The Quintic, the Icosahedron, and Elliptic Curves

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 $(x_1, \ldots , x_5)$ of a Brioschi quintic$$\begin{equation} X^5 + 10BX^3 + 45B^2X + B^2 = 0, {\tag{1}} \end{equation}$$

•

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 $x_1, x_2$ of a quadratic polynomial $X^2 + aX + b = 0$, namely

$$\begin{equation} x = \frac{-a \pm \sqrt {a^2 - 4b}}{2}. {\tag{2}} \end{equation}$$

Perhaps we are also familiar with Cardano’s formula (1545) for the roots of a cubic polynomial $X^3 + aX + b = 0$,

$$\begin{equation} x = c - \frac{a}{3c}, \quad \quad c = \sqrt [3]{-\frac{b}{2} + \sqrt {\frac{b^2}{4} + \frac{a^3}{27}}}. {\tag{3}} \end{equation}$$

There 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 ($+$, $-$, $\cdot$, $/$) and extraction of $n$th roots.

It is also commonly known that Ruffini (1799) and Abel (1824) showed that there is no such radical formula for the roots $x_i$ 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

$$\begin{equation} P = X^n + a_1 X^{n-1} + a_2 X^{n-2} + \cdots + a_{n-1} X + a_n {\tag{4}} \end{equation}$$

a finite group $G$, the Galois group of $P$ ,which is a certain subgroup of the group $S_n$ of permutations of the roots $x_1, \ldots , x_n$ of $P$. Galois showed that there is a radical formula for $x_i(a_1, \ldots , a_n)$ if and only if $G$ 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 $S_5$, 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 $P$ as in 4, one considers the collection of all polynomials of degree $n$ as a topological space:

$$\begin{equation*} \operatorname {Poly}_n = \{\mathrm{polynomials}\; X^n + a_1 X^{n-1}+ \cdots + a_{n-1}X + a_n,a_1, \ldots , a_n \in \mathbb{C} \}. \end{equation*}$$

To each polynomial $P \in \operatorname {Poly}_n$ we may associate the unordered set $\{x_1, \ldots , x_n\}$ of its roots, so that we have a covering space

$$\begin{align*} \operatorname {Roots}& \twoheadrightarrow \operatorname {Poly}_n \\ \{x_1, \ldots , x_n\} & \mapsto \prod _{i=1}^n (X - x_i) \end{align*}$$

which is branched over the discriminant locus $D \subset \operatorname {Poly}_n$ of polynomials with multiple roots, and is an $S_n$-principal bundle over the complement $\mathcal{P}_n = \operatorname {Poly}_n \setminus D$.

If we start at some fixed basepoint polynomial $P_0 \in \mathcal{P}_n$, and move along a path in $\mathcal{P}_n$, the roots of the polynomial move around (see Figure 1). If we loop back to $P_0$, then they will have undergone a permutation. We have established a monodromy map

$$\begin{equation} \pi _1(\mathcal{P}_n, P_0) \rightarrow S_n {\tag{5}} \end{equation}$$

which in fact classifies the covering space $\mathcal{P}_n$.

(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 $S_n$. But, for example, the 1-parameter Brioschi family of quintics

$$\begin{equation*} P_B (X) = X^5 + 10B X^3 + 45B^2X + B^2, \quad B \in \mathbb{C} \end{equation*}$$

has monodromy group $A_5$ (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

$$\begin{equation} X^5 + a_1 X^4 + a_2 X^3 + a_3X^2 + a_4 X + a_5=0 {\tag{6}} \end{equation}$$

reduces to finding the roots of the Brioschi quintic⁠¹ $P_B$ for a certain $B \in \mathbb{C}$.

¹

Using the Brioschi quintic as normal form is one trick which Klein missed, as he instead used a different form — the Bring-Jerrard quintic — where the maximal number of coefficients $a_i$ in 6 are zero.

✖

Figure 1.

When the Brioschi parameter loops around $B=0$ along the blue loop $\sigma$ as shown on the left, the roots undergo the cyclic permutation $(14352)$ as shown on the right. In a similar way, when $B$ loops around $B=-1/1728$ along the red loop $\gamma$, the roots undergo the cyclic permutation $(153)$. These permutations generate $A_5$.

