On Klein’s icosahedral solution of the quintic

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On Klein’s icosahedral solution of the quintic Oliver Nash February 5, 2012 Abstract We present an exposition of the icosahedral solution of the quintic equation first described in the classic work [17]. Although we are heavily influenced by [17] we follow a slightly different approach which enables us to arrive at the solution more directly.

1

Introduction

In 1858, Hermite published a solution of the quintic equation using modular functions [12]. His work received considerable attention at the time and shortly afterward Kronecker and Brioschi also published solutions, but it was not till Klein’s seminal work [17] in 1884 that a comprehensive study of the ideas was provided. Although there is no modern work covering all of the material in [17], there are several noteworthy presentations of some of the main ideas. These include an old classic of Dickson [4] as well Slodowy’s article [28] and the helpful introduction he provides in the reprinted edition [18] of [17]. In addition Klein’s solution is discussed in both [31], [32] as well as [15]. Finally the geometry is outlined briefly in [22] and a very detailed study of a slightly different approach is presented in the book [26]. Perhaps surprisingly, we believe there is room for a further exposition of the quintic’s icosahedral solution. For one thing, all existing discussions arrive at the solution of the quintic indirectly as a result of first studying quintic resolvents of the icosahedral field extension. Even Klein admits that he arrives at the solution ‘somewhat incidentally’1 and each of the accounts listed above, except [22] and [26], exactly follow in Klein’s footsteps. In addition, we believe the icosahedral solution deserves a short, self-contained account. 1

His words in the original German are ‘gewissermassen zuf¨alligerweise’.

1

We thus follow Klein closely but take a direct approach to the solution of the quintic, bypassing the study of resolvents of the icosahedral field extension. In fact our approach is closely related to Gordon’s work [9] and indeed Klein discusses the connection but, having already achieved his goal by other means, he contents himself with an outline. The direct approach enables us to present the solution rather more concisely than elsewhere and we hope this may render it more accessible; part of our motivation for writing these notes was provided by [20]. In addition our derivation of the icosahedral invariant of a quintic produces a different expression than that which appears elsewhere and which is more useful for certain purposes (for example our formula can be easily evaluated along the Bring curve). Finally it is worth highlighting the geometry that connects the quintic and the icosahedron. Using a radical transformation, a quintic can always be put in the form y 5 + 5Îąy 2 + 5βy + Îł = of ordered roots of such P0. ThePvector 2 a quintic lies on the quadric surface yi = yi = 0 in P4 and the reduced Galois group A5 acts on the two families of lines in this doubly-ruled surface by permuting coordinates. The A5 actions on these families, parameterized by P1 , are conjugate to the action of the group of rotations of an icosahedron on its circumsphere and the quintic thus defines a point in the quotients — the icosahedral invariants of a quintic. We discuss this in detail below but first we collect those facts about the icosahedron that we will need.

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The icosahedron

Given an icosahedron, we can identify its circumsphere S with the extended complex plane, and so also with P1 , using the usual stereographic projection: (x, y, z) 7→ x+iy . Orienting our icosahedron appropriately, the 12 vertices 1−z have complex coordinates: 0, ν ( + −1 ), ∞, ν ( 2 + −2 )

ν = 0, 1, . . . , 4

(1)

where = e2Ď€i/5 . Projecting radially from the centre, we can regard the edges and faces of the icosahedron as subsets of S. With the sole exception of figure 2 below, we shall always regard the faces and edges as subsets of S ' C âˆŞ ∞. The picture of the icosahedron we should have in mind is thus:

2

Figure 1: The icosahedron, projected radially onto its circumsphere. We may inscribe a tetrahedron in an icosahedron by placing a tetrahedral vertex at the centre of 4 of the 20 icosahedral faces as shown:

Figure 2: The icosahedron with inscribed tetrahedron. Note that for each icosahedral vertex, exactly one of the 5 icosahedral faces to which it belongs has a tetrahedral vertex at its centre. If we pick an axis 3

joining two antipodal icosahedral vertices, we can consider the 5 inscribed tetrahedra obtained by rotating this configuration through 2πν/5 for ν = 0, 1, . . . , 4. None of these tetrahedra have any vertices in common and so each of the 20 faces of the icosahedron are labeled by a number ν ∈ {0, 1, . . . , 4}. Figure 3 below exhibits such a numbering after stereographic projection.

2

3

4 1

0

3 2

4 0

1

0 2

3

1

1

4

3 2

4

0

Figure 3: Tetrahedral face numbering of the icosahedron under stereographic projection (the outer radial lines meet at the vertex at infinity). The symmetry group Γ of the icosahedron acts transitively on the set of 20 faces with stabilizer of order 3 at each face and so has order 60. The symmetry group also acts faithfully2 on the set of 5 tetrahedra as constructed above and so we obtain an embedding Γ ,→ S5 . Since A5 is the only subgroup of S5 of order 60 we must thus have: Γ ' A5 It will be useful later to have explicit generators for Γ. Thus let S be a rotation through 2π/5 about the axis of symmetry joining the antipodal 2

Consider how many rotations simultaneously stabilize at least three of the tetrahedra (either of figures 2 or 3 may help).

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vertex pair 0, ∞ and let T be a rotation through Ď€ about the axis of symmetry joining the midpoints of the antipodal edge pair [0, + −1 ], [∞, 2 + −2 ]. Using the face numbering in figure 3, S, T correspond to the permutations: S = (01234) T = (12)(34)

(2)

We note in passing that since these two permutations generate A5 we can use this to see that the action on tetrahedra is faithful. In any case we thus have generators for Γ.3 In addition, under the embedding of symmetry groups: Γ ,→ P SL(2, C) associated to the identification of the circumsphere of the icosahedron with P1 we have: 3 0 S= 0 2 (3) 1 −( − 4 ) 2 − 3 T =√ 2 − 3 − 4 5 Having pinned down the symmetry group and its generators, we turn our attention to the branched covering: P1 → P1 /Γ We wish to construct an explicit isomorphism P1 /Γ ' P1 . We thus study the invariant elements in the homogeneous coordinate ring: C[P1 ]Γ . Consider first the possible stabilizer subgroups for the action of Γ on P1 . The action is free except on the three exceptional orbits that are the sets of vertices, edge midpoints and face centres where it has stabilizer subgroups of order ν3 = 5, ν1 = 2, ν2 = 3 respectively (corresponding to the rotations about axes of symmetry through antipodal pairs of these points). Each of these exceptional orbits is the divisor of an invariant homogeneous polynomial. Using (1) we can calculate the polynomial corresponding to the vertices directly. We obtain: f (z1 , z2 ) = z1 z2 (z110 + 11z15 z25 − z210 )

(4)

We could also calculate the polynomials corresponding to the edge midpoints and face centres directly but this would be somewhat cumbersome. Instead we calculate the Hessian and Jacobian covariants of f (see, e.g., [5], [26]) and 3

In fact ST is a rotation about a face centre and we obtain the presentation of Γ as a triangle group: Γ '< S, T | T 2 , (ST )3 , S 5 > that Hamilton used for his Icosian Calculus.

