An Algebraic Construction of Hyperbolic Planes over a Euclidean Ordered Field
11 To my daughters Vianne and Natalie AN ALGEBRAIC CONSTRUCTION OF HYPERBOLIC PLANES OVER A EUCLIDEAN ORDERED FIELD
Nicholas Phat Nguyen Abstract. Using concepts and techniques of bilinear algebra, we construct hyperbolic planes over a euclidean ordered field that satisfy all the Hilbert axioms of incidence, order and congruence for a basic plane geometry, but for which the hyperbolic version of the parallel axiom holds rather than the classical Euclidean parallel postulate. 1. INTRODUCTION. Most readers are probably familiar with the Poincare model of a hyperbolic plane, based on either the upper half-plane or the interior of the unit disc in the standard Euclidian plane (which can also be thought of as the space of complex numbers). Such a hyperbolic plane is a 2-dimensional real manifold (or a 1-dimensional complex manifold) where a suitable concept of lines could be defined so that the hyperbolic version of the Euclidean parallel axiom holds: for any point not on a given line, there are two or more lines (in fact an infinite number of lines) passing through that point and not intersecting with the given line. We can construct a hyperbolic plane using a similar Poincare model applied to an affine plane over an euclidean ordered field. See [1] at chapter 7. Such a construction gives us a model that satisfies all the basic axioms of incidence, order and congruence for lines, Email: [email protected] segments and angles for a basic Hilbert plane geometry, but for which the hyperbolic version of the parallel axiom holds rather than the classical Euclidean parallel postulate. In the following, we will outline an alternative construction of such a Poincare model over any euclidean ordered field using methods from bilinear algebra (i.e. the study of symmetric bilinear forms or quadratic forms over a field). Such an algebraic construction provides new methods that could help in the exploration of general hyperbolic geometry. The construction also does not rely on any visual verification for its definitions or its proofs, and gives us more precise information, such as a natural parametrization of the set of lines through a point that are not parallel to a given line. Recall that an ordered field is a field endowed with a total ordering relation compatible with the additive and multiplicative operations of the field. An ordered field could be defined by giving a total ordering ≤ that satisfies the following basic propertie s associated with the usual ordering on the real line: • if a ≤ b then a + c ≤ b + c for any c • if 0 ≤ a and 0 ≤ b then 0 ≤ a b An ordered field is said to be euclidean if every positive number is a square. Examples of euclidean fields include the field of real numbers R, the field of all algebraic numbers contained in R, and the field of all real constructive numbers (real numbers that can be constructed from the rational numbers using ruler and compass constructions). Given any ordered field, there are always ordered extensions of that field which are euclidian. So there are infinitely many euclidean ordered fields beyond the subfields of R cited in our examples above. Let K be an ordered field, and consider the n-dimensional vector space K n of all n-tuples with coefficients in K. The standard dot product x.y = x y + … + x n y n is a symmetric bilinear form on K n . This symmetric bilinear form is non-degenerate, in fact anisotropic in the sense that x.x is never zero unless the vector x itself is zero. The fact that the standard dot product is anisotropic for any K n is a characteristic property of ordered field, as discovered by Emil Artin and Otto Schreier (a field can be ordered if and only if it is formally real). In the following, we will start out with a vector space E of dimension 2 over a euclidean ordered field K. We will assume that E has a symmetric bilinear form (x | y) that is positive definite, meaning that the product (x | x) is always > 0 unless x is the zero vector. Such a positive definite form is obviously anisotropic. The standard dot product on K is a prime example. Accordingly, we will use the dot product notation for the bilinear form on E to emphasize the analogy. Before we begin, we note here a potential difference in terminology. A regular quadratic space of dimension 2 is either anisotropic or isometric to a standard space known as an Artinian or hyperbolic plane which has two linearly independent lines of isotropic vectors. Relative to a basis formed of the two