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 $(x_1, \ldots , x_5)$ of roots of a Brioschi quintic $P_B$, and the set of points on the icosahedron!

To understand this, recall that the icosahedron

$$\begin{equation*} I \subset \mathbb{R}^3, \end{equation*}$$

that most enigmatic of the Platonic solids, has 30 edges, 20 equilateral faces, and 12 vertices. If we make the identification $\mathbb{R}^3 = \mathbb{C} \times \mathbb{R}$, we can take the vertices to be situated at:

$$\begin{equation} (0, \pm 1), \quad \pm \frac{1}{\sqrt {5}}(-2\zeta ^i, 1), \quad \zeta = e^{\frac{2 \pi i}{5}}. {\tag{7}} \end{equation}$$

Consider the group $G$ of rotational symmetries of $I$. Each nonidentity $g \in G$ is a rotation about an axis through an antipodal pair of edge midpoints, face midpoints, or vertices of $I$, with order 2, 3, and 5 respectively. In fact, $G$ is naturally isomorphic to $A_5$, 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 $I$! See Figure 2.

Figure 2.

One of the 5 inscribed octahedra in the icosahedron. On the right, Kepler’s view of planetary orbits as inscribed Platonic solids from Mysterium Cosmographicum (1596).

 

This natural isomorphism between $G$ and $A_5$ gives us a nice way to see that $A_5$ is simple (and hence not solvable). This is because $G$ is simple: if a normal subgroup $N$ contains a rotation about an axis through a vertex $v$, 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 $N$ contains a rotation about $v$, it must be all of $G$, and similarly for edge midpoints and face midpoints. So, $G$ is simple, and therefore $A_5$ is simple.

Invariant Polynomials

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

To write down such functions, we need to keep in mind that $S^2$ has the structure of a Riemann surface, since we can identify it with the complex projective plane

$$\begin{equation*} \mathbb{CP}^1 = \{ \text{1-dimensional linear subspaces of $\mathbb{C}^2$} \} \end{equation*}$$

via stereographic projection from the north pole,

$$\begin{align*} S^2 & \xrightarrow {\cong } \mathbb{C} \cup \{\infty \} \\ (a,b,c) & \mapsto \frac{a+bi}{1-c}, \end{align*}$$

followed by the identification

$$\begin{align*} \mathbb{C} \cup \{\infty \} & \xrightarrow {\cong } \mathbb{CP}^1 \\ z & \mapsto \mathbb{C}(z, 1) \\ \infty & \mapsto \mathbb{C}(1, 0). \end{align*}$$

In what follows, I will freely use these identifications; my preference is to use the $S^2$ picture because I want the visual image of the icosahedron in $\mathbb{R}^3$ to be front and center.

Under this identification, the $\mathrm{SO}(3)$ rotation action on $S^2$ translates into an $\mathrm{SO}(3)$ action on $\mathbb{CP}^1$, which can be explained as arising from the natural action of its double cover $\mathrm{SU}(2)$ on $\mathbb{C}^2$. We define the binary icosahedral group $\hat{G} \subset \mathrm{SU}(2)$ as the double cover of the icosahedral group $G \subset \mathrm{SO}(3)$. So, we seek $\hat{G}$-invariant homogenous polynomials on $\mathbb{C}^2$.

We have the vertex polynomial $V$ (of degree 12, vanishing on the 1-dimensional subspaces corresponding to icosahedron vertices), the face polynomial $F$ (of degree 20, vanishing on the 1-dimensional subspaces corresponding to icosahedron face midpoints), and the edge polynomial $E$ (of degree 30, vanishing on the 1-dimensional subspaces corresponding to icosahedron edge midpoints):

$$\begin{align} V &= uv(u^{10} + 11 u^5 v^5 - v^{10}) {\tag{8}}\\ F &= -u^{20} + 228(u^{15}v^5-u^5v^{15}) - 494u^{10}v^{10} - v^{20} {\tag{9}}\\ E &= u^{30} + 522(u^{25}v^5 - u^5v^{25}) - 1005(u^{20}v^{10} - u^{10}v^{20}) + v^{30} {\tag{10}} \end{align}$$

