FFinite Euclidean and Non-Euclidean Geometries
R. De Vogelaere Department of MathematicsUniversity of California, Berkeley, CA a r X i v : . [ m a t h . M G ] S e p Foreword
The author of this monograph was my father, Professor Ren´e De Vogelaere. He receivedhis PhD in Mathematics in 1948 from the University Louvain, Belgium. Shortly after grad-uation, he immigrated to Canada and taught at l’Universit´e Laval in Quebec, followed byNotre Dame in South Bend, Indiana and then the University of California, Berkeley, wherehe spent most of his career. He studied and taught a wide range of subjects, including differ-ential equations, numerical analysis, number theory, group theory, and Euclidean geometry,to mention a few.Georges Lemaˆıtre, the founder of the “Big Bang” theory, was my father’s thesis advisorand lifelong mentor. He was often a guest in our home, and at these meetings he encouragedmy father to study astronomy and planetary motion. After earning his doctorate degree,Professor Lemaˆıtre spent a year working with Arthur Eddington. Professor Eddington postu-lated that there were a finite number of protons in the universe. This is known as Eddington’snumber.Ren´e spent much of his career modeling the continuous world with discrete, finite num-bers. In the late 70’s he asked himself: what if the world was discrete rather than continuous?Would the proofs found in different mathematical branches still work? That is when his re-search in finite geometry began, culminating in this monograph, to which he dedicated thelast 10 years of his life. In my family, I was the only one who had studied math at thegraduate level, and so I was uniquely qualified to share in the excitement of his discoveriesand the number of theorems he was able to prove. He was like an archeologist having founda new field of dinosaur bones—discovering something new, then examining and documentingit. He taught classes on his findings, and wrote many papers (see the bibliography). He didnot publish his book; there were too many exciting theorems to prove, which were muchmore interesting to him than working with a publisher.Upon his passing in 1991, I inherited his unfinished book. I worked with a good friendand past classmate, Michael Thwaites, to try to compile the book written in LaTeX. But lifewas busy with family and work. It wasn’t easy stepping into my father’s shoes to completethis very involved task. Throughout the following 25 years, I looked for a way to preservethe book and disseminate its knowledge. Eventually technology and the right person cametogether. One late evening I was discussing my father’s Finite Geometry book with WilliamGilpin. He has just completed his PhD in Physics from Stanford University. He knew LaTeXvery well, and was able to assemble all the files. He also knew of the Cornell’s arXiv andrecommended posting it there. It is the perfect place to store Professor De Vogelaere’smagnum opus.I would like to thank:- My mother, Elisabeth De Vogelaere, who made it possible for my father to dedicatehis career to mathematics, which he loved- Arthur Eddington, for inspiring my father- Georges Lemaˆıtre, for inspiring and teaching my father- The University of California, Berkeley, for providing the facilities and allowing himtime to do his research, as well as for archiving the work he did over the 43 years ofhis career.- Michael Thwaites, for helping me get started on the book, and for encouraging me tocontinue the work- My wife Cynthia Haines, for carefully keeping and storing the computer disks, files andpapers all these years- My daughter Beth, for finding William Gilpin- William Gilpin, for his extreme generosity of time to rapidly assemble the book andfor facilitating having it stored at Cornell’s arXiv.- My siblings, Helene, Andrew, and Gabrielle for their patience and faith that this wouldhappen!- And Cornell’s arXiv, for being there to disseminate knowledge. Charles De VogelaereMountain View, CAAugust 2019 ontents
CONTENTS
ONTENTS p . . . . . . . . . . . . . . . . . . . . 1902.1.90 Answer to exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1952.1.91 Relation between Synthetic and Algebraic Finite Projective Geometry. 1972.2 Algebraic Model of Finite Projective Geometry. . . . . . . . . . . . . . . . . 1982.2.0 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 CONTENTS p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2012.2.4 Finite vector calculus and simple applications. . . . . . . . . . . . . . 2042.2.5 Anharmonic ratio, harmonic quatern, equiharmonic quatern. . . . . . 2062.2.6 Projectivity of lines and involution on a line. . . . . . . . . . . . . . . 2112.2.7 Collineation, central collineation, homology and elation. . . . . . . . . 2152.2.8 Correlations, polarity. . . . . . . . . . . . . . . . . . . . . . . . . . . 2182.2.9 Conics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2202.2.10 The general conic. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2242.2.11 The Theorem of Pascal and Brianchon. . . . . . . . . . . . . . . . . . 2262.2.12 The Theorems of Steiner, Kirkman, Cayley and Salmon. . . . . . . . 2312.2.13 B´ezier Curves for drawing Conics, Cubics, . . . . . . . . . . . . . . . . . 2352.2.14 Projectivity determined by a conic. . . . . . . . . . . . . . . . . . . . 2392.2.15 Cubics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2402.2.16 Other models for projective geometry. . . . . . . . . . . . . . . . . . . 2412.2.17 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2432.3 Geometric Models on Regular Pythagorean Polyhedra. . . . . . . . . . . . . 2432.3.0 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2432.3.1 The selector. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2442.3.2 The tetrahedron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2472.3.3 The cube. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2482.3.4 The dodecahedron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2502.3.5 Difference Sets with a Difference. . . . . . . . . . . . . . . . . . . . . 2522.3.6 Generalization of the Selector Function for higher dimension. . . . . . 2552.3.7 The conics on the dodecahedron. . . . . . . . . . . . . . . . . . . . . 2592.3.8 The truncated dodecahedron. . . . . . . . . . . . . . . . . . . . . . . 273 ∗ ∗ ∗ ∗ ONTENTS
CONTENTS p. The Hyperbolic Case. . . . . . . 4194.2.2 Trigonometry in a Finite Field for q = p e . The Hyperbolic Case. . . . 4224.2.1 Trigonometry in a Finite Field for p. The Hyperbolic Case. . . . . . . 4254.2.2 Trigonometry in a Finite Field for q = p e . The Hyperbolic Case. . . . 4284.2.3 Trigonometry in a Finite Field for q = p e . The Elliptic Case. . . . . . 4304.2.4 Periodicity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4404.2.5 Orthogonality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4414.2.6 Conics in sympathic geometry. . . . . . . . . . . . . . . . . . . . . . . 4424.2.7 Regular polygons and Constructibility. . . . . . . . . . . . . . . . . . 4434.2.8 Constructibility of the second degree. . . . . . . . . . . . . . . . . . . 4454.4 Contrast with classical Euclidean Geometry. . . . . . . . . . . . . . . . . . . 4454.4.0 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4454.3 Parabolic-Euclidean or Cartesian Geometry. . . . . . . . . . . . . . . . . . . 4464.3.0 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4464.3.1 Fundamental Definitions. . . . . . . . . . . . . . . . . . . . . . . . . . 4474.3.2 The Geometry of the Triangle in Galilean Geometry. . . . . . . . . . 4494.3.3 The symmetric functions. . . . . . . . . . . . . . . . . . . . . . . . . 4514.5 Transformation associated to the Cartesian geometry. . . . . . . . . . . . . . 4524.5.0 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4524.5.1 The geometry of the triangle, the standard form. . . . . . . . . . . . 4534.5.2 The cubic γ a of Gabrielle. . . . . . . . . . . . . . . . . . . . . . . . . 4574.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4634.6.1 Problems for Affine Geometry. . . . . . . . . . . . . . . . . . . . . . . 4644.6.2 Problems for Involutive Geometry. . . . . . . . . . . . . . . . . . . . 4644.90 Answers to problems and miscellaneous notes. . . . . . . . . . . . . . . . . . 465 ONTENTS V of a triangle. . . . . . . . . . . . . . . . . . . . . . . . . 4905.1.7 An alternate definition of the center V of a triangle. . . . . . . . . . . 4925.1.8 Intersections of the 4 circumcircles. . . . . . . . . . . . . . . . . . . . 4945.1.9 Other results in the geometry of the triangle. . . . . . . . . . . . . . . 4975.1.10 Circumcircle of a triangle with at least one ideal vertex. . . . . . . . . 4995.1.11 The parabola in polar geometry. . . . . . . . . . . . . . . . . . . . . . 5005.1.12 Representation of polar geometry on the dodecahedron. . . . . . . . . 5045.2 Finite Non-Euclidean Geometry. . . . . . . . . . . . . . . . . . . . . . . . . . 5105.2.0 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5105.2.1 Trigonometry for the general triangle. . . . . . . . . . . . . . . . . . . 5105.2.2 Trigonometry for the right triangle. . . . . . . . . . . . . . . . . . . . 5135.2.3 Trigonometry for other triangles . . . . . . . . . . . . . . . . . . . . . 5135.3 Tri-Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5145.3.1 The primitive case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5145.3.2 The case of 1 root. Inverse geometry. . . . . . . . . . . . . . . . . . . 5205.3.4 The case of a double root and a single root. . . . . . . . . . . . . . . 5245.3.5 The case of a triple root. Solar geometry. . . . . . . . . . . . . . . . . 5265.3.6 The case of 3 distinct roots. . . . . . . . . . . . . . . . . . . . . . . . 5285.3.7 Conjecture. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5305.3.8 Notes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5335.3.9 On the tetrahedron. . . . . . . . . . . . . . . . . . . . . . . . . . . . 535 CONTENTS
ONTENTS k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6848.5.2 Dimension 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6858.5.3 Dimension 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691 cn, sd ) and theWeierstrass elliptic curve . . . . . . . . . . . . . . . . . . . . . . . . . 7219.4 Complete elliptic integrals of the first and second kind. . . . . . . . . . . . . 7239.5 P-adic functions, polynomials, orthogonal polynomials. . . . . . . . . . . . . 7299.5.1 Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . . . . . . 7339.5.2 Integration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7379.6 P-adic field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7379.6.1 Generalities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7379.6.2 Extension to the half argument. . . . . . . . . . . . . . . . . . . . . . 7439.6.3 The logarithm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7459.6.4 P-adic Geometry and Related Finite Geometries. . . . . . . . . . . . 7484 CONTENTS
10 DIFFERENTIAL EQUATIONS AND FINITE MECHANICS 751
11 COMPUTER IMPLEMENTATION 763 hapter 0Preface
Purpose
The purpose of this book and of others that are in progress is to give an exposition ofGeometry from a point of view which in some sense complements Klein’s Erlangen program.The emphasis is on extending the classical Euclidean geometry to the finite case, but it goesway beyond that.
Plan
In this preface, after a brief introduction, which gives the main theme, and was presented insome details at the first Berkeley Logic Colloquium of Fall 1989, I present the main results,according to a synthetic view of the subject, rather that chronologically. First, some varia-tion on the axiomatic treatment of projective geometry, then new results on quaternioniangeometry, then results in geometry over the reals which are generalized over arbitrary fields,then those which depend on properties of finite fields, then results in finite mechanics. Therole of the computer, which was essential for these inquires is briefly surveyed. The method-ology to obtain illustrations by drawings is described. The interaction between Teaching andResearch is then given. I end with a table which enumerates enclosed additional materialwhich constitutes a small but representative part of what I have written.
Introduction
My inquiry started with rethinking Geometry, by examining first, what could be preservedamong the properties of Euclidean geometry when the field of reals is replaced by a finitefield. This led me to a separation of the notions concerned with the distance between 2 pointsand the angle between an ordered pair of lines, into two sets, those concerned with equalityand those concerned with measure. Properties relating to equality are valid for a Pappiangeometry, whatever the underlying field, those pertaining to measure require specifying thefield.I have also come to the conclusion that the more fruitful approach to the axiomatic of Eu-clidean geometry is to reduce it to that of Projective geometry followed by a preferenceof certain elements, namely the isotropic points on the ideal line. This preference can be156
CHAPTER 0. PREFACE presented alternately by choosing 2 points relatively to a triangle of coordinates, namelythe barycenter and the orthocenter. The barycenter is used to define the ideal line, theorthocenter is then used to define the fundamental involution of this line, for which theisotropic points are the (imaginary) fixed points. This program extends to all non-Euclideangeometries.The preference method, which I call the “Berkeley Program”, can be considered as the syn-thetic equivalent of the group theoretical relations between geometries, as advocated in FelixKlein’s Erlangen program.When I refer to Euclidean geometry, I always mean that the set of points and lines of thegeometry of Euclid have been completed by the ideal line and the ideal points on that line.
Axiomatic of projective geometry
Projective Geometry
Axiomatic.
The approach, used by Artzy, has the advantage of giving the equivalence between the syn-thetic axioms and the algebraic axioms, at each stage of the axiomatic development: forperspective planes, Veblen-Wedderburn planes, Moufang planes, Desarguesian planes, Pap-pian planes, ordered planes, and finally, projective planes. I have revised it, to give a uniformtreatment (particularly lacking at the intermediate step of the Veblen-Wedderburn plane, inwhich, for instance, vectors are introduced by Artzy and others, to prove commutativity ofaddition) and by giving, for all proofs, explicit, rather than implicit constructions, togetherwith drawings.
Notation.
The Theorems of Desargues, Pappus and Pascal play an important role in synthetic proofs inProjective geometry. A notation has been introduced for the repeated use of these theoremsand their converse, in an efficient and unambiguous way. A notation for configurations hasbeen introduced, which further helps in distinguishing non isomorphic configurations.
Desarguesian geometry
Quaternionian Geometry.
With Relative Preference of 2 Points.
A quaternionian plane is a well known, particularly important, example of a Desarguesianplane. I have introduced in it, the relative preference of 2 points, the barycenter and thecobarycenter and have obtained several Theorems, which in the sub-projective planes ofthe geometry correspond to Theorems in involutive geometry which are associated with thecircumcircle and with the point of Lemoine. But these Theorems cannot be consideredas simple generalizations. For instance, in the involution on the ideal line, defined by thecircumcircular polarity, which corresponds to a circumcircle, the direction of a side and that7of the comedian, which generalizes an altitude, are not corresponding elements, althoughthese correspond to each other, in the sub-projective planes. Moreover, what I call theLemoine polarity degenerates in the sub-projective planes into all the lines through the pointof Lemoine. The proofs given are all algebraic. These investigations are just the beginningof what should become a very rich field of inquiries.
Finite Quaternionian Geometry.
The Theorems in quaternionian geometry were conjectured using a geometry whose pointsand lines are represented by 3 homogeneous coordinates in the ring of finite quaternions over Z p . In the corresponding plane, the axioms of allignment are not allways satisfied. If theyare, Theorems and proofs for the quaternionian plane extend to the finite case. Pappian geometry over arbitrary fields
Pappian Geometry.
This can be considered as a projective geometry over an arbitrary field.
On Steiner’s Theorem.
Pappus’ Theorem is one of the fundamental axioms of Projective geometry. If the 3 pointson one of the lines are permuted, we obtain 6 Pappian lines which pass 3 by 3 through 2points, this is the Theorem of Jakob Steiner. By duality, we can obtain from these, 6 pointson 2 lines. That these 2 lines are the same as the original ones is a new Theorem. Detailedcomputer analysis of the mapping in special cases leads to conjectures in which twin primesappear to play a role.
Generalization of Wu’s Theorem.
I obtained some 80 new Theorems in Pappian geometry, generalizing a Theorem, in projectivegeometry, of Wen-Tsen Wu, related to conics through 6 Pascal points of 6 points on a conic,I have obtained a computer proof for all of these Theorems by means of a single program,which includes convincing checks, and then succeeded in obtaining a synthetic proof for eachof these Theorems, using several different patterns and approaches including duality andsymmetry. These proofs have benefited from the projective geometry notation. Drawingshave been made for a large number of these Theorems which have suggested 2 new Theoremsand a (solid) Conjecture. Many of the Theorems can be considered as Theorems in Euclideangeometry, (only one of which was known, the Theorem of Brianchon-Poncelet), others canbe considered as Theorems in Affine or in Galilean geometry.
Generalization of Euclidean Theorems.
The Theorems, given for involutive geometry, can be considered, alternately, as Theorems inPappian geometry, because they involve only the preference of 2 elements of the projectiveplane and not additional axioms.
Involutive Geometry.
I call involutive plane, a Pappian plane in which I prefer 2 points relative to a triangle, M ,the barycenter and M , the orthocenter. M allows for the definition of the ideal line, M allows, subsequently, for the definition of the fundamental involution on that line. Generalization of Theorems in Euclidean and Minkowskian Geometry over Arbitrary Fields. CHAPTER 0. PREFACE
In this, which constitues the more extensive part of my research, I have generalized, whenthe involution is elliptic, a very large number of Theorems in Euclidean geometry, namelythose which are characterized by not using the measure of distance and of angles and notinvolving elements whose construction leads to more than one solution. When the funda-mental involution is hyperbolic, each of the Theorems gives a corresponding Theorem in thegeometry of Hermann Minkowski.
Symmetry and Duality.
The barycenter and orthocenter have a symmetric role for many Theorems of Euclideangeometry, the line of Euler and the circle of Brianchon-Poncelet being the simpler examples.This has been systematically exploited, to almost double the number of Theorems knownin that part of Euclidean geometry which involves congruence and not measure. Dualitycan also be extended to Euclidean geometry by associating to M and M , the ideal line andthe orthic line and vice-versa. This also has been systematically exploited to help me, inobtaining constructions of new elements, and should be helpful in future constructions. Notation.
A set of notations was introduced, to allow for a compact description of some 1006 defini-tions, 1073 conclusions and for the corresponding proofs. The counts correspond to one formof counting, other forms give higher numbers. All these Theorems are valid for any Pappianplane and give directly both statement and new proofs in both Euclidean and Minkowskiangeometry.
The Geometry of the Triangle.
During the period 1870 to 1900, there was an explosion of results in what has been calledthe geometry of the triangle, prepared by Theorems due to Leonhard Euler, Jean Poncelet,Charles Brianchon, Emile Lemoine and others. The synthesis of the subject was never suc-cessfully accomplished, not only because of the wealth of Theorems, but because of thedifficulty of insuring that elements defined differently were in fact, in general, distinct. Theproofs, used in involutive geometry, not only throw a new light on the reason for the explo-sive number of results for the geometry of the triangle but also gives a exhaustive syntheticview of the subject.
Diophantine Equations.
Because an algebraic expression of the homogeneous coordinates of points and lines andthe coefficients for conics is given in terms of polynomials in 3 variables, a large number ofparticular results on diophantine equations in 3 variables are implicitly obtained in theseinvestigations.
Construction with the Ruler only.
In all of the classical investigations, the most extensive one being that of Henri Lebesgue,the impression is given that the compass is indispensable for most constructions in geometry.More than half of the Theorems for which a count is given above, can be characterized asusing the ruler only. Implicit, in this part of my Research, is, that many constructions, whichusually or by necessity were assumed to require the compass, in fact need the ruler only, thesimplest one is that for constructing the perpendicular to a line. The more remarkable oneis that the circles of Apollonius can be constructed with the ruler only. These are defined asthe circles which have as diameter the intersections of the bisectrices of an angle of a trianglewith the opposite sides. It is this reduction to construction with the ruler alone, which allowsfor the straigthforward proofs which constitutes a major success of these investigations.9
Construction with the Ruler and Compass.
The construction with compass can be envisioned as follows. Given M and M , by findingthe intersection of 2 circles centered at 2 of the vertices of a triangle with the adjacent sidesand by constructions with the ruler, we can construct the bissectrices of these angles, theincenter (center of the inscribed circle) and the point of Joseph Gergonne (the common inter-section of the lines through a vertex of the triangle and the point of tangency of the inscribedcircle with the opposite side). From these, a very large number of other points, lines andcircles can be constructed with the ruler only, for instance, the point of Karl Feuerbach, theexcribed circles and the circles of Spieker. One can therefore, in the framework of involutivegeomatry, prefer instead of M and M , the incenter I and the point of Gergonne J . Startingfrom I and J , we can construct M and M , using the ruler alone. This allows to extend theproof methodology considerably, allowing the generalization to arbitrary fields of Theoremsinvolving elements whose construction, in the classical case, would requires the compass. Cubics.
Very little has been written on the construction of cubics by the ruler. Starting with thework of Herman Grassmann of R. Tucker and of Ian Barbilian, I have obtaining a few resultsin this direction, one of which, incidentally, gives a illustration of the procedure of construc-tion with the compass as I am envisioning it, which is much simpler than those involvingbissectrices.
Galilean geometry.
When the fundamental involution is parabolic and when the field is the field of reals, thegeometry is called Galilean, because its group is the group of Galilean transformations ofclassical mechanics. Extending to the Pappian case and starting from the definitions andconclusions of involutive geometry, I have made appropriate modifications to obtain Theo-rems which are valid in Galilean geometry, but I have not yet completed the careful checkthat is required to insure the essential accuracy. Again a very large number of Theoremshave been obtained, which are new, even in the case of the field of reals.
Polar Geometry.
The extension, to n dimensions, can be obtained using an appropriate adaptation of thealgebra of Herman Grassmann. A first set of Theorems has been obtained in the case of 3dimensions, again for a Pappian space over arbitrary fields, in which preference is given toone plane, the ideal plane and one quadric. These Theorems generalize Theorems on thetetrahedron due to E. Prouhet, Carmelo Intrigila and Joseph Neuberg. The special case ofthe orthogonal tetrahedron has also been studied in a way which puts in evidence the reasonsbehind many of the Theorems obtained in this case. Non-Euclidean Geometry.
The beginning of the preference approach to obtain new results in non-Euclidean geometrywas started in January 1982. The confluence, in the case of a finite field, of the geometries ofJanos Bolyai and of Nikolai Lobachevsky was then explored. A new point, called the centerof a triangle was discovered and its properties were proven.0
CHAPTER 0. PREFACE
Pappian geometry over finite fields
The Case of Finite Fields.
All the results given for involutive geometry and in the following sections are true, irrespectiveof fields. In what follows, we describe results for finite fields.
Projective Geometry.
Representation on Pythagorean and Archimedean solids.
Fernand Lemay has shown how to represent the projective planes corresponding to theGalois fields, 2, 3 and 5 respectively on the tetrahedron, the cube (or octahedron) and thedodecahedron (or icosahedron). I have shown, that if we choose instead of the Pythagoreansolids, the Archimedean ones, the results extend to 2 and the 5-gonal antiprism and to3 and the truncated dodecahedron. I have studied also the corresponding representationsof the conics on the dodecahedron. This is useful for the representation on it of the finitenon-Euclidean Geometry associated with GF (5). Involutive Geometry.
Partial Ordering.
In the case of finite fields, ordering and therefore the notions of limits and continuity are notpresent. By using Farey sets or, alternately, by using a symmetry property of the continuedfraction algorithm, I have introduced partial ordering in Z p . If only, the properties of orderhave to be preserved which are related to the additive inverse and multiplicative inverse,then a Theorem of Mertens allows me to estimate the cardinality of the ordered subset of Z p by .61 p , when p is large. The cardinality is decreased logarithmicaly, by a factor 2, for eachadditional operation of addition and multiplication, for which order needs to be preserved. Orthogonal polynomials.
Orthogonal polynomials can be defined in a straightforward way in Z p . For those I havestudied, it turns out, that the classical scaling used in defining the classical orthogonal poly-nomials, there is a symmetry which is exibited in each case, with the exception of those ofCharles Hermite. In this case, by using an alternate scaling, with different expressions forthe polynomials of even and odd degree, symmetry can also be obtained. Finite Trigonometry.
Ones the measure of angles between an ordered pair of non ideal lines and the measure of thesquare of the distance between two ordinary points has been defined, it is straightforwardto obtain the trigonometric functions in Z p . There are in fact, for each prime p , two sets oftrigonometric functions, one corresponding to the circular ones, one to the hyperbolic ones.The proofs required, depend on the existence of primitive roots, in the case correspondingto Minkowskian geometry, and on a generalization to the Galois field GF ( p ) in the casecorresponding to Euclidean geometry. Finite Riccati Functions.
The functions of Vincenzo Riccati, which are generalization of the trigonometric functionshave been defined and studied in the finite case. They enable the definition of a Riccatigeometry. An invariant defines distances, the addition formulas, which correspond to multi-plication of associated Toeplitz matrices, define addition of angles. This again should be afruitful field of inquiry.
Finite Elliptic Functions.
After I conjectured that the Theorem of Poncelet on polygons inscribed to a conic and cir-1cumscribed to an other conic extended to the finite case, I knew that Finite Elliptic functionscould be defined in the finite case, because I had learned from Georges Lemaˆıtre the rela-tion between Theorems on elliptic functions and the Theorem of Poncelet. The functions Idefined, correspond to the functions sn , cn and dn of Karl Jacobi. After I found that JohnTate had defined the Weierstrass type of finite elliptic functions I established the relationbetween the 2. Construction with the compass.
In the case of finite fields, the points I and J will only exist if 2 and therefore all angles ofthe triangle are even. Prefering I and J instead of M and M , insures that the triangle iseven. Isotropic Geometry.
Many of the Theorems in involutive and polar geometry do not apply to the case of fieldsof characteristic 2, because the diagonal points of a complete quadrilateral are collinear, be-cause every conics has all its tangents incident to a single point and because in the algebraicformulations, 2, which occurs in many of the algebraic expressions involved in correspondingproofs of involutive geometry is to be replaced by 0. I call isotropic plane, a Pappian plane,with field of characteristic 2 and with the relative preference of 2 points, M , the barycenterand, O , the center. The orthocenter does not exist when the characteristic is 2 becauseeach line can be considered as perpendicular to itself. The difference sets of J. Singer, calledselectors by Fernand Lemay, were an essential tool in these investigations. In an honor The-sis, Mark Spector, now a Graduate Student in Physics at M.I.T. wrote a program to checkthe consistency of the notation in the statements of the Theorems and the accuracy of theproofs. He obtained new results. My results on cubics are not retained in his honors Thesis.Some of the results in isotropic geometry were anticipated by the work of J. W. Archbold,Lawrence Graves, T. G. Ostrom and D. W. Crowe. Finite mechanics and simplectic integration
I was asked to participate in a discussion, Spring 1988, at Los Alamos, on the field of sim-plectic integration which I originated in 1955. Simplectic integration methods are methodsof numerical integration which preserve the properties of canonical or simplectic transforma-tions. It then occured to me, that these methods were precisely what was needed to extendto the finite case the solution of problems in Mechanics. I had searched for a solution to thisproblem since I obtained, as first example, the solution, using finite elliptic functions, forthe motion in Z p of the pendulum with large amplitude, as well as the polygonal harmonicmotion, whose study was suggested by a Theorem of John Casey, and led to an equationsimilar to Kepler’s equation.More specifically, whenever the classical Hamiltonian describing a motion has no singulari-ties, a set of difference equations can be produced whose solutions at successive steps havethe properties associated with simplectic transformations. To confirm the solidity of this ap-proach, I studied, in detail, the bifurcation properties for one particular Hamiltonian. Thestudy can be made in a more complete fashion than in the classical case and requires a much2 CHAPTER 0. PREFACE simpler analysis using the p -adic analysis of Kurt Hensel. The role of the computer for conjectures and verification
The computer was an essential tool in the conjecture part of the Research described above,in the verification of the order of the statements and to insure the consistency of the notationused in the statements of the Theorems as well as in the verification of the proofs. In partic-ular, the Theorem refered to in the Steiner section was conjectured from examples from finitegeometry. All of the Theorems generalizing Wu’s Theorem were conjectured by examining,in detail, one appropriately chosen example, for a single finite field. Many Theorems in in-volutive geometry and all the Theorems in quaternionian geometry were so conjectured andthe methodology used was such that almost all conjectures could be proven. The remainingones could easily be disposed of, by a counterexample or algebraically. The only exceptionare the conjectures, indicated in the section on Steiner’s Theorem, which refer to twin primes.
Illustrations by drawings
Responding to natural requests for figures which illustrate the many Theorems obtained, Ihave also prepared a large number of drawings. These have been done for the case of the fieldof reals and therefore in the framework of classical Euclidean geometry. These are created bymeans of a VMS-BASIC program, which constructs a POSTSCRIPT file, for any set a data,including points, lines, conics and cubics. The position of the labels of points and lines canbe adjusted by adding the appropriate information to the data file in order to position thelabels properly. One such illustration was chosen by George Bergman, for this years posteron “Graduate opportunities in Mathematics for minority and women students”.
Interaction between research and teaching
These 2 obligations are for me very closely intertwined, my specific contributions to teachingare given in a separate document. The conjecture aspect of my research was exclusivelydependent on VMS-BASIC programs which were a natural extension of programs which Iwrote for my classes. Many of the proofs are dependent on material contained in notes Iprepared for students while teaching courses not related to my original specialty of Numeri-cal Analysis and of Ordinary Differential Equations.Many results have been presented in courses, a few, in Computation Mathematics, (Math.100), Abstract Algebra (Math. 113) and Number Theory (Math. 115), a large number, ina seminar on Geometry, 2 years ago, and in Foundations of Geometry (Math. 255), Fall 1989.3
Notes and publication
The scope of the results and their constant interaction during the years made it impracticalto publish incrementally without slowing down considerably the pace of the inquiry. I haveonly given a brief overview in 1983 and in 1986.
Finite Euclidean and non-Euclidean Geometry with application to the finitePendulum and the polygonal harmonic Motion. A first step to finite Cosmology.
The Big Bang and Georges Lemaˆıtre, Proc. Symp. in honor of 50 years after hisinitiation of Big-Bang Cosmology, Louvain-la-Neuve, Belgium, October 1983., D.Reidel Publ. Co, Leyden, the Netherlands. 341-355.
G´eom´etrie Euclidienne finie. Le cas p premier impair.
La Gazette des Sci-ences Math´ematiques du Qu´ebec, Vol. 10, Mai 1986.
Basic Discoveries in Mathematics using a Computer.
Symposium on Mathe-matics and Computers, Stanford, August 1986.
A short guide to the reader.
The reader may want to start directly with Chapter II and to read sections of the introductoryChapter as needed. He may perhaps wish to read the section on a model of finite Euclideangeometry with the framework of classical geometry, if he wishes to be more confortable aboutthe generalization of the Euclidean notions to the finite case. If at some stage the readerswants a more tourough axiomatic treatment it will want to read the section on axiomatic ofthe first Chapter.Chapter II is written in terms of finite projective geometry associated to the prime p ,but, except in obvious places, all definitions and Theorem apply to Pappian planes overarbitrary fields. Among the new results, included in this Chapter, are, a Theorem relatedto the Steiner-Pappus Theorem, considerations on a “general conic”, a description of theTheorems of Steiner, Kirkmanm Cayley and Salmon in terms of permutation maps. Afterdescribing the representation of the finite projective planes for p = 2, 3 and 5 on Pythagoriansolids, the generalization to the projective plane of order p on the truncated dodecahedronis given as well as that of the plane of order on the antiprism. Difference sets involving nonprimitive polynomials are studied which allow a definition of the notion of distance for affineas well as other planes.Attention is also drawn to B´ezier curves, which have not yet entered the classical reper-toire of Projective Geometry. These are used extensively in the computer drawing of curvesand surfaces.One of the reason for the historical delay of extended the Euclidean notions associatedwith distance between points and angle between lines is the lack of early distinction betweenequality and measure. Equality is a simpler notion which can be dealt with over arbitraryfields, while measure requires greater care. This is examplified by the comment on finiteprojective geometries by O’Hara and Ward, p. 289.4 CHAPTER 0. PREFACE
Their analytic treatment involves the theory of numbers, and, in particular thetheory of numerical congruences; it may be assumed that the synthetic treatmentof them is correspondingly complicated.It is my fondest hope that some of the material on finite geometry will be assimilatedto form the basis of renewal of the teaching of geometry at the high school level, combinedwith a well-thought related use of computers at that level. hapter 1MAIN HISTORICALDEVELOPMENTS
In this chapter, I give the main historical developments in Mathematics which have a bear-ing on the generalization of Euclidean Geometry to the finite case and to non EuclideanGeometries.What could be consider as the first contribution to Mathematics which covers number theory,geometry and trigonometry is a tablet in the Plimpton collection, this is briefly describedand discussed in a note at the end of the Chapter. The key to the treatment of geometry andits use of continuity dates from the discovery of the irrationals by the school of Pythagoras.This is commented upon to suggest an alternative which is consistent with finite Euclideangeometry. I thought it would be handy for many readers to have at hand the definitions andpostulates of Euclid, as well as a brief description of his 13 books, if only to see how we havetravelled in getting a more precise description of concepts and theorems in geometry. Dis-tances play an essential, if independent role, in the development of geometry, until recently,after some comments on the subject, I give some post Euclidean theorems involving distnaceson the sides of a triangle due to Menelaus and Ceva. The geometry of the triangle, whichhas played an important historical role, is illustrated by theorems due to Euler, Brianchonand Poncelet, Feuerbach, Lemoine and Schr¨oter.I then review quickly some of the major developments in projective geometry due to Menaech-mus, Apollonius, Desargues, Pascal, MacLaurin, Carnot, Poncelet, Gergonne and Chasles.In the next section, I start the process of going back from projective, to affine, to involutive,to Euclidean geometry.I then review the algebraization of geometry starting with Descartes and Poncelet and endingwith James Singer, who spured by a paper of Veblen and MacLagan-Wedderburn, introducedthe notion of difference sets which allows the representation of every point and line in a finitePappian plane by an integer, allowing an easy determination of incidence, without coordi-natization.This is followed by a section on trigonometry which gives the Lambert formulas valid in thecase of finite fields. 256
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
The section on algebra is for the reader which has been away from the subject for some time.It includes algorithms to solve linear diophantine equations and to obtain the representationof numbers as sum of 2 squares, the definition of primitive roots and the application to theextraction of square roots in a finite field, contrasting with the solution of the school ofPythagoras.The section on Farey sets includes original material on partial ordering of distances, whichat least suggest that the essential notion of ordering in the classical case can be extended tothe finite case.Definition of complex and quaternion integers, loops, groups, Veblen-Wedderburn systemsand ternary rings are given as a preparation for the section on axiomatic. The importantrelevant contributions of Klein, Gauss, Weierstrass, Riemann, Hermite and Lindenbaum arethen recalled.The subject of elliptic functions and the application of geometry to mechanics has lost, atthe present time, the great interest it had during last century. Because this too generalizesto the finite case and because this is not now part of the Mathematics curriculum, I havea long section introducing one of its components, the motion of the pendulum to introduceelliptic integrals, the elliptic functions of Jacobi as well as his theta functions, ending withthe connection given first by Lagrange between spherical trigonometry and elliptic functions.To add credibility to the existence of non Euclidean geometries, models were divised to givemodels within the framework of Euclidean geometry. The next section gives a model of finiteEuclidean geometry also within this framework. It can be used as an introduction to thesubject.The axiomatic of geometry in the next section is done using a uniform treatment, and explicitconstructions. It includes a plane which is, like the Moufang plane, intermediate betweenthe Veblen-Wedderburn plane and the Desaguesian plane. The geometry of Lenz-Barlottiof type I.1 discovered by Veblen and MacLagan-Wedderburn and studied by Hughes is anexample of this intermediate plane.
Introduction.
Besides estimating areas and volumes, the Babylonians had a definite interest in so calledPythagorian triples, integers a, b and c such that a = b + c . In tablet 322 of the Plimpton library collection from Columbia University, dated 1900 to1600 B.C., a table gives, with 4 errors, and in hexadesimal notation, 15 values of a, b, and ( ac ) = sec ( B ) , corresponding to angles varying fairly regularly from near 45 • to near 32 • . (See Note 1.13.2).It is still debated if their interest was purely arithmetical or was connected with geometry(See Note 1.13.1). .1. BEFORE EUCLID. That the ratio of the length of the sides of a triangle is equal to the ratio of 2 integers wasfirst contradicted by the counterexample of an isosceles right triangle A , A , A , with rightangle at A and with sides a and hypothenuse a . The theorem of Pythagoras states that a = a + a = 2 a , a > a > . (1)If a and a are positive integers, it follows from the fact that the square of an odd integeris odd and that of an even integer is even, and from (1), that a and therefore a is even,therefore a = 2 a and a = 2 a , a > a > . (2)The argument can be repeated indefinitely and an infinite sequence of decreasing positiveintegers is obtained, a > a > . . . > a n > . . . > . (3)But this contradicts the fact that only a finite number of positive integers exist which areless than a . Geometrically, the proof follows from the following figure: (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) (cid:0)(cid:0)(cid:0)(cid:0)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64) (cid:64)(cid:64)(cid:64)(cid:64) a a a a This argument has been refined through the ages, by a careful construction of the inte-gers, see for instance the Appendix by Professor A. Morse in Professor J. Kelley’s book onTopology, by an analysis of their divisibility properties (see the Theorem of Aryabatha) andby their ordering properties (the well ordering axiom of the integers). What is implicit inthe geometry considered by the Greeks, after Pythagoras, is that the circle with center A and radius a meets the line through A and A at a point, but this assumption is not madeexplicitely. From it follows the existence of points on the line corresponding to the irrational √ √ , . . . , √ , . . . , more generally, √ p, for p prime, eventually this lead Euclid to consider that the set of points on each lineforms a continuous set.Moreover the theorem of Pythagoras assumes the axiom on parallels of Euclid.In finite affine geometry, I will keep the axiom of parallels but assume that the number ofpoints on each line is finite. In finite Euclidean Geometry most of the notions of ordinaryEuclidean geometry are preserved, the measure of angles presents no difficulties and themeasure of distances requires the introduction of one irrational. On the other hand circlesmeet half of the lines through their center in 2 points and the other half in no point and √ CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
The greek geometer Euclid (300 B.C) constructed a careful theory of geometry based on theprimary notions of points, lines and planes and on a set of axioms, the last one being theaxiom on parallels.His first 3 books are devoted to a study of the triangle, of the circle and of similitude.I will list here the definitions, postulates and common notions as translated by Heath, p.153 to 155:
Definitions.
0. A point is that which has no parts.1. A line is breadthless length.2. The extremities of a line are points.3. A straight line is a line which lies evenly with the points on itself.4. A surface is that which has length and breath only.5. The extremities of a surface are lines.6. A plane surface is a surface which lies evenly with the straight lines on itself.7. A plane angle is the inclination to one another of two lines in a plane which meet oneanother and do not lie in a straight line.8. And when the lines containing the angle are straight, the angle is called rectilinear .9. When a straight line set up on a straight line makes the adjacent angles equal to oneanother, each of the equal angles is right , and the straight line standing on the otheris called a perpendicular to that on which it stands.10. An obtuse angle is greater than the right angle.11. An acute angle is an angle less than a right angle.12. A boundary is that which is an extremity of anything.13. A figure is that which is contained by any boundary or boundaries.14. A circle is a plane figure contained by one line such that all the straight lines fallingupon it from one point among those lying within the figure are equal to one another.15. And the point is called the center of the circle . .2. EUCLIDEAN GEOMETRY. diameter of the circle is any straight line through the center and terminated in bothdirections by the circumference of the circle, and such a straight line also bisects thecircle .17. A semicircle is the figure contained by the diameter and the circumference cut off byit. And the center of the semicircle is the same as that of the circle.18. Rectilineal figures are those which are contained by straight lines, trilateral figuresbeing those contained by three, quadrilateral those contained by four, and multilateralthose contained by more than four straight lines.19. Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal.20. Further, of trilateral figures, a right-angled triangle is that which has a right angle,an obtuse-angled triangle that which has an obtuse angle, and an acute-angled triangle that which has three angles acute.21. Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilat-eral but not right-angled; and a rhomboid that which has opposites sides and anglesequal to one another but is neither equilateral or right-angled. And let quadrilateralsother than these be called trapezia .22.
Parallel straight lines are straight lines which, being in the same plane and beingproduced indefinitely in both directions, do not meet one another in either direction.
Postulates.
Let the following be postulated.0. To draw a straight line from one point to any point.1. To produce a finite straight line continuously in a straight line.2. To desribe a circle with any center and distance.3. That all right angles are equal to one another.4. That, if a straight line falling on two straight lines make the interior angles on thesame side less than two right angles, the two straight lines, if produced indefinitely,meet on that side on which are the angles less than the two right angles.0
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Common notions. Things which are equal to the same thing are also equal to one another. If equals be added to equals, the wholes are equal. If equals be subtracted from equals, the remainders are equal. Things which coincide with one another are equal to one another. The whole is greater than the part.
Short description of the Books of Euclid.
The work of Euclid consists of 13 books which contain propositions which are either theoremsproving properties of geometrical figures or theorems concerned with proving that certainfigures can be constructed. It also consists of a study of integers, rationals and reals.- Book 1 is devoted mainly to congruent figures, area of triangles and culminates withthe Theorem of Pythagoras (Proposition 47).- Book 2 is concerned with construction of which the following is typical, determine P on AB such that AP = AB.BP .- Book 3 studies in detail circles, tangent to circles, tangent circles.- Book 4 constructs polygons inscribed and outscribed to circles.- Book 5 gives the theory of proportions.- Book 6 applies the theory of proportions to geometrical figures.- Book 7 studies integers, their greatest common divisor (Proposition 2) and their leastcommon multiple (Proposition 34).- Book 8 studies proportional numbers.- Book 9 studies geometrical progression, in Proposition 20, the proof that the numberof primes is infinite is given.- Book 10 studies the commensurables and incommensurables.- Book 11 is on 3 dimensional or solid geometry.- Book 12 studies similar figures in solid geometry.- Book 13 studies properties of pentagons and decagons as well as the regular solids. .2. EUCLIDEAN GEOMETRY. Comment.
These definitions, postulates and axioms have been discussed since the time of Euclid. Thereader is urged to study some of these discussion, for instance those in the book of Heath.Already Proclus (see Paul van Eecke) criticizes Postulate 5, and claim that it should beproven. Let me only observe here that except for the notion of being on the same side andthe notion of continuity, which are absent from finite Euclidean geometry, in some sense allof the definitions and postulates given above are valid in finite Euclidean geometry. It shouldbe stressed that the expression “produced indefinitely“ (eis apeiron) cannot be translatedby “to infinity” (see Heath, p. 190).Heath observes also (p. 234) that Euclid implies that “straight lines and circles determineby their intersections other points in addition to those given“ and that “the existence of suchpoints of intersection must be postulated”. He concludes that “the deficiency can only bemade good by the Principle of Continuity“ and proceed by giving the axioms of Killing.We will see that the alternate route of finite Euclidean geometry disposes of the problemquite differently and that some figures cannot always be constructed.It will also be seen that the great emphasis given to distance between points and angles oftwo straight lines and their equality are notions which we will derive from more basic notionsand that following the point of view adopted since the 19-th century no attempt will be madeto define points and lines, as in Euclid, but we will give instead properties that they possess.In this connection the critique of Laurent, H., 1906, p.69 is of interest:Euclid and Legendre have imagined that the word ‘distance’ has a mean-ing and they believed that the proofs using superposition have a ‘logical’ value.Moreover few of the present day geometers have observed that Legendre andEuclid have erred. And that is, I believe, one of the more curious psychologicalphenomenons that for more than two thousand years one does geometry withoutrealizing that its fundamental propositions have no sense from a ‘logical’ pointof view .I will now state a few theorems which play an important role in Part II. a few of whichare not in Euclid or Legendre. Definition.
The altitude through A is the line through A which is perpendicular to A A . The foot ofthe altitude through A is the point H on the altitude and on A A . Theorem.
The altitudes through A , A and A are concurrent in H. Euclide et Legendre se sont figur´e que le mot ‘distance’ avait un sens et ils ont cru que les d´emonstrationspar superposition avaient une valeur ‘logique’. D’ailleurs peu de g´eom`etres aujourd’hui s’apercoivent queLegendre et Euclide ont divagu´e. Et c’est l`a, `a mon avis, un des ph´enom`enes psychologiques les pluscurieux que, depuis plus de deux milles ans, on fait de le g´eometrie sans s’apercevoir que ses propositionsfondamentales n’ont aucun sens au point de vue ‘logique’. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition.
The point H is called the orthocenter.
Theorem.
Let M be the mid-point of A A , let M be the mid-point of A A and let M be the mid-pointof A A , then A M , A M and A M are concurrent in M. Definition.
The point M is called the barycenter or center of mass . Definition. m is the mediatrix of A A if m passes through M and is perpendicular to A A . Theorem. [Euclid, Book 4, Proposition 5.]
The mediatrices m , m and m are concurrent in O. Definition.
The point O is called the center of the circumcircle of the triangle A A A . Theorem. [Euler]
The points
H, M and O are on the same line e. Definition.
The line e is called the line of Euler . Comment.
The usual proof of 1.2.1 is geometric. The proof given by Euler is entirely algebraic. It isbased on an expression for the distances of
HG, HO and OG in terms of the sides of thetriangle. Let a, b and c be the sides of the triangle. Let p = a + b + c, q = bc + ca + ab, r = abc, (the symmetric functions of a, b and c ).The area A is given by AA = ( − p + 4 ppq − pr ) .Euler obtains HM HM = rrAA − ( pp − q ) ,HO HO = rrAA − ( pp − q ) ,M O M O = rrAA − ( pp − q ) . I use here the notation of Euler and of mathematicians before the middle of the 19th century, namely AA for A.A. .2. EUCLIDEAN GEOMETRY.
M O = HM = HO and HO = HM + M O therefore the points
H, M and O are collinear.If I is the center of the inscribed circle, Euler determines also HI, GI and
IO.
Theorem. [Euclid, Book 3, Propositions 35 and 36.]
If 2 lines through M, not on a circle meet that circle, the first one in A and B, the secondone in C and D, then | M A || M B | = | M C || M D | . Theorem.
Let A A A be a triangle and H H H be the feet of the perpendiculars from the vertices tothe opposite sides. then A H bisects the angle H H H . Proof: If H is the orthocenter, the quadrangle HH A H can be inscribed in a circleand therefore the angles A H H and H A A are equal. Similarly the angles H H A and A A H are equal, but the angles H A A and A A H are equal because there sides areperpendicular, therefore A H H and A A H are equal. Definition.
The triangle H H H is called the orthic triangle. Introduction.
The following theorems give a metric characterization of three points on the sides of a trianglewhich are collinear or which are such that the line joining these points to the opposite vertexare concurrent. For these theorems, an orientation is provided on each of the sides andtherefore the distances have a sign. The theorems are as follows:
Theorem. [Menelaus] If X is on a , X is on a and X is on a , then the points X , X and X are collinear iff | A X || A X || A X | = | A X || A X || A X | . Theorem. [Ceva] If X is on a , X is on a and X is on a , then the lines A X , A X and A X areconcurrent iff | A X || A X || A X | = −| A X || A X || A X | . The following theoremis a direct consequence of the theorem of Ceva.Theorem . . . seeCoxeter, I believe I saw it later ???4
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem.
Let X be a point not on the sides of a triangle A A A , let X , X , X be the intersectionof XA with A A , of XA with A A , and of XA with A A , , let Y , Y , Y be the otherintersection of the circle through X , X and X with the sides of the triangle, then A Y ,A Y and A Y are concurrent. Proof: If we eliminate | A i X j | from the relation of Ceva and from the relations | A X || A Y | = | A X || A Y || A X || A Y | = | A X || A Y || A X || A Y | = | A X || A Y | obtained from Theorem 1.2.1, we obtain | A Y || A Y || A Y | = −| A Y || A Y || A Y | . Therefore by the Theorem of Ceva, the lines A Y , A Y and A Y are concurrent. Inroduction.
The geometry of the triangle has its origin in the following theorems.
Theorem.
The 3 medians of a triangle meet at a point called the barycenter or, in mechanics, the centerof mass.
Theorem.
The 3 altitudes of a triangle meet at a point called the orthocenter.
Theorem.
The 3 mediatrices of a triangle meet at a point which is the center of the circumcircle.
Theorem.
The 3 bisectrices of a triangle meet at a point which is the center of the inscribed circle.
Theorem. [Euler]
The points
H, G and O are on a line, called the line of Euler, moreover | HG = 2 | GO | and | HO | = 3 | GO | . The proof of Euler is algebraic. He determines the distance
HG, GO and HO in termsof the length of the sides of the triangle. Other distances are also determined in the samepaper. .2. EUCLIDEAN GEOMETRY. Theorem. [Brianchon and Poncelet]
The mid-points of the sides of a triangle, the feet of the altitudes and the mid-points of thesegments joining the orthocenter to the vertices of the triangle are on a circle, called thecircle of Brianchon-Poncelet. It is also called the 9 point circle or the circle of Feuerbach,who discovered it independently, and improperly the circle of Euler.
Theorem. [Feuerbach]
The circle of Brianchon-Poncelet is tangent to the inscribed circle and to the three excribedcircles, the point of tangency for the inscribed circle is called the point of Feuerbach.
The proof given by Feuerbach is algebraic and trigonometric in character. It expressesdistances in terms of the length of the sides and of the trigonometric functions of the anglesof the triangle.
Introduction.
An interesting development of Euclidean geometry occured during the 19-th century, knownunder the name of the geometry of the triangle. The activity in this area was most intenseduring the period 1870-1900. A large number of elementary results were obtained especiallyin Belgium and France, but also in England, Germany and elsewhere. Strictly speaking,the Theorem of Euler of 1.2.3 can be considered as the first important new result in thisconnection since Euclid. Others which prepared the way were the theorems of Brianchon-Poncelet of 1.2.3 and the Theorem of Feuerbach of 1.2.3. A few theorems will be extractedfrom the long list.
Theorem. [Schr¨oter] If a × b denotes the point on a and b and A × B denotes the line through A and B ,Let F := ( M × H ) × ( M × H ) ,F := ( M × H ) × ( M × H ) ,F := ( M × H ) × ( M × H ) .G := ( M × M ) × ( H × H ) ,G := ( M × M ) × ( H × H ) ,G := ( M × M ) × ( H × H ) . F , F and F are on the line e of Euler. A × G , A × G and A × G are parallel and are perpendicular to e. A , F , G and G are collinear, and so are A , F , G and G as well as A , F , G and G . G , G , G are the vertices of a triangle conjugate to the circle of Brianchon-Poncelet. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS M × G , M × G and M × G pass through the same point S. H × G , H × G and H × G pass through the same point S (cid:48) . S and S (cid:48) are on the circle of Brianchon-Poncelet. S and S (cid:48) are on the polar of H with respect to the triangle A , A , A .S and S (cid:48) are called the points of Schr¨oter. The proof of this Theorem published by Schr¨oter in “Les Nouvelles Annales de Math´ematiques”in 1864, was obtained by several people. The published proof is that of a student of Sainte-Barbe, L. Lacachie. The Theorem is generalized to Projective Geometry in III.D8.1,D8.2,C8.0.It is stated in finite involutive geometry in III. ?? . The projective geometry has its source in the discovery of the conic sections, the ellipse,the parabola and the hyperbola, which is ascribed by Proclus to the Greek mathematicianMenaechmus, a pupil of Plato and Eudoxus. The conic sections were studied by Aristaeusthe Elder, Euclid, Archimedes, Pappus of Alexandria and finally by Apollonius of Perga.The conics are defined as the intersection of a (circular) cone by a plane not passing throughits vertex. If we make a cut of the cone with a plane through the vertex we obtain two lines c and c . Line a is the cut of a plane giving an hyperbola, line b is the cut of a plane giving a parabola, line c gives an ellipse and line d gives the special case of a circle. V (cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16) c (cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80) c ad (cid:65)(cid:65)(cid:65)(cid:65)(cid:65)(cid:65)(cid:65)(cid:65)(cid:65)(cid:65) b (cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80) c Among the many contributions of Pappus I will cite the discovery that the anharmonicratio of 4 points is unchanged after projection, where the anharmonic ratio of A , B , C and D is dist ( C,A ) dist ( D,Bdist ( C,B ) dis ( D,A ) . This is a fundamental property in geometry.The important notion of point at infinity can be traced to Kepler, in 1604, and Desargues,in 1639 (see Heath, I, p. 193). This leads to the notion of the extended Euclidian plane whichcontains besides the ordinary points, the directions, each one is what is what is common tothe set of parallel line, and the set of all directions, or line at infinity. .2. PROJECTIVE GEOMETRY. Introduction.
The extensive study of the conics by Apollonius was eventually taken up again by Pascal.One of his many new results is Theorem 2.2.11 which allows the construction 2.2.11 and1.2.2 of a conic using the ruler only. The second construction is attributed to MacLaurin.But the two constructions are closely related to each other as will be seen. The Theorem ofPascal was generalized to n dimension by Arthur Buchheim in 1984. Notation.
I will introduce in III. ?? detailed notations which allow a compact description of construc-tions. For instance, a := A × A means that the line a is defined as the line through the 2 points A and A . Theorem. [Pascal]
Given the points A , A , A , A , A and A . Let P be the point common to A × A and A × A , let P be the point common to A × A and A × A , let P be the point common to A × A and A × A . A necessary and sufficientcondition for A , A , A , A , A and A to be on the same conic is that P , P and P becollinear. This theorem leads to 2 construction of conics.
Construction.[Pascal]
Given 5 points A , A , A , A and A . To each line through A corresponds a point A onthe conic. a := A × A , a := A × A , a := A × A , a := A × A ,P := a × a , a is an arbitrary line through A ,P := a × a , e := P × P , P := e × a ,a := A × P , A := a × a . Construction. [MacLaurin] If the sides of a triangle pass through three fixed points, and two vertices trace straight lines,the third vertex will trace a conic through two of the given points.The proof follows from Pascal’s Theorem. The construction can be given in the followingexplicit form: A , A , A , A , A are 5 given points.To each line l through P will correspond a point A on the conic. a := A × A , a := A × A , a := A × A , a := A × A ,P := a × a , P := l × a , P := l × a , as stated by Braikenridge CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS a := A × P , a := A × P , A := a × a . The triangle is { P , P , A } , P is on a , P is on a , P × P passes through P , P × A passes through A , P × A passes through A . Comment.
Pascal would not have easily accepted a finite geometry. Indeed in his “Pens´ees”, he says(p. 567),that there are no geometers which do not believe that space is infinitely divisible.Also discussing both the infinitely large and the infinitely small, he writes (p. 564)In one word, whatever the motion, whatever the number, whatever the space,whatever the time, there is always one which is larger and one which is smaller,in such a way they they sustain each other between nothing and infinity, beingalways infinitely removed from those extremes. All these truths cannot be proven,and still they are the foundations and the principles of geometry.
A contemporary of Poncelet, Carnot obtained many results of which the following is in theline of Manelaus and Ceva applied to conics.
Theorem. [Carnot]
If a conic cuts the side A × B of a triangle { A, B, C } at C and C , and similarly the side B × C cut the conic at A and A and the side C × A at B and B , then the orienteddistances satisfy AC .AC .BA .BA .CB .CB = AB .AB .BC .BC .CA .CB This is generalized to curves of degree n . Theorem.
Let A B C be a triangle and X be a point not on its sides,Let A × X meet A × A at X , A × X meet A × A at X and A × X meet A × A at X . Let Y be a point on A × A , Y be a point on A × A and Y be a point on A × A , then a necessary and sufficient condition for X , X , X , Y , Y , Y to be on the same conicis that the lines A × Y , A × Y , A × Y be concurrent. This is a consequence of the Theorem of Carnot.
The work of Poncelet done while a prisoner of Russia at the end of Napoleon’s campaign,was fundamental in isolating those properties of Euclidean geometry which are independent Eves p.358 .3. RELATION BETWEEN PROJECTIVE AND EUCLIDEAN GEOMETRY.
Theorem.
If a n sided polygon is inscribed in a conic and outscribed to an other conic, then if with startfrom any point on the first conic and draw a tangent to the second, then obtain the otherintersection with the first conic and repeat the construction, the new polygon closes after n steps. There are many proofs of this Theorem. The proofs which are done using the theory ofelliptic functions, suggested to me that the Jacobi elliptic functions could be generalized tothe finite case.
Gergonne was the first to recognize the property of duality which plays a fundamental rolein projective geometry. Chasles greatest contribution to projective geometry, according to Coolidge is the study ofthe cross ratio also called anharmonic ratio. Projective geometry is concerned only with those properties in geometry which are preservedunder projection. Euclidean, as well as non Euclidean geometry can be derived from projec-tive geometry. The connection through transformation groups will be described in section1.6.11.The first connection goes back to the work of Poncelet, but it is has been deemphasizedin the teaching of the subject, except for the first step (affine geometry). I will presentlysummarize this approach. Terms which are unknown to the reader, will be defined in thelater Chapters.In projective geometry, no line is distinguished from any other, no point is similarly dis-tinguished. The main notions are those of incidence, perspectivity, projectivity, involution Coxeter, p.13 p.96. See also Coxeter, p.165. G13.TEX [MPAP], September 9, 2019 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS and polarity, the last notion leading naturally to conics. Euclidean geometry can be consid-ered as derived from projective geometry by choosing some elements in it and distinguishingthem from all others. I will proceed in 3 steps.
Introduction.
In this first step one line is distinguished. This line is called the ideal line, or line at infinity.When we do so, we obtain the so called affine geometry. Points fall now into two categories,the ordinary points, which are not on the ideal line and the ideal points which are. Lines fallin two categories, the ideal line and the others which we can call ordinary. From the basicnotion of parallelism follow the derived notions of parallelogram, equality of vectors on thesame line or on parallel lines, trapeze or rhombus, mid-point, barycenter, center of a conic,area of triangles.
Definition.
Two distinct ordinary lines are parallel iff their common point is an ideal point.
Definition. A vector (cid:45) B, C is an ordered pair of points.
Definition.
If the lines B × C and D × E are parallel, the vectors (cid:45) B, C and (cid:45)
D, E are equal iff the lines B × D and C × E are also parallel. Definition. If B, C, D and E are on the same line, the vectors (cid:45) B, C and (cid:45)
D, E are equal iff there exists2 points F and G on a parallel line, such that (cid:45) B, C = (cid:45) F, G and (cid:45)
F, G = (cid:45) D, E.
This definitionhas, of course, to be justified. It can be replaced by: (cid:45)
B, C and (cid:45)
D, E are equal iff thereexists a parabolic projectivity, with the fixed point being the ideal point on the line, whichassociates C to B and E to D. Definition.
The center of a conic is the pole of the ideal line, in the polarity whose fixed points are theconic.
Definition.
Two points are conjugate iff one is on the polar of the other. .3. RELATION BETWEEN PROJECTIVE AND EUCLIDEAN GEOMETRY. Theorem.
Conjugate points on a given line determine an involution.
Definition. A parabola is a conic tangent to the ideal line. The point of tangency is called the directionof the parabola . Example.
The parabola y = 4 cx, in homogeneous coordinates is Y = 4 cXZ. (1)Its intersection with Z = 0 is Y = 0 . The parabola is tangent to Z = 0 at (1,0,0). Definition.
The focus of a parabola is the intersection of the ordinary tangents to the parabola fromthe isotropic points. The directrix of the parabola is the polar of the focus. The axis of theparabola is the line through the focus and the direction of the parabola. The vertex of theparabola is the point of the parabola on its axis.
Example.
The tangent to the parabola at ( X , Y ,
1) is2 c X − Y Y + 2 c X Z = 0 . (2)It passes through the isotropic point (1 , i,
0) if 2 c = Y i, hence because of (1), X = − c. The tangent is therefore X + Y i − c = 0 . The tangent from the other isotropic point is X − Y i − c = 0 . They both intersect at ( c, , . The polar is obtained by substituting in (2) this point for ( X , Y , , this gives X = − c Z. The axis is Y = 0 , the vertex is (0,0,1). Comment.
The terminology can be changed by accepting as points and lines only those which areordinary. An ideal point is renamed a direction. We obtain in this way, something which iscloser to the terminology used by Euclid.
Introduction.
The second step consists in considering the involutions on the ideal line. Among all theinvolutions we can distinguish one of them and call it the fundamental involution. Threecases are possible, the involution may have 2 fixed ideal points, in which case it is calledhyperbolic, one fixed point in which case it is called parabolic and no fixed point, in whichcase it is called elliptic. If we extend the projective geometry to the complex case, these2
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS ideal points then exist, but are not real.The elliptic case, which leads to Euclidean Geometry and the hyperbolic case which leadsto the Geometry of Minkowski can be studied together. The parabolic case, which leads tothe Galilean Geometry is studied separately.Using the fundamental involution, either elliptic or hyperbolic, we can introduce the basicnotion of perpendicularity and from it follow the derived notion of right triangle, rectangle,altitude, orthocenter, circle, equal segment, isosceles and equilateral triangles, center ofcircumcircle, Euler line, circle of Brianchon-Poncelet.In the alternate second step, one involution with 2 real fixed points is distinguished. It isonly if we stay with real projective geometry as opposed to complex projective geometry thatthe hyperbolic involutive geometry is distinct from the elliptic involutive geometry. Stayingwith real projective geometry, the notions which are introduced can be given the same nameas in the elliptic involutive geometry, the definitions may differ slightly, but properties arequite analogous.
Definition.
When the fundamental involution has no real fixed points, I will call the geometry ellipticinvolutive geometry .When the fundamental involution has no real fixed points, I will call the geometry hyperbolicinvolutive geometry . Definition.
The fixed points of the fundamental involution are called isotropic points . Any ordinary linethrough an isotropic point is called an isotropic line . Strictly speaking, the ideal points arethose on the ideal line which are not isotropic, and the ordinary lines are those which arenot isotropic.
Definition.
Two lines are perpendicular iff their ideal points are pairs of the fundamental involution.
Definition.
A conic is a circle iff the involution that the conic determines on the ideal line is the funda-mental involution.
Theorem.
A conic which passes through the 2 isotropic points is a circle .
Definition. A segment [ AB ] is an unordered pair of points. .4. ANALYTIC GEOMETRY. Definition.
The segment [ AB ] and the segment [ CD ] are equal iff the point E constructed in such a waythat ACDE is a parallelogram, is such that E and B are on the same circle centered at A. Definition.
The center of a circle is the intersection of the tangents to the circle at the isotropic points.
Comment.
A geometry could also be constructed in which the correspondence on the ideal line associatesevery point to one of them. This corresponds, using algebra, to the transformation T ( x ) = ax + bcx − a , aa + bc = 0 . This is the parabolic involutive geometry.Before leaving the subject of involutive geometry, I would like to make the followingobservation, which will be useful to understand terminology in non-Euclidean geometry.The step to construct non-Euclidean geometry from projective geometry, which correspondto involutive geometry, is to choose a particular conic as ideal, or set of ideal points. In viewof the fact that a line conic can degenerate in the set of lines passing through either oneor the other of 2 points, we can observe that the ideal in the involutive geometry is such adegenerate conic. This analogy will be pursued to define, using the ideal conic, notions innon-Euclidean geometry which are related to notions of Euclidean geometry and will help inan economy of terminology, but nothing more.
The prime motivation of Descartes when he wrote, “La G´eom´etrie” appears to have been along standing problem, the determination of the locus of Pappus. In present day notation, given lines l i and angles α i , the problem is to determine the locusof a point C and its α i projections U i on l i , such that the angle of C × U i with l i is α i , andfor instance, with i = 0,1,2, 3, such that | CU | | CU | = k | CU | | CU | . (1)Descartes chooses as axis l and u := C × U , he chooses also some orientation whichallows him to associate to the points on these axis, some real number. If x := | U , A | ,y := | C, U | , a i := | A , A i | , i = 1 , , , if X i are the intersection of l i with a , then the pre-scribed angles imply by similarity | U X i || U A i | = b i e , | CU i || CX i | = c i e , for some b i , c i and unit of distance e .The distances | CU i | are linear functions of x and y and therefore replacing in (1) gives theequation of a conic through A . By symmetry, the conic passes through A , B and B . G14.TEX [MPAP], September 9, 2019 p. 8 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Indeed, | CU | = y, , U A = x, U X = b e x, | CX | = | CU | + | U X | = y + b e , CU = ( y + b e ) c e , | U A | = x + a , U X = ( x + a ) b e , CX = y + ( x + a ) b e ,CU = ( y + ( x + a ) b e ) c e , CU = ( y + ( x + a ) b e ) c e . Nowhere, in his work are the axis or arrows on them indicated specifically or are the axischosen at a right angle, except if convenient to solve the problem at hand.
Using modern terminology, the problem posed by Descartes, was to construct an algebraicstructure which is isomorphic to Euclidean geometry. More precisely the problem is to ob-tain algebraic elements P (cid:48) which are in one to one correspondence with points P, algebraicelements l (cid:48) which are in one to one correspondence with lines l, an algebraic relation P (cid:48) · l (cid:48) = 0associated to the incidence relation in geometry, P is on l or l is through P, written P · l = 0 , such that if l (cid:48) corresponds to l and P (cid:48) to P, P (cid:48) · l (cid:48) = 0 if and only if P · l = 0 . Similar correspondences have to be given for perpendicularity, equality of angles and seg-ments, measure of angle and segments, etc. Descartes’ solution is to choose 2 lines xx and yy in the Euclidean plane and to associate, if these are perpendicular, to a point P the 2real numbers x and y which are the distances from P to yy and to xx.P (cid:48) = δ ( P ) = ( x, y ) . This correspondence is not one to one. If x, y (cid:54) = 0 , there are four points which will givethe same pair ( x, y ) . To solve this problem a sign must be associated to the distances, cor-responding to an orientation on the lines xx and yy. Usually, with xx horizontal and yy vertical, x is positive to the right of yy, y is positive above xx. The distance between ( x, y ) and ( x (cid:48) , y (cid:48) ) of the points P and Q is given by (cid:112) ( x (cid:48) − x ) + ( y (cid:48) − y ) . To represent the lines, several choices are possible, one such choice, is the pair [ m, b ] , where b is the (oriented) slope and b it the distance from the intersection of the line with yy, theso called y intercept. In this case, if ( x, y ) corresponds to the point P and [ m, b ] to the line l, ( x, y ) is on [ m, b ] if and only if y = mx + b. Perpendicularity of [ m, b ] and [ n, c ] is defined by m n = − . The difficulty of this representation is that lines perpendicular to xx do not have a (finite)slope. Reversing the role of xx and yy does not help.An other representation of lines that can be chosen, is to take the pair { l , l } of the distances l and l from the origin to the intersection of the line l with xx and yy,l (cid:48) = δ ( l ) = { l , l } , with the incidence property represented by the relation l x + l y − l l = 0 . In particular, the points ( l ,
0) and (0 , l ) are on l (cid:48) . The perpendicularity property of l (cid:48) and m (cid:48) is represented by the bilinear relation l m + l m = 0 . This again is not suitable because this representation fails for lines through the origin. .4. ANALYTIC GEOMETRY. a, b, c ] which areobtained from l , l and l l by multiplication by some arbitrary non zero real k. [ a, b, c ] = k [ l , l , l l ] . For a line through the origin c = 0 , ba is the slope, A line parallel to yy is represented by(1 , , c ) where − c is the x intercept.The incidence property is the familiar linear relation ax + by + c = 0 . But it is important to realize that the correspondence is not one to one. The line is rep-resented by the set of all triples corresponding to all the possible value of k, a so calledequivalence class, the numbers a, b, c are called the homogeneous coordinates of the line.Perpendicularity of [ a, b, c ] and [ a (cid:48) , b (cid:48) , c (cid:48) ] is represented by aa (cid:48) + bb (cid:48) + cc (cid:48) = 0 . By analogy, one could represent points by a triple ( x, y,
1) or by any equivalent set (
X, Y, Z ) = k ( x, y, , k (cid:54) = 0 . This implies that Z = k (cid:54) = 0 and X = kx, Y = ky or x = XZ , y = YZ . The incidence property is then aX + bY + cZ = 0 . (2)( X, Y, Z ) are the so called homogeneous coordinates of an algebraic point.
Poncelet was one of the first to take full advantage of the fact that parallel lines define adirection, which can be called the point at infinity and that all the points at infinity can beconsidered to be on a line, the line at infinity. This constitutes the decisive step towards thedevelopment of projective geometry.The algebraic points (
X, Y, , with X and Y not both 0, correspond to these new geometricpoints, They are all on the line [0,0,1] which is the line at infinity. The distance between thealgebraic points ( X, Y, Z ) and ( X (cid:48) , Y (cid:48) , Z (cid:48) ) , with Z and Z (cid:48) (cid:54) = 0 , is given by (cid:113) ( X (cid:48) Z (cid:48) − XZ ) + ( Y (cid:48) Z (cid:48) − YZ ) . Comment.
The extension of the Euclidian plane by adding the points at infinity and a line at infinityis distinct from the extension of the complex plane, in which to all the points x + iy, x and y real (and i = -1), we add 1 point at infinity. In a complex plane, all lines pass throughthe point at infinity.It is not the place here to review all the other basic formulas of analytic geometry.However, there is an important consequence of the isomorphism between synthetic geometryand analytic geometry, which is implicit in the work of Poncelet and is associated to theproperties of circles, which was the basis of Poncelet’s method to obtain properties for conicsin general.The equation of a circle of center ( a, b ) and radius R is( x − + ( y − b ) = R , or in homogeneous coordinates,( X − aZ ) + ( Y − bZ ) = R Z . The points on the circle and on the line at infinity Z = 0 satisfy6 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS X + Y = 0 , which has no real solution. The introduction of complex numbers, whose use had becomestandard by the time of Poncelet, suggested the definition of a complex analytic geometry,with elements( X, Y, Z ) = k ( X, Y, Z ) , k, X, Y, Z complex, k (cid:54) = 0 and not all X, Y, Z equal to zero,and with elements( a, b, c ) = k (cid:48) ( a, b, c ) , k (cid:48) , a, b, c complex, k (cid:48) (cid:54) = 0 and not all a, b, c equal to zero, Theincidence property being again (1).The complex elements which are not real correspond to new points and lines in syntheticgeometry, the complex points and the complex lines. In this structure, (1 , i,
0) and (1 , − i, Introduction.
Inspired by the paper of Veblen and MacLagan-Wedderburn of 1907, Singer introduced inOctober 1934 (Singer, 1938, Baumert, 1971) the important concept of cyclic difference setswhich allows for an arithmetization of projective geometry which is as close to the syntheticpoint of view as is possible. With this notion, it becomes possible to label points andhyperplanes in N dimensional projective geometry of order p k . With it, in the plane, it isnot only trivial to determine all the points on a line, and lines incident to a point but alsothe lines through 2 points and points on 2 lines.Completely independently, one of my first students at the “Universit´e Laval”, Quebec City,made the important discovery that the regular polyhedra can be used as models for finitegeometries associated with 2, 3 and 5. Then, he introduced the nomenclature of selector(s´electeur) for the notion of cyclic difference sets, to construct an appropriate numberingof the points and lines on the polyhedra. The definition of selector function and selectorcorrelation is implicit in his work.The notion of cyclic difference sets makes duality explicit through the correlation, which isthe polarity when p ≥ , introduced by Fernand Lemay.After defining selector and selector function, I associate with them points and lines in theprojective plane, represented by integers and give Singer’s results which prove the existenceof selectors using the notion of primitive polynomials, 1.4.4.1.4.4 is a special case of what is needed to determine when an irreducible polynomial is aprimitive polynomial. Baumert, p. 101 .4. ANALYTIC GEOMETRY. Definition.
Given a power q = p k of a prime p , a selector or difference set is a subset of q + 1 distinctintegers, such that their q ( q + 1) differences modulo n := q + q + 1 are all of the integersfrom 1 to q + q . Example. [Singer.]
The following are selectors with q = p k :For p = 2 : 0, 1, 3, modulo 7.For p = 3 : 0, 1, 3, 9, modulo 13.For q = 2 : 0, 1, 4, 14, 16, modulo 21.For p = 5 : 0, 1, 3, 8, 12, 18, modulo 31.For p = 7 : 0, 1, 3, 13, 32, 36, 43, 52, modulo 57.For q = 2 : 0, 1, 3, 7, 15, 31, 36, 54, 63, modulo 73.For q = 3 : 0, 1, 3, 9, 27, 49, 56, 61, 77, 81, modulo 91.For q = 11 : 0, 0, 1, 3, 12, 20, 34, 38, 81, 88, 94,104, 109 modulo 133. Theorem. If s i , i = 0 to p is a selector then, for any j , s (cid:48) i = a + ks i + j , is also a selector. The indices are computed modulo q + 1 and the selector numbers, modulo n. Definition. If a = 1 and k = − , the selector s (cid:48) i := 1 − s i is called the complementary selector or co-selector of s i . The selectors obtained using k = 2 , , are called respectively bi-selector , semi-selector . Example.
0. For q = 4 , other selectors are 10, 12, 17, 18, 21 and 0, 1, 6, 8, 18.1. For p = 7 , ifthe selector is 0, 1, 7,24,36,38,49,54,thenthe co-selector is 0, 1, 4, 9,20,22,34,51,the bi-selector is 0, 1, 5,27,34,37,43,45,the semi-selector is 0, 1, 9,11,14,35,39,51. Definition.
The selector function f is the function from Z n to Z n f ( s j − s i ) = s i , i (cid:54) = j. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem. f ( j − i ) − i = f ( i − j ) − j. Example.
For p = 3 , and n = 13, the selector function associated with the selector 0,1,3,9 is i f ( i ) 0 1 0 − − − Definition.
Given a selector, points in the projective plane associated with q = p k , with n = q + q + 1 elements are integers in Z n , and lines are integers in Z n followed by ∗ , with the incidencedefined by a is on b ∗ iff f ( a + b ) = 0 or a + b = 0 . Theorem. a × b = ( f ( b − a ) − a ) ∗ . a ∗ × b ∗ = f ( b − a ) − a. a on b ∗ ⇒ b on a ∗ . The Statements immediately reflect the duality in projective geometry.
Example.
For p = 3 , and the selector 0,1,3,9, the lines and the points on them are lines ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ points on ∗ Theorem.
In the projective geometry associated with q = p k and the selector { s , s , . . . , s q } i + s , i + s , . . . , i + s p are the q + 1 points on the line − i ∗ , the addition being done modulo n = q + q + 1 . Definition. [Singer] P is a primitive polynomial in the Galois Field GF ( p k ) iff P is of degree k and I p k − is thesmallest power of I, modulo P , which is identical to 1. .4. ANALYTIC GEOMETRY. Example. I + I + 1 = 0 is primitive in GF (2 ) . With 2 ≡ I = I + 1, I = I + I, I = I + I + 1, I = I + 1, I = 1.It is well known that Theorem.
A primitive polynomial always exists.
Theorem. P is a primitive polynomial of degree m over the Galois field GF ( q ) , iff P is an irreduciblepolynomial of degree m over GF ( q ) and for a given primitive root ρ of GF ( q m ) , P ( ρ ) = 0 . Theorem. [Singer]
For each value of q = p k , a selector can be obtained by choosing a primitive polynomial ofdegree 3 over GF ( q ) . It is, with 0, the set of exponents of I such that the coefficient of I is0. Example.
For p = 3 , P = I − I + 1 , I = I − , I = I − I, I = I − I + 1 , I = I + I + 1 ,I = I − I − , I = I + 1 , I = I + 1 , I = I + I, I = I + I − , I = I − , I = 1 . Therefore the selector is 0, 1, 3, 9.
Theorem.
Let the primitive polynomial be P := I + bI − c and the generator be G := I + g, let g (cid:48) := 3 g + b, h (cid:48) = g + bg + c, h = h (cid:48) ,let J ( n ) := h − n G − n +1 ∗ G − n , then
0. 0. G = I + 2 gI + g , G = 3 gI + (3 g − b ) I + ( g + c ) , G − = hI − ghI + ( g + b ) h, G − = g (cid:48) h I + (1 − g (cid:48) gh ) hI + ( − gh + g (cid:48) ( g + b ) h ) .
1. 0. G n +3 = 3 gG n +2 − g (cid:48) G n +1 + h (cid:48) G n , G n = h ( g (cid:48) G n +1 − gG n +2 + G n +3 ) .
2. 0. J (0) = I , J (1) = gI + I, J (2) = ( g − b ) I + 2 gI + 1 .
3. 0. J ( n +3) = 3 gJ ( n +2) − g (cid:48) J ( n +1) + h (cid:48) J ( n ) , J ( n ) = h ( g (cid:48) J ( n +1) − gJ ( n +2) + J ( n +3) ) . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
In other words the 4 term recurrence relation is the same for the points associated to G n (1.) as for the lines associated to J ( n ) (3.).Proof: 0.0. is immediate. G = ( I + g ) , or because of P we get 0.1. Eliminating 1, I and I from G , G and G gives G = 3 gG − g (cid:48) G + h (cid:48) . Multiplying by G n gives 1.0 hence 1.1.From this recurrence relation is it easy to get 0.2. and 0.3. J ( n ) := h − n G − n ∗ G − n , this giveseasily 2.0., 2.1, 2.2. We should be careful not to scale.The definition of J ( n ) implies J ( n +3) = h − n − G − n − ∗ G − n − = h − n − G − n − ∗ ( g (cid:48) G − n − − gG − n − + G − n )= 3 gJ ( n +2) − h − n − G − n ∗ G − n − = 3 gJ ( n +2) − h − n − G − n × ( g (cid:48) G − n − − gG − n + G − n +1 )= 3 gJ ( n +2) − g (cid:48) J ( n +1) + h (cid:48) J ( n ) . Example. p = 5 , g = − , b = − , c = 2 , g (cid:48) = 1 , h (cid:48) = 1 , h = 1 ,P = I − I − , G = − G − G + 1 . J (3) = − J (2) − J (1) + 1 .i G i J ( i ) i G i J ( i ) i G i J ( i ) − , ,
2) [3 , ,
1] 9 (2 , ,
3) [0 , ,
2] 20 (4 , ,
4) [3 , , − , ,
3) [4 , ,
1] 10 (0 , ,
3) [3 , ,
0] 21 (4 , ,
0) [4 , , , ,
1) [1 , ,
0] 11 (2 , ,
4) [1 , ,
2] 22 (1 , ,
3) [4 , , , ,
3) [3 , ,
0] 12 (0 , ,
1) [1 , ,
0] 23 (4 , ,
1) [0 , , , ,
4) [0 , ,
1] 13 (3 , ,
3) [1 , ,
3] 24 (4 , ,
1) [0 , , , ,
4) [3 , ,
4] 14 (4 , ,
0) [4 , ,
4] 25 (3 , ,
1) [4 , , , ,
0) [0 , ,
0] 15 (3 , ,
3) [1 , ,
3] 26 (2 , ,
4) [1 , , , ,
0) [2 , ,
2] 16 (1 , ,
0) [1 , ,
1] 27 (4 , ,
1) [0 , , , ,
4) [1 , ,
2] 17 (0 , ,
2) [2 , ,
0] 28 (2 , ,
1) [3 , , , ,
1) [2 , ,
1] 18 (2 , ,
1) [3 , ,
2] 29 (1 , ,
2) [3 , , , ,
0) [4 , ,
4] 19 (4 , ,
2) [1 , ,
4] 30 (1 , ,
3) [4 , ,
1] The selector is0,1,4,10,12,17. Line 1 ∗ is incident to points -1=30, 0,3,9,11 and 16. The first known table of trigonometric functions corresponds crd ( α ) = 2 sin ( α )and to α = 0 to 90 o step 15 (cid:48) , using two sexadesimal places. for instance, crd (36 o ) =2 sin (18 o ) =; 37 , , . (See . . . ).The trigonometric functions were first defined as ratios of the sides of a triangle by Rh¨aticus,who constructed 10 place tables for sin, cos, tan, cot, sec and cosec, in increments of 10 (cid:48)(cid:48) , G15.TEX [MPAP], September 9, 2019 .5. TRIGONOMETRY AND SPHERICAL TRIGONOMETRY. sin, with first second and third difference. They were edited byPiticus.
Lambert gives, in 1770 (I, 190-191), the values of the trigonometric function sine for argu-ments in units π .These require s √ , s √ , s √ , s p = √ s , s m = √ − s . His table can then be rewritten as follows: sin (1) = − s s p + s p + s s s s s − s s − s ,sin (2) = s s s m − s − ,sin (3) = s s s − s m ,sin (4) = s s p − s s s ,sin (5) = s s − s ,sin (6) = s − ,sin (7) = s s m + s m − s s s s s − s s s ,sin (8) = − s s m + s s s ,sin (9) = − s s s s p ,sin (10) = sin (11) = s s p − s p + s s s s s − s s − s ,sin (12) = s s m,sin (13) = − s s m + s m + s s s s s s s s ,sin (14) = s s s p − s ,sin (15) = s ,sin (16) = s s p + s s − s ,sin (17) = s s m + s m + s s s − s s s s − s ,sin (18) = s ,sin (19) = s s p + s p − s s s s s s s − s ,sin (20) = s ,sin (21) = s s − s s p ,sin (22) = s s s m + s ,sin (23) = s s m − s m + s s s s s s s s ,sin (24) = s s p,sin (25) = s s s ,sin (26) = s s s p + s − ,sin (27) = s s s s m ,sin (28) = s s m + s s s ,sin (29) = s s p + s p + s s s − s s − s s s ,sin (30) = 1 . These tables are given here, because they can be used in the case of finite fields forappropriate values of p.
The first appearance of a spherical triangle is in book I of Menelaus’ treatise Sphaerica,known through its translation into Arabic. In it appears the first time a study of spherical2
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS triangles and of the formula for a spherical triangle
ABC with points
L, M, N on the sidescorresponding to IV.. . . ? sin ( AN ) sin ( BL ) sin ( CM ) = − sin ( N B ) sin ( LC ) sin ( M A ) . The law of cosine for a spherical triangle was given by al-Battani, it will be generalized tofinite non-Euclidean geometry in IV.. . . 2.0.The formula, for a spherical right triangle, called Geber’s Theorem, will be generalized inIV . . . 1.1.
Introduction.
This section uses extensively, material learned from Professor George Lemaˆıtre, in his classon Analytical Mechanics, given to first year students in Engineering and in Mathematics andPhysics, University of Louvain, Belgium, 1942. We first determine the differential equationfor the pendulum 6.1.3. using the Theorem of Toricelli 6.1.1. , we then define the ellipticintegral of the first kind and the elliptic functions of Jacobi 6.1.5., we then derive the Landentransformation which relates elliptic functions with different parameters 6.1.10., use it toobtain the Theorem of Gauss which determines the complete elliptic integrals of the firstkind from the arithmetico-geometric mean of its 2 parameters 6.1.14. and obtain the additionformulas for the these functions 6.1.16. using the Theorem of Jacobi on pendular motionswhich differ by their initial condition 6.1.7. We also derive the Theorem of Poncelet on theexistence of infinitely many polynomials inscribed in one conic and circumscribed to another6.1.9. We state, without proof, the results on the imaginary period of the elliptic functions ofJacobi 6.1.19. and 6.1.20. A Theorem of Lagrange is then given which relates identities forspherical trigonometry and those for elliptic function 6.1.23. Finally we state the definitionsand some results on the theta functions. Using this approach, the algebra is considerablysimplified by using geometrical and mechanical considerations.
Theorem. [Toricelli]
If a mass moves in a uniform gravitational field its velocity v is related to its height h by0. v = (cid:112) g ( h − h ) , where g is the gravitational constant and h is a constant, corresponding to the height atwhich the velocity would be 0. Proof: The laws of Newtonian mechanics laws imply the conservation of energy. In thiscase the total energy is the sum of the kinetic energy mv and the potential energy mgh, therefore mv + mgh = mgh , for some h . Definition. A circulatory pendular motion is the motion of a mass m restricted to stay on a verticalfrictionless circular track, whose total energy allows the mass to reach with positive velocity .5. TRIGONOMETRY AND SPHERICAL TRIGONOMETRY. oscilatory pendular motion is one for which the totalenergy is such that the highest point on the circle is not reached. The mass in this caseoscillates back and forth. The following Theorem gives the equation satisfied by a pendularmotion. Theorem.
If a mass m moves on a vertical circle of radius R, with lowest point A, highest point B andcenter O, its position M at time t, can be defined by φ ( t ) = ∠ ( AOM ) which satisfies0. Dφ = (cid:112) a − c sin φ, where1. a := 2 gh R , c = gR , for some h . Proof: If the height is measured from
A,h ( t ) = R − Rcos (2 φ ( t )) = 2 Rsin φ ( t ) , the Theorem of Toricelli gives RD (2 φ )( t ) = v ( t ) = (cid:112) gh − gRsin φ ( t ) , hence 0. The motion is circulatory if h > R or a > c, it is oscilatory if 0 < h , R or c > a. Notation. k := ca , b := a − c , k (cid:48) := ba , Definition. If a = 1 , and we express t in terms of φ ( t ) , t = (cid:82) φ ( t )0 1 √ − k sin . The integral 0. is called the incomplete elliptic integral of the firstkind . Its inverse function φ is usually noted1. am ( t ) , the amplitude function ,The functions2. sn := sin ◦ am, cn := cos ◦ am, dn := √ − k sn , are called the elliptic functions of Jacobi .3. K := (cid:82) π √ − k sin . is called the complete integral of the first kind , it gives half theperiod, Ka , for the circular pendulum. The functions which generalize tan, cosec, . . . are4. ns := sn , nc := cn , nd := dn , sc := sncn , cd := cndn , ds := dnsn , cs := cndn , dc := dncn , sd := sndn . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem.
If 0. s := sn ( t ) , c = cn ( t ) , d = dn ( t ) and1. s := sn ( t ) , c = cn ( t ) , d = dn ( t ) , we have2. sn + cn = 1 , dn + k sn = 1 , dn − k cn = k (cid:48) . − k s s = c + s d = c + s d . Theorem. [Jacobi]
Let M ( t ) describes a pendular motion, Given the circle γ which has the line r at height h as radical axis and is tangent to AM ( t ) , if N ( t ) M ( t ) remains tangent to that circle, then N ( t ) also describes a pendular motion, with N ( t ) = A. Proof: With the abbreviation M = M ( t ) , N = N ( t ) , let N M meets r at D, let M (cid:48) , N (cid:48) be the projections of M and N on r, let T be the point of tangency of M N with γ ,0. DM DN = DT , therefore1. DTND = DMDT = DT − DMND − DT = MTNT = (cid:113) DTND DMDT = (cid:113) DMND = (cid:113) M (cid:48) MN (cid:48) N When t is replaced by t + (cid:15) ,2. v M v N = lim M ( t + (cid:15) ) − M ( t ) N ( t + (cid:15) ) − N ( t ) = lim M ( t ) TN ( t + (cid:15) ) T = MTNT , because the triangles T, M, M ( t + (cid:15) ) and T, N, N ( t + (cid:15) ) are similar, because ∠ ( T, N, N ( t + (cid:15) ) = ∠ ( T, M ( t + (cid:15) ) , M ) as well as ∠ ( M ( t + (cid:15) ) , T, M ) = ∠ ( N ( t + (cid:15) ) , T, N ) . Therefore3. v M v N = (cid:113) M (cid:48) MN (cid:48) N . The Theorem of Toricelli asserts that v M = √ gM (cid:48) M , this implies, as we have just seen, v N = √ gN (cid:48) N , therefore N describes the same pendular motion with a difference in theorigin of the independent variable. Corollary. If M = B and N = A, the line M ( t ) × N ( t ) passes through a fixed point L on the verticalthrough O called point of Landen .Moreover, if b := BL and a := LA, we have v M v N = ba and h = a a − b . This follows at once from from 6.1.7.2. and 6.1.7.1. .5. TRIGONOMETRY AND SPHERICAL TRIGONOMETRY. Theorem. [Poncelet
Given 2 conics θ and γ , if a polygon P i , i = 0 to n, P n = P , is such that P i is on θ and P i × P i +1 is tangent to γ , then there exists infinitely many such polygons.Any such polygon is obtained by choosing Q on θ drawing a tangent Q Q to γ , with Q on θ and successively Q i , such that Q i is on θ and Q i − × Q i is tangent to γ , the Theoremasserts that Q n = Q . The proof follows at once from 6.1.7. after using projections which transform the circle θ and the circle γ into the given conics. The Theorem is satisfied if the circle have 2 pointsin common or not. Theorem. If M ( t ) describes a circular pendular motion, then the mid-point M ( t ) of M ( t ) and M ( t + K ) describes also a circular pendular motion. More precisely, M ( t ) is on a cicle with diameter LO, with LA = a, LB = b, and if φ ( t ) = ∠ ( O, L, M ( t ) , t = (cid:82) φ ( t )0 Dφ ∆ = (cid:82) φ ( t )0 Dφ ∆ . where1. ∆ := a cos φ + b sin φ and ∆ := a cos φ + b sin φ , where the relation between φ and φ is given by2. tan ( φ − φ ) = k (cid:48) tanφ , or3. sin (2 φ − φ ) = k sinφ , with4. a := ( a + b ) , b := √ ab, c := ( a − b ) , therefore5. a = a + c , b = a − c , c = 2 √ a c . Proof: First, it follows from the Theorem of Toricelli that the velocity v A at A and v B at B satisfy v A = √ gh = 2 Ra, v B = √ gh − R = √ R a − c R = 2 Rb, therefore
BLLA = ba . If P is the projection of L on BM and Q the projection of L on AM,LM = LP + LQ = a cos φ + b sin φ = ∆ .LQ = LM cos ( φ − φ ) = acosφ. We can proceed algebraically. Differentiating 2. gives a (1 + tan ( φ − φ ))( Dφ − Dφ ) = b (1 + tan φ ) Dφ, or a (1 + tan ( φ − φ )) Dφ = ( a (1 + tan ( φ − φ ) + b (1 + tan φ )) Dφ = ( a + b + b a tan φ + btan φ ) Dφ = ( a + b )(1 + ba tan φ ) Dφ = ( a + b )(1 + tanφtan ( φ − φ )) Dφ, or acos ( φ − φ ) Dφ = 2 a cos (2 φ − φ ) cosφcos ( φ − φ ) Dφ, or6
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS Dφ acosφcos ( φ − φ ) = Dφ a cos (2 φ − φ ) , or because LM = ∆ Dφ ∆ = Dφ . We can also proceed using kinematics.The velocity at M is v M = 2 RDφ = 2 R ∆ , If we project the velocity vector on a perpendicualr to
LM,LM Dφ = v M cos (2 φ − φ ) = 2 Rcos (2 φ − φ )∆ φ. Therefore Dφ ∆ = Dφ Rcos (2 φ − φ ) = a R Dφ ∆ = Dφ . Definition.
The transformation from φ to φ is called the forward Landen transformation . The trans-formation from φ to φ is called the backward Landen transformation . Comment.
The formulas 3. and 1. are the formulas which are used to compute t from φ ( t ). Theformulas 4. and 2. are used to compute φ ( t ) from t . Theorem. [Gauss]
Given a > b > , let0. a i +1 := ( a i + b i ) , b i +1 := √ a i b i , then the sequence a i and b i have a common limit a ∞ . The sequence a i is monotonicallydecreasing and the sequence b i is monotonically increasing. Proof: Because a i > a , b i +1 > b i , it follows that the sequence a i is bounded below by b , the sequence b i is bounded above by a , therefore both have a limit a ∞ and b ∞ . Taking the limit of 0. gives at once a ∞ = b ∞ . Theorem.
For the complete integrals we have Ka = (cid:82) π √ a cos + b sin = π a ∞ . Proof: If φ ( K ) = π , then φ ( K ) = π , therefore1. K = (cid:82) π Dφ ∆ = (cid:82) π Dφ = (cid:82) π Dφ ∆ + (cid:82) π π Dφ ∆ = (cid:82) π Dφ ∆ = (cid:82) π Dφ n ∆ n = (cid:82) π a ∞ = π a ∞ . .5. TRIGONOMETRY AND SPHERICAL TRIGONOMETRY. Lemma. c = c cn ( t + t ) + d s sn ( t + t ) , d = d dn ( t + t ) + k s c sn ( t + t ) . Proof: We use the Theorem 6.1.7. of Jacobi. Let R be the radius of θ and O its center,let r be the radius of γ and O (cid:48) its center, let s := OO (cid:48) . Let
A, N, M (cid:48) , M be the positionof the mass at time 0, t , t , t + t . The lines A × M (cid:48) and N × M are tangent to the same circle γ at T (cid:48) and T. Let X be the intersection of O × M and O (cid:48) × T, φ := ∠ ( A, O, N ) ,
2. 2 φ (cid:48) := ∠ ( A, O, M ) , we have ∠ ( N, O, M ) = 2( φ (cid:48) − φ ) , ∠ ( M, X, T ) = φ (cid:48) − φ, ∠ ( T, O (cid:48) , O ) = φ (cid:48) + φ .If we project M OO (cid:48) on O (cid:48) T,r = Rcos ( φ (cid:48) − φ ) scos ( φ (cid:48) + φ ) , or3. r = ( R + s ) cosφcosφ (cid:48) + ( R − s ) sinφsinφ (cid:48) .φ = amt , φ (cid:48) = am ( t + t ) ,sinφ (cid:48) = sn ( t + t ) , cosφ (cid:48) = cn ( t + t ) ,sinφ = sn t = s ,cosφ = cn t = c , when t = 0, cos ( ∠ ( A, B, M (cid:48) ) = cn t = c = BM (cid:48) AB = O (cid:48) T (cid:48) AO (cid:48) = rR + s ,the ratio of the velocities is v M (cid:48) v A = dn t dn = d = T M (cid:48) AT = O (cid:48) BAO (cid:48) = R − sR + s , substituting in 2. gives 0.The proof of 1. is left as an exercise. Theorem. [Jacobi] sn u cn u dn u + sn u cn u dn u sn ( u + u ) = 1 − k sn u sn u . cn u cn u − sn u dn u sn u dn u cn ( u + u ) = 1 − k sn u sn u . dn u dn u − k sn u sn u cn u cn u dn ( u + u ) = 1 − k sn u sn u . Proof: Let w = − k s s . Let s , s , . . . denote sn u , sn u , . . . , define S and C such that sn ( u + u ) = Sw, cn ( u + u ) = Cw.
The 6.1.15.0. gives c = c Cw + d s Sw or3. c Cw = − d s Sw + c , S w + C w = 1 , eliminating C gives the second degree equation in Sw :8 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS ( c + d s ( Sw ) − s c d ( Sw ) + c − c = 0 , one quarter of the discriminant is s c d − ( c − c )( c + d s )= s c d − c c + c − s c d + s c d = c ( c − c + s d ) = c s d , therefore Sw = ( s c d ± c d s ) w. One sign correspond to one tangent from M to γ , the other to the other tangent,therefore one corresponds to the addition, the other to the subtration formula. Fromthe special case k = 0 , follows that, by continuity, the + sign should be used. Thisgives 0., 1. follows from 3, 2. is left as an exercise. Corollary. sn ( u + K ) = cd ( u ) , cn ( u + K ) = − k (cid:48) sd ( u ) , dn ( u + K ) = k (cid:48) nd ( u ) . sn ( u + 2 K ) = − sn ( u ) , cn ( u + 2 K ) = − cn ( u ) , dn ( u + 2 K ) = dn ( u ) . sn ( u + 4 K ) = sn ( u ) , cn ( u + 4 K ) = cn ( u ) , dn ( u + 4 K ) = dn ( u ) . Definition. K (cid:48) ( k ) = K ( k (cid:48) ) . Theorem.
0. 0. ksn ◦ I + iK (cid:48) = sn, ikcn ◦ I + iK (cid:48) = ds, idn ◦ I + iK (cid:48) = cs,
1. 0. sn ◦ I + 2 iK (cid:48) = sn, cn ◦ I + 2 iK (cid:48) = − cn, dn ◦ I + 2 iK (cid:48) = − dn, Theorem. sn has periods K and iK (cid:48) and pole ± iK (cid:48) , cn has periods K and iK (cid:48) and pole ± iK (cid:48) , dn has periods K and iK (cid:48) and pole ± iK (cid:48) . Theorem. k = 0 ⇒ sn = sin, cn = cos, dn = 1 , k = 1 ⇒ sn = tanh, cn = sech, dn = sech. .5. TRIGONOMETRY AND SPHERICAL TRIGONOMETRY. Theorem. [Lagrange]
From the addition formulas of elliptic functions we can derive those for a spherical triangleas follows. Let u + u + u = 2 K, define sina := − snu , cosa := − cnu ,sinb := − snu , cosb := − cnu ,sinc := − snu , cosc := − cnu ,sinA := − ksnu , cosA := − dnu ,sinB := − ksnu , cosB := − dnu ,sinC := − ksnu , cosC := − dnu , then to any formula for elliptic functions of u , u , u , corresponds a formula for aspherical triangle with angles A, B, C and sides a, b, c.
For instance, sinAsina = sinBsinb = sinCsinc = k. cosa = cosbcosc + sinbsinccosA, cosA = − cosBcosC + sinBsinCcosa, sinBcotA = cosccosB + sinccota. Proof. 2. follows from the definition. 3. follows from c = c cn ( t + t ) + d s sn ( t + t ) after interchanging t and t and using6. 0. sn ( t + t ) = sn (2 K − t − t ) = sn t = s , cn ( t + t ) = − cn (2 K − t − t ) = − cn t = − c , dn ( t + t ) = dn (2 K − t − t ) = dn t = d , similarly, 4. follows from c = c cn ( t + t ) + d s sn ( t + t )after interchanging t and t and using 6 and 5. from sn t dn t = cn t sn ( t + t ) − sn t dn t cn ( t + t )after division by sn t . Definition.
Given the parameter q, called the nome,0. q := e − π K (cid:48) K , the functions1. θ := 2 q (cid:80) ∞ n =0 ( − n q n ( n +1) sin (2 n + 1) I θ := 2 q (cid:80) ∞ n =0 q n ( n +1) cos (2 n + 1) I θ := 1 + 2 (cid:80) ∞ n =1 q n cos nI θ := 1 + 2 (cid:80) ∞ n =1 ( − n q n cos nI are the theta functions of Jacobi .0 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition.
The functions, with v = π I K ) θ s := Kθ ◦ vDθ (0) , θ c := θ ◦ vθ (0) , θ d := θ ◦ vθ (0) , θ n := θ ◦ vθ (0) , are called the theta functions of Neville . Theorem. If p, q denote any of s, c, d, n,pq = θ p θ q . For instance sn = θ s θ n = Kθ ◦ vDθ (0) . θ (0) θ ◦ v . Theorem.
The Landen transformation replaces the parameter q, by q . References.
Jacobi, Fundamenta Nava Theoriae Funktionum Ellipticarum, 1829.Legendre, Trait´e des fonctions elliptiques et des int´egrales elliptiques. III, 1828.Gauss, Ostwald Klasiker?Landen John, Phil. Trans. 1771, 308.Abel, Oeuvres, 591.Bartky, Numerical Calculation of generalized complete integrals, Rev. of Modern Physics,1938, Vol. 10, 264-Lemaitre G. Calcul des integrales elliptiques, Bull. Ac. Roy. Belge, Classe des Sciences, Vol.33, 1947, 200-211.Fettis, Math. of Comp., 1965.Appell, Cours de Mechanique,
Notes ( Dy ) = C ( y − A )( y + B ) , ( Dz ) = C ( z − A )( z + B ) ,z = d ( y + y ) , l (cid:54) = 0 , d > . The equations are compatible iff (l in the beginning of next expres.?) d (1 − y ) C ( y + A )( y + B ) = C ( d ( y + ly ) + A )( d ( y + ly ) + A )this requires √ l to be a root of one of the factore of the second member, let it be the secondfactor, this implies d l + B = 0 , then, the second factor becomes, d ( y + ly ) + B = d (( y + ly ) − K ) = d ( y − ly ) therefore √ l is a double root of the second memeber and C ( y + ( A + B ) y + A B ) = d C ( y + (2 l + A ) y + l ) , therefore .6. ALGEBRA, MODULAR ARITHMETIC. C = d C , A B = B d , A + B = A − B d , For real transformations, A B > , if j = sign ( B ) and j = sgn ( B ) ,B = 4 j d √ A B , A = d ( A + B + 2 j √ A B = j ( (cid:112) | A | + j j (cid:112) | B | ) . If we want A B ¿0 then j = j . Geometry can be handled synthetically, with little or no reference to algebra. But it wasdiscovered little by little that an underlying algebraic structure lurks behind geometry. Ifwe deal with a geometry with a finite number of points on each line, we have to deal withan underlying algebraic structure which involves a finite number of integers. Such structurepresented itself in connection with application of mathematics to astronomy (and astrology),in studying the relative motion of sun and moon and the relative motion of the planets,mainly Jupiter. If the smallest unit of time used is t, the period of the sun around the earth,is s.t, the position of the sun is the same after 2 revolutions hence 2 s is equivalent to s and2 s + 1 is equivalent to s + 1 as well as 1. This led to the notion of working modulo s. Definition. p is a prime iff p is an integer larger than 1, which is only divisible by 1 and p. Restriction.
In the sequel, it is always assumed that p is odd. Introduction.
Although much of what I will do can be generalized, to the case of powers of primes, I will,for simplicity, restrict myself to the case of a prime p. Definition.
The integers modulo p are the integers x satisfying0 ≤ x < p. The set of these integers is denoted Z p . The operations modulo p are defined in terms ofthe operations on the integers as follows: G16.TEX [MPAP], September 9, 2019 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition.
0. If x and y are integers modulo p, addition modulo p, denoted + p is defined as the leastnon negative remainder of the division of the integer x + y by p. Multiplication modulo p, denoted . p , is defined as the least non negative remainder ofthe division of x.y by p. Subtraction modulo p, denoted − p , is defined as the inverse operation of addition, c + p b = a = ⇒ a − p b = c. Division modulo p, denoted / p , is defined as the inverse operation of addition, c. p b = a = ⇒ a/ p b = c, provided b (cid:54) = 0. Convention.
As I will not use simultaneously 2 different primes, and as it will usually be clear from thecontext that the addition, multiplication, . . . , are done modulo p, I will replace + p by + , . . . . An alternate notation, useful when several different moduli are used, is to use a + b ≡ c (mod p ) . Example.
We have, 0 + , , . Modulo 7: 5 + 4 = 2 , − , . , / , . Modulo 7: the inverses of 1 through 6 are respectively 1 , , , , , . Modulo 11: 9 + 5 = 3 , − , . , / , . . Comment.
Addition, subtraction and multiplication are easy to perform, moreover hand calculatorsand languages for microprocessors have functions which allow easy computations. Divisionrequires either a table of inverses or the inverses can be obtained, for large primes p, usingthe Euclid-Aryabatha algorithm.
10 11
Algorithm. [Euclid]
Let a ≥ b > . We determine in succession a := a, a := b, q , a , q , . . . , a n = 0 (cid:51) . a j − := a j q j + a j +1 , ≤ a j +1 < a j . To appreciate this contribution of the Hindus, Aryabatha lived at the end of the fifth Century, while anequivalent algorithm was only developed in the Western World by Bachet de Meziriac in 1624. Pulverizing a is meant to convey what we would now express by finding the inverse of a modulo n. .6. ALGEBRA, MODULAR ARITHMETIC. Algorithm. (Pulverizer of Aryabatha)
Given q , q , . . . , q n − , determine b n − := 0 , b n − := 1 , . b j − := b j q j + b j +1 , for j = n − , . . . , . Algorithm. (Continued fraction algorithm)
Given q , q , . . . , q n − , determine c := 0 , c := 1 , d := 1 , d := 0 , c j +1 := c j q j + c j − , for j = 1 , . . . , n − . d j +1 := d j q j + d j − , for j = 1 , . . . , n − . Algorithm. u j := c j + d j .v j := c j c j +1 + d j d j +1 . Example.
Let a = 10672 and b = 4147 , ↑ b = ↓ c = d = a = = .
2+ 2378 b = c = d = a = 4147 = 2378 .
1+ 1769 b = 39 c = 2 d = 1 a = 2378 = 1769 .
1+ 609 b = 29 c = 3 d = 1 a = 1769 = 609 .
2+ 551 b = 10 c = 5 d = 2 a = 609 = 551 .
1+ 58 b = 9 c = 13 d = 5 a = 551 = 58 .
9+ 29 b = c = 18 d = 7 a = 58 = 29 .
2+ 0 ↑ b = c = d = a = ↓ c = d = n = , · − ·
68 = 29. u = 35249 , u = 155873 , v = 74124 . The bold-faced number are initial values, the italicized numbers are final values. Notice that all a ’s have to be computed before the b ’s are computed, but this is not so for the c ’s and the d ’s.For instance, for the line starting with a , b = q .b + b = 1 . .c = c .q + c = 3 . , d = d .q + d = 1 . . Observe that 175 and 68 are obtained in 2 different ways.
Definition.
The greatest common divisor of a and b is the largest positive integer which divides a and b ,it is denoted ( a, b ) . Theorem. The algorithm terminates in a finite number of steps. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
1. ( a, b ) = ( a , a ) = ( a , a ) = . . . = ( a n − , a n ) = a n − . b j a j − − b j − a j = ( − n − j ( a, b ) , in particular, b b − b a = ( − n ( a, b ) . b j < a j . in particular, b < b, b < a. Theorem. a /a = q + 1 / ( q + 1 / ( q + . . . + 1 /q n − )) . b < a = ⇒ c i < c i +1 , i = 0 , . . . , n − . a i c i +1 + a i +1 c i = a, a i d i +1 + a i +1 d i = b. b.c i ≡ ( − i +1 a i (mod a ) , a.d i ≡ ( − i a i (mod b ) . If a ≤ b < a then q = 1 and c = c . Definition.
The second member in 1.6.2.0. is called a terminating continued fraction . Theorem. [Symmetry property] If ( a, b ) = 1 , b < a , and we repeat the algorithm with a (cid:48) := a and b (cid:48) := ± b − (mod a ) , b (cid:48) ≤ a , then this algorithm terminates in the same number n of stepsand a (cid:48) j = c n − j , c (cid:48) j = a n − j , q (cid:48) j = q n − j . In particular, if b ≡ − a ) and b < a , then n = 2 n (cid:48) + 1 is odd and c j = a n − j , q j = q n − j and a n (cid:48) + a n (cid:48) +1 = a. Example. i a i q i c i i a i q i c i c (cid:48) j q (cid:48) j a (cid:48) j j + 1 ≡ .c (cid:48) j q (cid:48) j a (cid:48) j j + 4 = 65 . Theorem. [Euler]
Every integer whose prime factors to an odd power are congruent to 1 modulo 4, can bewritten as a sum of 2 squares and vice-versa. .6. ALGEBRA, MODULAR ARITHMETIC. Example.
13 = 2 + 3 ,
52 = 4 + 6 .
585 = 9 + 24 = 12 + 21 . Definition. n is a quadratic residue of p iff there exist a integer x such that x is congruent to n modulo p. We write, with Gauss, n R p. n is a non residue of p, if there are no integer whose squareis congruent to n modulo p, and we write n N p. Theorem.
The product of 2 quadratic residues or of 2 non residues is a quadratic residue. The productof a quadratic residue by a non residue is a non residue.
Theorem. [Fermat] If a is not divisible by p then a p − ≡ p ) . Definition. g is a primitive root of p iff a i ≡ p ) and 0 < i < p = ⇒ i = p − . In other words, p − g which is congruent to 1 modulo p . Notation. [Euler] φ ( n ) denotes the number of integers betweem 1 and p, relatively prime to p . Theorem. [Gauss] There are φ ( p − primitive roots of p . If g is a primitive root of p , all primitive roots are g i with ( i, p −
1) = 1 . Example.
For p = 13 , i g i − − − − − − = 6, 2 = − = − . The easiest method to obtain all inverses moudo p is to first obtain a primitive root andthen to use g i . ( g p − − i ) − = 1.6 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem. If δ is a primitive root of p, the square root of an integer can be unambiguously defined if wechose a particular primitive root. It is sufficient to choose a or aδ , with 0 ≤ a < p − . Examples.
Modulo 5, δ can be chosen equal to 2 or 3, with δ = 3 , we have i √ i δ δ . Modulo 7, δ = 3 can be chosen equal to 3 or 5, with δ = 5 , we have i √ i δ δ δ Theorem. p ≡ ⇔ − p and p odd.1. p ≡ , − ⇔ p and p odd. Theorem. √ is rational in the field Z . This follows at once from the following figure and the fact that the mid-point of thesegment joining (1 ,
0) to (0 ,
1) is ( − , −
8) when p = 17 . This figure originates with thegeometric construction corresponding to the proof by the school of Pythagoras that there isno rational number whose square is 2. In fact, √ ± , when p = 17 . In the case of realnumbers, a corresponding figure corresponds to the geometric interpretation of the classicalproof of the irrationality of √
2, the squares becoming smaller and smaller. I suggest that thereader reflects on this, from a geometric point of view, together with the atomic structureof our Universe. (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) (cid:0)(cid:0)(cid:0)(cid:0)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64)(cid:64) (cid:64)(cid:64)(cid:64)(cid:64) (cid:113)(cid:113) (cid:113) (-8,-8) (8,8) .6. ALGEBRA, MODULAR ARITHMETIC. Introduction.
There has been, historically, a constant interplay between geometry and diophantine equa-tions, the former suggesting problems of the latter kind which also indicate the interest ofhaving problems in geometry solved using integers only. As evidence I will give just one suchproblem considered by Euler . Definition.
The median problem consists in constructing a triangle with integer sides and medians.
Theorem. If a i are the length of the sides and g i are twice the length of the medians, then
0. 2 a i +1 + 2 a i − = a i + g i . Proof: a + a − a a cos ( A ) = a , a + a − a a cos ( A ) = g , eliminating the terms involving the angle gives2 a + 2 a = a + g . Theorem. [Euler]
The solution of the preceding problem can be expressed in terms of 2 parameters a and b ,using C = (4 ab ) , D = (9 a + b )( a + b ), F = 2(3 a + b )(3 a − b ), a = 2 a ( D − F ) , a + a = 2 a ( C + D ) , a − a = 2 b ( C − D ) ,g = 2 b ( D + F ) , g + g = 6 a ( C − D ) , g − g = 2 b ( C + D ) , Theorem. [Euler]
An other solution can be obtained corresponding to a (cid:48) = b and b (cid:48) = 3 a for which a (cid:48) i = g i and g (cid:48) i = 3 a i . An example is provided with the pair (1,2) giving the pair (2,3) in the Example. In factwe have the following
Theorem.
If both a and b are not divisible by 3, then | g i and 3 does not divide a therefore the precedingTheorems gives a solution and for this solution b (cid:48) is divisible by 3. Indeed, a ≡ b ≡ , C ≡ F ≡ , D ≡ − , a ≡ − a , − a ≡ a ≡ b, g i ≡ . It is thereforesufficient, if we use Theorem 1.6.4 to consider all pairs in which one of the integers in thepair is divisible by 3. Similarly we have the following Theorem. Opera Minora Collecta, II, (1778) 294-301, (1779) 362-365, (1782) 488-491 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem. If a is not divisible by 3 and b is divisible by 9, then | a i , | /a i , | g i and the solution is thesame as that obtained from a (cid:48) = b/ and b (cid:48) = a . Proof:If 9 | b , then modulo 27, C ≡ , D ≡ F ≡ , therefore 9 | a (0) but 27 | /a (0) while 27 | g ( i ). Example.
The solutions for the pairs ( a, b ) =(1,3), (1,2), (2,3), (1,6), (3,1), (3,5) are given by Euler.Except for the pair (1,2), They are ordered by increasing maximum values of a i . a ∗ ∗ b a a a g g g a ∗ ∗ ∗ b a a a g g g ∗ are solutions only in a geometry with complex coordinates because a i +1 + a i − < a i for some i . The other degenerate solutions are obtained by observing that,in Euler’s proof, other solutions are obtained when b = a or 9 a . Introduction.
The basic idea is the following, a subset T ( n ), n >
0, of the set Z p can be placed into one toone correspondance with the set H n of irreducible rationals whose numerator, in modulus,and denominator are not larger than n , provided 2 n − n + 1 < p . The ordering ≤ in H n ∈ Q induces an ordering in T ( n ) such that a ≤ b and b ≤ c = ⇒ a ≤ c and − b ≤ − a .If, morever, 0 ≤ a then b − ≤ a − . If order is to be preserved, when we do one additionor one multiplication, we have to use T ( n ) := T ( n (cid:48) ) instead of T ( n ), with n = 2 n (cid:48) . Thisinsures that the sum or product of 2 elements in T ( n ) is in T ( n ). T , T are defined as thesets T ( n ), T ( n ) corresponding to the largest n . This can be repeated for a finite numberof additions and multiplications provided p is large enough. .6. ALGEBRA, MODULAR ARITHMETIC. H n is related to the Farey set F n which is its subset in [0,1]. Farey sets have been used, forinstance, by my colleague and friend Professor R. Sherman Lehman to factor medium sizednumbers.The cardinality of the partially ordered set is estimated in 1.6.5.The complement ( Z p − H n − {±√− } ) can be partitioned into 4 sets (cid:15) , − (cid:15) , λ and − λ whichmight play the role of the sets of smallest elements and the sets of largest elements as given in1.6.5 to 1.6.5.Given an integer k , we can determine the corresponding irreducible rationals,or in which of the small or large set k belongs, using algorithm 1.6.5, which depends on thesymmetry Theorem 1.6.2. We end by contrasting with the notion of continuity in the set ofreal numbers. Definition. A Farey set F n is the set of irreducible rationals a i b i , in ascending order, between 0 and 1,whose numerator and denominator do not exceed n. A Haros set H n is the set of irreducible rationals a i b i , in ascending order, between − n and n ,whose numerator, in modulus, and denominator do not exceed n. Theorem. [Haros] If a i b i and a i +1 b i +1 are any 2 successive rationals of a Farey set F n , then a i +1 b i − a i b i +1 = 1 .
1. The numerators and denominators of 2 successive rationals are relatively prime.2. a i b i = a i − + a i +1 b i − + b i +1 . The set F n can be constructed starting from and by inserting rationals using formula while the resulting numerators and denominators of the second member are not largerthan n. For a proof see Hardy and Wright, p. 23 to 26.The set H n can be deduced from F n , by multiplicative symmetry with respect to 1 and thenby additive symmetry with respect to 0. It can also be obtained from − n and n using formula2, but reduction is required and the termination condition is not as simple as for the set F n . Definition.
A set S is partially ordered by ≤ iff, with a, b, c ∈ S ,0. a ≤ a for all a in S, a ≤ b and b ≤ a = ⇒ a = b. a ≤ b and b ≤ c = ⇒ a ≤ c. But, for any 2 distinct elements a and b in S, we need not have a ≤ b or b ≤ a. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Notation. a < b if a ≤ b and a (cid:54) = b. Definition.
We define the set T ( n ) by:0. The set T ( n ) := { a i b i , a i and b i relatively prime, | a i | ≤ n, < b i ≤ n } . Theorem. If < n and n ( n −
1) + 1 < p or equivalently if < n < √ p − ,0. there is a bijection between the irreducible rationals in H n and the elements in thesubset T ( n ) ∈ Z p . If the order in T ( n ) is that induced by the order in H n ∈ Q and if x, y ∈ T ( n ) , then the set H n is partialy ordered, x < y ⇒ − y < − x,
2. 0 < x < y ⇒ < /y < /x. Proof: It is sufficient to prove, that under the given hypothesis, if a i b i and a j b j are any2 distinct elements in Q, they correspond to distinct elements of T ( n ) in Z p . Indeed, if rs ≡ tu then ru − ts ≡ p, but | ru − ts | ≤ n + ( n − = 2 n ( n −
1) + 1 < p, hence, by hypothesis, rs = tu . The bound cannot be improved for T , because, nn − ≡ − n − n if n + ( n − = p, whose positive root is √ p − , and the sequence of primes of the form m +12 is infinite. Definition.
For a given p, let n p be the largest positive integer such that0. 2 n p ( n p −
1) + 1 < p, then0. T := T ( n p ) , T := T ([ (cid:113) n p ]) . Theorem. If x, y, x (cid:48) , y (cid:48) ∈ T , x.y, x + y ∈ T ,1. 0 < x (cid:48) , x < y ⇒ x.x (cid:48) < y.x (cid:48) . x ≤ y, x (cid:48) ≤ y (cid:48) ⇒ x + x (cid:48) ≤ y + y (cid:48) . .6. ALGEBRA, MODULAR ARITHMETIC. | a | , | b | , | c | , | d | ≤ m := [ (cid:113) n p ] then | ad + bc | ≤ m , | ac | ≤ m , and | bd | ≤ m , therefore if x, y, x (cid:48) , y (cid:48) ∈ T , x + x (cid:48) and y + y (cid:48) ∈ T (2 m ) = T ( n p ) = T . Of course, for multiplication only, we could replace 2 m by m . Example.
In this, and in other examples, I have chosen as representative of an element in Z p , thatwhich is in modulus less than p .0. For p = 31 , n = 4 , T = T (4) is − < − < − < < < − < < − < < < − < < < − < − < < − < < − < − < < < . Indeed, the Farey set F is < < < < < < , the values in Z are0 < < − < − < < − < , their inverses are4 > > > − > − > .T = T (1) is − < < . For one multiplication we could use T (cid:48) = T (2) (2 = 4) which is − < − < < < − < < . For p = 617 , n = 18 , the positive elements of T are240 < − < < < − < < − < − < − < − < < − < − < < < < − < − < < < − < − < < < − < − < − < − < − < − < < − < < < − < − < − < − < − < − < − < < − < − < < < − < − < < − < − < < − < < < − < − < < < − < < < < < < < < − < − < < − < < < < < < < − < − < < < − < − < < < − < − < < < < − < < < < < − < < − < − < < − < < − < < < − < < − < − < − < − < − < − < − < < < − < < − < − < < − < − < − < < − < < − < − < − < − < − < − < < − < − < < − < < < < < − < − < < < < − < − < < < < < − < − < < < − < − < < − < − < − < < < − < < − < − < < < − < < − < − < < − < < < − < < − < − < < < − < − < < − < < − < < − < < < < < < < < < < . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
The positive elements in T = T (3) are206 < − < − < < − < < . Theorem (Mertens). (cid:80) nb =1 ( φ ( b )) = n π + O ( nlog ( n )) , where the last notation implies that the error divided by nlog ( n ) is bounded as n tends toinfinity. Theorem.
The number of terms in T is of the order of π p + O ( p log ( p )) , or approximately . p. This follows at once from the fact that the number of irreducible rationals with denomi-nator b is φ ( b ) , from T = 4 (cid:80) nb =2 ( φ ( b )) + 3 from p = 2 n + O ( n ), from φ (1) = 1 and fromthe Theorem of Mertens.For p = 31 ,
23 = . p, for p = 617 ,
405 = . p. The following Theorem gives a method to determine if a given integer in Z p is in T . Algorithm. [Modified continued fraction]
Given a := p, let n := n p , < a := a < p , c := 0 , d = 1 , c := 1 , d := 0 , i := 1l: q i := a i − /a i , a i +1 = a i − − a i q i ,c i +1 = c i − + c i q i , d i +1 = d i − + d i q i ,if a i +1 ≥ c i +1 then begin i := i + 1; goto l end,if a i < n then a ≡ ( − i a i +1 c i +1 (mod p ) ∈ T ,if c i +1 < n then a ≡ ( − i c i a i (mod p ) ∈ T ,if a i a i +1 > c i c i +1 then a ∈ ( − i +1 λ,if a i a i +1 < c i c i +1 then a ∈ ( − i (cid:15),if a i a i +1 = c i c i +1 then a = − /a . i is therefore the largest index for which a i ≥ c i .We observe that if we start with a (cid:48) := a and b (cid:48) := ± b − (mod a ), the + sign is to be chosenwhen n is even, and that by the symmetry property, when the algorithm stops, c (cid:48) j ≥ a (cid:48) j , c (cid:48) j +1 < a (cid:48) j +1 , therefore j = i + 1 and we have consistent conditions. Example.
For p = 31 , n = 4 ,
0. if a = 14 , the continued fraction algorithm gives .6. ALGEBRA, MODULAR ARITHMETIC. i a i q i c i i a i q i c i i a i q i c i c (cid:48) j q (cid:48) j a (cid:48) j j c (cid:48) j q (cid:48) j a (cid:48) j j c (cid:48) j q (cid:48) j a (cid:48) j j
1. For a = 14 , i = 2 , . − . − , | − | ≤ , ≡ − (mod 31) , which is in T .
2. For a = 12 , i = 3 , . − . > , ≡ (mod 31) , which is not in T . But12 ∈ − (cid:15) and 13 ∈ − λ .3. For a = 6 , i = 1 , . − . − , | − | > , ≡ − (mod 31) , which is in T . But6 ∈ λ and 5 ∈ − (cid:15) .4. In conclusion, λ = { , − } , (cid:15) = {− , − } . Example.
For p = 617 , the elements in λ and, below them, their inverse in (cid:15) , are given below. Thosein − λ and − (cid:15) are obtained by replacing x by − x. λ : 19 20 21 23 24 25 26 28 29 30 32 53 58 (cid:15) : 65 216 -235 161 180 -74 -261 -22 -234 144 135 163 -117 λ : 64 80 85 91 92 -98 -99 106 107 -118 -119 128 129 (cid:15) : -241 54 -196 278 -114 -170 -268 -227 173 183 -140 188 -110 λ : 159 160 162 187 -195 -197 -198 -199 213 214 215 -228 -229 (cid:15) : 260 27 -179 33 -212 -166 -134 31 -84 -222 -66 46 -97 λ : -242 -243 251 252 259 -272 277 -293 -294 -296 -297 -298 -299 (cid:15) : 283 -292 59 -71 81 93 -49 219 149 -148 -295 147 130 Definition.
Let x ∈| T , let a i and c i be defined as in 1.6.5, with b replaced by x and let a i +1 and c i +1 be the next pair, let a (cid:48) i and c (cid:48) i , a (cid:48) i +1 and c (cid:48) i +1 , be the corresponding quadruple for b (cid:48) := ± x − , the sign so chosen that b (cid:48) < a/
2, if0. c (cid:48) i +1 = a i , a (cid:48) i +1 = c i , c (cid:48) i = a i +1 , a (cid:48) i = c i +1 , then1. 0. a i a i +1 < c i c i +1 and i even ⇒ x ∈ (cid:15) ,1. a i a i +1 < c i c i +1 and i odd ⇒ x ∈ − (cid:15) ,2. 0. a i a i +1 > c i c i +1 and i + 1 even ⇒ x ∈ λ ,1. a i a i +1 > c i c i +1 and i + 1 odd ⇒ x ∈ − λ .and we have the partial ordering of these sets by << , − λ << − (cid:15) << << (cid:15) << λ. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem. x ∈ (cid:15) ⇒ − x ∈ − (cid:15), /x ∈ λ, − /x ∈ − λ. For a given p , all integers in the set [0 , p − are either in the set T ( n p ) or in one ofthe sets (cid:15) , − (cid:15), λ, or − λ, with the exception of ±√− when p ≡ − . We leave the proof as an exercise.
Theorem.
If for all (cid:15) ∈ (0 , (cid:15) ) ∃ δ ( (cid:15) ) > (cid:51) x − δ ( (cid:15) ) < x < x + δ ( (cid:15) ) ⇒ | f ( x ) − f ( x ) | < (cid:15) , then f is continuous at x . Indeed for the continuity criterium, we can choose δ ( (cid:15) ) = δ ( (cid:15) ) for (cid:15) ≥ (cid:15) . Comment.
The preceding Theorem is implicit in most text. In the older texts, it is alluded to by addingin the definition of continuity the phrase “however small is (cid:15) ”. If we choose (cid:15) = 10 − , say,and assume that for a given f and x , the hypothesis of the preceding Theorem is satisfied,it follows that the continuity at x depends only on the value of the function in the interval( x − − , x + 10 − ) . If we now try to give an example from the world we live in, nomeaning can be given to physical objects which have distances from each other less than (cid:15) . The definition of continuity gives therefore problems of interpretation in Atomic Physics.The same is true is Cosmology when the distances are of the order of the dimension of theUniverse. Continuity requires the notion of ordered set. We need to apply the more generalconcept of partialy ordered set, to allow for a criterium which test values which are small,but not too small, or large but not too large. This is what is achieved using Farey sets.
Introduction.
Hamilton introduced the notion of quaternions, to try to generalize the notion of complexnumber for application to 3 dimensional geometry.The elements are of the form a + b i + c j + d k , with a, b, c, d not all zero, with j . k = − k . j = i , k . i = − i . k = j , i . j = − j . i = k , i . i = j . j = k . k = 0 , and real numbers commute with i , j , k , addition of quaternions is commutative and isdistributive over multiplication. Definition.
Given a prime p and a non quadratic residue d, the set of complex integers C p is the set a + bδ, a, b ∈ Z p , δ = d. The operations are those of addition ,( a + b δ ) + ( a + b δ ) = a + b δ, where a := a + a (mod p ) , b := b + b (mod p ) . .6. ALGEBRA, MODULAR ARITHMETIC. multiplication ,( a + b δ ) . ( a + b δ ) = a + b δ, where a := a .a + b .b .d (mod p ) , b := a .b + a .b (mod p ) . This is entirely similar to the introduction of complex numbers, δ = d replacing i = − . Example.
For p = 5 and d = 2 , (1 + δ ) + (1 + 3 δ ) = 2 + 4 δ, (1 + δ ) . (1 + 3 δ ) = 2 + 4 δ. Definition. A quaternion integer is a quaternion with coefficients in Z p . Theorem. If p ≡ , , the quaternion integers are isomorphic to 2 by 2 matrices over Z p . The isomorphisms is deduced from the correspondance ∼ (cid:18) (cid:19) , i ∼ (cid:18) bb − (cid:19) , j ∼ (cid:18) − b b (cid:19) , k ∼ (cid:18) − (cid:19) , with b = −
2. For instance, for p = 11 , b = 3, for p = 17 , b = 7. Theorem. The quaternions form a skew field (or division ring). The quaternion integers form a non commutative ring with unity for which if a rightinverse exists then it is also a left inverse.
Definition. A loop ( L, +) is a non empty set of elements L together with a binary operation “ + ” suchthat, if l , l , l , are elements in L ,0. l + l is a well defined element of L.
1. There exists a neutral element e ∈ L, such that e + l = l + e = l . l + x = l has a unique solution x ∈ L , denoted x = l (cid:96) l (or x = l \ l , for ( L, . )),3. y + l = l has a unique solution y ∈ L denoted y = l (cid:97) l , (or y = l / l , for ( L, . ))6
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition. A group ( G, . ) is a non empty set of elements G together with an operation . such that0. If g and g are any elements of G, g .g is a well defined element of G.
1. The operation is associative , or for any elements g , g , g of G, ( g .g ) .g = g . ( g .g ) .
2. There exists a neutral element e in G, such that for all elements g ∈ G, e.g = g.e = g.
3. Every element g of G has an inverse, written g − , such that g.g − = g − .g = e. Notation.
If the operation is noted + instead of . , the neutral element is called a zero and is noted 0. Comment. ( G, +) or ( G, . ) is often abbreviated as G, if the operation is clear from the context. Theorem.
In a group, the neutral element is unique and in element has only one inverse.
Definition.
A group ( G, +) is abelian or commutative iff for every element g and g of G,g .g = g .g . Notation.
In a group ( G, +) , we define0. g = e, .g = g, ( n + 1) .g = n.g + g and ( − n ) .g = − ( n.g ) where n is any positive integer.In a group ( G, . ) , we use instead of 0 .g, .g and n.g, g g and g n , where n is any positive ornegative integer. Definition. A cyclic group ( G, . ) is a group for which there exist an element g, called a generator of thegroup such that every element if G is of the form g n . ( n.g if the operation is +). .6. ALGEBRA, MODULAR ARITHMETIC. Examples.
0. ( Z, +) is a cyclic group, 1 and − Z p , +) , p prime, is a cyclic group, every element different from 0 is a generator.2. ( Z n , +) , n composite, is an abelian group which is not cyclic.3. ( Z p − { } , . ) , p prime, is a cyclic group, any primitive root is a generator. Definition. A Veblem-Wederburn system (Σ , + , · ), is a set Σ, containing at least the elements 0 and 1which is such that for a, b , c ∈ Σ,0. Σ is closed under the binary operations “ + ” and ” · ” ,1. (Σ,+) is an abelian group,
2. (Σ − { } , · ) is a loop,
3. ( a + b ) · c = a · c + b · c ,4. a · , is right distributive, ( a + b ) · c = a · c + b · c .6. a (cid:54) = b ⇒ x · a = x · b + c has a unique solution. Definition. A division ring is a Veblen-Wederburn system which is left distributive. Definition. A alternative division ring is a division ring for which for all elements a (cid:54) = 0, the right inverse a R and left inverse a L are equal, so that we can write it as a − and such that for all b in theset ( a · b ) · b − = b − · ( b · a ) = a. Theorem.
In an alternative division ring ,( b · a ) · a = b · a , a · ( a · b ) = a · b. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition.
The
Cayley numbers or octaves consist of ( p + qe , + , · ), with (see also Stevenson p. 379)0. p and q are quaternions over the reals,1. ( p + qe ) + ( p (cid:48) + q (cid:48) e ) = ( p + p (cid:48) ) + ( q + q (cid:48) ) e ,
2. ( p + qe ) · ( p (cid:48) + q (cid:48) e ) = ( pp (cid:48) − q (cid:48) q + q (cid:48) p + qp (cid:48) e , Comment.
With l and l (cid:48) denoting i or j or k , e · l = − l · e = − le , l = ( l · e ) = − , e · ( le ) = − ( le ) · e = l , ( le ) · l (cid:48) = − ( ll (cid:48) e ) , l (cid:54) = l (cid:48) ⇒ ( le ) · ( l (cid:48) e ) = − ll (cid:48) . Definition.
The conjugate of an octave o = p + qe is defined by o = p − qe , the norm of an octave is defined by N ( o ) = o · o . Theorem. If o = p + qe , then0. o = o , o · o (cid:48) = o (cid:48) · o , N ( o ) = N ( o ) = N ( p ) + N ( q ) , N ( o ) = 0 iff o = 0 , N ( o · o (cid:48) ) = N ( o ) N ( o (cid:48) ) . Theorem. The octaves is an alternative division ring which is non associative.
For instance, ( i · j ) · e = ke , and i · ( j · e ) = − ke . .6. ALGEBRA, MODULAR ARITHMETIC. Definition. A ternary ring (Σ , ∗ ) is a set of elements Σ with at least 2 distinct elements 0 and 1, togetherwith an ternary operation “ ∗ ” such that if a , a , a , a are elements in Σ, then0. a ∗ a ∗ a is a well defined element of Σ,1. a ∗ ∗ a = a ,
2. 0 ∗ a ∗ a = a ,
3. 1 ∗ a ∗ a , a ∗ ∗ a , a (cid:54) = a ⇒ x ∗ a ∗ a = x ∗ a ∗ a , has a unique solution x ∈ Σ , a ∗ a ∗ y = a has a unique solution y ∈ Σ , a (cid:54) = a ⇒ a ∗ x ∗ y = a and a ∗ x ∗ y = a have a unique solution ( x, y ), x ∈ Σ ,y ∈ Σ . Theorem. a (cid:54) = 0 ⇒ ∃ a R (cid:51) a · a R = 1 , a R is called the right inverse of a .1. a (cid:54) = 0 ⇒ ∃ a L (cid:51) a L · a = 1 , a L is called the left inverse of a . Definition.
The addition in a ternary ring is defined by a + b := a ∗ ∗ b, the multiplication in a ternary ring is defined by a · b := a ∗ b ∗ . Theorem.
In a ternary ring (Σ , ∗ ),0. (Σ,+) is a loop with neutral element 0.1. (Σ − { } , · ) is a loop with neutral element 1 and a · · a = 0 . .0 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
The approach which has dominated the non axiomatic study of geometry during the last onehundred years has been influenced, almost exclusively , by the celebrated Inaugural addressgiven by Felix Klein, when he became Professor of the Faculty of Philosophy of Universityof Erlangen and a member of its senate in 1872. In it , Klein states that Geometriesare characterized by a subgroup of the projective group, with, for instance, the group ofcongruences characterizing the Euclidean Geometry. The success of this approach to thestudy of Geometry has been such that in may very well have led to the decline of thesynthetic Research and Teaching. It is hopped that this work, with its underlying program,which I call the Berkeley program, will revitalize the subject from the high school level on. Definition. A function f from a set D to a set R is a set of ordered pairs ( d , r ) , d in D, r in R, suchthat if to pairs have the same first elements, they have the same second element. We write r = f ( d ) . Definition.
The domain of a function is the set D (cid:48) which is the union of all the first elements of thepairs, the range of a function is the set R (cid:48) which is the union of all the second pairs. Definition. A function is one to one or bijective iff for every pair ( d , r ) ( d , r ) such that r = r then d = d . Theorem.
If a function is one to one, the set of pairs ( r , d ) is a function f − from R to D. Definition.
The function f − is called the inverse of f. Definition.
Given 2 functions f and g such that the domain of g is a subset of the range of f, the composition g ◦ f is the function ( d , g ( d )) . Diedonn´e characterizes it as a “ligne de partage des eaux” in the reedition of the French translation Abhandlungen, p.460-497 .7. THE REAL NUMBERS. Theorem.
The composition is associative. In other words, ( h ◦ g ) ◦ f = h ◦ ( g ◦ f ) . One of the most extensive type of problems in Euclidean Geometry is the constructibilityof geometric figures using the ruler and the compass. The construction of regular poly-gons lead Gauss, in his celebrated Disquisitiones Arithmeticae of 1801, to the study of rootsof cyclotomic polynomials and his discovery that the regular polygon with 17 sides is soconstructible. More generally, this is the case whenever the number of sides has the form2 n (cid:81) F i j j , where the F j ’s are Fermat primes (of the form 2 k + 1) . In so doing Gauss intro-duced, for the special case of cyclotomic equations, the method, which could be describedas baby Galois Theory, which was generalized by Galois to the case of general polynomialequations. But in his case Gauss gives explicitely the various subgroups required to analyzecompletely the solution to the problem.The general problem of constructibility has been extensively studied, I will mention onlyhere the work Emile Lemoine (1902), of Henri Lebesgue (1950) and of A. S. Smogorzhevskii(1961). In finite geometry, it would appear at first that the ruler is sufficient for all con-structions because any point in the plane can be obtained from 4 points forming a completequadrangle. But this interpretation should be rejected in favor of that which implies thatthe construction of geometric figures should be given completely independently of the prime,or power of prime, which characterizes the finite Euclidean geometry. The impression isgiven that with the ruler very little can be constructed. One of the consequences of theresults of Chapter 3, is to demonstrate, that both in the finite and classical case many morepoints, lines, circles, . . . can be constructed with the ruler than heretofore assumed, and thatit is a useful exercise to reduce the problem of construction with the compass to that of afew points obtained with it and then the ruler alone. This is pursued extensively startingwith the construction, first of the center of the inscribed circle. Of note is that the circle ofApollonius can be constructed with the ruler alone. In his Introduction to the History of Mathematics, Eves ascribed the beginning the arith-metization of analysis by Weierstrass and his followers to the problem presented by the Gauss gives, in n. 366, the polygons with number of sides less than 300, constructible with rule andcompass, namely, 2 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . G17.TEX [MPAP], September 9, 2019 p.426 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS existence (Riemann, 1874) of a continuous curve having no tangents at any of its points andthat (Riemann) of a function which is continuous for all irrational values and discontinuousfor all rational values in its domain of definition.
Introduction.
We have seen that the Pythagoreans discovered that if we want any circle centered at theorigin and passing to a point with rational coordinates to intersect always the x axis, ir-rationals have to be introduced. If that was all that was desired, it would be sufficient toconstruct first the extension field Q ( √
2) = { u + v √ } , where u and v are in Q then Q ( √ √
5) = { u v √ } , where u v Q ( √ , . . . . The successive integers 5, 13, 17 are all the primescongruent to 1 modulo 4, because of the result of Euler . . . and because all we would needwas to obtain the square root of integers which can be written as a sum of 2 squares.But, in fact we would like that circles centered at the origin, through a point with coordi-nates in one of these extension fields also intersect the x axis at a number in our system.This requires the introduction of algebraic numbers: Definition. An algebraic number is one which can be obtained as the real solution of a polynomial withinteger coefficients.A transcendental number is a real number which is not algebraic. Example. √ x − , An outstanding problems of the last part of the 19-th century was the following, is π , whichis the limit of the ratio of the length of a regular polygon with n sides to the diameter,algebraic or not.The proof that it was not algebraic was first give by Lindemann in 1882, using an earlierresult of Hermite of 1973, that e is not algebraic. This section uses extensively, material learned from George Lemaˆıtre, in his class on Analyt-ical Mechanics, given to first year students in Engineering and in Mathematics and Physics,University of Louvain, Belgium, 1942 and from de la Vall´ee Poussin in his class on elliptic .8. THE PENDULUM AND THE ELLIPTIC FUNCTIONS.
Theorem. [Toricelli]
If a mass moves in a uniform gravitational field, its velocity v is related to its height h by v = (cid:112) g ( h − h ) , where g is the gravitational constant and h is a constant, corresponding to the height atwhich the velocity would be 0. Proof: The laws of Newtonian mechanics laws imply the conservation of energy. In thiscase the total energy is the sum of the kinetic energy mv and the potential energy mgh, therefore mv + mgh = mgh , for some h . Definition. A circulatory pendular motion is the motion of a mass m restricted to stay on a verticalfrictionless circular track, whose total energy allows the mass to reach with positive velocitythe highest point on the circle. An oscilatory pendular motion is one for which the totalenergy is such that the highest point on the circle is not reached. The mass in this caseoscillates back and forth. The following Theorem gives the equation satisfied by a pendularmotion. Theorem.
If a mass m moves on a vertical circle of radius R, with lowest point A, highest point B andcenter O, its position M at time t, can be defined by CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS φ ( t ) = ∠ ( AOM ) which satisfies Dφ = (cid:112) a − c sin φ, where a := gh R , c = gR , for some h . D φ = − g R sin ◦ (2 φ ) . Proof: If the height of the mass is measured from
A,h ( t ) = R − Rcos (2 φ ( t )) = 2 Rsin φ ( t ) , the Theorem of Toricelli gives RD (2 φ )( t ) = v ( t ) = (cid:112) gh − gRsin φ ( t ) , hence 0.The motion is circulatory if h > R or a > c, it is oscilatory if 0 < h < R or c > a.
2, follows by squaring 0 and taking the derivative.
Notation. k := ca , b := a − c , k (cid:48) := ba , m := k . Theorem. [Jacobi]
Let M ( t ) describes a pendular motion. Given the circle γ which has the line r at height h as radical axis and is tangent to AM ( t ) , if N ( t ) M ( t ) remains tangent to that circle, then N ( t ) describes the same pendular motion, with N ( t ) = A. Proof: With the abbreviation M = M ( t ) , N = N ( t ) , let N M meets r at D, let M (cid:48) , N (cid:48) be the projections of M and N on r, let T be the point of tangency of M N with γ ,0. DM DN = DT , therefore1. DTND = DMDT = DT − DMND − DT = MTNT = (cid:113) DTND DMDT = (cid:113) DMND = (cid:113) M (cid:48) MN (cid:48) N When t is replaced by t + (cid:15) ,2. v M v N = lim M ( t + (cid:15) ) − M ( t ) N ( t + (cid:15) ) − N ( t ) = lim M ( t ) TN ( t + (cid:15) ) T = MTNT , because the triangles T, M, M ( t + (cid:15) ) and T, N, N ( t + (cid:15) ) are similar, because ∠ ( T, N, N ( t + (cid:15) ) = ∠ ( T, M ( t + (cid:15) ) , M ) as well as ∠ ( M ( t + (cid:15) ) , T, M ) = ∠ ( N ( t + (cid:15) ) , T, N ) . Therefore3. v M v N = (cid:113) M (cid:48) MN (cid:48) N . The Theorem of Toricelli asserts that v M = √ gM (cid:48) M , this implies, as we have just seen, v N = √ gN (cid:48) N , therefore N describes the same pendular motion with a difference in theorigin of the independent variable. .8. THE PENDULUM AND THE ELLIPTIC FUNCTIONS. Corollary. If M = B and N = A, the line M ( t ) × N ( t ) passes through a fixed point L on the verticalthrough O .Moreover, if b := BL and a := LA, we have v M v N = ba and h = a a − b . This follows at once from from 1.8.1.2, and 1.
Definition.
The point L of the preceding Corollary is called point of Landen . Theorem. [Poncelet]
Given 2 conics θ and γ , if a polygon P i , i = 0 to n, P n = P , is such that P i is on θ and P i × P i +1 is tangent to γ , then there exists infinitely many such polygons.Any such polygon is obtained by choosing Q on θ drawing a tangent Q Q to γ , with Q on θ and successively Q i , such that Q i is on θ and Q i − × Q i is tangent to γ , the Theoremasserts that Q n = Q . The proof follows at once from 1.8.1, after using projections which transform the circle θ and the circle γ into the given conics.The Theorem is satisfied if the circle have 2 points in common or not. Theorem. If M ( t ) describes a circular pendular motion, then the mid-point M ( t ) of M ( t ) and M ( t + K ) describes also a circular pendular motion. More precisely, M ( t ) is on a circle with diameter LO, with LA = a, LB = b, and if φ ( t ) = ∠ ( O, L, M ( t ) , t = (cid:82) φ ( t )0 Dφ ∆ = (cid:82) φ ( t )0 Dφ ∆ . where
1. ∆ := a cos φ + b sin φ and ∆ := a cos φ + b sin φ , where the relation between φ and φ is given by tan ( φ − φ ) = k (cid:48) tanφ , or sin (2 φ − φ ) = k sinφ , with k (cid:48) := ba , k := c a ,5. a := ( a + b ) , b := √ ab, c := ( a − b ) , therefore a = a + c , b = a − c , c = 2 √ a c . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Proof: First, it follows from the Theorem of Toricelli that the velocity v A at A and v B at B satisfy v A = √ gh = 2 Ra, v B = √ gh − R = √ R a − c R = 2 Rb, therefore
BLLA = ba . If P is the projection of L on BM and Q the projection of L on AM,LM = LP + LQ = a cos φ + b sin φ = ∆ .LQ = LM cos ( φ − φ ) = acosφ. We can proceed algebraically. Differentiating 2. gives a (1 + tan ( φ − φ ))( Dφ − Dφ ) = b (1 + tan φ ) Dφ, or a (1 + tan ( φ − φ )) Dφ = ( a (1 + tan ( φ − φ ) + b (1 + tan φ )) Dφ = ( a + b + b a tan φ + btan φ ) Dφ = ( a + b )(1 + ba tan φ ) Dφ = ( a + b )(1 + tanφtan ( φ − φ )) Dφ, or acos ( φ − φ ) Dφ = 2 a cos (2 φ − φ ) cosφcos ( φ − φ ) Dφ, or Dφ acosφcos ( φ − φ ) = Dφ a cos (2 φ − φ ) , or because LM = ∆ Dφ ∆ = Dφ . We can also proceed using kinematics.The velocity at M is v M = 2 RDφ = 2 R ∆ , If we project the velocity vector on a perpendicualr to
LM,LM Dφ = v M cos (2 φ − φ ) = 2 Rcos (2 φ − φ )∆ φ. Therefore Dφ ∆ = Dφ Rcos (2 φ − φ ) = a R Dφ ∆ = Dφ . Definition.
The transformation from φ to φ is called the forward Landen transformation .The transformation from φ to φ is called the backward Landen transformation .These transformations have also been applied to the integrals of the second kind and ofthe third kind. Comment.
The formulas 3. and 1. are the formulas which are used to compute t from φ ( t ). Theformulas 4. and 2. are used to compute φ ( t ) from t . Comment
Given the first order differential equations,( Dy ) = C ( y + A )( y + B ) , ( Dz ) = C ( z + A )( z + B ) , with .8. THE PENDULUM AND THE ELLIPTIC FUNCTIONS. z = d ( y + ly ) , l (cid:54) = 0 , d > . These equations are compatible iff d (1 − ly ) C ( y + A )( y + B ) = C ( d ( y + ly ) + A )( d ( y + ly ) + B )this requires √ l to be a root of one of the factors of the second member, let it be the secondfactor, this implies d l + B = 0 , then, the second factor becomes, d ( y + ly ) + B = d (( y + ly ) − l ) = d ( y − ly ) ,therefore √ l is a double root of the second member and C ( y + ( A + B ) y + A B ) = d C ( y + (2 l + A d ) y + l ) , therefore C = d C , A B = B d , A + B = A − B d , For real transformations, A B > , if j = sign ( B ) and j = sign ( B ) ,B = 4 j d √ A B , A = d ( A + B + 2 j √ A B = j d ( (cid:112) | A | + j j (cid:112) | B | ) . If we want A B > j = j . Introduction.
Gauss began his investigations after he showed that the length of the lemniscate could becomputed from the arithmetico geometric mean of √ r = cos (2 θ ), in polar coordinates. A quarter of its length is given by the integral (cid:90) dr √ − r , which is easily deduced from the general formula for the square of the arc length in polarcoordinates, ds = dr + r ( dθ ) . Gauss observed that to 9 decimal places the integral was 1.311028777 and so is π/ agm ( √ , ,where agm ( a, b ) denotes the arithmetico geometric mean of 2 numbers, defined below. Theorem. [Gauss]
Given a > b > , let a i +1 := ( a i + b i ) , b i +1 := √ a i b i , The sequences a i and b i have a common limit a ∞ . The sequence a i is monotonically decreasing and the sequence b i is monotonically in-creasing. Proof: Because a i > a , b i +1 > b i , it follows that the sequence a i is bounded below by b , the sequence b i is bounded above by a , therefore both have a limit a ∞ and b ∞ . Taking the limit of 0. gives at once a ∞ = b ∞ . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition. a ∞ is called the arithmetico-geometric mean of a and b . Example.
With a = √ b = 1, a = 1 . , b = 1 . a = 1 . , b = 1 . a = 1 . , b = 1 . Definition. If a = 1 , and we express t in terms of φ ( t ) , t = (cid:82) φ ( t )0 1 √ − k sin . This integral is called the incomplete elliptic integral of the firstkind . Its inverse function φ is usually noted1. am := φ , the amplitude function ,2. K := (cid:82) π √ − k sin is called the complete integral of the first kind , it gives half theperiod, Ka , for the circular pendulum. Theorem.
0. For the circulatory pendulum, the angle 2 φ between the lowest position of the mass andthat at time t is given by φ = am ( at ). The coordinates are R sin (2 φ ) , R − R cos (2 φ ).1. For the oscillatory pendulum, if the highest point is 2 Rsin ( α ) above the lowest point,the angle 2 θ between the lowest position of the mass and that at time t is given by sinθ = sinφ sinα where φ is given by φ = am ( at, sin α ). Theorem.
For the complete integrals we have Ka = (cid:82) π √ a cos + b sin = π a ∞ . Proof: If φ ( K ) = π , then φ ( K ) = π , therefore1. K = (cid:82) π Dφ ∆ = (cid:82) π Dφ = (cid:82) π Dφ ∆ + (cid:82) π π Dφ ∆ = (cid:82) π Dφ ∆ = (cid:82) π Dφ n ∆ n = (cid:82) π a ∞ = π a ∞ . .8. THE PENDULUM AND THE ELLIPTIC FUNCTIONS. Definition.
The functions0. sn := sin ◦ am, cn := cos ◦ am, dn := √ − k sn , are called the elliptic functions of Jacobi .The functions which generalize tan, cosec, . . . are1. ns := sn , nc := cn , nd := dn , sc := sncn , cd := cndn , ds := dnsn , cs := cndn , dc := dncn , sd := sndn . Theorem. If s := sn ( t ) , c = cn ( t ) , d = dn ( t ) and s := sn ( t ) , c = cn ( t ) , d = dn ( t ) , we have sn + cn = 1 , dn + k sn = 1 , dn − k cn = k (cid:48) .
3. 1 − k s s = c + s d = c + s d . Lemma. c = c cn ( t + t ) + d s sn ( t + t ) , d = d dn ( t + t ) + k s c sn ( t + t ) . Proof: We use the Theorem 1.8.1 of Jacobi. Let R be the radius of θ and O its center,let r be the radius of γ and O (cid:48) its center, let s := OO (cid:48) . Let
A, N, M (cid:48) , M be the position ofthe mass at time 0, t , t , t + t . The lines A × M (cid:48) and N × M are tangent to the same circle γ at T (cid:48) and T. Let X be the intersection of O × M and O (cid:48) × T, φ := ∠ ( A, O, N ) ,
2. 2 φ (cid:48) := ∠ ( A, O, M ) , we have ∠ ( N, O, M ) = 2( φ (cid:48) − φ ) , ∠ ( M, X, T ) = φ (cid:48) − φ, ∠ ( T, O (cid:48) , O ) = φ (cid:48) + φ .If we project M OO (cid:48) on O (cid:48) T,r = Rcos ( φ (cid:48) − φ ) scos ( φ (cid:48) + φ ) , or3. r = ( R + s ) cosφcosφ (cid:48) + ( R − s ) sinφsinφ (cid:48) .φ = amt , φ (cid:48) = am ( t + t ) ,sinφ (cid:48) = sn ( t + t ) , cosφ (cid:48) = cn ( t + t ) ,sinφ = sn t = s ,cosφ = cn t = c , CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS when t = 0, cos ( ∠ ( A, B, M (cid:48) ) = cn t = c = BM (cid:48) AB = O (cid:48) T (cid:48) AO (cid:48) = rR + s ,the ratio of the velocities is v M (cid:48) v A = dn t dn = d = T M (cid:48) AT = O (cid:48) BAO (cid:48) = R − sR + s , substituting in 2. gives 0.The proof of 1. is left as an exercise. Theorem. [Jacobi] sn u cn u dn u + sn u cn u dn u sn ( u + u ) = 1 − k sn u sn u . cn u cn u − sn u dn u sn u dn u cn ( u + u ) = 1 − k sn u sn u . dn u dn u − k sn u sn u cn u cn u dn ( u + u ) = 1 − k sn u sn u . Proof: Let w = − k s s . Let s , s , . . . denote sn u , sn u , . . . , define S and C such that sn ( u + u ) = Sw, cn ( u + u ) = Cw.
The 1.8.3.0. gives c = c Cw + d s Sw or3. c Cw = − d s Sw + c , S w + C w = 1 , eliminating C gives the second degree equation in Sw :( c + d s ( Sw ) − s c d ( Sw ) + c − c = 0 , one quarter of the discriminant is s c d − ( c − c )( c + d s )= s c d − c c + c − s c d + s c d = c ( c − c + s d ) = c s d , therefore Sw = ( s c d ± c d s ) w. One sign correspond to one tangent from M to γ , the other to the other tangent, thereforeone corresponds to the addition, the other to the subtration formula. From the special case k = 0 , follows that, by continuity, the + sign should be used. This gives 0., 1. follows from3, 2. is left as an exercise. Corollary. sn ( u + K ) = cd ( u ) , cn ( u + K ) = − k (cid:48) sd ( u ) , dn ( u + K ) = k (cid:48) nd ( u ) . sn ( u + 2 K ) = − sn ( u ) , cn ( u + 2 K ) = − cn ( u ) , dn ( u + 2 K ) = dn ( u ) . sn ( u + 4 K ) = sn ( u ) , cn ( u + 4 K ) = cn ( u ) , dn ( u + 4 K ) = dn ( u ) . Definition. K (cid:48) ( k ) = K ( k (cid:48) ) . .8. THE PENDULUM AND THE ELLIPTIC FUNCTIONS. Theorem.
0. 0. ksn ◦ I + iK (cid:48) = sn, ikcn ◦ I + iK (cid:48) = ds, idn ◦ I + iK (cid:48) = cs,
1. 0. sn ◦ I + 2 iK (cid:48) = sn, cn ◦ I + 2 iK (cid:48) = − cn, dn ◦ I + 2 iK (cid:48) = − dn, Theorem. sn has periods K and iK (cid:48) and pole ± iK (cid:48) , cn has periods K and iK (cid:48) and pole ± iK (cid:48) , dn has periods K and iK (cid:48) and pole ± iK (cid:48) . Theorem. k = 0 ⇒ sn = sin, cn = cos, dn = 1 , k = 1 ⇒ sn = tanh, cn = sech, dn = sech. Definition.
Given the parameter q, called the nome,0. q := e − π K (cid:48) K , the functions1. θ := 2 q (cid:80) ∞ n =0 ( − n q n ( n +1) sin (2 n + 1) I θ := 2 q (cid:80) ∞ n =0 q n ( n +1) cos (2 n + 1) I θ := 1 + 2 (cid:80) ∞ n =1 q n cos nI θ := 1 + 2 (cid:80) ∞ n =1 ( − n q n cos nI are the theta functions of Jacobi . Definition.
The functions, with v = π I K θ s := K θ ◦ vDθ (0) , θ c := θ ◦ vθ (0) , θ d := θ ◦ vθ (0) , θ n := θ ◦ vθ (0) , are called the theta functions of Neville .2 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem. If p, q denote any of s, c, d, n,pq = θ p θ q . For instance sn = θ s θ n = Kθ ◦ vDθ (0) . θ (0) θ ◦ v . Theorem.
The Landen transformation replaces the parameter q, by q . Theorem. [Lagrange]
From the addition formulas of elliptic functions, we can derive those for a spherical triangleas follows. Let u + u + u = 2 K, define sina := − snu , cosa := − cnu ,sinb := − snu , cosb := − cnu ,sinc := − snu , cosc := − cnu ,sinA := − k snu , cosA := − dnu ,sinB := − k snu , cosB := − dnu ,sinC := − k snu , cosC := − dnu , then to any formula for elliptic functions of u , u , u , corresponds a formula for aspherical triangle with angles A, B, C and sides a, b, c.
For instance, sinAsina = sinBsinb = sinCsinc = k. cosa = cosb cosc + sinb sinc cosA, cosA = − cosB cosC + sinB sinC cosa, sinB cotA = cosc cosB + sinc cota. Proof. 2. follows from the definition. 3. follows from c = c cn ( u + u ) + d s sn ( u + u ) after interchanging u and u and using6. 0. sn ( u + u ) = sn (2 K − u − u ) = sn u = s , cn ( u + u ) = − cn (2 K − u − u ) = − cn u = − c , dn ( u + u ) = dn (2 K − u − u ) = dn u = d , similarly, 4. follows from c = c cn ( u + u ) + d s sn ( u + u )after interchanging u and u and using 6, and 5. from sn u dn u = cn u sn ( u + u ) − sn u dn u cn ( u + u )after division by sn u . .8. THE PENDULUM AND THE ELLIPTIC FUNCTIONS. Introduction.
Because it is not germane in this context, I will only mention briefly the important contri-bution of Weierstrass, which proved that all doubly periodic meromorphic functions can beexpressed in terms of one of them, the p function. The addition formulas for this functionanf for the Jacobi functions and many other properties generalize to the finite case (DeVogelaere, 1983)..
0. Abel, Niels Henrik,
Oeuvres Compl`etes , Nouv. ed., publi´ees aux frais de l’Etat norv´egienpar L. Sylow et S. Lie, Christiania, Grondahl & Son, 1881, Vol. 1,2.1. Appell, Paul Emile & Dautheville, S.,
Pr´ecis de M´ecanique Rationnelle , Paris, Gauthier-Villars, 1924, 721 pp.2. Bartky,
Numerical Calculation of Generalized Complete Integrals , Rev. of ModernPhysics, 1938, Vol. 10, 264-3. Cayley, Arthur,
On the Addition of Elliptic Functions , Messenger of Mathematics, Vol.14, 1884, 56-61.4. Cayley, Arthur,
Note sur l’Addition des Fonctions Elliptiques , Crelle J., Vol. 41, 57-65.5. De Vogelaere, Ren´e,
Finite Euclidean and non-Euclidean Geometry with applicationto the Finite Pendulum and the Polygonal Harmonic Motion. A First Step to FiniteCosmology . The Big Bang and Georges Lemaˆıtre, Proc. Symp. in honor of 50 yearsafter his initiation of Big-Bang Cosmology, Louvain-la-Neuve, Belgium, October 1983.,D. Reidel Publ. Co, Leyden, the Netherlands. 341-355.6. Eisenstein, Ferdinand Gotthold Max,
Mathematische Abhandlungen , besonders ausdem Gebiete der hoheren Arithmetik und den elliptischen Functionen, Mit einer Vorredevon C.F. Gauss. (Reprografischer Nachdruk der Ausg., Berlin 1847.) Hildesheim, G.Olms, 1967.7. Emch,
An Application of Elliptic Functions , Annals of Mathematics, Ser. 2, Vol. 2,1901. III.4.4.0.8. Fettis, Henri E., Math. of Comp., 1965.9. Gauss, Carl Friedrich, Ostwald Klassiker der Exakten Wissenschaften, Nr 3.10. Jacobi, Karl Gustav Jakob,
Fundamenta Nova Theoriae Funktionum Ellipticarum ,1829.11. Landen, John, Phil. Trans. 1771, 308.4
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
12. Lagrange, Joseph Louis, comte,
Oeuvres , publi´ees par les soins de m. J.-A. Serret, sousles auspices de Son Excellence le ministre de l’instruction publique, Paris, Gauthier-Villars, 1867-92.13. Legendre, Adrien Marie,
Trait´e des Fonctions Elliptiques et des Int´egrales Euleriennes ,avec des tables pour en faciliter le calcul num´erique, Paris, Huzard-Courcier, Vol. 1-3,1825-1828.14. Lemaˆıtre, Georges,
Calcul des Int´egrales Elliptiques , Bull. Ac. Roy. Belge, Classe desSciences, Vol. 33, 1947, 200-211.15. Weierstrass, Karl,
Mathematische Werke , Hildesheim, G. Olms, New York, JohnsonReprint 1967.
0. Abramowitz, Milton & Stegun, Irene A. Edit.,
Handbook of Mathematical Functions ,U.S. Dept of Commerce, Nat. Bur. of Stand., Appl. Math, Ser., number 55, 1964,1046 pp.1. Adams, Edwin Plimpton, Ed.,
Smithsonian Mathematical Formulae and Tables ofElliptic Functions , under the direction of Sir George Greenhill, 3d reprint, City ofWashington, 1957, Its Smithsonian miscellaneous collections, v.74, no.1, SmithsonianInstitution, Publication 2672.2. Halphen, Georges Henri,
Trait´e des Fonctions Elliptiques et de leurs Applications ,Paris, Gauthier-Villars, 1886-91.3. Hancock, Harris,
Lectures on the Theory of Elliptic Functions , v. 1. 1st ed., 1stthousand, New York, J. Wiley, 1910. Dover Publ., 1958.4. Jahnke, Eugen & Emde, Fritz,
Tables of Functions with Formulae and Curves , 4th ed.,New York, Dover Publications, 1945.5. Jahnke, Eugen & Emde, Fritz & Losch, Friedrich,
Tables of Higher Functions , 6th ed.rev. by Losch, New York, McGraw-Hill, 1960.6. Jordan, Camille,
Fonctions Elliptiques , New York, Springer-Verlag, 1981.7. King, Louis Vessot,
On the Direct Numerical Calculations of Elliptic Functions andIntegrals , Cambridge, Eng., Univ. Press, 1924.8. Lang, Serge,
Elliptic functions , 2nd ed., New York, Springer-Verlag, 1987, Graduatetexts in mathematics, 112.9. Mittag-Leffler, Magnus Gustaf,
An Introduction to the Theory of Elliptic Functions ,Lancaster, Pa., 1923, Hamburg, Germany, Lutcke & Wulff. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY.
Elliptic Functions, a primer , Prepared for publication by W. J.Langford, 1st d. Oxford, New York, Pergamon Press, 1971.11. Neville, Eric Harold,
Jacobian Elliptic Functions , Oxford, The Clarendon Pr., 1944.12. Oberhettinger, Fritz Wilhelm & Magnus, Wilhelm,
Anwendung der elliptischen Funk-tionen in Physik und Technik , Berlin, Springer, 1949, Die Grundlehren der mathema-tischen Wissenschaften in Einzeldarstellungen, Bd. 55.13. Riemann, Bernhard,
Elliptische Functionen , Mit zusatzen herausgegeben von HermannStahl . . . Leipzig, B. G. Teubner, 1899.14. Schuler, Max & Gebelein, H.,
Eight and Nine Place Tables of Elliptical Functions basedon Jacobi Parameter q , with an English text by Lauritz S. Larsen, Berlin, Springer-Verlag, 1955, XXIV+296 pp.15. Spenceley, G.W. & Spenceley, R.M., Smithsonian Elliptic Functions Tables , Washing-ton, Smithsonian Institution 1947, Smithsonian miscellaneous collections v. 109.16. Sturm, Charles Fran¸cois,
Cours d’Analyse de l’Ecole Polytechnique , revu et corrig´e parE. Prouhet et augment´e de la Th´eorie ´el´ementaire des fonctions elliptiques, par H.Laurent, 14. ed., rev. et mise au courant du nouveau programme de la licence, par A.de Saint-Germain, Paris, Gauthier-Villars, 1909.17. Tannery, Jules & Molk, Jules,
El´ements de la th´eorie des fonctions elliptiques , Paris,Gauthier-Villars, Vol. 1-4, 1893-1902.18. T¨olke, Friedrich,
Praktische Funktionenlehre , Berlin, Springer, Vol. I to VI ab. Vol. 3and 4 deal with the elliptic functions of Jacobi.19. Tricomi, Francesco Giacomo,
Elliptische Funktionen , ¨ubers. und bearb. von Maximil-ian Krafft. Leipzig, Akademische Verlagsgesellschaft Geest & Portig, 1948, Mathematikund ihre Anwendungen in Physik und Technik, Bd. 20.20. Whittaker, Edmund Taylor & Watson, G. N.,
A Course of Modern Analysis , an intro-duction to the general theory of infinite processes and of analytic functions, Cambridge,Eng., Univ. Pr., 1963, 606 pp., (1927).
The purpose of this section is to give an informal introduction to finite Euclidean geometryfor those familiar with classical Euclidean geometry and analytic geometry.The definitions of points and lines will be given in terms of equivalence classes. The Theoremswill be derived from these definitions or can be derived from the classical Theorems. I will6
CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS restrict myself to the 2 dimensional case and will not attempt to give the most general results.In particular, I will assume that distances are defined in only one way.In this restricted framework there is one finite geometry for each prime integer p. p is assumedto be larger than 2, non degenerate circles require p larger than 3. The examples correspondto small p. The reader is encouraged to think of the implications when p is very large, forinstance of the order of 10 say, and is looking at points with coordinates of the order of10 to 10 . Notation.
A point P (cid:48) in Euclidean geometry will be denoted by its cartesian coordinates (( x, y )) givenbetween double parenthesis. A line l (cid:48) will be denoted by the coefficients [[ a, b, c ]] of itsequation ax + by + c = 0 , given between double brackets. These coefficients are not unique. They can be replaced by[[ ka, kb, kc ]]where k is any real number different from 0.For the points P and lines l in finite geometry, I will use the same notation with singleparenthesis and single brackets. Definition.
Given a prime p, if x is an integer, x mod p denotes the smallest positive remainder of thedivision of x by p. For instance, 28 mod
13 = 2 , − mod
11 = 6 . We observe that if x and y are non negative integers less than p, for any integers l and m , x + lp mod p = x, y + mp mod p = y. Definition.
Let x and y be integers. For any integers l and m, the points(( x + lp, y + mp ))are called equivalent points . A set of equivalent points is called a point ( x, y ) in finitegeometry.Let a, b, c be integers, a and b not both zero. For any integers l, m and n, the lines[[ k ( a + lp ) , k ( b + mp ) , k ( c + np )]] , k (cid:54) = 0 , are called equivalent lines . A set of equivalent lines is called a line in finite geometry.If P = ( x + mp, y + np ) is on l = [ a + k (cid:48) p, b + m (cid:48) p, c + n (cid:48) p ] , then( a + k (cid:48) p )( x + mp ) + ( b + m (cid:48) p )( y + np ) + ( c + n (cid:48) p ) = 0and therefore( ax + by + c ) mod p = 0 , this is 1.9.1 below. This method of reducing modulo p allows us to extend many of theproperties of Euclidean geometry to the finite case. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Example.
