On Klein’s icosahedral solution of the quintic

Oliver Nash

February 5, 2012

Abstract

We present an exposition of the icosahedral solution of the quinticequation first described in the classic work [17]. Although we areheavily influenced by [17] we follow a slightly different approach whichenables us to arrive at the solution more directly.

1 Introduction

In 1858, Hermite published a solution of the quintic equation using modularfunctions [12]. His work received considerable attention at the time andshortly afterward Kronecker and Brioschi also published solutions, but itwas not till Klein’s seminal work [17] in 1884 that a comprehensive study ofthe 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. Theseinclude an old classic of Dickson [4] as well Slodowy’s article [28] and thehelpful introduction he provides in the reprinted edition [18] of [17]. Inaddition Klein’s solution is discussed in both [31], [32] as well as [15]. Finallythe geometry is outlined briefly in [22] and a very detailed study of a slightlydifferent approach is presented in the book [26].

Perhaps surprisingly, we believe there is room for a further expositionof the quintic’s icosahedral solution. For one thing, all existing discussionsarrive at the solution of the quintic indirectly as a result of first studyingquintic resolvents of the icosahedral field extension. Even Klein admits thathe arrives at the solution ‘somewhat incidentally’1 and each of the accountslisted above, except [22] and [26], exactly follow in Klein’s footsteps. Inaddition, we believe the icosahedral solution deserves a short, self-containedaccount.

1His words in the original German are ‘gewissermassen zufalligerweise’.

1

We thus follow Klein closely but take a direct approach to the solution ofthe quintic, bypassing the study of resolvents of the icosahedral field exten-sion. In fact our approach is closely related to Gordon’s work [9] and indeedKlein discusses the connection but, having already achieved his goal by othermeans, he contents himself with an outline.

The direct approach enables us to present the solution rather more con-cisely than elsewhere and we hope this may render it more accessible; partof our motivation for writing these notes was provided by [20]. In additionour derivation of the icosahedral invariant of a quintic produces a differentexpression than that which appears elsewhere and which is more useful forcertain purposes (for example our formula can be easily evaluated along theBring curve).

Finally it is worth highlighting the geometry that connects the quinticand the icosahedron. Using a radical transformation, a quintic can always beput in the form y5 +5αy2 +5βy + γ = 0. The vector of ordered roots of sucha quintic lies on the quadric surface

∑yi =

∑y2

i = 0 in P4 and the reducedGalois group A5 acts on the two families of lines in this doubly-ruled surfaceby permuting coordinates. The A5 actions on these families, parameterizedby P1, are conjugate to the action of the group of rotations of an icosahedronon its circumsphere and the quintic thus defines a point in the quotients —the icosahedral invariants of a quintic. We discuss this in detail below butfirst we collect those facts about the icosahedron that we will need.

2 The icosahedron

Given an icosahedron, we can identify its circumsphere S with the extendedcomplex plane, and so also with P1, using the usual stereographic projection:(x, y, z) 7→ x+iy

1−z. Orienting our icosahedron appropriately, the 12 vertices

have complex coordinates:

0, εν(ε + ε−1), ∞, εν(ε2 + ε−2) ν = 0, 1, . . . , 4 (1)

where ε = e2πi/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 ∪ ∞. Thepicture of the icosahedron we should have in mind is thus:

2

Figure 1: The icosahedron, projected radially onto itscircumsphere.

We may inscribe a tetrahedron in an icosahedron by placing a tetrahedralvertex 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 facesto 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 inscribedtetrahedra obtained by rotating this configuration through 2πν/5 for ν =0, 1, . . . , 4. None of these tetrahedra have any vertices in common and so eachof 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.

0

00

0

1

1

1

12

2

2

2

3

3

3

3

4

4

4

4

Figure 3: Tetrahedral face numbering of the icosahedron understereographic projection (the outer radial lines meet at the vertex

at infinity).

The symmetry group Γ of the icosahedron acts transitively on the setof 20 faces with stabilizer of order 3 at each face and so has order 60. Thesymmetry group also acts faithfully2 on the set of 5 tetrahedra as constructedabove and so we obtain an embedding Γ ↪→ S5. Since A5 is the only subgroupof S5 of order 60 we must thus have:

Γ ' A5

It will be useful later to have explicit generators for Γ. Thus let S bea rotation through 2π/5 about the axis of symmetry joining the antipodal

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

4

vertex pair 0,∞ and let T be a rotation through π about the axis of symmetryjoining the midpoints of the antipodal edge pair [0, ε+ε−1], [∞, ε2+ε−2]. Usingthe 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 usethis to see that the action on tetrahedra is faithful. In any case we thus havegenerators for Γ.3 In addition, under the embedding of symmetry groups:Γ ↪→ PSL(2, C) associated to the identification of the circumsphere of theicosahedron with P1 we have:

S =

[ε3 00 ε2

]T =

1√5

[−(ε− ε4) ε2 − ε3

ε2 − ε3 ε− ε4

] (3)