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obtain4 : H(z1 , z2 ) = −(z120 + z220 ) + 228(z115 z25 − z15 z215 ) − 494z110 z210 T (z1 , z2 ) = (z130 + z230 ) + 522(z125 z25 − z15 z225 ) − 10005(z120 z210 + z110 z220 )

(5)

In view of their degrees and the fact that there are no other exceptional orbits, the divisors of these invariant polynomials must be the the face centres and edge midpoints respectively. Next note that since the divisor of any Γ-invariant polynomial p is a sum of Γ-orbits, repeated according to multiplicity, p must be of the form: Y p = f e1 H e2 T e3 pj (6) j

where 0 ≤ ei < νi for i = 1, 2, 3 and deg(pj ) = 60. In particular a degree-60 Γ-invariant polynomial vanishes on a unique orbit. Now the space of degree-60 Γ-invariant polynomials on P1 corresponds to degree-1 polynomials on P1 /Γ and so forms a 2-dimensional space. In fact it is easy to see that any set of three elements must be linearly dependent directly. Suppose that f1 , f2 , f3 are degree-60 Γ-invariant polynomials on P1 . Let [ui , vi ] be a zero of fi for i = 1, 2. We suppose that f1 , f2 are both non-zero, that neither is a multiple of the other so that the orbits containing [u1 , v1 ], [u2 , v2 ] are distinct and consider the polynomial (defined up to scale): g = f2 (u1 , v1 )f3 (u2 , v2 )f1 + f3 (u1 , v1 )f1 (u2 , v2 )f2 − f2 (u1 , v1 )f1 (u2 , v2 )f3 We must then have g = 0 for if not we would have an degree-60 Γ-invariant polynomial vanishing on both the orbit containing [u1 , v1 ] as well as the orbit containing [u2 , v2 ]. This is the required linear relationship. In particular we must have a linear relationship between f 5 , T 2 , H 3 . Using the above construction we obtain cf 5 − T 2 − H 3 = 0 for some constant c. Expanding and comparing coefficients of z160 we find c = 1728 and thus obtain the important syzygy: H 3 + T 2 = 1728f 5

(7)

Moving on from polynomials, we consider Γ-invariant rational functions. Because Γ has no non-trivial characters such a function r must be of the form r = p/q where p, q are Γ-invariant homogeneous polynomials of the 4

We think it worthwhile following the notation of [17] as closely as possible to aid the reader who wishes to compare. It is unfortunate that we must thus use T to denote both the rotation mentioned above as well as the invariant polynomial introduced here but we trust that context will protect us from confusion.

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same degree. Furthermore we can assume ei = 0 in (6) for both p, q since 0 0 0 the only solution of deg(f e1 H e2 T e3 ) ≡ deg(f e1 H e2 T e3 ) mod 60 is ei = e0i . If f1 , f2 are a basis for the degree-60 Γ-invariant homogeneous polynomials we can thus write: Y aj f1 + bj f2 r= cj f1 + dj f2 j = R(f1 /f2 ) Q where R is the rational function z 7→ j (aj z + bj )/(cj z + dj ). We see that, as expected, the space of Γ-invariant rational functions is simply the space of rational functions in f1 /f2 for any basis f1 , f2 as above. Following Klein we make the choice of basis such that the rational function f1 /f2 sends the vertices, edge midpoints and face centres to ∞, 1, 0, respectively, i.e., we use the rational function: H3 I= (8) 1728f 5 and we can summarise by saying that we have an isomorphism of function fields5 : C(P1 )Γ ' C(I) and a corresponding equivalence: I : P1 /Γ ' P1 Lastly we note a property of f, H, T , that we shall need later: they are invariant under composition of antipodal map z 7→ −1/¯ z and reflection z 7→ z¯ since the product of these two orientation-reversing icosahedral symmetries is necessarily a rotation. In particular: f (z2 , −z1 ) = f (z1 , z2 )

(9)

and similarly for H, T . If we were now to follow the usual approach to the icosahedral solution of the quintic, we would next study the quintic resolvents, of the Galois extension C(P1 ) ⊃ C(P1 )Γ . These are obtained by taking subgroups of index 5 in the Galois group A5 corresponding to the tetrahedron. However, as we have already mentioned, we follow a slightly different approach and so immediately turn our attention to the solution of the quintic. We first quickly gather the few facts about Tschirnhaus transformations that we need. 5

In fact we can do better than birational equivalence; indeed we have an isomorphism of graded algebras C[P1 ]Γ ' C[f, H, T ]/(1728f 5 −T 2 −H 3 ) where f, H, T are given weights 12, 30, 20 respectively. However we do not need this.

