# The Quintic, the Icosahedron, and Elliptic Curves

Communicated by *Notices* Associate Editor William McCallum

There is a remarkable relationship between the roots of a quintic polynomial, the icosahedron, and elliptic curves. This discovery is principally due to Felix Klein (1878), but Klein’s marvellous book 9 misses a trick or two, and doesn’t tell the whole story. The purpose of this article is to present this relationship in a fresh, engaging, and concise way. We will see that there is a direct correspondence between:

- •
“Evenly ordered” roots of a Brioschi quintic1

- •
Points on the icosahedron, and

- •
Elliptic curves equipped with a primitive basis for their 5-torsion, up to isomorphism.

Moreover, this correspondence gives us a very efficient direct method to actually *calculate* the roots of a general quintic! For this, we’ll need some tools both new and old, such as Cremona and Thongjunthug’s complex arithmetic geometric mean 3, and the Rogers-Ramanujan continued fraction 512. These tools are not found in Klein’s book, as they had not been invented yet!

If you are impatient, skip to the end to see the algorithm.

If not, join me on a mathematical carpet ride through the mathematics of the last four centuries. Along the way we will marvel at Kepler’s Platonic model of the solar system from 1597, witness Gauss’ excitement in his diary entry from 1799, and experience the atmosphere in Trinity College Hall during the wonderful moment Ramanujan burst onto the scene in 1913.

For the approach I present here, I have learnt the most from Klein’s book itself together with the new introduction and commentary by Slodowy 9, as well as 25811.

## Arnold’s Topological Proof of the Unsolvability of the Quintic

We are all familiar with the formula for the roots of a quadratic polynomial namely ,

2Perhaps we are also familiar with Cardano’s formula (1545) for the roots of a cubic polynomial ,

3There is a similar formula for the roots of a quartic polynomial, due to Ferrari (1545). We say formulas like 2 and 3 express the roots of a polynomial *in terms of radicals*, since the only ingredients necessary are the usual algebraic operations ( , , , and extraction of ) roots. th

It is also commonly known that Ruffini (1799) and Abel (1824) showed that there is no such radical formula for the roots of a general *quintic* equation. The standard modern way to understand these results is the algebraic framework of Galois (1832). Namely, we associate to a *specific* polynomial

a finite group the ,*Galois group* of ,which is a certain subgroup of the group of permutations of the roots of Galois showed that there is a radical formula for . if and only if is a *solvable* group (a tower built from iteratively stacking finite cyclic groups on top of each other, i.e., “built from epicycles” as Ptolemy might have put it). Now, the Galois group of a general quintic is which is not solvable. Therefore, there is no radical formula for the roots of a general quintic. ,

Galois’ approach is elegant but requires a semester’s worth of abstract algebra to understand. In 1963, the Russian mathematician Vladimir Arnold gave an alternative *topological* proof of the unsolvability of the quintic in a series of lectures to high school kids in Moscow. In Arnold’s approach, instead of focusing on finding an algebraic formula for the roots of a *specific* polynomial as in 4, one considers the collection of *all* polynomials of degree as a *topological space*:

To each polynomial we may associate the unordered set of its roots, so that we have a *covering space*

which is branched over the discriminant locus of polynomials with multiple roots, and is an bundle over the complement -principal .

If we start at some fixed basepoint polynomial and move along a path in , the roots of the polynomial move around (see Figure ,1). If we loop back to then they will have undergone a permutation. We have established a ,*monodromy map*

which in fact classifies the covering space .

(Note how this approach is more geometric than that of Galois. Instead of caring only about the *permutations* of the roots, we also care about the *journey* they undertook to accomplish that permutation.)

Arnold’s insight was to show that if there is a radical formula for the roots of a general polynomial, then the “dance of the roots” cannot be overly complex, in the sense that the image of the monodromy map must be a solvable subgroup of But, for example, the 1-parameter Brioschi family of quintics .

has monodromy group (see Figure 1), which is certainly not solvable since it is simple, as we will see by relating it to the icosahedron in the next section. Hence the unsolvability of the quintic.

In fact, after some algebraic manipulations involving at most two square roots 2, Theorem 6.6, finding the roots of the general quintic

6reduces to finding the roots of the Brioschi quinticFootnote^{1} for a certain .