To see that these polynomials are indeed $\hat{G}$-invariant (instead of picking up phase factors when acting with $g \in \hat{G}$), it helps to realize that $G$ is generated by $R_z$ and $R_y$, the rotations about the $z-$ and $y-$axis by angles $2 \pi / 5$ and $\pi$ respectively. We can take their preimages in $\mathrm{SU}(2)$ to be

$$\begin{equation*} \hat{R}_z = \begin{pmatrix} e^{\frac{\pi i}{5}} & 0 \\ 0 & e^{-\frac{\pi i}{5}} \end{pmatrix}, \quad \hat{R}_y = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{equation*}$$

which similarly generate $\hat{G}$, and which act on homogenous polynomials in $u, v$ as:

$$\begin{equation} \hat{R}_z\colon u \mapsto e^{-\frac{\pi i}{5}}u, v \mapsto e^{\frac{\pi i}{5}} v, \quad \hat{R}_y \colon u \mapsto v, v \mapsto -u {\tag{11}} \end{equation}$$

It is clear that our polynomials 8–10 are invariant under these transformations, so they are indeed $\hat{G}$-invariant. In fact, they generate the algebra of $\hat{G}$-invariant homogenous polynomials on $\mathbb{C}^2$.

Figure 3.

The round icosahedron and its stereographic projection. The edge midpoint $\alpha$ and face midpoint $\beta$, together with their stereographic projections $a$ and $b$, have been marked for later use.

We can play the same game with our 5 inscribed octahedra. Let $O_i$ be the vertex polynomial of the $i$th inscribed octahedron. It has degree 5 and vanishes at the 1-dimensional subspaces of $\mathbb{C}^2$ corresponding to the 8 vertices of $O_i$. We compute

$$\begin{equation*} O_1 = (u^2 + v^2)(u^2 - 2nuv - v^2)(u^2 - 2muv - v^2) \end{equation*}$$

where $m = \zeta + \zeta ^4=\frac{1}{\phi }$ and $n = \zeta ^2 + \zeta ^3 = -\phi$, with the other $O_i, i=2\ldots 5$ obtained by simply rotating $O_1$ around the $z$-axis using the $\hat{R}$-action in 11.

The Icosahedron and the Quintic

When we rotate the icosahedron, the 5 inscribed octahedra and hence their vertex polynomials $O_i$ undergo an (even) permutation. Consider the quintic

$$\begin{equation*} \prod _{i=1}^5 (X - O_i) \end{equation*}$$

whose coefficients are symmetric polynomials in the octahedral polynomials $O_i$ and hence $\hat{G}$-invariant polynomials on $\mathbb{C}^2$. If we multiply it out, we obtain a Brioschi-type quintic

$$\begin{equation*} X^5 - 10V X^3 + 45 V^2 X - E \end{equation*}$$

as the reader will verify! (The coefficients of $X^4$ and $X^2$ are invariant polynomials of degree $6$ and $18$ respectively and hence must vanish, while the coefficients of $X^3$, $X$ and $1$ are of degree 12, 24, and 30 and hence must be multiples of $V$, $V^2$, and $E$ respectively.)

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

$$\begin{equation*} x_i = - \frac{V^2}{E} O_i, \quad i = 1\ldots 5. \end{equation*}$$

They are of degree zero in $u$ and $v$. Therefore, away from the edge midpoints, they are well-defined complex-valued functions on $S^2$. In other words, to each point $z$ on the “round icosahedron” $S^2$ (excluding edge midpoints), we can associate an ordered tuple $(x_1(z), \ldots , x_5(z))$ of 5 complex numbers! We map these 5 numbers to the quintic

$$\begin{equation} P_{\{x_i\}} = \prod _{i=1}^5 (X - x_i(z)) {\tag{12}} \end{equation}$$

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

$$\begin{equation*} P_B = X^5 + 10BX^3 + 45B^2 X + B^2 \end{equation*}$$

with Brioschi parameter $B(z) = -V^5/E^2$! See Figure 4. In summary, we have:

Theorem 1.