linearly independent isotropic vectors, the matrix of the bilinear form is a 2 by 2 symmetric matrix that has zeros in the diagonal and a non-zero number in the cross diagonal. The literature on bilinear algebra more commonly refers to such a space as a hyperbolic plane, but in order to avoid confusion, we will refer to such a space as an Artinian plane, which is another name that is sometimes used for it. The late geometer Marcel Berger championed the name “Artinian plane” instead of “hyperbolic plane.” He regarded the term “hyperbolic plane” often used in bilinear algebra as undesirable, because the term hyperbolic plane can of course also mean a hyperbolic manifold of dimension 2 or a space that satisfies the axioms of hyperbolic plane geometry. See [5] at paragraph 13.1.4.4. 2. THE SPACE OF CYCLES AND THE CYCLE PAIRING PRODUCT. Consider the set 𝓕 of all functions p from E to K of the form p(X) = aX.X + b.X + c, where X and b are vectors in E, and a and c are elements in the field K. The set 𝓕 can naturally be endowed with the If we regard the coefficients of the vector X with respect to a fixed basis of E as variables, the function p is at most a second-degree polynomial in 2 variables. Because the ordered field K has structure of a K-vector space of dimension 2 + 2 = 4, being parametrized by the vector b and the elements a and c. We will refer to such a function p (if it is not the constant zero function) as a 2-cycle, 1-cycle, or 0-cycle depending on whether the degree of p is 2 (coefficient a is nonzero), 1 (a is zero but b is nonzero, or 0 (both a and b are zero). Aside from the natural structure of a vector space over K, we can also endow 𝓕 with a symmetric bilinear form < _ , _ > as follows. Given p = aX.X + b.X + c and p* = a*X.X + b*.X + c*, we define
as b.b* – – 𝓕 with this scalar product is isometric to the orthogonal sum of E and an Artinian plane. We will refer to this fundamental scalar product on 𝓕 as the cycle pairing or cycle product. The reader can easily verify the following properties from the formula for the cycle pairing product. • All 0-cycles are isotropic. • No 1-cycle is isotropic. • A 2-cycle p(X) = aX.X + b.X + c is isotropic if and only if b.b - 4ac = 0, or equivalently, if p(X) = a(X + b/2a).(X + b/2a). We will informally refer to such an isotropic 2-cycle as a zero circle centered at the point ( – b/2a), in analogy with the familiar circle equation in the Euclidean plane R . For each point u in the plane E, we write q(u) = X.X – characteristic 0 and therefore an infinite number of elements, such a function p is a zero function if and only if a, b, and c are all zero. simplify notation, we will assume that p has unit norm, which is possible by scaling because the field K is euclidean. We define 𝓗 to be the subset of E consisting of all points u in E such that
= – 𝓗 as follows. Any such line is a nonempty subset of 𝓗 consisting of all points u in 𝓗 such that
2
p is a normalized zero circle whose cycle product with p has the opposite sign and whose cycle product with m is also zero. Therefore, the zero points of m in E not orthogonal to p can be divided into two subsets of equal size. That means the hyperbolic line defined by m must have an infinite number of points. ■ Note that if a nonisotropic cycle m defines a line in 𝓗 , then any multiple µm of m ( µ ≠0) defines the same line. The complement of such a line in 𝓗 can naturally be divided into two sides, each side consisting all points u such that = a 4. THE HYPERBOLIC PLANE – ORDER AXIOMS. We need to define what it means for a point to lie between two other points on the same line. Let u, v and w be three distinct collinear points in 𝓗 . The fact that these three points are collinear means first of all that the four cycles p, q(u), q(v) and q(w) are not linearly independent. Otherwise they would generate the entire space 𝓕 of cocycles, and in that case there would be no nonisotropic cycle orthogonal to all of them. We claim that q(u), q(v) and q(w) must be linearly independent, because otherwise we would have an equation aq(u) + bq(v) + cq(w) = 0 with abc ≠ 0. Two of the three coefficient numbers must have the same sign. Let’s say a and b have the same sign. If we now take cycle product of both sides of the equation with q(w), the right hand side is obviously zero. On the left hand side we have a + 2yz is strictly positive, then t is a point in 𝓗 , and we are done. If not, then the reflected 2-cycle q(t) – 𝓗 lying on both M and L, and also lying between r and s. ■ Given a point u on a line L, we