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Tschirnhaus and the canonical equation

A common approach when solving the polynomial equation: xn + a1 xn−1 + · · · + an = 0

(10)

is to begin by making the affine substitution y = x − a1 /n and so eliminate the term of degree n − 1. This substitution is a special case of the so-called Tschirnhaus transformation [33] in which y is allowed to be a polynomial expression q in x. If αi are the roots of (10), the coefficients of the transformed equation: Y (y − q(αi )) = 0 i

are polynomials in the ai by Sn -invariance (or Newton’s identities). Using a Tschirnhaus transformation we can simultaneously eliminate further terms in the original polynomial. For example if n ≥ 3 and a1 = 0, it is easy to check that the substitution: y = x 2 + b1 x + b2 simultaneously eliminates the terms of degree n − 1 and n − 2 provided the coefficients b1 , b2 satisfy the auxiliary polynomial conditions [26]: p2 b21

b2 − p2 /n = 0 + 2p3 b1 + (p4 − p22 /n) = 0

P where pj = i αij are the power sums of the roots. Thus, provided we are willing to allow ourselves the auxiliary square root necessary to solve the above quadratic for b1 , we may take the general form of the quintic to be: y 5 + 5αy 2 + 5βy + γ = 0

(11)

(We include the factors of 5 for consistency with [17].) In fact it is possible to simultaneously eliminate the terms of degrees n−1, n − 2 and n − 3 (where the coefficients of the substitution are determined by polynomials of degree strictly less than n). Thus, as first shown by Bring [2] and subsequently by Jerrard [16], the general quintic can be reduced to the so-called Bring-Jerrard form: y5 + y + γ = 0 However it is not in this form that the quintic most easily reveals its icosahedral connections and so, except for section 8.1 and appendix A, we shall take the quintic in the form (11). 8

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The space of roots and the icosahedral invariant

Given the quintic (11), if we exclude the trivial equation with α = β = γ = 0 we may regard the roots yi as homogeneous coordinates of a point in P4 . More precisely, since the roots are unordered we obtain an orbit of S5 in P4 , where S5 acts by permuting coordinates. Furthermore because the quintic lacks terms of degree 4 and 3, this orbit lies in the non-singular S5 -invariant quadric surface: n o X X Q = [y0 , . . . , y4 ] ∈ P4 | yi = yi2 = 0 Now the quadric surface is ruled by two P1 -families of lines, the (reduced) Galois group acts on these and the quintic thus defines a point in the quotient of each family. Each quotient is isomorphic to the map P1 → P1 /Γ of section 2 and the points the quintic obtained are the icosahedral invariants of the quintic. We shall describe this explicitly in section 5 but first we discuss the geometry slightly more abstractly. P Thus let V = {(y0 , . . . , y4 ) ∈ C5 | yi = 0} and recall [34] that the space of lines in P(V ) is naturally the Klein quadric: K = [ω] ∈ P(∧2 V ) | ω ∧ ω = 0 The line determined by a point P [ω]2 ∈ K is L[ω] = P({v ∈ V | v ∧ ω = 0}). Now the quadratic form q = i yi restricted to V provides an isomorphism V ' V ∗ and so also an isomorphism: q : ∧2 V ' (∧2 V )∗ Combining this with the natural isomorphism provided by the wedge product: ∧ : ∧2 V ' (∧2 V )∗ ⊗ ∧4 V we thus have an isomorphism: ∧2 V ' ∧2 V ⊗ ∧4 V Furthermore this isomorphism is equivariant with respect to the orthogonal group O(V ) = O(V, q). If we now fix an orientation on V we have an SO(V )-equivariant automorphism of ∧2 V . This is of course simply the (complex) Hodge ?-operator and in this setting is self-inverse. Thus q provides a decomposition: ∧2 V = ∧+ ⊕ ∧− 9

of ∧2 V into the positive and negative eigenspaces of ?; this is a decomposition into irreducible SO(V ) spaces. K meets the two planes P(∧± ) in two nonsingular conics C ± and the natural map (intersection of lines) provides an SO(V )-equivariant isomorphism: C+ × C− ' Q

(12)

This is the well known double ruling of the non-singular quadric surface. Now the representation of S5 on ∧4 V is the sign representation and so in view of the above it is desirable to restrict our representation S5 → O(V ) to A5 → SO(V ). To do this we must be able to associate an A5 -orbit to our quintic rather than a full S5 orbit. We achieve this by supplying a square root ∇ of the discriminant: Y ∇2 = 3125 (yi − yj )2 i<j 5

= 108α γ − 135α4 β 2 + 90α2 βγ 2 − 320αβ 3 γ + 256β 5 + γ 4

(13)

in addition to α, β, γ. Thus, given a quintic, which for simplicity we assume has full Galois group S5 , together with ∇ we obtain an orbit of A5 in each of the conics C ± . Since these actions are non-trivial and A5 is simple, we must have icosahedral actions. We thus obtain two points: Z ± ∈ C ± /A5 ' P1 Note, for the sake of definiteness, that we are using the unique isomorphism C ± /A5 ' P1 sending the orbits with order 60/νi to 1, 0, ∞ respectively for νi = 2, 3, 5. We thus have a (possibly infinite) numerical invariant Z ± associated to our quintic. By implicitly employing the isomorphism C ± ' P1 above, we are using the fact that a non-singular conic is rational. In fact these conics have natural rational parameterizations if we supply V with just a little extra structure (a complex spin structure). Specifically we take two 2-dimensional complex vectors spaces S ± with structure group SL(2, C) and fix an isomorphism: S+ ⊗ S− ' V

(14)

commuting with the SL(2, C) × SL(2, C) and SO(4, C) actions. Using this extra data we can represent everything in terms of S ± . In particular ∧± ' Sym2 S ± and the symmetric square of S ± in Sym2 S ± is (the cone over) the conic C ± . We thus naturally have C ± ' P(S ± ). Using this, the double ruling (12) becomes the more familiar Segre embedding: P(S + ) × P(S − ) ' Q 10

The corresponding statement for homogeneous coordinate rings is that we can express C[Q] as a Segre product: C[Q] ' ⊕n≥0 C[P(S + )]n ⊗ C[P(S − )]n

(15)

where C[P(S ± )]n = H 0 (P(S ± ), O(n)) is the degree n summand of the graded ring C[P(S ± )]. In the following section, where we actually calculate Z ± in terms of α, β, γ, ∇, we give a more concrete, explicit construction of Z ± .