## Enter the Icosahedron

Now for a wonderful fact: we will show that there is a natural correspondence between the set of “evenly ordered” d5-tuples of roots of a Brioschi quintic and the set of points on the icosahedron! ,

To understand this, recall that the icosahedron

that most enigmatic of the Platonic solids, has 30 edges, 20 equilateral faces, and 12 vertices. If we make the identification we can take the vertices to be situated at: ,

7Consider the group of rotational symmetries of Each nonidentity . is a rotation about an axis through an antipodal pair of edge midpoints, face midpoints, or vertices of with order 2, 3, and 5 respectively. In fact, , is naturally isomorphic to the group of even permutations of 5 things. What are these 5 things that are being evenly permuted when we rotate the icosahedron? They are the 5 inscribed octahedra which have their vertices on the edge midpoints of , See Figure !2.

This natural isomorphism between and gives us a nice way to see that is simple (and hence not solvable). This is because is simple: if a normal subgroup contains a rotation about an axis through a vertex then it contains rotations about ,*all* axes passing through vertices (since it is closed under conjugation). But a rotation about an edge midpoint equals the product of the rotations about the three vertices in a triangle adjacent to it, while a rotation about a face midpoint equals the product of the rotations about the two vertices in an edge adjacent to it. So if contains a rotation about it must be all of , and similarly for edge midpoints and face midpoints. So, , is simple, and therefore is simple.

## Invariant Polynomials

Let denote the 2-dimensional unit sphere. Since is a group of rotational symmetries, it acts on Our goal in this section is to understand the quotient space . which we can think of as the “moduli space” of points on the “round icosahedron” (the soccer ball version of , obtained by inflating , outward onto the sphere see Figure ;3). We are going to need a toolbox of functions on -invariant .

To write down such functions, we need to keep in mind that has the structure of a Riemann surface, since we can identify it with the complex projective plane

via stereographic projection from the north pole,

followed by the identification

In what follows, I will freely use these identifications; my preference is to use the

Under this identification, the *binary icosahedral group*

We have the vertex polynomial

To see that these polynomials are indeed

which similarly generate

It is clear that our polynomials 8–10 are invariant under these transformations, so they are indeed

We can play the same game with our 5 inscribed octahedra. Let

where

## The Icosahedron and the Quintic

When we rotate the icosahedron, the 5 inscribed octahedra and hence their vertex polynomials

whose coefficients are symmetric polynomials in the octahedral polynomials

as the reader will verify! (The coefficients of

Let us make this correspondence between points on the icosahedron and ordered roots of a Brioschi quintic clear and precise. Consider the rationalized octahedral functions

They are of degree zero in *ordered* tuple

thereby forgetting their ordering. If we multiply out this quintic, we find that it is a Brioschi quintic

with Brioschi parameter

The only subtlety here is the notion of an “evenly” ordered set of roots of a Brioschi quintic *even* if, as

Theorem 1 tells us that in order to find the roots of a Brioschi quintic

## A Polynomial Relation

But first, let’s return to the subject of invariant polynomials on the icosahedron for a moment, as there is an important relation between

This reminds us of modular forms, where the Eisenstein series satisfies

This is the first clue that the icosahedron has something to do with moduli spaces of elliptic curves. But for now, what we need to get out of 13 is that it tells us that

so that

## Enter Elliptic Curves

Expressing a number “in terms of radicals” implies having access to the roots of unity, i.e., the *elliptic curve*, i.e., the set

also forms an abelian group (once one works projectively). It was natural to speculate that, while the roots of a quintic could not be expressed in terms of

In this way,

the solution of the general equation of degree 5 is made to depend upon the equations for the division of periods of the elliptic functions, as they used to say.

Hermite and Kiepert worked with the Bring-Jerrard form of the quintic, and their final expression for the roots of the quintic in terms of the 5-torsion points of an elliptic curve is, to this humble author, a bit convoluted and indirect. I will present my own streamlined and modernized form of Klein’s approach (1878) 9. We will see that the 60 evenly ordered 5-tuples of roots *bases* (not points themselves) for the 5-torsion points of an elliptic curve!

## Moduli Spaces of Elliptic Curves

In the previous section we found that the icosahedron is a 60-sheeted

Besides the icosahedron, there is another 60-sheeted

Recall that an elliptic curve

Let

for some matrix

We say that *primitive basis* for the 5-torsion of

for the set of equivalence classes (“moduli space”) of pairs *equivalent* if there is an isomorphism

Write

Here,

is the *principal congruence subgroup of level 5*. (The congruence relation is done independently entrywise, so the requirement is that

## Permutation Wizardry

Now,

The magic is that ^{2}.

The vertices of an inscribed octahedron

On the other hand,

which also consists of 6 things. So, let us identify the six vertex axes of the icosahedron in

In this way

Let us pre- and post-compose this isomorphism with the natural isomorphisms

## An Isomorphism of Covering Spaces

The isomorphism