Let p = 7 . The line a = [[1 , − , − a = [[1 , − , − , a =[[1 , − , , a = [[1 , − , . The line b = [[1 , , − b = [[1 , , − , b = [[1 , , − , b =[[1 , , − , b = [[1 , , − . The intersection P = ((9 , a and b is equivalent to the points all labelled Q, ((2 , , ((16 , , ((2 , , ((9 , , ((16 , P is in thedomain 0 ≤ x, y < p, namely ((2 , . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS ^y. . Qb2 . . . . . . Qb3 . . . . . . Qb4 .. a2 . . b2 . . . a1 . . b3 . . . a . .a2 . . . . . b2 a1 . . . . . b3 a . . .. b . . . . a1 . b2 . . . . a . b3 . .. . . b . a1 . . . . b2 . a . . . . b3__________________________. . . . a1 b . | . . . . a b2 . . . . .|b1 . . a1 . . . | b . . a . . . b2 . . a0|. . Qb1 . . . . | . . Pb . . . . . . Qb2 .|. a1 . . b1 . . | . a . . b . . . a0 . .|a1 . . . . . b1| a . . . . . b a0 . . .|. b0 . . . . a | . b1 . . . . a0 . b . .|. . . b0 . a . | . . . b1 . a0 . . . . bx >
Equivalence of points and lines . Fig.0a, p = 7 .In Fig.0a, I have not given those lines which are equivalent to a but have a different slope,if R is any point on such a line which is in the lower right square it is either Q or a pointlabelled a or a . In finite geometry, we do not distinguish points labelled a from those labelled a or the pointslabelled b and b from those labelled b. We have therefore Fig.0b below. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. ^y__________________________. . . . a b . ||b . . a . . . ||. . Qb . . . . ||. a . . b . . ||a . . . . . b ||. b . . . . a ||. . . b . a . | x > Points and lines in finite Euclidean geometry . Fig.0b, p = 7 .The point Q = (2 ,
4) is on the lines a = [1 , ,
2] = [4 , ,
1] and b = [1 , ,
4] = [2 , ,
1] inthe finite Euclidean geometry associated with p = 7 . Observe that from one point on a, the others are obtained by moving one to the right andone up, for b, we move 2 to the right and one down. Observe also what happens at theboundary using, if needed Fig. 0a.If we attempt to use the equivalence method when p is not a prime, the situation for 6points is typical. If a = [[1 , , − , b = [[1 , , c = [[1 , , , the points P = ((2 , Q = ((5 , a and b, while the lines a and c or their equivalenthave no point in common with coordinates reduced modulo 6.00 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS ^ya . b . . c. a c . . b. . bP . . c. . c a . b. . b . a c. . c . . bQ x > . Fig.0c . Comment.
When giving numerical examples, it is convenient to assume that x, y, a, b, c are non negativeintegers less than p, and that the right most of the triplet a, b, c which is non 0 is chosento be 1. This is always possible because, if c is for instance different from 0, then we canchoose k in such a way that, kc mod p = 1 . This property requires p to be a prime and wasknown, together with an algorithm to obtain k, by the Indian Astromomer-MathematicianAryabatha, 5-th Century A. D. as well as by the Chinese, the date of the invention of theiralgorithm, called the chiu-i, or search for 1 is not known.For instance, I will use, when p = 11 , [4 , ,
1] instead of [5 , , , [4 , ,
0] insteadof [2 , ,
0] and [1 , ,
0] instead of [7 , , . The main advantage of this convention is that it insures a unique representation of the linesin finite geometry.Because all computations for finite Euclidean geometry have to be done modulo p , it isuseful to have ready a table of multiples of p, of inverses modulo p and of squares modulo p. Two such tables are given, the others should be completed. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. p = 7 ,i p i i − i p = 11 ,i p i i − i p = 13 ,i p i i i p = 17 ,i p i i i i
13 14 15 16 p i i i p = 19 ,i p i i i i
13 14 15 16 17 18 p i i i Theorem.
A point ( x, y ) is on a line [ a, b, c ] if and only if ( ax + by + c ) mod p = 0 . For instance, with p = 11 , from ((12 , , − , , it follows that(1 ,
2) is on line [5 , ,
4] or [4 , , . Theorem.
There are p points and p + p lines. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem.
Moreover, if A = ( A , A ) and B = ( B , B ) , the line a through A and B is a = [ A − B , B − A , A B − B A ] . For instance, for p = 11 , if A = (9 ,
8) and B = (8 , , a = [2 , , . Theorem. l = [ l , l , l ] and m = [ m , m , m ] , let d := l m − l m , if d is different from0, then the point P common to l and m is P = ( l m − l m d , l m − l m d ) . For instance, with p = 11 , if l = [2 , ,
1] and m = [9 , , , d = 5 , mod
11 = 9 , P = (1 . , .
9) = (9 , . Definition.
If 2 lines have no points in common, they are called parallel .This will occur if d = 0 , because of 1.9.1.For instance, with p = 11 , a = [2 , ,
1] is parallel to b = [5 , , . The following figure gives also a representation of points in finite geometry. The rep-resentative which is chosen is that with integer coordinates, non negative and less than p (0 ≤ x, y < p ) . The reader is asked to ignore for now the information at the left of the figure. The possiblepoints are indicated with “.”, a named point has its name just to the right of it. All pointson a line a are indicated by replacing “.” by ” a ”. If 2 lines have a point in common, one ofthe 2 lines is chosen. The other could be indicated by the reader, if he so desires. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Example. . ^ybDb . . . . . c . b . . a. . . . c . . . . bA . .. . c . . . . aB . . b .. . . . . a . . . . . cC. b . a . . . . . cD . .. a b . . . . c . . . .. . . b . c . . . . a .. . . c b . . . a . . .cDa c . . . b a . . . . .. . . . a . b . . . c .. . a . . . . b c . . . x >
Points, lines and parallels . Fig. 1, p = 11 .The points are A = (8 , , B = (6 , , C = (10 , , the line a = [10 , ,
1] is the line through A and B, it passes through the points (0,5), (2,6), (4,7), (6,8), (8,9), (10,10), (1,0), (3,1),(5,2), (7,3) and (9,4). The line b = [9 , ,
1] is the line through A and C. The line c = [3 , , C, has no points in common with a and therefore is parallel to a. The line d, which is not indicated on the picture, is parallel to b and passes through B. The point D ison c and d .As an exercise determine the coordinates of D and the other points of d. Notation.
To ease description of constructions in geometry, I have introduced the notation A × B, forthe line l through the distinct points A and B, and l × m , for the point C common to thedistinct lines l and m. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Comment. If p is very large, and the unit used for the representation is very small, the Angstr¨om =10 − cm, say, the points on a line will appear as we imagine them in the classical case. Butit is clear that they are not connected. Connectedness is a property in classical Euclideangeometry, which has no counterpart in the finite case. Moreover, the finite case should, whenfully understood, give a better model for a world which is atomic, whatever the smallestparticle is and which is finite, whatever the size of the universe is. Introduction.
Parallels have been defined in 1.9.1. In this section, I will give properties of parallel linesand define parallelograms. It is appropriate at this stage to define distances between points.In the finite case, the square of a distance is the appropriate basic concept, if we do notwant to introduce “imaginaries”. Properties of the parallelogram allow us then to derive aconstruction for the mid-point of a segment. The barycenter will be define in section 1.9.4.
Theorem.
Given a line l and a point P not on l, there exists a unique line m through P parallel to l. Moreover, if P = ( P , P ) and l = [ l , l , l ] then m = [ l , l , − ( P l + P l )] , Definition.
Given 3 points
A, B, C not on the same line, let c be the line through C parallel to the line a through A and B, let d be the line through A parallel to the line b trough B and C, { A,B, C, D } is called a parallelogram . The lines A × C and B × D are called diagonals , theirintersection is called the center of the parallelogram. Comment.
In Euclidean geometry, opposites sides of a parallelogram are equal. To generalize, I observefirst, that distances in Euclidean geometry are always considered positive. This is consistentwith the distance of AB equal to the distance of BA.
But when working modulo p, we cannotintroduce positive numbers, keeping the requirement that the product and sum of positiveintegers modulo p is positive. Also not every integer modulo p has a square root hence weuse the square of the distance instead. To use a terminology reminiscent of that used inEuclid’s time, I will say the square on AB, for the square of the distance between A and B. Definition.
Given 2 points A = ( A , A ) and B = ( B , B ) , the square on AB , denoted ( AB ) is( B − A ) + ( B − A ) mod p. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. p = 19 , if A = (8 , , B = (12 , , the square on AB is( AB ) = (16 + 4) mod
19 = 1 . But, for p = 13 , the square on AB, where A = (1 ,
2) and B = (2 ,
7) is (1 + 25) mod
13 = 0 , therefore the square can bezero for distinct points A and B. See 1.9.5.
Definition. { AC } and { BD } are equal if the square on AC equals the square on BD.
I write AC = BD .For instance, with p = 19 , if A = (7 , , B = (11 , , C = (9 ,
7) and D = (9 , AC ) = 1 , ( BD ) = 1 , ( AB ) = 1 , ( CD ) = 17 . Therefore AC = BD . Definition.
A point M on the line through A and C such that the square on AM is equal to the squareon CM is called the mid-point of AC.
Theorem.
In a parallelogram { A, B, C, D, } with A × B parallel to C × D and A × D parallel to B × C, the square on AB is equal to the square on CD, the square on AD is equal to the square on BC.
The center M is the midpoint of the diagonals A × C and B × D. Moreover, if A = ( A , A ) , B = ( A + B , A + B ) , D = ( A + C , A + C ) , then C = ( A + B + C , A + B + C ) , M = ( A + B + C , A + B + C ) , ( AB ) = ( CD ) = B + B , ( AD ) = ( BC ) = C + C , ( AM ) = ( M C ) = (( B + C ) + ( B + C ) ) . Example.
The given points are A = (7 , , B = (9 , , C = (11 , . D = (9 , , M = (9 , . ( AB ) = ( CD ) = 1 , ( AD ) = ( BC ) = 8 . ( AM ) = ( M C ) = 5 , ( BM ) = ( M D ) = 9 . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS . ^y. . c . . . . . . . . . . . d a . b . ..Df c . . . d . . b . . . . . a . . . . .. . . . . . . . . . . . . a . d . . b c. . . . . . d . . b . . a . . . . . c .. . . . . . . . . . . a . . . . d c . b. . . . . . . d . . bB. . . . . c . . .. b . . . . . . . a . . . . . c . d . .. . . . . . . . dA. . b . . c . . . . ..Dm . b . . . . a . . .M. . c . . . . d .dDb . . . . . a . . d . . cC. . . . . . .. . . b . a . . . . . c . . . . . . . d. . . . a . . . . . dD. . b . . . . . .. d . a b . . . . c . . . . . . . . . .. . a . . . . . c . . d . . b . . . . .. a d . . b . c . . . . . . . . . . . .. . . . . . c . . . . . d . . b . . . a. . . d . c b . . . . . . . . . . . a .cDa . . . c . . . . . . . . d . . b a . .. . . c d . . b . . . . . . . . a . . . x >
Parallelogram and mid-point of A B . Fig. 2, p = 19 . Introduction.
The perpendicularity of lines is defined. Theorem 1.9.3 follows from the correspondingtheorem in classical geometry and from the analytical property of perpendicular lines adaptedmodulo p .An application giving the orthocenter of a triangle is also given. Definition.
Two lines l = [ l , l , l ] and m = [ m , m , m ] are perpendicular iff l m + l m mod p = 0 . For instance, with p = 11 , l = [7 , ,
1] and m = [3 , ,
1] are perpendicular.
Theorem.
If 2 lines l and m are perpendicular to the same line a, then they are parallel. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. p = 11 , l = [9 , ,
1] and m = [1 , ,
1] are perpendicular to a = [6 , ,
1] and are parallel.
Theorem.
Given a triangle
A, B, C with sides a, b and c, if p is the perpendicular from A to a, q isthe perpendicular from B to b and r the perpendicular from C to c, then the three lines p, q and r have a point H in common. Definition.
The lines p, q and r of Theorem 1.9.3 are called altitudes, the point H is called the orthocenter of the triangle { A, B, C } . Example.
The given points are A = (8 , , B = (4 , , C = (3 , , the sides are a = [1 , , , b = [6 , , ,c = [10 , , , the altitudes are p = [1 , , , q = [2 , , , r = [10 , ,
1] and the orthocenter is H = (7 , . As an exercise, indicate on the figure one of the sides and compare how points are derivedfrom each other with those of the perpendicular line, c and r are the easiest, b and q themore difficult.08 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS . ^y.Dc r q . . . p . . . . .. p . . . . . . . q . r.Db . . . . qB . p . . r .. q p . . . . . . r . .. . . . . . . . qH . . .pDp . . p q . . r . . . .qDq . . . . . r . . pA . q. . . . p r . q . . . ..Da . . q rC . . . . . p .rDr . . r . p . . . . q .. . r . . . q . . . . p x >
The orthocenter H of { A, B, C } . Fig. 3, p = 11 . Introduction.
Having the notion of distance, we can define a circle. Having a diameter { A, B } we candefine the tangent at A as the perpendicular to A × B. The medians and barycenter aredefined and the relation between the center of a circumcircle and the mediatrices of the sidesis given. The proofs depend on the following Theorem:Given a prime p, there exists a circle C (cid:48) of radius r (cid:48) in Euclidean geometry which containsrepresentatives P (cid:48) of each point P of a circle C of radius r := r (cid:48) mod p in finite Euclideangeometry.For instance, when p = 19 , the circle C of radius 1 contains the points (0 , , (3 , , (2 , , and therefore the points (7 , , (3 , − , (7 , − , ( − , , ( − , , ( − , − , ( − , − , (4 , , (2 , − , (4 , − , ( − , , ( − , , ( − , − , ( − , − , (1 , , (0 , − , ( − ,
0) altogether 20 points.0 + 75 = 60 + 45 = 21 + 72 = 75 , therefore, in Euclidean Geometry,((0 , , (60 , , ((21 , C (cid:48) of radius r (cid:48) = 75 , moeover .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. r (cid:48) mod
19 = − , mod
19 = 3 , mod
19 = 7 , mod
19 = 2 , mod
19 = − . Appropriate change of signs give the other points, for instance ((-60,45))corresponds to (-3,-7) is also on C (cid:48) .We can also replace in the Theorem just quoted the radius r (cid:48) by the radius square r (cid:48) = r (cid:48) . The following solutions are especially attractive, because the points on the circle in theEuclidian plane are also the representatives in the finite Euclidean plane. p r (cid:48) points5 1 (0 , , , , , , , , (3 , , , (3 , , , (4 , Definition.
Given a point A and an integer d, the points P such that the square on P A is equal to d areon a circle of center A and radius square d. Notation.
From here on, it is often more convenient to have the origin at the center of the figure. Wewill then replace the condition0 ≤ x, y, a, b, c < p by − p < x, y, a, b, c < p . Theorem.
For a circle centered at the origin,if ( x,
0) ( x (cid:54) = 0) is a point, so are ( − x, , (0 , x ) , (0 , − x ) , if ( x, x ) ( x (cid:54) = 0) is a point, so are ( x, − x ) , ( − x, x ) , ( − x, − x ) , if ( x, y ) is a point ( x (cid:54) = y , both non zero), so are ( y, x ) , ( − x, y ) , ( y, − x ) , ( x, − y ) , ( − y, x ) , ( − x, − y ) , ( − y, − x ) . Example. . ^y. . . . . . . . . . . . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS . . c . . . . . . . c .. . . . . c . c . . . .. . . . . . . . . . . .. . . c . . . . . c . .. . . . . . .A . . . . . x >. . . c . . . . . c . .. . . . . . . . . . . .. . . . . c . c . . . .. . c . . . . . . . c .. . . . . . . . . . . .
Circle of center A . Fig. 4, p = 11 . A = (0 , , the points labelled c are on a circle with center A and with radius square10. The line [0,1,0] through A has no point in common with the circle, the line [1,-3,0] has2 points in common with the circle, (3,1) and (3,-1), the line [1,-1,0] has also 2 points incommon with the circle, (4,4) and (-4,-4). Exercise.
Indicate on the Fig. 4 by r the points on a circle with radius square 3. Theorem. If p + 1 is divisible by 4, there are p + 1 points on the circle. Otherwise, there are p − pointson the circle. Definition.
If a line t through a point P on a circle has no other points in common with the circle, it iscalled a tangent to the circle. Theorem.
If a line l through a point P of a circle is not tangent to it, it intersects the circle at an otherpoint Q. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Definition.
A line through the center of a circle is called a diameter . Theorem.
Half of the diameters have 2 points in common with the circle, half of them of no points incommon with the circle.
Theorem.
The tangent at a point A of a circle is perpendicular to the diameter passing through A. Example.
Given the point A = (6 ,
6) and the radius square 5, the points labeled c are on the circlecentered at A. The tangent t at P = (4 ,
7) is [9 , , . The point C = (2 ,
3) is ont the tangent.The line d = [3 , ,
1] is a diameter through P. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS . ^y. t . . . . . c . . d .dDd d . . . . t . . . . .. . . d . . c . c . . t. . . . . tP . . . c . .tDt . . c . . . dA . . t c. . . . t c . . . d . .. . . . . . c . c t . d. . d tC . . . . . . . .. . . . d . . c t . . .. . t . . . d . . . . .. . . . . . . t d . . . x >
Circle, tangent and diameter . Fig. 5, p = 11 . Exercise.
Determine the other point Q on d and the circle and the tangent at Q. Theorem. If B and C are points on a diameter of a circle and on the circle and A is an other point ofthe circle, A × B is perpendicular to A × C. Definition.
The medians of a triangle are the lines joining the vertices to the mid-points of the oppositeside.
Theorem.
The medians of a triangle have a point G in common. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Definition.
The point G of 1.9.4 is called the barycenter of the triangle. Definition.
The anti-complementary triangle { D, E, F } has its side E × F through A parallel to B × C ,and similarly for E and F. Theorem.
The mid-points
M, N, O of the sides of the triangle { A, B, C } are also the mid-points of { AD } , { BE } and { CF } . Example. .Da ^y. . . . o m n . . . . . . . . . . . . .. . . . n . . o . . . . . . . m . . . .o . n . . . m . . . o . . . . . . . . .. . . . . . . . . . . . . o . . m . . n. . . . . . . mA. . . . . . . . oFnE. ..Dc . . . . . . . . . . . . . . n . m . o. . . o . . . . m . . . . n . . . . . .. . . . . . oO. . . . nN. . . . . . m .. . . . . . . . . oG. . . . . . . . . .m . . . . . . n . . . . o . . . . . . m. . . . . nB. . . . mM. . . . oC. . . .. m . n . . . . . . . . . . . . . . o .. n o . . . . . . . . m . . . . . . . ..Db . m . . o . . . . . . . . . . . . n .. . . . . . . . o . . . m . . . n . . .. . . m . . . . . . . o . . n . . . . .n . . . . . . . . . . . n mDo . . . . .. . . . m . . . . . n . . . . . . o . .. o . . . . . . n . . . . . m . . . . . x >
Medians and barycenter . Fig. 6a, p = 19 .The given points are A = (6 , , B = (4 , , C = (14 , . The anti-complementary points are D = (12 , , E = (16 , , F = (15 , . The mid-points are M = (12 , , N = (15 , , O = (16 , . The medians are m = [16 , , , n = [8 , , , o = [13 , , . The barycenter is G = (8 , . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Exercise.
Indicate on Fig. 6a, the line F × D through 2 mid-points and observe that F × D is parallelto C × A. Definition.
The mediatrix of AB is the line through the mid-point of AB perpendicular to A × B. Theorem.
The mediatrices of the sides of a triangle pass through the center of the circumcircle of thetriangle.
Example. .Da$ ^y. . q . r . . c . . p . . c . . . . . .. . . . . . q r . . p . . . . . . . . .rDr . . . . . . . . . rZ. . . . . . . . .qDq . . . . . . . . . p . . r q . . . . .. . . . . . . cA. . p . . c . . rF.Eq ..Dc . . q . . . . c . p . c . . . . . . r. . . r . . c q . . p . . . c . . . . .. . c . . . rO. . . p qN. . . . . . c .. . . . . . . . . r p . . . . q . . . .. . . . . . . . . . p . r . . . . . . q. . . . q cB. . . . pM. . . . cC. . . .. . . . . . . . q . p . . . . . . . r .. . r . . . . . . . p . q . . . . . . ..Db . . . . c . . . . p . . . . c q . . .. q . . . . . . r . p . . . . . . . . .. . . . . q . . . . p r . . . . . . . .. . c . . . . . . q p . . .Dr . . . c .. . . . . . c . . . p . . q c . . r . .pDp r . . . . . . c . p . c . . . . q . . x >
Mediatrices and center of circumcircle . Fig. 6b, p = 19 .The given points are the same as in Example 1.9.4. The mediatrices p = [2 , , , q =[13 , , , r = [13 , ,
0] pass through the center Z = (9 ,
16) of the circumcircle C of thetriangle { A, B, C } . .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Exercise.
Determine the radius square of the circle and check that ( AZ ) = ( BZ ) = ( CZ ) . Check that if Y is some point on q, ( AY ) = ( CY ) . Theorem. If A, B, C and D are points on a circle and AB is parallel to CD then the square on AC equals the square on BD and the square on AD equals the square on BC.
Introduction.
It is now time to explain the points located at the left of each figure. In classical geometry, theplane can be extended to contain elements which are not points but have similar properties.For instance, all lines which are parallel to a given line l have no points in common, but theyall have the same direction. A direction is also called a point at infinity or an ideal point.If we extend the Euclidean plane in this way we see that 2 points ideal or not determine aunique line, with the exception of 2 ideal points. To have no exceptions, we also introduce theline at infinity or ideal line, which contains all ideal points. This extended Euclidean plane,which is unfortunately not part of high school education, is a first step to the understandingof projective geometry. Other notions which are known to those familiar with complexEuclidean geometry are the isotropic points, the isotropic lines and their properties. Thesenotions also extend to the finite case and, with the definition of distance used, give rise toreal points when the prime is of the form 4 k + 1. The distance between points which are notboth ordinary is not defined.To represent points we will now use, as for lines, 3 coordinates, not all 0, and ( x, y, z )will not be considered distinct from ( kx, ky, kz ) , k (cid:54) = 0.The ordinary points ( x, y ) will also be noted ( x, y,
1) or ( kx, ky, k ) . Definition.
The ideal line is the line [0 , , , the ideal points or directions are the points ( P , P , . Definition.
A point P = ( P , P , P ) is on a line l = [ l , l , l ] if P l + P l + P l mod p = 0 . Theorem.
All p + 1 ideal points are on the ideal line. There are p + p + 1 ordinary and ideal pointsand p + p + 1 ordinary and ideal lines. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem.
If 2 lines l and m are parallel, they have an ideal point in common or have the same direction. Moreover, if l = [ l , l , l ] this point is D l = ( l , − l ,
0) and if m = [ m , m , m ] , then d := l m − l m = 0 . Definition.
If 2 lines l and m are perpendicular , their ideal points or directions are said to be perpendic-ular .Moreover, if the direction of l is D l = ( l , − l , m is D m = ( l , l , . Comment.
The ideal points are represented to the left of the figures. (1,0,0) is at the top the otherpoints are from the bottom up (0,1,0), (1,1,0), (2,1,0), (3,1,0), . . . . Example.
In Fig. 1, the point D a = (2 , ,
0) is the ideal point on a = [10 , , , and c = [3 , , , thepoint D b = (10 , ,
0) is the ideal point on b = [9 , ,
1] and d. In Fig. 2, the points D a = (1 , , , D b = (9 , , , D f = (17 , , , D m = (10 , ,
0) arerespectively the ideal points on a = [5 , , , b = [3 , , , f = [17 , , , m = [14 , , . m is the mediatrix of AC.
In Fig. 3, D a = (2 , ,
0) and D p = (5 , , , D b = (8 , ,
0) and D q = (4 , , , D c =(10 , ,
0) and D r = (1 , ,
0) are the direction of pairs of perpendicular lines.In Fig. 5, D d = (9 , ,
0) is the direction of the diameter d = [3 , , . D t = [6 , ,
0] is thedirection perpendicular to Dd and of the tangent t = [9 , , . Definition.
The isotropic points are the ideal points ( i, ,
1) and ( − i, ,
1) where i is a solution of i +1 = 0 . Theorem.
The isotropic points exist if p is of the form k + 1 (or p is congruent to 4 modulo 1), theydo not, otherwise. The proof of this result goes back to Euler.For instance, if p = 5 , i = 2 , if p = 13 , i = 5 , if p = 17 , i = 4 . Definition.
In the extended Euclidean plane, ( X , X , X ) is on the circle with center ( C , C , C ) and radius square R if0. ( X − C X ) + ( X − C X ) = R X . If X = 1 , we obtain the usual equation. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Theorem.
When the isotropic points exist, they are on each of the circles.
Indeed, if X = i, X = 1 and X = 0 , i + 1 = 0 . Definition.
The isotropic lines are any ordinary line passing through an isotropic point.
Theorem.
The isotropic lines are perpendicular to themselves.
Theorem.
The isotropic lines through the center of a circle are tangent to that circle at the isotropicpoint.
Theorem. If A and B are ordinary points on the same isotropic line, the square on AB is 0. Indeed, if A = ( A , A ,
1) and B = ( B , B , , the line A × B which is [ A − B , B − A , A B − A B ] passes through ( i, ,
0) if( A − B ) i = A − B . But the square on AB is ( A − B ) + ( A − B ) = ( A − B ) ( i + 1) = 0 . Comment.
Because of 1.9.5, when p is congruent to 1 modulo 4, it is possible for the square on AB tobe 0 for distinct points A and B. Example.
The circle C of center A = (8 ,
8) passes through P = (4 ,
10) and through the isotropicpoints J = (4 , ,
0) and K = (13 , , . The isotropic lines through A are j = [5 , ,
1] and k = [14 , , . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS i ^yi . . . . . . j . . . k . . . . . .i . . j . . . . . . . . . . . k . .i . k . . . . . c . c . . . . . j .cK . . . . . k . . . . . j . . . . .i . . . . . . c j . k c . . . . . .i . . . j . . . . . . . . . k . . .i k . . . cP. . . . . . . c . . . ji . . c . k . . . . . . . j . c . .i . . . . . . . . kA. . . . . . . .i . . c . j . . . . . . . k . c . .i j . . . c . . . . . . . c . . . ki . . . k . . . . . . . . . j . . .cJ . . . . . . c k . j c . . . . . .i . . . . . j . . . . . k . . . . .i . j . . . . . c . c . . . . . k .i . . k . . . . . . . . . . . j . .i . . . . . . k . . . j . . . . . . x >
Ideal line, isotropic points and isotropic lines . Fig. 7, p = 17 . Introduction.
The definition of angles is the most difficult aspect of finite geometry. To approach thesubject, I will give a construction which obtains points on a circle which are equidistant. Ifwe obtain in this way all 2 q = p + 1 or p − p = 11 , Construction.
Given a diameter of a circle with the points A and A q on it and an other point A on thecircle, I will construct A , A , . . . , as follows.Let C be the center of the circle, A is such that A q × A is parallel to C × A , and suchthat A × A is perpendicular to C × A ( A × A is parallel to the tangent at A ). Givensome point A j different from A , A j + 1 is such that A × A j + 1 is parallel to A × A j and A q × A j + 1 is perpendicular to A × A j . Using j = 2 ,
3, . . . , we obtain A , A , . . . . .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Theorem.
The square on A j A j + 1 is equal to the square on A A . The proof for j = 1 is as follows, let C = (0 , , A = ( r,
0) and A q = ( − r,
0) and A j = ( x j , y j ) , the square on A A is( x − r ) + y = 2 r ( r − x ) ,A × A = [ − y , x − r, ry ] , A q × A = [ − y , x + r, − ry ] ,C × A = [ y , − x , , parallelism requires0. y ( x + r ) − x y = 0 , perpendicularity requires1. y y + x ( x − r ) = 0 , therefore,( A A ) = ( x − x ) + ( y − y ) =2( r − x x − y y ) = 2 r ( r − x ) = ( a A ) , because of 1. Multiplying 0, by x ( x − r ) and 1, by y ( x + r ) and subtracting gives x y ( x + r )( x − r ) = x y y y or because x y is different from 0 , x + y = r .A is therefore on the circle. Theorem.
Given the construction 1.9.6,
0. ( A j A j + k ) = ( A A k ) ,
1. ( j + m − k − l ) (mod 2 q ) = 0 implies A j × A k is parallel to A l × A m . Definition.
Assume that the construction 1.9.6 gives all 2 q points of the circle, let the direction of A q A j be I j and that of the tangent at A q be I q , the set I , I , . . . , I q define a scale on the idealline. Definition.
Let the lines l, m, have directions I l , I m , the angle between l and m is given by ( m − l )(mod 2 q ) . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem.
The sum of the angles of a triangle is 0 (mod 2 q ) . Indeed, if the directions of the sides a, b, c are I a , I b , I c , the angles are ( c − b ) (mod 2 q ) , ( a − c ) (mod 2 q ) and ( b − a ) (mod 2 q ) . Theorem.
If 2 angles of a triangle are even, the third angle is even.
Definition.
A triangle is called even if 2 of its angles and therefore all its angles are even.
Example.
The points A = (10 , , A = (1 , , A = (2 , , A = (5 , , A = (8 , , A = (9 , ,A = (0 , , A = (9 , , A = (8 , , A = (5 , , A a = (2 , , A b = (1 ,
8) are on a circlecentered at C with radius square 3. These points have been obtained from A , A and A by the construction 1.9.6The angles can be determined using the scale defined by the ideal points I = (1 , , ,I = (7 , , , I = (5 , , , I = (1 , , , I = (9 , , , I = (8 , , , I = (0 , , ,I = (3 , , , I = (2 , , , I = (10 , , , I a = (6 , , , I b = (4 , , . If i + l = j + k then A ( i ) A ( j ) is parallel to A ( k ) A ( l ) , for instance, b or A A is parallel to c or A A . .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. .I0 ^ycI9 . . . . . .A3 c b . . ..I4 . . .Aa . . . . c bA8 . ..I5 . .Ab . . . . . . c bA7 ..I1 . . . . . . . . . c b.Ia b . . . . . . . . . c.I2 cA6 b . . . .C . . . . .A0.Ib . c b . . . . . . . ..I7 . . c b . . . . . . ..I8 . .A1 . c b . . . . .A5 ..I3 . . .A2 . c b . . .A4 . ..I6 . . . . . cA9 b . . . . x > Angles and equidistant points on a circle . Fig. 8, p = 11 . Exercise.
Obtain using the construction 1.9.6, the point A from the point A . Theorem.
If the angle of 2 lines l and m is even there are two lines b and b which form an equalangle with l and m. The lines b and b are perpendicular. Definition.
The lines b and b of 1.9.6 are called bisectrices. Theorem.
If a triangle is even, there exist 4 points C , C , C , C , which are on 3 bisectrices, eachpassing by a different vertex of the triangle. More precisely, if the 3 bisectrices d, e, f, which pass respectively Through A , A , A , are such that22 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
0. ( angle ( d, a ) + angle ( e, b ) + angle ( f, c )) (mod 2 q ) = q then d, e and f have a point in common. Definition.
The 4 points C , C , C , C are called center of the tangent circles . Theorem.
There exist a circle with center C i tangent to each of the sides of the triangle. .9. MODEL OF FINITE EUCLIDEAN GEOMETRY IN CLASSICAL EUCLIDEAN GEOMETRY. Example. . ^y. r . c . . . . .B0 . . . a . . . rc . . . . cA1 . . . . . . . . . . .. . . . . . . c . . . . . . a . . .r . . . . . a . . c . . . . . . . .. . .B1 . . . . . . . c . . . a . .. . . . . . r a . . . . r c . . . .. . . . . . r . . . . . r . . c a .. . . . . . . . a . . . . . . . . cA0a . c . . . . . . . . . . . . . . a x >. . . . c . . . . a . . . . . . . .B2. a . . . . c . . . . . . . . . . .. r . . . . . . c . a . . . . . . rr . aA2 . . . . r . c . . . . . . .. . . . . . . r . . . a c . . . . .. . . a . . . . . .I. . . . c . . .. . . . . . . r . . . r a . . . c .. c . . a . . . r . r . . . . . . .
Bissectrices and inscribed circle . Fig.9, p = 17 .The given triangle is A = (8 , , , A = ( − , , , A = ( − , − . Its sides are a =[2 , , , b = [ − , , , c = [5 , − , . The bisectrices meet the circle at B = ( − , , B =( − , , B = (8 , − . They are d = [8 , , , e = [ − , , , f = [ − , ,
1] and have the point I = (0 , −
6) in common. The tangent circle r has radius square 5. Its points of contact with a, b, c are respectively (2,-5), (-3,3), (1,-4). Only a and c are given on the Figure, not toclutter it. The other centers of tangent circles are (-1,4), (3,-3) and (-2,5). Exercise.
Determine that b is tangent to the circle r and that ( − ,
4) is indeed a center of a tangentcircle.
Definition. If r = 1 and the construction 1.9.6 gives all 2 q points A j = ( x j , y j ) of the circle with radiussquare 1. I define s in (2 j ) := y j , c os (2 j ) := x j . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Comment.
Because, in general, several points A can be chosen, there are several distinct but relatedtrigonometric functions sine and cosine. Each corresponds to a different choice of the unitangle. This is similar to the real case in which many different units are used, those withangles in radians, degrees, grades, for instance. Comment.
I will develop the properties of the trigonometric functions and obtain functions which canbe considered as an analogue of the hyperbolic functions. An efficient method to obtainthem for large p will also be given. Example.
For p = 11 ,i A i angle ( i ) − i A (cid:48) i − , −
3) 36 ◦
87 (( − , − − , −
4) 73 ◦
74 (( , ))3 (0 ,
5) 110 ◦
61 (( , − ))4 (3 , −
4) 147 ◦
48 (( − , ))5 (4 , −
3) 184 ◦
35 (( , ))6 ( − ,
0) 221 ◦
22 (( − , − ))If A i = ( − , − , then 4 + 3 = 5 , cos ( i ) = − = − sin ( i ) = − = − . For p = 13 ,i A i angle ( i ) − i A (cid:48) i − , −
4) 53 ◦
13 (( − , − − , −
3) 106 ◦
26 (( − , ))3 (0 ,
5) 159 ◦
39 (( , − ))4 (4 , −
3) 212 ◦
52 (( − , − ))5 (3 , −
4) 265 ◦
65 (( , ))6 ( − ,
0) 318 ◦
78 (( , − ))For p = 17 ,i A i angle ( i ) − i A (cid:48) i − , − ◦
125 ((8 , , −
1) 7 ◦
250 ((8 , − − ,
7) 21 ◦
375 (( , − ))5 (7 , −
4) 35 ◦
625 (( , − ))7 ( − ,
8) 49 ◦
875 (( , − ))9111315 .10. AXIOMATIC Exercise.
Continue the last table obtaining the missing values.
Exercise.
Obtain trigonometric functions for p = 11 and check the familiar identities0. sin ( x ) + cos ( y ) = 1 , sin ( x + y ) = sin ( x ) cos ( y ) + sin ( y ) cos ( x ) , cos ( x + y ) = cos ( x ) cos ( y ) − sin ( x ) sin ( y ) , Notation.
In a finite field there is no ambiguity in defining π := 2 q. The axiomatic study of Geometry has a long history, starting with Euclid. Among the mainearlier contributors are Giovanni Saccheri (1667-1733), Karl Gauss (1777-1855), Janos Bolyai(1802-1860), Nikolai Ivanovich Lobachevsky (1792-1856), de Tilly (1837-1906) , Pieri, CarlMenger, Oswald Veblen (1880-1960), William Young (1863-1942), Julius Dedekind, FrederigoEnriques (1871-1946), I. Schur, David Hilbert (1862-1943), Marshall Hall and Alfred Tarski(1901-1983).To obtain a clear understanding of the relation between the synthetic and the algebraicpoint of view, an important step was the realization of the connection between the Axiomof Pappus and the commutativity of multiplication, first considered by Schur, in 1898, thenby Hilbert in 1899 (p. 71), by Artin in 1957 and many others, see Artzy (1965), Hartshorne(1967).A detailed history of the developments concerning Finite Geometry can be obtained fromthe monumental work of Dembowsky, 1968 and Pickert (Chapter 12).For some authors, the word projective geometry as moved away from its original meaning,to become a synonym of incidence geometry. I will not follow that practice.What follows can be used to obtain a justification of the relation between the syntheticand algebraic axioms of Chapter II. With the exception of the proof of associativity andcommutativity of addition, I have borrowed heavily from Artzy’s book, which contains proofsnot given here, increasing the formalism to prepare for eventual computarization.The axioms will progress from those ofthe perspective plane, with (Σ , + , · ) a ternary ring, ( A ∗ B ∗ C ) and (Σ , +), (Σ − { } , · ) are G19.TEX [MPAP], September 9, 2019 Blumenthal considers than in the paper of 1892, de Tilly makes a fundamental contribution by intro-ducing n-point relations to characterize a space metrically. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS loops,to Veblen-Wedderburn plane, with (Σ , + , · ) a quasifield, (linear, right distributivity) and(Σ , +) an Abelian group,to Moufang plane, with (Σ , + , · ) an alternative division ring (left distributivity, right andleft inverse property),to Desarguesian plane, with (Σ , + , · ) a skew field, (associativity of multiplication),to Pappian plane, with (Σ , + , · ) a field, ( commutativity of multiplication),to Separable Pappian plane, with (Σ , + , · ) an ordered field,to Continuous Pappian plane with (Σ , + , · ) the field of reals.The definitions of Desargues and Pappus configurations, given in Chapter II, will not berepeated here. Introduction.
Marshall Hall and D.T. Perkins independently succeeded to construct an algebraic structure,called ternary ring, 1.10.1 to coordinatize 1.10.1 the perspective plane1.10.1. Theorem 1.10.1shows that the first 4 conditions of the definition of a ternary ring are associated with theincidence property 1.10.1.3 and the others with Theorem 1.10.1. Theorem 1.10.1 proves thatthe set of the ternary ring is a loop under addition and multiplication.
Axioms. [Of Allignment]
Given a set of elements called points and a set of elements called lines with the relation of incidence , such that0. 2 points are incident to one and only one line.1. 2 lines are incident to one and only one point.2. there exists at least 4 points, any 3 of which are not collinear,we say that the axioms of allignment are satisfied.The terminology is that of Seidenberg, 1962, p. 56.
Definition. A perspective plane is a set of points and lines satisfying the axioms of alignment. It is alsocalled a rudimentary projective plane . (Artzy, p. 201.) Theorem.
Duality is satisfied in a perspective plane.
Menger gives a self dual set of equivalent axioms. .10. AXIOMATIC
Definition.
Given a point P and 2 lines a and b not incident to P, a perspectivity Π( P, a, b ) is thecorrespondance between A i ι a and B i ι b, with B i := ( P × A i ) × b. Π − ( P, a, b ) := Π(
P, b, a ) is the inverse correspondance which associates to B i ,A i = ( P × B i ) × a. I will also use the notation Π(
P, A i , B i ).A projectivity is a perspectivity or the composition of 2 or more perspectivities. Theorem. Π is a bijection. Definition.
Given a line m , we say that l is m-parallel to l (cid:48) iff l, l (cid:48) and m are incident and we write l // m l (cid:48) and I ml := l × m . I ml is called the m-direction of l . Definition.
Given a line m , 2 points A and A (cid:48) , not on m and a point B neither on m nor on a := A × A (cid:48) , the translation T m,BAA (cid:48) is the transformation which associates to I, I if I ιm and to points P neitheron a , nor on m the point P (cid:48) := ( P × I mA × A (cid:48) ) × ( A (cid:48) × I mA × P ) , and to C ι a , C (cid:48) := ( I mB × C × B (cid:48) ) × a, where B (cid:48) := T m,BAA (cid:48) ( B ) . Definition. [Marshall Hall] (Σ , ∗ ) is a ternary ring iff Σ is a set and ∗ is an operation which associates to an orderedtriple in the set an element in the set satisfying the following properties0. A ∗ ∗ C = C,
1. 0 ∗ B ∗ C = C,
2. 1 ∗ B ∗ B, A ∗ ∗ A, A ∗ B ∗ X = D has a unique solution X, B (cid:54) = B = ⇒ X ∗ B ∗ C = X ∗ B ∗ C has a unique solution X, A (cid:54) = A = ⇒ A × X × X (cid:48) = D and A × X × X (cid:48) = D have a unique solution(X,X’) . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem. X ∗ ∗ has a unique solution X . Proof: 1.10.1.0 implies X ∗ ∗ , (cid:54) = 0, the Theorem follows from 1.10.1.5. Definition.
A perspective plane can be coordinatized as follows, (Fig. 20a’)H0.0. Q , Q , Q , U , 4 points, no 3 of which are collinear,D0.0. q := Q × Q , q := Q × Q , m := q := Q × Q ,D0.1. v := Q × U , i := Q × U , V := i × q , I := v × q ,D0.2. u := V × I .Let Σ be the set of points on q , distinct from Q . Define 0 := Q , 1 := V .The point Q is represented by ( ∞ ), ∞ being a new symbol.The points Q on q , distinct from Q are represented by the element Q in Σ, placed betweenparenthesis, Q = (Q).A point P not on q is represented by a pair of elements ( P , P ) in Σ defined by (Fig. 20a”) P := ((((( P × Q ) × i ) × Q ) × v ) × Q ) × q ,P := ((( P × Q ) × v ) × Q ) × q . In particular, if a point A is on q , then its second coordinate A = 0, we represent itsfirst coordinate by A , if a point C is on q , then its first coordinate C = 0, we represent itssecond coordinate by C . Points on v have first coordinate 1.The line q is represented by [ ∞ ],a line l through Q distinct from q is represented by [ A ], with a × q = ( A, , a line m not through Q is represented by the pair M , M ], where ( M ) is the representationof the point m × q and where the point m × q on q is (0 , M ).Let P := (( A, × Q ) × (( B ) × (0 , C )) = ( A, Y ) . Y is a function of A , B and C whichwe denote by Y := A ∗ B ∗ C . Theorem.
There is a bijection between the points ( A, on q , (0 , A ) on q and ( A ) on q . Proof: We use the perspectivity Π( Q , q , i ) , followed by Π( Q , i, q ) , or Π( Q , q , u ) , followed by Π( Q , u, q ) . Comment. If P = ( P , P ), all points on P × Q have the same first coordinate, P , in particular,( P × Q ) × q = ( P , P × Q have the same second coordinate, P , inparticular, ( P × Q ) × q = (0 , P ).In Euclidean Geometry, if q is the x axis, q is the y axis q is the ideal line and U = (1 , , , ( A, B ) corresponds to (
A, B, , ( A ) to (1 , A,
0) which is the direction of lineswith slope A, ( ∞ ) to (0 , ,
0) which is the direction of y axis. The slope of the line joiningthe origin to ( A, B,
1) is BA . .10. AXIOMATIC Theorem.
The incidence, noted “ ι ” satisfies ,0. ( Q ) ι [ ∞ ],1. ( P ) ι [ P , P ],2. ( P , P ) ι [ P ]),3. ( P , P ) ι [ M , M ] iff P = P ∗ M ∗ M . Theorem.
0. ( R , R ) ι ( Q × ( A, ⇒ R = A ,1. ( S , S ) ι ( Q × (0 , C )) = ⇒ S = C ,2. X := v × ( Q × ( B )) = ⇒ X = (1 , B ),3. ( Y , Y ) ι ( Q × V ) = ⇒ Y = Y . Theorem.
The pespective plane as coordinatized in satisfies the properties of a ternary ring. Inparticular the unique solution X of A ∗ B ∗ X = D is the second coordinate of the point (( A, D ) × ( B )) × q , with B (cid:54) = B , the unique solution X of X ∗ B ∗ C = X ∗ B ∗ C is the first coordinateof the point ((0 , C ) × ( B )) × ((0 , C ) × ( B )) , with B (cid:54) = B , the unique solution ( X, X (cid:48) ) of A × X × X (cid:48) = D and A × X × X (cid:48) = D is given by X := (( A , D ) × ( A , D )) × q , X (cid:48) := (( A , D ) × ( A , D )) × q , Proof: For 0. to 3. of 1.10.1, we consider the points( R , R ) := (( A, × Q ) × ((0) × (0 , C )) = ( A, A ∗ ∗ C ) = ( A, C ),( S , S ) := (( Q × Q ) × (( B ) × (0 , C )) = (0 , ∗ B ∗ C ) = (0 , C ), X := ((1 , × Q ) × (( B ) × (0 , , ∗ B ∗
0) = (1 , B ) , ( Y , Y ) := (( A, × Q ) × (( V ) × (0 , A, A ∗ ∗
0) = (
A, A ). Theorem.
The pespective plane satisfies also the properties: X ∗ B ∗ C = D has a unique solution, the first coordinate of the point ((0 , D ) × q ) × (( B ) × (0 , C )) , A ∗ X ∗ C = D has a unique solution, the coordinate of the point (( A, D ) × (0 , C )) × q , Proof: For 0, (
X, Y ) ι [ B, C ] , ( X, Y ) ι [0 , D ] , therefore Y = X ∗ B ∗ C = X ∗ ∗ D = D .For 1, ( A, D ) ι [ X, C ] therefore D = A ∗ X ∗ C .30 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Example. Q = (0 , , Q = (0) , Q = ( ∞ ) , U = (1 , V = (1) , I = (1 , u = [1 , z ] with 1 ∗ ∗ z = 0 .q = [0 , , q = [0] , q = [ ∞ ] , v = [1] , i = [1 , , Let (see Fig. 20a’)D0.3, J := u × q j := U × Q , W := j × q , w := J × Q ,D0.4. T := v × w , t := V × W, R := t × q , r := T × Q , S := r × q , then, with S = (S), J = (0 , S ), j = [0 , , W = (0 , , w = [0 , S ] , T = (1 , S ) , t = [1 , , r = [ S, , R = ( y,
0) with y ∗ ∗ . Definition.
The dual coordinatization can also be chosen. I will use the subscript d to indicate the dualrepresentation,The notation in the preceding example is chosen to allow the dual coordinatization using aselements of Σ the lines through a given point ( ∞ ) . We choose ( ∞ ) d as Q . The line q isrepresented by [ ∞ ] d , the line l := Q × ( L ,
0) is represented by [ l ] d = [ L ] , the line n notthrough Q is represented by [ n , n ] d , with n := ((((( l × q ) × I ) × q × V ) × q ) × Q ,n := ((( l × q × V ) × q ) × Q . in this case q = [0 , d , q = [0] d , q = [ ∞ ] d , u = [1 , d , , w = [0 , d , i = [1 , d , but j = [0 , R ] d , with t × q = ( R, d and t = [1 , R ] d . The representation of points is done duallyas in 1.10.1, with N represented by ( N , N ) d , with N × Q = [ N ] d and N × Q = [0 , N ] d . Theorem.
0. (Σ , +) is a loop, with 0 as neutral element,
1. (Σ − { } , · ) is a loop with 1 as neutral element. Proof: For the addition, the neutral element property follows from 1.10.1.1 with B = 1and from .3. The solution property follows from 1.10.1.4 and .6 with B = 1.For the multiplication, the neutral element property follows from 1.10.1.3 and 4. The solutionproperty follows from 1.10.1.4 and .5 with C = 0. Theorem.
If the number of elements in Σ is a small number n ,0. If n = 2,3,4,5, there is only one perspective plane, If n = 6, there is no perspective plane, If n = 14,21,22,30,33,38,42,46,54,57,62,66,69,70,77, 78,86,93,94,. . . , If n = 10, there is no perspective plane, .10. AXIOMATIC
0, is easily settled, see II1, originates with the problem of the 36 officers, Euler (1782), was settled by Tarry (1900),2, depends on the next Theorem,3, has a long history, and was finaly proven, using computers, by Lam, Thiel and Swiercz(1989), see also Lam (1991).
Theorem. [Bruck and Ryser] If n ≡ , mod x, y such that x + y = n then there are noperpective plane of order n . Notation. A + B := A ∗ ∗ B , A · B := A ∗ B ∗ , A + ( A (cid:96) B ) = B, ( B (cid:97) A ) + A = B. When A (cid:54) = 0 , A · ( A \ B ) = B, ( B/A ) · A = B. In the Euclidean case, the line joining the point ( A, A + B ) to the point (0 , B ) has slope1 and the slope of the line joining Q to ( A, A.B ) is C = A.B . Definition.
A ternary ring (Σ , ∗ ) is linear iff for every A , B , C in the set ( A ∗ B ∗ ∗ ∗ C = A ∗ B ∗ C. Theorem.
If a ternary ring is linear then A ∗ B ∗ C = A · B + C. Axiom. [Fano]
The diagonal points of every quadrangle are not collinear.
Axiom. [N-Fano]
The diagonal points of every quadrangle are collinear.
Definition. A Fano plane is a perspective plane which satisfies the N-Fano axiom.
Theorem.
In a Fano plane A + A = 0 .Proof: For the quadrangle Q = (0 , , X A = ( A, , Y A = (0 , A ) , A A, A ) , q , therefore the third diagonal point is V = ( Q × A × q , therefore ( A, A ∗ ∗ A ) coincides with X A and A + A = 0 . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Exercise.
0. Prove that in a Fano plane ( A ∗ B ) ∗ ( A · B ) = 0 .
1. Determine a subset of quadrangles with collinear diagonal points which justify the pre-ceding property in a perspective plane.2. Same question for the property A + A = 0 . Definition.
Two triangles { AP Q } and { A (cid:48) P (cid:48) Q (cid:48) } are m -parallel iff A × P // m A (cid:48) × P (cid:48) , A × Q // m A (cid:48) × Q (cid:48) , P × Q // m P (cid:48) × Q (cid:48) . Theorem. If A ι l , l (cid:48) := A (cid:48) × I ml and P ι l then P (cid:48) ι l (cid:48) . In general, a line n not through A is not transformed into a line. For this to be so, if P ι n and
Q ι n , we want P (cid:48) := T mAB ( P ) and Q (cid:48) := T mAA (cid:48) ( Q ) to be collinear with I mP × Q . Thissuggest the following Definition. Axiom. [Of Desargues]
In a perspective plane, given any 2 triangles { A i , a i } and { B i , b i } ,let c i := A i × B i , and C i := a i × b i , incidence ( c i , C ) = ⇒ incidence ( C i , c ) . C is called the center , c is called the axis of the configuration.I write Desargues ( C, { A i } , { B i } ; (cid:104) C i (cid:105) , c ) . Axiom. [Elated Desargues]
The
Elated Desargues axiom is the special case when we restrict Desargues’ axiom to thecase when the axis c passes through the center C of the configuration. More specifically, C ι c ,and for the 2 triangles { A i } and { B i } ,let C i := ( A i +1 × A i − ) × ( B i +1 × B i − ) ,c i := ( A i × B i ) , c i ι C , i = 0 , , , incidence( A × A j , B × B j , c ) , j = 1 , , = ⇒ incidence( A × A , B × B , c ) . We writeElated-Desargues ( C, { A i } , { B i } ; (cid:104) C i (cid:105) , c ) . The terminology comes from that in projective geometry, which calls elation, a collineationwith an axis of fixed point and a center of fixed lines, with the center on the axis. This axiomis also called the minor Desargues axiom, see for instance Artzy, p. 210.
Theorem.
Given 2 triangles { A i } and { B i } , let C i := ( A i +1 × A i − ) × ( B i +1 × B i − ) , C i := A i × B i , and C := c × c , (cid:104) C i , c (cid:105) and C ι c = ⇒ c ι C. We write
Elated-Desargues − ( c, { A i } , { B i } ; (cid:104) c , c , c (cid:105) , C ) Proof: Desargues ( C , { A , B , C } , { A , B , C } ; (cid:104) B , A , C (cid:105) , c ) . .10. AXIOMATIC Definition. A Veblen-Wedderburn plane is a perspective plane for which the elated Desargues axiom issatisfied on a specific line of the plane.
Comment.
In all the construction that follow, H0.0 and .1, D0.0 to .4, of 1.10.1 and 1.10.1 will beassumed, but not all these constructions are necessarily required.
Lemma. [For the linearity property.]
H1.0. X A = ( A, , Y C := (0 , C ) , ( B ) , (See Fig. 21a)D1.0. j b := Q × B, j (cid:48) b := Y C × B, j := Q × V, j (cid:48) := Y C × V, D1.1. x := X A × Q , K := x × j b , k := K × Q , L := k × j , D1.2. K (cid:48) := x × j (cid:48) b , c := L × Q , L (cid:48) := c × j (cid:48) , k (cid:48) := L (cid:48) × K (cid:48) , C1.0. Q ι k (cid:48) . Moreover, K = ( A, A · B ) , L = ( A · B, A · B ) , K (cid:48) = ( A, A ∗ B ∗ C ) , L (cid:48) = ( A · B, A · B + C ) , C1.0 = ⇒ A ∗ B ∗ C = A · B + C. Proof:Elated-Desargues( Q , { Q , K, L } , { Y C , K (cid:48) , L (cid:48) } ; (cid:104) Q , V, B (cid:105) , q ) = ⇒ Q ι k (cid:48) . Lemma. [For the additive associativity law]
H1.0. X A , X B , Y C , (See Fig. 21b)D1.0. a := X A × Q , D1.1. b := X B × Q , B := b × i, x := B × Q , Y := x × q , D1.2. i := Y × V, A := i × a, x := A × Q , D := x × i, D1.3. d := D × Q , i := Y C × V, D := d × i , D1.4. B := i × b, x := B × Q , Y := x × q , D1.5. i := Y × V, A := i × a, x := A × D , D1.6. e := A × B , e := A × B , E := e × e , C1.0.
E ι q , C1.1. Q ι x , Moreover, B = ( B, B ) , Y B = (0 , B ) , A = ( A, A + B ) , D = ( A + B, A + B ) ,B = ( B, B + C ) , Y = (0 , B + C ) , A = ( A, A + ( B + C )) ,D = ( A + B, (( A + B ) + C ) ,C1.1. = ⇒ A + ( B + C ) = ( A + B ) + C. Proof: Elated-Desargues( Q , { A , B , Y } ; { A , B , Y } ; (cid:104) Q , V, E (cid:105) , q ) , variant due to Michael Sullivan, October 24, 1989. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS = ⇒ Elated-Desargues − ( q , { A , B , D } ; { A , B , D } ; (cid:104) V, Q , E (cid:105) , Q )= ⇒ Q ι x . Corollary.
If 2 m -parallelograms { A j } and { B j } , j = 0 , , , , are such that A k × B k // m A × B , k = 1 , , the same is true for k = 3 . (See Fig. 21e) The parallelograms for which the proof is given in the Lemma are { A , Y B , Y , A } and { D , B , B , D } . Lemma. [For the right distributive law]
H1.0. X A = ( A, , Y = (0 , B ) , ( C ) , (See Fig. 21c)D1.0. x := X A × Q , D1.1. x := Q × Y , B := x × i, i := Y × V, A := i × x, D1.2. x := A × Q , F := x × i, f := F × Q , c := Q × C, D1.3. b := B × Q , B := b × c , F := f × c , D1.4. x := B × Q , Y := x × q , e := Y × A , e := B × F , D1.5. E := e × e , D1.6. c := Y × C, A := c × x, x := A × F , C1.0.
E ι q .C1.1. Q ι x . Moreover, B = ( B, B ) , A = ( A, A + B ) , F = ( A + B, A + B ) , B = ( B, B · C ) ,F = ( A + B, ( A + B ) · C ) , Y = (0 , B · C ) , A = ( A, A ∗ C ∗ ( B · C )) , andC1.1 = ⇒ ( A + B ) · C = A · C + B · C. Proof:Elated-Desargues( Q , { Y , A , Y } , { B , F , B } ; (cid:104) E, Q , V (cid:105) , q ) = ⇒ E ι q . Elated-Desargues − ( q , { A , A , Y } , { F , F , B } ; (cid:104) E, C, Q (cid:105) , Q ) = ⇒ Q ι x . Finally, from C1.0 follows ( A + B ) · C = A ∗ C ∗ ( B · C ) , but by linearity, the second memberequals A · C + B · C. Exercise.
Determine the identity corresponding to C1.0 or to the m -parallelism of Y × A and B × F . Lemma. [For the commutativity law]
H1.0. A , B , (See Fig. 21d)D1.0. a := A × Q , A := a × q , a := A × B , D1.1. b := B × Q , B := b × q , b := A × B, D := b × a , D1.2. x := A × Q , Y A := x × q , b := Y A × B, D1.3. y := B × Q , B := y × b , x := B × D, y := A × Q , D1.4. x := B × Q , Y B := x × q , a := Y B × A, A := a × y , C1.0. Q ι x , C1.1. A ι x , .10. AXIOMATIC A = ( X A , Y A ) , and B = ( X B , Y B ) , then A = ( X A , Y A + Y B ) , B = ( X B , Y B + Y A ) , C1.0 and .1 = ⇒ Y A + Y B = Y B + Y A .Proof:Elated-Desargues( B, { Q , Y A , A } , { B , B , D } ; (cid:104) Q , A, Q (cid:105) , q ) = ⇒ Q ι x . Elated-Desargues( A, { Q , Y B , B } , { A , A , D } ; (cid:104) Q , B, Q (cid:105) , q ) = ⇒ A ι x . therefore A and B have the same second coordinate Y .Because A ι a ι ( A ) , Y A = X A · A, by construction and because of linearity, A = ( X A , X A ∗ A ∗ Y B ) = ( X A , X A · A + Y B ) similarly Y B = X B · B, and B = ( X B , X B · B + Y A ) . Corollary.
If we make the same constructions as in the lemma with A = B = J, then Q ι ( A × B ) . Lemma. [Addition an Negation in Veblen-Wedderburn planes.]
H0.0. Y A , Y B , ( F ig. e ) D1.0. i := Y A × V, i := Y B × V, D1.1. x := Y A × Q , A := x × i, a := A × Q , A := a × i , D1.2. x := Y B × Q , B := x × i, b := B × Q , B := b × i , D1.3. x := A × B , C1.0. Q ι x , D2.0. U := x × v, c := U × Q , A := c × q , D2.1. c − := Y A × I, A − := c − × q , Moreover,If Y A = (0 , A ) and Y B = (0 , B ) , then A = ( A, A ) , B = ( B, B ) , A = ( A, A + B ) , B =( B, B + A ) ,U = (1 , A ) , A = ( A ) , A − = ( − A ) . Theorem.
In a Veblen-Wedderburn plane, the ternary ring (Σ,*) is a quasifield in the terminology ofDembowski (p. 129):0. (Σ , ∗ ) is linear, a ∗ b ∗ c = a · b + c ,1. ( Σ ,+) is an abelian group,
2. ( Σ − { } , · ) is a loop,
3. ( Σ , ∗ ) = (Σ , + , · ) is right distributive, ( a + b ) · c = a · c + b · c .4. a (cid:54) = b = ⇒ x · a = x · b + c has a unique solution. Theorem.
In a Veblen-Wedderburn plane with ideal line m , T m,BAA (cid:48) ( C ), C ι A × A (cid:48) , is independent of B .We can therefore use T mAA (cid:48) as notation for a translation.36 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition. m -equality is defined by [ A, A (cid:48) ] = m [ P, P (cid:48) ] iff P (cid:48) = T mAA (cid:48) ( P ) . Theorem.
In a Veblen-Wedderburn plane we can use systematically 3 coordinates as follows( Q ) is equivalent to (0 , , P ) is equivalent to (1 , P , , ( P , P ) is equivalent to ( P , P , , [ q ] is equivalent to [0 , , , [ M ] is equivalent to [1 , , − M ],[ M , M ] is equivalent to [ M , − , M ] . A point ( P , P , P ) is incident to the line [ l , l , l ] iff P l + P l + P l = 0 . Proof: In the general case, because of linearity, a point ( P , P ) is incident to the line [ M , M ] if P = P · M + M , which we can rewrite P · M + P · ( −
1) + 1 · M . The other correspondances can be verified using 1.10.1.
Theorem.
In a Veblen-Wedderburn plane with ideal line m , m -equality is an equivalence relation. Definition. A Moufang plane is a Veblen-Wedderburn plane in which the elated Desargues axiom issatisfied for every line in the plane. (See Fig. 3f ).
Theorem.
Duality is satisfied in a Moufang plane.
Definition.
The
C-Desargues Configuration is a Desargues Configuration, for which 2 correspondingsides intersect on the line joining the other vertices. The point of intersection will be under-lined.
Lemma.
The Elated-Desargues Configuration for all lines in the planes implies the C-Desargues Con-figuration. .10. AXIOMATIC
Proof: (See Fig 3f.) To prove C-Desargues ( C, { A i } , { B , B , B } ; (cid:104) C i (cid:105) , c ) , we apply Elated-Desargues − ( c , { A , B , C } , { A , B , C } ; (cid:104) B , A , C (cid:105) , C ) . Definition.
The is a Desargues Configuration, for which the vertex of 1 tri-angle is on the side of the other, this vertex will be underlined.
Lemma.
The Elated-Desargues Configuration for all lines in the planes implies the 1-Desargues Con-figuration.
Proof: (See Fig 3b.) To prove 1-Desargues ( C, { A i } , { B i } ; (cid:104) C , C , C (cid:105) , c ) , we apply Elated-Desargues ( B , { A , C , C } , { C, B , B } ; (cid:104) C , A , A (cid:105) , a ) . Theorem.
In a Moufang plane the C-Desargues Theorem is true .1. the 1-Desargues Theorem is true . Lemma. [For the left distributive law]
H1.0. X A = ( A, , ( B ) , ( C ) , (See Fig. 22a)D1.0. a := X A × Q , c := Q × C, U := c × u, x := U × Q , D1.1. Y := x × q , b := Y × B, U := b × u, D1.2. A := a × c , x := A × Q , Y := x × q , D1.3. b := Y × B, A := b × a, d := U × Q , C1.0. A ι d. Moreover, U = (1 , C ) , Y = (0 , C ) , U = (1 , ∗ B ∗ C ) , A = ( A, A · C ) , Y = (0 , A · C ) , A =( A, A ∗ B ∗ ( A · C ) , C1.0 = ⇒ A · B ) + ( A · C ) = A ∗ B ∗ ( A · C ) = A · (1 ∗ B ∗ C ) = A · ((1 · B ) + C ) = A · ( B + C ) . Proof:C-Desargues ( Q , { Y , U , U } , { Y , A , A } ; (cid:104) Q , B, Q (cid:105) , q ) = ⇒ (( U A × A ) × ( U × U )) ι ( Y × Y ) . Lemma. [For the inverse property]
H1.0. A, (See Fig. 22b)D1.0. a := X A × Q , A := a × j, A := a × i, D1.1. a := A × Q , A := a × u, a := A × Q , A := a × u, D1.2. a := A × Q , A := a × j, D1.4. a := A × Q , a := A × Q , A := a × a , D1.5. d := A × A , d := A × A , E := d × d , CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
C1.0.
E ι q , C1.1. A ι i. Moreover, A = ( A, , A = ( A, A ) , A = (1 , A ) , A = ( A L , , A = (1 , A R ) ,A = ( A L , A R ) , C1.1 = ⇒ A L = A R . Proof:1-Desargues ( Q , { A , A , A } , { A , A , U } ; (cid:104) Q , Q , E (cid:105) , q ) = ⇒ E ι q . − ( Q , { A , A , A } , { A , A , U } ; (cid:104) Q , Q , E (cid:105) , q ) = ⇒ A ι i. Notation. If B (cid:54) = 0 , we write B − = B R . Lemma. [For the right inverse property]
H1.0. X A , B, (See Fig. 22c)D1.0. a := X A × Q , A := a × i, b := Q × B, C := j × b, D1.1. c := C × Q , C := c × i, x C × Q , U := x × u, D1.2. A := a × b, x := A × Q , AB := x × i, ab := AB × Q , D1.3. b (cid:48) := U × Q , AB := ab × b (cid:48) , x := A × AB , D1.4. d := U × A , e := U × A , S := d × e, C1.0.
S ι q , C1.1. Q ι x . Moreover, A = ( A, A ) , C = ( B − , , C = B − , B − ) , U = (1 , B − ) , A = ( A, A · B ) , AB =( A · B, A · B ) , AB = ( A · B, ( A · B ) · B − ) , C1.1 = ⇒ ( A · B ) · B − = A. Proof:1-Desargues( Q , { U , A , C } , { U , A , C } ; (cid:104) Q , Q , S (cid:105) , q ) = ⇒ S ι q . − ( Q , { U , A , AB } , { U, A , AB } ; (cid:104) Q , Q , S (cid:105) , q ) = ⇒ Q ι x . Lemma. [For the left inverse property]
H1.0. X A , B, (See Fig. 22d)D1.2. b := Q × B, U := b × u, x := U × Q , D1.3. a := X A × Q , A := a × j, b (cid:48) := Q × A , U := b (cid:48) × u, D1.4. x := U × Q , C := x × i, c := C × Q , C := c × x , D1.5. A := a × b, x := A × Q , U := x × u, D1.6. ab := U × Q , D1.7. r := U × A , r := U × C , R := r × r , C1.0.
R ι q , C1.1. C ι ab, Moreover, A = ( A, , A = ( A, A · B ) , U = (1 , A − ) , U = (1 , B ) , U = (1 , A · B ) , C = ( A − , A − ) ,C = ( A − , A − · ( A · B )) . C1.1 = ⇒ A − · ( A · B ) = B. .10. AXIOMATIC Proof:1-Desargues( Q , { A , U , A } , { U , C , U } ; (cid:104) R, Q , Q (cid:105) , q )= ⇒ − ( q , { U, U , A } , { C , C , U } ; (cid:104) Q , R, Q (cid:105) , Q ) = ⇒ C ι ab. Theorem.
With the coordinatization of the plane as given in the ternary ring (Σ , + , · ) is left distributive, or A · ( B + C ) = A · B + A · C .1. B (cid:54) = 0 = ⇒ B R = B L = B − , ( A · B ) · B − = B − · ( B · A ) = A for all A . In other words, (Σ , + , · ) is an alternative division ring. Definition. A Desarguesian plane is a plane in which the Desargues Axiom is always satisfied.
Theorem.
Duality is satisfied in a Desarguesian plane.
Comment.
Instead of the Axiom of Desargues one can use the equivalent axiom of Reidemeister (SeeTheorem II.2.1.8 and Klingenberg, 1955).
Lemma. [For Associativity]
H1.0. X A , B, C, (See Fig. 23.)D1.0. b := Q × B, c := Q × C, U := b × u, x := U × Q , D1.1. D := x × i, d := D × Q , D := d × c, x := D × Q , D1.2. U := x × u, bc := U × Q , D1.3. a := X A × Q , A := a × b, x := A × Q , AB := x × i, D1.4. ab := AB × Q , AB ab × c, x := AB × Q , A := x × a, D1.5. r := U × D , r := A × AB , R := r × r , C1.0. A ι bc, Moreover, A = ( A, A · B ) , AB = ( A · B, A · B ) , AB = ( A · B, ( A · B ) · C ) ,A = ( A, A · ( B · C )) , U = (1 , B ) , D = ( B, B ) , D = ( B, B · C ) , U = (1 , B · C ) . C1.0 = ⇒ A · ( B · C ) = ( A · B ) · C. Proof:Desargues( Q , { D , D , U } , { AB , AB , A } ; (cid:104) R, Q , Q (cid:105) , q )= ⇒ Desargues − ( q , { U , U , D } , { A , A , AB } ; (cid:104) R, Q , Q (cid:105) , Q ) = ⇒ A ι bc. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem.
With the coordinatization of the plane as given in 1.10.1 ,0. (Σ , · ) is associative, A · ( B · C ) = ( A · B ) · C. In other words, ( Σ , + , · ) is a skew field. Theorem.
If a Desarguesian plane we use the coordinates of 1.10.2, we can make them homogeneousby multiplying the coordinates of points to the left by the same element in the set Σ , andthose of lines to the right by the same element in the set Σ . Associativity of multiplication is essential to allow for the left equivalence of points andthe right equivalence of lines.
Axiom. [Of Pappus]
In a perspective plane: If A i are 3 distinct points on a line a and B i are 3 distinct points ona line b and C i := ( A i +1 × B i − ) × ( A i − × B i +1 then incidence ( C i ) .I write Pappus ( { A i } , { B i } ; { C i } ) . Definition. A Pappian plane is a plane in which the Pappus Axiom is always satisfied.
Comment.
There are other axioms which are equivalent to that of Pappus. The Fundamental axiomand Axiom A (See Seidenberg, p. 25 and Chapter IV). The Fundamental axiom states thatthere is at most one projectivity which associates 3 given distinct collinear points into 3given distinct collinear. Axiom A states that if a projectivity which associates a line l into adistinct line l (cid:48) leaves l × l (cid:48) invariant then it is a perspectivity. Theorem.
Duality is satisfied in a Pappian plane.
Theorem.
A Pappian plane is a Desarguesian plane. .10. AXIOMATIC
Lemma. [For Commutativity]
H1.0. ( A ) , ( B ) , (See Fig. 24)D1.0. a := Q × A, b := Q × B, U := a × u, D1.1. x := U × Q , C := x × i, c := C × Q , C := c × b, D1.2. U := b × u, D1.3. x := U × Q , D := x × i, d := D × Q , D := d × a, D1.4. x := C × D , C1.0. D ι x , Moreover, U = (1 , A ) , U = (1 , B ) , C = ( A, A ) , D = ( B, B ) , C = ( A, AB ) , D = ( B, BA ) , C1.0 = ⇒ A · B = B · A. Proof:Pappus( (cid:104) D , C , Q (cid:105) , (cid:104) U , U , Q (cid:105) ; (cid:104) C , D , Q (cid:105) ) = ⇒ D ι x . Theorem.
With the coordinatization of the plane as given in 1.10.1 ,0. (Σ , · ) is commutative, a · b = b · a. In other words, (Σ , + , · ) is a field. Theorem.
The field of a Pappus-Fano plane has characteristic 2. Vice-versa if a field has characteristic2, the corresponding Pappian plane satisfies the axiom N-Fano.
Proof: We have seen than in a Fano plane A + A = 0 , for all A ∈ Σ , therefore thecharacteristic of the field is 2. To prove the converse,we choose as coordinates of the verticesof the quadrangle A = (1 , , , A = (0 , , , A = (0 , , and M = (1 , , , the diagonalelements are M = (0 , , , M = (1 , , , M = (1 , , , which are collinear iff . Axiom. [Of separation]
In a perspective plane, if A i , i = 0 , , , , are distinct points on the same line:0. There are at least 4 points on a line.1. σ ( A , A | A , A ) = ⇒ σ ( A , A | A , A ) and σ ( A , A | A , A )
2. only one of the relations σ ( A , A | A , A ) , σ ( A , A | A , A ) , σ ( A , A | A , A ) holds.3. σ ( A , A | A , A ) and σ ( A , A | A , A ) = ⇒ σ ( A , A | A , A ) . Π( P, A j , A (cid:48) j ) , j = 0 , , , , and σ ( A , A | A , A ) = ⇒ σ ( A (cid:48) , A (cid:48) | A (cid:48) , A (cid:48) ) . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition. A separable Pappian plane is a Pappian plane in which the separation axioms are satisfied. Theorem. σ ( A , A | A , A ) = ⇒ σ ( A , A | A , A ) , σ ( A , A | A , A ) , σ ( A , A | A , A ) , σ ( A , A | A , A ) , σ ( A , A | A , A ) σ ( A , A | A , A ) , σ ( A , A | A , A ) . σ ( A , A | A , A ) a ndσ ( A , A | A , A ) = ⇒ σ ( A , A | A , A ) , Notation.
When we use 1.10.6.3 or 1.10.6.1, I will underline the element in each quadruple of pointwhich is distinct, to ease the application of the axiom and write, for instance σ ( A , A | A , A ) and σ ( A , A | A , A ) = ⇒ σ ( A , A | A , A ) , or σ ( A , A | A , A ) and σ ( A , A | A , A ) = ⇒ σ ( A , A | A , A ) . Theorem.
In a Pappus-Fano plane, given a harmonic quadrangle
A, B, C, D , (See Fig. 2a”),
P, R | U, V ,where P , R are diagonal points and U , V are the intersection with P × R of the sides of thequadrangle which are not incident to P or R . Proof: Π( C, { P, U, R, V } , { D, Q, B, V } ) , Π( A, { D, Q, B, V } , { R, U, P, V } ) , therefore P, R | U, V = ⇒ R, P | U, V , while
P, U | R, V = ⇒ R, U | P, V , P, V | U, R = ⇒ R, V | U, P , thelast 2 conclusions are contradicted by 1.10.6.2.
Corollary. ( O, ∞| A, − A ) . Definition.
Given A i , i = 0 , , on a line a , a segment seg ( A , A \ A ) is the set of points A ι a suchthat σ ( A , A | A, A ) . Lemma. If A i ∈ Σ and σ ( A , A | A , A ), σ ( P + A , P + A | P + A , P + A ) ,1. P (cid:54) = 0 = ⇒ σ ( P · A , P · A | P · A , P · A ) . More generally, if Π is a projectivity which associates to X , ( A · X + B ) · ( C · X + D ) , A · C · D (cid:54) = 0, A · D (cid:54) = B · C ,then σ (Π( A ) , Π( A ) , Π( A ) , Π( A )) . .10. AXIOMATIC A i is replaced by ∞ and we use ∞ + A = ∞ and with A (cid:54) = 0, ∞ · A = ∞ . Proof: Π( V, q , p ) ◦ Π( Q , p, q ) transforms (0 , A i ) into ( P, P + A i ) into (0 , P + A i ) . Π( Q , q , i ) ◦ Π( Q , i, Q × ( P )) ◦ Π( Q , i, Q × ( P )) transforms (0 , A i ) into ( A i , A i ) into ( A i , P · A i ) into (0 , P · A i ) . The rest of the proof is left as an exercise.
Lemma.
In a separable Pappian plane, the characteristic is not 2.
Proof: If the characteristic was 2 and A is different from 0, 1 and ∞ ,either σ (0 , | A, ∞ ) or σ (0 , A |∞ , or σ (0 , ∞| , A ) .In the first case, adding 1 or A gives σ (1 , | A + 1 , ∞ ) or σ ( A, A + 1 | , ∞ ) , combining gives σ (1 , A | , ∞ ) which contradicts σ (0 , | A, ∞ ) . In the second case we add 1 or A and in thethird case we add 1 or A + 1 and proceed similarly to show contradiction. Definition. P is positive , or P > , iff σ (0 , ∞| − , P ) . P is negative , or P < , iff − P > or iff σ (0 , ∞| − , − P ) or iff σ (0 , ∞| , P ) . Theorem. > .1. A, B ∈ Σ , A >
B > ⇒ A + B > . A ∈ Σ, either A = 0 or A > − A > A, B ∈ Σ , A >
B > ⇒ A · B > . Proof:For 0, we use Corollary 1.10.6.For 1,
A > ⇒ σ (0 , ∞| − , A ) by the projectivity which associates to X , A − X − ,4. σ ( A − , ∞| A, − ,B > ⇒ σ (0 , ∞|− , B ) = ⇒ (adding A ) σ ( A, ∞| A − , A + B ) = ⇒ (combining with 4.) σ ( − , A + B | A, ∞ ) , with σ (0 , ∞| − , A ) = ⇒ σ (0 , A + B | A, ∞ ) , with σ (0 , ∞|− , A ) = ⇒ σ ( − , A + B | , ∞ ) = ⇒ A + B > . For 2, by the definition of
A > or − A > , it follows that A is not 0. A > and − A > are also mutually exclusive, otherwize A + ( − A ) = 0 would be positive. If A = − , then − A = 1 > . It remains to examine for a given A distinct from 0 and -1, the 3 possibilities, σ (0 , − | A, ∞ ) and σ (0 , ∞| − ,
1) = ⇒ σ (1 , A | , ∞ ) = ⇒ A < .σ (0 , A |∞ , − and σ (0 , ∞| − ,
1) = ⇒ σ (1 , A | , ∞ ) = ⇒ A < .σ (0 , ∞| − , A ) = ⇒ A > . For 3, σ (0 , ∞| − , B ) = ⇒ σ (0 , ∞| − A, A · B ) , A > ⇒ not σ (0 , ∞| − , − A ) , therefore either σ (0 , − |∞ , − A ) or σ ( − , ∞| , − A ) . Inthe first case, σ (0 , ∞| − A, A · B ) and σ (0 , − |∞ , − A ) = ⇒ σ ( − , A · B | , ∞ ) . In the second case, σ (0 , ∞| − A, A · B ) , and σ ( − , ∞| , − A ) = ⇒ σ ( − , A · B | , ∞ ) . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem.
With the coordinatization of the separable Pappian plane as given in 1.10.1 ,0. (Σ , + , · ) is an ordered field. Axiom. [Of continuity]
Let
S ⊂ ( seg ( A, C \ B ) , S non empty, ∃ L and U (cid:51) all P ∈ S , σ ( AP | LU ) = ⇒ ∃ G and H (cid:51) σ ( LP | GU ) and σ ( U P | HL ) . Definition. A Continuous Pappian or Classical Projective Plane is a separable plane for which the con-tinuity axiom is satisfied.
Theorem.
The field associated to a Continuous Pappian plane is the real field R . Introduction.
We have seen that we can coordinatize the various perspective planes by ternary rings whichhave special properties. The converse is also true. If a ternary rings has appropriate prop-erties there exists a plane as defined above which is isomorphic to it. More specifically:
Theorem.
There is an isomorphism between perspective planes and ternary rings (Σ , ∗ ). Veblem-Wedderburn planes and ternary rings with the properties 1.90.2. Moufang planes and alternative division rings. Desarguesian planes and skew fields. Papian planes and fields. .10. AXIOMATIC
Definition. A Moulton plane (1902) is the set of points in the Euclidean plane coordinatized with Carte-sian coordinates and the lines,0. the ideal line, [0,0,1],1. the lines [ m, − , n ], m ≤ ,
2. the lines consisting of two parts, first, the subset of [ m, − , n ], m > , which is in thelower half plane or on the ideal line, second, the subset of [ m/ , − , n ], m > , whichis in the upper half plane. Theorem. The Moulton plane is a perspective plane. The Moulton plane is not a Veblen-Wedderburn plane.
Proof: See Artzy, p. 210.
Definition. A is defined like a quaternion plane with ij = − ji = k replaced by ij = − ji = 2 k . Theorem. The 2-Q plane is a Veblen-Wedderburn plane. The 2-Q plane is not a Moufang plane.
Proof: See Artzy, p. 226.
Definition. A Cayleyian plane is defined like a quaternion plane, using Cayley numbers instead of quater-nions.
Theorem. The Cayleyian plane is a Moufang plane. The Cayleyian plane is not a Desarguesian plane.
Proof: See Artzy, p. 226.
Definition. A quaternion plane is defined using quaternions as coordinates instead of real numbers. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Theorem. The quaternion plane is a Desarguesian plane. The quaternion plane is not a Pappian plane.
Proof: See Artzy, p. 226.
Definition. A finite Pappian plane is a Pappian plane for which the number of points on one line isfinite. The field associated to it is therefore a finite field which is necessarily a Galois field GF ( p k ) with p prime, the number of points being p k + 1 . Theorem. The finite Pappian plane is a Pappian plane. The finite Pappian plane is not a Separable Pappian plane.
Proof: See Artzy, p. 210.
Definition.
If the field is the field of rationals, the Pappian plane is called the
Rational Pappian plane . Theorem. The Rational Pappian plane is a Separable Pappian plane. The Rational Pappian plane is not a Continous Pappian plane.
Proof: See Artzy, p. 210.
Exercise.
Give a synthetic definition of a0. The rational Pappian plane.1. The quaternion plane.2. The Cayleyian plane.3. 2-Q plane.It is clear how to proceed for the rational plane, immitating the definition of the rationalnumbers as equivalence classes of the integers. It is not known to me how to solve the otherexercises. .10. AXIOMATIC
Introduction.
For collineations, correlations and polarities in finite planes, see Dembowski, section 3.3 andChapter 4.
Definition.
Given 1.10.1,we say that the vectors AA (cid:48) and P P (cid:48) are m -equal and we write AA (cid:48) = m P P (cid:48) . Definition.
In a Veblen-Wedderburn plane with ideal line m , the elements of the set V are the equivalenceclasses of m − equal vectors and the addition of vectors is defined by P P + Q Q := P P ,where0. Q = P = ⇒ P = Q , P , Q , Q non collinear = ⇒ P is the point defined by P P = m Q Q ,2. if P , Q , Q collinear and X is not on Q × Q = ⇒ P is the point defined by P P = m Q Q , P P = m Q Q . Theorem.
The addition of vectors is well defined and ( V ,+) is an abelian group. This follows from the fact that any vector is equivalent to a vector (0 , A, B ) for some A and B Theorem.
The translations in a Veblen-Wedderburn plane with ideal line m are collineations, in otherwords, the image of points P on a fixed line l are points P (cid:48) on a line l (cid:48) . Each collineation isan elation with axis m and center m × ( A × A (cid:48) ). Theorem.
For any line n , the n -translations in a Moufang plane are collineations, in other words, theimage of points P on a fixed line l are points P (cid:48) on a line l (cid:48) . Each collineation is an elationwith axis n and center n × ( A × A (cid:48) ).48 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
Definition.
In a in Veblen-Wedderburn Plane the pre correlation configuration is defined as follows, (SeeFig. 25)Hy0. { Q i } , u ι Q , i, a, b, a (cid:48) ι Q ,De. U := u × a, D := d × a, x := U × Q , C := x × i, De. c := C × Q , C := c × b, U := b × u, x := U × Q , De. D := x × i, d := D × Q , D := d × a, x := D × Q , De. D (cid:48) := a (cid:48) × x , d (cid:48) := D (cid:48) × Q , D (cid:48) := d (cid:48) × i, x (cid:48) := D (cid:48) × Q , De. U (cid:48) := x (cid:48) × u, b (cid:48) := U (cid:48) × Q , B (cid:48) := b (cid:48) × q , U (cid:48) := a (cid:48) × u, De. x (cid:48) := U (cid:48) × Q , C (cid:48) := x (cid:48) × i, c (cid:48) := C (cid:48) × Q , C (cid:48) := c (cid:48) × b (cid:48) , De. x (cid:48) := C × Q , Hy1. C (cid:48) ι x (cid:48) , Let ( A ) = a × q , ( B ) = b × q , ( A (cid:48) ) = a (cid:48) × q , ( B (cid:48) ) = b (cid:48) × q , then D = ( B, B · A, D (cid:48) = ( B (cid:48) , B (cid:48) · A (cid:48) ) , C = ( A, A · B, C (cid:48) = ( A (cid:48) , A (cid:48) · B (cid:48) ) , If b (cid:48) or B (cid:48) ischosen in such a way that B · A = B (cid:48) · A (cid:48) , the configuration requires A · B = A (cid:48) · B (cid:48) . Thisdefines a correspondance γ between X = A · B and X (cid:48) = B · A . Exercise.
If we associate to ( Q ) , [ Q ] and to ( P , P ) , [ P γ, P γ ] , is the correspondance is a correlation?If not which of the axioms given below are required for the correspondance to be a correlation. Definition. A three net associated to the 3 points A , B , C in a perspective plane is the set of points P in the plane and the set of lines P × A, P × B, P × C . Theorem.
The coordinates of the lines of the three net associated to the points (0), (1), ( ∞ ) are [0 , P ],[1 , P ], [ P ], where P := ((( P × (0)) × v ) × Q ) × q , P := ((((( P × (1)) × q )) × (0)) × v ) × Q ) × q , P := ((((( P × ( ∞ )) × i ) × (0)) × v ) × Q ) × q . Lemma.
Let Y A = (0 , A ) , Y B = (0 , B ) , then((( Q × (1)) × ( Y A × (0))) × ( ∞ )) × ((( Q × ( ∞ )) × ( Y B × (0))) × (1)) = ( A, A + B ) . Definition.
Given Q (cid:48) = ( F, F + G ) , the F-G-sum of A and B , A ⊕ B is defined by ((( Q (cid:48) × (1)) × ( Y A × (0))) × ( ∞ )) × ((( Q (cid:48) × ( ∞ )) × ( Y B × (0))) × (1)) = ( X, A ⊕ B ) . .10. AXIOMATIC Theorem. A ⊕ B = ( A (cid:97) G ) + ( F (cid:96) B ) .X = A (cid:97) G .Proof: X A := (( Q (cid:48) × (1)) × ( Y A × (0)) = ( X, A ) and X + G = A , therefore X = A (cid:97) G . F B := (( Q (cid:48) × ( ∞ )) × ( Y B × (0)) = ( F, B ) , if Y Z := (( F, B ) × (1)) × q = (0 , Z ) , then C = F + Z and Z = F (cid:96) C, finally X Y := ( X A × ( ∞ )) × ( F B × (1) = ( X A × ( ∞ )) × ( Y Z × (1) = ( X, A ⊕ B ) , therefore ( A ⊕ B ) = X + Z, substituting for X and Z gives the Theorem. Theorem. (Σ , ⊕ ) is a loop.The neutral element is F + G .The solutions of A ⊕ B = C are given by A = ( C (cid:97) ( F (cid:96) B )) + G, B = F + (( A (cid:97) G ) (cid:96) C ) . Proof: The solutions follow directly from the preceding Theorem, the neutral elementproperty follows from ( F + G ) ⊕ H = (( F + G ) (cid:97) G ) + ( F (cid:96) H ) = F + ( F (cid:96) H ) = H,H ⊕ ( F + G ) = ( H (cid:97) G ) + ( F (cid:96) ( F + G )) = ( H (cid:97) G ) + G = H . Theorem.
The coordinates of the lines of the three net associated to the points Q , Q , Q are [ P , , P ], [ P ], where P := ((( P × (0)) × v ) × Q ) × q , P := ( P × Q ) × q , P := ((((( P × ( ∞ )) × i ) × (0)) × v ) × Q ) × q . Exercise.
Determine Theorems analogous to those associated with (0), (1) and ( ∞ ) . See Artzy, p. 206and p.210, 15. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
0. Artin, Emil,
Geometric Algebra , New York, Interscience Publishers, 1957. Inter-science tracts in pure and applied mathematics, no. 3.1. Artzy, Rafael,
Linear Geometry , Reading Mass., Addison-Wesley, 1965, 273 pp.2. Bolyai, Farkas,
Tentamen Juventutem Studiosam ein Elementa Mathiseos Parae, in-troducendi , Maros-Vasarhely, 1829, see Smith D. E. p. 375.3. Bolyai, Janos,
The Science Absolute of Space Independent of the Truth and Falsity ofEuclid’s Axiom XI , transl. by Dr George Brus Halstead, Austin, Texas, The Neomon,Vol. 3, 71 pp, 1886.4. Bolyai, Janos,
Appendix, the theory of space , with introduction, comments, and ad-denda, edited by Ferenc Karteszi, supplement by Barna Szenassy, Amsterdam, NewYork, North-Holland, New York, Sole distributors for the U.S.A. and Canada, Else-vier Science Pub. Co., 1987, North-Holland mathematics studies, 138.5. Bruck R. H. and Ryser H. J.,
The non existence of certain finite projective planes ,Can. J. Math., Vol. 1, 1949, 88-93.6. Dedekind, Julius Wilhelm,
Stetigkeit und Irrationalen Zahlen , 1872, see Smith D. E.,p. 35. (I,9,p.1)7. Dembowski, Peter,
Finite Geometries , Ergebnisse der Mathematik und ihrer Grenzge-biete, Band 44, Springer, New-York, 1968, 375 pp.8. Enriques, Federigo,
Lezioni di Geometria Proiettiva , Bologna, 1904, French Transl.,Paris 1930.9. Enriques, Federigo,
Lessons in Projective Geometry , transl. from the Italian by HaroldR. Phalen. Annandale-on-Hudson, N.Y., printed by the translator, [1932?].10. Euler, Leonhard,
Recherches sur une nouvelle esp`ece de quarr´es magiques , Verh. Zeeuwsch.Genootsch. Wetensch. Vlissengen, Vol. 9, 1782, 85-239.11. Fano, Gino,
Sui Postulati Fondamentali della Geometria Proiettiva , Giorn. di mat.,Vol. 30, 1892, 106-132. (PG( n, p ))12. Hall, Marshall, Jr,
Projective Planes , Trans. Amer. Math. Soc., Vol. 54, 1943,229-277.13. Hartshorne, Robin C.,
Foundation of Projective Geometry , N. Y. Benjamin, 1967,161 pp.14. Hilbert, David,
Grundlagen der Geometrie , 1899, tr. by E. J. Townsend, La Salle, Ill.,Open Court Publ. Cp., 1962, 143 pp. .10. AXIOMATIC
15. Hilbert, David,
The Foundations of Geometry , authorized transl. by E.J. Townsend. . . Chicago, The Open court publishing company; London, K. Paul, Trench, Trubner& co., ltd., 1902.16. Klingenberg, Wilhelm,
Beweis des Desargueschen Satzes aus der Reidemeisterfigur undVerwandte S¨atze . Abh. Math. Sem. Hamburg, Vol. 19, 1955, 158-175.17. Klingenberg, Wilhelm,
Grundlagen der Geometrie , Mannheim, Bibliographisches Insti-tut, 1971, B. I.-Hochschulskripten 746-746a.18. Lam, C. W. H.,
The Search for a Finite Projective Planes of Order 10 , Amer. Math.Monthly, Vo. 98, 1991, 305-318.19. Lam, C. W. H., Thiel, L. H. & Swiercz S.,
The non existence of finite projective planesof order 10 , Can. J. of Mat., Vol. 41, 1989, 1117-1123.20. Lobachevskii, Nikolai Ivanovich, see Norden A.,
Elementare Einfuhrung in die LobachewskischeGeometrie , Berlin, VEB Deuscher Verlag der Wissenschaften, 1958, 259 pp.21. Lobachevskii, Nikolai Ivanovich,
Geometrical Researches on the Theory of Parallels ,translated from the original by George Bruce Halsted, Austin, University of Texas,1891.22. Menger, Karl,
Untersunchungen ¨uber allgemeine Metrik , Math. Ann. Vol. 100, 1928,75-163.23. Moufang, Ruth,
Alternatievk¨orper und der Satz vom Vollst¨andigen Vierseit , Abh.Math. Sem. Hamburg, Vol. 9, 1933, 207-222.24. Pickert. Gunter,
Projektive Ebenen , Berlin, Springer, 1955, 343 pp.25. Pieri,
Un Sistema di Postulati per la Geometria Proiettiva , Rev. Math´em. Torino,Vol 6, 1896. See also Atti Torino, 1904, 1906.26. Pieri,
I Principii della Geometria di Posizione, composti in Sistema Logico Deduttivo ,Mem. della Reale Acad. delle Scienze di Torino, serie 2, Vol.48, 1899, 1-62.27. Reidemeister, Kurt,
Grundlagen der Geometrie , Berlin, Springer, Grundl. der math.Wissens. in Einz., Vol. 32, 1968, (1930),28. Saccheri, Giovanni Girolamo,
Euclides ab omni Naevo Vindicatus , Milan, 1732. tr.George Halstead, London Open Court Pr. 1920, 246 pp. See St¨ackel.29. Schur, Friedrich,
Grundlagen der Geometrie , mit 63 figuren im text. Leipzig, Berlin,B. G. Teubner, 1909.30. Schur, Issai,
Gesammelte Abhandlungen , Hrsg. von Alfred Brauer u. Hans Rohrbach,Berlin, Heidelberg, New York: Springer, 1973. CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
31. Stackel, Paul Gustav,
Die Theorie der Parallellinien von Euklid bis auf Gauss, eineUrkundensammlung zur Vorgeschichte der nichteuklidischen Geometrie, in Gemein-schaft mit Friedrich Engel , hrsg. von Paul Stackel, New York, Johnson Reprint Corp.,1968, Bibliotheca mathematica Teubneriana, Bd. 41.32. Tilly, Joseph Marie de,
Essai sur les Principes fondamentaux de G´eom´etrie et deM´ecanique , Bruxelles, Mayolez, 1879, 192 pp. Also, M´em. Soc. science phys. etnatur. de Bordeaux, Vol III, Ser. 2, cahier 1.33. Tilly, Joseph Marie de,
Essai de G´eom´etrie Analytique G´en´erale , Bruxelles, 1892.34. Tarry, G.,
Le probl`eme des 36 officiers , C. R. Assoc. Franc. Av. Sci., Vol. 1 (1900),122-123, Vol. 2 (1901), 170-203.35. Veblen, Oswald & Wedderburn, Joseph Henri MacLagan,
Non-Desarguesian and non-Pascalian Geometries , Trans. Amer. Math. Soc., Vol. 8, 1907, 279-388.36. Veblen, Oswald & Young, John,
Projective Geometry , Wesley, Boston, I, 1910, II,1918. Geometry is to be the support of the description of phenomenon in the real world. I willbriefly review Newton’s laws and 2 results to be generalized, the central force theorem ofHamilton and the motion of the pendulum.
Introduction.
Among many of the contribution of Kepler those which perpetuate his name are his 3 lawsof Mechanics and his equation discovered from 1605 to 1621. The first and third law are inAstronomi Nova, the second law and his equation in section V of his Epitome. . . .
We alsoknow that an ellipse can be generated by moving a segment of length a + b with one point onan axis and the other point on a perpendicular axis. 1.11.1.2 shows that the angle of the lineis also the eccentric anomaly.? .11. MECHANICS. Theorem.
If a point P ( x, y ) is restricted to move on an ellipse with major axis 2 a, minor axis 2 b andeccentricity e, and origin at a focus, x = a ( cos ◦ E − e ) , y = b sin ◦ E, where E is the excentric anomaly.If E (0) = 0 , then x (0) = ae, y (0) = 0 . Let I be the identity function, the motion preserves area iff Kepler’s equation1. I = E − esin ◦ E is satisfied.If v := ∠ ( A, F, P ) , called true anomaly then tanv = bsinEa ( cosE − e ) . Finally, if the line through P makes an angle E with AF, and intersect the major axisat L and the minor axis at M, P L = b, P M = a. Let A be twice the area (0 , , ( a, , ( x, y ) along the ellipse, divided by ab. Let T be twice the area of the triangle (0 , , ( x, y ) , ( x (cid:48) , y (cid:48) ) divided by ab, then T = xy (cid:48) − x (cid:48) y = ( cos ◦ E − e ) sin ◦ E (cid:48) − ( cos ◦ E (cid:48) − e ) sin ◦ E = sin ◦ ( E (cid:48) − E ) − e ( sin ◦ E (cid:48) − sin ◦ E ) , If E (cid:48) = E + ∆ E, and ∆ E is small, then ∆ A = (1 − ecos ◦ E )∆ E. Integrating gives6. A = E − e sin ◦ E. Therefore, if the area A is a linear function, with a proper choice of the unit of time, A = I and we have 1. Vice-versa, if 1. is satisfied then comparing 6, and 1, gives A = I and the area is proportional to the time. Theorem. [Hamilton]
Assuming Newton’s law, if a mass is to move on an ellipse, under a force passing through afixed point (central force), this force is proportional to the distance to the center and inversely proportional tothe cube of the distance to the polar of the center of force.54 CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS the relation between the eccentric anomaly E and the time t is given by aE ( t )+ c sin ( E ( t )) = C t.
Consider a conic with major axis of length 2 a on the x axis, with minor axis of length2 b and with center at ( c, d ) , the parametric representation is x = c + a cos ◦ E, y = d + b sin ◦ E. The acceleration is D x = − a cos ◦ E ( DE ) − a sin ◦ ED E, D y = − b sin ◦ E ( DE ) + b cos ◦ ED E, If we accept Newton’s law, the acceleration has to be in the direction of the force, if theforce is f ◦ E g ◦ E, where4. ( g ◦ E ) = ( c + a cos ◦ E ) + ( d + b sin ◦ E ) ,g ◦ E being the distance to the center of force,5.0. D x = f ◦ E ( c + a cos ◦ E ) , D y = f ◦ E ( d + b sin ◦ E ) , − a cos ◦ E ( DE ) − a sin ◦ ED E = f ◦ E ( c + a cos ◦ E ) , − b sin ◦ E ( DE ) + b cos ◦ ED E = f ◦ E ( d + b sin ◦ E ) , hence equating 3.0 and 5.0 as well as 3.1 and 5.1 we get 6.0 and 6.1, the combinations ( − b sin ◦ E ) 6 . . + ( a cos ◦ E ) 6 . . and ( − b cos ◦ E ) 6 . . − ( a sin ◦ E ) 6 . . give7.0. ab D E = f ◦ E ( ad cos ◦ E − bc sin ◦ E ) , ab ( DE ) = − f ◦ E ( ad sin ◦ E + bc cos ◦ E + ab ) . Taking the derivative of this equation and subtracting DE times 7.0. gives8. D ( f ◦ E )( ad sin ◦ E + bc cos ◦ E + ab ) + 3 f ◦ E ( ad cos ◦ E − bc sin ◦ E ) DE = 0 . Integrating gives9. f ◦ E ( ad sin ◦ E + bc cos ◦ E + ab ) = − aC for some constant C , but the polar of the origin is the line b cx + a dy − b c − a d + a b = 0 , therefore the distance of ( x, y ) to it is proportional to b c ( c + acos ◦ E ) + a d ( d + bsin ◦ E ) − b c − a d + a b or to .11. MECHANICS. bc cos ◦ E + ad sin ◦ E + ab, hence part 1 of the theorem.Replacing in 7.1. f ◦ E by its value gives ( DE ) = C ( ad sin ◦ E + bc cos ◦ E + ab ) , therefore C must be positive. Let C = C , then ( ad sin ◦ E + bc cos ◦ E + ab ) DE = C, and we obtain a generalization of Kepler’s equation − ad cos ◦ E + bc sin ◦ E + ab E = CI,
Let e and A be such that bc = ab e cos ( A ) , ad = ab e sin ( A ) , then e = ( ca ) + ( db ) , tan ( A ) = adbc . Let F = E − A and M = CI − ab A, then e sin ( F ) + F = M. Comment.
If the center of the conic is the center of force, c = d = 0 , f ◦ E is a constant and the forceis proportional to the distance. If the center of force is on the conic, 1.11.3.7 becomes f ◦ E ( ab ) (1 − cos ◦ ( E − E )) = − aC , when the center of force is c + acos ( E ) , d + bsin ( E ) . When the conic is a circle, g ◦ E = 2 a (1 − cos ◦ ( E − E )) , therefore, the force is inversely proportional to the 5-th power of the distance.? Comment. g ◦ E if h (( c + a cos ◦ E ) + ( d + b sin ◦ E ) ) = ( a d sin ◦ E + b c cos ◦ E + a b ) expanding will give terms in cos , sin cos, sin, cos and . The coefficient of sin cos must be , hence cd = 0 . Let d = 0 , the sin term disappears and the coefficients of , cos ◦ E, cos ◦ E give h ( c + b ) = a b ,h (2 ac ) = 2 b ac,h ( a − b ) = b c , CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS hence h = b and a = b + c or e := ca = (cid:113) − ba . Definition.
Given a curve ( x, y ) , the hodograph of the curve is the curve ( Dx, Dy ) . Comment.
The concept was first introduced by M¨obius (Mechanik des Himmels, (1843)), the name waschosen by Hamilton when he gave, independently, the definition in the Proc. Roy. IrishAcad., Vol. 3, (1845-1847) pp. 344-353.
Theorem.
If the force is central, and the center is chosen as the origin, the hodograph of the hodographis the original curve.
Indeed, the hodograph is ( D x, D y ) = f ◦ E ( x, y ) . Theorem.
If the central force obeys Newton’s law, the hodograph of the ellipse, 0.0. is the circle b (( Dx ) + ( Dy ) ) + 2 a e CDy − C = 0 , The proof is straightforward, the verification using Dx = − a sin ◦ EDE, Dy = b cos ◦ EDE and 1.11.3.9, .16 is even simpler.If the equation of the circle is ( − Rsin ( G ) , − k + Rcos ( G )) , equating to ( − asin ◦ EDE, bcos ◦ EDE ) for E = 0 and π and therefore G = 0 and π, gives bC = ab ( − k + R )(1 + E ) , − bC = ab ( − k − R )(1 − e ) , therefore − k + R = Ca (1+ e ) ) , k + R = Ca (1 − e ) , hence R = C ( ab ) , k = Re. moreover cos ( G ) = e + b a cos ◦ E e cos ◦ E . Introduction.
The generalization of classical mechanics to finite geometry turned out to be a thorny task. .11. MECHANICS.
Lemma. If x , y is a solution of x ( p + 1) − yp = 1 , all solutions are given by x = x + kp, y = y + k ( p + 1) , or x ≡ x (mod p ) , y ≡ y (mod p + 1) . Definition.
Kepler’s equation associated to the prime p is given by ( e sin ◦ E )( p + 1) − ( E − M ) p = 1 . This definition can be justified as follows, first when p is very large, we get the classicalKepler equation. Moreover from Lemma 10.4.1. all solutions are such that e sin ◦ E are equalmodulo p and E − M are equal modulo p + 1 which are precisely the congruence relations for e, sin ◦ E and for E − M. Example.
For p = 101 , . . . Theorem. (Of the circular hodograph of Hamilton). . . .
Starting with the work of Edwin P. Hubble,(1934) there had been mounting observationalastronomical evidence that the Universe is finite. This lead, Monseigneur Georges Lemaˆıtreto his hypothesis of the Primeval Atom and Sir Arthur Eddington to a possible a prioridetermination of the cosmical number N = 3 . . = 2 . . In his article published in1944, in the Proc. of the Camb. Phil. Soc., he first describes the number “picturesquely asthe number of protons and electrons in the universe“ and “interprets it by the considerationof a distribution of hydrogen in equilibrium at zero temperature, because the presence of thematter produces a curvature in space, the curvature causes the space to close when the numberof particles contained in it reaches the total N ”.If the work of Eddington would be reexamined today, protons and electron would probablybe replaced by quarks, if it were to be reexamined at some time in the future some otherparticles might play the fundamental role. In any case the lectures of Lemaˆıtre and the workof Eddington have been a primary motivation for my work on finite Euclidean and non-Euclidean geometry. As will be examined in more details when application will be made tothe finite pendulum, some elementary particle occupies a position and the possible positionsare discrete, they do this at a certain time, but again the time is not a continuous functionbut a discrete monotonic function. The fact that there are no infinitesimals in finite geometry CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS may very well be related to the uncertainty principle of Heisenberg (1927).H. Pierre Noyes and ANPA
All the earlier proofs in Mathematics were constructive, these proofs not only showed theexistence of objects, for instance the existence of the orthocenter of a triangle, where the 3perpendiculars from the vertex to the opposite sides meet, but also how to construct that point,by giving an explicit construction for a perpendicular to a line from a point outside it. Littleby little mathematicians have used more and more proofs using non constructive arguments,which show the existence of the object in question, without giving a method of construction.Such proofs are essential when no finite construction is possible, and are considered by manyas intellectually superior to a constructive proof when this one is possible. In finite geometry,it is desirable to limit oneself to constructive proofs, although this is not always possible, ata given point in time. I will give 2 examples later, the proof of Aryabatha’s theorem and theproof of the existence of primitive roots. Because in a finite geometry it is not easy to rely ontools such as the straightedge or the compass to experiment for the purpose of conjecturingtheorems, it is useful if not necessary to rely on computer experiments. Moreover, althoughthe simpler algorithm were for centuries given in the vernacular language, see for instancethe description of the so called Chinese remainder theorem by Ch’in Chiui-Shao, in UlrichLibbrecht’s translation, often the description avoids special cases or is ambiguous. Carefuldescription of algorithms started to appear with the advent of computers. The first formula oriented language was FORTRAN which evolved to FORTRAN 4 thenFORTRAN 77. It was developed enpirically. ALGOL was developed in 1958 and its syntaxcarefully defined in 1960 using the Backus normal form to attempt to define a priori an algo-rithmic language with a carefully constructed block structure. Its immediate successors wereALGOL 68 and PASCAL. APL was developped by Iverson to describe carfully the logic ofthe hardware of computers. It was magistrally adapted for the programming of Mathematicalproblems. LISP and its family of languages were developed when a list structure is required.BASIC was created at Dartmouth, to allow all undergraduates to learn programming in afriendly environment. It is the language which has evolved the most since its early days es-pecially by a small group at the Digital Equipment Corporation. This is the language whichI found most useful to discover mathematical conjectures because of the flexibility it offersin changing the program while in core and in examining easily, when needed, intermediateresults without prior planning. MAXIMA and its family of languages, MABEL, MATHE-MAICA and other recently developed languages are sure to play a more and more importantrole in discoveries.Elsewhere, I will describe some of the BASIC programs, that I have written to investigatenew areas of Mathematics, as well as the style used in the program descriptions and in theirdocumentation and use. Already in 1957, Lemaˆıtre used precise descriptions to communicate by letter with a person doing hiscalcultations on a EUCLID mechanical calculator. .13. NOTES.
Besides estimating areas and volumes, the Babylonians had a definite interest in so calledPythagorian triples, integers a, b and c such that a = b + c . It is still debated if their inter-est was purely arithmetical or was connected with geometry. On the one hand Neugebauer,states“It is easy to show that geometrical concepts play a very secondary part in Babylonean alge-bra, however extensively a geometrical terminology is used.” (p. 41)However, more recent discoveries, let him state (p.46), that these “contributions lie in thedirection of geometry”. One tablet computes the radius r of a circle which circumscribesan isosceles triangle of sides , and . An other tablet gives the regular hexagon, andfrom this the approximation √ ) can be deduced. . . . ( √ , . . . π = 3;7 , ) , . . . ” . He also describes, with Sachs, the data contained in tablet 322 of the Plimpton library col-lection from Columbia University (see Neugebauer and Sachs, vii and 38-41) as clearly in-dicating a relationship with right triangles “with angles varying regularly between almost 45degrees to almost 31 degrees”, while Bruins interpretation of the same table is purely al-gebraic. In fact the variation although monotonic is not that regular and the last trianglecorresponds to 31.84 degrees.FreibergThe tablet, dated 1900 to 1600 B.C., gives, with 4 errors, and in hexadesimal notation 15values of a, b, and ( ac ) = sec ( B ) where B is the angle opposite b,from or to or . Where the values between brackets are reconstructed values and 56 should be corrected to 28.
The tablet gives in hexadesimal notation columns I, II, III and IV, except for the line labelled11a in column IV. diff. is the difference between the numbers in column IV. The numbers inthe second line give, in hexadesimal notation uv and vu , for instance
2; 24 = 2 + = . CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
IV III II Iv u a c b B ( ac ) dif f. . . ,
0; 25 , . . − . , , ,
0; 25 , , , . . − . , , ,
0; 25 , , . . − . , , ,
0; 25 , , , . . − . ,
0; 26 , , . . − . , ,
0; 27 , . . − . , ,
0; 27 , , , . . − . ,
0; 28 , , , . . − . ,
0; 28 , ,
10 40 81 8161 6480 4961 37 . . − . , ,
0; 29 , , , ,
11 1 2 5 4 3 36 . . − . , a
64 125 19721 16000 11529 35 . . − . , , ,
0; 30 , , ,
12 25 48 2929 2400 1679 34 . . − . , ,
0; 31 , ,
13 8 15 289 240 161 33 . . − . , ,
0; 32 ,
14 27 50 3229 2700 1771 33 . . − . , , ,
0; 32 , ,
15 5 9 106 90 56 31 . . − . ∗
1; 48 ,
0; 33 , , There are 2 interpretations for the method of obtaining this table. The method of Neuge-bauer and Sachs, assumes the knowledge of the formulae a = u + v , b = u − v , c = 2 uv. It was proven later that all integer solutions of a = b + c , can be obtained from theseformulae and that the values of a, b and c are relatively prime if u and v are relatively primeand not both odd. They observe that u and v are always regular, it is, have only 2, 3 and 5 asdivisors, this implies that the reciprocals have a finite representation if we use hexadesimalnotation.This point of view is confirmed if we observe that u and v are precisely all the regular num-bers, which are relatively prime, satisfying .13. NOTES. ( √ − u < v < u < = 125 , except for the added pair, 11a, u = 125 , v = 64 . The first condition corresponds to requiringthat the triangle has an angle B opposite b less than 45 degrees. In this range, only one pairis such that u and v are both odd. This is the pair u = 9 , v = 5, which gives a = 106 , b = 56 ,c = 90 . The values a = 53 , b = 45 , c = 28 , could have been obtained with u = 7 and v = 2 , but these numbers are not both regular. It is interesting that one of the errors occurs for thispair, a being divided by 2 but not b. The other point of view is presented by Bruins which claims that a and b are obtained froma subset of tables of reciprocals, which we could write uv and vu , giving the values of a and b, because of ( uv + vu ) = ( uv − vu ) + 2 , after removing the common factors, which are necessarily 2, 3 or 5. This would give thetable for monotonically varying values of ac . We have given the corresponding hexadesimal values of uv and vu on alternate lines.Condition 0. adds credibility to the point of view of Neugebauer and would strengthen thegeometrical content of the table. A hope to get a deciding clue from one of the errors in thetable is not easily fulfilled. Indeed the second line gives for a and b, (64 + 27) + 2 ∗ ∗ . This requires several errors, first to add beforesquaring, then to add ∗ ∗ , which is explained by Gillings by the use of u + v = ( u + v ) − uv with − replaced by + and v = 64 replaced by v = . An other explanation, only slightlyless farfetched is to observe that, if we use Bruins approach, both numbers
2; 22 , , and
0; 25 , , have to be divided 3 times by 5, (or multiplied by 12 in hexadesimal notation).This gives for a, , , in base . If we assume that the scribe wrote instead , , , , using a large space, rather than a small one, and multiplies by 12 twice more, we get 3,12,1.An other explanation could start by explaining why the scribe computed instead of (64(= 60 + 4)) + (27(= 24 + 3)) = 4825 , (100(= 60 + 40)) + (39(= 36 + 3)) = 11521 . The argument could be decided if other tablets which continue this table are found. The tablePlimpt.tab, gives the values for angles less than 31.5 degrees, using criteria 0.There is an other minor controversy in the literature concerning the fact that the 1 in columnIV is visible or not in the tablet. If the opinion is taken, which is contrary to Neugebauer,that 1 is not there, column I is then ( bc ) = tan ( angle opposite b ) = ( ( uv − vu )) instead of ( ac ) = ( ( uv + vu )) . CHAPTER IFINITE PROJECTIVE CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS
GEOMETRY
Example.
Modulo 7, the inverses of 1 through 6 are respectively , , , , , . Answer to ??.
Notes for section on axiomatic, Pieri (coxeter, p. 12), Menger (Coxeter, p. 14) Dedekind(Coxeter, p 22), Enriques (Coxeter, p.22)The following does not work, leave for examination of other types, 1 where the triangleshave sides through Q and Q may give something, see also Pickert p. 74,75,80 Axiom. [2-point Desargues]
The is the special case when we restrict Desargues’ axiom to thecase when the center C of the configuration is one of 2 given points Q or Q of the givenaxis c . More specifically, C ι c , and for the 2 triangles { A i } and { B i } , ? let C i := ( A i +1 × A i − ) × ( B i +1 × B i − ) ,c i := ( A i × B i ) , c i ι C , i = 0 , , , incidence( A × A j , B × B j , c ) , j = 1 , , = ⇒ incidence( A × A , B × B , c ) . We write2-point-Desargues ( C, { A i } , { B i } ; (cid:104) C i (cid:105) , c ) . Theorem.
Given 2 triangles { A i } and { B i } , let C i := ( A i +1 × A i − ) × ( B i +1 × B i − ) , C i := A i × B i , and C := c × c , (cid:104) C i , c (cid:105) and C ι c = ⇒ c ι C. We write − ( c, { A i } , { B i } ; (cid:104) c , c , c (cid:105) , C ) Proof: 2-point-Desargues ( C , { A , B , C } , { A , B , C } ; (cid:104) B , A , C (cid:105) , c ) . Definition. A linear associative plane is a perspective plane for which the 2-point Desargues axiom issatisfied for 2 specific points on a specific line of the plane.If the line is q and the points are Q and Q , we have .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. Theorem.
In a linear associative plane, the ternary ring (Σ,*) is a . . . , more specifically: (Σ , ∗ ) is linear, a ∗ b ∗ c = a · b + c ,1. ( Σ ,+) is a group,
2. ( Σ − { } , · ) is a loop,
3. ( Σ , ∗ ) = (Σ , + , · ) is right distributive, ( a + b ) · c = a · c + b · c .4. a (cid:54) = b = ⇒ x · a = x · b + c has a unique solution. before CHAPTER 1. MAIN HISTORICAL DEVELOPMENTS hapter 2FINITE PROJECTIVE GEOMETRY
In Section 1, I give the axiomatic definition of synthetic projective geometry. In Section 2,I give an algebraic model of projective geometry. Although I will use, whenever possible asynthetic proof, I will use extensively an algebraic proof to proceed more expeditiously, if notmore elegantly. The reader is encouraged to replace these by the more satisfying syntheticproofs. In Section 3, I discuss the geometric model of the projective plane of order 2, 3 and5, discovered by Fernand Lemay and relate each model to classical configurations.
Projective Geometry implies usually that when we write down the equivalent algebraic axioms,the underlying field is the field of reals. Most of the properties that I will discuss in thisChapter and in the next one are valid whatever the field chosen. To deal with a set ofAxioms which characterize the plane, in a simpler setting, I will assume instead that thefield is finite. See 2.1.3. Most properties generalize to any field.
The objects or elements of plane projective geometry are points and lines. The relationbetween points and lines is called incidence. A point and a line are incident if and only ifthe point is on the line or if the line passes through the point.Identifiers are sequences of letters and digits, starting with a letter. If the first letter is alower case letter, the identifier will denote a line. If the first letter is an upper case letter,the identifier will denote a point. If the line ab is constructed as the line through the points A and B, we write ab := A × B. If the point A is constructed as the point on both a and a , we write G20.TEX [MPAP], September 9, 2019
CHAPTER 2. FINITE PROJECTIVE GEOMETRY A := a × a . The symbol “ := ” pronounced “is defined as” indicates a definition of a new point or of anew line. The symbol “ × ” will be justified in 2.2.2. A · ab = 0 , or A ι ab ,is an abbreviation for the statement “the point A is on the line ab ”. A · ab (cid:54) = 0 or A ι − ab ,is an abbreviation for the statement “the point A is not on the line ab ”. A = B, x = y, are abbreviations for “the points A and B or the lines x and y ”, all previously defined, “areidentical”. { A, B, C } or { a, b, c } denotes a triangle with vertices A, B and C or sides a, b and c. For Projective Geometry over fields we will use the following Axioms.
Of incidence and existence or of allignment :0. Given 2 distinct points, there exists one and only one line incident to, or passingthrough, the 2 points.1. Given 2 distinct lines, there exists one and only one point incident to, or on, the 2lines.2. There exists at least 4 points, any 3 of which are not collinear.
Of Pappus:
3. Let A , A , A be distinct points on a, let B , B , B be distinct points on b. Let C be the intersection of A × B and A × B or C := ( A × B ) × ( A × B ) . Similarly, let C := ( A × B ) × ( A × B ) , C := ( A × B ) × ( A × B ) , then the points C , C , C are collinear. (Fig. 1a) Notation.
The subscript i is usually restricted to the set { } and addition is then done modulo 3. Iwrite Pappus ( (cid:104) A i (cid:105) , (cid:104) B i (cid:105) ; (cid:104) C i (cid:105) ) or more generallyPappus ( (cid:104) A i (cid:105) [ , a ] , (cid:104) B i (cid:105) [ , b ]; (cid:104) C i (cid:105) [ , c ][ , X ]) .where “ (cid:104) X i (cid:105) ” indicate that the points X i are collinear, where the brackets indicate that whatis between them need not be given, and where X , if written, is the intersection of a and b .The axiom is trivially satisfied if X is one of the points A i or B i . If the axiom is used inproofs, it is always assumed that the points A i and B i are distinct from X . .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. Any plane satisfying the allignment axioms and the axiom of Pappus is called a
Pappianplane . For Projective Geometry over a specific field we will add one axiom or a set ofassociated axioms, for instance, for finite Projective Geometry over a Z kp , we add On the line l there are exactly p k + 1 points, p a prime. Exercise.
Write down the appropriate existence axiom associated with the fields,0. R , classical Projective Geometry ,1. C , complex Projective Geometry,2. Q , rational Projective Geometry. Theorem. Each line is incident to exactly p k + 1 points. Each point is incident to exactly p k + 1 lines. There are exactly p k + p k + 1 points and lines. The proof is left as an exercise.
Corollary.
There exists at least 4 lines, any 3 of which are not incident.
Comment.
If, contrary to 2.1.2.2, there is only one point P not on the line l, the geometry reduces to l, to a { pencil } of p + 1 lines through P, to P and to a set of p + 1 points on l. The axiom ofPappus is satisfied vacuously because no 2 distinct lines contain 3 points each.
Definition.
The line through C , C and C , in the axiom of Pappus, is called the Pappus line.68
CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Notation.
I introduce in the next Chapter a detailed notation for algebraic projective geometry. Anincomplete notation for the synthetic approach will now be introduced. The purpose is toformalize the Theorems, without the details of the approach of Russell and Whitehead. (cid:104) X i (cid:105) or ( (cid:104) X i (cid:105) , x ) indicates that the points X i are collinear and distinct, on x , (cid:104) x i (cid:105) or ( (cid:104) x i (cid:105) , X ) indicates that the lines x i are incident and distinct, through X , { X i } indicates that the points X i are distinct and not collinear, in other words forma triangle and similarly for the sides, { x i } . incidence ( A, B, C [ , l ]) or incidence ( A j [ , l ]) , j ∈ { , , . . . , k } , k ≤ ,is used to state that the points A, B, C or the points A j are on the same line l . “ [ , l ] ”indicates that the name of the line need not be given explicitely. incidence ( a, b, c [ , L ]) or incidence ( a j [ , L ]) is the corresponding statement for lines a, b, c or a j incident to the point L .No. Pappus ( (cid:104) A i (cid:105) , (cid:104) B i (cid:105) ; (cid:104) C i (cid:105) ) and the corresponding axioms can be written,in greater detail, as follows.Hy0. (cid:104) A i (cid:105) . Hy1. (cid:104) B i (cid:105) . De. C i := ( A i +1 × B i − ) × ( A i − × B i +1 ) . Co. (cid:104) C i (cid:105) . “No” is an abbreviation for “nomenclature” or “notation”, “Hy”, for “hypothesis”, “De”,for “Definition”, “Co” for “conclusion”.Notice that the order of the points is important.The reciprocal, P appus − ( (cid:104) A i (cid:105) , (cid:104) C i (cid:105) ; (cid:104) B i (cid:105) ) exchanges Hy1. and Co. and follows fromPappus ( (cid:104) A i (cid:105) , (cid:104) C i (cid:105) ; (cid:104) B i (cid:105) ) .In a statement, different letters indicate different elements with no special relationshipbetween them except as stated in the hypothesises “Hy”. Theorem.
Pappus ( (cid:104) A i (cid:105) , (cid:104) B i (cid:105) ; (cid:104) C i (cid:105) ) = ⇒ Pappus ( (cid:104) A , B , C (cid:105) , (cid:104) B , C , A (cid:105) ; (cid:104) C , A , B (cid:105) ) . Theorem. [Desargues]
Hy0. { A i } , { B i } , De0. c i := A i × B i , Hy1.
C ι c i , De1. a i := A i +1 × A i − , De2. b i := B i +1 × B i − , De3. C i := a i × b i , Co. ( (cid:104) C i (cid:105) , c ) . No.
Desargues ( C, { A i } [ , { a i } ] , { B i } [ , { b i } ]; (cid:104) C i (cid:105) [ , (cid:104) c i (cid:105) ][ , c ]) . .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. This is the notation for the following statements.
Given two triangles { A , A , A } and { B , B , B } , such that the lines A × B , A × B and A × B have a point C in common. Let C := ( A × A ) × ( B × B ) , C := ( A × A ) × ( B × B ) ,C := ( A × A ) × ( B × B ) . Then C , C , C are incident to the same line c (Fig. 3a). It is assumed that the trianglesare distinct and that the lines c i are distinct.This theorem can be proven using the incidence axioms in 3 dimensions. In 2 dimensions,it can be taken as an axiom or it can be derived from the axiom of Pappus, see 2.1.8. But theaxiom of Pappus does not derive from the incidence axioms and the Theorem of Desarguestaken as axiom. Theorem.
The axiom of incidence and the axiom of Pappus 2.1.2.4. imply the Theorem of Desargues.
See 2.1.8.
Introduction.
One of the characteristics of synthetic geometry is to start from a set of points and lines,to construct from them new points and lines and to extract known sets which have knownproperties. Hence, it is useful to describe some of the important sets, which are called con-figurations . We have seen 2 such configurations. In that of Pappus, we have 9 points and 9lines. In that of Desargues, we have 10 points and 10 lines. I will define here the completequadrangle and the complete quadrilateral configuration, the special Desargues configuration,as well as closely related configurations. To characterize the configuration further, I will usethe following notation:
Notation. ∗ ∗ , (11) indicates that each of the 10 points are incident to 3 lines, that each of the 10 lines areincident to 3 points and that the construction requires 11 independent data elements (2 fora given point or line, 1 for a point on a given line or a line through a given point). Or ∗ ∗ ∗ ∗ , indicates that 3 points are incident to 6 lines, that 8 points are incident to 3 lines and that12 lines are incident to 3 points and that 3 lines are incident to 2 points. The order chosenis that of decreasing number of incident elements.The notation does not uniquely define the configuration but is a useful tool. Definition. A confined configuration is a configuration in the description of which “ ∗ CHAPTER 2. FINITE PROJECTIVE GEOMETRY configuration to confined configuration and will use the adjective “non confined” otherwize.A self dual type configuration is one for which the information to the left of “ & ” is thesame as that to the right.It should not be confused with the notion of self dual configuration that will be introducedlater. A self dual configuration is a self dual type configuration but not vice-versa.
Theorem.
The configuration of Pappus is of type 9 ∗ ∗ , (10) . It can be viewed as a degeneratecase of that of Pascal. See 2.2.11. Hence the alternate name
Pappus-Pascal hexagon : If thealternate points of the hexagon A , A , A , A , A , A are on 2 lines, the three pairs ofopposites sides of the hexagon meet in 3 collinear points P , P and P . The correspondence between this notation and that used in the Theorem of Pappus is: A , A , A , A , A , A , P , P , P ,B , A , B , A , B , A , C , C , C . Theorem.
The configuration of Desargues is of type 10 ∗ ∗ , (11) . It can also be viewed as consisting of 2 pentagons which are inscribed one into the other.The points P , P , P , P , P and the points Q , Q , Q , Q , Q being such that P is on Q × Q , P is on Q × Q , P is on Q × Q , P is on Q × Q and P is on Q × Q . Q ison P × P , Q is on P × P , Q is on P × P , Q is on P × P and Q is on P × P . The correspondence between this notation and that used in the Theorem of Desargues is: P , P , P , P , P , Q , Q , Q , Q , Q ,B , A , B , A , C , C , B , C, A , C . Definition. A complete quadrangle is a configuration consisting of 4 points A , A , A , A , no 3 ofwhich are on the same line and of the 6 lines through each pair of points: a := A × A ,a := A × A , a := A × A , a := A × A , a := A × A , a := A × A . (Fig. 2a)It is of type ∗ ∗ , (8) . Definition.
The 3 points D := a × a , D := a × a and D := a × a are called the diagonal points of the complete quadrangle.The lines d i joining the diagonal points are called diagonal lines .These form, together with the quadrangle configuration, the completed non confined quad-rangle configuration. See Fig. 2a’.
Definition.
Given a complete quadrangle, a conic2 pseudo non confined configuration is the sub config-uration consisting of 3 of the points and the 3 lines joining these points to the 4-th one. It .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. is of type ∗ ∗ ∗ ∗ . (8)See 2.2.11. Definition.
Given a complete quadrangle, a completed quadrangle configuration is the configurationconsisting of the complete quadrangle, the diagonal points and the lines joining the diagonalpoints.
Theorem. If p = 2 the completed quadrangle configuration is of type7 ∗ ∗ , (8) . See 2.1.13 and 2.2.111. If p > , it is of type3 ∗ ∗ ∗ ∗ Definition. A complete n -angle is a configuration consisting of n points no 3 of which are on the sameline and of the n ( n − lines through each pair of points. Theorem.
A complete 5-angle does not exist if p < . Indeed, on the line through 2 of the points, wemust have 3 other points which are the intersection with the 3 pairs of lines through theother 3 points. We must have therefore at least 5 points on each line.
Exercise.
For which value of p does a complete n -angle exist for n > Definition. A complete quadrilateral is a configuration consisting of 4 lines a , a , a , a , no 3 of whichare incident to the same point and of the 6 points through each pair of lines: A := a × a ,A := a × a , A := a × a , A := a × a , A := a × a , A := a × a . (Fig. 2b)It is of type ∗ ∗ , (8) . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Definition.
The 3 lines A × A , A × A and A × A are called the diagonal lines of the completequadrilateral.The points joining the diagonal lines are called diagonal points . These together with the com-plete quadrilateral configuration form the completed quadrilateral non confined configuration (Fig. 2b’). Definition.
The special Desargues configuration , consists of 13 points and 13 lines obtained as follows. A , A , A , C is a complete quadrilateral, a := A × A , a := A × A , a := A × A ,c := C × A , c := C × A , c := C × A ,B := a × c , B := a × c , B := a × c ,b := B × B , b := B × B , b := B × B ,C := a × b , C := a × b , C := a × b ,r := A × C , r := A × C , r := A × C ,R := r × r , R := r × r , R := r × r ,c := C × C . (Fig. 3e’)This configuration is also called the quadrangle-quadrilateral configuration . The quad-rangle is { R i , C } or { c i , r i } , the quadrilateral is { b i , c } or { C i , B i . } The diagonal points are A i and the diagonal lines, a i . Comment.
The dual construction can be obtained with the upper case letters exchanged for the lowercase ones except for the exchange of B i and r i and b i and R i .This configuration plays an essential role in Euclidean Geometry. An example consist ofa triangle { A i } , C the barycenter, a i , the sides, c i , the medians, B i , the mid-points, b i ,the sides of the complementary triangle, C i , the directions of the sides, r i , the sides of theanticomplementary triangle, R i , its vertices, c, the ideal line. Definition.
Given a complete quadrangle-quadrilateral configuration, a conic3 pseudo non confined con-figuration is the sub configuration consisting of the quadrangle { R i , C } and the quadrilateral { b i , c } . It is of type ∗ ∗ ∗ ∗ (8).See 2.2.11. Theorem. [Special Desargues] C is on c, R is on c , R is on c , R is on c . .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. If p = 3 the special Desargues configuration is of type13 ∗ ∗ (8)See 2.1.6 If p > , it is of type9 ∗ ∗ ∗ ∗ , (8) . If we exclude r , r , r , R , R , R , we obtain a special case of the Desargues configu-ration in which P is on A × A , P is on A × A and P is on A × A . The proof will be given in section 2.1.8.
Definition. c is called the polar of C with respect to the triangle { A , A , A } . C is called the pole of c with respect to the triangle . Notation.
Part of Definition 2.1.6 and Theorem 2.1.6 can be noted as follows.No.
Special Desargues ( C, A i ; C i , c ) . De0. a i := A i +1 × A i − . De1. B i := a i × ( C × A i ) . De2. C i := a i × ( B i +1 × B i − ) . Co. ( (cid:104) C i (cid:105) , c ) . Exercise.
Construct the configuration starting from R i , C, and prove the 4 incidence properties corre-sponding to 2.1.6 in this construction. Exercise.
For p = 3 , prove that B i is on r i and C is on c. See also 2.2.11.For a connection between conics and the quadrangle-quadrilateral configuration, when p = 3 , see 2.2.11. Introduction.
There exist 2 other configurations of type ∗ ∗ , these will be constructed and defined.Many special cases of Desargues configurations will be defined, as well as the extended specialDesargues configuration and the dodecahedral configuration. I end by making some commentson the complete triangle in the more general case of the perspective plane. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Definition.
H0.0. A i , M, d ,H0.1. A ι d ,D0.0. a i := A i +1 × A i − ,D1.0. d := M × A , d := M × A ,D1.1. B i := d i × a i ,D1.2. mm := B × B , MA := mm × a ,D1.3. eul := M × MA , C := eul × d ,D1.4. c := B × C , c := B × C ,D1.5. C := d × c , C := d × c , c := C × C ,thenC0.0. B ι c , (Fig. 1c’)This defines the extended 2-Pappus Configuration . Definition.
The is the subset of the extended 2-Pappus Configurationconsisting of the point A i , B i , C i and of the lines a i , b i , c i (Fig. 1c). Theorem.
The extended 2-Pappus Configuration is of type3 ∗ ∗ ∗ ∗ , (9) . The 2-Pappus Pseudo Configuration is of type9 ∗ ∗ , (9) . The proof is left as an exercise.
Definition.
H0.0. A i , M, c ,H0.1. A ι c ,D0.0. a i := A i +1 × A i − ,D1.0. X := c × a ,D1.1. ma := M × A , ma := M × A , B := ma × a , B := ma × a ,D1.2. x := X × B , X := a × x , x := X × M, C := x × c ,D1.3. b := B × B , C := b × c , c := A × C ,D1.4. c := A × C , C := c × c ,D1.5. b := B × C , b := B × C , B := b × b ,thenC1.0. B ι a , (Fig. 1d’)This defines the extended 1-Pappus Configuration . Definition.
The is the subset of the extended 1-Pappus Configurationconsisting of the point A i , B i , C i and of the lines a i , b i , c i , (Fig. 1d). .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. Theorem.
The extended 1-Pappus Configuration is of type1 ∗ ∗ ∗ ∗ ∗ , (9) . The 1-Pappus Pseudo Configuration is of type9 ∗ ∗ , (9) . Definition.
There are many special cases of the Desargues configuration.0. 1-Desargues ( { A i } , { B , B , B } , (cid:104) C i (cid:105) ) , in which B ι a (Fig. 3b).1. 2-Desargues ( { A , A , A } , { B i } , (cid:104) C i (cid:105) ) , in which A ι b and A ι b (Fig. 3c).2. 1-1-Desargues ( { A , A , A } , { B , B , B } , (cid:104) C i (cid:105) ) , in which A ι b and B ι a (Fig. 3d).3. 3-Desargues ( { A i } , { B i } , (cid:104) C i (cid:105) ) , in which B i ι a i (Fig. 3e).4. C-Desargues ( { A i } , { B i } , (cid:104) C , C , C (cid:105) ) , in which C ι c (Fig. 3f ).5. C-1-Desargues ( { A i } , { B i } , (cid:104) C , C , C (cid:105) ) , in which B ι a and C ι c (Fig. 3g).6. Elated-Desargues ( C, { A i } , { B , B , B } , (cid:104) C i (cid:105) , c ) , in which C ι c (Fig. 3h).In each case the additional incident point(s) is (are) underlined.
Definition.
The extended special Desargues or extended quadrangle-quadrilateral configuration , consistsof 25 points and 25 lines, those of 2.1.6 and P Q i := p i +1 × q i − , QP i := q i +1 × p i − ,QR i := q i × r i , P R i := p × r i ,pq i := P i +1 × Q i − , qp i := Q i +1 × P i − ,qr i := Q i × R i , pr i := P × R i . Theorem.
All 25 points are on the 6 lines p i , r i of the quadrangle { Q i , P } . All 25 lines are on the 6points P i , R i of the quadrilateral { q i , p } . If p = 5 the extended special Desargues configuration is of type10 ∗ ∗ ∗ ∗ . If p > ∗ ∗ ∗ ∗ ∗ ∗ , (8) . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Definition.
The conical points and lines of the extended quadrangle-quadrilateral configuration are the 6points and 6 lines AF i := a i +1 × pr i − , F A i := pr i +1 × a i − ,af i := A i +1 × P R i − , f a i := P R i +1 × A i − , Theorem. AF i · pq i +1 = F A i · qp i − = 0 . Proof: To show that AF · pq = 0, we can use the dual of Desargues’ theorem applied to { p , p, p } = { R , Q , R } and { a , p , r } = { Q , P , A } with axial points A , R , A on the axis a and therefore central lines QR , P Q , a on the center AF . Comment.
We will see in 2.2.11 that the conical points are points on a conic. The conic thereforeappears in a natural way for p = 5 , in which case there are exactly 25 + 6 points and lines.(The Pascal line of N , M , N , M , N , M is R , R , R . ) Although, in some sense, the conicexits already for p = 2 and p = 3 , see 2.1.6, 2.2.11. Definition.
In view of 2.3.4, we define as the dodecahedral configuration , the configuration obtained byadding the 6 conical points to the extended special Desargues configuration.
Theorem. If p = 5 , the dodecahedral configuration is of type25 ∗ ∗ . If p > , the dodecahedral configuration is of type13 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ . Proof: The first part follows from 2.1.7 and 2.1.7. For p = 5 , all the points and lines ofthe dodecahedral configuration are distinct and are all the points and lines of the correspond-ing finite projective geometry. Any of the 6 conical points can be chosen to construct theextended special Desargues configuration. Moreover, pq i contains also P R i , QP i and F A i +1 ; qp i contains P R i , P Q i and AF i − ; qr i contains QP i , F A i , QR i +1 and QR i − ; pr i contains P Q i , QR i , F A i − and AF i +1 ; f a i contains QP i +1 QR i and AF i ; af i contains P Q i − , QR i and F A i . We leave, as an exercise, the proof of the following Theorem and the generalization ofthe definitions given therein. .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY.
Theorem.
The dodecahedral configuration can be continued indefinitely.
Starting with A = (1,0,0), A = (0,1,0), A = (0,0,1) and P = (1 , , , the coordinatesof the points and lines obtained by replacing lower case letter by the corresponding upper caseletter are the same, e.g. p = [1 , , . These are A = (1 , , , R = [0 , , − , P = (0 , , , Q = ( − , , ,P Q = ( − , , , QP = ( − , , ,QR = (2 , , , P R = ( − , , ,AF = (2 , , − , F A = (2 , − , , More points are
P G i := p i +1 × g i − , GP i := g i +1 × p i − ,AG i := a i +1 × g i − , GA i := g i +1 × a i − ,QRQR i := qr i +1 × qr i − , P QQP i := P Q i × QP i , and the lines are defined similarly, e.g. pg i := P i +1 × G i − . We have
P G = ( − , , , GP = ( − , , ,AG = (2 , , , GA = (2 , , ,QRQR = ( − , , , P QQP = (3 , , . and we have GP i · qp i +1 = P G i · pq i − = 0 ,GA i · pq i = AG i · qp i = 0 ,QRQR i · r i = 0 . Besides the conic { AF i , F A i } = 2( X + X + X ) + 5( X X + X X + X X ) = 0 , there are many more, such as { P Q i , QP i } = ( X + X + X ) + 6( X X + X X + X X ) = 0 , { P G i , GP i } = ( X + X + X ) + 11( X X + X X + X X ) = 0 , { AG i , GA i } = 2( X + X + X ) − X X + X X + X X ) = 0 . Comment.
We started with the special Desargues configuration with 13 points (and lines) which are allof the points when p = 3 , the extended special Desargues configuration consists of adding18 points and lines which are 31 distinct points and lines when p = . It would appearthat we could extend the construction in such a way that we get from the configuration with31 points a configuration with 57 points which would be all distinct when p = 7 , of 133points which would be all distinct when p = 11 , . . . . But this is not possible. For p = 7 , (1 , , − × (2 , − , gives (1,2,3) and by symmetry we get 5 other points but the points(0,1,3) give by symmetry (3 , ,
1) = (1 , , − which has already been constructed. Moreover,the point (1 , , − gives by symmetry ( − , ,
2) = (1 , , − hence for p = 7 , the same point.It is therefore not clear how to proceed in a systematic way. This may be related to the factthat there are only 5 regular polyhedra which are associated to p = 2 , CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Exercise.
Rewrite the statement of Theorem 2.1.6. in the form of a necessary and sufficient conditionfor A , A , A to be collinear, given that A , A and A are collinear. Exercise.
Let ω satisfy ω + ω + 1 = 0 .Let P = (0 , , − , Q = (0 , , − ω ) , R = (0 , , − ω ) ,P = ( − , , , Q = ( − ω, , , R = ( − ω , , ,P = (1 , − , , Q = (1 , − ω, , R = (1 , − ω , , then, with p = [1 , , , q = [1 , ω , ω ] , r = [1 , ω, ω ] ,p = [1 , , , q [0] = [1 , ω, ω ] , r = [1 , ω , ω ] , ω = 1 ,
1. incidence ( P i , p ) , incidence ( Q i , q ) , incidence ( R i , r ) ,2. incidence ( P i , Q i , R i , p i ) ,3. incidence ( P i , Q i +1 , R i − , q i ) ,incidence ( P i , Q i − , R i +1 , r i ) ,
4. the configuration is therefore of type ∗ ∗ . This configuration is that of the 9 inflection points of the cubic, X + X + X + kX X X = 0 . Comment.
Let { A , A , A , A } be a complete quadrangle and D , D , D be the diagonal points, severalsituation are possible in a perspective plane (See I).0. The diagonal points are always collinear, in this case, we have the N-Fano Configura-tion, N-Fano ( { A,B,C,D } ; (cid:104) P, Q, R (cid:105) ) .
1. The diagonal points are never collinear,in this case, we have the Fano Configuration,Fano ( { A,B,C,D } ; { P, Q, R } ) .
2. The diagonal points are sometimes collinear,in this case, we have either the pseudo con-figuration, ( { A,B,C,D } , (cid:104) P, Q, R (cid:105) ) or the pseudo configuration, ( { A,B,C,D } , { P, Q, R } ) . Notice the “;” in the first 2 cases.
Proof of the Theorem of Desargues.
Proof: The proof that Theorem 2.1.5 follows from the axioms of incidence and of Pappuswill now be given. .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY.
Cronheim (1953) showed that the proof reduces to 2 cases. In the first one, a permutationof the indices 0, 1, 2 is chosen in such a way that A ι − b and B ι − a . In the second one,except perhaps for an exchange of A i and B i , B i ι a i . In the first case (Hessenberg, 1905), letHe1.0. A ι − b , B ι − a ,De1.0. d := A × B , D := d × c , e := D × C , E := e × b , De1.1. f := D × C , F := f × a , G := a × b , g := F × G. De2.0. X := d × a , Y := d × b , Z := d × g ,The Pappus-Pascal hexagon D, A , A , A , B , C = ⇒ G, C and F are collinear.The Pappus-Pascal hexagon D, B , B , B , A , C = ⇒ G, C and E are collinear.Hence A , F, E , D, B , G is a Pappus-Pascal hexagon and C , C and C are collinear. Itis easy to verify that, because of He1.0, X is distinct from D, A , B , A , C , A , that Y isdistinct from D, B , A , B , C , B and that Z is distinct from A , D, B , E, G, F. This will be abbreviated as follows.Pr1.0. Pappus ( (cid:104) D, A , B (cid:105) , d, (cid:104) A , C , A (cid:105) , a ; (cid:104) G, C, F (cid:105) , X ) , Pr1.1. Pappus ( (cid:104) D, B , A (cid:105) , d, (cid:104) B , C , B (cid:105) , b ; (cid:104) G, C, E (cid:105) , Y ) , Pr1.2. ( (cid:104) E, G, F (cid:105) , g ) , Pr1.3. Pappus ( (cid:104) A , D, B (cid:105) , d, (cid:104) E, G, F (cid:105) , g ; (cid:104) C i (cid:105) , Z ) , Pr1.4. (cid:104) C i (cid:105) . In the second case (Cronheim, 1953), we have the 3-Desargues configuration (Fig. 3e),letDe3.0. r := A × C , R := c × r , R := c × r , De4.0. X := c × b ,Pr3.0. Pappus( (cid:104) C, A , B (cid:105) , c , (cid:104) C , B , B (cid:105) , b ; (cid:104) C , A , R (cid:105) , r , X ) , Pr3.1. Pappus( (cid:104)
C, A , B (cid:105) , c , (cid:104) C , B , B (cid:105) , b ; (cid:104) C , A , R (cid:105) , r , X ) , Pr3.2. Pappus( (cid:104) R , B , A (cid:105) , c , (cid:104) B , R , A (cid:105) , c ; (cid:104) C , C , C (cid:105) , c, C ) . Exercise.
Prove the Theorem of Cronheim, on the reduction to 2 cases, refered to in 2.1.8.
Theorem. [Dual of Pappus]
If the alternate sides of the hexagon { a , a , a , a , a , a } pass through 2 points, three pairsof opposites points of the hexagon are on 3 lines p , p and p which pass through the samepoint. See 2.1.10. (Fig. 1b)We write dual-Pappus ( (cid:104) a , a , a (cid:105) , (cid:104) a , a , a (cid:105) , (cid:104) p , p , p (cid:105) ) . Proof: Let A := a × a , A := a × a , A := a × a , A := a × a , A := a × a ,A := a × a . Let B := a × a , B := a × a , p := A × A , p := A × A , p := A × A , B := p × p . By hypothesis, B · a = B · a = 0 . B , A , A , B , A , A is a Pappus-Pascal hexagon,therefore A , B and A are collinear, in other words p passes through B . Proc. Amer. Math. Soc., , 219-221. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Definition.
The preceding configuration is a degenerate form of that of Brianchon. I will call it the
Pappus-Brianchon hexagon . The point common to p , p and p is called the Pappus point . Proof of the special Desargues Theorem.
The proof of Theorem 2.1.6 is as follows: 0. is a direct consequence of 2.1.8. 1. follows fromthe Axiom of Pappus 2.1.2.4. applied to the points P , A , R and P , A , R , proving that Q , P and P are collinear. Exercise.
The proof 2.1.8 of Theorem 2.1.6 is only given in the general case. Describe all the exceptionalcases and give a proof for each case.
Definition.
The
Reidemeister configuration consists of 11 points A , A , A , B , B , B , B , B , B , B , B , and 15 lines, a , a , a , b , b , b , b , b , b , b , b , b , b , b , b :Let A , A , A be a triangle, a := A × A , a := A × A , a := A × A , let b , b , b be 3 lines through A distinct from a and a , let B , B be points on b not on a , b := A × B , b := A × B , b := A × B ,b := A × B , B := b × b , B := b × b , B := b × b , B := b × b ,b := A × B , b := A × B , b := A × B , b := A × B , B := b × b ,B := b × b , b := B × B . (Fig. 11a) Lemma.
Let c := B × B , c := B × B , C := c × c , thenincidence ( C , A , A ) .Proof:Desargues ( A , B B B , B B B ; C A A , a ) . Theorem. [Reidemeister] A · b = 0 . The Reidemeister configuration is of type3 ∗ ∗ ∗ ∗ . Proof: After using the preceding Lemma, we useDesargues ( a , { c , b , b } , { c , b , b } ; (cid:104) b , b , b (cid:105) , A ) . .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. Theorem.
Let c := B × B ,c := B × B ,c := B × B , then incidence ( c i , C ) .Proof:Desargues − ( a , { b , b , c } , { b , b , c } ; (cid:104) c , c , c (cid:105) , C ) , Theorem.
Let c := B × B , c := B × B ,c := B × B , c := B × B ,c := B × B , c := B × B ,C i := c i × c i ,then incidence ( C i , c ) .Proof: Using the preceding Theorem,Desargues ( C, { B , B , B } , { B , B , B } ; (cid:104) C i (cid:105) , c ) . Exercise.
Let c (cid:48) := B × B , c (cid:48) := B × B ,c (cid:48) := B × B , c (cid:48) := B × B ,c (cid:48) := B × B , c (cid:48) := B × B , thenincidence ( C i , C (cid:48) i +1 , C (cid:48) i − ) . Definition.
The extended Reidemeister configuration consists of the points A , A , A , B jj , j = 0,1,2,3, B , B , B , B , C , C i , C (cid:48) i , i = 0,1,2, and of the lines a , a , a , b ij , i = 0,1,2, j =0,1,2,3, c ik , c (cid:48) ik , c (cid:48) i , , i, = 0,1,2, k = 0,1, c , see Fig. 11f. Exercise.
Prove0. that for given A , A , A , B , B , the correspondance between B and B is aprojectivity with center AEul on A × A (See 2.2.6). CHAPTER 2. FINITE PROJECTIVE GEOMETRY
1. The lines b and b coincide if the point B is on the conic through B tangent at A to A × A and tangent at A to A × A , represented by the matrix − −
2. that if we permute cyclically A , A , A , , then0. the lines B × B and the 2 other corresponding lines pass through the same point K .1. the lines A × B and the 2 other corresponding lines pass through the same point P .The same is true for the lines A × B and the 2 other corresponding lines, giving P .This configuration, see Fig. 26b, which I will call the K-Reidemeister configuration ispart of the Hexal configuration studied in Chapter III, with the correspondance A i B B b b b b b b c c (cid:48) A i M M ma ma ma ma ma ma eul mM a B B b b M aa M a cc cc Introduction.
If we permute in all possible way the 6 points of the Pappus configuration we obtain 6 Pappuslines. I prove in Theorem 2.1.9. that these pass 3 by 3 through 2 points. We obtain thereforea dual configuration, which therefore determines 6 Pappus points, which are 3 by 3 on 2 lines.I prove in Theorem 2.1.9 that these lines are the 2 original ones of the Pappus configuration.The points are not, in general the same. The proof of the first Theorem is synthetic, I haveno synthetic proof of the second Theorem. The algebraic proof uses a notation introduced inChapter III. Special cases of this configuration have been studied, but because some of theresults are still at the conjecture stage, these will not be discussed here, others are given asexercises.The term “rotate the points M , M , M “ means that we take the even permutations of M ,M , M , namely M , M , M and M , M , M .The notation is explained, in details, in Chapter III. Theorem. [Steiner (Pappus)] If we fix the points M , M , M on d and rotate the points M , M , M on d, we obtainthe 3 Pappus lines m , m , m . These pass through the point D . Similarly if we reverse theorder of the points of d and rotate, we obtain 3 other Pappus lines, m , m , m . These passthrough the point D . In detail, let H0. M i , M i , Steiner, Werke, I, p. 451 .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY.
D0. a i := M i × M i , D1. b i := M i +1 × M i − , b i := M i +1 × M i − , D2. L i := b i × b i , D3. N i := a i × b i , N i := a i × b i , D4. m := L × L , D5. m := N × N , m = N × N , D6. D := m × m , D7. Q i := a i +1 × a i − , D8. P i := b i +1 × b i − , P i := b i +1 × b i − , D9. m i := P i × P i , D10. D := m × m , then C0. m .L = 0( ∗ ) . C1. m .N = m · N = 0( ∗ ) . C2.
D.m = 0 . C3. m i · Q i = 0 . C4. D · m = 0( ∗ ) . See Fig. 9,Proof: A synthetic proof is as follows, C0, C1, C3, are direct consequences of Pappus’theorem applied toPappus( (cid:104) M , M , M (cid:105) , d, (cid:104) M , M , M , (cid:105) , d ; (cid:104) L , L , L (cid:105) , m ) Pappus( (cid:104) M , M , M (cid:105) , d, (cid:104) M , M , M , (cid:105) , d ; (cid:104) N , N , N (cid:105) , m ) Pappus( (cid:104) M , M , M (cid:105) , d, (cid:104) M , M , M , (cid:105) , d ; (cid:104) N , N , N (cid:105) , m ) Pappus( (cid:104) M , M , M (cid:105) , d, (cid:104) M , M , M , (cid:105) , d ; (cid:104) L , L , L (cid:105) , m ) M , M , M or M , M , M and M , M , M , or M , M , M or M , M , M . Thetriangles L i , N i , N i have m as axis of perspectivity for i = 1 and 2 therefore they have acenter of perspectivity D, by Desargues. I also note that for i = 2 and 0 the axis is m andfor i = 0 and 1 the axis is m . Hence C2. Symmetrically we get C4.For an algebraic proof, useful because of 2.1.9, letH0. M = (0 , , − , M = ( − , , , M = (1 , − , H1. M = (0 , m , − m ) , M = ( − m , , m ) , M = ( m , − m , . thenP0. a = [1 , , . P1. b = [ m , m , m ] , b = [ m , m , m ] . P2. L = ( m − m m , m ( m − m ) , − m ( m − m )) . P3. N = (0 , m , − m ) , N = (0 , m , − m ) . P4. m = [ m ( m + m ) , m ( m + m ) , m ( m + m )] , P5. m = [ m m , m m , m m ] , m = [ m m , m m , m m ] , P6. D = ( m m ( m − m m ) , m m ( m − m m ) , m m ( m − m m )) . P7. Q = (1 , , . P8. P = ( m ( m − m ) , m ( m − m ) , m ( m − m )) ,P = ( m ( m − m ) , m ( m − m ) , m ( m − m )) . P9. m = [0 , m ( m − m ) , − m ( m − m )] , P10. D = ( m m ( m − m )( m − m ) , m m ( m − m )( m − m ) , m m ( m − m )( m − m )) . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Definition.
The configuration of Theorem 2.1.9 which consists of 26 points and 17 lines is called the extended Pappus configuration .It is of type ∗ ∗ ∗ ∗ ∗ . (10)It can also be viewed, because of the synthetic proof as a multiple Desargues configuration,with 3 triangles perspective from D and 3 triangles perspective from D (cid:48) in which the axis ofone are the concurrent lines of the other. Definition. [Steiner]
The sub-configuration consisting of the points L i , N i , N i , Q i , P i , P i , D , D and of the lines a i , b i , b i , m i , m i , is called the Steiner configuration . It is of type ∗ ∗ . (10) Comment.
Part of a dual of the extended configuration is described in sections 1, 3 and 4 of the involutivegeometry of the triangle. The relation between the notations is as follows: d d M i M i a i b i b i L i N i N i M M ma i ma i A i M aa i M aa i mM a i cc i cc i m m m D Q i P i P i m i DK P P pp a i pap i pap i P ap i pap In particular, K is the point of Lemoine. On the other hand there is the followingcorrespondence m i and the dual of abr i , D and the dual of Ste, which passes through
BRa and
Abr.
Theorem.
If we make the dual construction starting with m , m , m on D and m i on D, the points M a i dual of m i is on the original line d and those M a i dual of m i is on the original line d :C5. M a i · d = 0 . C6.
M a i · d = 0 . See Fig. 10,Proof: An algebraic proof is as follows. P’0. A = (2 m m ( m − m ) , m ( m + m )( m − m ) , m ( m + m )( m − m )) .A = ( m m ( m − m ) , m ( m + m m − m m ) , m m ( m − m )) ,A = ( m m ( m − m ) , m m ( m − m ) , − m ( m − m m + m m )) , P’1. B = ( m m ( m − m ) , m m ( m − m ) , m ( m − m m + m m )) ,B = ( − m ( m + m m − m m ) , m m ( m − m ) , m m ( m − m )) ,B = ( m ( m + m )( m − m ) , m m ( m − m ) , m ( m − m )( m + m )) ,B = ( m m ( m − m ) , − m ( m − m m + m m ) , m m ( m − m )) ,B = ( m ( m − m )( m + m ) , m ( m − m )( m + m ) , m m ( m − m )) ,B = ( m ( m + m m − m m ) , m m ( m − m ) , m m ( m − m )) . P’2. l = [2 m m + 2 m m − m m − m m − m m − m m The reader will want to wait to check these algebraic manipulations until the notation has been explained. .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. − m m + m m m ( m + 2 m + 2 m ) , m ( m m − m m + m m − m m − m m + m m + 3 m m m ) , m ( m m − m m + m m − m m − m m + m m + 3 m m m ) ,l = [ m ( s − m m m ) , m (3 m + 4 m m − m m − m m + m m + m m − m m ) , − ( m + m )( m m + m m + m m − m m + m m + m m )+ 3 m m m )] ,l = [ m ( s − m m m ) , − ( m + m )( m m + m m + m m − m m + m m + m m )+ 3 m m m ) , m (3 m + 4 m m − m m − m m + m m + m m − m m )] , P’3. n = [ − ( m + m )( m m + m m + m m − m m + m m + m m )+ 3 m m m ) , m (3 m + 4 m m − m m − m m + m m + m m − m m ) , m ( s − m m m ) ,n = [( m + m )( − ( m m + m m + m m ) + 2( m m + m m + m m ) − m m m ) , m ( s − m m m ) , m (3 m − m m + 4 m m − m m + m m + m m − m m )] ,l = ( l , l , l )( m , m , m ) .n = ( l , l , l )( m , m , m ) .n = ( l , l , l )( m , m , m ) .n = ( l , l , l )( m , m , m ) . P’4.
M a = (2 m − m − m , m − m − m , m − m − m ) . P’5.
M a = (2 m − m − m , m − m − m , m − m − m ) ,M a = (2 m − m − m , m − m − m , m − m − m ) . P’7. q = [ − m ( m ( m + m )+ m ( m + m ) − m m − m m − m m + m m m (5 m − m − m )) , − m m ( − ( m m + m m + m m ) + 2( m m + m m + m m ) − m m m ) , m m ( − m m + m m + m m + ( m m + m m + m m )+ 3 m m m )] , P’8. p = [ − m m ( s − m m m ) , − m ( m + m )( − ( m m + m m + m m ) + 2( m m + m m + m m ) − m m m ) , − m ( m m − m m + 3 m m + m m + 4 m m − m m m (2 m + 5 m ))] ,p = [ − m m ( s − m m m ) , − m ( m m − m m + 3 m m + m m + 4 m m − m m m (2 m + 5 m )) , m ( m + m )( − m m + m m + m m ) + ( m m + m m + m m )+ 3 m m m )] . P’9.
M a = ( m m (2 m m − m m − m m ) , m m (2 m m − m m − m m ) , s is the symmetric function in m i , namely, m (m +m )+m (m +m )+m (m +m ) . CHAPTER 2. FINITE PROJECTIVE GEOMETRY m m (2 m m − m m − m m ) ,M a = ( m m (2 m m − m m − m m ) , m m (2 m m − m m − m m ) , m m (2 m m − m m − m m )) ,M a = ( m m (2 m m − m m − m m ) , m m (2 m m − m m − m m ) , m m (2 m m − m m − m m )) . Comment.
Continuing 2.1.9 we have the following relation between the above notation and that in theinvolutive geometry of the triangle. d d M i M i pp pap K, P, P P ap i a i b i b i kpa , tpa , tpa tpa , tpa , kpa tpa , kpa , tpa L i N i N i T tp , T kp , T kp T kp , T tp , T kp T kp , T kp , T tp m i D m i Dapa i M apa , apa , apa MQ i P i P i T tp , T kp , T kp T kp , T kp , T tp T kp , T tp , T kp Definition.
The mapping which associates to the points M i and M i , the points M a i and M a i , is calledthe Pappus-dual-Pappus mapping . Exercise. If a , a and a have a point in common, prove that the elements defined in 2.1.9 andtheir dual defined in 2.1.9 determine a self-dual configuration and the points M a i and M a i coincide, as a set, with the points M i and M i . If p > , there are 29 points and 29 lines.If p = 5 , there are 25 points and 25 lines, the type is ∗ ∗ ∗ ∗ ∗ ∗ . If p = 7 , it is of type ∗ ∗ ∗ ∗ ∗ ∗ . If p > , it is of type ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ . The configuration is therefore distinct from the extended special Desargues configuration of2.1.7.Prove that the 6 points and lines left over are also on a conic, as in 2.1.7. .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY.
Introduction.
This important concept, prepared by the work of Maurolycus and Poncelet, was introduced byJoseph Diaz Gergonne. We observe that if we join Theorem 2.1.8 to the axioms 2.1.2 and toTheorems 2.1.4, and then exchange the words line and point, we obtain the same statementsin some other order. Therefore in any result obtained, we can exchange the words line andpoint.
Definition.
The method of obtaining from a result an other result by exchange of the words line and pointis called duality . In particular, the Theorem of Desargues 2.1.5, becomes:
Theorem. [Dual of Desargues’ Theorem]
Given two triangles { a , a , a } and { b , b , b , } such that the points a × b , a × b and a × b are on the same line c. Let c := ( a × a ) × ( b × b ) , c := ( a × a ) × ( b × b ) ,c := ( a × a ) × ( b × b ) . Then c , c , c are incident to the same point C. Fig. 3a)This is the dual of Theorem 2.1.5.
Comment.
Fig. 1a and 1b are dual of each other, so are Fig. 2a and 2b, Fig. 2a’ and 2b’, Fig. 9 and10.Fig. 3a, Fig. 3e, Fig. 3h are self dual.
Theorem.
If 2 quadrangles { A , A , A , A } and { A (cid:48) , A (cid:48) , A (cid:48) , A (cid:48) } are such that none of their pointsand none of their lines coincide and are such that 5 of their corresponding lines are on thesame line p, then the 6-th pair of lines intersect on p. Proof: Using the notation 2.1.6 and “ (cid:48) ” for the second quadrangle, let B k := a k × a (cid:48) k , k =0 to 5 and let B , B , B , B , B , be all on the line p. Theorem 2.1.10, dual of Desargues canbe applied to the triangles { A , A , A } and { A (cid:48) , A (cid:48) , A (cid:48) } then to the triangles { A , A , A } and { A (cid:48) , A (cid:48) , A (cid:48) } . The consequence is that the lines A × A (cid:48) , A × A (cid:48) , A × A (cid:48) have a point P in common which is also on A × A (cid:48) . Therefore the Theorem of Desargues can be applied tothe triangles { A , A , A } and { A (cid:48) , A (cid:48) , A (cid:48) } which implies that the lines a and a (cid:48) intersecton the line p. Or using the synthetic notation, let b j := A j × A (cid:48) j , j = 0 to 3Desargues − p, { a , a , a } , { A , A , A } , { a (cid:48) , a (cid:48) , a (cid:48) } ; { A (cid:48) , A (cid:48) , A (cid:48) } ; (cid:104) b , b , b (cid:105) , P ) ,Desargues − p, { a , a , a } , { A , A , A } , { a (cid:48) , a (cid:48) , a (cid:48) } ; { A (cid:48) , A (cid:48) , A (cid:48) } ; (cid:104) b , b , b (cid:105) , , Q ) , = ⇒ P = ( A × A (cid:48) ) × ( A × A (cid:48) ) = Q, = ⇒ Desargues( P, { A , A , A } , { a , a , a } , { A (cid:48) , A (cid:48) , A (cid:48) } , { a (cid:48) , a (cid:48) , a (cid:48) } ; (cid:104) B , B , B (cid:105) , p ), CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Definitions.
The quadrangles of Theorem 2.1.11 are said to be homologic. p is called the axis and P the center of the homology . Corollary.
If two complete quadrangles with no points and lines in common are such that K := a × a is on a (cid:48) and a (cid:48) , L := a × a is on a (cid:48) and a (cid:48) ,M := a × a (cid:48) is on K × L, then N := a × a (cid:48) is also on K × L. Construction.
Given three points
K, L, M on a line p, choose arbitrarily a point A not on p and a point A on A × K distinct from A and K. Define A := ( A × L ) × ( A × M ) , A := ( A × K ) × ( A × L ) ,N := ( A × A ) × ( K × L ) . See Fig. 2a”.It follows from 2.1.9 that N is independent of the choice of A and A . Definition. N is called the harmonic conjugate of M with respect to K and L. Theorem.
If each line has q + 1 points on it, let l ( n, q ) denote the number of points on a complete n -angle, let l ∗ ( n, q ) denote the number of points not on a complete n-angle, let L ( n, q ) denotethe number of complete n -angles,0. l ( n, q ) = n n − q − n ( n − n − n +68 . l ∗ ( n, q ) = q − ( n + 1) n − q + ( n − n − n +7 n − . L ( n + 1 , q ) = n +1 L ( n, q ) l ∗ ( n, q ) . l ( n + 1 , q ) − l ( n, q ) = nq − ( n − n − n + 2) . Proof. l ( n + 1 , q ) is obtained from l ( n, q ) by adding points on each of the n lines throughthe new point A n and through one of the old points A say, plus the new point itself. On eachof the lines A n × A, we have q + 1 points from which we have to subtract the points A and A n as well as the points on the ( n − n − lines through each pair of the old points, A excluded.This gives n ( q + 1 − ( n − n − −
2) + 1 = nq − ( n − n + 4 n − . Using l (1 , q ) = 1 ,
0. follows by induction, 1. follows from l ∗ ( n, q ) = q + q + 1 − l ( n, q ) ,
2. follows from the fact that to each complete n -angle and each point not on its sides isassociated a complete ( n + 1) -angle each being counted n + 1 times. .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. Exercise. l ( I, q ) is a polynomial of degree 4, its successive forward differences at 0 are 1, q − , − . l ∗ ( + x, q ) = l ∗ ( − x, q ) l ∗ ( n, I ) is a quadratic function. Its discriminant is − ( n − n − n − n + 2) , itssuccessive forward differences at 0 are 4, − , − . The discriminant is negative if n > . Table. n l ( n, q ) l ∗ ( n, q ) discr. L ( n, q )0 0 q + q + 1 − q + 1) q q + q + 12 q + 1 q q ( q + 1)( q + q + 1)3 3 q ( q − q ( q + 1)( q + q + 1)4 6 q − q − q −
3) 1 q ( q − q − q − q − q + 21 − q ( q − q − q − q − q −
547 21 q − q − Exercise.
Complete the last 3 lines of the preceding table.
Definition. A collineation consists of a one to one function γ from the set of points of the plane ontoitself, such that all points on a line have their image also on a line and of the induced function γ (cid:48) from the set of lines of the plane onto itself. Definition. A correlation consists of a one to one function ρ from the set of lines of the plane onto itself,such that all lines through a point have their image also through a point and of the inducedfunction ρ (cid:48) from the set of points of the plane onto itself. Theorem.
If the geometry is of prime order, a collineation or a correlation is determined by the imageof a complete quadrangle onto a complete quadrangle or quadrilateral. (See 2.2.7) CHAPTER 2. FINITE PROJECTIVE GEOMETRY p . Introduction.
There is a well known, see for instance Stevenson, p. 72, or Dembowski, p. 144, 14. thatthere is, up to isomorphism, only one plane satisfying the incidence axioms, the axiom ofPappus and the finite field axiom 2.1.3. In the general case, the proof will require a fullknowledge of the material not only of section 1, but also of the existence of fundamentalprojectivities of order p − and p + 1 . The axiom of Pappus is not required for p ≤ , asproven by MacInnes in 1907 for p = 2 , p = 7 , see Bose and Nair, 1941, Hall1953, 1954b, Pierce, 1953, Pickert, 1955. Theorem.
For p = 2 , There exists, up to isomorphism, only one plane satisfying the incidence axioms. The diagonal points of a complete quadrangle configuration are collinear.
Proof: Assume that line [3] contains the points (0), (1) and (2). Let (3) be an otherpoint. Define line [0] as the line through (1) and (3), we abbreviate this as [0] := (1) × (3) . Similarly, [1] := (0) × (3) , [2] := (2) × (3) . Let the third point on [0] be (5), on [1] be (4) andon [2] be (6). Let [4] := (4) × (6) , [5] := (5) × (6) , [6] := (4) × (5) . The incidence propertiesimply (0) is on [5], which we abbreviate (0) · [5] = 0 , similarly (1) · [4] = 0 and (2) · [6] = 0 . This completes the incidence tables:line : Points on line Point : lines through Point
Theorem.
For p = 3 , there exists, up to isomorphism, only one plane satisfying the incidence axioms. Proof: Assume that the line [4] contains the points (0), (1), (2) and (3). Let (4) be apoint not on [4]. Let [0] := (1) × (4) , [1] := (0) × (4) , [2] := (3) × (4) , [3] := (2) × (4) . Let (7)and (10) be the other points on [0]. Let [7] := (0) × (10) , [10] := (0) × (7) , [9] := (2) × (10) , [11] := (2) × (7) , [8] := (3) × (10) , [12] := (3) × (7) . Let (9) := [2] × [10] , (11) := [2] × [7] , (8) := [3] × [10] , (12) := [3] × [7] , (5) := [1] × [9] , (6) := [1] × [8] . Let [5] := (1) × (9) , [6] := (1) × (8) . .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. At this stage with have the following incidence table:line : Points on line Point : lines through Point
It remains to complete the table using the incidence axioms:Line [8] contains (3), (6) and (10), but (3) is already on line [2] with (4) hence (4) cannotbe on [8]. Similarly (3) excludes (9), (11), (0), (1), (2), (6), (7); (6) excludes (5) and (10)excludes (12). The only point left is (8).Line [9] contains (2), (5) and (10), (2) excludes (4), (8), (12), (0), (1), (3), (7) and (10)excludes (11), (3), (6), only (9) remains.Line [5] contains (1) and (9), (1) excludes (4), (7), (10), (0), (2), (3), (8) and (10) excludes(11), (5), only (6) and (12) remain.Line [6] contains (1) and (8), (1) excludes (4), (7), (10), (0), (2), (3), (6), (9), (12), only(5) and (11) remain.Line [11] contains (2) and (7), (2) excludes (4), (8), (12), (0), (1), (3), (5), (9), (10), only(6) and (11) remain.Line [12] contains (3) and (7), (3) excludes (4), (9), (11), (0), (1), (2), (6), (8), (10), only(5) and (12) remain.This completes the incidence tables:line : Points on line Point : lines through Point
Exercise. D and D , are harmonic conjugates to d and d . Comes from Steiner for conics. Exercise.