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Calculating the invariant

We wish to calculate Z ± in terms of α, β, γ, ∇ (which we now regard as parameters). This is mostly just an exercise in classical invariant theory for the groups S5 and A5 . The rings of invariants of P(V ) are well known: C[P(V )]S5 ' C[q, α, β, γ] C[P(V )]A5 ' C[q, α, β, γ, ∇]/(∇2 − D(q, α, β, γ)) where D is the discriminant of the quintic y 5 + qy 3 + 5αy 2 + 5βy + γ. The rings of invariants of Q for its embedding in P(V ) are: C[Q]S5 ' C[α, β, γ] C[Q]A5 ' C[α, β, γ, ∇]/(∇2 − D(α, β, γ))

(16)

To see that these hold, let G denote S5 or A5 and take G-invariants of the exact sequence defining the homogeneous coordinate ring of Q. A priori we get the half-exact sequence: 0 → (q)G → C[P(V )]G → C[Q]G If we can show that the map to C[Q]G is in fact surjective then (16) follow trivially. Demonstrating surjectivity amounts to showing that any function d : G → (q) such that d(gh) = d(g) + g · d(h) for all g, h ∈ G is of the form d(g) = dˆ−P g · dˆ for some dˆ ∈ (q). This is easily seen to be the case by taking dˆ = 1/|G| g∈G d(g).6 Note also that in view of the relation ∇2 = D(α, β, γ), any A5 -invariant polynomial h on Q can be written uniquely in the form: h = hs + ha ∇ 6

Of course what we’re really showing is that the group cohomology H 1 (G, (q)) = 0.

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for unique polynomials hs , ha in Îą, β, Îł determined by: hs = (h + h∗ )/2 ha ∇ = (h − h∗ )/2

(17)

where h∗ is the polynomial obtained by transforming h under any odd permutation. In view of (15) and (16), any S5 -invariant element of the ring C[P(S + )]n ⊗ C[P(S − )]n can be written uniquely as a polynomial in Îą, β, Îł. We shall make repeated use of this essential fact below. We now introduce coordinates to make all this explicit. We first change basis from y0 , . . . , y4 to the (Fourier-dual) basis in which S is diagonal: X kj yj (18) pk = j

where as before = e2Ď€i/5 . Our subspace V now has equation p0 = 0 and our quadratic form q, restricted to V , is defined in terms of the pk by 25q = 10(p1 p4 + p2 p3 ). We let S + have coordinates Îť1 , Îť2 , S − have coordinates Âľ1 , Âľ2 and define the map (14) via the bilinear map: p1 = 5Îť1 Âľ1

p2 = −5Îť2 Âľ1

p3 = 5Îť1 Âľ2

p4 = 5Îť2 Âľ2

(19)

Thus [Îť1 , Îť2 ] are homogeneous coordinates on C + and [Âľ1 , Âľ2 ] homogeneous coordinates on C − . The A5 -action on the Îťi is given by the formulae (2), (3) and the dual action on the Âľi is obtained by replacing with 2 in the same formulae. The S5 -action of course interchanges the Îťi and Âľi , indeed the odd permutation R = (1243) acts as ([Îť1 , Îť2 ], [Âľ1 , Âľ2 ]) 7→ ([Âľ2 , âˆ’Âľ1 ], [Îť1 , Îť2 ]). Recalling our formulae for f, H, T from section 2 we define: f1 = f (Îť1 , Îť2 ) f2 = f (Âľ1 , Âľ2 ) and similarly we define H1 , H2 and T1 , T2 . Note that by (9), R interchanges f1 , f2 and similarly for the Hi , Ti . From now on we also denote Z Âą by Z1 , Z2 . In this notation we now have: Z1 =

H13 1728f15

We wish to express this in terms of Îą, β, Îł, ∇. We thus write: Z1 =

H13 f25 1728(f1 f2 )5 12

(20)

Now f1 f2 fixed by R and so is S5 -invariant. It thus lies in C[α, β, γ]. Since it is of degree 12, it must be a linear combination of α4 , β 3 , αβγ. To fix the coefficients we compare leading coefficients as polynomials in λi , µi . It is straightforward to verify that: α = − λ31 µ21 µ2 − λ21 λ2 µ32 − λ1 λ22 µ31 + λ32 µ1 µ22 β = − λ41 µ1 µ32 + λ31 λ2 µ41 + 3λ21 λ22 µ21 µ22 − λ1 λ32 µ42 + λ42 µ31 µ2 γ = − λ51 (µ51 + µ52 ) + 10λ41 λ2 µ31 µ22 − 10λ31 λ22 µ1 µ42 − 10λ21 λ32 µ41 µ2 − 10λ1 λ42 µ21 µ32 + λ52 (µ51 − µ52 ) 4 3 The coefficient of λ12 1 in f1 f2 is 0 whereas the same coefficients in α , β , αβγ 8 4 3 9 8 4 3 9 are µ1 µ2 , −µ1 µ2 , −µ1 µ2 − µ1 µ2 respectively. From this we see that we must have f1 f2 = A(α4 − β 3 + αβγ) for some constant A. Furthermore, upon 11 4 3 noting that the coefficient of λ11 1 λ2 µ1 µ2 in f1 f2 is 1 whereas it is 0 in α , β and 1 in αβγ we learn that A = 1. In other words we obtain:

f1 f2 = α4 − β 3 + αβγ This deals with the denominator in (20); we turn our attention to the numerator. Decomposing the numerator of (20) using (17) and recalling that our odd permutation R interchanges the f1 , f2 as well as H1 , H2 , we get: H13 f25 + H23 f15 H13 f25 − H23 f15 + 2 2 = p + ∇q

H13 f25 =

(21)

where p, q are polynomials in α, β, γ. We could now attempt to calculate p, q by the same means that we calculated f1 f2 above but this would be an enormous amount of work since p, q have degrees 60, 50 respectively. Instead, recall that we have the syzygies: Ti2 = 123 fi5 − Hi3 Multiplying these together and rearranging we obtain: 2p = H13 f25 + H23 f15 = 123 (f1 f2 )5 + 12−3 (H1 H2 )3 − 12−3 (T1 T2 )2 We will thus have the required expression for p in terms of α, β, γ as soon as we express H1 H2 and T1 T2 in these terms. To do this we do follow the same procedure as that used to find f1 f2 above and (admittedly with somewhat more effort) we obtain: H1 H2 = γ 4 + 40α2 βγ 2 − 192α5 γ − 120αβ 3 γ + 640α4 β 2 − 144β 5 T1 T2 = γ 6 + 60α2 βγ 4 + 576α5 γ 3 − 180αβ 3 γ 3 + 648β 5 γ 2 − 2760α4 β 2 γ 2 + 7200α7 βγ − 1728α10 + 9360α3 β 4 γ − 2080α6 β 3 − 16200α2 β 6 13