can divide the complement of u in L into two sides as follows. Let 𝓁 be a nonisotropic cycle defining the line L. The cycles p, q(u) and 𝓁 generate a regular subspace of dimension 3 in 𝓕 . The orthogonal complement of that subspace is a one-dimensional subspace generated by a nonisotropic cycle n. For any point v ≠ u on L, we must have must be > 0, and hence it is necessary and sufficient that λ > 0. We will call any orthogonal transformation T a proper transformation if it fixes p and has the property that for any u in 𝓗 , we have T(q(u)) = λ q(v) for some λ > 0. Such a proper transformation clearly induces a bijection of to itself, and gives us a transformation of 𝓗 that maps lines to lines and preserves the betweenness relationship. Any orthogonal transformation of 𝓕 can be written as a product of a finite number of basic transformations called reflections. (This is known as the Cartan – Dieudonne theorem.) Each nonisotropic cycle of 𝓕 defines such a reflection as follows. Let x be a nonisotropic cycle of 𝓕 , then by orthogonal decomposition each cycle v of 𝓕 can be expressed uniquely as a sum v = ax + y, where y is a cycle in 𝓕 orthogonal to x. The reflection defined by x maps v to v’ = – ax + y. In other words, the reflection leaves invariant the orthogonal complement of x and maps x to – x. Such a reflection is also called a hyperplane reflection along x or across the orthogonal complement of x. Note that all nonzero scalar multiples of the same nonisotropic cycle define the same reflection. It follows that any transformation T leaving the cycle p invariant can be expressed as a product of reflections defined by nonisotropic cycles that are orthogonal to p. Proposition 2: Any reflection defined by a nonisotropic cycle that is orthogonal to p and that has positive norm is a proper transformation. Proof. Let t = aX.X + b.X + c be a cycle orthogonal to p such that its norm 1, 1], whose orthogonal complement is generated by a nonisotropic cycle n of positive norm. We can assume by scaling that n has norm 1. The reflection defined by n induces a congruence transformation of order 2 that maps L into L and fixes the point u. Moreover, for any other point v on L, this transformation maps v to w such that for some positive coefficient λ , we have λ q(w) = q(v) – λ q(w)> = – 6. THE HYPERBOLIC PLANE – PARALLEL AXIOM. We can now consider the parallel axiom for our hyperbolic plane. (Parallel Axiom – Hyperbolic Version) Let L be a line in 𝓗 and let u be a point not on L. There are two or more lines passing through u that do not intersect with L. We will show that there are in fact infinitely many lines passing through the point u that do not intersect with the line L. Lines that do not intersect are also said to be parallel. In general, let 𝓁 and 𝓂 be nonisotropic cycles that define two distinct lines in 𝓗 . We claim that the lines defined by 𝓁 and 𝓂 do not intersect if and only if the 2-dimensional cycle subspace generated by 𝓁 and 𝓂 is isotropic. Indeed, we will show that the lines defined by 𝓁 and 𝓂 intersect if and only if the space generated by 𝓁 and 𝓂 is anisotropic, or equivalently, that its orthogonal complement is an Artinian plane. It is clear that if these lines intersect at a point w, then q(w) is cycle orthogonal to both 𝓁 and 𝓂 , and hence belongs to their orthogonal complement. That orthogonal complement has dimension 2 and also includes p. The orthogonal complement is therefore generated by p and q(w). It is regular and isotropic, and hence is an Artinian plane. Now consider the converse. We want to show that if the orthogonal complement to 𝓁 and 𝓂 is an Artinian plane, the two lines must intersect in 𝓗 . Such an Artinian plane must contain two isotropic lines not orthogonal to p. These isotropic lines therefore are multiples of the normalized 2-cycles q(v) and q(w) for two points v and w in E. It is straight-forward to verify that we must have q(w) = q(v) – 𝓗 . That point is the intersection of the two lines defined by 𝓁 and 𝓂 . Our claim is now proved. Let D be the orthogonal complement of p in the space 𝓕 of all cycles. D is a space of dimension 3, isometric to the quadratic space [1, 1, – complement in 𝓕 of the Artinian plane generated by p and q(u). The nonzero vectors in that subspace define lines that pass through the point u, with scalar multiples of the same vector representing the same line. Now look at the projective plane P(D) associated to D. In that projective plane, the lines passing through the point u correspond to points on the projective line 𝒰 defined by