Having pinned down the symmetry group and its generators, we turn ourattention to the branched covering:

P1 → P1/Γ

We wish to construct an explicit isomorphism P1/Γ ' P1. We thus study theinvariant 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 ofvertices, edge midpoints and face centres where it has stabilizer subgroupsof order ν3 = 5, ν1 = 2, ν2 = 3 respectively (corresponding to the rotationsabout axes of symmetry through antipodal pairs of these points). Each ofthese exceptional orbits is the divisor of an invariant homogeneous polyno-mial. Using (1) we can calculate the polynomial corresponding to the verticesdirectly. We obtain:

f(z1, z2) = z1z2(z101 + 11z5

1z52 − z10

2 ) (4)

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

3In fact ST is a rotation about a face centre and we obtain the presentation of Γ as atriangle group: Γ '< S, T | T 2, (ST )3, S5 > that Hamilton used for his Icosian Calculus.

5

obtain4:

H(z1, z2) = −(z201 + z20

2 ) + 228(z151 z5

2 − z51z

152 )− 494z10

1 z102

T (z1, z2) = (z301 + z30

2 ) + 522(z251 z5

2 − z51z

252 )− 10005(z20

1 z102 + z10

1 z202 )

(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 andedge midpoints respectively.

Next note that since the divisor of any Γ-invariant polynomial p is a sumof Γ-orbits, repeated according to multiplicity, p must be of the form:

p = f e1He2T e3

∏j

pj (6)

where 0 ≤ ei < νi for i = 1, 2, 3 and deg(pj) = 60. In particular a degree-60Γ-invariant polynomial vanishes on a unique orbit.

Now the space of degree-60 Γ-invariant polynomials on P1 corresponds todegree-1 polynomials on P1/Γ and so forms a 2-dimensional space. In factit is easy to see that any set of three elements must be linearly dependentdirectly. 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 bothnon-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-invariantpolynomial vanishing on both the orbit containing [u1, v1] as well as the orbitcontaining [u2, v2]. This is the required linear relationship.

In particular we must have a linear relationship between f 5, T 2, H3. Usingthe above construction we obtain cf 5 − T 2 − H3 = 0 for some constant c.Expanding and comparing coefficients of z60

1 we find c = 1728 and thus obtainthe important syzygy:

H3 + 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 theform r = p/q where p, q are Γ-invariant homogeneous polynomials of the

4We think it worthwhile following the notation of [17] as closely as possible to aid thereader who wishes to compare. It is unfortunate that we must thus use T to denote boththe rotation mentioned above as well as the invariant polynomial introduced here but wetrust that context will protect us from confusion.

6

same degree. Furthermore we can assume ei = 0 in (6) for both p, q sincethe only solution of deg(f e1He2T e3) ≡ deg(f e′

1He′2T e′

3) mod 60 is ei = e′i. Iff1, f2 are a basis for the degree-60 Γ-invariant homogeneous polynomials wecan thus write:

r =∏

j

ajf1 + bjf2

cjf1 + djf2

= R(f1/f2)

where R is the rational function z 7→∏

j(ajz + bj)/(cjz + dj). We see that,as expected, the space of Γ-invariant rational functions is simply the spaceof rational functions in f1/f2 for any basis f1, f2 as above. Following Kleinwe make the choice of basis such that the rational function f1/f2 sends thevertices, edge midpoints and face centres to ∞, 1, 0, respectively, i.e., we usethe rational function:

I =H3

1728f 5(8)

and we can summarise by saying that we have an isomorphism of functionfields5:

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 areinvariant under composition of antipodal map z 7→ −1/z and reflection z 7→ zsince the product of these two orientation-reversing icosahedral symmetriesis 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 Galoisextension C(P1) ⊃ C(P1)Γ. These are obtained by taking subgroups of index5 in the Galois group A5 corresponding to the tetrahedron. However, aswe have already mentioned, we follow a slightly different approach and soimmediately turn our attention to the solution of the quintic. We first quicklygather the few facts about Tschirnhaus transformations that we need.

5In fact we can do better than birational equivalence; indeed we have an isomorphismof graded algebras C[P1]Γ ' C[f,H, T ]/(1728f5−T 2−H3) where f,H, T are given weights12, 30, 20 respectively. However we do not need this.

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3 Tschirnhaus and the canonical equation

A common approach when solving the polynomial equation:

xn + a1xn−1 + · · ·+ an = 0 (10)

is to begin by making the affine substitution y = x− a1/n and so eliminatethe term of degree n− 1. This substitution is a special case of the so-calledTschirnhaus transformation [33] in which y is allowed to be a polynomial ex-pression q in x. If αi are the roots of (10), the coefficients of the transformedequation: ∏

i

(y − q(αi)) = 0

are polynomials in the ai by Sn-invariance (or Newton’s identities).Using a Tschirnhaus transformation we can simultaneously eliminate fur-

ther terms in the original polynomial. For example if n ≥ 3 and a1 = 0, it iseasy to check that the substitution:

y = x2 + b1x + b2

simultaneously eliminates the terms of degree n− 1 and n− 2 provided thecoefficients b1, b2 satisfy the auxiliary polynomial conditions [26]:

b2 − p2/n = 0

p2b21 + 2p3b1 + (p4 − p2

2/n) = 0

where pj =∑

i αji are the power sums of the roots.