It remains only to calculate q. This time the trick we use is to note that as well as (21) above, we have H23 f15 = p − ∇q and so: (H1 H2 )3 (f1 f2 )5 = p2 − ∇2 q 2 It follows that taking our above polynomial expressions for H1 H2 , f1 f2 , ∇2 = D(Îą, β, Îł), p we must find a factorization of ((H1 H2 )3 (f1 f2 )5 −p2 )/D(Îą, β, Îł). From this we determine7 : 2q = Âą (−8Îą5 Îł − 40Îą4 β 2 + 10Îą2 βγ 2 + 45ιβ 3 Îł − 81β 5 − Îł 4 )¡ (64Îą10 + 40Îą7 βγ − 160Îą6 β 3 + Îą5 Îł 3 − 5Îą4 β 2 Îł 2 + 5Îą3 β 4 Îł − 25Îą2 β 6 − β 5 Îł 2 ) The two signs corresponding to the two invariants: Z1 , Z2 . With this formula in hand we have achieved our goal of expressing Zi in terms of Îą, β, Îł, ∇. Our derivation and formulae for Zi in terms of Îą, β, Îł, ∇ differ from Klein’s ([17], pg. 213) and indeed from any other accounts we have come across which all reproduce Klein’s formulae exactly. In fact our approach is more closely connected to Gordon’s work [9]. Klein does outline the connection but he leaves it unfinished as he has already obtained his formulae for Zi by other means. Although the formulae for Zi in [17] have the advantage of being more concise than ours, they have the disadvantage that Zi is represented as a rather complicated rational expression (rather than in terms polynomials) and that Zi cannot be evaluated along the Bring curve Îą = 0 easily.

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Obtaining the roots

The invariants Zi defined above are obtained by applying the icosahedral function I to the coordinates Îťi , Âľi . These coordinates determine a point in the projective space of vectors of roots P(V ), and so if we could invert the icosahedral function I and express Îťi , Âľi in terms of the invariants Zi , we would be able to solve the quintic. We will show how to invert I in section 7 below but first we fill in the details of how an inverse for I yields a solution of the quintic. We begin by using (18), (19) to express roots in terms of Îťi , Âľi : yν = 4ν Îť1 Âľ1 − 3ν Îť2 Âľ1 + 2ν Îť1 Âľ2 + ν Îť2 Âľ2 7

(22)

That q itself turns out to factorize like this is unexpected, at least to the author.

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Now suppose [λ1 , λ2 ] is a solution of the icosahedral equation: I(λ1 , λ2 ) = Z1

(23)

The first thing to notice is that we do not also need to find an inverse for Z2 . Concretely, the idea is that if we can find A5 -invariant forms that are linear in µi then we can use these to eliminate the µi in (22) and so express yν in terms of just α, β, γ and our solution [λ1 , λ2 ] of (23). We thus enlarge the ring of invariant polynomials we are studying from the Segre product (15) to the full tensor product C[P(S + )] ⊗ C[P(S − )]. The two A5 -invariant forms linear in µi of lowest degree in λi are: N1 = (7λ51 λ22 + λ72 )µ1 + (−λ71 + 7λ21 λ52 )µ2 8 5 3 10 10 3 5 8 13 M1 = (λ13 1 − 39λ1 λ2 − 26λ1 λ2 )µ1 + (−26λ1 λ2 + 39λ1 λ2 + λ2 )µ2

(24)

There are a number of ways to derive these expressions. We follow Gordon [9] and use transvectants. We thus recall (see for example [3] or [5]) that if f, g are two homogeneous polynomials in λ1 , λ2 then the rth transvectant of f, g is given by: r X (−1)i ∂rg ∂rf (f, g)r = r−i i i i!(r − i)! ∂λr−i 1 ∂λ2 ∂λ1 ∂λ2 i=0

We extend this to homogeneous polynomials in both λi , µi using bilinearity, i.e., if: X j b−j fij λi1 λa−i f= 2 µ1 µ2 i,j

g=

X

l d−l gkl λk1 λc−k 2 µ1 µ2

k,l

then we define the (r, s)-transvectant: X j b−j k c−k l d−l fij gkl (λi1 λa−i (f, g)r,s = 2 , λ1 λ2 )r (µ1 µ2 , µ1 µ2 )s i,j,k,l

Using these transvectants, we can construct the invariants we need. Indeed it is straightforward to verify that: (α, β)0,3 = 6N1 ((α, α)0,2 , N1 )0,1 = 8M1 as required. 15

As a brief aside we comment on the geometry of N1 , M1 . Since they are linear in the Âľi they correspond to A5 -equivariant branched covers P(S + ) → P(S − ) of degrees 7, 13 respectively. From the Riemann-Hurwitz formula, the ramification index of these maps is 12, 24 respectively and so the branch locus in P(S − ) must be the set of vertices of an icosahedron in P(S − ) in each case (since they are stable under the A5 action). For N1 , there are 12 doublyramified points in P(S + ) which must also be the vertices of an icosahedron. There are 5 remaining unramified points in the fibre over each of the 12 points in the branch locus. These 60 points are the A5 -orbit containing the point Îť1 /Îť2 = 71/5 , i.e., the orbit with icosahedral invariant 64/189. It is interesting to wonder if it is possible to characterise the configuration of these points geometrically. The set of vertices of an Archimedean solid with icosahedral symmetry might seem like a plausible candidate but in fact this does not work as can be seen easily since the point 71/5 lies outside the field of coefficients of these solids (though the points are remarkably close to the vertices of the truncated dodecahedron). Similarly there are two distinguished orbits for M1 . If it is possible to come up with a geometric characterisation of these various orbits then it should also apply to the other two A5 -equivariant branched covers (also identified by Gordon in [9]) which have degrees 17 and 23. Returning to the task at hand we solve the 2 Ă— 2 system (24) and express Âľi in terms of M1 , N1 and using (22) obtain: yν = H1−1 bν M1 + H1−1 cν N1

(25)