The map

$$\begin{equation*} z \mapsto (x_1 (z), \ldots , x_5(z)) \end{equation*}$$

is an $A_5$-equivariant bijection between points on the round icosahedron (minus edge midpoints) and evenly ordered (defined below) roots of Brioschi quintics $P_B$ with Brioschi parameter $B(z) = -V^5 / E^2$. In other words, it is an explicit equivariant isomorphism of covering spaces:

$$\begin{equation*} \vcenter{\img[][195pt][76pt][{\renewcommand{\arraystretch}{1} \setlength{\unitlength}{1.0pt} \setlength{\extrarowheight}{0.0pt}\begin{tikzpicture}[xscale=4, yscale=2] \node(A) at (0,0) {$S^2 \setminus\{\text{edge midpoints}\}$}; \node(B) at (1,0) {$\operatorname{EvOrRoots}$}; \node(C) at (0,-1) {$S^2 \setminus\{\infty\}$}; \node(D) at (1,-1) {$\text{Brioschi quintics}$}; \draw[->] (A) -- node[above]{$\cong$} (B); \draw[->>] (A) -- node[left]{$B$} (C); \draw[->] (C) -- node[below]{$\cong$} (D); \draw[->>] (B) -- node[right]{$P$} (D); \end{tikzpicture}}]{Images/img3fb2962dee59206fac13d29d499ebf72.svg}} \end{equation*}$$

Figure 4.

The Brioschi parameter $B(z)$ on $S^2$. The absolute value maps to the brightness, while the argument maps to the hue. Note the zeros of $B$ at the vertices of the icosahedron, and the poles at the edge midpoints.

The only subtlety here is the notion of an “evenly” ordered set of roots of a Brioschi quintic $P_B$. We need this because there are 60 points on a generic orbit $A_5 \cdot z$ in the icosahedron, while there are 120 ways to order the 5 roots $x_1, \ldots , x_5$ of the quintic $P_B$. We call an ordering of the roots of a generic Brioschi quintic $P_B$ even if, as $B$ approaches $0$ through positive real values (which implies that $z$ approaches a vertex of the icosahedron) and track the roots continuously, the roots end up in an even permutation of the numbering shown in Figure 1.

Theorem 1 tells us that in order to find the roots of a Brioschi quintic $P_B$ for some Brioschi parameter $B \in \mathbb{C}$, we need to find a point $z$ on the icosahedron such that $-V^5(z) / E^2(z) = B$. Then our 5 ordered roots will be given by the 5 octahedral numbers $x_i(z)$. Actually calculating $z$ in terms of $B$ is hard (we need to solve a polynomial equation of degree 60), but it can be done efficiently using elliptic curves, as we will see shortly!

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 $V$, $E$, and $F$ which we will need later. The point is that these are polynomials on $\mathbb{C}^2$, and three polynomials on a 2-dimensional space must satisfy a relation. The first opportunity for this relation to occur is in degree 60, and indeed we have:

$$\begin{equation} E^2 + F^3 = 1728 V^5 {\tag{13}} \end{equation}$$

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

$$\begin{equation*} E_6^2 - E_4^3 = 1728 \Delta . \end{equation*}$$

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

$$\begin{equation*} B(z) = - \frac{V^5}{E^2} = -\frac{V^5}{1728 V^5 - F^3} \end{equation*}$$

so that $B$ sends vertices to $0$, edge midpoints to $\infty$, and face midpoints to $-1/1728$. This uniquely characterizes it as a holomorphic map from $S^2$ to $S^2$.