Thus, provided we are willing to allow ourselves the auxiliary square rootnecessary to solve the above quadratic for b1, we may take the general formof the quintic to be:

y5 + 5αy2 + 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 bypolynomials 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 theso-called Bring-Jerrard form:

y5 + y + γ = 0

However it is not in this form that the quintic most easily reveals its icosa-hedral connections and so, except for section 8.1 and appendix A, we shalltake the quintic in the form (11).

8

4 The space of roots and the icosahedral in-

variant

Given the quintic (11), if we exclude the trivial equation with α = β = γ = 0we 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 quinticlacks terms of degree 4 and 3, this orbit lies in the non-singular S5-invariantquadric surface:

Q ={

[y0, . . . , y4] ∈ P4 |∑

yi =∑

y2i = 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 quotientof each family. Each quotient is isomorphic to the map P1 → P1/Γ of section2 and the points the quintic obtained are the icosahedral invariants of thequintic. We shall describe this explicitly in section 5 but first we discuss thegeometry slightly more abstractly.

Thus let V = {(y0, . . . , y4) ∈ C5 |∑

yi = 0} and recall [34] that thespace of lines in P(V ) is naturally the Klein quadric:

K ={[ω] ∈ P(∧2V ) | ω ∧ ω = 0

}The line determined by a point [ω] ∈ K is L[ω] = P({v ∈ V | v ∧ ω = 0}).Now the quadratic form q =

∑i y

2i restricted to V provides an isomorphism

V ' V ∗ and so also an isomorphism:

q : ∧2V ' (∧2V )∗

Combining this with the natural isomorphism provided by the wedge product:

∧ : ∧2V ' (∧2V )∗ ⊗ ∧4V

we thus have an isomorphism:

∧2V ' ∧2V ⊗ ∧4V

Furthermore this isomorphism is equivariant with respect to the orthogo-nal group O(V ) = O(V, q). If we now fix an orientation on V we have anSO(V )-equivariant automorphism of ∧2V . This is of course simply the (com-plex) Hodge ?-operator and in this setting is self-inverse. Thus q provides adecomposition:

∧2V = ∧+ ⊕ ∧−

9

of ∧2V into the positive and negative eigenspaces of ?; this is a decompositioninto irreducible SO(V ) spaces. K meets the two planes P(∧±) in two non-singular conics C± and the natural map (intersection of lines) provides anSO(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 ∧4V is the sign representation and so in

view of the above it is desirable to restrict our representation S5 → O(V ) toA5 → SO(V ). To do this we must be able to associate an A5-orbit to ourquintic rather than a full S5 orbit. We achieve this by supplying a squareroot ∇ of the discriminant:

∇2 = 3125∏i<j

(yi − yj)2

= 108α5γ − 135α4β2 + 90α2βγ2 − 320αβ3γ + 256β5 + γ4 (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 icosahedralactions. We thus obtain two points:

Z± ∈ C±/A5 ' P1

Note, for the sake of definiteness, that we are using the unique isomorphismC±/A5 ' P1 sending the orbits with order 60/νi to 1, 0,∞ respectively forνi = 2, 3, 5. We thus have a (possibly infinite) numerical invariant Z± asso-ciated to our quintic.

By implicitly employing the isomorphism C± ' P1 above, we are usingthe fact that a non-singular conic is rational. In fact these conics have naturalrational parameterizations if we supply V with just a little extra structure(a complex spin structure). Specifically we take two 2-dimensional complexvectors 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 thisextra 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) theconic 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 wecan express C[Q] as a Segre product:

C[Q] ' ⊕n≥0C[P(S+)]n ⊗ C[P(S−)]n (15)

where C[P(S±)]n = H0(P(S±),O(n)) is the degree n summand of the gradedring C[P(S±)].

In the following section, where we actually calculate Z± in terms ofα, β, γ,∇, we give a more concrete, explicit construction of Z±.

5 Calculating the invariant

We wish to calculate Z± in terms of α, β, γ,∇ (which we now regard asparameters). This is mostly just an exercise in classical invariant theory forthe 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 y5 + qy3 + 5αy2 + 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 theexact sequence defining the homogeneous coordinate ring of Q. A priori weget 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) followtrivially. Demonstrating surjectivity amounts to showing that any functiond : G → (q) such that d(gh) = d(g) + g · d(h) for all g, h ∈ G is of the formd(g) = d− g · d for some d ∈ (q). This is easily seen to be the case by takingd = 1/|G|

∑g∈G d(g).6

Note also that in view of the relation ∇2 = D(α, β, γ), any A5-invariantpolynomial h on Q can be written uniquely in the form:

h = hs + ha∇6Of course what we’re really showing is that the group cohomology H1(G, (q)) = 0.