the 2-dimensisonal subspace of cycles orthogonal to q(u). Let 𝓁 be the cycle that defines the given line L in 𝓗 , and let the point Λ be the image of 𝓁 in P(D). If a cycle 𝓂 defines a line passing through u, its image in P(D) is a point μ on the line 𝒰 , and the projective line joining Λ and μ represents the 2-dimensional subspace generated by 𝓁 and 𝓂 . Such a subspace is isotropic if and only if the line joining Λ and μ intersects the projective conic defined by the cycle pairing in D. That cycle pairing conic is isomorphic to the projective conic X + Y – Z = 0, and therefore has an infinite number of points. For each point on the conic, consider the line joining Λ and that point. That line will intersect the line 𝒰 in a point because any two lines in a projective plane intersect. In light of our foregoing discussion, that point represents a line through u that is parallel to L. Because the point Λ is not on the conic, the lines joining Λ with points on 𝒰 and intersecting with the cycle pairing conic would correspond roughly to half of that conic. Except for the two tangents to the conic drawn from Λ , each line through Λ that intersects with the conic will do so in exactly two distinct points. (In terms of bilinear algebra, such a line represents an Artinian plane, and therefore has exactly two isotropic points.) Accordingly, there are an infinite number of lines through u that do not intersect with L. ■ . REFERENCES Hartshorne, R. (2000). Geometry: Euclid and Beyond. New York: Springer-Verlag. 2. Lang, S. (2002). Algebra. Revised Third Edition, Springer Graduate Texts in Mathematics Vol. 211. Berlin, Heidelberg, New York and Tokyo: Springer-Verlag. Scharlau, W. (1985). Quadratic and Hermitian Forms. Grundlehren der Mathematischen Wissenschaften, vol. 270. Berlin, Heidelberg, New York and Tokyo: Springer-Verlag. 4.. The left hand side is > 0 by definition of 𝓗 . For the right hand side, note that
= 4u.v – – – – v).(u – v) is < 0 by hypothesis. So the right hand side a
is < 0. This would be a contradiction, so the cycles p, q(u) and q(v) must be linearly independent. The subspace of 𝓕 generated by the cycles p, q(u) and q(v) therefore has dimension 3. Moreover we can readily check that the cycle pairing on this subspace has discriminant < 0, which means the cycle pairing on this subspace is regular. Therefore the orthogonal complement of this subspace in 𝓕 is a one-dimensional subspace generated by a nonisotropic cycle m. By construction m is orthogonal to p, q(u) and q(v), and therefore gives us a line in 𝓗 that contains u and v. This line is unique, since any such line by definition must correspond to a cycle that is orthogonal to p, q(u) and q(v), and therefore that cycle must be proportional to the cycle m. Proportional cycles obviously define the same line. ■ (Incidence Axiom 2) Every line contains at least two points. Proposition 1 tells us that every line in fact contains an infinite number of points. ■ (Incidence Axiom 3) There are at least three points that are not collinear. We can chose a basis s and t in E such that i.s and i.t are < 0. For any point u in 𝓗 , the points u + s, u + t are also in 𝓗 . Moreover, the points u, u + s, and u + t are not all collinear. Otherwise we would have an nonisotropic cycle m that is orthogonal to p, q(u), q(u + s) and q(u + t). But we can check by straight-forward computations that these four cycles generate the entire space 𝓕 of all cycles. The cycle pairing is nondegenerate, so the only vector in 𝓕 cycle orthogonal to all these four cycles is the zero vector. ■
+ b
≠ 0 because it is the sum of two numbers of the same sign. (Recall that for any two different points x and y in E,
= – – y).(x – y) is < 0.) Accordingly, the linear dependence of the four cycles p, q(u), q(v) and q(w) implies that we must have an equation p = aq(u) + bq(v) + cq(w) = 0 with abc ≠ 0. We claim that the three coefficient numbers cannot all have the same sign. Otherwise, by taking cycle product of both sides with p, we deduce that all three numbers must be positive. On the other hand, by taking the cycle product of both sides with q(w), we see that the left hand side is strictly positive (by definition of the set 𝓗 ) while the right hand side is strictly negative because
and
