March 3, 2024

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:

and perhaps ending here:

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’:

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 $X_0(11)$:

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

Posted at March 3, 2024 9:43 PM UTC

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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.

Layer 1

Posted by: Simon Burton on March 10, 2024 9:13 PM | Permalink | Reply to this

Re: Modular Curves and Monstrous Moonshine

This looks somewhat similar to a dessin d’infant. Do you think there is a connection?

Posted by: Leo Alonso on March 12, 2024 2:02 PM | Permalink | Reply to this

Re: Modular Curves and Monstrous Moonshine

Yes I think this is a dessin. When we project this down to the Riemann sphere, which is $X_0(1)$, the green dots are sent to zero, and the blue dots are sent to one. This projection map corresponds to forgetting the extra structure on the elliptic curve.

Posted by: Simon Burton on March 12, 2024 4:00 PM | Permalink | Reply to this

Re: Modular Curves and Monstrous Moonshine

Yes, this is the dessin I was talking about in my blog article:

More precisely, I think there’s a branched covering

$X_0(n)→X_0^+(n)$

And James and I think this should arise by taking the dessin for $X_0(n)$ and modding out by some symmetries [namely, the Fricke involution].

But now James believes the Fricke involution isn’t a symmetry of the dessin — and Simon’s picture seems to confirm that, because his picture seems to have no $\mathbb{Z}/2$ symmetry.

It’s sort of a bummer, especially since the formula for the Fricke involution is so simple. Maybe we should refine the dessin to get one with manifest $\mathbb{Z}/2$ symmetry.

I’m tempted to draw an edge from the green dot in the middle to the blue dot above it, and an edge from that blue dot to the green dot on top! That would give the picture a $\mathbb{Z}/2$ rotational symmetry. But I don’t know if that makes sense—I’m thinking purely visually right now.

Posted by: John Baez on March 12, 2024 6:48 PM | Permalink | Reply to this

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