11

for unique polynomials hs, ha in α, β, γ determined by:

hs = (h + h∗)/2

ha∇ = (h− h∗)/2(17)

where h∗ is the polynomial obtained by transforming h under any odd per-mutation.

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 α, β, γ. We shall makerepeated use of this essential fact below. We now introduce coordinates tomake all this explicit.

We first change basis from y0, . . . , y4 to the (Fourier-dual) basis in whichS is diagonal:

pk =∑

j

εkjyj (18)

where as before ε = e2πi/5. Our subspace V now has equation p0 = 0 andour quadratic form q, restricted to V , is defined in terms of the pk by 25q =10(p1p4 + p2p3). We let S+ have coordinates λ1, λ2, S− have coordinatesµ1, µ2 and define the map (14) via the bilinear map:

p1 = 5λ1µ1 p2 = −5λ2µ1 p3 = 5λ1µ2 p4 = 5λ2µ2 (19)

Thus [λ1, λ2] are homogeneous coordinates on C+ and [µ1, µ2] homogeneouscoordinates on C−. The A5-action on the λi is given by the formulae (2), (3)and the dual action on the µi is obtained by replacing ε with ε2 in the sameformulae. The S5-action of course interchanges the λi and µi, indeed the oddpermutation R = (1243) acts as ([λ1, λ2], [µ1, µ2]) 7→ ([µ2,−µ1], [λ1, λ2]).

Recalling our formulae for f, H, T from section 2 we define:

f1 = f(λ1, λ2) f2 = f(µ1, µ2)

and similarly we define H1, H2 and T1, T2. Note that by (9), R interchangesf1, f2 and similarly for the Hi, Ti. From now on we also denote Z± by Z1, Z2.In this notation we now have:

Z1 =H3

1

1728f 51

We wish to express this in terms of α, β, γ,∇. We thus write:

Z1 =H3

1f52

1728(f1f2)5(20)

12

Now f1f2 fixed by R and so is S5-invariant. It thus lies in C[α, β, γ]. Sinceit is of degree 12, it must be a linear combination of α4, β3, αβγ. To fixthe coefficients we compare leading coefficients as polynomials in λi, µi. It isstraightforward to verify that:

α =− λ31µ

21µ2 − λ2

1λ2µ32 − λ1λ

22µ

31 + λ3

2µ1µ22

β =− λ41µ1µ

32 + λ3

1λ2µ41 + 3λ2

1λ22µ

21µ

22 − λ1λ

32µ

42 + λ4

2µ31µ2

γ =− λ51(µ

51 + µ5

2) + 10λ41λ2µ

31µ

22 − 10λ3

1λ22µ1µ

42−

10λ21λ

32µ

41µ2 − 10λ1λ

42µ

21µ

32 + λ5

2(µ51 − µ5

2)

The coefficient of λ121 in f1f2 is 0 whereas the same coefficients in α4, β3, αβγ

are µ81µ

42,−µ3

1µ92,−µ8

1µ42 − µ3

1µ92 respectively. From this we see that we must

have f1f2 = A(α4 − β3 + αβγ) for some constant A. Furthermore, uponnoting that the coefficient of λ11

1 λ2µ111 µ2 in f1f2 is 1 whereas it is 0 in α4, β3

and 1 in αβγ we learn that A = 1. In other words we obtain:

f1f2 = α4 − β3 + αβγ

This deals with the denominator in (20); we turn our attention to the nu-merator.

Decomposing the numerator of (20) using (17) and recalling that our oddpermutation R interchanges the f1, f2 as well as H1, H2, we get:

H31f

52 =

H31f

52 + H3

2f51

2+

H31f

52 −H3

2f51

2= p +∇q (21)

where p, q are polynomials in α, β, γ.We could now attempt to calculate p, q by the same means that we cal-

culated f1f2 above but this would be an enormous amount of work since p, qhave degrees 60, 50 respectively. Instead, recall that we have the syzygies:

T 2i = 123f 5

i −H3i

Multiplying these together and rearranging we obtain:

2p = H31f

52 + H3

2f51 = 123(f1f2)

5 + 12−3(H1H2)3 − 12−3(T1T2)

2

We will thus have the required expression for p in terms of α, β, γ as soon aswe express H1H2 and T1T2 in these terms. To do this we do follow the sameprocedure as that used to find f1f2 above and (admittedly with somewhatmore effort) we obtain:

H1H2 = γ4 + 40α2βγ2 − 192α5γ − 120αβ3γ + 640α4β2 − 144β5

T1T2 = γ6 + 60α2βγ4 + 576α5γ3 − 180αβ3γ3 + 648β5γ2 − 2760α4β2γ2+

7200α7βγ − 1728α10 + 9360α3β4γ − 2080α6β3 − 16200α2β6

13

It remains only to calculate q. This time the trick we use is to note thatas well as (21) above, we have H3

2f51 = p−∇q and so:

(H1H2)3(f1f2)

5 = p2 −∇2q2

It follows that taking our above polynomial expressions for H1H2, f1f2,∇2 =D(α, β, γ), p we must find a factorization of ((H1H2)

3(f1f2)5−p2)/D(α, β, γ).