Let. . . complete this, change notation for BM in D2. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
D0. AM i := a i +1 × m i +1 , D2. BM i := m i +1 × b i − , D3. ab i := BM i +1 × AM i − , thenC0. P i .ab i = 0 . Proof: ab is the axis of perspectivity of the triangles N , with center of perspectivity M . Comment.
A configuration associated to antipolarity in 3 dimensions implies a configuration of 20 pointsand 22 lines in 2 dimensional geometry, see VI.6.1.5.INTEGRATE THEREDx. ce i := M M a i +1 × M M a i − ,M M a i − , Dy.
P Q i := ce i × ce i , Px. ce = [ m ( m − m ) , m ( m − m ) , m ( m − m )] ,ce = [ m ( m − m ) , m ( m − m ) , m ( m − m )] , Py.
P Q = (0 , m ( m − m ) , − m ( m − m )) . Examples.
In the following examples we can replace γ and γ (cid:48) by ρ and ρ (cid:48) . ρ composed with ρ (cid:48) gives acollineation σ . Properties and special cases of collineations and correlations will be discussedin 2.1.12 and 2.2.8. In these examples, the complete quadrangle in the domain is always(0), (1), (6), (12). t ( i ) denotes the smallest positive integer such that ( γ t )( i ) = i. C = C indicates that the function γ of the collineation C corresponds to the function γ of thecollineation C composed with itself 3 times. The examples will be used in 1.8.12.For p = 5 , C i γ ( i ) 1 6 21 11 26 16 12 22 7 17 27 2 3 0 4 5 γ (cid:48) ( i ) 5 10 18 26 14 22 0 2 1 4 3 9 17 30 13 21 t ( i ) 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 30 20 10 25 15 18 13 8 28 23 24 29 9 14 19 γ (cid:48) ( i ) 7 11 20 24 28 8 29 25 16 12 6 23 15 27 19 t ( i ) 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 C i γ ( i ) 1 6 16 26 11 21 12 27 17 7 22 2 5 4 0 3 γ (cid:48) ( i ) 5 10 22 14 26 18 0 3 4 1 2 9 21 13 30 17 t ( i ) 24 24 24 6 24 24 24 6 24 24 24 24 24 1 6 24 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 30 15 25 10 20 18 23 28 8 13 24 19 14 9 29 γ (cid:48) ( i ) 7 28 24 20 11 8 12 16 25 29 6 19 27 15 23 t ( i ) 6 24 24 24 24 24 24 6 24 24 24 24 24 24 6 .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. C = C i γ ( i ) 6 12 30 24 2 18 5 19 15 27 23 16 21 11 1 26 γ (cid:48) ( i ) 18 2 12 30 6 24 5 14 26 10 22 1 8 13 23 28 t ( i ) 12 12 12 3 12 12 12 3 12 12 12 12 12 1 3 12 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 29 3 13 22 20 25 28 14 17 4 8 10 0 7 9 γ (cid:48) ( i ) 3 27 25 11 9 4 21 7 29 15 0 20 19 17 16 t ( i ) 3 12 12 12 12 12 12 3 12 12 12 12 12 12 3 C = C i γ ( i ) 12 5 29 8 16 25 21 10 3 19 28 30 18 2 6 24 γ (cid:48) ( i ) 24 22 21 23 0 25 18 30 6 2 12 10 4 13 16 27 t ( i ) 8 8 8 2 8 8 8 2 8 8 8 8 8 1 2 8 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 9 26 4 23 20 13 14 0 15 11 17 22 1 27 7 γ (cid:48) ( i ) 14 19 29 9 1 26 8 3 15 17 5 11 20 28 7 t ( i ) 2 8 8 8 8 8 8 2 8 8 8 8 8 8 2 C = C i γ ( i ) 5 21 9 17 30 13 18 22 26 10 14 29 25 16 12 8 γ (cid:48) ( i ) 25 12 8 16 5 29 24 23 0 22 21 2 26 13 7 19 t ( i ) 6 6 6 3 6 6 6 3 6 6 6 6 6 1 3 6 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 7 24 11 28 20 4 0 1 3 2 15 23 6 19 27 γ (cid:48) ( i ) 30 20 15 1 10 6 4 14 17 28 18 9 11 27 3 t ( i ) 3 6 6 6 6 6 6 3 6 6 6 6 6 6 3 C = C i γ ( i ) 18 25 27 3 9 11 13 28 8 23 1 7 4 29 21 15 γ (cid:48) ( i ) 15 8 26 3 24 17 29 7 18 21 4 12 0 13 14 11 t ( i ) 4 4 4 1 4 4 4 1 4 4 4 4 4 1 1 4 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 19 17 16 0 20 2 6 12 24 30 26 14 5 22 10 γ (cid:48) ( i ) 16 9 28 2 22 5 6 23 27 19 25 10 1 20 30 t ( i ) 1 4 4 4 4 4 4 1 4 4 4 4 4 4 1 C = C i γ ( i ) 13 4 10 24 27 16 11 0 15 14 12 19 2 7 25 26 γ (cid:48) ( i ) 28 26 0 30 29 27 17 14 25 4 6 8 18 13 23 1 t ( i ) 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 3 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 22 3 29 6 20 30 5 21 17 9 8 1 18 28 23 γ (cid:48) ( i ) 3 10 19 12 21 24 5 7 20 11 15 22 2 9 16 t ( i ) 3 3 3 3 3 3 3 3 3 3 3 3 3 3 394 CHAPTER 2. FINITE PROJECTIVE GEOMETRY C = C i γ ( i ) 16 30 14 3 23 7 29 5 8 12 25 28 9 22 2 15 γ (cid:48) ( i ) 11 18 25 3 27 9 20 7 28 5 24 0 15 13 14 12 t ( i ) 2 2 2 1 2 2 2 1 2 2 2 2 2 1 1 2 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 0 17 19 18 20 27 13 4 24 10 26 21 11 6 1 γ (cid:48) ( i ) 16 21 1 26 6 17 29 23 10 2 19 4 8 22 30 t ( i ) 1 2 2 2 2 2 2 1 2 2 2 2 2 2 1 C i γ ( i ) 0 1 3 5 2 4 12 14 11 13 15 17 19 16 18 20 γ (cid:48) ( i ) 10 26 14 18 5 22 6 7 0 8 9 1 25 30 15 20 t ( i ) 4 5 20 20 20 20 1 1 4 4 4 5 20 20 20 20 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 22 24 21 23 25 27 29 26 28 30 7 9 6 8 10 γ (cid:48) ( i ) 21 19 28 3 12 11 2 29 17 23 16 13 27 24 4 t ( i ) 5 20 20 20 20 5 20 20 20 20 5 20 20 20 20 C = C i γ ( i ) 0 1 5 4 3 2 19 18 17 16 20 24 23 22 21 25 γ (cid:48) ( i ) 9 16 15 28 22 2 6 7 10 0 8 26 23 4 20 12 t ( i ) 2 5 10 10 10 10 1 1 2 2 2 5 10 10 10 10 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 29 28 27 26 30 9 8 7 6 10 14 13 12 11 15 γ (cid:48) ( i ) 11 3 27 18 25 1 14 24 19 29 21 30 13 17 5 t ( i ) 5 10 10 10 10 5 10 10 10 10 5 10 10 10 10 C = C i γ ( i ) 0 1 2 3 4 5 26 27 28 29 30 6 7 8 9 10 γ (cid:48) ( i ) 0 11 12 13 14 15 6 7 8 9 10 21 24 22 25 23 t ( i ) 1 5 5 5 5 5 1 1 1 1 1 5 5 5 5 5 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 γ (cid:48) ( i ) 26 28 30 27 29 16 20 19 18 17 1 5 4 3 2 t ( i ) 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 C i γ ( i ) 0 1 5 4 3 2 26 29 27 30 28 11 14 12 15 13 γ (cid:48) ( i ) 0 11 15 14 13 12 6 8 10 7 9 16 19 17 20 18 t ( i ) 1 4 4 4 4 4 1 4 4 4 4 4 4 4 4 4 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 21 24 22 25 23 6 9 7 10 8 16 19 17 20 18 γ (cid:48) ( i ) 21 22 23 24 25 1 4 2 5 3 26 30 29 28 27 t ( i ) 4 4 4 4 4 4 4 4 4 4 1 2 2 2 2 .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. C i γ ( i ) 0 6 9 7 10 8 12 15 13 11 14 24 23 22 21 25 γ (cid:48) ( i ) 5 26 18 10 22 14 1 4 3 0 2 6 23 15 27 19 t ( i ) 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 18 19 20 16 17 30 27 29 26 28 2 5 3 1 4 γ (cid:48) ( i ) 16 12 8 29 25 11 20 24 28 7 21 9 17 30 13 t ( i ) 1 5 5 5 5 5 5 5 5 5 5 5 5 5 5 C i γ ( i ) 11 7 24 28 20 5 2 17 29 23 10 13 12 15 14 0 γ (cid:48) ( i ) 15 27 6 19 23 5 30 1 25 20 10 0 12 11 14 13 t ( i ) 4 4 4 4 4 1 4 4 4 4 1 4 1 4 1 4 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 25 27 18 3 9 19 22 4 30 8 26 1 21 16 6 γ (cid:48) ( i ) 29 7 18 21 4 28 22 9 2 16 26 17 3 8 24 t ( i ) 4 4 1 4 4 4 1 4 4 4 1 4 4 4 4 C i γ ( i ) 10 25 15 1 30 20 11 23 17 29 2 9 0 6 8 7 γ (cid:48) ( i ) 29 27 30 0 28 26 7 15 4 18 21 17 23 10 11 2 t ( i ) 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 18 22 14 5 26 3 21 28 12 19 27 24 4 16 13 γ (cid:48) ( i ) 12 5 16 25 8 22 6 13 3 20 1 19 24 9 14 t ( i ) 31 31 31 31 31 31 31 31 31 31 31 31 31 31 31 C i γ ( i ) 13 17 30 21 9 5 24 27 16 4 10 15 12 0 14 11 γ (cid:48) ( i ) 13 17 30 21 9 5 24 27 16 4 10 15 12 0 14 11 t ( i ) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 γ ( i ) 8 1 18 28 23 3 22 20 6 29 26 7 19 25 2 γ (cid:48) ( i ) 8 1 18 28 23 3 22 20 6 29 26 7 19 25 2 t ( i ) 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer to 2.1.4.
First, prove that there are exactly p + 1 lines through P, more generally through any point noton l. Then, prove that on any line m distinct from l and not incident to both P and Q, thereare exactly p + 1 points. If Q is not on the line join Q to all the points on l and determinethe intersection with m. Then, for P × Q determine a point on an other line through P whichis not on l and repeat the argument just given. To count the points, observe that any point CHAPTER 2. FINITE PROJECTIVE GEOMETRY different from P is on a line through P. There are exactly p + 1 such lines and on each thereare p points distinct from P hence altogether ( p + 1) p + 1 points. Answer to 2.1.7.
Given the hexagon { A , A , A , A , A , A , } such that the alternate vertices A , A and A are collinear. The necessary and sufficient condition for A , A and A to be collinear isthat the points P , P and P be collinear. The necessary condition follows using 1.5.1. onthe hexagon { A , P , A , P , A , P } . Answer to 2.1.6.
The construction is r := P × Q , r := P × Q , r := P × Q ,p := Q × Q , p := Q × Q , p := Q × Q ,A := p × r , A := p × r , A := p × r ,a := A × A , a := A × A , a := A × A ,P := a × r , P := a × r , P := a × r ,q := P × P , q := P × P , q := P × P ,R := a × q , R := a × q , R := a × q ,p := R × R . We have to prove R is on p and R i is on p i .. . . ..This gives the configuration p on R i ; a i on P i , R i , A i +1 , A i − ; p i on A i , R i , Q i +1 , Q i − ,q i on R i , P i +1 , P i − ; r i on P, A i , P i , Q i . Similarly for lower case and upper case exchanged.If p = 3 , q i and p must contain a fourth point which is one of the 13 known point. Bynecessity P is on p and Q i is on q i . See 2.1.13
Answer to 2.1.8 and 2.1.8.
I will not repeat the computations of Chapter III. b = b if m m = − m − m , therefore if M is on the conic X − X X = 0 , which is representedby the matrix (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − − − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) .If B = ( a, m , m ) , then b = [ m , , − a ] , b = [ m, − a, , B = ( a, m , m ) , B =( a, m , m ) , b = [ m , − a, , b = [ m , , − a ] , B = ( a, m , m ) , hence B · b , hence2.1.8. B × B = [ m m − m , − a ( m − m ) , a ( M − − m )] , line incident to (0 , m − m , m − m ) = AEul , hence 2.1.8.0. 1, and 2, follow from Chapter III. .1. SYNTHETIC FINITE PROJECTIVE GEOMETRY. Introduction.
I start with affine geometry by choosing a particular line m as the ideal line and 2 ordinarylines x and y which intersect at O , as well as an ordinary point M on neither x nor y .I will first associate to ordinary points on the line integers from 0 to p − , by definingthe successor. I will then define addition of points on the line and prove commutativityusing the axiom of Pappus. I will then define multiplication of points on the line and provecommutativity using the axiom of Pappus. It remains to prove the distributivity law. Definition.
Let Y := x × m , M := ( Y × M ) × y.A := O,A i +1 := ((( A i × M ) × m ) × M ) × x, for i = 0 ,
1, . . . until A n = A .A i +1 is called the successor of A . Theorem. n = p .Proof: The parabolic projectivity associates to A , A , σ with fixed point Y which as-sociates to A , A , associates to A i , A i +1 . By definition σ n = (cid:15) , the identity mapping. If n < p , any other ordinary point on the line distinct from A i has therefore the same period n , after exausting all points in the line it follws that n must divide p , therefore n = p . Wecould also give a group theory proof of this Theorem and use the Theorem of Lagrange. Definition.
Given 2 points A and B on x , the addition of the 2 points, C := A + B is defined as follows,. . . .. Theorem.
The addition is commutative, in other words, A + B = B + A . Let u := Y × M , A := ( M × B ) × m, A := ( M × A ) × m,C := ( A × M ) × ( X × B ) , C := ( A × M ) × ( X × B ) . The axiom of Pappus applied to the points A , A , X on m and B , B , M on u impliesthat A × B , B × A intersect on the line A × B, therefore C = D or A + B = B + A. Definition.
Given 2 points A and B on x , the multiplication of the 2 points, C := A · B is defined asfollows,Let X := y × m, M := ( M × X ) × x, Z := ( M × M ) × m,B (cid:48) := ( B × Z ) × y, A (cid:48)(cid:48) := ( A × M ) × m, C := ( A (cid:48)(cid:48) × B (cid:48) ) × x. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem.
The multiplication is commutative, in other words, A · B = B · A . Proof: We have by definition A (cid:48) := ( A × Z ) × y, B (cid:48)(cid:48) := ( B × M ) × m, D := ( B (cid:48)(cid:48) × A (cid:48) ) × x, The axiom of Pappus appliedto the points A (cid:48) , B (cid:48) , M (cid:48) on y and A (cid:48)(cid:48) , B (cid:48)(cid:48) , Z on m implies that A (cid:48)(cid:48) × B (cid:48) , B (cid:48)(cid:48) × A (cid:48) intersecton the line A × B, therefore C = D or A · B = B · A. Theorem.
The distributive law applies, in other words A · ( B + C ) = ( A · B ) + ( A · C ) . Proof: Let Z be a point on m distinct from X and Y .Let B (cid:48) := ( B × Z ) × y, C (cid:48) := ( C × Z ) × y, BpC (cid:48) := ( BpC × Z ) × y, Let A (cid:48)(cid:48) be the direction of M × A or ( M × A ) × m . B × AtB , B × AtB and B × AtB have the same direction therefore, if U := ( B × Y ) × ( ZtC × X, then OB = AtCU , therefore
AtCAtB + C = OAtB , therefore. . . ..
In the general descriptions, I will from time to time give, between braces, information tothe reader with advanced knowledge. This information is not required for the reader withoutprior knowledge, and may be explained in later sections or not. In the next paragraph, thereare several examples of such use of braces. To construct finite Euclidean geometries I willuse a model which depends on the { field of } integers modulo p. The properties of the integers Z are assumed. The model will be constructed in 4 steps.In the first step, described in this section, I will not distinguish between points and directions,and use the well known algebraization of the finite projective plane { associated with a Galoisfield, corresponding to the prime p } , see also I.3.In the second step, (Section 8) I will introduce an ideal line { which plays the role of line atinfinity in the Euclidean plane } , the notion of parallelism and of mid-points.In the third step (III.1), I will introduce the notion of perpendicularity { associated to aninvolution on the ideal line } . { All these steps are valid in any field. } In the fourth step (III.2 and 3), I will introduce measure of angles and of distances, togetherwith a finite trigonometry. G21.TEX [MPAP], September 9, 2019 .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY.
Definition.
A point is represented by an ordered triple of integers modulo p, placed between parenthesis .Not all 3 integers can be simultaneously 0. Two triples are equivalent iff one of them can bederived from the other using multiplication, modulo p, by an integer which is not zero modulo p. Example.
If p = 3, there are 13 points:(0,0,1), (0,1,0), (0,1,1), (0,1,2), (1,0,0), (1,0,1),(1,0,2), (1,1,0), (1,1,1), (1,1,2), (1,2,0), (1,2,1), (1,2,2).(2,2,0) is the same as (1,1,0), (2,1,2) is the same as (1,2,1).
Convention.
When I will compute numerically, I will always choose the representation of triples in sucha way that the first non zero integer in the triple is 1. This representation will be called the normal representation . When I perform algebraic manipulations, I will multiply by the mostconvenient expression, to simplify the components or, if appropriate, to make the symmetryevident.
Notation.
A more compact notation for the triples is to use a single integer, as follows, (0) for (0 , , , ( i + 1) for (0 , , i ) , ≤ i < p, (( i + 1) p + j + 1) for (1 , i, j ) , ≤ i, j < p. When there is no ambiguity, I will often drop the parenthesis.
Exercise.
Justify the Notation 2.2.1 and therefore check that there are p + p + 1 points in the projectivegeometry associated to p. Definition.
A line is represented by an ordered triple of integers, modulo p, placed between brackets .Again, not all 3 integers can be simultaneously equal to 0, and 2 triples which can be obtainedfrom each other by multiplication of each integer by the same non zero integer modulo p areconsidered equal. Notation.
The notation [0] for [0,0,1], . . . , similar to 2.2.1 will be used for lines. I will, also drop thebracket around the single integer, if there is no ambiguity. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Definition.
The point P = ( P , P , P ) and the line l = [ l , l , l ] are incident , or P is on l or l goesthrough P , iff P · l := P .l + P .l + P .l = 0 (mod p )) .P and l are not incident iff P · l (cid:54) = 0 . Example.
For p = 5 , (1,0,1) is on [1,2,4], (1,2,3) is on [1,4,2].The points (5) = (0,1,4), (10) = (1,0,4), (14) = (1,1,3), (18) = (1,2,2), (22) = (1,3,1),(26) = (1,4,0) are the 6 points on the line [12] = [1,1,1]. Definition.
I recall the definition of the cross product of 2 three dimensional vectors . X ∗ Y := ( X , X , X ) ∗ ( Y , Y , Y ) :=( X Y − X Y , X Y − X Y , X Y − X Y ) Notation.
When I use the cross product of 2 vectors and then normalize using the convention 2.2.1, Iwill use the symbol “ × ” , which recalls the symbol “ ∗ “, instead of that symbol. The resultis unique, if I compute numerically. It is not unique, if I proceed algebraically. In this case,equality implies that an appropriate scaling has been used on either side of the equation oron both sides. See Chapter V, for some examples. Theorem. P × Q is the line through the distinct points P and Q.p × q is the point on the distinct lines p and q. This follows from ( P × Q ) · P = ( P × Q ) · Q = 0 or ( p × q ) · p = ( p × q ) · q = 0 . Example.
For p = 5 , [1 , ,
3] := (1 , , × (1 , , and (1 , ,
3) := [1 , , × [1 , , . Theorem. k A ∗ B − k C ∗ D is a line incident to A × B and C × D. The lines k A ∗ B − k C ∗ D, k C ∗ D − k E ∗ F and k E ∗ F − k A ∗ B , are incident. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Example.
For any p , let A = (0,1,1), B = (1,2,1), C = (2,1,1), D = (1,3,1), E = (2,4,1), F =(4,3,1), then A ∗ B = [ − , , − , C ∗ D = [ − , − , , E ∗ F = [1 , , − . If k = k = k = 1, we obtain the lines [1 , , − , [ − , − , , [2 , , − incident to (4,1,1). Comment.
The algebraic method allows the representation of points or lines by a single symbol. Thismethod which was well used in 19-th Century text, see for instance Salmon, 1879, ChapterXIV, has somehow fallen in disfavor. p . Introduction.
After proving that the algebraic model satisfies the axioms of finite projective geometry, Igive construction of points on a line whose coordinates have a simple algebraic relationship.These could be used as a tool for the construction of points whose coordinates are known interms of points contructed earlier. The notation O + kM used in Theorem 2.2.2 is partiallyjustified in section 2.2.4. Theorem.
Each line l contains exactly p + 1 points, each point P is on exactly p + 1 lines, thereforeThe model satisfies axiom 2.1.2.3 and its dual. Proof: We want to find the points ( x, y, z ) on the line [ a, b, c ] . At least one of the 3integers, a, b or c is different from 0, let it be c, in this case x and y cannot both be 0. Given x and y we can solve ax + by + cz = 0 for the integer z, using the algorithm of Euclid-Aryabatha, z := − ( a x + b y ) /c. (See I. ?? )If x = 1 , to each value of y from 0 to p − corresponds a value of z, namely − ( a + b y ) /c. If x = 0 and y = 1 , we obtain one value of z, namely − b/c. Therefore we obtain altogether p + 1 points.Exchanging brackets and parenthesis gives the dual property. Theorem.
The model satisfies the axiom of Pappus.
Proof: I will give the proof in the special case in which the lines are [0] = [0,0,1] and [1]= [0,1,0].The general case can be deduced from general considerations on projectivity or can be provendirectly. This direct proof is left as an exercise.We choose A = (1 , a , , A = (1 , a , , A = (1 , a , and B = (1 , , b ) , B = (1 , , b ) ,B = (1 , , b ) , with a a a b b b (cid:54) = 0 . Then C = ( a b − a b , a a ( b − b ) , b b ( a − a )) , CHAPTER 2. FINITE PROJECTIVE GEOMETRY C = ( a b − a b , a a ( b − b ) , b b ( a − a )) ,C = ( a b − a b , a a ( b − b ) , b b ( a − a )) . It is easy to verify that a b C + a b C + a b C = 0 , therefore the points C , C and C arecollinear as will be seen shortly, in 2.2.4.The special cases, where A = (0 , , or B = (0 , , or a or b = 0 , can be verified easily. Theorem.
The algebraic model satisfies the axioms of finite projective geometry and therefore itcan be used to prove all the theorems of finite projective geometry.
Definition.
Given a triangle { a , a , a } , and 2 arbitrary lines, x and y, the Pappus line of x and y, isthe line z associated to the application of the axiom of Pappus to the intersection with x and y of the lines a , a and a . Theorem. If a = [1 , , , a = [0 , , , a = [0 , , , if x = [x , x , x ] and y = [y , y , y ] then thePappus line of x and y is z = [ x y ( x y + x y ) , x y ( x y + x y ) , x y ( x y + x y )] . The proof is left as an exercise. Hint: One of the 3 points on z is ( x y y − x x y , x y ( x y − x y ) , x y ( x y − x y )) . Comment.
Definition 2.2.3 may be new, it was suggested by one of the construction in a triangle of thepoint of Lemoine from the barycenter and orthocenter. See 4.2.12. The operation of deriving z from x and y is commutative but is not associative. Exercise.
Verify that if p = 2 and p = 4 the diagonal points of a complete quadrangle are collinear.For p = 2 , choose one such quadrangle. For p = 4 , the coordinates of the points are u + vξ ,where u, v ∈ Z and ξ + ξ + 1 = 0 . Choose a quadrangle which is not in the subspace v = 0 . Exercise.
Prove algebraically the 2 cases of the Theorem of Desargues.
Definition.
Let the coordinates of the distinct points O = (o ,o ,o ) and M = (m ,m ,m ) be normalized, O + x M is the point on O × M whose coordinates are (o +x m ,o +x m ,o +x m ). .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Comment.
The following Theorem relates specific constructions in projective geometry to algebraic op-erations, nothing is claimed as to the projective properties of the operation “+”, these willrequire the introduction of preferences associated with affine and Euclidean geometries. Inthis Theorem we have not used the notation “ ∗ “ which appears in section 2.2.4, this thereason why the notation O + kM used in Theorem 2.2.2 is only partially justified in section2.2.4. Theorem. [Baker]
Let
A, B, C, E be points on the line a := O × M such that A = O + a M, B = O + b M, C = O + c M, E = O + M, the following constructions gives points O + x M, A (cid:48) for x = − a, D for x = a+b, D (cid:48) for x = a+b+c, L for x = ab, I for x = a − . Let P be a point not on a and let Q be a point on A × P, distinct from A and P . Let q := O × Q, U := q × ( P × M ) ,p := O × P, V := p × ( Q × M ) ,A (cid:48) := ( U × V ) × a, then A (cid:48) = O + (-a) A. Let pb := P × B, R := pb × ( O × Q ) ,b := R × M , pa := A × P, S := pa × b,c := Q × M,T := pb × c,D := ( S × T ) × a, then D = O + (a+b) M. Let R (cid:48) := ( O × T ) × b,T (cid:48) := ( C × R (cid:48) ) × c,D (cid:48) := ( S × T (cid:48) ) × a, then D (cid:48) = O + (a+b+c) M. Let J := pa × ( E × R ) ,K := pb × ( J × M ) ,L := ( Q × K ) × a, then L = O + ab M. CHAPTER 2. FINITE PROJECTIVE GEOMETRY Let G := ( P × E ) × c,H = ( Q × E ) × p,I = ( G × H ) × a, then I = O + a − M. Proof: Choose the coordinate system such that O = (1 , , , M = (0 , , , P = (0 , , , then, for some a, b, c and q (cid:54) = A = (1,a,0), B = (1,b,0), C = (1,c,0), E = (1,1,0), Q = (1,a,q).For 0, we have a = [0,0,1], q = [0,q, − a], U = (0,a,q), p = [0,1,0], V = (1,0,q), A (cid:48) =(1,-a,0).For 1, we have pb = [b, − ,0], R = (a,ab,bq), b = [bq,0, − a], pa = [a, − S = (a,a ,bq), c = [q,0, − T = (1,b,q), D = (1,a+b,0).For 2, we have R (cid:48) = (a,b ,bq), T (cid:48) = (b,bc+b − ac,bq), D (cid:48) = (1 , a + b + c, . For 3, we have J = (a(b − ),a (b − ),bq(a − )), K = (a(b − ),ab(b − ),bq(a − )), L = (1,ab,0).For 4, we have G = (1,1,q), H = ( − a,0,1), I = (a,1,0). Exercise.
Give constructions0. associated with the associativity, commutativity and distributivity rules (a + b) + c =a + (b + c), (a b) c = a (b c), b + a = a + b,b a = a b and a (b + c) = a b + a c.1. for F = O + ab − M. Introduction.
The following properties generalize, to the finite case, well known properties of vector calculus.Capital letters will represent points, lower case letters will represent lines, the role of pointsand lines can be interchanged because of duality. I have chosen to give at once the relationswhich directly apply to geometry rather than those which correspond to vector calculus. Thesewould be obtained if all lower case letters are replaced by upper case letters. 1 or 2 of Theorem2.2.4 justify the representation of any point on the line A ∗ B by kA + lB . Theorem. A ∗ B = − B ∗ A. ( A ∗ B ) ∗ c = ( A · c ) B − ( B · c ) A. a ∗ ( B ∗ C ) = ( a · C ) B − ( a · B ) C. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. ( A ∗ B ) · C = ( B ∗ C ) · A = ( C ∗ A ) · B = − ( B ∗ A ) · C = − ( C ∗ B ) · A = − ( A ∗ C ) · B. ( A ∗ B ) · ( c ∗ d ) = ( A · c )( B · d ) − ( A · d )( B · c ) . ( A ∗ B ) ∗ ( C ∗ D ) = ( A · ( C ∗ D )) B − ( B · ( C ∗ D )) A. ( C ∗ A ) ∗ ( A ∗ B ) = (( A ∗ B ) · C ) A. (( A ∗ B ) · C ) P = (( B ∗ C ) · P ) A + (( C ∗ A ) · P ) B + (( A ∗ B ) · P ) C. ( A ∗ B ) ∗ C + ( B ∗ C ) ∗ A + ( C ∗ A ) ∗ B = 0 . Proof: The proof of 0 is immediate, the proof of 1. follows from the computation of anyof the components of the triples on both sides, for the 0-th component, ( A B − A B ) c − ( A B − A B ) c = ( A c + A c ) B − ( B c + B c ) A , adding and subtracting A B c gives the 0-th component of the second member of 1. 2,follows from 0 and 1. 3, give various expressions of the 3 by 3 determinant constructed withthe 3 triples as the 3 columns, namely A B C + A B C + A B C − A B C − A B C − A B C . for 4, ( A ∗ B ) · ( C ∗ D ) = (( C ∗ D ) ∗ A ) · B = ( A · C )( D · B ) − ( A · D ) · ( C · B ) . because of 3 and then 1.6, follows from 1.7, follows from 0, 1 and 2 applied to ( B ∗ A ) ∗ ( C ∗ P ) .
8, follows from 1.
Theorem.
A, B and C are collinear iff ( A ∗ B ) · C = 0 . Proof: This an immediate consequence of the fact that ( A ∗ B ) · C = 0 iff C is on the line A ∗ B. Theorem. [Fano] If p is odd and { A, B, C, D } is a complete quadrangle, the intersections b ∗ b , c ∗ c , d ∗ d of the opposite sides are not collinear. Proof: ( b ∗ b ) ∗ ( c ∗ c ) ∗ ( d ∗ d )= 2(( B ∗ C ) · D )(( C ∗ D ) · A )(( D ∗ A ) · B )(( A ∗ B ) · C ) (cid:54) = 0 , by repeated use of 2.2.4.0 to .3. Comment. p = 2 is excluded, because in this case, the preceding Theorem is false, in fact every completequadrangle has its diagonal collinear. I leave as an exercise the determination of where theabove theory breaks down. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Comment.
In algebraic manipulations, although A i = a i +1 × a i − , we can not use this expression whenthe sum of two or more terms is involved, because the scaling to go from “ ∗ ” to “ × ” isdifferent for each index i . It is therefore essential to do these algebraic manipulations using“ ∗ “. For the various proof, a i will always denote A i +1 ∗ A i − and use will often been madeof the following Theorem: Theorem. If a i := A i +1 ∗ A i − and t := ( A ∗ A ) · A , then a i +1 ∗ a i − = tA i . Again, i = 0 , , Proof: The conclusion follows from 2.2.4.7 and from 2.2.4.3.
Comment.
The identity 2.2.4.8 is the fundamental identity in Lie algebras. The set of points or the setof lines form a Lie algebra, if we use as multiplication “ ∗ ”. See, for instance, Cohn, Liegroups. Notation. det ( A, B, C ) will denote ( A ∗ B ) · C = ( B ∗ C ) · A = ( C ∗ A ) · B = . Theorem. det ( A, C, E ) det ( B, D, E ) det ( A, B, F ) det ( C, D, F )= det ( A, C, F ) det ( B, D, F ) det ( A, B, E ) det ( C, D, E ) iff ((( A × E ) × ( D × C )) × (( E × B ) × ( C × F ))) · (( B × D ) × ( A × F )) = 0 .Proof: By 2.2.4.1 and .2, the second equation is equivalent to (( A × E ) × ( D × C )) · ( C × F ) (( B × D ) × ( A × F )) · ( E × B ) =(( A × E ) × ( D × C )) · ( E × B ) (( B × D ) × ( A × F )) · ( C × F ) , by 2.2.4.3 this is equivalent to (( D × C ) × ( C × F )) · ( A × E ) (( E × B ) × ( B × D )) · ( A × F ) =(( E × B ) × ( A × E )) · ( D × C ) (( A × F ) × ( C × F )) · ( B × D ) , by 2.2.4.6 this is equivalent to ( det ( C, F, D ) C ) · ( A × E ) ( det ( B, D, E ) B ) · ( A × F ) =( det ( A, B, E ) E ) · ( D × C ) ( det ( A, F, C ) F ) · ( B × D ) . This can be considered as an algebraic form of Pascal’s Theorem. for the order
A, E, B, D, C, F.
Convention.
In this section, I will use the convention that if point is on the line [0,1,0], ( ∞ , , denotes the point (1 , , . .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Definition.
Given 4 points on the line [0,1,0], A = ( m , , , A = ( m , , , A = ( m , , , A = ( m , , .The anharmonic ratio is defined byanhr ( A , A , A , A ) := anhr (m , m , m , m ) := (m − m )(m − m )(m − m )(m − m ) . If m i = ∞ then the 2 factors containing m i are dropped, e.g.if m = ∞ then anhr ( A , A , A , A ) := anhr ( ∞ , m , m , m ) := m − m m − m . Lemma.
Let a , b , c and d be such that ad − bc (cid:54) = 0, if t (m) := a m+ bc m+ d thenanhr(m , m , m , m ) = anhr( t (m ) , t (m ) , t (m ) , t (m )). If we project 4 points A i on a onto 4 points B i on b from the point B , it is easy to see thateach coordinate of B is a linear functions of m , therefore the ratio of 2 of some specificatecoordinates of B are functions of the form t (m) . This justifies the following 2 Theorems. Theorem.
Given 4 points B i on a line b , the anharmonic ratio of the 4 points is the anharmonic ratioof the 4 ratios obtained by dividing the j - th coordinate of B i by the k - th coordinate forappropriate j (cid:54) = k . Theorem.
If 4 points B i are obtained by successive projections from 4 points A i , then anhr( B , B , B , B )= anhr( A , A , A , A ). Theorem. If r :=anhr( A , A , A , A ) , then if we permute the points in all possible way we obtain, ingeneral 6 different values of the anharmonic ratio: For the anharmonic ratio is A , A , A , A A , A , A , A A , A , A , A A , A , A , A rA , A , A , A A , A , A , A A , A , A , A A , A , A , A r A , A , A , A A , A , A , A A , A , A , A A , A , A , A − rA , A , A , A A , A , A , A A , A , A , A A , A , A , A − r A , A , A , A A , A , A , A A , A , A , A A , A , A , A r − r A , A , A , A A , A , A , A A , A , A , A A , A , A , A rr − Theorem.
There are 3 cases for which the 6 values are not distinct:0. 0 , ∞ , , , ∞ , , when 2 points are identical.1. − , − , , , , . v, v , v , v, v, v , with v − v + 1 = 0 or v = √− v is real, if p ≡ CHAPTER 2. FINITE PROJECTIVE GEOMETRY
For instance, if p = 7 , then v = − or 3, if p = 19 , then v = − or 8. Theorem.
Given a complete quadrangle
A, B, C, D , the intersection of 2 of lines through oppositevertices and the line through 2 diagonal points make a harmonic quatern with these diagonalpoints. More precisely, let E := ( A × B ) × ( C × D ) and F := ( A × D ) × ( B × C ) be 2 ofthe diagonal points, let a := E × F , G := a × ( B × D ) and H := a × ( A × C ), then r :=anhr( E, F, G, H ) = − Let I be the third diagonal point, projecting the 4 points from B on A × C and these from D on a gives r = anhr ( A, C, I, H ) = anhr ( F, E, G, H ) = r , therefore r = 1 but we do nothave case 0, r = 1 , therefore r = − which is case 1. Definition.
In the special case of the preceding Theorem:Case 1, we say that A , A are harmonic conjugate of A , A , or that A , A , A , A form a harmonic quatern .Case 2, we say that A , A , A , A form a equiharmonic quatern . Definition.
The pre-equiharmonic non confined configuration is defined as follows:Given a complete quadrangle
A, B, F, K, determine q := A × B, p := A × F, b := B × F, r := A × K, f := B × K,H := p × f, J := b × r, choose C on q, distinct from A and B, determine c := C × K, P := b × c, R := p × c, g := C × F, Q := f × g, L := g × r,d := P × L, h := Q × R, D := d × h. Theorem.
Given D · q = 0 . J · h = 0 ⇒ H · d = 0 . The geometric condition J · h = 0 is equivalent to A, B, C, D is an equiharmonic quatern.
Proof: Let A = (0 , , , B = (0 , , , F = (1 , , , K = (1 , , . We have q = [1 , , ,p = [0 , , , b = [ − , − , , r = [ − , , , f = [0 , − , , H = (1 , , , J = (1 , , . Let C = (0,c,1), then c = [1 − c, − ,c], g = [c,1, − c], P = (1 − c,1,2 − c), R = (c,0,c − Q =(c − L = (c,c,c+1), d = [c − c+1,2c − , − c ], h = [c − c,2c − , − c ], D = (0,c ,2c − J · h = 0 or H · d = 0 are equivalent to c − c+1 = 0. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Definition.
Given 2.2.5 and J · h = 0 , the pre-equiharmonic configuration is then called an equiharmonicconfiguration . Theorem.
A pre-equiharmonic configuration is of type10 × × × × , unless it is equiharmonic, in which case, it is of type12 × × . The sets of 4 points on each of the 9 lines is an equiharmonic quatern.
Proof: The configuration, with projections as given below is as follows. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Points: lines: from A : q, p, r, q : A, B, C, D,B : q, b, f, p : A, F, R, H, P ( Q, H ) C : q, c, g, r : A, J, K, L, P ( Q, K ) D : q, h, d, b : F, B, P, J, R ( L, J ) P : b, c, d, f : H, B, K, Q, R ( L, K ) R : p, c, h, c : R, K, C, P, H ( J, K ) H : p, f, d, g : F, Q, C, L, H ( J, L ) K : r, f, c, d : L, H, P, D, K ( F, H ) Q : f, g, h, h : J, Q, R, D, K ( F, R ) L : r, g, d,J : r, b, h,F : p, b, g, If we project
A, B, C, D from P on p we get A, F, R, H ; if we project A, B, C, D from Q on p we get A, H, F, R ; therefore r = − r . To establish the results for the other sets, it is sufficientto project from a point, those of the line q. The points on each line have been arrangedcorrespondingly. For instance, for f, the point of projection is R and the lines are p, c and h ; if the point of projection is L, the order is K, B, Q, H, (the second point corresponds to A or B for p and r the others are obtained circularly). Exercise.
Prove that the configuration of 2.1.7 is equiharmonic.
Exercise.
Study the configuration which starts with P , P , P , P and P on P × P . Constructs l := P × P , l := P × P , l := P × P , l := P × P , l := P × P , l := P × P ,P := l × l , l := P × P , P := l × l , l := P × P , P := l × l .Using a coordinate system such that P = (1 , , , P = (0 , , , P = (0 , , and P =(1 , , , determine an algebraic condition involving the coordinates of P for P to be on l .Prove that if P is on l , the configuration is of type × × . Exercise.
Study the configuration which starts with A , A , B , B and A on A × A , constructs d := A × B , d := A × B , d := A × B , d := A × B , d := A × B , d := A × B ,a := A × A , b := B × B , P := a × b, C := d × d , C := d × d , C := d × d ,a := C × C , d := A × C , d := A × C , B := d × d . Determine a geometric condition on A for P, A , A , A to be an equiharmonic quatern,prove that in this case P , B , B , B is also a equiharmonic quatern. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Introduction.
In the next section we will study algebraically the isomorphisms of the plane into itself. Thespecial case of the mapping of a line of the plane into a line will be defined here. Thejustification will follow from the general definition. Such a mapping is called a projectivity.Special cases will be studied and appropriate constructions will be given. The notion ofamicable projectivities, which are at the basis of the definition of equality of angles is alsointroduced. The concept of harmonic conjugates is due to LaHire . The term projectivitywill be used here only for correspondances between points on lines not for correspondance ofa plane with itself as done by some authors. Theorem 2.2.6 gives a construction, when 2points A and B are fixed, and D corresponds to C. Convention.
For simplicity, I will assume that the line is [0,0,1], the last component of all the points is0, I will therefore only write the first 2 components.
Definition.
The mapping which associates to the point ( x , x ) the point ( y , y ) given by ( y ) = ( ab )( x ) , ( y ) = ( cd )( x ) with ad − bc (cid:54) =
0, is called a projectivity . Theorem. If C is the intersection of c and A × B, the point D such that A, B, C and D form a harmonicquatern is given by D := B · cA + A · cB. Definition. D is called the harmonic conjugate of C with respect to A and B. Theorem. If K = (1 , k, , L = (1 , l,
0) and M = (1 , m,
0) then N, the conjugate of M withrespect to K and L, is given by N = (2 m − l − k, k m + l m − k l, . If N is the harmonic conjugate of M with respect to K and L ,then M is the harmonic conjugate of N with respect to K and L, Coxeter, p.16 CHAPTER 2. FINITE PROJECTIVE GEOMETRY K is the harmonic conjugate of L with respect to M and N , and L is the harmonic conjugate of L with respect to M and N. Theorem. If A := (1 , , , B := (0 , , , C := (1 , k, , D := (1 , l, , the projectivity on c := A × B which associates A to A, B to B and C to D, can be constructed by choosing a line b through a, dictinct from c, a line a through b distinct from c, a point P on a not on b or c, then S := ( P × C ) × b, T := ( P × D ) × b. The mapping N of M := (1 , m,
0) is obtained by the construction Q := ( M × S ) × a, N := ( Q × T ) × c and N = ( k, lm, . The proof is left as an exercise.
Theorem.
Let u := lk , φ ( M ) = (1 , um,
0) and φ j ( M ) = (1 , u j m, . If u is a primitive root of p, theprojectivity has order p − . The proof is left as an exercise.
Theorem. If K, L and M are distinct, N is distinct from M. Proof. The last theorem would imply that m (2 m − l − k ) = km + lm − klor ( m − l )( k − m ) =0 . Theorem. If K and M are exchanged and L is replaced by N, then N is replaced by L. Indeed, n (2 m − k − l ) = k m + l m − k l can be written l (2 k − m − n ) = m k + n k − m n. The following theorem gives a construction of a projectivity on a line in which B i corre-sponds to A i , i = 0 , , . Theorem.
Given A i , B i , i = 0 , , , u, such that the A i are distinct and the B i aredistinct. Choose the line s (cid:54) = u and the point S, with S · u (cid:54) = 0, S · s (cid:54) = 0.Construct C i := ( S × A i ) × s, D j := ( B × C j ) × ( C × B j ) , j = 1 , , d := D × D , then for any A l on u , construct C l := ( S × A l ) × s, D l := ( B × C l ) × d, B l := ( C × D l ) × u. The mapping which associates B l to A l is a projectivity. Theorem.
Given A = (1 , , , A = (1 , a , , A = (1 , a ,
0) and B = (0 , , , B = (1 , b , ,B = (1 , b , , then the projectivity which is defined in the preceding theorem and associates .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. A i , B i , i = 0 , , , associates to A j = (1 , a j , , the point B j = (( a − a ) a j , ( a b − a b ) a j − a a ( b − b ) , , j > . The proof is left as an exercise.
Theorem.
Let g = a b − a b a − a ) and h = a a b − b a − a , The projectivity, which associates to (1 , u, , (1 , g − hu ) , is an involution iff g = 0 , is an hyperbolic projectivity if g − h is a quadratic residue modulo p, is an elliptic projectivity if g − h is a non residue and is a parabolic projectivity if h = g , the fixed point being g. Proof. If we eliminate u from u = 2 g − hu and u = 2 g − hu , we get g ( u − g u + h ) . Ifthis relation is to be satisfied for all u , it is necessary that g = 0 . The condition is sufficientbecause if φ ( u ) := hu , then φ ◦ φ is the identity. Theorem.
Given 3 distinct points A , A and A on the line a and 3 distinct points B , B and B onthe line b, let A = r A + r A and B = s B + s B , thenif A j = t A + t A , B j = φ ( A j ) := s t r B + s t r B is a projectivity which associates A j to B j for all j. Proof. The correspondance clearly associates A j to B j for j = 0 , using t = 0 , for j = 1 using t = 0 and for j = 2 using t = r and t = r . The proof that it is a projectivity is leftas an exercise.
Theorem. If the lines a and b of the preceeding Theorem coincide, there exists constants f , f , f and f such that B j = ( f t + f t ) A + ( f t + f t ) A . If B = b A + b A and B = b A + b A , then f = s b r , f = s b r , f = s b r , f = s b r . The values t and t for which A j is a fixed point, in other words, for which A j = B j satisfy f t − ( f − f ) t t − f t = 0 . The projectivity is hyperbolic, parabolic or elliptic if( f − f ) + 4 f f is positive, zero or negative.14 CHAPTER 2. FINITE PROJECTIVE GEOMETRY The projectivity is an involution iff f + f = 0 . The proof is left as an exercise.
Definition.
Using the notation of 2.2.6 and of 2.2.6 with primes used for an other projectivity, we saythat iff there exists a constant k different from0 such that f (cid:48) = kf , f (cid:48) = kf , f (cid:48) − f (cid:48) = k ( f − f ) . Theorem.
Two amicable projectivities are either both hyperbolic, or both parabolic or both elliptic. Ifthey are both hyperbolic, they have the same fixed points.
Example.
For p = 5 , a projectivity φ associates to A = (5) , A = (10) , A = (14) , (18) , (22) , (26) , B = (26) , B = (5) , B = (18) , (22) , (10) , (14) .r = r = 1 , s = 1 , s = − , b = − , b = 1 , b = 1 , b = 0 , f = − , f = − , f = 1 ,f = 0 . A second projectivity φ (cid:48) associates to A (cid:48) = (5) , A (cid:48) = (10) , A (cid:48) = (14) , (18) , (22) , (26) , B (cid:48) = (18) , B (cid:48) = (14) , B (cid:48) = (10) , (5) , (26) , (22) .r (cid:48) = r (cid:48) = 1 , s (cid:48) = − , s (cid:48) = 2 , b (cid:48) = 2 , b (cid:48) = 1 , b (cid:48) = 1 , b (cid:48) = 1 , f (cid:48) = − , f (cid:48) = 2 , f (cid:48) = − ,f (cid:48) = 2 . φ (cid:48) is an involution an f (cid:48) + f (cid:48) = 0 . φ and φ (cid:48) are sympathic, with k = − . The fixedpoints are complex and correspond to t = 1 and t = 1 + √− or t = 1 − √− . Comment.
The definition 2.2.6 will be used in III.1.3. to define equality of angles.
Theorem. If x is one of the coordinates, the projectivity takes the form F ( x ) = a + bxc + dx and the fixed points are the roots of dx + ( c − b ) x − a = 0 . Exercise.
Prove that the following construction defines a projectivity on u in which A i +1 corresponds to A i , the points A to A being given. Let l f is a line through A distinct from u, E is a point on l f distinct from A , F is a point on l f distinct from A and E, l d is a line through A distinctfrom u, D := l f × l d , D is a point on A × D distinct from A and D , E := ( A × E ) × ( A × F ) , E := ( A × F ) × ( E × ( l d × ( A × D ))) , A i +1 := ((((( A i × D ) × l d ) × E ) × l e ) × F ) × i,i = 4 , . . . . The preceding construction is less efficient than that in 2.2.6. G22.TEX [MPAP], September 9, 2019 .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY.
Introduction.
Collineations, { which are isomorphisms of the plane onto itself } have been defined in 2.1.12.They will now be studied algebraically. The point mapping which associates points to pointsis represented by a non singular matrix, and so is the line mapping which associates lines tolines. Two matrices which can be obtained from each other by multiplication, modulo p, byan integer different from 0 correspond to the same collineation. Theorem.
Given 2 complete quadrangles A j and B j , j = 0,1,2,3,Let a i := A i +1 ∗ A i − and b i := B i +1 ∗ B i − , let A = r A + r A + r A , B = s B + s B + s B q i := s i r i , u i := q i +1 q i − , then, up to a proportionality constant, q i = b i · B a i · A . Moreover, the mapping γ defined by B l := γ ( A l ) := q ( a · A l ) B + q ( a · A l ) B + q ( a · A l ) B is the point mapping of a collineation which associates to A j , B j for j = 0 to 3.1. the mapping γ (cid:48) defined by γ (cid:48) ( a l ) := u ( A · a l ) b + u ( A · a l ) b + u ( A · a l ) b is the corresponding line mapping. Proof: By hypothesis, r (cid:54) = 0 , because A is not on A × A , similarly, r , r , as well as s , s and s are (cid:54) = 0 , therefore, q , q and q are well defined and (cid:54) = 0 . a · A = ( A ∗ A ) · ( r A + r A + r A ) = r det ( A , A , A ) , similarly, a i · A = r i det ( A , A , A ) , b i · B = r i det ( B , B , B ) , hence the alternate expression for q i .0. follows from 2.2.4 by observing that ( A i ∗ A ) ∗ a i = r i +1 A i +1 − r i − A i − . The details are left as an exercise.If M and N are any 2 points on a l and a l = M ∗ N,γ (cid:48) ( a l ) = γ (cid:48) ( M ∗ N ) = γ ( M ) ∗ γ ( N )= q q ( a · M a · N − a · M a · N ) b + . . . = q q (( a ∗ a ) . ( M ∗ N )) b + . . . , = u t ( A · a l ) b + . . . , because of 2.2.4. Dividing by t, we get 1. Theorem.
If a collineation transforms each of the points of a complete quadrangle into itself, everypoint is transformed into itself.
Definition.
The collineation of 2.2.7 is called the identity collineation (cid:15) . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Comment.
Theorem 2.2.6 for 1 dimensional sets and Theorem 2.2.7 for 2 dimensional sets generalizeby induction to n dimensions. Example.
For p = 5, let A = (0) = (0 , , , A = (1) = (0 , , , A = (6) = (1 , , and A = (12) =(1 , , , let B = (1) , B = (6) , B = (12) , B = (3) = (0 , , , to obtain the point mapping γ which associates to A j , B j , a = [0 , , − , a = [0 , − , , a = [ − , , , b = [0 , − , ,b = [ − , , , b = [0 , , − .q = − , q = − , q = 2 , u = − , u = − , u = 2 , therefore γ XYZ = − − − XYZ = − X + 2 Y − X + Z − X ,γ (cid:48) xyz = − − − xyz = y − z − x − y + z . Comment.
The following mapping can be used in certain cases but not in all cases: φ ( M ) := q ( M · A ) φ ( A ) + q ( M · B ) φ ( B ) + q ( M · C ) φ ( C ) , with q = A · P ( φ ( B ) ∗ φ ( C )) φ ( P ) , q = B · P ( φ ( C ) ∗ φ ( A )) φ ( P ) ,q = C · P ( φ ( A ) ∗ φ ( B )) φ ( P ) . Indeed, one of the scalar product A · P or B · P or C · P can be 0, and cases exists for whichwhatever permutation of the 4 points A, B, C and P is used, the same difficulty occurs. Theorem. can be rewritten using matrix notation. Let a be a matrix whose rows are the compo-nents of the sides of the triangle A , A , A , a i,j := a j,i , etc.Let Q be the matrix Q i,i := q i , Q i,j := 0 for i (cid:54) = j, let U i,i := q i +1 q i − , U i,j := 0 , i (cid:54) = j. Let A l and B l be column vectors,then M := B Q a T defines the collineation B l = M A l and M (cid:48) := b U A T gives b l = M (cid:48) a l . Moreover M (cid:48) = M − is the adjoint matrix. Example.
Let A = (7) , A = (15) , A = (19) , A = (28) ,B = (27) , B = (3) , B = (10) , B = (14) . B = − − , Q = − − , a T = − − , .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. M = B Q a T = − − − − , b = − − − − − − , U = − , A T = −
11 2 − , m = b U A T = − − − , Definition.
A collineation is called a central collineation if the collineation transforms every point of agiven line into itself, and it is not the identity.The line is called the axis of the central collineation . Theorem.
Let a collineation be given by 2 complete quadrangles with 2 fixed points A and B and 2other pairs A , B and A , B , the necessary and sufficient condition for this collineation tobe a central collineation is that A × A , B × B and A × A have a point in common. Theorem.
In a central collineation, if B l corresponds to A l and is distinct from A l , then A l × B l passesthrough a fixed point F. Definition. F is called the center of the central collineation . Definition.
A central collineation is a homology iff its center is not on its axis.A central collineation is an elation iff its center is on its axis.
Comment.
Theorem 2.2.7 or 2.2.7 could serve as an alternate definition of collineation.oreo
Exercise.
Characterize the matrix of a central collineation, and of an elation. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Notation.
When matrices are used to represent collineations correlations it is convenient to have anotation for the inverse matrix scaled by a convenient non zero factor, meaning that eachentry is multiplied by that factor, N I will be used. Introduction.
Correlations have been defined in 2.1.12. Their algebraic study follows directly from that ofcollineations. Their importance is due to their intimate relation with conics as will be seenin 2.2.9.
Definition.
The mapping which associates to the point ( m ) , the line [ m ], for all m = 0 top k + p k is calleda basic duality . It will be denoted by δ . Theorem.
The mapping δ is a correlation. Theorem.
Given a point collineation γ and the corresponding line collineation γ (cid:48) , then the mapping ρ := δ ◦ γ is a point correlation, and the corresponding line correlation is ρ (cid:48) := δ ◦ γ (cid:48) . In particular,if Q = γ ( P ) and q = γ (cid:48) ( p ) , then ρ ( P ) = Q and ρ (cid:48) ( p ) = q . Theorem.
Given a complete quadrangle A j and a complete quadrilateral b j , j = 0 , , , , Let a i := A i +1 ∗ A i − and B i = b i +1 ∗ b i − ,A = r A + r A + r A , b = s b + s b + s b q i := s i r i , u i := q i +1 q i − , then q i := B i · b a i · A . Moreover, the correlation which associates to A j , a j , j = 0 to 3, is given by the point to line mapping b l := ρ ( A l ) := q ( a .A l ) b + q ( a .A l ) b + q ( a .A l ) b ,1. and the line to point mapping ρ (cid:48) ( a l ) := u ( A .a l ) B + u ( A .a l ) B + u ( A .a l ) B . The proof follows from 2.2.6 and from 2.2.8. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY.
Example.
For p = 5, the correlation ρ defined by ρ (0) = [0] , ρ (1) = [1] , ρ (6) = [12] , ρ (12) = [19] , = (1,2,3), implies a = [0 , , , a = [0 , , , a = [1 , , and ρ (cid:48) ( a ) = (1 , , − , ρ (cid:48) ( a ) = (1 , − , , ρ (cid:48) ( a ) = ( − , , .q = 2 , q = 1 , q = 1 , u = 1 , u = 2 , u = 2 , therefore ρ ( X, Y, Z ) = [ − X, − X − Y, − X − Z ] , ρ (cid:48) [ x, y, z ] = (2 x − y − z, y, z ) . Theorem.
Using the notation of can be written in matrix notation.Let Q i,i := Q i and Q i,j := 0 for i (cid:54) = j, let R i,i := q i +1 q i − and R i,j := 0 for i (cid:54) = j, then N := b R a T defines a correlation b l = N A l , and N (cid:48) := B V A T determines B l = N (cid:48) a l . Moreover N (cid:48) = N − T is the adjoint matrix. Definition. A polarity is a correlation which satisfies ρ (cid:48) ◦ ρ = (cid:15) .In this case ρ ( P ) is called the polar of P and ρ (cid:48) ( p ) is called the pole of p. Example.
The correlation which associates to A = (0) , B = (1) , C = (6) and P = (13) the lines [11],[7], [2] and [15], is a polarity and ρ ( X, Y, Z ) = [ Y + Z, X + Z, X + Y ] , ρ (cid:48) [ x, y, z ] = ( − x + y + z, x − y + z, x + y − z ) . Theorem. If M is a matrix associated to a correlation, then this correlation is a polarity iff the matrixis symmetric, in other words iff M = M T . Comment.
Theorem 2.2.8 or 2.2.8 could serve as an alternate definition of correlations.
Definition. A degenerate line correlation ρ d corresponds to a function which associates to the set ofpoints in the plane, lines which are obtained by multiplying the vector associated to the pointto the left by the matrix D = − U U U − U − U U . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem. If U is the point ( U , U , U ) , then D associates to the point V = ( V , V , V ) , the line U × V .In the correlation, the image of all points are lines through the point U and therefore alllines have U has their image. The matrix corresponding to ρ (cid:48) d is therefore, U U U U U U U U U . Exercise.
Prove (Seidenberg, p.193-196)0. that a linear transformation is the product of 2 polarities.1. that the set of fixed point and fixed lines of a linear transformation form a self dualconfiguration.
Introduction.
The following definition was first given by von Staudt. The connection between polarity andconics was anticipated already by Apollonius and clearly understood by La Hire.
Definition.
Given a polarity ρ with inverse ρ (cid:48) , a conic is the set of points P such that P · ρ ( P ) = 0 . and the set of lines p such that p · ρ (cid:48) ( p ) = 0 . In other words it is the set of points which are on their polar and the set of lines which areon their pole.If the polarity corresponds to a symmetric matrix a b b b a b b b a , the equation of the corresponding point conic is a X + a X + a X + 2( b X X + b X X + b X X ) = 0 . Theorem. G23.TEX [MPAP], September 9, 2019 .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY.
1. the conic through A, B, C, D and E is given by k [ A × B ] ×× [ C × D ] = k [ A × D ] ×× [ B × C ] , with k = [ A × D ] · E . [ B × C ] · E, k = [ A × B ] · E . [ C × D ] · E. Example.
Given the data of 2.2.2, the conic through A, B, C, D and E is X − X − X + 5 X X − X X = 0 . Exercise.
Prove that a conic has p + 1 points in a finite projective plane associate with p .If we join one point P to the p others we obtain p lines through P therefore the left overline is the tangent at P . Comment.
For p = 3 , the conic has 4 points, hence it cannot be constructed by giving 5 points, but itcan be constructed if we give 4 points and a tangent at one of these points or 3 points andthe tangents at 2 of these points. See 2.1.6.For p = 2 , a conic can be constructed using 3 non collinear points and the tangents at 2 ofthese points. Theorem.
The pole of [1,1,1] with respect to the conic b X X + b X X + b X X ρ + ( X + X + X )( u X + u X + u X ) = 0,is( b ( − b + b + b ) + 2 u b − u ( b + b − b ) − u ( b − b + b ) , . . . , . . . ) . Theorem.
The pole of [1,1,1] with respect to the conic c X X + c X X + c X X + u X + u X + u X = 0 , is ( c ( − c + c + c ) − u c − u c + 4 u u , . . . , . . . ) . Example.
For p = 13 , if b = 1 , b = 6 , b = 2 , u = − , u = 4 , u = 2 , then c = − , c = 2 , c = 5 , and the pole of [1,1,1] is (1,6,3) = (95). CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem.
Given the conic a X + a X + a X b X X + b X X + b X X = 0 . and a point ( P , P , P ) , with P (cid:54) = 0 , on the conic, all the points are given by X = a P u − (2 a P + b P ) uv − ( a P + b P + b P ) v ,X = − ( b P + a P + b P ) u − (2 a P + b P ) uv + a P v ,X = a P u + b P uv + a P v , using the p + 1 values of the homogeneous pair ( u, v ) . Proof: The points ( v, u, on [0,0,1] joined to P is the line l = [ − P u, P v, P u − P v ] .X is on l iff P X v = P X u − P X u + P X v, substituting in the equation of the conic, if A is the coefficient of X and B that of X , using the property of the products of the roots ofthe equations gives P X = A, P X = B, this gives X and X , substituting in l gives X . Theorem. [Chasles]
Given the configuration of Desargues there exists a conic such that A i is the pole of b i := B i +1 × B i − and vice-versa.Clearly B i is also the pole of a i := A i +1 × A i − . Proof: Let A = (1 , , , A = (0 , , , A = (0 , , , C = (1 , , , c = ( c , c , c ) and B = ( b, , . We have C = (0 , c , − c ) , C = ( − c , , c ) , C = ( c , − c , , b = [ c , c , − bc − c ] , b = [ c , − bc − c , c ] , B = ( c , ( b − c + c , c ) , B = ( c , c , ( b − c + c ) . The transformation whichassociates to a i , k i B i , with k := c c , k := c , k := c is bc c c c c c c c (( b − c + c ) c c c c c c c (( b − c + c ) c , is a line to point polarity because the representative matrix is symmetric. Its inverse caneasily be obtained by determining b . This is left as an exercise.
Notation. If u = [ u , u , u ] and v = [ v , v , v ] are 2 lines, then u ×× v = ( u X + u X + u X )( v X + v X + v X ) . Definition.
Given 2 conics α and β if there exist integers k and l and lines u and v such that kα + lβ = u ×× v, then v is called the radical axis with respect to u of α and β . Lemma. If N is a symmetric matrix and A and B are 2 vectors, then A · ( N B ) = B · ( N A ). .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Theorem.
A conic or the corresponding polarity determines an involution on every line, by associatingto each point its conjugate on that line. Moreover if A and B are conjugates as well as A and B , if A l = t A + t A its conjugate B l is given by B l = (( A .B ) t + ( A .B ) t ) A − (( A .B ) t + ( A .B ) t ) A . This property follows from the notion of conjugates and from B l = ( A ∗ A ) ∗ ( t N A + t N A ) , with B = N A and B = N A . The Lemma confirms the involutive property.
Example.
For p = 5 , starting with A (0) = (6) , A (1) = 1 , A (2) = 0 , A (3) = 12 , the quadrangle-quadrilateral configuration is a , = [6] , a , = [1] , a , = [0] , a , = [5] , a , = [10] , a , = [26] , D = (2) , D = (7) ,D = (11) , d = [30] , d = [27] , d = [15] , A , = (5) , A , = (10) , A , = (26) , A , = (24) ,A , = (17) , A , = (13) , a = [24] , a = [17] , a = [13] , a = [12] . Example.
The points and lines of the extended quadrangle-quadrilateral configuration are those of 2.2.9and B , = (9) , B , = (16) , B , = (3) , B , = (21) , B , = (4) , B , = (8) , B , = (15) ,B , = (27) , B , = (30) , B , = (18) , B , = (22) , B , = (14) , b , = [4] , b , = [8] ,b , = [21] , b , = [3] , b , = [9] , b , = [16] , b , = [2] , b , = [7] , b , = [11] , b , = [18] ,b , = [22] , b , = [14] . Example.
The conical points and lines of the extended quadrangle- quadrilateral configuration are the 6points and 6 lines C , = (20) , C , = (29) , C , = (19) , C , = (28) , C , = (23) , C , = (25) ,c , = [20] , c , = [29] , c , = [19] , c , = [28] , c , = [23] , c , = [25] . Definition. A degenerate conic is a set of points and lines represented by an equation corresponding toa singular 3 by 3 symmetric matrix. Exercise.
Describe all the types of degenerate conics.
Exercise.
The number of conics, degenerate or not is ( q + q + 1)( q + 1) , The number of degenerate conics are CHAPTER 2. FINITE PROJECTIVE GEOMETRY line ×× line q + q + 1 ,line ×× line ( q + q + 1) q ( q + 1) non real line ×× its conjugate ( q + q + 1) q ( q − , The number of non degenerate conics is q − q . Table. q line ×× line non real line ×× its conjugate line ×× line
21 78 210 465 1596 8778 non degenerate conics
28 234 1008 3100 16758 160930 all conics
63 364 1365 3906 19608 177156
Introduction.
There is a more general connection between correlations and conics, which leads to the conceptof a general conic, which is one of 4 types, the points of a conic of von Staudt and the linesof an other conic of von Staudt. It has p + 1 points and p + 1 lines; a degenerate conicconsisting of p + 1 points on 2 distinct lines and p + 1 lines through 2 distinct points; adegenerate conic consisting of p + 1 points on 1 line and of p + 1 lines through 1 point; andfinally the degenerate conic consisting of 1 point and 1 line. In the last case, in complexprojective geometry, all the complex points are on a pair of complex conjugate lines and allthe complex lines are through a pair of complex conjugate points. To every correlation isassociated a general point conic and a general line conic. Definition. A general conic consists of a point conic which is the set of points in a correlation whichare on their image and of a line conic which is the set of lines in a correlation which are ontheir image. Theorem. If N is the matrix associated to a correlation, the equation of the point conic is X T N X = 0 , where X is the vector ( X , X , X ) . The equation of the line conic is x T N − x = 0 , where x is the vector ( x , x , x ) . .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Theorem.
Let A be the most general antisymmetric matrix, A = − w vw − u − v u , all the correlations associated to N + A define the same point conic. Definition.
Given a matrix N , its symmetric part N S is defined by N S := N + N T , and its antisymmetric part N A by N A := N − N T . Theorem.
Given a correlation ρ, ρ (cid:48) .If T is on the point conic, then ρ ( T ) is on the line conic.If t is on the line conic, then ρ (cid:48) ( t ) is on the point conic.The general conic degenerates if det ( N ) = 0 . The center corresponds to the vector which is the homogeneous solution of N S C = 0 , andthe central line, to that of ( N S ) − c = 0 . Definition.
Given a general conic, the tangent t at the point T of the point conic is defined by t := N S T. The contact T of a line t which belongs to a line conic is defined by T := ( N S ) − t. Theorem.
If the correlation is a polarity then the tangent at a point T of a point conic is on thecorresponding line conic. Similarly, the contact of a line t of line conic is on the correspondingpoint conic. Theorem.
If a conic is non degenerate, the necessary and sufficient condition for the set of tangents toa point conic to coincide with the set of lines on the line conic is that the correlation be apolarity.
Proof: Let N be the matrix associated to the correlation. The line conic is x T N − x = 0 . The tangents t = N S X to the point conic are on t T ( N S ) − N ( N S ) − t = 0 . If ρ is a polarity, N = N S and N − = N S − N ( N S ) − . Vice-versa,if N − = ( N S ) − N ( N S ) − then N S = N ( N S ) − N . Therefore, using N = N S + N A , N A = − N A ( N S ) − N A , CHAPTER 2. FINITE PROJECTIVE GEOMETRY transposing, − N A = − N A ( N S ) − N A, because N A T = − N A , therefore N A = 0 , N is symmetric and therefore ρ is a polarity. Introduction.
A fundamental theorem associated to conics was discovered by Blaise Pascal. It allows con-struction of any point on a conic given by 5 points and in particular the other intersectionof a line through one point of a conic. See I, . . . .There is a general principle of linear construction that if a point or line is uniquelydefine, that point or line can be obtained by a linear construction. The points of intersectionof a conic with a general line are not uniquely defined and therefore do not admit a linearconstruction on the other hand if the line passes to a known point of the conic, the otherintersection of the line and the conic is uniquelly defined. The Pascal construction of 2.2.11is a solution to this problem which follows from the following Theorem.
Theorem [Pascal].
If 6 points A , A , A , A , A , A are on a conic and the Pascal points are defined as P := ( A × A ) × ( A × A ) ,P := ( A × A ) × ( A × A ) ,P := ( A × A ) × ( A × A ) , then the points P , P , P are collinear (Pascal, 1639, Lemma 1 and 3).There are “degenerate” forms of this theorem in which 2 consecutive points coincide andthe cord is replaced by the tangent at these points for instance if the tangent at A is t , thePascal points are P := t × ( A × A ) ,P := ( A × A ) × ( A × A ) ,P := ( A × A ) × ( A × A ) , and the points P , P , P are collinear .Proof: The Theorem of Pascal will now be proven in the 4 cases, 6 points, 5 points andthe tangents at one of them, 4 points and the tangents at 2 of them and finally 3 points andtheir tangents. In each case, the coordinates will be chosen to simplify the algebra. See also2.2.2.0. Let the 6 points of the conic be A , C , A , B , A , B . Choose the coordinates suchthat A = (1 , , , A = (0 , , and A = (0 , , , choose the barycenter M = (1 , , at the intersection of A × B and A × B , let the line A × B be [0 , r, − s ] . Because the conic passes through A i , it has an equation of the form0. uX X + vX X + wX X = 0 .A × B = A × M = [1 , , − , A × B = A × M = [1 , − , , therefore, B = ( u + w, − v, u + w ) , B = ( u + v, u + v, − w ) ,C = ( − urs, s ( vr + ws ) , r ( vr + ws )) , hence the Pascal points are P = ( A × C ) × ( B × A ) = ( s ( u + w ) , − vs, − vr ) , .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. P = ( C × A ) × ( A × B ) = ( urs, urs, − r ( vr + ws )) ,P = ( A × B ) × ( B × A ) = ( w, − ( u + v ) , w ) . which are all on [ v ( ru + rv + sw ) , w ( su + rv + sw ) , su ( u + v + w )] .
1. Let the 5 points of the conic be A , A , B , A , B , and let the tangent t be chosen at A . With the coordinates chosen as above and the conic again of the form 0.0, the tangentis [0 , w, v ] and the Pascal points are P = t × ( B × A ) = ( u + w, − v, w ) ,P = ( A × A ) × ( A × B ) = (1 , , ,P = ( A × B ) × ( B × A ) = ( w, − ( u + v ) , w ) . which all are on [ − w, w, u + v + w ] .
2. Let the 4 points be A , A , A and B and the tangents be t at A and t at A . Choose the coordinates as above, except for M on A × B = [0 , , − , then r = s = 1 .B = ( − u, v + w, v + w ) , the tangents are t = [ w, , u ] and t = [ v, u, . The Pascal points are P = ( A × A ) × t = ( − u, v, ,P = ( A × A ) × ( A × B ) = (0 , , ,P = t × ( A × B ) = ( − u, v + w, w ) , which all are on [ v, u, − u ] .
3. Let the points be A , A and A and the tangents be those at these points, using again0.0. as the equation of the conic, the tangents are t = [0 , w, v ] , t = [ w, , u ] , t = [ v, u, . the Pascal points are P = t × ( A × A ) = (0 , v, − w ) ,P = t × ( A × A ) = ( − u, , w ) ,P = t × ( A × A ) = ( u, − v, , which are all on [ vw, wu, uv ] .
4. The other cases, 4 tangents and 2 points of contact, 5 tangents and 1 point of contact,6 tangents, can be proven by duality.
Theorem [Pascal].
The reciprocal of the preceding Theorem is true. In other words, if the Pascal points P i are collinear, the 6 points A k are on a conic. The proof is left as an exercise.
Notation.
The property that 6 points are on a conic γ will be noted incidenceconic ( A, B, C, D, E, F [ , γ ]) or incidenceconic ( A k [ , γ ]) . Similar notation will be used for degenerate or for dual forms, for instance incidenceconic ( A, t, B, C, D, E ) , where t is the tangent at A. CHAPTER 2. FINITE PROJECTIVE GEOMETRY incidenceconic ( a, b, c, d, e, f ) where a, b, c, d, e, f are 6 tangents to the conic.Theorem 2.2.11 will be denoted as follows.No. P ascal ( A k [ , a k ]; (cid:104) P i [ , p ] (cid:105) ) Hy. incidenceconic ( A k ) . De. P i := ( A i × A i +1 ) × ( A i +3 × A i +4 ) Co. (cid:104) P i , p (cid:105) . If t k is the tangent at A k , A k is followed by t k .The Pascal line associated to the points A k will be denoted by p := Pascal ( A k ) . Definition.
The dual of the Theorem of Pascal is called the
Theorem of Brianchon . Brianchon discoveredthe Theorem before Gergonne discovered the important principle of duality.In the degenerate case of a triangle inscribed in a conic and of the triangle outscribed to theconic at these points (2.2.11.3), the line is called the
Pascal line of the triangle and the pointof the dual Theorem, the
Brianchon point of the triangle . von Staudt (1863) calls them, poleand polar of the triangle.
Theorem. [Generalization of von Staudt] If p is the Pascal line of the hexagon A , A , A , A , A , A inscribed in a conic γ and P isthe Brianchon point of the outscribed hexagon formed by the tangents at A j , then P is thepole of p . The proof follows at once from the properties of poles and polars.
Corollary. [von Staudt] If { A , A , A } is a triangle inscribed in a conic, then its Pascal line is the polar of itsBrianchon point. Theorem. [von Staudt]
If 2 triangles { A , A , A } and { B , B , B } are inscribed in a conic γ and are perspectivewith center C and axis c , and P , Q are their Brianchon pointa and p and q are theirPascal lines, tehn (cid:104) P, Q, C ; pq (cid:105) and (cid:104) p, q, c ; P Q (cid:105) , moreover, quatern(
P, Q, C, pq × c ) andquatern( p, q, c, P Q × C ). Notation. ( A i,j,k ) := det ( A i , A j , A k ) . Theorem.
If 6 points A k , k = 0 to 5, are on a conic then( A k +2 ,k +3 ,k +1 )( A k +3 ,k +4 ,k )( A k +4 ,k +5 ,k +1 )( A k +5 ,k,k +2 ) − ( A k +2 ,k +3 ,k )( A k +3 ,k +4 ,k +1 )( A k +4 ,k +5 ,k +1 )( A k +5 ,k,k +2 ) .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. A k +3 ,k +4 ,k +1 )( A k +4 ,k +5 ,k +2 )( A k +5 ,k,k +2 )( A k,k +1 ,k +3 ) − ( A k +3 ,k +4 ,k +1 )( A k +4 ,k +5 ,k +2 )( A k +5 ,k,k +3 )( A k,k +1 ,k +2 ) = 0 . The addition for the subscript is done modulo 6.
Proof: The Theorem will be proven for k = 0 . Let a k := A k × A k +1 . The Pascal pointsare P k := a k × a k +3 , we have ( P , , ) := det ( P , P , P ) = ( P ∗ P ) · P = 0 , but P = ( A ∗ A ) ∗ ( A ∗ A ) = det ( A , A , A ) A − det ( A , A , A ) A , = ( A , , ) A − ( A , , ) A , similarly, P = ( A , , ) A − ( A , , ) A ,P = ( A , , ) A − ( A , , ) A = ( A , , ) A − ( A , , ) A , therefore det ( P , P , P ) = ( A , , )( A , , )( A , , )( A , , ) − ( A , , )( A , , )( A , , )( A , , )+( A , , )( A , , )( A , , )( A , , ) − ( A , , )( A , , )( A , , )( A , , ) = 0 . (cid:3) Construction. point Pascal( A , A , A , A , A , A (cid:48) ; [ P , P , P , ] A ) is used as an abbreviation for the Pascal construction P := ( A × A ) × ( A × A ) ,P := ( A × A ) × ( A × A (cid:48) ) ,P := ( A × A ) × ( P × P ) ,A := ( A × A (cid:48) ) × ( P × A ) . It gives the point A on the conic through A to A on the line A × A (cid:48) . linePascal( a , a , a , a , a , a (cid:48) ; [ p , p , p , ] a ) is used for the dual construction. Theorem.
When p = 2, the points and lines of a conic2 configuration are the points and lines ofa conic. Proof: The Pascal points are the diagonal points of the complete quadrangle configurationwhich are collinear because of Theorem 2.1.13.
Theorem.
Let p = 3 , in a quadrangle quadrilateral configuration, Q i , P, q i , p (2.1.6), there is a conicwhose tangent at P is p and at Q i is p i . In other words the elements of a conic3 configuration (2.1.6 are the points and lines of aconic.
Proof: From 2.1.6, Q i is on q i , the Pascal-Brianchon theorem givesPascal ( P, p, Q i +1 , q i +1 , Q i − , q i − ; (cid:104) R , P i − , P i +1 , p (cid:105) ) . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem.
The conical points of Definition are on a conic and the conical lines are on a conic.
Proof: Pascal ( AF , F A , AF , F A , AF , F A ; (cid:104) R , R , R , p (cid:105) . Theorem.
The following points of the extended Pappus configuration are on a conic, 1 point on eachof the lines d, d, say M and M and the intersection with the lines joining the other pointssay a and a with the lines joining M or M with the other points on d or d. This gives the 18 conics M i , N i +1 , N i +1 , M i , N i − , N i − , M i , P i − , P i − , M i , P i +1 , P i +1 , M i +1 , N i +1 , L i − , M i − , N i − , L i − , M i +1 , Q i − , P i − , M i − , Q i +1 , P i − , M i − , Q i +1 , P i +1 , M i +1 , Q i − , P i − , M i − , L i +1 , N i +1 , M i +1 , L i − , N i − . Proof: This follows from Pascal’s Theorem applied to the points in the given order, orderwhich was chosen in such a way that the Pascal line was always m , containing P , P , Q and D. Exchanging N i +1 and N i − for 0. gives an other Pascal line m . The 9 Pappus linesare therefore the Pascal lines of the 18 conics.
Theorem.
The conics 0. and 1. have the same tangent at their common point.
Proof: The coefficients of conic 0. are, for i = 0 , a = m m m ,a = m m , a = m m , b = m m ( m + m ) , b = m ( m + m m ) ,b = m ( m + m m ) , This follows easily because M , M are on a , giving a , a and b , N , N are on a , thisgives a and b , N , N are on a , giving b . The coefficients of conic 1, for i = 0 are thesame except for a = m ( m − m m + m ) . The algebra is simplified by noting that theequation for P and P , gives after subtraction b from a and b , and for P and P , givesafter subtraction b from a and b . The Theorem follows at once. The tangent at M is [ m ( m + m − m m , m m , m m ] and at M is [ m + m − m , m , m ] . Exercise.
Study the configuration of all 18 conics associated to the extended Pappus configuration. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY.
Introduction.
The set of Theorems given here originates with the work of Steiner (1828, 1832 - Werke I,p.451). Proofs have been given using Pascal’s Theorem and Desargues Theorem in the planeor starting with properties of the configuration5 * 3 & 5 * 3 in three (Cremona, 1877) or four (Richmond, 1894, 1899, 1900, 1903)dimensions, subjected to a linear condition. An alternate approach starts with the work ofSylvester 1844 (Papers, I, p.92), 1862 (II, p265.) For a good summary, see Salmon, 1879,p. 379-383, Baker, II, (2d Ed. 1930), p. 219-236 and Friedrich Levi, 1929, p.192-199..The cyclic permutation notation allows the results to be given in a simple algebraic way andsuggests the related synthetic construction.
Definition.
Given 6 points A j , j = 0 to 5, on a conic, a conical hexagon abbreviated here by hexagon isa permutation h of 0 to 5. Given h this defines a specific Pascal line p ( h ) := P ascal ( A h (0) , A h (1) , A h (2) , A h (3) , A h (4) , A h (5) ) , A map will denote here a permutation which acts on h . Example.
Let h = [013524] = ( 012345013524 ) = (2354) . The ordered set of point associated to h is A A A A A A . The map σ = (135) associates to this set, the set (2354)(135) = (15)(234) =[053421] = h (cid:48) or A A A A A A , for instance, h (cid:48) (2) = hσ (2) = h (2) = 3 , h (cid:48) (3) = hσ (3) = h (5) = 4 . The multiplication of permutations is done from right to left.
Definition.
0. The
Steiner map is σ = (135),1. the Steiner conjugate map is γ = (35).2. the Kirkman map is κ = (021)(345),3. the Cayley-Salmon map is χ = (14),4. the Salmon map is λ = (2354).5. the line-Steiner maps are σ = (23) and σ = (45). Theorem.
Given r = (012345) and s = (05)(14)23) ,h = ( . . . ij . . . ) , r − hr = r − ( r . . . i − , j − . . . ) and s − hs = ( . . . s ( i ) , s ( j ) . . . ) have the samePascal line. The permutations r − k hr k and s − hs are called Pascal equivalent . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem.
0. (024), (042), (153) are Pascal equivalent to the Steiner map (135).1. (02), (04), (13), (15), (24) are Pascal equivalent to the Steiner conjugate map (35).2. (012)(354),(015)(243), (045)(132), (051)(234), (054)(123) are Pascal equivalent tothe Kirkman map (021)(345).3. (03), (25) are Pascal equivalent to the Cayley-Salmon map (14).4. (0132), (0215), (0451), (0534), (1243) are Pascal equivalent to the Salmon map (2354).5. (024), (042), (153) are Pascal equivalent to the line-Steiner maps (23) and (45).
Theorem. [Steiner (Pascal)] (cid:104) p ( h ) , p ( hσ ) , p ( hσ ); S ( h ) (cid:105) , S ( h ) is called the Steiner point of h .1. S ( hγ ) , called the Steiner conjugate point of h , is on the polar of S ( h ) with respect tothe conic. there are 10 pairs of conjugate Steiner points. See 2.1.9.Proof: Let h = [012345] = (). I will use here the abbreviations ij for the line A i × A j , ijkl for the Pascal point ( A i × A j × ( A k × A l ) .p ( h ) = P × P (cid:48) , with P = 0134 , P (cid:48) = 0523 ,p ( hκ ) = P × P (cid:48) , with P = 0125 , P (cid:48) = 1423 ,p ( hκ ) = P × P (cid:48) , with P = 2534 , P (cid:48) = 0514 , Let Q := ( P × P ) × ( P (cid:48) × P (cid:48) ) = 2514 ,Q := ( P × P ) × ( P (cid:48) × P (cid:48) ) = 3450 ,Q := ( P × P ) × ( P (cid:48) × P (cid:48) ) = 0123 , Pascal( A A A A A A ; (cid:104) Q , Q , Q (cid:105) ) , thereforeDesargues − ( { P , P , P } , { P (cid:48) , P (cid:48) , P (cid:48) } ; (cid:104) Q , Q , Q }(cid:105) , S ( h )) ,orDesargues − ( { , , } , { p , p , p }{ p × p , , } , { p , p , p } ; (cid:104) S ( e ) , S ( σ ) , S ( σ }(cid:105) , s ( e )) , Theorem. [Kirkman 1849, 1850] (cid:104) p ( h ) , p ( hκ ) , p ( hκ ); K ( h ) (cid:105) , K ( h ) is called the Kirkman point of h .1. there are 60 Kirkman points which are 3 by 3 on the 60 Pascal lines, giving a configu-ration of type
60 * 3 &
60 * 3 . Levi, p. 194 .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY.
Proof: Let h = [012345] = (). The proof, for i = 0 is as follows. p ( h ) = P × P (cid:48) , with P = 0134 , P (cid:48) = 0523 ,p ( hκ ) = P × P (cid:48) , with P = 0134 , P (cid:48) = 1245 ,p ( hκ ) = P × P (cid:48) , with P = 2534 , P (cid:48) = 0312 , Let Q := ( P × P ) × ( P (cid:48) × P (cid:48) ) = 0325 ,Q := ( P × P ) × ( P (cid:48) × P (cid:48) ) = 0145 ,Q := ( P × P ) × ( P (cid:48) × P (cid:48) ) = 1234 , Pascal( A A A A A A ; (cid:104) Q , Q , Q (cid:105) ) , thereforeDesargues − ( { P , P , P } , { P (cid:48) , P (cid:48) , P (cid:48) } ; (cid:104) Q , Q , Q }(cid:105) , S ( h )) ,orPascal(014235) = ⇒ (cid:104) , , p (cid:105) , Desargues − ( p , { , , } , { , , } , { , , } , { , p , } ; (cid:104) p , p , p }(cid:105) , K ( e )) , Theorem [Salmon]
If 2 triangles have their vertices on a conic, their sides are tangent to a conic .Proof:Desargues − ( { , , } , { , p , } , { , , } , { , p , } ; (cid:104) , K ( e ) , P (cid:105) ) . Exercise.
Prove (cid:104) p , p , p (cid:105) . Theorem. [Steiner] (cid:104) S ( h ) , S ( hσ ) , S ( hσ ); s ( h ) (cid:105) , s ( h ) is called the Steiner line of h ,1. S ( hσ σ ) ι s ( h ) , there are 15 Steiner lines s ( h ) . The proofs follows from 2.2.11.0 and from the fact that the Brianchon lines of the conicinscribed in the 2 triangles are Pascal lines of the original conic.
Theorem. [Cayley and Salmon] (cid:104) K ( hχ ) , K ( hχ ) , K ( hχ ); cs ( h ) (cid:105) , cs ( h ) is called the Cayley-Salmon line of h ,1. S ( h ) ι cs ( h ) , there are 20 Cayley-Salmon lines. The 60 Kirkman points, the 20 Steiner points, the 60 Pascal lines and the 20 Cayley-Salmon lines form a 80 ∗ ∗ (See Levi, p. 199). Salmon, p. 381 CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem.
18 Pascal points and 12 Pascal lines are used in the preceding Theorem and these are verticesand sides of 3 complete quadrilaterals.
Theorem. [Salmon] (cid:104) cs ( h ) , cs ( hλ ) , cs ( hλ ); Sa ( h ) (cid:105) , Sa ( h ) is called the Salmon point of h ,1. cs ( hλ ) ι Sa ( h ) , there are 15 Salmon points Sa ( h ) . Theorem.
In the preceding Theorem:0.
Each of the 24 Pascal lines occurs exactly twice. The Pascal points of h occur 4 times, the other 30 Pascal points occur twice. The 3 Pascal points of h, the 8 points S ( hλ i ) , K i ( hχ ) , i = 0 , , , ∗ ∗ ∗ , (11) . Example.
In all cases h = e = [012345] = () .
0. The Theorem of Steiner. S ( e ) ι S ( σ ) ι S ( σ ) ι S ( σ σ ) ι () = [012345] (23) = [013245] (45) = [012354] (23)(45) = [013254](135) = [032541] (1235) = [023541] (1345) = [032451] (12345) = [023451](153) = [052143] (1523) = [053142] (1453) = [042153] (14523) = [043152] (cid:104)(cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , (cid:105)(cid:105) . (Fig. 200b)1. The Theorem of Cayley-Salmon. S ( e ) ι K ( χ ) ι K ( σχ ) ι K ( σ χ ) ι () = [012345] (14) = [042315] (1435) = [042531] (1453) = [042153](135) = [032541] (024531) = [204153] (0241) = [204315] (024351) = [204531](153) = [052143] (043512) = [420531] (045312) = [420153] (0412) = [420315] (cid:104)(cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105)(cid:105) . (Fig. 200b’)2. The Theorem of Salmon.Add to Example 1, (Fig. 200b” and b4) .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. S ( λ ) ι K ( λ ) ι K ( λσχ ) ι K ( λσ χ ) ι (2354) = [013524] (12354) = [023514] (12345) = [023451] (123) = [023145](15)(234) = [053421] (031) = [302145] (03541) = [302514] (03451) = [302451](1423) = [043125] (02)(1345) = [230451] (02)(13) = [230145] (02)(1354) = [230514] (cid:104)(cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105)(cid:105) . ( F ig. b S ( λ ) ι K ( λ ) ι K ( λσχ ) ι K ( λσ χ ) ι (25)(34) = [015432] (134)(25) = [035412] (1325) = [035241] (13)(254) = [035124](14325) = [045231] (054231) = [503124] (052341) = [503412] (051)(23) = [503241](12543) = [025134] (0322)(15) = [350241] (031542) = [350124] (034152) = [350412] (cid:104)(cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105)(cid:105) . ( F ig. b S ( λ ) ι K ( λ ) ι K ( λσχ ) ι K ( λσ χ ) ι (2453) = [04253] (15324) = [054213] (15)(24) = [054321] (15243) = [054132](1245) = [024351] (0431)(25) = [405132] (041)(253) = [405213] (04251) = [405321](13)(245) = [034152] (05142) = [540321] (052)(143) = [540132] (0532)(14) = [540213] (cid:104)(cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105) , (cid:104) p , p , p (cid:105)(cid:105) . (Fig. 200b3) Exercise.
0. Give the geometric interpretation of the Theorems in this section.1. Determine the pseudo configuration associated to the Theorem of Salmon.
Introduction.
The drawing of curves is facilitated by the notion of B´ezier curves. These originate withthe work of de Casteljau at Citro¨en in 1959 and were popularized and generalized by B´ezier.To describe easily complicated curves in 2, 3, . . . dimensions, we start with a B´ezier poly-gon 2.2.13 to construct a parametric representation of points on the curve iteratively. Theassociated theory is briefly given here. The curve can be expressed in terms of the B´ezierpolygon by means of Bernstein polynomials (2.2.13), the derivatives and differences of thecurve can be similarly expressed and related to each other. The example for a curve whose i -th coordinates can be approximated by cubic polynomials is given in 2.2.13. Theorem.
Let P := ( w (1 − I ) , w I (1 − I ) , w I ) , w w w (cid:54) = 0 then P is the parametric equation of a conic, which passes through the points P (0) =(1 , , , P (1) = (0 , , , P ( ∞ ) = ( w , − w , w ) , the tangent t at P is[ w w I , − w w I (1 − I ) , w w (1 − I ) ] , CHAPTER 2. FINITE PROJECTIVE GEOMETRY in particular, the tangent at P (0) is [0 , , , at P (1) is [1 , , , (which meet at U =(0 , , P ( ∞ ) is[ w w , w w , w w ] , t meets t (0) at T = [ w (1 − I ) , w I, , in particular, T ( ∞ ) = [ − w , w , , the anharmonic ratio anhr ( U, P (0) , T ( ∞ ) , T ) = w − w w I. The proof starts with the observation that the coordinates P , P , P , satisfy the equation w P P = w w P , which is indeed the equation of a conic with the prescribed properties. The correspondingpolarity matrix is w − w w w . Notice that T (0) = (1 , , and is not undefined.The tangent can either be obtained from the polarity or using P × DP , where its direction DP = ( − w (1 − I ) , w (1 − I ) , w I ) .The last statement of the Theorem is associated with the four tangents Theorem of J.Steiner. It can be used as a method to draw conics. In the excellent language Postcript(see Reference Manual), a general method is given to draw curves based on rhe work of deCateljau and B´ezier as well as a method to draw ellipses using the Euclidean concepts ofrotation and scaling differently in the direction of its axis. This method does not allow todraw hyperbolas or parabolas and ignores the fact that a conic is a projective concept. Thefollowing gives a method which allows to draw conics using 3 points
A, B, C, and the tangents t A , t B at two of the 2 points.It is then generalized to other curves. Algorithm.
If the barycentric coordinates are chosen in such a way that A = (1 , , , B = (0 , , ,t A × t B = (0 , , and C = (1 , , , then the points on the conic are given by P of thepreceding Theorem, with, for instance, w = 2 , w = − , w = 2 . In the case of a finite field,we compute P for each element of the field or for an appropriate subset of it. In the caseof the field of reals, we can compute P for tan ( πt ) , t = 0 to 1, avoiding / , a section ofthe conic can be obtained by appropriately limiting the set { t } , joining the successive pointsby segments will automatically give the asymptotes for an hyperbola, which is appropriatebecause their directions are indeed points in the Euclidean plane, as we prefer to consider it(in its extended form). An other approach is to limit the domain of P to [0 , to obtain onesection of the conic and to replace w by − w , which is equivalent to compose P with I I − , to obtain the complement, see Farin, p.185.For some of the Theorems, see Farin.In what follows, the superscript of B, P and P are indices and not exponents. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Definition.
The
Bernstein polynomials are B ni := (cid:18) ni (cid:19) I i (1 − I ) n − i , ≤ i ≤ n .By convention B n − = B nn +1 := 0 . In particular, B = (1 − I ) , B = 2 I (1 − I ) , B = I . Theorem. B = 1 , B ni = (1 − I ) B n − i + IB n − i − , DB ni = n ( B n − i − − B n − i ) . (cid:80) nj =0 B nj = 1 . Definition. A weighted point P is a set of 3 non homogeneous coordinates which are not all 0. Wecan add weighted points and multiply by scalars, but two weighted points which differ by amultiplicative constant are not equivalent. I will use the notation P for the equivalent point. Definition. [de Casteljau]
Given n + 1 weighted points P , P , . . . , P n , called the B´ezier polygon , define P i := P i , ≤ i ≤ n, P ji := (1 − I ) P j − i + I P j − i +1 , ≤ j ≤ n, ≤ i ≤ n − j. P n := P n , The curve P n is called the de Casteljau curve of order n .The same curve is also called the B´ezier curve . Theorem. P n (0) = P , P n (1) = P n .1. P ji = (cid:80) jk =0 P i + k B jk , ≤ j ≤ n, ≤ i ≤ n − j, in particular, P n = (cid:80) nk =0 P k B nk , D P n = n (cid:80) n − k =0 ( P k +1 − P k ) B n − k . Definition. ∆ Q k = Q k +1 − Q k , ∆ r +1 Q k = ∆ r Q k +1 − ∆ r Q k , CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem. ∆ Q = (cid:80) rk =0 ( − r − j (cid:18) rj (cid:19) Q i + k . Theorem. D r P n = n !( n − r )! (cid:80) n − rk =0 ∆ r P k B n − rk .D r P n = n !( n − r )! ∆ r P n − r . In particular , D P n = n ( P n − − P n − ) . Curves with cubic parametrization.
For n = 3 , P = P (1 − I ) + 3 P (1 − I ) I + 3 P (1 − I ) I + P I .D P (0) = P − P ,D P (1) = P − P . In other words the direction of the tangents at the end points is that of the line joining theend points to the nearest point.If the cubic associated with the i -th coordinate of the curve P is f = c + c I + c I + c I ,then the i -th coordinate a j of the B´ezier polygon P j is given by a = c , a = c + b , a = a + ( c + c ) , a = c + c + c + c . Indeed, a (1 − I ) + 3 a (1 − I ) I + 3 a (1 − I ) I + a I = f. These last formulas allows for the determination of the weighted points P i of the cubic(approximation) given the 3 non homogeneous coordinates of the parametrized curve. If thecubic associated with the i -th coordinate reduces to a linear function then P i = P ( i/ ,i = 0 , , , .It is often convenient to choose − and 1 for the end points instead of 0 and 1 bymeans of a change of variable. If g = d + d I + d I + d I is the new polynomial, f = c + c I + c I + c I = g ◦ φ, with φ = 2 I − . In this case, we obtain the symmetric formulas, a = d − d + d − d , a = d − d − d + d , a = d + d − d − d , a = d + d + d + d , Example.
For the curve ( I − I , − I , , for the first coordinate, d = d = 0 , d = 1 , , d = − ,therefore a = 2 , a = − , a = , a = − . for the second coordinate, d = d = 0 ,d = 1 , , d = − , therefore a = 0 , a = , a = , a = 0 . Therefore the B´ezier polygon is P = (2 , , , P = ( − , , , P = ( , , , P = ( − , , . This gives the Cartesian coordinates of the following points on the curve associated with i/ , i = 0 to 20:2.000,0.00; 1.287,0.19; 0.736,0.36; 0.329,0.51; 0.048,0.64; -0.125,0.75, -0.208,0.84, -0.219,0.91,-0.176,0.96, -0.097,0.99; 0.000,1.00; 0.097,0.99; 0.176,0.96; 0.219,0.91; 0.208,0.84; 0.125,0.75;-0.048,0.64; -0.329,0.51; -0.736,0.36; -1.287,0.19; -2.000,0.00.The complement of the curve using i/ i/ − is .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Problem.
Given a curve in the plane, what are the condition for a representation of the 3 non-homogeneous coordinates by polynomials of degree n . For conics, we have seen that n = 2 . Definition.
Joining 2 distinct points of a conic, is to determine the line through the 2 points. Joining apoint of a conic to itself is to determine the tangent to the conic at that point.
Example.
For p = 3 , The conic X + 2 Y Z = 0 has the points (0), (1), (13), (17), (25) and (29).The tangents at ( X , Y , Z ) is [ X , Z , Y ] . The tangent at (0) is [1] and the tangent at (1) is [0].These points joined to (0) give the lines [6], [26], [16], [21], [11]. These points joined to (1)give [0], [8], [10], [7], [9].These lines determine on the ideal line [12], the projectivity which associates to(26), (5), (14), (18), (22), (26), the points (5), (26), (18), (22), (10), (14).This is precisely the projectivity φ of 2.2.6. Theorem.
Let N be a symmetric matrix associated to a conic. P is on the conic if P · N P = 0 . If P is on the conic and C is not, the other point on the conic, if any, is P + yC , with y = − C · NPC · NC . Proof: ( P + yC ) · N ( P + yC ) = 0 , or P · N P + yC · N P + yP · N C + y C · N C = 0 , but P · N P = 0 and C · N C (cid:54) = 0 and N is symmetric, therefore C · N P = P · N C, hence y = − C · NPC · NC . Theorem.
Let l be a line and A, B be 2 points on the line but not on the conic associated with thesymmetric matrix N ,let a := A · N A, b := B · N B, c := A · N B = B · N A, let C be an arbitrary point on the line, C = A + kB. Let P and P be 2 distinct points on the conic, let a = ( P · A ∗ P ) , b := ( P · B ∗ P ) , CHAPTER 2. FINITE PROJECTIVE GEOMETRY c := ( A · B ∗ P ) , c := ( A · B ∗ P ) , d := A · N P , d := B · N P . If C × P meets the conicat Q and P × Q meets l at D, then D = (( aa ) + 2( ca − d c ) k + ( ba + 2 d c ) k ) B − (( ab − d c ) + 2( cb − d c ) k + bb k A .1. The correspondance berween C and D is a projectivity. Proof: Q = ( C · N C ) P − C · N P ) C,D = ( A ∗ B ) ∗ ( P ∗ Q ) = ( A · P ∗ Q ) B − ( B · P ∗ Q ) A = ( Q · A ∗ P ) B − ( Q · B ∗ P ) A = (( C · N C )( P · A ∗ P ) − C · N P )( C · A ∗ P )) B − (( C · N C )( P · B ∗ P ) − C · N P )( C · B ∗ P )) A, but C · N C = a + 2 kc + k b therefore D = (( a + 2 ck + bk a − d + kd )( − c k )) B − (( a + 2 kc + bk b − d + d k )( c )) A Theorem.
If the line l is [1 , ,
1] then the conic X + Y + kZ = 0 determines on l the involution ηη (1 , Y, − − Y ) = (1 , f ( Y ) , − − f ( Y )) , with f ( Y ) = − ( k )+ kYk +(1+ k ) Y . Proof. The point (1 , Y, − − Y ) on l has the polar [1 , Y, − k (1 + Y )] , which meets l at ( Y + k (1 + Y ) , − − k (1 + Y ) , − Y ) = (1 , f ( Y ) , − − f ( Y )) . Notation.
In this section, the cubic is denoted by γ , ( I, i ) will denote an inflection point and thecorresponding tangent, ( A, a ) a point on the cubic and its tangent, Theorem.
Given I, there exists 3 ( A i , a i ) such that a i · I = 0 and A i are collinear. Theorem.
Let B j , j = 0 to 5 be on γ and a conic θ , if C j is the third point on B j × B j +3 , then C j arecollinear. Corollary.
Given ( A i , a i ) , i = 0 to 2, let B i be the other point on a i then B i are collinear. .2. ALGEBRAIC MODEL OF FINITE PROJECTIVE GEOMETRY. Corollary. If B k , k = 0 to 3 are on γ . Let a conic θ l meet γ also at C l, and C l, , C l, × C l, passesthrough a fixed point D of γ . Theorem.
The third point on I × I is an inflection point. Theorem.
Given ( A l , a l ) , l = 0 ,
1, and B l is the other point on a l , ( A × A ) × ( B × B ) is on thecubic. Theorem.
The anharmonic ratio of the 4 tangents through A distinct from a is constant. Introduction.
Many models can be derived from the model given in section 1. This is most easily ac-complished by starting with a correspondence between points in the plane and adjusting forspecial cases. One such correspondence is ( x , x , x ) to ( x , x , x ) , and will be studied insome detail. It assumes some given triangle { A , A , A } , whose vertices have coordinates (1 , , , (0 , , , (0 , , . Definition.
In inversive geometry, the “points” are the points ( x , x , x ) , with x x x (cid:54) = 0 together withthe lines [0 , x , x ] , [ x , , x ] , [ x , x , , the “lines” are the point conics a x x + a x x + a x x = 0 . degenerate or not. A “point” is on a “line” , which is a non degenerate point conic, iff itbelongs to it or is tangent to it. If the point conic degenerates in 2 lines, one which is a sideof the triangle and the other passes through the opposite vertex, then the “points” who belongto it are the two lines and the points on the line through the opposite vertex but not on thesides of triangle. If the point conic degenerates in 2 lines, which are 2 sides of the triangle,the “points” which belong to it are the lines through the common vertex. Example.
The “line” x x + 2 x x + 3 x x = 0 belongs to the “points” ( − x x , (2 x +3 x ) x , (2 x +3 x ) x ) , (2 x +3 x ) x x (cid:54) = 0 and to the “points” [0 , , , [3 , , , [2 , , tangent respectively at A , A and A . G24.TEX [MPAP], September 9, 2019 CHAPTER 2. FINITE PROJECTIVE GEOMETRY
The “line” x x + 3 x x = 0 belongs to the “points” ( x , , − , x (cid:54) = 0 and to the “points”[1,0,0] and [0,3,2].The “line” x x = 0 belongs to the “points” [0 , x , x ] . Theorem.
The model satisfies the axioms of projective geometry.
This is most easily seen if we associate to the point P = ( x , x , x ) , x x x (cid:54) = 0 the“point” P (cid:48) = ( x , x , x ) or ( x x , x x , x x ) , to the point Q = (0 , x , x ) , the “point” Q (cid:48) = [0 , x , − x ] , to the point Q = ( x , , x ) , the “point” Q (cid:48) = [ x , , − x ] , to the point Q = ( x , x , , the “point” Q (cid:48) = [ x , x , , and to the line l = [ a , a , a ] , the line l (cid:48) ,a x x + a x x + a x x = 0 . Indeed if P · l = 0 , P (cid:48) is on l (cid:48) and if Q · l = 0 , a x + a x = 0 , while the tangent to l (cid:48) at A is [0 , a , a ] = [0 , x , − x ] . Theorem.
The “lines” are the conics through the vertices A , A , A . Theorem.
The “conics” are the quartics0. b x x + b x x + b x x +( c x + c x + c x ) x x x = 0 . The quartic has double points (or nodes) at the vertices A , A , A . The branches through A are real if and only if c > b b , the branches through A are real if and only if c > b b , the branches through A are real if and only if c > b b . Vice versa if a quartic as double points at A , A and A it is of the form 0. Theorem.
If the quartic has double points with real branches at A , A and A , the tangents P (cid:48) P (cid:48) at A ,P (cid:48) P (cid:48) at A and P (cid:48) P (cid:48) at A are such that if K (cid:48) is the tangent to the conic ( A , ( A , P (cid:48) ) , ( A , P (cid:48) )) , if K (cid:48) is the tangent to the conic ( A , ( A , P (cid:48) ) , ( A , P (cid:48) )) , and if K (cid:48) is the tangent to the conic( A , ( A , P (cid:48) ) , ( A , P (cid:48) )) , then there is a conic through A , A , A with tangents K (cid:48) , K (cid:48) , K (cid:48) . This is a direct consequence of the Theorem of Pascal associated to the model.
Theorem.
If a quartic has double points with real branches at A , A and A , then the 6 tangents atthese points belong to the same line conic. Proof: Let the tangents at A , A , A be [0 , , z ] , [0 , , z (cid:48) ] , [ x, , , [ x (cid:48) , , , [1 , y, , [1 , y (cid:48) , , the tangents [0 , , z ] at A satisfy b z + c z + b , therefore zz (cid:48) = b b , similarly, yy (cid:48) = b b and xx (cid:48) = b b . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. On the other hand, applying Brianchon’s theorem to these tangents gives the Brianchon lines [0 , x (cid:48) y, − , [ − , , y (cid:48) z ] , [ z (cid:48) x, − , and these belong to the same point if x (cid:48) yy (cid:48) zz (cid:48) x = 1 . This conjecture was most strongly confirmed by a computer program and proven within anhour.
Theorem. If b = 0 , then the quartic degenerates in the side A × A and a cubic with double point at A passing through A and A . The conic c c yz + b c zx + b c xy = 0 plays, in the invertible geometry, the role of the“line” tangent at the “point” [1,0,0]. Theorem. [Jones]
Let n be even. If an n -gon is inscribed in a conic and n -1 sides meet a line at fixed points,then the n -th side also meets the line at a fixed point and dually. Theorem. [Jones]
The preceding Theorem, when n = 4 is equivalent to Pascal’s Theorem. Completely independently, one of my first student at the “Universit´e Laval”, Quebec City,made the important discovery that the regular polyhedra can be used as models for finitegeometries associated with 2, 3 and 5. Then, he introduced the nomenclature of selector(s´electeur) for the notion of cyclic difference sets, introduced by J. Singer, in 1938, to la-bel points and hyperplanes in N dimensional projective geometry of order p k (See Baumert,1971) and to construct an appropriate numbering of the points and lines on the polyhedra.Except for the fundamental contribution of Singer, the introduction of selector polarity (pre-pared by the use of f ( a + b ) instead of f ( a − b ) in the definition of incidence), the introductionof auto-polars and those on the conics for the dodecahedron, all the results in this section aredue to Fernand Lemay.Clearly we have only to study the tetrahedron, the cube and the dodecahedron, because theoctahedron is dual to the cube and the icosahedron is dual to the dodecahedron. G25.TEX [MPAP], September 9, 2019 CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Introduction.
The important concept of the cyclic difference sets allows for an arithmetization of projectivegeometry which is as close to the synthetic point of view as is possible. With it, it is notonly trivial to determine all the points on a line, and lines incident to a point, but also thelines through 2 points and points on 2 lines. This concept makes duality explicit through thecorrelation, which is a polarity when p ≥ . The definitions of selector function and selectorcorrelation is implicit in Lemay’s work.
Definition. A difference set associated to q = p k is a set of q + 1 integers { s , s , . . . , s q } such that the q + q diferences s i − s j , i (cid:54) = j modulo n := q + q + 1 are distinct and different from 0. Whenapplied to Geometry, I will prefer the terminology of Lemay and use the synonym selector .The elements of the selector are called selector numbers . Theorem. [Singer]
For any q = p k there exists difference sets. Theorem. If { s i } , i = 0 to q, is a difference set and k is relatively prime to n, then { s (cid:48) i = a + ks i +1 } , is also a difference set.The indices are computed modulo q + 1 and the selector numbers, modulo n. Using 0, we can always find a selector for which 0 and 1 are selector numbers.
Example. [Singer]
The following are difference sets associated with q = p k : For p = 2 : {
0, 1, 3 } modulo 7.For p = 3 : {
0, 1, 3, 9 } modulo 13.For q = 2 : {
0, 1, 4, 14, 16 } modulo 21.For p = 5 : {
0, 1, 3, 8, 12, 18 } modulo 31.For p = 7 : {
0, 1, 3, 13, 32, 36, 43, 52 } modulo 57.For q = 2 : {
0, 1, 3, 7, 15, 31, 36, 54, 63 } modulo 73.For q = 3 : {
0, 1, 3, 9, 27, 49, 56, 61, 77, 81 } modulo 91.For q = 11 : {
0, 1, 3, 12, 20, 34, 38, 81, 88, 94, 104, 109 } modulo 133. Definition. If a = 1 and k = − , the selector s (cid:48) i := 1 − s i is called the complementary selector or co-selector of s i . The selectors obtained using k = 2 , , − , − are called respectively bi-selector , semi-selector , co-bi-selector , co-semi-selector . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. Example.
For q = 4 , other selectors are { , , , , } , { , , , , } and { , , , , } . For p = 7 , ifthe selector is { , , , , , , , } , thenthe co-selector is { , , , , , , , } ,the bi-selector is { , , , , , , , } ,the co-bi-selector is { , , , , , , , } ,the semi-selector is { , , , , , , , } ,the co-semi-selector is { , , , , , , , } . Program.
All selectors derived by multiplication from one of them are given in [113]MODP30.
Definition.
The selector function f associated to the selector { s i } is the function from Z n to Z n f (0) = 0 , f ( s j − s i ) = s i , i (cid:54) = j. Example.
For p = 2 , the selector function associated with { , , } mod is i f ( i ) 0 0 1 0 3 3 1 For p = 3 , the selector function associated with { , , , } mod is i f ( i ) 0 0 1 0 − − − For p = 5 , the selector function associated with { , , , , , } mod is i f ( i ) 0 0 1 0 8 3 12 1 0 3 8 1 0 18 18 3 i
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 f ( i ) 18 1 0 12 12 18 12 8 8 18 8 12 3 3 1 Theorem. f ( j − i ) − i = f ( i − j ) − j (mod n ) . Points, lines and incidence in the 2 dimensional geometry associated with q = p k and n := q + q + 1 are defined as follows. Definition.
The points are elements of the set {
0, 1, . . . , n − } ,The lines are elements of the set {
0, 1, . . . , n − } .A point a is incident to a line b iff f ( a + b ) = 0 . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Notation.
The points are denoted by a lower case letter or by an integer in Z n . The lines are denotedby a lower case letter or by an integer in Z n followed by an asterix. The line incident to thepoints a and b is denoted a × b, the point incident to the lines a ∗ and b ∗ is denoted a ∗ × b ∗ . Theorem.
Given a selector { s j } associated with q = p k and the corresponding selector function f : The q + 1 points incident to or on the line i ∗ are s j − i mod n. The q + 1 lines incident to or on the point i are ( s j − i mod n ) ∗ . a (cid:54) = b = ⇒ a × b = ( f ( b − a ) − a ) ∗ . a (cid:54) = b = ⇒ a ∗ × b ∗ = f ( b − a ) − a. a on b ∗ iff b on a ∗ . The statements in the preceding Theorem reflect the duality in projective geometry.
Definition.
The selector polarity is the correlation which associates to the point i the line i ∗ . The points x which are on x ∗ are called auto-polars .The name “polarity” is appropriate because of 2.3.1.4.The selector polarity and the auto-polars play an important role in a natural way of labelingthe elements of the Pythagorean solids. Theorem.
The auto-polars are given by a i = s i , modulo n. Indeed we should have for an auto-polar x, x = s i − x. Definition. A primitive polynomial of degree over GF ( q ) , is an irreducible polynomial P of degree 3such that I k (cid:54) = 1 for k = 1 to q − , where I is the identity function and the constant polynomial 1.The multiplication is done modulo P and polynomials which differ by a multiplicative constant (cid:54) = q are equivalent. Theorem. [Singer]
For each value of q = p k a selector can be obtained by choosing a primitive polynomial ofdegree 3 over GF ( q ) . The selector is the set of exponents of I between 0 and q − .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. Example.
For p = 3 , P = I − I + 1 ,I = 1 , I = I, I = I , I = I − , I = I − I, I = I − I + 1 , I = I + I + 1 ,I = I − I − , I = I + 1 , I = I + 1 , I = I + I, I = I + I − , I = I − and wehave I = 1 . Therefore the selector is { , , , } . Introduction.
I have found useful to introduce the adjectives vertex, edge and in later sections, face, todistinguish points and lines which have different representation in the Pythagorean solids.
Definition.
The points in the tetrahedron model consist of0. The 4 vertex-points , which are the 4 vertices (or the opposite planes or the line throughthe center C of the tetrahedron perpendicular to one of the 4 planes).1. The 3 edge-points , which are the pairs of orthogonal edges, (or the mid-points of 3 nonorthogonal edges or the line through these points and the center C ).The lines in the tetrahedron model consist of2. The 6 edge-lines , which are incident to the 2 vertex-points and to the edge-point onthem.3. The tetrahedron-line , which is incident to the 3 edge-points. Theorem.
The model satisfies the axioms of projective geometry for p = 2 . Theorem.
With the selector { } mod , the 3 points, 0, 4 and 5 are auto-polars. It is thereforenatural to associate them to the 3 edge-points. These points are on the line 3 ∗ , it is naturalto associate it to the tetrahedral line. Any of the vertex-points can be chosen as the polar 3 of3 ∗ . We will choose the 3 adjacent edge-lines as 0 ∗ , ∗ and 5 ∗ such that 0 · ∗ = 4 · ∗ = 5 · ∗ = 0 . The other vertex-points are the third point on 0 ∗ , ∗ and 5 ∗ , therefore 2 is on the line and2 ∗ is the line orthogonal to the line associated to 5, similarly for 1 and 1 ∗ , to 0 and 6 and 6 ∗ , to 4.48 CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Figure. (cid:20)(cid:20)(cid:20)(cid:20)(cid:20)(cid:20)(cid:20) (cid:84)(cid:84)(cid:84)(cid:84)(cid:84)(cid:84)(cid:84) (cid:113) A (cid:97) P (cid:113) A (cid:113) P (cid:113) P (cid:97) A (cid:97) P Theorem.
A complete quadrangle configuration consists of the 4 vertex-points A , A , A , P and the6 edge-lines a = A × A , a = A × A , a = A × A , p = P × A , p = P × A ,p = P × A . It has the 3 edge-points P i = p i × a i as its diagonal points, and these are onthe tetrahedron-line p. Exercise.
0. For q = 2 , determine the primitive polynomial giving the selector { , , } .1. Determine the correspondence between the selector notation and the homogeneous co-ordinates for points and lines. Note that these are not the same.2. The correspondence i to i ∗ is a polarity whose fixed points are on a line. Determinethe matrix representation and the equation satisfied by the fixed points.3. Determine the degenerate conic through 0, 1, 2 and 5 with tangent ∗ at , its matrixrepresentation and its equation in homogeneous coordinates.4. Determine all the non degenerate conics. Convention.
In what follows we identify elements of the cube, which are symmetric with respect to its cen-ter C, for instance, the parallel faces. There are therefore 3 independent faces, 4 independentvertices and 6 independent edges. Definition.
The points in the cube model consist of0. The 3 face-points , which are the square faces or their centers or the lines joining C tothese points.1. The 4 vertex-points , which are the vertices or the lines joining C to these vertices. .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA.
2. The 6 edge-points , which are the edges, or the mid-points of the edges or the linesjoining C to these points.The lines in the cube model consist of3. The 3 face-lines , corresponding to a face f, which are incident to the 2 face-points andto the 2 edge-points in the plane through C parallel to f.
4. The 4 vertex-lines , corresponding to a vertex V, which are ncident to the vertex-points V and to the 3 edge-points not adjacent to V.
5. The 6 edge-lines , corresponding to an edge e, which are incident to the face-pointperpendicular to e, to the 2 vertex-points and the edge-point on e. Theorem.
The cube model satisfies the axioms of projective geometry for p = 3 . Theorem.
With the selector 2.3.1 for p = 3, the auto-polars are 0, 7, 8 and 11. If we examine thequadrangle-quadrilateral configuration, we observe that p and q i are the lines which requirea 4-th point, it is easy to verify that, with p = 3 , P is on p and Q i is on q i . Moreover0 ∗ · P = 0 , Q i = 7 , , . Hence r i := P × Q i = 9 , , p i := Q i +1 × Q i − = 5 , , A i := r i × p i = 4 , , a i := A i +1 × A i − = A i ; P i := a i × r i = p i ; q i := P i +1 × P i − = Q i ; R i := a i × q i = r i ; p := R × R = p. Theorem.
Because of 2.3.3, the vertex-points are auto-polars, we can choose them as 0, 7, 8 and 11, theother elements of the cube follow from 2.3.3. The edge-points are R i and P i , the face-pointsare A i . Figure. (cid:0)(cid:0)(cid:0) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) (cid:113) Q (cid:97) R (cid:113) Q (cid:113) P (cid:97) P (cid:113) Q (cid:97) P (cid:98) A (cid:97) R (cid:113) (cid:97) (cid:113) (cid:97) (cid:98) A (cid:97) R (cid:97) P (cid:98) A (cid:113) (cid:97) CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Exercise.
0. For p = 3 , determine the primitive polynomial giving the selector { , , , } .1. Determine the correspondence between the selector notation and the homogeneous co-ordinates for points and lines. Note that these are not the same.2. The correspondence i to i ∗ is a polarity whose fixed points are on a line. Determinethe matrix representation and the equation satisfied by the fixed points.3. Determine the degenerate conic through 0, 1, 2 and 5 with tangent ∗ at , its matrixrepresentation and its the equation in homogeneous coordinates. Hint: use 2.2.11.4. Determine all the conics. Convention.
In what follows we identify elements of the dodecahedron which are symmetric with respectto its center C, for instance, the parallel faces. There are therefore 6 independent faces, independent vertices and independent edges. Definition.
The points in the dodecahedron model consist of0. The 6 face-points , which are the pentagonal faces or their center or the lines joining C to these points.1. The 10 vertex-points , which are the vertices or the lines joining C to these vertices.2. The 15 edge-points , which are the edges, or the mid-points of the edges or the linesjoining C to these points.The lines in the dodecahedron model consist of3. The 6 face-lines , which are incident to the corresponding face-point F and to the 5edge-points in the plane through C perpendicular to CF.
4. The 10 vertex-lines , corresponding to a vertex V , which are incident to the 3 edge-points in the plane through C perpendicular to CV and to the 3 vertex-points whichjoined to V form an edge.5. The 15 edge-lines , corresponding to an edge E, which are incident to the 2 face-points,the 2 vertex-points and the 2 edge-points in the plane through C and E. Theorem.
The dodecahedron model satisfies the axioms of projective geometry for p = 5 . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. Example.
For p = 5 , the selector function associated with the selector { , , , , , } is i f ( i ) 0 0 1 0 8 3 12 1 0 3 8 1 0 − −
13 3 type f e e e f v f v e f v v e v e si
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 f ( i ) −
13 1 0 12 12 −
13 12 8 8 −
13 8 12 3 3 1 type f f e v v v e e v e e e e v e
The auto-polars are { , , , , , } .The “type” is explained in the following Theorem. Theorem.
A natural labeling of the points of the dodecahedron and of the dodecahedral configuration,associated with the selector 2.3.1, for p = 5, can be obtained as follows. If we examine thedodecahedron configuration,
F A i · f a i = AF i · af i = 0 , it is therefore natural to choose F A i and AF i as the auto-polars, but this cannot be donearbitrarily. Let us choose any 3 of them as F A i ,
0, 16 and 17. To obtain P and A i , we canproceed as follows. pq i = F A i × F A i +1 = 18 , , P Q i = pq i ; P R i = pq i × F A i − = 14 , , qp i = P R i × P Q i = 25 , , AF i = QP i × QP i +1 = 6 , , a i = F A i +1 × AF i − = 23 , , A i = a i ; p = P R × P R = 29 . We therefore choose P = 29 and A i = 23 , , . We obtain, according to 2.1.6, 2.1.7 and2.1.7: a i = A i , r i = 20 , , ,P i = 11 , , , q i = 19 , , , R i = r i , p i = P i , p = 29 ,P Q i = pq i = 18 , , , QP i = qp i = 25 , , ,QR i = qr i = 12 , , , P R i = pr i = 14 , , ,AF i = af i = 6 , , , F A i = f a i = 0 , , . Therefore the face-points are
F A i , AF i ; the vertex-points are p i , q i , r i ; the edge-points are A i , P Q i , QP i , QR i , P R i . Figure.
Q221QR1 o o PQ127 FA0 15R2 o 0 P010 11 CHAPTER 2. FINITE PROJECTIVE GEOMETRY o QP2 A1 o .A0 30 26 PR223 o P QP0 R0 o 2o 29 o 25 o 20 .P2 FA2 QP1 QR2 AF1 Q124 17 28 FA1 22 4 7R1 16 Q0o 5 19 .PQ0 A2 PQ218 QR0 8 1 PR1 1812 P1 3. AF2 13 AF0 .9 o 67 PR0 242 14 2311 o 1015 2721
Comment.
We observe that in the dodecahedron,
F A i are adjacent and if AF i are constructed as in2.3.4, that these are not. Moreover, F A i , F A i +1 and AF i +1 are adjacent; AF i , AF i − , F A i − are not. Therefore AF i +1 and AF i − are not adjacent to F A i +1 and AF i and therefore areadjacent to F A i and F A i − . There is therefore a consistent way to define adjacency of conicalpoints, if, given 3 of them named F A i , the 3 others are labelled according to the construction2.3.4. Definition.
If 3 conical points are labelled
F A i and the 3 others, AF i , are labelled according to theconstruction 2.3.4. The triples F A i ; F A i , F A i +1 , AF i +1 ; AF i +1 , AF i − , F A i and AF i +1 ,AF i − , F A i − are adjacent and the other triples are not adjacent .The notion of “adjacent” and “not adjacent” can be interchanged. Introduction.
After introducing distances in n dimensional affine geometry and the associated selector, itoccured to me that we could consider other difference sets for sets associated to p k by choosingpolynomials which are not irreducible. I discuss here briefly the extension to difference setsappropriate to the study of geometries in 2 and higher dimensions. .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. Definition. A difference set associated to q = p k and to a polynomial of degree 3 with one root, is a set of q integers { s , s , . . . , s q − } such that the q − q differences s i − s j , i (cid:54) = j modulo n = q − are distinct and different from 0 modulo q + 1 .A difference set associated to q = p k and to a polynomial of degree 3 with two roots, is a setof q − integers { s , s , . . . , s q − } such that the q − q + 2= ( q − q − differences s i − s j , i (cid:54) = j modulo n = q − q are distinct and different from0 modulo q and modulo q − .When applied to Geometry, I will prefer the terminology of Lemay and use the synonym selector . The elements of the selector are called selector numbers . Theorem.
There exists always a polynomial P of degree 3 with one root or 2 roots such that I is agenerator of the multiplicative group of polynomials, of degree at most 2, with coefficientsin Z p , normalized to have the coefficient of the highest power 1, which are relatively primeto P . The selector numbers are the powers of I modulo P which are polynomials of degreeat most 1. The proof can be adapted easily from that of the irreducible case and is left as an exercise.
Example.
For p = 3 , P = I + I + 1 ,I = 1 , I = I, I = I , I = I + 1 ,I = I + I, I = I − I − , I = I − I + 1 , I = I + 1 , and we have I = 1 . Therefore the selector is 0, 1, 3.
Example.
The following are difference sets associated with q = p k : For p = 3 : I + I + 1 , root 1, selector 0,1,3 (mod 8) .I + I − I − , roots 2,2,1, selector 0,1 (mod 6) . For p = 5 : I − I − , root 2, selector 0,1,3,11,20 (mod 24) .I − I − , roots 3,3,4, selector 0,1,3,14 (mod 20) .For p = 7 : I − I − , root 5, selector 0,1,7,11,29,34,46 (mod 48) .I − I − , roots 2,6,6, selector 0,1,3,11,16,20 (mod 42) . For q = 11 : I − I − , root 6,selector 0,1,3,28,38,46,67,90,101,107,116 (mod 120) .I − I − I − , roots 7,7,9,selector 0,1,9,15,36,38,43,62,94,107 (mod 110 . ) Definition.
The selector function f associated to the selector { s i } is the function from Z n to Z n f ( s j − s i ) = s i , i (cid:54) = j, for all other values f ( l ) = − . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Example.
For p = 5 , the selector function associated with {
0, 1, 3, 11, 20 } is i f ( i ) − − i
12 13 14 15 16 17 18 19 20 21 22 23 f ( i ) − − Theorem. If the defining polynomial has 1 root then the selector has n := p elements, the selectorfunction has p − − p − p + 1 . If the defining polynomial has 2 distinct roots then the selector has n := p − p ( p −
1) elements and is − p − p and p − . Theorem. If f ( i − j ) (cid:54) = − f ( j − i ) − i = f ( i − j ) − j. Points, lines and incidence in the 2 dimensional geometry associated with q = p k and n := q + q + 1 are defined as follows. Definition.
The points are elements of the set {
0, 1, . . . , n-1 } , the lines are elements of the set {
0, 1,. . . , n − } , a point a is incident to a line b iff f ( a + b ) = 0 . Notation.
The points are denoted by a lower case letter or by an integer in Z n . The lines are denotedby a lower case letter or by an integer in Z n followed by an asterix. The line incident to thepoints a and b is denoted a × b, the point incident to the lines a ∗ and b ∗ is denoted a ∗ × b ∗ . We leave as an exercise to state and prove Theorems analogous to those in Section 2.3.1.
Definition.
The dual affine plane , is a Pappian plane in which we prefer the “special“ points which arethose on a line l and a point P not on l and the “special” lines which are those through P and the line l .Th dual affine geometry can be studied by associating with it a polynomial which has 1root. I give here some examples of Desargues, Pappus and Pascal configurations.I illustrate Pappus and Desargues configurations using the notation of 2.1.2 and of 2.1.5and give the points on a conic obtained using Pascal’s construction. .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. Example.
For p = 11 , with the polynomial I − I − , we havePappus ( (cid:104) , , (cid:105) , ∗ , (cid:104) , , , (cid:105) , ∗ ; (cid:104) , , (cid:105) , ∗ ) ,with A i +1 × B i − = (56 ∗ , ∗ , ∗ ) and A i − × B i +1 = (28 ∗ , ∗ , ∗ ) . Desargues (98 , { , , } , { ∗ , ∗ , ∗} , { , , } , { ∗ , ∗ , ∗} ; (cid:104) , , (cid:105) , (cid:104) ∗ , ∗ , ∗(cid:105) , ∗ ) , Desargues (111 , { , , } , { ∗ , ∗ , ∗} , { , , } , { ∗ , ∗ , ∗} ; (cid:104) , , (cid:105) , (cid:104) ∗ , ∗ , ∗(cid:105) , ∗ ) ,From Pascal’s construction we obtain the following points are on a conic: 9,10,33,51,58,60,74,77,79,87,96,98. Exercise.
Define a geometry corresponding to a polynomial which has 2 roots.
Introduction.
I will briefly stae one result for dimensions 3 and 4 concerning defining polynomials associ-ated to the non irreducible case and illustrate for dimnesions 3, 4 and 5.
Theorem.
If the P i denotes a primitive polynomial of degree i. For k = 3 , the defining polynomials P can have the following form, P , P P , P P , there are p + p + p + p + 1 , p − , p − p polynomials relatively prime to P, in theserespective cases. For k = 4 , the defining polynomials P can have the following form, P , P P , P P , P P . there are p + p + p + p + p + 1 , ( p − p + 1) , p − , p − p polynomials relativelyprime to P, in these respective cases. Proof: The polynomials in the sets are those which are relatively prime to the definingpolynomial. There are p k homogeneous polynomials of degree k. If, for instance, k = 4 andthe defining polynomial P is P P , there are p + p + 1 polynomials which are multiple of P and p + 1 , which are multiples of P , hence p + p + p + p + 1 − ( p + p + 1) − ( p − polynomials relatively prime to P. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Example. a , a , . . . a k represent I k +1 − a I k − a I k − − a k . k p period def. pol. | sel. | roots of def. pol. , , , −−
26 1 , , , , , , ,
25 156 1 , , , −−
124 1 , , , , , , ,
47 400 0 , , , −−
342 0 , , , , , , ,
511 1464 0 , , , −− , , , , , , ,
14 3 121 2 , , , , −−
104 0 , , , , I + I − I − I + I + 1)80 0 , , , , , , , , ,
25 781 4 , , , , −−
744 2 , , , , I + I + 2)( I + 2 I − I + 2)624 2 , , , , , , , , ,
37 2801 3 , , , , −− , , , , I + 2 I − I − I − I − , , , , , , , , , , , , , −− , , , , I + I − )( I − I − I − )14640 0 , , , , , , , , ,
313 30941 8 , , , , −− , , , , I − I + 6)( I − I + I + 2)28560 2 , , , , , , , , , −−
242 1 , , , , , , , , , , , .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. Definition.
Given a selector s, the selector function associates to the integers in the set Z n a set of p + 1 integers or p integers obtained as follows, s ( j ) ∈ f i iff sel ( l ) − sel ( j ) = i for some l. Theorem. f ( i ) is the set of points on the line i ∗ × ∗ . f ( i ) − j, where we subtract j from each element in the set, is the set of points in( i + j ) ∗ × j ∗ , equivalently f ( i − j ) − j, is the set of points in i ∗ × j ∗ . a ∗ × b ∗ × c ∗ = (( a − i ) ∗ × ( b − i ) ∗ × ( c − i ) ∗ ) − i. Theorem. If the defining polynomial is primitive, then | s | = p k − p − ,
1. if i ≡ / p k − p − , | f ( i ) | = p + 1 . If the defining polynomial has one root, then | s | = p k ,
1. if i (cid:54) = 0 , | f ( i ) | = p, If the defining polynomial has a double root, then | s | = p k − ,
1. if i ≡ / p, p − , | f ( i ) | = p,
2. if i ≡ p and i (cid:54) = 0 , | f ( i ) | = p − , Example. k = 3 , p = 3 , defining polynomial I − I − I − I − I − I − I + 1) , selector: { , , , , , , , , , , , , } CHAPTER 2. FINITE PROJECTIVE GEOMETRY selector function: − − − − k = 3 , p = 3 , defining polynomial I − I − I − I − I − ( I + I − , selector: { , , , , , , , , } selector function: − − − − − − k = 3 , p = 3 , defining polynomial I − I − I − . selector: { , , , , , , , } selector function: − − − − − −
11 0 1 14 7 14 19 21 13 1 2 15 19 0 2 192 0 2 19 8 − − − − − − −
14 0 15 21 10 4 14 15 16 − − − Example.
In the case of Example 2.3.6.0. If we denote by i † , the lines ∗ × i ∗ , these lines, which aresets of 4 points can all be obtained from † = { , , , } , † = { , , , } , † = { , , , } and † = { , , , } by adding an integer modulo n . † + 0 = 1 † , † , † , † + 1 = 39 † , † , † , † + 9 = 6 † , † , † , † + 15 = 34 † , † , † , † + 0 = 2 † , † , † , † + 2 = 22 † , † , † , † + 37 = 3 † , † , † , .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. † + 24 = 13 † , † , † , † + 0 = 4 † , † , † , † + 4 = 17 † , † , † , † + 21 = 12 † , † , † , † + 33 = 7 † , † , † , † + 0 = 10 † , † , † . Introduction.
The reader may want to skip this section until he has become familiar with conics. In it, wesummarize the various types and sub-types of conics as they relate to the representation ofthe finite projective plane, for p = 5 , on the dodecahedron. We will see later, IV.1.12. thatthe dodecahedron can also be used to represent the finite polar and the finite non-Euclideangeometry, for p = 5 . Definition.
The conics are all of the same type if the classification into face-points, vertex-points andedge-points is the same. The conics are of the same sub-type if they can be derived from eachother using any of the 60 collineations which exchange face-points.
Notation.
In the following Theorem we use the notation“60 f f f vve, C (6,9,17;11,29;22), 30 C (9,16,17;7,13;15).“to indicate that we have 60 conics with 3 face-points, 2 vertex-lines and 1 edge-line. Theseare of the sub-type C and C . An example of a conic of a given sub-type is provided inparenthesis, “;” separates points of a different classification, these points are given in theorder, face-point, vertex-point, edge-point, and in the same classification in increasing order.A pictorial representation of the sub-types is given in Figure 2.3.7. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem.
The 31 . . . . = 3100 conics are of the following type and sub-type.1 f f f f f f, A (0 , , , , , . f f f f ee, B (6 , , ,
17; 2 , . f f f vve, C , ,
17; 11 ,
29; 22) , C , ,
17; 7 ,
13; 15) . f f f vee, D , ,
17; 7; 1 , , D , ,
17; 5; 1 , , D , , , ; 5; 1 , , D , ,
6; 13; 18 , . f f vvvv, E ,
4; 5 , , , , E ,
4; 7 , , , . f f vvve, F (0 ,
4; 11 , ,
29; 8) . f f vvee, G ,
4; 11 ,
20; 1 , , G ,
4; 5 ,
13; 3 , , G ,
4; 7 ,
21; 18 , , G ,
4; 19 ,
21; 12 , , G ,
4; 11 ,
29; 1 , , G ,
4; 5 ,
11; 14 , , G ,
4; 21 ,
29; 1 , , G ,
4; 5 ,
21; 2 , . f f veee, H ,
4; 11; 8 , , , H ,
4; 13; 8 , , , H ,
4; 21; 2 , , , H ,
4; 21; 1 , , . f f eeee, I ,
4; 2 , , , , I ,
4; 2 , , , , I ,
4; 15 , , , , I ,
4; 1 , , , . f vvvvv, J , , , , , J , , , , . f vvvve, K , , ,
19; 27) , K , , ,
11; 26) . f vvvee, L , ,
20; 1 , , L , ,
13; 27 , , L , ,
24; 12 , , L , ,
20; 1 , , L , , , , , L , ,
13; 12 , . f vveee, M ,
21; 2 , , , M ,
20; 25 , , , M ,
21; 2 , , , M ,
24; 2 , , , M ,
20; 1 , , , M ,
29; 12 , , , M ,
29; 8 , , , M ,
21; 1 , , , M ,
21; 3 , , . f veeee, N , , , , N , , , , N , , , , N , , , , N , , , , N , , , , N , , , , N , , , , . f eeeee, O , , , , , O , , , , . vvvvvv, P (5 , , , , , . vvvvee, Q , , ,
29; 15 , , Q , ,
13; 21; 1 , . vvveee, R , ,
29; 3 , , , R , ,
20; 23 , , , R , ,
29; 14 , , , R , ,
24; 3 , , , R , ,
29; 3 , , , R , ,
29; 1 , , . vveeee, S ,
20; 3 , , , , S ,
20; 2 , , , , S ,
20; 1 , , , , S ,
20; 1 , , , , S ,
21; 18 , , , , S ,
21; 2 , , , , S ,
21; 3 , , , , S ,
21; 8 , , , , S ,
21; 2 , , , . veeeee, T , , , , , T , , , , . T , , , , . eeeeee, U , , , , , , U , , , , , . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. The proof of the decomposition into types was done using a computer program which took22 minutes to run an an IBM PC.
Figure.
The pictorial representation of a conic of a given sub-type on the dodecahedron is as follows.
A: the 6 faces.B: f . f.. . . .. .o of f. . .. .C1: . o . C2: f . f. . .f . . . .o . o o. . f. . . . .. f . f . . o .. . . . .. .. oD1: o D2: . . .f . f . .f. . .. . . . . .. . o o ofo o . f . f .. . . . . .. .. . . . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
D3: . . . D4: . . .. . o of f f f f. . . .o o . .. . . . . .. . . . . .. o . . o .. . . .. . . .E1: o . o E2: o o. . . .. . .. .. . . f . f .. o . . . .. f . f . . .o o. o .. .. .F: . o o. .. .. .. o .. f . f .. . .. .. o. oG1: G2: o . o. .. . . .. o . . .. f . f . . . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. o . o. o .. . . f . f .. . . . .o . . .. .. oG3: G4:o o o .. . . .o . o . . .. f . f . . f . f .. . . . . .. . . .. . . oo .o .G5: G6: o . .. oo . . o. . . . .. f . f . . .o o .. o .. . . f . f .. . . . .. . . .. .o .G7: G8: o . .. o. . . .. . . . o. f . f . o .. . o. . o. . . f . f .. o . . .. . . . CHAPTER 2. FINITE PROJECTIVE GEOMETRY . .H1: . o . H2: . o o. . . .. . . .. . . o. o . . . .. f . f . . f . f .. . . . . .o o o .. . . .o . . .H3: . . . H4: . . .. . o .. o . .o . o .o . . . . o. f . f . . f . f .. . . . . .. . . .. . . oI1: I2: . . .. .o o o o. . . . .. f . f . o o. . .. . .o o . f . f .. . . . .. .. . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. oI3: I4:. . . .. o . oo . . . . o. f . f . . f . f .. . o . . .o . . o. . . .J1: . . . J2: .. . o. . . .o o o o. . f. o . . .o . o o . o. . . . .. .. .fK1: f K2: . . .o o. . . .. o . f. .. . . . .. o .. . .o . o . oo . . . o. .o .L1: o . o L2: . . .. . . .f f. . . .. . o oo . o . o . CHAPTER 2. FINITE PROJECTIVE GEOMETRY . . . o . o. o . . . .. . . .. . . .L3: L4: o . oo o. o o .. . . f. .o o . . .. o .. . .. . . . .. .f o . .. .. .L5: o . o L6: . . .. . . .f f. . . .. . . o. . . . o .. o o o . .. . . . . .. o p .. . o .M1: o M2: o. .. . . . . o o .. . o of fo o . .o . o . . .. . o .M3: . o . M4: . o . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. . . . .f f. . . .. . . .o . o . . .. . . o . o. . . . . .. . . .o o . .. . o .M5: o . o M6: o . o. o . of f. . . .. . . .. o . . . o. . . . . .. . o . . .. . o .. . . .M7: . . . M8: . . .o . . .f f. . . .o . . o. o . . o oo . . . . .o . . . . o. . . .. . o .M9: . . . M10: . o .. . . .f f. . . .. . . .o o o o . . CHAPTER 2. FINITE PROJECTIVE GEOMETRY . . . . o o. . o . .. . . .. o . ooN1: . . . N2: . .o . . .f o o o. .o o . f o .. o . . . .. . . . .. .. . .o .. .N3: . . . N4: o o .. . . .f. o . .. o o oo . . . . .. . . . f o .. . . . . .. o . .. . . .o . . .N5: . . . N6: . . .. . o of f. . . .o . . .. o o . . .. . . . . . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. o . . . o .. . . o. . . .. . o .o . o .N7: . . . N8: . . .. . . .f f. . . .o . . .o . . o . o. . . . . .. o o o o .. . . .. . . .oO1: . O2: o. .. o o . o . . o. . . .fo o . .. o . . . .. . o ofP: o .. .. o .o . o. o .. .. o CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Q1: . Q2:. .. o o. o o . . o .o o . . .. . . o .o . o . oo .. .R1: . . . R2: o. . o. o o .o o . .. .. o . . .. o o . o. . . . .o o. .R3: . o . R4: o . o. . . .o o . .. . . .o o o . . .. . . . o .. . . . o .. . o o. . . .R5: . . . R6: . . .. o o . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. o o o o. . . o. o o o . .. . . . .. . . . . .o . . .. . . o. .S1: S2:. . . .. . o oo o o . o .. . . . . .o o o . o .. . o o. . . .o . . oS3: S4:. . . .o . . .. o . o o .. . . . . .. o . . o oo . . .. . . .o . o .oS5: . . . S6: .. . .. o o .o o o o. .o . o o o. . . . . . CHAPTER 2. FINITE PROJECTIVE GEOMETRY . . . . .o o. .S7: . . . S8: . o .o o o oo o o o. . . .. . . . . .. . . . o .o . o. .. .. .S9: o oo .o . .. . .o . .. o. .. .T1: . . . T2: o o .o o o .. . . .. . . oo . o . . o. o . . . o .. o . . . .. . . .. . . .T3: . . .o . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. . .. .. o o. o. . oo .. .U1: . . . U2: . o .o o . .. . . .. . o oo . o . . .. . . . o .. . . o . oo o . .. . . .
The family of types of conics was determined interactively using a computer program.
Introduction.
After defining convex uniform polyhedra, whose notion may go back to Archimedes and werefully studied by Kepler, we will show that one of them, the truncated dodecahedron can beused as a model for the finite projective plane of order . Definition.
A polyhedron with regular faces, in Euclidean 3-space is uniform if it has symmetry operationstaking a given vertex into any other vertex, otherwize it is non-uniform . If, in addition, allfaces are congruent, the polyhedra is regular . Theorem. [Euclid]
There are 5 convex regular polyhedra.
Notation. [See Johnson.]
In the following Theorem, we use the following notation, developped by several Mathemati-cians. { n } denotes a regular polygon with n sides, ( n.q.n.q ) denotes a vertex with adjoiningfaces successively with n, q, n, q sides, < n.q > denotes an edge ajoining a face with n sidesand one with q sides. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem. [Kepler]
Besides regular prisms and antiprisms, there are 13 convex uniform, non-regular polyhedra.These are
Name Faces Vertices Edges
Cuboctahedron { } , { } . . .
4) 24 < . >Icosidodecahedron { } , { } . . .
5) 60 < . >T runcated tetrahedron { } , { } . ) 12 < . >, < . >T runcated octahedron { } , { } . ) 24 < . >, < . >T runcated cube { } , { } . ) 24 < . >, < . >T runcated icosahedron { } , { } . ) 60 < . >, < . >T runcated dodecahedron { } , { } . ) 60 < . >, < . >Rhombicuboctahedron { } , { } . ) 24 < . >, < . >Rhombicosidodecahedron { } , { } , . . .
4) 60 < . >, < . > { } T runcated cuboctahedron { } , { } , . .
8) 24 < . >, < . >, { } < . >T runcated { } , { } , . .
10) 60 < . >, < . >,icosidodecahedron { } < . >Snubcuboctahedron { } , { } .
4) 36 < . >, < . >Snubicosidodecahedron { } , { } .
5) 90 < . >, < . >n − gonal prism n { } , { n } n (4 .n ) n < . >, n < .n >n − gonal antiprism n { } , { n } n (3 .n ) 2 n < . >, n < .n > Theorem. [N. W. Johnson]
There are 92 convex non-uniform regular-faced polyhedra.
The fact that all vertices are of the same type does not insure uniformity, as the exampleof the elongated square gyrobicupola of J. C. P. Miller shows. This non-uniform polyhedrahas the same characteristics as the rhombicuboctahedron, but has the part below the 8 squaresturned 45 degrees.Before discussing the truncated dodecahedron as a model for the Pappian plane associatedwith , I will discuss the pentagonal antiprism as a model for the Pappian plane associatedwith . Notation.
I identify elements which are symmetrical with respect to the center of the antiprism. Forthe pentagonal antiprism, with i = 0 , , , , , I will denote by t i , the 5 triangular faces, by v i , the 5 vertices, by e i , the 5 pentagonal-triangular edges, by f i , the 5 triangular-triangularedges and by p, pentagonal face. We have altogether 21 elements to represents the 21 pointsin the plane associated with . Theorem.
For q = 2 , The selector is { , , , , } . The corresponding selector function f is, and the representation of the points on theantiprism are .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. i f ( i ) 0 0 14 1 0 16 16 14 14 16 4 repr. of points v e v f f t t v v e e repr. of lines v t v f f e e v v t t i
11 12 13 14 15 16 17 18 19 20 f ( i ) 14 4 1 0 1 0 4 4 16 1 repr. of points v f f p t t e f e t repr. of lines v f f p e e t f t e The incidence properties are e ∗ i ι v i , e i ± , t i ± , t ∗ i ι v i , t i ± , f i ± , f ∗ i ι v i , e i ± , f i ± , v ∗ i ι p, v i , e i , t i , f i , p ι v i . Proof: I leave as an exercise the determination of the fundamental polynomial and thecorresponding selector.The selector function follows easily from its definition.The selector polarity which associates i to i ∗ has the fixed points 7,8,11,0 and 2 on the line ∗ . I will associates to 14 and to ∗ the pentagonal face and its incident points or lines to v i . Starting from that, one of the possible solution is given in 1. Notice that I use the samecorrespondance between e i , v i and f i for the points and the lines but exchange t i and e i toget the corresponding points and lines. Figure.
The corresponding drawing for the Projective plane over is given page . . . . Notation.
For the truncated dodecahedron, I will denote by t, a triangular face, by d, a decagonal face,by v, a vertex, by e, a < , > edge and by u, a < . > edge. The lower case notationis used indifferently for points and lines, the upper case notation for points. Lemma. If x Y denotes the number of points of type Y incident to a line of type x, then2 | f Y , | d Y , | t Y for Y (cid:54) = T, | t T − . Proof: For instance, there are 10 T -points, each is adjacent to 10 lines; on the otherhand, the 30 e -lines are adjacent to 30 e T triangles, the 15 f -lines are adjacent to 15 f T triangular-points, . . . . This implies e T + 15 f T + 30 v T + 10 t T + 6 d T = 100 , which gives, modulo , f T = 0 , modulo ,d T = 0 , modulo , t T = 1 . CHAPTER 2. FINITE PROJECTIVE GEOMETRY
Theorem.
For q = 3 , a primitive polynomial is I − I − (cid:15), with (cid:15) = 1 + α, an 8-th root of unity and α = − . The powers of (cid:15) are 1, 1 + α, − α, − α, − , − − α, α, − α. The corresponding selector is { , , , , , , , , , } . The corresponding selector function is { , , , , , , , , , } . The letters refer to the type. The last row gives the conjugate, for instance, 61 is theconjugate of 77.
Proof: To retrieve the primitive polynomial associated with S, the selector 1, because ∈ S, I = βI + γ, β and γ are chosen in such a way that I has no term of second degree.The computations are facilitated by preparing first a table giving g ( i ) (cid:51) (cid:15) i = (cid:15) g ( i ) , ≤ i ≤ t, and by use of the convention (cid:15) − = 0 . The conjugates are obtained when α is replaced by − α . Heuristics.
The truncated dodecahedron has 182 faces, vertices and edges. using symmetry with respectto the center we expect that a model can be found for the projective geometry of order , with 91 points and with 10 points on each line . We will solve simultaneously the followingproblems, discover appropriate incidence properties, associate to the vertices, integers from0 to 90 to take advantage of the selector and determine a fundamental projectivity on a line .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. to prepare for a representation of finite Euclidean geometry. I will describe here some of thesteps which have led me to the solution given in 2.3.8 to 2.3.8.The auto-correlates should be the points of a conic γ . I will choose this conic as a circlein the corresponding Euclidean plane. The intersection of the lines ×
70 = 77 ∗ and of ×
28 = 72 ∗ , which is 75, is chosen as the center of the circle. The points on the polar ∗ are 2, 6, 16, 17, 19, 25, 43, 65, 72, 77.To obtain a fundamental projectivity, we want to choose 2 points, A, B, on the circle andproject from them any point X on the circle onto ∗ , giving X A and X B , X A correspondsto X B , we want to choose A and B such that the projectivity is of order 10. A trial gave aprojectivity of order 5, it was then easy to obtain one of order 10 using A = 0 and B = 50 . The computations start as follows: × ∗ × ∗ = 77 × ∗ with 0 as the other pointon γ . × ∗ × ∗ = 65 × ∗ with 50 as the other point on γ . ×
50 = 50 ∗ × ∗ = 6 × ∗ with 46 as the other point on γ . ×
46 = 31 ∗ × ∗ = 25 × ∗ with 84 as the other point on γ . . . . . Hence the projectivity 2.3.8 and the equidistant points 0, 50, 46, 84, 47, 70, 86, 28, 76,59 on γ .With one d -face chosen as ∗ , the 10 t -faces are subdivided into 2 sets, those adjacentto the d-face and those which are not. The vertices of the pentagonal points 0, 46, 47, 86,76 are chosen for the successive triangles adjacent to the d -face. The diametrically oppositepoint, e. g. 70 of 0 is chosen for the triangle not adjacent to the d -face but adjacent to thetriangle 0.Because ×
46 = 3 , ×
47 = 45 , . . . , ×
70 = 77 , ×
86 = 6 , I chose the pentagonalside 3 for the e -point between the t -points 0 and 46, . . . , the diameter 77 for the e -pointbetween the t -points 0 and 70, . . . . Because | Y , we chose these 10 e -points as incident to ∗ . These consideration suggest Definition 2.3.8 and Theorems 2.3.8 and 2.3.8.
Definition.
The points in the truncated dodecahedron model consist of0. The 10 triangular face-points T.
1. The 6 decagonal face-points D.
2. The 30 vertex-points V.
3. The 30 triangular-decagonal edge-points U.
4. The 15 decagonal-decagonal edge-points E. The lines in the truncated dodecahedron model consist of0. The
10 triangular face-lines t . Each is incident to itself as a point, to the 3 adjacent < . > edge-points E, and to the 6 vertex-points V which are the vertices of thetriangle adjacent to the 3 edge-points which are not themselves adjacent to these points.For instance, for t = 0 , the incident points are T ) , E ) , E ) , E ) , V ) , V ) , V ) , V ) , V ) , V ) :78 CHAPTER 2. FINITE PROJECTIVE GEOMETRY o 61 27 o. 1 77 .56 o o . . o o 81o _0. . .. o 3 . .. . .. o o .. 49 9 .
1. The d. Each is incident to its 5 < . > edges U , and the 5 < . > edges E adjacent to its 5 adjacent triangles.For instance 75( d ) is incident to U ) , U ) , U ) , U ) , U ) and E ) , E ) , E ) , E ) , E ) : .o72.. o .. 19 .o . . . . o77 o 17 25 o 2. _75 .. .o 65 43 o. .. .16 o o 6. .
2. The
30 vertex-lines v. Each is incident to the < . > edge E adjacent to it andto the vertex at the other end of it, to the 2 triangular points T and T adjacent tothe other edges E and E , to the 2 < . > edges E and E opposite to E or E belonging to the same decagon as v , to the vertices adjacent to E or E closest to v ,to the < . > edges U belonging to the same decagon as T or T and the triangleopposite E .For instance, 9( v ) is adjacent to E ) , V ) , T ) , T ) , E ) , E ) , V ) , V ) , U ) , U ) : .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. . .. .. o 47 . . 0 o .. ._9 .52 o o 83. o 72 . .. o 68 .40 85. o . . o .o 18 82 o. .
3. The
30 triangular-decagonal edge-lines u. Each is incident to the adjacent decagonalpoint D , to the < . > edge E adjacent to the triangle adjacent to u and to the < . > edges U adjacent to E , to the vertices in the same decagons D and D as E opposite the vertex adjacent to E and the same triangle as u , and to the < . > edges U and U adjacent to the triangle adjacent to D and D or D not adjacentto these decagons and to the vetices adjacent to D and the < . > edges of D adjacent to U or U .For instance, 17( t ) is adjacent to D ) , E ) , T ) , T ) , V ) , V ) , U ) , U ) , V ) , V ) : . .32 o o 44. .o 75. .74 o . _17 . o 60. . . . .. . .. o 77 . .10 o . o 6439 o o 83. . . .. .
4. The
15 decagonal-decagonal edge-lines e. Each is incident to the 2 decagonal points,the 2 triangular points, the 2 < . > edges, the 2 vertices, the < . > edge, whosecenter in the the equatorial plane through e and the < . > edge perpendicular tothat plane.For instance, 1( e ) is incident to D ) , D ) , T ) , T ) , U ) , U ) , V ) , V ) , E ) , E ) :80 CHAPTER 2. FINITE PROJECTIVE GEOMETRY . o .. 90 .. .o 55. .. 26 .. . o . .o 0. o 48 . 2. _1 . . o .. o 8 .o 76. . o . .. 60 .. .o 80. .. .. (o) .(90)
Theorem.
The truncated docecahedron model satisfies the axioms for q = 3 . Figure.
The corresponding drawing for the Projective plane over is given page . . . . Theorem.
A fundamental projectivity on line 75 ∗ is(77 , , , , , , , , , . The elements are alternately of type v and u . Exercise.
Given the selector function f of 2.3.8 and the 6 dodecagonal faces, 4, 55, 75, 78, 80, 89,reconstruct the preceding figure using the following rules, which are first examplified,0. D ) × D ) = 5 ∗ ( e ) , .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. E ) × D ) = 60 ∗ ( u ) , E ) × E ) = 0 ∗ ( t ) , T ) × E ) = 56 ∗ ( v ) , f (89 − − .The above rules are clearly redundant.Determine alternate rules, for instance the rule corresponding to 2 triangular faces or 2vertices adjacent to the same decagonal-decagonal edge. Slightly more ambitious is to dermineall the possible rules. Theorem.
There are several configurations which represent a projective plane of order 3. The quad-rangle consists of 4 triangular face-points, the diagonal points, of 3 decagonal-decagonaledge-points, the quadrilateral, of 6 vertex-points. All the other points on the truncateddodecahedron represent complex points, 6 on each of the 13 lines.
The first example is associated with the primitive polynomial 2.3.8.0. . .o o47 . o . 59. 20 .. .. .o 9 18 o. . . .. o 10 . 1. o 0 o 90 76 o . o .. o 82 .. . . .o 81 71 o. .
The conjugates are given in table 2.3.8.2.A second example is as follows CHAPTER 2. FINITE PROJECTIVE GEOMETRY
82 10o o o. 90 .. .. .86 o . . o 84. . . .. o 8 . 2. o 1 o o o. o 48 . 7 54. . . .. .o 46 o 70. .o 80. .. .. (o) .(90) on the t -line , the conjugates are V ) and E ) , E ) and E ) , V ) and V ) .on the e -line ∗ , the conjugates are T ) and T ) , U ) and U ) , D ) and D ) . on the v -line ∗ , the conjugates are V ) and U ) , V ) and U ) , U ) and U ) . Proof: For the conjugates we use the Pascal construction to determine the 6-th point onthe line on a conic through 4 real points and 1 complex point.
Exercise.
For q = 2 ,
0. determine the primitive polynomial giving the selector 0, 1, 4, 14, 16.1. Determine the correspondance between the selector notation and the homogeneous co-ordinates for points and lines. Note that these are not the same.2. The correspondance i to i ∗ is a polarity whose fixed points are on a line. Determinethe matrix representation, the polar of ( X, Y, Z ) and the equation satisfied by the fixedpoints.3. Determine the fundamental projectivity on the line ∗ using a point conic which hasno points on ∗ . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA.
4. Illustrate Pascal’s Theorem.
Exercise.
0. Explore the usefulness of the truncated cuboctahedron less the hexagonal faces and the < . > edges as a model for the projective geometry of order 7.1. Show that the 14-gonal antiprism can be used as a model for the projective geometry oforder 7. More generally,2. Show that the n -gonal antiprism can be used as a model for the projective geometry oforder q = p k when p ≡ − , with n = q + q .
3. Show that the n -gonal antiprism can be used as a model for the projective geometry oforder q = p k when q ≡ , with n = q + q . Finally,4. Show that the n -gonal prism can be used as a model for the projective geometry of order q = p k when q ≡ − , with n = q + q .
5. is there a general theory when using prisms or antiprisms?
Exercise.
For q = 2 .
0. to 4. Answer question similar to those of 2.3.85. Show that the -gonal antiprism can be used as a model for the projective geometry oforder . More generally,6. Show that the n -gonal antiprism can be used as a model for the projective geometry oforder q = 2 k , with n = q + q . Answer to 2.3.2.
0. For q = 2 , the primitive polynomial giving the selector 0, 1, 3, is I + I + 1 . The auto-correlates are 0 11 2 7 8.The selector function is i f ( i ) 0 14 1 0 16 16 14 14 16 type F V F V T T V F F T E F i
12 13 14 15 16 17 18 19 20 f ( i ) 4 1 0 1 0 4 4 16 1 type T E P V E T E V E CHAPTER 2. FINITE PROJECTIVE GEOMETRY
1. The correspondence between the selector notation and the homogeneous coordinates forpoints and lines is i I i i ∗ ∗ : 1 , , , I ∗ : 0 , , , I ∗ : 0 , , , I + 1 5 ∗ : 2 , , , I + I ∗ : 0 , , , I + I + 1 4 ∗ : 3 , , , I + 1 2 ∗ : 1 , , .
2. The matrix representation is M = , M − = . and the equation satisfied by thefixed points is ( X + X ) = 0.3. The degenerate conic through 0, 1, 2 and 5 with tangent ∗ at 5, is represented by thematrix N = . The polar of 0 is ∗ , of 1 is ∗ , of 2 is ∗ , of 4 is ∗ , of 5 is ∗ of 6 is ∗ and of 3 isundefined. The equation in homogeneous coordinates is X ( X + X ) = 0 .
4. A circle with center 14 can be constructed as follows. I first observe that a directionmust be orthogonal to itself. Indeed, if 0 is a direction, the others form an angle 1,2,3,4 mod , we cannot play favorites and must choose 0. If A = 1 , C × A and thereforethe tangent has direction 0, A × A i +1 has direction i mod or are the points 0, 7, 8,2, 11.It is natural to choose the pentagonal face-point as 14, and the edge-points on thepentagon as 0, 8, 11, 7, 2. The points on the circle 1, 6, 3, 15, 19 are chosen asthe vertex-points opposite the corresponding edge-point, 1 opposite 0, 6 opposite 8, . . . .This gives the types, with subscripts indicated in 0. and the definition:The points are represented on the 5-anti-prism as follows. The pentagonal face-point,P, the 5 triangular face-points, T i , the 5 vertex-points, V i , the 5 triangular-triangularedge-points, E i , the 5 pentagonal-triangular edge-points F i . The lines are represented on the 5-anti-prism as follows. The pentagonal face-line, f,which is incident to F i , the 5 triangular face-lines, t i , which are incident to F i , F i , T i +1 ,T i − , E i +2 , E i − . If f is the pentagonal edge of t i and V, V (cid:48) are on f , F i is on it, T i +1 ( T i − ) share V ( V (cid:48) ) , E i +2 ( E i − ) are on an edge through V ( V (cid:48) ) not on t i the 5 vertex-lines, v i , which are incident to F i , V i +2 , V i − , E i +1 , E i − . If t is the face with v i on its pentagonal edge these are allthe vertices, and edge-points on it distinct from v i . the 5 triangular-triangular edge-lines, e i , which are incident to F i , T i +2 , T i − , V i +1 , .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. V i − . V i +1 and V i − are on the same edge as e i , the line which joins the center C of theantiprism to E i is parallel to the edge containing F i , T i +2 and T i − are the triangularfaces which are not adjacent to E i or F i . the 5 pentagonal-triangular edge-lines. f i , which are incident to P, T i , V i , E i , F i . T i isadjacent to f i , V i is opposite f i , E i joined to the center of the antiprism is parallel to T i . Answer to 2.3.3.
For p = 3 ,
0. The primitive polynomial giving the selector 0, 1, 3, 9 is I − I − .
1. The correspondence between the selector notation and the homogeneous coordinates forpoints and lines is i I i i ∗ ∗ : 1 , , , , I ∗ : 0 , , , , I ∗ : 0 , , , , I + 1 7 ∗ : 2 , , , , I + I ∗ : 0 , , , , I + I + 1 4 ∗ : 5 , , , , I + 2 I + 1 1 ∗ : 3 , , , , I + I + 2 6 ∗ : 3 , , , , I + 1 2 ∗ : 1 , , , , I + 2 11 ∗ : 2 , , , I + 2 I ∗ : 0 , , , , I + 2 I + 2 5 ∗ : 4 , , , , I + 2 8 ∗ : 1 , , , .
2. The matrix representation of the polarity i to i ∗ is M = , M − = . The equation satisfied by the fixed points is X + X + 2 X X = 0 .
3. The degenerate conic through 0, 1, 2 and 5 with tangent ∗ at , is obtained by con-structing the quadrangle-quadrilateral configuration starting with P = 5 and Q i = { , , } . We obtain q i = { ∗ , ∗ , ∗ } , which are the tangents at Q i . The matrix repre-sentation is N = with equation X X + X X + X X = 0 . We can check that the polar of
10 = 3 ∗ × ∗ is ∗ = 0 × . Answer to ??. CHAPTER 2. FINITE PROJECTIVE GEOMETRY
0. For q = 2 , the primitive polynomial giving the selector 0, 1, 4, 14, 16 is I − I − I − (cid:15) , with (cid:15) + (cid:15) + 1 = 0 .
1. The correspondence between the selector notation and the homogeneous coordinates areas follows, i ∗ has the homogeneous coordinates associated with I i .i I i i ∗ ∗ I ∗ I ∗ I + I + (cid:15) ∗ I + (cid:15) ∗ I + (cid:15) I ∗ I + (cid:15) I + 118 ∗ I + 1 15 ∗ I + (cid:15) ∗ I + (cid:15) I + (cid:15) ∗ I + (cid:15)I + 1 9 ∗ I + (cid:15) ∗ I + (cid:15)I + (cid:15) ∗ I + I + (cid:15) ∗ I + 1 2 ∗ I + I ∗ I + (cid:15) ∗ I + (cid:15)I ∗ I + (cid:15)I + (cid:15) ∗ I + (cid:15) I + (cid:15) ∗ I + I + 1 7 ∗ To obtain the last column, for row 9, [1 , (cid:15) , (cid:15) ] = (1 , , × (1 , (cid:15),
0) = 20 ×
17 = 5 ∗ .
2. The correspondence i to i ∗ is a polarity whose fixed points are on a line. The matrixrepresentation is obtained by using the image of 4 points.0 = (0,0,1), M (0) = 0 ∗ = [1 , , , M (1) = 1 ∗ = [1 , , , M (2) = 2 ∗ = [0 , , ,
18 = (1 , (cid:15), (cid:15) ) , M (18) = 18 ∗ = [1 , (cid:15) , . The first 3 conditions give the polarity matrix asThe last condition gives β(cid:15) + α(cid:15) = 1 , γ + β(cid:15) = (cid:15) , γ = 1 . Hence γ = 1 , β = 1 , α = 1 . Therefore M = , M − = . Note that M is real and could have been obtained from the reality and non singularityconditions, giving directly α = β = γ = 1 . The polar of ( X , X , X ) is [ X + X , X + X , X ] . .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA. The fixed points ( X , X , X ) satisfy X = 0 corresponding to ∗ .
3. A point conic with no points on 14 is 1, 3, 4, 5,13,the corresponding line conic is 15,19,10,16, 8.Projecting from 1 and 3, 1, 3, 5,13, 4,we get the fundamental projectivity, 8, 2,11, 0, 7 on ∗ .
4. To illustrate Pascal’s Theorem, because there are only 5 points on a conic, we need touse the degenerate case. The conic through 0, 1, 2 and the conjugate points 9 and 18is The last condition gives β(cid:15) + α(cid:15) = 1 , γ + β(cid:15) = (cid:15) , γ = 1 . Hence γ = 1 , β = 1 , α = 1 . Therefore M = , M − = . Note that M is real and could have been obtained from the reality and non singularityconditions, giving directly α = β = γ = 1 . The polar of ( X , X , X ) is [ X + X , X + X , X ] . The fixed points ( X, X , X ) satisfy X = 0 corresponding to ∗ .
5. A point conic with no points on 14 is 1, 3, 4, 5,13, The tangents at (0,0,1), (0,1,0), (1,0,0), (1 , (cid:15) , (cid:15) ) , (1 , (cid:15), (cid:15) ) are [1,1,0], [1,0,1], [0,1,1], [1 , (cid:15) , (cid:15) ) , (1 , (cid:15), (cid:15) ] , or ∗ , ∗ , ∗ , ∗ , ∗ . On the other hand, using Pascal’s Theorem,the tangent at 0 is given by ((((0 × × (9 × × ((18 × × (1 × × (2 × ×
0= (((0 ∗ × ∗ ) × (4 ∗ × ∗ )) × ∗ ) ×
0= (((14 ×
17) = 8 ∗ ) × ∗ or13) × ∗ . Answer to ??.
For q = 57 , choose the auto-correlates as point on a circle although 0 is on the circle drawas it is the center. With the succession of points X i ,x i = 0 × X i , , , , , , ,X i , , , , , , ,y i +1 = X i − × X i +1 , , , , , , ,y i +2 = X i − × X i +2 , , , , , , ,y i +3 = X i − × X i +3 , , , , , , ,y i +1 × x i , , , , , , ,y i +2 × x i , , , , , , ,y i +3 × x i , , , , , , . This gives all the points in the projective plane of order 7. We observe CHAPTER 2. FINITE PROJECTIVE GEOMETRY ∗ ∗ ∗ ∗ ∗ ∗ ∗
36 36 36 36 36 36 3616 35 ,
30 18 ,
26 50 , ,
44 22 8 ,
28 14 , ,
20 34 17 ,
40 2 , ,
54 31 ,
39 7 6 , ,
33 5 , , ,
49 56 ,
47 48 ,
38 2437 ,
11 12 ,
19 9 ,
55 5335 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ,
18 50 ,
30 29 , , ,
44 10 , ,
23 2 40 ,
20 34 , ,
25 7 , , , ,
45 51 21 , ,
38 24 ,
48 15 ,
47 569 ,
19 53 ,
37 55 ,
11 1218 ∗ ∗ ∗ ∗ ∗ ∗ ∗
52 52 52 52 52 52 5218 35 ,
50 16 ,
29 26 , ,
14 8 22 ,
10 28 , ,
40 41 20 ,
23 2 , ,
39 31 ,
27 6 7 , ,
45 21 ,
33 5 51 , ,
47 56 ,
15 24 ,
49 4853 ,
55 12 , ,
19 37
Answer to ??.