Enter Elliptic Curves

Expressing a number “in terms of radicals” implies having access to the roots of unity, i.e., the $n$-torsion points (points $p$ such that $np = 0$) in the nonzero complex numbers $\mathbb{C}^*$, thought of as an additive abelian group. In the nineteenth century, mathematicians discovered that the set of points on an elliptic curve, i.e., the set $E$ of complex solutions to a cubic equation of the form

$$\begin{equation} y^2 = 4x^3 - g_2 x - g_3 {\tag{14}} \end{equation}$$

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 $n$-torsion points on the circle, perhaps they could be expressed in terms of the $5$-torsion points of an elliptic curve somehow associated with the quintic. Remarkably, this is precisely what Hermite (1858) and Kiepert (1878) managed to do! To quote McKean and Moll 10:

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 $(x_1, \ldots , x_5)$ of a Brioschi quintic directly correspond to the 60 equivalence classes of primitive 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 $A_5$ equivariant covering space of $S^2$ via the Brioschi map $B$. And moreover, we found that away from the edge midpoints, this covering space is explicitly isomorphic to the covering space of evenly ordered roots of Brioschi quintics.

Besides the icosahedron, there is another 60-sheeted $A_5$-equivariant covering space of $S^2$ occurring in nature: the moduli space $X(5)$ of elliptic curves equipped with a primitive basis for their 5-torsion!

Recall that an elliptic curve $E \subset \mathbb{CP}^2$ given by a cubic equation as in 14 identifies holomorphically with $\mathbb{C}/\Lambda$, the quotient of $\mathbb{C}$ by some rank 2 lattice $\Lambda \subset \mathbb{C}$. So, topologically, $E$ looks like a doughnut. Moreover, under this identification the addition operation on $E$ is just the standard addition in $\mathbb{C} / \Lambda$. Therefore,

$$\begin{equation*} \text{5-torsion points of $E$} \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z} / 5 \mathbb{Z}. \end{equation*}$$

Let $p,q$ be 5-torsion points in $\mathbb{C} / \Lambda$ and let $\omega _1, \omega _2 \in \mathbb{C}, \operatorname {Im}(w_1/w_2) > 0$ be generators of $\Lambda$. Since the equivalence classes $[\omega _1/5], [\omega _2/5]$ generate the 5-torsion of $\mathbb{C} / \Lambda$, we can write

$$\begin{align*} p &= a_1 [\omega _1/5] + a_2 [\omega _2/5] \\ q &= b_1 [\omega _1/5] + b_2 [\omega _2/5] \end{align*}$$

for some matrix

$$\begin{equation*} \gamma \in \begin{pmatrix} a_1 & a_2 \\ b_1 & b_2 \end{pmatrix} \in \mathrm{GL}_2(\mathbb{Z}, \mathbb{Z}/5\mathbb{Z}). \end{equation*}$$

We say that $(p,q)$ is a primitive basis for the 5-torsion of $E$ if $\det \gamma \equiv 1 \mod 5$. Write

$$\begin{equation*} X(5) = \{ (E, (p, q)) \} / \sim \end{equation*}$$

for the set of equivalence classes (“moduli space”) of pairs $(E, (p,q))$ where $E$ is a complex elliptic curve and $(p,q)$ is a primitive basis for its 5-torsion (see 4). Two such pairs $(E, (p,q))$ and $(E', (p',q'))$ are equivalent if there is an isomorphism $E \rightarrow E'$ which carries $(p,q) \mapsto (p', q')$.

Write $X(1)$ for the “vanilla” moduli space of elliptic curves (no extra torsion information tagged on), and $\mathbb{H} = \{ \tau \in \mathbb{C} : \operatorname {Im}(\tau ) > 0 \}$ for the upper half plane. Thinking of an elliptic curve as a quotient $\mathbb{C} / \mathbb{Z} \oplus \mathbb{Z} \tau$ for some $\tau \in \mathbb{H}$, we can identify these moduli spaces as:

$$\begin{equation*} X(1) \cong \mathbb{H}^* \big / SL(2, \mathbb{Z}), \quad X(5) \cong \mathbb{H}^* \big / \Gamma (5). \end{equation*}$$

Here, $\mathbb{H}^* = \mathbb{H} \cup \mathbb{Q} \cup \{i \infty \}$ is the extended upper half plane (the extra “cusp” points $\mathbb{Q} \cup \{i \infty \}$ are needed to get a compact moduli space; they contribute a single point to the quotient in $X(1)$ and 12 points in $X(5)$) and

$$\begin{equation*} \Gamma (5) = \left\{\,\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{SL}(2, \mathbb{Z}) : \begin{pmatrix} a & b \\ c & d \end{pmatrix} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} , \operatorname {mod} 5 \,\right\} \end{equation*}$$

is the principal congruence subgroup of level 5. (The congruence relation is done independently entrywise, so the requirement is that $a \equiv 1 \pmod {5}$, $b \equiv 0 \pmod {5}$, $c\equiv 0 \pmod {5}$ and $d \equiv 1 \pmod {5})$.