From this we determine7:

2q = ± (−8α5γ − 40α4β2 + 10α2βγ2 + 45αβ3γ − 81β5 − γ4)·(64α10 + 40α7βγ − 160α6β3 + α5γ3−5α4β2γ2 + 5α3β4γ − 25α2β6 − β5γ2)

The two signs corresponding to the two invariants: Z1, Z2. With this formulain hand we have achieved our goal of expressing Zi in terms of α, β, γ,∇.

Our derivation and formulae for Zi in terms of α, β, γ,∇ differ fromKlein’s ([17], pg. 213) and indeed from any other accounts we have comeacross which all reproduce Klein’s formulae exactly. In fact our approach ismore closely connected to Gordon’s work [9]. Klein does outline the connec-tion but he leaves it unfinished as he has already obtained his formulae forZi by other means. Although the formulae for Zi in [17] have the advantageof being more concise than ours, they have the disadvantage that Zi is rep-resented as a rather complicated rational expression (rather than in termspolynomials) and that Zi cannot be evaluated along the Bring curve α = 0easily.

6 Obtaining the roots

The invariants Zi defined above are obtained by applying the icosahedralfunction I to the coordinates λi, µi. These coordinates determine a point inthe projective space of vectors of roots P(V ), and so if we could invert theicosahedral function I and express λi, µi in terms of the invariants Zi, wewould be able to solve the quintic. We will show how to invert I in section 7below but first we fill in the details of how an inverse for I yields a solutionof the quintic.

We begin by using (18), (19) to express roots in terms of λi, µi:

yν = ε4νλ1µ1 − ε3νλ2µ1 + ε2νλ1µ2 + ενλ2µ2 (22)

7That q itself turns out to factorize like this is unexpected, at least to the author.

14

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 linearin µi then we can use these to eliminate the µi in (22) and so express yν interms of just α, β, γ and our solution [λ1, λ2] of (23). We thus enlarge thering 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 formslinear in µi of lowest degree in λi are:

N1 = (7λ51λ

22 + λ7

2)µ1 + (−λ71 + 7λ2

1λ52)µ2

M1 = (λ131 − 39λ8

1λ52 − 26λ3

1λ102 )µ1 + (−26λ10

1 λ32 + 39λ5

1λ82 + λ13

2 )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 iff, g are two homogeneous polynomials in λ1, λ2 then the rth transvectant off, g is given by:

(f, g)r =r∑

i=0

(−1)i

i!(r − i)!

∂rf

∂λr−i1 ∂λi

2

∂rg

∂λi1∂λr−i

2

We extend this to homogeneous polynomials in both λi, µi using bilinearity,i.e., if:

f =∑i,j

fijλi1λ

a−i2 µj

1µb−j2

g =∑k,l

gklλk1λ

c−k2 µl

1µd−l2

then we define the (r, s)-transvectant:

(f, g)r,s =∑i,j,k,l

fijgkl(λi1λ

a−i2 , λk

1λc−k2 )r(µ

j1µ

b−j2 , µl

1µd−l2 )s

Using these transvectants, we can construct the invariants we need. Indeedit 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 arelinear in the µi 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 branchlocus in P(S−) must be the set of vertices of an icosahedron in P(S−) in eachcase (since they are stable under the A5 action). For N1, there are 12 doubly-ramified 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 12points in the branch locus. These 60 points are the A5-orbit containingthe point λ1/λ2 = 71/5, i.e., the orbit with icosahedral invariant 64/189.It is interesting to wonder if it is possible to characterise the configurationof these points geometrically. The set of vertices of an Archimedean solidwith icosahedral symmetry might seem like a plausible candidate but in factthis does not work as can be seen easily since the point 71/5 lies outsidethe field of coefficients of these solids (though the points are remarkablyclose to the vertices of the truncated dodecahedron). Similarly there are twodistinguished orbits for M1. If it is possible to come up with a geometriccharacterisation of these various orbits then it should also apply to the othertwo A5-equivariant branched covers (also identified by Gordon in [9]) whichhave degrees 17 and 23.