For q = ,
36 : 0 37 38 40 44 52 18 27 68 1 ∗ ∗ ∗ ∗ ∗ ∗ × ∗ : 0 1 3 7 15 31 36 54 63 0 0 0 1 3 3136 ×
37 = 37 ∗ : 17 26 36 37 39 43 51 67 72 72 51 67 72 72 2636 ×
38 = 38 ∗ : 16 25 35 36 38 42 50 66 71 35 71 66 71 50 7136 ×
40 = 40 ∗ : 14 23 33 34 36 40 48 64 69 14 33 69 34 69 6936 ×
44 = 44 ∗ : 10 19 29 30 32 36 44 60 65 30 60 29 29 32 1036 ×
52 = 52 ∗ : 2 11 21 22 24 28 36 52 57 2 28 24 52 11 236 ×
18 = 18 ∗ : 13 18 36 45 55 56 58 62 70 62 70 56 13 70 5836 ×
27 = 27 ∗ : 4 9 27 36 46 47 49 53 61 53 4 47 61 27 4936 ×
68 = 68 ∗ : 5 6 8 12 20 36 41 59 68 6 12 8 5 59 68 Conic with no point on 36: 2, 4, 5, 6,13,28,31,46,63line conic: 29,59,31, 9,18,43,28,35,64.Fundamental projectivity: from 2 and 5 on the conic, the points2, 5, 6,31,13,28, 4,46,63 give the points on ∗ : .3. GEOMETRIC MODELS ON REGULAR PYTHAGOREAN POLYHEDRA.
38, 0,68,27,52,37,40,18,44.empty CHAPTER 2. FINITE PROJECTIVE GEOMETRY hapter 3FINITE PRE INVOLUTIVEGEOMETRY
In the geometry of Euclid, not every pair of lines have a point in common, namely the paral-lel ones. I call Euclidean Geometry, that geometry which consists in completing the plane ofEuclid by the ideal points and the lines of Euclid by the ideal line. To each set of parallel linescorrespond its direction or point at infinity or ideal point. The line at infinity or ideal line isincident to all ideal points. Figures Pl and St may help the reader to visualize. In Fig. Pl,projecting the line b on the line c from the point P establishes a one to one correspondancebetween the points on these lines, if we include the ideal point C i , on c , corresponding to B i and the ideal point B ∞ , on b , corresponding to C ∞ . Replacing lines b and c by planes,perpendicular to the plane P of figure establishes a one to one correspondance between a linethrough C ∞ perpendicular to P and the ideal line through B ∞ .This led to the concept of perspectivity, which I have schematized in Fig. St. In it, the shad-ing, corresponds to the method used by Chinese artists to represent distances in paintings.The tiling corresponds to the method used by Western painters. Johannes Vermeer’s use ofperspective in his paintings was so accurate as to allow P. T. A. Swillens to reconstruct, fromthe size of a chair, in the painting, not only the size of the rooms, but also to estimate theheight of the artist.Affine geometry is obtained from Euclidean geometry by discarding the notions associatedwith congruences of figures, projective geometry is obtained by discarding the notion of paral-lelism, thereby making the properties of any point or line in the plane indistinguishable fromthat of any other.I will describe at a later time, how I was lead to the discovery of finite Euclidean geometryand to the extension of many of the properties of Euclidean geometry. While working out aproof for these results, it occured to me, that the results can be placed in the framework of fi-nite projective geometry. I will, as I proceed, make the connection with the results in classical CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Euclidean geometry. The results can be considered as proceeding from an, apparently new,configuration consisting of 14 points and 13 lines. This configuration is defined starting froman ordered complete 5-angle, A , A , A , M and M , in which the first 3 points can be rotatedand the last 2 points interchanged. In other words the configuration is the same if we replace A , A and A by A , A and A and independently M by M and M by M. In involutivegeometry, (the Euclidean geometry without measure of angles and distances), we define alti-tudes and their intersection, the orthocenter, we define medians and their intersection, thebarycenter. In the generalization to projective geometry, the orthocenter and the barycenterbecome two arbitrary points, whose role is interchangeable. The proofs are constructive, andthe only construction required are those of lines through 2 given points and of points at theintersection of two given lines, but these constructions must be valid for all p . They do notinvolve the construction of an arbitrary line through a given point, as required to obtain, forinstance, an arbitrary point on a conic, by the construction of Pascal or of MacLaurin.No special relations will be assumed here between the points obtained during the construc-tion. The special relation M on the polar of M with respect to the triangle A , A , A willbe studied in Chapter IV,in the section on Cartesian geometry and the special case where M and M are respectivelyon the polar of M and M with respect to the triangle will also be discussed elsewhere.The beginning of a synthetic proof is given in section 4.3. Synthetic proofs are highly desir-able and are from my point of view more elegant, but require much more time to develop.The constructions and statements are given in a compact found using a notation which willnow be explained. Introduction.
In the preceding Chapter, I have introduced a notation for points, lines, incidence and state-ments. Additional notation is given here for conics, for points on conics and tangents toconics and a notation which allows to describe at once 3 points or 6 points associated to atriangle.
Notation.
The identifier for a point conic will be a lower case Greek letter or an identifier starting witha lower case preceded by a backward quote “ ‘ ”. The identifier for a line conic will be anupper case Greek letter or an identifier starting with an upper case preceded by a backwardquote “ ‘ ”.The subscript i , will have the values , and . Hence A i denotes 3 points A , A and A . If subscripts involve the letter i and addition, the addition is done modulo 3, for instance, a i := A i +1 × A i +2 is equivalent to .1. AN OVERVIEW OF THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. a := A × A , a := A × A , a := A × A . It represents the construction of the sides a , a and a of a triangle with vertices A , A and A . To indicate that a conic γ is constructed as that conic which passes through the 5 distinctpoints P , P , P , P , and P , we write γ := conic ( P , P , P , P , P ) . To indicate that a conic γ is constructed as that conic whose tangent at P is a and at P is a and passes also by P , we write γ := conic (( P , a ) , ( P , a ) , P ) . or γ := conic ( P , a , P , a , P ) . When 3 lines x i are concurrent, the intersection X can be obtained using any of the threepairs. I have chosen, arbitrarily, X := x × x ( ∗ ) , as a reminder that 2 other definitions of X could have been chosen. In the special case, x = x , the other choice X := x × x , will be used. “ (*) ” denotes therefore not only a Definition but also a Theorem or Conclusion.A similar notation will be used for conics. X · γ = 0 and X i · γ = 0 , are the notations corresponding to the point X is on theconic γ and the triple X , X , X is on the conic γ . P = P ole ( p, α ), is the notation for P is the pole of p with respect to the conic α . γ is a circle = 0 or γ is a cocircle is either an hypothesis, to indicate a prefered conicfrom which all other circles are defined or a Conclusion, X = Center ( γ ) and X = Cocenter ( γ ) is an abbreviation for X is the center of theconic γ (not necessarily a circle) and X is the cocenter of the conic γ , in other words X, ( X ) is the polar of m ( m ) with respect to γ . See section 4.3.3. Example.
With this notation, the special configuration of Desargues of 0.4.6. can be defined by a i := A i +1 × A i − , r i := P × A i ,P i := a i × r i , q i := P i +1 × P i − ,R i := a i × q i , p i := A i × R i ,Q i := p i +1 × p i − , p := R × R ( ∗ ) , and the conclusions of the special Desargues Theorem are implied by the last Definition-Conclusion and by the Conclusion, Q i · r i = 0 . Let P = ( p , p , p ) , and A = (1 , , , A = (0 , , , A = (0 , , , then a = [1 , , , r = [0 , p , − p ] ,P = (0 , p , p ) , q = [ − p p , p p , p p ] ,R = (0 , p , − p ) , p = [0 , p , p ] ,Q = ( − p , p , p ) , p = [ p p , p p , p p ] . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Example.
For p = 3 , prove that if A = (1 , , , A = (0 , , , A = (0 , , and P = (1 , , then theother elements of the quadrangle quadrilateral configuration II.2.1.6 are P = (0 , , , Q = ( − , , , R = (0 , , − , . . . , and a = [1 , , , p = [1 , , ,p = [0 , , , q = [ − , , , r = [0 , , − , . . . and thatthe conic of II.2.2.11 is X + X + X = 0 . Theorem.
With the above notation, the polar p can be obtained algebraically from the pole P or thepole P from the polar p using the first or the second formula: pA i = P ∗ A i , P a i = p ∗ a i , pA i = ( P · a i ) A i − ( A i · a i ) P, P a i = ( p · A i ) a i − ( a i · A i ) p, p i = ( P · a i +1 )( P · a i − ) A i +1 ∗ A i − + ( P · a i +1 )( A i − · a i − ) P ∗ A i +1 − ( P · a i − )( A i +1 · a i +1 ) P ∗ A i − ,P i = ( p · A i +1 )( p · A i − ) a i +1 ∗ a i − + ( p · A i +1 )( a i − · A i − ) p ∗ a i +1 − ( p · A i − )( a i +1 · A i +1 ) p ∗ a i − . P a i = ( P · a i +1 ) A i +1 − ( P · a i − ) A i − ,P A i = ( p · A i +1 ) a i +1 − ( p · A i − ) a i − . p = P · a a + P · a a + P · a a , P = p · A A + p · A A + p · A A . Proof: Only the first part of 2 to 4 needs to be proven, because of duality. To obtain 2,we use P ∗ P = 0 and A i +1 ∗ P = − P ∗ A i +1 . To obtain 3, we recall that a i = A i +1 ∗ A i − , weuse A i ∗ a j = 0 when i (cid:54) = j and A i ∗ a i = ( A ∗ A ) · A = t, then divide by t and by P · a i (cid:54) = 0 .To obtain 4, we use p = R i +1 ∗ R i − . We divide by ( P · a i )( P · a i +1 )( P · a i − (cid:54) = 0 , and obtain p = A i +1 ∗ P · a i A i − + A i − ∗ P · a i +1 A i + A i ∗ P · a i − A i +1 , or p = P · a i a i + P · a i +1 a i +1 + P · a i − a i − . Example.
For p = 13 , A i = (36(1 , , , , , , , , , P = (68(1 , , ,a i = [175(1 , , , , , , , , , r i = [77 , , ,S i = (143 , , , p i = [108 , , , R i = (48 , , ,p = a + a + a = 2 a + a + 4 a = [124] , p i = [140 , , ,P i = (51 , , , P = A + A + A = 6 A + 3 A + 11 A = (68) . .1. AN OVERVIEW OF THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Definition. An hexal complete 5-angle configuration , is a configuration which starts with an ordered setof 5 points A , A , A , M and M .
In the configuration obtained from it, if a point X is constructed, 5 other points are ob-tained. X is obtained by replacing A , A , A by A , A , A ; X is obtained by replacing thesame points by A , A , A and the points X i are obtained by exchanging in the constructionof X i , M and M .
The same holds for lines. The first letter has a macron placed above it inthe naming of the construction which exchanges M and M . { In the group of permutation onthe 5 points of the complete 5-angle, the figure is invariant under the cyclic group generatedby the permutation ( A A A M M )( A A A M M ) } . In special cases, several of these elements or all of the elements may coincide.
Comment.
We know from II.1.5.6. that a complete 5-angle requires p ≥ , therefore, the definition andresults that follow are non vacuous only if p ≥
5. We introduce here a terminology inspiredfrom corresponding terms in Euclidean geometry. In some instances, the correspondencewill be made explicitly. For instance, the line m which will be constructed corresponds tothe ideal line or line at infinity in Euclidean geometry, we will therefore call m the idealline. In the symmetry which exchanges M and M , to m corresponds m , which will be calledthe coideal line. { m corresponds to the orthic axis. } The conic θ which will be constructedcorresponds to the circumcircle and the conic γ to the circle of Brianchon-Poncelet also calledthe nine-point circle. Definition. θ and any conic δ, ( δ ) such that there exists a radical axis u , ( u ) with respect to m ( m ) iscalled a circle ( cocircle ) and u is called the radical ( coradical ) axis of θ and δ, ( δ ) .Algebraically, we have, for some integers k , k and k ,k δ + k θ = k ( m ) ×× ( u ) , where θ , m, δ and u are expressed exactly as in the corresponding expressions P0.7, P1.19,1.20, . . . , below.A triangle consists of its vertices and its sides. When we want to be specific we will useeither or both, for instance the given triangle can be written as { A i } or { a i } or { A i , a i } .To each of the section of this Theorem corresponds a sequence of theorems in Euclideangeometry which will be given in the corresponding sections of Chapter IV.We will give separately the construction of the various points and lines of the hexal con-figuration ( zetetic part ) and the proof that the construction satisfies the given properties( poristic part ). CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Introduction.
As the generalization proceeds, the 4 points on the line of Euler, become 10 points on itsgeneralization. The 9 (or 12) points on the circle of Brianchon-Poncelet (also called circleof Euler) become 20 points on the corresponding conic. New results, which will be given inpart III, are further consequences.The definitions are numbered starting with D, the conclusions are numbered stating with C,the proofs, which consist of the algebraic expressions of the various points, lines and conics,which can easily be checked and from which the conclusions can easily be verified, have anumber corresponding to the definition, starting with P.The numbering in this overview is the same as the number in the complete theory, given inChapter 5 and 6.
Theorem.
If we derive a point X and a line x by a given construction from A i , M and M , with thecoordinates as given in G0.0 and G0.1, below, and the point X and line x are obtain by thesame construction interchange M and M , X = ( f ( m , m , m ) , f ( m , m , m ) , f ( m , m , m )) ,x = [ g ( m , m , m ) , g ( m , m , m ) , g ( m , m , m )] , = ⇒ X = ( m f ( m − , m − , m − ) , m f ( m − , m − , m − ) , m f ( m − , m − , m − )) ,x = [ m − g ( m − , m − , m − ) , m − g ( m − , m − , m − ) , m − g ( m − , m − , m − )] . Proof: The point collineation C = q q
00 0 q , associates to (1,1,1), ( q , q , q ) , andto ( m , m , m ) , ( r , r , r ) , if r i = q i m i . In the new system of coordinates, X = ( q f ( q − r , q − r , q − r ) , q f ( q − r , q − r , q − r ) , q f ( q − r , q − r , q − r )) . Exchanging q i and r i and then replacing q i by 1 and r i by m i is equivalent to substituting m i for q i and 1 for r i , which gives X . x is obtained similarly.The line collineation is q − q −
00 0 q − . Theorem.
Given a complete 5-angle, 5 distinct points, no 3 of which are on the same line, A , A , A ,M and M , A i are called the vertices, M is called the barycenter and M , the orthocenter. .1. AN OVERVIEW OF THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES.
1. The ideal line and the orthic line. See Fig. 1,H0.0. A i , H0.1.
M, M ,D0.0. a i := A i +1 × A i − , D0.1. ma i := M × A i , ma i := M × A i , D0.2. M i := ma i × a i , M i := ma i × a i , D0.3. mm i := M i +1 × M i − , mm i := M i +1 × M i − , D0.4.
M A i := a i × mm i , M A i := a i × mm i , D0.7. m := M A × M A ( ∗ ) , m := M A × M A ( ∗ ) . The nomenclature:N0.0. a i are the sides .N0.3. ma i are the medians , ma i are the comedians or ma i are the altitudes , ma i are the coaltitudes ,N0.4. M i are the mid-points of the sides. M i are the feet or the feet of the altitudes,N0.5. ( M i , mm i ) is the complementary triangle , ( M i , mm i ) is the orthic triangle ,N0.6. M A i are the directions of the sides ,N0.8. m is the ideal line corresponding to the line at infinity, m is the coideal line or the orthic line , which is the polarof M with respect to the triangle.Proof:P0.0. a = [1 , , , P0.1. ma = [0 , , − , ma = [0 , − m , m ] , P0.2. M = (0 , , , M = (0 , m , m ) , P0.3. mm = [ − , , , mm = [ − m m , m m , m m ] , P0.4.
M A = (0 , , − , M A = (0 , m , − m ) , P0.7. m = [1 , , , m = [ m m , m m , m m ] ,
2. The line of Euler and the circle of Brianchon-Poncelet. See Fig. 2, 2b.LetD1.0. eul := M × M D1.20. γ := conic ( M , M , M , M , M )( ∗ ) , thenC1.1 γ is a circle, γ is a cocircle = 0 . The nomenclature:N1.0. eul is the line of Euler .N1.11. γ is the circle of Brianchon-Poncelet . In Euclidean geometry, the circle ofBrianchon-Poncelet, is also called the circle of 9 points or circle of Feuerbach or,improperly, the circle of Euler. It passes through the midpoints of the sides, the feet ofthe altitudes and the midpoints of the segment joining the vertices to the orthocenter.The Definition-Conclusion D1.20. corresponds to the first part of the Theorem ofBrianchon-Poncelet. CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Proof:P1.0. eul = [ m − m , m − m , m − m ] , P1.20. γ : m m X + m m X + m m X − m ( m + m ) X X − m ( m + m ) X X − m ( m + m ) X X = 0 ,γ − : m ( m − m ) x + m ( m − m ) x + m ( m − m ) x − m m ( m (3 s − m ) + m m ) x x − m m ( m (3 s − m ) + m m ) x x − m m ( m (3 s − m ) + m m ) x x .
3. The circumcircle. See Fig. 4, 4b.LetD1.6.
Imm i := m × mm i , Imm i := m × mm i , D1.7. ta i := A i × Imm i , D1.19. θ := conic ( A , ta , A , ta , A ) , thenC1.2. Imm i · ta i = 0 . H1.1. θ is a circle = θ is a cocircle = 0 . The nomenclature:N1.4.
Imm i are the directions of the antiparallels a i with respectto the sides a i +1 and a i − . N1.5. ta i are the tangents at A i to the circumcircle ,N1.10. θ is the circumcircle .Proof:P1.6. Imm = ( m ( m − m ) , − m ( m + m ) , m ( m + m )) ,Imm = ( m ( m − m ) , − m ( m + m ) , m ( m + m )) , P1.7. ta = [0 , m ( m + m ) , m ( m + m )] , P1.19. θ : m ( m + m ) X X + m ( m + m ) X X + m ( m + m ) X X = 0 , θ + γ = ( m ) ×× ( m ) .θ − : m ( m + m ) x + m ( m + m ) x + m ( m + m ) x − m m ( m + m )( m + m ) x x − m m ( m + m )( m + m ) x x − m m ( m + m )( m + m ) x x = 0 ,
4. The point of Lemoine. See Fig. 3.LetD1.2.
M aa i := ma i +1 × ma i − , M aa i := ma i − × ma i +1 , D1.3. mM a i := M aa i × M aa i , D1.4. K := mM a × mM a ( ∗ ) , D1.8. T i := ta i +1 × ta i − , D12.1. at i := A i × T i , The nomenclature:N1.5. ( T i , ta i ) is the tangential triangle , .1. AN OVERVIEW OF THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. N12.1. at i are the symmedians , of d’Ocagne,N1.2. K is the point of Lemoine , also called point of Grebe or of Lhuillier.Proof:P1.2. M aa = ( m , m , m ) , M aa = ( m , m , m ) , P1.3. mM a = [ q , m ( m − m ) , − m ( m − m )] , P1.4. K = ( m ( m + m ) , m ( m + m ) , m ( m + m )) , P1.8. T = ( − m ( m + m ) , m ( m + m ) , m ( m + m )) , P12.1. at = [0 , m ( m + m ) , − m ( m + m )] , ∗ ∗ ∗ ∗ configuration. Introduction.
It would be desirable to have a synthetic proof of the sequence of Theorems given in thisand in the following Chapters. In many instances, it is not difficult to obtain it, using thestandard Theorems of projective geometry, mainly those of Pappus, Desargues and Pascal.In other cases, the proof is less obvious. Theorem 4.3.1., which can be considered as thestarting point, has a first part which required additional constructions. The proof impliesthe validity of the extension of all the Theorems to finite projective geometries associated toGalois fields of order p k , p > and to the projective geometries associated to the field ofrationals, the field of reals, the field of complex numbers, the real p-adic field, the complexp-adic field, . . . . For the second part, the proof is synthetic. Theorem.
Let A , A , A , M and M be a complete 5-angle, see Fig. 0, a i := A i +1 × A i − ma i := M × A i , ma i := M × A i M i := ma i × a i , M i := ma i × a i ,M aa i := ma i +1 × ma i − ,cc i := A i × M aa i ,P := cc × cc ( ∗ ) ,CA i := cc i × a i ,caa i := CA i +1 × CA i − ,c i := M i +1 × M i − ,CC i := a i × c i p := CC × CC ( ∗ ) ,γ := conic( M i , M , M )( ∗ ). then CC i · caa i = 0 . The configuration involves the 14 points A i , M i , M i , C i , M and M and the 13 lines a i ,ma i , ma i , c i and p. Proof: For the first part, (see Fig. 0’)dual-Pappus ( (cid:104) ma , ma , ma (cid:105) , (cid:104) ma , ma , ma (cid:105) ; (cid:104) cc , cc , cc (cid:105) , P ) ,therefore cc i are incident to P ,Desargues ( P, { A i } , { CA i } ; (cid:104) CC i (cid:105) , p ) , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY therefore caa i × a i are incident to p ,Desargues − ( cc , { ma , cc , a } , { ma , cc , a } ; (cid:104) caa , c , a (cid:105) , CC ) therefore caa , c and a are incident to CC .For the second part, the Theorem of Pascal implies that the points M , M , M , M , M ,M , are on a conic, because the points C , C and C are collinear.The conic may degenerate in two lines. This will occur if, for instance, M is on M × M and in this case M is on M × M . Indeed,
Theorem.
Let A , A , A , M and M be a complete 5-angle such that M , M and M are collinearthen M , M and M are collinear. Proof: A synthetic proof is as follows. Let E := M × ( M × M ) , the Theorem of Pappusapplied to A j M M j and A M M for j = 1 and implies that M , E, M j are collinear,therefore M M and M are collinear. Theorem. If m = [1 , ,
1] and m = [ m , m , m ] , then with respect to the line conic a x + a x + a x + 2( a x x + a x x + a x x ) = 0the pole of m is( a + a + a , a + a + a , a + a + a )and the pole of m is( a m m + a m m + a m m , a m m + a m m + a m m ,a m m + a m m + a m m ) . Introduction.
I will now give a construction associated to a conic inscribed in a triangle. The degeneratecase of the Theorem of Brianchon implies that if
J J i are the points of contact on A i +1 × A i − , then the lines A i × J J i pass through a point J. We can choose arbitrarily a point I or itspolar i. The construction in Theorem 3.12 determines a pair of points M and M which in thecase of Euclidean geometry will correspond to the barycenter and to the orthocenter. As willbe seen later, the function which associates M, M to J, I is not one to one. It is thereforenecessary to start with this construction if we want to extend to projective geometry that partof the geometry of the triangle which is related to the inscribed circles. In this case, Part 0.should precede Part 1.
Theorem.
Given a complete 5-angle, 5 distinct points, no 3 of which are on the same line, A , A , A ,J and I, A i are called the vertices, .1. AN OVERVIEW OF THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. J, is the point of Gergonne and I, is the center of the inscribed circle .
0. The barycenter and orthocenter derived from the point of Gergonne and the center ofthe inscribed circle.LetH0.0. A i , (See Fig. 20b)H0.2. J, I,
D0.0. a i := A i +1 × A i − , D0.8. ja i := J × A i , D0.9.
J J i := ja i × a i , D0.10. j i := J J i +1 × J J i − , D0.11.
J a i := j i × a i , D0.23’. ji i := J J i × I, D0.26.
J ia i := ji i +1 × a i − , J ia i := ji i − × a i +1 , D0.27. jia i := J ia i +1 × J a i − , jia i := J ia i − × J a i +1 , D0.28.
J ai i := jia i +1 × j i − , J ai i := jia i − × j i +1 , D20.0.
J i i := jai i +1 × jai i − , D20.22. ι := conic ( J J , J J , J J , J i , J i )( ∗ ) , D0.5’. m i := J ai i × J ai i , D0.6.
M M i := m i +1 × m i − , D0.1’. ma i := A i × M M i , D0.4’.
M A i := m i × a i , D0.7. m := M A × M A ( ∗ ) , D0.H. M := ma × ma ( ∗ ) , D0.25’.
IM a i := m × ji i , D0.1’. ma i := A i × Ima i , D0.H. M := ma × ma ( ∗ ) , thenC0.2. a i · ι = 0 . C0.5. m i · A i = 0 . C20.3. ι is a circle = 0 . C20.4. I = Center ( ι ) . The nomenclature:N20.3. ι is the inscribed circle ,N0.12. J J i are the Gergonnian points , these are the points of contact of the inscribed circlewith the sides of the triangle.
J a i is the pole of ja i with respect to the inscribed circle . J i i is the point of the inscribed circle diametrically opposite to J J i . Again, M is the barycenter and M is the orthocenter.Proof:For a synthetic proof see section . . . G272.tex.Let A = (1 , , , A = (0 , , , A = (0 , , , J = ( j , j , j ) , I = ( i , i , i ) ,m is constructed in such a way that I is the pole of m with respect to ι , therefore if the line m is chosen to be [1 , , , then CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY I = ( j ( j + j ) , j ( j + j ) , j ( j + j )) , therefore there is no loss of generality if we set1. i := j ( j + j ) , i := j ( j + j ) , i := j ( j + j ) . We will use the abbreviations forsymmetric functions of j , j , j using “p“ instead of “s” as used for the symmetricfunctions of m , m , m . For instance,2. p = j j + j j + j j . P0.0. a = [1 , , . P0.8. ja = [0 , j , − j ] . P0.9.
J J = (0 , j , j ) . P0.10. j = [ − j j , j j , j j ] . P0.11.
J a = (0 , j , − j ) . P0.23. ji = [ j j ( j − j ) , j j ( j + j ) , − j j ( j + j )] . P0.26.
J ia = ( j ( j − j ) , j ( j + j ) , , J ia = ( j ( j − j ) , , j ( j + j )) . P0.27. jia = [ j j ( j + j ) , j j ( j + j ) , j j ( j − j )] ,jia = [ j j ( j + j ) , j j ( j − j ) , j j ( j + j )] . P0.28.
J ai = ( j ( j + j ) , − j j , j j ) , J ai = ( j ( j + j ) , j j , − j j ) . P20.0.
J i = ( j ( j + j ) , j j , j j ) . P20.22. ι : j j X + j j X + j j X X − j j j ( j X X + j X X + j X X ) = 0 .j − θ + ι = − i ×× [ j j , j j , j j ] .ι − : j j x x + j j x x + j j x x = 0 . P0.25.
IM a = ( j ( j + j ) , j ( j − p ) , j ( j − p )) . P0.12. ma = [0 , j ( j − p ) , − j ( j − p )] . P0.16. M = ( j ( j − p )( j − p ) , j ( j − p )( j − p ) , j ( j − p )( j − p )) . For m i , M M i , ma i , M A i , m and M, see 3.7.The following relations are useful in the derivation of some of the formulas either given aboveor given below:0. m = j ( j − p )( j − p ) m + m = − j ( j + j ) ( j − p ) , m ( m + m ) = ( j ( j + j )) jp, with3. jp = − ( j − p )( j − p )( j − p ) m − m = − ( j − j )( j − p )( p + j j ) m ( m − m ) = − j ( j − j )( p + j j )( j − p ) s = 4 j j j p , s + m = j ( ... ) , j m − j m = ( j − j )( j − p ) p , m m = − j j ( j − p ) jp. G27.TEX [MPAP], September 9, 2019 .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES.
Section . . . contains a synthesis of a very large number of Theorems in Euclidean Geometry,using the presentation introduced in section i. This is followed by a proof given also as insection 1.The set of Theorems includes some which are always valid, some which are valid when thegiven triangle has a tangent circle and some which are valid when the point of Steven exits.In the second case I indicate that the definitions and Theorem are meaningful by labeling thesection with (J). In the third case I label the section with (Mu), if neither case apply I labelthe section with (M). Definitions and conclusions contained in sections without (M), (J) or(Mu) are always meaningful.I start with a triangle { A i } .In case (M), I choose the barycenter M and the orthocenter M .In case (Mu), I choose the barycenter M = (m , m , m ) and the point of Steven, M u = ( √ m , √ m , √ m ) . This assumes that k m , k m and k m are quadratic residuesfor some k . I then determine the orthocenter from M and M u .In case (J), I assume that the triangle has a tangent circle and derive the orthocenter fromthe barycenter and the point of Gergonne J . The point J , if it exist is such that the con-structions obtained by use J instead of J give eventually M instead of M and vice-versa. Ifthe J does not exist these constructions are meaningless. The construction in the rightmostcolumn of the sections marked with (J) should therefore be ignored.At the end of section 0, whatever the variant, the ideal line and orthic line have been con-structed as well as the medians and altitudes, mid-points, the feet, and the complemantaryand anticomplematary triangles.In section 1, we construct the line eul of Euler, the point K of Lemoine, the circumcircle θ and the circle γ of Brianchon-Poncelet. Hypothesis . . .Because in finite geometry If the section starts with (M), (J) or (Mu), it is only to Inthis Chapter, I will give systematically most of the results which generalize the known resultsof the geometry of the triangle in classical Euclidean geometry. In 5.1. and in 5.4. thecorresponding constructions can be done with the ruler alone. In 5.5. the correspondingconstructions in classical Euclidian geometry would require also the compass. ( CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem.
Given a complete 5-angle, 5 distinct points, not 3 of which are on the same line, A , A ,A , M and M ,
The vertices A i are those of a triangle, M is the barycenter and M is theorthocenter . Proof of Theorem 5.1.1.
The algebraic proof will be summarized by giving the coordinates of the points and linesconstructed in 5.1.1. The incidence properties follow from straightforward computation ofscalar products or substitution in the equation of the conics.For triples, the coordinates of the 0-th subscript will be given. The coordinates of subscript1 and 2 are obtained by applying the mapping ρ and ρ to it. ρ is defined as follows,we substitute m , m , m for m , m , m in each of the components, and rotate these, the0-th coordinate becoming the first, the first coordinate becoming the second and the secondcoordinate becoming the 0-th. For instance, from em of 5.0. we get em = [ s + m , m − m , − ( s + m )] and em = [ − ( s + m ) , s + m , m − m ] . The hypothesis imply, m (cid:54) = 0 , m (cid:54) = 0 , m (cid:54) = 0 , m (cid:54) = m , m (cid:54) = m and m (cid:54) = m . I will use the usual abbreviations for the symmetric functions, s := m + m + m , s := m m + m m + m m ,s := m + m + m , etc. and q := m − m m , q := m − m m , q := m − m m , and the identity ( m + m )( m + m )( m + m ) = s + 2 s . The dual of the reciprocal of all elements have also been included.
Comment.
To determine in succession the homogeneous coordinates, we have used the definition. Tocheck the results, if for instance x := P × Q, we can simply verify x · P = x · Q = 0 . Theconstruction asserts implicitly that for x := P × Q, in general, P and Q are distinct, in otherwords for some value of p and some M , P and Q are distinct. It may of course happen thatfor a particular example P = Q, x, for this example,are not all 0, this means that some alternate construction, using for instance one of theconclusions, will determine x, in the other case, x can not be constructed. For instance, for p = 37 and M = (202) = (1 , , , Ste = AA = (880) , but stAA = (472) = (1 , , is well defined but cannot be obtained using Ste × AA . On the hand, for any p, if M =(1 , p − , p − , F = M = (0 , , and f m = (0 , , and is therefore undefined. .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Comment.
The determination of a conic, with known intersections X , X with a , Y1, Y2 with b and Z , Z with a , can be obtained easily. Introduction.
The following pairs of point can not be obtained by the construction involving only intersectionof known lines or lines through known points, they are sufficiently important to be defined.
Definition.
I, I (cid:48) = m × γ, I, I (cid:48) = m × γ .The first pair corresponds to the isotropic points , the second pair to the co-isotropic points . Theorem.
With the definitions of Theorem 5.1, we have
I, I (cid:48) = ( m ( m + m ) , − m ( m + jσ ) , − m ( m − jσ )) ,j = +1 or − , σ := √− s ,I, I (cid:48) = ( m ( m + m ) , − m m − jτ, − m m + jτ ) , where j := +1 or − , τ := √− m m m s . Definition. F is a focus of a non degenerate conic iff both F × I and F × I are tangent to the conic. Theorem.
If the conic is not a parabola, there are 4 foci, real or complex.
Definition.
Two directions IA and IB are perpendicular iff one direction is on the polar of the otherwith respect to any circle. We will write IA ⊥ IB.
Theorem.
Let ( X , X , X ) be an ideal point, the perpendicular direction is( m ( m X − m X ) , m ( m X − m X ) , m ( m X − m X )) , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem. ( X , X , X ) and ( Y , Y , Y ) are perpendicular directions if m m X Y + m m X Y + m m X Y = 0 . Theorem.
The following are perpendicular directions. Let D0.
Imeul = ( s − m , s − m , s − m ) , thenC0. M A i ⊥ Ima i . C1. Im i ⊥ Im i . C2. I ⊥ I, I (cid:48) ⊥ I (cid:48) . C3.
EU L ⊥ Imeul.
See also C12.4, C12.5, C16.7,
Exercise.
Construct
Imeul of the preceding Theorem.
Introduction.
Besides the properties given in Theorem 5.1.1., many other properties of Euclidean geometrygeneralize to projective geometry. These will now be stated. The numeration started inTheorem 5.1.1. is continued.
Theorem.
Given the hypothesis of Theorem 5.1.1. and the points and lines defined (or constructed) inthat Theorem.
Notation.
To make some of the algebraic expression less cumbersome, we have often used the symmetricfunctions s := m + m + m ,s := m m + m m + m m ,s := m + m + m ,s := m ( m + m ) + m ( m + m ) + m ( m + m ) . and similarly in the equations for conics other symmetric functions. We have also used, attimes, .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. q := m − m m , q := m − m m , q := m − m m , and the following identities in the calculations: m q + m q + m q = 0 andm q + m q + m q = 0 . Proof of Theorem 5.4.1..
The proof is given in the same way as the proof of 5.1.1.
Proof of Theorem 5.4.3., P.15.16..
The details of the proof to obtain the equation of the circle of Brocard will now be given. Theequation of a conic which has the radical axis m with the circle γ is0. m ( m + m ) X X + m ( m + m ) X X + m ( m + m ) X X +( X + X + X )( u x + u X + u X ) = 0 .u , u and u are determined in such a way that the conic passes through Br i . For Br we have X + X + X = 4 m m + m m + m m , and for the first line of 0. (4 m m + m m + m m )( m m ( m + m )( m + m )) . Hence we have to solve m m u + m ( m + m ) u + m ( m + m ) u + m m ( m + m )( m + m ) = 0 ,m ( m + m ) u + 2 m m u + m ( m + m ) u + m m ( m + m )( m + m ) = 0 ,m ( m + m ) u + m ( m + m ) u + 2 m m u + m m ( m + m )( m + m ) = 0 . Replacing u , u and u in terms of v , v and v , given by v u m m , v u m m , v u m m , weget m m v m ( m + m ) v m ( m + m ) v m + m )( m + m ) = 0 ,m ( m + m ) v m m v m ( m + m ) v m + m )( m + m ) = 0 ,m ( m + m ) v m ( m + m ) v m m v m + m )( m + m ) = 0 . The determinant is D = − s + 2 s . The numerator for v is E ( m + m )( m + m ) with E = ( s − s ) . Hence the solution for u . To obtain P15.16., we have to determine u + u + m ( m + m ) = m ( m + m )( D + Es + Em m ) D = m ( m + m )( − s + s + Em m ) D , because Es = s s − s s . But ( s − s ) s = 3 s − s , hence the equationfor PUb.To obtain the relation between θ and β , knowing that Aθ + β = B ( m ) ×× ( lem ) , it is easy to obtain A and B, for instance, K.β gives A (3 m m m ( m + m )( m + m )( m + m ))= B (2 s m m m ( m + m )( m + m )( m + m )) therefore with B = 1 , A = 2 s . ADD PERPENDICULARITY, e.g.
IiI i ⊥ Iai i . Cross refer. at end of G2705 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Introduction.
I will now give a construction associated to a conic inscribed in a triangle. The degeneratecase of the Theorem of Brianchon implies that if
J J i are the points of contact on A i +1 A i − , then the lines A i × J J i pass through a point J. We can choose arbitrarily a point I or its polar i. The construction in Theorem 5.5.1. determines a pair of points M and M which in thecase of Euclidean geometry will correspond to the barycenter and to the orthocenter. As willbe seen later, the function which associates M, M to J, I is not one to one. It is thereforenecessary to start with this construction if we want to extend to projective geometry thatpart of the geometry of the triangle which is related to the inscribed circles. Part 0. shouldtherefore precede Part 1. of Theorem 5.1.1. Part 20. given next follows Part 19. of Theorem5.1.1.IN THE NEXT SECTION REVERSE THE ORDER. FIRST SHOW THAT THE pointdiametrically opposed to
J J is on the line m × j × J J , then that m × j is on the line j × a and ij × a STUDY from which it follows that m × j defines m with A and can be obtained from J J i . Heuristics.
Before giving the construction I will look back at Euclidean geometry and determine propertieswhich have guided me in the construction given below. Let I be the center of the circle ι inscribed in the triangle ( A , A , A ) , let J J i be the point of contact with a i , let m i be theparallel to a i through A i . First, if
J a := j × a , J ia := a × ji , and J ai := j × m , then J a , J ia and J ai arecollinear because they are the Pascal points of the hexagon with cords or tangents j , a , j ,a , ji , jai . If we start from A i , J and I we can therefore construct J J i , J a , J ia , J ai and m , hence M A := a × m , similarly we can construct M A and therefore the ideal line m := M A × M A . (The construction below is a variant which uses a “symmetric” point J ai also on m . ) From m we can derive the barycenter M as the polar of m with respect tothe triangle { A i } . Next, the conic through
J J i with tangent a , a can be defined as a circle,the altitude ma can be obtained as parallel to I × J J and therefore the orthocenter M canbe constructed.Finally, let jai := J ai × J J and J i := jai × ji , I claim that
J i is on the inscribedcircle. Indeed, first the triangles ( J J i − , A i , J J i +1 ) are isosceles triangles, then, for i = 0 , the triangle ( J J , A , J ai ) which is similar to the triangle ( J J , A , J J ) is therefore anisosceles triangle and | A , J ai | = | A , J J | = | A , J J | . Therefore angle ( A , J J , J ai ) = ( π − angle ( J J , A , J ai )) = angle ( A , A , A ) = angle ( A , A , I ) , therefore j is parallelto A × I and therefore perpendicular to j , it follows that ( J i , J J ) is a diameter.We can therefore construct J i on ι . .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Proof of Theorem 5.5.2.. Asyntheticproof of . . . isasf ollows.P ascal (cid:48) sT heoremgivesct ( ji , a , j , a , j , jai ?) = ( J ia , J a ) ⇒ J ai ⇒ jai ⇒ J i .ct ( ji , a , j , a , j , jai ?) = ( J ia , J a ) ⇒ J ai ⇒ jai ⇒ J i , hence 0.0 and 0.2. ct ( j , a , jai , jai , a , j ) = ( J ai , A , J ai ) , which are therefore collinear, hence 0.3. ct ( jai , j , a , j , jai , tangent ( J i )) =( J ai , J ai , M A ) , hence 0.4. Im i is the pole of ji i , therefore i is the polar of I, hence 0.5.The coordinates of the various points are easy to derive. Let A = (1 , , , A =(0 , , , A = (0 , , , J = ( j , j , j ) , I = ( i , i , i ) ,m is constructed in such a way that I is the pole of m with respect to ‘ ι , therefore ifthe line m is chosen to be [1 , , , then1. I = ( j ( j + j ) , j ( j + j ) , j ( j + j )) , therefore there is no loss of generality if we set2. i := j ( j + j ) , i := j ( j + j ) , i := j ( j + j ) . We will use the abbreviations for symmetric functions of j , j , j using “p” instead of“s” as used for the symmetric functions of m , m , m . For instance,3. p = j j + j j + j j . We have also expressed the coordinates in terms of i , i and i . The symmetric functions of i , i and i use “o” instead of “s”. The expression of j , j and j in terms of i , i and i is given by4. j = ( o − i )( o − i ) ip , . . . , where ip = 2( o − i )( o − i )( o − i ) . This alternate notation has the advantage thatthe information on the associate construction for the excribed circles is obtained byreplacing either i by − i , or i by − i , or i by − i . Proof of Theorem 5.5.2., P21.8.
The proof or the preceding theorem is straightforward, I will only give details for the deter-mination of π : Let C be the symmetric matrix associated to the polarity of pi , let M be thematrix whose i-th column are the coordinates of m i , let J be the matrix whose i-th columnare the coordinates of M na i , let K be a diagonal matrix of unknown scaling factors k , k ,k . CJ = MK or C = MKJ − expresses the fact that m i is the polar of M na i . J − = p j j − p j j − p j j − p p j j − p j j − p j j − p p . The problem is now reduced to a set ot 3 homogeneous equations in the unknowns k , k , k , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY which express the symmetry of C , namely, after simplification, − j j k + j j k + j ( j − j ) k = 0 ,j ( j − j ) k − j j k + j j k = 0 ,j j k + j ( j − j ) k − j j k = 0 , giving k = j j , k = j j , k = j j . Comment.
The following alternate definition for Nagel’s point which is clearly more clumsy: oi := O × I,Ioi := i × oi,nm := M × Ioi,N := nm × mi, We have oi = [ j j ( j − j )( j + j )( j + j ) , j j ( j − j )( j + j )( j + j ) ,j j ( j − j )( j + j )( j + j )] ,Ioi = ( j ( j + j )( p − j p ) , . . . ) ,nm = [ j ( j − j )( j − p ) , . . . ], done backward from N. We also have mj = [ j − j , j − j , j − j ] , Comment.
An alternate method to obtain quickly the relation between the barycentric coordinates of thepoint of Gergonne and of the orthocenter is as follows.Let n = m ( m + m ) , . . . , we known that are circles are n x x + . . . − ( X + X + X )( u X + . . . ) = 0 . the line equation of the inscribed circle is j j x x + . . . = 0 to express that it is a circle we can use A = 2 adjoint ( B ) , where A is the polarity matrix associated to the general circle and B the matrix associatedto (2). The constant is arbitrary and reflect the chosen scaling. A = − u n − u − u n − u − u n − u − u − u n − u − u n − u − u n − u − u − u , B = j j j j j j j j j j j j . This gives at once u = ( j j ) and n = 2 j j j + u + u = ( j ( j + j )) . Comment.
To obtain the points of contact of the outscribed circle ι , Let J be the corresponding pointof Gergonne [ g , g , G ] . We have g ( g + G ) = − j ( j + j ) ,g ( G + g ) = j ( j + j ) , .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. G ( g + g ) = j ( j + j ) , adding 2 equations and subtracting the third gives g G = p , G g = − j j , g g = − j j , with p = j j + j j + j j . Hence with an appropriate constant of proportionality, ( g , g , G ) = ( − j j j , j p , j p ) . Therefore the points of contact with a , a and a are N a = (0 , j , j ) , N ai = ( j j , , − p ) , N ai = ( j j , − p , . Theorem.
The isogonal transformation of J is isog ( J ) = ( j ( j + j ) , j ( j + j ) , j ( j + j ) ) , − j j ( j + j )( p j )) , Proof: x = [ − j j , j j , j j ] ,X = ( j ( j + j ) , j ( j + j ) , − j ( j + j )) ,Y = ( j ( j + j ) , j ( j + j ) , j j ) ,y = [ j j ( j + j )( j + j ) , j j ( j j − j j + 3 j j + j ) ,Z = ( j ( j + j − j j − j j ) , j ( j + j ) , j ( j + j ) ) ,z = [0 , j ( j + j ) , − j ( j + j ) ] , hence the Theorem. Definition.
Many other constructions can be easily derived from the following operation called the dualconstruction . Instead of the quintuple A i , M, M , consider instead the quintuple A i , M,M (cid:48) := T mm.
The construction associated to every point X = ( X , X , X ) , a point X (cid:48) whose coordinatesare the reciprocal X (cid:48) = ( X X , X X , X X ) and to every line x = ( x , x , x ) the reciprocal x (cid:48) = ( x x , x x , x x ) . A few of the dual points and lines are not new but most are and lead easily to the constructionof important points and lines. See for instance the exercise on the line of Longchamps. Wehave ma (cid:48) i = ma i , M (cid:48) i = M i , m (cid:48) i = m i , Im (cid:48) i = Im i , AC (cid:48) i = AC i , i (cid:48) = i, O (cid:48) = K, oa (cid:48) i = at i ,S (cid:48) = S, Ima (cid:48) i = Ima i , K (cid:48) = O, ok (cid:48) = ok. An example is given in ex3.3.
Corollary.
We can now summarize incidence properties associated with the historically important lineof Euler and circle of Brianchon- Poncelet.
0. The following 14 points are on the line of Euler: the barycenter M, the orthocenter M , the point
P P of D3.3, the center EE and cocenter EE of the circle of Brianchon-Poncelet, the center O and cocenter O of the circumcircle, the points Am, Am of D7.9,the points D i of D8.4, the center G and the cocenter G of the orthocentroidal circle. CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
1. The following 24 points are on the conic of Brianchon-Poncelet: the midpoints M i , thefeet M i , the Euler points E i , E i , the points F i and F i of D6.2, points of Schr¨oter S and S, the points of Feuerbach, F e and
F e i .The complete set of incidence properties are given in detail in section 5.7. Comment.
Given the algebraic coordinates of a point it is sometimes difficult to obtain a constructionstarting from M and M .
One additional tool is provided by using homologies. We will givehere an example, which allows the easy construction of other points on the line of Euler.
Definition. A barycentric homology is a homology with center M and axis m. Example.
One such homology and its inverse is D = , D − = − − − . Theorem.
The transforms of the 14 points on the line of Euler, given in 5.10.0 are as follows: D ( M ) = M, D ( M ) = O, D ( P P ) = ( q + q , q + q , q + q ) , D ( EE ) = (3 s − m , s − m , s − m ) , D ( EE ) = (3 s − m m , s − m m , s − m m ) , D ( O ) = EE, D ( O ) = ( s − m ( m + m ) , s − m ( m + m ) , s − m ( m + m )) , D ( Am ) = M , D ( Am ) = ( s − s − m s , s − s − m s , s − s − m s ) , D ( D ) = ( m ( m + m ) − m m , m ( m + m ) − m m , m ( m + m ) − m m ) , D ( G ) = (5 s − m , s − m , s − m ) , D ( G ) = ( s − s − m s , s − s − m s , s − s − m s ) , Exercise.
Complete the table for the inverse transform, D T-1T ( M ) = M, D T-1T ( M ) = ( s − m , s − m , s − m ) . Observe that D T-1T ( M ) .m = 0 . .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Introduction. . . .
Theorem. If m + m , m + m and m + m are all quadratic residues or or non quadratic residue,then both the dual of the inscribed circle and the symmetric of the inscribed circle are real.Moreover, if i and j are the dual of I and J and if I and J are the symmetric of I and J then i = [ √ m + m , √ m + m , √ m + m ] , j = [( − i + i + i ) − , ( i − i + i ) − , ( i + i − i ) − ] and2. I = ( m i , m i , m i ) , J = ( m j , m j , m j ) , Proof:For the symmetric case, I × ( a × ( A × J )) , A × M A and m are concurrent, moreover I is the pole of m with respect to ι .Therefore,4. J ( I m ( m + m ) + I m m ) = J ( I m m + I m ( m + m )) , and in view of P0.15,5. I = J J m m + J J m m . This relation and the 2 others obtained by circularitygive − I m m + I m m + I m m = 2 J − m Using 4, we get I
20 ( m − m ) m − I
21 ( m + m ) m + I
22 ( m + m ) m = 0 , as well as 2 other similar equations. These equations are compatible and give using theminors ( I m ) = m + m , ( I m ) = m + m , ( I /m ) 2 = m + m . For the dual case, it follows from .1 and .6 (G2722), that the coordinates of I are proportionalto (cid:112) m ( m + m ) , . . . , those of the dual are obtained by replacing m by m m , . . . . Theorem. If m , m and m are all quadratic residues or all non quadratic residues, then the dual ofthe symmetric of the inscribed circle is real. Morover if j and i are the dual of the symmetricof J and I, then14 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
0. 0. i = [ (cid:113) m + m m , (cid:113) m + m m , (cid:113) m + m m ] , and j = [( m ( − m i + m i + m i )) − , ( m ( m i − m i + m i )) − , ( m ( m i + m i − m i )) − ] . Example.
For p = 29 , if M = (60) and M = (258) = (1 , , , (1 , m + m m + m , m + m m + m ) = (1 , − , − , with achoice of the square roots, i = 1, i = -12, i = -14, hence i = [538] and j = [1 , − , −
9] =[833] . Moreover I = (1 , , −
2) = (144) , and J = (1 , − ,
7) = (472) . ( m , m , m ) =(1 , , , with a choice of the square roots, ( √ m , √ m , √ m ) = (1 , , , hence i = [1 , − ,
3] =[816] and j = [1 , , −
14] = [248] . then J = (164) , I = (448) , I = (144) , J = (472) , and i = [538] , j = [833] , i = [816] , j = [248] . Introduction.
The incidence properties of points, lines and conics will now be summarized. There areseveral reasons for doing this. First, having so many elements, it is difficult to keep in onesmind at any one time all of the properties given above. Second, it is important to insurethat the elements obtained are in general distinct. Third, it is important to obtain from theelements defined any incidence properties not already discovered. For this purpose, I createa program, which, for given examples, determine all incidence properties, by comparison, itwas possible to eliminate a few incidence properties which were peculiar to a given example,for the others attempting an algebraic proof determined if the incidence property was indeedgeneral. Quite a few new Theorems were obtained in this way. They have been indicated by(
Theorem.
The incidence properties are as follows:
Proof of Theorem 6.1.1.Exercise.
Construct the vertical tangent of the parabola of Kiepert and prove that it is[ m ( m − m )( s − m )( s − m ) , m ( m − m )( s − m )( s − m ) ,m ( m − m )( s − m )( s − m )] . Exercise.
Construct the conic of Jerabek (Vigarie, N.99), m ( m − m ) X X + m ( m − m ) X X + m ( m − m ) X X = 0 . .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Comment.
There exist a large number of conditional theorems. For instance, if s + 12 s = 0 then G · i = G · i = 0 . An example is provided by p = 29 , m = 1 , m = 6 , m = 11 , corresponding to J = 94 ,I = 315 . Exercise.
The line of Simson.LetD.0. Y i := X × Im i , D.1. y := Y × Y , H.0. X · θ = 0 , thenC.0. Y · y = 0 . P.0. Y = () , P.1. y = [(( m + m )( X + X ) − m X ) X X , (( m + m )( X + X ) − m X ) X X , (( m + m )( X + X ) − m X ) X X ] . Exercise.
The excribed circles.
Let iii[i] := radical axis(iota[i+1],iota[i-1]),theniii[i]\cdot En = 0.Ex g277, iii[] = [647,435,847],Ex4.0, iii[] = [873,651,964],Ex5.0, iii[] = [723,837,965].The conic of Neuberg. (Mathesis, Ser.2, Vol.6, p. 95).P40.2. ‘Neuberg: m_0m_1m_2(m_0 X_0^2 + m_1 X_1^2 + m_2 X_2^2)+ s11( m_0(m_1+m_2)X_1X_2 + m_1(m_2+m_0)X_2X_0+ m_2(m_0+m_1)X_0X_1) = 0.(Bastin, Mathesis, p.97)
Exercise. (Neuberg, see Casey, no 80,81,82)The barycenter of the triangle { Ste, BRa, Abr } (see D16.5,14,16) is M . BRa × Abr =[ m m ( m − m )( m − m ) , m m ( m − m )( m − m ) , m m ( m − m )( m − m )] . Exercise.
Some points on the circumcircle.Construct the points
M iqm i on θ and ac i distinct from A i and the points M iqm i on θ and CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY on ac i distinct from A i . Answer to (partial).
M iqm [0] = ( m ( m + m ) , m ( m − m ) , m ( m − m )) ,M iqm [0] = ( m + m , m − m , m − m ) , Example. p = 29 , A i = (30 , , , M = (60) , M = (215) ,M iqm i = (545 , , , M iqm i = (115 , , . Exercise.
The point of Miquel.Given an arbitrary line which does not pass through the vertices and is neither the ideal orcoideal line, q = [ q , q , q ] , Let Q i := q × a i . Determine µiq i := conic ( I (cid:48) , I (cid:48)(cid:48) , A i , Q i +1 , Q i − ) ,µiq i := conic ( I (cid:48) , I (cid:48)(cid:48) , A i , Q i +1 , Q i − ) , Construct the point
M iq which is on µiq i and θ, the point M iq which is on µiq i and θ, the circle µq of M iquel which passes through the center of µiq i and the cocircle µq which passes through the center of µiq i . The following special cases are of interest: q = m, in which case M iq = M iq, which we denote
M iqi,q = e, we then denote the point and copoint of Miquel by M iqe and
M iqe,q = m i , giving M iqm i of Exercise ¡...¿ above, q = m i , giving M iqm i of Exercise ¡...¿ above. Answer to (Partial) miq = ( q − q q − q m ( m + m ) X X + ... )+ ( X + X + X )( m ( m + m ) q q − q X + m ( m + m ) q q − q .miq = ( m q − m q m q − m q m ( m + m ) X X + ... ) + ( m m X + m m X + m m X )(( m + m ) q m q − m q X + ( m + m ) q m q − m q X ) = 0 .M iq = ( m ( m + m ) q q q − q q − q , m ( m + m ) q q q − q q − q , m ( m + m ) q q q − q q − q ,M iq = (( m + m ) q q m q − m q m q − m q , ( m + m ) q q m q − m q m q − m q , ( m + m ) q q m q − m q m q − m q ,M iqi = (( m + m )( m − m )( m − m ) , ( m + m )( m − m )( m − m ) , ( m + m )( m − m )( m − m )) ,M iqe = ( m ( m + m )( m − m )( m − m )( m − m + m )( m + m − m ) , m ( m + .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. m )( m − m )( m − m )( m − m + m )( m + m − m ) ,m ( m + m )( m − m )( m − m )( m − m + m )( m + m − m )) , M iqe = (( m + m )( m − m )( m − m )(2 m m − m ( m + m ))(2 m m − m ( m + m )) , ( m + m )( m − m )( m − m )(2 m m − m ( m + m ))(2 m m − m ( m + m )) , ( m + m )( m − m )( m − m )(2 m m − m ( m + m ))(2 m m − m ( m + m ))) . Exercise. (Sondat, See Mathesis, Ser. 2, Vol.6, pp. 81-83)Let B · µiq = 0 , letD.0. b := B × Q , b := Q × B , D.1. B := muiq × b − Q , B := muiq × b − Q , D.2. b := B × B , D.3. ab i := A i × B i , D.4. S := ab × ab , Let S · muiq = 0 , S · µiq = 0 , D.5. sa := S × A , sa := S × A , D.6. T := sa × sa , D.7. sa := T × A , D.8. S := muiq × sa − A , D.9. σ := conic ( S i , S, T ) , C.0. Q [0] · b = 0 . C.1. S · ab [0] = 0 . C.2. S · θ = 0 . C.3. σisacircle,
C.4. T = M ==¿ center( σ ) · q = 0 . ¡...¿ double check the above. Exercise.
Construct the point common to the circumcircle and the circle through A i , M i +1 and M i − . (See also the transformation of Hamilton. Partial Answer to a) The center is E i ,x i := E i × Imm i ,y i := O × Im i ,Z i := x i × y i ,z i := A i × Z i ,z i corresponds to the perpendicular to O × E i , hence contains the desired intersection HH i .x = [ m m ( s + m ) , − m (2 m s + m ( m + m )) , − m (2 m + 3 m m + m s )] ,y = [ − ( s + m ) , m + m , m + m ] ,Z = (( m − m )( m + m ) , m ( s m )( m +2 m ) , − m ( s m ) s , z = [0 , m s , m ( m +2 m )] , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
The rest of the construction is that of Pascal: aa i := A i +1 × AA i − ,ZZ i := aa i × z i ,zz i := Im i +1 × ZZ i − ,Y i := a i +1 × zz i − ,yy i := Y i +1 × AA i − ,M iqm i := yy i +1 × z i . Exercise.
0. Complete a section on the conic of Nagel, with ν := conic ( J J , J J , J J , N a , N a ) ,ν i := conic ( . . . ) ,
1. Give a construction for the other intersection of the conic with
J a i and N a i .
2. Give a construction for the center of the conic.3. Are there some other points on this conic which have already been constructed or thatyou can construct?
Partial Answer to µ = j j j ( X + X + X ) − j ( j + j ) X X − . . . = 0 , ¡...¿ not checkedother intersection with J a = ( p + j ) , j j , j j ) , other intersection with N a = ( p + j j , j j j , j j j ) , center (. . . ?) ( j ( j ( j − j )+ j ( j − j )+ j j (2 j +3 j +3 j )+ j j ( j +3 j )+ j j (3 j + j )) , . . . ) Exercise.
The circles of Lemoine-Tucker.D.0. X i := K − xA i , x is some integer,D.1. x i := X i +1 × X i − , D.2. XX i := x i +1 × a i − , D.3. XX i := x i − × a i +1 , D.4. ξ := conic ( XX , XX , XX , XX , XX ) , thenC.0. x i · Im i = 0 . C.1. ξ · XX = 0 . C.2. ξ is a circle.D.0., can be replaced by a construction which start with a point X on the symmedian at , the parallel through X to the side a or a intersect the symmedians at or at at X or X . Proof.P.0. X = ( m ( m + m ) + x, m ( m + m ) , m ( m + m )) , P.1. x = [ m ( m + m ) , m ( m + m ) , − s − m m + x ] , P.2. XX = ( s + m m − x, m ( m + m ) , , P.3. XX = ( s + m m − x, , m ( m + m )) , .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. P.4. (2 s − x ) θ − ( m ) ×× ( u ) = 0 , with u = m m ( m + m )( m + m )( s + m m − x ) , . . . . Comment.
The following are special cases: x = 0 gives the first circle of Lemoine lambda1, x = s gives the second circle of Lemoine lambda2, x = . . . gives the circle of Taylor, x = 2 s gives the degenerate circle ( i ) ×× ( i ) ,x = gives θ. Exercise.
Definition.
Given a conic θ and a point K not on the conic, an inscribed polygon A i , i = 0 , ...n − is a harmonic polygon if ( A i − , A i , A i +1 , A (cid:48) i ) is harmonic for all i , where ka i := K × A i ,A (cid:48) i := θ × ka i − A i ,k := polar ( K ) ,B i := polar ( ka i ) ,K is called the point of Lemoine of the polygon, k is called the line of Lemoine. Theorem. If A i , i = 0 to n − A (cid:48) i , i = 0 to n − Construction.
Given
K, A , A , construct k, ka , B , ka , B , for i = 1 to n − begin ka i := polarA i , B i := ka i × k, c i := A i − × B i , A i +1 := θ × c i − A i − , end Construction.
Details.H.0. x ( A − B ) = y ( A − B ) + B A − A B H.1. x + y = 1 , thenC.0. y (( A − B ) +( A − B ) )+2 y ( A − B )( B A − A B )+( B A − A B ) − ( A − B ) = 0 , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
C.1. y = − A B )( B A − A B ) / (( A − B ) + ( A − B ) ) − A . CHECK THE ABOVE ¡...¿
Example.
For p = 31 , let K = (0 , − , , A = (1 , , , A = () , then A i = (1 , , , Exercise.
Complete a section on polars of the vertices with respect to the conic of Brianchon-Poncelet.0. Give an explicit construction for the tangents to gamma at the mid-points and at thefeet.1. Give an explicit construction for the polar pp i of A i with respect to γ.
2. Verify that the intersections
P P i := pp i × A i are collinear on pp. This result can be used as the starting point for special results in the geometry of the tetrahe-dron. (An other approach is suggested by the theorem ¡...¿). The lines a i and pp correspondto the ideal lines in the four faces of a tetrahedron whose opposite vertices are perpendicular.The tetrahedron so obtained have the additional properties that A i × M i are concurrent aswell as A i × M i . Comment.
An other model of projective geometry within projective geometry is suggested by the follow-ing.Associate to the point ( X , X , X ) , the point ( X X , X X , X X ) , associate to the line [ a , a , a ] , the conic a X X + a X X + a X X = . The ideal is the conic X X + X X + X X = , and the coideal is m X X + m X X + m X X = 0 . Some care has to be exercised because if, for instance, two of the coordinates X , X , X are0, the image is not defined.In the following definition “Point’” and “Line” is used for the new objects which have theproperties of “point” and “line” defined above. Definition.
Given a triangle ( A , A , A ) , ( a , a , a ) , the Points are- the points not on the sides of the triangle, .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. - the line through the vertices, (including a , a , a ),the Lines are- the conics through A0, A1 and A2, including the degenerate conics which consist ofone side and a line through the opposite vertex.A Point is on a Line if- . . .If two of the points are the isotropic points, the lines become the circles passing througha given point. A large number of properties of circles as well as properties of projectivegeometry can be obtained by pursuing this approach. In particular a study of the quarticswhich are associated to the circles is of interest.An early reference on circular triangles is by Miquel, J. de Liouville, Vol. 9, 1844, p. 24.Special cases. 2 Special cases are of interest. Notation. P :== p × p does not denote an actual construction, but a construction in which p or p are assumed to be known.“==” was suggested by the mode of drawing using dashed lines rather than continuous ones.In the example below, D is not known, hence we can not construct A × D. The following problem is of interest.
Exercise.
Given 2 conics with 3 points in common, determine by a linear construction the fourth pointon both conics.One solution is the following.Let
A, B, C be the known points and D be the unknown point. Let E and F be on the firstconic γ, U and V on the second conic γ (cid:48) . Determine first by the Pascal’s constructionpoint Pascal ( U, V, C, B, A, E ; E (cid:48) ) , and point Pascal ( A, B, C, U, V, E ; E (cid:48) ) , E (cid:48) on γ (cid:48) and A × E , F (cid:48) on γ (cid:48) and B × F ,let K :== ( D × A ) × ( C × B ) , L := ( A × E ) × ( B × F ) , M :== ( E × C ) × ( F × D ) ,M (cid:48) :== ( E (cid:48) × C ) × ( F (cid:48) × D ) , then Pascal( D, A, E, C, B, F ; K, L, M ) , and Pascal( D, A, E (cid:48) , C, B, F (cid:48) ; K, L, M (cid:48) ) . This impliesincidence ( L, M, M (cid:48) ) . Using Desargues − ( (cid:104) L, M (cid:48) , M (cid:105) , { C, E, E (cid:48) } , { D, F, F (cid:48) } ; G ) , it follows that D is incident to c × G , with G := ( E × F ) × ( E (cid:48) × F (cid:48) ) , the triangles { C, E, E (cid:48) } and { D, F, F (cid:48) } being perspective. D follows from point Pascal ( B, A, E, F, C, G ; D ) . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Construction.
The complete construction is the following: P := ( U × V ) × ( B × A ) , P := ( V × C ) × ( A × E ) , P := ( C × B ) × ( P × P , E (cid:48) :=( A × E ) × ( P × U ) ,P (cid:48) := ( V × C ) × ( B × F ) , P (cid:48) := ( C × A ) × ( P × P (cid:48) ) , F (cid:48) := ( C × A (cid:48) ) × ( P (cid:48) × U ) ,G := ( E × F ) × ( E (cid:48) × F (cid:48) ) ,Q := ( B × A ) × ( F × C ) , Q := ( A × E ) × ( C × G ) , Q := ( E × F ) × ( Q × Q ) ,D := ( C × G ) × ( Q × B ) . An other solution is the following
Theorem.
Let t A and t B be the tangents to the first conic at A and at B, let t (cid:48) A and t (cid:48) B the tangents to the second conic at A and B,O := t A × t B , O (cid:48) := t (cid:48) A × t (cid:48) B , oo (cid:48) := O × O (cid:48) , ab := A × B, BA (cid:48) := t B × t (cid:48) A , AB (cid:48) := t A × t (cid:48) B ,ab (cid:48) := BA (cid:48) × AB (cid:48) , E := ab × ab (cid:48) , cd := C × E, bc := B × C, F := bc × oo (cid:48) , ad := A × F,D := ad × cd, then D is on conic( A, t A , B, t B , C ) and conic( A, t (cid:48) A , B, t (cid:48) B , C ) .Proof: Assume that A and B are the isotropic points then the 2 conics are circles. O and O (cid:48) are their centers. cd ⊥ oo (cid:48) . Therefore ( A, B, D, ab × oo (cid:48) ) is a harmonic quatern. bc and ad meet on oo (cid:48) . This can be checked using A = (1 , i, , B = (1 , − i, , C = (0 , , ,D = (0 , − , , oo (cid:48) = [1 , , , bc = [ − i, − , , ad = [ i, − , − , F = (0 , , . Introduction.
Cubics have extensively studied by Newton, MacLaurin, Gergonne, Plucker, Salmon, . . . .I will give here a few properties, many of which generalize to higher degree curves, most ofthem taken from Salmon, 1979, sections 29 to 31 and 148 to 159?:
Theorem.
All cubics which pass trough 8 fixed points pass also through a ninth.
Definition.
If 9 points are on a one parameter family of non degenerate cubics, we say that they form a cubic configuration.
This configuration is not confined.
Theorem. [MacLaurin]
Let A , to A be 8 points of a cubic, such that A , A , A , A , A , A , are on a conic α and A , A , A , A , A , A , are on a conic β then ( A × A ) × ( A × A ) is on the cubic and the9 points form a cubic configuration. .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Proof: This follows when the preceding Theorem is applied to the degenerate cubics con-sisting of the conic α and the line A × A and the conic β and the line A × A . Corollary.
If 2 lines meet a cubic at points B ,i and B ,i then the 3 points B ,i on the cubic and on B ,i × B ,i are collinear, or equivalently If 6 points B j , j = 0 to 5, are on a cubic and 2 of the points C i := ( B i × B i +1 ) × ( B i +3 × B i +4 ) are on the cubic then the third point is on the cubic and the 9 points form a cubicconfiguration. Proof: This follows when the preceding Theorem is applied to the degenerate conics α through B ,i and B ,i and beta through B i, and B i, .The alternate form corollary gives Pappus’ Theorem when the cubic degenerates into 3lines. Notation.
I will write C B , B , B , B , B , B ; C , C , C ) . Theorem. [Salmon]
The 3 parameter family of cubics through the 6 points A i and B i , which are not on a conicis Σ i =0 , , s i ( A i × B i ) ×× ( A i +1 × A i − ) ×× ( B i +1 × B i − )= ( A i +1 × B i − ) ×× ( A i − × B i ) ×× ( A i × A i +1 ) . Proof: It is easy to verify that each of the points is on each of the 4 degenerate cubicsand that these are independent.There are many alternate forms possible, I have chosen the above one which displays a usefulsymmetry property.
Definition.
The tangential point of a point C on a cubic is the third intersection of the tangent at C with the cubic. Corollary.
If 3 points of a cubic are on a line a , their tangential points are on a line s . This follows from the degenerate case B ,i = B ,i . Definition.
The line s is called the satellite of the line a . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Notation.
Given 2 points A and B on a cubic, the third point on the cubic and the line A × B is denoted A (cid:63) B . Theorem.
Given 6 lines a i and b i and their 9 intersections i j := a i × b j ,0. These 9 points form a cubic configuration. If C is a point on 11 × , the cubic of the family through the points i j and C aresuch that if we define the following points, C i := i + 1 , i + 1 (cid:63) i − , i − , D i := i + 1 , i − (cid:63) i − , i + 1 ,E i := i, i + 1 (cid:63) i − , i, E i := i + 1 , i (cid:63) i, i − ,F i := i, i (cid:63) i + 1 , i − , F i := i, i (cid:63) i − , i + 1 ,Cc i := C i +1 (cid:63) C i − , Cd i := C i +1 (cid:63) D i − , Dc i := D i +1 (cid:63) C i − ,CF i := C i (cid:63) F i , Cf i := C i +1 (cid:63) F i − , F c i := F i +1 (cid:63) C i − ,CF i := C i (cid:63) F i , Cf i := C i +1 (cid:63) F i − , F c i := F i +1 (cid:63) C i − ,CF i := C i (cid:63) F i , Cf i := C i +1 (cid:63) F i − , F c i := F i +1 (cid:63) C i − ,DE i := D i (cid:63) E i , De i := D i +1 (cid:63) E i − , Ed i := E i +1 (cid:63) D i − ,DE i := D i (cid:63) E i , De i := D i +1 (cid:63) E i − , Ed i := E i +1 (cid:63) D i − ,DF i := D i (cid:63) F i , DF i := D i (cid:63) F i ,Ee i := E i +1 (cid:63) E i − , EF i := E i (cid:63) F i , Ef i := E i +1 (cid:63) F i − ,F e i := F i +1 (cid:63) E i − , Ef i := E i +1 (cid:63) F i − , F e i := F i +1 (cid:63) E i − ,Ee i := E i +1 (cid:63) E i − , EF i := E i (cid:63) F i , Ef i := E i +1 (cid:63) F i − ,F e i := F i +1 (cid:63) E i − , Ef i := E i +1 (cid:63) F i − , F e i := F i +1 (cid:63) E i − ,F F i := F i (cid:63) F i ,C (cid:48) i := C i (cid:63) C i , D (cid:48) i := D i (cid:63) D i , E (cid:48) i := E i (cid:63) E i , E (cid:48) i := E i (cid:63) E i , F (cid:48) i := F i (cid:63) F i , F (cid:48) i := F i (cid:63) F i , K i := C i (cid:63) i + 1 , i − , K i := C i (cid:63) i − , i + 1 , L i := D i +1 (cid:63) i − , i − , L i := D i − (cid:63) i + 1 , i + 1 , M i := E i (cid:63) i, i , M i := E i (cid:63) i, i , N i := E i (cid:63) i − , i + 1 , N i := E i (cid:63) i + 1 , i − , P i := F i +1 (cid:63) i + 1 , i , P i := F i +1 (cid:63) i, i + 1 , Q i := F i − (cid:63) i, i − , Q i := F i − (cid:63) i − , i , .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. We have the following table for the operation (cid:63) between points on the cubic: (cid:63)
00 11 22 12 20 01 21 02 10 C C C D D D C D
22 11 K CF CF K CF CF C (cid:48) Cc Cc Cd Dc C D CF K CF CF K CF Cc C (cid:48) Cc Dc Cd C
11 00 D CF CF K CF CF K Cc Cc C (cid:48) Cd Dc D C L L F e Ef F e Ef Dc Cd D (cid:48) (cid:48) (cid:48) D L C L Ef F e Ef F e Cd Dc (cid:48) D (cid:48) (cid:48) D L L C F e Ef F e Ef Dc Cd
22 11 (cid:48) (cid:48) D (cid:48) E M DE DE F
01 20 N EF EF (cid:48) E E DE Ed De E DE M DE F EF N EF E (cid:48) E De DE Ed E DE DE M
20 12 F EF EF N E E (cid:48) Ed De DE E M DE DE N EF EF F
10 02 12 (cid:48) E E DE Ed De E DE M DE EF N EF F E (cid:48) E De DE Ed E DE DE M EF EF N
02 21 F E E (cid:48) Ed De DE F F e Ef Dc Cd E P Q CF F c Cf DF F F F Ef F e Cd Dc Q E P Cf CF F c F DF F F F e Ef Dc Cd P Q E F c Cf CF F F DF F F e Ef E P Q Dc Cd CF F c Cf DF F F F Ef F e Q E P Cd Dc Cf CF F c F DF F F F e Ef P Q E Dc Cd F c Cf CF F F DF (cid:63) E E E E E E F F F F F F E E (cid:48) Ee Ee (cid:48) C C EF Ef F e Ef F e E Ee E (cid:48) Ee C (cid:48) C F e EF Ef F e Ef E Ee Ee E (cid:48) C C (cid:48) Ef F e EF Ef F e E (cid:48) C C E (cid:48) Ee Ee Ef F e EF Ef F e E C (cid:48) C Ee E (cid:48) Ee F e Ef F e EF Ef E C C (cid:48) Ee Ee E (cid:48) Ef F e Ef F e EF F EF F e Ef F e Ef F (cid:48) (cid:48) (cid:48) F F D D F F e EF Ef Ef F e (cid:48) F (cid:48) (cid:48) D F F D F F e Ef EF F e Ef
10 02 (cid:48) (cid:48) F (cid:48) D D F F F F e Ef EF F e Ef F F D D ovF (cid:48) (cid:48) (cid:48) F Ef F e Ef EF F e D F F D (cid:48) F (cid:48) (cid:48) F F e Ef F e Ef EF D D F F (cid:48) (cid:48) F (cid:48) CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Proof: α D C D
21 20 10 11 ; 00 12 22) ρα D E F
01 00 20 21 ; 10 22 02) ρ α C E F
11 10 00 01 ; 20 02 12) α C E E
21 11 22 12 ; C
02 01) σα C E E
12 11 22 21 ; C
20 10) βα C F F
11 21 12 22 ; D
02 01) σβα C F F
11 12 21 22 ; D
20 10) α C C D F
20 00 ; Cd
10 22) σα C C D F
02 00 ; Cd
01 22)021021 α C C D F
10 00 ; Dc
20 11) σ α C C D F
01 00 ; Dc
02 11) α C C F C
10 11 ; CF
12 22) σα C C F C
01 11 ; CF
21 22)021021 α C C F C
20 22 ; CF
21 11) σ α C C F C
02 22 ; CF
12 11) α C D E E
11 21 ; DE
01 12) σα C D E E
11 12 ; DE
10 21)021021 α C D E E
22 12 ; DE
02 21) σ α C D E E
22 21 ; DE
20 12) α C E F E
02 01 ; EF
00 20) σα C E F E
20 10 ; EF
00 02)021021 α C E F E
01 02 ; EF
00 10) σ α C E F E
10 20 ; EF
00 01) α C E F F
11 12 ; Ef
10 01) σα C E F F
11 21 ; Ef
01 10) α C F E F
22 20 ;
F e
21 11) σα C F E F
22 02 ;
F e
12 11) α C C E
20 21 21 11 ; 21 (cid:48)
01 22) σα C C E
02 12 12 11 ; 12 (cid:48)
10 22)012210 α C F F
22 21 21 11 ; 21 (cid:48)
01 20) σα C F F
22 12 12 11 ; 12 (cid:48)
10 02)210102 α C E E
01 00 00 10 ; 00 (cid:48)
20 02)120102 α C D D
01 00 00 20 ; 00 (cid:48)
10 02) α C C
10 11 C
02 22 ; CF
12 00) σα C C
01 11 C
20 22 ; CF
21 00)210012 α C D
10 11 F
22 02 ;
F e
12 20) σ α C D
01 11 F
22 20 ;
F e
21 02)012102 α C F
11 10 D
02 22 ; Ef
12 01) σ α C F
11 01 D
20 22 ; Ef
21 10)102210 α C E
02 01 E
10 20 ; EF
00 12) σ α C E
20 10 E
01 02 ; EF
00 21)120201 α C E
22 20 E
11 01 ; DE
21 12) σ α C E
22 02 E
11 10 ; DE
12 21)102120 α C F
01 02 F
10 20 ; Dc
00 11)201210 α C F
02 01 F
20 10 ; Cd
00 22) .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES.
Definition.
Given 6 lines a i and b i , among the 15 intersections we choose the following 9,D1.0. A i := a i +1 × a i − , D1.1. B i := b i +1 × b i − , D1.2. E i := a i × b i , the non confined configuration consisting of these 9 points and 6 lines each containing 3 ofthe points is called a Grassmann configuration.
It is noted ( { A i } , { B i } , { E i } ) . Theorem. [Grassmann]
Given 2 triangles { A i , a i } and { B i , b i } , the locus of the points X is a cubic, if X is such thatthe points obtained by finding the intersections of the lines joining X to the vertices of oneof the triangles and the corresponding sides of the second triangle, namely ( X × A i ) × b i , arecollinear. Theorem.
Let
D2.0. aB i := A i +1 × B i − , aB i := A i − × B i +1 , D2.1. AB i := aB i × aB i , D2.2. abE i := AB i +1 × E i − , abE i := AB i − × E i +1 , D2.3. DE i := abE i × abE i , D2.4. de i := DE i × E i , D2.5. ab i := A i × B i , D2.6. D i := de i × ab i , D2.6. ba i := AB i +1 × AB i − , D2.8. abd i := D i × AB i , D2.9. C i := ba i × abd i , D4.0. ae i := A i × E i , be i := B i × E i , D4.1. ce i := C i × E i , D4.2. aab i := A i × AB i , bab i := B i × AB i , D4.3. F i := be i × aab i , F i := ae i × bab i , D4.4. f i := F i +1 × F i − , f i := F i +1 × F i − , D4.5. A (cid:48) i := ce i × f i , B (cid:48) i := ce i × f i , D4.6. aC i := A i +1 × C i − , aC i := A i − × C i +1 , D4.6. bC i := B i +1 × C i − , bC i := B i − × C i +1 , D4.7. CF i := bC i × bC i , CF i := aC i × aC i , D4.8. cD i := C i +1 × D i − , cD i := C i − × D i +1 , D4.9. aF i := A i +1 × F i − , aF i := A i − × F i +1 , D4.10. Cd i := cD i × aF i , Dc i := cD i × aF i , D4.11. aD i := A i +1 × D i − , aD i := A i − × D i +1 , D4.12. eF i := E i +1 × F i − , eF i := E i − × F i +1 , D4.13. Ef i := aD i × eF i , F e i := aD i × eF i , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
D4.14. bD i := B i +1 × D i − , bD i := B i − × D i +1 , D4.15. f E i := E i +1 × F i − , f E i := E i − × F i +1 , D4.16. Ef i := bD i × f E i , F e i := bD i × f E i , D4.17. ef i := E i × F i , ef i := E i × F i , D5.0. a (cid:48) D i := A (cid:48) i +1 × D i − , a (cid:48) D i := A (cid:48) i − × D i +1 , D5.1. abe i := AB i × E i , D5.2. M i := a (cid:48) D i × abe i , D5.3. dAB i := D i +1 × AB i − , dAB i := D i − × AB i +1 , D5.4. a (cid:48) E i := A (cid:48) i +1 × E i − , a (cid:48) E i := A (cid:48) i − × E i +1 , D5.5. L i := dAB i × a (cid:48) E i , L i := dAB i × a (cid:48) E i , D5.6. f B i := F i +1 × B i − , f B i := F i − × B i +1 , D5.7. f A i := F i +1 × A i − , f A i := F i − × A i +1 , D5.8. P i := f B i × f A i , Q i := f B i × f A i , D5.9. ac i := A i × C i , bc i := B i × C i , D5.10. bde i := B i × DE i , ade i := A i × DE i , D5.11. K i := ac i × bde i , K i := bc i × ade i , D5.12. dE i := D i +1 × E i − , eD i := E i +1 × D i − , D5.13. bK i := B i +1 × K i − , kB i := B i − × K i +1 , D5.14. Ed i := eD i × bK i , De i := dE i × kB i , D6.0. cL i := C i +1 × L i − , cL i := C i − × L i +1 , D6.1. C (cid:48) i := cL i × cL i , D6.2. kf i := K i × F i , kf i := K i × F i , D6.3. D (cid:48) i := kf i × kf i , D6.4. a (cid:48) f i := A (cid:48) i × F i , a (cid:48) f i := A (cid:48) i × F i , D6.5. df i := D i × F i , df i := D i × F i , D6.6. DF i := a (cid:48) f i × df i , DF i := a (cid:48) f i × df i , D6.7. f C i := F i +1 × C i − , f C i := F i − × C i +1 , D6.8. f C i := F i +1 × C i − , f C i := F i − × C i +1 , D6.9. f DE i := F i +1 × DE i − , f DE i := F i +1 × DE i − , D6.10. aM i := A i +1 × M i − , bM i := B i +1 × M i − , D6.11.
F c i := f C i × f DE i , F c i := f C i × f DE i , D6.12. Cf i := f C i × aM i , Cf i := f C i × bM i , D6.13. c i := C i +1 × C i − , ll i := L i × L i , D6.14. Cc i := c i × ll i , D6.15. ccL i := Cc i +1 × L i − , ccL i := Cc i − × L i +1 , D6.16. AB (cid:48) i := ccL i × ccL i , D6.17. lQ i := L i +1 × Q i − , pL i := P i +1 × L i − , D6.18. F (cid:48) i := lQ i × pL i , D6.19. f f i := F i × F i , dab (cid:48) i := D i × AB (cid:48) i , D6.20.
F F i := f f i × dab (cid:48) i , D7.0. a (cid:48) i := A i × A (cid:48) i , D7.1. b (cid:48) i := B i × A (cid:48) i , D7.2. c (cid:48) i := C i × C (cid:48) i , D7.3. d (cid:48) i := D i × D (cid:48) i , D7.4. ab (cid:48) i := AB i × AB (cid:48) i , .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. D7.5. e (cid:48) i := E i × AB (cid:48) i , D7.6. f (cid:48) i := F i × F (cid:48) i , D7.7. f (cid:48) i := F i × F (cid:48) i , then C2.0. C i ι e i , C2.1. X i = Y i ,C2.2. A (cid:48) i = B (cid:48) i , C2.3. A i ι ef i , B i ι ef i , C2.4. A (cid:48) i +1 × D i − = M i , C2.5. A (cid:48) i +1 × E i − = L i , C2.6. E i +1 × A (cid:48) i − = M i , C2.7. F (cid:48) i = F (cid:48) i , Theorem.
Given a Grassmann configuration, ( { A i } , { B i } , { E i } ) , the points C i , AB i , D i , F i , F i , Cd i , Dc i , CF i , CF i , DE i , Ef i , F e i , Ef i , F e i , are onthe cubic γ through A i , B i , E i . ( A i a (cid:48) i ) , ( B i b (cid:48) i ) , ( C i c (cid:48) i ) , ( D i d (cid:48) i ) , ( AB i ab (cid:48) i ) , ( E i e (cid:48) i ) , ( F i f (cid:48) i ) , ( F i f (cid:48) i ) ι γ. E i (cid:63) E i = AB i (cid:63) AB i .3. The points A i , B i , AB i , are on a cubic configuration. (cid:104) A (cid:48) i +1 , A (cid:48) i − , AB (cid:48) i , (cid:105) ,5. (cid:104) B (cid:48) i +1 , B (cid:48) i − , AB (cid:48) i , (cid:105) ,6. (cid:104) AB (cid:48) i +1 , AB (cid:48) i − , C (cid:48) i , (cid:105) ,7. (cid:104) AB (cid:48) i , C (cid:48) i , D (cid:48) i , (cid:105) ,8. (cid:104) A (cid:48) i , AB (cid:48) i , F (cid:48) i , (cid:105) ,Proof: To prove that AB is on the cubic, we have to prove, because of 3.2.10, (cid:104) ( AB × A ) × b , ( AB × A ) × b , ( AB × A ) × b , (cid:105) , but the second point is B and the third is B and both are on b . The rest of the proof follows from 6.0.2 given below. Comment.
The preceding Theorem was conjectured in the process of construction the third point on aGrassmann cubic and the line through 2 points of the cubic, using intersection of conics andlines, (Fig. hd.c) using the Theorem of Grassmann, (Henry White, 1925, p. 109, Fig. 27.)For instance, the conic through B = B , C = B , A (cid:48) = B , D = ( B × C (cid:48) = B ) × ( C × B (cid:48) = B ) and X, intersects A × X at the third point A (cid:63) X . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem.
We have the following table for the operation (cid:63) between points on a Grassmann cubic: (cid:63) AB AB AB A A A B B B AB AB (cid:48) C C F B B F A A AB C AB (cid:48) C B F B A F A AB C C AB (cid:48) B B F A A F A F B B A (cid:48) E E D AB AB A B F B E A (cid:48) E AB D AB A B B F E E A (cid:48) AB AB D B F A A D AB AB A (cid:48) E E B A F A AB D AB E A (cid:48) E B A A F AB AB D E E A (cid:48) (cid:63) AB AB AB A A A B B B C C C D D D C D AB AB K CF CF K CF CF C (cid:48) Cc Cc AB Cd Dc C AB D AB CF K CF CF K CF Cc C (cid:48) Cc Dc AB Cd C AB AB D CF CF K CF CF K Cc Cc C (cid:48) Cd Dc AB D C L L B F e Ef A F e Ef AB Dc Cd D (cid:48) AB (cid:48) AB (cid:48) D L C L Ef B F e Ef A F e Cd AB Dc AB (cid:48) D (cid:48) AB (cid:48) D L L C F e Ef B F e Ef A Dc Cd AB AB (cid:48) AB (cid:48) D (cid:48) E M DE DE F A A F B B A (cid:48) E E DE Ed De E DE M DE A F A B F B E A (cid:48) E De DE Ed E DE DE M A A F B B F E E A (cid:48) Ed De DE F A F e Ef AB Dc Cd E P Q CF F c Cf DF F F F Ef A F e Cd AB Dc Q E P Cf CF F c F DF F F F e Ef A Dc Cd AB P Q E F c Cf CF F F DF F B F e Ef E P Q AB Dc Cd CF F c Cf DF F F F Ef B F e Q E P Cd AB Dc Cf CF F c F DF F F F e Ef B P Q E Dc Cd AB F c Cf CF F F DF (cid:63) E E E F F F F F F E AB (cid:48) C C B Ef F e A Ef F e E C AB (cid:48) C F e B Ef F e A Ef E C C AB (cid:48) Ef F e B Ef F e A F B F e Ef F (cid:48) A (cid:48) A (cid:48) F F D D F Ef B F e A (cid:48) F (cid:48) A (cid:48) D F F D F F e Ef B A (cid:48) A (cid:48) F (cid:48) D D F F F A F e Ef F F D D F (cid:48) A (cid:48) A (cid:48) F Ef A F e D F F D A (cid:48) F (cid:48) A (cid:48) F F e Ef A D D F F A (cid:48) A (cid:48) F (cid:48) .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Proof: α D C D B A B AB ; AB A AB ) ρα D E F A AB A B ; B AB B ) ρ α C E F AB B AB A ; A B A ) α C E E B AB AB A ; C B A ) βα C F F AB B A AB ; D B A ) σβα C F F AB A B AB ; D A B ) α C C D A F A AB ; Cd B AB ) σα C C D B F B AB ; Cd A AB )021021 α C C D B F B AB ; Dc A AB ) σ α C C D A F A AB ; Dc B AB ) α C C F AB C B AB ; CF A AB ) σα C C F AB C A AB ; CF B AB )021021 α C C F AB C A AB ; CF B AB ) σ α C C F AB C B AB ; CF A AB ) α C D E A E AB B ; DE A A )021021 α C D E B E AB A ; DE B B ) α C E F A E B A ; B AB A ) σα C E F B E A B ; A AB B )021021 α C E F B E A B ; A AB B ) σ α C E F A E B A ; B AB A ) α C E F AB F AB A ; Ef B A ) σα C E F AB F AB B ; Ef A B ) α C F E B F AB A ; F e B AB ) σα C F E A F AB B ; F e A AB ) σα C C E B A A AB ; A (cid:48) B AB ) α (cid:48) C A A B B B A ; A (cid:48) AB E )012210 α C F F AB B B AB ; A (cid:48) A A ) σα C F F AB A A AB ; A (cid:48) B B )210102 α C E E A AB AB B ; AB (cid:48) A B )120102 α C D D A AB AB A ; AB (cid:48) B B )210012 α C D B AB F AB B ; F e A A ) σ α C D A AB F AB A ; F e B B )012102 α C F AB B D B AB ; Ef A A ) σ α C F AB A D A AB ; Ef B B )120201 α C E AB A E AB A ; DE B A ) α C AB E A D A (cid:48) A ; M A B ) α C AB D B A (cid:48) E A ; L A B )? α C AB D B A (cid:48) E A ; L A B ) α C F B B A F A ; P E AB )? α C F B B A F A ; Q E AB ) α C C C AB C L AB ; C (cid:48) D AB )? α C C C AB C L AB ; C (cid:48) D AB ) α C F F B F F A ; F (cid:48) E AB ) α C A C E DE B B ; K A (cid:48) D )? α C B C E DE A A ; K A (cid:48) D ) α C AB AB AB L Cc C ; AB (cid:48) C D )? α C AB AB AB L Cc C ; AB (cid:48) C D ) α C AB AB C Cc C AB ; AB (cid:48) AB C ) α C C C AB L L AB ; Cc D D ) α C F F D L P A ; F (cid:48) F AB ) α C D F E F A (cid:48) B ; DF B A )? α C D F E F A (cid:48) A ; DF A B ) CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY α C F C AB F DE E ; F c AB B )? α C F C AB F DE E ; F c AB A ) α C C F E M A AB ; Cf B AB )? α C C F E M B AB ; Cf A AB ) α C F F D L Q A ; F (cid:48) F AB )? α C F F D L P A ; F (cid:48) F AB ) α C D D C K F B ; D (cid:48) AB A )? α C D D C K F A ; D (cid:48) AB A ) α C E D B B K C ; Ed A E )? α C E D B B K C ; De A E ) α C C F A B AB (cid:48) AB ; CF AB D )? α C C F B A AB (cid:48) AB ; CF AB D ) α C F F B D AB (cid:48) AB ; F F AB A ) Theorem.
Given a Grassmann configuration ( { A i } , { B i } , { E i } ) , THe tangential points at A i and B i are the same, in the above case A (cid:48) i , it will be added to the configuration after a semi colon. Lemma.
If ( { A i } , { B i } , { E i } , { A (cid:48) i } ) , is a Grassmann configuration so is ( { A i } , { B i } , { E i } ) , , where AB i := A i +1 (cid:63) B i − , and C i := AB i +1 (cid:63) AB i − . ( { F i } , { F i } , { A (cid:48) i } ) , , where F i := AB i (cid:63) A i , and F i := AB i (cid:63) B i , This follows at once from grom 6.0.2.
Theorem.
Given a Grassmann configuration ( { A i } , { B i } , { E i } ; { AB i } ) , the following are also Grass-mann configurations: ( { AB i } , { E i } , { C i } ; { AB (cid:48) i } ) , ( { F i } , { F i } , { A (cid:48) i } ; { F (cid:48) i } ) , ( { DE i } , { C i } , { C i (cid:63) AB (cid:48) i } ) , ( { D i } , { A (cid:48) i } , { AB (cid:48) i } ; { d (cid:48) i } ) . Proof: This follows by repeated applications of the Lemma 3.2.10. .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES.
Definition.
In involutive geometry I will give the name of
Grassmannian cubic to the special case wherethe 6 lines are m i and m i . Theorem.
The correspondence between the elements as given above and those of involutive geometryis as follows A i = CF i B i = CF i E i = DE i AB i = C i = Cc i F i = K i F i = K i M M i M M i A i AT i F i F i Ef i = Cf i Ef i = Cf i F e i = F c i F e i = F c i A (cid:48) i = M i L i = Cd i L i = Dc i P i Q i Ef i Ef i F e i F e i M M (cid:48) i L i L i P i Q i a i b i aB i aB i ab i e i ae i be i ba i mm i mm i c i c i a i aeU L i nm i nm i eul Theorem. The Grassmann cubic passes through the points
M M i , M M i , A i , D i . Its equation is m ( m + m ) X X (( m + m X + ( m + m ) X ) m ( m + m ) X X (( m + m X + ( m + m ) X ) m ( m + m ) X X (( m + m X + ( m + m ) X ) + ( s +2 s ) X X X = 0 Proof: Using 3.2.9 on the points A i and AT i , we obtain the given form, to determine thecoefficients g i of X X (( m + m X + ( m + m ) X ) , . . . and g of X X X we impose thecondition that the cubic passes through M M i , this gives the system of equations . . . . Theorem.
Let
D2.0. aB i := A i +1 × B i − , aB i := A i − × B i +1 , D2.1. AB i := aB i × aB i , D2.2. abE i := AB i +1 × E i − , abE i := AB i − × E i +1 , D2.3. DE i := abE i × abE i , D2.4. de i := DE i × E i , D2.5. ab i := A i × B i , D2.6. D i := de i × ab i , D2.6. ba i := AB i +1 × AB i − , D2.8. abd i := D i × AB i , D2.9. C i := ba i × abd i , D4.0. ae i := A i × E i , be i := B i × E i , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
D4.1. ce i := C i × E i , D4.2. aab i := A i × AB i , bab i := B i × AB i , D4.3. F i := be i × aab i , F i := ae i × bab i , D4.4. f i := F i +1 × F i − , f i := F i +1 × F i − , D4.5. A (cid:48) i := ce i × f i , B (cid:48) i := ce i × f i , D4.6. aC i := A i +1 × C i − , aC i := A i − × C i +1 , D4.6. bC i := B i +1 × C i − , bC i := B i − × C i +1 , D4.7. CF i := bC i × bC i , CF i := aC i × aC i , D4.8. cD i := C i +1 × D i − , cD i := C i − × D i +1 , D4.9. aF i := A i +1 × F i − , aF i := A i − × F i +1 , D4.10. Cd i := cD i × aF i , Dc i := cD i × aF i , D4.11. aD i := A i +1 × D i − , aD i := A i − × D i +1 , D4.12. eF i := E i +1 × F i − , eF i := E i − × F i +1 , D4.13. Ef i := aD i × eF i , F e i := aD i × eF i , D4.14. bD i := B i +1 × D i − , bD i := B i − × D i +1 , D4.15. f E i := E i +1 × F i − , f E i := E i − × F i +1 , D4.16. Ef i := bD i × f E i , F e i := bD i × f E i , D4.17. ef i := E i × F i , ef i := E i × F i , D5.0. a (cid:48) D i := A (cid:48) i +1 × D i − , a (cid:48) D i := A (cid:48) i − × D i +1 , D5.1. abe i := AB i × E i , D5.2. M i := a (cid:48) D i × abe i , D5.3. dAB i := D i +1 × AB i − , dAB i := D i − × AB i +1 , D5.4. a (cid:48) E i := A (cid:48) i +1 × E i − , a (cid:48) E i := A (cid:48) i − × E i +1 , D5.5. L i := dAB i × a (cid:48) E i , L i := dAB i × a (cid:48) E i , D5.6. f B i := F i +1 × B i − , f B i := F i − × B i +1 , D5.7. f A i := F i +1 × A i − , f A i := F i − × A i +1 , D5.8. P i := f B i × f A i , Q i := f B i × f A i , D5.9. ac i := A i × C i , bc i := B i × C i , D5.10. bde i := B i × DE i , ade i := A i × DE i , D5.11. K i := ac i × bde i , K i := bc i × ade i , D5.12. dE i := D i +1 × E i − , eD i := E i +1 × E i − , D5.13. bK i := B i +1 × K i − , kB i := B i − × K i +1 , D5.14. Ed i := eD i × bK i , De i := dE i × kB i , D6.0. cL i := C i +1 × L i − , cL i := C i − × L i +1 , D6.1. C (cid:48) i := cL i × cL i , D6.2. kf i := K i × F i , kf i := K i × F i , D6.3. D (cid:48) i := kf i × kf i , D6.4. a (cid:48) f i := A (cid:48) i × F i , a (cid:48) f i := A (cid:48) i × F i , D6.5. df i := D i × F i , df i := D i × F i , D6.6. DF i := a (cid:48) f i × df i , DF i := a (cid:48) f i × df i , D6.7. f C i := F i +1 × C i − , f C i := F i − × C i +1 , D6.8. f C i := F i +1 × C i − , f C i := F i − × C i +1 , D6.9. f DE i := F i +1 × DE i − , f DE i := F i +1 × DE i − , D6.10. aM i := A i +1 × M i − , bM i := B i +1 × M i − , D6.11.
F c i := f C i × f DE i , F c i := f C i × f DE i , .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. D6.12. Cf i := f C i × aM i , Cf i := f C i × bM i , D6.13. c i := C i +1 × C i − , ll i := L i × L i , D6.14. Cc i := c i × ll i , D6.15. ccL i := Cc i +1 × L i − , ccL i := Cc i − × L i +1 , D6.16. AB (cid:48) i := ccL i × ccL i , D6.17. lQ i := L i +1 × Q i − , pL i := P i +1 × L i − , D6.18. F (cid:48) i := lQ i × pL i , D6.19. f f i := F i × F i , dab (cid:48) i := D i × AB (cid:48) i , D6.20.
F F i := f f i × dab (cid:48) i , D7.0. a (cid:48) i := A i × A (cid:48) i , D7.1. b (cid:48) i := B i × A (cid:48) i , D7.2. c (cid:48) i := C i × C (cid:48) i , D7.3. d (cid:48) i := D i × D (cid:48) i , D7.4. ab (cid:48) i := AB i × AB (cid:48) i , D7.5. e (cid:48) i := E i × AB (cid:48) i , D7.6. f (cid:48) i := F i × F (cid:48) i , D7.7. f (cid:48) i := F i × F (cid:48) i , then C2.0. C i ι e i , C2.1. X i = Y i ,C2.2. A (cid:48) i = B (cid:48) i , C2.3. A i ι ef i , B i ι ef i , C2.4. A (cid:48) i +1 × D i − = M i , C2.5. A (cid:48) i +1 × E i − = L i , C2.6. E i +1 × A (cid:48) i − = M i , C2.7. F (cid:48) i = F (cid:48) i ,Proof: H0.0. m = [0 , , , m = [0 , m , m , D1.0.
M M = ( − , , , M M = ( m , − m , − m D0.10. A = (1 , , D0.10. a = [1 , , D.. ma = [0 , , − , ma = [0 , m , − m , D.. eul = [ m − m , − ( m − m , − ( m − m , D.. y = [ m − m , − ( m m , − ( m m , D.. y = [ m − m , m m , m m , D.. AT = (0 , m m , − ( m − m , D.. k = [ m m , m m , m m , D.. tAM = [ s m , m m , m m , D.. tAM = [ s
11 + m m , m m m , m m m , D.. F = ( s
11 + m m , − m s m , − m s m , D.. F = ( m s m , − ( s
11 + m m , − ( s
11 + m m , D.. f f = [ m m − m , s
11 + m m , m s m , D.. D = ( m m q , − m m − m m m , m m m m − m , D.. eul = [ m − m , m − m , m − m , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
D.. aAT = [0 , m m , m m , D.. f = [( m m q , − ( m m s
11 + m m , − ( m m s
11 + m m , D..
M M (cid:48) = ( m m − m m m m m , − ( m m m m q , ( m m m m q , D.. = [( m − m s
11 + m m , m m − m , m q − m q m m m − m , D.. = [ s
11 + m m , m s m , , D..
F e = ( m s m m q − m q m m m − m , − ( s
11 + m m m q − m q m m m − m , ( s
11 + m m m q − m q − m m m − m , D.. = [] , D.. = [] , D.. Ef = ( m s m m q − m q − m m m − m , ( s
11 + m m m q − m q m m m − m , − ( s
11 + m m m q − m q − m m m − m , D.. = [] , D.. = [] , D..
F e = (( s
11 + m m m q − m q − m m m − m , ( s
11 + m m m q − m q − m m m − m , − ( s
11 + m m m q − m q m m m − m , D.. = [] , D.. = [] , D.. Ef = (( s
11 + m m m q − m q m m m − m , m s m m q − m q − m m m − m , − m s m m q − m q m m m − m , D.. = [] , D.. = [] , D.. Ed = ( m m m ( m m m − m , − m q m m m
2) + ( m m m m , s
21 + 2 s , D.. = [] , D.. = [] , D.. De = ( m m m ( m − m m m , s
21 + 2 s , m q m m m
2) + ( m m m m , D.. = [] , D.. = [] , D.. L = (( m m q m + m m m m m , m m m m − m m + m m m m m , − ( m m m − m s
21 + 2 s , D.. = [] , D.. = [] , D.. L = (( m m q m + m m m m m , ( m − m m m s
21 +2 s , − m m − m m m m + m m m m m , D.. = [] , D.. = [] , D.. P = ( m m m ( m − m m m , ( s
21 + 2 s q , − m q m + m m m m m , D.. = [] , D.. = [] , D.. Q = ( m m m ( m m m − m , − m q m + m m m m m , − ( s s q , .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Theorem.
We have the following table for the operation (cid:63) between points on the Grassmannian cubic: (cid:63) AT AT AT M M M M M M M M M M M M A A A D D D AT D AT AT F M M M M F M M M M M M (cid:48) A A AT L L AT AT D AT M M F M M M M F M M A M M (cid:48) A L AT L AT AT AT D M M M M F M M M M F A A M M (cid:48) L L AT M M F M M M M M M (cid:48) A A D AT AT F M M M M M M Ef F e M M M M F M M A M M (cid:48) A AT D AT M M F M M F e M M Ef M M M M M M F A A M M (cid:48) AT AT D M M M M F Ef F e M M M M F M M M M D AT AT M M (cid:48) A A F M M M M M M Ef F e M M M M F M M AT D AT A M M (cid:48) A M M F M M F e M M Ef M M M M M M F AT AT D A A M M (cid:48) M M M M F Ef F e M M A M M (cid:48) A A F M M M M F M M M M D AT AT A De Ed A A M M (cid:48) A M M F M M M M F M M AT D AT Ed A De A A A M M (cid:48) M M M M F M M M M F AT AT D De Ed A D AT L L M M F e Ef M M F e Ef A Ed De D (cid:48) D D D L AT L Ef M M F e Ef M M F e De A Ed D D (cid:48) D D L L AT F e Ef M M F e Ef M M Ed De A D D D (cid:48) F M M F e Ef AT L L A P Q M M F e Ef F D F F F Ef M M F e L AT L Q A P Ef M M F e F F D F F F e Ef M M L L AT P Q A F e Ef M M F F F D F M M F e Ef A P Q AT L L M M F e Ef F D F F F Ef M M F e Q A P L AT L Ef M M F e F F D F F F e Ef M M P Q A L L AT F e Ef M M F F F D (cid:63) F F F F F F F F (cid:48) M M (cid:48) M M (cid:48) D (cid:48) D D F M M (cid:48) F (cid:48) M M (cid:48) D D (cid:48) D F M M (cid:48) M M (cid:48) F (cid:48) D D D (cid:48) F D (cid:48) D D F (cid:48) M M (cid:48) M M (cid:48) F D D (cid:48) D M M (cid:48) F (cid:48) M M (cid:48) F D D D (cid:48) M M (cid:48) M M (cid:48) F (cid:48) CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Proof: .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. α D C D M M M M M M AT ; AT M M AT ) ρα D E F M M AT M M M M ; M M AT M M ) ρ α C E F AT M M AT M M ; M M M M M M ) α C E A M M AT AT M M ; AT M M M M ) βα C F F AT M M M M AT ; D M M M M ) σβα C F F AT M M M M AT ; D M M M M ) α C C D M M F M M AT ; L M M AT ) σα C C D M M F M M AT ; L M M AT )021021 α C C D M M F M M AT ; L M M AT ) σ α C C D M M F M M AT ; L M M AT ) α C C F AT AT M M AT ; M M M M AT ) σα C C F AT AT M M AT ; M M M M AT )021021 α C C F AT AT M M AT ; M M M M AT ) σ α C C F AT AT M M AT ; M M M M AT ) α C D A M M A AT M M ; A M M M M )021021 α C D A M M A AT M M ; A M M M M ) α C E F M M A M M M M ; M M AT M M ) σα C E F M M A M M M M ; M M AT M M )021021 α C E F M M A M M M M ; M M AT M M ) σ α C E F M M A M M M M ; M M AT M M ) α C E F AT F AT M M ; Ef M M M M ) σα C E F AT F AT M M ; Ef M M M M ) α C F A M M F AT M M ; F e M M AT ) σα C F A M M F AT M M ; F e M M AT ) σα C C A M M M M M M AT ; A (cid:48) M M AT ) α (cid:48) C M M M M M M M M M M M M ; A (cid:48) AT A )012210 α C F F AT M M M M AT ; A (cid:48) M M M M ) σα C F F AT M M M M AT ; A (cid:48) M M M M )210102 α C E A M M AT AT M M ; AB (cid:48) M M M M )120102 α C D D M M AT AT M M ; AB (cid:48) M M M M )210012 α C D M M AT F AT M M ; F e M M M M ) σ α C D M M AT F AT M M ; F e M M M M )012102 α C F AT M M D M M AT ; Ef M M M M ) σ α C F AT M M D M M AT ; Ef M M M M )120201 α C E AT M M A AT M M ; A M M M M ) α C AT A M M D A (cid:48) M M ; M M M M M ) α C AT D M M A (cid:48) A M M ; L M M M M )? α C AT D M M A (cid:48) A M M ; L M M M M ) α C F M M M M M M F M M ; P A AT )? α C F M M M M M M F M M ; Q A AT ) α C C AT AT AT L AT ; C (cid:48) D AT )? α C C AT AT AT L AT ; C (cid:48) D AT ) α C F F M M F F M M ; F (cid:48) A AT ) α C M M AT A A M M M M ; F A (cid:48) D )? α C M M AT A A M M M M ; ovF A (cid:48) D ) α C AT AT AT L Cc AT ; AB (cid:48) AT D )? α C AT AT AT L Cc AT ; AB (cid:48) AT D ) α C AT AT AT Cc AT AT ; AB (cid:48) AT AT ) α C C AT AT L L AT ; Cc D D ) α C F F D L P M M ; F (cid:48) F AT ) α C D F A F A (cid:48) M M ; DF M M M M )? α C D F A F A (cid:48) M M ; DF M M M M ) CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY α C F AT AT F A A ; F c AT M M )? α C F AT AT F A A ; F c AT M M ) α C C F A M M M AT ; M M M M AT )? α C C F A M M M AT ; M M M M AT ) α C F F D L Q M M ; F (cid:48) F AT )? α C F F D L P M M ; F (cid:48) F AT ) α C D D AT F F M M ; D (cid:48) AT M M )? α C D D AT ovF F M M ; D (cid:48) AT M M ) α C E D M M M M F AT ; Ed M M A )? α C E D M M M M F AT ; A M M A ) α C C F M M M M AB (cid:48) AT ; M M AT D )? α C C F M M M M AB (cid:48) AT ; M M AT D ) α C F F M M D AB (cid:48) AT ; F F AT M M ) Exercise.
Study the Grassmannian cubic when the 6 lines are mm i and mm i . .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. In involutive geometry I will give the name of
Grassmannian cubic to the special case wherethe 6 lines are mm i and mm i . Theorem.
The correspondence between the elements as given above and those of involutive geometryis as follows A i B i E i AB i C i = Cc i M i M i EU L i D i Aeul i a i b i aB i aB i ab i e i ae i be i ba i mm i mm i c i c i a i aeU L i nm i nm i eul Theorem. The Grassmann cubic passes through the points M i , M i , EU L i , D i . Its equation is ? g X ( − X + X + X )( − m m X + m m X + m m X ) + g X ( X − X + X )( m m X − m m X + m m X )+ g X ( X + X − X )( m m X + m m X − m m X ) = 8 m m m ( m X + m X − m X )( − m X + m X + m X )( m X − m X + m X ) where g = ( m − m )( s − m ( m + m − m m )) , . . .Proof: Using 3.2.9 on the points M i and M i , we obtain the given form, to determine g i we impose the condition that the cubic passes through EU L i , with EU L = ( − m ( m − m ) , m ( m − m ) , m ( m − m )) , this gives the system of equations m m ( m − m ) s + 4 m m ( m − m ) s = ( m − m )( m − m )( m + m ) , m m ( m − m ) s + 4 m m ( m − m ) s = ( m − m )( m − m )( m + m ) , m m ( m − m ) s + 4 m m ( m − m ) s = ( m − m )( m − m )( m + m ) , the s i are proportional to ( m − m )( s − m ( m + m − m m )) and the constant ofproprtionality is easily determined by substitution into one of the equations.Verify that (0 , m − m , − ( m − m )) is on the cubic. CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY Lemma. m ( m − m ) q + m ( m − m ) q = ( m − m ) q . m ( m − m ) q q + m ( m − m ) q q + m ( m − m ) q q = − s s ( m − m )( m − m )( m − m ) . m m ( m − m )( m − m ) q + m m ( m − m )( m − m ) q + m m ( m − m )( m − m ) q = − ( q q q ) . Definition.
Let A i and Q be a complete quadrilateral, the family of cubics associated to A i and Q are thecubics through A i , Q i and tangent at A i to aq i , with q , the polar of Q with respect to { A i } , Q i := a i × q, aq i := A i × Q i . Theorem. If Q = ( T , T , T ) , the Tucker family of cubics is( T X X + T X X + T X X )( T T X + T T X + T T X )= kT T T X X X . Any point R distinct from A i and Q is on one and only one of these cubics, notedTucker( Q )( R ) . Theorem. If R = ( R , R , R ) is on Tucker( M ) , so are isobaric ( R ) or ( R , R , R ) , ( R , R , R ) , ( R , R , R ) , semi reciprocal ( R ) or ( R , R , R ) , ( R , R , R ) , ( R , R , R ) , reciprocal ( R ) or ( R R , R R , R R ) , ( R R , R R , R R ) , ( R R , R R , R R ) , iso reciprocal ( R ) or ( R R , R R , R R ) , ( R R , R R , R R ) , ( R R , R R , R R ) . Theorem.
The following are special cases of Tucker cubics: k = 1 / , for R · a i = 0 , T ucker ( Q )( R ) = a ×× a ×× a .k = 0 , for R · m = 0 or on conic ( Q ) := conic ( A , aq , A , aq , A ) , where aq i := A i × Q i , Tucker( Q )( R ) = conic ( Q ) ×× m. k = 1, for R · aq i = 0, Tucker( Q )( R ) = aq ×× aq ×× aq .k = 9 , Tucker( Q )( Q ) . Finally the constant k is the same for Tucker( Q )( R ) and for Tucker( R )( Q ) . Tucker, Messenger of Mathematics, Ser. 2, Vol. 17, 1887-1888, p. 103 .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES.
Theorem.
0. conic ( K ) = θ.
1. Tucker ( M )( M ) is incident to A i , M A i , M , P O , P O , M AI, P, P , Atm i ,
2. Tucker ( M )( M ) is incident to A i , M A i , M, Atm i , T mm, T mm, T mm.
3. Tucker ( M )( K ) is incident to A i , M A i , K, Br i , Br, Br
4. Tucker ( M )( K ) is incident to A i , Im i , K, Br i .
5. Tucker ( M )( O ) is incident to A i , M A i , , LEM ,6. Tucker ( M )( O ) is incident to . . . Theorem.
In the finite case, there are p + 1 such cubics, each has besides the 6 vertices A i and Q i of thecomplete quadrilateral, a number of points which is a multiple of 6 except when k = and1, when it is 3( p − , k = 0 , when it is 2 p − − (cid:18) − p (cid:19) , k = 9 when it is p − − (cid:18) − p (cid:19) . (cid:18) − p (cid:19) is the Jacobi symbol = 1 when p = 1 (mod 6) and = -1 when p = 5 (mod 6) . Construction of the cubic of Tucker ( M )( M ) by the ruler only. H0.0. A i , See Fig. t and t’H0.1.
M, M ,
D0.0 to .5, construct a i , ma i , ma i , M i , M i , mm i , M A i , mm i , m i , D1.2, D3.0, 3.1, D4.12 and D4.26 construct
M aa i , M aa i , cc i , cc i , M M b i ,M M b i , mn i , mn i , P O, P O. We then proceed as followsD80.0. Aa i := M , P O, P O,
D80.0. aaA i := A i +1 × Aa i − , aaA i := A i − × Aa i +1 , D80.0. Ad i := aaA i × aaA i , D80.0. maaA i := M A i +1 × Aa i − , maaA i := M A i − × Aa i +1 , D80.0. Ab i := maaA i × maaA i , D80.0. aab i := A i +1 × Ab i − , aab i := A i − × Ab i +1 , D80.0. Ac i := aab i × aab i , D80.1. aaaa i := Aa i +1 × Aa i − , acac i := Ac i +1 × Ac i − , D80.1. Ba i := aaaa i × acac i , D80.1. aaac i := Aa i +1 × Ac i − , aaac i := Aa i − × Ac i +1 , D80.1. Bc i := aaac i × aaac i , D80.1. abA i := A i +1 × Ba i − , abA i := A i − × Ba i +1 , D80.1. Bd i := abA i × abA i , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
D80.1. mabA i := M A i +1 × Ba i − , mabA i := M A i − × Ba i +1 , D80.1. Bb i := mabA i × mabA i , D80.2. baba i := Ba i +1 × Ba i − , bcbc i := Bc i +1 × Bc i − , D80.2. Ca i := baba i × bcbc i , D80.2. babc i := Ba i +1 × Bc i − , babc i := Ba i − × Bc i +1 , D80.2. Cc i := babc i × babc i , D80.2. acA i := A i +1 × Ca i − , acA i := A i − × Ca i +1 , D80.2. Cd i := acA i × acA i , D80.2. macA i := M A i +1 × Ca i − , macA i := M A i − × Ca i +1 , D80.2. Cb i := macA i × macA i , D80.3. ‘ T ucker := cubic ( A i , mm i , M A , M A , M ) , C80.0.
M A · ‘ T ucker = 0 , C80.0.
P O · ‘ T ucker = 0 , P O · ‘ T ucker = 0 , C80.0. iOK · ‘ T ucker = 0 , C80.0. Ba · iOK = 0 , C80.0. ( Ba i × Aa i ) · ‘ T ucker = 0 , at Aa i ?C80.0. ( Bb i × Ab i ) · ‘ T ucker = 0 , at Ab i ?C80.0. ( Bc i − × Ad i ) · ‘ T ucker = 0 , at Ad i ?C80.0. ( Bd i +1 × Ac i ) · ‘ T ucker = 0 , at Ac i ?C80.1. Ab i , Ac i , Ad i · ‘ T ucker = 0 , C80.1. Ba i , Bb i , Bc i , Bd i · ‘ T ucker = 0 , C80.1. Ca i , Cb i , Cc i , Cd i · ‘ T ucker = 0 , C80.2. Ad i · tmm i = 0 , C80.2. Ac i · mIA i = 0 , C80.2. Ab i · mAM i = 0 , C80.2. Ac i · ( M A i × Ad i ) = 0 , C80.2. Ba i · pOL i = 0 , we can continue indefinitely. The cubic of Tucker ( M )( M ) . I have determined all the intersections of the following lines with the cubic of Tucker ( M )( M ) . mm i : A i , A i , M A i ,a i : A i +1 , A i − , M A i ,ma i : A i , M , M N a i ,mn i : A i , P O, M N a i +1 ,mn i : A i , P O, M N a i − ,mIA i : A i , M AI, Atm i ,cc i : A i , P, Atm i +1 ,cc i : A i , P , Atm i +1 ,m : M A i ,maM i : M A i , M , Atm i ,aaM i : M A i +1 , P O, Atm i − ,aaM i : M A i − , P O, Atm i +1 , .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. tmm i : M A i , P , M N a i − ,tmm i : M A i , P , M N a i +1 ,mpo : M , P O, P AMmpo : M , P O, , P AM ı P OK : M , M AI, : M , P, : M , P ,pOL : P O, P O, P Ol, : M A i , M N A i , M AI,pmai : P, M AI, , P AMpmai : P , M AI, , P AMpp : P, P , P Ol,
Notation.
The correspondance between the notation used here and that used in EUC. is as follows: aaA i aaA i maaA i maaA i mn , ma , mn mn , mn , ma aaM , maM , aaM aaM , aaM , maM Vigari´e (Mathesis. S´erie 1, Vol. 9, 1889, Suppl. pp. 1-26 gives the distances to the sides δ a ,δ b , δ c , the normal coordinates x, y, z which are proportional to these, and or the barycentriccoordinates α, β , γ, which are proportional to ax, by, cz, where a, b, c are the lengths of thesides.These are given in terms of a, b, c and the trigonometric functions of the angles of thetriangle. To obtain our barycentric coordinates it is sufficient to replace in α, β and γ,a by m ( m + m ) , or a by a j j ( j + j ) ,b by m ( m + m ) , or b by a j j ( j + j ) ,c by m ( m + m ) , or c by a j j ( j + j ) , and to replace the trigonometric functions as follows: sinA by sa , cosA by c m + m a , tanA by tm ,sinB by sa , cosB by c m + m a , tanB by tm ,sinC by sa , cosC by c m + m a , tanC by tm , where p = j j + j j + j j ,j = ( p − j )( p − j )( p − j ) ,s = m + m + m ( m + m )( m + m )( m + m ) = p ( j ( j + j )( j + j )( j + j ) ) ,c = m m m ( m + m )( m + m )( m + m ) , c = j ( j + j )( j + j )( j + j ) ,t = sc . (Vigari´e’s notation is here given between quotes.Twice the area “2S” by a a a s , m + m + m = 4 j j j p , m m m = j j j jp , (”2 S ”) = m m m ( m + m + m ) = (2 j j j jp ) p , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY the radius of the inscribed circle “r” by r = ( j j j ) /p . Moreover a + b + c = 2 j p , b + c − a = j j ( j + j ) , b − c = m ( m − m ) ,b + c = j ( p + j j ) ,b + c − a = 2 m m . The coordinates are given for the following points, I give first Vigari´e’s notation and underit my notation.
G K H O H o Ω Ω I I a I b I c M K { M } O M AI Br Br I I I I O I c ( ) ν Γ ν ‘ ν (cid:48) b ν (cid:48) c Γ (cid:48) a Γ (cid:48) b Γ (cid:48) c EE En EE N JI o J δ J ρ N R ρ ρ (cid:48)
V WT ar Ste BRa BraV W P P D D Z A B C T bb T nn Bro Br Br Br A B C A B C Br Br Br Br Br Br
58 59 60 61 62 63 64 δ δ mm i m m i j i mf i m aia at i lem o ok
68 69 Σ Σ (cid:48) Σ (cid:48)(cid:48) Σ IO KH o HH o bbr eul In our notation we have, ” D ” = ( m m m m m m , . . . ) , ” I ” = ( j j ( j + j )( j + j ) , . . . )” J δ ” = ( j j ( j + j )( j + j ) , j j ( j + j )( j + j ) , j j ( j + j )( j + j )) , ” J ρ ” = ( j j ( j + j )( j + j ) , j j ( j + j )( j + j ) , j j ( j + j )( j + j )) , ” P
2” = inverse (” P ”)” P
2” = (( s + m m )( s + m m ) , . . . )”41” = ( m m ( m + m )( m + m ) , . . . ) . The equations are given for the following lines, m i , m i , X i +1 × X i − , where X i = a i × ai i ,m i ] , ı , I i +1 × I i − , at i , “65”, “66”, ok, “line of Brocard” := Br × Br , e, “70”, “71”, “72”.The equations of the circles. θ , ι , ι i , γ , ”78” , polarcircle : m m X + m m X + m m X = 0 , couldonlyf indtheobviousI, andI [ i ] onit ”79” , anticomplementaryof ”78” :( X X X m ( m + m ) X m ( m + m ) X m ( m + m ) X − m ( m + m ) X X m ( m + m ) X X m ( m + m ) X X
1) = 0 , m ( m + m ) X + m ( m + m ) X + m ( m + m ) X + 2( m m X X m m X X m m X X
1) = 0 β, ,“81” family of Circles ofTucker“, λ , lambda , ”Circle of Taylor ‘Tay?“, ”Circles of Neuberg“ (D35.4), ”86“ ”Circlesof M’Cay“ ”87“:‘alphap[i] ”Circles of Apolonius“, ”88“: family of ”Circles of Schoute“,The equations are given for other curves,The conic of Brocard (D36.19), the conic of Lemoine (D36.7), also mentioned by Neu-berg, m´emoire sur le t´etra`edre, p.5 VI, iM := I × M, ı M := I × M .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. i (cid:48) M := I (cid:48) × M, ı (cid:48) := I (cid:48) × M. . . − · iM = . . . − · i (cid:48) M = . . . − · i = . . . − · i (cid:48) M .Hence the foci of . . . are M and K andthe cofoci of . . . are M and K. M K = (5 s − m m , s − m m , s − m m ) ,M K = ( m (5 s − m ) , m (5 s − m ) , m (5 s − m )) ,. . . − : ( s + 3 m m ) x x + ( s + 3 m m ) x x + ( s + 3 m m ) x x = 0 .. . . − : (5 m m − m + m + m ) x x + . . . = 0 . points of contact = (( s + 3 m m )( s + 3 m m ) , . . . ).These are the feet of the symedians of A M A . points of contact = ((5 m m + m − m + m )(5 m m + m + m − m ) , . . . ).the conic K (D36.2),the conic of Simmons (92),the conic of Steiner (S36.3),the ”hyperbola“ of Kiepert (D16.19),the first parabolas of Artzt (D36.8),The second parabolas of Artzt (96):Artzt2:The parabolas of Brocard (97):Brocard1:Brocard2:Focus(Kiepert2).theta = 0.The conic of Jerabek (99):Jerabek = inverse(e)Jerabek: m ( m − m ) X X + m ( m − m ) X X + m ( m − m ) X X = 0 . The conic centrally associated to a point (99’):conic(X). Given X = ( X , X , X ) , Let X i := a i × ( A i × X ) , conic(X) := conic( X i , × poleof i ),conic(X): ( − X + X + X ) X X X + . . . − X X X X X + . . . ) = 0 . I := conic(I),”I“ no point on it The conic I (100):supplementary( θ ) = I,I: ( j j ) (( j + j )( j + j )) X + . . . -2( ( j ( j + j )) j j ( j + j )( j + j ) X X + . . . ) = 0 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY or ( p + p p )( X + X + X ) − j ( j + j ) X X + j ( j + j ) X X + j ( j + j ) X X ) = 0 the conics of Simson (D16.18),m1m2 X1X2 + m2m0 X2X0 + m0m1 X0X1= 0 no point on it Introduction.
The cubic of 17 points is defined without explicit reference by Vigari´e. It can be defined asthe cubic through the vertices of a triangle, its midpoints and the midpoints Mma i betweenthe vertices and the feet. The other 8 points are the barycenter, orthocenter, center of theoutscribed circle, point of Lemoine, and the 4 centers of the tangent circles. Other pointsand tangent on it will also be given. In particular, KLL i , Flor, ART M , are on the cubicand at i is the tangent at A i , mf i is the tangent at M i , mk is the tangent at M, ok is thetangent at K. Definition.
The cubic of 17 points is defined by”cubic17 := cubic(A i ,M i ,Mma i ). Theorem. O · “cubic17 = M · ”cubic17 = M · “cubic17 = K · ”cubic17= I · cubic17 = I i · “cubic17 = 0. K LL i · ”cubic17 = 0. A RTM · “cubic17 = 0. at i · ”cubic17 = 0,Proof.“cubic17: m ( m + m ) X X ( X − X ) + m ( m + m ) X X ( X − X )+ m ( m + m ) X X ( X − X ) = 0 . .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Theorem.
M M O K F lor ART M C a C b C c C dM K O M M ART M F lor C bM C a M ART M K MO F lor K O ? K O MF lor ART M ? O ? MART M F lor K ? M MC a C b M ? ? C b C aC c ART M C d ? ? M ? ? A i M i M ma i KLL i M M i A i KLL i M ma i M M ma i C d i A i C b i O KLL i M i C c i A i K A i M ma i M i C d i F lor C d i C b i M ma i C e i ART M C c i KLL i C f i M i C a C f i C c i ? KLL i C b C b i C e i C d i C g i C c C d i C b i ? ? A i M i M ma i KLL i A i K M M OM i M O K ART MM ma i M K C c i MKLL i O ART M M C aA i − M i − M ma i − KLL i − A i +1 M i A i KLL i M ma i M i +1 A i M ma i M i C d i M ma i +1 KLL i M i C c i A i KLL i +1 M ma i C d i C b i M i Exercise.
Construct the tangents to ”cubic17 at Mma i and at M . Exercise.
Give properties of ” cubic17 := cubic(A i , M i , M ma i ). Partial answer to is [( m − m s − m , ( m m s − m , − ( m m s − m . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
The tangent at M is [ m m m − m s − m , m m m − m s − m ,m m m − m s − m . A ⊕ A = K, A ⊕ A = M , A ⊕ M = M, A ⊕ Mma = M , A ⊕ Mma = K LL ,A ⊕ K LL = O, M ⊕ M = K, M ⊕ M = Mma ,M ⊕ K LL = A RTM, M ⊕ K LL = C17a ,O ⊕ O = Flor, whereFlor = (( m m s − m s − m , ( m m s − m s − m , ( m m s − m s − m ,tangent at M is [ m m m − m s − m , . . . ] . M × M = ( m m m s − m s − m s − m ) , . . . ) ,M × F lor = [( m − m s − m , ( m − m s − m , ( m − m s − m ,M × ART M = [( m − m s − m s − m ) , . . . ] ,M ⊕ ART M = K,K × F lor = [( m − m m m m m s − m , . . . ] ,C a = ( m m m s − m s − m s − m ) , . . . ) , C b = (( s − m s − m )( s − m ) , . . . ) , C c = (( m m s − m s − m , . . . ) , C d = (( s − m )( s − m )( s − s − s m , . . . ) , C b = (2 m m m s − m s − m , s s − m s − m ) ,s s − m s − m )) , C c = (( m m s − m s − m , m s s − m , m s s − m ,C d = ( m s , ( m m s − m , ( m m s − m ,C e = (( s − m )( s − m )( s − s − s , − m m m s − m )( s − m − s m m + m )+2 s , − m m m s − m )( s − m − s
21 + 2 m m + m ) + 2 s ,C f = ( s s − m )( s − m ) , m m m s − m s − m ) , − m m m s − m s − m )) ,C g = (( m m s s − m s − m s − m s − s
21 + 2 s
111 + 2 m s − m ))(( s − s
21 + 2 s
111 + 2 m s − m ))) , − m s − s − s s − s
21 +2 s
111 + 2 m s − m ))( s s − m + m m )) , − m s − s − s s − s
21 + 2 s
111 + 2 m s − m ))( s s − m + m m )) , Answer to O · ” cubic17 = M · ” cubic17 = M · ” cubic17 = K · ” cubic17= 0.KLL i · ” cubic17 = 0. at i · ” cubic17 = 0, ” cubic17: m ( m + m ) X X ( m X − m X ) + m ( m + m ) X X ( m X − m X ) + m ( m + m ) X X ( m X − m X ) = 0 . .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. cubic21 Introduction.
In posthumously published works of Dan Barbilian, also known in his native RoumanianCountry as the poet Eon Barbu, the following Theorem is proven. The loci of the pseudocenters of the isotropic cubics which pass through the vertices of a complete quadrilateraland 2 of its diagonal elements is a circle. I observed that in the case where the isotropicpoints are the fixed points of the involution determined by the 3 pairs of opposite sides of thequadrilateral, the third diagonal point is also on the cubics. It is this family of cubics whichwill be studied now, to which I will give the name of the Poet-Mathematician Barbilian.
Definition. An isotropic cubic is a cubic which passes through the isotropic points.The pseudo center of an isotropic cubic is the intersection of its tangents at the isotropicpoints. Theorem. [Barbilian]
The family of isotropic cubic through the vertices B j of a complete quadrangle and 2 of itsdiagonal points A := ( B × B ) × ( B × B ) and A := ( B × B ) × ( B × B ) has a circle asthe locus of the pseudo centers. This circle is the
Miquel circle of the complete quadrangleand the 2 diagonal points.I remind the reader that this circle passes through the center of the circumcircles of thetriangles { B , B , A } , { B , B , A } , { B , B , A } , { B , B , A } . See g334 Definition.
The isotropic cubics through the vertices of a triangle, the feet and the orthocenter will becalled
Barbilian cubics.
Corollary.
The family of Barbilian cubics has a circle as the locus of its pseudo centers.
In this case, B = A , B = M , B = M , B = M and the circles circumscribed to { B , B , A } , { B , B , A } pass through the point of Miquel, M . The isotropic points are also called circular points. Barbilian calls a pseudo center, a pseudo focus. CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem.
The Miquel circle of B j , A and A is the circle of Brianchon-Poncelet. Theorem.
The following are degenerate Barbilian cubics.0. ‘ Aam ×× ma , its equation is ( m + m ) m X X ( m X − m X ) − m m X X ( m X − m X ) − m m X X ( m X − m X ) = 0 . ‘ M ma ×× a , its equation is m X X ( m X − m X ) − m X X ( m X − m X ) = 0 . Proof: 0 and 1, follow from the definition of the circles ‘Aam i and ‘Mma i given in sectionD11.1 and .2. Theorem. The cubics through A i , M i , M , are a m X X ( m X − m X ) + a m X X ( m X − m X )+ a m X X ( m X − m X ) = 0 . A necessary and condition for the cubics through A i , M i , M , to be Barbilian cubics,is a + a + a = 0 .Proof: It is easy to verify 0. For 1, any Barbilian cubic is a linear combination of thedegenerate cubics given in the preceding Theorem and this satisfy the given condition. More details on Recall that the isotropic points are ( m ( m + m ) , − m m − jτ, − m m + jτ ) , with j = ± and τ = − m m m s . Theorem.
In homogeneous Cartesian coordinates (
X, Y, Z ) , with A = (0 , h, , A = ( b, , , A = ( c, ,
1) and isotropic points ( ± j, , The coordinates of the sides, feet, orthocenter and altitudes are a = [0 , , , a = [ h, c, − ch ] , a = [ h, b, − bh ] ,M = (0 , , , M = ( c ( h + bc ) , hc ( c − b ) , h + c ) , M = ( b ( h + bc, hb ( b − .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. c ) , h + b ) ,M = (0 , bc, − h ) ,m = [1 , , m = [ c, − h, − bc ], m = [ b, − h, − bc ].1. The circle through A , A , M , M is αm : X + Y + bcZ − ( b + c ) ZX = 0 . The circle through A , M , M , M is µa : h ( X + Y ) − hbcZ + ( bc − h ) XY = 0 . The Barbilian cubics are kαm ×× X + lµa ×× Y = 0.4. The pseudo center is, with d = k ( b + c ) + l ( bc − h ),( kd, − lhd, k + l h )) . This is a parametric equation of the circle of Brianchon-Poncelet :2 h ( X + Y ) − ( h − bc ) XY − h ( b + c ) ZX = 0 . The transformation from barycentric to Cartesian coordinates is b ch and its inverse is h ( c − b ) c − b − h − c chh b − bh , where the barycentric coordinates of the orthocenter are given by m = bc ( c − b ) h + bc , m = − c, m = b ,provided h = m m s m . The barycentric coordinates of the pseudo center are easily derived using the value of h and of m d = km ( m − m ) − l ( s − m m s ).The details of the proof is left to the reader. Answer to
0, is straigthforward.For 1, αm is u [ h, b, − hb ] ×× [ h, c, − hc ] = v [ c, − h, − bc ] ×× [ b, − h, − bc ] . With j = − , u = ( cj − h )( bj − h ) , v = ( hj + b )( hj + c ) = − u. After dividing by h + bc we get the equation 1.For 2, µa is u [ h, b, − hb ] ×× [ c, − h, − bc ] = v [ h, c, − hc ] ×× [ b, − h, − bc ] .u = ( hj + c )( bj − h ) , v = ( hj + b )( cj − h ) = u. After dividing by c − b we get the equation 2.For 4, the tangent at ( j, , − k + 2 lhj, kj + 2 lh, k ( b + c ) + l ( bc − h )] . The tangent at the other isotropic point is obtained by replacing j by − j . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Their intersection is 4, after dividing by 4 j . It is easy to verify that 5, is the equation of acircle through M i , and that the pseudo center is on it for all values of k and l .For 5, the coordinates of vertices A i give the coefficients of the matrix. The transform of( m , m , m ) is ( m b + m c, hm , s ). Comparing with (0 , bc, − h ) gives 5. 6, is straightfor-ward. For 7, we need to related the cubics in Cartesian and barycentgric coordinates.If weuse k’ and l’ for the barycentric case, comparison of the coefficients of X X k (cid:48) m m = kc (2 bc − bc − c ) = kc ( b − c ) , and l (cid:48) m m = lh ( hc − hbc ) = lh ( c − b ) . Usingproportionality we can therefore write 7.The pseudo center will be (0,0,1) if d = 0 , this gives k = h − bc = m m ( s + m ) m , l = m − m .Substituting, we get, after division by m m m − , k (cid:48) = s + m , l (cid:48) = ( m − m ) s , hence a = − ( m + m )( s + m ) , a = (2 m − m ) s + m m , a = (2 m − m ) s + m m . To checkthis independently, we should verify that M × I is tangent to the cubic for these values of k (cid:48) and l (cid:48) . (0 , , × ( m ( m + m ) , − m m − jτ, − m m + jτ ) = [ m m + jτ, m ( m + m ) , . Theorem.
Given an Barbilian cubic Γ , there exists a line l and a circumscribed conic φ such that Γ = θ ×× l + m ×× φ .More specifically, with l arbitrary, l = [ l , l − a , l + a ], φ = b m X X + b m X X + b m X X = 0, with b = − m a + m a − ( m + m ) l , b = − m a − ( m + m ) l , b = m a − ( m + m ) l . Proof: Identification of the coefficients of X X and X X gives a m = ( m + m ) l + b , − a m = ( m + m ) l + b , subtracting gives, a = l − l , and similarly a = l − l , and a = l − l . By substitution, we obtain b and similarly b and b , using a + a + a = 0. Definition. l is called a radical axis of Γ, φ is called the corresponding radical conic of Γ. θ could be replaced by an other circle. Theorem. The non trivial ideal point is ( a , a , a ).1. The tangent at the non trivial ideal point, or asymptote is [ m a a ( m a − m a ) , m a a ( m a − m a ) , m a a ( m a − m a )] . Proof: This follows from the fact that the non trivial ideal point is m × l . The tangentis obtain by taking the partial derivatives respectively with respect to X , X and X at( a , a , a ). The first one is m a a ( m a − m a ) − m a a ( m a − m a ) = − m a a ( m a − m a ) . .2. THE GEOMETRY OF THE HEXAL COMPLETE 5-ANGLES. Comment.
Special Barbilian cubics can be obtained by combining the equations of Theorem 3.2.17 Forinstance, βa m , m and m . βa
1, by using on the equations 0, the multipliers m m , m m and m m . βa
2, by using on the equations 1, equal multipliers.
Theorem.
The Barbilian cubic βa m ( m − m ) X X ( m X − m X ) + m ( m − m ) X X ( m X − m X )+ m ( m − m ) X X ( m X − m X ) = 0 has the following properties: A radical axis is − mai , with mai = [ m , m , m ] . The corresponding radical conic has the equation m ( m + m ) X X + m ( m + m ) X X + m ( m + m ) X X . The non trivial ideal point is
M K = ( m − m , m − m , m − m ) . The asymptote is [ m ( m − m )( m − m )( s − m s ) , m ( m − m )( m − m )( s − m s ) , m ( m − m )( m − m )( s − m s )] . The tangent at A i is mka i , with mka = [0 , m − m , − ( m − m )] . Theorem.
The Barbilian cubic βa m ( m − m ) X X ( m X − m X ) + m ( m − m ) X X ( m X − m X )+ m ( m − m ) X X ( m X − m X ) = 0 has the following properties: A radical axis is − m .1. The corresponding radical conic is m m m ‘ Steiner, with ‘ Steiner = m X X + m X X + m X X = 0 , The non trivial ideal point is
EU L with
EU L = ( m ( m − m ) , m ( m − m ) , m ( m − m )) . The asymptote is m .4. The tangent at A i is mka i , with mka = [0 , m − m , − ( m − m )] . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem.
The Barbilian cubic βa m ( s − m ) X X ( m X − m X )+ m ( s − m ) X X ( m X − m X ) + m ( s − m ) X X ( m X − m X ) = 0 has the following properties: A radical axis is eul with eul = [ m − m , m − m , m − m ] . The corresponding radical conic is m ( m − m ) X X + m ( m − m ) X X + m ( m − m ) X X = 0,2. The non trivial ideal point is
Ieul , where
Ieul = ( s − m , s − m , s − m ) . The asymptote is
M km × Ieul, [ m ( m − m )( s − m )( s − m ) , m ( m − m )( s − m )( s − m ) , m ( m − m )( s − m )( s − m )] . The tangent at A i is Several mappings are defined and these allow an algebraic definition of many of thepoints, these will be given here as theorems for the points already defined and as definitionfor the others.reciprocal( X , X , X ) := ( X X , X X , X X ) , reciprocal( x , x , x ) := ( x x , x x , x x ) , inverse( X , X , X ) := ( m ( m + m ) X X , m ( m + m ) X X , m ( m + m ) X X ) , complementary( X , X , X ) := ( X + X , X + X , X + X ) , [Nagel, 1885]anticomplementary( X , X , X ) := ( − X + X + X , X − X + X , X + X − X ) , [inversetransformation of de Longchamps, 1886]supplementary( X , X , X ) := . . . algebraically associated( X , X , X ) := (( − X , X , X ) , ( X , − X , X ) , ( X , X , − X )) , Brocardian( X , X , X ) := (( X X , X X , X X ) , ( X X , X X , X X )) , isobaric( X , X , X ) := (( X , X , X ) , ( X , X , X )) , semi reciprocal( X , X , X ) := (( X , X , X ) , ( X , X , X ) , ( X , X , X )) , associated( X , X , X ) := ( X − X , X − X , X − X ) . Theorem. K = inverse ( M ) , O = inverse ( M ) = complementary ( M )2. ( Br , Br
2) = brocardian ( K ) , ? 3. I i = algebraically associated ( I ) , N = anticomplementary ( I ) , J = reciprocal ( N ) , .3. FINITE PROJECTIVE GEOMETRY. N i = algegraically associated ( N ) , J i = reciprocal ( N i ) . Definition.
The following are the definition of other points.0. H := reciprocal ( M )1. I c := complementary ( I ) , I := reciprocal ( I ) ,
3. “Center of equal parallels” := anticomplememtary(I ) ,
4. (” J δ ” , ” J ρ ”) := Brocardian(I). Exercise.
Define in terms of the above functions as many points as you can in Theorem . . . . Exercise.
Determine, for many of the points of Definition . . . a linear construction and determine theirbarycentric coordinates.
The Theorems given here are deduced from Theorems of Involutive Geometry.
Theorem.
Given 6 points A i and B i , forming an hexagon inscribed to a conic α and outscribed to another conic β . Let C be the point common to A i × B i . Let T i be the vertices of the tangentsto α at A i . The lines B i × T i have a point D in common. The line C × D passes through the pole of with respect to the triangle { A i } of theDesargues line of the perspective triangles { T i } and { B i } . The Theorem generalizes a Theorem of Kimberling 3.4.6 and 3.4.6 using 3.4.6.58
CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem.
Given the special Desargues configuration with the points A i on the lines of the triangle { M M , M M , M M } with center of perspectivity M . Let m be an arbitrary line and M A i be its intersection with the side a i of the triangle { A , A , A } , if T M a i is the intersectionof the line M M i +1 M A i − and the line M M i − M A i +1 , then the lines joining the points A i to the T M a i have a point ART M in common.
The Theorem generalizes a Theorem of Kimberling 3.4.3 assuming that the excenters arereplaced by the vertices of the anti complimentary triangle and the direction of the altitudesare replaced by the intersections of the orthic line with the side of the triangle.
Definition.
The point
ART M is called the point of Luke.
I will now describe the Theorems of involutive Geometry in the traditional way, refering forto proofs of the corresponding sections of the hexal configuration.Starting with affine geometry, we obtain an involutive geometry, if we choose among allthe possible involutions on the ideal line, a particular one, called fundamental involution.We could also start directly from projective geometry and choose among all the possibleinvolutions one involution on one of all the possible lines.This involution can be given in many ways,0. by 2 points, the fixed points of the involution,1. by 2 pairs of corresponding points on a line,2. by a polarity and a line which does not belong to its line conic,3. by an hexal complete 5-angle, . . . . See II.3.The definitions will be given in terms of the fundamental involution. Because this involutioncan be elliptic or hyperbolic, there are 2 distinct types of real involutive geometries, ellipticand hyperbolic. I will study them together and give theorems, in the hyperbolic case, whichin some cases can be used as an alternate definition of the concepts. Such theorems will benoted with (H. D.) . When the additional notions of measure of distance and angles willhave been introduced, the elliptic involutive geometry will become the Euclidean Geometryand the hyperbolic one, will be that of Minkowski. A third geometry which correspondsto the confluence of the 2 fixed points of the involution will be considered later, it is theparabolic (involutive) geometry which becomes the Galilean geometry.Among the many ways of starting I will give one. It is a good Exercise to ask students totry other approaches.I will choose one line m as the ideal line and a conic, given by 5 points (again an other set .4. FINITE INVOLUTIVE GEOMETRY. A on the conic, determine its tangent t A ,the parallel tangent t B , by a construction which is the dual of that of finding the secondpoint of intersection of a line with the conic, and the point of contact B . A × B is a diameter.Perpendicular directions are obtsined as follows. If I p is an ideal point, we determine thesecond intersection P of A × I p with the conic, the perpendicular direction is then ( P × B ) × m. We can therefore construct the altitudes and therefore the orthocenter.
Definition.
Starting with an affine geometry associated to p , a particular involution on the ideal linewill be called the fundamental involution . Definition.
If the fundamental involution is hyperbolic, its fixed points are called isotropic points , theother points on the ideal line will be called ideal points or directions . (In a hyperbolicinvolutive geometry, the isotropic points are no more called ideal points). The lines throughthe isotropic points, distinct from the ideal line, are called isotropic lines .The lines which are neither ideal or isotropic are called ordinary lines , the points whichare not ideal or isotropic points are called ordinary points , ordinary lines or points willabbreviated from now on by lines or points. On an ordinary line, there are p ordinary pointsand one ideal point. Definition.
Corresponding pairs of points in the fundamental involution are called perpendicular idealpoints or perpendicular directions . 2 lines whose ideal points are perpendicular directions arecalled perpendicular lines .Some obvious results follow from these definitions and from those of the corresponding affinegeometry. For instance: Theorem.
All the lines perpendicular to a given line are parallel.
Definition.
If the involution defined by a conic on the ideal line is the same as the fundamental involution,the corresponding conic is called a circle and the corresponding polarity is called a circularity .60
CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem. (H. D)
In a hyperbolic involutive geometry, a necessary and sufficient condition for a conic to be acircle is that it passes through the isotropic points.
Definition.
The center of a conic is the pole of the ideal line in the corresponding polarity. (See II.2.3.0).
Theorem. (H. D)
In a hyperbolic geometry, the center of a circle is the intersection of its isotropic tangents.
Definition. A diameter of a conic is a line passing through its center (See II.2.3.1). Definition. A mediatrix of 2 points A and B on a line l , which is not an isotropic line, is the lineperpendicular to l through the mid-point of A B . (See II.6.2.6)
Example.
In the examples of involutive and Euclidean geometry, I will make one of 2 choices for theideal line and for the defining circle.0. In the first choice,0.[1 , ,
1] is the ideal line, as in affine geometry.1. X + X + k X = 0, k (cid:54) = − , is the defining circle, − (1 + 2 k ) N p for the elliptic case, − (1 + 2 k ) R p for the hyperbolic case.2. Let δ := − − k .If k (cid:54) = −
1, the isotropic points (1 , y, − − y ) correspond to the roots y and y of3. (1 + k ) y + 2 ky + (1 + k ) = 0 . or with4. k (cid:48) = k k ) , to the roots of5. y + k (cid:48) y + 1 = 0.Therefore y = − k + δ k and y = − k − δ k . If k = − , the isotropic points are (0,1,-1) and (1,0,-1).The polar of ( X , X , X
2) is [ X , X , kX X , X , − X − X
1) is( kX k ) X , − (1 + k ) x − kX , X − X k = − , the conic 0.1. is tangent to the ideal line. .4. FINITE INVOLUTIVE GEOMETRY. , ,
1] is the ideal line, as in Euclidean geometry.1. kX + X = X , k (cid:54) = 0, the defining circle, − k N p for the elliptic case, − k R p for the hyperbolic case. δ := − k. If k = 0, the conic 1.1. is tangent to the ideal line.The isotropic points are (1 , δ,
0) and (1 , − δ, X , X , X
2) is [ kX , X , − X X , X ,
0) is ( X , − kX , Definition.
In a triangle { A i } , the altitude ma i from A i is the line through A i which is perpendicular tothe opposite side a i := A i +1 × A i − (C0.1,N0.3). Theorem.
The altitudes ma i of a triangle are concurrent at a point M . (D0.12) Definition.
The point M is called the orthocenter of the triangle.(N0.2) Theorem.
The necessary and sufficient condition for a triangle to be a right triangle at A i is that itsorthocenter M coincides with A i . Theorem.
The necessary and sufficient condition for a triangle to be an isosceles triangle is that theorthocenter be on the altitude from A i and distinct from the center of mass. Theorem.
The necessary and sufficient condition for a triangle to be an equilateral triangle is that theorthocenter and the barycenter coincide.
Introduction.
We are now ready to give a large number of results of finite involutive geometry associatedto a scalene triangle whose vertices A , A , A and whose sides a , a , a are ordinary.62 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem II.6.2.7. determines a point M, the center of mass, at the intersection of the medians A M , A M , A M , the points M i being the mid-points of pairs of vertices.Theorem 3.1. determines a point M , the orthocenter, at the intersection of the altitudes A M , A M , A M , the points M i being the feet of the altitudes.In a scalene triangle, M and M are distinct, are distinct from the vertices and are notcollinear with any of the vertices. A large number of results can therefore be obtained asdirect consequences of rephrasing the results of Theorem 3.6. and 4.0.Similar results can be obtain for right triangles, for isosceles triangle and for equilateraltriangles. These will be left as exercises.These results were in fact the starting point of our study of finite Euclidean geometry, asexplained in section . . . .All references will be to Theorems 3.6. and 4.0. unless explicitely indicated. Definition.
The ideal points
M A i of a triangle are the ideal points on its sides.The orthic points M A i of a triangle are the points on the corresponding sides a i of thetriangle and mm i of the orthic triangle (D0.13, N0.6). See Fig. 1. Theorem.
The orthic points
M A i are on the orthic line m (D0.14*). Definition.
The triangle M i is called the complementary triangle. Its sides are denoted mm i . The triangle M i is called the orthic triangle . Its sides are denoted mm i (D0.18, N0.5). Definition.
The orthic line m of a triangle is the polar of its orthocenter with respect to the triangle(N0.8).Its direction EU L is called the orthic direction (N1.1).
Definition.
The line eul := M × M is called the line of Euler (D1.0, N1.0). Theorem.
The mid-points M i at the intersection of the medians ma i with the sides a i and the feet M i of the altitudes ma i are on a circle γ (D1.20, C1.4). See Fig. 2. Definition.
The circle γ is called the circle of Brianchon-Poncelet (N1.11). .4. FINITE INVOLUTIVE GEOMETRY. Theorem. If M aa i ( M aa i ) is the intersection of the median ma i +1 ( ma i − ) with the altitude ma i − ( ma i +1 ) , then the lines mM a i joining M aa i and M aa i have a point K in common (D1.2,D1.3, D1.4*). See Fig. 3. Definition.
The point K is called the point of Lemoine (N1.2). Definition.
The circumcircle θ of a triangle { A , A , A } is the circle passing through the vertices of thetriangle (D1.19, H1.1, N1.10). Theorem.
The line ta i through the vertex A i parallel to the side m i of the orthic triangle is the tangentat A i to the circumcircle (D1.7, D1.19*). See Fig. 4. Definition.
The triangle with sides ta i is called the tangential triangle. Its vertices are denoted by T i (D1.8, N1.5). Definition.
The mixed triangles are the triangles with respective sides c i := M i +1 × M i − and c i := M i +1 × M i − (D1.13, N1.6).The mixed feet are the points CC i , ( CC i ) on the side of the given triangle and the corre-sponding side c i , ( c i ) of the mixed triangle (D1.14, N1.7). Dee Fig. 5. Theorem.
The mixed feet CC i , ( CC i ) of a mixed triangle are collinear on the line p ( p ) (D1.15*). Definition.
The line p and p are called the mixed lines of a triangle (N1.8). Theorem.
The mixed lines p and p of a triangle meet at the point P P which is on the line of Euler (D1.16, C1.0).
Definition.
P P is called the mixed center of the triangle (N1.9).64
CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Definition.
The intersection
IM a i of a median with the orthic line is called a medorthic point (D0.15,N0.9). Definition.
The intersection of the lines me i ( me i ) joining the medorthic points IM a i +1 ( IM a i − ) tothe foot M i − ( M i +1 ) are called the points of Euler EE i ( Eulerian points E i ) (D5.0, D5.1).Fig. 6. Theorem.
The points of Euler EE i are the mid-points of the segment joining the orthocenter M to thevertex A i (D5.1, C5.3). Theorem.
The points of Euler EE i are on the circle of Brianchon-Poncelet (C5.5). The Eulerian point EE i is on the median ma i as well as on the circle of Brianchon-Poncelet (C5.0, C5.5). Theorem.
The lines em i joining the mid-points M i to the Eulerian points EE i are concurrent at a point EE.
The lines em i joining the feet M i to the Eulerian points EE i are concurrent at a point EE (D5.2, D5.3*). EE is on the line of Euler and is the center of the circle of Brianchon-Poncelet. EE is on the line of Euler and is the pole of the orthic line with respect to the circle ofBrianchon-Poncelet (C5.1, C5.4, N5.1). Definition.
The mediatrix mf i is the line through through the mid-point M i perpendicular to the corre-sponding side a i (D6.0, N6.0). Theorem.
The vertex T i of the tangential triangle is on the mediatrix mf i (D6.0, C6.8, N6.0). The mediatrices mf i are concurrent at a point O. The diameters mf i of the circle of Brianchon-Poncelet which pass through the feet of thealtitudes pass through the same point O (D6.4*, N6.1). Definition. O is the circumcenter or center of the circumcircle (N6.1). .4. FINITE INVOLUTIVE GEOMETRY. Theorem.
The circumcenter O is on the line of Euler.The point O is on the line of Euler (C6.1). Definition. An equilateral conic is a conic whose ideal points are harmonic conjugates of the isotropicpoints.An coequilateral conic is a conic whose points on the orthic line are harmonic conjugates ofthe coisotropic points.We leave as an exercise the pproof of the following Theorem and Corollary. Theorem.
If an conic passes through the vertices of the triangle it is equilateral if and only if it passes through the orthocenter.Its center is on the circle of Brianchon-Poncelet. it is coequilateral if and only if it passes through the barycenter.Its cocenter is on the circle of Brianchon-Poncelet. Corollary.
A conic a X + a X + a X + b X X + b X X + b X X = 0,0. is equilateral if and only if m m ( a + a − b ) + m m ( a + a − b ) + m m ( a + a − b ) = 0.1. it is coequilateral if and only if m ( m a + m a − m m b ) + m ( m a + m a − m m b ) + m ( m a + m a − m m b ) = 0. Definition.
The conic of Kiepert is the conic circumscribed to the triangle passing through the barycenterand the orthocenter. (D3.8.)The conic of Jerabek is the conic circumscribed to the triangle passing through the orthocenterand the point of Lemoine. (D36.16.)These are therefore equilateral. (C3.3 and C36.). The center of one conic is the cocenter ofthe other and these are on the circle of Brianchon-Poncelet (C8.9 and C36.18.)
Definition.
The circle through the vertices T i of the tangential triangle is the circle of Neff . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem. The circle of Neff is a cocircle. The ortic line is the radical axis of the circle of Neff and both the circumcircle and thecircle of Brianchon-Poncelet.
Definition. A triangle of Neff is a triangle whose orthocenter is on the conic. Exercise.
Prove that in a triangle of Neff, one of the sides of the tangential triangle is a diameter ofthe circle of Neff. Determine other conditions for this to happen. xxx
Definition.
The points
EU L i at the intersection of the corresponding sides of the complementary triangleand of the orthic triangle are called the complorthic points (D8.0, N8.0).The lines aeU L i joining the complorthic points are called complorthic lines (D8.3, N8.1).The triangle whose vertices are the complorthic points is called the complorthic triangle (N8.2). Definition.
The intersections of corresponding sides of the mixed triangles are the mixed points D i (D8.4,N8.4). Theorem. The mixed points D i and D i are on the line of Euler (C8.2).1. The vertex A i and the mixed point D i are on the complorthic line aeU L i . (C8.1, C8.3).2. The lines nm i joining the mid-points of the sides to corresponding complorthic points EU L i are concurrent in a point S. The lines nm i joining the feet of the altitudes to the corresponding complorthic points EU L i are concurrent in a point S (D8.1, D8.2*). Definition.
The points S and S are the point and copoint of Schr¨oter (N8.3). .4. FINITE INVOLUTIVE GEOMETRY. Theorem. S is the point of Miquel of the quadrangle a i +1 , a i − , ma i +1 , ma i − . S is therefore also onthe circles through A i , M , M i +1 , M i − of center E i and on the circles through A i +1 , A i − ,M i of center M i . (See .) Theorem. The points of Schr¨oter are on the circle of Brianchon-Poncelet (C8.8).1.
The first point of Schr¨oter S, the Eulerian point EE i and the mixed point D i are onthe same line s i The second point of Schr¨oter S, the point of Euler point EE i and the mixed point D i are on the same line s i . (D8.5, C8.4). Theorem.
The conic through the barycenter M , the orthocenter M and the feet Gm i of the perpen-dicular iM A i from M to the corresponding altitude ma i are on a circle ‘ omicron (D10.3,D10.4, D10.7, C10.7). This circle passes also through the perpendiculars Gm i which are the feet of the perpendicu-lars gm i from M to the corresponding median ma i . { M, M } is a diameter whose mid-pointis G (C10.1, C10.8). See Fig. 9. Definition. ‘ omicron is called the orthocentroidal circle (N10.2). Theorem.
If we join . . . .(D6.1, D10.1, D10.2, D10.3*).
Definition.
The G is the center of the orthocentroidal circle ‘ omicron (N10.13). Theorem.
The line be i is parallel to the median ma i (D10.5, N10.0, C10.2). Theorem.
The 3 circles, the circumcircle θ , the circle γ of Brianchon-Poncelet and the orthocentroidalcircle ‘ omicron have the same radical axis m. (C1.5, C10.9) CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Definition. An orthocentric quadrangle is . . . .An example is provided by the circumcentral orthocentric quadrangle (N10.1). Theorem.
The line tm i through the mid-point M i parallel to the side m i of the orthic triangle is tangentat M i to the circle of Brianchon-Poncelet. (D12.0, C12.11) Definition.
The line at i joining the vertex A i of the triangle to the vertex T i of the tangential triangleare called the symmedians (D12.1, N12.0). Theorem.
The symmedians at i are concurrent at a point K (C12.6). Theorem.
The point K of Lemoine, the first point S of Schr¨oter and the point G are collinear on theline gk. The point K of Lemoine, the second point S of Schr¨oter and the point G are collinear onthe line gk (D12.2, C12.7). Definition. Te tangential point AM a i ( AM a i ) is the intersection od the parallel am i +1 ( am i − ) through A i − ( A i +1 ) to the altitude ma i and the parallel am i − ( am i +1 ) through A i to ma i +1 ( ma i − )(D6.8, D14.4, N14.0). Definition.
The tangential circle χa i ( χa i ) is the circle though the vertices A i +1 and A i − tangent at A i − ( A i +1 ) to the side a i +1 ( a i − ) (D14.13, C14.8, C14.5). Theorem.
The tangential circle χa i ( χa i ) passes through the tangential point AM a i +1 ( AM a i − ) (D14.13). Definition.
The parallels of Lemoine kk i are the lines through the point K of Lemoine parallel to thesides of the triangle (D15.0, N15.0). See Fig. 13. .4. FINITE INVOLUTIVE GEOMETRY. Definition.
The vertices Br i of the first triangle of Brocard are the intersections of the mediatrices mf i with the parallels of Lemoine kk i (D15.1, N15.1). Theorem.
The lines br i joining the vertices of a triangle A i to the corresponding vertex Br i of thefirst triangle of Brocard are concurrent at a point BR (D15.2, D15.3*). Theorem.
The lines br i ( br i ) joining the vertices A i − ( A i +1 ) to the vertices Br i +1 ( Br i − ) of the firsttriangle of Brocard are concurrent at a point Br ( Br (D15.4, D15.5*). Definition.
The point Br ( Br ) is called the first (second) point of Brocard (N15.4). Definition. d2trbrThe points Br i at the intersection of the parallel ok i to the side a i through the center O of the circumcircle and the perpendicular km i to a i through the point K of Lemoine arethe vertices of the second triangle of Brocard (D13.4, D13.3, D15.6, N15.2). Theorem.
The lines br i joining corresponding vertices Br i and Br i of the first and second triangleof Brocard are concurrent at a point Bro (D15.9, D15.10*).
Definition.
The cross tangential line mf f i is the line through the tangential points AM a i +1 and AM a i − (D15.7, N15.6). Definition.
The vertices of the third triangle of Brocard Br i are the intersections of the cross tangentialline mf f i and the corresponding symmedian at i (D15.8, N15.3). Theorem.
The vertices Br i , Br i and Br i of the first, second and third triangle of Brocard, the firstand second point of Brocard Br Br , the center O of the circumcircle and the point K of Lemoine are on a circle β with center Bro , the mid-point of { K, O } (D15.18, C15.17,C15.18, C15.12, C15.13, C15.7). CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Definition.
The circle β is called the circle of Brocard (N15.6). Definition.
The conics of Tarry ‘
T arry [ i ] are the conics through the barycenter M and through 2 vertices,tangent there to the side through the third vertex, A i . (N19.0.) Theorem.
Let
Apt , ( Apt ) be the intersection of the line through A parallel to the median ma ( ma )with the line through A ( A ) parallel to the median ma and circularly for Apt , ( Apt ), Apt , ( Apt ), then the line Apt × ( Apt ) is the tangent common to ‘ T arry and ‘ T arry with Apt and Apt as point of contact. (D19.7, C19.0, D33.7, C33.5, C33.6.)From the coordinates associated with the symmetric Theorem using M instead of M , itis easy to solve the problem of C. Bindschelder, El. Math. 1990, p. 56. Definition.
The line of Schr¨oter , pap is . . . . (N4.1) It is tangent to the conics of Steiner, Lemoine andSimmons, (P. de Lepiney, Math. 1922-133)(C36.7) !dont have def. of Lemoine and Sim-mons, these are of the form b x x + b x x + b x x = 0 , with b m ( m − m ) + b m ( m − m )+ b m ( m − m ) = 0 . ! M K.Center (‘ Lemoine ) = 0 , M K.Center (‘ Simmons ) =0 , ?? Theorem. [Kimberling] The lines joining the vertices T i of the tangential triangle to the second intersection B i of the medians ma i with the circumcircle θ are concurent at a point CK . (C47.0.)1. The lines joining the vertices T i of the tangential triangle to the second intersection B i of the altitudes ma i with the circumcircle are concurent at a point C K . (C47.0.) Definition.
The points CK and CK just defined is called respectively the point and copoint of Kimberling .(N47.0.) Theorem. [Kimberling]
The point and copoint of Kimberling are on the line of Euler. (C47.4) See 3.3.0 .4. FINITE INVOLUTIVE GEOMETRY.
Theorem.
0. Desargues( M, { A i } , { B i } , ee ). (D47.21)1. Desargues( M , { A i } , { B i } , ee ). (D47.21)2. Desargues( CK, { T i } , { B i } , m ). (C47.6)3. Desargues( M , { A i } , { B i } , m ). (C47.6) Theorem. [Sekigichi]
The set of points on a triangle at which the sum of the distances to the sides is equal to thearithmetic mean of the lengths of the altitudes is a segment of a line through the barycenter. (Amer. Math. Monthly, 1981, 349 and 1984, 257.)
Definition.
The line defined in the preceding Theorem is called the line of Sekiguchi . Theorem.
The line of Sekiguchi is perpendicular to the line ok joining the center O of the circumcircleto the point K of Lemoine. The segment [ A , Sek ] is equal to the segment M , A , (D18.27), the segment [ M , Set ] is equal to the segment A , M , (D18.28), the line sek joining Sek and Set has the direc-tion of O × K (C18.23). Definition.
The triangle of Nagel , { N a i } has as its vertices the point of contact of the i -th exscribedcircles with the i -th side .(N21.0.) Definition.
The conic of Feuerbach is the conic through the vertices of the triangle, the point of Gergonne J and the incenter I . (N20.6.) Theorem. [Feuerbach]
The conic of Feuerbach is an equilateral hyperbola, it passes through the orthocenter and thepoint of Nagel, it is tangent at I to the line through I and O (Th´ebault), it has the point ofFeuerbach as its center. (D20.23., C20.14, C20.15, C20.17,C23.8.) See also Neuberg, Math.1922-51-90.72 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Theorem. [Kimberling] If Kim is the intersection of the lines from the center I and I of the excribed circles on theexterior bissectrix through A perpendicular respectively to the sides a and a , then the line kimc joining Kim to A and the similarly obtained lines kimc and kimc have a point Kim in common. (D21.30.) See 3.3.0
Definition.
The point
Kim is called the excribed point of Kimberling. (N21.5.)
Theorem.
The point of Kimberling is on the conic of Feuerbach. (C21.11.)
Theorem. If Kid is the intersection of the lines from the center I and I of the excribed circles on theexterior bissectrix through A and respectively the points M A and M on the orthic line m and the sides a and a , then the line kidc joining Kid to A and the similarly obtainedlines kidc and kidc have a point Kid in common. (D21.26.)
Definition.
The point
Kid is called the excribed orthic point. (N21.4.)
Theorem.
The barycentric coordinates of the incenter I are proportional to the lengths of the sides ofthe triangle. Exercise.
Prove that the point En is the centroid of a wire of uniform density forming the sides of thetriangle A i . See C. J. Bradley, Math. Gazette, 1989, p. 44. for the latter. Definition. [Mandart]
The conic of Nagel is the conic tangent at the vertices of the triangle of Nagel to the sidesof the triangle.(N27.0)
Theorem. [Mandart and Neuberg]
The center of the conic of Nagel is on the conic of Feuerbach.
C27.1. (Math. 1922-125)
Definition. [Mandart]
The cercle of Nagel is the circle circumscribed to the triangle of Nagel.(Math. 1922-125) .4. FINITE INVOLUTIVE GEOMETRY.
Theorem.
The complimentary point En of the incenter I is the center of gravity of the perimeter of thetriangle. (See Math. 1889, Suppl. p. 8, ) Introduction.
This section discusses is some detail the notion of equality of angles in involutive geometry.
Definition. A sympathic projectivity is one which is amicable with the fundamental involution. (II,1.5.10) Theorem. (H. D.)
If the involutive geometry is hyperbolic, a sympathic projectivity has 2 fixed points, theisotropic points.
Theorem.
The sympathic projectivities form an Abelian group under composition.Moreover, using 1.0.10.0.4., if f b ( y ) = − byb + k (cid:48) + y , then f b ◦ f b = f b , with b − b b k (cid:48) + b b . Proof. f b ( f b ( y )) = ( − ( b b k (cid:48) ) + ( − b b y ( b k (cid:48) ( b k (cid:48) ) − b b k (cid:48) ) y ) , dividing numerator and denominator by b b k (cid:48) gives the conclusion of the Theorem.See also . . . , Section 7. Example.
The method of obtaining sympathic projectivities will be studied in section . . . . It will beseen that all are powers of a sympathic projectivity S which is of order p-1 in the hyperboliccase and of power p+1 in the elliptic case. This generating projectivity is not unique, choos-ing one of these as fundamental sympathic projectivity will constitute the next step towardsEuclidean geometry, the sympathic geometry. The fundamental involution is S ( p − ) or S ( p +12 ) . With p = 7, (elliptic), we will choose k = 0, δ = 6,The sympathic projectivities are S i , i = 0 to 7, with S (1 , j, − − j ) = (2 − j, j, − j ) ,S (0 , ,
6) = (1 , , , or74 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY S, = , ( 7 , , , , , , , , , , , , , , ,
32) = S ,S = ( 7 , , , , , , , , , , , , , , ,
7) = S ,S = ( 7 , , , , , , , , , , , , , , ,
38) = S . The fundamental involution is S , = , ( 7 , , , , , , , , , , , , , , . The isotropic points are (1 , δ, − − δ ) , (1 , − δ, − δ )With p = 7 , (hyperbolic), we will choose k = 1 , δ = 2 , The sympathic projectivities are S i , i = 0 to 5, with S = (26 , , , , , , , , , , , , , ,
32) = S ,S = (26 , , , , , , , , , , , , , ,
14) = S , The fundamental involution is S = (26 , , , , , , , , , , , , , , . The isotropic points are (26) = (1,2,4) and (38) = (1,4,2).anti. . . .
Definition. An angle is an ordered pair { a, b } of ordinary lines a and b . Definition.
