A Brief Account of Klein's Icosahedral Extensions
aa r X i v : . [ m a t h . NA ] F e b A Brief Account ofKlein’s Icosahedral Extensions
Leonardo Solanilla, Erick S. Barreto and Viviana Morales ∗ February 2021
Abstract
We present an alternative relatively easy way to understand anddetermine the zeros of a quintic whose Galois group is isomorphicto the group of rotational symmetries of a regular icosahedron. Theextensive algebraic procedures of Klein in his famous
Vorlesungen ¨uberdas Ikosaeder und die Aufl¨osung der Gleichungen vom f¨unften Grade are here shortened via Heymann’s theory of transformations. Also,we give a complete explanation of the so-called icosahedral equationand its solution in terms of Gaussian hypergeometric functions. Asan innovative element, we construct this solution by using algebraictransformations of hypergeometric series. Within this framework, wedevelop a practical algorithm to compute the zeros of the quintic.
Keywords (MSC 2020):
Computational methods for problems pertain-ing to field theory (12-08); Polynomials in real and complex fields: locationof zeros (algebraic theorems) (12D10); Real polynomials: location of zeros(26C10); Zeros of polynomials, rational functions, and other analytic func-tions of one complex variable (30C15); Numerical computation of solutionsto single equations (65H05). ∗ Departamento de Matem´aticas y Estad´ıstica, Universidad del Tolima, BarrioSanta Elena, Ibagu´e, Tolima, Colombia; [email protected], [email protected],[email protected] Introduction
Let f ∈ K [ X ] be an irreducible quintic with coefficients in a field K ≤ C .Let L be the splitting field of f , K ≤ L ≤ C , and let Gal( K ≤ L ) denote theGalois group of L over K . It is a well-known fact that f is solvable by radicalsif and only if Gal( K ≤ L ) is a subgroup of the Frobenius group F , i.e. ifand only if Gal( K ≤ L ) is isomorphic (up to conjugacy) to F of order 20,to the dihedral group D of order 10, or to the cyclic group Z / Z . Since theGalois group of certain quintics is A (isomorphic to the group of symmetriesof a regular icosahedron) and this group is not solvable, the solutions to thistype of quintic cannot be expressed by field operations and radicals from thepolynomial coefficients. We will say that L is an icosahedral extension of K if Gal( K ≤ L ) = A . We assume the reader already knows the basics ofGalois Theory and quintics, as they are given e.g. in Cox [5] and the earliestworks of Abel [1] and Ruffini [12].In this paper we give a method to solve any quintic whose splitting field isicosahedral. Furthermore, we also provide a numerical implementation of themethod. Our approach is based on Klein’s [10] foundational paper, as wellas on the more recent renditions of Heymann [8], Shurman [13] and Nash[11]. For the numerical part, we have been inspired by Cox, Little & O’Shea[6] and Trott [15]. Some incomplete sketchy elementary ideas on the subjectcan be found in [2] and [3].Tschirnhaus transformations [16] are key ingredients of Klein’s method. Asusual, given a general quintic x + c x + c x + c x + c x + c , the firstdegree transformation z = x + ( c /
5) eliminates the term of 4th degree andreduces the problem to a quintic z + pz + qz + rz + s . After a furtherquadratic transformation y = z − az − b , we obtain the so-called principalquintic y + 5 αy + 5 βy + γ = 0 . (1)The quadratic transformation extends the coefficient field K to K ( δ ), where δ = 9 q p + 3 p − rp . For simplicity’s sake, we will denote this new field bythe same letter K .From a purely field-theoretical approach, our main task consists of proving1he following theorem, that we will call Klein’s theorem. C. f.
Slodowy [14].
Theorem 1.