Here H1 appears as it is the determinant of the matrix which we invert and the coefficients bν , cν are given by: bν cν = 3 5 8 13 4ν âˆ’Îť71 + 7Îť21 Îť52 26Îť10 3ν 2ν ν 1 Îť2 − 39Îť1 Îť2 − Îť2 Îť1 − Îť2 Îť1 + Îť2 8 5 3 10 −7Îť51 Îť22 − Îť72 Îť13 1 − 39Îť1 Îť2 − 26Îť1 Îť2 Now in (25) we can calculate H1 , bν , cν using our solution of (23). We deal with M1 , N1 turning them into forms of the same degree in Îťi and Âľi which we can thus express in terms of Îą, β, Îł, ∇. We thus rewrite (25) as: bν f1 M1 f2 cν T1 N1 f12 T2 yν = ¡ ¡ + H1 f1 f2 H1 f12 T1 T2 The methods described in section 5 then allow us to calculate: M1 f2 = (11Îą3 β + 2β 2 Îł − ιγ 2 )/2 − âˆ‡Îą/2 N1 f12 T2 = r + ∇s 16

where: 2r = α2 γ 5 − αβ 2 γ 4 + 53α4 βγ 3 + 64α7 γ 2 − 7β 4 γ 3 − 225α3 β 3 γ 2 − 12α6 β 2 γ + 216α9 β + 717α2 β 5 γ − 464α5 β 4 − 720αβ 7 2s = − α2 γ 3 + 3αβ 2 γ 2 − 9β 4 γ − 4α4 βγ − 8α7 − 80α3 β 3 and since we already have formulae for f1 f2 and T1 T2 we have the required expression for yν in terms of α, β, γ, ∇, λ1 , λ2 . It thus remains only to show how to solve (23) to find λ1 , λ2 .

7

Solving the icosahedral equation

We wish to invert the equation: I(z) = Z

(26)

(In this section we regard I as a function of the single variable z = z1 /z2 .) This problem was essentially solved by Schwarz in 1873 when he determined the list parameters for which the hypergeometric differential equation has finite monodromy. We thus need only cast our problem in the right form and appeal to general theory. The solution can be written in terms of Schwarz s-functions (as discussed in [23] for example). We must identify domains of injectivity for I (equivalently, fundamental domains for the action of Γ on P1 ). We thus note that if r is any reflection about a plane of symmetry of the icosahedron then: I ◦ r = I¯ for the reasons explained when discussing (9). Since there is a plane of symmetry through any edge of the icosahedron as well as a plane of symmetry through each of the altitudes of any face of the icosahedron, it follows that the edges and altitudes of the faces of the icosahedron constitute the preimage of RP1 = R ∪ ∞ under I. The altitudes divide each face into six triangles with angles π/2, π/3, π/5 = π/νi . I sends the vertices of each triangle to 0, 1, ∞ and is injective on the interior. It maps three of them biholomorphically to upper half space H + and three of them biholomorphically to lower half space H − , according to whether their vertices are sent to 0, 1, ∞ in anti-clockwise or clockwise order respectively. Subdividing faces like this, figure 1 becomes:

17

Figure 4: Icosahedral tiling of sphere. I maps the interior of each light and dark triangle biholomorphically onto the upper and lower half-planes respectively. Under stereographic projection, the subdivision of the face with vertices 0, + −1 , 2 + 1 is: Îľ2+1

h

0

Îľ+Îľ-1

t

Figure 5: Domains of injectivity for I under stereographic projection. The points h, t are the images of the face centre and edge midpoint respectively. We shall construct an inverse for the restriction of I to the interior of the triangle with vertices 0, t, h. I maps the interior biholomorphically to the 18

lower half-plane H − and the boundary homeomorphically to RP1 , sending the points 0, t, h to ∞, 1, 0 respectively. Now the equation (26) we wish to solve is a degree 60 polynomial but it has the special property that its monodromy lies in P SL(2, C), i.e., the 60 branches of a local inverse are related by a Mobius transformation. Since the Schwarzian derivative is invariant under Mobius transformations, the Schwarzian derivative of the inverse is independent of the branch; moreover it is easily calculated. This yields a differential equation for the inverse which, using standard methods, may be transformed to Gauss’s hypergeometric differential equation. It then follows that an inverse may be expressed as a ratio of linearly independent solutions to this equation, i.e., as a Schwarz s-function. Indeed, as shown in [23], the inverse we seek is determined up the scale by the angles in the triangle, π/νi , and is given by: s : H − −→ C Z 7−→ C

Z c−1 2 F1 (a0 , b0 ; c0 ; Z −1 ) −1 ) 2 F1 (a, b; c; Z

Here 2 F1 is the analytic continuation of Gauss’s hypergeometric series to C − [1, ∞), Z c−1 is defined using the principle branch of log on C − (−∞, 0]. The values of the constants are: a = 12 1 − ν13 + ν12 − ν11 b = 12 1 − ν13 − ν12 − ν11 c = 1 − ν13 a0 = a − c + 1

b0 = b − c + 1

c0 = 2 − c

To determine C we compare the coefficients of Z on both sides of the power series identity: Z · H 3 (z1 , z2 ) = f 5 (z1 , z2 ) with z1 = CZ 1−c 2 F1 (a0 , b0 ; c0 ; Z) and z2 = 2 F1 (a, b; c; Z). This yields C = 1728−1/5 and so s is completely determined.8 In summary then, for im Z < 0, our inverse is: s(Z) =

31 11 6 −1 ) 2 F1 ( 60 , 60 ; 5 ; Z 19 1 4 1/5 (1728Z) 2 F1 ( 60 , − 60 ; 5 ; Z −1 )

8

An alternate means of determining C would be to impose the condition s(1) = t and to use Gauss’s expression for 2 F1 (a, b, c, 1) in terms q of Γ-functions. This yields C = t·

61 Γ( 54 )Γ( 41 60 )Γ( 60 ) 49 . Since )Γ( Γ( 65 )Γ( 29 60 60 ) −1/5

know C = 1728

t = 2 (cos(π/10) − cos(π/5)) =

we obtain a Γ-function identity.

19

√ 5+ 5 2

−

√ 1+ 5 2

and we already

Using a similar expression for im(Z) > 0, we could extend this function to the open set H + âˆŞ H − âˆŞ (0, 1) so that we would have an inverse for the restriction of I to the interior of the triangle with vertices 0, + −1 , h. (Indeed using Schwarz reflection, we can extend s across any of the edges of the triangle and of course this process generates the monodromy A5 .)

8

Further properties and parting words

Our focus in these notes has been to present the icosahedral solution of the quintic as concisely as possible, subject to the conditions of remaining as explicit as [17] and as self-contained as possible. As a result we have been forced us to omit discussion of many related matters. We comment briefly on some of these here.