Two angles { a, b } and { a , b } are equal and we write { a, b } = { a , b } , if the ideal points on these lines, A, B , A , B are such that there exists a sympathicprojectivity which associates A to B and A to B . Compare with Coxeter, p. 9 and p.125). Notation.
In view of 2.3., we will also use { A, B } = { A , B } instead of { a, b } = { a , b } , where A , B , A , B are the ideal points on a , b , a , b . Example.
For p = 5, starting with Example II.1.5.12. if φ (cid:48) is used to to define the fundamentalinvolution, then φ is a sympathic projectivity. We have the equality of angles { (10),(5) } = { (5),(26) } = { (26),(14) } = { (14),(18) } = { (18),(22) } = { (22),(10) } and of angles { (10),(14) } = { (5),(18) } = { (26),(22) } = { (14),(10) } . .4. FINITE INVOLUTIVE GEOMETRY. Theorem. If a and b are perpendicular, then { a, b } = { b, a } . If c and d are also perpendicular, then { a, b } = { ( c, d } . Definition. If a and b are perpendicular, the angle { a, b } is called a right angle . Definition. If a and b are not parallel and c , through a × b , is such that { a, c } = { c, a } , c is called a bisectrix of { a, b } . If a bisectrix exist, we say that the angle { a, b } can be bisected . Theorem.
If the ideal points on a and b are (1 , a , − − a ) and (1 , b , − − b ) , then the ideal point (1 , z, − − z ) on the bisectrix c of { a, b } satisfies the second degree equation
0. ( k (cid:48) + a + b ) z − a b − z − ( a + b + k (cid:48) a b ) = 0 , with k (cid:48) = k k ) . The discriminant of 0. is t (cid:48) = ( a + k (cid:48) a + 1)( b + k (cid:48) b + 1)2. Moreover,0. if t (cid:48) (cid:54) = 0 is a quadratic residue, the bisectrices are real and perpendicular to eachother,1. if t (cid:48) is a non residue, there are no real bisectrices,2. if a or b is an isotropic line, t (cid:48) = 0 , the bisectrices coincide with the isotropic line,3. if both a and b are isotropic, the bisectrices are undefined,4. if a and b are parallel, the bisectrices do not exist but the directions given by 0. arethat of a and of the perpendicular to a . Proof: Let the sympathic projectivity which associates to the ideal points on a and c theideal points on c and b , have the form f ( x ) = a (cid:48) + b (cid:48) xc (cid:48) + d (cid:48) x , then z ( c (cid:48) + d (cid:48) a ) = a (cid:48) + b (cid:48) a ,b ( c (cid:48) + d (cid:48) z ) = a (cid:48) + b (cid:48) z. Because of 0.0.10., d (cid:48) = − a (cid:48) = 1 + k, c (cid:48) − b (cid:48) = 2 k. Substituting and multiplying the first equation by ( c − b ) , the second by ( c − a ) andadding, we obtain the equation 0.If a corresponds to an isotropic point, a + k (cid:48) a + 1 = 0 , t (cid:48) = 0 , the roots of 0. are a b − k (cid:48) + a + b = a a b − a k (cid:48) a b a = a . If a = b ,
0. can be written ( z − a )(( k (cid:48) + 2 a ) z + (2 + k (cid:48) a ) = 0 . The perpendicularity followsfrom 1.9.6.76
CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Example.
For p = 7, hyperbolic case, let the ideal point on “a“ be (32) and on “b” be (20), a = 3 ,b = 1, k = k (cid:48) = 1 ,
0. is 5 z - 4z = 0, with roots 0 and 5 giving the points (14) and (44). If a = b = 0 , one root is 0, the other is 5. Definition.
The angle between distinct non isotropic lines is even if and only if the angle can be bisected.
Theorem.
Under the hypothesis of Theorem 0.2.7., an angle is even if a + k (cid:48) a + 1 and b + k (cid:48) b + 1 are both quadratic residues or both non residues. The proof follows at once from . . .
Theorem.
The relation “even” is a equivalence relation.
Again this follows from . . . .
Definition.
The sum of two angles . . .. . . circle, angle at the center, rotation. congruence (translation composed with rotation)
Theorem.
Any congruence can be written as the composition of a translation rotation and a translation.
Definition. A segment [ A, B ] is an unordered pair of ordinary points A and B .. . . not on the same isotropic line? Definition.
Two segments are equal iff Theorem. If [ A, B ] = [
B, C ] and C is on A × B, then either A = C or B is the mid-point of [ A, C ] . . . . equality of segments on parallel line iff equal in the affine sense or AB = CD in affinesense or BA = CD .4. FINITE INVOLUTIVE GEOMETRY. Theorem. { A, B } = { C, D } implies [ A, B ] = [
C, D ] . equal. on non parallel segment using translation and circle may have to use tangent tocircledef. of congruence. Definition. A right triangle is a triangle with 2 perpendicular sides. If a and a are perpendicular wesay that the triangle is a right triangle at A . Theorem.
A necessary and sufficient condition for a triangle to be a right triangle at A is that m = m = 0 . Exercise.
If we start with A i , M and M in involutive geometry, we cannot derive the properties ofthe right triangles. Other elements have to be prefered. Make an appropriate choice andconstruct enough elements to determine θ and γ. Answer to 3.4.11.
To obtain the coordinates, we replace m , m , m by 1 , (cid:15)m , (cid:15)m , and when the coordinates contain terms of different order of (cid:15), we neglect the terms of higherorder.For instance, q = 1 , q = − m , q = − m ,θ : m ( m + m ) X X + m ( m + m ) X X + m ( m + m ) X X = 0 , becomes θ : ( m + m ) X X + m X X + m X X = 0 , Of course many points or lines will coincide and some of the construction which are invalidmust be replaced by other constructions but the coordinates do not have to be rewritten.For instance, A = M M = IM a , , m = m = mm = ta , ma = mk, M A = T Aa , eul = ma , EU L = Imm , ta = m, Aat = M . We can start, for instance, with A i , M and K = ( m + m , m , m ) , on mm , we construct,as usual, ma i , M i , mm i , M A i , m i , M M i , m. Then mk := M × K, mk = [ m − m , − m , m ] ,at i := K × A i , at = [0 , m , − m ] , at = [ m , , − ( m + m )] ,Aat i := at i × a i , Aat = ( m + m , , m ) ,Iat i := m × at i , Iat = ( m + m , − (2 m + m ) , m ) , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY ta i := A i × Iat i , ta = [0 , m , m ] , ta = [ m , , m + m ] ,T Aa := ta × a , T Aa [0] = (0 , m , − m ) . In general, the construction cannot be done for all 3 elements, but can done simultaneouslyfor the elements with index 1 and 2, we can use j for 1 and − j for 2. θ = conic ( A , ta , A , A , M M ) ,θ : ( m + m ) X X + m X X + m X X = 0 ,γ := conic ( M i , Aat , A ) ,γ : m X + m X − ( m + m ) X X − m X X − m X X = 0 . When we start with J and M , the triangle is a right triangle if I × J // a and I × J // a .Moreover, j = p , m = j ( j + j )( j − j ) , m = j ( j + j )( j − j ) . The usual constructiongives M = Aat , then at := A × Aat , K := at × mm . Definition. An isosceles triangle { A i } at A is a triangle whose angles A A A and A A A are equal.What about right isosceles? Theorem.
A necessary and sufficient condition for a triangle to be an isosceles triangle at A is that m = m . Theorem. If { ABC } is an isosceles triangle at A , then the sides AB and AC are equal. Theorem. If { ABC } is an isosceles triangle, then the angle ( A × B, A × C ) is even. Definition.
A triangle { ABC } is an equilateral triangle iff it is isosceles at B and C . Theorem.
A necessary and sufficient condition for a triangle to be an equilateral triangle at A is that m = m = m . Theorem.
If a triangle is equilateral, then all its angles
ABC, BCA and
CAB are equal and all itssides are equal.
Definition.
A triangle which is neither a right triangle nor an isosceles triangle, and therefore not anequilateral triangle is called a scalene triangle . .4. FINITE INVOLUTIVE GEOMETRY. Theorem.
A necessary and sufficient condition for a triangle to be a scalene triangle is that m , m and m be distinct. Definition.
A triangle which is an isosceles triangle at A but is not an equilateral triangle is called a proper isosceles triangle. Theorem.
If a triangle is equilateral, then all its angles
ABC, BCA and
CAB are equal and all itssides are equal.
The following Definitions and Theorem are only meaningfull in Minkowskian Geometry. Definition. An isotropic triangle is a triangle with one isotropic side. Definition. A doubly isotropic triangle is a triangle with 2 isotropic sides Theorem. A necessary and sufficient condition for a triangle to be isotropic is that the barycenterbe on the complementary triangle. A necessary and sufficient condition for a triangle to be doubly isotropic is that thebarycenter be one of the vertices of the complementary triangle.
Theorem. If a [0] is an isotropic line, then
0. m + m = 0,1. the circumcircle degenerates into a and the line [0,m +m ,-(m -m )], Theorem. If a and a are isotropic lines, then
0. m = 1, m , m = -1,1. the circumcircle degenerates into a and a . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Introduction.
There are many other types of triangles that can be defined. I will give here 2 exampleswhich allow the constructions of configurations of the type 9 * 3 & 9 * 3, distinct from thatof Pappus. For the first one, if we choose B = M , B = M and C = M , the constructionof Section 19 gives C = M am , C = T ara , B = tara × tarb . from P19.7, follows that B · a = 0 iff q = 0 . This suggest the definition of a triangle of Tarry and the constructionof the 1-Pappus configuration.A similar approach determines the construction of the 2-Pappus configuration.
Definition.
The is the set of points1-Pappus( A i , B i , C i ),such that { A i } , { B i } , { C i } are 3 triangles and incidence( A i , C i , C i − ), incidence( B i +1 , B i − , C i ). Definition.
The is the set of points2-Pappus( A i , B i , C i ),such that { A i } , { B i } , { C i } are 3 triangles and incidence( A i , B i , C i ), incidence( B i , C i +1 , C i − ). Definition. A triangle of Tarry is a triangle which is not a right triangle whose point of Tarry is welldefined and coincides with one of the vertices of the triangle. Theorem.
A necessary and sufficient condition for a triangle to be a triangle of Tarry at A is that q = 0 . The proof follows at once from m i (cid:54) = 0 and from P16.3.Moreover, if q = q = 0 , then the point of Tarry is undefined. Corollary.
A necessary and sufficient condition for a triangle to be a triangle of Tarry at A is that theorthocenter be on the conic of Tarry. Theorem. M · ‘ T arry [0] ⇒ A i , T arb M M , M am M T ara ). .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. Theorem.
In Involutive Geometry, A = ( x, y, , A = (0 , , , A = (1 , , , M = (1 + x, y, ,M = ( xy, x (1 − x ) , y ) is a triangle of Tarry iff u − (1 + 2 y ) u + y = 0 and x − x + u = 0 . Proof: Assuming that m = [0 , ,
1] and X + X = X is a circle, a trivial computationdetermines M and M as given. The conic of Tarry is a ×× a + ka ×× a = 0 , where k is determined in such a way that M · ‘ T arry = 0 , this gives k = 1. To insure that M is on the conic of Tarry gives after division by u := x (1 − x ) ,u + ( y − u )( u − y ) = 0 , a simple discussion determines that 0 < y ≤ √ . Definition. An Eulerian triangle is a triangle for which the line of Euler is parallel to one of its sides.
Theorem.
A necessary and sufficient condition for a triangle to be an Eulerian triangle for side a isthat m = m + m . The proof follows at once from P1.17.
Theorem. eul // a ⇒ A i , M M M , M T arc T arc ). (bissectrices) CHAPTER IIFINITE PROJECTIVEGEOMETRY
Answer to 2.2.3.
Let A = (1 , , , A = (0 , , , A = (0 , , , C = (1 , , , B = ( b , , , B = (1 , b , , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY B = (1 , , b ) , by hypothesis, b i (cid:54) = 1 , b (cid:54) = 0 , b b (cid:54) = 1 and 2 − b − b − b + b b b (cid:54) = 0 (because of { B i } .a = [1 , , , b = [1 − b b , − (1 − b ) , − (1 − b )] ,c = [0 , , − , C = (0 , − b , − (1 − b )) ,c = [(1 − b )(1 − b ) , (1 − b )(1 − b ) , (1 − b )(1 − b )] , d = [0 , b , − ,D = ( b , , b ) , e = [ b (1 − b ) , b (1 − b ) , − b − b + b b ] ,E = (1 − b + b b − b b − b b + b b b , − b + b − b b − b + b b + b b b ,b (2 − b − b − b + b b b )) , f = [1 − b b , − b (1 − b ) , − b (1 − b )] , F = ( b (1 − b ) , , − b b ) , G = (1 − b , − b b , ,g = [1 − b b , − (1 − b ) , − b (1 − b )] ,X = (0 , , b ) , Y = (1 − b − b + b b b , − (1 − b ) , − b (1 − b )) ,Z = (1 − b + b − b b , − b b , b (1 − b )) . Exercise. [Pappus]
Define α := ( A ∗ A ) · A , β := ( B ∗ B ) · B ,α , := ( A ∗ A ) · B , β , := ( B ∗ B ) · A ,α , := ( A ∗ A ) · B , β , := ( B ∗ B ) · A ,α , := ( A ∗ A ) · B , β , := ( B ∗ B ) · A , Using 2.3.17.0, C = ( A ∗ B ) ∗ ( A ∗ B = (( A ∗ B ) · B ) B − (( B ∗ A ) · B ) A = α , B − β , A , similarly, C = α , B − β , A ,C = α , B − β , A , therefore( C ∗ C ) · C = βα , α , α , − αβ , β , β , + α , α , β , β , − β , α , α , β , + α , β , α , β , − β , β , α , α , + α , β , β , α , − β , α , β , α , = βα , α , α , − αβ , β , β , .Therefore, if the points A , A , A and the points B , B , B are collinear, α = 0 and β = 0 , therefore ( C ∗ C ) · C = 0 and the points C , C , C are collinear by 2.3.18. Exercise. (Harmonic quatern). a = [0 , , , let A = (0 , , , A × K = [ k, − , , let B = (1 , k, , l (cid:54) = 0 and 1. B × L = [ l, − , k − l ] , A × M = [ m, − , , D = ( l − k, ml − mk, l − m ) , D × K =[ k ( l − m ) , m − l, ( l − k )( m − k )] , A × L = [ l, − , , C = ( m − k, lm − lk, l − m ) , B × C =[2 kl − km − lm, m − k − l, ( k − l )( k − m )] , N = (2 m − l − k, km + lm − kl, . Exercise. (Projectivity). Choose b = [0 , , , a = [1 , ,
0] and P = (0 , , . c = [0 , , , S = (1 , , − k ) ,T = (1 , , − l ) , Q = (0 , m, k ) , N = [ k, ml, . .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. Exercise. (Projectivity with 3 pairs). C = (1 , , − , C j = (1 , a j , − − a j ) , j > , with obviousnotation, D j = (1 , − a j b j , − − a j ) , j = 1 , , t = ( a a ( b − b ) + a b − a b , a − a , a b − a b ) ,D j = ( a − a , a b − a b − a a ( b − b ) , ( a − a )(1 + a j )) , j > , hence B j given above. Answer to 2.6.14.
In part, the coefficients of A and A in A l and B l must be proportional, therefore, f t + f t f t + f t = t t , this gives 1. Answer to 2.3.7.
For p = 2 , if A [] = ((1 , , , (1 , , , (1 , , , (1 , , , and the diagonal points are B i ,A × A = [0 , , , A × A = [1 , , , B = (0 , , . A × A = [1 , , , A × A = [0 , , ,B = (0 , , . A × A = [0 , , , A × A = [1 , , , B = (0 , , . The diagonal points areon [1,0,0]. For p = 4 , the coordinates are 0, 1, x, y = 1 + x. The addition and multiplicationtables are + 0 1 x y · x y x y y x x yx x y x x y y y x y y x If A [] = ((1 , , , (1 , , , (1 , x, , (1 , y, ,A × A = [0 , , , A × A = [ x, , , B = (1 , , x ) . A × A = [ x, , x ] , A × A = [0 , , y ] ,B = (1 , , x ) . A × A = [0 , , , A × A = [ y, , y ] , B = (0 , , . The diagonal points areon [ x, , . Answer to 2.5.10.
In part, C = ( c, − c, , a = (1 , c, − c − , b = [1 , , , B = ( c − c + 1 , − c + 1 , . A geometriccondition is a · C = 0 or b · B = 0 . The configuration is then of type6 ∗ ∗ ∗ ∗ ∗ . Notes.
On 2.2.2:Commutativity implies that if J (cid:48) := ( B × P ) × ( E × Q ) , J = ( b ( a − , b ( a − , a ( b − ,K (cid:48) := ( A × P ) × ( J (cid:48) × M ) , K = ( b ( a − , ab ( a − , a ( b − , then L · ( J (cid:48) × K (cid:48) ) = 0 . The construction D (cid:48)(cid:48) := ( R × T (cid:48) ) × a, with D” = (1,b+c)84 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY is related to the associative property( a + b ) + c = a + ( b + c ) . Before2.2.9
Theorem. . . . describe the degenerate conic, perhaps in 2.10.8 . . . determine the collineation which leavea general conic fixed also special case when it is a conic.
Examples.
For p = 5, C0 N = − −
12 0 0 , N I = − − − . Point conic and its mapping 0 1, 1 6, 10 27, 14 4, 23 8, 27 29,Line conic and its mapping 1 10, 4 14, 6 0, 8 1, 27 23, 29 27,Points on line conic and tangent, 0 16, 1 0, 10 22, 14 4, 23 14, 27 29,Lines on line conic and contact, 1 7, 4 14, 6 5, 8 13, 27 29, 29 27.The equation of the point conic is X + 2YZ + ZX = 0.The equation of the line conic is − z + yz - zx + xy = 0. C1 N = − − − − , N I = − −
12 2 1 . Point conic and its mapping 0 1, 1 6, 10 22, 12 5, 18 25, 29 9,Line conic and its mapping 1 10, 5 18, 6 0, 9 1, 22 12, 25 29, C2 N = −
21 1 0 − , N I = − − − − − . Point conic and its mapping 0 6, 5 18, 6 5, 15 26, 19 22, 21 25,22 28, 23 14, 24 17, 25 4, 27 10,The center is (23), the points are on [21] or [2]. .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. Y + YZ + 2ZX + 2XY = 0.The equation of the line conic is y - 2 z -yz - 2zx + xy = 0. C3 Point conic and its mapping 7 10, 9 19, 11 30, 12 18, 29 27, 30 7,Line conic and its mapping 7 30, 10 12, 18 29, 19 9, 27 11, 30 7, C4 Point conic and its mapping 11 29, 14 12, 17 24, 19 28, 26 15, 27 23,Line conic and its mapping 12 26, 15 19, 23 14, 24 17, 28 11, 29 27, C5 Point conic and its mapping 2 27, 3 3, 8 8, 10 1, 16 19, 18 16,Line conic and its mapping 1 8, 3 3, 8 18, 16 16, 19 2, 27 10, C6 Point conic and its mapping 9 14, 10 12, 15 26, 18 29, 23 21, 24 17,Line conic and its mapping 12 18, 14 23, 17 10, 21 24, 26 15, 29 9, C7 Point conic and its mapping 0 16, 3 3, 8 8, 11 28, 16 0, 28 11,Line conic and its mapping 0 11, 3 3, 8 28, 11 0, 16 16, 28 8, C8 N = , N I = − − . Point conic and its mapping 17 24, the center is (17).86
CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Line conic and its mapping 24 17, the central line is [24].The equation of the point conic is, with δ = 2,(X - (2 + 2 δ )Y - (2 + δ )Z) (X - (2 − δ )Y - (2 − δ )Z) = 0.The equation of the line conic is (x + (2 + 2 δ )y + (1 − δ )z) (x + (2 − δ )y + (1 + 2 δ )z)= 0. C9 Point conic and its mapping 2 5, 5 2, 15 25, 20 30, 26 14, 29 11,Line conic and its mapping 2 15, 5 2, 11 26, 14 20, 25 29, 30 5,
C10
Point conic and its mapping 3 3, 4 4, 8 28, 9 29, 13 8, 14 9,Line conic and its mapping 3 13, 4 14, 8 8, 9 9, 28 4, 29 3,
C11
Point conic and its mapping 2 5, 5 2, 12 14, 15 13, 17 24, 20 23,Line conic and its mapping 2 15, 5 12, 13 17, 14 20, 23 2, 24 5,
C12 N = − − − − − , N I = − − . Point conic and its mapping 1 6, 7 15, 8 13, 15 25, 16 18, 19 16,Line conic and its mapping 6 1, 13 15, 15 19, 16 16, 18 8, 25 7,
C13 N = , N I = − − − − − − . Point conic and its mapping 0 11, 1 7, 6 2, 13 15, 17 27, 24 30,Line conic and its mapping 2 6, 7 1, 11 0, 15 13, 27 17, 30 24,The equation of the point conic is YZ + ZX + XY = 0.The equation of the line conic is x + y + z - 2yz - 2zx - 2xy = 0. C14 .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. N = − − − , N I = − −
21 1 1 . Point conic and its mapping 2 15, 9 29, 15 7, 16 18, 22 21, 28 4,The points are on [19].Line conic and its mapping 4 28, 7 15, 15 2, 18 16, 21 22, 29 9,The lines pass through (25).
C15 N = N I = − − − − − − Point conic and its mapping 19 28, 20 23, 23 20, 25 29, 28 19, 29 25,Line conic and its mapping 19 28, 20 23, 23 20, 25 29, 28 19, 29 25,The equation of the point conic is X + Y + Z + Y Z + ZX + XY = 0 . The equation of the line conic is x + y + z + yz + zx + xy = 0 . Answer to 2.2.9.
For p = 13 , points on the conic are, (0,1,1) = (2), (0,1,2) = (3), (1,0,1) = (15), (1,0,4) =(18), (1,1,0) = (27), (1,2,0) = (40).The point conic is2, 3, 15, 18, 27, 35, 40, 51,133,135,146,151,158,168,the line conic is111, 83,156,121,179, 22, 98,148,112,129, 86, 25,165,166.The representative matrix is − − − − − − (0,1,2) = (13) and (1,0,12) = (26) are on [1,1,1], the polars are [1,6,5] = [97] and [0,1,11] =[12]. Hence the pole of [1,1,1] is (1,6,3) = (95).1 . − . − . − . ( −
3) = −
17 = − . ( −
3) + 8 . − . ( − − ( − .
5) = 67 = 2 , . . − ( − . ( − − . ( −
6) = 27 = 1 , ( − , ,
1) = (1 , , . Answer to 2.2.9. A × B = [1 , − , , C × D = [2 , , − , A × D = [2 , − , , B × C = [1 , , − , k = 1 . ,k = − . , therefore the conic is, after dividing by 3,( X − X + X )(2 X + X − X ) + (2 X − X + X )( X + X − X ) = 0 , which gives twice the result of the Example.88 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
Answer to 2.3.2.
0. For q = 2 , the primitive polynomial giving the selector 0, 1, 3, is I + I + 1 . The auto-correlates are 0 11 2 7 8.The selector function is i f ( i ) 0 14 1 0 16 16 14 14 16 type F V F V T T V F F T E F i
12 13 14 15 16 17 18 19 20 f ( i ) 4 1 0 1 0 4 4 16 1 type T E P V E T E V E
1. The correspondence between the selector notation and the homogeneous coordinatesfor points and lines is i I i i ∗ ∗ : 1 , , , I ∗ : 0 , , , I ∗ : 0 , , , I + 1 5 ∗ : 2 , , , I + I ∗ : 0 , , , I + I + 1 4 ∗ : 3 , , , I + 1 2 ∗ : 1 , , .
2. The matrix representation is M = , M − = . and the equation satisfied by thefixed points is ( X + X ) = 0.3. The degenerate conic through 0, 1, 2 and 5 with tangent 5 ∗ at 5, is represented by thematrix N = . The polar of 0 is 0 ∗ , of 1 is 0 ∗ , of 2 is 5 ∗ , of 4 is 4 ∗ , of 5 is 5 ∗ of 6 is 6 ∗ and of 3 isundefined. The equation in homogeneous coordinates is X ( X + X ) = 0 .
4. A circle with center 14 can be constructed as follows. I first observe that a directionmust be orthogonal to itself. Indeed, if 0 is a direction, the others form an angle 1,2,3,4 mod , we cannot play favorites and must choose 0. If A = 1 , C × A and thereforethe tangent has direction 0, A × A i +1 has direction i mod .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. T i , the 5 vertex-points, V i , the 5 triangular-triangularedge-points, E i , the 5 pentagonal-triangular edge-points F i . The lines are represented on the 5-anti-prism as follows. The pentagonal face-line, f,which is incident to F i , the 5 triangular face-lines, t i , which are incident to F i , F i , T i +1 ,T i − , E i +2 , E i − . If f is the pentagonal edge of t i and V, V (cid:48) are on f , F i is on it, T i +1 ( T i − ) share V ( V (cid:48) ) , E i +2 ( E i − ) are on an edge through V ( V (cid:48) ) not on t i the 5 vertex-lines, v i , which are incident to F i , V i +2 , V i − , E i +1 , E i − . If t is the face with v i on its pentagonal edge these are allthe vertices, and edge-points on it distinct from v i . the 5 triangular-triangular edge-lines, e i , which are incident to F i , T i +2 , T i − , V i +1 ,V i − . V i +1 and V i − are on the same edge as e i , the line which joins the center C of theantiprism to E i is parallel to the edge containing F i , T i +2 and T i − are the triangularfaces which are not adjacent to E i or F i . the 5 pentagonal-triangular edge-lines. f i , which are incident to P, T i , V i , E i , F i . T i isadjacent to f i , V i is opposite f i , E i joined to the center of the antiprism is parallel to T i . Answer to 2.3.3.
For p = 3 ,
0. The primitive polynomial giving the selector 0, 1, 3, 9 is I − I − .
1. The correspondence between the selector notation and the homogeneous coordinatesfor points and lines is i I i i ∗ ∗ : 1 , , , , I ∗ : 0 , , , , I ∗ : 0 , , , , I + 1 7 ∗ : 2 , , , , I + I ∗ : 0 , , , , I + I + 1 4 ∗ : 5 , , , , I + 2 I + 1 1 ∗ : 3 , , , , I + I + 2 6 ∗ : 3 , , , , I + 1 2 ∗ : 1 , , , , I + 2 11 ∗ : 2 , , , I + 2 I ∗ : 0 , , , , I + 2 I + 2 5 ∗ : 4 , , , , I + 2 8 ∗ : 1 , , , .
2. The matrix representation of the polarity i to i ∗ is M = , M − = . The equation satisfied by the fixed points is X + X + 2 X X = 0 . CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY
3. The degenerate conic through 0, 1, 2 and 5 with tangent 4 ∗ at 5, is obtained by con-structing the quadrangle-quadrilateral configuration starting with P = 5 and Q i = { , , } . We obtain q i = { ∗ , ∗ , ∗ } , which are the tangents at Q i . The matrix repre-sentation is N = with equation X X + X X + X X = 0 . We can check that the polar of 10 = 3 ∗ × ∗ is 9 ∗ = 0 × . Answer to 2.3.8.
0. For q = 2 , the primitive polynomial giving the selector 0, 1, 4, 14, 16 is I − I − I − (cid:15) , with (cid:15) + (cid:15) + 1 = 0 .
1. The correspondence between the selector notation and the homogeneous coordinatesare as follows, i ∗ has the homogeneous coordinates associated with I i .i I i i ∗ ∗ I ∗ I ∗ I + I + (cid:15) ∗ I + (cid:15) ∗ I + (cid:15) I ∗ I + (cid:15) I + 118 ∗ I + 1 15 ∗ I + (cid:15) ∗ I + (cid:15) I + (cid:15) ∗ I + (cid:15)I + 1 9 ∗ I + (cid:15) ∗ I + (cid:15)I + (cid:15) ∗ I + I + (cid:15) ∗ I + 1 2 ∗ I + I ∗ I + (cid:15) ∗ I + (cid:15)I ∗ I + (cid:15)I + (cid:15) ∗ I + (cid:15) I + (cid:15) ∗ I + I + 1 7 ∗ To obtain the last column, for row 9, [1 , (cid:15) , (cid:15) ] = (1 , , × (1 , (cid:15),
0) = 20 ×
17 = 5 ∗ .
2. The correspondence i to i ∗ is a polarity whose fixed points are on a line. The matrixrepresentation is obtained by using the image of 4 points.0 = (0,0,1), M (0) = 0 ∗ = [1 , , , .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. M (1) = 1 ∗ = [1 , , , M (2) = 2 ∗ = [0 , , ,
18 = (1 , (cid:15), (cid:15) ) , M (18) = 18 ∗ = [1 , (cid:15) , . The first 3 conditions give the polarity matrix asThe last condition gives β(cid:15) + α(cid:15) = 1 , γ + β(cid:15) = (cid:15) , γ = 1 . Hence γ = 1 , β = 1 , α = 1 . Therefore M = , M − = . Note that M is real and could have been obtained from the reality and non singularityconditions, giving directly α = β = γ = 1 . The polar of ( X , X , X ) is [ X + X , X + X , X ] . The fixed points ( X , X , X ) satisfy X = 0 corresponding to 14 ∗ .
3. A point conic with no points on 14 is 1, 3, 4, 5,13,the corresponding line conic is 15,19,10,16, 8.Projecting from 1 and 3, 1, 3, 5,13, 4,we get the fundamental projectivity, 8, 2,11, 0, 7 on 14 ∗ .
4. To illustrate Pascal’s Theorem, because there are only 5 points on a conic, we need touse the degenerate case. The conic through 0, 1, 2 and the conjugate points 9 and 18is The last condition gives β(cid:15) + α(cid:15) = 1 , γ + β(cid:15) = (cid:15) , γ = 1 . Hence γ = 1 , β = 1 , α = 1 . Therefore M = , M − = . Note that M is real and could have been obtained from the reality and non singularityconditions, giving directly α = β = γ = 1 . The polar of ( X , X , X ) is [ X + X , X + X , X ] . The fixed points (
X, X , X ) satisfy X = 0 corresponding to 14 ∗ .
5. A point conic with no points on 14 is 1, 3, 4, 5,13, The tangents at (0,0,1), (0,1,0), (1,0,0), (1 , (cid:15) , (cid:15) ) , (1 , (cid:15), (cid:15) ) are [1,1,0], [1,0,1], [0,1,1],[1 , (cid:15) , (cid:15) ) , (1 , (cid:15), (cid:15) ] , or 1 ∗ , ∗ , ∗ , ∗ , ∗ . On the other hand, using Pascal’s Theorem,the tangent at 0 is given by((((0 × × (9 × × ((18 × × (1 × × (2 × ×
0= (((0 ∗ × ∗ ) × (4 ∗ × ∗ )) × ∗ ) ×
0= (((14 ×
17) = 8 ∗ ) × ∗ or13) × ∗ . Answer to 2.3.8.
For q = 57 , choose the auto-correlates as point on a circle although 0 is on the circle drawas it is the center. With the succession of points X i , CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY x i = 0 × X i , , , , , , ,X i , , , , , , ,y i +1 = X i − × X i +1 , , , , , , ,y i +2 = X i − × X i +2 , , , , , , ,y i +3 = X i − × X i +3 , , , , , , ,y i +1 × x i , , , , , , ,y i +2 × x i , , , , , , ,y i +3 × x i , , , , , , . This gives all the points in the projective plane of order 7. We observe16 ∗ ∗ ∗ ∗ ∗ ∗ ∗
36 36 36 36 36 36 3616 35 ,
30 18 ,
26 50 , ,
44 22 8 ,
28 14 , ,
20 34 17 ,
40 2 , ,
54 31 ,
39 7 6 , ,
33 5 , , ,
49 56 ,
47 48 ,
38 2437 ,
11 12 ,
19 9 ,
55 5335 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ,
18 50 ,
30 29 , , ,
44 10 , ,
23 2 40 ,
20 34 , ,
25 7 , , , ,
45 51 21 , ,
38 24 ,
48 15 ,
47 569 ,
19 53 ,
37 55 ,
11 1218 ∗ ∗ ∗ ∗ ∗ ∗ ∗
52 52 52 52 52 52 5218 35 ,
50 16 ,
29 26 , ,
14 8 22 ,
10 28 , ,
40 41 20 ,
23 2 , ,
39 31 ,
27 6 7 , ,
45 21 ,
33 5 51 , ,
47 56 ,
15 24 ,
49 4853 ,
55 12 , ,
19 37
Answer to 2.3.8.
For q = 2 , .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. ∗ ∗ ∗ ∗ ∗ ∗ × ∗ : 0 1 3 7 15 31 36 54 63 0 0 0 1 3 3136 ×
37 = 37 ∗ : 17 26 36 37 39 43 51 67 72 72 51 67 72 72 2636 ×
38 = 38 ∗ : 16 25 35 36 38 42 50 66 71 35 71 66 71 50 7136 ×
40 = 40 ∗ : 14 23 33 34 36 40 48 64 69 14 33 69 34 69 6936 ×
44 = 44 ∗ : 10 19 29 30 32 36 44 60 65 30 60 29 29 32 1036 ×
52 = 52 ∗ : 2 11 21 22 24 28 36 52 57 2 28 24 52 11 236 ×
18 = 18 ∗ : 13 18 36 45 55 56 58 62 70 62 70 56 13 70 5836 ×
27 = 27 ∗ : 4 9 27 36 46 47 49 53 61 53 4 47 61 27 4936 ×
68 = 68 ∗ : 5 6 8 12 20 36 41 59 68 6 12 8 5 59 68Conic with no point on 36: 2, 4, 5, 6,13,28,31,46,63line conic: 29,59,31, 9,18,43,28,35,64.Fundamental projectivity: from 2 and 5 on the conic, the points2, 5, 6,31,13,28, 4,46,63 give the points on 36 ∗ :38, 0,68,27,52,37,40,18,44.94 CHAPTER 3. FINITE PRE INVOLUTIVE GEOMETRY hapter 4FINITE INVOLUTIVE SYMPATHICAND GALILEAN GEOMETRY
In part II, I have given a construction of a finite projective geometry associated to a prime p. In it, there is no notion of parallelism, equality of segments or of angles, perpendicularity,etc . I have then obtained the well known finite affine geometry. In it, we have the notion ofparallel lines, equality of segments on a given line or on parallel lines, but we have no circles,no notion of equality on non parallel lines, no perpendicularity, etc . It is the purpose ofPart III to construct a finite Euclidean geometry in which these notions as well as measureof angles and distances can be obtained.In the first step, which I will call involutive geometry , I choose an involution on theideal line. This involution either is elliptic, in which case it has no real fixed points oris hyperbolic, in which case it has 2 real fixed points. The elliptic case resembles morethe standard Euclidean geometry, while the hyperbolic case is easier to deal with, but theproperties of both geometries go hand in hand. In it we define circles and perpendicularity. Aprinciple of compensation, which is not evident in the classical case, makes its appearance.For instance, if we consider the lines through the center of a circle, half of them do notintersect the circle, but the other half do and then at two points. As an other example, notall triangles have an inscribed circle, only roughly one in 4 has, but these have 4 inscribedcircles. In the involutive geometry, I also define the equality of angles and the equality ofsegments.In the second step, I will introduce the sympathic geometry , in which we have the notionof measure of angle. The algebraic development suggests a finite trigonometry. In fact 2 suchtrigonometries are required for each prime, corresponding to the elliptic and to the hyperboliccase. The trigonometry for the elliptic case is obtained easily from the notion of primitiveroots associated to p. The trigonometry for the hyperbolic case, requires a generalization.In the last step, I introduce the notion of measure of distances and obtain the finiteEuclidean geometry . G30.TEX [MPAP], September 9, 2019
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
The Theorems in finite Euclidean Geometry fall also in several categories. The first one, . . .The theorems are a direct consequence of . . . .The proof follows by assuming like in section . . . that m corresponds to the line at infinityand . . . . The reference in parenthesis is to the section ¡?¿ in Theorem . . . . Theorem.
0. Let M × H A × A C
0, . . . , then the points C C C p.
1. Let H M A A D
0, . . . , then the points D D D q.
2. The intersection P of p and q is on the line eul of Euler.Proof: Use AA1, 3.0, 3.1, with H M , M M , C C , D C , p = p and q = p. . . . Involves problems of the second degree, bisectrices, inscribed circles for even triangles. Introduction.
It is not my intention to devlop here the extensive theory on circles for involutive geometryover arbitrary fields. I will simply give an example which illustrates how the problem canbe approached effectively.
Definition.
Let θ be a defining circle and m , the ideal line, any circle γ can be written as γ = θ + ( m ) ×× ( r ) , where ( r ) = [ r , r , r ] is a given constant times the radical axis with θ . The γ is defined by the point, with coordinates r , r and r . I willwrite ( r ) := ( r , r , r ) for that representation. θ is represented by the origin. A degeneratecircle ( m ) ×× ( r ) is represented by the direction of r . Exercise.
What is the representation of tangent circles. .1. FINITE INVOLUTIVE GEOMETRY.
Lemma. If γ and γ are circles, represented by ( r ) and ( r ) , then the family of circles throughtheir intersections is represented by ( r ) + k ( r ) ,with k an arbitrary element in the field together with ∞ , where ∞ represents γ . The additionis that of vectors in 3 dimensions and the multiplication by k the scalar multiplication. I will also denote the family by γ + kγ .One can also use the homogeneous representation, k γ + k γ .Strictly speaking, this is the representation used in the proofs, although I have used the nonhomogeneous representation to simplify the writing. Theorem. [Bundle]
Let γ j , j = 0 to , be 4 circles, if there is a circle α which passes through the intersection ofthe circles γ and γ , as well as the intersection of γ and γ , then there is a circle β passingthrough the intersections of γ and γ , as well as those of γ and γ . This is the so called bundle Theorem. Proof: If r j ) is the representation of γ j . The family through the first 2 circles is repre-sented by ( r ) + k ( r ) and that through the last 2 circles by( r ) + l ( r ) , the hypothesis concerning the circle α implies( r ) + k ( r ) = u (( r ) + l ( r ) , which can be rewritten( r ) + u ( r ) = − k (( r ) + ul ( r ) ), which gives the conclusion concerning the circle β. Introduction.
The parabola, ellipse and hyperbola have already be defined in affine geometry. Here westudy their properties in involutive geometry.The parabola.The ellipse and hyperbola.If we assume that the isotropic points are ( δ, ,
0) and ( − δ, , , where δ = d = N p, wewill see that by an appropriate . . . transformation, these can be reduced to X A + X B = X . Recall also that i = − Definition.
The isotropic tangents are isotropic lines tangent to the conic. The foci are the intersectionof 2 isotropic tangents through 2 different isotropic points. see Dembosky, p. 256 CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem.
Given the conic X A + X B = X . D0. C = A + Bd, thenC0. The point polarity is B A
00 0 − AB C1. The line polarity is A B
00 0 − C2. The isotropic tangents through ( δ, ,
0) are(1 , − δ, √ C ) and (1 , − δ, − sqrtC )C3. The foci areC3.0. ( −√ C, , , ( √ C, , , C3.1. (0 , −√ C, δ ) , (0 , √ C, δ ) , C4.0. C =R p ⇒ the foci C.3.0. are real, the foci C.3.1. are not.C4.1. C =N p ⇒ the foci C.3.1. are real, the foci C.3.0. are not. Theorem.
Given the conic
D0. X A + X B = X . H1.0. A =R p, B =R p, D1.0. a = √ A, b = √ B, H1.1. A =N p, B =N p, D1.1. a = (cid:113) Ad , b = (cid:113) Bd , H1.2. A =R p, B =N p, D1.2. a = √ A, b = (cid:113) Bd , H1.3. A =N p, B =R p, .1. FINITE INVOLUTIVE GEOMETRY. a = (cid:113) Ad , b = √ B, then the conic takes the form C1.0. x a + y b = 1 , C1.1. d x a + d y b = 1 , C1.2. x a + d y b = 1 , C1.3. d x a + y b = 1 , Theorem.
H0.0. p = − mod andAB = Rp, orH0.1. p = 1 mod andAB = N p, then
C0. the conic is an ellipse,
Theorem.
H0.0. p ≡ and AB R p, or H0.1. p ≡ − and AB N p, then C0. the conic is a hyperbola.The ideal points on it are
C1.0. ( ab i, , , ( − ab i, , , C1.1. ( ab , , , ( − ab , , . Introduction.Notation.
A pair of reals between parenthesis will denote the Cartesian coordinates of a point. Wecannot choose a pair of reals between brackets to denote the x and y intercept of a line inthe Cartesian plane, because we have then no way to represents lines through the origin. Wewill therefore use triplets, with the last non zero coordinate normalized to 1. CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem.
If we choose as x axis the line [0,1,0] and as y axis [1,0,0], then we have the correspondence: C ( i, j,
1) = ( i, j ) ,C [ i, j, k ] = [ ik , jk , , k (cid:54) = 0 ,C [ i, j,
0] = [ ij , , , j (cid:54) = 0 ,C [ i, ,
0] = [1 , , . Theorem.
Given a triangle whose vertices have the Cartesian coordinates(0 , a ) , ( b, , ( c, , a (cid:54) = 0 , b (cid:54) = c.
0. The point whose barycentric coordinates are ( q , q , q , with q q q (cid:54) = 0 , corresponds to the point whose Cartesian coordinates are( bq cq q q q , aq q q q ) . REDO 1. IN view of the preceding theorem1. The line, distinct from the ideal line, whose barycentriccoordinates are[ l , l , l bl − cl l − l , bl − cl c − b ) l − cl bl , if bl − cl c − b ) l − cl bl (cid:54) = 0 , it corresponds to0 , l − l c − b ) l − cl bl , and (( c − b ) l − cl bl
2) = 0 , it corresponds to1 , ,
2. The values of the coordinates of the orthocenter are m bc ( b − c ) , m c ( a + bc ) , m − b ( a + bc ) . Definition. The following mapping associates to the non ideal points in the finite Euclidean plane asso-ciated to p , points in the classical Euclidean plane. T ( i, j ) = ( i + kp, j + lp ) , where k and l are any integers. Theorem.
Let d = i j − i j , then ( i , j x ( i , j
2) = [ j − j d , i − i d , , d (cid:54) = 0 , ( i , j x ( i , j
2) = [ j − j i − i , , , d = 0 , i (cid:54) = i , ( i , j x ( i , j
2) = [1 , , , d = 0 , i i , j (cid:54) = j . For the following see ..[1,135]/cartes .1. FINITE INVOLUTIVE GEOMETRY. Example.
For p = 13, let the circles beCr: x + y = r . The points on the circles areC1: (1 , , ( − , , (0 , , (0 , − , (6 , , ( − , , (6 , − , ( − , − , (2 , , ( − , , (2 , − , ( − , − , C2: (2 , , ( − , , (0 , , (0 , − , (4 , , ( − , , (4 , − , ( − , − , (1 , , ( − , , (1 , − , ( − , − , C3: (3 , , ( − , , (0 , , (0 , − , (6 , , ( − , , (6 , − , ( − , − , (5 , , ( − , , (5 , − , ( − , − , C4: (4 , , ( − , , (0 , , (0 , − , (5 , , ( − , , (5 , − , ( − , − , (2 , , ( − , , (2 , − , ( − , − , C5: (5 , , ( − , , (0 , , (0 , − , (4 , , ( − , , (4 , − , ( − , − , (3 , , ( − , , (3 , − , ( − , − , C6: (6 , , ( − , , (0 , , (0 , − , (3 , , ( − , , (3 , − , ( − , − , (1 , , ( − , , (1 , − , ( − , − , The isotropic lines through the origin contain the points:i0: (0 , , (1 , − , (2 , , (3 , − , (4 , , (5 , , (6 , − , ( − , , ( − , − , ( − , , ( − , − , ( − , − , ( − , , i1: (0 , , (1 , , (2 , − , (3 , , (4 , − , (5 , − , (6 , , ( − , − , ( − , , ( − , − , ( − , , ( − , , ( − , − , If we join the origin to the points (1,k) and (1,l) we obtain perpendicular directions, withk,l = 0,oo; 1,-1; 2,6; 3,4; -2,-6; -3,-4.For p = 13, let the circles beCr: x − xy + y = r . The points on the circles areC1: (0 , , (1 , , (1 , , (4 , , (4 , − , (6 , , (6 , − , C2: (0 , , (1 , − , (1 , − , (2 , , (2 , − , (5 , , (5 , − , C3: (0 , , (1 , , (1 , , (3 , , (3 , , (5 , , (5 , , C4: (0 , , (2 , , (2 , − , (3 , , (3 , , (4 , , (4 , − , C5: (0 , , (4 , , (4 , , (5 , , (5 , , (6 , , (6 , , C6: (0 , , (2 , , (2 , − , (3 , − , (3 , − , (6 , , (6 , − , as well as the points symmetric with respect to the origin.If we join the origin to the points (1,k) and (1,l) we obtain perpendicular directions, withk,l = 0,-4; 1,-1; 2,-5; 3,oo; 4,-2, 5,-6; 6,-3.For p = 11, let the circles beCr: ex x − xy + y = r . The points on the circles areC1: (0 , , (1 , , (4 , , CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
C2: (0 , , (2 , − , (3 , , C3: (0 , , (1 , , (1 , , C4: (0 , , (4 , , (5 , , C5: (0 , , (2 , , (2 , − , as well as the points symmetric with respect to the diagonals, ( i, j ) here means ( i, j ) , ( j, i ) , ( − i, − j ) , ( − j, − i ) . The isotropic points arel0: ex(0 , , (1 , , (1 , , (2 , − , (2 , − , (3 , − , (3 , , (4,1),(4,5),(5,-2),(5,4), l ex (0,0),(1,-3),(1,-4),(2,3),(2,5),(3,2),(3,-1),(4 , − , (4 , − , (5 , , (5 , − , If we join the origin to the points (1 , k ) and (1 , l ) we obtain perpendicular directions, with k, l = 0,-5; 1,-1; 2, ∞ ; 3,5; 4, -2.For p = 11, let the circles beCr: ex x + y = r . The points on the circles areC1: (0,1), (3,5),C2: (0,2), (1,5),C3: (0,3), (2,4),C4: (0,4), (1,2),C5: (0,5), (3,4),as well as the points symmetric with respect to the 2 axis and the diagonals.( i, j ) here means ( i, j ) , ( i, − j ) , ( j, i ) , ( j, − i ) , ( − i, − j ) , ( − i, j ) , ( − j, − i ) , ( − j, i ) . If we join the origin to the points (1 , k ) and (1 , l ) we obtain perpendicular directions, with k, l = 0 , ∞ ; 1,-1; 2,5; 3,-4; 4,-3; -2,-5. Introduction.Theorem.
To the point ( x, y ) , in classical geometry, on a circle centered at the origin and of radius r, corresponds, if r is not congruent to 0 modulo p , the point ( x/rmodp, y/rmodp ) on a circleof radius 1 in the finite geometry associated to p .Vice-versa, given a point P = ( x, y ) on a circle of radius 1 in the finite geometry associatedto p , we can always find a point on a circle in the classical geometry which is one of therepresentatives of P , in the mapping given in . . . .The proof is left to the reader. The first part is trivial, the second part is not trivial. Seealso [135]FINPYT.BAS .1. FINITE INVOLUTIVE GEOMETRY. Example.
For p = 13,(2,6) for r = 1 is associated to (15,20) for r = 25.For p = 29,(5,11) for r = 1 is associated to (24,18) for r = 30.(8,13) for r = 1 is associated to (108,45) for r = 117.(6,9) for r = 1 is associated to (180,96) for r = 204. Theorem.
There exist a circle of radius u in R which contains all the representatives of a circle in Z p . Indeed, for the radius 1, for instance, if one of the representatives is on the circle x + y = r and if s = 1 /r , then u = r i ( s i + k i p ) , for all i,by finite induction, if u = r ( s + pk ) = r ( s + pk ) , then r k − r k = ( r s − r s ) /p, this gives k = a + r k (cid:48) and with s (cid:48) = ( s + a p ) /r ,u = r r ( s (cid:48) + pk (cid:48) ) , . . . . Example.
For p = 29, for r = 1, we start withpoint in Z r i s i k − k = 3 , a = − , s (cid:48) = − ,u = 5 . − k (cid:48) ) , k (cid:48) − k = 3 , a = 1 , s (cid:48) = 5 ,u = 5 . . k (cid:48) ) , hence the suitable circle in R with smallest radii has radius u = 1625 and contains thepoints in R in Z -1300,-975 5,11;-1500,-625 8,13;-1560,-455 6,9.04 CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Answer to 1.13.1.
For the second part.The problem can be restated successively as follows, given a solution of0. x + y = z , there exist i, j, k such that ( x + ip ) + ( y + jp ) = ( z + kp ) , or there exist u and v such that u − v = x, uv = y, u + v = z, eliminating v from the first 2 equations and using 0., gives u = r + x , v = r − x , r + x need not be a quadratic residue, therefore we use instead u = b r + x , v = b r − x , c = 1 /b, this gives u and v , x + ip = ( u − v ) c, y + ip = 2 uvc, z + ip = ( u + v ) c. A more careful discussion will show that signs may have to be changed and the role of x and y interchanged.For instance, for p = 13 and 2 + 6 = 1 , b = -2, c = 6, u = ( − −
2) = 6 , v = (6)( −
2) =1 , hence x + ip = 35 , − ( y + jp ) = 72 , z + kp = 222 . For p = 17, and x = 4, y = 6, z = 1, b = 3, c = 6, u = (6)(3) = 1 , v = (12)(3) = 2 , hence after interchange of x and y , x + ip = 72 , y + jp = 210 , z + kp = 222 . For p = 19 and 3 = 1 , b = 2, c = 10, u = (2)(2) = 2 , v = ( − , hence − ( x + ip ) = 320 , − ( y + jp ) = 240 , z + kp = 400 . (AFTER INVOLUTIVE GEOMETRY) Comment.
For the following theorem, I will not give a linear construction, although one could be given.The theorem is a generalization of the Theorem of Miquel and can be further generalized inthe context of Gaussian geometry.
Notation. If u and v are 2 lines and ξ is a conic, ξ − u ×× v = 0,is equivalent to ξ ( X ) − ( u · X ) ( v · X ) = 0.This should be moved before the definition of circles. Theorem.
The radical axis of the 2 circles µ j := θ − m ×× u j , j = 0 , is u − u . Indeed, µ ( X ) − µ ( X ) = − ( m · X ) (( u − u ) · X ) = 0 therefore µ = µ − m ×× ( u − u ) . .1. FINITE INVOLUTIVE GEOMETRY. Theorem.
The radical axis of each pair of 3 circles are concurrent .Proof: Let the 3 circles be µ i := θ − m ×× u i , u i are the radical axis of these circles with θ .the 3 radical axis are u − u , u − u , u − u , but u − u = ( u − u ) + ( u − u ), thereforeone of the axis passes through the intersections of the other 2. Theorem. [Miquel]
H0. N i · a i = 0 , H1. N i · m (cid:54) = 0 , D0. µ i := circle ( A i , N i +1 , N i − ) , D1.
M iquel := ( µ × µ ) − N , then C0.
M iquel × µ = 0 . The nomenclature.N0. Miquel is called the point of Miquel associated to N i . Proof. Let N = (0 , , q ) , N = ( q , , , N = (1 , q , . H1. implies that 1 + q i (cid:54) = 0 . If the equation of µ a circle is θ − m ×× u = 0 , with u = [ u , , u , , u , ] ,m = [1 , , θ = m (cid:48) X X + m (cid:48) X X + m (cid:48) X X m = [1 , , m (cid:48) = m m m , . . . ).D0. implies, for i = 0 ,u = [0 , m (cid:48) q , q m (cid:48) q ] . Let
M iquel = ( X , X , X ). If the 3 circles have a point in common, it is on the intersectionof the 3 radical axis u i , it is therefore necessary that( u i − u i +1 ) · M iquel = 0 , i = 0 , , , therefore M iquel = ( u − u ) × ( u − u ) , this gives after simplification, M iquel = ( m (cid:48) q ( − q m (cid:48) q + q q m (cid:48) q + m (cid:48) q , . . . ) . It remains to verify that
M iquel belongs to µ i . First, u i · M iquel = m (cid:48) m (cid:48) m (cid:48) q q q q (1 + q )(1 + q )) , second, m · M iquel = − q m (cid:48) m (cid:48) (1+ q ) + . . . + m (cid:48) m (cid:48) q q (1+ q )(1+ q ) + . . . .It is straigthforward to verify that the product of these two expressions is precisely m (cid:48) X X + m (cid:48) X X + m (cid:48) X X . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem. [Miquel]
H0. n := N × N , n · N = 0 . D0. n = [ n , n , n ] , then C0.
M iquel · θ = 0 . C1. q = − n n , q = − n n , q = − n n . C2.
M iqnel = ( n n m (cid:48) n − n , n n m (cid:48) n − n , n n m (cid:48) n − n ) . The condition that the points N i be collinear is precisely 1 + q q q = 0 , but in this case θ = 0 as follows from the expression u i · M iquel.
It is straightforward to verify C1 and C2.
Corollary. The circles µf i circumscribed to A i , M A i +1 , M A i − have a point F ock in common.1.
F ock is in the circumcircle θ . µf i = θ + m ×× [0 , m m ( m + m ) m − m , − m m ( m + m ) m − m ].3. F ock = ( m ( m + m )( m − m )( m − m ) , m ( m + m )( m − m )( m − m ) , m ( m + m )( m − m )( m − m )). This is the special case when n is the orthic line m = [m ,m ,m ].The point F ock had been constructed before (D38.9) and proven to be on θ (C38.4). Theorem. [Miquel]
D0. N i,j := midpoint ( A i , N j ) , D1. n i,j := mediatrix ( A i , N j ) , D2. C i := n i,i +1 × n i,i − , D3. φ := circle ( C , C , C ) , then C0. O · φ = 0 . Proof:P0. N , = (1 + 2 q , , , N , = (2 + q , q , . P1. n , = [ m m , m − (1 + 2 q ) m , − (1 + 2 q )( m m ,n , = [ − q ( m m , (2 + q )( m m , (2 + q ) m − q m . P2. C = ( m q + q q ) m q )(1 + q ) m q )(1 + q ) m , ( m m q ) m q q − m , ( m m − q q ) m q ) m . P3. φ : . . . Problem.
The following question suggests itself. Let ν := circle ( N , N , N ) . What relation exists between all circles ν having the same point of Miquel?Same question in the case for which the point of Miquel is on θ . .1. FINITE INVOLUTIVE GEOMETRY. Theorem. [Simson and Wallace]
H0. X · θ = 0 , D0. n i := X × Im i , D1. N i := n i × a i , D2. n := N × N , then C0. N · n = 0( ∗ ) . C1. ( W × N i ) ⊥ a i . ?Proof:P0. n = [ m X − m X , ( m m X + m X , − m X − ( m m X ] . P1. N = (0 , m X + ( m m X , m X + ( m m X ) . P2. n = [ X X ( − m X + ( m m X + X )) ,X X ( − m X + ( m m X + X )) ,X X ( − m X + ( m m X + X ))] . To obtain the last expression we use in each coordinate the relation H0, m m m X X + m m m X X + m m m X X = 0 . MAY WANT TO REFER HERE TO THE FOLLOWING BUT MOVE IT AS APPLI-CATION OF PARABOLAS.
Theorem.
The set of lines having the same point X of Miquel are on a line parabola :C0. mup − ( X ) : X u u − u /m (cid:48) = X u u − u /m (cid:48) = X u u − u /m (cid:48) .µp ( X ) : ( X X m (cid:48) U ) + ( X X m (cid:48) U ) + ( X X m (cid:48) U ) − X X X ( X m (cid:48) m (cid:48) U U + X m (cid:48) m (cid:48) U U + X m (cid:48) m (cid:48) U U ) . C1. a i · µp ( X ) = 0 . C2.
The line of Simson and Wallace is the tangent at the vertex.
C3.
The point of Miquel is its focus.
Proof. C2 of Theorem . . . gives C0.
Introduction.
The conic of Kiepert has been constructed in 5.4.1.D3.8. .Kiepert showed that, in the classical case, if V i is a point on the mediatrix mf i such that angle ( A , A , V ) = angle ( A , A , V = angle ( A , A , V ) , then v i := A i × V i have a point V in common which is on a hyperbola, now known as the CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY hyperbola of Kiepert. After proving this Theorem in the finite case, I will consider severalspecial cases of interest, which can be obtained either by a linear or by a second degreeconstruction. In the latter case, if the angle is π , the point is called the point of Vectem, towhich is associated a special chapter of the classical theory of the geometry of the triangle.The cases when the angle is π and π are also discussed and a new property is obtained. Theorem.
Let H0.0. X · θ = 0 . G0.0. X = ( X , X , X ) . D1.0. x A × X, P1.0. x X , , − X ] , D1.1. V := x × mf , P1.1. V = (( m m X , ( m m X − ( m − m X , ( m m X ) . D1.2. v := A × V , P1.2. v = [0 , ( m m X , ( m − m X − ( m m X ] . D1.3. x A × V , P1.3. x m m X − ( m − m X , − ( m m X , . D1.4. x M a × X, P1.3. x − X − X , X , X ] . D1.5. Y = x × x . P1.4. Y = (( m m X , ( m m X − ( m − m X , ( m − m X + ( m m X )D2.0. x A × X, P2.0. x , X , − X ] . D2.1. X x × m, P2.1. X X + X , − X , − X ) . D2.2. x A × X , P2.2. x X , X + X , . .1. FINITE INVOLUTIVE GEOMETRY. V := mf × x , P2.3. V = (( m m X + X ) , ( m m X , m X + ( m m X ) . D2.4. v := A × v , P2.4. v = [2 m X + ( m m X ) , , − ( m m X + X )] . D3.0. y A × Y, P3.0. y , Y , − Y ] . D3.1. Y y × m, P3.1. Y Y + Y , − Y , − Y ) . D3.2. y A × Y , P3.2. y Y , , Y + Y ] . D3.3. V := mf × y , P3.3. V = (( m m Y + Y ) , ( m m Y + 2 m Y , − ( m m Y ) . D3.4. v := A × V , P3.5. v m m Y + 2 m Y , − ( m m Y + Y ) , . D4.0. V = v × v , P4.0. V = ( u ( X + X ) , ( m m m − m m − m X − m m m X + uX , m m m m m X + uX ) , where u := ( m m m m m m . then C0.0. V · v = 0 . C0.1. V · κiepert = 0 . The construction is based on angle ( X, A , A ) = angle ( A , A , Y ) = angle ( X, A , A ) = angle ( A , A , V ) , implying the parallelism of A × X and A × V and symmetrically for V . For P4.0., after replacing Y , Y and Y by their values from P1.4., the equation for θ is usedto express X X in terms of X X and X X . Exercise.
To complete the proof of x.x.1., the 2 special case X = A and X = A should be considered.This is left as an exercise.In the first case x ta at A, in the second case x ta at A . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Exercise.
Proceed in the inverse order and construct X from V. Prove that if V is on κiepert then X is on θ . Exercise.
Study the projectivity which associates to ( X , X , X ) , the point ( V , V , V ) , as given byP4.0. without assuming that ( X , X , X ) is on θ . Determine 4 points and their images andconstruct any of these points if they have not been constructed in this book.The following are special cases. X = A , α = 0 gives V = M.σ = π gives V = M .σ = π gives the point of Vectem (see below). σ = π gives the equilateral point (see below) σ = π gives the hexagonal point (see below) σ = angle ( A i − , A i , A i +1 ) gives V = A i . Other angles give V = T ar. (5.4.1.D16.3.), V = Br . (5.4.1.D15.3.) V = Br . (5.4.1.D15.3.)and V = En. (5.4.1.D21.10)D5.0.
M am i := ma i +1 × ma i − , M am i := ma i +1 × ma i − , D5.1. Ae i := a i × e, D5.2. mae i := Ae i +1 × M am i − , mae i := Ae i +1 × M am i − , D5.3.
M M a i := mae i × a i , M M a i := mae i × a i , D5.4. mm := M M a × M M a , mm := M M a × M M a , then C5.0. n i · Kiepert . C5.1. mm · Kiepert mm · Kiepert . C5.2. mm · K = mm · K = 0 . C5.3. ] S is the center of Kiepert1, S is the cocenter .The nomenclature:Proof.P5.0. M am = ( m , m , m , M am = ( m , m , m , P5.1. Ae = (0 , m − m , m − m , P5.2. mae = [ m m − m , m m − m , m m − m ,mae = [ m m − m , m m − m , m m − m , .1. FINITE INVOLUTIVE GEOMETRY. M M a = (0 , m m − m , m m − m ,M M a = (0 , m m − m , m m − m , P5.4. mm = [ m m − m , m m − m , m m − m ,mm = [ m m m − m , m m m − m , m m m − m , The tangent at
T ar is[ m m q ( m − m , m m q ( m − m , m m q ( m − m . The tangent at Br m ( m m ( m − m , m ( m m ( m − m , m ( m m ( m − m . The tangent at Br m m m m ( m − m , m m m m ( m − m , m m m m ( m − m . Example.
With p = 13 , A [] = (14 , , , M = (28) , M = (44) , M am = (41 , , , M am =(31 , , , Ae = (4 , , , mae = [138 , , , mae = [151 , , , M M a =(9 , , , M M a = (7 , , , mm = [146] , mm = [136] ,V = ( asin ( A − α ) , bsin ( B − α ) , csin ( C − α ) ) . Theorem. If V σ = ( V , V , V ) is associated with the angle σ , then V − σ = (( m − m m m m m V + 2 m m m m − m V + 2 m m − m m m V , m m − m m m V + ( m − m m m m m V + 2 m m m m − m V ) + 2 m m m m − m V , m m − m m m V + ( m − m m m m m V ) . Proof: v i := V × A i , v = [0 , V , − V ] .V i := v i × mf i ,V = (( m m V − V ) , ( m − m V , ( m − m V ) .va i = A i +1 × V i − ,va = [( m m V − V ) , , ( m − m V ] .va i = A i − × V i +1 ,va = [( m m V − V ) , ( m − m V , .V a i = m × va i ,V a = (( m − m V , m V − ( m m V , − ( m m V − V )) .V a i = m × va i ,V a = (( m − m V , ( m m V − V ) , − m V + ( m m V )) .vb i := V a i × A i ,V b = [0 , ( m m V − V ) , m V − ( m m V ] .vb i := V a i × A i ,vb = [0 , m V − ( m m V , ( m m V − V )] . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY V − σi := vb i +1 × vb i − ,V − σ = (( m m V − V ) , m V − ( m m V , − m V + ( m m V ) .v − σi := V − σi × A i ,v − σ = [0 , m V − ( m m V , m V − ( m m V ] .V − σ = v − σ × v − σ ,V − σ · v − σ = 0( ∗ ) . For the determination of V − σ , I have multiplied the components by m − m V σ is on κiepert to eliminate V V . Every component is thendivisible by V . Example. p = 11 , M = (1 , , ,V σ = (1 , − , − , V − σ = (1 , , ,V i = { (1 , − , , (1 , , − , (1 , − , − , } V − i = { (1 , , , (1 , , , (1 , , , } the sides of these triangles are vv i = { [1 , , , [0 , , , [1 , , , } vv − i = { [1 , − , , [1 , − , − , [0 , , − } . vv i × vv − i = { (1 , , − , (1 , − , − , (1 , − , , } which have [1,4,7] in common. Exercise . x.x.x. defines a projectivity which fixes the conic of κiepert.
Determine other propertiesof this projectivity.
Exercise.
Construct V π − σ and V π + σ and obtain properties involving these points and V σ , V − σ andlines derived from these. Introduction.
In classical Euclidean Geometry, the construction of the point of Vectem starts with thatof squares on the sides of the triangle, outside of it. In the finite case, there is ambiguityand the squares need not exist. It is easy to determine the intersections of the circle κ with the perpendicular through A to a . This leads to the expression for X , given below.To insure the consistency associated to the outside condition of the classical case I havestarted with that definition for X , , chosen X , on κ and the perpendicular at A to a in such a way that X , × X , is parallel to a . X , , X , and X , , X , , are definedusing the symmetry operation ρ , defined in section ?.?.?. Because α is obtained in section?.?.?. using a square root operation the definitions can be repeated using − α instead of α , the corresponding elements are denoted with a superscript -. These have been givenexplicitely. The conclusions have not been written explicitely. To each conclusion (and .1. FINITE INVOLUTIVE GEOMETRY. − and α by − α .Explicit expression for distances and area, if needed, are given uding the same notationas in the conclusions, replacing C by F.One could also proceed by first choosing one of the intersections of κ with A × Ima as X . and constructing all the other points. For instance, X , by ( A × Ima ) × X , × M A ,X , by ( A × Ima ) × ( ma × ( A × X , )) , etc. Theorem.
H0.0. X , := ( m m m , α − m m , − m m , X i,i := ρ i X , , X , := ( m m m , − m m , α − m m , X i,i := ρ i X , , H0.1. X − , := ( m m m , − α − m m , − m m , X − i,i := ρ i X − , , X − , := ( m m m , − m m , − α − m m , X − i,i := ρ i X − , , D0.1. x i,j,k := A i × X j,k , i (cid:54) = k. x − i,j,k := A i × X − j,k , i (cid:54) = k. D0.2. V i := x i +1 ,i − ,i × x i − ,i +1 ,i , V − i := x − i +1 ,i − ,i × x − i − ,i +1 ,i , D0.3. W i := x i +1 ,i,i +1 × x i − ,i,i − , W − i := x − i +1 ,i,i +1 × x − i − ,i,i − , D0.4. U i := x i +1 ,i − ,i +1 × x i − ,i +1 ,i − , U − i := x − i +1 ,i − ,i +1 × x − i − ,i +1 ,i − , D0.5. v i := A i vV i , v − i := A i × V − i , D0.6. w i := X i +1 ,i − × X i − ,i +1 , w − i := X − i +1 ,i − × X − i − ,i +1 , D0.7. V := v × v , V − := v − × v − , D0.8. v := V × V − . D1.0. Ix i,j,k := m × x i,j,k , j (cid:54) = k. D1.1. Iv i = m × v i , D1.2. Iw i = m × w i , then C0.0. ( X i +1 ,i × X i − ,i ) · M A i = 0 . C0.1. V · v = 0( ∗ ) . C0.2. U i · ma i = 0 . C0.3. W i · w i = 0 . C0.4. W i · v i = 0 . C0.5. V − i · w i = 0 . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
C0.6. x i +1 ,i,i +1 ??? x i − ,i,i − . C0.7. v i · w i . C0.8. dist ( A i +1 , X i +1 ,i ) = dist ( A i − , X i − ,i ) = dist ( A , A − ) . C0.9. dist ( A i +1 , X i,i +1 ) = dist ( A i − , X i,i − ) . C0.10. dist ( A i , V i ) = dist ( V i +1 , V i − ) . C0.11.
Area ( A i , X i,i +1 , X i,i − ) = Area ( A , A , A ) . P0.1. x , , = [0 , m , m m .x , , = [0 , m m , m .x , , = [0 , α − m m , m m .x , , = [0 , m m , α − m m .x , , = [0 , m m − α, m m m .x , , = [0 , m m m , m m − α ] . P0.2. V = ( m m m , α − m m , α − m m . P0.3. W = ( m m m α − m m m s + m , m m α − m m , m m α − m m . P0.4. U = ( − m m m , m α − m m , m α − m m . P0.5. v = [0 , m m − α, α − m m . P0.6. w = [2 m m α, m m m s − α ) , m m m s − α )] . P0.7. V = (( α − m m α − m m , ( α − m m α − m m , ( α − m m α − m m . P0.8. v = [( m − m m s − m m , ( m − m m s − m m , ( m − m m s − m m . F0.6. dist ( A , X , ) = 2 α − m m − m m m .dist ( A , X , ) = 2 α − m m − m m m . P1.0. Ix , , = ( m , − ( m m , m .Ix , , = ( m , m , − ( m m .Ix , , = ( α − m m m , m m , m m − α ) .Ix , , = ( m m m − α, α − m m , − m m .Ix , , = ( α − m m m , m m m , m m − α ) .Ix , , = ( m m α, m m − α, − m m m . P1.1. Iv = ( m m m − α, α − m m , − m m . P1.2. Iw = ( m m − m α, m m m s − (2 m m α, m m m α − m m s )) . The nomenclature: .1. FINITE INVOLUTIVE GEOMETRY.
Theorem.
0. 2
Area ( A , A , X , ) = − m m .
1. 2
Area ( A , A , X , ) = α − m m . dist ( V , V ) = Comment.
The isotropic points are real if − α is a quadratic residue (5.5.2.) if p ≡ p − π = p − π with the sides, this is consistent with thefact that X i,j are integers. If α is imaginary and p ≡ − , then π = p + 1 is divisibleby 4 and the same situation exist. The equilateral and hexagonal points.
Let β = α √ ,H0.0. V ei = κ i +1 × κ i − ,V e = ( m m m , β − m m , β − m m . D0.0. vea i := V i +1 × A i − ,vea = [ m m m , m m − β, .vea i := V i − × A i +1 ,vea = [ m m m , , m m − β ] . D0.1.
V ea i := vea i +1 × M A i − ,V ea = ( β − m m , m m m , m m β )) .V ea i := vea i − × M A i +1 ,V ea = ( β − m m , − m m − β, m m m . D0.2. veb i := V ea i +1 × M i − ,veb = [ m m m , − m m m , β + m m − m .veb i := V ea i − × M i +1 ,veb = [ m m m , β − m m − m , − m m m . D0.3.
V eb i := veb i × vea i ,V eb = ( β − m m , β − m m , m m m .V eb i := veb i × vea i ,V eb = ( β − m m , m m m , β − m m . D0.4. vec i := vV eb i × A i ,vec = [0 , − m m m , β − m m .vec i := vV eb i × A i ,vec = [0 , β − m m , − m m m . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
D0.5. V hi := vec i +1 × vec i − ,V h = ( m m m , β − m m , β − m m . D1.0 vve i := V ei +1 × V ei − ,vve = [ β + m m , β − m m m , β − m m m . D1.1. vvh i := V hi +1 × V hi − ,vvh = [ m m − β, β − m m m , β − m m m . D1.2.
V eh i := vve i × vvh i ,V eh = (0 , − (2 β − m m m , β − m ∗ ( m m . D1.3. veh := V eh × V eh ,veh = [(2 β − m m m β − m m m , (2 β − m m m β − m m m , (2 β − m m m β − m m m . then C0.0. V ei · mf i = 0 . C0.1. V e · v e = 0( ∗ ) . C0.2. V h · κiepert = 0C0.3. V h · v h = 0( ∗ ) . C0.4. V e · κiepert = 0C0.5. V eh i · a i = 0 . C0.6.
V eh · veh = 0( ∗ ) . The nomenclature:N0.0. V ei are the equilateral points , such that angle ( V ei , A i +1 , A i − ) = π .V e is therefore on κ i +1 and κ i − . N0.1. V hi are the hexagonal points , such that angle ( V hi , A i +1 , A i − ) = π .V h is therefore the barycenter of the equilateral triangle ( V ei , A i +1 , A i − ) . Answer to x.x.4. X = A gives V = A ,X = A gives V = (( m m m m , m m m , m m m ,X = A gives V = M.X = ( m m , − ( m − m , m − m
2) gives V = A . .1. FINITE INVOLUTIVE GEOMETRY. Introduction.
When p = 5 , it is natural to try to represent involutive geometry on the dodecahedron. Themost natural choice, for the ideal line, in the hyperbolic case is an edge-line. We can choosetwo face-points as the isotropic points. In the elliptic case, the simplest choice for the idealline appears to be a vertex-line. The fundamental involution associates to a vertex-point anedge-point. Definition.
In the case of hyperbolic involutive geometry, the isotropic points are chosen as 2 face-points.
Theorem.
With the chosen fundamental involution, the perpendicular direction of a vertex-point is avertex point and to an edge-point is an edge-point.
Example.
If the isotropic points are 0 and 4, the perpendicular directions are 10 and 24 as well as 23and 26.
Theorem.
There are 100 circles in a hyperbolic involutive geometry.Number of center sub − types2 f B, G D , H . f B, G D , H . v C , D G , I . v C , D G , I . v B, H G , G . s A, E E , I . s B, I F, H . s C , D D , G . s C , D D , G . Proof: For the type f f f f ss, out of 15 quadruples only 6 contain 2 given ones, thereforethe number of conics must be divided by . For the type f f f xxx, out of 20 triples only 4 contain 2 given ones, hence the number ofconics must be divided by = 5 . For the type f f xxxx, out of 15 pairs, only one is the given one, therefore, the number ofconics is to be divided by 15.As a check there are 25 ∗ ∗ = 100 conics through 2 given points.More precisely the conics are1 of type f f f f f f, sub-type A. CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
12 of type f f f f ss, sub-type B.
12 of type f f f vvs, C
1, 6 of sub-type C .
24 of type f f f vss, D , D , D D . f f vvvv, E , E . f f vvvs, sub-type F.
24 of type f f vvss, G G G G .
12 of type f f vsss, H , H H H . f f ssss, I , I , I , and 1 of sub-type I . The centers and their relationship to the conic have been determined using the program[130]DODECA.
Theorem.
In the case of elliptic involutive geometry, if a vertex-line is chosen as the ideal line, thereexists an elliptic projectivity which associates, alternately, to a vertex-point, an edge-pointand to an edge-point, a vertex-point.
Definition.
The projectivity of Theorem 4.1.11 is chosen as the fundamental projectivity . Example.
We can choose as ideal line 5 ∗ and as fundamental projectivity (7,13,23,27,29,26). Theorem.
Given a center, there are 4 circles with 6 ordinary points on them. 2 have a diameter inthe direction of a ideal face-point and 2 have a diameter in the direction of the other idealface-point.
Theorem.
There are 100 circles in an elliptic involutive geometry.
Number of center sub − types3 f H , M S , G . f H , M S , G . v A, P ; U , U . v I , S H , F. s I , G L , M . s I , G L , M . s H , S G , G . Proof: The proof was done using the program [130] EUCLID5 and the interactive program[130] DODECA. The semi colon separates the circles whose diameter have a different idealface-points. .2. FINITE SYMPATHIC GEOMETRY.
See Example 1.10 . . . Measure of angles, separate from measure of distances,2 triangles having the same angles are similar, their sides are not equal.For measure of distances we can do it starting from a unit (2 ordinary distinct points) on alllines which have the same parity (even or odd), the other parity requires an other unit, thetwo become connected as a subset of sympathic projectivity which is Euclidean geometry.Although we could have subordinated measure of angles to measure of distance we prefer todo the reverse. p. The Hyperbolic Case.
Introduction.
The trigonometry associated to the finite Euclidean plane with real isotropic points willfirst be defined and studied in this section for the finite field Z p . Theorems 1.4. and 1.6.determine sin (1) and cos (1) from which all other values can be obtained using the additionformulas. In section 2, definitions and results will be extended, for the finite field associatedto p e , with proofs left as an exercise. Definition.
Given the sets Z of the integers, Z p of the integers modulo p, Z p − of the integers modulo p − , let δ be a square root of a non quadratic residue of p, I define π as follows π := p − . Therefore π is an integer.The problem addressed here is to construct 2 functions s ine or sin and c osine or cos with domain Z and range { Z p , δZ p } which satisfy:The Theorem of Pythagoras,0. 0. sin ( x ) + cos ( x ) = 1 , The addition formulas,1. sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y ) , .cos ( x + y ) = cos ( x ) cos ( y ) − sin ( x ) sin ( y ) . The periodicity property1. 0. sin (2 π + x ) = sin ( x ) , cos (2 π + x ) = cos ( x ) , The symmetry properties G34.TEX [MPAP], September 9, 2019 CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
2. 0. sin ( π + x ) = − sin ( x ) , cos ( π + x ) = − cos ( x ) , .sin ( − x ) = − sin ( x ) , cos ( − x ) = cos ( x ) , .sin ( π − x ) = sin ( x ) , cos ( π − x ) = − cos ( x ) , .sin ( π − x ) = cos ( x ) , cos ( π − x ) = sin ( x ) , .sin ( π + x ) = cos ( x ) , cos ( π + x ) = − sin ( x ) , and such that3. 0. sin (0) = 0 , cos (0) = 1 , .sin ( π ) = 1 , cos ( π ) = 0 , .cos ( x ) (cid:54) = 0 for 0 < x < π . Theorem.
Let g be a primitive root of p, γ := √ g, i := γ p − , e ( j ) := γ j , sin ( j ) = i ( e ( j ) − e ( − j )) , cos ( j ) = ( e ( j ) + e ( − j )) , then i = − andsatisfy 1.1.0.0. to 1.1.3.2.. Proof. Because g is a primitive root, i = g p − = − . From the definition of sin ( j ) and cos ( j ) follows cos ( j ) + isin ( j ) = e ( j ) , cos ( j ) − isin ( j ) = e ( j ) , therefore cos ( j ) + sin ( j ) = 1 , hence 1.1.0.0.From the exponentiation properties follows e ( j + k ) = γ j + k = ( cos ( j ) + isin ( j ))( cos ( k ) + isin ( k ))= ( cos ( j ) cos ( k ) − sin ( j ) sin ( k )) + i ( cos ( j ) sin ( k ) + sin ( j ) cos ( k )) , hence 1.1.0.1. Because of 1. and 5., e ( π ) = γ p − = i, e ( π ) = − i, hence 1.1.3.1.0. implies that π is the smallest exponent of g which gives − π is the smallest exponent of γ which gives +i or -i,therefore 1.1.3.2. The proof of all other properties is left as an exercise. Theorem.
Assume p ≡ . Let g be a primitive root of p, .2. FINITE SYMPATHIC GEOMETRY. i := g p − , δ := γ = sqr ( g ) , g (cid:48) := − g p − , then sin (1) = i g (cid:48) − δ, cos (1) = g (cid:48) +12 δ. Proof: gg (cid:48) = − g p − = 1 , i = g p − = − .δ − = δ/g = g (cid:48) δ, hence sin (1) = δ − δ − i = − i − g (cid:48) δ,cos (1) = δ + δ − = g (cid:48) δ. Theorem.
Assume p ≡ − . Let g be a primitive root of p, δ := i or δ := − , g (cid:48) := − g p − , then sin (1) = ( g − g (cid:48) , cos (1) = ( g + 1) g (cid:48) δ. Proof: gg (cid:48) = g p − = − /δ , therefore γg (cid:48) = 1 /δ,γ − = g (cid:48) δ and γ = gg (cid:48) δ, hence sin (1) = γ − γ − i = ( g − g (cid:48) ,cos (1) = γ + γ − = ( g + 1) g (cid:48) δ. Example.
For p = 13 , g = δ = 2 , i = − , g (cid:48) = − , then i sin ( i ) cos ( i ) tan ( i )0 0 1 01 − δ δ − − δ δ − − −
45 4 δ − δ −
26 1 0 ∞ For p = 11 , g = 2 , g (cid:48) = − , δ = − , then i sin ( i ) cos ( i ) tan ( i )0 0 1 01 − δ − δ δ − δ δ − δ δ − δ ∞ CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem.
Given a trigonometric table of sin and cos, all other φ ( p − tables can be obtained by using sin ( e ) ( j ) = sin ( je ) , cos ( e ) ( j ) = cos ( je ) , ( e, p −
1) = 1 , with < e < p − . Proof: We know that there are φ ( p −
1) primitive roots. If g e is an other primitive root,then g ( e ) = g e , ( e, p −
1) = 1 ,δ ( e ) = g e − δ, for p ≡ , i ( e ) = g e p − , g (cid:48) ( e ) = − g e p − for p ≡ − , g (cid:48) ( e ) = − g e p − . Substituting in 2.1.3. and 2.1.4. gives the theorem.Replacing δ by − δ gives tables for which sin ( π ) = − . Example.
For p = 13, g = 2, e = 5 , , ,g ( e ) = g e = 6 , − , − ,δ ( e ) = 4 δ, − δ, δ,i ( e ) = g e = − , , ,g (cid:48) ( e ) = − g e = − , , sin ( e ) (1) = 4 δ, − δ, δ,cos ( e ) (1) = − δ, δ, − δ. For e = 5 , sin (5) (1) = 5 . δ ( e ) = 1 δ ( e ) = 4 δ, cos (5) (1) = − δ ( e ) = 6 δ ( e ) = − δ. The tables are: i sin ( i ) cos ( i ) tan ( i )0 0 1 01 4 δ − δ − − − − δ − δ
14 2 − − δ δ
66 1 0 ∞ q = p e . The HyperbolicCase. Introduction.
After recalling the definition of Galois fields, for p , I will state the Theorems which generalize2.1.2., 2.1.3 and 2.1.4. .2. FINITE SYMPATHIC GEOMETRY.
Definition.
Let n be a non quadratic residue, the set of elements in the Galois field p , GF ( p ) , arethe polynomials of degree 0 or 1, for which addition is performed modulo p and multi-plication is performed modulo I − n. More specifically ( uI + v ) + ( u (cid:48) I + v (cid:48) ) =( u + u (cid:48) mod p ) I + ( v + v (cid:48) mod p, ( uI + v ) . ( u (cid:48) I + v (cid:48) ) = ( uv (cid:48) + u (cid:48) v mod p ) I + ( vv (cid:48) + nuu (cid:48) mod p ) . Moreover ( uI + v ) − = − uI + vv − nu . More generally, if P is a primitive polynomial of degree n, i.e. a polynomial which has nofactors with coefficients in Z p , the set of elements in the Galois field p e , GF ( p e ) , are the poly-nomials of degree less than e, for which addition is performed modulo p and multiplicationis performed modulo P. Notation. uI + v will be written u.v or up + v.tI + uI + v will be written t.u.v or tp + up + v, . . . . Example.
Let q = 5 , n = 3 , g = I + 1 = 1 . , then g − = − . . , g = 2 . − . , g = 1 . − . , g = 0 . − . ,g = 0 . − . , hence − g = g − . Theorem. p e . The proof as well as the proof of the other theorems in this section are left as exercises.
Theorem.
Assume q = p e ≡ . Let g be a primitive root of p e , i := g q − , δ := sqr ( g ) , g (cid:48) := − g q − , then sin (1) = i ( g (cid:48) − δ , cos (1) = ( g (cid:48) + 1) δ . Theorem.
Assume q = p e ≡ − . Let g be a primitive root of p e , CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY g (cid:48) = − g q − , δ = − , g − = − g q − , then sin (1) = ( g − g (cid:48) , cos (1) = ( g + 1) g (cid:48) δ . Example.
For q = 5 , n = 3 , δ = g = 6 , i = g = 3 , g (cid:48) = − g = 17 ,sin (1) = 2 . δ = 14 δ, cos (1) = 4 . δ = 24 δ.sin (2) = ( − . . (2 . − − . − . , cos (1) = (2 . − . (1 .
1) = 1 . ,cos (2) = 2 cos (1) − . − . . − . . This gives the Table: k sin ( k ) cos ( k )0 0 11 14 δ δ δ δ δ δ δ δ δ δ
10 14 1911 24 δ δ
12 1 0
Exercise.
Verify the following and construct the corresponding trigonometric table.0. For q = 13 , n = − , δ = g = 15 , i = g = 8 , g (cid:48) = − g = − , g = 35 ,sin (1) = 110 δ, cos (1) = 18 δ,
1. For q = 7 , n = 3 , δ = − , g = 8 ,sin (1) = 3 . .δ = 25 δ, cos (1) = 2 . δ = 18 δ,
2. For q = 11 , δ = 13 ,sin (1) = 0 . δ = 2 δ, cos (1) = 8 . δ = 89 δ,
3. For q = 13 , δ = 15 ,sin (1) = 11 . δ = 143 δ, cos (1) = 3 . δ = 40 δ,
4. For q = 17 , δ = 20 ,sin (1) = 11 . δ = 203 δ, cos (1) = 7 . δ = 124 δ,
5. For q = 5 , δ = 9 ,sin (1) = 3 . . δ = 90 δ, cos (1) = 4 . . δ = 121 δ,sin (2) = 87 , cos (2) = 110 . .2. FINITE SYMPATHIC GEOMETRY. Introduction.
The trigonometry associated with the finite Euclidean plane with real isotropic points willfirst be defined and studied in this section for the finite field Z p . Theorems 4.2.1 and 4.2.1determine sin (1) and cos (1) from which all other values can be obtained using the additionformulas of 4.2.1.In section 4.2.2, definitions and results will be extended, for the finite field associated with p e , with proofs left as an exercise.The trigonometry associated with the finite Euclidean plane with no real isotropic pointswill obtained in section 4.2.3, sin (1) and cos (1) will be determined, for the general case p e in 4.2.3 and 4.2.3. Definition.
Given the sets Z of the integers, Z p of the integers modulo p,Z p − of the integers modulo p − , let δ be a square root of a non quadratic residue of p.π := p − . The problem addressed here is to construct 2 functions sine or sin and cosine or cos withdomain Z and range { Z p ∪ δZ p } which satisfy:The Theorem of Pythagoras,0. sin ( x ) + cos ( x ) = 1 . The addition formulas,1. 0 . sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y ) , cos ( x + y ) = cos ( x ) cos ( y ) − sin ( x ) sin ( y ) . The periodicity property2. sin (2 π + x ) = sin ( x ) , cos (2 π + x ) = cos ( x ) , The symmetry properties3. 0 . sin ( π + x ) = − sin ( x ) , cos ( π + x ) = − cos ( x ) , . sin ( − x ) = − sin ( x ) , cos ( − x ) = cos ( x ) , . sin ( π − x ) = sin ( x ) , cos ( π − x ) = − cos ( x ) , . sin ( π − x ) = cos ( x ) , cos ( π − x ) = sin ( x ) , . sin ( π + x ) = cos ( x ) , cos ( π + x ) = − sin ( x ) , and such that G35.TEX [MPAP], September 9, 2019 CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
4. 0 . sin (0) = 0 , cos (0) = 1 , . sin ( π ) = 1 , cos ( π ) = 0 , . sin ( x ) (cid:54) = ± < x < π . Theorem.
Let g be a primitive root of p, γ := √ g, ι := γ p − , e ( j ) := γ j , then ι = − sin ( j ) = e ( j ) − e ( − j )2 ι , cos ( j ) = e ( j )+ e ( − j )2 , satisfy g is a primitive root, ι = g p − = − . From the definition of sin(j) and cos(j) follows cos ( j ) + ιsin ( j ) = e ( j ) , cos ( j ) − ιsin ( j ) = e ( j ) − , therefore cos ( j ) + sin ( j ) = 1 , hence 4.2.1.0.From the exponentiation properties follows e ( j + k ) = γ ( j + k ) = ( cos ( j ) + ιsin ( j ))( cos ( k ) + ιsin ( k ))= ( cos ( j ) cos ( k ) − sin ( j ) sin ( k )) + ι ( cos ( j ) sin ( k ) − sin ( j ) cos ( k )) , hence 4.2.1.1.From 5 and 4.2.1.1 follows 4.2.1.3.0, implies that π is the smallest exponent of g which gives -1, hence π is the smallest exponentof γ which gives + ι or − ι, therefore 4.2.1.4.2. The proof of all other properties is left as anexercise. The next 2 Theorems give sin (1) and cos (1) first when ι ∈ Z p , then when this isnot the case. Theorem.[Hyperbolic case]
Assume p ≡ . Let g be a primitive root of p ,1. i := ι := g p − , δ := γ = √ g, g (cid:48) := − g p − , then sin (1) = i g (cid:48) − δ, and cos (1) = g (cid:48) +12 δ. Proof: gg (cid:48) = − g p − = 1 , i = g p − = − . δ − = δg − = g (cid:48) δ ,hence sin (1) = δ − δ − i = − i − g (cid:48) δ, and cos (1) = δ + δ − = g (cid:48) δ . .2. FINITE SYMPATHIC GEOMETRY. Theorem.[Hyperbolic case]
Assume p ≡ − . Let g be a primitive root of p, δ := ι or δ := g p − = − , g (cid:48) := − g p − , then sin (1) = ( g − g (cid:48) , cos (1) = ( g +1) g (cid:48) δ. Proof: gg (cid:48) = g p − = − δ − . Because γ := √ g, by taking square roots, γg (cid:48) = δ − ,γ − = g (cid:48) δ and γ = gg (cid:48) δ ,hence sin (1) = γ − γ − ι = ( g − g (cid:48) , cos (1) = γ + γ − = ( g +1) g (cid:48) δ . Example.
For p = 13 , g = δ = 2 , i = − , g (cid:48) = − , then j sin ( j ) cos ( j ) tan ( j )0 0 1 01 − δ δ − − δ δ − − −
45 4 δ − δ −
26 1 0 ∞ For p = 11, g = 2 , g (cid:48) = − , δ = − . then j sin ( j ) cos ( j ) tan ( j )0 0 1 01 − δ − δ δ − δ δ − δ δ − δ ∞ Theorem.
Given a trigonometric table of sin and cos, all other φ ( p − tables can be obtained by using sin ( e ) ( j ) = sin ( je ) , cos ( e ) ( j ) = cos ( je ) , ( e, p −
1) = 1 , with 0 < e < p − . Proof: We know that there are φ ( p −
1) primitive roots.If g ( e ) is an other primitive root, then g ( e ) e = g e , ( e, p −
1) = 1 , δ ( e ) = g e − δ ,for p ≡ , i e = g e p − , g (cid:48) e = − g e p − ,for p ≡ − , g (cid:48) e = − g e p − . Substituting in 2.1.3. and 2.1.4. gives the theorem.Replacing δ by − δ gives tables for which sin ( π ) = − . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Example.
For p = 13 , g = 2 ,e g ( e ) = g e − − δ e δ − δ δi e = g e − g (cid:48) e = − g e − sin ( e ) (1) 4 δ − δ δcos ( e ) (1) − δ δ − δ For e = 5 , sin (5) (1) = 5 . δ e = 1 δ e = 4 δ, cos (5) (1) = − δ e = 6 δ e = − δ .The tables are: j sin ( j ) cos ( j ) tan ( j )0 0 1 01 4 δ − δ − − − − δ − δ
14 2 − − δ δ
66 1 0 ∞ q = p e . The HyperbolicCase.
Introduction.
After recalling the definition of Galois fields, I will generalize Theorems 4.2.1, 4.2.1 and 4.2.1.
Definition.
Let n be a non quadratic residue, the set of elements in the Galois field GF ( p ) , associatedwith p , are the polynomials of degree 0 or 1, for which addition is performed modulo p andmultiplication is performed modulo P := I − n. More specifically( uI + v ) + ( u (cid:48) I + v (cid:48) ) = ( u + u (cid:48) mod p ) I + ( v + v (cid:48) mod p ) , ( uI + v ) · ( u (cid:48) I + v (cid:48) ) = ( uv (cid:48) + u (cid:48) v mod p ) I + ( vv (cid:48) + nuu (cid:48) mod p ) . Moreover ( uI + v ) − = − uI + vv − nu . More generally, if P is a primitive polynomial of degree n, i.e. a polynomial which hasno factors with coefficients in Z p , the set of elements in the Galois field GF ( p e ) , are thepolynomials of degree less than e , for which addition and multiplication is performed modulo P. Notation. uI + v will be written u.v or up + v, tI + uI + v will be written t.u.v or tp + up + v . .2. FINITE SYMPATHIC GEOMETRY. Example.
Let q = 5 , n = 3 , g = I + 1 = 1 . , then g − = − . . , g = 2 . − . , g = 1 . − . , g = 0 . − . ,g = 0 . − . , hence − g = g − . Theorem.
Theorems and p replaced by q := p e . Example.
For q = 5 , n = 3 , δ = g = 6 , i = g = 3 , g (cid:48) = − g = 17 ,sin (1) = 2 . δ = 14 δ, cos (1) = 4 . δ = 24 δ . sin (2) = ( − . · (2 . − − . − . , cos (1) = (2 . − · (1 .
1) = 1 . , cos (2) =2 cos (1) − . − . . − . . This gives the Table: k sin ( k ) cos ( k )0 0 11 14 δ δ δ δ δ δ δ δ δ δ
10 14 1911 24 δ δ
12 1 0
Exercise.
Verify the following and construct the corresponding trigonometric table.0. For q = 13 , n = − , δ = g = 15 , i = g = 8 , g (cid:48) = − g = − , g = 35 ,sin (1) = 110 δ , cos (1) = 18 δ ,1. For q = 7 , n = 3 , δ = − , g = 8 , sin (1) = 3 . .δ = 25 δ,cos (1) = 2 . δ = 18 δ,
2. For q = 11 , δ = 13 , sin (1) = 0 . δ = 2 δ , cos (1) = 8 . δ = 89 δ ,3. For q = 13 , δ = 15 , sin (1) = 11 . δ = 143 δ , cos (1) = 3 . δ = 40 δ ,4. For q = 17 , δ = 20 , sin (1) = 11 . δ = 203 δ , cos (1) = 7 . δ = 124 δ ,30 CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
5. For q = 5 , δ = 9 , sin (1) = 3 . . δ = 90 δ , cos (1) = 4 . . δ = 121 δ, sin (2) = 87 ,cos (2) = 110 . q = p e . The Elliptic Case.
Notation. ( GF ( q ) , + , . ) is a finite field with q = p e elements,( GF ( q ) b , + , . ) for the corresponding extension field GF ( q )( β ) , with β = b, where b is a nonquadratic residue modulo p. Convention.
I will heretofore assume that p is a given odd prime. The sets G b and G b depend on q, wecould indicate that dependence by writing G b,q for G b and G b,q for G b . Definition.
Let G b = GF ( q ) ∪ {∞} . The operation ◦ is defined by0. ∞ ◦ a = a, a ∈ G b , − a ◦ a = a ◦ − a = ∞ , a ∈ GF ( q ) , a ◦ a (cid:48) = a.a (cid:48) + ba + a (cid:48) , a and a (cid:48) ∈ GF ( q ) , a + a (cid:48) (cid:54) = 0 . To avoid confusion with the power notation in GF ( q ) , the k -th power in G b precedesthe exponent with “ o ”. For instance,3. a o = ∞ , a o = a, a ok = a ◦ a o ( k − . Theorem. If (cid:18) bp (cid:19) = − , in other words, if b is non quadratic residue modulo p, then { G b , o } is an Abelian group, ∞ is the neutral element, the inverse of a ∈ GF ( q ) is − a, r ◦ s = t ⇒ r ◦ ( − t ) = − s. Proof: The associativity property follows from( a ◦ a (cid:48) ) ◦ a (cid:48)(cid:48) = a.a (cid:48) .a (cid:48)(cid:48) + b ( a + a (cid:48) + a (cid:48)(cid:48) ) a (cid:48) .a (cid:48)(cid:48) + a (cid:48)(cid:48) .a + a.a (cid:48) + b = a ◦ ( a (cid:48) ◦ a (cid:48)(cid:48) ) , if a (cid:48) (cid:54) = − a and a ◦ a (cid:48) (cid:54) = − a (cid:48)(cid:48) , and from the special cases,( a ◦ − a ) ◦ a (cid:48) = a (cid:48) = a ◦ ( − a ◦ a (cid:48) ) , ( ∞ ◦ a ) ◦ a (cid:48) = a ◦ a (cid:48) = ∞ ◦ ( a ◦ a (cid:48) ) . .2. FINITE SYMPATHIC GEOMETRY. Example.
With p = 13 , b g g o g o g o g o g o g o g o g o g o g o g o g o g o − − − − − − ∞ − − − ∞ − − − − − − ∞ Comment. If p = 2 , (cid:18) bp (cid:19) = − p odd. Definition. If (cid:18) bp (cid:19) = − β = √ b , then G b = { } ∪ { r + βr − β , r ∈ GF ( q ) } . The elements in G b are distinct and G b is a subset of GF ( q ) b . The operation of multipli-cation in GF ( q ) b induces one in the set G b . Theorem. ( G b , . ) and ( G b , o ) are isomorphic, with the correspondance
0. 1 ∈ G b , corresponds to ∞ ∈ G b , r + βr − β ∈ G b corresponds to r ∈ G b . Proof: r + βr − β . s + βs − β = ( rs + b )+( r + s ) β ( rs + b ) − ( r + s ) β = ( r ◦ s + βr ◦ s − β ) , if r + s (cid:54) = 0 . If s = − r, r + βr − β . s + βs − β = ( r + β )( − r + β )( − r + β )( − r − β ) = 1 , Theorem. G b,p is an Abelian group, of order p + 1 .
1. ( r + βr − β ) p +1 ≡ p ) for any r ∈ G b,p . Lemma. If A is a cyclic group of order q + 1 , the number of elements of order d, where d | q + 1 , is φ ( d ) and q + 1 = Σ d | q +1 φ ( d ) . Lemma. [Gauss] If A is an abelian group of order q which is not cyclic then there exists a divisor d of q suchthat the number of solutions of x d = e is larger than d . Herstein, p. 76, 39 CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Lemma.
The polynomial R d := ( r + β ) d − ( r − β ) d has at most d roots in G d . Proof: Dividing by β, we obtain a polynomial in Z p of degree d − , which has thereforeat most d − z ∈ Z p or d roots in G d ( ∞ being a root). Example.
With p = 13 , b = 2 , if S d is the set of roots of R d d = 1 , S = {∞} .d = 2 , S = { } ∪ S ,d = 7 , S = {± , ± , ± } ∪ S ,d = 14 , S = {± , ± , ± } ∪ S ∪ S . Theorem. ( G b,p , . ) is a cyclic group of order p + 1 . ( G b,p , o ) is a cyclic group of order p + 1 . Example. G , , G , . Theorem.[Elliptic case]
Given q = p e ≡ . Let b be a non quadratic residue, r b be a generator of G b , i := − , β := b, r := r o p − b , then sin (1) = r + br − b , cos (1) = − rir − b β. Proof: Let σ = r b + βr b − β , then σ p +1 = 1 and 0 < i < p + 1 ⇒ σ i (cid:54) = 1 .ρ = σ ⇒ ρ p +1) = 1 and 0 < i < p + 1) ⇒ ρ i (cid:54) = 1 . If we take square roots twice, ρ p +12 = ± i, we want ρ p +12 = σ p +14 = i, then ρ = cos (1) + i sin (1) , and ρ p +1) = cos (2( p + 1)) + isin (2( p + 1)) = 1 . If r = r o p − b , then ρ p − = σ p − = r + βr − β , or i = ρ p +12 = ρρ p − = ρ r + βr − β , or cos (1) + i sin (1) = ρ = i r − βr + β , cos (1) − i sin (1) = ρ − = − i r + βr − β , .2. FINITE SYMPATHIC GEOMETRY. cos (1) = − i r β r − β , sin (1) = r + β r − β . Example. q = 13 , i = 5, b = β = 6 , r b = 1, r = 5 , r = − , sin (1) = 3 , cos (1) = − β.k sin ( k ) cos ( k ) tan ( k ) atan ( kβ )0 0 1 0 β
01 3 − β − β −
32 2 β − β
43 5 − β − β − β β − β β − − β β
27 1 0 β ∞ − q b i r sin (1) cos (1)5 2 2 2 − − β
17 4 3 6 − − β
29 12 2 7 − − β
37 6 2 5 6 − β
41 9 2 − − − β Theorem.[Elliptic case]
Given q = p e ≡ − . Let b be a non quadratic residue, r b be a generator of G b , ι := − , δ := ι, β := b, r := r o p +14 b , then cos (1) = r b √ b − r b δ, sin (1) = rcos (1) r b . Proof: The proof proceeds at first as in 4.2.3. r + βr − β = i, therefore r = − βi, this establishesthe relationship between the sign for the square root of -1 and b.cos (2) = ( σ + σ − ) = r b + br b − b and sin (2) = i ( σ − σ − ) = r b r b − b . cos (1) = 1 + cos (2) ⇒ cos (1) = r b √ r b − b , moreover r b b − r b = − cos (1) , sin (1) follows from 2 sin (1) cos (1) = sin (2) = rcos (1) r b , insuring the consis-tency between the signs of sin (1) and cos (1) to insure that sin ( π ) = 1 . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Example. q = 11 , δ = − , b = 2 , r b = 1 , r = − , cos (1) = δ, sin (1) = − δ . k sin ( k ) cos ( k ) tan ( k ) atan ( kβ )1 − δ δ − − − − −
23 4 δ δ − − − −
55 1 δ − δ − ∞ − q b r b r sin (1) cos (1)3 2 1 1 δ δ δ − δ
19 2 1 6 6 δ δ
23 5 1 8 4 δ − δ
31 3 1 − − δ δ
43 3 5 − − δ δ
47 5 4 18 3 δ − δ. Theorem.
If as usual, π := p + 1 in the elliptic case or π := p − in the hyperbolic case, then
0. 3 | π ⇒ sin ( π ) = , cos ( π ) = √ .
1. 4 | π ⇒ cos ( π ) = √ .
2. 5 | π ⇒ cos ( π ) = √ , cos ( π ) = √ − ,sin ( π ) = √ − √ , sin ( π ) = √ √ . In the classical case, there is no ambiguity of sign, because 0 < x < π = ⇒ sin ( x ) , cos ( x ) > . This is not the case in a finite field, the formulas can only give the trigonometric functionsup to the sign, or alternately one of the values of √ √ √ (cid:112) ± √ cos ( π ), cos ( π ), cos ( π ), sin ( π ) and sin ( π ). Example. p = 11, elliptic case, sin (2) = − , cos (2) = − ⇒ √ , with δ = − , cos (3) = 4 δ = ⇒ √ − δ. p = 11, hyperbolic case, cos (2) = 4 , cos (4) = − ⇒ √ , with γ = − , sin (2) = − γ = ⇒ √ − γ, sin (4) = γ = ⇒ √− γ .p = 19, elliptic case, cos (5) = − δ = ⇒ √ δ, with δ = 2 , cos (4) = − ⇒ cos (8) = 7 = ⇒ √ − ,sin (4) = 4 = ⇒ √ − , sin (8) = 3 = ⇒ √− − . .2. FINITE SYMPATHIC GEOMETRY. Definition.Lemma. If r b is a generator of G b and b (cid:48) := br b , then kr b is a generator of G bk , G b (cid:48) . Theorem. There exists always fundamental roots. There exist φ ( p + 1) fundamental roots associated with p. Example.
6, 7 and 8 are the fundamental roots for p = 13 ,
6, 7 and 12 are the fundamental roots for p = 17 ,
3, 11, 18 and 27 are the fundamental roots for p = 29 ,
6, 14, 15, 18, 19, 20, 23, 24 and 32 are the fundamental roots for p = 37 ,
12, 13, 28, 29, 30 and 35 are the fundamental roots for p = 41 . Theorem.
Given an involution, I ( x ) = ax + bcx − a , aa + bc (cid:54) = 0 , an amicable projectivity, in other words aprojectivity with the same fixed points, real or complex, is given by T ( x ) = ( a + f ) x + bcx − a + f , where f = ± (cid:112) ( aa + bc ) /d ) . Proof: If (cid:18) dp (cid:19) = − F d is a fundamental projectivity: F d = y + dy +1 . In view of 4.2.3, we have F d = y ◦ Comment. ( r + βr − β ) e ≡ , for e | p + 1 = ⇒ ( r + β ) e − ( r − β ) e ≡ , dividing by β , we obtain a polynomial in Z p of degree e − , which has therefore at most e − z in Z p or e roots in G ( ∞ being a root). We want to show that ( G , . ) andtherefore ( G, ∞ ) are cyclic groups. Theorem.
Let r = 1 , r = r, r i +1 = r i ◦ r, CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY x i ≡ /r i ( ∈ G ) , u i +1 ≡ ru i + su i − , u = 0 , u = 1 ,v i +1 ≡ rv i + sv i − , v = 2 , v = r,
3. 4 s ≡ d − r , α = r + √ d and β = r −√ d , then x i +1 ≡ rx i +1 dx i + r (mod p ) , x = 0 , r = α + β, √ d = α − β, s = − αβ. u i = ( α i − β i ) / √ d, v i = α i + β i ,
8. 2 u i + j = u i v j + v i u j , v i + j = v i v j + bu i u j , u i +1 ( bu i + rv i ) − v i +1 ( ru i + v i ) = 0 , x i v i = u i . Proof: ( ru i + su i − ) = ( α + β )( α i − β i ) − αβ ( α i − − β i − ) √ d = α i +1 − β i +1 √ d = u i +1 . Substituting in 2. with j = 1 after multiplication by 2 gives(( u i v + v i u )( bu i + rv i ) − ( v i v + bu i u )( ru i + v i )= u i ( bv − rbu ) + u i v i ( bu + rv − rv − bu ) + v i ( ru − v ) = 0 . Morover, ru i + v i du i + rv i = ( α + β )( α i − β i )+( α − β )( α i + β i )(( α − β )( α i − β i )+( α + β )( α i + β i )) √ d = u i +1 v i +1 = x i +1 . Example.
For p = 7 , elliptic case, ι = − ,A k = ( cos ( k ) , sin ( k ) , x + y = z . If we define it as a circle and z = 0 as the ideal line, the isotropic points are not real and wehave a Euclidean geometry.The center of the circle, which is the pole of z = 0 is (0,0,1). There are 8 real points on thecircle, A = (1 , , , A = ( − , − , , A = (0 , , , A = (2 , − , ,A = ( − , , , A = (2 , , , A = (0 , − , , A = ( − , , . The distances on the lines a k = O × A k , k = 0 ,
1, 2, 3, are real. a = [0 , , , a = [2 , − , , a = [1 , , , a = [1 , , . The other lines through O intersect the circle at complex points: A = (2 ι, ι, , A = (4 ι, ι, , A = ( − ι, ι, ,A = ( − ι, ι, , A = ( − ι, − ι, , A = ( − ι, − ι, ,A = (4 ι, − ι, , A = (2 ι, − ι, . .2. FINITE SYMPATHIC GEOMETRY. a k +1 = O × A k +1 , k = 0 ,
1, 2 , 3, a = [ − , , , a = [1 , − , , a = [1 , , , a = [2 , , . If B = (1 , ,
1) is a real point on a , B is on a circle x + y = 5 z . this circle intersects a k +1 at real points and a k at complex points. The distances betweenpoints on a are multiples of ι, because √ ι. The same is true on the lines a , a , a . Definition.
The smallest j such that u j ≡ p ) is called the rank of apparition of p. Hence
Theorem.
For a fixed r there are φ ( p + 1) values of s in [1 , p − for which the rank of apparition is p + 1 . More generally, there are φ ( e ) values of s in [1 , p − for which the rank of apparitionis e, e divides p + 1 , e > . Theorem. If b is a fundamental root modulo p , then b = N p, − b = N p. Comment.
For p = 17, 11 N p, 7 N p, but 7 is not a fundamental root. Theorem.
For a given p, the sets S h = { cos ( j ) , j = 1 , . . . , p − } , in the hyperbolic case and S e = { cos ( j ) , j = 1 , . . . , p − } , in the elliptic case areindependent of the choice of the primitive root or of the fundamental root. Example. p = 11 , S h = { , , , } , S e = { , , , , } p = 29 , S h = { , , , , , , , , , , , , } ,S e = { , , , , , , , , , , , , , } . Theorem. Let sin ( j ) ∈ { Z p − { } − { }} ,if p ≡ − , then sin ( j ) sin ( k ) (cid:54) = 1 , for all k, if p ≡ , then sin ( j ) sin ( k ) = 1 for some k. CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY sin ( j ) γ ∈ { Z p − { }} if p ≡ , then sin ( j ) sin ( k ) (cid:54) = 1 , for all k, if p ≡ − , then sin ( j ) sin ( k ) = 1 , for some k. .2. FINITE SYMPATHIC GEOMETRY. Example. . . .
Give examples of associated fundamental sympathic projectivities, see 1.10.If T ( r ) = r + dr +1 . For p = 7 , d = 3 , (5 is the other choice) r ∞ T ( r ) 1 3 2 4 5 0 6 ∞ . . . talk about transformations such as r = 2 s leading to the form used in 1.10 S ( s ) = s + b − s . Example.
For p = 13 , (see g35.bas .5.) A = (0 , , , A j = (1 , j, , j = 0 , . . . , q − , A × A j = [0 , , , d j = dist ( A, A j ) : cos ( d j ) = j √ j , sin ( d j ) = √ j ,j (cid:112) j sin ( d j ) cos ( d j ) d j . . . .tan ( dist ( A j , A k )) = k − j jk tan ( dist ( A, A k ) = − k tan ( dist ( A j , A l )) = tan ( dist ( A j , A k ) + dist ( A k , A l )) . Indeed, the second member is ( k − j )(1+ lk )+( l − k )(1+ jk )(1+ jk )(1+ lk ) − ( k − j )( l − k ) = ( l − j )(1+ k )(1+ lj )(1+ k ) = tan ( dist ( A j , A l )).and tan ( dist ( A j , A l )) = tan ( dist ( A j , A ) + dist ( A, A l ))Indeed, the second member is j − l kl = l − j kl = tan ( dist ( A j , A l )). j \ k
312 1112 1012 412 512
912 812 712 112 212
112 412 1112 512 612
212 512 112 612 712
812 1112 712 612 112
712 1012 612 512 1112 A
612 912 512 412 1012 1112
Example. p = 13 , h, the correspondence between the point A j = (1 , j,
0) and d ( j ) is j − − ∞ − − − d ( j ) 0 1 2 3 4 5 6 7 8 9 10 11 i = ± d (5) = d ( −
5) = ∞ .elliptic case: jδ − − − ∞ − − − d ( j ) 0 1 2 3 4 5 6 7 8 9 10 11 12 1340 CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY tan ( dist ( A j , A k )) = d ( k ) − d ( j )1+ d ( j ) d ( k ) . Definition.
Let f be a function, g (0) is arbitrary, g ( i + 1) = ai + g ( i ) + f ( i ) + f ( i + 1) , where a is such that g ( T ) = g (0) , and we write g = T f.
Theorem. If f is a periodic function with period T, then0. g is periodic.1. f odd ⇒ geven. Example. i f ( i ) 1 − − − − − g ( i ) 0 − − − − a = 0 . Example. i f ( i ) 0 4 7 7 4 0 g ( i + ) 0 9 − h ( i ) − ai − − h ( i ) 0 − − a = − − . Definition.
Let f be a function. Let g be defined by g ( i + ) = u ( f i , f i +1 ) , where u is symmetric in its arguments,we write g = U f.
Let h be defined by h (0) is arbitrary, h ( i + 1) = ai + h ( i ) + g ( i + ) ,h ( T ) = h (0) . .2. FINITE SYMPATHIC GEOMETRY. h = M U f.
Theorem.
If be a periodic function with period T, then0. h is a periodic function with period T. Theorem. If p ≡ − , choose the elliptic case and q = p +12 , If p ≡ , choose the hyperbolic case and q = p − ,
0. The trigonometric functions sin and cos are orthogonal.1. ( . . . .. ..) ( . . . sin ( ij ) . . . ) , i, j = 1 to q − ( . . . .. ..)is orthogonal, symmetric and SS = q I. ( . . . ..s . . . .. ) C = ( scos ( ij )( − i s ) , i, j = 0 to q + 1 , ( . . . ( − j s.. ) with s = , is orthogonal, symmetric and CC = q I. Example. p = 7 , Elliptic case, q = 4 , ( − − − − − − − , ( − − S = (10 − , C = ( − − − − − − , ( − − − − − p = 13 , hyperboliccase, q = 6 , ( − sssss − − −
6) ( s − − − s )(220 − −
2) ( s − − − s ) S = (10 − , C = ( s − − s )(2 − −
2) ( s s )( − − − −
6) ( s − − − s )( − − ss − ss − s − s = 2 δ, δ = 5 . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem. X i = acos (2 i ) , X i = bsin (2 i ) , X i = 1 .X i = aδ cos (2 i + 1) , X i = bδ sin (2 i + 1) , X i = 1 . X i = acos (2 i ) , X i = bδ sin (2 i ) , X i = 1 .X i = aδ cos (2 i + 1) , X i = bsin (2 i + 1) , X i = 1 . Example.
Let p = 11 , a = 1 , b = 2 , δ = i, i = − X + X = 1 :(1 , , ( − , , ( − , , (0 , , (5 , , (3 , , ( − , , (3 , − , (5 , − , (0 , − , ( − , − , ( − , − . . . . gives − X − X = 1 :(1 , , (4 , − , ( − , , (3 , , ( − , − , ( − , , ( − , − , ( − , , (3 , − , ( − , − , (4 , , (1 , − . In the hyperbolic case, . . . gives X − X = 1 :(1 , , (4 , , ( − , − , (2 , − , ( − , , ( − , , ( − , − , (2 , , ( − , , (4 , − . The asymptotic directions are (5,1,0) and ( − , , − X + X = 1 :(5 , − , (2 , − , (0 , , ( − , − , ( − , − , ( − , , ( − , , (0 , − , (2 , , (5 , . The asymptotic directions are (5,1,0) and ( − , , . Example.
Let p = 13 , a = 1 , b = 2 , δ = 2 , i = 5 . In the elliptic case, . . . gives X + X = 1 :(1 , , ( − , − , (5 , − , (3 , , ( − , , ( − , − , (4 , − , ( − , − , (4 , , ( − , , ( − , − , (3 , − , (5 , , ( − , . . . . gives2 X + X = 1 :(3 , , ( − , − , (5 , , (0 , , ( − , , (1 , − , ( − , , ( − , − , (1 , , ( − , − , (0 , − , (5 , − , ( − , , (3 , − . In the hyperbolic case, . . . gives X + X = 1 :(1 , , ( − , , ( − , − , (0 , , (6 , − , (2 , , ( − , , (2 , − , (6 , , (0 , − , ( − , , ( − , − , the asymptotic directions are (4,1,0) and ( − , , . .2. FINITE SYMPATHIC GEOMETRY. X + X = 1 :(4 , − , (6 , − , ( − , − , (2 , − , ( − , − , ( − , − , ( − , , ( − , , (2 , , ( − , , (6 , , (4 , , the asymptotic directions are (4,1,0) and ( − , , . Definition. A regular polygon . . .because the angles are multiples of rp − or rp +1 .The only regular polygons are those whose number of sides is a divisor of p − p +1 . If we give the unit angle then we can define “convex polygons” and “star polygons”,find appropriate names.The constructibility by rule and compass in Euclidean geometry corresponds here to thosewhich demand the solution of equations of the first and second degree.The work of Gauss on cyclotomic polynomials extend immediately to the finite case becauseof Theorem . . . on trigonometric functions.
Theorem.
0. For a regular polygon of n sides to exist, n must divide . . .1. For a regular polygon to be constructible using only equations of the second degree, n must have the form i p i p i . . . p ikk , where i is a non negative integer, i , i , . . . ik, are 0or 1 and p j are primes of the form k + 1 , namely, 3, 5, 17, 257, 65537, . . . .All square roots are integers except perhaps the last one. Theorem.
For triangles. cos ( r ) = , sin ( r ) = (cid:113) . Example. p = 5, elliptic case, cos ( r ) = 3 , sin ( r ) = √ p = 7 , hyperbolic case, cos ( r ) = 4 , sin ( r ) = √ ,p = 23 , elliptic case, cos ( r ) = 12 , sin ( r ) = 8 . Theorem.
For hexagons, we first obtain the triangle and then use cos ( r ) = sin ( r ) , sin ( r ) = cos ( r ) . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Example. p = 23 , elliptic case, cos ( r ) = 8 , sin ( r ) = 12 . Theorem.
For pentagons. The polynomial to solve is x − x + 1 = 0 ,cos ( r ) = x , sin ( r ) = (cid:113) − x . Example. p = 11 , hyperbolic case, γ = 8 .x (cid:113) = = 4 , x − = − ,cos r ) = 2 , cos r ) = 4 ,sin r ) = √− γ, sin r ) = √− − γ .The choice of 1 or 2 is arbitrary as is the choice of the sign of the coefficient of γ. The secondcase corresponds to sin ( r ) of the trigonometric table, trig.tab.The first, corresponds to sin ( r ) of the same table. p = 19 , elliptic case, δ = 10 . x = = 5 ,cos r ) = = − cos ( r ) of trig.tab. sin r ) = √ sin ( r ) . Theorem.
For decagons, we first obtain the pentagon and then use cos ( r ) = (cid:113) (1+ cos r , sin ( r ) = sin r cos ( r . Example. p = 19 , elliptic case, cos r ) = √− cos ( r ) , sin r ) = = − . Theorem.
For 17 sided polygons. The polynomials to solve are in succession: u + u + 4 = 0 , of which we choose 1 root, v − uv − ,v (cid:48) = − v − v ,w − vw + v (cid:48) = 0 ,cos ( rp + j ) = w , sin ( rp + j ) = (cid:112) − ( w ) . .4. CONTRAST WITH CLASSICAL EUCLIDEAN GEOMETRY. Example. p = 67 , elliptic case. u = − √ = 16 ,v = u + √ u +42 = ( √ = = 28 ,v (cid:48) = 61 = − ,w = v + √ v − v (cid:48) = (28+ √ = 15 ,cos r ) = = −
26 = cos ( r )of the table obtained using the program trig.bas. sin r ) = √ − − . The other choices for the roots of the above equations lead, with the right choice of sign,to cos ( kr ) , k = 1,2,3,4,5,6,7. From these all the other angles can ge obtained using thetrigonometry identities . . . . p = 137 , hyperbolic case, u = − = 23 , v = = 52 ,v (cid:48) = − − = 63 , w = = 58 ,cos r ) = 29 = cos ( r ) of the table obtained using the program trig.bas. Introduction.
In this section we examine the problems which correspond or require the intersection of aconic or of a circle with a line when one of the intersections is not known.
To contrast the notions within finite Euclidean geometry with those of Euclidean geometry,with have the following summary:In finite Euclidean geometry (of the elliptic type),The following properties are different in finite Euclidean geometry:0. There are p points on each line.1. There are p + 1 lines through every point.2. There are p + 1 points and p + 1 tangents for each circle.3. There are even and odd angles, the even ones can be bisected, the odd ones cannot.4. There are even triangles, for which there are 4 inscribed circles, the others have noinscribed circles.5. Angles can be expressed as integers, addition of angles is done modulo p + 1 . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
6. line through the center of a circle does not necessarily have an intersection with thecircle. The angle between any two lines, through the center, which have an intersectionis even.7. Regular polygons exist only if the number of vertices is a divisor of p + 1 .
8. Distances can be expressed either as integers or as integers times an irrational, theaddition of distances on the same line is done by adding the integers modulo p. Thesquare of the irrational is an integer which is not a square. For instance, for p = 7 , theirrational can be chosen to be √ .
9. Trigonometric functions sin and cos can expressed like the distances. The cosine ofan even angle is always an integer. The cosine of an odd angle is an integer times anirrational. If p ≡ , the sine of an even angle is an integer that of an oddangles is an integer times an irrational, the reverse is true if p ≡ − .
10. Ordering cannot be introduced. This is replaced by partial ordering.Among the properties which are similar, we have the following: Incidence, parallelism,equiangularity , equidistance, perpendicularity, congruence (of figures), similarity, the barycen-ter, the orthocenter, the circumcircle, the theorem of Pythagoras.The constructibility of regular polygons (if they exist in the finite case), for instance ifwe replace the field Z p by the field ( Z p , √ , we always have regular octogons, if we replaceby the field ( Z p , √ p i , p i being all the primes, the constructible polygons always exist. If wereplace the field Z p by the field of algebraic numbers, it is those which are roots of somepolynomial, then all regular polygons exist.A similar discussion can be made if we start from the field Q of the rationals. With Q wecan only construct squares, extended using . . . .If A is the field of algebraic numbers, . . . .The length of the circle as a limit of the length of polygons only make sense if we start with Q. The implication of the transcendence of . . . in A is not a number in A. . . . The field of algebraic numbers, the rational case and the existence or non existence ofregular polygons. The Euclidean geometry can be obtained from the projective geometry by choosing an ap-propriate set of elements, namely the ideal line and 2 complex conjugate points on the line,the isotropic points. Similarly, for the hyperbolic geometry, one choose one real conic as theideal. In the Cartesian geometry, we choose a line, the ideal line and a point on that line, theisotropic point. Definitions and properties in this geometry will be stated. A construction, G39.TEX [MPAP], September 9, 2019 .3. PARABOLIC-EUCLIDEAN OR CARTESIAN GEOMETRY. i, in thatplane is chosen, called the ideal line. A specific point, I, on that line is also chosen, calledthe isotropic point. Because a line is chosen, we can use all the concepts of affine geometry.In particular, 2 lines are called parallel if they have the same ideal point. The mid-point of2 points A, B is the harmonic conjugate with respect to
A, B of the ideal point on A × B. A parabola is a conic tangent to the ideal line.I now will define new concepts in the Cartesian geometry. To focus the attention ona specific set of properties, I have chosen properties which have been inspired by thoseassociated to the geometry of the triangle. Because we want properties which are true inany field, it is not appropriate to derive them by a limiting procedure. I have therefore statedand proven them independently from the corresponding properties in Euclidean geometryand have indicated the correspondence by giving the same name as that of the correspondingelement in Euclidean geometry, but without giving the justification.The configuration should give theorems in 2 ways , using . . . .The equality of angles is associated with a parabolic projectivity (with 2 coincident fixedpoints).Recall the construction of a parabolic projectivity on i, let I , I I , corresponding to I , is obtained as follows,given A, choose B on A × I,C := ( A × I × ( B × I , c := I × C,D := ( B × I × c, a := A × D, then I a × i. Measure of distances and of angles.
The measure of angles and distances play a fundamental role in the geometry of Euclidand in the study of non Euclidean geometry by de Tilly. On the other hand, starting fromprojective geometry, these notions are derived notions. The appropriate definitions for themeasure of distances and of angles will be given first in the case of a real field using a modelon the Euclidean plane with given perpendicular axis x and y through a point O and theline l with equation y = 1 . This will justify the name of Cartesian geometry. Definition.
In in affine geometry let us choose one point on the ideal line as a double isotropic point.This point will be called the isotropic point or sun . The associated geometry will be called parabolic-euclidean or Cartesian . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Definition.
Any line through the sun is called an isotropic line or solar axis . A parabola with the sunas ideal point is called a parcircle . Comment.
There is a configuration which is a special case of the hexal configuration which allows thestudy of the geometry of the triangle in the Cartesian geometry. Indeed it is sufficient tochoose M to be the isotropic point. Example. x i +1 := x i + 1 mod p is such a projectivity.With p = 5 ,X = { , , , , , , . . . } . Hence both angles and distances can be represented by an integer modulo p. The circle is replaced by a parcircle,
Notation.
Let π be the parabolic projectivity with the ideal point as fixed point:0. pi = { ( x, x + 1) } . Theorem. pi i = { ( x, x + i ) } , i ∈ R, therefore, if the coordinates of the point P are x P and y P and if the parallels to the lines a and b through O meet l at A and B, we are justified to give the following Definition. dist ( P, Q ) = y Q − y P .angle ( a, b ) := x B − x A . We have
Theorem.
Given any 3 points
A, B and C and any 3 lines a, b and c,dist ( A, C ) + dist ( C, B ) + dist ( B, A ) = 0 ,angle ( a, c ) + angle ( c, b ) + angle ( b, a ) = 0 . Comment.
Because, in both instances, the notion of measure are associated with the coordinates of a1 dimensional set of points, both measure of distances and of angles can be given a sign. Ofcourse what we obtain is not a metric but a semi metric because .3. PARABOLIC-EUCLIDEAN OR CARTESIAN GEOMETRY. | dist ( A, B ) | ≤ | dist ( A, C ) | + | dist ( C, B ) | , but dist ( P, Q ) = 0 does not imply P = Q but only y P = y Q . In the case of a Gaussian field, if [0,0,1] is the ideal line and (0,1,0) is the isotropic point,we can give
Definition. dist (( A , A , , ( B , B , B − A mod p k ,angle ([ a , − , a , [ b , − , b b − a mod p k , Theorem.
The set of points Q such that angle AQB is constant is a parcircle . Theorem.
The set of points Q such that angle QA = angle AQO isosceles triangle) is a set of 2 lines(which can coincide), A × S and B × S, such that O is equidistant in the Euclidean sensefrom these 2 lines. Proof: A (cid:48) := ( B × O ) × ( A × S ) , B (cid:48) := ( A × O ) × ( B × S ) ,I i × ( B × O ) , I i × ( A × B ) = i × ( A (cid:48) × B (cid:48) ) ,I i × ( A × O ) , then ( I , I
2) = ( I , I
3) and
ABO is an isosceles triangle. (
Ii, Ij ) denotes the angle of anypair of lines through Ii and Ij on i. Any other point D on B × S is such that ADO is an isosceles triangle.
Definition. AO = BO either if A, B and I are collinear, or if A (cid:48) B (cid:48) is parallel to AB where A (cid:48) = ( B × O ) × ( A × S ) andB (cid:48) = ( A × O ) × ( B × S ) . Definition.
Two lines are antiparallel if . . . .
Definition.
A line in a triangle is a symmedian if . . . . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Comment.
We could also define measure of distance and angles dualy .( A, B ) = (
C, D ) if (( I × A ) , ( I × B )) and (( I × C ) , ( I × D )) are corresponding pairs in anparabolic projectivity with fixed line i. If we use as model in the Euclidean plane the line at infinity as the ideal line and the pointin the direction of the x axis as the sun all points on a line through the sun are equidistantfrom points on another line through the sun the measure of distances between the pointscan be chosen as the measure of the distances between the lines. Therefore, the distancebetween A := ( a , a ,
1) and B := ( b , b ,
1) is b − a . The angle between a := [1 , − a ,
0] and b := [1 , − a ,
0] is a − a . a corresponds to X = a Y, if α is the angle with the y axis in the clockwise direction, tan ( α ) = a , the “sun” angle isdoubled if the tangent is doubled. Definition.
The line of Euler is . . . .
Definition.
The circumparcircle
Definition.
The first circle of Lemoine.
Definition.
The second circle of Lemoine.
Definition.
The circle of Brocard.
The Brianchon-Poncelet-Feurbach theorem becomes Theorem.
Given a triangle { A i , a i } and the parcircle ι tangent to a i . Let M be any point not on theside of the triangle or on i, Let M i := ( M × A i ) × a i , the parcircle γ through M i is tangentto the parcircle ι .By duality, let m be a line not through A i or I, let m i := ( m × a i ) × A i , the parcircle tangentto m i is tangent to the parcircle ι . .3. PARABOLIC-EUCLIDEAN OR CARTESIAN GEOMETRY. Theorem.
The symmetric functions can be expressed in terms of s and s , . More precisely H0. s = 0 , b := s , c := s , then C2. s = − b, C3 s = − c, s c, C4 s = b , s = − b , s = 2 b , s = 0 , C5 s = bc, s = − bc, s = − bc, s = 5 bc, s = − bc, C6 s = c , s = b + 3 c , s = − c , s = 3 c , s = − b − c , s = 2 b − c ,s = − b + 3 c . C7 s = 0 , s = b c, s = − b c, s = − b c, s = 2 b c, s = 3 b c, s = − b c,s = 7 b c. C8 s = bc , s = − bc , s = − bc , s = b + 4 bc ,s = 5 bc , s = − b − bc , s = − bc , s = 2 b + 2 bc , s = − b + 8 bc ,s = 2 b − bc , C9 s = c , s = − c , s = b c + 3 c , s = 3 c , s = − b c − c , s = − b c − c ,s = 3 b c +6 c , s = 2 b c − c , s = − b c − c , s = − b c +3 c , s = 9 b c − c ,s = − b c + 3 c , C10 s = 0 , s = b c , s = − b c , s = − b c ,s = b + 5 b c , s = 2 b c , s = 3 b c , s = − b − b c , s = − b c ,s = 2 b + 6 b c , s = 7 b c , s = − b + b c , s = 2 b − b c , s = − b + 6 b c , C11 s = bc , s = − bc , s = − bc , s = bc +4 bc , s = 5 bc , s = − bc − bc ,s = − bc − bc , s = − bc , s = 2 bc + 2 bc , s = 3 bc + 11 bc , s = − bc +8 bc , s = − bc − bc , s = 2 bc − bc , s = 7 bc + 5 bc , s , = − bc + 11 bc ,s = 11 bc − bc , C12 s = c , s = − c , s = b c + 3 c , s = 3 c , s = − b c − c , s = − b c − c , s = b + 6 b c + 3 c , s = 2 b c − c , s = 3 b c + 6 c , s = − b − b c − c , s = − b c + 3 c , s = − b c − c , s = 2 b + 8 b c − c ,s = 9 b c − c , s = − b − b c +6 c , s , , = − b c +3 c , s , = 2 b − b c − c ,s , = − b + 24 b c − c , s = 2 b − b c + 3 c . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem.
Given a triangle A i a point M not on the sides of the triangle and a point M on the polar m of M with respect to the triangle. γ is a parcircle, θ is a parcircle, χ i and χ i are parcircles. Proof: We will use the abbreviation s
11 = m m m m m m m − m m m − m m m − m m m + m + m m − s .s + m m − ( m + m ) , . . . . s + m = m m , . . . .2 s − m m m m − m , ( m − m = − ( s + 3 m m . COMPARE
M mm and jia, M mm and jia . Theorem.
The conic m X X + m X X + m X X = 0 passes through M, A i , and ZZ i = ( m , − ( m − m , m − m , the tangent at A i is ac i , the tangent at M is mai = [ m , m , m , the tangent at ZZ i is [( m − m , m m , m m . Such transformation must preserve measure of angles and distances.
Theorem.
The transformations associated to the Cartesian geometry are represented by unit uppertriangular matrices in the Euclidean The following are subgroups of these transformationThe translations v w . The shears .5. TRANSFORMATION ASSOCIATED TO THE CARTESIAN GEOMETRY. u v , and the special shears u
00 1 00 0 1 .Indeed, Definition.Theorem.
Given a point P, the set of points whose polars with respect to a triangle pass through P are on a conic through the vertices of the triangle and vice-versa.Proof. The pole of [ q , q , q ] is ( q q , q q , q q ) . It is on line ( a , a , a ) if a q q + a q q + a q q = 0 . Definition.
Given a triangle, the point of Lemoine of a conic through the vertices of the triangle is thepoint P of the preceding Theorem. Definition.
Given a triangle, the line of Lemoine of a conic tangent to the triangle is the set of pointswhose polars with respect to the triangle are tangent to the conic.
Corollary.
The point of Lemoine of the circumcircle is the classical point of Lemoine. The point ofLemoine of the conic of Simsom m m X X + . . . = 0 is T mm = ( m m , m m , m m ) , The line of Lemoine of the inscribed conic is [ j j , j j , j j ] , it is the line through J a i . Introduction.
In this section, we do give only a representative set of Theorems using a form similar to thatfound in works on Geometry since Euclid. Many more Theorems can be deduced from thecompact form given in section 9.5. The vertices of the triangle are denoted by A , A , A , its sides by a , a , a . Definition.
A Fano line p of a point P is the line through the diagonal points of the quadrangle A , A ,A , P.54
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Definition.
The cocenter M of a triangle is the Fano point of the ideal line m. (D0.1., .2., .12.) Construction.
Given a triangle A i , the ideal line m and the center M , we can obtain a conic as follows. Thetangents to the conic θ at the vertices of the triangle are be constructed using m i = M x A i . Any point on the conic and on a given line through one of its points, are be obtainedusing the construction of Pascal.
Definition.
The conic θ constructed in 9.6.3. is by definition a circle the circumcircle of the triangle. (D1.12., H1.0.) Definition.
The Euler line of a triangle is the line eul through the cocenter and the center of the triangle.(D1.0.)
Definition.
The central parallels kk i are the lines parallel to the sides of a triangle passing through thecenter M . (D1.1.)
Theorem.
The central parallels kk i intersect the sides a i +1 and a i − at points K A i − and K A i +1 which are on a circle λ . (D1.2., D2.11., C2.1.,C2.2.) Definition.
The circle λ is called central circle. Theorem.
The circumcircle and the central circle are tangent at a point
LO. (C23.0.)
Definition.
The central points M i of a triangle are the intersection of a tangent at a vertex with theopposite sides. (D0.11.) .5. TRANSFORMATION ASSOCIATED TO THE CARTESIAN GEOMETRY. Definition.
The central line m of a triangle is the Fano line of its center M . (D0.12.)
Definition.
The associated circles α i are the circles through the center M of the triangle and its vertices A i +1 and A i − . (D3.6., C3.1.) Definition.
The center-vertex circles κ c i and κ c i are the circles centered at one vertex of a trianglepassing through an other vertex. (D4.12., C4.0.) Definition.
The bissectrices i i of a triangle are the lines through a vertex A i such that the lines formsequal angles with the sides of the triangle passing through A i . Theorem.
The bissectrices have a point I in common. (9.5.5., D0.3.)
Definition.
The bissector is the point I common to the bissectrices i i . (D.0.3.) The bissector line i is theFano line of the bissector. (D20.1) Comment.
The sides of a triangle do not have a point in common, therefore, there is no circle tangentto its sides.
Definition.
The circles of Apollonius α p i are the circles centered at a central point M i through a oppositevertex A i . (D5.12., C5.0., C5.1.) They have a common tangent with the circumcircle (C5.3.).The point of contact with the side a i is on the bissectrix through A i . (C22.1.) Theorem.
The circles of Apollonius have the same radical axis, l mm, which is the common tangent ofthe circumcircle and the central circle. (C6.2., C2.6.)
Definition.
The sun
M I is the direction of the bissector line. (D24.0.)56
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Definition.
Any parabola, i.e. a conic tangent to the ideal line, whose ideal point is the sun
M I is calleda sun-parabola . Comment.
In the isotropic geometry, the center of a parabola is an ideal point which is not necessarilyits ideal point.
Definition.
The center-cocenter conic γ is the conic through the vertices of the triangle, its center andits cocenter. (D7.10.) Theorem.
The center-cocenter conic is a sun-parabola. (C24.1.)
Definition.
The tangential circles χ t i ( χ t i ) are the circles tangent to a i +1 ( a i − ) at A i − ( A i +1 ) passingthrough A i +1 ( A i − ) . (D7.8., C7.0.) Theorem.
The other intersections
K L i and K L i with the tangential circles and the sides of thetriangles are on a conic ξ which is a sun-parabola. (D3.1., D7.9., C7.2., D24.2.) Theorem.
The cocircumcircle θ is the conic through the vertices of the triangle for which the tangentsare parallel to the opposite side. (D1.12., D0.1., C1.0.) Definition.
Let
Eul and
E ul be the points of contact of the circumcircle and of the co-circumcircle withthe line of Euler, the conic ι through these points and circumscribed to the triangle is calledthe bissector conic . (D20.19., C20.2., C20.3.) Theorem.
The center of the bissector conic is the bissector point. (C20.7.)
Comment. .5. TRANSFORMATION ASSOCIATED TO THE CARTESIAN GEOMETRY. γ a of Gabrielle. Introduction.
This section and the related section 11. was conceived after my daugther asked when I wouldname a Theorem for her. It concerns a general construction which starts from a parabola andconstructs points on a cubic of which several are assoiated to the geometry of the triangle.
Definition.
Let x = (x0,x1,x2) be any line of the dual of the sun-parabola Γ , 0. ( m m x x m m x x m m x x . Let k k i = [m1+m2,m2,m1]. The following constructs points X = (X0,X1,X2) of the curve γ a called the cubic of Gabrielle: D1. X i := xxk k i , D2. x i := X i xA i , D3. X := x xx . D4. γa := { X } . Definition.
A parametric representation of a curve, with constraint arbitrary point are given in termsof 3 homogeneous parameters subjected to an homogeneous relation R between these 3parameters.
Theorem.
The curve γ a is a point cubic, with axis 0. df = [ m ( m m , m ( m m , m ( m m ] . It contains the points A i , M i , M, M , LM. A parametric representation, with constraint9.7.1.0., is 1. ( x x m m m x x
2) + m m x ,x x m m m x x
0) + m m x ,x x m m m x x
1) + m m x , Its equation in homogeneous coordinates is P4. γa : m X X + X ) + m X X + X ) + m X X + X ) = 0 . Proof: Definition 9.7.1. gives P1. X = ( m x m x , m x m m x , m x m m x , P2. x = [0 , m x m m x , m x m m x , P3. X = (( m x m m x m x m m x , ( m x m m x m x m m x , ( m x m m x m x m m x , if we multiply all coordinates by x2 and use 9.7.1.0. we get 9.7.3.1. By a long algebraicverification it can be shown that the equation P . is satisfied by 1. CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem.
A parametric representation, with constraint 0. x x x x x m x m x , x x m x m x , x x m x m x . T hepoint .isthepointonthecubicγaandtheline [ x , x , x throughM distinctf romM. T hereisonelinemiwhereM isatriplepoint.
Proof: Let (X0,X1,X2) be any point on the line [x0,x1,x2] passing through M, eliminating X γ a and X x X x X x gives ( X X ( X m x x x
2) + m x ) + X m x x x
2) + m x )) = 0or X x x m x m x , X x x m x m x , because of 0., by symmetry we get 1. X1+X2 = 0, gives the point M and because (X1+X2)is a double factor this point has to be counted twice, hence 2., the point M is a node . Thepoint 1. coincides with M iff the 3 coordinates are equal, the first 2 give after eliminationof x , ( m m x = ( m m x , hence x = i2+i0, x2 = i0+i1, by symmetry x0 =i1+i2, hence 3. Theorem.
If the point of contact of a line [x0,x1,x2] through LM with the cubic γ a is (X0,X1,X2),then 0. X = m x x m x m x , X = m x x m x m x ,X = m x x m x m x . We have 1. m m x m m x m m x and 2. x X x X x X . Eliminating X between 2. and the equation of the cubic gives ( m X m X m x m x m x X + m x m x m x X ) = 0 , becauseof . The first factor corresponds to the point
LM, the other factor has a double root which gives0.
Definition.
The cubic χ of Charles is the cubic through the points M i , M i , LM i . Theorem.
Let 0. a := m m m , a m m + m + m m , a m m + m + m m ,a m m + m + m m , then 2. χ : a ( X + X + X )+ a X X X X
2) + a X X X X
0) + a X X X X
1) = 0 . T hetangentat ( X , X , X is [ aX + a X + a X , aX + a X + a X ,aX + a X + a X ] , T heotherpoint ( Y , Y , Y onthetangent [ x , x , x at ( X , X , X , isobtainedbyeliminatingZ f rom ., where .5. TRANSFORMATION ASSOCIATED TO THE CARTESIAN GEOMETRY. X , X , X isreplacedby ( Z , Z , Z andx Z x Z x Z .T hecoef f icientof Z isY X andthatof Z isY X . Proof: For 4. we observe that the elimination should lead to the equation ( X Z X Z ( Y Z Y Z
2) = 0 . An illustration of 4. is given by 12.4.
Conjecture.
Given 9 points A i , B i , C i , on a cubic such that A i , B i C i and ( A , B , C ) , ( A , B , C ) arecollinear, then ( A , B , C ) are collinear. Corollary.
If 3 points A i are on a cubic, the third point C i on the tangent to the cubic at A i are alsocollinear. Example.
For q = 16, i0 = 1, i1 = (cid:15) , i2 = (cid:15) , m , m (cid:15), m (cid:15) , H0.0. M = 253, E0.10. M = 184, H0.1. A i = 2 , , , H0.2. I = 130 , E0.0. a i = 0 ∗ , ∗ , ∗ , E0.1. m i = 253 ∗ , ∗ , ∗ , E . .m i = 179 ∗ , ∗ , ∗ , E0.2. M i = 136 , , , E . .M i = 15 , , , E0.3. i i = 125 ∗ , ∗ , ∗ , E0.4. I i = 238 , , , E0.5. im i = 44 ∗ , ∗ , ∗ ,im i = 4 ∗ , ∗ , ∗ , E0.6.
T m i = 194 , , ,T m i = 3 , , , E0.7. tm i = 61 ∗ , ∗ , ∗ ,tm i = 271 ∗ , ∗ , ∗ , E0.8. IA i = 94 , , ,IA i = 101 , , , E0.12. m = 137 ∗ , m = 189 ∗ , E1.0. eul = 20 , E1.1. kk i = 152 ∗ , ∗ , ∗ , kk i = 258 ∗ , ∗ , ∗ , E1.2. KA i = 234 , , , K A i = 195 , , ,KA i = 126 , , , K A i = 180 , , , E1.3. kl i = 203 ∗ , ∗ , ∗ , k l i = 126 ∗ , ∗ , ∗ ,kl i = 15 ∗ , ∗ , ∗ , k l i = 127 ∗ , ∗ , ∗ , E1.4. B i = 147 , , , B i = 101 , , , CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
E1.5. bb i =E1.6. Eul i = 116 , , , E1.7. Ba i = Ba i =E1.8. tB i =E1.9. KK i = 147 , , , K K i = 101 , , , E1.10. eul i = 88 ∗ , ∗ , ∗ , E1.11.
T T =E1.12. θ = 0 , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , E2.0. tim i = 118 ∗ , ∗ , ∗ , t im i = 182 ∗ , ∗ , ∗ , E2.1. LI i = 63 , , , L I i = 181 , , , E2.2. li i = 192 ∗ , ∗ , ∗ , l i i = 13 ∗ , ∗ , ∗ , E2.3.
Atm i = 211 , , , A tm i = 151 , , , E2.4. lt i = 125 ∗ , ∗ , ∗ , l t i = 125 ∗ , ∗ , ∗ , E2.5. LM = 155 , L M = 163 , E2.6. LT i = 238 , , , L T i = 238 , , , E2.7. lm = 98 ∗ , l m = 162 ∗ , E2.8.
LM M = 96 , L M M = 174 ,LM M = 118 , L M M = 265 , E2.9. tKKL i = tKKL i =E2.10. lmm = 83 ∗ , l mm = 92 ∗ ,lm m = 49 ∗ , l m m = 110 ∗ , E3.0. ka i = 255 ∗ , ∗ , ∗ , k a i = 233 ∗ , ∗ , ∗ ,ka i = 193 ∗ , ∗ , ∗ , ka i = 254 ∗ , ∗ , ∗ , E6.13. Γ = 0 , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , E7.1.
T M i = 171 , T mi = 225 , E8.0. dt i = 139 ∗ , ∗ , ∗ ,dt i = 91 ∗ , ∗ , ∗ , E8.1. Du i = 116 , , ,Du i = 90 , , , E9.0. Eb i = 133 , , , E b i = 150 , , , .5. TRANSFORMATION ASSOCIATED TO THE CARTESIAN GEOMETRY. Eb i = 189 , , , E b i = 9 , , , E9.2. ed i = 226 ∗ , ∗ , ∗ , e d i = 183 ∗ , ∗ , ∗ ,ed i = 128 ∗ , ∗ , ∗ , e d i = 158 ∗ , ∗ , ∗ , E11.19. Dh i = 177 , , , D h i = 133 , , ,Dh i = 83 , , , D h i = 253 , , , E11.20. di i = 180 , , , d i i = 180 , , ,di i = 204 , , , d i i = 204 , , , E11.21. Dj i = 100 , , , D j i = 100 , , , E11.22. dk i = 27 ∗ , ∗ , ∗ , d k i = 92 ∗ , ∗ , ∗ ,dk i = 23 ∗ , ∗ , ∗ , E11.23. du i = 88 ∗ , ∗ , ∗ ,du i = 114 ∗ , ∗ , ∗ , E11.24. Dl i = 101 , , ,Dl i = 101 , , , E11.25. Dm i = 204 , , ,Dm i = 14 , , , E11.26. Dn i = 30 , , ,Dn i = 16 , , , E11.27. dn = 33 ∗ ,dn = 100 ∗ , E11.28. Do = 200 ,Do = 174 , E11.29. dp = 102 ∗ ,dp = 55 ∗ , E11.30. Dq = 178 , E11.31. dr = 26 ∗ , E11.32. γa = 0 , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ ,γ a = 0 , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , E12.0.
N a i = 89 , , , N a i = 272 , , , E12.1. na i = 7 ∗ , ∗ , ∗ , n a i = 204 ∗ , ∗ , ∗ , E12.2.
N b i = 248 , , , N b i = 29 , , , E12.3. nc i = 8 ∗ , ∗ , ∗ , n c i = 158 ∗ , ∗ , ∗ , E12.4. lM M i = 192 ∗ , ∗ , ∗ , l M M i = 114 ∗ , ∗ , ∗ , E12.5. nd i = 263 ∗ , ∗ , ∗ , n d i = 103 ∗ , ∗ , ∗ , CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
E12.6.
N e i = 124 , , , N e i = 208 , , , E12.7. nf i = 80 ∗ , ∗ , ∗ , E12.8. ng i = 182 ∗ , ∗ , ∗ , n g i = 118 ∗ , ∗ , ∗ , E12.9.
N h i = 106 , , , N h i = 186 , , , E12.10. lI = 170 ∗ , l I = 52 ∗ , E12.11.
N i i = 98 , , , N i i = 170 , , ,N i i = 86 , , , N i i = 218 , , , E12.12. nj i = 179 ∗ , ∗ , ∗ , n j i = 253 ∗ , ∗ , ∗ , E12.13. nk i = 32 ∗ , ∗ , ∗ , n k i = 221 ∗ , ∗ , ∗ ,nk i = 252 ∗ , ∗ , ∗ , n k i = 213 ∗ , ∗ , ∗ , E12.14.
N l i = 95 , , , N l i = 83 , , ,N l i = 157 , , , N l i = 20 , , , E12.15. nl = 99 ∗ , n l = 150 ∗ ,nl = 119 ∗ , n l = 161 ∗ , E12.16. nm i = 268 ∗ , ∗ , ∗ , n m i = 98 ∗ , ∗ , ∗ , E12.17. np i = 47 ∗ , ∗ , ∗ , n p i = 265 ∗ , ∗ , ∗ ,np i = 76 ∗ , ∗ , ∗ , n p i = 140 ∗ , ∗ , ∗ , E12.18. nq i = 24 ∗ , ∗ , ∗ , n q i = 48 ∗ , ∗ , ∗ ,nq i = 146 ∗ , ∗ , ∗ , n q i = 82 ∗ , ∗ , ∗ , E12.19.
N r i = 157 , , , N r i = 15 , , ,N r i = 134 , , , N r i = 198 , , , E12.20. nr = 252 ∗ , n r = 218 ∗ ,nr = 202 ∗ , n r = 191 ∗ , E12. .N D i = 256 , , , E12. .χ = 13 , , , , , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , , , , , , , , , , , , , , , , , , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , , , , , , , , , , , , , ( thesearethetangentsandtheotherpointonthetangent )E19.4. lI = 170 ∗ , l I = 52 ∗ , E19.5. LJ = 11 , E22.0. iA i = 179 ∗ , ∗ , ∗ , E22.1. IA = 230 , E22.2. ab = 224 ∗ , E22.6. Ia i = 15 , , , E22.7. ia = 112 ∗ , E25.0. Dk = 176 , .6. PROBLEMS dl = 100 ∗ , Given the center C of a circle and one of its points A, the point X on any line x through A (or y through C ) can be obtained by construction y through C (or x through A ) such thatthe angle XAC = angle BCX. The above as to be reviewed. A construction of a point on agiven tangent (or radius) follows. There must be a simpler way. Let the given points be A ,A , A , let the center be C, let the radius-tangent be t, Mt := t x m, find A on the circleand A x Mt, find A on the circle and A x Mt, let Y := ( A x A ) x ( A x A ) , (C x Y)x t is the point of contact with the circle. To find the bissectrix of an angle A − A A weuse the above construction with the tangent-radius Cx ( mx ( A − xA ) the ?? point X on thecircle is also on the bissectrix. Notation.
Angles and directions will be denoted by an upper case letter and a lower case letter under-lined.
Theorem.
If the angle of the direction of the sides b i is nb i , then the angle of the direction of the tangentis d i +1 + d i +2 − d i m odq + 1 . Theorem.
If the direction of a i is a i , the angles at A i are A i = a i +1 − a i − m odq + 1. Theorem.
0. The angle of the direction of the center of χ i isc i = a i + A i − , thatof thecenterof χ i is c i = a i − A i +1 .
1. The center of χ i is ( A i +1 xM ) xa i +1 , c i thatof χ i is ( A i − xM ) xa i − , c i . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY
Theorem. If m is the ideal line, and A = ( A , A , A ), B = ( B , B , B ) then the mid-point A + B of A and B is A + B = ( m · B ) A + ( m · A ) B .1. the symmetric 2 B − A of A with respect to B is2 B − A = 2( m · A ) B − ( m · B ) A . Theorem.
The mid-points of the diagonals of a complete quadrilateral are collinear.
D37.5, C37.15.(020, Chou and Schelter 1986, p. 18)
Definition.
The line of the preceding Theorem is called the mid-line of the complete quadrilateral.
Theorem.
Given a complete 5-lines, the mid-lines of the 5 complete quadrilaterals obtained by sup-pressiong any of the 5 lines have a point in common. (025, Chou and Schelter 1986, p.19)
Theorem.
Given a triangle A i , and a point M , let M j be the symmetric of M j − with respect to A j − , then M i +3 is the symmetric of M i with respect to M M i +1 , vertex of the anticomplementarytriangle of A i .1. M = M .2. M i , M i +3 , M i +1 , M i +4 are parallelograms. Theorem.
The perpendicular direction to ( IX , IX , IX ) is ( m ( m − m ) I + m ( m + m )( I − I ) , m ( m − m ) I + m ( m + m )( I − I ) , m ( m − m ) I + m ( m + m )( I − I )) . Theorem. [Buterfly Theorem]
If a quadrangle is inscribed in a circle with cent O , then a diagonal point, D , is the midpointof the intersection with the other sides of a perpendicular through D to O × D .(041, Chou(1984), p.269.) .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. Answer to 4.6.1.
Let the lines be a i , m and m (cid:48) = [ m (cid:48) , m (cid:48) , m (cid:48) ] .The midlines are l = [ s − m , s − m , s − m ] ,l = [ s (cid:48) − m (cid:48) , s (cid:48) − m (cid:48) , s (cid:48) − m (cid:48) ] , with s (cid:48) = m (cid:48) + m (cid:48) + m (cid:48) , M A + M (cid:48) A = ( m (cid:48) ( m − m ) − m ( m (cid:48) − m (cid:48) ) , m (cid:48) ( m − m ) , m ( m (cid:48) − m (cid:48) )) .M (cid:48) A + M A = ( m ( m (cid:48) − m (cid:48) ) − m (cid:48) ( m − m ) , m ( m (cid:48) − m (cid:48) ) , m (cid:48) ( m − m )) .l = [ m m m (cid:48) ( s (cid:48) − m (cid:48) ) − m (cid:48) m (cid:48) m ( s − m ) , m (cid:48) m (2 m (cid:48) ( m − m ) − m (cid:48) ( s − m )) − m m (cid:48) (2 m ( m (cid:48) − m (cid:48) ) − m ( s (cid:48) − m (cid:48) )) , m (cid:48) m (2 m (cid:48) ( m − m )+ m (cid:48) ( s − m )) − m m (cid:48) (2 m ( m (cid:48) − m (cid:48) ) + m ( s (cid:48) − m (cid:48) ))] , The common point is P = ( m (cid:48) ( m − m ) − m ( m (cid:48) − m (cid:48) ) , m (cid:48) ( m − m ) − m ( m (cid:48) − m (cid:48) ) , m (cid:48) ( m − m ) − m ( m (cid:48) − m (cid:48) )) . Answer to 4.6.1.
Let M = ( m , m , m ) , with m + m + m = 1 . M = 2 A − M = (2 − m , − m , − m ) , M = 2 A − M = ( − m , m , m ) , M = 2 A − M = (2 − m , − − m , − m ) , M = 2 A − M = ( m , m , − m ) , M = 2 A − M = ( − m , − m , − m ) , M = 2 A − M = ( m , m , m ) . Answer to 4.6.2.
Let 3 of the points be A i , let D := (0 , , x ) , be on a , then the 4-th point is ( y, , x ) , with y = − m ( m + m ) xm ( m + m )+ m ( m + m ) x . O = ( m + m , m + m , m + m ) ,D × O = [( m + m ) x − ( m + m ) , − ( m + m ) x, ( m + m )] , its direction is (( m + m )( m x + m ) , m ( m + m ) x − s − m m ) , m ( m + m ) + x ( s + m m )) .The direction perpendicular to D × O is ( m ( m − m )( m + m )( m x + m ) + m ( m + m )( m ( m + m ) x − s − m m ) − m ( m + m ) + x ( s + m m )) , m ( m − m ) m ( m + m ) x − s − m m ) + m ( m + m )( m ( m + m ) + x ( s + m m ) − ( m + m )( m x + m )) , m ( m − m ) m ( m + m ) + x ( s + m m ) + m ( m + m )(( m + m )( m x + m ) − m ( m + m ) x − s − m m ))) .. . . . CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY.90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES.
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY.90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES.
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY.90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES.
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY.90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES.
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY.90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES.
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY.90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES.
CHAPTER 4. FINITE INVOLUTIVE SYMPATHIC AND GALILEAN GEOMETRY hapter 5FINITE NON-EUCLIDEANGEOMETRY
In Chapter IV, Finite Euclidean geometry was constructed. In it, we have seen that theangles can be given as integers. In the finite hyperbolic Euclidean geometry, the angles canbe represented by elements in Z p − and in finite elliptic Euclidean geometry by elements in Z p +1 . The distances can, in either case, be represented elements in Z p or by δ times anelement in Z p , where δ is such that δ is a non quadratic residue in Z p . I made many attempts to define angles and distances for a geometry which can be consideredas the finite form of non-Euclidean geometry. The clue was finely provided by the work ofLaguerre. I will show that using this definition, both angles and distances can be treated sym-metrically, or to use a mathematical terminology, that we have duality between the notionsof angle of 2 lines and distance of 2 points.For those familiar with non-Euclidean geometry, in the classical case, there is a distinc-tion between the hyperbolic non-Euclidean geometry of Lobatchevski and the elliptic non-Euclidean geometry of Bolyai. The axioms, in a form already familiar to Saccheri, are:there exists a triangle whose sum of interior angles is equal to (Euclidean), smaller than(Lobatchevski) or greater than (Bolyai) 180 degrees.In the hyperbolic case, the set of lines through a point P not on a line l is subdivided into2 sets, those which intersect l and those who do not. If we assume continuity, there are 2lines which form the boundary of either set and are called parallels. The simplest model isobtained by starting with the 2 dimensional projective plane and choosing a given conic asideal. We define as points those inside the conic and as lines the portion of the lines of theprojective plane inside the conic. The parallels to l from a point P not on l are those whichpass through the intersection of l with the conic.In the elliptic case, there are no parallels, the lines always intersect. The simplest modelis obtained by choosing a sphere in 3 dimensional Euclidean geometry. We define as linesthe great circles of a sphere and as points the points of the sphere, identifying each point withits antipode.In the finite case, there is no distinction between the elliptic and the hyperbolic case. CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Indeed in finite projective geometry, the inside or the outside of a conic cannot be defined.Instead, for some lines there are no parallels and for others the situation is analogous tothat described in the classical hyperbolic case. For those who like to refer to some geometricpicture, the image of the geometry on the sphere will be useful although imperfect. I willrefer to it from time to time. Again, although I would find it more satisfactory to proceedsynthetically, I will proceed algebraically to reach the goal more quickly.In finite Euclidean geometry, I proceeded from projective geometry in 3 steps, affine ge-ometry, involutive geometry and Euclidean geometry. Here I will proceed in 2 steps, polargeometry and non Euclidean geometry.In the involutive geometry, an involution on the ideal line is chosen, from which thenotions of perpendicularity and circles are derived. In finite projective geometry, no coniccan be distinguished from any other. To define finite non-Euclidean geometry, I proceed in 2steps. In the first step, I define the finite polar geometry by chosing, or better still, preferinga specific polarity, or equivalently a specific conic. From it, the notions of parallelism, circles,equality of segments, . . . , are derived. In the second step, I introduce the notions of measureof distances and measure of angles, in this case also, the ideal conic plays again an essentialrole.
After defining the geometry starting from a finite projective geometry in which a given polarityis preferred, I define elliptic, parabolic and hyperbolic points and lines. I then define circleswithout using the notion of distance, equidistance and the dual notion of equiangularity are,as in finite Euclidean geometry derived notion. After defining perpendicularity, I definespecial triangles using equidistance and right angles. I then proceed to define mid-points,medians and mediatrices and finally the circumcircles of a triangle. A new point, which Icall the center of a triangle is defined using 2 independent methods. This point also exists inclassical non-Euclidean geometry, but I have not found any reference in the literature. Theintersection of the circumcircles of a triangle are obtained and constructed. Various resultsobtained while studying the center of a triangle are derived. The circumcircle for the specialcase of a triangle with an ideal vertex is studied and finally the properties of the parabola aregiven in detail.
Definition.
Among all the conics in the plane, the chosen one is called the ideal conic or the ideal . Thepoints on the conic are called ideal points or parabolic points . The lines tangent to the conicare called ideal lines or parabolic lines . They could also be called isotropic, by analogy withthe Euclidean case, but I will not use this terminology.A line which intersects the ideal in 2 real points is called a hyperbolic line , a point which is .1. FINITE POLAR GEOMETRY. incident to 2 ideal lines is called a hyperbolic point .A line which does not contain ideal points or a point which is not on an ideal line is called elliptic .A point or a line is said to be an ordinary point if the point or the line is either elliptic orhyperbolic.Two points or two lines are said to be of the same type if they are both either elliptic orhyperbolic. Points of the same type are necessarily ordinary. Convention.
By convention, the conic chosen for the algebraic derivation is X · X = 0 or X + X + X = 0 . Example.
For p = 13 , the ideal points are 6, 9, 19, 22, 57, 62, 69, 76, 79, 118, 134, 141, 148, 153. Theorem.
The polar of A = ( A , A , A ) with respect to the ideal conic is a = [ A , A , A ] . Notation.
The polar of A will be denoted a, the pole of a, A .This notation should not be confused with the notation in section . . . on finite projectivegeometry. Theorem.
With j = +1 or −
1, the ideal points on the line a = [ a , a , a ] are if a + a (cid:54) = 0 , ( a + a , − a a + ja √ d, − a a − ja √ d ), where d = − ( a + a + a ) , if a + a = 0 and a .a (cid:54) = 0,(0 , a + ja √− , a − ja √− , if a = a = 0,(0 , , j √− . Example.
For p = 13 , let a = [124] = [1 , , , then d = 3 , √ d = 4 , the ideal points on a are ( − , − − , − −
6) = (1 , ,
3) = (134) and ( − , − , − , ,
0) = (118) . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Theorem.
The point A = ( A , A , A ) and the line a = [ A , A , A ] are parabolic, iff A · A = 0 , elliptic, iff − A · A is a non quadratic residue modulo p ,in other words,if there is no integer x such that x = − A · A, hyperbolic, iff − A · A is a quadratic residue modulo p . Example.
For p = 13 , (6) = (0,1,5) is parabolic, (172) = (1,12,2) is elliptic and (124) = (1,8,6) ishyperbolic. Theorem. There are p + 1 parabolic or ideal points, There are p ( p − elliptic points, There are p ( p +1)2 hyperbolic points. Proof: Each of the p + 1 parabolic line meets the other p parabolic lines in a hyperbolicpoint. Definition.
Two lines are parallel if they have an ideal point in common.Two points are parallel if they have an ideal line in common.
Example.
For p = 13 , (61) = (1,3,8) and (71) = (1,4,5) are parallel, they are on [134] = [1,9,3]. Theorem.
The intersections of the sides of a triangle with the polars of the opposite vertex with respectto any conic are collinear.
This follows at once from II.2.2.4.7 if we choose the coordinates in such a way that theconic is the ideal conic. .1. FINITE POLAR GEOMETRY.
Introduction.
There are 3 kinds of circles in polar geometry.A hyperbolic circle is a conic tangent to the ideal conic at 2 distinct points. Its center is theintersection of these tangents.An elliptic circle is a conic tangent to the ideal conic at 2 distinct complex conjugate points.A parabolic circle is one for which the two points of tangency coincide.I will give now the corresponding algebraic definition, when convention 5.1.1 is used.Having introduced the notion of circles, it is natural to define the notion of equidistancebetween points and equiangularity between lines. When measure of angles and distances willbe introduced, the compatibility of the 2 concepts equivalence and measure will be made clear.
Definition.
The circles of center C = ( C , C , C ) are the conics with equation X · X + k ( X · C ) = 0 . Definition.
The line c = [ C , C , C ] is called the central line of the circle. Theorem.
The central line is the polar of the center in the polarity associated to the circle as well asin the polarity associated to the ideal conic.
Definition.
A circle is called hyperbolic if its center is hyperbolic, elliptic, if its center is elliptic and parabolic if its center is a parabolic or ideal point.
Theorem.
The ordinary points on a circle are all either hyperbolic or elliptic.
Proof: If k is a quadratic residue modulo p, then − X · X is a quadratic residue and X is necessarily hyperbolic. If k is a non residue, then − X · X is a non residue and X isnecessarily elliptic. Theorem.
If a circle is hyperbolic, the lines through the center and the ideal points on the central lineare tangent to both the ideal conic and the circle. If the circle is parabolic, the center C isan ideal point and its polar is the tangent at the ideal point to both the ideal conic and thecircle.84 CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
All hyperbolic circles can be constructed using the degenerate form of Pascal’s construc-tion. The following Theorem allows the construction of parabolic circles and of many ellipticcircles.
Theorem.
For any circle of center C and central line c through a point X not on c, if I is an idealpoint on C × X and I is a distinct ideal point not on c and I × I meets c in X then X := ( X × X ) × ( C × I ) is also on the circle. Theorem.
If a circle of center C is not parabolic, let A and B be arbitrary points on the circle, let M and N be the other ideal points on C × A and C × B, then the central line, A × B and M × N pass through the same point. Example.
For p = 13 , (see g13.tab)0. One of the hyperbolic circles of center (124) has the equation x + y + z + ( x + 8 y + 6 z ) = 0 , or x + 5 y + 3 z + 5 yz − zx + 3 xy = 0 . It contains the ideal points 118 and 134 and the elliptic points 2, 7, 44, 46, 54, 56,105, 111, 135, 151, 158, 164.1. One of the elliptic circles with center (172) has equation x + y + z + ( x − y + 2 z ) = 0 , or x + 3 y + 6 z − yz + 4 zx − xy = 0 . It contains the elliptic points 7, 13, 15, 21, 41, 44, 70, 77, 98, 111, 116, 151, 156, 169.2. One of the parabolic circles with center (6) has the equation ( x + y + z − ( y + 5 z ) = 0 , or x + 2 z + 3 yz = 0 . It contains the ideal point 6 and the hyperbolic points 1, 33, 39, 81, 89, 100, 101, 109,110, 121, 129, 171, 177.
Definition.
Two circles are parallel if they have one ideal point in common. Two circles are concentric if they have the same center.
Theorem.
Two concentric circles have all their ideal points in common. One for the parabolic circles,2 for the hyperbolic circles. .1. FINITE POLAR GEOMETRY.
Definition.
The points A and B are equidistant from the point C iff there exists a circle of center C passing through both A and B. Theorem. A and B are equidistant from C iff( A · C ) ( B · B ) = ( B · C ) ( A · A ) . This suggest the more general definition:
Definition.
The distance between the points A and B is the same as the distance between the points C and D iff ( A · B ) ( C · C )( D · D ) = ( C · D ) ( A · A )( B · B ) . The angle between the lines a and b is the same as the angle between the lines c and d iff ( a · b ) ( c · c )( d · d ) = ( c · d ) ( a · a )( b · b ) . Definition.
The angle between a and b is a right angle iff a · b = 0 and the distance between A and B isa right distance iff A · B = 0 . Comment.
Although the distance between 2 points A and B has not yet been defined, I will by conventionuse the notation d ( A, B ) . This will be acceptable, in polar geometry, as long as the notationappears in both sides of an equality. I will later define the distance between 2 points andshow that it is consistent with 5.1.2. Theorem.
The notion of equidistance between pairs of points and the notion of equiangularity betweenpairs of lines is an equivalence relation, in other words, the relation isreflexive: d ( A, B ) = d ( B, A ) , symmetric: d ( A, B ) = d ( C, D ) = ⇒ d ( C, D ) = d ( A, B ) , transitive: d ( A, B ) = d ( C, D ) and d ( C, D ) = d ( E, F ) = ⇒ d ( A, B ) = d ( E, F ) . Definition.