are both < 0. Therefore exactly two of the three coefficients have the same sign. The remaining coefficient has a different sign, and we say that the point corresponding to that coefficient lies between the other two points. Our definition immediately implies the following two axioms: (Order Axiom 1) For three collinear points u, v and w, if w is between u and v, then w is also between v and u. (Order Axiom 2) For any three collinear points u, v and w, exactly one point is between the other two points. We now consider the following more difficult axioms. (Order Axiom 3) For any two points s and t, we can find at least three points u, v and w on the line passing through s and t such that s is between u and t, v is between s and t, and t is between s and w. We will later prove that for any two lines L and M in 𝓗 , there is a transformation of that maps the line L bijectively onto the line M and respecting the betweenness relationship of any three points. Assuming this, it is enough for us to show that Order Axiom 3 applies to one specific line. Consider the line defined by the 1-cycle j.X, where j is a nonzero vector in E such that i.j = 0. This line consists of all points in E of the form – > 0. On this line, there is a natural betweenness induced by the given ordering of K. That natural betweenness clearly satisfies Order Axiom 3. We will show that the natural betweenness relationship on this particular line is the same as the betweenness relationship defined above. Let u, v and w be three distinct points on this line. Under our definition, to say that v lies between u and w means there is an equation p = xq(u) + yq(v) + zq(w) where x, y and z are nonzero numbers such that x and z have the same sign. Let u = – – – – – – – – – x + B y + C z = 0 From linear algebra, we know that up to a common factor, x and z are proportional to (C – B ) and (B – A ). That means the numbers x and z have the same sign if and only if we have (C > B > A ) or (C < B < A ). But these conditions on the numbers A, B and C mean that v lies between u and w in the natural ordering on that line. ■ (Order Axiom 4 – Pasch ’ s Axiom) Let u, v, w be three distinct points in 𝓗 , and let M be a line in 𝓗 that does not pass through any of these three points. If M passes through a point between u and v, then M also passes through a point either between u and w, or between v and w. Let m be a nonisotropic cycle defining the line M. We claim that in order for the line M to pass through a point between r and s, it is necessary and sufficient that the numbers
= 0. Such a monic quadratic equation must have a root x in the euclidean field K because the constant term 2yz
is strictly negative. Indeed, the number yz is > 0 because y and z are nonzero numbers of the same sign, and we know from earlier computation that
is a strictly negative number for any two points r and s in 𝓗 . The 2-cycle xp + yq(r) + zq(s) therefore is isotropic and must be equal to α q(t) for some vector t in E and some nonzero coefficient α . By construction q(t) is cycle orthogonal to both 𝓁 and m. If
/(
.
) It is clear that d(u, v) = d(v, u) ≥ 0 and d(u, v) = 0 if and only if u = v. We will refer to the function d(u, v) as the quasi-distance of u and v. Let T be a congruence transformation that maps u and v to u’ and v’ respectively. We have T(q(u)) = λq ( u’ ) and T(q(v)) = µq( v’ ). It follows readily that d(u, v) = d( u’’ , v’ ). So the quasi-distance function is invariant under the congruence transformations. Proposition 3: (a) For any two distinct points u and v in 𝓗 , there is a reflection that maps u to v and that maps other points on the line through u and v to points on the same line. (b) For any two lines L and M in 𝓗 , there is a reflection that maps L to M. (c) Let u be a point on the line L and v be a point on the line M, there is a congruence transformation that maps L to M and u to v. (d) Let R = [uv, ∞) and S = [xy, ∞) be two rays in 𝓗 . There is a congruence transformation mapping u to x and the ray R to the ray S. Proof. Let λ be the number > 0 such that
= < λq (v), p>. The cycle r = q(u) – λq (v) has norm – λ
which is > 0. Moreover,
because T is an orthogonal transformation fixing the cycle p. Let 𝓁 be a cycle corresponding to the line L. Any congruence transformation T that fixes the points of L must send 𝓁 to a scalar multiple of itself. Because T is orthogonal, T( 𝓁 ) is either 𝓁 or –𝓁 . In the first case, T is the identity transformation. In the second case, T is the reflection defined by 𝓁 . That reflection maps 𝓁 to –𝓁 , and leaves invariant any cycle orthogonal to 𝓁 . Accordingly, such a reflection fixes all points of L and exchanges the two sides of L, and it is uniquely determined by these two properties. That proves (a). The space generated by p, q(u) and 𝓁 is a regular 3-dimensional subspace isometric to [1, –