Returning to the task at hand we solve the 2× 2 system (24) and expressµi in terms of M1, N1 and using (22) obtain:

yν = H−11 bνM1 + H−1

1 cνN1 (25)

Here H1 appears as it is the determinant of the matrix which we invert andthe coefficients bν , cν are given by:[

bν cν

]=[

ε4νλ1 − ε3νλ2 ε2νλ1 + ενλ2

] [−λ7

1 + 7λ21λ

52 26λ10

1 λ32 − 39λ5

1λ82 − λ13

2

−7λ51λ

22 − λ7

2 λ131 − 39λ8

1λ52 − 26λ3

1λ102

]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 λi and µi

which we can thus express in terms of α, β, γ,∇. We thus rewrite (25) as:

yν =bνf1

H1

· M1f2

f1f2

+cνT1

H1f 21

· N1f21 T2

T1T2

The methods described in section 5 then allow us to calculate:

M1f2 = (11α3β + 2β2γ − αγ2)/2−∇α/2

N1f21 T2 = r +∇s

16

where:

2r = α2γ5 − αβ2γ4 + 53α4βγ3 + 64α7γ2 − 7β4γ3 − 225α3β3γ2−12α6β2γ + 216α9β + 717α2β5γ − 464α5β4 − 720αβ7

2s = − α2γ3 + 3αβ2γ2 − 9β4γ − 4α4βγ − 8α7 − 80α3β3

and since we already have formulae for f1f2 and T1T2 we have the requiredexpression 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 determinedthe list parameters for which the hypergeometric differential equation hasfinite monodromy. We thus need only cast our problem in the right form andappeal to general theory. The solution can be written in terms of Schwarzs-functions (as discussed in [23] for example).

We must identify domains of injectivity for I (equivalently, fundamentaldomains for the action of Γ on P1). We thus note that if r is any reflectionabout a plane of symmetry of the icosahedron then:

I ◦ r = I

for the reasons explained when discussing (9). Since there is a plane ofsymmetry through any edge of the icosahedron as well as a plane of symmetrythrough each of the altitudes of any face of the icosahedron, it follows that theedges and altitudes of the faces of the icosahedron constitute the preimage ofRP1 = R∪∞ under I. The altitudes divide each face into six triangles withangles 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 toupper half space H+ and three of them biholomorphically to lower half spaceH−, according to whether their vertices are sent to 0, 1,∞ in anti-clockwiseor clockwise order respectively. Subdividing faces like this, figure 1 becomes:

17

Figure 4: Icosahedral tiling of sphere. I maps the interior of eachlight and dark triangle biholomorphically onto the upper and

lower half-planes respectively.

Under stereographic projection, the subdivision of the face with vertices0, ε + ε−1, ε2 + 1 is:

0 t

h

ε+ε-1

ε2+1

Figure 5: Domains of injectivity for I under stereographicprojection. 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 thetriangle with vertices 0, t, h. I maps the interior biholomorphically to the

18

lower half-plane H− and the boundary homeomorphically to RP1, sendingthe points 0, t, h to ∞, 1, 0 respectively.

Now the equation (26) we wish to solve is a degree 60 polynomial but ithas the special property that its monodromy lies in PSL(2, C), i.e., the 60branches of a local inverse are related by a Mobius transformation. Sincethe Schwarzian derivative is invariant under Mobius transformations, theSchwarzian derivative of the inverse is independent of the branch; moreoverit is easily calculated. This yields a differential equation for the inverse which,using standard methods, may be transformed to Gauss’s hypergeometric dif-ferential equation. It then follows that an inverse may be expressed as aratio of linearly independent solutions to this equation, i.e., as a Schwarzs-function.

Indeed, as shown in [23], the inverse we seek is determined up the scaleby the angles in the triangle, π/νi, and is given by:

s : H− −→ C

Z 7−→ CZc−1

2F1(a′, b′; c′; Z−1)

2F1(a, b; c; Z−1)

Here 2F1 is the analytic continuation of Gauss’s hypergeometric series toC− [1,∞), Zc−1 is defined using the principle branch of log on C− (−∞, 0].The values of the constants are:

a = 12

(1− 1

ν3+ 1

ν2− 1

ν1

)b = 1

2

(1− 1

ν3− 1

ν2− 1

ν1

)c = 1− 1

ν3

a′ = a− c + 1 b′ = b− c + 1 c′ = 2− c

To determine C we compare the coefficients of Z on both sides of thepower series identity:

Z ·H3(z1, z2) = f 5(z1, z2)

with z1 = CZ1−c2F1(a

′, b′; c′; Z) and z2 = 2F1(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) =2F1(

3160

, 1160

; 65; Z−1)

(1728Z)1/52F1(

1960

,− 160

; 45; Z−1)

8An alternate means of determining C would be to impose the condition s(1) = tand to use Gauss’s expression for 2F1(a, b, c, 1) in terms of Γ-functions. This yields C =

t · Γ( 45 )Γ( 41

60 )Γ( 6160 )

Γ( 65 )Γ( 29

60 )Γ( 4960 )

. Since t = 2 (cos(π/10)− cos(π/5)) =√

5+√

52 − 1+

√5

2 and we already

know C = 1728−1/5 we obtain a Γ-function identity.