Let K be a subfield of the complex numbers containing δ andthe fifth roots of unity. We suppose that L is a Galois extension of K with Gal( K ≤ L ) = A , the alternating group of five letters. Then, there exists a J ∈ K such that the solution Y of the icosahedral equation q ( Y ) = J (Section2 below) yields L = K ( Y ) , that is, L is obtained from K by adjoining Y . Notably, along the proof we will develop a practical numerical method forfinding the zeros of our principal quintic.Since Klein’s theorem essentially asserts that the solution of the quintic re-duces to the solution of the icosahedral equation, in Section 2 we introducethis equation together with a method of solution. Gaussian hypergeometricfunctions provide the means to determine such a solution. Our approach iscomprehensive and we provide full details. Although this solution has beendescribed barely in Nash [11], we use a simpler procedure based on the al-gebraic transformations of the hypergeometric series. We devote Section 3to the algebraic treatment of the principal quintic. We have closely followedthe theory of Heymann’s [8] resolvents. They allow us to show the way thesolution of the quintic effectively amounts to the solution of the icosahedralequation. With this, Klein’s theorem is proved. In Section 4 we rewritethe previous sections as a useful algorithm that produces as outputs the fivesolutions of an icosahedral quintic. In the end, we draw some conclusionsregarding the main features of our method.
The solution of a Klein’s quintic relies heavily on the geometry of the icosa-hedron.
Icosahedral symmetry
The icosahedron is the surface of the regular polyhedron or Platonic solidhaving 12 vertices, 30 edges and 20 triangular faces (the ancient Greek num-ber ǫικoσι , e´ıkosi , means twenty). Once it is inscribed in a sphere, we projectradially from the center to obtain the image of the icosahedron on the sphere.Then, we project the circumsphere onto the complex plane via the stereo-graphic projection to get a plane image of the icosahedral surface. As it2s usual, the extended complex plane or Riemann sphere S is convenientlyidentified with the complex projective space P .The icosahedral rotation group I has order 60 and is isomorphic to A , thegroup of the even permutations of five letters. It comprises all sphericaltransformations or rotations under which the icosahedron remains invariant.By virtue of the identification S ≃ P , the elements of I can be understood asM¨obius transformations or, alternatively, as homogeneous forms. As well, A is isomorphic to the projective special linear group PSL(2 , P → P / I . Invariant forms of 60th degree
Klein (1884) describes an inverse of this covering in relation to polynomialsor homogeneous forms. For instance, the vertices of the icosahedron areprecisely the zeros of the invariant polynomial f = f ( z, w ) = zw ( z + 11 z w − w ) . (2)Similarly, the Hessian determinant of f produces the invariant form H = H ( z, w ) = − ( z + w ) + 228( z w − z w ) − z w , (3)which vanishes at the centers of the icosahedron faces. Also, the Jacobiandeterminant of f, H provides the 30th degree invariant T = T ( z, w ) = ( z + w )+522( z w − z w ) − z w + z w ) , (4)vanishing at the midpoints of the edges. We should standardize these in-variant forms in the 60th degree in order to deal with the spherical orbitsunder the action of I . Once we do so, the set O of the invariant forms of60th degree is easily furnished with a structure of vector space of dimensiontwo over C . Among other choices, the forms H and f form a basis for O .Remarkably enough, T = 1728 f − H in this basis.It is not hard to establish an identification between the homogeneous poly-nomials in O with the orbits in P / I , where they vanish.3 he isomorphism P ≃ P / I The canonical projection P → P / I takes a point ( z, w ) to its orbit [ z, w ].It corresponds to a