8.1

Bring’s curve and Kepler’s great dodecahedron

We mentioned in section 3 that it is possible to reduce the general quintic to the so-called Bring-Jerrard form: y5 + y + γ = 0 but that we would work with the quintic in the form (11). We did this because we were following [17], because (11) is more general and because it is easy to bring out the icosahedral connection using the A5 actions on the lines in the doubly-ruled quadric surface. However there is an appealing way to connect the icosahedron with the quintic in Bring-Jerrard form which is worth mentioning. The construction below is described in [11]. Firstly note that the family of quintics in Bring-Jerrard form is the P P P smooth genus 4 curve B cut out of P4 by the equations yi = yi2 = yi3 = 0. This is known as the Bring curve and has automorphism group S5 corresponding to the general Galois group. The branched covering B → B/A5 ' P1 allows us to define an invariant as before and connects with our work above. Secondly, starting with an icosahedron in R3 we form Kepler’s great dodecahedron GD . This regular solid, which self-intersects in R3 , has one face for each vertex of the icosahedron. It is formed by spanning the five neighbouring vertices of each vertex of the icosahedron with a regular pentagon and then dismissing the original icosahedron. GD thus has the same 12 vertices and 30 edges as the icosahedron but only 12 faces. Projection onto the common circumsphere S yields a triple covering GD → S with a double branching at the 12 vertices and after identifying S with P1 provides GD with 20

a complex structure. Evidently GD has Euler characteristic −6 and so genus 4. In fact, as explained in [11], GD is isomorphic to the Bring curve. The isomorphism GD ' B can be used to bring out the relationship between the quintic and the icosahedron.

8.2

Modular curves, Ramanujan’s continued fraction and Shioda’s modular surface

From one point of view, the exceptional geometry of the quintic is a result of the exceptional isomorphism: A5 ' P SL(2, 5) Corresponding to the exact sequence defining the level 5 principal congruence subgroup of the modular group: 0 → Γ(5) → P SL(2, Z) → P SL(2, 5) → 0 there is a factorization of the modular quotient: Iˆ

j5

j : H ∗ −→ X(5) −→ X(1) where H ∗ = H + ∪ QP1 is the upper half-plane together with the P SL(2, Z)orbit of ∞ and X(N ) is the compactified modular curve of level N . The curves X(5), X(1) are rational and the map Iˆ : X(5) → X(1) is a quotient by P SL(2, 5) and is thus an icosahedral quotient. We can use this to find an inverse for the icosahedral function I in terms of Jacobi ϑ-functions (provided we are willing to invert Klein’s j-invariant). Indeed the map j5 may be expressed as: P 5n2 +3n q ϑ1 (πτ ; q 5 ) 2/5 j5 (τ ) = q PZ 5n2 +n = q −3/5 (27) ϑ1 (2πτ ; q 5 ) Zq where q = eπτ i and we are using the ϑ-function notational conventions of [35]. Thus given Z as in section 7, if τ satisfies j(τ ) = 1728Z then z = j5 (τ ) is a solution to I(z) = Z. In fact there is another expression for j5 , it is none other than Ramanujan’s continued fraction: q 1/5 q

j5 (τ ) = 1+

q2

1+ 1+ 21

q3 1 + ···

Furthermore because we know that the icosahedral vertices, edge midpoints and face centres in X(5) lie above the points ∞, 1, 0 in X(1), we can calculate 1 the values of this continued fraction at those orbits in H ∗ which 1728 j maps to ∞, 1, 0. For example j(i) = 1728 and the corresponding edge midpoint equality: s √ √ 5+ 5 1+ 5 j5 (i) = t = − 2 2 is one of the identities that famously caught Hardy’s eye when Ramanujan first wrote to him. An excellent account of these results together with a proof of (27) can be found in [6]. This method of inverting I treats the icosahedral invariant Z of a quintic as a point in the moduli space of elliptic curves X(1). It is natural to wonder if the quintic in fact provides a geometric realization of the elliptic curve with j-invariant Z. Since X(5) is the moduli space of elliptic curves together with level 5 structure, inverting I would then correspond to finding a basis for the 5-torsion of such a curve in which the Weil pairing is in standard form. If this were the case we would expect a link with the elliptic curves embedded in P4 defined in [14] as an intersection of quadrics. The union of all of these curves is an irreducible surface (with singularities) whose normalization is Shioda’s modular surface (a compactification of the universal elliptic curve with level 5 structure).

8.3

Parting words

There is of course much more to say beyond even those remarks in sections 8.1 and 8.2 above. Other matters which we shall not discuss at all but which seem worth mentioning include the following: • The group P SL(2, 5) is one of a ‘trinity’of groups, the other two being P SL(2, 7) and P SL(2, 11). Higher degree equations can have these latter two groups as their Galois group and for these there exists a higher genus version of the icosahedral solution of the quintic. • The rational parameterization of the singularity T 2 + H 3 = 1728f 5 we have described can be used to find solutions of the Diophantine equation a2 + b3 + c5 = 0. See Beukers [1] for details. • We have worked exclusively over C. Needless to say, this is not necessary, the field Q(eπi/5 ) suffices for most of the above. Serre makes some helpful comments in this direction in his letter [27]. 22

• The icosahedral solution of the quintic is not usually the most efficient technique for finding the roots. More practical formulae appear in [30] for example.

A

An earlier solution

In addition to the techniques described above, there is another approach to the solution of the quintic discovered by Lambert9 [19] in 1758 and again by Eisenstein [7] in 1844. Consider the quintic in Bring-Jerrard form (up to a sign): y5 − y + γ = 0

(28)

Viewing y as a function of γ, we claim that the branch of y such that y(0) = 0 has power series: X 5k γ 4k+1 y(γ) = (29) k 4k + 1 k≥0 (This can also be expressed in terms of the generalized hypergeometric func 5γ 4 4 3 2 1 5 3 1 tion as: y(γ) = 5 F4 1, 5 , 5 , 5 , 5 ; 4 , 1, 4 , 2 ; 5( 4 ) γ.) This appealing result is established using analytic methods (Lagrange inversion) in [24] and [29] as well as [21]. However since this statement is really an identity of binomial coefficients it is desirable to have a combinatorial proof for the identity (28) satisfied by the generating function (29). Now the coefficients in (29) are a special case of the sequence: pk 1 p dk = (p − 1)k + 1 k which specializes to the Catalan numbers for p = 2. This sequence, considered long ago by Fuss [8], was studied in some detail in [13]. Just as various identities for the Catalan numbers can be established by observing that the they count (amongst many other things) certain lattice paths, so too can those identities for p dk which we seek for p = 5 be established by demonstrating that these coefficients count certain paths introduced in [13]. Although the results in [13] thus provide a combinatorial proof of the generating function identity (28), there is a more direct combinatorial proof 9

It should be pointed out that Lambert would not have been aware that his method provided a solution of the general quintic since the reduction the Bring-Jerrard form was not known in his time.