Permutation Wizardry

Now, $\Gamma (5)$ is a normal subgroup of $\Gamma \coloneq \mathrm{SL}(2, \mathbb{Z})$, and so we can form the quotient group

$$\begin{equation*} \Gamma / \Gamma (5) \cong \mathrm{PSL}(2, \mathbb{Z} / 5\mathbb{Z}). \end{equation*}$$

The magic is that $\mathrm{PSL}(2, \mathbb{Z} / 5\mathbb{Z})$ is isomorphic to $A_5$! We will need to understand this isomorphism explicitly in terms of the action of $A_5$ on the five inscribed octahedra⁠².

²

I learned this argument from David Speyer and Beren Gunsolus 13.

✖

The vertices of an inscribed octahedron $O_i$ are located at edge midpoints of the icosahedron $I$. Therefore, $O_i$ partitions the vertices of $I$ into pairs, and hence encodes a fixed-point free involution of the 12 vertices of $I$. This involution commutes with the map $p \rightarrow -p$, and so we conclude that each inscribed octahedron $O_i$ encodes (and is in fact encoded by) a fixed-point free involution $o_i$ of the 6 vertex axes of $I$.

On the other hand, $PSL(2, \mathbb{Z}/5 \mathbb{Z})$ acts naturally on projective space

$$\begin{equation*} \mathbb{P}^1(\mathbb{Z}/5\mathbb{Z}) = \{ 0, 1, 2, 3, 4, \infty \}, \end{equation*}$$

which also consists of 6 things. So, let us identify the six vertex axes of the icosahedron in $\mathbb{R}^3$ with $\mathbb{P}^1(\mathbb{Z}/5\mathbb{Z})$ in the natural way:

$$\begin{align*} \mathbb{R} (-2 \zeta ^k, 1) & \mapsto k, \quad k=0\ldots 4 \\ \mathbb{R} (0, 0, 1) &\mapsto \infty . \end{align*}$$

In this way $PSL(2, \mathbb{Z}/5\mathbb{Z}) \subset S_6$ acts by conjugation on the 5 octahedral involutions $o_i \in S_6$, and it turns out that it permutes them evenly, giving our isomorphism $PSL(2, \mathbb{Z}/5\mathbb{Z}) \cong A_5$.

Let us pre- and post-compose this isomorphism with the natural isomorphisms $\Gamma / \Gamma (5) \cong \mathrm{PSL}(2, \mathbb{Z} / 5\mathbb{Z})$ and $A_5 \cong G$ and record the result explicitly for later use.

Lemma 1.

The explicit isomorphism

$$\begin{align*} \Gamma / \Gamma (5) & \rightarrow G, \end{align*}$$

at the level of generators, sends

$$\begin{align*} T &\mapsto R_z \\ S &\mapsto R_\alpha , \end{align*}$$

where $T(\tau ) = \tau +1$ and $S(\tau )=-\frac{1}{\tau }$, and $R_z (z) =\zeta z$ and $R_\alpha (z) = \frac{nz + 1}{z-n}$ are rotations by $\frac{2\pi }{5}$ and $\pi$ counterclockwise about the positive z-axis and the axis through the edge midpoint $\alpha$ shown in Figure 3 respectively.

An Isomorphism of Covering Spaces

The isomorphism $PSL(2, \mathbb{Z}/5\mathbb{Z}) \cong A_5$ means that the forgetful map

$$\begin{equation*} \pi : X(5) \twoheadrightarrow X(1) \end{equation*}$$

is an $A_5$-equivariant 60-sheeted covering space of $X(1)$. Note that we can also make our own direct count of the sheets in the covering. Given an elliptic curve $E \in X(1)$, there are