map P → O , taking ( z, w ) to the form P ( z, w ), whichequals zero at ( z, w ). An arbitrary P ∈ O is written P = µH − λf , in thebasis H , f . Then, P = 0 implies H f = λµ . So, the covering can be written ( z, w ) ( λ, µ ). In non-homogeneous coor-dinates we have (cid:0) Y = zw , (cid:1) (cid:16) λµ = J, (cid:17) , or simply Y J .The inversion of the covering reveals interesting facts. First, we must restrictourselves to one of the possible branches or restricted domains. Then, weconstruct the inverse map [ z, w ] ( z, w ) by means of ( H ( z, w ) , f ( z, w )) ( z, w ) and thus J Y . In the subset of the normal orbits [ z, w ] ∈ P / I , it ispossible to choose a unique pair of holomorphic functions ( z, w ). However,the abnormal orbits introduce discontinuities corresponding to the vertices,edge midpoints and face midpoints. The nature of this inverse map is nextelucidated with the aid of Complex Analysis. Differential resolvents
Even more, we can construct an explicit inverse map J Y of the covering.For sure, H and f are connected through variable J and the constant ranktheorem reduces the problem to the equivalent form f ( z, w ) = k,H ( z, w ) = u, where k is constant and u is a complex variable. We recall that the jacobiandeterminant of the map P → P , ( z, w ) ( k, u ), equals − T , where T isthe icosahedral form (4). Also, the derivatives of z, w with respect to u canbe computed implicitly through dzdudwdu ! = − T ∂H∂w − ∂f∂w − ∂H∂z ∂f∂z (cid:18) (cid:19) . ddu (cid:18) ∂f∂z (cid:19) = ∂ f∂z · dzdu + ∂ f∂z∂w · dwdu and ddu (cid:18) ∂f∂w (cid:19) = ∂ f∂w · dwdu + ∂ f∂z∂w · dzdu , this yields the linear ordinary homogeneous equations T d zdu + T dTdu · dzdu + 11 Hz
400 = T d wdu + T dTdu · dwdu + 11 Hw
400 = 0 . (5)By using T = 12 f − H and setting u = H , we obtain(12 f − H ) d zdH − H dzdH + 11 Hz
400 = 0 (6)and a similar equation for w . Then, we suitably recover J = H / f andperform the change of variables H J to get dzdH = 3 JH · dzdJ and d zdH = 9 J H · d zdJ + 2 H · dzdH . The substitution of these derivatives in (6) produces the well-known hyper-geometric equation J (1 − J ) d zdJ +[ c − ( a + b +1) J ] dzdJ − abz = 0 , with a = 1160 , b = − , c = 23 , (7) and a corresponding equation for w . This equation has three regular singu-larities located respectively at J = 0 , , ∞ . They correspond to the branchpoints of the covering: J = 0 means H = 0, J = 1 means T = 0 and J = ∞ means f = 0. A general exposition of the nature of the solutionsto general hypergeometric equations similar to (7) near ordinary points andregular singularities can be found in Copson [4, Chapter X] and Hille [9,Section 6.1].The solutions to (7) can be expressed in terms of the hypergeometric series F ( a, b, c ; J ) = 1 + ∞ X n =1 ( a ) n ( b ) n (1) n ( c ) n J n , where ( p ) n = p ( p + 1) · · · ( p + n − a, b, c in (7), there arefundamental systems z, w of solutions at the singular points J = 0 , , ∞ . Inparticular, at J = ∞ , since a − b is not an integer, z ( J ) = 1 J F (cid:18) , , , J (cid:19) , w ( J ) = J F (cid:18) − , , , J (cid:19) are linearly independent for | J | >
1. It is a known fact that the functions z, w have an analytical continuation in the whole plane, save for the singularities.5 olving for Y Inasmuch as our main interest lies in the ratio of functions z, w and not inthe actual solutions to the differential equation, we consider the function s = s ( J ) = z ( J ) w ( J ) = F (cid:0) , , , J (cid:1) J F (cid:0) − , , , J (cid:1) . (8)With it, the solution to the icosahedral equation is accomplished by compos-ing s with a M¨obius transformation M ( s ) = M ( s ( J )) = Y ( J ) = Y. The spherical symmetry M rotates s to the correct value of Y , which satisfiesthe correct “initial or boundary conditions” at three convenient points. Wecan determine M assisted by Table 1. J ∞ s = s ( J ) ∞ − ( −