19

Using a similar expression for im(Z) > 0, we could extend this function to theopen set H+∪H−∪(0, 1) so that we would have an inverse for the restrictionof I to the interior of the triangle with vertices 0, ε + ε−1, h. (Indeed usingSchwarz reflection, we can extend s across any of the edges of the triangleand 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 thequintic as concisely as possible, subject to the conditions of remaining asexplicit as [17] and as self-contained as possible. As a result we have beenforced us to omit discussion of many related matters. We comment brieflyon 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 tothe so-called Bring-Jerrard form:

y5 + y + γ = 0

but that we would work with the quintic in the form (11). We did thisbecause we were following [17], because (11) is more general and because itis easy to bring out the icosahedral connection using the A5 actions on thelines in the doubly-ruled quadric surface. However there is an appealing wayto connect the icosahedron with the quintic in Bring-Jerrard form which isworth mentioning. The construction below is described in [11].

Firstly note that the family of quintics in Bring-Jerrard form is the smoothgenus 4 curve B cut out of P4 by the equations

∑yi =

∑y2

i =∑

y3i = 0.

This is known as the Bring curve and has automorphism group S5 correspond-ing 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 do-

decahedron GD. This regular solid, which self-intersects in R3, has one facefor each vertex of the icosahedron. It is formed by spanning the five neigh-bouring vertices of each vertex of the icosahedron with a regular pentagonand then dismissing the original icosahedron. GD thus has the same 12 ver-tices and 30 edges as the icosahedron but only 12 faces. Projection ontothe common circumsphere S yields a triple covering GD → S with a doublebranching 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 genus4. In fact, as explained in [11], GD is isomorphic to the Bring curve.

The isomorphism GD ' B can be used to bring out the relationshipbetween the quintic and the icosahedron.

8.2 Modular curves, Ramanujan’s continued fractionand Shioda’s modular surface

From one point of view, the exceptional geometry of the quintic is a resultof the exceptional isomorphism:

A5 ' PSL(2, 5)

Corresponding to the exact sequence defining the level 5 principal con-gruence subgroup of the modular group:

0 → Γ(5) → PSL(2, Z) → PSL(2, 5) → 0

there is a factorization of the modular quotient:

j : H∗ j5−→ X(5)I−→ X(1)

where H∗ = H+ ∪QP1 is the upper half-plane together with the PSL(2, Z)-orbit of ∞ and X(N) is the compactified modular curve of level N . Thecurves X(5), X(1) are rational and the map I : X(5) → X(1) is a quotientby PSL(2, 5) and is thus an icosahedral quotient. We can use this to find aninverse for the icosahedral function I in terms of Jacobi ϑ-functions (providedwe are willing to invert Klein’s j-invariant). Indeed the map j5 may beexpressed as:

j5(τ) = q2/5

∑Z q5n2+3n∑Z q5n2+n

= q−3/5 ϑ1(πτ ; q5)

ϑ1(2πτ ; q5)(27)

where q = eπτi 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 Ramanu-jan’s continued fraction:

j5(τ) =q1/5

1 +q

1 +q2

1 +q3

1 + · · ·

21

Furthermore because we know that the icosahedral vertices, edge midpointsand face centres in X(5) lie above the points∞, 1, 0 in X(1), we can calculatethe values of this continued fraction at those orbits in H∗ which 1

1728j maps

to ∞, 1, 0. For example j(i) = 1728 and the corresponding edge midpointequality:

j5(i) = t =

√5 +

√5

2− 1 +

√5

2

is one of the identities that famously caught Hardy’s eye when Ramanujanfirst wrote to him. An excellent account of these results together with a proofof (27) can be found in [6].

This method of inverting I treats the icosahedral invariant Z of a quinticas a point in the moduli space of elliptic curves X(1). It is natural to wonderif the quintic in fact provides a geometric realization of the elliptic curve withj-invariant Z. Since X(5) is the moduli space of elliptic curves together withlevel 5 structure, inverting I would then correspond to finding a basis for the5-torsion of such a curve in which the Weil pairing is in standard form. Ifthis were the case we would expect a link with the elliptic curves embeddedin P4 defined in [14] as an intersection of quadrics. The union of all of thesecurves is an irreducible surface (with singularities) whose normalization isShioda’s modular surface (a compactification of the universal elliptic curvewith level 5 structure).

8.3 Parting words

There is of course much more to say beyond even those remarks in sections8.1 and 8.2 above. Other matters which we shall not discuss at all but whichseem worth mentioning include the following:

• The group PSL(2, 5) is one of a ‘trinity’of groups, the other two beingPSL(2, 7) and PSL(2, 11). Higher degree equations can have theselatter two groups as their Galois group and for these there exists ahigher genus version of the icosahedral solution of the quintic.

• The rational parameterization of the singularity T 2 + H3 = 1728f 5

we have described can be used to find solutions of the Diophantineequation a2 + b3 + c5 = 0. See Beukers [1] for details.