23

presented in [10] based on an observation of Raney [25]. He noticed that if a1 , . . . , am is any sequence of integers that sum to 1, then exactly one of the m cyclic permutations of this sequence has all of its partial sums positive. With this in mind we consider the problem of counting sequences a0 , . . . , akp such that: • a0 + · · · + akp = 1 • All partial sums are positive • Each ai is either 1 or 1 − p Using Raney’s observation it is clear that the number of such sequences is p dk . The natural recursive structure of such sequences provided by concatenation of p such sequences, followed by a terminating value of 1−p then corresponds to the identity we seek. The interested reader will find details in [10].

References [1] F. Beukers. The diophantine equation axp + by q = cz r . Duke Mathematical Journal, 91(1), 1998. [2] E.S. Bring. Meletemata quaedam mathematematica circa transformationem aequationum algebraicarum. Lund University, Promotionschrift, 1786. [3] W. Crawley-Boevey. Lectures on representation theory and invariant theory. Erg¨anzungsreihe Sonderforschungsbereich 343 ’Diskrete Strukturen in der Mathematik’, 90-004, Bielefeld University, 1990. [4] L.E. Dickson. Modern Algebraic Theories. Ben J.H Sanborn & Co., 1926. [5] I.V. Dolgachev. Lectures on Invariant Theory. Number 296 in London Mathematical Society Lecture Note Series. Cambridge University Press, 2003. [6] W. Duke. Continued fractions and modular functions. Bulletin of the American Mathematical Society, 42(2):137–162, 2005. [7] F.G.M. Eisenstein. Allgemeine aufl¨osung der gleichungen von den ersten vier graden. J. Reine Angew. Math., 27:81–83, 1884.

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[8] N. Fuss. Solutio quaestionis, quot modis polygonum n laterum in polygona m laterum, per diagonales resolvi quaeat. Nova acta academiae scientiarum imperialis Petropolitanae, 9:243–251, 1791. [9] P. Gordon. Ueber die aufl¨osung der gleichungen vom f¨ unften grade. Mathematische Annalen, 13:375–404, 1869. [10] R.L. Graham, D.E. Knuth, and O. Patashnik. Concrete mathematics. Addison Wesley, 1994. [11] M.L. Green. On the analytic solution of the equation of the fifth degree. Compositio Mathematica, 37:233–241, 1978. [12] C. Hermite. Sur la r´esolution de l’´equation du cinqu`eme degr´e (1858). Oevres de Charles Hermite, 2:5–21, 2009. [13] P. Hilton and J. Pedersen. Catalan numbers, their generalization, and their uses. Mathematical Intelligencer, 13(2):64–75, 1991. [14] K. Hulek. Geometry of the horrocks-mumford bundle. In Proceedings of Symposium in Pure Math, volume 46, pages 69–85. American Mathematical Society, 1987. [15] B. Hunt. The geometry of some special arithmetic quotients. Springer, 1996. [16] G.B. Jerrard. Mathematical researches. William Strong, Bristol, 2, 1834. [17] F. Klein. Lectures on the icosahedron and the equation of the fifth degree (trans.). Cosimo Classics, 1884. [18] F. Klein and P. Slodowy. Vorlesungen u ¨ber das Ikosaeder und die Aufl¨osung der Gleichungen vom f¨ unften Grade. Birkh¨auser, 1993. [19] J.H. Lambert. Observationes variae in mathesin puram. Acta Helvetica, 3(1):128–168, 1758. [20] Mathoverflow. Do there exist modern expositions of klein’s icosahedron? http://mathoverflow.net/questions/9474, 2009. [21] Mathoverflow. What is lagrange inversion good for? mathoverflow.net/questions/32099/#32261, 2010.

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[22] H. McKean and V. Moll. Elliptic Curves. Cambridge University Press, 1997. 25

[23] Z. Nehari. Conformal mapping. McGraw-Hill, 1952. [24] S.J. Patterson. Eisenstein and the quintic equation. Historia mathematica, 17:132–140, 1990. [25] G.N. Raney. Functional composition patterns and power series reversion. Transactions of the American Mathematical Society, 94:441–451, 1960. [26] M.J. Schurman. Geometry of the quintic. Wiley Interscience, 1997. [27] J.P. Serre. Extensions icosa´edriques. S´eminaire de Th´eorie des Nombres de Bordeaux, 19(123):550–552, 1979. [28] P. Slodowy. Das ikosaeder und die gleichungen f¨ unften grades. Arithmetik und Geometrie. Vier Vorlesungen (Mathematischen Miniat¨ uren, Band 3), 1986. [29] J. Stillwell. Eisenstein’s footnote. Mathematical Intelligencer, 17(2):58– 62, 1995. [30] B. Sturmfels. Solving algebraic equations in terms of a-hypergeometric series. In Department Of Mathematics, Texas A&M University, College Station, TX 77843. (EM), pages 171–181, 1996. [31] G. Toth. Finite M¨obius groups, minimal immersions of spheres, and moduli. Springer, 2002. [32] G. Toth. Glimpses of Algebra and Geometry. Springer, second edition, 2002. [33] E. Tschirnhaus. Methodus auferendi omnes terminos intermedios ex data equatione. Acta Eruditorum, II, 1683. [34] R. S. Ward and R. O. Wells, Jr. Twistor geometry and field theory. Cambridge Monographs on Mathematical Physics. Cambridge University Press, Cambridge, 1990. [35] E.T. Whittaker and G.N. Watson. A Course of Modern Analysis. Cambridge University Press, 1927.

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