110 5 √ Y = M ( s ) ∞ − ( − Y .The first row contains three different values of J . The second row is calculatedfrom the first row by using the formula defining s ( J ) and some hypergeomet-ric identities that we will state shortly. The values of Y in the third row areobtained from J = H / f . For instance, J = ∞ implies f = 0 and thisis achieved through f ( Y ∞ = 0 ,
1) = 0. For the case J = 0 we realize that, atthe infinity, 1 J = H f (1 , Y = ∞ ) = ∞ . Finally, an easy calculation proves that J = H f (cid:16) − ( − , (cid:17) = 1 . s ( ∞ ) , s (0) are easily determined. s (1) is found with the aidof the following algebraic transformations of the Gaussian hypergeometricfunction, cf. Vidunas [17, Section 6.3]: F (cid:18) , , , ϕ ( x ) (cid:19) = (1 − x + 494 x + 228 x + x ) x − x , (9) F (cid:18) − , , , ϕ ( x ) (cid:19) = (1 − x + 494 x + 228 x + x ) − , (10)where ϕ ( x ) = 1728 x ( x − x − (1 − x + 494 x + 228 x + x ) . (11)The quotient of (9) into (10) provides a feasible expression for (8): s ( J ) = F (cid:0) , , , ϕ ( x ) (cid:1) J F (cid:0) − , , , ϕ ( x ) (cid:1) = (1 − x + 494 x + 228 x + x ) J (1 + 11 x − x ) . (12)Thus, we may take at convenience x = i to obtain ϕ ( i ) = 1 = 1 /J , that is J = 1. Therefore, s (1) = F (cid:0) , , , ϕ ( i ) (cid:1) (1) F (cid:0) − , , , ϕ ( i ) (cid:1) = (1 − i ) + 494( i ) + 228( i ) + ( i ) ) (1) (1 + 11( i ) − ( i ) )= − ( −
110 5 √ . Finally, we write Y = M ( s ) = a s + bc s + d , for some a , b , c , d ∈ C . Since s ( ∞ ) = 0 yields Y = 0, we must have b = 0and d = 0. Without loss of generality, we set a = 1. Then, s (0) = ∞ and Y = ∞ imply 1 / c = ∞ and so, c = 0. This leaves us with s (1) / d = − ( − and an elementary calculation gives d = √ Y is given by Y = M ( s ) = 1 √ z ( J ) w ( J ) = F (cid:0) , , , J (cid:1) √ · J × F (cid:0) − , , , J (cid:1) . (13)This non-homogeneous coordinate solves the problem.7 Proof of Klein’s Theorem
Heymann’s resolvents
The coefficients α, β, γ of the principal quintic y + 5 αy + 5 βy + γ = 0 belongto a quadratic Tschirnhaus extension K of the original coefficient field. Weremind that∆ = 3125 · (108 α γ − α β + 90 α βγ − αβ γ + 256 β + γ )is the so-called discriminant of the quintic. By hypothesis, K also containsthe fifth roots of unity and √ ∆, because Gal( K ≤ L ) = A .First and foremost, the immemorial symmetric relation yz = y + z transformsthis quintic into (1+5 α +5 β + γ ) z − α +4 β + γ ) z +5(3 α +6 β +2 γ ) z − α +4 β +2 γ ) z +5( β + γ ) z − γ. The fourth and third degree terms vanish whenever α = γ/ β = − γ/ y = η , z = η , h = − γ − and h = 1 − h , we obtain Heymann’s η -resolvents h η − η + 15 η − ,h η − η + 15 η − . We observe that h + h = 1 and η η = η + η . Many higher degreepolynomials in η , η are easily reduced to quadratic or linear polynomials bythese identities. For example, h η η = 10 η + η −
6. Such polynomials arecalled Heymann’s simultaneous resolvents.
The first form of the solution
We are interested in solutions of the form y = pη + qη , p, q ∈ C . (14)The substitution of this expression in (1) yields the simultaneous resolvent M η + N η + P η + Qη + R, M, N, P, Q and R are given by M = 5 p (2 h p + 2 h q + αh h ) ,N = 5 q (2 h q + 2 h p + αh h ) ,P = 5 p ( − h p + 10 h p q + 6 h pq + h q + 2 αh h q + βh h ) ,Q = 5 q ( − h q + 10 h q p + 6 h qp + h p + 2 αh h p + βh h ) ,R = 6( h p − h p q + 10 h q p + 10 h p q − h pq + h q ) + γh h . In order to find values of the coefficients α, β, γ , we require M = N = R = 0.With this, α = − p h − + q h − ) ,β = 3( p h − ( p − q ) − q h − (2 p − q )) ,γ = − p h − ( p − pq + 10 q ) + q h − (10 p − pq + q )) . These formulas provide a map ( p, q, h ) ( α, β, γ ). We also need certainmap ( α, β, γ ) ( p, q, h ). To do so, the previous equations for α and β produce h = − p ( p − q )3 α (2 p − q ) − β , h = − q ( p − q )3 α ( p − q ) + 2 β . By replacing these expressions in the equation for γ ,12 αr + 