The line b is perpendicular to the line a iff b passes through the pole A of a with respect tothe ideal conic, in other words, when a and b are conjugates with respect to the ideal conic. CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Theorem.
If the line b is perpendicular to a, then b · a = 0 . In other words, the angle between a and b is a right angle. This follows at once from 5.1.2.
Theorem.
The perpendicular h , h , h from the vertices A , A , A of a triangle to the opposite sideshave a point H in common. Moreover, h = ( A · A ) A − ( A · A ) A ,h = ( A · A ) A − ( A · A ) A ,h = ( A · A ) A − ( A · A ) A .H = ( A · A )( A · A ) A ∗ A + ( A · A )( A · A ) A ∗ A +( A · A )( A · A ) A ∗ A . Proof: h := A ∗ ( A ∗ A ) , is indeed a line through A perpendicular to A ∗ A . Theresults follow easily from II.2.2.4. Related results are obtained in 5.1.4. Definition. h i are called the altitudes of the triangle. The point H is called the orthocenter . Example.
For p = 13 , if A = (0) = (0 , , , A = (18) = (1 , , , A = (67) = (1 , , , then h = (27) = [1 , , , h = [1 , , , h = [1 , , and H = (171) = (1 , , . Comment.
Section 7 could be placed here, but then the motivation would be absent.
Definition. A right , double right , polar triangle is a triangle which has one, two or three right angles.A right sided or double right sided triangle is a triangle for which the distance between onepair or two pairs of vertices is a right distance. Examples.
For p = 13 : The triangle A = (8) = (0 , , , B = (17) = (1 , , , C = (36) = (1 , , , withsides [44] = [1,2,4], [150] = [1,10,6], [161] = [1,11,4] is a right triangle at B .The triangle A = (44) , B = (17) , C = (36) , with sides [44], [130] = [1,8,12], [161] is adouble right triangle at B and C . The triangle A = (44) , B = (17) , C = (161) , with sides[44], [17] [161] is a polar triangle. .1. FINITE POLAR GEOMETRY. Exchanging vertices and sides, we obtain, by duality, examples of right sided and double rightsided triangles.
Definition. A triangle is isosceles if 2 pairs of vertices are equidistant.A triangle is equilateral if all 3 pairs of vertices are equidistant. Theorem. If a triangle { ABC } is such that d ( A, B ) = d ( A, C ) , then d ( a, b ) = d ( a, c ) . If a triangle { ABC } is such that d ( A, B ) = d ( B, C ) = d ( A, C ) , then d ( a, b ) = d ( a, c ) = d ( b, c ) . Proof: The second part follows, by transitivity, from the first part. For the first part, letus set p = A · A, q = B · B, r = C · C, t = B · C, u = C · A, v = A · B. The hypothesis implies v r = u q = s. We want to prove that w = ( a · b ) c · c does not change when we exchange b and c or q and r as well as u and v . Using II.2.2.4.2and .3, a · b = ut − vr and c · c = pq − v , therefore w = ( u t + v r − tuvr )( pq − v )= pt s − t u v + psqr − s − ptqruv − tsuv. Example.
For p = 13 : The triangle A = (172) = (1 , , , B = (7) = (0 , , , C = (13) = (0 , , , with sides a = [14] = [1 , , , b = [182] = [1 , , , c = [74] = [1 , , is an isoscelestriangle. The triangle A = (172) , B = (7) , C = (15) = (1 , , , with sides a = [104] =[1 , , , b = [182] , c = [74] is an equilateral triangle. Theorem.
In a polar triangle, each vertex is the pole of the opposite side and the distance between thevertices is a right distance.
Definition.
Two triangles are dual of each other iff the sides of one are the polar of the vertices of theother.
Example.
The dual of the triangle of example 5.1.3 is, with p = 13 ,A = (173) = (1 , , , A = (53) = (1 , , , A = (1) = (0 , , . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Theorem.
A polar triangle is its own dual.
Theorem.
The altitudes of a triangle and of its dual coincide. The orthocenter of a triangle and of itsdual coincide.
The proof is left as an exercise.
Introduction.
For this section, the analogy with the model of the non-Euclidean geometry on the sphere isuseful. We recall that each point has 2 representations on the sphere, which are antipodes ofeach other. If we take 2 points A and B, let A (cid:48) and B (cid:48) be their antipodes, there are 2 pointson the great circle, (in the plane through the center of the sphere) which are equidistant from A and B, namely a point on the arc AB and a point on the arc A (cid:48) B, which is the antipode ofthe mid-point on the arc AB (cid:48) . But the analogy is not complete, in the finite case, it is onlywhen the points are of the same type that mid-points exist. There is about 1 chance in 2 thatthe points are not of the same type, there are then no mid-points, there is about 1 chancein 2 that they are of the same type, there are then 2 mid-points, this is an other example ofwhat I call the law of compensation.To simplify the algebra, I will introduce a scaling in 5.1.5. The scaling contains an arbitrarysign, which may be thought as corresponding to the 2 representations on the sphere. Thesystematic way which is chosen could be replaced by some other one. The choice is influencedby the choice of a primitive root of p and the sign of the square root and depends on the rule . . . . Having the concept of mid-points, we can consider those of the vertices of a triangle, ifthe vertices are scaled, we can define interior and exterior mid-points. Again, the choice isarbitrary and depends on the rule . . . .To each side correspond 2 medians, these meet 3 by 3 in 4 points corresponding to thebarycenter. Again the analogy with the geometry on the sphere is useful, the 4 barycenterscan be considered as corresponding to the triangles { ABC } , { A (cid:48) BC, } { AB (cid:48) C } , { ABC (cid:48) } .A similar treatment can be made for the mediatrices which meet 3 by 3 in 4 points, each isthe center of a circumcircle of the triangle { ABC } . But, again, the analogy with the geometry on the sphere is not complete. Given a triangle,there are about 3 chances in 4 that the 3 vertices are not all of the same type, in this casethere is no barycenter and no circumcircle. In about 1 chance out of 4, the 3 points are of thesame type, and there are 4 barycenters and 4 circumcircles. Again this is the compensation.If the vertices of the triangle are of the same type, the 4 lines joining a barycenter to thecorresponding center of a circumcircle can be considered as generalizations of the line ofEuler. It is natural to conjecture that these four lines are concurrent. This is indeed thecase. The surprise is that this point V is not the orthocenter. The coordinates of V are realeven if the vertices of the triangle are not of the same type. V must therefore be obtainablein an independent way. One such method is described in section 7. .1. FINITE POLAR GEOMETRY. I first recall the convention of I. ?? . Convention.
Given δ a specific square root of a specific non quadratic residue of p , we choose the squareroot a of a quadratic residue or the square root aδ , of a non residue in such a way that ≤ a < p − . Notation.
Using the preceding convention, a square root is uniquely defined. It is convenient to introducean other scaling for points and lines different from that given in II.2.2.1.If A = ( A , A , A and A is not an ideal point, A (cid:48) = A √− A · A . | A | = √− A · A is called the length of A. Either each component is an integer, oreach component is an integer divided by δ , in this last case we say that A (cid:48) is pure imaginary. Theorem. If A is hyperbolic, A (cid:48) is real, if A is elliptic, A (cid:48) is pure imaginary. Moreover A (cid:48) · A (cid:48) = − . Definition.
Given 2 points A and B of the same type, M on A × B is called a mid-point of [A,B] iff thedistances M A and
M B are equal.
Theorem.
The mid-points of [
A, B ] are M = A (cid:48) + B (cid:48) and M − = A (cid:48) − B (cid:48) . Proof: Because of 5.1.2, d ( M, A ) = d ( M, B ) if ( M · A (cid:48) ) = ( M · B (cid:48) ) , or if ( A (cid:48) · A (cid:48) ) + ( A (cid:48) · B (cid:48) ) + 2( A (cid:48) · A (cid:48) )( A (cid:48) · B (cid:48) )= ( B (cid:48) · B (cid:48) ) + ( B (cid:48) · A (cid:48) ) + 2( B (cid:48) · B (cid:48) )( B (cid:48) · A (cid:48) ) , which is satisfied because of A (cid:48) · A (cid:48) = B (cid:48) · B (cid:48) = − and A (cid:48) · B (cid:48) = B (cid:48) · A (cid:48) . The proof is similar for M − . Definition. M is called the interior mid-point, M − is called the exterior mid-point . Example.
For p = 13 , the mid-points of (44) = (1,2,4) and (164) = (1,11,7) are (115) = (1,7,10) and(124) = (1,8,6). Indeed | | = √ , | | = √ , hence, | || | = (cid:113) − − = √ , thereforethe mid-points are (1,2,4) + 2(1,11,7) = (3,24,18) = (1,8,6) and (1 , , − , ,
7) =90
CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY ( − , − , −
10) = (1 , , .With δ = 8 , A (cid:48) = Aδ , B (cid:48) = B (6 δ , therefore the interior mid-point is A + B = A − B =(1 , , and the exterior mid-point is A − B = A + 2 B = (1 , , . Definition. m is called a mediatrix of [ A, B ] iff m is perpendicular to A × B and passes through amid-point of [ A, B ] . Theorem. m := A (cid:48) − B (cid:48) passes through M = A (cid:48) + B (cid:48) and m − := A (cid:48) + B (cid:48) passes through M − = A (cid:48) − B (cid:48) . Proof: m is perpendicular to A × B because m · ( A ∗ B ) = m · ( A (cid:48) ∗ B (cid:48) ) = A (cid:48) · ( A (cid:48) ∗ B (cid:48) ) − B (cid:48) · ( A (cid:48) ∗ B (cid:48) ) = 0 .m passes through M, because m · M = ( A (cid:48) − B (cid:48) ) · ( A (cid:48) + B (cid:48) ) = A (cid:48) · A (cid:48) − B (cid:48) · B (cid:48) = − − ( −
1) = 0 . Theorem.
The set of points equidistant from A and B are on m or m (cid:48) . Definition.
In a triangle, the line joining a vertex to the interior (exterior) mid-points of the oppositeside is called an interior (exterior) median . Theorem.
If a triangle is isosceles, with d ( A , A ) = d ( A , A ) , then a median through A is also amediatrix. V of a triangle. Theorem.
Let M i and M − i be the interior and exterior mid-points of A i − and A i +1 , let n i and n − i bethe interior and exterior medians associated to A i . G := n × n ⇒ G · n = 0 . (*)1. G i := n i × n − i +1 ⇒ G i · n − i − = 0 (*). Let m i and m − i be the interior and exterior mediatrices of A i +1 A i − . O := m × m ⇒ O · m = 0 . (*)3. O i := m i × m − i +1 ⇒ O i · m − i − = 0 . (*)4. e j := O j × G j and V := e × e ⇒ V · e = V · e = 0 . (*) .1. FINITE POLAR GEOMETRY. The proof follows from Theorem 4.4.12. As in finite Euclidean geometry “*” indicatesthat there are equivalent definitions, for instance 0 could be written G := n × n and G · n = 0 . Definition.
By analogy with Euclidean geometry, the points G j are called the barycenters of the triangle.The points O j are called the centers of the circumcircles of the triangle.The lines e j are called the lines of Euler of the triangle . Definition. V is called the center of the triangle . Theorem.
Let A (cid:48) i be the normalized coordinates of the vertices of a triangle .0. The mid points are A (cid:48) i − + j i A (cid:48) i +1 , j i = +1 or − . The mediatrices are A (cid:48) i − − j i A (cid:48) i +1 . The medians n i are A (cid:48) i × ( A (cid:48) i − + j i A (cid:48) i +1 ) . Choosing j j j = 1 , the medians meet 3 by 3 at the 4 barycenters which are A (cid:48) + j A (cid:48) + j A (cid:48) . The mediatrices meet 3 by 3 at the 4 centers of circumcircles, which are A (cid:48) ∗ A (cid:48) + j A (cid:48) ∗ A (cid:48) + j A (cid:48) ∗ A (cid:48) . The Euler lines, joining G j to O j are, with d (cid:48) i = A (cid:48) i − · A (cid:48) i +1 , ( j d (cid:48) − d (cid:48) ) A (cid:48) + ( j d (cid:48) − j d (cid:48) ) A (cid:48) + ( d (cid:48) − j (cid:48) d (cid:48) ) A (cid:48) . The Euler lines intersect at V and V = d (cid:48) A (cid:48) ∗ A (cid:48) + d (cid:48) A (cid:48) ∗ A (cid:48) + d (cid:48) A (cid:48) ∗ A (cid:48) . Moreover, V = A · A A · A A ∗ A + A · A A · A A ∗ A + A · A A · A A ∗ A . V exists when the triangle is not a polar triangle. Proof: 0. and 1. follow from . . . . For 2., the intersection of n and n is n ∗ n = − A (cid:48) ( A (cid:48) · ( A (cid:48) ∗ A (cid:48) ))+ j A (cid:48) ( A (cid:48) · ( A (cid:48) ∗ A (cid:48) ))+ j j A (cid:48) ( A (cid:48) · ( A (cid:48) ∗ A (cid:48) ))= A (cid:48) + j j A (cid:48) + j A (cid:48) . The same point is obtained if j j = j or if j j j = 1 . There are 4 points corresponding to j = + or − and j = + or − . The proof of 3. is similar. The proof of 4. to 8. is left as exercises. CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Comment.
Reality requires, because A (cid:48) i = A i | A i | that the lengths | A i | be either all real or all imaginary,hence: Theorem.
The mid-points, mediatrices, medians, barycenter and center of circumcircles are real if andonly if the vertices are either all elliptic or all hyperbolic. V is always real. Example.
For p = 13 , the triangle A = (58) = (1 , , , A = (51) = (1 , , , A = (159) = (1 , , , has all its vertices hyperbolic.Let A (cid:48) = (6 , , , A (cid:48) = (6 , − , , A (cid:48) = (6 , , − . The mid-points of A and A are (14) =(1,0,0) and (13) = (0,1,12). All the mid-points are (14), (13); (115), (12); (139), (8).The mediatrices are [13], [14]; [12], [115]; [8], [139].The medians are [3], [126]; [91], [176]; [76], [161].The interior mediatrices [13], [12], [8] meet at O = (14) . The centers of the circumcircles are (56), (8), (12) and (14).The interior medians [3], [91], [76] meet at G = (33) . The barycenters are (152), (179), (106), and (33).The center of the triangle is V = (152) . V of a triangle. Notation.
From here on, the following notation will be used systematically:0. a i := A i +1 × A i − n i := A i +1 ∗ A i − a i , which means that A i +1 ∗ A i − = n i a i , defines n i ,n i is the normalization factor, see 2.3.2. and 2.3.11.2. l i := A i · A i , d i := A i +1 · A i − , t := ( A ∗ A ) · A . Similarly,5. N i := a i +1 ∗ a i − A i , which means that a i +1 ∗ a i − = N i A i , defines N i , L i := a i · a i , D i := a i +1 · a i − , T := ( a ∗ a ) · a . .1. FINITE POLAR GEOMETRY. Theorem. t = n n N = n n N = n n N . n i L i = n i a i · a i = l i +1 l i − − d i . n i +1 n i − D i = n i +1 n i − a i +1 · a i − = d i +1 d i − − d i l i . n n n T = t . and the dual relations T = N N n = N N n = N N n . N i l i = N i A i · A i = L i +1 L i − − D i . N i +1 N i − d i = N i +1 N i − A i +1 · A i − = D i +1 D i − − D i L i . N N N t = T . Theorem. a i +1 ∗ a i − = tA i , n i a i ∗ A i = d i − A i − − d i +1 A i +1 . n i a i ∗ A i +1 = l i +1 A i − − d i A i +1 . n i a i ∗ A i − = d i A i − − l i − A i +1 . The proof follows easily from 2.3.17. and from 4.6.0.
Example.
For p = 13 , with A [] = { (0) = (0 , , , (18) = (1 , , , (67) = (1 , , } , then a [] = { [173] = [1 , , , [53] = [1 , , , [1] = [0 , , } . l [] = [1 , , , d [] = [5 , , , n [] = [10 , , ,L [] = (11 , , , D [] = (3 , , , N [] = (1 , , ,t = [4] , T = (3) . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Theorem.
Let h be the polar of H with respect to the triangle. Let K i be the intersection of h and a i , let v i be the perpendicular at A i to A [ i ] × K i .Then v i have a point in common V .Moreover, if we define u i := d i +1 d i − − d i l i , we have h = u a + u a + u a . K i = u i − A i +1 − u i +1 A i − . v i = ( d i +1 l i +1 − d i − l i − ) A i − ( d i d i +1 l i − d i − l i − l i ) A i +1 + ( d i d i − l i − d i +1 l i +1 l i ) A i − . V = d l A ∗ A + d l A ∗ A + d l A ∗ A . Proof: Because of 5.1.3, H = d d a + d d a + d d a , after simplification, H · a = ( d d − d l )( d d − d l ) , using the definition 0, H · a = u u , and because of 2.3.20, after multiplication by u u u , we obtain 1. K i := H ∗ a i , gives 2, after division by t . A i ∗ K i = u i +1 a i +1 + u i − a i − , therefore, V i = A i ∗ ( A i ∗ K i ) , substituting and using 2.3.17.0., we get V i = ( u i − d i − − u i +1 d i +1 ) A i − u i − l i A i +1 + u i +1 l i A i − , replacing u i by its value, we get, 3, from which we obtain v ∗ v = d l A ∗ A + d l A ∗ A + d l A ∗ A , after dividing each term by ( d d l + d d l + d l l l − d l − d d l l ) . Introduction.
In this section we study the 4-th point of intersection of the 4 circumcircles of a triangle.The expression for these 6 points is given in 5.1.8. A construction in the case where thecenters of the 2 circles are given is described in 5.1.8. .1. FINITE POLAR GEOMETRY. Notation. C i denotes the circumcircle with center C i .X j,k denotes the intersection of C j and C k distinct from the vertices of the triangle A i , nor-malized to A (cid:48) i .a i := A (cid:48) i · A (cid:48) ,d := A (cid:48) ∗ ( A (cid:48) ∗ A (cid:48) ) . Theorem. The intersections of the circumcircles of a triangle A i are given, using X , = (1 + a ) A (cid:48) + ( a − a ) ( A (cid:48) − A (cid:48) ) ,X , = (1 + a ) A (cid:48) + ( a − a ) ( A (cid:48) − A (cid:48) ) ,X , = (1 + a ) A (cid:48) + ( a − a ) ( A (cid:48) − A (cid:48) ) ,X , = (1 − a ) A (cid:48) + ( a + a ) ( A (cid:48) + A (cid:48) ) ,X , = (1 − a ) A (cid:48) + ( a + a ) ( A (cid:48) + A (cid:48) ) ,X , = (1 − a ) A (cid:48) + ( a + a ) ( A (cid:48) + A (cid:48) ) . Proof: Let B i := A (cid:48) i +1 ∗ A (cid:48) i − , then C = − B + B + B ,C = B + B + B , the circles with these centers are kX = ( X · C ) and kX = ( X · C ) , they pass through the vertex A of the triangle if k − A · B ) = d and therefore also through the vertices A and A if k = − d . X is common to the 2 circlesif X · C = jX · C ( j = +1 or − ), j = +1 leads to the vertices A (cid:48) or A (cid:48) , j = − gives X · ( C + C ) = 2 X · ( B + B ) =2 X · ( A (cid:48) ∗ ( A (cid:48) − A (cid:48) )) = 0 . therefore, for some l, and with M − = A (cid:48) − A (cid:48) ,X = lA (cid:48) + M − , hence X = − l + ( M − ) + 2 lA (cid:48) · M − , and X · C = ( lA (cid:48) + M − ) · ( B + A ∗ M − ) = lA (cid:48) · B = ld.X is on the circles C if − d − l + ( M − ) + 2 lA (cid:48) · M − ) = l d , therefore l = − ( M − ) A (cid:48) · M − = − − a )2( a − a hence the expression for X , . The other points are derived similarly. Corollary.
The intersections X , and X , coincide if1 − a + a + a = 0 Salzbourg-Innsbruck 29-30.9.83 CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY and X , = X , = − A (cid:48) + A (cid:48) + A (cid:48) , with similar expressions for other pairs. The proof is straightforward.
Example.
With the triangle of 5.1.6, a i = 5,, X , = X , = X , = X , = X , = X , = Theorem.
Let M , , M , be the mid-points of a , . . . , in the algebraic order defined above, then the dual lines are the mediatrices,the dual of M , passes through M , , . . . the points M , , M , , M , are on o , the points M , , M , , M , are on o , the points M , , M , , M , are on o , the points M , , M , , M , are on o . the dual of o i is the center O i of one of the 4 circumcircles of the triangle A i . By duality, the mediatrices m , , m , , m , are on O , . . . .Let m (cid:48) , , m (cid:48) , , . . . be the medians A · M , , A · M , , . . . then the medians m (cid:48) , , m (cid:48) , , m (cid:48) , are on G , the medians m (cid:48) , , m (cid:48) , , m (cid:48) , are on G , the medians m (cid:48) , , m (cid:48) , , m (cid:48) , are on G , the medians m (cid:48) , , m (cid:48) , , m (cid:48) , are on G . Notation. u ∗ v is the vector ( u v − u v , u v − u v , u v − u v ) u ×× v is the vector ( u v + u v , u v + u v ,u v + u v ) u O v is the vector ( u v , u v , u v )) .1. FINITE POLAR GEOMETRY. Algorithm.
Given two circles through the points A i , with centers C and C , With j in the set { , } and i in the set { , , } , and addition within the indices done modulo 2, d j := C j O C j ,B i := a i +1 Xa i +2 ,L := d ∗ d ,f i := B i · L,G := f O f,s i := a i +1 a i +2 A i ,O := (cid:80) i =0 ( s i G i A i ) . Theorem. O is the 4-th point common to the two circles. Introduction.
The following results were obtained while searching for a construction of V, independent fromthe centers of mass and center of circumcircles. Theorem.
Let I be an ideal point on the line I × B . Let J := ( B · B ) I − I · B ) B, then J is the other ideal point on I × B. Example.
For p = 13 , Let I = (22) = (1 , , and B = (4) = (0 , , , then J = − , ,
8) + 4(0 , ,
3) = (1 , ,
4) = (57) . The line I × J is [48] = [1 , , . Theorem.
Let a be an ordinary line and B an ordinary point not on a. Let I and K be the ideal points on a and J and L be the other ideal points on I × B and K × B, let c := J ∗ J J, then c = ( B · B ) a − a · B ) B. Proof: With a = I ∗ K,J = ( B · B ) I − I · B ) B, L = ( B · B ) K − K · B ) B, hence L ∗ J = ( B · B ) K ∗ I + 2( B · B )(( K · B ) I − ( I · B ) K ) ∗ B, = ( B · B )( − ( B · B ) a + 2(( K ∗ I ) ∗ B ) ∗ B ) because of 2.3.17.0., CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY = ( B · B )( − ( B · B ) a − a ∗ B ) ∗ B ) , = ( B · B )( − ( B · B ) a + 2( B · B ) a − a · B ) B ) because of 2.3.17.0., hence the Theorem. Example.
For p = 13 , let a = [139] = [1 , , and B = (4) , then c = − , , − , ,
3] = [1 , ,
9] =[88] . The ideal points are I = 22 , J = 57 , K = 76 , L = 79 . Definition. c as defined in the preceding theorem is called the conjugate of a with respect to B. Theorem.
The lines x i +1 A i +1 − x i − A i − + y i − n i − a i − − y i +1 a i +1 are concurrent at the point (cid:80) i ( y i +1 y i − t + ( x i d i +1 − x i − l i − ) y i +1 + ( x i d i − − x i +1 l i +1 ) y i − ) A i + (cid:80) i ( x i +1 x i − n i ) a i . Theorem.
Given a triangle A i with sides a i , let b i be the conjugate of a i with respect to A i . Assumethat the b i are not collinear. Let B i be the vertices of the triangle b i , then A i × B i are concurrent at W . A i × b i are concurrent at H. B i × b i are concurrent at W . Moreover, b i = l i A i +1 ∗ A i − − tA i . B i = − l i − l i +1 A i + 2 l i − d i − A i +1 +2 l i +1 d i +1 A i − + 4 tA i +1 ∗ A i − . A i ∗ B i = l i +1 d i +1 A i ∗ A i − − l i − d i − A i +1 ∗ A i +2 td i +1 A i +1 − td i − A i − . W = (cid:80) i (( − l i − l i +1 d i − d i +1 ) A i + 2 d i ( l i − d i − + l i +1 d i +1 A i +4 td i +1 d i − A i +1 ∗ A i − ) . A i × b i = d i +1 A i +1 − d i − A i − . H = d d A ∗ A + d d A ∗ A + d d A ∗ A . .1. FINITE POLAR GEOMETRY. With x i = 8 t d i + 2 l i +1 l i − d i +1 d i − − l l l d i and y i = 4 tl i d i ,B i × b i = x i +1 A i +1 − x i − A i − + y i − A i ∗ A i +1 − y i +1 A i − ∗ A i . W = Proof: 3., is immediate,Using 2.3.17.0 and 1.7.2., we obtain .4 after division by t,
5. and 6. after division by t , 7. after division by l i .
8. is immediate and is indeed the same as 1.3.3.
Example.
For p = 13 , using the same triangle as Example 1.7.4. B i = { , , ,
137 = (1 , , ,
61 = (1 , , } ,b i = {
180 = [1 , , ,
101 = [1 , , ,
80 = [1 , , } , W = 7 = (0 , , , H = 171 = (1 , , , W = 77 = (1 , , . Exercise.
Using the dual triangle, A i = { , , } , determine B i , b i and W , H and W . Introduction.
In the preceding section I have dealt with circumcircles through 3 ordinary points. I will nowdiscuss the case when 1 or more points are ideal points.
Theorem.
The only circle through 3 distinct ideal points is the ideal conic, X = 0. Theorem.
The only circle through 2 distinct ideal points A and B and through an ordinary point C is( c · C ) X − ( c · X ) C = 0 , where c := A × B. Example. p = 11 , A = (31) = (1 , , , B = (26) = (1 , , ,C = 54 = (1 , , . The circumcircle through
A, B and C is X + X + X = 4( X − X − X ) . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Theorem.
There are 2 circles through 1 ideal point A and through two distinct ordinary points B and C not collinear with A. Theorem.
Let X be a point not on the sides of a triangle A i . Let X i be the intersection with a i of theline A i × X , or X i := ( A i × X ) × a i , Let ξ be a circle through X i . Let Y i be the other intersection of a i with ξ ,let y i := A i × Y i , then the lines y i have a point Y in common. Comment.
The theorem 5.1.10 is analogous to theorem . . . in Euclidean geometry. It follows from itsgeneralization . . . to projective geometry.Because of the clear connection with the Theorem of Ceva, I have the following:
Definition.
The correspondence X to the various points Y associated to the several circumcircles through X i , is called the Ceva correspondence . I will ignore those points Y which happen to coincidewith a vertex of the triangle. Comment.
Clearly if Y is associated to X, X is associated to Y in a Ceva correspondence. But wecannot call this an involution because the correspondence is not one to one or bijective. Program.
The program NETR1.BAS determines the Ceva correspondence. It is illustrated in NETR1.HOM.
Introduction.
In this section I have defined a parabola for non Euclidean geometry and many of the re-lated elements of the parabola, by analogy with the definitions of Euclidean geometry. Byduality we essentially double the number of these elements, for instance to the focus in Eu-clidean geometry corresponds the focal point and the focal line. The basic equation is givenby 5.1.11.0. .1. FINITE POLAR GEOMETRY.
Definition. A parabola is a conic which is tangent at one point to the ideal conic and at one point only.This point is called the isotropic point of the parabola , the tangent is called the isotropic lineof the parabola . Definition. A focal tangent t of a parabola is an ideal tangent to the parabola which is not isotropic.A focal point F is an ideal point which is not isotropic. There are 2 focal points F and F which are either real or complex conjugate. Definition.
The focus F of a parabola is the intersection of the focal tangents. The focal line f of aparabola is the line through the focal points. Theorem.
The focus is not on the isotropic line. The focus is not an ideal point.
Proof: In the first case, through the focus we could draw 3 tangents to the parabola. Inthe second case, the parabola would be tangent at a second point to the ideal conic and wouldtherefore be a circle.
Definition.
The director D of a parabola is the pole of its focal line with respect to the parabola. Definition.
The axis a of a parabola is the line through its focus and its isotropic point.The axial point A of a parabola is the point on the focal line and on the isotropic line. Definition.
The vertex V of a parabola is the ordinary point on the parabola and its axis.The vertical line v of a parabola is the ordinary tangent through the axial point. Theorem.
A parabola with isotropic point I and focal tangent f is I · X )( f · X ) = t ( X · X ) , f · I (cid:54) = 0 , f · f (cid:54) = 0 , t (cid:54) = 0 , t (cid:54) = f · I . The polar of X ( I · X f + ( f · X I − tX . The pole of a is02 CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY ( f ∗ I ) · aI ∗ f + t ( a · f ) I + t ( a · I ) f + ( t − tI · f ) a. Proof: 0, follows from the general equation of a conic through the intersections of theideal and the lines I and f, see . . . and from 5.1.11.11.4.0, represents a degenerate conic corresponding to the lines I and f if t = 0 and to the lines I × F and I × F if t = f · I .The proof of the last fact is left as an exercise. 1 follows from 0. See . . . . 2 is obtained bychosing 2 points on a, a ∗ I and f ∗ a, and determining the intersection of the polars of theselines. ( f ∗ I ) · a is a factor of each term. Example.
For p = 13 , I = (1 , , , f = [1 , , , t = 4 , The parabola is x + 5 y )( x + y + 4 z ) = 4( x + y + z . The polar of X x , y , z is [ − x y z , x y − z , x − y − z . The pole of a = [ a , b , c is [ a c , b c , a b c . The director D is (1,-1,6).The axial point A is I ∗ f = (1 , , . With v = − , the ideal points on the parabola are I and F , v, − − v ) , F , − v, − v ) . The tangent at F is f v − v, − v, − − v ] = [1 , − v, v ] The ideal lines t , , and t , − , are tangent to the parabola at T , , − and T , − , , they meet at the focus F = (1 , − , − . The directrix is [1,4,4].The axis a is I ∗ F = [1 , , . The vertex V is (1,-2,-6).The tangent at the vertex is [1,4,1].The vertical line v is [1,4,1]. Theorem.
The polar of X is (cid:80) j A i,j X j , with A j,j = 2 I j f j − t,A j,k = I j f k + I k f j , j (cid:54) = k, The pole of a is (cid:80) k B j,k a k , with B j,j = t − t ( I · f − I j f j ) − ( I ∗ f ) j ,B j,k = t ( I j f k + I k f j ) − ( I ∗ f ) j ( I ∗ f ) k . The dual equation of the conic is2( I · x )( F · x ) = u ( x · x ) , The pole of x is ( I · x ) F + ( F · x ) I − ux and F and u follow from . . .if I j (cid:54) = f rac , u = ( I B , + I B , , − I I B , ) , .1. FINITE POLAR GEOMETRY. F j = u + B j,j I j , if I = 0, and I (cid:54) = 0 , u = − B , ,F = B , − B , I ,F = B , − B , I ,F = B · I , if I = (1 , , , u = − B , , F = B , − B , , F = B , , F = B , . Proof: The matrix A follow from (1), the matrix B is its adjoint divided by 2. Theorem.
The director is D = ( f · f ) I + ( t − f · I ) f. Proof: Replace a by f in 5.1.11.2. Exercise.
Complete the following sentences, for the parabola 5.1.11.0:0. The axis a is .1. The axial point A is .2. The vertex V is .3. The vertical line v is .4. The ideal point J is . Theorem. The vertical line v passes trough the vertex V. The director D , the pole f of the focal line f, the pole d of the directrix d are all onthe axis a. The directrix d, the polar F of the focus F, the polar D of the director D all passthrough the axial point A. The other ideal point J on the axis and the other ideal line j through the axial pointare incident. Answer 5.1.9. B i = { , , } , b i = { , , } ,W = (62) , H = (171) , W = (88) . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Introduction.
When p = 5 , the representation of polar geometry on the dodecahedron is suggested by thefact that the 6 faces form a conic which can be chosen as the ideal. Definition.
Using the dodecahedral representation, the conic which consists of the 6 face-points is the ideal conic . Theorem.
The 15 side-points are hyperbolic and the 10 vertex-points are elliptic.
This follows at once from the incidence definitions, II.2.3.4.
Theorem.
With the ideal conic of type A ,the 3 . 15 conics of type I , E E J , J , O , O P, U U Although this should be placed in the Chapter on non-Euclidean geometry, we have.
Theorem.
With a particular choice of unit, the radii of the various sub-types are as follows, U E π , U E π , P and I π . Example.
Computations relating to g434.PRN:If we use as primitive polynomial I − I − , we obtain the correspondence: .1. FINITE POLAR GEOMETRY. x ( y , y , y ) ( y ) [ z ]0 (0 , ,
1) (0) [6]1 (0 , ,
0) (1) [1]2 (1 , ,
0) (6) [11]3 (0 , ,
2) (3) [21]4 (1 , ,
0) (16) [16]5 (1 , ,
1) (22) [26]6 (1 , ,
4) (30) [7]7 (1 , ,
3) (9) [9]8 (0 , ,
3) (4) [8]9 (1 , ,
0) (21) [10]10 (1 , ,
4) (20) [0]11 (1 , ,
1) (7) [12]12 (0 , ,
1) (2) [24]13 (1 , ,
0) (11) [18]14 (1 , ,
2) (13) [30]15 (1 , ,
2) (23) [2]16 (1 , ,
4) (15) [17]17 (1 , ,
2) (8) [14]18 (0 , ,
4) (5) [28]19 (1 , ,
0) (26) [25]20 (1 , ,
3) (29) [4]21 (1 , ,
3) (14) [22]22 (1 , ,
2) (28) [29]23 (1 , ,
3) (19) [13]24 (1 , ,
1) (17) [20]25 (1 , ,
1) (12) [3]26 (1 , ,
2) (18) [27]27 (1 , ,
1) (27) [19]28 (1 , ,
3) (24) [23]29 (1 , ,
4) (25) [15]30 (1 , ,
4) (10) [5]
The second column is I x mod P, the third column is the representationof Chapter II, the fourth column is obtained as follows.The ideal conic passes through 0,4,6,9,16 and 17 and is therefore represented by the matrix Q = or the quadratic form Q ( x, y ) = x y + x y + x y + x y . This determines the polar of ( y .y , y ) as [ y + y , y , y ] , but the polar of x is x ∗ , thereforeif x = ( y .y , y ) and [ z ] = [ y + y , y , y ] then x ∗ = [ z ] . For instance, if x = 7 , ( y ) = (1 , , , [ z ] = [ − , ,
1] = [10] . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Example.
The hyperbolic circles have as center a edge-point.Those with center ∗ × ∗ and through the point u can be constructed as follows, let up = 0 × (4 × u ) , given any line v through 0, such that u · v (cid:54) = 0 , the Pascal constructiongives the other point on the conic and v using (((( v × × up ) × × u ) × v. This gives, with the radii determined below:0,4;;2,15,22,25 of type f f ssss and sub-type I , radius π . f f vvvv and sub-type E , radius π . f f vvvv and sub-type E . radius π . Before having obtained a synthetic construction of the parabolic and elliptic circles wehave used the algebraic definition.The algebraic definition is kx T Qx − ( x T QC ) = 0 , with C on the ideal conic, for parabolic circles and C a vertex-point for elliptic circles.For k = 0 , the circle degenerates in (a double) line, consisting of the points at distance π from C. Let C = 0 , the polar ∗ = [1 , , , hence the parabolic circles are k ( x − x + 2 x x ) + x = 0 . With k (cid:48) = k , the points are (0,0,1) = 0, and (1 , x , − k (cid:48) + x ) . This gives for, k = − , (1,0,4) = 30, (1,1,1) = 25, (1,4,1) = 27, (1,2,2) = 26, (1,3,2) = 15, in view of thetable above. Hence k = 4 , { } of type f sssss and sub-type O .k = 2 , { } of type f vvvvv and sub-type J .k = 3 , { } of type f vvvvv and sub-type J .k = 1 , { } of type f sssss and sub-type O . Let C = 5 , the polar ∗ = [1 , − , − , hence the circles are k ( x + x + 2 x x ) − ( x − x − x ) = 0 . The points are, (0 , , ± √− k and (1 , x , − k + 2( x − ± (cid:112) k − k ( x − x + 2) . This gives, for k = − , (0,1,3) = 8, (0,1,1) = 12, (1,4,2) = 22, (1,2,3) = 23, (1,3,2) = 15, (1 , ,
3) = 22 , in view of the table above.Hence, with the radii determined below: k = 4 , { ssssss and sub-type U , radius π .k = 2 , { } of type vvvvvv and sub-type P, radius π k = 3 , { } , radius 0. k = 1 , { } of type ssssss and sub-type U , radius π . To summarize, we see that, with the ideal conic of type A, the 3 . 15 conics of type I , E and E are hyperbolic circles,the 4 . 6 conics of type J , J , O , O are parabolic circles andthe 3 . 10 conics of type P, U and U are elliptic circles. .1. FINITE POLAR GEOMETRY. Exercise.
For a synthetic construction of the parabolic circles and some elliptic ones, we can useIV.1.2.7. This is a good exercise.
Example of Distances.
We recall the trigonometric tables for p = 5:With δ = 2, x sin ( x ) cos ( x ) x sin ( x ) cos ( x )0 0 1 0 0 11 2 δ δ − δ δ −
23 1 0 cos ( d ) 0 1 2 3 4 dπ
16 14 13 where cos ( d ( C, X )) = Q ( X,C ) Q ( X,X ) = kQ ( C,C ) . Hence the distances, recorded above. For instance, in the case of elliptic circles, for C = 5 = (1 , , and X = 8 = (0 , , , Q ( C, C ) = 2 , Q ( X, X ) = 1 , Q ( X, C ) = 1 ,cos ( d ( C, X )) = 3 , d ( C, X ) = π . Hencefor k = − , the radius is π , for k = 2 , cos ( d ( C, X )) = − , d ( C, X ) = π , for k = 3 , d ( C, X ) = 0 , for k = 1 , d ( C, X ) = π . In the case of hyperbolic circles with C = 8 and X = 5 , we have d (8 ,
5) = π , the other radiiare obtained directly, cos ( d (8 , ( − . = − , hence d (8 ,
2) = π ,cos ( d (8 , ( − . = 2 , hence d (8 ,
7) = π . Example.
Computations relating to g434.PRN: If we use as primitive polynomial I − I − , we obtainthe correspondence: CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY x ( y , y , y ) ( y ) [ z ]0 (0 , ,
1) (0) [6]1 (0 , ,
0) (1) [1]2 (1 , ,
0) (6) [11]3 (0 , ,
2) (3) [21]4 (1 , ,
0) (16) [16]5 (1 , ,
1) (22) [26]6 (1 , ,
4) (30) [7]7 (1 , ,
3) (9) [9]8 (0 , ,
3) (4) [8]9 (1 , ,
0) (21) [10]10 (1 , ,
4) (20) [0]11 (1 , ,
1) (7) [12]12 (0 , ,
1) (2) [24]13 (1 , ,
0) (11) [18]14 (1 , ,
2) (13) [30]15 (1 , ,
2) (23) [2]16 (1 , ,
4) (15) [17]17 (1 , ,
2) (8) [14]18 (0 , ,
4) (5) [28]19 (1 , ,
0) (26) [25]20 (1 , ,
3) (29) [4]21 (1 , ,
3) (14) [22]22 (1 , ,
2) (28) [29]23 (1 , ,
3) (19) [13]24 (1 , ,
1) (17) [20]25 (1 , ,
1) (12) [3]26 (1 , ,
2) (18) [27]27 (1 , ,
1) (27) [19]28 (1 , ,
3) (24) [23]29 (1 , ,
4) (25) [15]30 (1 , ,
4) (10) [5]
The second column is I x mod P, the third column is the representationof Chapter II, the fourth column is obtained as follows.The ideal conic passes through 0,4,6,9,16 and 17 and is therefore represented by the matrix Q = or the quadratic form Q ( x, y ) = x y + x y + x y + x y . This determines the polar of ( y .y , y ) as [ y + y , y , y ] , but the polar of x is x ∗ , thereforeif x = ( y .y , y ) and [ z ] = [ y + y , y , y ] then x ∗ = [ z ] . For instance, if x = 7 , ( y ) = (1 , , , [ z ] = [ − , ,
1] = [10] . .1. FINITE POLAR GEOMETRY. Example.
The hyperbolic circles have as center a edge-point.Those with center ∗ × ∗ and through the point u can be constructed as follows, let up = 0 × (4 × u ) , given any line v through 0, such that u · v (cid:54) = 0 , the Pascal constructiongives the other point on the conic and v using (((( v × × up ) × × u ) × v. This gives, with the radii determined below:0,4;;2,15,22,25 of type f f ssss and sub-type I , radius π . f f vvvv and sub-type E , radius π . f f vvvv and sub-type E . radius π . Before having obtained a synthetic construction of the parabolic and elliptic circles wehave used the algebraic definition.The algebraic definition is kx T Qx − ( x T QC ) = 0 , with C on the ideal conic, for parabolic circles and C a vertex-point for elliptic circles.For k = 0 , the circle degenerates in (a double) line, consisting of the points at distance π from C. Let C = 0 , the polar ∗ = [1 , , , hence the parabolic circles are k ( x − x + 2 x x ) + x = 0 . With k (cid:48) = k , the points are (0,0,1) = 0, and (1 , x , − k (cid:48) + x ) . This gives for, k = − , (1,0,4) = 30, (1,1,1) = 25, (1,4,1) = 27, (1,2,2) = 26, (1,3,2) = 15, in view of thetable above. Hence k = 4 , { } of type f sssss and sub-type O .k = 2 , { } of type f vvvvv and sub-type J .k = 3 , { } of type f vvvvv and sub-type J .k = 1 , { } of type f sssss and sub-type O . Let C = 5 , the polar ∗ = [1 , − , − , hence the circles are k ( x + x + 2 x x ) − ( x − x − x ) = 0 . The points are, (0 , , ± √− k and (1 , x , − k + 2( x − ± (cid:112) k − k ( x − x + 2) . This gives, for k = − , (0,1,3) = 8, (0,1,1) = 12, (1,4,2) = 22, (1,2,3) = 23, (1,3,2) = 15, (1 , ,
3) = 22 , in view of the table above.Hence, with the radii determined below: k = 4 , { ssssss and sub-type U , radius π .k = 2 , { } of type vvvvvv and sub-type P, radius π k = 3 , { } , radius 0. k = 1 , { } of type ssssss and sub-type U , radius π . To summarize, we see that, with the ideal conic of type A, the 3 . 15 conics of type I , E and E are hyperbolic circles,the 4 . 6 conics of type J , J , O , O are parabolic circles andthe 3 . 10 conics of type P, U and U are elliptic circles. CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Exercise.
For a synthetic construction of the parabolic circles and some elliptic ones, we can useIV.1.2.7. This is a good exercise.
Example of Distances.
We recall the trigonometric tables for p = 5:With δ = 2, x sin ( x ) cos ( x ) x sin ( x ) cos ( x )0 0 1 0 0 11 2 δ δ − δ δ −
23 1 0 cos ( d ) 0 1 2 3 4 dπ
16 14 13 where cos ( d ( C, X )) = Q ( X,C ) Q ( X,X ) = kQ ( C,C ) . Hence the distances, recorded above. For instance, in the case of elliptic circles, for C = 5 = (1 , , and X = 8 = (0 , , , Q ( C, C ) = 2 , Q ( X, X ) = 1 , Q ( X, C ) = 1 ,cos ( d ( C, X )) = 3 , d ( C, X ) = π . Hencefor k = − , the radius is π , for k = 2 , cos ( d ( C, X )) = − , d ( C, X ) = π , for k = 3 , d ( C, X ) = 0 , for k = 1 , d ( C, X ) = π . In the case of hyperbolic circles with C = 8 and X = 5 , we have d (8 ,
5) = π , the other radiiare obtained directly, cos ( d (8 , ( − . = -1, hence d (8 ,
2) = π ,cos ( d (8 , ( − . = 2, hence d (8 ,
7) = π . Introduction.
Spherical trigonometry refers to the relation between the measure of angles and arcs of atriangle on a sphere.The formulas of al-Battani (Albategnius, about 920 A.D.) and of Jabir ibn Aflah (Geber,about 1130 A.D.) have to be adapted to the finite case in which the sine of an angle in thefirst 2 quadrants cannot be considered as positive. There are several possible solutions. Oneof these will be given in Theorem 5.2.1.Let
A, B, C be the vertices of a triangle, a, b, c be its sides. .2. FINITE NON-EUCLIDEAN GEOMETRY.
The measure of the angle between b and c will be denoted A, . . . .The distance between the points B and C will be denoted a, . . . . Definition.
Let
A, B, C be 3 points on the sphere x + y + z = 1 , of center O = (0 , , , . The direction DA of A is the ideal point on OA.
The side a = { B, C } is a section of the circle C a which is the intersection of the sphere andthe plane O × B × C. The spherical distance of the side a , also denoted a is the angle betweenthe directions DB and DC.
The angle
BAC, also denoted A is the angle of the directions of the tangents at A to thecircles C b and C c . Theorem.
Between the trigonometric functions of the angles and sides of a general triangle we have therelations: | sina | sinA = | sinb | sinB = | sinc | sinC = r.
1. 0. cosA = cosBcosC + sinBsinCcosa, cosB = cosCcosA + sinCsinAcosb, cosC = cosAcosB + sinAsinBcosc.
2. 0. cosa = cosbcosc + | sinb || sinc | cosA, cosb = cosccosa + | sinc || sina | cosB, cosc = cosacosb + | sina || sinb | cosC.
3. 0. sinA = cos B − cos CsinBcosCcosc − sinCcosBcosb , sinB = ( cos C − cos AsinCcosAcosa − sinAcosCcosc , sinC = ( cos A − cos BsinAcosBcosb − sinBcosAcosa .
4. 0. | sina | = cos b − cos c | sinb | cosccosC −| sinc | cosbcosB , | sinb | = cos c − cos a | sinc | cosacosA −| sina | cosccosC , | sinc | = cos a − cos b | sina | cosbcosB −| sinb | cosacosA .
5. 0. cosA = sinBcosBcosc − sinCcosCcosbsinBcosCcosc − sinCcosBcosb , cosB = sinCcosCcosa − sinAcosAcoscsinCcosAcosa − sinAcosCcosc , cosC = sinAcosAcosb − sinBcosBcosasinAcosBcosb − sinBcosAcosa .
6. 0. cosa = | sinb | cosbcosC −| sinc | cosccosB | sinb | cosccosC −| sinc | cosbcosB , cosb = | sinc | cosccosA −| sina | cosacosC | sinc | cosacosA −| sina | cosccosC , cosc = | sina | cosacosB −| sinb | cosbcosA | sina | cosbcosB −| sinb | cosacosA . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Proof :Let the coordinates of the points A, B, C be ( A , A , A , , ( B , B , B , , ( C , C , C , . Those of
DA, DB and DC are ( A , A , A , , ( B , B , B , , ( C , C , C , . if A · B := A B + A B + A B and A · A := A A + A A + A A , by definition (see . . . ) cosa = B · C because B · B = C · C = 1 . The plane A × B × O is { A B − A B , A B − A B , A B − A B , } the tangent to the sphere at A is { A , A , A , − } and the ideal plane is { , , , , } therefore the direction of A × B is DAB = A · AB − A · BA , A · AB − A · BA , A · AB − A · BA , . Similarly the direction of A × C is DAC = A · AC − A · CA , A · AC − A · CA , A · AC − A · CA , .DAB · DAB = 1 − ( A · B ) = 1 − cos c = sin c, and DAC · DAC = sin b. Therefore cosA = B · C + A · BA · C − A · BA · C − A · CA · B | sinb || sinc | = cosa − cosccosb | sinb || sinc | , hence 2.0. sin Asin bsin c = (1 − cos A ) sin bsin c = sin bsin c − cos a − cos bcos c + 2 cosacosbcosc = 1 − cos a − cos b − cos c + 2 cosacosbcosc Therefore, if r := sqrt − cos a − cos b − cos c +2 cosacosbcoscsin asin bsin c then sinA | sina | = sinB | sinb | = sinC | sinc | = r. Simple algebraic manipulations give 3 to 6.If we eliminate cosB and cosC from 1.1 and 1.2, cosB = − sinCcosb + cosAsinBcoscsinA ,cosC = − sinBcosc + cosAsinCcosbsinA , substituting in 1.0. gives cosA = ( sinCcosb + cosAsinBcosc )( sinBcosC + cosAsinCcosb ) /sin A − sinBsinCcosacosA = ( sinccosb + cosAsinbcosc )( sinbcosc + cosAsinccosb ) /sin a − sinbsinccosacosBcosC − cosA = ( cosb − cosccosa )( cosc − cosbcosa ) − sin a ( cosa − cosbcosc ) sin a | sinb || sinc | = cosa ( − cos b − cos c +2 cosacosbcosc +1 − cos asin a | sinb || sinc | = cosasinBsinC. Hence 1.0. Echo Lake 22.7.84 .2. FINITE NON-EUCLIDEAN GEOMETRY.
Example.
For p = 13 , with δ = 2,let A = (0 , , , , B = (1 , , , , C = (6 , , , .cosa = 7 , cosb = 4 , cosc = 3 , | sina | = 2 , | sinb | = 5 δ, | sinc | = 3 δ.cosA = 2 , cosB = 4 δ, cosC = 2 δ, sinA = 6 , sinB = 2 δ, sinC = − δ. Theorem.
For a triangle with a right angle at A, let sinA = 1 , cosA = 0 , then we have the relations: | sinb | = | sina | sinB, | sinc | = | sina | sinC, cosBcosC = sinBsinCcosa, cosB = sinCcosb, cosC = sinBcosc. cosa = cosbcosc, Proof: 1 and 0.2 follow from 5.2.1.0.0 follows from 5.2.1.2.0.1 and 1.2 follow from 5.2.1.1 which gives 1.0, using 2.0.
Definition. An auto-dual triangle is a triangle such that A = a, B = b, C = c. Theorem.
If a triangle is auto dual, then cosA = cosBcosC sinBsinC , sinA = − sinB + sinC sinBsinC . Proof: 0 follows from Theorem 5.2.1. If we substitute cosA using 0 in sin A + cos A = 1 , we get sinA = + j sinB + sinC sinBsinC , j = +1 or − . replacing sinA and cosA by their expression in1.1, gives after multiplication by sinBsinC, sinBsinC = cos C − i ( sinBsinC + sin C ) , therefore j = − . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Notation. A = ( s, c ) , is an abbreviation for sinA = s, cosA = c. Example.
For p = 13 , let a = A = ( − , − δ ) , b = B = ( − , − , c = C = (3 , δ ) , we easily verify5.2.3.0 and .1: cosA = − δ = − . δ . − , sinA = − − − . − . Introduction.
To a given polynomial P of the third degree, we can associate a selector. The first case I willconsider is that when the polynomial has no integer roots or is primitive. To a given suchpolynomial corresponds a selector called the fundamental selector and a tri-geometry withnon-integer isotropic points and lines. To this fundamental selector we can associate others,see g25.prn, The semi-selector gives conics associated to the auto-polars, the co-selector andthe bi-selector are associated to the point-conics and line-conics through the isotropic points,the bi-selector and the co-selector to the point-conics and line-conics tangent to the isotropiclines. Examples indicated that the other selectors do not give lines or conics or, in general,cubics. It is an open question if they have any geometrical significance. Definition. If s is the selector, the selector function is a function from Z p + p +1 to Z p + p +1 given by0. f ( s j − s i ) := s i , i (cid:54) = j, f (0) = − . Theorem.
The selector for the c -lines is the co-selector of the lines. More precisely, s c ( i ) = 1 − s ( i ) . The selector function for c -lines is given by f c ( i ) = 1 − f ( − i ) . Theorem. a × b = ( f ( b − a ) − a ) ∗ . a ∗ × b ∗ = f ( b − a ) − a. .3. TRI-GEOMETRY a is on b ∗ iff f ( a + b ) = 0 or f ( a + b ) = − . the points on a ∗ are s ( i ) − a, i = 0 to p. acb = (1 − f ( b − a ) − b ) c . a c cb c = 1 − f ( b − a ) − b. a is on b c iff f ( − a − b ) = 1 . the points on a c are 1 − a − s ( i ) , i = 0 to p. Definition.
Let c ∗ := a × b . Let a = s ( i ) − c, let b = s ( j ) − c, the gap of a and b, written gap ( b, a ) := j − i mod p + 1 . Let c := a ∗ × b ∗ . Let a ∗ = s ( i ) − c, let b ∗ = s ( j ) − c, the gap of a ∗ and b ∗ , written gap ( b ∗ , a ∗ ) := j − i mod p + 1 . Theorem.
Let a be a point on b , let let a i be on b i , such that gap ( a i , a ) + gap ( b i , b ) = 0,the points a i are on a c -line through a tangent to b . Table.
The selector for some values of p and equivalent ones which are not complementary (obtainedby reversing the order are p = 3 ,
0: 0,1,3,9. 1: 0,1,4,6. p = 5 ,
0: 0,1,3,8,12,18. 1: 0,1,3,10,14,26. 2: 0,1,4,6,13,21.3: 0,1,4,10,12,17. 4: 0,1,8,11,13,17. p = 7 ,
0: 0,1,3,13,32,36,43,52. 1: 0,1,4,9,20,22,34,51.2: 0,1,4,12,14,30,37,52. 3: 0,1,5,7,17,35,38,49.4: 0,1,5,27,34,37,43,45. 5: 0,1,7,19,23,44,47,49. p = 11 ,
0: 0,1,3,12,20,34,38,81,88,94,104,109.1: 0,1,3,15,46,71,75,84,94,101,112,128.2: 0,1,3,17,21,58,65,73,100,105,111.3: 0,1,3,17,29,61,80,86,91,95,113,126.4: 0,1,4,12,21,26,45,+68,84,97,99,127.5: 0,1,4,16,50,71,73,81,90,95,101,108.6: 0,1,4,27,51,57,79,89,100,118,120,125.7: 0,1,5,12,15,31,33,39,56,76,85,98.8: 0,1,5,21,24,39,49,61,75,92,125,127.9: 0,1,5,24,44,71,74,80,105,112,120,122.10: 0,1,5,25,28,68,78,87,89,104,120,126.11: 0,1,6,18,39,68,79,82,98,102,124,126. CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
12: 0,1,8,21,33,36,47,52,70,74,76,124.13: 0,1,9,19,24,31,52,56,58,69,72,98.14: 0,1,15,18,20,24,31,52,60,85,95,107.15: 0,1,15,25,45,52,58,61,63,80,84,92.16: 0,1,16,21,24,49,51,58,62,68,80,94.17: 0,1,23,37,57,62,75,83,86,90,92,102.
Example.
Let p = 3 . If we use the selector 0,1,3,9 and use the representation on the cube (g25.prn),the complementary selector 0,1,5,11 gives the c -lines which can be classified as follows: 3 oftype V V SS,
F SSS,
F V SS,
F SV V,
V F F F,
Example.
Let p = 7 , P = I + 2 , The powers of I + 3 are: , , − , , , , − , − , − , ,
10 1 3 − , , − − , − − , − ,
15 1 2 1 , , , − , ,
20 1 0 − , − , − − , − − , ,
25 1 − − , , − , − − , ,
30 1 − , − , − , − − , − ,
35 1 − , , − , , ,
40 1 − , − , − , − , ,
45 1 − , − , − , − − , − ,
50 1 1 1 , , , − − , − ,
55 1 0 − , − , Example, p = 7 , P = I + 2 , ∗ : 0 1 7 24 36 38 49 541 ∗ : 0 6 23 35 37 48 53 56 .3. TRI-GEOMETRY ∗ : 0 17 29 31 42 47 50 5124 ∗ : 0 12 14 25 30 33 34 4036 ∗ : 0 2 13 18 21 22 28 4538 ∗ : 0 11 16 19 20 26 43 5549 ∗ : 0 5 8 9 15 32 44 4654 ∗ : 0 3 4 10 27 39 41 52The points 0,3,8,19,21,33,50,56 are on a c -line through 0.The points 0,5,16,18,30,47,53,54 are on a c -line through 0.The part proving that co-, bi- and semi-selectors are conics was proven before this date .The equation of the conics through 2 coordinate points was also obtained earlier. It remainsto prove that the 2 are identical. Lemma. If i is an element of the co-selector, the tangent is (1 − i ) ∗ . If i is an element of the bi-selector, the tangent is ( a − i ) ∗ , for some a. Proof:For 0, (1 − i ) ∗ is on i because f (1 − i ) = 0 if i is an element of the co-selector. It remainsto prove that it is the only point on (1 − i ) ∗ . For 2, by duality?
Theorem.
Let S be a selector .0. The points associated to the co-selector are on a conic which passes through theisotropic points. The points associated to the bi-selector are on a conic which is tangent to the isotropiclines. The points associated to the semi-selector are on a conic for which the isotropic triangleis a polar triangle. The conics of the same family are such that 2 distinct points determine a conic and 2distinct conics have exactly one point in common.
Proof: Le P = I + bI − c .For 0. Consider the selector associated to the line through G and G = I + g. Let G i = I + h. The corresponding point on the co-selector is G − i +1 . We obtain G − i +1 = ( g − h ) I − h ( g − h ) I + ( g ( b + h ) + c ) It is easy to check that that point is on the conic ( bg + c ) X + gX − gX X − X X and that the isotropic points are on this conic. CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Part 1, follows by duality in view of Lemma 3.2.10.1.For 2, because the line i ∗ is on the point i, the correspondance which associates i ∗ to i isa polarity and the points on their polars is a conic, the auto-polar conic. These points aresuch that f (2 i ) = 0 , where f is the selector function and therefore the solutions i are pointscorresponding to the semi-selector . In view of g142.prn, the symmetric matrix M whichrepresents the auto-conic satisfies for some values of u, v, w u gv ( g − b ) w v gw w = a b b b a b b b a g g g . The inverse of the last matrix is g − g − g . Multiplying the first matrix by this last matrix gives, because of the symmetry, u = 1 , v = 1 , w = 1 and with b = s ,M = − s . This matrix clearly associates to the pole (1 , − ( π + ρ ) , ρ ρ ) the polar, [ ρ , ρ , , because s = 0 . Answer to 5.3.1.0.
I fill in here some of the details:If ( I + g ) ∗ ( I + h ) − = uI + vI + w, then (( v + uh ) I + ( w + vh − ub ) + ( wh + uc ) = k ( I + g ) , therefore v = − uh, wh + uc = g ( w − uh − ub ) or u = g − h, v = − h ( g − h ) , w = g ( b + h ) + c. The conic through the isotropic points and through the points 0 and 1 is of the form a X + gX − X X + b X X + b X X . To insure that it passes through the isotropic points gives 3 linear equations for a , b and b . It is easiest to check a posteriori that ( bg + c ) X + gX − gX X − X X passes through the isotropic points, for instance through (1 , ρ , r ρ ) : g ( ρ ρ + ρ ρ + ρ ρ + ρ − ρ ρ ) + ( c − ρ ρ ρ ) = 0 . The point ( u, v, w ) is on the conic because ( bg + c )( g − h ) + gh ( g − h ) − g ( g − h ) w + h ( g − h ) w = ( g − h ) ( bg + c + gh − g ( b + h ) − c ) = 0 . Definition.
The mapping which associates to a point P corresponding to G k , the point Q correspondingto G − k is called the inversion mapping . .3. TRI-GEOMETRY Theorem. If P = I + bI − c, The inversion mapping T associates to ( x, y, z ) , ( X, Y, Z ) with X = bx + y − xz,Y = cx − yz,Z = ( bx − z ) + by − cxy.T ◦ T ( x, y, z )= ( c x + bcx y + b x z − cxyz − bxz + cy + by z + z ) . ( x, y, z ) ? Example. p = 7 , P = I + 2 , selector: 0 1 7 24 36 38 49 54selector function:0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23-1 0 36 54 54 49 1 0 49 49 54 38 24 36 24 49 38 7 36 38 38 36 36 124 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 470 24 38 54 36 7 24 7 49 24 24 1 0 1 0 54 24 54 7 38 49 36 49 748 49 50 51 52 53 54 55 561 0 7 7 54 1 0 38 1c-selector: 0 1 4 9 20 22 34 51c-selector fuction:0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23-1 0 20 1 0 4 51 51 1 0 51 9 22 9 20 51 4 34 4 1 0 1 0 3424 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 4734 9 51 34 51 22 4 20 34 1 0 22 22 20 20 22 51 20 9 34 22 34 20 448 49 50 51 52 53 54 55 569 9 1 0 9 4 4 22 1 ×
22 = 42 ∗ , c
22 = 39 c , ∗ is tangent at 12 to c , ∗ is tangent at 22 to c , c istangent at 12 to ∗ , c is tangent at 22 to ∗ , ∗ × ∗ = 6 , c c c = 28 . ×
28 = 30 ∗ , c
28 = 51 c , ∗ is tangent at 6 to c , ∗ is tangent at 28 to c , c is tangent at 6 to ∗ , c is tangent at 28 to ∗ , ∗ × ∗ = 14 , c c c = 50 . There appears to be no connection.the c -lines (conics through the isotropic points) are − mX + lX + 3 kX − mX X − lX X − kX X = 0 . the c -line with l = m = 0 is , ,
0; 1 , ,
0; 1 , ,
1; 1 , ,
6; 1 , ,
2; 1 , ,
5; 1 , ,
3; 1 , , or 12, 18, 19, 22, 27, 38, 40,52,the 57 others are all obtained by adding a constant, for instance, if we add 38 we get 50, 56,0, 3, 8, 19, 21, 33 or 1,1,1; 0,0,1; 1,4,5; 1,3,2; 1,2,4; 1,0,0; 1,6,1; 1,5,4which corresponds to k = m = 0 , if we add 19 we get 31, 37, 38, 41,46, 0, 2, 14, ( k = l = 0)20 CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY if we add 33 we get 45, 51, 52, 55, 3, 14, 16, 28, ( k = l = m = 1) Is there any significance to the fact that 19, 38 are , and . ?the c -points (conics tangent to isotropic lines) are kx + lx − kx − kx x − lx x − kx x = 0 . Example. p = 7 , P = I + 2 , selector, 0,1,7,24,36,38,49,54, x − x x : , , , , , , , , (0 , , , , , , , x − x x :
56, 0, 3, 8,19,21,33,50, − x − x x : I by I + 1 we obtain P = I + 3 I + 3 I + 3 , this gives the same selectors, P = I + 1 , and P = I + I − give the selectors(0,1,6,15,22,26,45,55), (0,1,3,13,32,36,43,52) and (0,1,5,7,17,35,38,49). Example.
For p = 3 and P = I + 2 , the auto-polar conics through two of the points (0,0,1), (0,1,0)and (1,0,0) are X − X X = 0 or x + 3 x x = 0 ,X − X X = 0 or x + 2 x x = 0 ,X − X X = 0 or x − x x = 0 . All 3 do not have 3 points or 3 lines in common.
Comment.
If 2 conics are in the same family and we known the tangents corresponding to the points ofone we can obtain those of the other. The sum of the corresponding representation of pointsand lines is a constant.
Program.
Examples can be studied using 130 \ TWODIM.BAS.
Introduction.
Let P = ( I + aI + b )( I + c ) , a − b N p. There is one isotropic point (1 , a, b ) and one isotropic line [ c , − c, . The isotropic point is not on the line otherwize, − c whould be a root of I + aI + b. The p + 1 ideal points are (0 , , c ) and (1 , x, c ( x − c )) . The p + 1 ideal lines are [0 , b, − a ] and [1 , x, − axb ] . In the case of the complex field, if P = ( I + 1) I, the c -lines are the circles through theorigin, it is therefore natural to call this geometry inverse geometry . .3. TRI-GEOMETRY Definition.
The pseudo-bi-selector is the set { s i } , The pseudo-semi-selector is the set { s i } , Example. p = 5 . With P = I − I − I − , a generator is I + 2 , its powers are , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , The lines are0 [1,0,0]: { } and (0,1,4)1 [1,2,4]: { } and (1,1,3) . . . ..The selector function is i f ( i ) 0 9 1 0 4 4 1 0 1 0 11 11 9 9 11 9 4 4 11 1 The isotropic line [1,1,1], the isotropic point is (1,0,3). The ideal lines are [ j ] = { j, j +6 , j + 12 , j + 18 } , for j = 0 to 5.I will now examine the case when Z p is replace by an infinite field R, for instance. p = 7 , P = I + I, G = I + 3 , , , , , , , , , , , , , , , , , , , , , , i I , i , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ I , ∗ I, ∗ , , , , , , , , , , , , , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ I , ∗ , ∗ , ∗ , ∗ , ∗ , , , , , , , , , , , , , , , , , , , , , ∗ , i I , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , i I , ∗ , , , , , , , , , , , , , , , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , i I , ∗ , ∗ , ∗ , ∗ , , , , , , , , , , , , , , , , , , , , , ∗ , ∗ , i I , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , i , , , , , , , , , , , , , , I , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , i The real isotropic point is denoted by I, the real isotropic line by i. × − − i , (0 , , × (1 , ,
5) = [1 , , . ∗ × ∗ = ( − − I , [1 , , × [1 , ,
1] = (1 , , . Observethat k I corresponds to IG (cid:48) I , with G (cid:48) = I + 3 (mod I + 1) . The selector is 0,1,7,11,29,34,46.The co-selector is 0,1,3,15,20,38,42.The pseudo-bi-selector is { , , , , , , , I } .The corresponding tangents to the bi-conic are { i, ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ } which is a member of the dual of the co-selector family .The pseudo-semi-selector is { , , , , , , I , I } .The corresponding tangents to the semi-conic are { ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , i , i } . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
The points I , I , are obtained from the ideal tangents i , i . The same can be checked if we add 1, . . . , to the values above, we get in this way 24hyperbolas, e.g. { , , , , , , I , I }{ ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , i , i } , and 24 ellipses, e.g. { , , , , , , , } , { ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ } . Hence, do we also have therefore the Theorem that a selector has p − even values and p +12 odd values? Comment.
In the case of the field R, every polynomial of degree 3 has necessarily one root. There is norestriction in assuming that it is P := I + I. In this case the isotropic points are (1,0,1),and the Euclidean isotropic points (1,i,0), (1,-i,0).
Theorem.
If the field is R and P := I + I, the transformation associated to k = − , transforms the lines into circles through the point (1,0,1). to k = 2 , transforms the lines into parabolas with focus (1,0,1). to k = , transforms the lines into equilateral hyperbolas with center (1,0,1). Proof:For 0, the conics which pass trough (1 , i, and (1 , − i, are circles. (1,0,1) is the thirdisotropic point.For 1, the conics are tangent to the isotropic line through (1 , i, and (1 , − i, which is theideal line. Because the focus of a parabola is at the intersection of the tangent through theEuclidean isotropic point, we have 1.For 2, because (1,0,1) is the pole of the opposite isotropic line which is the ideal line, (1,0,1)is the center of the conic. Because the points on the conic and the ideal line form a harmonicquatern with the pole (1 , i, and the intersection (1 , − i, with its polar, the correspondingdirections, which are those of the asymptotes to the hyperbola and therefore perpendicular.More explicitely: Theorem.
The transformation associated to the case k = − x, y,
1) or ( x, y ) , the point ( X, Y,
1) or (
X, Y ) , with ( xI + yI + 1)( XI + Y I + 1) = 1 (mod P ) . this gives X − x − x − + y , Y = − y ( x − + y . .3. TRI-GEOMETRY The point Q = ( X, Y ) and the point P = ( x, y ) are on a line through (1,0), the productof the distances to that point is 1, and the points P and Q are separated by (1,0). Proof:0, gives with I replaced by − I and I replaced by − I , − xX + x + X + yY = 0 , − xY − yX + y + Y = 0 . solving for X and Y gives easily 1.Moreover, X − Y = − x − y , Theorem.
The transformation associated to the case k = 2 associates to to the point ( x, y,
1) or ( x, y ) , the point ( X, Y,
1) or (
X, Y ) , with ( XI + Y I + 1) = ( xI + yI + 1) (mod P ) . this gives X − y − ( x − ,Y = − y ( x − . The line associated to a ( x −
1) + by + c is the parabola( − ab ( X −
1) + b ( a − b ) Y − ac ) = 4( a + b ) c ( c − b ( X − − abY ) . Proof: 0, gives with I replaced by − I and I replaced by − I , we obtain at once 1. For2, to simplify let us write the line as y = mx (cid:48) + d, with x (cid:48) := x − , m = − ab and d = − cb . Expressing y in terms of x (cid:48) gives, with X (cid:48) := X − , mX (cid:48) + ( m − Y = 2( m + 1) dx (cid:48) + 2 md and X (cid:48) + mY = − ( m + 1) x (cid:48) + d eliminating x (cid:48) gives 2. Theorem.
The transformation associated to the case k = associates to the point ( x, y,
1) or ( x, y ) , the point ( X, Y,
1) or (
X, Y ) , with ( XI + Y I + 1) = ( xI + yI + 1) (mod P ) . this gives x − Y − ( X − , y = − Y ( X − . The line associated to a ( x −
1) + by + c is the hyperbola a ( Y − ( X − ) − bY ( X −
1) + c = 0 . Proof: 0, gives with I replaced by − I and I replaced by − I , we obtain at once 1. CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY
Theorem. The lines joining the points associated to the selector and their inverse are tangent toa conic. 2 y + bz + xz = 0 or bX − Y − XZ = 0 . The lines joining the points associated to the co-selector and their inverse are tangentto a conic.
Proof: The points of the selector are I − h, their inverse is I + hI + h + b. The linethrough these points is [ − h − b, h, . Problem.
Complete a set of axioms of inverse geometry using an appropriate form of the axiom ofPappus:0. Given 2 distinct points, there exist one and only one line incident to, or passing through,the 2 points, or the points are parallel .1. Given 2 distinct lines, there exists one and only one point incident to, or on, the 2lines, or the lines are parallel .2. There exists at least one line l and two distinct points P and Q not incident to l.
3. On the line l there are exactly p points, p an odd prime .4. Given a line l and a point P not on the line, there exists one and only one line parallelto l through P.
5. Given a point P and a line l not through the point, there exists one and only one pointparallel to P on l. Introduction.
There is no ambiguity to call this also the case of 2 roots . Definition.
The selector function is a function from the set Z p ( p − − { p ) } − { p − } into Z p ( p − , with f ( i − j ) = i − j (mod p ( p − , for all i and j on [1 , , . .3. TRI-GEOMETRY Example. p = 5 The cyclic group is0,0,1 0,1,1 1,2,1 1,2,4 1,4,1 1,0,1 1,3,3 0,1,3 1,4,3 1,2,31,0,2 1,1,1 1,4,2 1,1,2 1,1,4 1,0,3 1,4,4 1,3,2 0,1,2 1,3,2.The lines are0 [1,0,0]: { }
19 [1,4,1]: { }
18 [1,2,0]: { }
17 [1,3,2]: { } . . . .The selector is i f ( i ) 0 18 18 1 0 18 7 7 7 1 0 1 If f ( j − i ) does not exist, then if j − i ≡ (mod 4) , the points are on a line through (1,4,0).If f ( j − i ) does not exist, then if j − i ≡ (mod 5) , the points are on a line through (1,0,0).Otherwize, the line if f ( j − i ) − i (mod 20) . There is no restriction in assuming that P := I − I . Definition.
The bi-isotropic point is I := (1 , − , , the isotropic point is I := (1 , , . The bi-isotropic line is i := [0 , , , the isotropic line is i := [1 , , . Theorem. The points associated to the co-selector are on a conic, the co-conic, which passesthrough the isotropic point I and is tangent to the isotropic line i at the co-isotropicpoint I . The points associated to the bi-selector are on a conic, the bi-conic, which is tangentto the co-isotropic line i and is tangent to the isotropic line i at the isotropic point I . The points associated to the semi-selector are on a conic, the semi-conic, which istangent to the isotropic line i at the co-isotropic point I and is such that the polarof the isotropic point I is the co-isotropic line i . Proof:For 0, The conic of 3.1.8.0. reduces to ( k − l ) Y + mZ + ( m − l ) Y Z + ( m − k ) ZX + ( k − l ) XY = 0 , which passes through I and for which [1,1,1] is the polar of (1 , − , . For 1,. . . .For 2,. . . . CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY Introduction.
In this case the . . .
There is one special point and a line belonging to each other. The specialpoint and the special lines are called respectively the isotropic point and the isotropic line .The other points on the isotropic line are called ideal points . The other lines through theisotropic point are called ideal line . The other points and lines are called ordinary . Theorem. There are p ideal points, p ideal lines. There are p ordinary points, p ordinary lines. The isotropic line belongs to 1 isotropic point and p ideal points. The isotropic point belongs to 1 isotropic line and p ideal lines. The ideal lines belong to . . . . The ordinary lines belong to . . . .
Comment.
In the parabolic-Euclidean or sun-geometry, among all points and lines of projective geometry,one point and a line through it are preferred, in this case this is also true, but if we representthis geometry in the Cartesian plane and choose the isotropic line as the line at infinity andthe isotropic point as the direction of the x axis, the c -lines are parabolas which have . . . . Itis therefore natural, by analogy to choose the names solar geometry and bi-solar geometry ,for the geometry in question. Lemma. X − X ( X + aX ) = 0and Y − Y ( Y + aY ) = 0implies ( X Y + X Y ) − X Y ( X Y + X Y + X Y + a ( X Y + X Y )) = 0 . Proof: The first member of 2. is the sum of the first member of 0 and 1 multipliedrespectively by Y and X . .3. TRI-GEOMETRY Theorem.
Let a = 0 , generators of T are I + 1 and I + 2 . The cyclic group of order p generated by I+bcorresponds to points on the conic X − X ( X + aX ) = 0where a := b . The cyclic group of order p generated by I + I + corresponds to points onthe conic X − X X = 0 . Proof:The point ( X , X , X ) corresponds to the polynomial X I + X I + X . The product of ( X I + X I + X ) and ( Y I + Y I + Y ) is ( X Y + X Y + X Y , X Y + X Y , X Y ) . The Theorem folllows at once from Lemma . . . . Definition.
The selector function is a function from Z p ×× Z p to Z p ×× Z p . . . . Definition.
The ideal lines can be represented by [ i ] , i ∈ Z p . The points on [ i ] are ( j, i + j mod p ) . Theorem. ( x, y ) × ( x (cid:48) , y (cid:48) ) = ( f ( x (cid:48) − x, y (cid:48) − y ) − ( x, y ) mod Z p ×× Z p ) . [ x, y ] × [ x (cid:48) , y (cid:48) ] = [ f ( x (cid:48) − x, y (cid:48) − y ) − ( x, y ) mod Z p ×× Z p ] . ( x, y ) · [ x (cid:48) , y (cid:48) ] iff f ( x (cid:48) − x, y (cid:48) − y ) = (0 , . Example.
For p = 3 , the selector function is x , , , , , , f ( x ) 0 , , , , , , With the exponents in the order exponent of b then exponent of a, points line , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , The points on the 3 ideal lines are CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY [0] = { } [1] = { } [2] = { } Example. If p = 7 and P = I then the group T is , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , A selector is e, a, b, ab , a b , a b , a b . points on [1,0,0].The conics are , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , The line is [0,1,0] with points (1 , , · (1 , , c ) = (1 , , c c ) )0 , , , , , , , , , , , , , , . Other points are on [0,0,1].They all have a contact of order 2 at (0,0,1) with tangent [1,0,0]. Definition.
If the roots are a, b, c,
0. The polynomial which has 2 of the roots corresponds to a point called isotropic point .1. The 3 lines through 2 of the 3 isotropic points are called isotropic lines .2. Any non isotropic line through an isotropic point is called an ideal line .3. Any non isotropic point on an isotropic line is called an ideal point . .3. TRI-GEOMETRY Example. p = 7 , a = 1 , b = 2 , c = 4 .0. The isotropic points are A = (1 , , , A = (1 , , , A = (1 , , .
1. The isotropic lines are a = [1 , , , a = [1 , , , a = [1 , , .
2. The generators of the group are α = (0,1,2) and β = (0 , , .
3. The ideal lines through A are [1 , ,
0] = S = { e, b , b , a , a b , a b } , [1 , ,
3] = bS , [0 , ,
3] = aS , [1 , ,
6] = abS , [1 , ,
5] = a S , [1 , ,
2] = a bS .
4. The ideal lines through A are [1 , ,
0] = S = { e, a , a , ab , a b , a b } , [1 , ,
6] = aS , [1 , ,
4] = bS , [1 , ,
3] = abS , [0 , ,
5] = b S , [1 , ,
5] = b aS .
5. The ideal lines through A are [1 , ,
0] = S = { e, a , a b , a b , a b, b , a b } , [1 , ,
5] = abS , [1 , ,
1] = aS , [1 , ,
6] = a bS , [0 , ,
6] = bS , [1 , ,
3] = ab S .
6. A selector is ( e, b , ab , a b , a b ) giving the points (0 , , , (1 , , , (1 , , , (1 , , , (1 , , on [110] , the ideal points on this line are (1,6,0), (1,6,5), (1,6,2).The 36 other lines are obtained by multiplication by any of the elements in the group,e.g. if we multiply to the left by a b , the points are (0,1,4), (1,6,4), (1,2,2), (1,3,6),(1,0,1), and the ideal points are (1,1,5), (1,5,0), (1,4,3). Notes.
We have therefore the following operations: l := P × Q, L := p × qR := P • Q, r := p • q ? earlier version CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY where • is done modulo a polynomial of degree 3. If the polynomial is primitive the propertiesare well known, what is probably new is what happens when the polynomial is not primitive.If it has 3 roots it makes sense to normalize to have the isotropic points at (1,0,0), (0,1,0)and (0,0,1), but I do not see how this can be done in view of the fact that an isotropic pointcorresponds to ( I − a )( I − b ) . Definition.
Given a polynomial of the third degree with 3 distinct roots, a line generator is a generatorof a cyclic group of order p − whose elements correspond to p − points of a line throughone of the isotropic points, the last point is an ideal point on the isotropic line which doesnot belong to the isotropic point. Definition.
Two line generators are said to be independent if they are associated to lines through distinctisotropic points.Why did I not worry about this when constructing an example and use simply distinctlines? Given a polynomial of the third degree with 3 distinct roots, there exists 2 independent linegenerators.
Comment.
I will choose the roots to be 0, 1 and -1. P = I − I. The isotropic points are A = (1 , , − , A = (1 , , , A = (1 , − , . The isotropic lines are a = [0 , , , a = [1 , , , a = [1 , − , . Conjecture.
With the choice just given, there exist an x such thatif y = − x + 1) , (1 , , x ) , (1 , , y ) , ( x + y + 1 , x + 1 , xy ) are line generators correspondingto lines [0 , , , [1 , − , , [1 , , − x + y +2 xy ] through A , A and A . .3. TRI-GEOMETRY Example. p = 3 , , , , , , , , , p = 5 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , p = 7 , roots , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , I − The lines are obtained form (0 , , , (0 , , , (0 , , , (0 , , , (0 , , or 0,0 4,3 2,4 3,4 5,5for instance, adding 2,3 modulo 6,6 , , , , , , or (1,5,4),(0,0,1),(1,5,3),(1,5,2),(1,5,5) on [1,4,0].The c -lines are all obtained from (0 , , , (0 , , , (1 , , , (1 , , , (1 , , or 0,0 5,5 3,1 1,2 2,1for instance, adding 3,2 modulo 6,6 gives , , , , , or (1,6,1),(1,6,4),(1,2,5),(1,1,2),(1,4,5). p = 11 , line generators: (1,0,1), (1,1,7), (1,10,2). p = 13 , line generators: (1,0,1), (1,1,9), (1,12,2). p = 17 , line generators: (1,0,2), (1,1,11), (1,16,4). Definition. A selector is a set of p − elements P ik Q jk which are on an ordinary line. Theorem.
Given 2 independent line generators P and Q, the isotropic lines are obtained as cosets ofthe cyclic groups generated by P, Q and P • Q .The ordinary lines are obtained by multiplication modulo P , P l Q m by the elements of aselector. We may want to put this in a section on triangular geometry .In this geometry we have ordinary, ideal and isotropic points, ordinary, ideal and isotropiclines, and c -lines. These are represented by conics through the isotropic points, the ideal c -lines are the degenerate conics consisting of an isotropic line and an ideal line to the opositeisotropic point. The isotropic c -lines are the degenerate conics consisting of two isotropic CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY lines. The lines and the c -lines can be interchanged. If the c -lines are considered as lines,then the lines are c -lines, in other words if we start with a geometry where we define the conicsthrough 3 given points as lines, the conics are represented by lines. Pascal’s Theorem givesthe following, consider a line l, with points P i on a i , and 3 other points P , P , P , this linecan be considered a c -conic, indeed, the c -lines through successive points are the degenerate c -lines or ideal c -lines, a = ( A × P ) + a , a = ( A × P ) + a , a = ( A × P ) + a ,a = ( A × P ) + a , a = ( A × P ) + a , a = ( A × P ) + a . The c -Pascal points are Q = ( A × P ) × ( A × P ) , Q = ( A × P ) × ( A × P ) , Q =( A × P ) × ( A × P ) . These points are on a conic, with A , A , A because the Pascal line for the sequence A ,Q , A , Q , A , Q , gives the Pascal points P , P , P . This can be used to study what couldbe called a bi-triangular geometry . Comment.
The c -lines can be deduced from the line by the transformation which associates, in general,to ( X , X , X ) , ( X X , X X , X X ) , a conic becomes then a quadric with double points,isolated or not in the case or a real field. A conic through A , A , but not through A , becomesa quadric which degenerates in a , a and a conic through A , A but not A . Comment.
We could choose as isotropic points, in a model of this geometry in the Euclidean plane, withCartesian coordinates, by choosing one of them at the origin, and the 2 others at the directionof the axis. The c -lines are then hyperbolas passing through the origin, with asymptotes inthe direction of the axis. Problem.
Study the axiomatic of the triangular geometry and obtain Theorems in it. Circles could beconics through 2 of the isotropic points.
Problem.
Study the axiomatic of the triangular bi-geometry and obtain Theorems in it.
Comment.
The analysis can be repeated in the form of Euclidean geometry by considering the non-homogeneous points ( x, y ) and the homogeneous lines [ a, b, c ] . This can be done directly orinfered from the cases 1, 2 and 4 above, with one of the isotropic lines playing the role of theline at infinity in Euclidean geometry. .3. TRI-GEOMETRY
In G45, I give a special case of the following Theorem valid when s = a = 0 , this generalizesthe Theorem, with b = s and c = s . It was obtained earlier.
Theorem.
The symmetric functions of the roots are s := ρ + ρ + ρ = a,s := ρ ρ + ρ ρ + ρ ρ = b,s := ρ ρ ρ = c,s := ρ + ρ + ρ = a − b,s := ρ ( ρ + ρ ) + ρ ( ρ + ρ ) + ρ ( ρ + ρ ) = ab − c,s := ρ + ρ + ρ = a ( a − b ) + 3 c,s := ac,s := b − ac,s := a ( ab − c ) − b ,s := a ( a − ab + 4 c ) + 2 b . Theorem. The conic which pass through the isotropic points is k (( b − ac ) X + bX + 3 X +2 aX X + 2( a − b ) X X − (3 c − ab ) X X )+ l ( bcX + 3 cX + aX +2 bX X − (3 c − ab ) X X + 2 acX X )+ m (3 c X + acX + ( a − b ) X − (3 c − ab ) X X + 2( b − ac ) X X + 2 bcX X ) = 0 , which is tangent to the isotropic lines is k (3 x + ( a + b ) x + acx − (3 c + ab ) x x + 2 bx x − ax x )+ l ( ax + a ( a − b ) + 3 c ) x + ( a − b ) cx − ( a ( ab + c ) − b ) x x − (3 c − ab ) x x − a − b ) x x )+ m (( a − b ) x + a ( a − ab + 4 c ) x + ( a ( a − b ) + 3 c ) cx +( a ( − a b + 3 b − ac ) − bc ) x x + ( a ( ab − c ) − b ) x x − ( a (2 a − b ) + 3 c ) x x ) = 0 . Proof:The degenerate conics through the isotropic points are α ( ρ X + ρ X + X )( ρ X + ρ X + X ))+ α ( ρ X + ρ X + X )( ρ X + ρ X + X ))+ α ( ρ X + ρ X + X )( ρ X + ρ X + X )) = 0 . If we choose, in succession, α = α = α = 1, α = ρ , α = ρ , α = ρ , and α = ρ ,α = ρ , α = ρ , CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY we obtain respectively the expressions whose coefficients are k, l and m. Similarly for the c -points we start with the degenerate conics tangent to the isotropic lines, which are α ( x − ( ρ + ρ ) x + ρ ρ x )( x − ( ρ + ρ ) x + ρ ρ x ))+ α ( x − ( ρ + ρ ) x + ρ ρ x )( x − ( ρ + ρ ) x + ρ ρ x ))+ α ( x − ( ρ + ρ ) x + ρ ρ x )( x − ( ρ + ρ ) x + ρ ρ x )) = 0 . The following Theorem was develppoed to prove the relation between the conics associatedto the co, bi and semi-selectors but were found not to be needed. It is now an answer to anexercise.
Answer to exercise .
Let s = 0, the conic through (0,1,0), (0,0,1)0. which passes through the isotropic points is s X − X X = 0 .
1. which is tangent to the isotropic lines is x − s x x + s x x = 0 . or ( s X + s X ) − s X X = 0 ,
2. which has the isotropic triangle as polar triangle is s X − s X X + 2 X X = 0 . Proof: Using3. ρ + ρ = − ρ , we can checkfor 0, ρ ρ ρ − ( ρ + ρ ) ρ ρ = 0 , for 1, ρ − ρ ρ ρ ρ + ( ρ ρ + ρ ρ + ρ ρ ) ρ = 0 , for 2, ρ ρ ρ = s − s − s − ρ − ρ ρ ρ . Exercise.
Let s = 0 . Determine the conic through (0,1,0), (0,0,1),0. which passes through the isotropic points,1. which is tangent to the isotropic lines,2. which has the isotropic triangle as polar triangle. .3. TRI-GEOMETRY
Comment.
To obtain the statements of the preceding Theorem, I will illustrate for the case 2. If theconic is represented by the symmetric matrix γ βγ αβ α .The condition that I is the pole of i gives µ ρ = 1 − γ ( ρ + ρ ) + βρ ρ ,µ ρ = γ + αρ ρ ,µ = β − α ( ρ + ρ ) . Eliminating µ from the first 2 and the last 2 equations gives − γ ( ρ + ρ ) + βρ ρ − γρ − αρ ρ ρ = 0 ,γ + αρ ρ − βρ + α ( ρ ρ + ρ ρ ) = 0 , or using 3, βρ ρ − αs = 0 ,γ + αs − βρ = 0 . Because the conic cannot depend on individual values of ρ , ρ , ρ , β = 0 and then α = s and γ = − αs . Example.
Let the roots be 0,1,2,3 and p = 5,the isotropic points are (1,-1,1,-1), (1,0,1,0), (1,1,-2,0), (1,2,2,0). P = I − I + I − I, therefore, I = I (mod P ) and I = I m odP . The cubic surface is given by (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k l m nZ + T Y + Z X + Y XX − Z T − Y Z − x YY + Z X + Y X + T Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . For instance, a point on I × I is ( u + v, − u, u + v, − u ) and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k l m nv v v u + v − uv v v u + v (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 , because 2 rows are equal.Similarly for a point on I × I , ( u + v, u + v, u − v, , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k l m n u − v u − v u + 2 v u + v − u + 3 v − u − v u − v u + v u − v u + 2 v u + v u − v (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 , because the sum of the last 3 rows is equal to . No other conditions are needed to obtain a family of cubics with 3 parameters because if theisotropic points are chosen as (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) the cubic is CHAPTER 5. FINITE NON-EUCLIDEAN GEOMETRY d X X X + d X X X + d X X X + d X X X = 0 . Example. For the semi-transformation, let p = 5 and P = I − I + I − I, ( X, Y, Z, T ) = ( x, y, z, t ) gives x = 2 XT + 2 Y Z + (2 XZ + Y ) y = 2 Y T + Z − (2 XZ + Y ) + X ,z = 2 ZT + (2 XZ + Y ) + 2 XY,t = T . A plane { k, l, m, n } is therefore transforned in the quadric represented by the symmetricmatrix l m k − l + m km k − l + m k lk − l + m k l mk l m n . The isotropic points are (1,-1,1,-1), (1,0,1,0),(1,1,-2,0), (1,2,2,0),and the corresponding isotropic planes are { } , { } , { -2,-1,2,1 } , { } .It is easy to check the latter are the polar of the former, independently from k, l, m, n. Itis easy to verify that the quadric which have the isotropic tetrahedron as polar tetrahedronform a 3 parameter family and that this generalizes to n dimensions.
Exercise.
Study the relation which exist between the correspondance between a pair of points and thepair obtained at the intersection of the tangents at the 2 points to the c -line through thesepoints and the intersection of the c -lines tangent to the line through the 2 points at the twopoints.Hint: Study first how to obtain from the point on any other line through a point and the c -line that line through the point which is tangent to the c -line. Definition.
Let l c be a line through P, . . . . hapter 6GENERALIZATION TO 3DIMENSIONS I will sketch here part of the generalization to 3 dimensions of what has been presented inthe preceding parts. It will be obvious how to generalize further to n dimensions. After abrief look at the history, I will review the application of Grassmann algebra to the incidenceproperties of the fundamental objects in 3 dimensions, the points, the lines and the planes.The finite polar geometry will be introduced in section 6.2. It is obtain by preferinga plane, the ideal plane, to which correspond the notions of affine geometry, parallelism,mid-points, equality of segments on parallel lines, and a quadric, the fundamental quadric,which, together with the ideal plane, allow for the definition of spheres and therefore equalityof distances between unordered pairs of points as well as orthogonality or more generallyequality of angles between ordered pair of lines.To illustrate properties in 3 dimensions, the geometry of the triangle in involutive geometrywill be generalized, in section 6.2.3 to the study of the general tetrahedron in finite polargeometry. In the classical case, the first work on the subject is that of Prouhet, this wasfollowed by important memoirs of Intrigila and Neuberg.We will see that a special case occurs very naturally, that of the orthogonal tetrahedron,studied in section 6.2.4. We will see that the success of the theory of this special case isexplained by the generalization to 3 dimension of the symmetry which exists in 2 dimensionswhen we exchange the barycenter and the orthocenter.The isodynamic tetrahedron is studied in section 6.2.5.The generalization of many other 3 dimensional and n dimensional concepts is left to thereader.This part ends with an introduction to the anti-polar geometry 6.1.5. CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Introduction.
In the classical case, the extension to 3 dimensions is already given by Euclid. The earlierdefinitions of conics derives from the circular cone in 3 dimensions. Of note is also the factthat the 2-dimensional Desargues’ theorem derives directly from the incidence properties in3 dimensions. Although the algebraic notation of analytic geometry, introduced by Descartesimmediately extends and at once suggests to go beyond the observable to 4 and to n dimen-sions, it is not suitable if we progress from finite polar geometry - where equality of distanceand angles are defined and are not primary notions - to finite 3 dimensional Euclidean ge-ometry. Instead, I will use the notation of exterior algebra introduced by Grassmann. Introduction.
After introducing in 6.0.2 the algebraic representation of points, lines and planes in Z p , Irecall the basic concepts and properties of the exterior algebra of Grassmann (6.0.2 to 6.0.2),I define the incidence relations (6.0.2) and derive the associated properties (6.0.2, 6.0.2 to6.0.2). Definition.
The points and planes in 3 dimensions will be represented using 4 homogeneous coordinates.(Not all coordinates are 0, and if all coordinates are multiplied modulo p by the same nonzero element in Z p , we obtain the same point or plane.)Points will be denoted by a capital letter and the coordinates will be placed between parenthe-sis. Planes will be denoted by a capital letter preceded by the symbol “ | “ or by a calligraphicletter and the coordinates will be placed between braces.The lines will be represented by 6 homogeneous coordinates [ l , l , l , l , l , l ] , such that l l + l l + l l = 0 . This part of the definition will be justified in 6.0.2.4 and 6.0.2.The normalization will again be such that the leftmost non zero coordinate is 1.
Example.
Let p = 7 ,P := (2 , , ,
1) = (1 , , , , l := [3 , , , , ,
5] = [1 , , , , , , Q := { , , , } = { , , , } are respectively a point, aline and a plane. Notation.
To define algebraically the incidence properties, I will use Grassmann algebra with exteriorproduct multiplication. If e , e , e , e are “unit” vectors, we write .0. INTRODUCTION. P := ( P , P , P , P ) := P e + P e + P e + P e . l := [ l , l , l , l , l , l ]:= l e ∨ e + l e ∨ e + l e ∨ e + l e ∨ e + l e ∨ e + l e ∨ e , with2. l l + l l + l l = 0 . Q := {Q , Q , Q , Q } := Q e ∨ e ∨ e + Q e ∨ e ∨ e + Q e ∨ e ∨ e + Q e ∨ e ∨ e . In each case not all coefficients are zero.The specific notation for l and Q will is justified in 6.0.2. For Q the order of the unitvectors is chosen in such a way that the last ones are consecutive, e , e , e , e . Ifcondition 2 is not satisfied, the 2-form will be denoted using an identifier starting witha lower case letter and followed by “ (cid:48) ”. If an identity is satisfied for the general 2-form l (cid:48) as well as for the line l, I will use the notation l (cid:48) (see for instance 6.0.2).I recall: Definition.
The exterior product is defined by using the usual rules of algebra, namely, commutativity,associativity, neutral element property and distributivity with the exception e i ∨ e j = − e j ∨ e i which gives, in particular, e i ∨ e i = 0 . Lemma. ( e i ∨ e j ) ∨ ( e k ∨ e l ) = ( e k ∨ e l ) ∨ ( e i ∨ e j ) . Theorem. P ∨ Q = − Q ∨ P, l (cid:48) ∨ m (cid:48) = m (cid:48) ∨ l (cid:48) . Corollary. P ∨ P = 0 . Definition.
Given any expression involving points, lines or planes using the Grassmann representation,the dual of an expression is obtained by replacing the coefficient by itself and e i ∨ . . . ∨ e i k − by j e i k ∨ . . . ∨ e i where i , . . . , i k − , i k , . . . i is a permutation of 0,1,2,3 and j = 1 if the permutation is even, − if the permutation is odd. CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Theorem. dual ( P ) = P e ∨ e ∨ e + P e ∨ e ∨ e + P e ∨ e ∨ e + P e ∨ e ∨ e . dual ( l (cid:48) ) = l e ∨ e + l e ∨ e + l e ∨ e + l e ∨ e + l e ∨ e + l e ∨ e . Because of the notation 6.0.2.1, duality, for a line, simply reverses the order of thecomponents of l. Notation. P · Q := Q · P := dual ( P ∨ Q ) , l (cid:48) ∧ P := P ∧ l (cid:48) := dual ( dual ( P ) ∨ dual ( l (cid:48) )) . P ∧ Q := Q ∧ P := dual ( dual ( P ) ∨ dual ( Q )) . Definition. A point P is incident to a line l iff P ∨ l = 0 . A point P is incident to a plane Q iff P ∨ Q = 0 . A line l is incident to a plane Q iff l ∧ Q = 0 . Theorem. P ∨ Q = ( P Q − P Q ) e ∨ e + ( P Q − P Q ) e ∨ e +( P Q − P Q ) e ∨ e + ( P Q − P Q ) e ∨ e +( P Q − P Q ) e ∨ e + ( P Q − P Q ) e ∨ e . P ∨ Q = ( P Q − P Q ) e ∨ e ( P Q − P Q ) + e ∨ e +( P Q − P Q ) e ∨ e + ( P Q − P Q ) e ∨ e +( P Q − P Q ) e ∨ e + ( P Q − P Q ) e ∨ e . P ∨ l (cid:48) = ( P l + P l + P l ) e ∨ e ∨ e +( − P l + P l − P l ) e ∨ e ∨ e +( − P l − P l + P l ) e ∨ e ∨ e +( − P l + P l − P l ) e ∨ e ∨ e . P ∨ l (cid:48) = ( P l + P l + P l ) e ∨ e ∨ e +( −P l + P l − P l ) e ∨ e ∨ e +( −P l − P l + P l ) e ∨ e ∨ e +( −P l + P l − P l ) e ∨ e ∨ e . l (cid:48) ∨ m (cid:48) = ( l m + l m + l m + l m + l m + l m ) e ∨ e ∨ e ∨ e . P ∨ Q = ( P Q + P Q + P Q + P Q ) e ∨ e ∨ e ∨ e . .0. INTRODUCTION. P ∨ ( P ∨ l (cid:48) ) = 0 . Q ∨ ( Q ∧ l (cid:48) ) = 0 .
8. 0. ( P ∨ l (cid:48) ) ∧ l (cid:48) = − ( l l + l l + l l ) P. ( P ∨ l ) ∧ l = 0 .
9. 0. ( Q ∧ l (cid:48) ) ∨ l (cid:48) = − ( l l + l l + l l ) Q . ( Q ∧ l ) ∨ l = 0 . The proof is straightforward or follows from duality.The condition 6.0.2.2 that a sextuple be a line is precisely chosen to insure 8.1 and 9.1.
Example.
For p = 7 , given P := (1 , , , , P := (1 , , , , P := (1 , , , , P := (1 , , , ,l := [1 , , , , , , l := [1 , , , , , , Q := { , , , } , Q := { , , , } , we can easily verify P and P are incident to l , P and P are incident to l ,P , P , P , l and l are incident to Q , P and l are incident to Q . Notation.
As for 2 dimensional finite projective geometry, we will make use of a compact notation,assuming that the elements are ordered as if the 4 or 6 normalized coordinates were formingan integer in base p. We have the correspondence (0) := (0 , , , , [0] := [0 , , , , , , (1) := (0 , , , , [1] := [0 , , , , , , ( p + 1) := (0 , , , , [ p + 1] := [0 , , , , , , ( p + p + 1) := (1 , , , , [ p + p + 1] := [0 , , , , , , [ p + p + p + 1] := [0 , , , , , , [ p + p + p + p + 1] := [1 , , , , , . Example.
Continuing Example 6.0.2, P = (180) , P = (65) , P = (107) , P = (58) , l = [7222] , l = [17509] , Q = { } , Q = { } . Theorem. P and Q are distinct iff P ∨ Q (cid:54) = 0 . Proof: By Corollary 6.0.2, if P and Q are not distinct, Q = kP, k (cid:54) = 0 and P ∨ Q = P ∨ kP = 0 . If P ∨ Q = 0 , let P be a coefficient of P different from 0, 6.0.2.0 gives P Q = P Q , P Q = P Q , P Q = P Q , therefore if Q = 0 then Q = Q = Q = 0 , and Q is not a point.If Q (cid:54) = 0 , I can, by homogeneity choose Q = P and then Q = P , Q = P and Q = P or Q = P. CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Theorem.
Given 2 distinct points P and Q, there exist one and only one line l = P ∨ Q incident to P and Q. Proof: Because of associativity, P ∨ ( P ∨ Q ) = 0 and ( P ∨ Q ) ∨ Q = 0 , therefore l = P ∨ Q is incident to both P and Q and l (cid:54) = 0 because P and Q are distinct.The line is unique. Let P ∨ Q (cid:54) = 0 , P ∨ l = Q ∨ l = 0 , l (cid:54) = 0 . Because P ∨ Q (cid:54) = 0 , one of the coordinates is different from 0, let it be P Q − P Q . Theorem 6.0.2.1 gives 4 equations associated to P ∨ l = 0 and 4 equationsassociated to Q ∨ l = 0 , the last equations are − P l + P l − P l = 0 − Q l + Q l − Q l = 0 . Multiplying the first by − Q and the second by P gives0. ( P Q − P Q ) l = ( P Q − P Q ) l , Similarly multiplying by − Q and P gives1. ( P Q − P Q ) l = ( P Q − P Q ) l , The third equation of each set gives similarly2. ( P Q − P Q ) l = ( P Q − P Q ) l , ( P Q − P Q ) l = ( P Q − P Q ) l , If we add the first equations for P ∨ l = 0 multiplied by − Q and − Q , we get4. ( P Q − P Q ) l = − Q P l + Q P l + Q P l + Q P l = 0 . Because l (cid:54) = 0 , the first parenthesis is different from 0, otherwise, it follows from 0,1, 2 and 3 then l = l = l = l = 0 , and from 4 that l = 0 . We can, because ofhomogeneity write l = P Q − P Q , it follows that l = P Q − P Q .l = P Q − P Q . and from 2 and 3, l = P Q − P Q ,l = P Q − P Q , Replacing in 4, gives ( P Q − P Q ) l = Q P ( P Q − P ) Q ) − Q P ( P Q − P Q )+ Q P ( P Q − P Q ) + Q P ( P Q − P Q )= ( P Q − P Q )( P Q − P Q ) . hence l = P Q − P Q . Therefore l = P ∨ Q. .0. INTRODUCTION. Theorem.
Given a point P and a line l, not incident to P, there exists one and only one plane Q = P ∨ l incident to P and l. Proof: Because of associativity, 6.0.2 and 6.0.2.4.1, P ∨ ( P ∨ l ) = ( P ∨ l ) ∧ l = 0 , therefore Q = P ∨ l is incident to both P and l and Q (cid:54) = 0 because P and l are not incident.The plane is unique, if P ∨ Q = l ∨ Q = 0 and Q (cid:54) = 0 , let Q be a coefficient of Q (cid:54) = 0 .P ∨ Q = 0 and l ∨ Q = 0 give P Q + P Q + P Q + P Q = 0 , Q l + Q l + Q l = 0 , −Q l + Q l − Q l = 0 , −Q l − Q l + Q l = 0 , −Q l + Q l − Q l = 0 . Multiplying the equations respectively by l ,
0, 0, − P and P and adding gives using homo-geneity, and the same argument used in the preceding Theorem, Q = P l + P l + P l , Q = − P l − P l + P l . Similarly, if we multiply respectively by l , , P , and − P and then add, Q = − P l + P l − P l , and if we multiply respectively by l , − P , P and 0 and then add, Q = − P l − P l + P l . Therefore Q = P ∨ l. Using duality, it is easy to deduce from 6.0.2 and 6.0.2.
Theorem.
Given 2 distinct planes P and Q , there exist one and only one line l = P ∧ Q incident to P and Q . Theorem.
Given a plane Q and a line l, not incident to Q , there exists one and only one point P = Q ∧ l incident to Q and l. Lemma. If l and m are lines, then( l m − l m )( l m + l m + l m ) + ( l m − l m )( l m − l m )= l m ( l m + l m + l m + l m + l m + l m ) . Lemma. If l = [0 , , , l , l , l ] and m = [ m , m , m , , , , then l and m have a point P in commoniff l ∨ m = 0 and P = (0 , m , m , m ) . Proof: The 4 conditions associated with P ∨ l = 0 give, because not all l , l and l canbe 0, P = 0 and P l + P l + P l = 0 . The 4 conditions associated with P ∨ m = 0 give, CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS because not all m , m and m can be 0, P = m , P = m , P = m , substituting in theremaining equation gives the equivalence with l ∨ m = 0 . Lemma. If l m (cid:54) = 0 and l m (cid:54) = l m , if P is on l and m , then l and m have a point P in commoniff l ∨ m = 0 and P = ( l m + l m + l m , l m − l m , l m − l m , l m − l m ) . Proof: The first component of P ∨ l = 0 and of P ∨ m = 0 implies P = k ( l m − l m ) , P = k ( l m − l m ) and P = k ( l m − l m ) . Similarly, the second components implies P = k ( l m − l m ) , P = k ( l m − l m ) and P = k ( l m − l m ) . Consistency implies ( l m − l m )( l m − l m ) = ( l m − l m )( l m − l m ) . or l m ( l m + l m + l m + l m ) + m ( − l l − l l )+ l ( − m m − m m )= l m ( l m + l m + l m + l m + l m + l m ) , because l and m are lines. Chosing k = ( l m − l m ) / ( l m − l m ) and k = − gives theexpression for P using Lemma 6.0.2. Theorem.
If 2 distinct lines l and m are such that l ∨ m = 0 , they are incident to a point noted l ×× m andto a plane noted l X m. Vice-versa, if 2 distinct lines are incident to the same point or the sameplane then l ∨ m = 0 . Moreover, if l = [ l , l , l , l , l , l ] and m = [ m , m , m , m , m , m ] , then one of the following will give the point l ×× m ( l m + l m + l m , l m − l m , l m − l m , l m − l m ) . ( l m − l m , l m + l m + l m , l m − l m , l m − l m ) , ( l m − l m , l m − l m , l m + l m + l m , l m − l m ) . ( l m − l m , l m − l m , l m − l m , l m + l m + l m ) . and the plane l X m { l m + l m + l m , l m − l m , l m − l m , l m − l m } . { l m − l m , l m + l m + l m , l m − l m , l m − l m } . { l m − l m , l m − l m , l m + l m + l m , l m − l m } . { l m − l m , l m − l m , l m − l m , l m + l m + l m } . The proof follows by a judicious application of the Lemmas.
Exercise. If l and m are lines and l ∨ m = 0 then l ∨ ( l ×× m ) = ( l ×× m ) ∨ m = 0 . .1. AFFINE GEOMETRY IN 3 DIMENSIONS. Exercise. If l ×× m and l X m are defined by 6.0.2.0 and .4, then ( l ×× m ) ∨ ( l X m ) = 0 . To define a 3 dimensional Euclidean geometry, I will start with a preferred plane I to whichare associated the notions of affine geometry. Just as in the case of the Pappian plane, wecan define the notions of parallelism, mid-point, equality of segments (ordered pair of points)on parallel lines. It is convenient to intoduce a matrix notation to express parallelism andin later sections polarity and orthogonality. Bold faced letters will be used for matrices. Thecoordinates of points, lines and planes will have associated with them vectors which will beconsidered as row vectors. Definition.
The preferred plane is called the ideal plane . There is no restriction in chosing the ideal plane I = { , , , } because the coordinates of I simply corresponds to those of the unit point(1,1,1,1) and can be considered as the polar of the unit point with respect to the tetrahedronof the coordinate system (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1). Definition.
The points in the ideal plane are called the ideal points , the lines in the ideal plane are calledthe ideal lines . Definition.
Two lines are parallel iff they are incident to I at the same point.Two planes are parallel iff they are incident to I on the same line.A plane Q and a line l are parallel iff the line Q ∧ I is incident to the point
I ∧ l. Definition.L := − − − − − − , P := − − − −
11 0 − − . CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Theorem. If l is a line then L l T is the direction of the line l .If P is a plane then P P T is the direction of the plane P . The proof follows from 6.0.2.3 and 1. Notice that P is obtained from the transpose of L by exchanging row i with row − i , i = 0,1,2. Theorem.
Let Q := {Q , Q , Q , Q } and l := [ l , l , l , l , l , l ] , The plane Q is parallel to the line l iff Q L l T = 0 , or − Q ( l + l + l ) + Q ( l − l + l ) + Q ( l + l − l ) + Q ( l − l + l ) = 0 . The proof follows from the property that the direction L l T of the line l is incident to theplane Q . Alternately, we can obtain the Theorem by introducing first, directional correspon-dance. Definition.
Let P := ( P , P , P , − ( P + P + P )) be an ideal point. Let m := [ m , m , m , m , m , m ] be an ideal line. These point and line can also be determinedby 3 well chosen coordinates. The coordinates of points are placed between double parenthesisand that of lines, between double brackets, while the point P as viewed as a point in the planeis denoted ( P ) and the line as [ m ] .One of the good choices is [[ m , − m , m ]] , indeed, in the ideal plane, m e + m e + m e is the dual of m e ∨ e + m e ∨ e + m e ∨ e , while the 3 chosen components give m e ∨ e − m e ∨ e + m e ∨ e . The other components are m = − m − m , m = m − m ,m = m + m . We have P T = U ( P ) T and [ m ] T = V m T , with U = − − − and V = − . The correspondence which associates to P = ( P , P , P , P ) in the 3 dimensional spacethe point ( P ) = ( P , P , P ) in the 2 dimensional plane I , and which associates to the line m = [ m , m , m , m , m , m ] , in the 3 dimensional space the line [ m ] = [[ m , − m , m ]] , inthe 2 dimensional plane I , is called the directional correspondence . Theorem.
The directional correspondence is a homomorphism from the 3 dimensional space onto theideal plane.
Theorem.V P = U T . .2. POLAR GEOMETRY IN 3 DIMENSIONS. Theorem. If P and m are in the ideal plane, P is on m, iff ( P ) is on [ m ] , iff( P ) · ( m ) = (( P , P , P )) · [[ m , − m , m ]] = P m − P m + P m = 0 . Alternate Proof of I l and line i Q of Theorem 6.1.1, correspond, the point ( I l ) and theline [ i Q ] in I . ( I l ) = (( l + l + l , − l + l − l , − l − l + l )) , [ i Q ] = [[ Q − Q , Q − Q , Q − Q ]] , Q is parallel to l iff − ( l + l + l )( Q − Q ) + ( − l + l − l )( Q − Q ) − ( − l − l + l )( Q − Q ) = 0 which is 6.1.1.0. Theorem.
The mid-point of two points A = ( a , a , a , a ) and B = ( b , b , b , b ) is( b + b + b + b ) A + ( a + a + a + a ) B. Notation.
The mid-point of A and B is denoted by A + B . Exercise.
Generalize the construction of the polar p of a point P with respect to a triangle to that ofthe polar P of a point P with respect to a tetrahedron and prove that if P = ( p , p , p , p ) then P = {P P P , P P P , P P P , P P P } . To define a polar geometry in 3 dimensions, I will start with an affine Geometry in 3 di-mansions and a preferred non degenerate quadric θ which is not tangent to the ideal plane I .Using the ideal plane and the prefered quadric we can define orthogonality, spheres, centers,equality of pairs of points and lines, . . . .The preferred quadric is represented by a symmetric 4 by 4 matrix F , which associates toa 4-vector representing a point (pole), a 4-vector representing a plane (polar). Its adjoint G gives the correspondance from polar to pole. From F , we can derive a 6 by 6 matrix H whichgives the correspondance between a line and its polar. From F we can also derive a polarityin the ideal plane represented by a 3 by 3 matrix J , giving the correspondance from pole topolar and its adjoint K , giving the correspondance from polar to pole from which we canderive perpendicularity between a line and a plane, the direction of the line giving the poleand the direction of the plane giving the polar. The 6 by 6 matrix J , derived from J , allows CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS for a direct check of the orthogonality of 2 lines and the 4 by 4 matrix K , derived from K ,allows for a direct check of the orthogonality of 2 planes. Introduction.
The properties of pole and polar are properties in Pappian Geometry. They are easily gener-alized by using a 4 dimension collinearity which transforms the fundamental quadric into anarbitrary quadric or by chosing a coordinate system with the four base points on the quadric.
Definition.
The preferred quadric is called the fundamental quadric . There is no restriction in chosingthe fundamental quadric θ as follows, because it simply assumes that the quadric passesthrough the base points (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1). Let0. F := n n n n n n n n n n n n , Θ := ( X , X , X , X F ( X , X , X , X T = n X X n X X n X X n X X n X X n X X . The condition of non degeneracy and non tangency are2. d := det ( F ) = n n + n n + n n − n n n n + n n n n + n n n n ) (cid:54) = 0 , t := n n n + n n n + n n n + n n n + 2 n n ( n + n )+ 2 n n ( n + n ) + 2 n n ( n + n ) − ( n n + n n + n n ) n (cid:54) = 0 , where n := n + n + n + n + n + n . The condition t (cid:54) = 0 will be verified in 6.2.1. Definition.
In polar geometry, the points in the ideal plane which are not on the quadric are called the ideal points , all lines in the ideal plane which are not tangent to the quadric are the ideallines . Definition.
The points in the ideal plane and the quadric are the isotropic points . The lines in the idealplane tangent to the quadric are the isotropic lines .If the isotropic points are real, the polar geometry is said to be hyperbolic , if there are no .2. POLAR GEOMETRY IN 3 DIMENSIONS. real isotropic points, it is said to be elliptic , if there is exactly one isotropic point, the quadricbeing tangent to the plane, the geometry is said to be parabolic .Again as for the involutive geometry, I will not study the parabolic case and will studytogether the elliptic and hyperbolic case.
Definition.
The polar of the point P = ( P , P , P , P ) , is the plane Q T := F P T = { n P + n P + n P , n P + n P + n P ,n P + n P + n P , n P + n P + n P } .P is called the pole of the plane Q . Theorem. P T = G Q T where G is the adjoint of F : G = n n n n ( n n − n n − n n ) n ( n n − n n − n n ) n ( n n − n n − n n ) n ( n n − n n − n n ) 2 n n n n ( n n − n n − n n ) n ( n n − n n − n n ) n ( n n − n n − n n ) n ( n n − n n − n n ) 2 n n n n ( n n − n n − n n ) n ( n n − n n − n n ) n ( n n − n n − n n ) n ( n n − n n − n n ) 2 n n n . Theorem. I is tangent to the fundamental quadric iff(1 , , , G { , , , } T = 0or t = 0 . where t is defined in 6.2.1 . t is simply the sum of all the elements of G . Theorem. If Q is on the polar P of P then its polar Q is incident to P. The proof is left as an exercise. The theorem justifies the following definition:
Definition.
A line m is a polar of a line l iff it is the line common to all the polars of the points of l. CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Theorem.
Let H := n n − n n n n − n n n n n n − n − n n n n − n n n n n n − n n n n n − n n n n − n n − n n n n n − n n − n n − n n n − n n n n − n n − n n − n − n n − n n n n − n n n n − n − n n − n n n n − n n n n − n n , then the polar m of l is given by m T = H l T . Moreover
H H = d I , where I is the identity matrix and d is the determinant of F given in 6.2.1.3. The proof is left as an exercise. As a hint, consider 2 points P and Q on l = P ∨ Q andtheir polar F P and F Q . Definition.
The center of a quadric is the pole of I , Example.
The pole of { } is (2 n n n , n ( n n − n n − n n ) , n ( n n − n n − n n ) ,n ( n n − n n − n n )) . The center of the fundamental quadric is (2 n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) . Introduction.
Prefering both an ideal plane and a fundamental quadric allows us to define othogonalityof lines and planes with lines and planes. After defining the polarity in the ideal plane,induced by the fundamental quadric, we use it to derive the 3 conditions which express theorthogonality of lines and planes and the condition which express the orthogonality of 2 linesor of 2 planes.
Definition. A line is orthogonal to a plane iff the polar of its ideal point is incident to the ideal line ofthe plane. A line is orthogonal to a line iff the polar of its ideal point is incident to the ideal .2. POLAR GEOMETRY IN 3 DIMENSIONS. point of the other line. A plane is orthogonal to a plane iff the ideal line of one is the polarof the ideal line of the other. Definition.
The ideal polarity is the polarity induced in the ideal plane I by the polarity defined by thequadric θ . Notation.
It is sometimes convenient to use an other notation for the elements of the fundamentalquadric. n ij = n ji is the coefficient of X i X j in the equation of the fundamental quadric, more specifi-cally, n := n , n := n , n := n , n := n , n := n and n := n . The elements of the matices K and of K are more easily expressed in terms of i ii and i ij , using ijkl for permutation of 0123, i ii = − ( n kl + n lj + n jk + 2( n lj n kl + n kl n jk + n jk n lj ) ,i ij = ( n ik − n il )( n jk − n jl ) + n kl (3 n ij + 2 n kl − n ) , with n := n + n + n + n + n + n . For instance, i = − ( n + n + n + 2( n n + n n + n n ,i = ( n − n )( n − n ) + n (3 n + 2 n − n ) , Theorem.
The point to line ideal polarity in given by the matrix J = U T F U = − n n − n − n n − n − n n − n − n − n n − n − n n − n − n n − n − n − n . The adjoint matrix K gives the line to point ideal polarity. K = i i i i i i i i i . Indeed given a point ( P ) in I , U ( P ) gives the coordinates of P is space, multiplication tothe left by F determines the polar plane P . P P gives the direction i P of P , muiltiplicationto the left by V gives the 3 coordinates ( i P ) of the direction in the ideal plane, using 6.1.1gives J .For the adjoint matrix, i = 4 n n − ( n − n − n ) = − ( n + n + n ) + 2( n n + n n + n n = − ( n + n + n + 2( n n + n n + n n .i = ( n − n − n )( n − n − n ) + 2 n ( n − n − n )= ( n − n − n )( n − n − n ) + 2 n ( n − n − n )= ( n − n )( n − n ) + n (3 n + 2 n − n ) . CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Theorem. det ( J ) = 2 t. The ideal polarity is not degenerate.
Theorem. If P is on the conic associated with the idea polarity then P is on the fundamental quadric. Theorem.
Let l a := l + l + l , l b := − l − l + l , l c := − l − l + l , l d := − l − l + l ,i P := P ∧ I , i Q := Q ∧ I , I l := l ∧ I , I m := m ∧ I . then [ l a , l b , l c , l d ] = I l = L l. A line l = [ l , l , l , l , l , l ] is orthogonal to the plane P = { P , P , P , P } iff0. [ i P ] = J ( I l ) , or if for some k (cid:54) = 0,1. k ( P − P ) = l a n + l b ( n − n ) + l c ( n − n ) + l d n ,k ( P − P ) = l a ( n − n ) + l b ( − n ) + l c ( n − n ) + l d n ,k ( P − P ) = l a ( n − n ) + l b ( n − n ) + l c ( − n ) + l d n . Other relations can be derived from these, e.g.2. k ( P − P ) = l a n + l b n + l c ( n − n ) + l d ( n − n ) ,k ( P − P ) = l a n + l b ( n − n ) + l c n + l d ( n − n ) , A line l = [ l , l , l , l , l , l ] is orthogonal to the line m = [ m , m , m , m , m , m ] iff0. ( I m ) J ( I l ) = 0 , or − (( l + l + l )( m + m − m ) + ( m + m + m )( l + l − l )) n − (( l + l + l )( m + m − m ) + ( m + m + m )( l + l − l )) n − (( l + l + l )( m + m − m ) + ( m + m + m )( l + l − l )) n + (( l + l − l )( m + m − m ) + ( m + m − m )( l + l − l )) n + (( l + l − l )( m + m − m ) + ( m + m − m )( l + l − l )) n + (( l + l − l )( m + m − m ) + ( m + m − m )( l + l − l )) n = 0 . or m J l = 0 , with J = − n − n − n + n − n − n + n n − n + n − n + n − n n − n − n + n − n − n + n − n − n − n + n n − n − n − n + n + n − n n − n + n − n − n + n − n − n + n − n n − n − n + n − n + n + n n − n − n n − n + n n − n − n n − n − n + n − n n + n − n n − n + n − n + n − n − n + n + n − n − n + n + n n + n − n − n − n + n + n n − n − n + n n − n + n n − n − n n − n + n − n + n + n − n . A plane P = { P , P , P , P } is orthogonal to the plane Q = { Q , Q , Q , Q } iff ( i Q ) K ( i P ) T = 0 , or Q K P T = 0 , .2. POLAR GEOMETRY IN 3 DIMENSIONS. K is the symmetric matrix K = i i i i i i i i i i i i i i i i . Example.
The pole of the line
I ∧ A = [[1 , , , is (( i , i , i )) , which gives I Q = ( i , i , i , − ( i + i + i )) , hence if f oot := ( A ∨ I Q ) ∧ A , then f oot = (0 , i , i , i ) , with i = − ( i + i + i ) = ( n − n − n )( n − n − n ) + 2 n ( n − n − n ) . Definition.
The defining quadric and any other which has the the same ideal polarity is called a sphere . Theorem.
All spheres degenerate or not are given byΦ := k Θ + k I ×× R . with not both k and k equal to 0 and R a plane, distinct from the ideal plane. A sphere can be reduced to a point or be degenerate in the ideal plane and an other plane,when k = 0 . Definition.
The plane R of the preceding Theorem is called the radical plane of the 2 spheres Θ and Φ . Exercise.
Give an example of a sphere which reduces to a single point.
Theorem.
Given 2 ordinary points A and B, on a sphere and the polar of the ideal point on A × B isincident to A × B at C, then C is independent of the sphere. C is called mid-point of ( A, B ) . Exercise.
Generalize the construction of the polar p of a point P with respect to a conic to that of thepolar P of a point P with respect to a quadric. Exercise.
Give a construction of the mid-point of 2 points. CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Introduction.
The study of the geometry of the triangle can be generalized in 3 dimensions to the study of thegeneral tetrahedron. A special case occurs very naturally, that of the orthogonal tetrahedron,studied in section 6.2.4.
Notation.
Let IV be the set { , , , } and V I be the set { , , , , , } , let d be a function from theset IV ×× IV to the set V I defined by d , = 0 , d , = 1 , d , = 2 ,d , = 3 , d , = 4 , d , = 5 ,d i,i is undefined, d i,j = d j,i . d − denotes the inverse function.Similarly, let e be a function from the set IV ×× IV ×× IV to the set V I defined by e , , = 0 , e , , = 1 , e , , = 2 , e , , = 3 ,e i,j,k unchanged when we permute indices and e i,j,k undefined if 2 indices are equal. e − denotes its inverse. Notation. a ×× b indicates first that the lines a and b have a point P in common and second define P. The plane through the lines a and b is similarly denoted by a X b. See 3.2.D.12. In this section,the indices i, j, k, l are in the set { , , , } , the indices u, v are in the set { , , , , , } . Unless indicated explicitely the indices i, j, k, l or u, v in a given statement are distinct. l i,j or l d i,j represent the same line, the second forms indicates explicitely the mapping usedto map the 2 dimensional array into a 1 dimensional array.The set I J := { (0 , , (0 , , (0 , , (1 , , (3 , , (2 , } . The set J I := { (1 , , (2 , , (3 , , (2 , , (1 , , (3 , } . The definitions only define the object u ( i, j ) , ( i, j ) ∈ I J and not that ( i, j ) is in the set I J unless indicated explicitely.If ( j, i ) ∈ J I , then u ( i, j ) = u ( j, i ) . If a definition is followed by ( ∗ ) , this means that one of several definitions can be used,those not used are Theorems, for instance in D1.12. O can be defined by f acealtitude , ×× f acealtitude , , and O ∨ f acealtitude , = 0 . A quadric is denoted by a greek letter, θ say,the point quadric is then denoted by Θ , the plane quadric by | Θ . Comment.
In this section, I will only give the expression of one of the points in a set, the others areobtained as follows, if a point P nu is defined symmetrically from A , A and A the point P nv is obtained as follows,let nu = n d i,j then nv = n d i +1 ,j +1 . where the addition witin the subscripts is done modulo 4.In particular, .2. POLAR GEOMETRY IN 3 DIMENSIONS. n = n , becomes n , = n ,n = n , becomes n , = n ,n = n , becomes n , = n ,n = n , becomes n , = n ,n = n , becomes n , = n ,n = n , becomes n , = n , If a line l u is defined non symmetrically in terms of A , A , A , A then l v is obtained bymeans of a permutation P of { } .If l = f ( n , n , n , n , n , n ) , then l = f ( n , n , n , n , n , n ) , l = f ( n , n , n , n , n , n ) .l = f ( n , n , n , n , n , n ) ,l = f ( n , n , n , n , n , n ) , l = f ( n , n , n , n , n , n ) . Notation. { A i } will denote the tetrahedron with vertices A i . If we want to indicate explicitely not onlythe vertices A i but also the edges a u and the faces A j we will use the more elaborate notation { A i , a u , A j } . Comment.
For the tetrahedron with vertices A , A , A , A , the algebra will be done assuming thesehave the coordinates to be (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1), and that the barycentricpoint M has coordinates (1,1,1,1). Theorem.
If the coordinates of a point P are ( m , m , m , m ) , m , m , m , m (cid:54) = 0 , those of the plane P , which is its polar with respect to the tetrahedron { A i } are { m − , m − , m − , m − } . The Euclidean geometry will be defined starting with the ideal plane I which is the polarof M with respect to the tetrahedron and starting from the quadric Θ : n X X + n X X + n X X + n X X + n X X + n X X = 0 . as one of the spheres. Prefering I and Θ allows us to define parallelism and orthogonality. Theorem.
Given
H0.0. A , A , A , A , H0.1.
M, M ,
H0.2. Θ . Let
D0.0. a i,j := A i ∨ A j , D0.1. A l := A i ∨ A j ∨ A k , D0.2. I := polar ( M ) with respect to the tetrahedron, D1.0. C := pole ( I ) , D1.1. euler := C ∨ M, CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
D1.2. AP i := pole ( A i ) , D1.3. med i := C ∨ A i , D1.4.
Imed i := I ∧ med i , D1.5. alt i := A i ∨ Imed i , D1.6.
F oot i := A i ∧ alt i , D1.7. ipa i,j := Imed i ∨ Imed j , ( i, j ) ∈ I J , D1.8. P erp i,j := ipa ( i, j ) ∨ A i , ( i, j ) ∈ I J , P erp j,i := ipa ( i, j ) ∨ A j , ( j, i ) ∈ J I , D1.9.
F acef oot i,j := P erp i,j ∧ a k,l , ( i, j ) ∈ I J or J I , D1.10. f acealtitude i,j := F acef oot i,j ∨ A i , ( i, j ) ∈ I J or J I , D1.11. O i := f acealtitude j,i ×× f acealtitude k,i , i, j, k distinct ( ∗ ) , D1.12.
M id i := F oot i + O i , D1.13. mid i := M id i ∨ Imed i , D1.14. H := mid ×× mid , ( ∗ ) D1.15. η :=quadric through alt i ( ∗ ) , then C1.0. M = C + H C1.1. H ∨ euler = 0 . C1.2.
F oot i j = F oot j i . C1.3. O i ∨ η = 0 . The nomenclature or alternate definitions: A i are the vertices , M is the barycenter ,N0.0. a u are the edges ,N0.1. A i are the faces ,The tetrahedron is ( A i , a u , A i ) , N0.2. I is the ideal plane ,N1.0. C is the center of the circumsphere ,N1.1. euler is the line of Euler ,N1.2. AP i is the pole of the face A i , N1.3. med i is the mediatrix of the face A i , N1.4.
Imed i is the ideal point on the mediatrix med i , N1.5. alt i is the altitude corresponding to A , N1.6.
F oot i is the foot of alt i , corresponding to A i , N1.7. ipa i,j is the direction of the planes perpendicular to a k,l , N1.8. P erp i,j , ( i, j ) ∈ I J , is the plane perpendicular to a k,l through A i , P erp j,i , ( j, i ) ∈ J I , is the plane perpendicular to a k,l through A j , N1.9.
F acef oot i,j , ( i, j ) ∈ I J or J I , is the face-foot in the face A j , on the edgeopposite A i , A i ∨ F acef oot i,j is perpendicular to a k,l , N1.10. f acealtitude i,j , ( i, j ) ∈ I J or J I , is the face-altitude in the face A j through the vertex A i , perpendicular to a k,l , N1.11. O i is the orthocenter of A i , N1.12. mid i is the perpendicular to A i through M i , N1.13. H is the center of the hyperboloid η. N1.14. η is the hyperboloid of Neuberg. .2. POLAR GEOMETRY IN 3 DIMENSIONS. Proof:G0.0. A = (1 , , , . G0.1. M = (1 , , , . G0.2.
Θ : n X X + n X X + n X X + n X X + n X X + n X X = 0 . P0.0. a = [1 , , , , , . P0.1. A = { , , , } . P0.2. I = { , , , } . P1.0. C = (2 n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n ( n − n − n ) + n n ( n − n − n ) + n n ( n − n − n ) . P1.1. euler = [2 n ( n n − n n )+( n n − n n )( n + n − n − n )+ n n ( n + n − n − n ) , n ( n n − n n ) + ( n n − n n )( n + n − n − n ) + n n ( n + n − n − n ) , n ( n n − n n ) + ( n n − n n )( n + n − n − n ) + n n ( n + n − n − n ) , n ( n n − n n ) + ( n n − n n )( n + n − n − n ) + n n ( n + n − n − n ) , n ( n n − n n ) + ( n n − n n )( n + n − n − n ) + n n ( n + n − n − n ) , n ( n n − n n ) + ( n n − n n )( n + n − n − n ) + n n ( n + n − n − n ) , P1.2. AP = (2 n n n , n ( n n − n n − n n ) , n ( n n − n n − n n ) ,n ( n n − n n − n n )) . P1.3. med = [ n ( n − n − n ) , n ( n − n − n ) , n ( n − n − n ) ,n ( n − n ) − n ( n − n ) , n ( n − n ) − n ( n − n ) ,n ( n − n ) − n ( n − n )] . P1.4.
Imed = ( n + n + n − n n + n n + n n ) , − n (3 n + 2 n − n ) − ( n − n )( n − n ) , − n (3 n + 2 n − n ) − ( n − n )( n − n ) , − n (3 n + 2 n − n ) − ( n − n )( n − n ))) . P1.5. alt = [ n (3 n + 2 n − n ) + ( n − n )( n − n ) ,n (3 n + 2 n − n ) + ( n − n )( n − n ) ,n (3 n + 2 n − n ) + ( n − n )( n − n ) , , , . P1.6.
F oot = (0 , n (3 n + 2 n − n ) + ( n − n )( n − n ) ,n (3 n + 2 n − n ) + ( n − n )( n − n ) ,n (3 n + 2 n − n ) + ( n − n )( n − n )] , P1.7. ipa = [2 n , n − n − n , − n + n − n , − n + n + n , − n + n − n , − n + n + n − n . P1.8. P erp , = { , − n + n + n − n , − n + n − n , n + n − n } , P erp , = { n − n − n + n , , − n + n − n , n + n − n } . P1.9.
F acef oot , = (0 , , − n + n + n , n − n + n ) ,F acef oot , = (0 , , − n + n + n , n − n + n ) . P1.10. f acealtitude , = [0 , − n + n + n , n − n + n , , , ,f acealtitude , = [0 , , , n − n − n , n − n + n , . P1.11. O = (0 , n − ( n − n ) , n − ( n − n ) , n − ( n − n ) ) . P1.12.
M id = (0 , n (3 n + n − n ) + ( n − n )( n − n + n − n ) ,n (3 n + n − n ) + ( n − n )( n − n + n − n ) ,n (3 n + n − n ) + ( n − n )( n − n + n − n )) , CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
P1.13. mid = [ n (3 n + n − n ) + ( n − n )( n − n + n − n ) ,n (3 n + n − n ) + ( n − n )( n − n + n − n ) ,n (3 n + n − n ) + ( n − n )( n − n + n − n ) , ( n − n − n + n )( n + n − n ) , ( n − n − n + n )( n + n − n ) , ( n − n − n + n )( n + n + n )] .P1.14. H = ( n n n + n n n + n n n + n n n + n n ( n − n − n )+ n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n n + n n n + n n n + n n ( n − n − n )+ n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n n + n n n + n n n + n n ( n − n − n )+ n n ( n − n − n ) + n n ( n − n − n ) , n n n + n n n + n n n + n n n + n n ( n − n − n )+ n n ( n − n − n ) + n n ( n − n − n )) . P1.15. η : r X X + r X X + r X X + r X X + r X X + r X X = 0 .r = ( n − n − n + n )( n (3 n + 2 n − n ) + ( n − n )( n − n ) ,r = ( n − n − n + n )( n (3 n + 2 n − n ) + ( n − n )( n − n ) ,r = ( n − n − n + n )( n (3 n + 2 n − n ) + ( n − n )( n − n ) ,r = ( n − n − n + n )( n (3 n + 2 n − n ) + ( n − n )( n − n ) ,r = ( n − n − n + n )( n (3 n + 2 n − n ) + ( n − n )( n − n ) ,r = ( n − n − n + n )( n (3 n + 2 n − n ) + ( n − n )( n − n ) . Details for the computation of P.15 are given in 6.2.3.
Comment.
A simple derivation for some of the points in the faces follows from a direct application ofthe results on the geometry of the triangle. Indeed, the circumcircle in A is on the one hand n X X + n X X + n X X = 0 and on the other hand m ( m + m ) X X + m ( m + m ) X X + m ( m + m ) X X = 0 , assuming the coordinates of the orthocenter A to be (0 , m , m , m ) . Comparing we get m m = n + n − n , m m = n + n − n , m m = n + n − n .m , m , m are proportional to ( m m )( m m ) , ( m m )( m m ) , ( m m )( m m ) , thereforeusing homogeneity m = n − ( n − n ) ,m = n − ( n − n ) ,m = n − ( n − n ) . This can therefore be used to derive all the elements in the plane directly from Theorem 2.6.of Chapter 2. The following are useful, m + m = n ( n + n − n ) ,m + m = n ( n + n − n ) ,m + m = n ( n + n − n ) (The notation is only valid in A , to have a notation for all faces, m , m , m should bereplaced by m , m , m . )For instance, to obtains F oot , .2. POLAR GEOMETRY IN 3 DIMENSIONS. ia := | A | I = [0 , , , , ,
1] = [[1 , , ,P ia := pole ( ia ) ∈ I = (( i , i , i )) = ( i , i , i , i ) , with i = − i − i − i = ( n − n − n )( n − n − n ) + 2 n ( n − n − n ) . hence F oot := ( A ∨ P ia ) A = (0 , i , i , i ) . Hence
F oot i,i = 0 ,f I J := F oot i j = F oot j i = ( n i,k − n i,l )( n j,k − n j,l ) + n k,l (3 n i,j + 2 n k,l − n ) for i (cid:54) = j, and i, j, k, l a permutation of 0,1,2,3.Hence the Theorem as well as C1.3.Similarly the center of the circumcircle ∈ A is (0 , n ( n + n − n ) , n ( n + n − n ) , n ( n + n − n )) . Theorem.
Let r i,j be the coordinate of X i X j in η .Let the coordinates of the feet F oot , F oot , F oot , F oot be (0 , f , f , f ) , ( f , , f , f ) , ( f , f , , f ) , ( f , f , f , , then r i,j = f k,l ( f i,k f j,l − f i,l f j,k ) , where i, j, k, l is an even permutation of f J I of f I J modulo p be denoted g I J . If all f ij (cid:54) = 0 , expressing the fact that the quadric contains the altitudes gives the equations(0) f r + f r + f r = 0 , (1) g r + g r + g r = 0 , (2) f r + f r + f r = 0 , (3) g r + g r + g r = 0 , (4) f r + f r + f r = 0 , (5) g r + g r + g r = 0 , (6) f r + f r + f r = 0 , (7) g r + g r + g r = 0 . Equations (0) and (7) are obtained by substituting in the equation of the quadric, X i by kA i + F oot i ,g , g , g are proportional to f f , f f , f f , and therefore to f − , f − , f − .We solve (0) with respect to r , and (3) with respect to r , in terms of r and r , (5) gives r , substitution in (6) gives an homogeneous equation in terms of r and r only, hence afterdivision by f f + f f r , = r = f ( f f − f f ) , r , = r = − f ( f f − f f ) . equations (0) give r , (5) gives r , (7) gives r and (1) gives r . Hence r , = r = − f ( f f − f f ) ,r , = r = f ( f f − f f ) ,r , = r = f ( f f − f f ) ,r , = r = f ( f f − f f ) , equations (2) and (4) can be used as a check. Summarizing the results gives the Theorem.Simplifying by a common factor, we obtain P1.15. Theorem.
Let
D2.0. Ia u := a u ∧ I , D2.1. P olara u := polar ( Ia u ) , then CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
C2.0. ipa u ∧ P olar − u = 0 . Nomenclature:N2.0. Ia u are the ideal points on the edges a u , N2.1. polara u is the equatorial plane perpendicular to a u . Proof.P2.0. Ia = (1 , − , , . P2.1. P olara = {− n , n , n − n , n − n } . Definition. A tetrahedron is orthogonal iff the 3 pairs of opposite sides are perpendicular. Lemma. a · a = 0 iff n + n = n + n ,a · a = 0 iff n = n + n ,a · a = 0 iff n = n + n ,a · a = 0 iff n = n + n . The first condition expresses the orthogonality of opposite sides, the other conditions theorthogonality of adjacent sides.
Theorem.
The tetrahedron is orthogonal iff the parameters of the circumsphere satisfy n + n = n + n = n + n . Proof. The perpendicularity of A ∨ A and A ∨ A implies, because of 6.2.2.1, with l j = m j = 0 , except for l = m = 1 , n − n = n − n or n + n = n + n , Similarly that of A ∨ A and A ∨ A implies n + n = n + n . Theorem.
Given an orthogonal tetrahedron whose adjacent sied are not orthogonal, let m = ( n + n − n ) − , m = ( n + n − n ) − ,m = ( n + n − n ) − , m = ( n + n − n ) − , then1. n = ( m + m ) m m , n = ( m + m ) m m , n = ( m + m ) m m ,n = ( m + m ) m m , n = ( m + m ) m m , n = ( m + m ) m m . Proof: The non othogonality of adjacent sides implies that the m j are well defined. Weobtain, because of the orthogonality of opposite sides, m − + m − = 2 n , m − + m − = 2 n , m − + m − = 2 n , .2. POLAR GEOMETRY IN 3 DIMENSIONS. we also obtain m − + m − = n − n + n + n = 2 n ,m − + m − = n − n + n + n = 2 n ,m − + m − = n + n + n + n − n = 2 n . If we multiply by m m m m we get 1. Comment.
The indices obey the following rules.Let n i,j be the coefficient of X i X j , we have n , = n , n , = n , n , = n , n , = n , n , = n , n , = n . The orthogonality takes the form, n , + n , = n , + n , = n , + n , .m i is the inverse of n i,j + n i,k − n j,k , where i, j, k are distinct.For instance, if l is distinct from i, j, k, m i is also the inverse of n i,j + n i,l − n j,l . Definition.
A orthogonal tetrahedron is called a special orthogonal tetrahedron at A i if 2 adjacent sidesthrough A i are also orthogonal. Theorem. If A ∨ A is orthogonal to A ∨ A and the tetrahedron is orthogonal then these lines areorthogonal to A ∨ A and n + n − n = n + n − n = n + n − n = 0 . Vice versa, if n + n − n = 0 and the tetrahedron is orthogonal, then it is special at A . Exercise.
Discuss the special cases0. n + n − n = 0 , n + n − n (cid:54) = 0 . . . .1. n + n − n = 0 , n + n − n = 0 n = 0 . Theorem.
The coordinates of the points lines and planes defined in 6.2.3 are
G0.0. A = (1 , , , , G0.1. M = (1 , , , , G0.2.
Θ : ( m + m ) m m X X + ( m + m ) m m X X + ( m + m ) m m X X + ( m + m ) m m X X + ( m + m ) m m X X + ( m + m ) m m X X = 0 . CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
P0.0. a = a , = [1 , , , , , , P0.1. A = { , , , } , P0.2. I = { , , , } , P1.0. C = ( − m + m + m + m , m − m + m + m , m + m − m + m ,m + m + m − m ) , P1.1. euler = [ m − m , m − m , m − m , m − m , m − m , m − m ] , P1.2. AP = ( m ( m + m )( m + m )( m + m ) , − m ( m + m )( m m + m m ) , − m ( m + m )( m m + m m ) , − m ( m + m )( m m + m m )) , P1.3. med = [ m + m , m + m , m + m , m − m , m − m , m − m ] , P1.4.
Imed = ( − ( m + m + m ) , m , m , m ) , P1.5. alt = [ m , m , m , , , , P1.6.
F oot = (0 , m , m , m ) , P1.7. ipa , = [ m + m , − m , − m , m , − m , P1.8. P erp , = P erp , = { , , − m , m } , P1.9.
F acef oot , = F acef oot , = ( m , m , , , P1.10. f acealtitude , = [0 , m , m , , , , f acealtitude , = [0 , , m , , m , , P1.11. O = (0 , m , m , m ) , P1.12.
M id = (0 , m , m , m ) , P1.13. mid = altitude ,P1.14. H = ( m , m , m , m ) , P1.15.
Hyperboloid : m m X X − m m X X − m m X X m m X X or m m X X − m m X X − m m X X m m X X . Exercise.
Construct a quadric generalizing the conic of Brianchon-Poncelet, and verify that its equationis m m m X + . . . − m m ( m + m ) X X + . . . = 0 . Determine points on this quadric by linear constructions which are in none of the faces. .2. POLAR GEOMETRY IN 3 DIMENSIONS.
Exercise.
Construct a quadric which generalizes the sphere of Prouhet, passing through the barycentersand orthocenters of the faces and verify that its equation is m + m ) m m X X + . . . − X + X + X + X )( m m m X + . . . ) = 0 . (Coolidge, Treatise, p. 237) Definition. A symmedian is a line joining a vertex to the point of Lemoine of the opposite face. Definition. An isodynamic tetrahedron is a tetrahedron in which 3 of the symmedians are concurrent. Theorem.
A tetrahedron is isodynamic iff n n = n n = n n . Proof:Let K i be the symmedian in the place A i , let k i := A i × K i ,K = (0 , n , n , n ) , K = ( n , , n , n ) ,K = ( n , n , , n ) , K = ( n , n , n , .k = [ n , n , n , , , , k = [ n , , , n , n , and k = [0 , n , , n , , n ] . k and k are coplanar if n n = n n , k and k are coplanar if n n = n n , hence thetheorem. Theorem.
In an isodynamic tetrahedron all 4 symmedians are concurrent.
Definition.
A tetrahedron is orthogonal iff opposite sides are perpendicular.
Lemma. a · a = 0 iff n + n = n + n ,a · a = 0 iff n = n + n ,a · a = 0 iff n = n + n ,a · a = 0 iff n = n + n . CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Theorem.
The tetrahedron is orthogonal iff the parameters of the circumsphere satisfy n + n = n + n = n + n . Proof: The perpendicularity of A ∨ A and A ∨ A implies (0 , , , − F (1 , − , , T = 0 , or ( n − n ) − ( n − n ) = 0 or n + n = n + n , Similarly that of A ∨ A and A ∨ A implies n + n = n + n . Theorem. If n + n − n (cid:54) = 0 , n + n − n (cid:54) = 0 , n + n − n (cid:54) = 0 , n + n − n (cid:54) = 0 , and the tetrahedronis orthogonal. Let m = ( n + n − n ) − , m = ( n + n − n ) − ,m = ( n + n − n ) − , m = ( n + n − n ) − , then n = ( m + m ) m m , n = ( m + m ) m m , n = ( m + m ) m m ,n = ( m + m ) m m , n = ( m + m ) m m , n = ( m + m ) m m . Proof: We obtain at once, m − + m − = 2 n , m − + m − = 2 n , m − + m − = 2 n , using 1.3.3., we also obtain m − + m − = n − n + n + n = 2 n ,m − + m − = n − n + n + n = 2 n ,m − + m − = n + n + n + n − n = 2 n . If we multiply by m m m m we get 1. Comment. ?? when hyp. of prec theorem replace (cid:54) = 0 by = 0 , see Theorem 3.7. Comment.
The indices obey the following rules.Let n , be the coefficient of X X , . . . ,we have n , = n , n , = n , n , = n ,n , = n , n , = n , n , = n . The orthogonality takes the form, n , + n , = n , + n , = n , + n , .mi is the inverse of n i,j + n i,k − n j,k , where i, j, k are distinct.For instance, if l is distinct from i, j, k, mi is also the inverse of n i,j + n i,l − n j,l . .2. POLAR GEOMETRY IN 3 DIMENSIONS. Definition.
A orthogonal tetrahedron is called a special orthogonal tetrahedron at A i if 2 adjacent sidesthrough A i are also orthogonal. Theorem. If A ∨ A is orthogonal to A ∨ A and the tetrahedron is orthogonal then these lines areorthogonal to A ∨ A and n + n − n = n + n − n = n + n − n = 0 . Vice versa, if n + n − n = 0 and the tetrahedron is orthogonal, then it is special at A . Exercise.
Discuss the special cases0. n + n − n = 0 , n + n − n (cid:54) = 0 . . . .1. n + n − n = 0 , n + n − n = 0 n = 0 . Theorem.
The coordinates of the points lines and planes defined in in 1.2.7. are
H0.0. A =(1 , , , , H0.1. M = (1 , , , , P0.0. a = a , = [1 , , , , , , P0.1. | { , , , } , P0.2. | I = { , , , } , P1.0. C = ( − m + m + m + m , m − m + m + m , m + m − m + m , m + m + m − m ) , P1.1. euler = [ m − m , m − m , m − m , m − m , m − m , m − m ] , P1.2. AP = ( m ( m + m )( m + m )( m + m ) , − m ( m + m )( m m + m m ) , − m ( m + m )( m m + m m ) , − m ( m + m )( m m + m m )) , P1.3. med = [ m + m , m + m , m + m , m − m , m − m , m − m ] , P1.4.
Imed = ( − ( m + m + m ) , m , m , m ) , P1.5. alt = [ m , m , m , , , , P1.6.
F oot = (0 , m , m , m ) , P1.7. ipa , = [ m + m , − m , − m , m , − m , P1.8. | P erp. , = { , , − m , m } , perp. to a , through A × A = same ?? P1.9.
F acef oot , = ( m , m , , , on a , = same, P1.10. f acealtitude , = [0 , m , m , , , , F acef oot , ∨ A [1 ,
0] = [0 , , m , , m , , F acef oot , ∨ A P1.11. O = (0 , m , m , m ) , P1.12.
M id = (0 , m , m , m ) , CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
P1.13. mid = [ m , m , m , , , , P1.14. H = ( m , m , m , m ) , P1.15.
Eta : m m X X − m m X X − m m X X m m X X m m X X − m m X X − m m X X m m X X . C oideal = { m m m , m m m , m m m , m m m } ,Cocenter = Barycenter,
Theorem.
If (0 , p , p , p ) is on the Euler line eul (0) then p ( m − m ) + p ( m − m ) + p ( m − m ) = 0 , P ∨ IC (0) intersects the Euler line eu ; at(( p ( m − m ) + p ( m − m ))( m + m + m ) ,p (( m − m )( m + m + m ) + m ( m − m )) + p m ( m − m ) ,p (( m − m )( m + m + m ) + m ( m − m )) + p m ( m − m ) ,p (( m − m )( m + m + m ) + m ( m − m )) + p (( m − m )( m + m + m ) + m ( m − m ))) , or more symmetrically , ( p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m ) ,p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m ) ,p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m ) ,p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m )) , Moreover if p = km + s − m , p = km + s − m , p = km + s − m , then the pointon e is(( k − m + s, ( k − m + s, ( k − m + s, ( k − m + s ) . In particular, M = ( m + m + m , m + m + m , m + m + m , m + m + m ) , P = ( s t − m t , s ( m − m m ) − m t ,s ( m − m m ) − m t ,s ( m − m m ) − m t ) , w itht = m m + m m + m m − m m − m m − m m ,O = ( Am = ( s − m , s − m , s − m , s − m ) ,G = ( s + 2 m , s + 2 m , s + 2 m , s + 2 m ) ,Am = () D = () D = () D = () G = () G = () Answer (partial). . . . ? The polar pp = [ − m m , m ( m + m ) , m ( m + m ] , . . . ? The intersection P P = (0 , − m ( m + m ) , m ( m + m )) , .2. POLAR GEOMETRY IN 3 DIMENSIONS. . . . ? pp = [ m m ( m + m )( m + m ) , m m ( m + m )( m + m ) , m m ( m + m )( m + m )] . Point “O“, intersection of perpendicular to faces through their barycenter(special case of . . . with k = 0 hence “ O ” = ( s − m , s − m , s − m , s − m ) . “Conjugate tetrahedron”, “ A (cid:48) ” = ( − s − m , s + 2 m , s + 2 m , s + 2 m ) , barycenter of faces of [ A (cid:48) [ i ]] are “ M (cid:48) ” = (0 , m , m , m ) , which are the orthocenters of the faces,Perpendiculars through M (cid:48) to the faces, (which are parallel to those of [A[]] meet at “ O (cid:48) ” = (3 s − m , s − m , s − m , s − m ) . Hence his theorem: Then he generalizes the circle of Brianchon-Poncelet and gives its centeras the midpoint of H and “ O ” I believe his “ O ” is my G. Orthocenter,barycenter and “ O ” are collinear Definition. A symmedian is a line joining a vertex to the point of Lemoine of the opposite face. Definition. An isodynamic tetrahedron is a tetrahedron in which 3 of the symmedians are concurrent. Theorem.
A tetrahedron is isodynamic iff n n = n n = n n . Proof: K = (0 , n , n , n ) , K = ( − n , , n , − n ) ,K = ( − n , − n , , n ) , K = () .k and k are coplanar if n n = n n , k and k are coplanar if n n = n n , hence thetheorem. Theorem.
In an isodynamic tetrahedron all 4 symmedians are concurrent.
In 3 dimensions start with A = (1 , , , , . . . , A = (0 , , , , and with M = (1 , , , and M = ( m , m , m , m ) . M corresponds to the barycenter, M to the intersection of thelines joining the orthocenter of the faces to the opposite vertex and A [] to the vertices of anorthogonal tetrahedron. See . . . . Theorem. Prove that the tetrahedron is orthogonal.. Theorem. Prove that the circumscribed quadric is given by ( m + m ) m m X X + . . . = 0 . CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS . Theorem. Construct a quadric generalizing the conic of Brianchon- Poncelet, and verifythat its equation is m m m X + . . . − m m ( m + m ) X X + . . . = 0 . Determine points on this quadric by linear constructions which are in none of the faces.. Theorem. Construct a quadric which generalizes the sphere of Prouhet, passing throughthe barycenters and orthocenters of the faces and verify that its equation is m + m ) m m X X + . . . − X + X + X + X )( m m m X + . . . ) = 0 . (Coolidge, Treatise, p. 237)Point “O“, intersection of perpendicular to faces through their barycenter (special case of . . . with k = 0 hence “ O ” = ( s − m , s − m , s − m , s − m ) “Conjugate tetrahedron”, “ A (cid:48) ” = ( − s − m , s + 2 m , s + 2 m , s + 2 m ) , barycenter of faces of [ A (cid:48) [ i ]] are “ M (cid:48) ” = (0 , m , m , m ) , which are the orthocenters of the faces,Perpendiculars through M (cid:48) to the faces, (which are parallel to those of [ A []] meet at “ O (cid:48) ” = (3 s − m , s − m , s − m , s − m ) , Hence his theorem:Then he generalizes the circle of Brianchon-Poncelet and gives its center as the midpoint of H and “ O ” I believe his “ O ” is my G .Orthocenter,barycenter and “ O ” are collinear. Definition.
Consider the 2-form [ l , l , l , l , l , l ] with0. l l + l l + l l (cid:54) = 0 , the point to plane antipolarity associates to a point P a plane P := l (cid:48) ∨ P, the plane to point antipolarity associates to a plane P a point P := l (cid:48) P . Theorem.
The point to plane antipolarity can be represented by an antisymmetric matrix P = l l l − l l − l − l − l l − l l − l The plane to point antipolarity is represented by the antisymmetric matrix Q = l l l − l l − l − l − l l − l l − l .2. POLAR GEOMETRY IN 3 DIMENSIONS. l l + l l + l l ) (cid:54) = 0 . Theorem. If P is associated to P in an antipolarity then P is associated to | P and | P is incident to P. The proof is left as an exercise.
Theorem.
Let d := l l + l l + l l , the planes associated in the antipolarity 6.1.5 to the points of a line m are all incidentto a line q and if L := l dual ( l ) T − d I , then L = − l l − l l l l l l l l l l l l l − l l − l l l l l l l l l l l l l − l l − l l l l l l l l l l l l − l l − l l l l l l l l l l l l l − l l − l l l l l l l l l l l l l − l l − l l q = L m. det ( L ) = − d . The proof is left as an exercise.
Example.
Let p = 29 , l (cid:48) = [1 , , , , ,
10] = [3644209] , E0. d = 1 . E2. L = − −
14 12 −
13 41 12 3 9 12 3 − −
14 5 3 −
14 1111 − −
14 12 −
14 413 11 − E3. m = 732541 , , , , , .q = 20154561 , , , , , . CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Theorem.
The antipolarity of vertices, faces and edges of a tetrahedron follows from what follows.Given
H.0. l (cid:48) = [ l , l , l , l , l , l ] , H.1. A i , let D0.0. a i,j := A i ∨ A j , D0.1. A i := a j,k ∨ A l , D1.0. B i := l (cid:48) ∨ A i , B i := l (cid:48) A i , D1.1. ba i := | B i | A i , ba i := B i ∨ A i , D1.2. N i,j := a k,l B j , N i,j := a i,j ∨ B j , D1.3. n i,j := B i ∨ A j , D1.4. b d ( i,j ) := | B i | B j , then C1.0. B i ∨ B j = 0 . C1.1. N i,j ∨ N j,i = 0 . C1.2. N i,j ∨ n i,j = 0 , C1.3. n i,j = | B j | A i . C1.4. b d ( i,j ) = B k ∨ B l . C1.5. b u = L a u . Proof.P1.0. B = { , − l , − l , − l } .B = (0 , − l , − l , − l ) . P1.1. ba = [0 , , , l , l , l ] .ba = [ l , l , l , , , . P1.2. N , = (0 , , l , l ) .N , = { , , l , l } . P1.3. n , = [0 , , , l , − l , . P1.4. b = [ − l l − l l , l l , l l , l l , l l , l ] . Corollary. If l is a line and the definitions of Theorem 6.1.5 hold then all the conclusions of 6.1.5 hold.Moreover b u = l for all u .The mapping is not one to one. The image of a point P is the plane P ∨ l, the image of theplane Q is the point Q ∨ l. Example.
Let p = 29 , l (cid:48) = [1 , , , , ,
10] = [3644209] , A = (871 , , , , E0.0. a = [732541 , , , , , . E0.1. A = { , , , } . E1.0. B = { , , , } .B = (149 , , , . E1.1. ba = [139 , , , , .2. POLAR GEOMETRY IN 3 DIMENSIONS. ba = [3634832 , , , . E1.2. N = −− −−
875 139337 883 −− −− . N = −−
11 34 37819 −−
883 145149 878 −− −− . E1.3. n = −−
639 56 26659374 −− −− −− . E1.4. b = [20154561 , , , , , . Example.
Let p = 29 , l = [3623186] = [1 , , , − , , , A = (871 , , , , E1.0 B = { , , , } .B = (148 , , , . E1.1. ba = [286 , , , ,ba = [3610443 , , , . E1.2. N = −−
16 32 14622 −−
875 127735 897 −− −− . N = −−
12 53 32028 −−
881 156744 878 −− −− .E1.3. n = −−
436 57 26537429 −− −− −− .E1.4. b u = [3623186] . Theorem.
An antipolarity can be determined as follows,Given 4 points A i , a line ba ∈ A = A ∨ A ∨ A , a line ba ∈ A := A ∨ A ∨ A and a point B on n , = A ∨ ( ba ×× a ) but not on (( ba ×× a ) ∨ A ) ×× ( ba ×× a ) ∨ A )) ∨ (( ba ×× a ) ∨ A ) ×× ( ba ×× a ) ∨ A ))(= B ∨ B ) . Proof. Let us choose the A i as basis for the coordinate system. ba = [0 , , , l , l , l ] determines l , l , l .ba = [0 , l , l , , , l ] determines after scaling l and l . CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS B = (0 , l , l , l ) , determines, after scaling l . Scaling the last component should check with l . Example.
Let p = 29 , A = (871 , , , . Let ba = [139] = [0 , , , , , − , ba = [220389] = [0 , , , , , − ,N , = ba ×× a = (9) , n , := N , ∨ A = [639] . Finally let B = (149) = (0 , , , on [639] = [0 , , , , − , but not on [20154561] =[1 , − , , , − , − . (For the details of the computations see Example 6.1.5.) ba gives l = 1 , l = 3 , l = − ,ba gives l = t, l = 8 t, l = − t, t is the scaling factor,hence l = t = = 3 , l = − .B gives l = u, l = 4 u, l = 3 u, hence l = u = = 8 .l = 3 . − is a check.Therefore l (cid:48) = [8 , , − , , , −
7] = [1 , , , , , . The associated construction is as follows.
Construction.
Given A , A , A , A , ba ∈ A , ba ∈ A ,N , := ba ×× a , n , := N , ∨ A ,N , := ba ×× a , n , := N , ∨ A ,N , := ba ×× a , n , := N , ∨ A .N , := ba ×× a , n , := N , ∨ A ,N , := ba ×× a , n , := N , ∨ A ,N , := ba ×× a , n , := N , ∨ A .B := n , ×× n , , n , := B ∨ A ,N , := n , ×× a ,B := n , ×× n , , n , := B ∨ A ,N , := n , ×× a . Given B on n , , notonB ∨ B (otherwize l (cid:48) is a line), n , := B ∨ A , N , := n , ×× a ,n , := B ∨ A , N , := n , ×× a .ba := N , ∨ N , ,N , := ba ×× a , n , := N , ∨ A ,B := n , ×× n , .B i are the antipoles of A i .ba := N , ∨ N , . B i := A i ∨ ba i . B i are the antipolars of A i . To complete the construction, N , := ba ×× a , n , := B ∨ A , we can check0. N , ∨ n , = 0 . .2. POLAR GEOMETRY IN 3 DIMENSIONS. Theorem.
In the geometry of the triangle if A is M , ba is m , and ba is an arbitrary line and B is anarbitrary point on ( ba × ( A × M )) × A , the configuration of 6.1.5 consisting of 20 points A, B and N , and of 22 lines a, ba, n , satisfies 6.1.5.0. Theorem.
D1.0. P a r := P ∨ a − r , D1.1.
P ab r := P a r b − r , D1.2. P := P ab ∨ P ab ∨ P ab , then C1.0.
P ∨ P = 0 , C1.1. P = P ∨ l (cid:48) . Example.
With p, l (cid:48) and A as in Example 6.1.5.Let P = (1742) = (1 , , , , then P a = { , , } , P ab = (2357 , , , P = { } . Let P = (5350) = (1 , , , , then P a = { , , , P ab = (5356 , , , P = { } . Exercise.
The antipolarity associates to a point quadric
Alpha , a plane quadric B eta, the points of oneare on the tangent of the other. Study this correspondance in detail. Case 0. p = 13 , Barycenter = 366 , n = 1 , , , , , , m = 1 , , , , The tetrahedron is orthogonal.
Barycenter = (366) I deal = { } A = (183 , , , a = [30941 , , , , , Center = (1244)
P oleof A = (1509 , , , mediatrix = [271483 , , , IC = (387 , , , altitude = [107836 , , , F oot = (49 , , , ipa = [85840 , , , , , P erp. = { , , , , , } = { , , , , , } CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
F acef oot = (12 , , , , , , , , , , f acealtitude = [26547 , , , , , , , , , , Orthocenters = (49 , , , M id = (49 , , , mid = [107836 , , , Orthocenter = (578) C oideal = { } Cocenter = (366)
Barycenter (check) = (366)
Hyperboloid : 1110035 , The coordinates of IC are (1,1,2,9), (1,3,4,5), (1,2,5,5), (1,2,4,6),those of F oot are (0,1,2,9), (1,0,4,5), (1,2,0,5), (1,2,4,0),those of the
Centerof Hyperboloid are (1 , , , , the hyperboloids are ( − r r X X + r X X + r X X − r X X + 3 r X X + (5 r r X X = 0 . Case 1. p = 13 , Barycenter = 1504 , n = 1 , , , , , , m = 1 , , , , The tetrahedron is orthogonal.The
Center is an ideal point.
Barycenter = (1504) I deal = { } A = (183 , , , a = [30941 , , , , , Center = (2368)
Case 2. p = 13 , Barycenter = 366 , n = 1 , , , , , , The
Center is an ideal point.
Barycenter = (366) I deal = { } A = (183 , , , a = [30941 , , , , , Center = (2248)
Case 3. p = 13 , Barycenter = 366 , n = 1 , , , , , , m = 1 , , , , The tetrahedron is orthogonal.
Barycenter = (366) I deal = { } .2. POLAR GEOMETRY IN 3 DIMENSIONS. A = (183 , , , a = [30941 , , , , , Center = (94)
P oleof A = (1156 , , , mediatrix = [208070 , , , IC = (1156 , , , altitude = [252838 , , , F oot = (115 , , , ipa = [269114 , , , , , P erp. = { , , , , , } = { , , , , , } F acef oot = (8 , , , , , , , , , , f acealtitude = [17759 , , , , , , , , , , Orthocenters = (115 , , , M id = (115 , , , mid = [252838 , , , − Orthocenter = (1589) C oideal = { } Cocenter = (299)
Barycenter (check) = (366)
Hyperboloid : 6100106 , Case 4. p = 13 , Barycenter = 366 , n = 1 , , , , , ,Barycenter = (366) I deal = { } A = (183 , , , a = [30941 , , , , , Center = (1833)
P oleof A = (281 , , , mediatrix = [207817 , , , IC = (1504 , , , altitude = [237459 , , , F oot = (108 , , , ipa = [269480 , , , , , P erp. = { , , , , , } = { , , , , , } F acef oot = (2 , , , , , , , , , , f acealtitude = [4577 , , , , , , , , , , Orthocenters = (115 , , , CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
M id = (112 , , , mid = [246391 , , , − Centerof hyperb. = (194) C oideal = {− } Cocenter = ( − Barycenter (check) = (366)
Hyperboloid : 9800114
The coordinates of
Center are (1 , , , , those of IC are (1 , , , , (1 , , , , (1 , , , , (1 , , , , those of F oot are (0 , , , , (1 , , , , (1 , , , , (1 , , , , those of orthocenters are (0 , , , , (1 , , , , (1 , , , , (1 , , , , those of the Centerof Hyperboloid are (1 , , , , the hyperboloid is X X − X X − X X − X X = 0 . Case 5. p = 17 , Barycenter = 614 , n = 1 , , , , , ,Barycenter = (614) I deal = 614 A = (307 , , , a = [88741 , , , , , Center = (2954)
P oleof A = (3872 , , , mediatrix = [174494 , , , IC = (4484 , , , altitude = [904299 , , , F oot = (184 , , , ipa = [170402 , , , , , P erp. = { , , , , , } = { , , , , , } F acef oot = (12 , , , , , , , , , , f acealtitude = [59263 , , , , , , , , , , Orthocenters = (0 , , , M id = (173 , , , mid = [850281 , , , Centerof hyperb. = (2687) C oideal = {− } Cocenter = ( − Barycenter (check) = (614)
Hyperboloid : 42149113
The coordinates of the
Center are (1 , , , , those of IC are (1 , , , , (1 , , , , (1 , , , , (1 , , , , .2. POLAR GEOMETRY IN 3 DIMENSIONS. those of F oot are (0 , , , , (1 , , , , (1 , , , , (1 , , , , those of orthocenters are (0 , , , , (1 , , , , (1 , , , , (1 , , , , those of the Centerof Hyperboloid are (1 , , , , the hyperboloid is X X + X X + 7 X X − X X − X X − X X = 0 . Case 6. p = 17 , Barycenter = 614 , n = 1 , , , , , ,Barycenter = (614) I deal = 614 A = (307 , , , a = [88741 , , , , , Center = (3114)
P oleof A = (3984 , , , mediatrix = [95281 , , , IC = (4564 , , , altitude = [1218731 , , , F oot = (248 , , , ipa = [1427515 , , , , , P erp. = { , , , , , } = { , , , , , } F acef oot = (15 , , , , , , , , , , f acealtitude = [74002 , , , , , , , , , , Orthocenters = (1 , , , M id = (61 , , , mid = [304633 , , , Centerof hyperb. = (2623) C oideal = {− } Cocenter = ( − Barycenter (check) = (614)
Hyperboloid
The coordinates of
Center are (1 , , , , those of Centerof hyperboloid are (1 , , , . Observe that the last 2 coordinates are exchanged.
Case 7. p = 17 , Barycenter = 614 , n = 1 , , , , , , THE QUADRIC IS DEGENERATE CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS
Case 8. p = 13 , Barycenter = 1165 , n = 1 , , , , , , m = 1 , , , , The tetrahedron is orthogonal.The
Center is an ideal point.