• We have worked exclusively over C. Needless to say, this is not neces-sary, the field Q(eπi/5) suffices for most of the above. Serre makes somehelpful comments in this direction in his letter [27].

22

• The icosahedral solution of the quintic is not usually the most efficienttechnique 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 tothe solution of the quintic discovered by Lambert9 [19] in 1758 and again byEisenstein [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) = 0has power series:

y(γ) =∑k≥0

(5k

k

)γ4k+1

4k + 1(29)

(This can also be expressed in terms of the generalized hypergeometric func-tion as: y(γ) = 5F4

(1, 4

5, 3

5, 2

5, 1

5; 5

4, 1, 3

4, 1

2; 5(5γ

4)4

)γ.)

This appealing result is established using analytic methods (Lagrange in-version) in [24] and [29] as well as [21]. However since this statement is reallyan identity of binomial coefficients it is desirable to have a combinatorialproof for the identity (28) satisfied by the generating function (29).

Now the coefficients in (29) are a special case of the sequence:

pdk =1

(p− 1)k + 1

(pk

k

)which specializes to the Catalan numbers for p = 2. This sequence, con-sidered long ago by Fuss [8], was studied in some detail in [13]. Just asvarious identities for the Catalan numbers can be established by observingthat the they count (amongst many other things) certain lattice paths, sotoo can those identities for pdk which we seek for p = 5 be established bydemonstrating that these coefficients count certain paths introduced in [13].

Although the results in [13] thus provide a combinatorial proof of thegenerating function identity (28), there is a more direct combinatorial proof

9It should be pointed out that Lambert would not have been aware that his methodprovided a solution of the general quintic since the reduction the Bring-Jerrard form wasnot known in his time.

23

presented in [10] based on an observation of Raney [25]. He noticed that ifa1, . . . , am is any sequence of integers that sum to 1, then exactly one of them 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 pdk.The natural recursive structure of such sequences provided by concatenationof p such sequences, followed by a terminating value of 1−p then correspondsto the identity we seek. The interested reader will find details in [10].

References

[1] F. Beukers. The diophantine equation axp + byq = czr. Duke Mathe-matical Journal, 91(1), 1998.

[2] E.S. Bring. Meletemata quaedam mathematematica circa transforma-tionem aequationum algebraicarum. Lund University, Promotionschrift,1786.

[3] W. Crawley-Boevey. Lectures on representation theory and invarianttheory. Erganzungsreihe Sonderforschungsbereich 343 ’Diskrete Struk-turen 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 LondonMathematical Society Lecture Note Series. Cambridge University Press,2003.

[6] W. Duke. Continued fractions and modular functions. Bulletin of theAmerican Mathematical Society, 42(2):137–162, 2005.

[7] F.G.M. Eisenstein. Allgemeine auflosung der gleichungen von den erstenvier graden. J. Reine Angew. Math., 27:81–83, 1884.

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[8] N. Fuss. Solutio quaestionis, quot modis polygonum n laterum in polyg-ona m laterum, per diagonales resolvi quaeat. Nova acta academiaescientiarum imperialis Petropolitanae, 9:243–251, 1791.

[9] P. Gordon. Ueber die auflosung der gleichungen vom funften 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 resolution de l’equation du cinqueme degre (1858).Oevres de Charles Hermite, 2:5–21, 2009.

[13] P. Hilton and J. Pedersen. Catalan numbers, their generalization, andtheir uses. Mathematical Intelligencer, 13(2):64–75, 1991.

[14] K. Hulek. Geometry of the horrocks-mumford bundle. In Proceedingsof Symposium in Pure Math, volume 46, pages 69–85. American Math-ematical 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 uber das Ikosaeder und dieAuflosung der Gleichungen vom funften Grade. Birkhauser, 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? http://

mathoverflow.net/questions/32099/#32261, 2010.

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

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[23] Z. Nehari. Conformal mapping. McGraw-Hill, 1952.

[24] S.J. Patterson. Eisenstein and the quintic equation. Historia mathemat-ica, 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 icosaedriques. Seminaire de Theorie des Nombresde Bordeaux, 19(123):550–552, 1979.

[28] P. Slodowy. Das ikosaeder und die gleichungen funften grades. Arith-metik und Geometrie. Vier Vorlesungen (Mathematischen Miniaturen,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-hypergeometricseries. In Department Of Mathematics, Texas A&M University, CollegeStation, TX 77843. (EM), pages 171–181, 1996.

[31] G. Toth. Finite Mobius groups, minimal immersions of spheres, andmoduli. Springer, 2002.

[32] G. Toth. Glimpses of Algebra and Geometry. Springer, second edition,2002.

[33] E. Tschirnhaus. Methodus auferendi omnes terminos intermedios exdata 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 Univer-sity Press, Cambridge, 1990.

[35] E.T. Whittaker and G.N. Watson. A Course of Modern Analysis. Cam-bridge University Press, 1927.

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