6 βs − γ = 0 , (15)where we have introduced the new variables r = ( p − q ) , s = p + q (thisvariable s is different from variable s ( J ) in Section 2). Clearly, r and s depend on each other. Now we replace h , h in the fundamental relation h + h = 1 and use (15) to obtain the central quadratic relation( α + αβγ − β )(12 r ) − (2 α γ + 11 α β + βγ )(12 r ) + ( αγ − β ) = 0 . (16)We notice that the quadratic discriminant needed to solve for r is108 α γ − α β + 90 α βγ − αβ γ + 256 β + γ = ∆3125 . Therefore, the solution of (16) does not extend our field K . Each value of r provides a value of s and so, values of p and q . Heymann’s parameters h , h are the solutions of the product-sum or trinomial-factoring relations h + h = 1 and h h = 432 βp q αγ − β ) r − γ . (17)9 ulerian resolvents and icosahedral invariants forms Heymann’s ideas recall some results of Euler (1764) on solvable quintics ofthe form (1). His solutions have the form y = ǫz − ǫ w, where ǫ denotes a fifth root of unity. In particular, ǫ = 1 yields y = z − w .Also, z, w are related to the polynomial coefficients through α = − zw , β = z w, γ = − ( z − w ) . Euler’s solution is not our solution. However, we can use the map ( α, β, γ ) ( p, q, h ) and give it a shot. First, we notice that the second degree coefficientof equation (16) vanishes with these expressions of α, β, γ . The remaininglinear equation gives r = w (7 z + w ) f , where f is the well-known icosahedral invariant form (2). From (15) it iseasy to find the corresponding value s = z ( − z + 39 z w + 26 w )6 f . With r and s we determine p and q . Then, from (17) we get4 h h = H f , (18)where H is the expression in (3). In other words, J = 4 h h is the icosahedralparameter.Now, from h + h = 1 and (18) we obtain h − h = p f − H f √ f = T f √ f , (19)where T is just but the invariant (4).The smooth way to compute the parameters h , h is through r and s : h = − s + √ r ) √ r α ( s + 3 √ r ) − β , h = 9( s − √ r ) √ r α ( s − √ r ) − β . h − h = κ ( r ) √ r = T f √ f , (20)where κ = κ ( r ) is a rational function of r and so, p rf = T κf . Since h − h ∈ K , neither √ r nor √ f introduces extra elements into field K . The second form of the solution
The main idea to find the solutions of the quintic runs as follows. Fromcoefficients α, β, γ of (1), we find r, s . With them, we determine p, q togetherwith h , h . Then, we pose the problem by forcing the solution to have alsoEuler’s form but not its original meaning. That is, y = pη + qη = z − w, η η = η + η . (21)The order pair ( z, w ) represents homogeneous coordinates of P and shouldbe found. The resolution for Heymann’s resolvents gives η = − √ ft − √ f and η = 2 √ ft + √ f , where t is the well-konwn octahedral form t ( z, w ) = z + 2 z w − z w − z w − zw + w . In this way, the quadratic term η η in (21) does not introduce new elementsto field K . Further, the solution y to the reduced quintic is obtained bysubstituting η , η in system (21), thus y = − sf + 2 t √ rft − f . We observe that r, s are fixed by the coefficients of the quintic and f isinvariant under the action of the icosahedral group I . The only quantitythat changes is t . Certainly, the action of I yields the five values t ν = ǫ ν z + 2 ǫ ν z w − ǫ ν z w − ǫ ν z w − ǫ ν zw + ǫ ν w , (22)11n which ǫ = e πi/ ∈ K and ν = 0 , , , ,
4. In brief, the solution set is (cid:26) y ν = − sf + 2 t ν √ rft ν − f : ν = 0 , , , , (cid:27) . (23)Now, from α, β, γ and the values p, q , we determine the icosahedral parameter J = 4 h h = H ( z, w )12 f ( z, w ) . (24)With this, the problem is reduced to determine a pair ( z, w ) from an orbit[ z, w ] ∈ P / I , i.e. a solution of the icosahedral equation. The corresponding Y = Y ( J ) is determined as in Section 2. Then, we replace f ( Y, , H ( Y, t ν ( Y,
1) in (23) to find the zeros of the principal quintic.In such a way, Klein’s theorem has been proved.