Barycenter = (1165) I deal = { } A = (183 , , , a = [30941 , , , , , Center = (501)
Case 9. p = 19 , Barycenter = 762 , n = 1 , , , , , ,Barycenter = (762) I deal = { } A = (381 , , , a = [137561 , , , , , Center = (5279)
P oleof A = (2484 , , , mediatrix = [2260013 , , , IC = (597 , , , altitude = [82689 , , , F oot = (12 , , , ipa = [881367 , , , , , P erp. = { , , , , , } = { , , , , , } F acef oot = (12 , , , , , , , , , , f acealtitude = [82689 , , , , , , , , , , Orthocenters = (203 , , , M id = (296 , , , mid = [233678 , , , Centerof hyperb. = (2557) C oideal = {− } Cocenter = ( − Barycenter (check) = (762)
Hyperboloid : 10870118
Case 10. p = 29 , Barycenter = 1742 , n = 1 , , , , , ,Barycenter = (1742) I deal = { } A = (871 , , , .2. POLAR GEOMETRY IN 3 DIMENSIONS. a = [732541 , , , , , Center = (7629)
P oleof A = (13904 , , , mediatrix = [5332964 , , , IC = (11875 , , , altitude = [13487988 , , , F oot = (553 , , , ipa = [8254786 , , , , , P erp. = { , , , , , } = { , , , , , } F acef oot = (17 , , , , , , , , , , f acealtitude = [415484 , , , , , , , , , , Orthocenters = (109 , , , M id = (512 , , , mid = [12490883 , , , Centerof hyperb. = (19403) C oideal = { } Cocenter = (759)
Barycenter (check) = (1742)
Hyperboloid : 5714192025
Case 11. p = 29 , Barycenter = 19403 , n = 1 , , , , , ,Barycenter = (19403) I deal = { } A = (871 , , , a = [732541 , , , , , Center = (759)
P oleof A = (13904 , , , mediatrix = [18523362 , , , IC = (13904 , , , altitude = [21121745 , , , F oot = (866 , , , ipa = [1759206 , , , , , P erp. = { , , , , , } = { , , , , , } F acef oot = (17 , , , , , , , , , , f acealtitude = [415484 , , , , , , , , , , Orthocenters = (527 , , , M id = (202 , , , CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS mid = [4930377 , , , Centerof hyperb. = (6873) C oideal = { } Cocenter = (17926)
Barycenter (check) = (19403)
Hyperboloid : 21181812216
Theorem.
If (0 , p , p , p ) is on the Euler line eul then p ( m − m ) + p ( m − m ) + p ( m − m ) = 0 , P ∨ IC (0) intersects the Euler line eul at(( p ( m − m ) + p ( m − m ))( m + m + m ) ,p (( m − m )( m + m + m ) + m ( m − m )) + p m ( m − m ) ,p (( m − m )( m + m + m ) + m ( m − m )) + p m ( m − m ) ,p (( m − m )( m + m + m ) + m ( m − m ))+ p (( m − m )( m + m + m ) + m ( m − m ))),or more symmetrically,( p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m ) ,p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m ) ,p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m ) ,p ( s ( m − m ) + m ( m − m ) + p ( s ( m − m ) + m ( m − m )) . Moreover, if p = km + s − m , p = km + s − m , p = km + s − m , then the point on eul is(( k − m + s, ( k − m + s, ( k − m + s, ( k − m + s ) . In particular, M = ( m + m + m , m + m + m , m + m + m , m + m + m ) ,P = ( s t − m t , s ( m − m m ) − m t , s ( m − m m ) − m t ,s ( m − m m ) − m t ) , with t = m m + m m + m m − m m − m m − m m ,O = ( Am = ( s − m , s − m , s − m , s − m ) ,G = ( s + 2 m , s + 2 m , s + 2 m , s + 2 m ) ,Am = ( D = () D = () D = () G = () G = () .90. ANSWERS TO PROBLEMS AND MISCELLANEOUS NOTES. Answer to ??.Answer (partial). . . . ? The polar pp = [ − m m , m ( m + m ) , m ( m + m ] , . . . ? The intersection P P = (0 , − m ( m + m ) , m ( m + m )) , . . . ? pp = [ m m ( m + m )( m + m ) , m m ( m + m )( m + m ) , m m ( m + m )( m + m )] . Point “O“, intersection of perpendicular to faces through their barycenter(special case of . . . with k = 0 hence “ O ” = ( s − m , s − m , s − m , s − m ) . “Conjugate tetrahedron”, “ A (cid:48) ” = ( − s − m , s + 2 m , s + 2 m , s + 2 m ) , barycenter of faces of [ A (cid:48) [ i ]] are “ M (cid:48) ” = (0 , m , m , m ) , which are the orthocenters of the faces,Perpendiculars through M (cid:48) to the faces, (which are parallel to those of [A[]] meet at “ O (cid:48) ” = (3 s − m , s − m , s − m , s − m ) . Hence his theorem: Then he generalizes the circle of Brianchon-Poncelet and gives its centeras the midpoint of H and “ O ” I believe his “ O ” is my G. Orthocenter,barycenter and “ O ” are collinear CHAPTER 6. GENERALIZATION TO 3 DIMENSIONS hapter 7QUATERNIONIAN GEOMETRY It is a classical result, (see Artin, Harsthorne) that if the coordinates which are used to definea projective geometry are elements of a non commutative division ring, then Desargues’Theorem is true, but Pappus’ Theorem is, in general, not true. More precisely, Pappus’theorem implies that the division ring or skew fielf is commutative. I will prove here detailledgeometric properties which justify the definitions of medians and circumcircular polarity ina quaternionian plane.The results were conjectured by taking the coordinates in the ring with unity associatedwith quaternions over the finite field Z p , p prime. This is not a division ring because a finitedivision ring is a field. In this geometry, not all points define a line and vice-versa. Thesituation is similar to that described by Kn¨uppel and Salow, for the case of a commutativering with unity. This generalization merits to be explored in detail.In involutive Geometry, we started with the triangle { A i , a i } , the barycenter M and theorthocenter M . We constructed the medians ma i , the altitudes ma i , the mid-points M i , thefeet M i , the complementary triangle ( M i , mm i ) , the orthic triangle { M i , mm i } , the idealpoints M A i , the orthic points M A i , the ideal line m, the orthic line m, the orthic directions, Imm i , the tangential triangle { T i , ta i } , the symmedians at i and the point of Lemoine K. Moreover the same tangential triangle T i can be obtained if we interchange the role of M and M .
I attempted the same construction for the Geometry over the quaternion skew field. In thiscase, however, the lines at i are not concurrent, in general, but form a triangle K i and thereexists a polarity in which K i is the pole of a i and therefore A i is the pole of at i . This polaritydegenerates into all the lines through K in the involutive Geometry.The plane corresponding to the involutive plane is defined by chosing a complete 5-anglein the quaternionian plane. 3 points are the vertices of the basic triangle, 1 is the barycenter,1 is the cobarycenter. We define the ideal line as the polar of the barycenter with respect to thetriangle and as comedians, the lines joining the vertices of the triangle to the cobarycenter.It can be shown that there is a polarity which associates to the vertices of the triangle 3 ofthe lines through them, corresponding to the tangential lines in Euclidean geometry, but, in CHAPTER 7. QUATERNIONIAN GEOMETRY general, the involution defined by this polarity on the ideal line, does not have the directionsof the sides and the direction of the comedians as corresponding points.
Notation.
Identifiers, starting with a lower case letter, will denote quaternions, q denotes the conjugateof q, q (cid:48) , the conjugate inverse, q n := q q = q q, q − n := ( q n ) − . Definition.
The elements and incidence in
Quaternionian geometry in 2 dimensions are defined as fol-lows.0. The points are ( q , q , q ) with right equivalence,1. The lines are [ l , l , l ] with right equivalence,2. A point P is incident to a line l iff P · l := (cid:80) i =0 P i l i = 0 . Condition 2 is consistent with equivalence and can also be written l · P = 0 . I prefered it, because the usual form (cid:80) i =0 P i l i = 0 implies (cid:80) i =0 l i P i = 0 . We remind the reader of the following
Theorem.
In any skew field, if a matrix A has a left inverse and a right inverse, these are equal. Proof: Let C be the left inverse of A and B be its right inverse, by associativity ofmatrices, C = C ( AB ) = ( CA ) B = B . Lemma. p i (cid:54) = q i , i = 1 , ⇒ (1 , p , p ) × (1 , q , q )= [( p (cid:48) − q (cid:48) ) − ( p (cid:48) p − q (cid:48) q )( p − q ) − , ( p − q ) − , − ( p − q ) − ] , (1 , p , p ) × (1 , p , q ) = [ p , − , , (0 , , × (1 , p ,
0) = [ p , − , ,3. (0 , , × (0 , ,
0) = [1 , , . .1. QUATERNIONIAN GEOMETRY OVER THE REALS. Proof: Let the line be [ x , x , x ] , we must have x + p x + p x = 0 , and x + q x + q x = 0 , subtracting gives ( p − q ) + x ( p − q ) x = 0 , hence x and x . x follows from substitution into the second equation. Theorem.
A quaternionian geometry is a perspective geometry.
Lemma.
Let a i and b i be different from 0. The points P := (0 , q , r ) , P := ( r , , q ) , P := ( q , r , q r (cid:48) )( q r (cid:48) )( q r (cid:48) ) = − r (cid:48) q , q (cid:48) r , − , [ − , r (cid:48) q , q (cid:48) r ] , [ r (cid:48) q , − , q (cid:48) r ] . Proof: Let y := [ y , y , −
1] := P × P , we have q y + r = 0 and r y + q = 0 , this gives the first form y . To verify that P is on y , weneed q (cid:48) r (cid:48) q + r q (cid:48) r = 0 . Lemma.
In a quaternionian geometry the Theorem of Desargues is satisfied.
Proof: We can always choose the coordinates of A i and C as follows A = (1 , , , A = (0 , , , A = (0 , , , C = (1 , , . It follows that a = [1 , , , a = [0 , , , a = [0 , , ,c = [0 , , − , c = [ − , , , c = [1 , − , , It follows that B i are B = ( q , , , B = (1 , q , , B = (1 , , q ) , therefore b = [0 , q − , − ( q − , b = [ − ( q − , q − , ,b = [ q − , − ( q − , ,B = ( q , , , B = (1 , q , , B = (1 , , q ) ,C = (0 , q − , q − , C = ( q − , q − , ,C = ( q − , q − , , the Theorem follows from Lemma 7.1.1. Theorem.
A quaternionian geometry is a Desarguesian geometry.
Notation.
To give an explicit way of indicating that a 3 by 3 matrix is a point or line collineation ora correlation from points to lines or from lines to point a parenthesis is used on the sideof the point and a bracket on the side of a line. This notation is used when we give the CHAPTER 7. QUATERNIONIAN GEOMETRY table of elements. It could also be used in connection with the bold face letter representingcollineation or correlation. This notation is only useful when we apply algebra to geometry.
Theorem. If C is a point collineation, the line collineation is C (cid:48) T . In particular, the point collineation which associates to A i , A i and to (1 , , , ( q , q , q ) is q q
00 0 q , and the line collineation is q (cid:48) q (cid:48)
00 0 q (cid:48) . Proof: If Q is the image of P , and m is the image of l , we want P · l = Σ P i l i = Σ C − ij Q j l i = Σ Q j C (cid:48) ij l i = Σ Q j m j = 0 , for all points P and incident lines l . This requires m j = Σ C (cid:48) Tji l i . Definition. A Hermitian matrix M is a matrix which is equal to its conjugate transpose. M defines a transformation from points to line, M − , the inverse, defines the transformation from lines to points. Theorem. If M is Hermitian and p = M P, q = M Q and P · q = 0 then Q · p = 0 . Proof: Q · p = (cid:80) i Q i p i = (cid:80) i (cid:80) j Q i M i,j P j = (cid:80) j (cid:80) i ( Q i M j,i ) P j = (cid:80) j (cid:80) i ( M j,i Q i ) P j = (cid:80) j q j P j = q · P = 0 . Theorem. The transformation defined by a Hermitian matrix is a polarity .1.
The columns of a polarity M are the polars of the points A i , with A = (1 , , A = (0 , , A = (0 , , The columns of a inverse polarity M − are the poles of the lines a i , with a = [1 , , a = [0 , , a = [0 , , .1. QUATERNIONIAN GEOMETRY OVER THE REALS. Definition.
Polar points are points incident to their polar.
Polar lines are lines incident to their pole.
Comment.
A polar line can contain infinitely many polar points. For instance, let the polarity be − − − − − − . P := (0 , , , has for polar p = [1 , , . A point Q := (0 , , q ) , on p has for polar [1 + q, q, . Q is a polar point if (cid:60) ( q ) = 0 . Therefore all the points (0 , , a i + a j + a k ) are polar points on [1,0,0]. Lemma.
Let c i := q i − q i q i +1 + q i +1 q i q i − = 2 Re ( q i − q i q i +1 ) , if q i q i (cid:54) = 0 then c = c = c and we define c := c . Proof: q q c = q c q = q ( q q q + q q q ) q = q q ( q q q ) + ( q q q ) q q = q q ( q q q + q q q ) = q q c . Theorem.
Let a i = a i ,b i := a i +1 a i − − q i q i ,r i := q i − q i +1 − a i q i , then a q q q a q q q a b r r r b r r r b = d E , where E is the identity matrix and d := a a a − a q q − a q q − a q q + 2 Re ( q q q ) . Moreover, b i = b i ,a i := b i +1 b i − − r i r i ,q i := r i − r i +1 − b i r i , Proof: For instance, a b + q r + q r = a a a − a q q + q q q − q a q + q q q − q a q = d and a r + q b + q r = a q q − a a q + q a a − q q q + q q q − q a q = 0 . The second part of the proof is obtained similarly or follows from 7.1.1.1. CHAPTER 7. QUATERNIONIAN GEOMETRY
The 2 parts of the following Lemma use different approaches to the problem of constructingpolarities.
Lemma. If all the components of the lines x i are non zero and the i -th component of x i is real,necessary and sufficient conditions for x i to be polars of A i are x − i +1 ,i − x i − ,i +1 = k i , k i real .1. k k k = 1 .
1. Let 3 points have coordinates P = ( a , q , q − ) , P = ( q − , a , q ) , P = ( q , q − , a ) , where a i = a i , then, if the norm of q q q = 1 and if the matrix P is obtained bymultiplying the column vectors P i respectively by , q n , q n , this matrix defines a polaritywhich associates the lines a i to the points P i . Proof: For part 0, the condition that the i -th component of x i is real can always be satisfiedby multiplying the components of x i by x ii . It remains to find real numbers which multiplied by x i give the columns of an Hermitianmatrix. Considering the elements x and x requires x = x k , k real, or more generally0. Multiplying the first and second column by k and k − requires condition 1, for the elementsin position 12 and 21 to be conjugates of each other.For the second part, the matrix is then a q q (cid:48) q q n a q n q q − q n q q − n a . Exercise.
State and prove the Theorem which extends the preceding Theorem to the case where someof the components of the vectors x i are 0, or some of the q i are 0. Lemma. u nj (cid:54) = 0 , v nj (cid:54) = 0 , j = 1 , , and d := u u − + v v − , e := u u − + v v − , ⇒ u − d v = u − e v and d n v n u n = e n v n u n . Proof: u − d v = u − v + u − v and u − e v = u − v + u − v . Lemma. ( u u u v v v ) n (cid:54) = 0 ,d i := u i +1 u − i − + v i +1 v − i − , d := d d d , .1. QUATERNIONIAN GEOMETRY OVER THE REALS. e i := u i − u − i +1 + v i − v − i +1 , e := e e e , ⇒ d n = e n . Proof: This follows from Lemma 7.1.1, taking norms and using the fact that the norm ofa product is the product of the norms.
Notation.
In what follows, I will use the same notation as in involutive Geometry, namely, l := P × Q, means that the line l is defined as the line incident to P and Q. If subscripts are used these have the values 0, 1 and 2 and the computation is done modulo3, P · l = 0 means that the point P is incident to the line l. When 3 lines intersect, this intersection can be defined in 3 ways, this has been indicated byusing (*) after the definition and implies a Theorem. σ := polarity (( M i , a i )) . implies that σ is the polarity which associates M to a , M to a and M to a . m = polar ( σ, M ) . implies that in the polarity σ , m is the polar of M. The labeling used is “H,“ for Hypothesis, “D”, for definitions, “C“, for conclusions,“N”, fornomenclature, “P“, for proofs, this labelling being consistent with that of the correspondingdefinitions. The example given is associated to the quaternions over Z , the labelling is “E”and is consistent with the corresponding definitions.Because any 3 pairs of points and lines do not necessarily define a polarity, if a polarity isdefined it implies a conclusion (or Theorem) I have therefore replaced “D“ by “DC”. The special configuration of Desargues.
With this notation, the special configuration of Desargues can be defined by a i := A i +1 × A i − , qa i = Q × A i ,Q i := a i × qa i , qq i := Q i +1 × Q i − ,QA i := a i × qq i , q i := A i × QA i ,QQ i := q i +1 × q i − , q := QA × QA ( ∗ ) , and the other conclusion of the special Desargues Theorem can be written, QQ i · qa i = 0 . Let Q and A i be Q = ( q , q , q ) , and A = (1 , , , A = (0 , , , A = (0 , , , then we have the following results, not obtained in the given order, A = (1 , , , a = [1 , , ,Q = ( q , q , q ) , q = [ q (cid:48) , q (cid:48) , q (cid:48) ] ,QA = (0 , q , − q ) , qa = [0 , q (cid:48) , − q (cid:48) ] ,Q = (0 , q , q ) , q = [0 , q (cid:48) , q (cid:48) ] ,QQ = ( − q , q , q ) , qq = [ − q (cid:48) , q (cid:48) , q (cid:48) ] , The self duality of the configuration corresponds to the replacement of points by lines CHAPTER 7. QUATERNIONIAN GEOMETRY where upper case letters are replaced by lower case letters and coordinates by their conjugateinverse.
Fundamental Hypothesis, Definitions and Conclusions.
The ideal line and the coideal line.GivenH0.0. A i , H0.1.
M, M ,
LetD1.0. a i := A i +1 × A i − , D1.1. ma i := M × A i , ma i := M × A i , D1.2. M i := ma i × a i , M i := ma i × a i , D1.3. eul = M × M ,
DC1.4. σ := polarity (( M i , a i )) , σ := polarity (( M i , a i )) , D2.0. mm i := M i +1 × M i − , mm i := M i +1 × M i − , D2.1.
M A i := a i × mm i , M A i := a i × mm i , D2.2. m i := A i × M A i , m i := A i × M A i , D2.3.
M M i := m i +1 × m i − , M M i := m i +1 × m i − , D2.4. m := M A × M A ( ∗ ) , m := M A × M A ( ∗ ) , D2.5.
Ima i := m × ma i , Ima i := m × ma i , D2.6.
IM a i := m × ma i , IM a i := m × ma i , D2.7. iM A i := M × M A i , ıM A i := M × M A i , thenC2.0. m = polar ( σ, M ) , m = polar ( σ, M ) . C2.1. mm i = polar ( σ, A i ) , mm i = polar ( σ, A i ) . C2.2. ma i = polar ( σ, M A i ) , ma i = polar ( σ, M A i ) . C2.3. iM A i = polar ( σ, Ima i ) , ıM A i = polar ( σ, Ima i ) . LetD3.0. mf i := M i × IM a i , mf i := M i × IM a i , D3.1. O := mf × mf ( ∗ ) , O := mf × mf ( ∗ ) , D3.2.
M f a i := a i +1 × mf i − , M f a i := a i +1 × mf i − ,M f a i := a i − × mf i +1 , M f a i := a i − × mf i +1 , D3.3. mf a i := M f a i +1 × A i − , mf a i := M f a i +1 × A i − ,mf a i := M f a i − × A i +1 , mf a i := M f a i − × A i +1 , D3.4.
M f m i := mf a i × m i , M f m i := mf a i × m i , thenC3.0. O · eul = O · eul = 0 . C3.1.
M f m i · mf a i = M f m i · mf a i = 0 . LetD4.0.
Imm i := m × mm i , Imm i := m × mm i , D4.1. ta i := A i × Imm i , D4.2. T i := ta i +1 × ta i − , D4.3. at i := A i × T i , D4.4. K i := at i +1 × at i − , .1. QUATERNIONIAN GEOMETRY OVER THE REALS. D4.5.
T Aa i := ta i × a i , D4.6. poK i := T aa i +1 × T aa i − , DC4.7. θ := polarity (( A i , ta i )) , DC4.8. λ := polarity (( A i , at i )) , thenC4.0. Imm i · ta i = 0 . C4.1. T i · mf i = 0 . C4.2. a i := polar ( θ, T i ) . C4.3. a i := polar ( λ, K i ) . The nomenclature:N0.0. A i are the vertices of the triangle,N0.1. M is the barycenter , M is the cobarycenter .N1.0. a i are the sides .N1.1. ma i are the medians , ma i are the comedians N1.2. M i are the mid-points of the sides. M i are the feet of the comedians N1.3. eul is the line of Euler ,N1.4. σ is the Steiner polarity . σ is the co-Steiner polarity .N2.0. { M i , mm i } is the complementary triangle , { M i , mm i } is the orthic triangle ,N2.1. M A i are the directions of the sides ,N2.2. { M M i , m i } is the anticomplementary triangle .N2.3. m is the ideal line corresponding to the line at infinity, m is the orthic line which is the polar of M with respect to the triangle.N2.4. Ima i are the directions of the medians . IM a i are the directions of the comedians .N3.0. mf i are the mediatrices ,N3.1. O is the center ,N3.2. M f m i are the trapezoidal points ,N4.0. Imm i are the directions of the antiparallels of a i with respect to thesides a i +1 and a i − . N4.1. ( T i , ta i ) is the tangential triangle ,N4.2. at i are the symmedians ,N4.3. K i is the triangle of Lemoine .N4.4. θ is the circumcircular polarity N4.5. λ is the Lemoine polarity . Theorem.
If we derive a point X and a line x by a given construction from A i , M and M , with thecoordinates as given in G0.0 and G0.1, below, and the point X and line x are obtain by thesame construction interchange M and M , X = ( f ( m , m , m ) , f ( m , m , m ) , f ( m , m , m )) ,x = [ g ( m , m , m ) , g ( m , m , m ) , g ( m , m , m )] , = ⇒ X = ( m f ( m − , m − , m − ) , m f ( m − , m − , m − ) , m f ( m − , m − , m − )) , CHAPTER 7. QUATERNIONIAN GEOMETRY x = [ m (cid:48) g ( m − , m − , m − ) , m (cid:48) g ( m − , m − , m − ) , m (cid:48) g ( m − , m − , m − )] . Proof: The point collineation C = q q
00 0 q , associates to (1,1,1), ( q , q , q ) , andto ( m , m , m ) , ( r , r , r ) , if r i = q i m i . In the new system of coordinates, X = ( q f ( q − r , q − r , q − r ) , q f ( q − r , q − r , q − r ) , q f ( q − r , q − r , q − r )) . Exchanging q i and r i and then replacing q i by 1 and r i by m i is equivalent to substituting m i for q i and 1 for r i , which gives X . x is obtained similarly.The line collineation is q (cid:48) q (cid:48)
00 0 q (cid:48) . Exercise.
Prove that if a point to line polarity [ P ) has its i, j -th element P ij = f ( m , m , m ) , then the i, j -th element of the polarity obtained by the same construction, after exchange of M and M , is P ij = m (cid:48) i f ( m − , m − , m − ) m − j . Similarly, for a line to point polarity ( P − ]( P − ] ij = g ( m , m , m ) , = ⇒ ( P − ] ij = m i g ( m − , m − , m − ) m j . Lemma. m − ( m + m )( m − m ) − = − ( m − + m − )( m − − m − ) − m − . This Lemma is useful in checking equivalent representations of coordinates of points andlines.
Notation. r i := ( m − i − + m − i ) − ( m − i +1 − m − i − ) ,s i := − ( m − i − + m − i ) − ( m − i + m − i +1 ) ,t i := s − i +1 s − i − ,f i := s i − s − i +1 s − i − ,g i := t i − t − i +1 t − i − . Lemma. s s s = − . norm ( t t t ) = 1 . s (cid:48) f s − = − f s . t (cid:48) f t − = − f t . .1. QUATERNIONIAN GEOMETRY OVER THE REALS. Proof: For 0, we use Lemma 7.1.1 and obtain 1, from the definition of t i . For 2, wesubstitute f by its definition and compare the terms of both sides of the equality which havethe same sign. Proof of 7.1.2 .LetG0.0. A = (1 , , , A = (0 , , , A = (0 , , , G0.1. M = (1 , , , M = ( m , m , m ) , thenP1.0. a = (1 , , , a = (0 , , , a = (0 , , , P1.1. ma = [0 , , − , ma = [0 , m (cid:48) , − m (cid:48) ] , P1.2. M = (0 , , , M = (0 , m , m ) , P1.3. eul = [1 , ( m − m ) − ( m − m ) , ( m − m ) − ( m − m )] , P1.4. S = − − − − − − , S − = . S = m − n − m (cid:48) m − − m (cid:48) m − − m (cid:48) m − m − n − m (cid:48) m − − m (cid:48) m − − m (cid:48) m − m − n , S − = m m m m m m m m m m m m . P2.0. mm = [1 , − , − , mm = [ m (cid:48) , − m (cid:48) , − m (cid:48) ] , P2.1.
M A = (0 , , − , M A = (0 , m , − m ) , P2.2. m = [0 , , , m = [0 , m (cid:48) , m (cid:48) ] , P2.3.
M M = (1 , − , − , M M = ( m , − m , − m ) , P2.4. m = [1 , , , m = [ m (cid:48) , m (cid:48) , m (cid:48) ] , P2.5.
Ima = (2 , − , − , Ima = (2 m , − m , − m ) , P2.6.
IM a = ( m + m , − m , − m ) , IM a = ( m ( m − + m − ) , − , − , P2.7. iM A = [2 , − , − , ıM A = [2 m (cid:48) , − m (cid:48) , − m (cid:48) ] , P3.0. mf = [( m + m ) (cid:48) ( m − m ) , , − ,mf = [ m (cid:48) ( m − + m − ) (cid:48) ( m (cid:48) − m (cid:48) ) , m (cid:48) , − m (cid:48) , , P3.1. O = ( m + m , m + m , m , m ) ,O = ( m ( m − + m − ) , m ( m − + m − ) , m ( m − + m − )) , P3.2.
M f a = (1 , , − ( m + m )( m − m ) − ) ,M f a = ( m , , − m ( m − + m − )( m − − m − ) − ) ,M f a = (1 , m (cid:48) ( m + m )( m − m ) − , ,M f a = ( m , m ( m − + m − )( m − − m − ) − , , P3.3. mf a = [( m + m ) (cid:48) ( m − m ) , , ,mf a = [ m (cid:48) ( m − + m − ) (cid:48) ( m (cid:48) − m (cid:48) ) , m (cid:48) , ,mf a = [( m + m ) (cid:48) ( m − m ) , , − ,mf a = [ m (cid:48) ( m − + m − ) (cid:48) ( m (cid:48) − m (cid:48) ) , , − m (cid:48) ] , P3.4.
M f m = (( m + m )( m − m ) − , − , ,M f m = ( m ( m − + m − )( m − − m − ) − , − m , m ) , CHAPTER 7. QUATERNIONIAN GEOMETRY
P4.0.
Imm = ( r , , s ) , Imm = ( − r , , s ) , P4.1. ta = [0 , , − s (cid:48) ] , P4.2. T = (1 , s , s − ) , P4.3. at = [0 , s (cid:48) , − s ] = [0 , , − t (cid:48) ] , P4.4. K = (1 , t , t − ) , P4.5.
T aa = (0 , , s ) , P4.6. poK = [ − , s (cid:48) , s ] , P4.7. T = f − f s − f − f s − s (cid:48) f − s f , T − = s s (cid:48) s s n s n s s − s n s s − n . P4.8. L = g − g t − g − g t − t (cid:48) g − t g , L − = t t (cid:48) t t n t n t t t − n t (cid:48) t − n . Details of proof:For P4.0, if the coordinates of
Imm are x , 1 and x , we have to solve x + 1 + x = 0 , − m − x + m − + m − x = 0 . Multiplying the equations to the left respectively by m − and -1, or by m − and 1 and addinggives x and x using the notation 7.1.2.For P4.7, it is easier to obtain T − first, the columns are T , T , T , multiplied to the rightby 1, s n , s − n . The matrix T is then obtained using Theorem 7.1.1, multiplying by − s − n . Theequivalence with the matrix whose columns are ta i can be verified using Lemma 7.1.2.2. Asimilar proof gives P4.8. It is trivialize by the notationb used for t . Theorem.
The product of the diagonal elements of T − and of L − is the same. This follows from Lemma 7.1.1.
Exercise.
Prove that the center of the circumcircular polarity is ( m (cid:48) ( m − + m − ) , m (cid:48) ( m − + m − ) , m (cid:48) ( m − + m − )) . Therefore, in general, it is distinct from O . From this follows, that, in general, mf i is notthe polar of M A i in the circumcircular polarity. .2. FINITE QUATERNIONIAN GEOMETRY. Definition.
Finite Quaternions over Z p are associative elements of the form q + q i + q j + q k , where q i are elements of Z p and i , j , k are such that i = j = − and k = ij = − ji . Theorem.i , j , k satisfy k = − , i = jk = − kj , j = ki = − ik . This follows at once from associativity.
Theorem.
Finite quaternions in Z p can be represented by 2 by 2 matrices over Z p .In particular, if j + j = −
1, then we can represent1 by (cid:18) (cid:19) , i by (cid:18) − (cid:19) , j by (cid:18) j j j − j (cid:19) , k by (cid:18) j − j − j − j (cid:19) . Comment. If p ≡ , we can find an interger j such that j = − , and choose j = 0 . Example. p = 5 , we can represent by (cid:18) (cid:19) , i by (cid:18) − (cid:19) , j by (cid:18) − (cid:19) , k by (cid:18) − − (cid:19) . p = 7 , we can represent by (cid:18) (cid:19) , i by (cid:18) − (cid:19) , j by (cid:18) − (cid:19) , k by (cid:18) − (cid:19) . Finite quaternions will be represented by an integer using the following notation.
Notation.
In the example the quaternion over Z p ,q = q + q i + q j + q k is represented by q = q + q p + q p + q p , ≤ q i < p. For instance, when p = 19, the representation of
18 + 3 i + 6 j is 2222, of
11 + 10 i + 4 j + 3 k is 22222, of
16 + 8 i + 18 j is 6666 and of
14 + 12 i + 13 j + 9 k is 66666. CHAPTER 7. QUATERNIONIAN GEOMETRY
Let p = 19 , G0.0. A = (1 , , , A = (0 , , , A = (0 , , , G0.1. M = (1 , , , M = (1 , , , thenE1.0. a = [1 , , , a = [0 , , , a = [0 , , , E1.1. ma i = [0,1,13827], [22219,0,1], [1,6378,0], ma i = [0,1,41987], [66657,0,1], [1,3333,0],E1.2. M i = (0,1,48176), (22219,0,1), (1,2222,0), M i = (0,1,21174), (70903,0,1), (1,6666,0),E1.3. eul = [1,35222,126587],E1.4 S = , S − = , S = , S − = , E2.0. mm i = [1,6378,22222], [2205,1,13827], [22219,48173,1], mm i = [1,3333,70894], [6653,1,41987], [66657,21177,1],E2.1. M A i = (0,1,82525), (115341,0,1), (1,5017,0), M A i = (0,1,116386), (66657,0,1), (1,573,0),E2.2. m i = [0,1,116874], [115341,0,1], [1,861,0], m i = [0,1,95573], [70903,0,1], [1,3906,0],E2.3. M M i = (1,5017,115338), (868,1,82525), (115341,116883,1), M M i = (1,573,70894), (3903,1,116386), (66657,95586,1),E2.4. m = [1,861,115338], m = [1,3906,66666],E2.5. Ima i = (1,6128,61279), (624,1,41443), (61290,123602,1), Ima i = (1,3906,35637), (2132,1,58383), (101928,116383,1),E2.6. IM a i = (1,31398,82872), (70470,1,745), (35569,2751,1), IM a i = (1,84862,112419), (112219,1,114203), (4535,17280,1),E2.7. iM A i = [2,6378,22222], [2205,2,13827], [22219,48173,2], ıM A i = [2,3333,70894], [6653,2,41987], [66657,21177,2],E3.0. mf i = [1,57399,96485], (22698,1,119282), (59539,116028,1), mf i = [1,17052,63592), (87814,1,43860), (49067,45624,1),E3.1. O = (1,39571,2622), O = (1,26376,18393).E3.2. M f a i = (1,0,59534), (57399,1,0), (0,119282,1), M f a i = (1,22693,0), (0,1,116019), (96498,0,1), M f a i = (1,0,49068), (17053,1,0), (0,43863,1), M f a i = (1,87803,0), (0,1,45633), (63575,0,1),E3.3. mf a i = (1,57399,0], [0,1,119282], [59539,0,1], mf a i = (1,0,96485], [22698,1,0], [0,116028,1], mf a i = (1,17052,0], [0,1,43860], [49067,0,1], mf a i = (1,0,63592], [87814,1,0], [0,45624,1],E3.4. M f m i = (1,57399,57265), (10093,1,49647), (92996,18940,1), .2. FINITE QUATERNIONIAN GEOMETRY. M f m i = (1,39191,23604), (90214,1,112020), (72715,69295,1),E3.5. iM A i = [1,6628,76281], [1112,1,7094], [76270,89247,1], ıM A i = [1,5096,35637], [3336,1,21174], [101928,79188,1],E4.0. Imm i = (1,101541,76547), (91854,1,115568),(74057,64703,1), Imm i = (1,36019,60652),(45706,1,21992), (63503,72857,1),E4.1. ta i = [0,1,19660], [64952,0,1], [1,51999,0],E4.2. T i = (1,115899,64951), (51988,1,114948), (39743,19651,1),E4.3. at i = [0,1,86571], [100052,0,1], [1,66787,0],E4.4. K i = (1,52716,100037), (66802,1,11323), (84095,86576,1),E4.5. T aa i = (0,1,114948), (39743,0,1), (1,115899,0),E4.6. poK i = [1,51999,39734], [115882,1,19660], [64952,1149331],E4.7. T = , T − = , E4.8. L = , L − = . Except for interchanges the computation of × is done as followswe normalize l to 1 l .P + l .P + P = 0 , l .Q + l .Q + Q = 0 ,Multiplying the first by P − .Q to the right and subtract from the second equation gives l ( Q − P P − Q ) + ( Q − P P − Q ) = 0 ,therefore if r := ( Q − P P − Q ) − and r = − ( Q − P P − Q ) , then l = the conjugate of r .r and l = − the conjugate of ( l ) p + p ) p − . The interchange is done as followsif P = 0 , then we exchange P i and Q i ,if after exchange, P = 0 , we consider all permutations sub0, sub1, sub2, of the subscripts 0,1 and 2 .Correspondance in Z between representation and quaternion. representation r i j k f or M ∗ M eul CHAPTER 7. QUATERNIONIAN GEOMETRY Ψ of Veblen-Wedderburn. Starting with the work of L. E. Dickson of 1905, non-Desarguesian planes of order 9 werediscovered by Veblen and Wedderburn in 1907, I will here consider only one of these whichis self dual, and for which non trivial polarities exists, and refer to the work of G. Zappa(1957), T. G. Ostrom (1964), D. R. Hughes (1957) and T. G. Room and P. B. Kirkpatrick(1971) for further reading.The synthetic definition used can be traced to Veblen and Wedderburn, who first considerpoints obtained by applying a transformation (see p. 383), later generalized by J. Singer.The notation is inspired by Room and Kirkpatrick (see Table 5.5.4) using the same methodI used for the finite plane reversing the indices for lines.An alternate definition, (5.6.1), is given by Room and Kirkpatrick.
Definition. A near-field ( N, + , ◦ ) is a set N with binary operations such that0. N is finite,1. ( N, +) is an Abelian group, with neutral element ( N − { } , ◦ ) is an group, with neutral element ◦ is right distributive over + , or ( ξ + η ) ◦ ζ = ξ ◦ ζ + η ◦ ζ, for all ξ, η, ζ ∈ N ξ ◦ , for all ξ ∈ N . Theorem.
In any near-field, ◦ ξ = 0 , for all ξ ∈ N .1. ξ ◦ η = 0 = ⇒ ξ = 0 or η = 0 . , − (cid:54) = 0 . Theorem.
In any near-field of order 9, { , , − } ≈ Z . ξ + ξ + ξ = 0 , for all ξ ∈ Q , .3. MINIQUATERNIONIAN PLANE Ψ OF VEBLEN-WEDDERBURN. − ◦ ξ = ξ ◦ ( −
1) = ξ, for all ξ ∈ Q , ( − ξ ) ◦ η = ξ ◦ ( − η ) = − ( ξ ◦ η ) , for all ξ, η ∈ Q , ( − ξ ) ◦ ( − η ) = ξ ◦ η, for all ξ, η ∈ Q , Given κ ∈ Q ∗ , λ = s − κr determines a one to one correspondance between the elements λ ∈ Q and the pairs ( r, s ), r, s ∈ Z . Q being an other near-field of order 9, the groups ( Q , +) and ( Q (cid:48) , +) are isomorphic. Besides GF(3 ) there is only one near-field of order 9, which is the smallest near-fieldwhich is not a field, (Zassenhaus, 1936). Exercise.
Determine the correspondance of 7.6.2.5.
Definition.
The miniquaternion set Q := { , ± , ± α, ± β, ± γ } with the operations of addition and mul-tiplications defined from, ξ + ξ + ξ = 0 for all ξ ∈ Q , α − β, α + 1 = γ,α = β = γ = αβγ = − .The set Q ∗ := {± α, ± β, ± γ } . Theorem. α − β = β − γ = γ − α = 1 , α + β + γ = 0 . βγ = − γβ = α, γα = − αγ = β, αβ = − βα = γ. the multiplication is right distributive, ( ρ + σ ) τ = ρτ + στ, for all ρ, σ, τ ∈ Q .3. { Q , + , . } is a near-field .4. { Q , + , . } is not a field , e. g. α ( α + β ) = α ( − γ ) = β , αα + αβ = − γ = α .5. + 1 − α − α β − β γ − γ − γ − β α − γ β − α − β − γ γ − α α − βα γ β − α − γ − β − − α − β − γ α − γ ββ α γ − γ − − β − α − β − γ − α γ β − αγ β α − β − α − − γ − γ − α − β − β α γ CHAPTER 7. QUATERNIONIAN GEOMETRY · − α − α β − β γ − γ − α − α β − β γ − γ − − − α α − β β − γ γα α − α − γ − γ − β β − α − α α − − γ γ β − ββ β − β − γ γ − α − α − β − β β γ − γ − − α αγ γ − γ β − β − α α − − γ − γ γ − β β α − α − Ψ . Definition.
With i, i (cid:48) ∈ { , , } , j ∈ { , , . . . , } , and the addition being performed modulo for thefirst element of a pair, and modulo , for the second element in the pair or for the element,if single, then the elements and incidence in the miniquaternionian plane Ψ are defined asfollows.0. The points P are ( j ) , ( i, j ) , ( i (cid:48) , j ) ,1. The lines l are [ j ] , [ i, j ] , [ i (cid:48) , j ] ,2. The incidence is defined by [ j ] := { ( − j ) , (1 − j ) , (3 − j ) , (9 − j ) , ( i, − j ) , ( i (cid:48) , − j ) } , [ i, j ] := { ( − j ) , ( i, − j ) , ( i, − j ) , ( i, − j ) , ( i (cid:48) , − j ) , ( i (cid:48) , − j ) , ( i (cid:48) + 1 , − j )( i (cid:48) + 1 , − j ) , ( i (cid:48) − , − j ) , ( i (cid:48) − , − j ) } , [ i (cid:48) , j ] := { ( − j ) , ( i (cid:48) , − j ) , ( i (cid:48) , − j ) , ( i (cid:48) , − j ) , ( i, − j ) , ( i, − j ) , ( i + 1 , − j )( i + 1 , − j ) , ( i − , − j ) , ( i − , − j ) } . Exercise. [ j ] is a subplane, [ i, j ] and [ i (cid:48) , j ] are copseudoplanes.0. Perform a similar representation of points, lines and incidence starting with a subplanewhich is a Fano plane.1. Determine similar representations for non Desarguesian geometries of order , usinga subplane of order 4, or of order 5 (651 = 31 · .
2. Determine other such representation for non Desarguesian geometries of higher order.
Theorem.
The same incidence relations obtain, if we interchange points and lines in .3. MINIQUATERNIONIAN PLANE Ψ OF VEBLEN-WEDDERBURN.
Theorem. [see Room and Kirkpatrick]
0. 0
The correspondance ( j ) to [ j ] and ( i, j ) to [ i, j ] and ( i (cid:48) , j ) to [ i (cid:48) , j ] is a polarity P ( J ∗ ). The 16 auto-poles are (0), (7), (8), (11), (0,8), (0,12), (1,4), (1,7), (2,10), (2,11),(0’,8), (0’,12), (1’,4), (1’,7), (2’,10), (2’,11).1. 0
The correspondance ( j ) to [ j ] and ( i, j ) to [ i (cid:48) , j ] and ( i (cid:48) , j ) to [ i, j ] is a polarity P ( J (cid:48)∗ ). The 22 auto-poles are (0), (7), (8), (11), (0,1), (0,3), (0,9), (1,1), (1,3), (1,9),(2,1), (2,3), (2,9), (0’,1), (0’,3), (0’,9), (1’,1), (1’,3), (1’,9), (2’,1), (2’,3), (2’,9).2 (0), (7), (8), (11), (0,1), (1,9), (2,3), (0’,9), (1’,3), (2’,1),(0), (7), (8), (11), (1,1), (2,9), (0,3), (2’,9), (0’,3), (1’,1),(0), (7), (8), (11), (2,1), (0,9), (1,3), (1’,9), (2’,3), (0’,1) are ovals.
Exercise.
0. Prove that the correspondance ( j ) to [ j ] and ( i, j ) to [( i + 1) (cid:48) , j ] and ( i (cid:48) , j ) to [ i − , j ] is a polarity P .1. Prove that the correspondance ( j ) to [ j ] and ( i, j ) to [( i − (cid:48) , j ] and ( i (cid:48) , j ) to [ i + 1 , j ] is a polarity P . Exercise.
0. Determine a configuration in 7.6.4.0.2, which gives an example were the Theorem ofPascal is satisfied and an other, in which it is not satisfied.1. Determine ovals which are subsets of 7.6.4.1.1.
Theorem.
The polar m of a point M with respect to a triangle is incident to that point. Indeed, we can always assume thast the triangle consists of the real points A = (0) , A = (1) , A = (2) , and that M = (5) = (1 , , . It follows that m = [4] = [1,1,1] which isincident to M . Exercise.
Check that the other points and lines of the polar construction are M i = (4) , (8) , (3) , M A i =(10) , (12) , (9) , M M i = (7) , (6) , (11) , a i = [12] , [1] , [0] , ma i = [9] , [8] , [11] , m i = [3] , [2] , [7] ,mm i = [6] , [10] , [5] . CHAPTER 7. QUATERNIONIAN GEOMETRY
Theorem. [see Room and Kirkpatrick] The planes obtained by taking the complete quadrangle associated with 3 real points A , A , A , and a point M which such that none of the lines M × A i are real are Fanoplanes associated with Z . There are ( . . .
24 = 5616 Fano planes that contain 3 real points.
Example.
The following is a Fano plane (0), (1), (2), (0,3), (2,1), (0’,12), (1,0), [12], [1], [0], [0’,0],[0,12], [2’,11], [0’,7].
Exercise.
Determine the Fano plane associated with (0), (1), (2), (0,7).
Comment.
The correspondance between the notation of Veblen-Wedderburn and Room-Kirkpatrick is
V eblen − Wedderburn a j b j c j d j e j f j g j Room − Kirkpatrick k j a j b j c j a (cid:48) j b (cid:48) j c (cid:48) j De V ogelaere [ − j ] [0 , − j ] [1 , − j ] [2 , − j ] [0 (cid:48) , − j ] [1 (cid:48) , − j ] [2 (cid:48) , − j ] V eblen − Wedderburn A j B j C j D j E j F j G j Room − Kirkpatrick K j A (cid:48) j C (cid:48) j B (cid:48) j A j C j B j De V ogelaere ( j ) (0 , j ) (1 , j ) (2 , j ) (0 (cid:48) , j ) (1 (cid:48) , j ) (2 (cid:48) , j ) Example. [Veblen-Wedderburn]
With the notation ( C (cid:104) c , c , c (cid:105) , { A , A , A }{ a , a , a } , { B , B , B }{ b , b , b } ; { C , C , C }{ d , d , d } ) , with d i := C i + j × C i − j , the following configuration shows that the Desargues axiom is not satisfied ((0) (cid:104) [0 , , [1 (cid:48) , , [2 , (cid:105) , { (0 , , (1 , , (1 (cid:48) , }{ [1 , , [0 (cid:48) , , [0 (cid:48) , } , { (2 , , (0 (cid:48) , (2 , }{ [0 , , [2 (cid:48) , , [10] } ; { (2 (cid:48) , , (0 , , (1 (cid:48) , }{ [2 , , [1 , , [0 , } ) . Definition.
The
Singer matrix G := . Its powers G k are the columns k, k + 1 , k + 2 of k = 0 1 2 3 4 5 6 7 8 9 10 11 121 0 0 1 0 1 1 1 − − −
10 1 0 1 1 1 − − − − − − Problem.
Can we characterize the plane Ψ using Theorem 7.6.4.0. .3. MINIQUATERNIONIAN PLANE Ψ OF VEBLEN-WEDDERBURN. move to g6a.tex:
Answer to 7.6.2. κ λ = 0 1 − α − α β − β γ − γα r − − − s − − − β r − − − s − − − γ r − − − s − − − Definition.
The elements and incidence in the miniquaternionian plane Ψ are defined as follows.0. The points are ( ξ , ξ , ξ ) with right equivalence,1.2. A point P is incident to a line l iff Definition. [Veblen-Wedderburn]
The points P are ( x, y, , ( x, , , (1 , , , the lines l are [1 , b, c ] , [0 , , c ] , [0 , , , and the incidence is P · l = 0 . Theorem. [Veblen-Wedderburn] [1 , b, c ] × [1 , b (cid:48) , c (cid:48) ] = ( − ( yb + c ) , y, , with y ( b − b (cid:48) ) = − ( c − c (cid:48) ) . [1 , b, c ] × [0 , , c (cid:48) ] = ( c (cid:48) b − c, − c (cid:48) , , [1 , b, c ] × [0 , ,
1] = ( − b, , , [0 , , c ] × [0 , , c (cid:48) ] = (1 , , , [0 , , c ] × [0 , ,
1] = (1 , , , Theorem. [Veblen-Wedderburn]
Let (a(b+c) = ab+ac)0. M := − − − . A := ( − , , , B := ( − γ, α, , C := ( β, − α, , D := ( − β, γ, , E := ( α, − γ, ,F := ( γ, − β, , G := ( − α, β, , CHAPTER 7. QUATERNIONIAN GEOMETRY A j := M j A , B j := M j B , . . . , for j = 0 to 12 ,3. a := [1 , , , b := [1 , α, , c := [1 , − α, , d := [1 , γ, , e := [1 , − γ, , f :=[1 , − β, , g := [1 , β, , then a = { A , A , A , A , B , C , D , E , F , G } ,b = { A , B , B , D , D , E , E , E , G , G } ,c = { A , C , C , E , E , F , F , G , G , G } ,d = { A , B , B , D , D , F , F , F , G , G } ,e = { A , B , B , B , C , C , E , E , F , F } ,f = { A , C , C , D , D , D , E , E , F , F } ,g = { A , B , B , C , C , C , D , D , G , G } , A = { a , a , a , a , b , c , d , e , f , g } ,B = { a , b , b , d , d , e , e , e , g , g } ,C = { a , c , c , e , e , f , f , g , g , g } ,D = { a , b , b , d , d , f , f , f , g , g } ,E = { a , b , b , b , c , c , e , e , f , f } ,F = { a , c , c , d , d , d , e , e , f , f } ,G = { a , b , b , c , c , c , d , d , g , g } , X j ι x k = ⇒ X j + l mod ι x k + l mod . M = − − −
11 1 1 , M = − − − , M = − − −
11 0 0 , M = − −
11 0 1 , M = − −
10 1 − − , M = − − − − , M = − − − , M = − − − , M = − −
10 0 − − − , M = − − , M = − − − −
11 1 0 , M = . Proof: [ x, y, z ] ι β ? iff x ( M k + M k ) + y ( M k + M k ) + z ( M k + M k ) + ( z M k + y M k + z M k ) β = 0 , . . . . Example. [Veblen-Wedderburn]
With the notation ( C (cid:104) c , c , c (cid:105) , { A , A , A }{ a , a , a } , { B , B , B }{ b , b , b } ; { C , C , C }{ d , d , d } ) , with d i := C i + j × C i − j , the following configuration shows that the Desargues axiom is not satisfied ( A (cid:104) b , f , d (cid:105) , { B , C , F }{ c , e , e } , { D , E , D }{ b , g , a } ; { G , B , F }{ d , c , b } ) . .3. MINIQUATERNIONIAN PLANE Ψ OF VEBLEN-WEDDERBURN.
Partial answer to 7.6.4.
For n = 7 , 2451 = 57.43, for n = 9 , 6643 = 91.73, for n = 11 , 14763 = 57.259, For n = 13 , 28731 = 3.9577. Answer to 7.6.4.
The other points are (1,1), (0’,12), (0’,0), the lines are [12], [1], [0], [1’,0], [0,12], [0,11]and the polar of (0,7) is [0’,6].
Answer to 7.6.4. (7) × (8) = [6] , [6] × [0] = (3) , (8) × (0) = [1] , [1] × [7] = (2) , (0) × (7) = [9] , [9] × [8] = (5) , (cid:104) (3) , (2) , (5); [11] (cid:105) . (7) × (8) = [6] , [6] × [0 ,
1] = (1 (cid:48) , , (8) × (0 ,
1) = [0 , , [0 , × [7] = (0 (cid:48) , , (0 , × (7) =[2 (cid:48) , , [2 (cid:48) , × [8] = (2 , , (2 , is not incident to (1 (cid:48) , × (0 (cid:48)
6) = [2 , . This has not been checked.From Dembowski, p. 129
Definition.
A linear ternary ring (Σ , + , · ) is called a cartesian field iff (Σ , +) is associative and is there-fore a group. Definition.
A cartesian field is called a quasifield iff the right distributivity law holds: ( x + y ) z = xz + yz. Artzy adds that xa = xb + c has a unique solution, but this is a property (28). This isVeblen-Wedderburn. Definition.
A quasifield is called a semifield iff the left distributivity law holds: z ( x + y ) = zx + zy. Definition.
A quasifield is called a nearfield iff (Σ , · ) is associative and is therefore a group. Definition.
A semifield is called a alternative field iff x y = x ( xy ) and xy = ( xy ) y. CHAPTER 7. QUATERNIONIAN GEOMETRY
Theorem. P is ( p, L ) transitive iff P is ( p, L ) Desarguesian. p is point, L is a line. Dembowski p.123,16 Let Q = (79) , Q = (80) , Q = (90) , and U = (81) then q = [79] , q = [90] , q = [80] ,v = [88] , i = [78] , V = (78) , I = (82) , j = [89] , W = (89) ,Points on q : q : ×
12 = [7] : 84(78 , , , , , , , , , , , , , , , , ×
25 = [61] : 82(78 , , , , , , , , , , , , , , , , ×
38 = [75] : 81(78 , , , , , , , , , , , , , , , , ×
51 = [4] : 87(86 , , , , , , , , , , , , , , , , ×
64 = [8] : 83(86 , , , , , , , , , , , , , , , , ×
77 = [68] : 88(86 , , , , , , , , , , , , , , , , Coordinates of points: (0) 64 ,
64 38 ,
80 77 ,
78 78 ,
51 38 ,
25 38 ,
64 51 ,
86 64 , ,
25 86 ,
77 77 ,
51 80 ,
77 12 51 ,
51 12 ,
80 64 , ,
77 12 ,
38 12 ,
51 77 ,
86 51 ,
38 77 ,
38 86 ,
64 64 , ,
64 25 77 ,
77 25 ,
80 51 ,
78 78 ,
64 25 ,
12 25 , ,
86 77 ,
12 64 ,
12 86 ,
51 51 ,
64 80 ,
51 38 25 , ,
80 38 ,
78 78 ,
12 77 ,
64 77 ,
25 12 ,
86 25 ,
64 12 , ,
38 38 ,
12 80 ,
38 51 12 ,
12 51 ,
80 25 ,
78 78 , ,
77 51 ,
12 38 ,
86 12 ,
77 38 ,
77 86 ,
25 25 ,
38 80 , ,
38 64 ,
80 12 ,
78 78 ,
25 64 ,
51 64 ,
38 25 , ,
51 25 ,
51 86 ,
12 12 ,
25 80 ,
12 77 78 80 , ,
78 78 ,
80 86 ,
78 78 ,
86 80 ,
86 86 86 , ,
80 80 , ∞ Coordinates of lines: [0] 78 ,
38 38 77 ,
78 12 ,
12 51 ,
77 86 ,
38 12 ,
86 12 , ,
77 77 ,
25 64 ,
12 80 ,
25 51 ,
80 78 ,
12 12 64 , ,
25 77 ,
64 86 ,
12 25 ,
86 25 ,
64 51 ,
64 64 ,
38 51 , ,
38 77 ,
80 78 ,
25 25 51 ,
78 38 ,
38 64 ,
51 86 , ,
86 38 ,
51 77 ,
51 51 ,
12 77 ,
38 80 ,
12 64 ,
80 78 , ,
78 51 ,
51 12 ,
38 86 ,
77 51 ,
86 51 ,
38 25 , ,
64 25 ,
51 80 ,
64 12 ,
80 78 ,
51 51 25 ,
78 64 , ,
25 86 ,
51 64 ,
86 64 ,
25 12 ,
25 25 ,
77 12 ,
64 80 , ,
80 78 ,
64 64 12 ,
78 77 ,
77 25 ,
12 86 ,
64 77 , ,
12 38 ,
12 12 ,
51 38 ,
77 80 ,
51 25 ,
80 78 , ∞ [80] 80 78 ,
78 86 86 ,
78 86 ,
86 80 ,
86 86 ,
80 78 , ,
78 80 , , , , , , , , , , , , , , , , , , , /inf ty ) B = A + α, ( A, B ) ι [ V, Y ] , V = (78) , (76) = (80 ,
12) = (0 , α ) , Y × V = (78) × (76) = [13] . [13] : 78(78) , , , , , , , , , , , , , , , , , , hence
12 = 80 + α,
51 = α + α = − α,
25 = 78 + α = 1 + α = γ,
64 = 25 + γ = γ + γ = − β,
38 = 86 + α = − α = β,
77 = 38 + α = β + α = − γ. .4. AXIOMATIC. ∞ − α − α β − β γ − γ
90 80 78 86 12 51 38 64 25 77[ α,
0] = (12) × (80 ,
80) = (12) × (79) = [51] , ( a, b ) ι [51] = ⇒ b = a · α. (42) = (78 ,
12) = (1 , α ) = ⇒ α = 1 × α, (45) = (12 , − ( α, −
1) = ⇒ − α × α, (23) = (64 ,
77) = ( − β, − γ ) = ⇒ − γ = − β × α, (28) = (51 ,
78) = ( − α,
1) = ⇒ − α × α, (35) = (86 ,
51) = ( − , − α ) = ⇒ − α = − × α, Using DATA 6,0, 6,4, 6,10, 6,12, 0,0, 1,0, 2,0, 3,0, 4,0, 5,0 DATA 6,0, 0,7, 0,8, 0,11,3,2, 3,10, 4,4, 4,6, 5,5, 5,12 gives the same multiplication table give left not right distibutivelaw with Q i = 79 , , , U = (83) , α = (12) , q = [87] = { , , , , , , . . . } ,q = [81] = { , , , , , , . . . } ,q = [79] = { , , , , , , . . . } , with case 7, data 79,81,87,83,12: ∞ = 87 , , , − , α = 12 , − α = 77 , β = 38 , − β = 51 , γ = 25 , − γ = 64 . This is a try for a section to be included in g19.tex between Moufang and Desargues.
Introduction.
The first example of a Veblen-Wedderburn plane was given in 1907 by Veblen and MacLagan-Wedderburn. It is associated to the algebraic structure of a nearfield, which is a skew fieldwhich lacks the left distributive law, hence is an other plane between the Veblen-Wedderburnplane and the Desarguesian plane.
Axiom. [Da] Given a Veblen-Wedderburn plane, 2 points Q and Q on the ideal line and an other point Q not on it, any 2 parallelograms A i and B i with directions Q and Q , with no sides incommon . . . ,???, such that A j and B j are perspective from Q for j = 0 To 2, imply that A and B are perspective from Q . Notation. Da ( { Q , Q , Q } , { A j } , { B j } ) . Definition. A Veblen-MacLagan plane is a Veblen-Wedderburn plane in which the axiom Da is satisfied. Da for Desargues leading to associativity of multiplication. CHAPTER 7. QUATERNIONIAN GEOMETRY
Lemma. [For Associativity]
H1.0. A , a , x, (See Fig. 2?.)D1.0. a := Q × A , a := Q × A , D1.1. A := a × a , A := a × a , D1.2. a := Q × A , a := Q × A , A := a × a , D2.0. a := Q × A , a := Q × A , a := Q × A , a := Q × A , D2.1. B := a × y,b := Q × B , b := Q × B , D2.1. B := b × b , B := b × b , D2.2. b := Q × B , b := Q × B , B := b × b , C1.0. B ι b , Moreover A = ( A, B ) , A = ( A (cid:48) , B ) , A = ( A, B (cid:48) ) , A = ( A (cid:48) , B (cid:48) ) , B = Proof: Da ( { Q , Q , Q } , { A j } , { B j } ) . Theorem.
In a Veblen-MacLagan plane, the ternary ring (Σ , ∗ ) is a nearfield. :0. (Σ , +) is an Abelian group, (Σ − { } , · ) is a group,
2. ( Σ , ∗ ) = (Σ , + , · ) is right distributive, ( a + b ) · c = a · c + b · c . Theorem. The Cayleyian plane is not a Veblen-MacLagan plane.replace Desarg.?
Definition. A miniquaternion plane . . . . Theorem. A miniquaternion plane is a Veblen-MacLagan plane. A miniquaternion plane is not a Moufang plane.
Tables.
The following are in an alternate notation the known table for p = 3 and a new table for p = 5. The other incidence are obtained by adding one to the subscripts of the lines andsubtracting one for the subscript of the points. Selectors for Ψ plane, when p = 3: (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , .4. AXIOMATIC. (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , Selectors for Ψ plane, when p = 5: (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [1 ] , [0 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [3 ] , (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [1 ] , [0 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [3 ] , (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [1 ] , [0 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [3 ] , (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [1 ] , [0 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [3 ] , (0 ) : [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [1 ] , [0 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [3 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [0 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [0 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [0 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [0 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , (1 ) : [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [0 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , (2 ) : [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [3 ] , [2 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [1 ] , (2 ) : [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [3 ] , [2 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [1 ] , (2 ) : [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [3 ] , [2 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [1 ] , (2 ) : [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [3 ] , [2 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [1 ] , (2 ) : [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [3 ] , [3 ] , [2 ] , [3 ] , [2 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [1 ] , (3 ) : [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [2 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , (3 ) : [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [2 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , (3 ) : [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [2 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , (3 ) : [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [2 ] , CHAPTER 7. QUATERNIONIAN GEOMETRY [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , (3 ) : [3 ] , [3 ] , [3 ] , [2 ] , [2 ] , [2 ] , [2 ] , [3 ] , [3 ] , [2 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , [0 ] , [0 ] , [0 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [1 ] , [0 ] , An abbreviated form is as follows:The array for the indices: i a i
24 29 14 1 20 b i c i
22 28 27 15 4 d i
16 19 8 10 9 e i
26 13 21 6 25
Selectors for the Ψ plane, when p = 5: (0 ) : [0 e ] , [0 e ] , [0 d ] , [1 e ] , [1 b ] , [1 e ] , [1 b ] , [0 c ] [0 c ] , [1 c ] , [2 b ] , [2 b ] , [2 a ] , [2 e ] , [2 a ] , [2 d ] , [2 a ] , [2 d ] , [3 d ] , [3 c ] , [3 d ] , [3 c ] , [2 a ] , [2 a ] , [3 b ] , (1 ) : [1 d ] , [1 d ] , [1 e ] , [0 b ] , [0 e ] , [0 b ] , [0 e ] , [1 b ] , [1 b ] , [0 c ] , [3 c ] , [3 c ] , [3 a ] , [3 d ] , [3 a ] , [3 e ] , [3 a ] , [3 e ] , [2 c ] , [2 d ] , [2 c ] , [2 d ] , [3 a ] , [3 a ] , [2 b ] , (2 ) : [2 e ] , [2 e ] , [2 d ] , [3 e ] , [3 b ] , [3 e ] , [3 b ] , [2 c ] [2 c ] , [3 c ] , [0 b ] , [0 b ] , [0 a ] , [0 e ] , [0 a ] , [0 d ] , [0 a ] , [0 d ] , [1 d ] , [1 c ] , [1 d ] , [1 c ] , [0 a ] , [0 a ] , [1 b ] , (3 ) : [3 d ] , [3 d ] , [3 e ] , [2 b ] , [2 e ] , [2 b ] , [2 e ] , [3 b ] , [3 b ] , [2 c ] , [1 c ] , [1 c ] , [1 a ] , [1 d ] , [1 a ] , [1 e ] , [1 a ] , [1 e ] , [0 c ] , [0 d ] , [0 c ] , [0 d ] , [1 a ] , [1 a ] , [0 b ] , I will attempt to generalize the results of quaternionian geometry to Desarguesian geometry.It is not clear to me now that polarities exist in general. Indded, we have seen that we canconstruct a system of homogeneous coordinates over a skew field, for which the incidenceproperty is Σ P i l i = 0 , with right equivalence for the lines l anf left equivalence for the points P . A line collineation can be represented by a matrix, m = C l l while a point collineation re-quires, P T = Q T C p , to allow for right and left equivalence. For a polarity, these equivalencesdo not appear to be compatible with a matrix transformation.It should be kept in mind that every skew field which is not a filed has a non trivial subfield .5. DESARGUESIAN GEOMETRY. generated by 1, which can be finite (Ore) or not. This implies that given 4 points forming acomplete quadrangle, there exist a Pappian subgeometry through these 4 points, the elementsof which are obtained from the linear constructions which start from these 4 points. Theorem.
In any skew field, if a matrix A has a left inverse and a right inverse, these are equal. Proof: Let C be the left inverse of A and B be its right inverse, by associativity ofmatrices, C = C ( AB ) = ( CA ) B = B . Theorem. IF C p is a point collineation, the line collineation C l is C p − . In particular, the point collineation which associates to A i , A i and to (1 , , , ( q , q , q ) is q q
00 0 q , and the line collineation is q − q −
00 0 q − . Proof: If Q is the image of P , and m is the image of l , we want P · l = Σ P i l i = Σ Q i C p ij C l jk m k = Σ Q i m i = 0 , for all points P and incident lines l iff C l = C p − . Notation.
In what follows, I will use the same notation as in involutive Geometry, namely, l := P × Q, means that the line l is defined as the line incident to P and Q. If subscripts are used these have the values 0, 1 and 2 and the computation is done modulo3, P · l = 0 means that the point P is incident to the line l. When 3 lines intersect, this intersection can be defined in 3 ways, this has been indicated byusing (*) after the definition and implies a Theorem.The labeling used is “H,” for Hypothesis, “D”, for definitions, “C”, for conclusions, “N”, fornomenclature, “P”, for proofs, this labelling being consistent with that of the correspondingdefinitions.
The special configuration of Desargues.
With this notation, the special configuration of Desargues can be defined by a i := A i +1 × A i − , qa i = Q × A i ,Q i := a i × qa i , qq i := Q i +1 × Q i − , CHAPTER 7. QUATERNIONIAN GEOMETRY QA i := a i × qq i , q i := A i × QA i ,QQ i := q i +1 × q i − , q := QA × QA ( ∗ ) , and the other conclusion of the special Desargues Theorem can be written, QQ i · qa i = 0 . Let Q and A i be Q = ( q , q , q ) , and A = (1 , , , A = (0 , , , A = (0 , , , then we have the following results, not obtained in the given order, A = (1 , , , a = [1 , , ,Q = ( q , q , q ) , q = [ q − , q − , q − ] ,QA = (0 , q , − q ) , qa = [0 , q − , − q − ] ,Q = (0 , q , q ) , q = [0 , q − , q − ] ,QQ = ( − q , q , q ) , qq = [ − q − , q − , q − ] , The self duality of the configuration corresponds to the replacement of points by lineswhere upper case letters are replaced by lower case letters and coordinates by their inverse.
Fundamental Hypothesis, Definitions and Conclusions.
The ideal line and the coideal line.GivenH0.0. A i , H0.1.
M, M ,
LetD1.0. a i := A i +1 × A i − , D1.1. ma i := M × A i , ma i := M × A i , D1.2. M i := ma i × a i , M i := ma i × a i , D1.3. eul = M × M ,
D2.0. mm i := M i +1 × M i − , mm i := M i +1 × M i − , D2.1.
M A i := a i × mm i , M A i := a i × mm i , D2.2. m i := A i × M A i , m i := A i × M A i , D2.3.
M M i := m i +1 × m i − , M M i := m i +1 × m i − , D2.4. m := M A × M A ( ∗ ) , m := M A × M A ( ∗ ) , D2.5.
Ima i := m × ma i , Ima i := m × ma i , D2.6.
IM a i := m × ma i , IM a i := m × ma i , D2.7. iM A i := M × M A i , ıM A i := M × M A i , LetD3.0. mf i := M i × IM a i , mf i := M i × IM a i , D3.1. O := mf × mf ( ∗ ) , O := mf × mf ( ∗ ) , D3.2.
M f a i := a i +1 × mf i − , M f a i := a i +1 × mf i − ,M f a i := a i − × mf i +1 , M f a i := a i − × mf i +1 , D3.3. mf a i := M f a i +1 × A i − , mf a i := M f a i +1 × A i − ,mf a i := M f a i − × A i +1 , mf a i := M f a i − × A i +1 , D3.4.
M f m i := mf a i × m i , M f m i := mf a i × m i , thenC3.0. O · eul = O · eul = 0 . C3.1.
M f m i · mf a i = M f m i · mf a i = 0 . .5. DESARGUESIAN GEOMETRY. LetD4.0.
Imm i := m × mm i , Imm i := m × mm i , D4.1. ta i := A i × Imm i , D4.2. T i := ta i +1 × ta i − , D4.3. at i := A i × T i , D4.4. K i := at i +1 × at i − , D4.5.
T Aa i := ta i × a i , D4.6. poK i := T aa i +1 × T aa i − , thenC4.0. Imm i · ta i = 0 . C4.1. T i · mf i = 0 . The nomenclature:N0.0. A i are the vertices of the triangle,N0.1. M is the barycenter , M is the orthocenter .N1.0. a i are the sides .N1.1. ma i are the medians , ma i are the altitudes N1.2. M i are the mid-points of the sides. M i are the feet of the altitudes N1.3. eul is the line of Euler ,N2.0. { M i , mm i } is the complementary triangle , { M i , mm i } is the orthic triangle ,N2.1. M A i are the directions of the sides ,N2.2. { M M i , m i } is the anticomplementary triangle .N2.3. m is the ideal line corresponding to the line at infinity, m is the orthic line which is the polar of M with respect to the triangle.N2.4. Ima i are the directions of the medians . IM a i are the directions of the altitudes .N3.0. mf i are the mediatrices ,N3.1. O is the center ,N3.2. M f m i are the trapezoidal points ,N4.0. Imm i are the directions of the antiparallels of a i with respect to thesides a i +1 and a i − . N4.1. ( T i , ta i ) is the tangential triangle ,N4.2. at i are the symmedians ,N4.3. K i is the triangle of Lemoine . Theorem.
If we derive a point X and a line x by a given construction from A i , M and M , with thecoordinates as given in G0.0 and G0.1, below, and the point X and line x are obtain by thesame construction interchange M and M , X = ( f ( m , m , m ) , f ( m , m , m ) , f ( m , m , m )) ,x = [ g ( m , m , m ) , g ( m , m , m ) , g ( m , m , m )] , = ⇒ X = ( f ( m − , m − , m − ) m , f ( m − , m − , m − ) m , f ( m − , m − , m − ) m ) ,x = [ m − g ( m − , m − , m − ) , m − g ( m − , m − , m − ) , m − g ( m − , m − , m − )] . CHAPTER 7. QUATERNIONIAN GEOMETRY
Proof: The point collineation C p = q q
00 0 q , associates to (1,1,1), ( q , q , q ) , and to ( m , m , m ) , ( r , r , r ) , if r i = m i q i . In the new system of coordinates, X = ( f ( q − r , q − r , q − r ) q , f ( q − r , q − r , q − r q ) , f ( q − r , q − r , q − r ) q ) . Exchanging q i and r i and then replacing q i by 1 and r i by m i is equivalent to substituting m i for q i and 1 for r i , which gives X . x is obtained similarly.The line collineation is q − q −
00 0 q − . Notation. a i := ( m − i +1 − m − i − )( m − i − + m − i ) − ,s i := − ( m − i + m − i +1 )( m − i + m − i − ) − ,t i := s i +2 s i +1 ,f i := s i − s − i +1 s − i − ,g i := t − i − t i +1 t i − . Proof of .LetG0.0. A = (1 , , , A = (0 , , , A = (0 , , , G0.1. M = (1 , , , M = ( m , m , m ) , thenP1.0. a = (1 , , , a = (0 , , , a = (0 , , , P1.1. ma = [0 , , − , ma = [0 , m − , − m − ] , P1.2. M = (0 , , , M = (0 , m , m ) , P1.3. eul = [1 , ( m − m ) − ( m − m ) , ( m − m ) − ( m − m )] , P2.0. mm = [1 , − , − , mm = [ m − , − m − , − m − ] , P2.1.
M A = (0 , , − , M A = (0 , m , − m ) , P2.2. m = [0 , , , m = [0 , m − , m − ] , P2.3.
M M = (1 , − , − , M M = ( m , − m , − m ) , P2.4. m = [1 , , , m = [ m − , m − , m − ] , P2.5.
Ima = (2 , − , − , Ima = (2 m , − m , − m ) , P2.6.
IM a = ( m + m , − m , − m ) , IM a = (( m − + m − ) m , − , − , P2.7. iM A = [2 , − , − , ıM A = [2 m − , − m − , − m − ] , P3.0. mf = [( m + m ) − ( m − m ) , , − ,mf = [ m − ( m − + m − ) − ( m − − m − ) , m − , − m − , , P3.1. O = ( m + m , m + m , m + m ) ,O = (( m − + m − ) m , ( m − + m − ) m , ( m − + m − ) m ) , P3.2.
M f a = (1 , , − ( m − m ) − ( m + m )) ,M f a = ( m , , ( m − − m − )( m − + m − ) − m ) ,M f a = (1 , ( m − m ) − ( m + m ) , ,M f a = ( m , ( m − − m − ) − ( m − + m − ) m , , P3.3. mf a = [( m + m ) − ( m − m ) , , , .5. DESARGUESIAN GEOMETRY. mf a = [ m − ( m − + m − ) − ( m − − m − ) , m − , ,mf a = [( m + m ) − ( m − m ) , , − ,mf a = [ m − ( m − + m − ) − ( m − − m − ) , , m − ] , P3.4.
M f m = (( m + m )( m − m ) − , − , ,M f m = (( m − − m − ) − ( m − + m − ) m , − m , m ) , P4.0.
Imm = ( a , , s ) , Imm = ( − a , , s ) , P4.1. ta = [0 , , − s − ] , P4.2. T = (1 , s , s − ) , ? P4.3. at = [0 , s − , − s ] = [0 , , − t ] , P4.4. K = (1 , t − , t ) , P4.5.
T aa = (0 , , s ) , P4.6. poK = [ − , s − , s ] , Details of proof:For P4.0, if the coordinates of
Imm are x , 1 and x , we have to solve x + 1 + x = 0 , − x m − + m − + x m − = 0 . Multiplying the equations to the right respectively by m − , and -1 or by m − and 1 and adding gives x and x using the notation 7.5.1. Definition.
Given M and a direction I x , the perpendicular direction I y is defined by the following constructionD5.0. b := A × I x , D5.1. B := b × a , D5.2. c := B × IM a , D5.3. C := c × ma , D5.4. d := C × A , D5.5. I y := d × m, Theorem. If I x = ( − − q, q,
1) and I y = (1 − r, r,
1) then r = . Proof:P5.0. b = [0 , − q − , , P5.1. B = (0 , q, , P5.2. c = [ x, − q − , , with x = m − ( m + m + m q − ) , P5.3. C = ( y, m , m ) , with y = ( m q − − m ) x − , P5.4. d = [ y − , , − m − ] , P5.5. I y = ( y, z, m ) , with z = − y − m ,Therefore r = − m − ( y + m ) = − m − ( m + ( m q − − m )( m + m + m q − ) − m ) . If the skew field we therefore have − m ( r + 1) m − (( m + m ) q + m ) = m − m q . − rm − ( m + m ) q − rm − m − m − m q − m − m = 0 .which is, in general, not an involution. CHAPTER 7. QUATERNIONIAN GEOMETRY
There are essentially 2 methods to algebraize a plane. The first one which start with the work ofDesargues coordinatized the plane using 2 coordinates, the difficulty of representing the ideal pointsor points at infinity can be dealt with by using 3 homogeneous coordinates. This approach hasbeen generalized to perspective planes, for which the only axioms are those of incidence, by usingas coordinates, elements of a ternary ring instead of elements in a field. This generalization wasgiven by Marshall Hall in 1943, but its origin can be found, for the case of nearfields, introducedby Dickson (1905), in the most remarkable paper of Veblen and MacLagan-Wedderburn in 1907 (p.380-382).In this paper they give, independently from Vahlen the first example of non Pappian Geometry.The indpendent result consited in showing that quaternions could be used as coordinates for such ageometry.The second approach, which can be used in finite planes, is to construct a difference set, of q = p k integers as a subset { , . . . , q + q } from which the points incident to each line, and the lines incidentto every point can be completely derived. This approach was fully examined for the finite Pappianplanes by J. Singer in 1938, but it again can be traced in the paper of 1907 (p. 383 and 385).Moreover, the generalization to non Desarguesian planes is given explicitely for a plane of order 9,called Ψ plane by Room and Kirkpatrick.I do prefer, when applying the notion of difference sets to geometry, to use, instead of it, theterminology of selector introduced by Fernand Lemay, in 1979. (See his most accessible paper of1983.)It is the second approach, that I am exploring in this paper, gives many of the results in the formof conjectures.We will see that to give the incidence properties for planes of the Ψ type and order p we have togive p selectors of p elements, and in a particular notation the points which are incident to p − lines from which all other incidences can be derived. The notation is such that the same incidencetables are valid for the points on any line, giving rise to a fundamental polarity.One of the advantages of the selector approach is to eliminate the need of addition and multiplicationtables in the particular nearfield which greatly simplifies the exploration of new properties with acomputer. Many of the planes are special case of Hughes planes, hence the tiltle of the section.I will assume that p is an odd prime. Definition. [Dickson] A left nearfield ( N, + , ◦ ) is a set N with binary operations such that0. N is finite,1. ( N, +) is an Abelian group, with neutral element ( N − { } , ◦ ) is an group, with neutral element ◦ is left distributive over + , or ζ ◦ ( ξ + η ) = ζ ◦ ξ + ζ ◦ η, for all ξ, η, ζ ∈ N .6. THE HUGHES PLANES. ◦ ξ = 0 , for all ξ ∈ N .For a right nearfield, the left distributive law is replaced by the right one and ◦ ξ = 0 , isreplaced by ξ ◦ . Theorem.
In any left nearfield, ξ ◦ for all ξ ∈ N .1. ξ ◦ η = 0 = ⇒ ξ = 0 or η = 0 . , − (cid:54) = 0 . Definition. [Dickson]
Let n be a non residue of p . A Dickson left nearfield ( N, + , ◦ ) is a set N with the operations ( a + b α ) + ( a + b α ) := (( a + a ) + ( b + b ) α ) , ( a + b α ) · ( a + b α ) := (( a a + e n b b ) + ( a b + e a b ) α ) , where e = +1 if a − n b is a quadratic residue of p and e = − otherwize.A Dickson right nearfield is obtained by the replacement of ( a b + e a b ) α ) , by ( a b + e a b ) α ) . Theorem.
A Dickson left nearfield is a left nearfield.
Definition.
Let β, γ , ξ and η are elements of a Dickson nearfield. In a Hughes plane , the points are the triples (1 , β, γ ) , (0 , , γ ) , (0 , , , and the lines are the triples [ η, ξ, − , [ η, − , , [1 , , A point ( P , P , P ) is incident to a line [ l , l , l ] if and only if P l + P l + P l = 0 .A point or a line is real if the coefficient of α in its coordinates are 0. A point or line is complex ,otherwize.The notation is used to indicate the close relationship with the corresponding coordinates in aternary ring, see for instance Artzy, p. 203-203,for the points: ( b, c ) = (1 , b, c ) , ( c ) = (0 , , c ) , ( ∞ ) = (0 , , , for the lines: [ x, y ] = [ y, x, − , [ y ] = [ y, , , [ ∞ ] = [1 , , indeed · y + b · x − c = 0 corresponds to c = b · x + y , giving the ternary ring conditions of incidence. Theorem. [Hughes]
A Hughes plane is of Lenz-Barlotti type I.1
See Hughes, Rosati and Dembowski, p. 247. The simplest case p = 3, is given by Veblen andMacLagan-Wedderburn p. 383, it is called in this case a Ψ plane by Room and Kirkpatrick. CHAPTER 7. QUATERNIONIAN GEOMETRY
Theorem.
A real line has p + 1 real and p − p complex points incident to it.A complex line has 1 real and p complex points incident to it. See, for instance, Room and Kirkpatrick.
Theorem.
The p selectors and the negative inverses of the fundamental selector modulo p + p + 1 form apartition of the set { , . . . , p + p } . Theorem.
The p + p + 1 points are partitioned into p + p + 1 real points and p − p classes. Each class consists of p + p + 1 points, which form by definitiona coplane. Starting with the work of L. E. Dickson of 1905, non-Desarguesian planes of order 9 werediscovered by Veblen and Wedderburn in 1907, I will here consider only one of these which is selfdual, and for which non trivial polarities exists, and refer to the work of G. Zappa (1957), T. G.Ostrom (1964), D. R. Hughes (1957) and T. G. Room and P. B. Kirkpatrick (1971) for furtherreading.The synthetic definition used can be traced to Veblen and Wedderburn, who first consider pointsobtained by aplying a transformation (see p. 383), later generalized by J. Singer. The notationis inspired by Room and Kirkpatrick (see Table 5.5.4) using the same method I used for the finiteplane reversing the indices for lines.An alternate definition, (5.6.1), is given by Room and Kirkpatrick.
Theorem.
In any left nearfield Q , of order 9, { , , − } ≈ Z . ξ + ξ + ξ = 0 , for all ξ ∈ Q , − ◦ ξ = ξ ◦ ( −
1) = ξ, for all ξ ∈ Q , ( − ξ ) ◦ η = ξ ◦ ( − η ) = − ( ξ ◦ η ) , for all ξ, η ∈ Q , ( − ξ ) ◦ ( − η ) = ξ ◦ η, for all ξ, η ∈ Q , Given κ ∈ Q ∗ := Q − , λ = s − κr determines a one to one correspondance between theelements λ ∈ Q and the pairs ( r, s ), r, s ∈ Z . Q being an other nearfield of order 9, the groups ( Q , +) and ( Q (cid:48) , +) are isomorphic. Besides GF(3 ) there is only one nearfield of order 9, which is the smallest nearfield which isnot a field, (Zassenhaus, 1936). .6. THE HUGHES PLANES. Exercise.
Determine the correspondance of 7.6.2.5.
Definition.
The left miniquaternions is the set Q := { , ± , ± α, ± β, ± γ } with the operations of addition andmultiplications defined from, ξ + ξ + ξ = 0 for all ξ ∈ Q , α − β, α + 1 = γ,α = β = γ = − αβγ = − .The set Q ∗ := {± α, ± β, ± γ } .For the right miniquaternions, we replace αβγ = 1 by αβγ = − . Theorem. α − β = β − γ = γ − α = 1 , α + β + γ = 0 . − βγ = γβ = α, − γα = αγ = β, − αβ = βα = γ. the multiplication is left distributive, τ ( ρ + σ ) = τ ρ + τ σ, for all ρ, σ, τ ∈ Q .3. { Q , + , . } is a left nearfield .4. { Q , + , . } is not a field , e. g. α ( α + β ) = α ( − γ ) = β , αα + αβ = − γ = α .5. + 1 − α − α β − β γ − γ − γ − β α − γ β − α − β − γ γ − α α − βα γ β − α − γ − β − − α − β − γ α − γ ββ α γ − γ − − β − α − β − γ − α γ β − αγ β α − β − α − − γ − γ − α − β − β α γ · − α − α β − β γ − γ − α − α β − β γ − γ − − − α α − β β − γ γα α − α − − γ γ β − β − α − α α − γ − γ − β ββ β − β γ − γ − − α α − β − β β − γ γ − α − αγ γ − γ − β β α − α − − γ − γ γ β − β − α α − For the right miniquaternions, we change the sign of the products in 1. and exchange rows andcolumns for the multiplication table, e.g. αβ = γ. CHAPTER 7. QUATERNIONIAN GEOMETRY
Definition. [Veblen-Wedderburn]
The points P are ( x, y, , ( x, , , (1 , , , the lines l are [1 , b, c ] , [0 , , c ] , [0 , , , and the incidence is P · l = 0 , where x, y, b and c are elements of a left nearfield. Theorem. [Veblen-Wedderburn]
With b , c , b (cid:48) , c (cid:48) in Q , [1 , b, c ] × [1 , b (cid:48) , c (cid:48) ] = ( − ( yb + c ) , y, , with y ( b − b (cid:48) ) = − ( c − c (cid:48) ) . [1 , b, c ] × [0 , , c (cid:48) ] = ( c (cid:48) b − c, − c (cid:48) , , [1 , b, c ] × [0 , ,
1] = ( − b, , , [0 , , c ] × [0 , , c (cid:48) ] = (1 , , , [0 , , c ] × [0 , ,
1] = (1 , , , Theorem. [Veblen-Wedderburn]
Let ( a ( b + c ) = a b + a c ) M := − − − , A := ( − , , , B := ( − γ, α, , C := ( β, − α, , D := ( − β, γ, , E := ( α, − γ, ,F := ( γ, − β, , G := ( − α, β, , A j := M j A , B j := M j B , . . . , for j = 1 to 12 ,3. a j := { M j X i } , X i ∈ a , and similarly for b j to g j .4. a := [1 , , , b := [1 , α, , c := [1 , − α, , d := [1 , γ, , e := [1 , − γ, , f := [1 , − β, , g := [1 , β, , then M is of order 13.6. a = { A , A , A , A , B , C , D , E , F , G } ,b = { A , B , B , D , D , E , E , E , G , G } ,c = { A , C , C , E , E , F , F , G , G , G } ,d = { A , B , B , D , D , F , F , F , G , G } ,e = { A , B , B , B , C , C , E , E , F , F } ,f = { A , C , C , D , D , D , E , E , F , F } ,g = { A , B , B , C , C , C , D , D , G , G } , A = { a , a , a , a , b , c , d , e , f , g } ,B = { a , b , b , d , d , e , e , e , g , g } ,C = { a , c , c , e , e , f , f , g , g , g } ,D = { a , b , b , d , d , f , f , f , g , g } , .6. THE HUGHES PLANES. E = { a , b , b , b , c , c , e , e , f , f } ,F = { a , c , c , d , d , d , e , e , f , f } ,G = { a , b , b , c , c , c , d , d , g , g } , X j ι x k = ⇒ X j + l mod ι x k + l mod . Given a , to g , it is easy to verify that A is on all these lines and determine B to G all on a and B on b , C on c , . . . .Having determined, the other points using 2, it is easy to verify which points are on b , . . . .The notation helps gretaly in justifyong that 2 points have one and only one line in common and 2liners have only one point in common. The notation can be made even more compact. See 7.6.4.The following are the powers of M . M = − − −
11 1 1 , M = − − − , M = − − −
11 0 0 , M = − −
11 0 1 , M = − −
10 1 − − , M = − − − − , M = − − − , M = − − − , M = − −
10 0 − − − , M = − − , M = − − − −
11 1 0 , M = . Example. [Veblen-Wedderburn]
With the notation non − Desargues ( C (cid:104) c , c , c (cid:105) , { A , A , A }{ a , a , a } , { B , B , B }{ b , b , b } ; { C , C , C }{ d , d , d } ) , with d i := C i + j × C i − j , the following configuration shows that the Desargues axiom is not satisfied non − Desargues ( A (cid:104) b , f , d (cid:105) , { B , C , F }{ c , e , e } , { D , E , D }{ b g , a } ; { G , B , F }{ d , c , b } ) . Ψ . Definition.
With i ∈ { , , } , i (cid:48) ∈ { (cid:48) , (cid:48) , (cid:48) } , j ∈ { , , . . . , } , and the addition being performed modulo forthe first element of a pair, and modulo , for the second element in the pair or for the element,if single, then the elements and incidence in the miniquaternionian plane Ψ are defined as follows.(See 7.6.40. The 91 points P are ( j ) , ( i, j ) , ( i (cid:48) , j ) ,1. The 91 lines l are [ j ] , [ i, j ] , [ i (cid:48) , j ] ,2. The incidence is defined by [ j ] := { ( − j ) , (1 − j ) , (3 − j ) , (9 − j ) , ( i, − j ) , ( i (cid:48) , − j ) } , [ i, j ] := { ( − j ) , ( i, − j ) , ( i, − j ) , ( i, − j ) , ( i (cid:48) + 1 , − j ) , ( i (cid:48) + 1 , − j ) , ( i (cid:48) − , − j )( i (cid:48) − , − j ) , ( i (cid:48) , − j ) , ( i (cid:48) , − j ) } , [ i (cid:48) , j ] := { ( − j ) , ( i (cid:48) , − j ) , ( i (cid:48) , − j ) , ( i (cid:48) , − j ) , ( i − , − j ) , ( i − , − j ) , ( i + 1 , − j )( i + 1 , − j ) , ( i, − j ) , ( i, − j ) } . giving the 10 points on each line. i and i (cid:48) in the same definiton correspond to the same integer, 0,03-’ or 1, 1’ . . . . CHAPTER 7. QUATERNIONIAN GEOMETRY
Exercise. [ j ] is a subplane, [ i, j ] and [ i (cid:48) , j ] are copseudoplanes.0. Perform a similar representation of points, lines and incidence starting with a subplane whichis a Fano plane.1. Determine similar representations for non Desarguesian geometries of order , using a sub-plane of order 4, or of order 5 (651 = 31 · .
2. Determine other such representation for non Desarguesian geometries of higher order.
Theorem.
The same incidence relations obtain, if we interchange points and lines in
Theorem. [see Room and Kirkpatrick]
0. 0
The correspondance ( j ) to [ j ] and ( i, j ) to [ i, j ] and ( i (cid:48) , j ) to [ i (cid:48) , j ] is a polarity P ( J ∗ ). The 16 auto-poles are (0), (7), (8), (11), (0,8), (0,12), (1,4), (1,7), (2,10), (2,11),(0’,8), (0’,12), (1’,4), (1’,7), (2’,10), (2’,11).1. 0
The correspondance ( j ) to [ j ] and ( i, j ) to [ i (cid:48) , j ] and ( i (cid:48) , j ) to [ i, j ] is a polarity P ( J (cid:48)∗ ). The 22 auto-poles are (0), (7), (8), (11), (0,1), (0,3), (0,9), (1,1), (1,3), (1,9), (2,1),(2,3), (2,9), (0’,1), (0’,3), (0’,9), (1’,1), (1’,3), (1’,9), (2’,1), (2’,3), (2’,9).2 (0), (7), (8), (11), (0,1), (1,9), (2,3), (0’,9), (1’,3), (2’,1),(0), (7), (8), (11), (1,1), (2,9), (0,3), (2’,9), (0’,3), (1’,1),(0), (7), (8), (11), (2,1), (0,9), (1,3), (1’,9), (2’,3), (0’,1) are ovals.
Exercise.
0. Prove that the correspondance ( j ) to [ j ] and ( i, j ) to [( i + 1) (cid:48) , j ] and ( i (cid:48) , j ) to [ i − , j ] is apolarity P .1. Prove that the correspondance ( j ) to [ j ] and ( i, j ) to [( i − (cid:48) , j ] and ( i (cid:48) , j ) to [ i + 1 , j ] is apolarity P . Exercise.
0. Determine a configuration in 7.6.4.0.2, which gives an example were the Theorem of Pascalis satisfied and an other, in which it is not satisfied.1. Determine ovals which are subsets of 7.6.4.1.1.
Theorem.
The polar m of a point M with respect to a triangle is incident to that point. Indeed, we can always assume thast the triangle consists of the real points A = (0) , A = (1) , A = (2) , and that M = (5) = (1 , , . It follows that m = [4] = [1,1,1] which is incident to M . .6. THE HUGHES PLANES. Exercise.
Check that the other points and lines of the polar construction are M i = (4) , (8) , (3) , M A i =(10) , (12) , (9) , M M i = (7) , (6) , (11) , a i = [12] , [1] , [0] , ma i = [9] , [8] , [11] , m i = [3] , [2] , [7] , mm i =[6] , [10] , [5] . Theorem. [see Room and Kirkpatrick] The planes obtained by taking the complete quadrangle associated with 3 real points A ,A , A , and a point M which such that none of the lines M × A i are real are Fano planesassociated with Z . There are ( . . .
24 = 5616 Fano planes that contain 3 real points.
Notation.
For a Fano subplane with 7 elements, I will use the notation associated with the selector , , andconstruction:Given a complete quadrangle , , , , ∗ := 0 × , ∗ := 1 × , ∗ := 2 × , ∗ := 2 × , ∗ := 0 × , ∗ := 1 × , ∗ × ∗ , ∗ × ∗ , ∗ × ∗ , ∗ := 3 x . The Fano plane propery implies ι ∗ .The configuration is denoted by F ano (0 , , , , , , , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ , ∗ ) . Example.
The following is a Fano plane configuration:
F ano ((0) , (1) , (2) , (2 , , (1 , , (0 , , (1 (cid:48) , , [0] , [1] , [0 , , [1 (cid:48) , , [1 (cid:48) , , [0 (cid:48) , , [12]) . Exercise.
Determine the Fano plane associated with (0), (1), (2), (0,7).
Comment.
The correspondance between the notation of Veblen-Wedderburn and Room-Kirkpatrick is
V eblen − Wedderburn a j b j c j d j e j f j g j Room − Kirkpatrick k j a j b j c j a (cid:48) j b (cid:48) j c (cid:48) j De V ogelaere [ − j ] [0 , − j ] [2 (cid:48) , − j ] [1 , − j ] [0 (cid:48) , − j ] [1 (cid:48) , − j ] [2 , − j ] V eblen − Wedderburn A j B j C j D j E j F j G j Room − Kirkpatrick K j A (cid:48) j C (cid:48) j B (cid:48) j A j C j B j De V ogelaere ( j ) (0 (cid:48) , j ) (2 , j ) (1 (cid:48) , j ) (0 , j ) (1 , j ) (2 (cid:48) , j ) Example. [Veblen-Wedderburn]
The example 7.6.3 becomes with the above notation ((0) (cid:104) [0 , , [1 (cid:48) , , [2 , (cid:105) , { (0 , , (1 , , (1 (cid:48) , }{ [1 , , [0 (cid:48) , , [0 (cid:48) , } , { (2 , , (0 (cid:48) , (2 , }{ [0 , , [2 (cid:48) , , [10] } ; { (2 (cid:48) , , (0 , , (1 (cid:48) , }{ [2 , , [1 , , [0 , } ) . Problem.
Can we characterize the plane Ψ using Theorem 7.6.4.0. CHAPTER 7. QUATERNIONIAN GEOMETRY
Definition.
The
Singer matrix G := . Its powers G k are the columns k, k + 1 , k + 2 of k = 0 1 2 3 4 5 6 7 8 9 10 11 121 0 0 1 0 1 1 1 − − −
10 1 0 1 1 1 − − − − − − move to g6a.tex: Answer to κ = λ = 0 1 − α − α β − β γ − γα r − − − s − − − β r − − − s − − − γ r − − − s − − − For − α, − β , − γ , change the sighof r . Definition.
The elements and incidence in the miniquaternionian plane Ψ are defined as follows.0. The points are ( ξ , ξ , ξ ) with right equivalence,1.2. A point P is incident to a line l iff .6. THE HUGHES PLANES. Partial answer to
For n = 7 , 2451 = 57.43, for n = 9 , 6643 = 91.73, for n = 11 , 14763 = 57.259, For n = 13 ,28731 = 3.9577. Answer to
We have
F ano ((0) , (1) , (2) , (1 (cid:48) , (2 , , (0 , , (1 (cid:48) , , [0] , [1] , [0 , , [2 (cid:48) , , [1 (cid:48) , , [0 , − , [12]) . Answer to (7) × (8) = [6] , [6] × [0] = (3) , (8) × (0) = [1] , [1] × [7] = (2) , (0) × (7) = [9] , [9] × [8] = (5) , (cid:104) (3) , (2) , (5); [11] (cid:105) . (7) × (8) = [6] , [6] × [0 ,
1] = (1 (cid:48) , , (8) × (0 ,
1) = [0 , , [0 , × [7] = (0 (cid:48) , , (0 , × (7) = [2 (cid:48) , , [2 (cid:48) , × [8] = (2 , , (2 , is not incident to (1 (cid:48) , × (0 (cid:48)
6) = [2 , . This has not been checked.From Dembowski, p. 129
Definition.
A linear ternary ring (Σ , + , · ) is called a cartesian field iff (Σ , +) is associative and is therefore agroup. Definition.
A cartesian field is called a quasifield iff the right distributivity law holds: ( x + y ) z = xz + yz. Artzy adds that xa = xb + c has a unique solution, but this is a property (28). This is Veblen-Wedderburn. Definition.
A quasifield is called a semifield iff the left distributivity law holds: z ( x + y ) = zx + zy. Definition.
A quasifield is called a nearfield iff (Σ , · ) is associative and is therefore a group. Definition.
A semifield is called a alternative field iff x y = x ( xy ) and xy = ( xy ) y. Theorem. P is ( p, L ) transitive iff P is ( p, L ) Desarguesian. p is point, L is a line. Dembowski p.123, 16Let Q = (79) , Q = (80) , Q = (90) , and U = (81) then q = [79] , q = [90] , q = [80] ,v = [88] , i = [78] , V = (78) , I = (82) , j = [89] , W = (89) ,Points on q : q : CHAPTER 7. QUATERNIONIAN GEOMETRY ×
12 = [7] : 84(78 , , , , , , , , , , , , , , , , ×
25 = [61] : 82(78 , , , , , , , , , , , , , , , , ×
38 = [75] : 81(78 , , , , , , , , , , , , , , , , ×
51 = [4] : 87(86 , , , , , , , , , , , , , , , , ×
64 = [8] : 83(86 , , , , , , , , , , , , , , , , ×
77 = [68] : 88(86 , , , , , , , , , , , , , , , , Coordinates of points: (0) 64 ,
64 38 ,
80 77 ,
78 78 ,
51 38 ,
25 38 ,
64 51 ,
86 64 , ,
25 86 ,
77 77 ,
51 80 ,
77 12 51 ,
51 12 ,
80 64 , ,
77 12 ,
38 12 ,
51 77 ,
86 51 ,
38 77 ,
38 86 ,
64 64 , ,
64 25 77 ,
77 25 ,
80 51 ,
78 78 ,
64 25 ,
12 25 , ,
86 77 ,
12 64 ,
12 86 ,
51 51 ,
64 80 ,
51 38 25 , ,
80 38 ,
78 78 ,
12 77 ,
64 77 ,
25 12 ,
86 25 ,
64 12 , ,
38 38 ,
12 80 ,
38 51 12 ,
12 51 ,
80 25 ,
78 78 , ,
77 51 ,
12 38 ,
86 12 ,
77 38 ,
77 86 ,
25 25 ,
38 80 , ,
38 64 ,
80 12 ,
78 78 ,
25 64 ,
51 64 ,
38 25 , ,
51 25 ,
51 86 ,
12 12 ,
25 80 ,
12 77 78 80 , ,
78 78 ,
80 86 ,
78 78 ,
86 80 ,
86 86 86 , ,
80 80 , ∞ Coordinates of lines: [0] 78 ,
38 38 77 ,
78 12 ,
12 51 ,
77 86 ,
38 12 ,
86 12 , ,
77 77 ,
25 64 ,
12 80 ,
25 51 ,
80 78 ,
12 12 64 , ,
25 77 ,
64 86 ,
12 25 ,
86 25 ,
64 51 ,
64 64 ,
38 51 , ,
38 77 ,
80 78 ,
25 25 51 ,
78 38 ,
38 64 ,
51 86 , ,
86 38 ,
51 77 ,
51 51 ,
12 77 ,
38 80 ,
12 64 ,
80 78 , ,
78 51 ,
51 12 ,
38 86 ,
77 51 ,
86 51 ,
38 25 , ,
64 25 ,
51 80 ,
64 12 ,
80 78 ,
51 51 25 ,
78 64 , ,
25 86 ,
51 64 ,
86 64 ,
25 12 ,
25 25 ,
77 12 ,
64 80 , ,
80 78 ,
64 64 12 ,
78 77 ,
77 25 ,
12 86 ,
64 77 , ,
12 38 ,
12 12 ,
51 38 ,
77 80 ,
51 25 ,
80 78 , ∞ [80] 80 78 ,
78 86 86 ,
78 86 ,
86 80 ,
86 86 ,
80 78 , ,
78 80 , , , , , , , , , , , , , , , , , , , /inf ty ) B = A + α, ( A, B ) ι [ V, Y ] , V = (78) , (76) = (80 ,
12) = (0 , α ) , Y × V = (78) × (76) = [13] . [13] : 78(78) , , , , , , , , , , , , , , , , , , hence
12 = 80 + α,
51 = α + α = − α,
25 = 78 + α = 1 + α = γ,
64 = 25 + γ = γ + γ = − β,
38 = 86 + α = − α = β,
77 = 38 + α = β + α = − γ. ∞ − α − α β − β γ − γ
90 80 78 86 12 51 38 64 25 77[ α,
0] = (12) × (80 ,
80) = (12) × (79) = [51] , ( a, b ) ι [51] = ⇒ b = a · α. (42) = (78 ,
12) = (1 , α ) = ⇒ α = 1 × α, (45) = (12 , − ( α, −
1) = ⇒ − α × α, (23) = (64 ,
77) = ( − β, − γ ) = ⇒ − γ = − β × α, (28) = (51 ,
78) = ( − α,
1) = ⇒ − α × α, (35) = (86 ,
51) = ( − , − α ) = ⇒ − α = − × α, Using DATA 6,0, 6,4, 6,10, 6,12, 0,0, 1,0, 2,0, 3,0, 4,0, 5,0 DATA 6,0, 0,7, 0,8, 0,11, 3,2,3,10, 4,4, 4,6, 5,5, 5,12 gives the same multiplication table give left not right distibutive law with Q i = 79 , , , U = (83) , α = (12) , q = [87] = { , , , , , , . . . } , .7. AXIOMATIC. q = [81] = { , , , , , , . . . } ,q = [79] = { , , , , , , . . . } , with case 7, data 79,81,87,83,12: ∞ = 87 , , , − , α = 12 , − α = 77 , β = 38 , − β = 51 , γ = 25 , − γ = 64 . This is a try for a section to be included in g19.tex between Moufang and Desargues.
Introduction.
The first example of a Veblen-Wedderburn plane was given in 1907 by Veblen and MacLagan-Wedderburn. It is associated to the algebraic structure of a nearfield, which is a skew field whichlacks the left distributive law, hence is an other plane between the Veblen-Wedderburn plane andthe Desarguesian plane.
Axiom. [Da] Given a Veblen-Wedderburn plane, 2 points Q and Q on the ideal line and an other point Q noton it, any 2 parallelograms A i and B i with directions Q and Q , with no sides in common . . . ,???,such that A j and B j are perspective from Q for j = 0 To 2, imply that A and B are perspectivefrom Q . Notation. Da ( { Q , Q , Q } , { A j } , { B j } ) . Definition. A Veblen-MacLagan plane is a Veblen-Wedderburn plane in which the axiom Da is satisfied.
Lemma. [For Associativity]
H1.0. A , a , x, (See Fig. 2?.)D1.0. a := Q × A , a := Q × A , D1.1. A := a × a , A := a × a , D1.2. a := Q × A , a := Q × A , A := a × a , D2.0. a := Q × A , a := Q × A , a := Q × A , a := Q × A , D2.1. B := a × y,b := Q × B , b := Q × B , D2.1. B := b × b , B := b × b , D2.2. b := Q × B , b := Q × B , B := b × b , C1.0. B ι b , Moreover A = ( A, B ) , A = ( A (cid:48) , B ) , A = ( A, B (cid:48) ) , A = ( A (cid:48) , B (cid:48) ) , B = Proof: Da ( { Q , Q , Q } , { A j } , { B j } ) . Da for Desargues leading to associativity of multiplication. CHAPTER 7. QUATERNIONIAN GEOMETRY
Theorem.
In a Veblen-MacLagan plane, the ternary ring (Σ , ∗ ) is a nearfield. :0. (Σ , +) is an Abelian group, (Σ − { } , · ) is a group,
2. ( Σ , ∗ ) = (Σ , + , · ) is right distributive, ( a + b ) · c = a · c + b · c . Theorem. The Cayleyian plane is not a Veblen-MacLagan plane. replace Desarg.?
Definition. A miniquaternion plane . . . . Theorem. A miniquaternion plane is a Veblen-MacLagan plane. A miniquaternion plane is not a Moufang plane.
Tables.
The following are in an alternate notation the known table for p = 3 and a new table for p = 5.The other incidence are obtained by adding one to the subscripts of the lines and subtracting onefor the subscript of the points.
1. Artzy, Rafael, Linear Geometry, Reading Mass., Addison-Wesley, 1965, 273 pp.2. Baumert, Leonard D., Cyclic Difference Sets, N. Y., Springer, 1971.3. Dembowski, Peter, Finite Geometries, Ergebnisse der Mathematik und ihrer Grenzgebiete,Band 44, Springer, New-York, 1968, 375 pp.4. Dickson, G¨ottingen Nachrichten, 1905, 358-394.5. Hall, Marshall, Projective Planes , Trans. Amer. Math. Soc., Vol. 54, 1943, 229-277.6. Hughes, D. R., A class of non-Desarguesian projective planes, Canad. J. of Math., Vol. 9,1957, 378-388. (I,9.p.1)7. Lemay, Fernand, Le dod´eca`edre et la g´eom´etrie projective d’ordre 5, p. 279-306 of Johnson,Norman L., Kallaher, Michael J., Long Calvin T., Edit., Finite Geometries, N. Y., MarcelDekker Inc. 1983. .8. BIBLIOGRAPHY.
8. Maclagan-Wedderburn, J. H.,
A theorem on finite algebras,
Trans. Amer. Math. Soc., Vol.6, 1905, 349. (A finite skew-field is a field)9. Moore, E. H., Mathematical Papers, Chicago Congress, 1893, 210-226. (Def. of GaloisFields.)10. Room, Thomas Gerald and Kirkpatrick P. B., Miniquaternion geometry; an introduction tothe study of projective planes, Cambridge [Eng.] University Press, 1971.11. Rosati, L. A., I gruppi di collineazioni dei pliani di Hughes. Boll. Un. Mat. Ital. Vol. 13,505-513.12. Singer, James, A Theorem in Finite Projective Geometry and some applications to numberTheory, Trans. Amer. Math. Soc., Vol. 43, 1938, 377-38513. Vahlen, Karl Theodor,
Abstrakte Geometrie, Untersuchungen ¨uber die Grundlagen der Euk-lidischen und nicht-Euklidischen Geometrie , Mit 92 abbildungen im text, 2., neubearb. aufl.Leipzig, S. Hirzel, 1940, Series title: Deutsche mathematik, im auftrage der Deutschenforschungsgemeinschaft, 2. beiheft.14. Veblen, Oswald & Bussey, W. H.,
Finite Projective Geometries , Trans. Amer. Math. Soc.,Vol. 7, 1906, 241-259. (On PG( n, p k ))15. Veblen, Oswald, and Wedderburn, Jospeh Henri MacLagan, Non-Desarguesian and non-Pascalian geometries, Trans. Amer. Math. Soc., Vol. 8, 1907, 379-388.16. Zappa, G. Sui gruppi di collineazioni dei paini di Hughes, Boll. Un. Mat. Ital. (3), Vol. 12,1957, p. 507-516. CHAPTER 7. QUATERNIONIAN GEOMETRY
Notation. u ij := q i q − j − r i r − j ,v ij := q i q − j + r i r − j ,s i := − v − i,i − v i,i +1 ,a i := − v − i − ,i u i − ,i +1 ,t i := s i +2 s i +1 ,f i := s i − s − i +1 s − i − ,g i := t − i − t i +1 t i − ,k i := q (cid:48) i − q i +1 ,l i := r (cid:48) i − r i +1 . Exercise.
Prove q − u r + q − u r + q − u r = 0 , associated with M · eul = 0 . The proof follows form substitution of u i j by their definition. Exercise.
Prove u u − u = − u u − u , associated with 2 equivalent forms of eul one for which the firstcoordinate is one and the other obtain by “rotation”, the second coordinate being one.Form the definition of u it follows by multiplication to the right or left by u − , that q q − u − − r r − u − = 1 ,u − q q − − u − r r − = 1 . Moreover, u = q q − − r r − = − q q − u r r − or u − = r r − u − q q − . If we substitute in the identity to prove, with both terms in the second member, u , u , u and u , by their definition, we get q q − u − q q − − q q − u − r r − − r r − u − q q − + r r − u − r r − − q q − r r − u − q q − q q − + q q − r r − u − q q − r r − + r r − r r − u − q q − q q − − r r − r r − u − q q − r r − = 0 , because terms 3 and 7 cancel, terms 1 and 5 as well as 4 and 8 give 1 and -1, terms 2 and 6 give0 by application of the identities given at the begginning of the proof. Lemma. norm ( s s s ) = 1 . norm ( t t t ) = 1 . s (cid:48) f s − = − f s . t g t = − g t − . .90. ANSWER TO PROBLEMS AND COMMENTS. Proof: For 0, we use Lemma 7.1.1 and obtain 1, from the definition of t i . For 2, we substitute f by its definition and compare the terms of both sides of the equality which have the same sign. Proof of .LetG0.0. A = (1 , , , A = (0 , , , A = (0 , , , G0.1. M = ( q , q , q ) , M = ( r , r , r ) , thenP1.0. a = (1 , , , a = (0 , , , a = (0 , , , P1.1. ma = [0 , q (cid:48) , − q (cid:48) ] , ma = [0 , r (cid:48) , − r (cid:48) ] , P1.2. M = (0 , q , q ) , M = (0 , r , r ) , P1.3. eul = [1 , − u (cid:48) u , − u (cid:48) u ] , P1.4. S = q − n − q (cid:48) q − − q (cid:48) q − − q (cid:48) q − q − n − q (cid:48) q − − q (cid:48) q − − q (cid:48) q − q − n , S − = q q q q q q q q q q q q . S = r − n − r (cid:48) r − − r (cid:48) r − − r (cid:48) r − r − n − r (cid:48) r − − r (cid:48) r − − r (cid:48) r − r − n , S − = r r r r r r r r r r r r . P2.0. mm = [ − q (cid:48) , q (cid:48) , q (cid:48) ] , mm = [ − r (cid:48) , r (cid:48) , r (cid:48) ] , P2.1.
M A = (0 , q , − q ) , M A = (0 , r , − r ) , P2.2. m = [0 , q (cid:48) , q (cid:48) ] , m = [0 , r (cid:48) , r (cid:48) ] , P2.3.
M M = ( − q , q , q ) , M M = ( − r , r , r ) , P2.4. m = [ q (cid:48) , q (cid:48) , q (cid:48) ] , m = [ r (cid:48) , r (cid:48) , r (cid:48) ] , P2.5.
Ima = ( − q , q , q ) , Ima = ( − r , r , r ) ,Ima = ( − q ( q − r + q − r ) , r , r ) , Ima = ( − r ( r − q + r − q ) , q , q ) , P2.6. iM A = [2 q (cid:48) , − q (cid:48) , − q (cid:48) ] , ıM A = [2 r (cid:48) , − r (cid:48) , − r (cid:48) ] , P3.0. mf = [ k v (cid:48) u , k − , − , mf = [ l v (cid:48) u , − l − , , P3.1. O = [] , O = [] , P3.2.
M f a = ( k , , − k (cid:48) v u − ) , M f a = ( l , , l (cid:48) v u − ) ,M f a = (1 , k (cid:48) v u (cid:48) , , M f a = (1 , − l (cid:48) v u (cid:48) , , P3.3. mf a = ( k v (cid:48) u , k − , , mf a = ( l v (cid:48) u , − l − , ,mf a = ( k v − u , , − , mf a = ( l v − u , , , P3.4.
M f m = ( k − v u − , − k , , M f m = ( l − v u − , l , − , P4.0.
Imm = ( a , , s ) , Imm = ( − a , , s ) , P4.1. ta = [0 , , − s (cid:48) ] , P4.2. T = (1 , s , s − ) , P4.3. at = [0 , s (cid:48) , − s ] = [0 , , − t ] , P4.4. K = (1 , t − , t ) , P4.5.
T aa = (0 , , s ) , P4.6. poK = [ − , s (cid:48) , s ] , P4.7. T = f − f s − f − f s − s (cid:48) f − s f , T − = s s (cid:48) s s n s − n s − s − s n s s − n . P4.8. L = g − g t g − g t − − t g − t (cid:48) g , L − = t (cid:48) t t − t − n t n t t t n t t n . CHAPTER 7. QUATERNIONIAN GEOMETRY
Proof:For P4.0, if the coordinates of
Imm are x , 1 and x , we have to solve q − x + q − + q − x = 0 , − r − x + r − + r − x = 0 . Multiplying the equations to the left respectively by q and − r , or by q and r and adding gives x and x using the notation 7.90.For P4.7, it is easier to obtain T − first, the columns are T , T , T , multiplied to the right by 1, s n , s − n . The matrix T is then obtained using Theorem 7.1.1, multiplying by − s − n . The equivalencewith the matrix whose columns ate ta i can be verified using Lemma 7.90.2. A similar proof givesP4.8. Theorem.
The product of the diagonal elements of T − and of L − is the same. This follows from Lemma 7.1.1.The correspondance between the definitions in EUC and here is as followsD0.0 D1.0 D0.1 D1.1 D0.2 D1.2 D1.0 D1.3 D36.12 DC1.4D0.3 D2.0 D0.4 D2.1 D0.5 D2.2 D0.6 D2.3 D0.7 D2.4D10.3 D2.5 D0.25 D2.6 D10.3 D2.7 D6.0 D3.0 D6.4 D3.1? D3.2 ? D3.3 D14.0?D3.4 D1.6 D4.0 D1.7 D4.1D1.8 D4.2 D12.1 D4.3 D1.2 D4.4 D1.4 D4.4 D1.9 D4.5D15.12?D4.6 D1.19 DC4.7 DC4.8 hapter 8FUNCTIONS OVER FINITEFIELDS
Notation.
The first notation is standard, the second is useful for Theorem 8.2.1.2.1. ( a ) i := (cid:81) i − ( a + i ) = a ( a + 1) . . . ( a + i − . [ a ] i := (cid:81) i − ( a + 2 i ) = a ( a + 2) . . . ( a + 2 i − . Notation.
The following notation, favored on the European continent, but seldom used elsewhere, is quiteuseful: n !! := 2 . . . . . . n. (2 n + 1)!! := 1 . . . . . . (2 n + 1) . Introduction.
In a finite field, we can define polynomials of degree up to p − . These are determined by theirvalues at i in Z p . If these are defined in the real field with rational coefficients, the definition andproperties automatically extend to the finite field.
Definition. A polynomial is a function a I p − + a I p − + . . . + a p − which associates to x ∈ Z p the integer a x p − + a x p − + . . . + a p − . The polynomial is of degree k iff a . . . a p − k − are congruent to modulo p and a p − k is not. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Theorem. [Lagrange]
Given k + 1 distinct integers x i (modulo p ), k < p, and given k + 1 integers f i , ∃ a polynomial P of degree k (cid:51) P ( x i ) = f i , i = 0 to k. Definition.
The derivative of the polynomial ( p − a I p − + ( p − a I p − + . . . + a p − . The main purpose of writing this Chapter is connected with interesting symmetry properties of theorthogonal polynomials, in Z p . In the classical theory there is a scaling factor which is arbitrairelychosen for each of the families of orthogonal polynomials. For some time now, the same scalingfactor, in each case, is universely used. When determining values of the Chebyshev polynomials forsome small values of p, I was struck by the symmetry properties given in 8.2.2 and 8.2.2. Theseproperties are dependent on the scaling factors and it turns out that the unanimously acceptedones are essentially the only ones giving this property. The same property has been found for thepolynomials of Legendre and of Laguerre. For the polynomial of Hermite this is not the case. I havesucceeded in obtaining some scaling, given in 8.2.5 for which a symmetry can be obtained. Thisscaling is given by expressions which are different for the even and for the odd Hermite polynomials,therefore the recurrence relation has a constant whose expression differs for even and odd indices.It is therefore possible to give an a-posteriori justification of the scaling factor for the classicalpolynomial, and there is some reason to introduce a different scaling for the Hermite polynomials.The case of the Jacobi polynomials with 2 parameters a and b is left as an exercise. With a = b ,again a scaling is required to obtain symmetry. Introduction.
For orthogonal polynomials, recurrence relations, differential equations and values of the coefficientsgeneralize automatically, from the classical case. Therefore, we have the definitions 8.2.1 and thetheorems 8.2.1 to 8.2.1.
Definition.
The polynomials of Chebyshev of the first ( T n ) and of the second kind ( U n ) , of Legendre ( P n ) , ofLaguerre ( L n ) and of Hermite ( H n ) are defined by the recurrence relations:0. T := 1 , T := I, T n +1 := 2(2 I − T n − T n − , U := 1 , U := 2 I, U n +1 := 2(2 I − U n − U n − , .2. ORTHOGONAL POLYNOMIALS OVER FINITE FIELDS. P ( a )0 := 1 , P ( a )1 := I, ( n + 2 a + 1) P ( a ) n +1 := (2 n + 2 a + 1) I P ( a ) n − n P ( a ) n − , a ≥ , n < p − − a, L := 1 , L := 1 − I, ( n + 1) L n +1 := (2 n + 1 − I ) IL n − nL n − , n < p − , L ( a )0 := 1 , L ( a )1 := a + 1 − I, ( n + 1) L ( a ) n +1 := (2 n + a + 1 − I ) L ( a ) n − ( n + a ) L ( a ) n − , n < p − , H := 1 , H := 2 I, H n +1 := 2 IH n − nH n − . The Legendre polynomial is P n := P (0) n and P ( a ) n := n ! a !( n + a )! P ( a,b ) n , where P ( a,b ) n , are the polyno-mials of Jacobi, scaled so that P ( a ) n (1) = 1 .L ( a ) n are the generalized Laguerre polynomials and L n = L (0) n .See for instance Handbook of Mathematical functions, p. 782. Theorem. If X n,j denotes the coefficient of I j in the polynomial X n , T n,n − j = n ( n − j ) ( − j ( n − j − j !( n − j )! , U n,n − j = n ( n − j ) ( − j ( n − j )! j !( n − j )! , P n,n − j = 2 ( − n ) ( − j (2 n − j )! j !( n − j !( n − j )! , P ( a )2 n, j = ( − ( n − j ) (cid:18) nj (cid:19) [2 a +2 n +1] j [2 j +1] n − j [2 a +2] n ,P ( a )2 n, j = ( − ( n − j ) (cid:18) nj (cid:19) [2 a +2 n +3] j [2 j +3] n − j [2 a +2] n , L n,j = ( − j n !( n − j )! j ! , L ( a ) n,j = ( − j ( n + a )!( n − j )!( a + j )! j ! , H n,n − j = n !2 ( n − j ) ( − j j !( n − j )! , See for instance Handbook of Mathematical functions, p. 775.
Theorem.
The polynomials of Chebyshev, of the first ( T n ) and of the second kind ( U n ) , of Legendre ( P n ) , ofLaguerre ( L n ) and of Hermite ( H n ) satisfy by the differential equations (1 − I ) D T n − IDT n + n T n = 0 , (1 − I D U n − IDU n + n ( n + 2) U n = 0 , CHAPTER 8. FUNCTIONS OVER FINITE FIELDS (1 − I D P ( a ) n − a + 1) I ) DP ( a ) n + n ( n + 2 a + 1) P ( a ) n = 0 , ID L n + (1 − I ) DL n + nL n = 0 . ID L ( a ) n + ( a + 1 − I ) DL ( a ) n + nL ( a ) n = 0 . D H n − IDH n + 2 nH n = 0 . See for instance Handbook of Mathematical functions, p. 781.
Theorem. T n (1) = 1 , DT n (1) = n . U n (1) = n + 1 , DU n (1) = n ( n +1)( n +2)3 . P ( a ) n (1) = 1 , DP ( a ) n (1) = − n ( n +2 a +1)2( a +1) . L n (0) = 1 , DL n (0) = − n. L ( a ) n = (cid:18) n + an (cid:19) , DL ( a ) n = − (cid:18) n + an − (cid:19) . H n (0) = ( − n (2 n )! n ! , DH n (0) = 0 . H n +1 (0) = 0 , DH n +1 (0) = ( − n +1 (2 n )! n ! . Comment.
It is easy to verify that, contrary to the classical case, the roots of the orthogonal polynomials arenot necessarily in Z p . For instance, T has a root in Z p iff R p or p ≡ ± . Program. [m130]FIN ORTHOG.HOM illustates the use of the program [m130]FIN ORTHOG.BAS, whichdetermines these various orthogonal polynomials. [m130]FIN ORTHOG.NOT are notes tracingsome of the steps leading to the conjectures proven here.
Theorem. T p + i,j = T p − i,j . T i +2 pk,j = − T i + pk,j = T i,j , j < p. Proof: T p + i,j = ( − ( p + i − j ) j ( ( p + i + j ) − ( p + i )( ( p + i − j ))! j ! = ( − ( p + i − j ) ( − ( p − i − j ) ( − ( p − i + j + i ) 2 j ( ( p − i + j − ( p − i )( ( p − i − j ))! j ! .2. ORTHOGONAL POLYNOMIALS OVER FINITE FIELDS. = ( − ( p − i − j ) j ( ( p − i + j − ( p − i )( ( p − i − j ))! j ! = T p − i,j . Example.
For p = 5 ,T = − T = T = 1 .T = − T = − T = T = + I.T = − T = − T = T = − I .T = − T = − T = T = +2 I − I .T = − T = − T = T = 1 + 2 I − I .T = T = 0 . Theorem. U p − i,j = U p − − i,j . U i +2 pk,j = − U i + pk,j = U i,j , j < p. Proof: U p − i,j = ( − ( ( p − i − j ) ( ( p − i + j ))!2 j ( ( p − i − j ))! j ! = ( − ( ( p − i − j ) ( − ( ( p − i − j − ) ( − ( 12 ( p − i + j − ( ( p − i + j − j ( ( p − i − j ))! j ! = ( − ( ( p − i − j ) j ( ( p − − i + j ))! (( p − − i − j ))! j ! = U p − − i,j . Example.
For p = 5 ,U = U = − U = − U = 1 .U = U = − U = − U = 2 I.U = U = − U = − U = − − I .U = U = − U = − U = I − I .U = − U == 1 − I + I .U = − U == 0 . Introduction.Theorem. P a ) p − − a − n = P ( a ) n , n ≤ p − − a. Proof:Let p (cid:48) = ( p − . The recurrence relations 8.2.1.2 imply CHAPTER 8. FUNCTIONS OVER FINITE FIELDS ( p (cid:48) + 1) P p (cid:48) +1 = − p (cid:48) P p (cid:48) − , hence P p (cid:48) +1 = P p (cid:48) − . They can also be written, ( n + 1) P p − n − = − (2 n + 1) P p − n − − nP p − n . Therefore, starting from P p (cid:48) and from P p (cid:48) − and P p (cid:48) +1 , we obtain by induction P p − − n = P n . Example.
F orp = 11 ,P = P = 1 ,P = P = I,P = P = 5 − I ,P = P = 4 I − I ,P = P = − − I + 3 I ,P = − I + 5 I + I , F orp = 13 ,P = P = 1 ,P = P = I,P = P = 6 − I ,P = P = 5 I − I ,P = P = 2 + 6 I + 6 I ,P = P = − I + I + 3 I ,P = − − I − I − I , Theorem. P p − − n = P n , n < p. ? P ( a )0 = 1 ,P ( a )1 = I,P ( a )2 = − a +3) I a +1) ,P ( a )3 = − − I +(2 a +5) I ( a +1)( a +2) ,P ( a )4 = − a +5) I +(2 a +5)(2 a +7) I ( a +1)( a +2) ,P ( a )5 = I − a +7) I +(2 a +7)(2 a +9) I ( a +1)( a +2) , Example.
F orp = 11 , a = 2 P (2)0 = 1 ,P (2)1 = I,P (2)2 = − I ,P (2)3 = 5 I − I ,P (2)4 = − I ,P (2)5 = I,P (2)6 = 1 , F orp = 13 , a = 2 P (2)0 = 1 ,P (2)1 = I,P (2)2 = 2 − I ,P (2)3 = 6 I − I ,P (2)4 = − − I − I ,P (2)5 = 6 I − I ,P (2)6 = 2 − I ,P (2)7 = I,P (2)8 = 1 , .2. ORTHOGONAL POLYNOMIALS OVER FINITE FIELDS. Theorem (La). L p − − i,j = ( − j L i + j,j , ≤ i, j, i + j < p. Proof:1. L n,j = ( − j j ! (cid:18) nj (cid:19) . See for instance, Handbook p.775.
Example.
For p = 7 ,L = 1 .L = 1 − I.L = 1 − I − I .L = 1 − I − I + I .L = 1 + 3 I + 3 I − I − I .L = 1 + 2 I − I + 3 I − I − I .L = 1 + I − I + I − I + I − I . Theorem (La). L ( a ) p − a − − i,j = ( − j + a L i + j,j ( a ) , ≤ i, j, i + j < p − a. L ( a ) j,j = − ( a − I p − a , < a, p − a ≤ j < p. L ( a ) i,j = 0 , a > , j < p − a ≤ i < p. The proof is left to the reader.
Example.
For p = 13 , a = 5 L (5)0 = 1 .L (5)1 = 6 − I.L (5)2 = − I − I .L (5)3 = 4 − I + 4 I + 2 I .L (5)4 = − − I + 5 I + 5 I + 6 I .L (5)5 = 5 − I − I − I − I + 4 I .L (5)6 = − I − I + 5 I + 5 I + 5 I − I .L (5)7 = − − I + 6 I + 2 I − I + 4 I + 5 I − I .L (5)8 = 2 I .L (5)9 = 2 I − I .L (5)10 = 2 I + I − I .L (5)11 = 2 I − I − I − I .L (5)12 = 2 I + 2 I + I − I − I . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Definition.
The scaled Hermite polynomials are defined by0. H s = 1 , H s = I, [ n ] H sn = a n IH sn − − ( n − H sn − , where a n = 1 if n is even and a n = [ n ] , the largest integer in n if n is odd. Example.
In the fields Q or R , H s = − + I ,H s = − I + I ,H s = − I + I ,H s = I − I + I ,H s = − + I − I + I ,H s = − I + I − I + I . Theorem. H s n (0) = ( − n (2 n − n )!! , DH s n (0) = 0 .H s n +1 (0) = 0 , DH s n +1 (0) = ( − n (2 n +1)!!(2 n )!! . Lemma. In Z p , p > , ( p − − . ( p − − i )! = ( − ( i +1) 1 i ! , ≤ i < p. (cid:18) p − − ij (cid:19) = ( − j (cid:18) i + jj (cid:19) , ≤ i, j, i + j < p. (cid:18) kp + ij (cid:19) = (cid:18) ij (cid:19) , j < p. ( p − − i )!! i !! = ( − k ( p − − k − i )!!( k + i − ≤ i < p − , < k + i < p. Proof: 0, is the well known Theorem of Wilson. 1, can be considered as a generalization. ( p − − i )! = ( − i ( p − . . . ( i + 1)= ( − i ( p − i ! = ( − ( i +1) 1 i ! . For 2, ( p − − i )!( p − − i − j )! j ! = ( − ( i +1) ( i + j )!( − i + j +1 i ! j ! = ( − j (cid:18) i + jj (cid:19) . .2. ORTHOGONAL POLYNOMIALS OVER FINITE FIELDS. Lemma.
Modulo p, p > , (( p − = ( − ( p − ( p − p − − . ( p − − i )!! i !! = ( − s ( p − , where s = i when i is even and s = (p-2-i) when i is odd. or where s = [ ([ p ] + 1 + i )] + [ ( p + 1)] . i is even, ( p − − i )!! i !! = ( p − i )!!( i − ip − i or − − ( i ) ( p − . if i is odd, ( p − − i )!! i !! = ( p − − i )!!( i + 2)!!( p − − ii +2 or − − ( p − − i ) (0)!! p − . p = 1 , , , and i =0 , , , , . Theorem.
For scaled Hermite H si + j,j = 0 , ≤ i, j, i odd.1. H sp − − i,j = H si + j,j , ≤ i, j, i even, j even, i + j < p. H sp − − i,j = H si + j,j , ≤ i, j, i even, j odd, i + j < p. The proof is left as an exercise.
Example.
For p = 11 ,H s = 1 .H s = I.H s = 5 + I H s = 4 I + I H s = − I − I H s = − I + 3 I − I H s = − − I − I + 2 I H s = 4 I + 3 I + I + 2 I H s = 54 I − I + 4 I − I H s = I + I − I + 2 I − I .H s = 1 + I − I + 2 I − I − I . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Problem.
The Jacobi polynomials can be defined by
P s ( a,b )0 := 1 , P ( a,b )1 := a − b +( a + b +2) I a +1) , n + 1)( n + a + b + 1)(2 n + a + b ) P ( a,b ) n +1 :=((2 n + a + b + 1)( a − b )+ (2 n + a + b )(2 n + a + b + 1)(2 n + a + b + 2) I ) P ( a,b ) n − n + a )( n + b )(2 n + a + b + 2) P ( a,b ) n − . Determine an appropriate scaling for the Jacobipolynomials that gives symmetry properties which generalize those of the special case where a = b . Ungar, gave recently the addition formulas associated with a generalization of the trigonometric andhyperbolic functions by Ricatti. This suggested the extension to the finite case. Section 1, is theTheorem of Ungar, the special case for 3 functions is given in 8.3.2, with the associated invariant8.3.2.2. The invariant defines the distances, addition, which in fact corresponds to the multiplicationof associated Toeplitz matrices gives the angles. For 3 dimensions we have 2 special cases, p ≡ and p ≡ − . In the latter case all non isotropic direction form a cycle. In theformer case, we can consider that the set of ( p − non isotropic directions corresponds to a directproduct of 2 cyclic groups of order p − , I conjecture (14) that there are always pairs of generatorswhich are closely related called special generators (13). This is extended to more than 3 functionsin 8.3.3. The connection with difference sets is given at the end of that chapter.
Theorem. [Ungar] If f is a solution of D n +1 f + a n D n f + . . . + a f = 0 , a n +1 = 1 , D n f (0) = 1 , D k f (0) = 0 , ≤ k < n, then f ( x + y ) = (cid:80) nm =0 a m +1 (cid:80) mk =0 D k f ( x ) D m − k f ( y ) . More generally, if D k f (0) = d k , ≤ k ≤ n, then (cid:80) nm =0 a m +1 (cid:80) mk =0 D k f ( x ) D m − k f ( y )= (cid:80) nm =0 a m +1 (cid:80) mk =0 d k D m − k f ( x + y ) . Proof: See Abraham Ungar, 1987. .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS.
Example. a m = 0 , ≤ m ≤ n, f = I n n ! , ( x + y ) n = (cid:80) nk =0 (cid:18) nk (cid:19) x k y n − k . n = 0 , a = 1 , a − = 0 , f = e I , e x + y = e x e y . n = 1 , a = 1 , a = 0 , a = 1 , f = sin, sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y ) . n = 1 , a = 1 , a = 0 , a = − , f = sinh, sinh ( x + y ) = sinh ( x ) cosh ( y ) + cosh ( x ) sinh ( y ) . a n +1 = 1 , a k = 0 , < k ≤ n, a = − j, where j = ± ,D i f = R ( jn, = (cid:80) ∞ k =0 I rk − i ( rk − i )! , where r = n + 1 , ( j = − R ( − n, ( x + y ) = R ( − n, ( x ) R ( − n,n ) ( y )+ R ( − n, ( x ) R ( − n,n − ( y ) + . . . . + R ( − n, ( x ) R ( − n,n ) ( y ) . ( j = − These are, with my notation, the functions of Vincenzo Ricatti. In particular, when n = 2 , ?we have the following Theorem. Theorem. If n is odd, then R ( − n, = R ( n. ( − I ) . R ( − n,j ) = ( − j R ( n.j ) ( − I ) . Theorem.
Let f = R (2 , , g = R (2 , h = R (2 , , then f ( x + y ) = f ( x ) h ( y ) + g ( x ) g ( y ) + h ( x ) f ( y ) ,g ( x + y ) = g ( x ) h ( y ) + h ( x ) g ( y ) + f ( x ) f ( y ) ,h ( x + y ) = h ( x ) h ( y ) + f ( x ) g ( y ) + g ( x ) f ( y ) , f + g + h − f gh = 1 . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS ( f ( x ) h ( y ) + g ( x ) g ( y ) + h ( x ) f ( y )) +( g ( x ) h ( y ) + h ( x ) g ( y ) + f ( x ) f ( y )) +( h ( x ) h ( y ) + f ( x ) g ( y ) + g ( x ) f ( y )) − f ( x ) h ( y ) + g ( x ) g ( y ) + h ( x ) f ( y ))( g ( x ) h ( y ) + h ( x ) g ( y ) + f ( x ) f ( y ))( h ( x ) h ( y ) + f ( x ) g ( y ) + g ( x ) f ( y ))= ( f ( x ) + g ( x ) + h ( x ) − f ( x ) g ( x ) h ( x ))( f ( y ) + g ( y ) + h ( y ) − f ( y ) g ( y ) h ( y )) . Proof: g = Df, h = Dg = D f, f = Dh = D g = D f,f, g and h satisfy the same differential equation, whose Wronskian is constant this gives det (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f g hg h fh f g (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − . Theorem.
The solution of 8.3.2, f, g, h is given by f = Ae I + Be βI + Ce β − I , g = Ae I + Bβe βI + Cβ − e βI , h = Ae I + Bβ − e βI + Cβ − e βI , where β + β + 1 = 0 , A = , B = β, C = β − . Corollary. f = e − I cos ( √ I ) ,g = e − I cos (( √ + π ) I ) ,h = e − I cos (( √ − π ) I ) , is a solution of D f = f. I examined the more general case starting from f , g , h and using the addition formulas, itappears that the period is always p − and that if f + g + h − f gh = 1 then we have f ( π ) = g ( π ) = 0 when p ≡ . Application to the case of 3 dimensional Affine geometry associated to p . Lemma.
Let .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS. T ( x, y, z ) := x + y + z − xyz, L ( x, y, z ) := x + y + z,S ( x, y, z ) := x + y + z − yz − zx − xy, then T ( x, y, z ) = L ( x, y, z ) S ( x, y, z ) . If − p or p ≡ , then the only points of S = 0 are( a, a, a ) . If − p or p ≡ , then S ( x, y, z ) = ( x + τ y + τ (cid:48) z )( x + τ (cid:48) y + τ z ) , with τ := ( − √− τ (cid:48) ) := ( − − √− . The number of lines through the origin on T ( x, y, z ) = 0 is3 p if p ≡ p + 2 if p ≡ − . Proof: The number of lines through the origin is the same as the number of points in a planenot through the origin which are on the ideal line or on S. If p ≡ , this gives ( p + 1) + 1 , if p ≡ − , this gives p + 1) − . Definition.
Let T be the set of points ( x, y, z ) (cid:51) x + y + z − xyz = 1 . Let the addition in T be defined by1. ( x, y, z ) + ( x (cid:48) , y (cid:48) , z (cid:48) ) := ( yy (cid:48) + xz (cid:48) + zx (cid:48) , xx (cid:48) + yz (cid:48) + zy (cid:48) , zz (cid:48) + xy (cid:48) + yx (cid:48) ) Theorem. ( T , +)is an Abelian group with neutral element (0 , , . ( x, y, z ) + ( x (cid:48) , y (cid:48) , z (cid:48) ) + ( x (cid:48)(cid:48) , y (cid:48)(cid:48) , z (cid:48)(cid:48) )= ( x ( x (cid:48) y (cid:48)(cid:48) + y (cid:48) x (cid:48)(cid:48) + z (cid:48) z (cid:48)(cid:48) ) + y ( x (cid:48) x (cid:48)(cid:48) + y (cid:48) z (cid:48)(cid:48) + z (cid:48) y (cid:48)(cid:48) ) + z ( x (cid:48) z (cid:48)(cid:48) + y (cid:48) y (cid:48)(cid:48) + z (cid:48) x (cid:48)(cid:48) ) ,x ( x (cid:48) z (cid:48)(cid:48) + y (cid:48) y (cid:48)(cid:48) + z (cid:48) x (cid:48)(cid:48) ) + y ( x (cid:48) y (cid:48)(cid:48) + y (cid:48) x (cid:48)(cid:48) + z (cid:48) z (cid:48)(cid:48) ) + z ( x (cid:48) x (cid:48)(cid:48) + y (cid:48) z (cid:48)(cid:48) + z (cid:48) y (cid:48)(cid:48) ) ,x ( x (cid:48) x (cid:48)(cid:48) + y (cid:48) z (cid:48)(cid:48) + z (cid:48) y (cid:48)(cid:48) ) + y ( x (cid:48) z (cid:48)(cid:48) + y (cid:48) y (cid:48)(cid:48) + z (cid:48) x (cid:48)(cid:48) ) + z ( x (cid:48) y (cid:48)(cid:48) + y (cid:48) x (cid:48)(cid:48) + z (cid:48) z (cid:48)(cid:48) )) . ( x, y, z ) + ( y − zx, x − yz, z − xy ) = (0 , , . Corollary. x, y, z ) = ( y + 2 zx, x + 2 yz, z + 2 xy ) . x, y, z ) = (3( x y + y z + z x ) , x z + y x + z y, xyz )) . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Theorem. If ( x n , y n , z n ) := n ( x, y, z ) then x n + y n + z n = ( x + y + z ) n . x k ( p − i + y k ( p − i + z k ( p − i = x i + y i + z i . Theorem.
Let u = − x + 2 sx − s + s y = ( s − x ± u ) , z = s − x − y, then( x, y, z ) ∈ T .Proof:Substitute z by s − x − y ∈ x + y + z − xyz − gives s y − s ( s − x ) y + (3 s x − s x + s −
1) = 0 , dividing by s, the discriminant is the second member of 0. (cid:3) Definition.
0. The distance d between 2 points ( x, y, z ) and ( x (cid:48) , y (cid:48) , z (cid:48) ) is given by d ( x, x (cid:48) ) := ( x (cid:48) − x ) + ( y (cid:48) − y ) + ( z (cid:48) − z ) − x (cid:48) − x )( y (cid:48) − y )( z (cid:48) − z ) .
1. If the distance between 2 distinct points is 0, the line incident to the 2 points is called isotropic . Theorem.
The isotropic lines are those on the surface T(x,y,z) = 0.
Lemma. d ( x, x (cid:48) ) = d (0 , x ) − d (0 , x (cid:48) ) − x ( x (cid:48) − y (cid:48) z (cid:48) ) + y ( y (cid:48) − z (cid:48) x (cid:48) ) + z ( z (cid:48) − x (cid:48) y (cid:48) ))+3( x (cid:48) ( x − yz ) + y (cid:48) ( y − zx ) + z (cid:48) ( z − xy )) . Theorem. d ( P, Q ) = − d ( Q, P ) . If P = (0 , , , , then P × Q is isotropic iff Q is on the line l joining P to (1,1,1,1) or on aline through P perpendicular to l. The ideal points on the surface satisfy x + y + z − xyz = 0 . .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS. Definition.
The normal to the surface T at (a,b,c) is [ a − bc, b − ca, c − ab ] . Notation. If p ≡ , δ := p − . Theorem. If p ≡ , ( T , +) ∼ C p − ×× C p − . Proof: The order of the group follows from Lemma 8.3.2?.
Lemma.
If an Abelian group is isomorphic to C q ×× C q if u and v are of order p and u i (cid:54) = v j for all i and j between 1 and q, u and v are generators of the group. Lemma. If p ≡ , g is a primitive root of p and ( a, b, c ) and ( a (cid:48) , b (cid:48) , c (cid:48) ) are obtained using b, c = g − a ± (cid:113) − g ag − a g − if their i -th iterates are distinct, 0 < i < p − , then (a,b,c) and (a’,b’,c’) are generators.In particular, if h = 1 then b, c = h − a ± √ ( h +3 a )( h − a )2 Proof: the fact that g is primitive insures that the sum of the components of the i -th iterate of ( a, b, c ) is g i , because these are distinct for i = 1 to p − , the Lemma follows. Definition.
If the pair ( a, b, c ) and ( b, a, c ) are pairs of generators of (T,+) then ( a, b, c ) is called a specialgenerator of T. Conjecture.
Given a primitive root of p ≡ , there exists always special generators ( a, b, c ) (cid:51) a + b + c = g (mod p ) . Theorem.
If ( a , b , c ) is a special generator, the period is0 1 2 . . . δ δ +1 δ +2 . . . 2 δ δ +1 2 δ +2 . . .0 a a . . . 1 c c . . . 0 b b . . .0 b b . . . 0 a a . . . 1 c c . . .1 c c . . . 0 b b . . . 0 a a . . . . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
For ( b , a , c ) the period, scaled again is0 1 2 . . . δ δ +1 δ +2 . . . 2 δ δ +1 2 δ +2 . . .0 b b . . . 0 a a . . . 1 c c . . .0 a a . . . 1 c c . . . 0 b b . . .1 c c . . . 0 b b . . . 0 a a . . . . Algorithm.
For a given p, we determine the smallest positive primitive root g, then for increasing values of a, we determine b and c using 8.3.2.0, if c ( delta ) = 1 we permute a, b, c, in the order b, a, c, c, b, a, a, c, b, c, a, b, b, c, a, unless p ≡ , in which case we try a new p, if c(delta) = 0, we save the period and permute,if c ( delta ) = 0 and the values of a ( i ) , b ( i ) , c ( i ) are not distinct from some i, from the correspondingsaved values, we permute again, if we exaust the permutations, we ignore this value of a. When wehave obtained ( a, b, c ) such that the first a(delta) = -1 and and the second = 1 we exchange. Example. p = 7 , g i = 1 , , , , , , T =0 1 2 3 4 5 6 7 8(0 , ,
1) (0 , ,
2) (0 , ,
4) (0 , ,
0) (0 , ,
0) (0 , ,
0) (1 , ,
0) (1 , ,
6) (1 , , , ,
1) (0 , ,
1) (0 , ,
1) (0 , ,
0) (0 , ,
0) (0 , ,
0) (1 , ,
0) (5 , ,
2) (2 , , , ,
3) (2 , ,
0) (2 , ,
3) (2 , ,
6) (2 , ,
5) (3 , ,
6) (3 , ,
3) (3 , ,
2) (3 , , , ,
1) (1 , ,
0) (2 , ,
3) (5 , ,
1) (5 , ,
2) (1 , ,
2) (3 , ,
3) (3 , ,
2) (2 , , , ,
4) (3 , ,
1) (4 , ,
0) (4 , ,
5) (4 , ,
3) (4 , ,
6) (5 , ,
5) (5 , ,
6) (5 , , , ,
5) (1 , ,
5) (1 , ,
0) (5 , ,
1) (5 , ,
2) (2 , ,
3) (3 , ,
3) (2 , ,
1) (1 , , , ,
3) (5 , ,
1) (5 , ,
2) (6 , ,
3) (6 , ,
5) (6 , ,
1) (6 , ,
6) (6 , ,
2) (6 , , , ,
2) (3 , ,
2) (2 , ,
5) (2 , ,
1) (1 , ,
2) (2 , ,
5) (3 , ,
3) (1 , ,
5) (3 , , When scaled the group is isomorphic to C ×× C , we have the equivalences,0,1,2; 3,4,5; 6,10,20; 7,13,22; 8,11,23; 9,12,21; 14,27,3; 15,24,33; 16,28,35; 17,25,30; 18,29,32;19,26,34.We have the table + 0 9 3 18 6 140 0 9 3 18 6 1416 16 7 15 19 8 17 p = 13 , g i = 1 , , , , , , , , , , , , the scaled period is .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS. (0 , ,
1) (7 , ,
6) (7 , ,
11) (11 , , , (1 , ,
0) (6 , ,
1) (11 , ,
9) (6 , , , (0 , ,
0) (1 , ,
7) (9 , ,
7) (10 , , , p = 19 , g i = 1 , , , , , , , , , , , , , , , , , , the scaled period is (0 , , , (3 , , , (15 , , , (8 , , , (7 , , , (11 , , , , , (8 , , , (4 , , , (18 , , , (8 , , , (9 , , , , , (9 , , , (1 , , , (13 , , , (5 , , , (0 , , Example.
The following are special generators, for the given primitive root, which is the smallest positive one: p g sp.gen. p g sp.gen. p g sp.gen. , ,
3) 283 3 (3 , , , , , ,
12) 307 5 (4 , , , , , ,
16) 313 10 (5 , , , , , ,
0) 331 3 (0 , ,
97) 673 5 (3 , , , ,
14) 337 10 (180 , , , , , ,
3) 349 2 (50 , ,
2) 709 2 (424 , , , ,
58) 367 6 (22 , ,
5) 727 5 (377 , , , ,
55) 373 2 (53 , , , , , ,
62) 379 2 (5 , , , , , ,
35) 397 5 (8 , , , , , ,
75) 409 21 (390 , ,
2) 757 3 (5 , , , ,
4) 421 2 (6 , , , , , , , , , , , ,
53) 439 15 (0 , , , , , ,
3) 457 13 (14 , ,
0) 823 3 (15 , , , ,
0) 463 3 (0 , , , , , , , , , , , , , , , , , , , , , , , ,
73) 541 2 (93 , , , , , , , , , , , , , , , , , ,
88) 577 5 (4 , , , , , ,
66) 601 7 (138 , ,
7) 967 5 (3 , , , , , , , , , , , , , , , , , , Lemma. If p ≡ − , s i := P i + P i + P ⇒ s i = s i . f i := P i + P i + P − ( P i P + P P i + P i P i ) ⇒ f i = f i . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Lemma. If p ≡ − , if g is a primitive root of p , Proof: Let . . .
To determine what happens for the solutions for i = 1 , , . . . p − :let g be a generator for p, we want To determine what happens for the solutions for i ≡ p − : a + b + c = 1 , a + b + c − ( bc + ca + ab ) = 1 , ⇒ ( a + b + c ) = 1 , bc + ca + ab = 0 , given a, b + c = 1 − a, bc = a ( a − ,b, c = − a ± √ ( a − − a )2 special solutions (1 , , , ( − , , ) , − , , ) = (0 , , , , − , ) = (1 , , , there should be p +1 − − = p − possible values of a . Notation. (cid:15) := (0 , , , α := (1 , , , β := (0 , , . Theorem. If p ≡ − , and 3 δ = p − , then ( T , +) ∼ C δ . If h is a generator of this group then h δ = (0 , ,
0) or (1 , , , in the former case, we willchoose g = h − otherwizewe will choose g = h. If g i = ( a, b, c ) , then g i + δ = ( c, a, b ) , g i +2 δ = ( b, c, a ) , g pi = ( b, a, c ) , g pi + δ = ( c, b, a ) ,g pi +2 δ = ( a, c, b ) .3. d ( P i , P p +( p +1) l +( p − kj + i ) = d ( P i , P p +1) l − ( p − kj + i ) . d ( P i , P ( p +1) l +( p − kj + i ) = d ( P i , P ( p +1) l − ( p − kj + i ) . Proof:If p ≡ − , to any of the p − line through the origin which does not pass through(1,1,1,1) and is not in the plane perpendicular to this last line, associate a point a, b, c, say let u := a + b + c − abc, u (cid:54) = 0 and u has a unique cube root v in Z p , therefore the point ( av , bv , cv ) ∈ T . ( a, b, c ) + (1 , ,
0) = ( c, a, b ) , ( a, b, c ) + (0 , ,
0) = ( b, c, a ) . Lemma.
Proof: φ ( p − = 2 φ (p-1) φ (p+1), Example. p = 11 , g = 7 , .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS. Lemma. ( a + b + c )( a + b + c − bc − ca − ab ) = a + b + c − abc. Given a and g, a primitive root of p, then b, c = g − a ± (cid:113) − g ag − a g − Proof: a + b + c = g and a + b + c − bc − ca − ab = g − ⇒ ( a + b + c ) = g and bc + ca + ab = 0 , therefore b + c = g − a and bc = g − g − − a ( g − a ) , hence b and c are roots of a quadratic equations,this gives 1. Theorem. If p ≡ − , A necessary condition for ( a, b, c ) in T to be a generator is that a + b + c be a primitive rootfor p. If ( a, b, c ) is a generator such that ( a, b, c ) δ = (1 , , , p ≡ , or p ≡
11 (mod 18) ⇒ ( b, c, a ) δ = (cid:15), ( c, a, b ) δ = β, ( c, b, a ) δ = (cid:15), ( a, c, b ) δ = α, ( b, a, c ) δ = β, p ≡ , or p ≡ ⇒ ( b, c, a ) δ = β , ( c, a, b ) δ = (cid:15) , ( c, b, a ) δ = α , ( a, c, b ) δ = (cid:15), ( b, a, c ) δ = β ,1. p ≡ , or p ≡
17 (mod 18) ⇒ ( b, c, a ) δ = α , ( c, a, b ) δ = α , ( c, b, a ) δ = β , ( a, c, b ) δ = β , ( b, a, c ) δ = β, p ( u, v, w ) = ( v, u, w ) . Definition.
Given a generator ( a, b, c ) and a non isotropic scaled direction ( u, v, w ) the corresponding angulardirection is the multiplier i such that i(a,b,c) = (u,v,w). Conjecture. angular direction ( P i + k , P i ) = i + angular direction ( P k , P )(mod p − . angular direction ( O, M i ) = i + angular direction (0 , M ) , where M i is the mid-point of ( P i , P i +1 ) .2. angulardirection ( O, N i ) = i + angulardirection (0 , N ) , where N i is the mid-point of ( P i − , P i +1 ) .3. angular direction ( P i , N i ) = i + angular direction ( P , N ) . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Example. p = 17 , generator (13 , , ,angular direction ((0 , , , (13 , , ,angular direction ((0 , , , (9 , , ,M = (15 , , , angular direction ( O, M ) = 60 ,N = (13 , , , angular direction ( O, N ) = 33 ,angular direction ( P , N ) = 40 . Corollary.
The coordinates of the normal to the surface T are those of − p ( a, b, c ) . Lemma.
If ( a, b, c ) is a generator and g i is a primitive root of p, then i, prime, ≡ ⇒ ( g i ) δ = α. ?1. i, prime, ≡ ⇒ ( g i ) δ = β. ? i is not a prime ⇒ ( g i ) δ = (cid:15). Example.
The following table gives generators ( a, b, c ) for the given values of p and g,α , (cid:15) , β − (cid:15) , α , β α , β , (cid:15) − α , (cid:15) , β α , α , α − β , β , β .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS. p = 11 , , , p = 5 , , , p = 17 , , , g = 2 , , , g = 2 , g = 3 , , , p = 29 , , , p = 23 , , , p = 53 , , , g = 2 , , , g = 5 , , , g = 2 , , , , , , , , , , ,
23 18 , , , , p = 47 , , , p = 41 , , ,
10 11 , , g = 5 , , , g = 6 , , , , , , ,
46 30 , , p = 71 , , , , ,
21 17 , , g = 7 , , , , ,
27 40 , ,
15 4 , , , ,
41 31 , ,
28 21 , , , , p = 59 , , , , , , , g = 2 , , , , , p = 83 , , , , , , , g = 2 , , ,
63 48 , , , , , ,
70 19 , , p = 89 , , , , , , , g = 3 , , , , ,
61 27 , ,
13 37 , , , ,
49 18 , ,
50 29 , , , ,
81 20 , , , ,
51 43 , , , ,
42 46 , , , ,
19 47 , , , , , , Example. p = 5 , T =0 1 2 3 4 5 6 7 8 9 10 110 0 1 − − − − − − − − − − − − − − − − − − −
212 13 14 15 16 17 18 19 20 21 22 23 − − − − − − − − − − − − − − − − − − − − The ideal points are (last coord. 0), A B C D E F G (0,1,-1), (1,0,-1), (1,-1,0), (1,1,1),(1,1,-2),(1,2,2),(1,-2,1).successive powers, A,G,E,F,B,C,0; B,F,E,G,A,C,0; C,C; D,D; E,D,E; F,G,D,F.1. p = 11 , g = 2 , T = CHAPTER 8. FUNCTIONS OVER FINITE FIELDS − − − − − −
40 3 3 − − − − − − − − − −
21 4 0 − − − − − −
217 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 − − − − − − − − − − −
44 3 − − − − − − − − − − − − − − − −
134 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 505 5 4 1 4 2 1 4 0 − − − − − − − − − − − − − − − − − − − . . .2. p = 17 , g = 3 , T = { (0 , , , (1 , , , (10 , , , (1 , , , (4 , , , (8 , , , (6 , , , (9 , , , (14 , , , (11 , , , (1 , , , (13 , , , (5 , , , (11 , , , (7 , , , (7 , , , (12 , , , (2 , , , (11 , , , (1 , , , (1 , , , (16 , , , (5 , , , (16 , , , (4 , , , (11 , , , (14 , , , (1 , , , (15 , , , (9 , , , (5 , , , (7 , , , (1 , , , (3 , , , (13 , , , (4 , , , . . . . } Example.
A generator associated to the given primitive root is such that ( a, b, c ) δ = α = (1 , , , with δ := p − . .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS. p g generator , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , − , , , , − , , − , , − , , , , , , − , , See [M130] RICATTI. for more.
Definition.
The set R is the set of elements0. ( x, y, z, t ) (cid:51) x, y, z, t ∈ Z p and − ( x − z ) + ( y − t ) + 4(( x + z ) yt − ( y + t ) xz ) = 1 , with addition1. ( x, y, z, t ) + ( x (cid:48) , y (cid:48) , z (cid:48) , t (cid:48) ) = ( xt (cid:48) + tx (cid:48) + yz (cid:48) + zy (cid:48) , xx (cid:48) + zz (cid:48) + yt (cid:48) + ty (cid:48) ,xy (cid:48) + yx (cid:48) + zt (cid:48) + tz (cid:48) , yy (cid:48) + tt (cid:48) + xz (cid:48) + zx (cid:48) ) . Theorem. ( R , +) is an Abelian group. Conjecture. If p ≡ p − . If p ≡ − p − . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Example.
0. If p = 3 , (1 , , , , is of period 8.1. If p = 5 , (1 , , , , is of period 4.2. If p = 7 , (3 , , , , is of period 48.3. If p = 11 , (3 , , , , is of period 120.4. If p = 13 , (1 , , , , is of period 12.5. If p = 17 , (15 , , , , is of period 16.6. If p = 19 , (12 , , , , is of period 360.7. If p = 23 , (2 , , , , is of period 528.8. If p = 29 , (15 , , , , is of period 28. Lemma. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z y xx z yy x z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x + y + z y xx + y + z z yx + y + z x z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y x z y x z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y x z − y y − x x − z z − y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z )(( z − y ) − ( x − z )( y − x )) . Lemma. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t z y xx t z yy x t zz y x t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z + t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z y x t z y x t z y x t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z + t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − z z − y y − xx − t t − z z − yy − x x − t t − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z + t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − z + y − x z − y y − xx − t + z − y t − z z − yy − x + t − z x − t t − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z + t )( − x + y − z − t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − y y − x − t − z z − y x − t t − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z + t )( − x + y − z + t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − y y − x t − y z − xx − z t − y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x + y + z + t )( − x + y − z + t )(( t − y ) + ( x − z ) ) . .3. ADDITION FORMULAS FOR FUNCTIONS ON A FINITE FIELDS. Definition. det ( x, y, z, t, u ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u t z y xx u t z yy x u t zz y x u tt z y x u (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Definition.
