### Modular Curves and Monstrous Moonshine

#### Posted by John Baez

Recently James Dolan and I have been playing around with modular curves — more specifically the curves $X_0(n)$ and $X^+_0(n)$, which I’ll explain below. Monstrous Moonshine says that when $p$ is prime, the curve $X^+_0(p)$ has genus zero iff $p$ divides the order of the Monster group, namely

$p = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 47, 59, 71$

Just for fun we’ve been looking at $n = 11$, among other cases. We used *dessins d’enfant* to draw a picture of $X_0(11)$, which seems to have genus $1$, so for $X^+_0(11)$ to have genus zero it seems we want the picture for $X_0(11)$ to have a visible two-fold symmetry. After all, the torus is a two-fold branched cover of the sphere, as shown by Greg Egan here:

But we’re not seeing that two-fold symmetry. So maybe we’re making some mistake!

Maybe you can help us, or maybe you’d just like a quick explanation of what we’re messing around with.

To get the modular curve $X_0(n)$, we start with the quotient space

$\mathrm{H}/\Gamma_0(n)$

where $\mathrm{H}$ is the complex upper half-plane and the discrete group

$\Gamma_0(n) = \left\{\; \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \; \Big\vert \; a,b,c,d \in \mathbb{Z}, \quad a d - b c = 1, \quad b = 0 \; mod \; N \; \right\} \subset SL(2,\mathbb{Z})$

acts on the upper half-plane by Möbius transformations

$z \mapsto \frac{a z + b}{c z + d}$

This is the moduli space of elliptic curves with some extra structure. As a Lie group an elliptic curve is just a torus, so it has a group of $n$-torsion points that looks like

$\mathbb{Z}\!/\!n \times \mathbb{Z}\!/\!n$

Thus, we can equip any elliptic curve with a *subgroup* of its $n$-torsion points that’s isomorphic to $\mathbb{Z}\!/\!n$, in various possible ways — and that’s the extra structure I’m talking about!

Then, to form $X_0(n)$ we compactify $\mathrm{H}/\Gamma_0(n)$ by adding finitely many extra points.

I wish I had the energy to explain how you can draw these modular curves $X_0(n)$ as triangulated surfaces using a trick Grothendieck came up with, called *dessins d’enfant*. It is really quite childish and fun, but it’s only easy to explain using lots of pictures, and I don’t have the time to draw those. The best I can do is refer you to a series of nice posts here on the $n$-Café, starting here:

- Tim Silverman, Pictures of modular curves (I).

and perhaps ending here:

- Tim Silverman, Pictures of modular curves (XI).

though this post declared itself the “antepenultimate”, so we may still get two more out of Tim someday!

Anyway, the normalizer of $\Gamma_0(n)$ in $SL(2,\mathbb{Z})$ is a slightly bigger discrete subgroup called $\Gamma_0^+(n)$. We can compactify

$\mathrm{H}/\Gamma^+_0(n)$

by adding finitely many extra points, and we get another modular curve, called $X^+_0(n)$.

$X^+_0(n)$ is a kind of quotient of $X_0(n)$. More precisely, I think there’s a branched covering

$X_0(n) \to X^+_0(n)$

And James and I think this should arise by taking the *dessin* for $X_0(n)$ and modding out by some symmetries.

We’re thinking about examples. When $n = 11$, it seems $X_0(11)$ has genus 1, and people tell us that $X_0^+(11)$ has genus 0. So we’re hoping to find a visible two-fold symmetry of the *dessin* for $X_0(11)$, such that when we mod out by it we get a curve of genus zero. Can you help?

Well, I just cheated a bit and discovered that this symmetry should be the ‘Fricke involution’:

- Wikipedia, Supersingular prime (moonshine theory).

But I bet there’s a way to get a lot of visual intuition for this symmetry using *dessins* and other tricks.

This paper might help if you want to see more details about $X_0(11)$:

- Tom Weston, The modular curves $X_0(11)$ and $X_1(11)$.

This paper about the group $PSL(2,11)$ is probably related. but I’m not completely sure how:

- Bertram Kostant, The graph of the truncated icosahedron and the last letter of Galois,
*Notices of the American Mathematical Society***42**(1995), 959–968.

## Re: Modular Curves and Monstrous Moonshine

I had a go at making a picture of $X_0(11)$. The edges are labelled with elements of the projective line over the field of order 11, and the opposite boundaries are identified so this lies on a torus. The green points are the (two) cusps.

I also asked James Dolan about this, and he is suggesting that $\Gamma_0^{+}(n)$ is the normalizer of $\Gamma_0(n)$ in SL$(2,\mathbb{R})$ not SL$(2,\mathbb{Z})$. So the Fricke involution is a bit more tricky than just finding a symmetry of this diagram.