The foregoing procedure forms a finite sequence of computer-implementableinstructions. 12 ead: coefficients α, β, γ of a reduced quintic with Galois group A ; Function f( z ) :Data: z , a complex number Result: inhomogeneous icosahedral form f ( z ) = f ( z, f ( z ) ← z ( z + 11 z − return ; Function t( z ) :input: z , a complex number output: inhomogeneous octahedral forms t ν ( z ) = t ν ( z, ǫ ← exp (cid:0) πi (cid:1) ; for ν ← to do t ν ( z ) ← ǫ ν z + 2 ǫ ν z − ǫ ν z − ǫ ν z − ǫ ν z + ǫ ν ; endreturn ; Function F( a, b, c ; z ) :Data: argument z ; parameters a, b, c Result:
Gaussian hypergeometric function F ( a, b, c ; z ) = F ( a, b, c ; z ) F ( a, b, c ; z ) ← ∞ X n =1 ( a ) n ( b ) n (1) n ( c ) n z n ; return ; /* initialization */ r ← roots of( α + αβγ − β )(12 r ) − (2 α γ + 11 α β + βγ )(12 r ) + ( αγ − β ) = 0; s ← solutions of 12 αr + 6 βs − γ = 0; p ← ( √ r + s ); q ← ( s − √ r ); J ← h h = 4 432 βp q αγ − β ) r − γ ; Y ← F (cid:0) , , , J (cid:1) √ · J × F (cid:0) − , , , J (cid:1) ; /* computation of zeros */ for ν ← to do y ν ← − sf ( Y ) + 2 t ν ( Y ) p rf ( Y ) t ν ( Y ) − f ( Y ) ; endWrite: zeros y ν , ν = 0 , , , ,
4, of the quintic.13 xample
We consider the following quintic with coefficients in Q [ i, √ ∆]: y + 5 iy − y + (1 − i ) = 0 . We know its Galois group is A . We have α = i, β = − , γ = 1 − i . Firstly,we conveniently pick the solution r = − . − . i of(12 r ) (cid:0) α + αβγ − β (cid:1) − (12 r ) (cid:0) α γ + 11 α β + βγ (cid:1) + (cid:0) αγ − β (cid:1) = 0 . The corresponding value s = 0 . − . i is obtained from 12 αr +6 βs − γ = 0. Next, p = ( √ r + s ) and q = ( s − √ r ) provide p = 0 . − . i and q = 0 . . i. With all this, we obtain immediately J = 4 h h = H f = − . − . i. A solution of this icosahedral equation is Y = Y ( − . − . i ) = F ( , ; ; J − ) √ J F ( − , ; ; J − )= 0 . . i. Afterwards, we calculate the values f ( Y ) and t ν ( Y ): f = − . − . i, t = 0 . − . i,t = − . . i, t = 0 . − . i,t = 0 . . i, t = − . − . i. Finally, we substitute these values in y ν = − sf ( Y ) + 2 t ν ( Y ) p rf ( Y ) t ν ( Y ) − f ( Y )to get the desired solutions y = 0 . − . i, y = − . . i,y = − . − . i, y = − . − . i,y = 1 . − . i. This example has been developed with the routines of the program
WolframMathematica , version 12.0.0.0. 14
Concluding remarks
We have constructed a simple algorithm for the evaluation of the zeros ofany quintic polynomial whose splitting field is icosahedral. The mathemat-ical machinery used to construct the method is not that arduous and moreabridged than the usual approaches employed to compute such zeros. Inparticular, the theory of Heymann [8] permits to rapidly establish a centralrelation between the solutions to the equation and the icosahedral invari-ants. Also, the algebraic transformations of hypergeometric series furnish analternative simplifying way to solve the icosahedral equation.Some other interesting questions arise from these results. They will be re-ported at a later date.
Acknowledgments
This research was partially funded by the
Comit´e Central de Investigaciones,Universidad del Tolima, Ibagu´e, Colombia , grant number 60120. We alsothank the
Facultad de Ciencias, Universidad del Tolima , for its logistic sup-port to craft the manuscript.
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