The set R is the set of elements0. ( x, y, z, t, u ) (cid:51) x, y, z, t, u ∈ Z p and det ( x, y, z, t, u ) = 1 with addition1. ( x, y, z, t, u ) + ( x (cid:48) , y (cid:48) , z (cid:48) , t (cid:48) , u (cid:48) )= ( xt (cid:48) + yu (cid:48) + zt (cid:48) + tz (cid:48) + uy (cid:48) , xy (cid:48) + yx (cid:48) + zu (cid:48) + tt (cid:48) + uz (cid:48) ,xz (cid:48) + yy (cid:48) + zx (cid:48) + tu (cid:48) + ut (cid:48) , xt (cid:48) + yz (cid:48) + zy (cid:48) + tx (cid:48) + uu (cid:48) , xu (cid:48) + yt (cid:48) + zz (cid:48) + ty (cid:48) + ux (cid:48) ) . Lemma. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u t z yx u t zy x u tz y x u (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( u − xt ) − ( tu − xz ) ∗ ( xu − yt )+ ( zu − xy ) ∗ ( x − yu ) + ( t − zu ) ∗ ( yu − zt ) − ( zt − yu ) ∗ ( xy − zu ) + ( z − yt )( y − xz )= u − x y − y t − z x − t z + x t + y z +2 x zu + 2 y xu + 2 z tu + 2 t uy − u xt − u yz − xyzt Theorem. det ( x, y, z, t, u ) = s (2( x + y + z + t + u ) − s ( x + y + z + t + u )+ x ( y (2 z + 2 t − u ) − zt + 2 tu + 2 uz ) + . . . − yztu − ztux − tuxy − uxyz − xyzt ) , with s = x + y + z + t + u. Proof: We use (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u t z y xx u t z yy x u t zz y x u tt z y x u (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t z y x u t z y x u t z y x u t z y x u (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and then the Lemma. Theorem. ( R , +) is an Abelian group. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS
Conjecture. If p ≡ p − . If p ≡ p − .2. If p ≡ ± p − .For examples see 8.4.1. To define distances in a sub geometry of affine k -dimensional geometry, we have to define a ho-mogeneous function f ( P ) of degree k. We can then either define the distance between 2 points P and Q by the k -th root of f ( Q − P ) or the hypercube between 2 points P and Q by f ( Q, P ) . I willnot discuss here the extension of a 2-dimensional distance to n -dimension as is done in Euclideangeometry.To define angles, we can associate to a point P, a k by k matrix by a bijection, if the set of thesematrices, which are of determinant 1, form a subset of an Abelian group under matrix multiplica-tion, with generator G , . . . G l , we can define then angular direction of a point associated to thematrix G i . . . G i l l by ( i , . . . , i l ) . We can also define f ( P ) as the determinant of the associated matrix. If in the 2 dimensional realaffine geometry, we associate to ( x, y ) the matrix (cid:18) y x − x y (cid:19) , then f ( x, y ) = x + y , the matrices of determinant 1 for an Abelian group with generator x = sin (1) ,y = cos (1) , and we obtain the -dimensional Euclidean distance and angle.If in the -dimensional real affine geometry, we associate to ( x, y ) the matrix (cid:18) y xx y (cid:19) , then f ( x, y ) = y − x , the matrices of determinant 1 for an Abelian group with generator x = sinh (1) , y = cosh (1) , and we obtain the -dimensional Minkowskian distance and angle.This will now be extended using the generalization of the hyperbolic functions by Ricatti. Definition. In k -dimensional affine geometry we define the Ricatti function as the function which associates tothe point P = ( P , . . . , P k − ) , the Toeplitz matrix T , defined by T i,j = P k − − i + j , ≤ i, j The matrix multiplication defines an addition (which is a convolution) for the points as follows, if T is associated to P and U , to Q, TU is associated to R with P ◦ Q := R i = (cid:80) j P j Q i − − j . For instance, k = 3 , ( P ◦ Q ) = P Q + P Q + P Q , ( P ◦ Q ) = P Q + P Q + P Q , ( P ◦ Q ) = P Q + P Q + P Q . k = 4 , ( P ◦ Q ) = P Q + P Q + P Q + P Q , ( P ◦ Q ) = P Q + P Q + P Q + P Q , ( P ◦ Q ) = P Q + P Q + P Q + P Q , ( P ◦ Q ) = P Q + P Q + P Q + P Q . Corollary. The set of matrices, associated to all the non ideal points of k -dimensional affine geometry withdeterminant 1, form an abelian group under matrix multiplication. Theorem. If p = k then P i = δ i, has period p . Moreover, the j -th iterate P ( j ) is such that P ( j ) i = δ j,i . Theorem. Let det ( . . . , z, y, x ) denote the determinant of the Toeplitz matrix associated with P = ( x, y, z, . . . ) , let s be the sum of the components of P, then k = 3 ,det ( zyx ) = x + y + z − xyz = s (( z − y ) − ( x − z )( y − x )) . k = 4 ,det ( tzyx ) = s ( − x + y − z + t )(( t − y ) + ( x − z ) ) . k = 5 ,det ( x, y, z, t, u ) = s (2( x + y + z + t + u ) − s ( x + y + z + t + u )+ x ( y (2 z + 2 t − u ) − zt + 2 tu + 2 uz ) + . . . − yztu − ztux − tuxy − uxyz − xyzt ) . In the following examples we have obtained what a cyclic generator of what appears to be thelongest period, without examining the details of the structure of the solution. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS Example. k = 4 .p period cyclic generator , , , , , , , , , , , , 11 120 (3 , , , , 13 12 (1 , , , , 17 16 (15 , , , , 19 360 (12 , , , , 23 528 (2 , , , , 29 28 (15 , , , . k = 5 .p period cyclic generator , , , , , , , , , , , , , , , 11 10 (4 , , , , . 13 28560 (3 , , , , , , , , , 19 18 (7 , , , , , 23 279840 (14 , , , , , 29 840 (13 , , , , , 31 30 (26 , , , , . k = 6 ,p period cyclic generator , , , , , , , , , , , , , , , , , , 11 120 (5 , , , , , , 13 12 (3 , , , , , , , , , , , 19 18 (2 , , , , , , 23 528 (3 , , , , , , 29 840 (10 , , , , , , 31 30 (1 , , , , , , k = 7 , .4. APPLICATION TO GEOMETRY. p period cyclic generator , , , , , , , , , , , , , , , , , , , , , , 11 1330 (0 , , , , , , , 13 168 (6 , , , , , , , 17 24137568 (3 , , , , , , , 19 47045880 (18 , , , , , , , 23 12166 (0 , , , , , , , 29 28 (26 , , , , , , , 31 (6) 887503680 (26 , , , , , , , 37 (3) 50652 (9 , , , , , , , 41 (2) 1680 (7 , , , , , , . k = 8 ,p period cyclic generator , , , , , , , , , , , , , , , , , , , , , , , , 11 120 (5 , , , , , , , , 13 168 (7 , , , , , , , , 17 16 (9 , , , , , , , , 19 360 (18 , , , , , , , , 23 528 (9 , , , , , , , , 29 840 (28 , , , , , , , , 31 960 (28 , , , , , , , , 37 1368 (0 , , , , , , , , 41 40 (16 , , , , , , , , k = 9 ,p period cyclic generator , , , , , , , , , , , , , , , , , , may not be largest period , , , , , , , , , 19 18 (2 , , , , , , , , , , , , , , , , , Example. Here we have written j when the maximum period is p j − , unless the number is underlined inwhich case the period is given. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS k \ p k k k k k k k 10 10 10 1 10 5 5 ≥ k k ≥ 12 6 3 4 ≥ ≥ k k k ≥ 16 4 − ≥ ≥ ≥ 16 ? ?19 18 5 3 3 ≥ 18 9 − ≥ 18 6 2 ? k \ p 43 47 53 59 61 67 71 73 79 83 89 973 1 2 2 2 1 1 2 1 1 2 2 14 2 2 1 2 1 2 2 1 2 2 1 15 4 4 4 2 1 4 1 4 2 4 2 46 1 2 2 2 1 1 2 1 1 2 2 17 1 6 3 6 6 3 1 6 3 2 6 28 2 2 2 2 2 2 2 1 2 2 1 19 3 6 2 6 3 3 2 1 3 6 2 310 4 4 4 2 1 4 1 4 2 4 2 411 2 5 5 5 ≥ 10 1 5 ≥ ≥ ≥ 10 1 512 2 2 2 2 1 2 2 1 2 2 2 113 6 4 1 ≥ ≥ ≥ ≥ ≥ 714 1 6 3 6 6 3 1 6 3 2 6 215 4 4 4 2 1 4 2 4 2 4 2 416 4 2 4 4 4 4 2 2 2 4 2 117 8 ? ? 8 ≥ 16 2 ? ? ? ≥ ≥ ≥ ≥ ≥ ≥ Moreover it appears that {\em k | 18 | 20 | 21 | 22 | 24 | 25 |26 | 27}| 3 k | 5 k | 3 104k | 11 5k | 3 k | 5 k |13 2k | 3 k| | | 7 6k | | | | |{\em k | 28 | 30 | 33 | 34 | 35 | 36 |38 | 39}| 7 12k | 3 8k | 3 22k | 17 $8kL\hti{-1}/$ |5 .k | 3 .k | 19 9k | 3 .k| | 5 4k | 11 40k | | 7 560k | || 13 .k .4. APPLICATION TO GEOMETRY. {\em k | 40 | 42 | 44 | 45 | 46 | 48 |49 | 50}| 5 3k | 3 .k | 11 15k | 3 .k | 23 11k | 3 5k |7 .k | 5 k| | 7 .k | | 5 .k | | | |{\em k | 51 | 52 | 54 | 55 | 56 | 57 |58 | 60}| $3\geq$ 200k | 13 3k | 3 | 11 5k | 7 8k | 3 .k |29 .k | 3 .k| 17 .k | | | | | 19 .k | |5 .k Conjecture. If k = 4 , then0. if p ≡ then the maximum period is p − . 1. if p ≡ − then the maximum period is p − . Conjecture. If k = 5 , 0. if p ≡ then the maximum period is p − , 1. if p ≡ then the maximum period is p − ,2. if p ≡ ± then the maximum period is p − .The above examples may lead to other conjectures perhaps for all k. Conjecture. Let p | /k. The maximum period is p e − , where e depends on p and k, e ( p i , p (cid:48) ) = e ( p i , p ”) if p (cid:48) ≡ p ” (mod p i ) . ( q , q ) = 1 ⇒ e ( k, q q ) = lcm ( e ( k, p ) , e ( k, p ) . e ( p i , p (cid:48) ) = order ( p (cid:48) ) ∈ Z p,. . In view of 0, we can define e u := e ( p i , u ) for u ∈ Z p i ,. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS e ( p i , 1) = 1 , e ( p i , p i − 1) = 2 , k = 5 , e = e = 4 ,k = 7 , e , e = 3 , e , e = 6 ,k = 2 , e , e = 2 ,k = 3 , e , e = 3 , e , e = 6 ,k = 11 , e , e , e , e = 5 , e , e , e , e = 10 ,k = 13 , e , e = 3 , e , e = 4 , e , e = 6 , e , e , e , e = 12 , k = 17 , e , e , e , e = 5 ,e , e , e , e = 10 , ? Theorem. If, for k = 3 , p ≡ , we construct a period associated to a generator and determine thecoplanar directions to the directions associated to 0 and 1, we obtain a difference sets For the set Z p ,. of the numbers from 0 to p relatively prime to p. The sets have p ( p − 1) elements. Proof: The proof is similar to that of Singer. In this case, the directions are the non isotropicones and 2 non isotropic directions determine exactly one plane, through the origin, which contains p + 1 directions.This Theorem extends to any dimension. We should check if these difference sets are alsoobtained by some other method. .4. APPLICATION TO GEOMETRY. Example. k = 3 , ([130 \ RIC.BAS] p, then diff. set then generator) p gen. dif f erence set (mod p − of p ( p − elements , , 3) 0 , , , , , , 4) 0 , , , , , , , , , , , , 13) 0 , , , , , , , , , , , , , , , , , , 12) 0 , , , , , , , , , , , , , , , , , , , , , , , , 7) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 17) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 47 (0 , , 37) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 53 (0 , , 20) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 59 (0 , , 11) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , CHAPTER 8. FUNCTIONS OVER FINITE FIELDS p gen. dif f erence set (mod p − of p ( p − elements 71 (0 , , 54) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 83 (0 , , 60) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 89 (0 , , 77) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 101 (0 , , 60) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 29) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , .4. APPLICATION TO GEOMETRY. p gen. dif f erence set (mod p − of p ( p − elements 131 (0 , , 18) 0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Introduction. It occured to me that just like in 3 dimensional Euclidean geometry, the geometry on the sphere canbe used as a model for the non euclidean geometry of elliptic type, in the same way the geometryon the surface T : x + y + z − xyz = 1 , can be used as a model for an other geometry, if p ≡ − . I will call this geometry, Ricatti geometry . It turns out that this geometry is more akinto an Euclidean geometry. It can be considered as starting from a dual affine geometry in whichwe prefer a line (the ideal line) and a point (the ideal point) which is not on the line. The linecorresponds to the intersection with t = 0 of the plane x + y + z = 0 , the point to the direction ofthe line through the origin and the point (1 , , . Definition. Given p ≡ − , the group ( T , +) is cyclic (8.3.2. We determine a generator ( a, b, c ) of thegroupt using in part 8.3.2. The points on T are labelled according to i ( a, b, c ) from 0 to p − . The lines are the set of points on T and a plane through the origin distinct from [1 , , , and whichdoes not contain the line from the origin to (1 , , , . The line through i and i + 1 is labelled − i ∗ . CHAPTER 8. FUNCTIONS OVER FINITE FIELDS Notation. Points are denoted by a lower case letter or integer mod p − , lines by the same followed by a “ ∗ ”. Definition. If 2 points do not determine a line they are called parallel .If 2 lines are not incident to a point they are called parallel . Definition. There is a correspondence between the point i and the line i ∗ , called polarity . Theorem. There are p − A line is incident to p points and a point to p lines. A point is parallel to p − p − There is duality in this geometry. Theorem. If i ∗ is incident to i , i , i , i and i , then ( i + j ) ∗ is incident to i − j, i − j, i − j, i − j and i − j. Definition. Let D be a difference set associated to the integers in Z p ,. , between 0 and p − , relatively primeto p, D = { d , d , . . . , d p − } . 0. The selector function is defined as follows, f ( k ( p + 1)) = − ,f ( d j − d i ) = d i . 1. With points represented by elements in Z p − and lines similarly represented but followed by“ ∗ , ” the incidence relation is defined by i is on j ∗ iff f ( i + j ) = 0 . Theorem. i is parallel to j or i ∗ is parallel to j ∗ iff f ( i − j ) = − , the line ( i × j ) ∗ incident to i and j , not parallel, is ( f ( i − j ) − j ) ∗ . the line k ∗ incident to i parallel to j ∗ is . . . Proof: For 2, we want k to be ≡ j (mod p + 1) such that f ( k + i ) = 0 , . . . . .4. APPLICATION TO GEOMETRY. Example. p = 5 , D = { , , , , } , i f ( i ) − − i 12 13 14 15 16 17 18 19 20 21 22 23 f ( i ) − − Examples of such differences sets are given in 8.4.1. Example. p = 5 , the computations are done mod . 0. The coordinates of the i -th point on T are a i , b i , c i , the distance between j and j + i is d i = (cid:113) a i + b i + ( c i − − a i b i ( c i − .i a i − − − − − − b i − − − − − c i − − − − − − − − d i i 12 13 14 15 16 17 18 19 20 21 22 23 a i − − − − − − − b i − − − − − − − − c i − − − − − d i 1. The line i ∗ is incident to the points − i, − i, − i, − i, − i. 2. The point i is parallel to i + 6 , i + 12 and i − . 3. The angle between lines j ∗ and ( j + i ) ∗ is d i . In particular, ∗ : 0 , , , , , is parallel to ∗ , ∗ , ∗ , ∗ : 23 , , , , , is parallel to ∗ , ∗ , ∗ , ∗ : 10 , , , , , ∗ : 8 , , , , , ∗ : 3 , , , , , ∗ : 1 , , , , . on ∗ , the distances are CHAPTER 8. FUNCTIONS OVER FINITE FIELDS The “circles” of radius r and center 0 are r points on , , , , , 174 7 , , , , , 222 18 , , 233 1 , , 60 0 , , , , , Example Let p = 11 ,i a i b i c i d i i 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 a i b i c i d i i 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 a i b i c i d i .4. APPLICATION TO GEOMETRY. i 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 a i b i c i d i i 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 a i b i c i d i i 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 a i b i c i d i i 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 a i b i c i d i i 112 113 114 115 116 117 118 119 120 a i b i c i d i The “circles” of radius r and center 0 are r points on , , , , , , , , , , , 118 (12)10 2 , , , , , , , , , , , 115 (12)2 4 , , , , , , , , , , , 106 (12)9 14 , , , , , , , , , , , 116 (12)3 3 , , , , , , , , , , , , , , 119 (15)8 1 , , , , , , , , , , , , , , 117 (15)4 17 , , , , , 108 (6)7 12 , , , , , 103 (6)5 24 , , , , , , , , 107 (9)6 13 , , , , , , , , 96 (9)0 0 , , , , , , , , , , , 110 (12) p = 17; circles of center 0 with given radius: CHAPTER 8. FUNCTIONS OVER FINITE FIELDS , , , , , , , , , , , , , , , , , 272 (18)1 : 29 , , , , , , , , , , , , , , , , , , , , , , , 278 (24)2 : 24 , , , , , , , , , , , 282 (12)3 : 1 , , , , , , , , , , , 268 (12)4 : 18 , , , , , , , , , , , , , , 262 (15)5 : 23 , , , , , , , , , , , 234 (12)6 : 3 , , , , , , , , , , , , , , , , , , , , 277 (21)7 : 2 , , , , , , , , , , , , , , 207 (15)8 : 5 , , , , , , , , , , , , , , , , , , , , , , , 276 (24)9 : 12 , , , , , , , , , , , , , , , , , , , , , , , 283 (24)10 : 81 , , , , , , , , , , , , , , 286 (15)11 : 11 , , , , , , , , , , , , , , , , , , , , 285 (21)12 : 54 , , , , , , , , , , , 265 (12)13 : 26 , , , , , , , , , , , , , , 270 (15)14 : 20 , , , , , , , , , , , 287 (12)15 : 6 , , , , , , , , , , , 264 (12)16 : 10 , , , , , , , , , , , , , , , , , , , , , , , 259 (24) The points common to the circles given above and noted i : if the radius is i are given below ifthere are points which are common with .4. APPLICATION TO GEOMETRY. center 1, radius 1: 30,61,63,74,90,112,119,134,146,148,157,160,162,163,191,196,203,206,217,236,246,252,267,2790: 112,160; 1: 162; 3: 236,267; 4: 74,134; 6: 157;7: 61,63,119; 8: 206; 9: 148,191,203; 10: 90,279;11: 217,246,252; 12: 146,196; 13: 163; 14: 30;center 1, radius 2: 25,48,68,121,152,187,224,254,264,270,276,2830: 48,224; 6: 121; 7: 68; 8: 276; 9: 283; 10: 254; 11: 187;13: 270; 14: 152; 15: 25,264;center 1, radius 3: 2,9,18,39,67,71,137,220,237,259,268,2692: 67,269; 3: 268; 4: 18,39; 6: 71; 7: 2,9; 10: 220; 11: 237;15: 137; 16: 259;center 1, radius 4: 19,40,75,79,88,101,107,110,114,126,135,175,194,261,263 2: 263; 5: 110; 6: 88,101; 7: 107; 8: 40,194; 9: 79,261; 10: 135;12: 75; 13: 114,175; 15: 19; 16: 126;center 1, radius 5: 24,60,93,104,111,125,140,143,166,181,214,2351: 60,111,235; 2: 24; 4: 125; 8: 104,140; 9: 166; 10: 181;13: 214; 16: 93,143;center 1, radius 6: 4,34,37,42,43,45,51,52,56,57,72,78,89,102,122,139,158,173,274,275,2781: 89,278; 2: 275; 4: 78; 5: 139; 6: 42,51,56,274; 7: 4,34,173;8: 45,57,122; 9: 158; 14: 52; 15: 102; 16: 37,43,72;center 1, radius 7: 3,5,8,10,35,62,64,69,92,108,120,154,174,199,2080: 64,208; 1: 62; 2: 120; 3: 8; 4: 174; 5: 92; 6: 3; 8: 5; 12: 108;13: 154; 14: 69; 15: 35; 16: 10,199;center 1, radius 8: 6,28,32,41,46,47,58,59,77,85,86,98,105,106,123,131,141,172,190,195,207,210,240,2770: 32,240; 1: 190,195; 2: 47; 4: 106; 5: 59; 6: 41,77,172,277;7: 207; 8: 46,58,85,105; 11: 131; 12: 123; 13: 28,210; 15: 6;16: 86,98,141;center 1, radius 9: 13,50,80,83,95,100,118,149,159,167,184,185,192,204,205,213,231,232,243,244,249,258,262,2840: 80,192; 1: 118,159,205; 3: 258; 4: 100,262; 5: 213; 6: 50;9: 184,204,231,243; 10: 284; 11: 167,232,244; 12: 149,185;13: 95,249; 15: 13; 16: 83;center 1, radius 10: 82,91,116,136,170,182,198,221,226,228,255,280,282,285,2872: 282; 3: 136; 7: 91,198; 9: 82; 11: 116,255,285; 13: 182;14: 280,287; 15: 221; 16: 170,226,228; CHAPTER 8. FUNCTIONS OVER FINITE FIELDS center 1, radius 11: 12,15,16,117,132,151,168,188,201,212,218,233,234,238,239,245,247,248,253,256,2860: 16,256; 1: 245; 2: 151,253; 5: 234; 8: 239; 9: 12,117,212,248;10: 286; 11: 15,233,238,247; 13: 188,201; 14: 218; 15: 168; 16: 132;center 1, radius 12: 55,76,109,124,147,150,165,179,186,197,230,2661: 147,266; 2: 186; 4: 109; 5: 124,165; 6: 55; 8: 76; 9: 230;10: 197; 11: 150; 13: 179;center 1, radius 13: 27,29,96,115,155,164,176,180,183,189,202,211,215,250,2710: 96,176; 1: 29,202; 5: 180; 8: 27,189; 9: 183; 10: 115; 11: 211;12: 164; 14: 250,271; 16: 155,215;center 1, radius 14: 21,22,31,53,70,153,219,223,251,272,281,00: 0,272; 1: 251; 2: 223; 3: 70,219; 7: 153; 8: 31; 10: 281;14: 21; 16: 22,53;center 1, radius 15: 7,14,20,26,36,66,103,138,169,222,242,2653: 66; 5: 103; 6: 36,138; 7: 7; 9: 242; 10: 169; 11: 14; 12: 265;13: 26; 14: 20,222;center 1, radius 16: 11,23,38,44,54,73,84,87,94,99,127,128,130,133,142,144,156,171,178,200,216,227,229,2600: 128,144; 1: 73,133,156,216; 3: 38; 4: 87,260; 5: 23,142; 6: 44;8: 84,130,171; 9: 94,99; 10: 227; 11: 11,200; 12: 54,178,229;16: 127;center 2, radius 1: 31,62,64,75,91,113,120,135,147,149,158,161,163,164,192,197,204,207,218,237,247,253,268,2800: 64,192; 1: 62,147,161; 2: 120,253; 3: 268; 4: 113; 7: 91,207;8: 31; 9: 158,204; 10: 135,197; 11: 237,247; 12: 75,149,164;13: 163; 14: 218,280;center 2, radius 2: 26,49,69,122,153,188,225,255,265,271,277,2846: 277; 7: 153; 8: 122; 9: 49; 10: 225,284; 11: 255; 12: 265;13: 26,188; 14: 69,271;center 2, radius 3: 3,10,19,40,68,72,138,221,238,260,269,2702: 269; 4: 260; 6: 3,138; 7: 68; 8: 40; 11: 238; 13: 270;15: 19,221; 16: 10,72;center 2, radius 4: 20,41,76,80,89,102,108,111,115,127,136,176,195,262,264 0: 80,176; 1: 89,111,195; 3: 136; 4: 262; 6: 41; 8: 76; 10: 115;12: 108; 14: 20; 15: 102,264; 16: 127;center 2, radius 5: 25,61,94,105,112,126,141,144,167,182,215,2360: 112,144; 3: 236; 7: 61; 8: 105; 9: 94; 11: 167; 13: 182; 15: 25;16: 126,141,215; .4. APPLICATION TO GEOMETRY. center 2, radius 6: 5,35,38,43,44,46,52,53,57,58,73,79,90,103,123,140,159,174,275,276,2791: 73,159; 2: 275; 3: 38; 4: 174; 5: 103; 6: 44;8: 5,46,57,58,140,276; 9: 79; 10: 90,279; 12: 123; 14: 52; 15: 35;16: 43,53;center 2, radius 7: 4,6,9,11,36,63,65,70,93,109,121,155,175,200,2093: 70; 4: 109; 6: 36,121; 7: 4,9,63; 8: 209; 11: 11,200; 13: 175;15: 6,65; 16: 93,155;center 2, radius 8: 7,29,33,42,47,48,59,60,78,86,87,99,106,107,124,132,142,173,191,196,208,211,241,2780: 48,208; 1: 29,60,278; 2: 47; 4: 78,87,106; 5: 59,124,142;6: 33,42; 7: 7,107,173; 9: 99,191; 11: 211; 12: 196; 15: 241;16: 86,132;center 2, radius 9: 14,51,81,84,96,101,119,150,160,168,185,186,193,205,206,214,232,233,244,245,250,259,263,2850: 96,160; 1: 205,245; 2: 186,263; 4: 193; 6: 51,101; 7: 119;8: 84,206; 10: 81; 11: 14,150,232,233,244,285; 12: 185; 13: 214;14: 250; 15: 168; 16: 259;center 2, radius 10: 83,92,117,137,171,183,199,222,227,229,256,281,283,286,00: 0,256; 5: 92; 8: 171; 9: 117,183,283; 10: 227,281,286; 12: 229;14: 222; 15: 137; 16: 83,199;center 2, radius 11: 13,16,17,118,133,152,169,189,202,213,219,234,235,239,240,246,248,249,254,257,2870: 16,240; 1: 118,133,202,235; 3: 17,219; 5: 213,234; 8: 189,239;9: 248,257; 10: 169,254; 11: 246; 13: 249; 14: 152,287; 15: 13;center 2, radius 12: 56,77,110,125,148,151,166,180,187,198,231,2672: 151; 3: 267; 4: 125; 5: 110,180; 6: 56,77; 7: 198;9: 148,166,231; 11: 187;center 2, radius 13: 28,30,97,116,156,165,177,181,184,190,203,212,216,251,2720: 272; 1: 156,190,216,251; 5: 165; 8: 97; 9: 184,203,212; 10: 181;11: 116; 13: 28; 14: 30; 16: 177;center 2, radius 14: 22,23,32,54,71,154,220,224,252,273,282,10: 32,224; 2: 282; 3: 1; 5: 23; 6: 71,273; 10: 220; 11: 252;12: 54; 13: 154; 16: 22;center 2, radius 15: 8,15,21,27,37,67,104,139,170,223,243,2661: 266; 2: 67,223; 3: 8; 5: 139; 8: 27,104; 9: 243; 11: 15; 14: 21;16: 37,170;center 2, radius 16: 12,24,39,45,55,74,85,88,95,100,128,129,131,134,143,145,157,172,179,201,217,228,230,2610: 128; 1: 145; 2: 24; 4: 39,74,100,134; 6: 55,88,157,172;8: 45,85; 9: 12,230,261; 11: 131,217; 13: 95,179,201;16: 129,143,228; CHAPTER 8. FUNCTIONS OVER FINITE FIELDS Introduction. On the surface T , for p ≡ − , we can define besides lines (intersection with a planethrough the origin), circles (pts equidistant using the cubic function from a given point), line-circle(set of tangents in space to the circles), podars (set of points where tangents in space intersect T ),mediatrices (set of points equidistant from 2 points). This section describes those curves. Definition. The circles are the set of points on T such that the cubic distance from a given point on T , calledthe center of the circle, is a given integer r, called the radius of the circle. Theorem. The circles of radius r and center (0 , , , are the points ( x, y, z ) which satisfy 0. and 1. or 0. and2. x + y + z − xyz = 1 . x + y + z − z + 3 z − − xyz + 3 xy = r . z − z − xy = − r . Definition. A line-circle is the set of lines tangent in space to a circle. Theorem. The line-circle associated to the circles in 8.4.3 have at ( x, y, z ) the direction (∆ x, ∆ y, ∆ z ) givenby ∆ x = xz + z (cid:48) y , ∆ y = − yz − z (cid:48) x , ∆ z = − xx + yy , where x = x − yz, y = y − zx, z = z − xy, z (cid:48) = 2 z − . Proof: the component of the direction satisfy ( x − yz )∆ x + ( y − zx )∆ y + ( z − xy )∆ z = 0 , − y ∆ x − x ∆ y + (2 z − z = 0 , hence 0. Definition. A podar is set of points where tangents in space intersect T. .4. APPLICATION TO GEOMETRY. Theorem. The coordinates of points on the podar associated to the circle in 8.4.3 are the points ( x + t ∆ x, y + t ∆ y, z + t ∆ z ) where t satisfies t = − x ∆ x + y ∆ y + z ∆ z (∆ x +∆ y +∆ z (∆ x + ∆ y + ∆ z )) . where ∆ x = ∆ x − ∆ y ∆ z, ∆ y = ∆ y − ∆ z ∆ x, ∆ z = ∆ z − ∆ x ∆ y. Proof. A point ( x + t ∆ x, y + t ∆ y, z + t ∆ z ) on the line ( x, y, z ) with direction (∆ x, ∆ y, ∆ z ) ison T if t satisfies the cubic equation, (∆ x + ∆ y + ∆ z )(∆ x + ∆ y + ∆ z ) t + 3( x ∆ x + y ∆ y + z ∆ z ) t + 3( x ∆ x + y ∆ y + z ∆ z ) t + ( x + y + z − xyz − 1) + 1 = 0 , the coefficient of t is 0 because ( x, y, z ) is on T , that of t is 0 because it is x ( xz + z (cid:48) y ) + y ( − yz − z (cid:48) x ) + z ( − xx + yy ) = 0 . Theorem. If the tangent k ∗ at i to the circle, centered at 0 of radius r, meets T at j, then the tangent( k +2 i ) ∗ at − i to the circle, centered at 0 of radius − r (mod p ) , meets T at j − i. If the tangent at i to the circle, centered at 0 of radius r, meets T at j parallel to i, then thetangent at − i to the circle, centered at 0 of radius − r (mod p ) , meets T at j − i parallel to − i. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS Example. Let p = 11 ,r = 1 ,circle podar 92 15 113 43 51 52 4 44 107 45 81 97 line − circle ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ r = 10 ,circle podar 101 93 87 31 118 98 62 63 71 61 117 82 line − circle ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ r = 2 circle podar 28 59 36 118 6 68 36 66 95 49 98 85 line − circle −− ∗ ∗ ∗ ∗ −− ∗ ∗ ∗ ∗ ∗ ∗ r = 9 circle 14 19 32 34 51 63 76 81 89 93 112 116 podar 113 16 113 43 48 42 100 48 56 102 43 20 line − circle ∗ ∗ ∗ ∗ ∗ −− ∗ ∗ ∗ ∗ −− r = 3 circle podar line − circle ∗ ∗ ∗ ∗ −− ∗ ∗ ∗ ∗ ∗ r = 3 circle 92 104 105 109 119 podar 38 83 84 40 80 line − circle ∗ ∗ ∗ ∗ ∗ r = 8 circle podar 82 62 114 115 94 35 94 54 65 74 line − circle ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ r = 8 circle 72 79 87 113 117 podar 72 25 33 74 3 line − circle −− ∗ ∗ ∗ ∗ r = 4 circle 17 62 67 82 84 108 podar 41 113 91 43 84 108 line − circle −− ∗ −− ∗ −− −− r = 7 circle 12 36 38 53 58 103 podar 12 36 119 77 109 7 line − circle −− −− ∗ −− ∗ −− r = 5 circle 24 42 49 59 85 95 97 102 107 podar 24 21 73 83 16 56 103 111 53 line − circle −− ∗ −− −− ∗ ∗ ∗ ∗ ∗ r = 6 circle 13 18 23 25 35 61 71 78 96 podar 79 27 29 106 86 85 95 57 96 line − circle ∗ ∗ ∗ ∗ ∗ −− −− ∗ −− .4. APPLICATION TO GEOMETRY. Definition. A mediatrix of 2 points is the set of points equidistant from them. Conjecture. The number of points on the mediatrix is ≡ , unless the points are i and i + k ( p − inwhich case it is . . . namely these points are for 0 and p − , ( p − k and ( p + 1) k − . When p = 11 , the multiples are 8, 12 and 16; when p = 17 , 12, 16, 20 and 24; when p = 23 , 0, 16, 20, 24, 28, 32. Example of mediatrices for p = 11 : For 0 and 3, 0 and 6, 0 and 9, no points.For 0 and 1: 16,22,29,55,66,92,99,105.For 0 and 2: 25,34,38,45,77,84,88,97.For 0 and 4: 1,3,6,7,8,9,15,31,52,72,93,109,115,116,117,118.For 0 and 5: 16,18,19,23,44,58,67,81,102,106,107,109.For 0 and 7: 2,5,25,48,49,78,79,102.For 0 and 8: 1,7,39,41,54,74,87,89.For 0 and 10: i and i − .For 0 and 11: 2,5,6,9,52,56,75,79.For 0 and 12: 25,28,35,39,63,64,68,69,93,97,104,107.For 0 and 13: 28,29,32,44,56,57,76,77,89,101,404,105.For 0 and 14: 3,11,15,29,43,91,105,119.For 0 and 15: 6,9,16,21,34,37,43,48,53,82,87,92,98,101,114,119.For 0 and 16: 22,37,45,64,72,91,99,114.For 0 and 17: 26,28,35,44,45,51,53,59.For 0 and 18: 3,4,7,11,14,15,29,32,42,57,65,73,81,96,106,109.For 0 and 19: 3,16,21,27,51,52,66,73,87,88,112,118. Definition. The horizon of a point P on T is the set of points on T and the tangent plane through P. Theorem. The coordinates of points on the tangent at P = ( x, y, z ) in the plane P through O and Q = ( x (cid:48) .y (cid:48) , z (cid:48) )which is also on T is( x + t ∆ x, y + t ∆ y, z + t ∆ z ) , where t satisfies t = − x ∆ x + y ∆ y + z ∆ z (∆ x +∆ y +∆ z (∆ x + ∆ y + ∆ z )) . where ∆ x = ∆ x − ∆ y ∆ z, ∆ y = ∆ y − ∆ z ∆ x, ∆ z = ∆ z − ∆ x ∆ y. Proof. The direction of the normal to P at P is ( a := yz (cid:48) − zy (cid:48) , b := zx (cid:48) − xz (cid:48) , c := xy (cid:48) − yx (cid:48) ).The direction (∆ x, ∆ y, ∆ z ) satisfies a ∆ x + b ∆ y + c ∆ z = 0 and x ∆ x + y ∆ y + z ∆ z = 0 , where x = x − yz, y = y − zx, z = z − xy, therefore ∆ x = y c − z b, ∆ y = z a − x c, ∆ z = x b − y a. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS A point ( x + t ∆ x, y + t ∆ y, z + t ∆ z ) on the line ( x, y, z ) with direction (∆ x, ∆ y, ∆ z ) is on T if t satisfies the cubic equation, (∆ x + ∆ y + ∆ z )(∆ x + ∆ y + ∆ z ) t + 3( x ∆ x + y ∆ y + z ∆ z ) t + 3( x ∆ x + y ∆ y + z ∆ z ) t +( x + y + z − xyz − 1) + 1 = 0 , the coefficient of t is 0 because ( x, y, z ) is on T, that of t is 0 because it is x ( y c − z b ) + y ( z a − x c ) + z ( x b − y a ) = 0 . Algorithm. To determine the horizon as 0 we determine for each point with z = 1 , on which the line × sel ( i ) itis located, if x + y = 0 , the point is the ideal point, if x = 0 or y = 0 , × sel ( i ) , then it corresponds to points parallel to it. This is implement in [ \ \ RIC.BAS]option 12.Proof: The horizon of P of 0 are the points on T and z = 1 or x + y − xy = 0 , those in theplane x = kt, y = lt satisfy ( k + l ) t − kl = 0 and t = 0 , twice. If k = 0 or l = 0 then t = 0 is atriple root, if k = − l, or x + y = 0 , then the point is an ideal point. In all other cases, t = klk + l ) . Example. For p = 11 , the points H on the horizon of 0 have their tangent t and the points Q on T for whichthe tangent is t ∗ given by H t ∗ Q ∗ , , , , , , , , , y = 0)41 ∗ , , , , , , , , , x = 0)6 28 ∗ , , , , , , , , , ∗ , , , , , , , , , −− , , , , , , , , ∗ , , , , , , , , , ∗ , , , , , , , , , ∗ , , , , , , , , , ∗ , , , , , , , , , ∗ , , , , , , , , , ∗ , , , , , , , , , ∞ ∗ , , , , , , , , , Conjecture. The points on the horizon of 0 are multiples of 3. Example. The horizon of 0 is for0. p = 5 ,selector −− horizon ∞ p = 11 ,selector −− horizon ∞ .4. APPLICATION TO GEOMETRY. p = 17 ,selector horizon 114 111 24 246 225 150 0 165 210 81 159 213 selector 191 195 251 259 282 −− horizon ∞ 93 0 141 120 108 p = 23 ,selector horizon selector 271 298 305 333 342 352 363 375 450 488 503 −− horizon 81 231 426 444 33 0 498 402 453 ∞ 294 408 Introduction. The selector function was introduced by Fernand Lemay to determine easily from the selector, pointson 2 lines, lines incident to 2 points, points on lines or lines incident to points. This notion isgeneralized to 3 and more? dimensions. Definition. defining polynomial Theorem. If the P i denotes a primitive polynomial of degree i, for k = 3 , the defining polynomials P can havethe following form, P , P P , P P , there are p + p + p + p + 1 , p − , p − p polynomials relatively prime to P, in these respectivecases.For k = 4 , the defining polynomials P can have the following form, P , P P , P P , P P . there are p + p + p + p + p + 1 , ( p − p + 1) , p − , p − p polynomials relatively prime to P, in these respective cases.Proof: The polynomials in the sets are those which are relatively prime to the defining polyno-mial. There are p k homogeneous polynomials of degree k. If, for instance, k = 4 and the definingpolynomial P is P P , there are p + p + 1 polynomials which are multiple of P and p + 1 , whichare multiples of P , hence p + p + p + p + 1 − ( p + p + 1) − ( p − 1) polynomials relatively primeto P. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS Example. a , a , . . . represents I k +1 − a I k − a I k − − . . . .k p period def.pol. sel. rootsof def.pol.orprim.pol. , , , −− 26 1 , , , , , , , 25 156 1 , , , −− 124 1 , , , , , , , 44 3 121 2 , , , , −− 104 0 , , , , I + I − I − I + I + 1)80 0 , , , , , , , , , 25 781 4 , , , , −− 744 2 , , , , I + I + 2)( I + 2 I − I + 2)624 2 , , , , , , , , , 37 2801 3 , , , , −− , , , , I + 2 I − I − I − I − , , , , , , , , , , , , , −− , , , , I − I + 6)( I − I + I + 2)28560 2 , , , , , , , , , −− 242 1 , , , , , , , , , , , Definition. Given a selector s, the selector function associates to the integers in the set Z n a set of p +1 integersor p integers obtained as follows, s ( j ) ∈ f i iff sel ( l ) − sel ( j ) = i for some l. Theorem. f ( i ) is the set of points on the line i ∗ × ∗ . 1. 0. f ( i ) − j, where we subtract j to each element is the set, is the set of points in ( i + j ) ∗ × j ∗ , equivalently 2. 1. f ( i − j ) − j, is the set of points in i ∗ × j ∗ . a ∗ × b ∗ × c ∗ = (( a − i ) ∗ × ( b − i ) ∗ × ( c − i ) ∗ ) − i. .4. APPLICATION TO GEOMETRY. Definition.Theorem.Theorem. If the defining polynomial is primitive, then 1. 0. | s | = p k − p − , 2. 1. if i (cid:54) = 0 , | f ( i ) | = p + 1 If the defining polynomial has one root, then 4. 0. | s | = p k , 5. 1. if i (cid:54) = 0 , | f ( i ) | = p, If the defining polynomial has one root, then 7. 0. | s | = p k − , 8. 1. if i ? , | f ( i ) | = p, 9. 2. if i ? , | f ( i ) | = p − , If the defining polynomial has one quadratic factor, then 11. 0. | s | = ( p k − p + 1)? , 12. 1. if i ? , | f ( i ) | = p, Example. k = 3 , p = 3 , defining polynomial I − I − I − I − . selector: 0 1 2 9 10 13 15 16 18 20 24 30 37selector function: − − − − CHAPTER 8. FUNCTIONS OVER FINITE FIELDS k = 3 , p = 3 , def iningpolynomialI − I − I − I − . selector: 0 1 2 8 11 18 20 22 23selector function: − − − − − − k = 3 , p = 3 , def iningpolynomialI − I − I − . selector: 0 1 2 4 14 15 19 21selector function: − − − − − − 11 0 1 14 7 14 19 21 13 1 2 15 19 0 2 192 0 2 19 8 − − − − − − − 14 0 15 21 10 4 14 15 16 − − − Example. In the case of Example 3.6.x.0. if we denote by i % , the lines ∗ × i ∗ , these lines, which are sets of4 points can all be obtained from % = { , , , } , % = { , , , } , % = { , , , } and % = { , , , } by adding an integer modulo n. % + 0 = 1 % , % , % , % + 1 = 39 % , % , % , % + 9 = 6 % , % , % , % + 15 = 34 % , % , % , % + 0 = 2 % , % , % , % + 2 = 22 % , % , % , % + 37 = 3 % , % , % , % + 24 = 13 % , % , % , % + 0 = 4 % , % , % , % + 4 = 17 % , % , % , % + 21 = 12 % , % , % , % + 33 = 7 % , % , % , % + 0 = 10 % , % , % . Definition.Conjecture. k . Introduction. If we choose the “sphere“ x + y + z − xyz = 1 in 3 dimension we do not obtain for a givenprime all periods as we do with the selector. We have to generalyze using what is derived fromdifferential equations with constant coefficients in which the coefficient of the k − -th derivative iszero to obtain a constant Wronskian. But, just as in the case of 2 dimensions, to obtain all sets oftrigonometric functions, corresponding to the circular and hyperbolic functions, for all p, we have .5. GENERALIZATION OF THE SPHERES IN RICCATI GEOMETRY. to introduce in 3 dimension a cubic non residue if there is any, . . . I will first recall same wellknown definitions and Theorems of linear differential equations. Definition. Given a linear differential equation D k x = C x + C Dx + . . . C k − D k − x, and k solutions y i of these equations, the Wronskian is the matrix of functions whose j -th row are the j -th derivativesof y i , for i = 0 to k − . Theorem. The functions y i are independent solution iff the determinant of the Wronskian is differentfrom 0 for a particular value of the independent variable. The determinant of the Wronskian is a constant function. If W (0) = E, then W ( x + y ) = W ( x ) W ( y ) . Theorem. If the linear differential equation 8.5.1.0. is such that C i are constant functions then any linearcombination of x and its derivatives is also a solution of 8.5.1.0. Comment. If we choose x such that its derivatives are 0 except the k − -th, chosen equal to 1, it is easy toobtain independent solutions using linear combination of x and its derivatives to insure W (0) = E. If det ( W ( t )) = 1 , then det ( W ( nt )) = 1 and the surface D i x ( nt ) , i = 0 to k − , can be chosen as a“sphere” in k -dimension and n as the angle between the directions joining the origin to the points ( D i x (( n + a ) t )) and ( D i x ( at )) . Notation. Given 2 solutions x and x (cid:48) of 8.5.1.0. and a parameter t let x i := { D j x ( it ) } and x (cid:48) i := { D j x (cid:48) ( it ) } ,y i,j := x i x (cid:48) j + x j x (cid:48) i , i (cid:54) = jy i,i := x i x (cid:48) i , Theorem. For k = 3 , let D x = C x + C Dx , with Dx (0) = 0 , Dx = x (0) = 0 , D x (0) = x (0) = 1 , then x − C x , x , x are independent CHAPTER 8. FUNCTIONS OVER FINITE FIELDS 1. their Wronskian is W = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − C x x x C x x x C x C x + C x x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) W (0) = E. 2. The distance from (0 , , to ( x , x , x ) is C x + C x + x − x x x − C x x − C x x + 2 C C x x + C x x . 3. The addition formulas are x (cid:48)(cid:48) = y , + y , − C y , ,x (cid:48)(cid:48) = y , + C y , ,x (cid:48)(cid:48) = y , + C y , + C y , . 4. The tangent plane at ( x , x , x ) is [3 C x − C x x − C x + 4 C C x x + C x , C x − C x x + 2 C C x − C x x + 2 C x x , x − C x x − C x x − C x ] . Definition. If 0. p ≡ , then ν = n is a non cubic residue and the functions are not necessary real,we therefore denote then by ξ i instead of x i and express ξ i in terms of a power of ν and aninteger x i as follows,1. x (3 i ) = x (3 i ) , ξ (3 i ) = x (3 i ) ν, ξ (3 i ) = x (3 i ) ν , ξ (3 i + 1) = x (3 i ) ν , ξ (3 i + 1) = x (3 i + 1) , ξ (3 i ) = x (3 i ) ν , ξ (3 i + 2) = x (3 i ) ν , ξ (3 i + 2) = x (3 i + 2) ν , ξ (3 i ) = x (3 i ) . Moreover C is replaced by C ν . The addition formulas become, for instance,2. x (3 i ) = x (1) x (3 i − 1) + x (1) x (3 i − n + x (1) x (3 i − − C x (1) x (3 i − n,x (3 i ) = x (1) x (3 i − 1) + x (1) x (3 i − 1) + C x (1) x (3 i − ,x (3 i ) = x (1) x (3 i − n + C ( x (1) x (3 i − 1) + x (1) x (3 i − n )+ C x (1) x (3 i − n. x (3 i + 1) = x (1) x (3 i ) + x (1) x (3 i ) n + x (1) x (3 i ) n − C x (1) x (3 i ) n,x (3 i + 1) = x (1) x (3 i ) + x (1) x (3 i ) n + C x (1) x (3 i ) ,x (3 i + 1) = x (1) x (3 i ) + C ( x (1) x (3 i ) + x (1) x (3 i ))+ C x (1) x (3 i ) n. x (3 i + 2) = x (1) x (3 i + 1) + x (1) x (3 i + 1) + x (1) x (3 i + 1) − C x (1) x (3 i + 1) ,x (3 i + 2) = ( x (1) x (3 i + 1) + x (1) x (3 i + 1) n ) + C x (1) x (3 i + 1) ,x (3 i + 2) = x (1) x (3 i + 1) n + C ( x (1) x (3 i + 1) + x (1) x (3 i + 1))+ C x (1) x (3 i + 1) n. .5. GENERALIZATION OF THE SPHERES IN RICCATI GEOMETRY. Theorem. (on the period special case for type 1 and 2 and p ≡ ) The period for type 0, 1 and 2 is respectively p + p + 1 , p − , p − p. Notation. The period in k -dimension, which depends on the type is denoted by π k . Theorem. (on the selector) Example. For k = 3 , (See \ RIC.BAS)0. p = 5 ,type period C , C x (1) , x (1) , x (1)0 31 1 , , , 11 24 1 , , , 32 20 1 , , , p = 7 , ν = 2 ,type period C , C x (1) , x (1) , x (1)0 57 1 , , , 61 48 1 , , , 42 42 3 , , , p = 11 ,type period C , C x (1) , x (1) , x (1)0 133 1 , , , 61 120 1 , , , 52 110 1 , , , p = 13 , ν = 2 ,type period C , C x (1) , x (1) , x (1)0 183 1 , , , 31 168 1 , , , 102 156 4 , , , p = 17 ,type period C , C x (1) , x (1) , x (1)0 307 1 , , , 21 288 1 , , , 32 272 1 , 12 0 , , p = 19 , ν = 2 ,type period C , C x (1) , x (1) , x (1)0 381 1 , , , 71 360 1 , , , 62 342 4 , , , CHAPTER 8. FUNCTIONS OVER FINITE FIELDS p = 23 ,type period C , C x (1) , x (1) , x (1)0 553 1 , , , 161 528 1 , , , 92 506 1 , , , p = 29 ,type period C , C x (1) , x (1) , x (1)0 871 1 , , , 11 840 1 , , , 132 812 1 , 10 0 , , p = 31 , ν = 3 ,type period C , C x (1) , x (1) , x (1)0 993 2 , , , 241 960 1 , , , 52 930 6 , , , Example. For k = 3 , (See [m130] WRONSKI.BAS)The table also includes the coordinates of a line,e.g., for p = 5 , type 0, ∗ = [2 , , . p = 5 , type 0, C = 1 , C = 3 ,i x x x l ∗ 11 14 28 29 19 20 2 23 24 8 0 10 27 15 4 5 i 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 x x x l ∗ selector: { , , , , , } selector function: − p = 5 , type 1, C = 1 , C = 0 ,i x x x i 12 13 14 15 16 17 18 19 20 21 22 23 x x x selector: { , , , , , −−} .5. GENERALIZATION OF THE SPHERES IN RICCATI GEOMETRY. selector function: − − − − p = 5 , type 2, C = 1 , C = 2 ,i x x x selector: { , , , , −− , −−} selector function: − − − − − − − − p = 7 , type 0, n = 5 , C = 1 , C = 0 ,i x x x i 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 x x x i 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 x x x selector: { , , , , , , , } selector function: − p = 7 , type 1, n = 2 , C = 1 , C = 1 ,i x x x i 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 x x x i 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 x x x selector: { , , , , , , , −−} selector function: CHAPTER 8. FUNCTIONS OVER FINITE FIELDS − − − − − − p = 7 , type 2, n = 5 , C = 3 , C = 3 ,i x x x i 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 x x x i 32 33 34 35 36 37 38 39 40 41 x x x selector: { , , , , , , −− , −−} selector function: − − − − − − − − − − − − Definition. 0. The direction dir ( i, j ) of 2 points i and j on the “sphere“ is the direction of the line associatedto the 2 points.1. A triangle ( i, j, k ) is isosceles iff j − i = k − j. 2. The planar direction pl ( i, j ) of 2 points i and j on the “sphere” is when C = 1 and C = 0 that of the normal to the plane passing through the origin, i and j. 3. The t plane t ∗ ( i ) at the point i is the plane through the origin parallel to the tangent planeat i. Theorem. dir ( i, j ) = dir ( i + k, j + k ) − k. if c = 1 and c = 0 then pl ( i, j ) = pl ( i + k, j + k ) + pk. if c = 1 and c = 0 then n ( i ) = − pi, For types 0 and 1, the correspondance i, n ( i ) is a bijection. For type 2, there are 2 p − n which are undefined because the length of the normalis 0 (ideal point). For types 0 and 1, t ∗ ( i ) + i = t ∗ ( j ) + j. .5. GENERALIZATION OF THE SPHERES IN RICCATI GEOMETRY. Corollary. If a triangle ( i, j, k ) is isosceles, then dir ( j, k ) = dir ( i, j ) + k − i Example. p = 5 , type 0, dir (0 , 1) = 21 , dir (1 , 2) = 22 , dir (0 , 2) = 5 , dir (0 , 3) = 25 , dir (0 , 4) = 17 .t ∗ (0) = 3 ∗ , t ∗ (1) = 2 ∗ .In the triangle { , , } , dir (2 , 4) = 5 , dir (4 , 0) = 17 , dir (2 , 4) = 7 = 5 + 2 .p = 5 , type 1, pl(0,1) = 8, pl(1,2) = 3, pl(2,3) = 22, pl(0,2) = 6, pl (1 , 2) = 1 .t ∗ (0) = 8 ∗ , t ∗ (1) = 7 ∗ .p = 11 , type 1, pl (0 , 1) = 40 , pl (1 , 2) = 29 , pl (2 , 3) = 18 . Theorem. For k = 4 , let D x = C x + C Dx + C D x , with0.1. Dx (0) = 0 , Dx = x (0) = 0 , D x (0) = x (0) = 0 , D x (0) = x (0) = 1 , then the functions x − C x − C x , x − C x , x and x are independent their Wronskian is W = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − C x − C x x − C x x x C x x − C x x x C x C x + C x x x C x C x + C x C x + C x + C x x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) W (0) = E. The addition formulas are x (cid:48)(cid:48) = y , + y , − C y , − C y , x (cid:48)(cid:48) = y , + y , − C y , − C y , x (cid:48)(cid:48) = y , + C y , + C y , x (cid:48)(cid:48) = y , + C ( y , + y , ) + C y , + C y , Example. CHAPTER 8. FUNCTIONS OVER FINITE FIELDS hapter 9FINITE ELLIPTIC FUNCTIONS The success of the study of the harmonic polygons of Casey (II.6.1), suggested the study of thepolygons of Poncelet. After having conjectured that the Theorem of Poncelet, as given in I.2.2.generalized to the finite case, and because one of the proof of this Theorem, in the classical case,is by means of elliptic functions, this suggested that these too could be generalized to the finitecase. Just as the additions properties were used to define the trigonometric functions, the sameproperties were generalized to the finite case. It was soon realized that the poles of the ellipticfunctions correspond to values, which in the finite case are outside of the finite field. The basicdefinitions and properties of section 1 do not give directly functions but an abelian group structureon a set E, whose elements are, in general, triplets of integers modulo p. In section 2, this structurewill be described as the direct product of the Klein 4-group and an abelian group which can be usedas seen in section 3 to define 3 functions which generalize, in the finite case, the functions sn, cn and dn of Jacobi.In this Chapter, j and j (cid:48) will denote +1 or − . Introduction. Given p and m different from 0 and 1, we will define in 3.1.1, the set E = E ( p, m ) and, in 3.1.7.,an operation “+” from E ×× E into E. The basic result that ( E, +) is an abelian group is given in3.1.15. Definition. Given s, c, d ∈ Z p . The elements of E are ( s, c, d ) such thatD0. s + c = 1 and d + m s = 1 . as well as, when − and − m are quadratic residues, ( ∞ , c ∞ , d ∞ ) , where c = − and d = − m. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Notation. i := √− , m := 1 − m, k := √ m, k := √ m . Theorem. H0. ( s, c, d ) , ( s , c , d ) , ( s , c , d ) ∈ E, H1. j = +1 or − , then C0. d − m c = m . C1. c + m s = d . C2. c + s d = d + m s c = 1 − m s . C3. m (1 − c )(1 + c ) = (1 − d )(1 + d ) . C4. ( c + d )(1 + jd ) = (1 + j c )( j m + d + m c ) . C5. m ( c + d )(1 − j c ) = (1 − j d )( j m + d + m c ) C6. d − m s c = d + c ( d − . C7. d d + m m s s = m − m c c . C8. ( d s c + d s c )( d d − m s s c c ) = ( s c d + s c d )( d d + mm s s ) . Proof: Each of the identities can easily be verified using Definition 1.1. If C3, is writtenC3’ m (1 − j c )(1 + j c ) = (1 − j d )(1 + j d ) , then C5, follows from C4. Lemma. H0. m s s = 1 , then C0. d = − m s c , d = − m s c , C1. ( s c d ) = ( s c d ) , C2. ( c c ) = ( d s d s ) , C3. ( d d ) = ( m s c s c ) , C4. s s (cid:54) = 0 . Lemma. H0. m s s = 1 , H1. s c d = − j s c d , then C0. c c = j d s d s , C1. d d = j m s c s c . C2. c = 0 ⇒ d = 0 and c (cid:54) = 0 .c = 0 ⇒ d = 0 and c (cid:54) = 0 . Proof: If c = c = 0 then s = s = 1 hence m = 1 , which is excluded. Lemma. H0. ( s c d ) = ( s c d ) , then C0. s = j s or m s s = 1 . .1. THE JACOBI FUNCTIONS. Definition. The addition is defined as follows:Let D = 1 − m s s . If D (cid:54) = 0 , thenD0. ( s , c , d ) + ( s , c , d ) = ( s c d + s c d D , c c − d s d s D , d d − m s c s c D ) , If D = 0 , s c d = s c d , c (cid:54) = 0 and c (cid:54) = 0 , thenD1. ( s , c , d ) + ( s , c , d ) = ( ∞ , c ∞ , d ∞ ) , where c = c s d and d = d s c , If D = 0 , s c d = − s c d , c (cid:54) = 0 and c (cid:54) = 0 , thenD2.0. ( s , c , d ) + ( s , c , d ) = ( s − s s c d , c + c c c , d + d d d ) , If D = 0 , s c d = j s c d , c = 0 and c (cid:54) = 0 , thenD2.1. ( s , c , d ) + ( s , c , d ) = ( ∞ , c ∞ , d ∞ ) , where c = − d s c and d = d m s c . If D = 0 , s c d = j s c d , c (cid:54) = 0 and c = 0 , thenD2.2. ( s , c , d ) + ( s , c , d ) = ( ∞ , c ∞ , d ∞ ) , where c = − d s c and d = d m s c . If s (cid:54) = 0 , thenD3.0. ( ∞ , c ∞ , d ∞ ) + ( s , c , d )= ( s , c , d ) + ( ∞ , c ∞ , d ∞ ) = ( − cdm s , dd m s , cc m s ) . If s = 0 , thenD3.1. ( ∞ , c ∞ , d ∞ ) + (0 , c , d )= (0 , c , d ) + ( ∞ , c ∞ , d ∞ ) = ( ∞ , c d ∞ , d c ∞ ) . D4. ( ∞ , c ∞ , d ∞ ) + ( ∞ , c ∞ , d ∞ ) = (0 , d d m , c c ) . Example. With p = 11 , m = 3 , ( − ) = ( − ) = − ,E = { (0 , , , (0 , , − , (0 , − , , (0 , − , − , (1 , , , (1 , , − , ( − , , , ( − , , − , (5 , , , (5 , , − , (5 , − , , (5 , − , − , ( − , , , ( − , , − , ( − , − , , ( − , − , − } . If the elements of E in the above order are abbreviated 0,1,2, . . . ,15, then the addition table is CHAPTER 9. FINITE ELLIPTIC FUNCTIONS + 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 151 1 0 3 2 7 6 5 4 13 12 15 14 9 8 11 102 2 3 0 1 6 7 4 5 14 15 12 13 10 11 8 93 3 2 1 0 5 4 7 6 11 10 9 8 15 14 13 124 4 7 6 5 2 1 0 3 10 13 14 9 8 15 12 115 5 6 7 4 1 2 3 0 9 14 13 10 11 12 15 86 6 5 4 7 0 3 2 1 12 11 8 15 14 9 10 137 7 4 5 6 3 0 1 2 15 8 11 12 13 10 9 148 8 13 14 11 10 9 12 15 4 1 2 5 0 7 6 39 9 12 15 10 13 14 11 8 1 6 7 2 5 0 3 410 10 15 12 9 14 13 8 11 2 7 6 1 4 3 0 511 11 14 13 8 9 10 15 12 5 2 1 4 3 6 7 012 12 9 10 15 8 11 14 13 0 5 4 3 6 1 2 713 13 8 11 14 15 12 9 10 7 0 3 6 1 4 5 214 14 11 8 13 12 15 10 9 6 3 0 7 2 5 4 115 15 10 9 12 11 8 13 14 3 4 5 0 7 2 1 6 Example. With p = 13 , m = 3 , ( − ) = ( − ) = 1 ,E = { (0 , , , (0 , , − , (0 , − , , (0 , − , − , ( ∞ , ∞ , ∞ ) , ( ∞ , ∞ , − ∞ ) , ( ∞ , − ∞ , ∞ ) , ( ∞ , − ∞ , − ∞ ) , (6 , , , (6 , , − , (6 , − , , (6 , − , − , ( − , , , ( − , , − , ( − , − , , ( − , − , − } . If the elements of E in the above order are abbreviated 0,1,2, . . . , 15, then the addition table is + 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 151 1 0 3 2 6 7 4 5 13 12 15 14 9 8 11 102 2 3 0 1 5 4 7 6 14 15 12 13 10 11 8 93 3 2 1 0 7 6 5 4 11 10 9 8 15 14 13 124 4 6 5 7 3 1 2 0 12 14 13 15 11 9 10 85 5 7 4 6 1 3 0 2 10 8 11 9 13 15 12 146 6 4 7 5 2 0 3 1 9 11 8 10 14 12 15 137 7 5 6 4 0 2 1 3 15 13 14 12 8 10 9 118 8 13 14 11 12 10 9 15 7 1 2 4 0 5 6 39 9 12 15 10 14 8 11 13 1 4 7 2 6 0 3 510 10 15 12 9 13 11 8 14 2 7 4 1 5 3 0 611 11 14 13 8 15 9 10 12 4 2 1 7 3 6 5 012 12 9 10 15 11 13 14 8 0 6 5 3 4 1 2 713 13 8 11 14 9 15 12 10 5 0 3 6 1 7 4 214 14 11 8 13 10 12 15 9 6 3 0 5 2 4 7 115 15 10 9 12 8 14 13 11 3 5 6 0 7 2 1 4 Theorem. C0. ( s , c , d ) + ( j (cid:48) s , jc , j j (cid:48) d ) = (0 , j, j j (cid:48) ) . C1. ( ∞ , c ∞ , d ∞ ) + ( ∞ , j c ∞ , j (cid:48) d ∞ ) = (0 , − j (cid:48) , − j ) . C2. − ( s , c , d ) = ( − s , c , d ) . .1. THE JACOBI FUNCTIONS. C3. − ( ∞ , c ∞ , d ∞ ) = ( ∞ , − c ∞ , − d ∞ ) . C4. ( s , c , d ) + (0 , j, j j (cid:48) ) = ( js , j (cid:48) c , j j (cid:48) d ) . C5. ( ∞ , c ∞ , d ∞ ) + (0 , j (cid:48) , j ) = ( ∞ , jc ∞ , j (cid:48) d ∞ ) . Theorem. H0. ( s , c , d ) + ( s , c , d ) = ( s , c , d ) , then C0. ( s , c , d ) + ( − s , − c , d ) = ( − s , − c , d ) . C1. ( s , c , d ) + ( s , − c , − d ) = ( s , − c , − d ) . C2. ( s , c , d ) + ( − s , c , − d ) = ( − s , c , − d ) . Notation. We will use the notation, which is customary in abelian groups with addition as operation symbol, n ( s , c , d ) = ( n − s , c , d ) + ( s , c , d ) , n ∈ Z using induction starting with n = 0 or n = − . Theorem. C0. n ( − s , − c , − d ) = j n ( s , c , d ) , with j = ( − n . Theorem. D0. D := 1 − m s , then C0. 2( s, c, d ) = ( s c dD , c − s d D , d − m s c D ) . Theorem. ( E, +) is an abelian group. Its order is divisible by 4. The proof, although tedious, is straigthforward. The closure follows from the definition .6.Associativity follows, non trivially from .6. The neutral element is (0 , , . The additive inverseelement of ( s, c, d ) is given by 1.10. C2. and C3. Definition. The group ( E, +) is called the Jacobian elliptic group associated to the prime p and the integer m ∈ Z p . Corollary. The following constitute special cases. For m = 0 , the elements of the group are ( sink, cosk, and ( sink, cosk, − , and the addition formulas reduce to ( sink, cosk, j ) + ( sinl, cosl, j ) =( sin ( j k + j l ) , cos ( j k + j l ) , j j ) , j and j are +1 or − . CHAPTER 9. FINITE ELLIPTIC FUNCTIONS For m = 1 , the elements of the group are ( tanhk, cosechk, cosechk ) , ( tanhk, cosechk, − cosechk ) and if c = − , ( ∞ , c ∞ , c ∞ ) , ( ∞ , − c ∞ , c ∞ ) , ( ∞ , c ∞ , − c ∞ ) , ( ∞ , − c ∞ , − c ∞ ) . and the addition formulas correspond to tanhk + tanhk = tanhk + tanhk tanhk tanhk .cosech ( k + k ) = cosechk cosechk tanhk tanhk . Comment. To remove some of the mystery associated with some of the formulas just given, assume that thefinite field is replaced by the field of reals. For instance, D4, is obtained by replacing in D0, c by i s, d by i k s, c by i s , d by i k s and letting s and s tend to infinity. Introduction. It can be shown that ( E, +) is isomorphic to the direct product of the Klein 4-group and the groupE associated to the finite Weierstrass p function introduced by Professor Tate and that the kernelof a homomorphism between the 2 groups is the subgroup of ( E, +) of elements with s = 0 . A lessprecise form of this Theorem is given in 2.1. and is illustrated by the examples given in this sectionand prepares for the definition of finite Jacobi elliptic functions. In many cases the generator ofthe larger group allows the inclusion of one of the generators of the Klein 4-group. Theorem. ( E, +) is isomorphic to Z ×× Z n or to Z ×× Z n . Example. In the example the generators of the factor groups will be given. The additional information in thesecond column will be explained in the Chapter on isomorphisms and homomorphisms. .1. THE JACOBI FUNCTIONS. p m E is isomorphic generatorto Z i ×× Z n of Z i and Z n Z ×× Z (0 , , − , (0 , − , Z ×× Z (0 , , − , (0 , − , m (cid:48) (2) Z ×× Z (0 , , − , (1 , , m j (2) (cid:48)(cid:48) (0 , , − , ( ∞ , ∞ , ∞ )7 3 Z ×× Z (0 , , − , (0 , − , Z ×× Z (0 , − , , (2 , , m (cid:48)(cid:48) (2) (cid:48)(cid:48) (0 , , − , (1 , , m (cid:48) (4) (cid:48)(cid:48) (0 , , − , (1 , , Z ×× Z (0 , , − , (2 , − , Z ×× Z (0 , − , , (3 , , m (cid:48) (9) (cid:48)(cid:48) (0 , , − , (1 , , m (cid:48)(cid:48) (5) (cid:48)(cid:48) (0 , , − , (1 , , Z ×× Z (0 , , − , (3 , , (cid:48)(cid:48) (0 , , − , (5 , , (cid:48)(cid:48) (0 , − , , (5 , , − Z ×× Z (0 , , − , (5 , , m (cid:48)(cid:48) (3) (cid:48)(cid:48) (0 , − , , (3 , , m (cid:48) (3) (cid:48)(cid:48) (0 , , − , (3 , , m (cid:48) (2) Z ×× Z (0 , , − , (1 , , m j (2) = m (cid:48)(cid:48) (12) (cid:48)(cid:48) (0 , , − , ( ∞ , ∞ , ∞ )4 = m (cid:48) (10) = m (cid:48)(cid:48) (10) Z ×× Z ( ∞ , ∞ , ∞ ) , (1 , , m j (4) (cid:48)(cid:48) ( ∞ , ∞ , ∞ ) , (1 , , Z ×× Z (0 , , − , (2 , , m j (6) (cid:48)(cid:48) (0 , − , , (6 , , m j (11) Z ×× Z (0 , , − , (6 , , m j (9) (cid:48)(cid:48) (0 , , − , (6 , , m (cid:48)(cid:48) (3) (cid:48)(cid:48) (0 , , − , (2 , , m (cid:48) (5) (cid:48)(cid:48) (0 , , − , (2 , , m j (7) Z ×× Z (0 , − , , (2 , , − m (cid:48) (2) Z ×× Z ( ∞ , ∞ , ∞ ) , (1 , , m (cid:48)(cid:48) (2) = m j (9) (cid:48)(cid:48) ( ∞ , ∞ , ∞ ) , (1 , , m j (2) = m (cid:48)(cid:48) (16) (cid:48)(cid:48) ( ∞ , ∞ , ∞ ) , (1 , , Z ×× Z (0 , , − , (3 , , m j (6) (cid:48)(cid:48) (0 , , − , (4 , , Z ×× Z (0 , , − , (3 , , m j (13) (cid:48)(cid:48) (0 , , − , (6 , , m (cid:48)(cid:48) (4) = m j (5) (cid:48)(cid:48) (0 , , − , (6 , , m j (4) = m (cid:48) (5) (cid:48)(cid:48) (0 , , − , (4 , , Z ×× Z (0 , , − , (4 , − , m j (7) (cid:48)(cid:48) (0 , , − , (3 , , − Z ×× Z (0 , , − , (4 , , (cid:48)(cid:48) (0 , , − , (4 , , m (cid:48) (3) = m j (8) (cid:48)(cid:48) (0 , , − , (3 , , m j (3) = m (cid:48)(cid:48) (8) (cid:48)(cid:48) (0 , , − , (3 , , CHAPTER 9. FINITE ELLIPTIC FUNCTIONS p m E is isomorphic generatorto Z i ×× Z n of Z i and Z n 19 12 Z ×× Z (0 , , − , (3 , , − Z ×× Z (0 , , − , (7 , , m (cid:48) (3) (cid:48)(cid:48) (0 , , − , (3 , , m (cid:48)(cid:48) (11) (cid:48)(cid:48) (0 , − , , (2 , , (cid:48)(cid:48) (0 , , − , (2 , , m (cid:48)(cid:48) (4) (cid:48)(cid:48) (0 , − , , (4 , , m (cid:48) (4) (cid:48)(cid:48) (0 , , − , (4 , , Z ×× Z (0 , , − , (4 , − , (cid:48)(cid:48) (0 , , − , (3 , − , (cid:48)(cid:48) (0 , , − , (2 , , Z ×× Z (0 , − , , (7 , , m (cid:48) (16) (cid:48)(cid:48) (0 , , − , (2 , , m (cid:48)(cid:48) (6) (cid:48)(cid:48) (0 , , − , (3 , , (cid:48)(cid:48) (0 , , − , (4 , , m (cid:48) (9) (cid:48)(cid:48) (0 , , − , (3 , , m (cid:48)(cid:48) (9) (cid:48)(cid:48) (0 , − , , (7 , , Z ×× Z (0 , , − , (2 , , Z ×× Z (0 , − , , (4 , , m (cid:48)(cid:48) (4) (cid:48)(cid:48) (0 , , − , (8 , , m (cid:48) (6) (cid:48)(cid:48) (0 , , − , (11 , , Z ×× Z (0 , , − , (4 , , (cid:48)(cid:48) (0 , , − , (4 , − , (cid:48)(cid:48) (0 , , − , (9 , − , Z ×× Z (0 , − , , (11 , , m (cid:48)(cid:48) (2) (cid:48)(cid:48) (0 , , − , (9 , , m (cid:48) (12) (cid:48)(cid:48) (0 , , − , (10 , , (cid:48)(cid:48) (0 , − , , (8 , , m (cid:48)(cid:48) (3) (cid:48)(cid:48) (0 , , − , (10 , , m (cid:48) (8) (cid:48)(cid:48) (0 , , − , (11 , , (cid:48)(cid:48) (0 , − , , (9 , , m (cid:48)(cid:48) (13) (cid:48)(cid:48) (0 , , − , (8 , , m (cid:48) (16) (cid:48)(cid:48) (0 , , − , (9 , , Z ×× Z (0 , , − , (4 , , (cid:48)(cid:48) (0 , , − , (8 , − , (cid:48)(cid:48) (0 , , − , (8 , − , Z ×× Z (0 , − , , (4 , , m (cid:48)(cid:48) (9) (cid:48)(cid:48) (0 , , − , (10 , , m (cid:48) (18) (cid:48)(cid:48) (0 , , − , (4 , , Definition. Given a prime p and an integer m in Z p , n and 4 n. If wechoose a generator g := ( s , c , d ) of this group, we obtain by successive addition n g = n ( s , c , d )= ( s n , c n , d n ) . The finite Jacobi elliptic functions sn , cn and dn , scd are defined by .1. THE JACOBI FUNCTIONS. sn ( n ) := s n , cn ( n ) := c n , dn ( n ) := d n , scd ( n ) := ( s n , c n , d n ) . The period is denoted by 4 K. Example. For p = 11 , m = 3 , K = 2 , — For p = 13 , m = 3 , K = 2 ,i sn ( i ) cn ( i ) dn ( i ) i sn ( i ) cn ( i ) dn ( i )0 0 1 1 0 0 1 11 − − ∞ − ∞ − ∞ − − − − − − 64 0 − − − 15 5 − − − − − ∞ ∞ ∞ − − Definition. ns := sn , nc := cn , nd := dn , sc := sncn , cd := cndn , ds := dnsn , cs := cnsn , dc := dncn , sd := sndn . The notation is due to Glaisher, Glaisher, J.W.L., On elliptic functions, Messenger of Mathe-matics, Vol. 11, 1881, 81-95. Introduction. The formulas given in this section are for the most part the same as in the real case. Theorem 9.1.4gives the addition formulas. Theorem 9.1.4, which may be new is needed to prove the additionformula for the Jacobi Zeta function . Theorem 9.1.4 is given for sake of completeness. It is clearlyless elegant than 9.1.4. Lemma. − ms s = c + d s = d + ms c . Theorem. sn ( u ) cn ( v ) dn ( v ) − sn ( v ) cn ( u ) dn ( u )= (1 − msn ( u ) sn v )( sn ( u ) − sn ( v ) . cn ( u ) cn ( v ) − sn ( u ) sn ( v ) dn ( u ) dn ( v = (1 − msn ( u ) sn v )(1 − sn ( u ) − sn ( v ) . dn ( u ) dn ( v ) − m sn ( u ) sn ( v ) cn ( u ) cn ( v = (1 − msn ( u ) sn v )(1 − msn ( u ) − msn ( v ) + msn ( u ) sn ( v ) . CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Theorem. sn ( u + v ) = sn ( u ) − sn ( v ) sn ( u ) cn ( v ) dn ( v ) − sn ( v ) cn ( u ) dn ( u ) . cn ( u + v ) = − sn ( u ) − sn ( v ) cn ( u ) cn ( v )+ sn ( u ) sn ( v ) dn ( u ) dn ( v ) . dn ( u + v ) = − msn ( u ) − msn ( v )+ msn ( u ) sn ( v ) dn ( u ) dn ( v )+ msn ( u ) sn ( v ) cn ( u ) cn ( v ) . cn ( u + v ) = sn ( u ) cn ( u ) dn ( v ) − sn ( v ) cn ( v ) dn ( u ) sn ( u ) cn ( v ) dn ( v ) − sn ( v ) cn ( u ) dn ( u ) , for u (cid:54) = ( v ) . dn ( u + v ) = sn ( u ) dn ( u ) cn ( v ) − sn ( v ) dn ( v ) cn ( u ) sn ( u ) cn ( v ) dn ( v ) − sn ( v ) cn ( u ) dn ( u ) , for u (cid:54) = ( v ) . Formulas 0., 3. and 4. are due to Cayley (1884). Theorem. − mcn ( u ) cn ( v ) cn ( u + v ) + dn ( u ) dn ( v ) dn ( u + v ) = 1 − m. dn ( v ) dn ( u + v ) + mcn ( u ) sn ( v ) sn ( u + v ) = dn ( u ) . sn ( v ) dn ( u ) sn ( u + v ) + cn ( v ) cn ( u + v ) = cn ( u ) . Theorem. sn ( u + v + w )( sn ( v ) sn ( u + w ) − sn ( w ) sn ( u + v ))= sn ( u )( sn ( v ) sn ( u + v ) − sn ( w ) sn ( u + w )) . sn ( a − a ) sn ( a − a ) sn ( a − a ) − sn ( a − a ) sn ( a − a ) sn ( a − a )+ sn ( a − a ) sn ( a − a ) sn ( a − a ) − sn ( a − a ) sn ( a − a ) sn ( a − a ) = 0 . Proof: If we write u = a − a , v = a − a , w = a − a , then u + v = a − a ,u + w = a − a , u + v + w = a − a and we obtain 0, from 1. To prove 1, let us introduce thenotation s := sna , s := sna , s := sna , s := sna . and similarly for c i and d i . Let B := ( s − s )( s − s )( s − s )( s c d + s c d )( s c d + s c d )( s c d + s c d ) . Let B , B , B be obtained by adding 1, 2, 3 modulo 4 to each digit, using 9.1.4.0 in 1. andreducing to the same denominator, we have to prove that B − B + B − B = 0 . Using .0.D0., B = ( s ( s − s ) + s ( s − s ) + s ( s − s ))( s s s c d (1 − s )(1 − ms )+ s s s c d (1 − s )(1 − ms ) .1. THE JACOBI FUNCTIONS. + s s s c d (1 − s )(1 − ms )+ s s s c d (1 − s )(1 − ms )+ s c d c d c d + s s c d c d c d + s s c d c d c d + s s c d c d c d )therefore B − B + B − B =( s s s c d ( s s ( s − s )( m − m ) + . . . + s s s (1 + m − − m ) + . . . + s s (1 − 1) + . . . ) + . . . )+( s c d c d c d ( s s s (1 − 1) + . . . + s s s ( − . . . + s s s ( − . . . ) + . . . ) = 0 . The given terms come from s ( s − s ) s s s c d ms in B and from the term in B corresponding to the term s ( s − s ) s s s c d ms in B , the term in B corresponding to − s s s s s c d ( − − m ) s in B and the term in B , to − s s s s s c d ( − − m ) s in B , the term in B corresponding to − s s s s s c d in B and the term in B , to − s s s s s c d in B , the term in B corresponding to − s s s s c d c d c d in B and theterm in B , to − s s s s c d c d c d in B , the term − s s s c d c d c d in B and the term in B , to s s s s c d c d c d in B and the term in B corresponding to − s s s s c d c d c d in B and the term in B , to − s s s s c d c d c d in B , The reduction involves 4(6 . . . . . . 2) + 4(3 + 6 + 3)2 terms, which exausts the list of 4(6(4 . B − B + B − B . Comment. Formula 9.1.4.0, should be compared with the formula of Jacobi, (Crelle Vol. 15) sn ( u + v + w ) sn ( u )(1 − msn ( v ) sn ( w ) sn ( u + v ) sn ( u + w ))= sn ( u + v ) sn ( u + w ) − sn ( v ) sn ( w ) . Formulas 9.1.4.0. to 1. should also be compared with the formulas of Glaisher (1881) and of Cayley(Crell Vol. 41), Corollary. sn ( u + 1) = sn (1)( sn (1) sn (2) − sn ( u − sn ( u )) sn (1) sn ( u ) − sn (2) sn ( u − , u = 3 , . . . . Proof: Use 9.1.4.0. with u = v = 1 and w = u − , Theorem. cn ( u + v + w ) = sn ( u ) dn ( v ) dn ( w )( cn ( v ) cn ( u + v ) − cn ( w ) cn ( u + w )) − dn ( u )( sn ( v ) cn ( v ) dn ( w ) − sn ( w ) cn ( w ) dn ( v )) dn ( u )( sn ( v ) dn ( w ) cn ( u + w ) − sn ( w ) dn ( v ) cn ( u + v )) . sd ( a − a ) cn ( a − a ) − sd ( a − a ) cn ( a − a )= sd ( a − a )( cn ( a − a ) cn ( a − a ) − cn ( a − a ) cn ( a − a )) − cn ( a − a )( cn ( a − a ) sd ( a − a ) − sd ( a − a ) cn ( a − a )) CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Proof: One proof is to derive first 9.1.4.1. using the same method as in 9.1.4, the other is toset a = 0 and derive the corresponding formula using 9.1.4. Theorem. sn (2 u ) = sn ( u ) cn ( u ) dn ( u )1 − msn ( u ) . cn (2 u ) = cn ( u ) − sn ( u ) dn ( u )1 − msn ( u ) = cn ( u ) − sn ( u ) dn ( u ) cn ( u )+ sn ( u ) dn ( u ) . dn (2 u ) = dn ( u ) − msn ( u ) cn ( u )1 − msn ( u ) . = dn ( u )+ cn ( u )( dn ( u ) − − msn ( u ) dn ( u ) − cn ( u )( dn ( u ) − . Theorem. D0. s := (cid:113) − c d , D1. c := (cid:113) d d , D2. d := s ( c + d )(1+ c )(1+ d ) s c , then C0. d = c + d c , C1. 2( s , c , d ) = ( s, c, d ) . Proof. C0. follows directly from D0. to D1. It is not used to define d to insure that 2( s , c , d )is ( s, c, d ) not ( − s, c, d ) . The formulas can be derived starting from d − ms c = d (1 − ms ) . Expressing c and d in terms of s gives m (1 + d ) s − ms + 1 − d = 0 , hence s = m + j √ m − m (1 − d ) m (1+ d ) , where j = +1 or − , hence s = − jc d . therefore c = jc + d d and d = m d + jmc d = jc + d jc . It remains to verify, by substitution, for c and s. For c, c − s d = jc ( jc + d )(1+ d )(1+ jc ) , therefore j = 1 . For s, ms = 1 + m (1 − c ) (1+ d ) = 1 + − c d − d c = c + d )(1+ d )(1+ c ) = s c d s , s c d = (cid:113) − j c j c j c + d d = s ( c + d )(1+ c )(1+ d ) = j s ( jc + d )(1+ j c )(1+ d ) ,j = +1 or − .1. THE JACOBI FUNCTIONS. Example. p = 19 , m = 2 , δ = 2,Let ( s, c, d ) = (4 , , ,s = 2 , c = − , d = − , therefore s = δ , c = 3 δ or − δ , d = 4 or − . Theorem. If dn ( u ) (cid:54) = − , then at u ,0. sn ◦ I = (cid:113) − cn dn . cn ◦ I = (cid:113) cn + dn dn . dn ◦ I = (cid:113) − m + mcn + dn dn . Theorem. If sn ( u ) = 0 , cn ( u ) = 1 and dn ( u ) = − , then0. sn ( u ) = (cid:113) m cn ( u ) = (cid:113) m − m . dn ( u ) = 0 . Theorem. If sn ( u ) = 0 , cn ( u ) = − and dn ( u ) = − , then0. sn ( u ) = ∞ . cn ( u ) = √− ∞ . dn ( u ) = √− m ∞ . Conjecture. 0. (1 − m ) R p ⇒ scd (2 K ) = (1 , , √ − m ) . − p and − m R p ⇒ scd ( K ) = ( √− ∞ , √− m ∞ , ∞ ) and scd (2 K ) = (0 , − , − . m R pand ( m − 1) R p ⇒ scd ( K ) = ( (cid:113) m , (cid:113) − m , 0) and scd (2 K ) = (0 , , − . Conjecture. If (1 − m ) R p then sn ( K − u ) = cd ( u ) . CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Introduction. Definitions 9.1.6 are inspired by the relation which exist, in the real case, between the Jacobi Zetafunction , the θ functions and the Weierstrass ζ function. See Handbook p.578, 16.34 and p.650,18.10.7. Definition. The function u is defined by: u (1) := 0 ,u ( i + 1) := u ( i ) − msn (1) sn ( i ) sn ( i + 1) . Definition. The Jacobi Zeta function Z is defined by Z (1) := − u ( K ) K .Z ( i ) := u ( i ) + Z (1) i, i (cid:54) = 1 . Theorem. Z ( u + v ) = Z ( u ) + Z ( v ) − msn ( u ) sn ( v ) sn ( u + v ) . Z ( u + v ) = Z ( u ) + Z ( v ) + msd ( u )( cn ( v ) cn ( u + v ) − cn ( u )) Z ( u + v ) = Z ( u ) + Z ( v ) + sc ( u )( dn ( v ) dn ( u + v ) − dn ( u ))Proof of 0. The formula is true, by definition, for v = 1 . It follows by induction on v and from . . . Z ( u + v + 1) = Z ( u + v ) + Z (1) − msn (1) sn ( u + v ) sn ( u + v + 1)= Z ( u ) + Z ( v ) + Z (1) − msn ( u ) sn ( v ) sn ( u + v ) − msn (1) sn ( u + v ) sn ( u + v + 1)= Z ( u )+ Z ( v +1)+ msn (1) sn ( v ) sn ( v +1) − msn ( u ) sn ( v ) sn ( u + v ) − msn (1) sn ( u + v ) sn ( u + v + 1) = Z ( u ) + Z ( v + 1) − msn ( u ) sn ( v + 1) sn ( u + v + 1) . The proof of 1. and 2. is left as an exercise. Hint: Use 9.1.4. Theorem. Z ( K − u ) = − Z ( u ) + msn ( u ) cd ( u ) . Z ( K ) = m sn ( K ) cd ( K ) , if K is even.2. Z ( K ) = 0 . Z ( K + u ) = − Z ( K − u ) . Z (2 K − u ) = − Z (2 K + u ) . Z (2 K + u ) = Z ( u ) . Proof: 0, follows from the additional formula for Z eta ( K − u ) and from . . . 9.1.5 See Example3.1.1. and \ 130 elliptic.bas .1. THE JACOBI FUNCTIONS. Definition. z ( u ) := Z ( u ) + cn ( u ) ds ( u )1. z ( u ) := Z ( u ) − dn ( u ) sc ( u ) . z ( u ) := Z ( u ) + msn ( u ) cd ( u ) . z ( u ) := Z ( u ) . Several examples of Jacobian elliptic functions follow. p = 5 m = 3 δ = 2 p = 5 m = 2 δ = 2 p = 5 m = 4 δ = 2 . , − , δ ) . δ, , δ ) . δ, δ, , − , 1) 1 (1 , , 2) 1 ( ∞ , ∞ , ∞ )2 K = 1 2 (0 , − , 1) 2 (0 , − , − − , , 2) 3 ( ∞ , − ∞ , − ∞ )2 K = 2 2 K = 2 p = 7 m = 3 δ = 3 p = 7 m = 2 δ = 3 p = 7 m = 4 δ = 3 . , − , δ ) . δ, , δ ) . δ, δ, , − , 1) 1 (2 , , 0) 1 (1 , , K = 1 2 (0 , , − 1) 2 (0 , − , − , , 0) 3 ( − , , K = 2 2 K = 2 p = 7 m = 6 δ = 3 p = 7 m = 5 δ = 3 . , δ, δ ) . δ, , δ )1 (1 , , 3) 1 (2 , − , , − , 1) 2 ( − , , − − , , 3) 3 (0 , − , − K = 2 4 (2 , , − − , − , K = 3 p = 11 m = 5 δ = 2 p = 11 m = 8 δ = 2 p = 11 m = 9 δ = 2 . − δ, , δ ) . , δ, δ ) . δ, δ, , , 0) 1 (1 , , 2) 1 (1 , , , , − 1) 2 (0 , − , 1) 2 (0 , − , − , , 0) 3 ( − , , 2) 3 ( − , , K = 2 2 K = 2 2 K = 2 p = 11 m = 2 δ = 2 p = 11 m = 6 δ = 2 p = 11 m = 10 δ = 2 . δ, , − δ ) . , δ, δ ) . δ, δ, , , 4) 1 (5 , , 4) 1 (5 , , − − , − , − 4) 2 (5 , − , 4) 2 (5 , , , − , − 1) 3 (0 , − , 1) 3 (0 , , − , − , − 4) 4 ( − , − , 4) 4 ( − , , − , , 4) 5 ( − , , 4) 5 ( − , , − K = 3 2 K = 3 2 K = 3 CHAPTER 9. FINITE ELLIPTIC FUNCTIONS p = 11 m = 3 δ = 2 p = 11 m = 4 δ = 2 p = 11 m = 7 δ = 2 . δ, δ, . δ, δ, . δ, δ, − , , − 5) 1 (3 , , 3) 1 ( − , , , , 3) 2 (5 , , 0) 2 (1 , , − , − , − 5) 3 (3 , , − 3) 3 ( − , − , , − , 1) 4 (0 , , − 1) 4 (0 , − , , − , − 5) 5 ( − , , − 3) 5 (3 , − , − , , 3) 6 ( − , , 0) 6 ( − , , , , − 5) 7 ( − , , 3) 7 (3 , , K = 4 2 K = 4 2 K = 4 p = 13 m = 2 δ = 2 p = 13 m = 12 δ = 2 p = 13 m = 4 δ = 2 . − δ, , δ ) . δ, δ, . δ, − , δ )1 (1 , , 5) 1 ( ∞ , ∞ , ∞ ) 1 (1 , , , − , 1) 2 (0 , − , − 1) 2 (0 , − , − , , 5) 3 ( ∞ , − ∞ , − ∞ ) 3 ( − , , K = 2 2 K = 2 2 K = 2 p = 13 m = 10 δ = 2 p = 13 m = 6 δ = 2 p = 13 m = 8 δ = 2 . , δ, δ ) . − , δ, δ ) . δ, − , − δ )1 (1 , , 2) 1 (2 , , 4) 1 (6 , , , − , 1) 2 (2 , − , 4) 2 (6 , , − − , , 2) 3 (0 , − , 1) 3 (0 , , − K = 2 4 ( − , − , 4) 4 ( − , , − − , , 4) 5 ( − , , K = 3 2 K = 3 p = 13 m = 3 δ = 2 p = 13 m = 5 δ = 2 p = 13 m = 9 δ = 2 . − δ, , δ ) . δ, , δ ) . δ, δ, , , 6) 1 ( − , , 4) 1 (2 , , ∞ , − ∞ , − ∞ ) 2 (1 , , ∞ , − ∞ , − ∞ )3 ( − , − , − 6) 3 ( − , − , 4) 3 ( − , − , − , − , − 1) 4 (0 , − , 1) 4 (0 , − , − , − , − 6) 5 (6 , − , 4) 5 (2 , − , − ∞ , ∞ , ∞ ) 6 ( − , , ∞ , ∞ , ∞ )7 ( − , , 6) 7 (6 , , 4) 7 ( − , , K = 4 2 K = 4 2 K = 4 p = 13 m = 11 δ = 2 p = 13 m = 7 δ = 2 . δ, − , δ ) . − δ, , δ )1 (2 , , 3) 1 (2 , , − , , − 4) 2 (6 , − , , − , 3) 3 (6 , − , − , − , 1) 4 (2 , , − , − , 3) 5 (0 , , − − , , − 4) 6 ( − , , − , , 3) 7 ( − , − , − K = 4 8 ( − , − , − , , − K = 5 .1. THE JACOBI FUNCTIONS. Theorem. H0. (cid:18) − mp (cid:19) = 1 , then C0. (1 , , k ) ∈ E, C1. (1 , , k ) + (1 , , k ) = (0 , − , . Definition. Let e = ( s, c, d ) , s (cid:54) = ∞ ,sin (2 e ) := 2 s c, cos (2 e ) := c − s . Can this be justified?This is done better using sn = sin ◦ am, cn = cos ◦ am. Theorem?. sin (2 e + 2 e ) = . . . . Theorem. [Landen] Let H0. e := ( s , c , d ) ∈ E, H1. s c d (cid:54) = 0 , s (cid:54) = oo, H2. e := ( s , c , d ) := ( s , c , d ) + (1 , , k ) , H3. l (2 e ) := sin (2 e ) cos (2 e ) − cos (2 e ) sin (2 e ) sin (2 e ) − sin (2 e ) , then C0. l ( e ) = − k k .l or l ( p, m ) is the Landen constant associated to p and m. Proof.P0. e = ( cd , − k sd , k d ) . P1. l ( e ) = k ( s − c )+ c − k s d + k = − k k . Comment. We can replace in the above Theorem (1 , , k ) by ( − , , k ) , this gives the same constant l. We can also replace k by − k , this gives the constant l = l . Theorem. If k is real, there exists an isomorphism φ (cid:48) between the elliptic group associated to m and thatassociated to m (cid:48) = mm − ,φ (cid:48) ( s, c, d ) := ( k sd , cd , d ) φ (cid:48) ( s, c, 0) := ( ∞ , ck s ) ∞ , k s ∞ ) ,φ (cid:48) ( ∞ , c ∞ , d ∞ ) := ( k d , cd , . CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Corollary. If k is real the order of the group associated to m and to mm − are the same. Theorem. [Jacobi] If k is real, there exists an isomorphism φ (cid:48)(cid:48) between the elliptic group associated to m and thatassociated to m (cid:48)(cid:48) = m ,φ (cid:48)(cid:48) ( s, c, d ) := ( k s, d, c ) φ (cid:48)(cid:48) ( ∞ , c ∞ , d ∞ ) := ( ∞ , dk ∞ , ck ∞ ) . Corollary. If k is real the order of the group associated to m and to m are the same. Theorem. [Jacobi] If p ≡ , there exists an isomorphism φ between the elliptic group associated to m andthat associated to m j = m ,φ ( s, c, d ) := ( √− sc , c , dc ) ,φ ( s, , d ) := ( ∞ , c √− s ) ∞ , d √− s ) ∞ ) ,φ ( ∞ , c ∞ , d ∞ ) := ( √− c , , dc ) . Corollary. If p ≡ , the order of the group associated to m and to m are the same. Theorem. [Gauss] If k is real, there exist a homomorphism φ G from the elliptic group associated to m and thatassociated to m G = k (1+ k ) ,φ G ( s, c, d ) := ( (1+ k ) sD , c dD , D − , where D = 1 + k s ,φ G ( s, c, d ) := ( ∞ , c d (1+ k ) s ∞ , k s ) ∞ ) , if ks = 0 .φ G ( ∞ , c ∞ , d ∞ ) := (0 , c d, − . CHECK cdThe kernel of the homomorphism is { (0,1,1), (0,-1,-1) } .The image of the homomorphism is a subgroup of index 2. Corollary. If k is real the order of the group associated to m and to k (1+ k ) are the same. Theorem. [Landen] If k is real, there exist a homomorphism φ L from the elliptic group associated to m and thatassociated to m L = ( − k k ) , .1. THE JACOBI FUNCTIONS. φ L ( s, c, d ) := ( (1+ k ) s cd , k m ( d − k ) d , − k m d + k d ) ,φ L ( s, c, 0) := ( ∞ , − km s c ∞ , k (1+ k s c ) ∞ ) ,φ L ( ∞ , c ∞ , d ∞ ) := ( ∞ , d m c ) ∞ , d (1+ k ) 2 c ) ∞ ) , The kernel of the homomorphism is { (0,1,1), (0,-1,1) } .The image of the homomorphism is a subgroup of index 2. Corollary. If k is real the order of the group associated to m and + o ( − k k ) are the same. Definition. The amplitude function is defined by sin ◦ am = sn, cos ◦ am = cn. The example below gives sin (2 k ) , cos (2 k ) under sin and cos. Theorem. C0. D am = dn, C1. D sin = cos, D cos = − sin. C2. D sn = cn dn, D cn = − sn dn, D dn = − m sn cn. C3. D (2 am ) = − m sin ◦ (2 am )Proof: Using the derivative of composition of functions, D sn = D ( sin ◦ am ) = cos ◦ amD am = cn dn. The other relations in C2, follow from sn + cn = 1 and dn + m sn = 1 . C3, follows from D (2 am ) = 2 D dn = − m sn cn = − m sin ◦ amcos ◦ am = − m sin ◦ (2 am ) . Comment. The derivatives will have to be defined in a separate section. Somehow the connection with p -adicfunctions will have to be involved.If | h | < , then sin ( x + h ) = sin ( x ) cos ( h ) + cos ( x ) sin ( h ) ,sin ( x + h ) − sin ( x ) = sin ( x ) ( cos ( h ) − 1) + cos ( x ) sin ( h ) , but sin ( h ) = h + o ( h ) and cos ( h ) − h + o ( h ) , hence sin ( x + h ) − sin ( x ) h = cos ( x ) + o (1) and D sin = cos. For the elliptic functions, we have, see for instance Handbook, l. c., p. 575, 16.22.1 to .3 and am ( h ) = h − h m + h m (4 + m ) − . . . ,sn ( h ) = h + o ( h ) ,cn ( h ) = 1 + o ( h ) ,dn ( h ) = 1 + o ( h ) ,am ( h ) = h + o ( h ) . Hence sn ( x + h ) − sn ( x ) = sn ( x )( cn ( h ) dn ( h ) − 1) + cn ( x ) dn ( x ) sn ( h ) = h ( cn ( x ) dn ( x )) + o ( h ) , therefore D sn = cn dn.D ( sin ◦ am ) = cos ◦ amD am = cn dn, therefore D am = dn. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Definition. Let us associated to e = ( s, c, d ) the point P ( e ) = ( sin ( e ) , cos ( e ) , . The set of points P ( e + j e ) , j = 0 , , . . . are the vertices of a polygon called the polygon of Poncelet . Theorem. [Poncelet] The sides P ( e + j e ) × P ( e + j e + e ) are tangent to a circle. Proof. Let e := e + j e, e := e + e,P := P ( e ) , P := P ( e ) , then P = (2 s c , c − s , ,e = ( s , c , d ) ,e = ( s c d + s c dD , c c − d s d s D , d d − m s c s c D ) , with D = 1 − m s s ,P = (2 ( s c d + s c d )( c c − d s d s ) D , ( c c − d s d s ) − ( s c d + s c d ) D , ,P × P = [ . . . ] . Corollary. [Landen] The lines P ... × P ... pass through a fixed point L := (0 , l, called the point of Landen . This is a special case of the Theorem of Poncelet, when( s , c , d ) = (1 , , k ) . Construction. Determination of Poncelet’s polygons.Let the outscribed circle θ be X + ( X − d ) = S , let the inscribed circle γ be R ( t + t ) = ( c t + t ) , Given a point P = ( P , P , P ) on θ and a tangent t = ( t , t , t ) to γ through P, the other tangent u = ( u , u , u ) is given by u = t ( P − R P ) , u = t ( P c − R ( P + P ) , u = − P u + P u P . Given a tangent t to γ and a point on it and θ , the other point Q = ( Q , Q , Q ) common to t and θ is given by Q = 1 , Q = 2 t d − t t t + t , Q = − Q t + t t . .3. THE WEIERESTRASS FUNCTIONS. Definition. Given g, a non residue of p, or (cid:18) gp (cid:19) = − , then C p,g = C p is the set of pairs ( a, b ) , a, b ∈ Z p such that( a, b ) + ( c, d ) = ( a + b, c + d ) , ( a, b ) . ( c, d ) = ( a c + b d g, a d + b c ) . We could also write ( a, b ) as a + bγ , with γ = g. Definition. Given s, c, d ∈ C p , we can repeat definition . . . . Definition. S, C, D. Theorem. H0. δ = d, D0. e := ( s δ, c δ, d ) , e := ( s , c , d ) ,e s δ, c δ, d ) , D1. D := 1 − m d s s , D2. s := s c d + s c d D , c := c c − s s d d D ,d := d d − m d s s c c D , Hence replace md by m (cid:48) D3. D := 1 − m d s s ,s := ( s c d + s c d ) dD , c := ( c c s + s d d ) dD ,d := d d − m d s s c c D , H1. D (cid:54) = 0 , D (cid:54) = 0 , then C0. e + e = ( s δ, c δ, d ) , C1. e + e s , c , d ) . C2. 2 n e ∈ E, (2 n + 1) e ∈ S. Definition. ( a, b ) > b = 0 and 0 < a < p or if 0 < b < p . Comment. All that has been said above can be repeated. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Introduction. The modern work on elliptic curves starts and ends with the elliptic curves of Weierstrass. I referthe reader to Lang S. Theorem. Let D0. e := − m , D1. g = 4 m − m +13 , g = ( m + 1)( m − m − . D2. ∆ := g − g , J := g ∆ , D3. pn := e + s , Dpn := − cds , D4. e := − m , D5. g (cid:48) := m − m + 1 , g (cid:48) := (2 m − m − m − , D6. ∆ (cid:48) := g (cid:48) − g (cid:48) , J (cid:48) = g (cid:48) ∆ (cid:48) , D7. qn := e + c − c , Dqn := − d (1+ c ) s , then C0. Dpn = 4 pn − g pn − g . C1. Dqn = 4 qn − g (cid:48) qn − g (cid:48) . C2. ∆ = 3(12 m ( m − , J = ( m − m + 1) ( m ( m − ) . C3. ∆ (cid:48) = 256 m ( m − , J (cid:48) = (16 m − m + 1) m ( m − . The proof is straithforward. Substituting D2 in C0, multiplying by s and expressing c and d in terms of s gives a polynomial of the second degree in s . The coefficients of 1, s and s give inturn, e , g and g . Substituting D5 in C1, multiplying by (1 − c ) gives similarly a polynomial of the second degree inx := 1-c. The coefficients of 1, x and x give in turn e , g (cid:48) and g (cid:48) .pn corresponds to the Weierstrass P function and Dpn to its derivative.The formulas correspond to those of real elliptic functions with the ratio of the period ω and thecomplete elliptic integral K set to 1.(See for instance Handbook for Mathematical functions, p649, 18.9.1, 2,3,4,5,8,9 and 11). Example. With p = 7 , m = 2 , then e = − , g = 3 , g = − , with p = 7 , m = 6 , then e = 0 , g = 3 , g = 0 . with p = 19 , k = 2 , then e = 18 , g = 4 , g = 0 ,e = 17, g (cid:48) = 6 , g (cid:48) = − ,sn cn dn sin cos am pn Dpn qn Dqn ∞ ∞ ∞ ∞ − − − − 83 7 3 − − − − − − − − 55 0 1 − ∞ ∞ ∞ ∞ .3. THE WEIERESTRASS FUNCTIONS. Jacobi Z function = 0, -3, -2, 2, 3, 0Weierstrass ζ function = ∞ , 1, 6, -6, -1, ∞ .with p = 19 , k = 3 , then e = 5 , g = 3 , g = − ,e = 3 , g (cid:48) = 9 , g (cid:48) = 5 ,sn cn dn sin cos am pn Dpn qn Dqn ∞ ∞ ∞ ∞ − − − − − − − − − − − − − 19 3 − − − − − − 86 1 0 6 3 − − − − − − − − − 38 0 1 1 4 2 0 ∞ ∞ ∞ ∞ Jacobi Z function = 0, -7, 0, 7, 0, -7, 0, 7, 0.Weierstrass ζ function = ∞ , 6, 0, -6, ∞ , 6, 0, -6, ∞ ..If 2 T is the period for the Jacobi functions, T is the period of pn, which is even and of Dpn which is odd. See the next section for the last 2 columns.RERUN LAST EXAMPLE using ..[130]/ELLIPT Theorem. Let a := e . D0. ( pn , Dpn 3) := ( pn , Dpn 1) + ( pn , Dpn , D1. Q := ( pn − a )( pn − , Q (cid:48) := ( pn − a − pn − a − , then C0. pn − a = 4( pn − a )( pn − a )C1. pn − a = ( ( pn − a )( pn − a ) − m ( pn − a ) Dpn pn − a ) Dpn ) C2. Dpn Q ( Q − m )( Dpn Dpn − QQ (cid:48) )( Dpn Dpn − mQ ) Q ( pn − a ) Dpn pn − a ) Dpn . RECHECK THIS (the line above) If pn (cid:54) = pn then C3.0. pn Dpn − Dpn pn − pn ) − ( pn pn . C4.0. Dpn pn Dpn − Dpn pn Dpn − pn Dpn pn − pn . If pn pn and Dpn − Dpn then C3.1. pn ∞ . C4.1. Dpn ∞ . If pn pn and Dpn Dpn then C3.2. pn − pn pn − g Dpn ) , C4.2. Dpn (3 pn − g )( pn − pn Dpn − Dpn . If pn pn and Dpn Dpn then C3.3. pn ∞ . C4.3. Dpn ∞ . The proof of C3 and C4 follow from the addition formulas for the Jacobi functions. That of C2and C3 is analoguous of the formulas for the real case (Handbook p.635, 18.4.1, 18.4.2) CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Theorem. If H0. ( pn , Dpn 1) + ( pn , Dpn 2) = ( pn , Dpn , then C0. ( pn , Dpn , ( pn , Dpn , ( pn , − Dpn 3) are collinear.This follows at once from 3.10.3. and is the geometric interpretation of C3. Theorem. If D0. pn ( ti, t ) := t − pn ( i ) , D1. Dpn ( ti, t ) := t − Dpn ( i ) , D2. g ( t ) := t − g , D3. g ( t ) := t − g , then C0. Dpn ( ti, t ) = 4 pn ( ti, t ) − g ( t ) pn ( ti, t ) − g (3) . The proof follows at once from . . . .ellinv.tab gives a table of the invariants g ( t ) and g ( t ) for p = 19 and m = 2 to 18. Theorem. Making explicit the dependence of g and g on m , C0. g m + 1 , t ) = g − m, t ) = g m +1 , − t ) = g − m, − t ) . C1. g m + 1 , t ) = − g − m, t ) = − g m + 1 , − t ) = g − m, − t ) . C2. g (cid:48) ( m + 1 , t ) = g (cid:48) ( − m, t ) = g (cid:48) ( m + 1 , − t ) = g (cid:48) ( − m, − t ) . C3. g (cid:48) ( m + 1 , t ) = − g (cid:48) ( − m, t ) = − g (cid:48) ( m + 1 , − t ) = g (cid:48) ( − m, − t ) . The sections 3.10.7. to 10.12. were inspired from the formulas on complex elliptic functions.(See for instance, Handbook, p.635 18.4.3.,18.4.8)In the classical case, the Weierstrass p function is defined in such a way that the constant term inthe Maclaurin expension is 0. The Weierstrass ζ function is defined as its integral and the constantterm is again chosen as zero, which is natural if we want ζ to be an odd function. ζ is not periodic.In the finite case, we have chosen pn and ζ using the same definition in terms of the Jacobi functionsas in the real case, but now ζ as an odd periodic function. This should be contrasted with theclassical case in which the Weierstrass function is not periodic. Theorem 3.10.9. gives an interestingproperty of ζ (1) . Definition. The Weierstrass ζ function is defined by ζ ( u ) =Z( u ) + cn ( u ) dn ( u ) sn ( u ) . Definition. The function u is defined as follows:Given u (1) = 0 , D0.0. pn ( i ) (cid:54) = pn ( j ) ,u ( i + j ) := u ( i ) + u ( j ) + Dpn ( i ) − Dpn ( j )2( pn ( i ) − pn ( j )) . .3. THE WEIERESTRASS FUNCTIONS. D0.1. pn ( i ) = pn ( j ) andDpn ( i ) (cid:54) = Dpn ( j ) , u ( i + j ) = ∞ . D1.0. Dpn ( i ) (cid:54) = 0 , u (2 i ) := 2 u ( i ) + pn ( i ) − g Dpn ( i ) . D1.1. Dpn ( i ) = 0 , pn ( i ) = g (2) , u (2 i ) = ∞ . D2.0. u (0) = u ( T ) = ∞ .T is the period . . . . Theorem. There exist a constant ζ (1) such that u ( j ) + j ζ (1) = u ( T − j ) + ( T − j ) ζ (1) , j = 1 to T − . Proof. . . . ? Theorem. The Weierstrass ζ function is related to the u function by ζ ( j ) := u ( j ) + j ζ (1) . Comment. The definitions and theorems can be repeated replacing respectively pn, Dpn, u, ζ by qn, Dqn, v,ζ (cid:48) , but we have the additional property of the next Theorem. Theorem. ζ (cid:48) ( T ) = 0. Proof. . . . ? Theorem. ζ and ζ (cid:48) are odd functions and their period is either T or T. Notation. card ( X ) denotes the cardinality of the set X . Theorem. If p ≡ − , then card ( g , g 3) + card ( g , − g 3) = 2( p + 1) . If p ≡ , then card ( g , g 3) = card ( g , − g . Corollary. If p ≡ − , then card ( g , 0) = p + 1 . Comment. Examples can be obtained using P.BAS see P.HOM. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Definition. The function Ke is defined by − p < Ke ( m ) ≡ K ( m ) < p, Ke ( m ) is even. Theorem. [Hasse conjectured by Artin] − √ p < Ke ( m ) < √ p. Conjecture. Given an integer x in the range − √ p < x < √ p. then there exist a pair ( g , g such that the corresponding Weierstrass elliptic curve W has card p + 1 + x and the corresponding group is abelian. This has been verified up to p = 47 .1. If the cardinality of W is divisible by 4 and W is not abelian, then there exist a J or a J isomorphic to W . If e + e + e = 0 , e i − e k are all non quadratic residue, and j (cid:48) (cid:54) = 0 , then the elliptic groupis isomorphic to C l +2 ×× C for some l. This has been verified up to p = 97 . See g7622, Example.. Comment. If for a given m we obtain J m ) and J m ) and the corresponding W W (cid:48) , card ( W 2) = card ( W (cid:48) ) but W W (cid:48) are not necessarily isomorphic. E.g. p = 17 , J − 2) = C ×× C ,J − 2) = C . Comment Excluding ( g , g 3) = (0 , , if j (cid:48) (cid:54) = 0 , g , g g , − g . What is the connection between the structure if any?None except that concerning cardinality. E.g. p = 31 , W , ∼ C , W , − ∼ C ×× C ,W , ∼ C , W , − ∼ C ×× C . This should be integrated with 3.1. .3. THE WEIERESTRASS FUNCTIONS. Introduction. The usual correspondance in the real field between the functions sn, cn and dn of Jacobi and P,DP of Weierstrass should be modified to insure an isomorphism between the 3 dimensional ellipticcurve associated to ( sn, cn, dn ) and the 2 dimensional elliptic curve associated to ( P, DP ) . Thisrequires in fact to associate, in the real case to sn ( t ) , P (2 t ) . . Theorem. The curve ( P, P (cid:48) ) has a singularity when m = 0 and m = 1 , When m = 0 , the singularity is ( − , , because − is a double root of 4 p - g p - g , the regular solution when P (cid:48) = 0 is ( , . When m = 1 , the singularity is ( , , because is a double root of 4 p − g p − g , the regularsolution when P (cid:48) = 0 is ( − , . Definition. Let ( s, c, d ) in E, if m = 0 we add the restriction d = 1, if m = 1 , we add the restriction c = d. Let e := − m , e := e + 1, e := e + m,0. T ( s, c, d ) := ( e + d − c , ( c + d )(1+ d ) s ( c − ) ,if s (cid:54) = f rac , ∞ . 1. 0. T (0 , , 1) := ( ∞ , ∞ ) , T (0 , , − 1) := ( e , , m (cid:54) = 0 , , item 2. T (0 , − , 1) := ( e , , item3. T (0 , − , − 1) := ( e , , (when m = 0 , d = 1 , when m = 1 , d = − , T ( ∞ , c ∞ , d ∞ ) := ( e − dc , ( c + d ) dc ) , ( m (cid:54) = 0 , for m = 1 and c = √− , T ( ∞ , c ∞ , c ∞ ) = ( − , c )) . e = m − , e = − m . . . . D0. gives s = − c d , c = (cid:113) c + d d ,d s ( c + d )(1+ c )(1+ d ) s c . Theorem. Let a := p (cid:48) p − e )( p − e − , then0. a (cid:54) = 0 , − ⇒ T − ( p, p (cid:48) ) = ( s, c, d ) , where c := − a a , s := c − a , d := (1 − c )( p − e ) − . a = − ⇒ T − ( p, p (cid:48) ) = ( ∞ , a ∞ , a ( e − p ) ∞ ) . 2. 0. T − ( e [3] , 0) = (0 , − , − , for m (cid:54) = 0 . T − ( e [3]+1 , 0) = (0 , − , , for m (cid:54) = 1 . T − ( e [3]+ m, 0) = (0 , , − , for m (cid:54) = 0 , . T − ( ∞ , ∞ ) = (0 , , . Proof: If a (cid:54) = 0 , − , solving (1 − c )( p − ) = 1 + d and sp (cid:48) = − c + d )( p − e ) for c gives c − a s . Hence ( c − = a (1 − c ) , but c (cid:54) = 1 , − c = a (1 + c ) , hence c, s = c − a = − a a and d. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Theorem. Let g := 4 m − m +13 and g := ( m − m − m + 1) , let T ( s, cd ) = ( p, p (cid:48) ) , then p (cid:48) = 4 p − g p − g . Definition. The bijection defined by 9.3.3 and justified by 9.3.3 defines an [ isomorphism between the 3 dimen-sional elliptic curve of . . . and the 2 dimensional elliptic curve 9.3.3.0. Definition. The invariant j of sec-tell32a.0. is j := 2 j (cid:48) , with j (cid:48) := g g − g . Theorem. g − g = ( m ( m − . j = 2 m − m +1) ( m ( m − . If j is given and M is a solution of 1, the other solutions are − m, m , − m , − m , m − m . Example. For p = 11 , let u := m ( m − , j (cid:48) = − u = 1 , − , − m = − , , , − , − , K ( j ) = 4 or − .j (cid:48) = 4 corresponds to u = 2 , − m = 2 , − , − K ( j ) = 0 .j (cid:48) = ∞ corresponds to u = 0 or m = 0 , K ( j ) = 1 or − .j (cid:48) = 0 corresponds to u = − m = − δ , giving K ( j ) = 0 . Theorem. Given e , e and e such thatH.0. e + e + e = 0 , H.1. e i are distinct,Let m := e − e e − e ,e − e = d,g e − e e ) ,g e e e . thenC.0. ( d p ) = 1 ⇒ J m ) ∼ W g , g . C.1. ( d p ) = − ⇒ ( p + 1 − | J m ) | ) − ( − p )( p + 1 − | W g , g | ) = 0 . Proof: Let c = d , e (cid:48) i = c e i , then e (cid:48) = e (cid:48) + m and e (cid:48) = e (cid:48) + 1. . . . .3. THE WEIERESTRASS FUNCTIONS. ( cn, sd ) and the Weierstrass elliptic curve Definition. Let m := 1 − m, e := 2 − m . T ( cn, sd ) := ( e + cn − cn , − sd m + m cn (1 − cn ) , if cn (cid:54) = 1 . T (1 , 0) := ( ∞ , ∞ ) , T ( ∞ , sd ) := ( e − , − msd ) , If − m m = b thenT ( b, ∞ ) = ( e + b − b , . Theorem. Let a := p − e + 1 , then a (cid:54) = 0 , cn := a − a ,m + mcn (cid:54) = 0 thenT − ( p, p (cid:48) ) = ( cn, sd ) , where sd := P (cid:48) (1 − cn ) − m + mcn ) ,m + mcn = 0 then T − ( p, p (cid:48) ) = ( cn, ∞ ) , T − ( ∞ , ∞ ) = (1 , . a = 0 ⇒ T − ( p, p (cid:48) ) = ( ∞ , p (cid:48) − m ) . T ( − , 0) = ( e , . If T − ( x, 0) = ( y, ∞ ) and x (cid:54) = e then m ( m − 1) = d and x = e or e = − e ± d. Theorem. Let g := (16 m − m + 1) and g := (2 m − m − m − , let T ( cn, sd ) = ( P, P (cid:48) ) , then P (cid:48) = 4 P − g P − g . Definition. The bijection defined by 9.3.4 and justified by 9.3.4 defines an isomorphism between the 2 dimen-sional elliptic curve of . . . and the 2 dimensional elliptic curve sec-tell32a.0. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Theorem. Using 9.3.3, g − g = 256 m ( m − . j = 16 (16 m − m +1) m ( m − . Corollary. Let c := (cid:113) − m , p = e + c − c , using e from 9.3.4 and g , g from sec-tjacweia, then p − g p − g = − .. . . DOUBLE CHECK THIS. The proof follows from the fact that the denominator m + m cn in 9.3.4.0. cannot be zero. Theorem. Given e , e and e such that H.0. e + e + e = 0 , H.1. − ( e − e e ) = f , H.2. e i are distinct,Let m := (1 + d ) ,d = 2 − m e ,g e − e e ) ,g e e e . thenC.0. ( d p ) = 1 ⇒ J m ) ∼ W g , g . C.1. ( d p ) = − ⇒ ( p + 1 − | J m ) | ) − ( − − p )( p + 1 − | W g , g | ) = 0 . Proof: Let e (cid:48) i = c e i . We have a J m ) if e (cid:48) = 2 − m and e (cid:48) − e (cid:48) = 4 d = 4 (cid:112) m ( m − .4. COMPLETE ELLIPTIC INTEGRALS OF THE FIRST AND SECOND KIND. Example. p e , e , e g , g j (cid:48) m (cid:48) structure 13 1 , , , − C ×× C 29 1 , , − , − 12 13 3 C ×× C 37 1 , , − , − − C ×× C , , 20 0 , − − C ×× C 41 1 , , − , − − C ×× C 53 1 , , − , − − C ×× C , , − , − 21 3 C ×× C 61 1 , , − , 17 15 − C ×× C , , 31 7 , − C ×× C , , 33 0 , − 29 0 − C ×× C 73 1 , , 64 0 , C ×× C , , 57 15 , − 11 13 − C ×× C , , − , − − C ×× C 89 1 , , 75 20 , − 16 44 − C ×× C , , 63 23 , − − − C ×× C , , 59 13 , − C ×× C 97 1 , , 88 1 , − C ×× C , , 78 14 , − − − C ×× C , , 61 0 , C ×× C , , 58 15 , − 11 7 − C ×× C . . . .. . . . , , − , 10 21 C ×× C . . . Introduction. Several conjectures appear to justify the terminology of complete elliptic integrals of the first andsecond kind for the functions K and E defined below. The definitions are inspired from thedefinitions in the real and complex fields. Their importance is associated with Conjecture 9.4. Byconvention in this section I will use q := [ p ] = p − . Again I denotes the identity function (I(i) = i), D denotes the derivative operator and f ◦ g denotesthe function which is obtained by composition from the functions f and g. Definition. K j := ( (2 j − j !!) , (mod p ) , ≤ j ≤ q. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS E (cid:48) j := K j j j − , j = 0 to q, E := 1 , E j := − K j j − , j = 1 to q, Definition. K := (cid:80) qj =0 ( K j I j ) , E (cid:48) := (cid:80) qj =0 ( E (cid:48) j I j ) , E := (cid:80) qj =0 ( E j I j ) , Example. For p = 11 ,j K E (cid:48) E D B C K (cid:48)(cid:48) all [ j ]0 1 0 1 − − − − − − − − − − − − − − − 44 3 5 − − − − − − − − p = 13 ,j K E (cid:48) E D B C K (cid:48)(cid:48) all [ j ]0 1 0 1 − − − − − − − − − − − − − − − − − − − − − − − Lemma. K j = K q − j , ≤ j ≤ q. E (cid:48) j = E (cid:48) q +1 − j , < j ≤ q. E (cid:48) j = j − j K j − , < j ≤ q. E j +1 = (2 j − j − j E j , < j ≤ q. 4. 2 jK j − (2 j + 1) K j − = 2 jE j , < j ≤ q. 5. 2 jE j + E (cid:48) j = 0 , ≤ j ≤ q. 6. 2( j + 1) E (cid:48) j +1 − jE (cid:48) j − E j = 0 , ≤ j < q. The proof follows at once from Lemma 3.0.8.4 (g730) and from 9.4. .4. COMPLETE ELLIPTIC INTEGRALS OF THE FIRST AND SECOND KIND. Theorem. (see KE.NOT) K ( m ) = ( m p ) K ( m ) , < m < p. Proof. K ( m ) = m q K ( m ) , because of 9.4.0 and m q = ( m p ) by the Theorem of Euler. See for instanceAdams and Goldstein, p. 107. Corollary. p ≡ ⇒ K ( − 1) = 0 . Theorem. 0. 2 IDK + K − DE (cid:48) = 0 . 1. 2(1 − I ) DK − K − DE = 0 . 2. 2 IDE + E (cid:48) = 0 . 3. 2(1 − I ) DE (cid:48) − E = 0 . Proof: This follow immediately from the definitions, for instance, the coefficient of I j is2 jK j + K j − j + 1) E (cid:48) j +1 = 0 f or ≤ j < q, and that of I q = 0 because 2 q + 1 = p = 0 . Corollary. K (0) = 2 DE (cid:48) (0) = E (0) = 1 . DK (0) − DE (0) = . E (cid:48) (0) = 0 . K (1) = − DE (1) = 2 DE (cid:48) (1) − DK (1) = − E (cid:48) (1) . E (1) = 0 . Theorem. 0. 4 D ( I (1 − I ) DK ) − K := 0 . 1. 4 ID ((1 − I ) DE (cid:48) ) + E (cid:48) := 0 . 2. 4(1 − I ) D ( IDE ) + E := 0 . K = E + E (cid:48) . This derives from 9.4 by elimination, for instance, eliminating E’ from 2 and 3 gives4(1 − I ) D ( IDE ) = − − I ) DE (cid:48) = − E ID ((1 − I ) DE (cid:48) ) = 2 IDE = − E (cid:48) . 1, times I gives4 I (1 − I ) DK − IK − IDE = 0 , CHAPTER 9. FINITE ELLIPTIC FUNCTIONS using 2 and 0 gives4 I (1 − I ) DK − IK + 2 E (cid:48) = 0 , D ( I (1 − I ) DK ) − D ( IK ) + 2 IDK + K = 0 . Finally, it follows from 9.4.0 and 1 that D ( K − E − E (cid:48) ) = 0 , but K (0) = E (0) + E (cid:48) (0) = 1 , hence3. Definition. K (cid:48)(cid:48) j := ((2 j − j − ... ) j ! (mod p ) , K (cid:48)(cid:48) := (cid:80) q j =0 ( K q − j I ( q − j ) ) . Lemma. p ≡ ⇒ K (cid:48)(cid:48) iseven, p ≡ − ⇒ K (cid:48)(cid:48) isodd. 1. ( q − j + 2)( q − j + 3) K (cid:48)(cid:48) q − j +2 = (2 q − j + 1) K (cid:48)(cid:48) q − j , < j ≤ q. D ((1 − I ) DK (cid:48)(cid:48) ) − K (cid:48)(cid:48) = 0 . cK (cid:48)(cid:48) are, for arbitrary constant c, the only solutions of 2 .To prove 3., we substitute (cid:80) qj =0 ( X q − j I ( q − j ) ) , in 2, this gives − (2 q + 1) X q ) I ( q ) − (2 q − X q − ) I ( q − )+ (cid:80) qj =2 (( q − j + 2)( q − j + 3) X q − j +2 − (2 q − j + 1) X q − j ) I q − j = 0 . The coefficient of I q is zero because 2 q + 1 = 0 , hence X q is arbitrary. The coefficient of I ( q − must be zero, therefore X q − = X q − = . . . = 0 . Lemma. K ◦ (1 − I ) satisfies E ◦ (1 − I ) satisfies K ◦ ( I + ) satisfies K (cid:48)(cid:48) = sK (cid:48)(cid:48) ◦ ( − I ) . K = sK ◦ (1 − I ) . E = sE (cid:48) ◦ (1 − I ) . For instance, K ◦ (1 − I ) = 4( D ( I (1 − I ) DK ) ◦ (1 − I )= − D ((1 − I ) IDK ◦ (1 − I ))= 4 D ((1 − I ) ID ( K ◦ (1 − I ))) . Hence 0. 3, from 9.4.0 and s = ( − p − . K = cK (cid:48)(cid:48) ◦ ( I − ) = scK (cid:48)(cid:48) ◦ ( − I ) = sK ◦ (1 − I ) , hence 4. Finally, E ◦ (1 − I ) and E (cid:48) satisfy the same differentialequation, of second order, moreover E (1) = sE (cid:48) (0) = 0 and DE (1) = − sDE (cid:48) (0) because of . . . andof K (1) = sK (0) hence 5. .4. COMPLETE ELLIPTIC INTEGRALS OF THE FIRST AND SECOND KIND. Corollary. K ( m ) = sK (1 − m ) , with s = ( − q .1. K ( mm − ) = ( m − m ) ) K ( m ) , < m < p − . E ( m ) + sE (1 − m ) = K ( m )3. E ( m ) K ( m ) + E (1 − m ) K (1 − m ) = 1 . m R p, k = m ⇒ K ( k k ) 2) = K ( m ) , < m < p − . 5. 1 − m R p, k = 1 − m ⇒ K (((1 − k )(1 + k )) = K ( m ) , < m < p − . Corollary. ( − k (cid:80) jk =0 ( (2 j − j )!! ) (cid:18) jk (cid:19) = ( − q ( (2 k − k )!! ) . Exchanging k and j, 1. follows from Lemma 3.0.x. of g730. Theorem 7. Corollary. p ≡ − ⇒ K ( p +12 ) = 0 , p ≡ ⇒ K ( p +12 ) = 2 E ( p +12 ) . Conjecture. With the exception of p = 7 , K (3) = 3(= − ,m (cid:54) = 0 , , | K ( m ) | < p ⇒ K ( m ) ≡ p + 1 (mod 4) . Corollary. p ≡ , ⇒ K ( m ) (cid:54) = 0 . Example. For p = 11 ,m K E (cid:48) E D B C K (cid:48)(cid:48) EK all ( m )0 1 0 1 1 1 1 0 11 − − − − − − − − ∞ − − − − − − − − − − − ∞ − − − − − − − − − − − − − 110 0 − − − − ∞ CHAPTER 9. FINITE ELLIPTIC FUNCTIONS For p = 13 ,m K E (cid:48) E D B C K (cid:48)(cid:48) EK all ( m )0 1 0 1 1 1 1 1 11 1 1 0 6 − − − − − − − − − − − − − − − − − − − − − − − − 68 2 − − − − − − − − − − − − − − − − − − − − − − − Definition. By analogy with the case of the real or complex field,0. D j := K j +1 − E j +1 , ≤ j < q, D q := 0 . B j := K j − D j , ≤ j ≤ q. C j := D j +1 − B j +1 , ≤ j < q, C q := 0 . D := (cid:80) qj =0 ( D j I j ) , B := (cid:80) qj =0 ( B j I j ) , C := (cid:80) q j =0 ( C j I j ) . Theorem. D j = E j +1 , ≤ j < q. B j = E q − j , ≤ j ≤ q. D = E (cid:48) I . IB = IE + ( I − E (cid:48) . I C = (2 − I ) E (cid:48) − IE. I C = 2 E (cid:48) − IK. ? .5. P-ADIC FUNCTIONS, POLYNOMIALS, ORTHOGONAL POLYNOMIALS. Comment. In a p -adic field, we can define polynomials of degree up to p − . These are determined by theirvalues at i in Z p . If these are defined in the real field with rational coefficient, the definition andproperties are automatically extended to the p -adic field. For orthogonal polynomials, recurrencerelations, differential equations and values of the coefficients generalize automatically. Therefore,we have the definitions 1. and the theorems 2. and 3. Definition. The polynomials of Chebyshev of the first ( T n ) and of the second kind ( U n ) , of Legendre ( P n ) , ofLaguerre ( L n ) and of Hermite ( H n ) are defined by the differential equations:0. (1 − I D T n − IDT n + n T n ≡ ,T n (0) ≡ , DT n (0) ≡ . 1. (1 − I D U n − IDU n + n ( n + 2) U n ≡ ,U n (0) ≡ , DU n (0) ≡ . 2. (1 − I D P n − IDP n + n ( n + 1) P n ≡ ,P n (0) ≡ , DP n (0) ≡ . ID L n + (1 − I ) DL n + n ≡ . D H n − IH n + 2 n ≡ . Theorem. If X n,j denotes the coefficient of I j in the polynomial X n , then T n,n − j ≡ n ( n − j ) ( − j ( n − j − j !( n − j )! , U n,n − j ≡ n ( n − j ) ( − j ( n − j )! j !( n − j )! ) , P n,n − j ≡ ( − n ) ( − j (2 n − j )! j !( n − j )!( n − j )! , L n,j ≡ ( − j n !( n − m )!( m !) , H n,n − j ≡ n !2 ( n − j ) ( − j j !( n − j )! , See for instance Handbook of Mathematical functions, p. 775. Theorem. T , T ≡ I, T n +1 ≡ I − T n − T n − , U ≡ , U ≡ I, U n +1 ≡ I − U n − U n − , P ≡ , P ≡ I, ( n + 1) P n +1 ≡ − (2 n + 1) IP n − nP n − , CHAPTER 9. FINITE ELLIPTIC FUNCTIONS P ≡ , P ≡ I, ( n + 1) P n +1 ≡ (2 n + 1 − I ) P n − nP n − , H ≡ , H ≡ I, H n +1 ≡ IH n − nH n − . See for instance Handbook of Mathematical functions, p. 782. Theorem (T). T i +2 pk,j = − T i + pk,j = T i,j , j < p. Proof: T p + i,j ≡ ( − p + i − j j p + i + j − p + i ( p + i − j )! j ! ≡ ( − p + i − j ( − p − i − j ( − p − i + j + i j ( p − i + j − )! p − i ( p − i − j )! j ! ≡ ( − p − i − j j p − i + j − )! p − i ( p − i − j )! j ! ≡ T p − i,j . Theorem (U). U i +2 pk,j = − U i + pk,j = U i,j , j < p. U p − i,j ≡ ( − p − i − j ( p − i + j )!2 j ( p − i − j )! j ! U p − i,j ≡ ( − p − i − j ( − p − i − j − 3. ( − p − i + j − ( p − i + j − )!2 j ( p − i − j )! j ! U p − − i,j ≡ ( − p − i − j j ( p − − i + j )!( p − − i − j )! j ! ≡ U p − − i,j . Theorem (Le). P p − − n = P n , n < p. Proof: The polynomials can be defined by the recurrence relations, P = 1 , P = I, ( n + 1) P n +1 = (2 n + 1) IP n − nP n − , n < p − . The last equation is valid for n + 1 = p and therefore P p can be considered as 0 as far as the proofof the theorem is concerned. They satisfy the Rodrigues’ formula P n = n n ! D n ( I − n Therefore P p − = p − ( p − )!) D ( p − ( − I ( p − ) = ( − p − ( p − p − ( p − !) , but 2 ( p − = 1 , ( p − − , and( p − !) = ( − ( p − ) ( p − p − i = − p + i , hence P p − = 1 . .5. P-ADIC FUNCTIONS, POLYNOMIALS, ORTHOGONAL POLYNOMIALS. By convention we can write P p = P − = 0 , if we replace in the recurrence relation n by p − n − n + 1) P p − n − = (2 n + 1) P p − n − − nP p − n , and therefore, by induction, P p − − n = P n . Example. p = 11 , see orthog, 120. P = 1 ,P = I,P = 5 − I ,P = 4 I − I ,P = − − I + 3 I ,P = − I + 5 I + I ,P = − − I + 3 I ,P = 4 I − I P = 5 − I ,P = I,P = 1 . For p = 13 ,P = P = 1 ,P = P = I,P = P = 6 − I ,P = P = 5 I − I ,P = P = 2 + 6 I + 6 I ,P = P = − I + I + 3 I ,P = − − I − I − I , Theorem (La).Theorem (H).Definition. The scaled Hermite polynomials are defined by0. H = 1 , H = I, 2. [ n ] H n = a n H n − − n − H n − , where a n = 1 if n is even and a n = [ n ] , the largest integer in n if n is odd. Example. H = − + I ,H = − I + I ,H = − I + I ,H = I − I + I , CHAPTER 9. FINITE ELLIPTIC FUNCTIONS H = − + I − I + I ,H = − I + I − I + I . Lemma. Modulo p, p > , 0. ( p − ≡ − . 1. ( p − − i )! ≡ ( − ( i +1) 1 i ! , ≤ i < p. (cid:18) p − − ij (cid:19) ≡ ( − j (cid:18) i + jj (cid:19) , ≤ i, j, i + j < p. (cid:18) kp + ij (cid:19) ≡ (cid:18) ij (cid:19) , j < p. 4. ( p − − i )!! i !! ≡ ( − k ( p − − k − i )!!( k + i − ≤ i < p − , < k + i < p. Proof: 0. is the well known Theorem of Wilson. 1, can be considered as a generalization.( p − − i ) ≡ ( − i ( p − . . . ( i + 1) ≡ ( − i ( p − i ! ≡ ( − i +1 1 i ! . For 2. ( p − − i )!( p − − i − j )! j ! ≡ ( − ( i +1) ( i + j )!( − i + j +1 i ! j ! ≡ ( − j (cid:18) i + jj (cid:19) . Lemma. Modulo p, p > , 0. (( p − ≡ ( − ( p − )1. ( p − p − ≡ − . 2. 0.( p − − i )!! i !!( − s ( p − , where s = i w ≡ n i is evenand s = p − − i when i is odd. 1. or where s = [ [ p ]+1+ i ] + [ p +14 ] . 0, and 1 are well known and given for completeness. 2, if i is even,( p − − i )!! i !!( p − i )!!( i − i p − i ) or − − ( i ) ( p − . if i is odd,( p − − i )!! i !!( p − − i )!!( i + 2)!!( p − − ii +2 or − ≡ ( − ( p − − i )) (0)!! p − . p = 1 , , , i = 0 , , , , . .5. P-ADIC FUNCTIONS, POLYNOMIALS, ORTHOGONAL POLYNOMIALS. Theorem (La). L p − − i,j ≡ ( − j L i + j,j , ≤ i, j, i + j < p. Proof:For 0, L n,j = ( − j (cid:18) nj (cid:19) j ! . See for instance, Handbook p.775. Lemma. (cid:80) ((2 j − k )!! j ) !!(2 k − j ! k !( j − k ) !)= ( − ( [ p ] − k ) , j = k to [ p ] . This is needed for g761, . . . . Not yet proven.The expression which is summed in the first member can be replaced by ( (2 j )!(2 k !) ) k ! j !( j − k )!2 j − k ) . Introduction. . Connection with p -adic fields. In p -adic fields, introduced by Kurt Hensel, trigonometric functionsare defined. The connection between these and those obtained in finite fields has to be explored.To that effect, 2 programs have been written, padic.bas and sin.bas. The first program obtains thefunctions sin and cos for arguments which are congruent to 0 modulo p. For instance to 7 Example. x sin ( x ) cos ( x ) . . . . . . . . . . . . . . . . . . . . . . 01 0 . . CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Example. In base 7 and 7 , we have for the elliptic case y sin ( y ) cos ( y ) sin ( y ) cos ( y )0 0 1 0 . 000 1 . 000 0 . 000 1 . . 126 2 . 406 2 . 653 2 . . 053 0 . 514 1 . 000 0 . . 054 5 . 143 2 . 653 5 . . 332 6 . 606 0 . 000 6 . . 444 5 . 352 . . . .. 12 6 0 6 . 631 0 . . 030 2 . . 101 1 . . 332 = − . 434 and 6 . 606 = − . , we get the first clue for the relation between the functions in the p -adic field and in base 7 . .5. P-ADIC FUNCTIONS, POLYNOMIALS, ORTHOGONAL POLYNOMIALS. For 13 n ,x sin ( x ) cos ( x )0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 0 . 10 0 11 8 4 3 5 1 . . 11 0 . 11 0 10 6 7 3 10 1 . . 12 0 . 12 0 11 3 2 11 4 1 . . . . π . . π . . π . . Example. p,first e,s%? 13,2,7c% = 21 7 2 7. 0 2. 02 2 7 2. 0 7. 03 1 0 1. 0 0. 04 2 6 2. 0 6. 0p,first e,s%? 169,2,137c% = 411 137 41 137. 0 41. 02 80 150 80. 0 150. 03 1 91 1. 0 91. 04 2 45 2. 0 45. 05 163 50 163. 0 50. 06 13 168 13. 0 168. 07 58 37 58. 0 37. 08 11 162 11. 0 162. 09 168 65 168. 0 65. 010 76 98 76. 0 98. 011 149 28 149. 0 28. 012 143 1 143. 0 1. 013 85 54 85. 0 54. 014 67 33 67. 0 33. 015 1 117 1. 0 117. 016 15 97 15. 0 97. 017 46 63 46. 0 63. 0 CHAPTER 9. FINITE ELLIPTIC FUNCTIONS 18 39 168 39. 0 168. 019 110 24 110. 0 24. 020 24 110 24. 0 110. 021 168 39 168. 0 39. 022 63 46 63. 0 46. 023 97 15 97. 0 15. 024 117 1 117. 0 1. 025 33 67 33. 0 67. 026 54 85 54. 0 85. 027 1 143 1. 0 143. 028 28 149 28. 0 149. 029 98 76 98. 0 76. 030 65 168 65. 0 168. 031 162 11 162. 0 11. 032 37 58 37. 0 58. 033 168 13 168. 0 13. 034 50 163 50. 0 163. 035 45 2 45. 0 2. 036 91 1 91. 0 1. 037 150 80 150. 0 80. 038 41 137 41. 0 137. 039 1 0 1. 0 0. 040 41 32 41. 0 32. 0 Comment. Using s = x, s i +1 = s i x i − i (2 i +1) , and arcsin ( x ) = s + s + . . . ,arcsin ( . , , , , , , 6) = . , , , , , , sin ( . , , , , , , 5) = . , , , , , , . , , , , , , π correspond to arcsin ( . , , 6) = . , , p to . , , , , , , sin ( α ) and cos ( α ) , obtained using 115a n2adic,determine sin ( iα ) and cos ( iα ).For p = 13 , using sin ( α ) ≡ mod , we get (mod p ) all distinct values except(0 , , , , , , 11) 169 ± . The non zero numbers in the parenthesis are the non residues.This indicates that the connection is tenuous. Theorem. If s (1) ≡ g (mod p n ) , where g is a primitive root of p, then . . .Let s c = 6 . . . . , c s (cid:113) = 2 . . . . , given ξ , determine s = sin ( ξ ) , c = cos ( ξ ) , in the p -adic field, ( ξ ≡ p ) , .6. P-ADIC FIELD. determine, s ( i ) = sin ( iξ ) cp ( i ) + cos ( iξ ) sp ( i ) , c ( i ) = cos ( iξ ) cp ( i ) − sin ( iξ ) sp ( i ) . ( ξ was { control H − )? Definition. Int j i sin = cos (2 j ) − cos (2 i ) . M id j i sin = sin (2 i + 1) + sin (2 i + 3) + . . . + sin (2 j − . T rap j i sin = sin (2) + sin (2 i + 2) + . . . + sin (2 j − 2) + sin (2 j ) . Simpson j i sin = sin (2 i ) + 4 sin (2 i + 2) + 2 sin (2 i + 4) + 4 sin (2 i + 6)+ . . . + 4 sin (2 j − 2) + sin (2 j ) . Theorem. Int j i sin = − sin (1) M id j i sin = − tan (1) T rap j i sin = − sin (2)2+ cos (2) Simpson j i sin . Notation. Writing x = x + x p + x p + x p + . . . in the form x = x .x x x . . . , we will also write x ≡ x (mod p ) , x ≡ x .x (mod p ) , . . . .We have, for instance,0. = 0. 7 6 6 6 6 6 6. Example. p = 5 , up to p , = 3 . , indeed, 2 3 = 6 ≡ , . 2) = 2 . 13 = 26 ≡ ) , . . . . − / . , − . / . , − . / . . Definition. In the p -adic field, the exponetial, logarithmic and trigonometric functions are defined by: exp ( x ) := 1 + x + x + x + . . . , for | x | < . . . . log ( x ) := x − x + x + . . . , for | x | < . CHAPTER 9. FINITE ELLIPTIC FUNCTIONS sin ( x ) := x − x + x − . . . , for | x | ≤ p − cos ( x ) := 1 − x + x − . . . , for | x | ≤ p − . Example. p = 5 , x = 0 . , . x = 0 . x = 0 . x / . − x = 0 . x / . . x = 0 . x / . . − x = 0 . x /. 44 = 0 . x /. . x /. . − x / . . x /. . x / . . x /. . exp (0 . 1) = 1 . log (1 . 1) = 0 . x = 0 . . − x / . . − x = 0 . x /. 44 = 0 . x / . − x /. . − x / . . x /. . sin (0 . 1) = 0 . cos (1 . 1) = 1 . Example. For p = 13 ,x sin ( x ) cos ( x )0 . . , . . . , . . . , . . . , . . . , . . . , . . . , . . . , . . . , . . . , . . 10 0 . 10 0 11 8 4 3 5 , . . 11 0 . 11 0 10 6 7 3 10 , . . 12 0 . 12 0 11 3 2 11 4 , . . . , . . 11 1 0 . 11 1 10 4 10 9 1 , . . . , . . 10 2 0 . 10 2 11 12 2 1 8 , . . . , . Definition. The Chebyshev polynomials are defined by the recurrence relation T i +1 ( x ) := − T i − ( x ) + 2 xT i ( x ) , i = 1 , , . . . , with .6. P-ADIC FIELD. T ( x ) := 1 , T ( x ) := x. Definition. For p ≡ , a root c T i is called a primitive root modulo p , if all the roots of T i ,c , c , . . . , c i − can be obtained from it using the addition formulas, c := c , c := 4 c − c , c i +1 := − c i − + 2 c i − (2 c − , i = 2 , . . . , i − . Example. For p = 13 , the roots of T = 4 x − x are 0, 2 and − . − c c = 2 then c = 0 , c = − . For p = 17 , T ( x ) = 8 x − x + 1 , which has the primitive roots 4 , − , , − . Notation. For p ≡ , the roots of T p − will be denoted, cos ( α ) , cos (3 α ) , . . . , cos ( p − α ) . with α = π p − . We will also define cos (0 α ) := 1 ,cos (2 kα ) := − cos ((2 k − α ) + 2 c cos ((2 k − α ) ,k = 1 , . . . , p − .sin ( kα ) = cos (( p − − k ) α ) . Example. For p = 13 , α = π , δ = 2 , (the lines k = , , . . . will be explained in 1.1, 1.2), k sin ( kα ) cos ( kα ) CHAPTER 9. FINITE ELLIPTIC FUNCTIONS . . . δ . δ . . . δ . δ . . . δ . δ . . . δ . δ . . . δ . δ . . . δ . 11 4 5 6 10 6 5 δ . . 12 12 12 12 12 12 12 . δ . 11 4 5 6 10 6 5 δ . . . δ . δ . . . 11 4 5 6 10 6 5 δ . δ . 12 12 12 12 12 12 12 0 . . 11 4 5 6 10 6 5 δ . δ 10 11 . . . δ . δ 11 6 . . . δ . δ 12 0 . . sin ( α ) = , cos ( α ) = √ , are sufficient to obtain the entries 0, 1, 2, . . . ,12, in the table.We have therefore a first method of obtaining tables of trigonometric functions in a finite field .We choose x, such that | x | = p − , therefore | kx | ≤ p − . ( If | x | < p − , primitivity is not insured.See 1.3.)We compute sin ( k x ) and cos ( i x ) by Maclaurin series, (see 0.1.) and use the addition formulas sin ( k ( α + x )) = sin ( kpα ) cos ( kx )+ cos ( kpα ) sin ( kx ) , cos ( k ( α + x )) = cos ( kpα ) cos ( kx ) − sin ( kpα ) sin ( kx ) , where kp is k (mod p ) . Example. For p = 13 , and x = 0. 1,(The lines k = , , . . . willbeexplainedin . .k sin ( k ( α + x )) cos ( k ( α + x )) .6. P-ADIC FIELD. . δ . δ . . 10 8 7 2 10 10 6 . δ . 10 12 0 3 3 5 7 δ . . . δ . δ . . 10 12 10 6 3 9 6 . δ . 11 8 8 0 12 4 5 δ . . 11 6 10 6 9 12 0 . δ . δ . . 12 1 7 5 3 7 5 . δ . δ . . 12 4 1 11 8 4 20 . . δ . 11 2 9 10 12 1 2 δ . . . . δ . 12 5 8 7 11 2 6 δ . . . . δ . δ . 12 7 9 4 3 11 2 0 . . . 12 8 7 3 11 6 6 δ . δ 10 11 . . . . δ . δ 11 6 . . . . δ . . 12 0 11 3 2 11 4 1 . . . δ . 12 5 5 0 2 9 110 . . . . . δ . δ . . 11 2 1 9 11 6 8 7 . . . δ . δ . . . 11 11 9 10 7 2 105 . . 11 2 0 10 2 0 5 δ . δ . . . . . 11 2 0 2 12 9 0 δ . 11 4 4 1 12 8 3 δ . . 11 9 3 11 12 4 10 11 . . . 11 8 0 12 9 2 12 δ . δ . . . 12 5 12 7 1 5 711 . . δ . δ . . . 11 0 6 6 2 8 60 . . δ . δ A second method to obtain trigonometric tables in finite fields is to start with sin ( α ) such that x .x is a primitive root for p and cos ( α ) = (cid:112) − sin ( α ) and use the addition formulas to obtainin succession sin ( kα ) and cos ( kα ) for k = 2 , , . . . . For instance, we have the following Example. For q = 13 and sin ( α ) = 7. 8 0 4, then cos ( α ) = 2.10 8 7, the period is 12 . and CHAPTER 9. FINITE ELLIPTIC FUNCTIONS k sin ( kα ) cos ( kα )1 7 . . 10 8 72 2 . . . . 10 12 104 2 . . 11 6 105 7 . . 12 1 76 0 . . 12 4 17 6 . . . . . 12 7 9 0 . . . . . . 12 0 11 1 . . . . . . 11 1 10 1 . . . . . . 10 2 11 1 . . . . Theorem. If sinh ( x ) = x + x + x + . . . , | x | ≤ p − , and cosh ( x ) = 1 + x + x + . . . , | x | ≤ p − , then sin ( xi ) = isinh ( x ) , cos ( xi ) = cosh ( x ) , Example. With p = 11 ,sin (0 . i ) = 0. 6 5 810 3 8 0 i, cos (0 . i ) = 1. 0 7 9 5 7 0 2. sin (0 . i ) =0. 1 0 2 9 0 9 6 i , cos (0 . i ) =1. 0 6 5 0 5 9 6. Example. For p = 11 and x = . 1, Using 6.1.13 and sin ( α ) = 9. 3 5 9 2 2 410 and cos ( α ) = 5. 7 6 3 0 4 6 9 i,sin ( α ) = 2.10 5 5 6 5 6 6 i and cos ( α ) = 4. 919 3 8 7 9 4of 1.6, we obtain sin (( α +12 ) i ) = 9. 6 910 9 2 1 0 i, cos (( α +12 ) i ) = 5. 8 610 0 0 7 2, sin (( α + 1) i ) = 2. 3 5 7 8 5 5 2 i,cos (( α + 1) i ) = 4. 0 1 210 6 810.The table can be computed from the first values the second are used here as a check: k sin ( k ( α + x ) / cos ( k ( α + x ) / .6. P-ADIC FIELD. . . . . . . i . . i . . . . i . . i . . . . i . . i . . . . i . . i . . . . i . . i . 10 3 4 6 3 9 90 . . . i . . i . . . . i . . i . . . 10 8 9 6 0 8 7 0 . i . . i . . . . i . . i . . . . 10 5 5 8 1 1 10 i . . 10 0 9 7 9 9 0 i . . . . i . . 10 9 3 7 3 9 5 i . . . . i . . i . . . . i . . i . . . . i . . i . . . . 10 4 2 0 9 10 0 i . . i . 10 2 6 8 4 8 89 . . . i . . i . . . . i . . i . 10 2 3 7 6 3 02 . . 10 4 9 7 9 9 0 0 . i . . i . . . . i . . 10 8 5 1 5 10 4 i . . . . i . . i . Introduction. The tables of trigonometric functions can be extended to the half arguments. These are requiredfor the angles in finite Euclidean geometry. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Theorem. If g is a primitive root of p, and δ = g, then c (cid:48) = cos ( α ) δ − is a primitive root of S (cid:48) p − = T p − ◦ ( δI ) , where I is the identity function.Indeed, T n = T n ◦ (2 I − . The other roots are denoted by c (cid:48) , c (cid:48) , . . . . Example. For p = 13 , g = 2, S (cid:48) = 256 I − I + 36 I − ,c (cid:48) = cos ( α ) δ − i = , , . . . , . Comment. The method given at the end of section 0.6. enables to complete the table of Example 0.7. Alter-nately, if g is a primitive root for p , p ≡ , we know that g is a primitive root for p e ,e = 3 , , . . . . If δ = g, sin ( α/ 2) = δ (cid:113) − cos ( α )2 g and cos ( α ) = δ (cid:113) cos ( α )2 g . Example. For p = 13 ,sin ( α ) δ − √ (3. 7 7 412 310 4) = 9. 8 211 7 1 5 9 δ, cos ( α ) δ − √ (4.1211 1 7 2 9 1) = 2. 3712121211 1 δ. One of the signs of the square roots can be chosen arbitrarily, the other must be chosen in such away that sin ( α ) = 2 g sin ( α ) cos ( α ) . Theorem. For p ≡ − , with δ = − ,cos ( α/ δ − is a primitive root of V p − = ( T p − I − ) ◦ ( δI ) .sin ( α/ is a primitive root of U p − = ( T p − I − ) ◦ √ − I . Example. For p = 11 , δ = − , α = π ,V = 16 I + 20 I + 5 , with roots 5. 7 6 3 0 4 6 9 and 2.10 5 5 6 5 6 6, U = 16 I − I + 1 , with roots 9. 3 5 9 2 2 410 and 4. 910 3 8 7 9 4. Hence, k sin ( kα ) cos ( kα ) .6. P-ADIC FIELD. . . . . δ . 10 5 5 6 5 6 6 δ . . . 10 5 5 6 5 6 6 δ . δ . . . . δ . . . δ . 10 5 5 6 5 6 6 δ . . . δ . , . 10 10 10 10 10 10 10 , Tables. These can be found in the Handbook for Mathematical functions. Table 22.3 gives T and V and,by a simple transformation, S (cid:48) . Table 22.5 gives U.U and V can be obtained by recurebces: U = 1 , U = 4 I − , U i +2 = 2(2 I − U i − U i − .V = 1 , V = 4 I + 3 , V i +2 = 2(2 I + 1) V i − V i − . We have p = 5 , g = 2 , S (cid:48) = 4 I − ,c s c p = 7 , U = 4 I − , s V = 4 I + 3 , c p = 11 , U = 16 I − I + 1 , s V = 16 I + 20 I + 5 , c p = 13 , g = 2 , S (cid:48) = 256 I − I + 36 I − ,c (cid:48) = s c (cid:48) = c p = 17 , g = 3 , S (cid:48) = 10368 I − I + 1440 I − I + 1 ,c (cid:48) = s c (cid:48) = c p = 19 , U = 256 I − I + 240 I − I + 1 ,s V = 256 I + 576 I + 432 I + 120 I + 9 ,c Comment. The following values may be useful, √− = 2. 1 2 1 3 4 2 3 0 3 2 √− = 5. 5 1 0 5 5 1 0 1 8 8, √− = 4. 210 5121612 813 314, √− =12. 112 1181615 3 92425. Definition. The exponentiaonal function and the logarithmic function are defined by the following p -adic ex-pansion. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS D.0. exp ( x ) = 1 + x + . . . + n ! x n + . . . , | x | < . D.1. log (1 + x ) = x − x + . . . + ( − n n x n + . . . , | x | < . The classical theorem is (See for instance Koblitz, 1977.) Theorem. . . . Motivation. For p = 5, log ( − 4) = log (1 − 5) = 0 . ,log (1 . 1) = log (1 + 5) = 0 . ,log ( − . 1) = log (1 − 10) = 0 . . If we want log ( x y ) = log ( x ) + log ( y ) to hold, we have 3 equations to determine log ( − , log (2) and log (3) . log (1 . 3) and log ( − . 4) can be used as check. This gives log ( − 1) = 0 ,log (2) = 0 . ,log (3) = 0 . . Clearly we now have a function which is not a bijection, for instance, log (1 . . . This suggest that we can extend the range of definition of the logarithm function. The equation x p − ≡ p ) has p − p − , therefore the equation x p − = 1 has p − p -adic field, with first digit 1,2, . . . , p − . In general the roots are 1 , x , x , . . . , − x , − x , − . Algorithm. If g is a primitive root in Z p , the corresponding primitive root g (cid:48) ∈ . . . can be obtained by Newton’smethod: y := g,y = y − p − ( y + y p − for i = 1 to n . Theorem. Algorithm 9.6.3 determines the first n digits of g (cid:48) . Theorem. Given a prime p, a primitive root g ∈ Z p and x, H0. | x | = 1 , D0. x ind ( int ( x )) , where ind is the index function in Z p associated to g, D1. y := x/g (cid:48) x − z := y − y + . . . + n ( − y ) n then C0. | z | < . C1. log p,g ( x ) = { x , z } .6. P-ADIC FIELD. Example. For p = 5 , x . . (All the roots are 1, 2 .1213423, − . . , − . p = 7 , x . , x . . For p = 11 , x x x x p = 13 , x x x x x p = 17 , x x x x x x x p = 19 , x x x x x x x x p = 23 , x x x x x x x x x x p − Definition. Given a primitive root g,log p,g ( x ) = { i , x , log p ( x ) } ,i ∈ ( Z, +) , x ∈ ( Z p − , +) , log p ( x ) ∈ . . . . where | x | = p − i , g x = int ( x ) , and log p ( x ) is the p -adic logarithm. i characteristic , x , the index and log p ( x ) , the mantissa. Example. For p = 5 and g = 2 ,log , (1 . { , . } , log , (2 . { , . } ,log , (3 . { , . } , log , (4 . { , . } . Theorem. H0. | x | , | y | = 1 . C0. log p,g ( x ∗ y ) = log p,g ( x ) + log p,g ( y ) . CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Problem. Extend the definition to allow | x | , | y | to be anything using the relation C.0. and the idea ofmantissa. Introduction. Some thought on finite geometry for different powers of p and the p -adic geometry.Given p, if we take the points on a line, with coordinates of the form x . . . . these are indistin-guishable if we use the p -adic valuation to the precision 1.If we consider the points x .x , these are distinguishable to the precision 1 but not to the precision p . They can be considered, if p is 5 say, to have a color associated to the various digits of x x . Comment. The trigonometric functions work as follows.In the hyperbolic case,for p, the period is 2( p − , for p , the period is 2 p ( p − , . . . .The factor 2 corresponds to the fact, that just as in Euclidean geometry the total angle is 2 π, lineswhich form an angle π correspond to the same direction.For the hyperbolic case, there are 2 real isotropic points,There are on the ideal line p + 1 points, p − p -geometry, p − p -geometry, of which p − p are not included in the preceding set.The hyperbolic trigonometry associated to p is presumably for these p − p directions . . . .For the elliptic case, the period is 2( p + 1) as one would expect for the p -geometry.For the p case, the period of the trigonometric functions is p + p. I do not yet understand howthis comes into the picture. Definition. p -ADIC GEOMETRYLet p ≡ − z = 0 be the ideal line, let x + y = z be a circle.There are no solutions of x + y = 0 , therefore the isotropic points are not real. .6. P-ADIC FIELD. If z (cid:54) = 0 , then we can normalize using z = 1 , consider x + y = 1 . Let | x | ≥ | y | . If | x | > p | x | and work modulo p ) Lemma. If s = sin ( α ) is a root of . . . then sin ( pα ) = sin ( α ) , cos ( pα ) = cos ( α ) . Theorem. Let s = sin ( α ) be a root of . . . .If | sin ( β ) − sin ( α ) | < then lim n →∞ sin ( p n β ) = sin ( α ) . If | x | = 1 , there are solutions, x = x .x x . . . . Let | x | be a primitive solution of the polynomial. . . .let x . be . . . .To x corresponds the pair ( x, y ) = ( sin ( α ) , cos ( α )) , Let X ( x ) = sin ( pα ) , then the sequence x, X ( x ) , X x ) , . . . convergences to x (cid:48) . Moreover x (cid:48) is a rootof . . . Example. p = 11 , let x = sin ( α ) = 3 , X(3. 0 0 0 0 0 0 0) = 3. 0 2 8 011 4 6, cos ( pα ) = 10 . , X(3. 0 2 0 0 0 0 0) = 3. 0 2 6 0 71212,X(3. 0 2 6 0 0 0 0) = 3. 0 2 6 7 612 2,X(3. 0 2 6 7 0 0 0) = 3. 0 2 6 71211 2,X(3. 0 2 6 712 0 0) = 3. 0 2 6 71212 1,X(3. 0 2 6 71212 0) = 3. 0 2 6 71212 2. CHAPTER 9. FINITE ELLIPTIC FUNCTIONS Example. p = 11 , elliptic case, g = − sin, with sin ( k = (2 l + 1) α ) δ − , sin ( k = 2 lα )1 4 . 000 4 . 100 4 . 200 4 . 300 4 . 400 4 . 500 4 . . 325 5 . 112 5 . 665 5 . 450 5 . 245 5 . 034 5 . . 350 2 . 241 2 . 115 2 . 030 2 . 624 2 . 506 2 . . 063 1 . 031 1 . 033 1 . 064 1 . 050 1 . 000 1 . . 632 2 . 322 2 . 033 2 . 451 2 . 130 2 . 522 2 . . 524 5 . 430 5 . 303 5 . 200 5 . 130 5 . 026 5 . . . 566 4 . 550 4 . 540 4 . 536 4 . 524 4 . 511 4 . . 500 4 . 510 4 . 520 4 . 530 4 . 540 4 . 550 4 . . 034 5 . 013 5 . 061 5 . 040 5 . 026 5 . 005 5 . . 506 2 . 562 2 . 556 2 . 543 2 . 530 2 . 524 2 . . 000 1 . 000 1 . 000 1 . 000 1 . 000 1 . 000 1 . . 522 2 . 562 2 . 533 2 . 504 2 . 544 2 . 515 2 . . 026 5 . 013 5 . 000 5 . 063 5 . 050 5 . 044 5 . . . 511 4 . 510 4 . 516 4 . 515 4 . 514 4 . 513 4 . . 510 4 . 511 4 . 512 4 . 513 4 . 514 4 . 515 4 . . 013 5 . 011 5 . 016 5 . 014 5 . 012 5 . 010 5 . . 562 2 . 561 2 . 560 2 . 566 2 . 565 2 . 564 2 . . 000 1 . 000 1 . 000 1 . 000 1 . 000 1 . 000 1 . . 562 2 . 566 2 . 563 2 . 560 2 . 564 2 . 561 2 . . 013 5 . 012 5 . 011 5 . 010 5 . 016 5 . 015 5 . . . 510 4 . 510 4 . 510 4 . 510 4 . 510 4 . 510 4 . hapter 10DIFFERENTIAL EQUATIONS ANDFINITE MECHANICS In the context of Finite Geometry, we should examine the subject of Differential Equations, theirapproximation and the application to finite mechanics.I will describe the first success associated with the harmonic polygonal motion, then . . . Introduction. In classical Euclidean geometry as well as in finite Euclidean geometry I define the harmonicpolygonal motion as the motion which associates to linearly increasing time successive points of theharmonic polygon. I will determine, for the classical case, the differential equation of the motion, byconsidering first points which are close to each other 10.1.1. I will then prove that this equation issatisfied when points are not close to each other 10.1.1. The equation bears ressemblance with theequation of Kepler. Because the method uses derivatives of functions of the trigonometric functionsonly and in view of the method of Hensel for p -adic functions, the result extend automatically tothe finite case. Definition. Given a conic0. A ( E ) = ( acos ( E ) , bsin ( E ) , , and the point of Lemoine K = ( q, , given the correspond harmonic polygon of Casey (g2734, p. . . . ), A i , we define the harmonic polygonal motion by1. A ( E ( t + i h )) = A i . CHAPTER 10. DIFFERENTIAL EQUATIONS AND FINITE MECHANICS Theorem. Let r = qa , if h is small, the motion satisfies the differential equation C DE = 1 − r cos ( E ) , for some constant of integration C. Proof: The polar k of K is2. k = [ qa , , − , the polar a ( E ) of A ( E ) is3. a ( E ) = [ cos ( E ) a , sin ( E ) b , − , it meets k at4. B ( E ) = ( a sin ( E ) , b ( q − acos ( E )) , qsin ( E )) . the condition that A i − × A i +1 passes through B ( E ( t )) gives5. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a sin ( E ( t )) b ( q − acos ( E ( t ))) qsin ( E ( t )) acos ( E ( t − h )) bsin ( E ( t − h )) 1 acos ( E ( t + h )) bsin ( E ( t + h )) 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Let6. s ( t ) = ( E ( t + h ) + E ( t − h )) , d ( t ) = ( E ( t + h ) − E ( t − h )) , then to the second order in h, with k = h ,s = E + kD E, d = hDE, cos ( s ) = cos ( E ) − ksin ( E ) D E, sin ( s ) = sin ( E ) + kcos ( E ) D E, cos ( d ) = 1 − k ( DE ) . Replacing the determinant by that obtained by using instead of the last 2 lines their halfsum and their half difference, gives after division of the first row and first column by a andthe second column by b, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) sin ( E ) r − cos ( E ) rsin ( E ) cos ( s ) cos ( d ) sin ( s ) cos ( d ) 1 − sin ( s ) sin ( d ) cos ( s ) sin ( d ) 0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 . Dividing the last row by sin ( d ) and expanding with respect to the first row gives, afterchanging sign,8. sin ( E ) cos ( s ) + ( r − cos ( E )) sin ( s ) − rsin ( E ) cos ( d ) = 0 , or, after using 7 and dividing by k, 9. (1 − rcos ( E )) D E = rsin ( E )( DE ) , integrating gives 1, with some appropriate constant C, 10. (1 − ecos ( E )) DE = C. This is tantalizing close to Kepler’s equation. Theorem. The harmonic polygonal motion associated to the point of Lemoine ( r a, , is described on theellipse by the differential equation C DE = 1 − r cos ( E ) . Proof: We have to show that if we take the derivative of the relation between 3 pointsequidistant in time, namely 10.1.1.8, this derivative is 0 if the differential equation 0. issatisfied. We can assume that C = 1 . From 0 and from 10.1.1.6. follows1. Ds = 1 − r ( cos ( E ( t + h )) + cos ( E ( t − h )))= 1 − rcos ( s ) cos ( d ) , Dd = − r ( cos ( E ( t + h )) − cos ( E ( t − h )))= rsin ( s ) sin ( d ) . Taking the derivative of 10.1.1.8, gives cos ( s ) cos ( E )(1 − rcos ( E )) − sin ( s ) sin ( E )(1 − rcos ( s ) cos ( d ))+ sin ( s ) sin ( E )(1 − rcos ( E )) + cos ( s )( r − cos ( E ))(1 − rcos ( s ) cos ( d )) − rcos ( d ) cos ( E )(1 − rcos ( E )) + r sin ( s ) sin ( E ) sin ( d ) . We would like to prove that this expression is identically zero. 0, gives3. rcos ( d ) = cos ( s ) + sin ( s ) r − cos ( E ) sin ( E ) , substituting in the expression gives cos ( s ) cos ( E )(1 − rcos ( E )) − rsin ( s ) sin ( E ) cos ( E ) + cos ( s )( r − cos ( E ))+ r sin ( s ) sin ( E )+( sin ( s ) cos ( s ) sin ( E ) − rcos ( s ) + cos ( s ) cos ( E ) − cos ( E )+ rcos ( E ))( cos ( s ) + sin ( s ) r − cos ( E ) sin ( E ) ) − sin ( s ) sin ( E )( cos ( s ) + sin ( s )( r − cos ( E ))) /sin ( E )) . The coefficient of r is sin ( s ) sin ( E ) + ( cos ( E ) − cos ( s )) sin ( s ) sin ( E ) − sin ( s ) sin ( E ) = sin ( s )( sin ( E )+ cos ( E ) − cos ( s ) − sin ( s ) sin ( E ) = 0 . The coefficient of r is − cos ( s ) cos ( E ) − sin ( s ) sin ( E ) cos ( E ) + cos ( s )+( cos ( s ) − sin ( s ) cos ( E ) sin ( E ) ( cos ( E ) − cos ( s ))+ sin ( s )( sin ( s ) cos ( s ) sin ( E ) + cos ( s ) cos ( E ) − cos ( E )) /sin ( E ) − sin ( s ) sin ( E )( cos ( s ) − sin ( s ) cos ( E ) /sin ( E )) sin ( s ) /sin ( E )= sin ( s )( − cos ( E ) + cos ( E ) cos ( s ) − cos ( E ) + sin ( s ) cos ( s ) sin ( E )+ cos ( s ) cos ( E ) − cos ( s ) sin ( s ) sin ( E ) + 2 sin ( s ) cos ( E )) /sin ( E ) − cos ( s ) cos ( E ) − sin ( s ) sin ( E ) cos ( E ) + cos ( s ) + cos ( s ) cos ( E ) − cos ( s )= sin ( s )( cos ( E )( − cos ( E ) + cos ( s ) − cos ( s ) + 2 sin ( s )) /sin ( E ) − sin ( s ) sin ( E ) cos ( E ) − sin ( s ) sin ( E ) cos ( E ) + cos ( s ) sin ( s )= sin ( s ) cos ( E ) sin ( E ) − sin ( s ) cos ( s ) − sin ( s ) sin ( E ) cos ( E ) + cos ( s ) sin ( s ) = 0 . The term independent of r is cos ( s ) cos ( E ) − cos ( s ) cos ( E )+( sin ( s ) cos ( s ) sin ( E ) + cos ( s ) cos ( E ) − cos ( E ))( cos ( s ) − sin ( s ) cos ( E ) /sin ( E ) − sin ( s ) sin ( E )( cos ( s ) − sin ( s ) cos ( E ) /sin ( E )) CHAPTER 10. DIFFERENTIAL EQUATIONS AND FINITE MECHANICS = ( cos ( s ) − sin ( s ) cos ( E ) /sin ( E ))( sin ( s ) cos ( s ) sin ( E ) + cos ( s ) cos ( E ) − cos ( E ) − sin ( s ) cos ( s ) sin ( E )+ sin ( s ) cos ( E )) = 0 . Theorem. Let e (cid:48) = 1 − e , then e (cid:48) tan ( e (cid:48) M ) = (1 + e ) tan ( E ) . Theorem. If t ( M ) = tan ( E ) and k = e +1 e − , t (cid:48) i = √− kt ( M i ), then t (cid:48) = t (cid:48) + t (cid:48) − t (cid:48) t (cid:48) . Example. For p = 13 and e = 2 , k = 3 and M t ( M ) 0 1 7 11 8 4 ∞ t (cid:48) M − − ∞ − − − E p = 13 and e = 3 , k = 2 and M t ( M ) 0 2 12 4 8 7 3 ∞ 10 6 5 9 1 11 E Programs. The programs pl.bas and planet.bas . . . Exercise. Prove that the acceleration is − r cosEC ( − ( a cosE, b sinE, 0) + r ( a cos (2 E ) , b sin (2 E ) , Introduction. The parabola has been studied in g33. Galileo Galilei was the first to show that the motion of aparticle in a uniform gravitional field is a parabola. (Love, p.45) The result extend to the finitecase. Theorem. In both the infinite and finite cases, the solution of mD x = 0 and mD y = − mg is x ( t ) = v t, y ( t ) = − gt + v t, or y = ax + bx, with a := − g v , b := v v . Proof: Comparing the equation in the form( x + b a ) = ya + ( b a ) . with the standard equation y = 4 cx shows that the vertex V and the directrix d are V = ( b a , ( b a ) ) ,d : y = v + v g = v g corresponding to the Torricelli law. Example. For p = 7 , g = 1 and v = v = 4 , then a = − b = 1, y ( x ) = − x + x = 1 − x − ,x − − − y − − − − z t − − − Introduction. I have made many attempts to generalize Kepler’s equation or the simple planetary motion tothe finite case. In section . . . , I examine the use of p -adic function to obtain a solution in theneighbourhood of a circular motion. Definition. The circular motion is defined by x ( t ) = cos ( t ) , y ( t ) = sin ( t ) ,Dx ( t ) = − sin ( t ) , Dy ( t ) = cos ( t ) . This assumes that the unit of distance is chosen as the radius of the circle and the unit of time ischosen in such a way that the period is 2 π . CHAPTER 10. DIFFERENTIAL EQUATIONS AND FINITE MECHANICS To approximate the solution of differential equations it is important to insure that essential proper-ties are preserved. In particular, for conservative systems, the same should hold. In this connection,I developed in 1956 a method of first order and a method of second order which are contact trans-formations and therefore preserve the essential properties of conservative systems. These will beapplied to the finite case. Algorithm. The first order algorithm is defined by Theorem.Algorithm. The second order algorithm for the solution of the differential equation D x = f ◦ x , x (0) = x , Dx (0) = Dx , is defined by x i +1 = x i + h Dx i + h f i , Dx i +1 = Dx i + h ( f i +1 + f i ) , where f i := f ( x i ) . Definition. A mapping is reversible iff Theorem. Given the Algorithm 10.2.1, the mapping is reversible. Proof: If we solve for x i and Dx i , we get Dx i = Dx i +1 − h ( f i +1 + f i ) , x i = x i +1 − h Dx i − h f i , = x i +1 − h Dx i +1 + h ( f i + f i +1 ) . Definition. A mapping is symplectic iff Theorem. The mapping defined in algorithm 10.2.1 is symplectic. Proof: Example. Let x and f be one dimensional, let f ( x ) = − x − x , let Dx = 0 , we have the following solutions, for h = 1 and various initial conditions ( x (0) , Dx (0) =0). p = 11 ,i x, Dx ) i , , − , − , − , − , − , , , , − , − , − , − , − − , − , − , , , , − , − , − − , − − , − − , − , − , p = 13 ,i x, Dx ) i , , − , , , , − , − , − , , , − , − , − , − , − , , , − , − , , , , − , − , , − , , − , , − − , − , Theorem. If we apply the mapping 10.2.1 to D x = − x, x (0) = 0 , Dx (0) = 1 , we obtain, up to a scaling factor the trigonometric functions. Example. With p = 13 and h = 1 ,i x, Dx ) i , , − , − − , − − , , , , − δx, Dx )(( i ) = ( sin, cos )( − i ) , if sin (1) = 3 and cos (1) = 3 δ, with δ = 2 . Program. [130] PENDUL(um) The parabola has been studied in g33. Galileo Galilei was the first to show that the motion of aparticle in a uniform gravitional field is a parabola. (Love, p.45) The result extend to the finitecase. CHAPTER 10. DIFFERENTIAL EQUATIONS AND FINITE MECHANICS Theorem. In both the infinite and finite cases, the solution of mD x = 0 and mD y = − mg is x ( t ) = v t, y ( t ) = − gt + v t, or y ( x ) = ax + bx, with a := g v , b := v v . Proof: Comparing the equation in the form( x − b a ) ) = − ( y − b a ) ) a with the standard equation y = 4 cx shows that the vertex V and the directrix d are V = ( b a ) , b a ) ) ,d : y = v + v g = v g ) corresponding to the Torricelli law. Example. For p = 7 , g = 1 and v v , y(x) = − x + x, x − − − y − − − − z t − − − Introduction. I have made many attempts to generalize Kepler’s equation or the simple planetary motion tothe finite case. In section . . . , I examine the use of p -adic function to obtain a solution in theneighbourhood of a circular motion. Definition. The circular motion is defined by x ( t ) = cos ( t ) , y ( t ) = sin ( t ) ,Dx ( t ) = − sin ( t ) , Dy ( t ) = cos ( t ) . This assumes that the unit of distance is chosen as the radius of the circle and the unit of time ischosen in such a way that the period is 2 π . Introduction. To approximate the solution of differential equations it is important to insure that essential proper-ties are preserved. In particular, for conservative systems, the same should hold. In this connection,I developed in 1956 a method of first order and a method of second order which are contact trans-formations and therefore preserve the essential properties of conservative systems. These will beapplied to the finite case. Algorithm. The first order algorithm is defined by Theorem.Algorithm. The second order algorithm for the solution of the differential equation D x = f ◦ x , x (0) = x , Dx (0) = Dx , is defined by x i +1 = x i + h Dx i + h f i , Dx i +1 = Dx i + h ( f i +1 + f i ) , where f i := f ( x i ) . Definition. A mapping is reversible iff Theorem. Given the Algorithm 4.1.3., the mapping is reversible. Proof: If we solve for x i and Dx i , we get Dx i = Dx i +1 − h ( f i +1 + f i ) , x i = x i +1 − h Dx i − h f i , = x i +1 − h Dx i +1 + h ( f i + f i +1 ) . Definition. A mapping is iff Theorem. The mapping defined in algorithm 4.1.3. is Proof: CHAPTER 10. DIFFERENTIAL EQUATIONS AND FINITE MECHANICS Example. Let x and f be one dimensional, let f ( x ) = − x − x , let Dx = 0 , we have the folowing solutions p = 11 ,i x, Dx ) i , , − , − , − , − , − , , , − , − , − , − − , − , − , − , , , − , − , − − , − − , − − , − , − , , , , p = 13 ,i x, Dx ) i , , − , , , − , − , − , , − , − , − , − , − , , − , − , , , , − , − , − , − , − , − , − − , − − , − , − , − , , − , , − − , − Theorem. If we apply the mapping 0.3. to D x = − x, x (0) = 0 , Dx (0) = 1 , we obtain, up to a scaling factor the trigonometric functions. Example. With p = 13 and h = 1 ,i x, Dx ) i , , − , − − , − − , , , , − δx, Dx )(( i ) = ( sin, cos )( − i ) , if sin (1) = 3 and cos (1) = 3 δ, with δ = 2 . Program. [130] PENDUL(um) Introduction. I will first give a non constructive proof of the existence of primitive roots and the give a construc-tion. The first proof insures that the construction is always successful. Theorem. d = ord p ( x ) , ( d, p ) = g, < l < d ⇒ ord p ( x l ) = dg , d = ord p ( x ) , ≤ i, j < d, x i ≡ x j (mod p ) ⇒ i = j. If d | p − then x d ≡ p ) has φ ( d ) solutions of order d. Hint 2.25.3. In particular, there are φ ( p − primitive roots of p. d = ord p ( z ) , e = ord p ( y ) , ( d, e ) = 1 ⇒ ord p ( z.y ) = d.e. What follows is a Theorem which gives a constructive method ofdetermining primitive roots or more generally of solutions of d = ord ( x ) , where d | p − . The construction is inspired by Gauss, 1801, section 55. Theorem. Let Π nj =1 p i j j be a prime factorization of q − . Let a ( q − j p j − (cid:54) = 1 (mod q ) and a ( q − j ≡ q ) , for j = 1 , , . . . n, then p kj = ord q ( a q − Pj j ) , ≤ k j ≤ i j . Let P j = p i j ,j let h j ≡ a q − Pj j (mod q ) , then, in particular, p j = ord q ( h j ) . Let h k j j = a q − ikjj j (mod q ) , ≤ k j < i j , then 2. Π nj =1 h k j j = ord q (Π nj =1 h k j j ) . Let h = Π nj =1 h j (mod q ) , then, in particular, q − ord q ( h ) , q is prime, h is a primitive root of q. CHAPTER 10. DIFFERENTIAL EQUATIONS AND FINITE MECHANICS hapter 11COMPUTER IMPLEMENTATION One of the tradition of Mathematicians is to discover properties by working on special cases or ex-amples, this is especially so at the beginning of many branches of Mathematics, geometry, numbertheory, algebra, . . . . This was certainly the tradition kept up by Euler, see . . . , by Gauss, see . . . .In Euclidean geometry, the special cases were obtained by drawing a reasonably accurate figure,in number theory by numerical computation, and in algebra by algebraic manipulations. All threecan now be done accurately and with great speed using computers and these are now becomingmore and more available to every one.Depending on our training or, I believe, on the structure of our individual brain, such experimen-tation is almost essentail for many to obtain a thourough understanding of basic concepts.To help in the understanding of the material given above and, I hope, to help the reader in thediscovery of new properties, it is becoming essentail to provide him with the tools to realize quicklycomputer programs.When the subject matter is well settled and the experimentation is not at the basic level, a higherlevel non interactive language such as FORTRAN, ALGOL, PASCAL, PL1, ADA, is an excel-lent choice. When this is not the case, an interactive language such as BASIC or APL is by farpreferable. BASIC, BASIC+, BASIC+ extended.Hardware, operating system, files, interaction, language, compiler, interpreter. CHAPTER 11. COMPUTER IMPLEMENTATION EFERENCES 1. 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