Such a great colloquium last week by Peter Sarnak, this year’s Hilldale Lecturer, on his paper with Bourgain and Gamburd. My only complaint is that he promised to talk about the mapping class group and then barely did! So I thought I’d jot down what their work has to do with mapping class groups and spaces of branched covers.

Let E be a genus 1 Riemann surface — that is, a torus — and O a point of E. Then pi_1(E-O) is just a free group on two generators, whose commutator is (the conjugacy class of) a little loop around the puncture. If G is a group, a G-cover of E branched only at O is thus a map from pi_1(E-O) to G, which is to say a pair (a,b) of elements of G. Well, such a pair considered up to conjugacy, since we didn’t specify a basepoint for our pi_1. And actually, we might as well just think about the *surjective* maps, which is to say the *connected* G-covers.

Let’s focus on the case G = SL_2(Z). And more specifically on those maps where the puncture class is sent to a matrix of trace -2. Here’s an example: we can take

You can check that in this case the puncture class has trace -2; that is, it is the negative of a unipotent matrix. Actually, I gotta be honest, these matrices don’t generate SL_2(Z); they generate a finite-index subgroup H of SL_2(Z), its commutator.

Write S for the set of all conjugacy classes of pairs (a,b) of matrices which generate H and have commutator with trace -2. It turns out that this set is the set of integral points of an affine surface called the *Markoff surface*: namely, if we take x = Tr(a)/3, y = Tr(b)/3, and z = Tr(ab)/3, then the three traces obey the relation

and indeed every solution to this equation corresponds to an element of S.

So the integral points on the Markoff surface are acted on by an infinite discrete group. Which if you just look at the equation seems like kind of a ridiculous miracle. But in the setting of H-covers is very natural. Because there’s a natural group acting on S: namely, the mapping class group Γ of type (1,1). This group’s whole purpose in life is to act on the fundamental group of a once-punctured torus! (For readers unfamiliar with mapping class groups, I highly recommend Benson Farb and Dan Margalit’s wonderful textbook.) So you start with a surjection from pi_1(E-O) to H, you compose with the action of Γ, and you get a new homomorphism. The action of Γ on pi_1(E-O) is only outer, but that’s OK, because we’re only keeping track of conjugacy classes of homomorphisms from pi_1(E-O) to H.

So Γ acts on S; and now the lovely theorem is that this action is *transitive.*

I don’t want to make this mapping class group business sound more abstract than it is. Γ isn’t a mystery group; it acts on H_1(E-O), a free abelian group of rank 2, which gives a map from Γ to SL_2(Z), which turns out to be an isomorphism. What’s more, the action of Γ on pairs (a,b) is completely explicit; the standard unipotent generators of SL_2(Z) map to the moves

(a,b) -> (ab,b)

(a,b) -> (a,ab)

(Sanity check: each of these transformations preserves the conjugacy class of the commutator of a and b.)

Sarnak, being a number theorist, is interested in *strong approximation*: are the integral solutions of the Markoff equation dense in the adelic solutions? In particular, if I have a solution to the Markoff equation over F_p — which is to say, a pair (a,b) in SL_2(F_p) with the right commutator — can I lift it to a solution over Z?

Suppose I have a pair (a,b) which lifts to a pair (a,b). We know (a,b) = g(a_0,b_0) for some g in Γ. Thus (a,b) = g(a_0,b_0). In other words, if strong approximation is true, Γ acts transitively on the set S_p of Markoff solutions mod p. And this is precisely what Bourgain, Gamburd, and Sarnak conjecture. (In fact, they conjecture more: that the Cayley-Schreier graph of this action is an *expander*, which is kind of a quantitative version of an action being transitive.) One reason to believe this: if we replace F_p with C, we replace S with the SL_2(C) character variety of pi_1(E-O), and Goldman showed long ago that the action of mapping class groups on complex character varieties of fundamental groups was ergodic; it mixes everything around very nicely.

Again, I emphasize that this is on its face a question of pure combinatorial group theory. You want to know if you can get from any pair of elements in SL_2(F_p) with negative-unipotent commutator to any other via the two moves above. You can set this up on your computer and check that it holds for lots and lots of p (they did.) But it’s not clear how to prove this transitivity for all p!

They’re not quite there yet. But what they can prove is that the action of Γ on S_p has a very big orbit, and has no very small orbits.

Now that G is the finite group SL_2(F_p), we’re in my favorite situation, that of Hurwitz spaces. The mapping class group Γ is best seen as the fundamental group of the moduli stack M_{1,1} of elliptic curves. So an action of Γ on the finite set S_p is just a cover H_p of M_{1,1}. It is nothing but the Hurwitz space parametrizing maps (f: X -> E) where E is an elliptic curve and f an SL_2(F_p)-cover branched only at the origin. What Bourgain, Gamburd, and Sarnak conjecture is that H_p is connected.

If you like, this is a moduli space of curves with nonabelian level structure as in deJong and Pikaart. Or, if you prefer (and if H_p is actually connected) it is a noncongruence modular curve corresponding to the stabilizer of an element of S_p in Γ = SL_2(Z). This stabilizer is in general going to be a noncongruence subgroup, except it *is *a congruence subgroup in the more general sense of Thurston.

This seems like an interesting family of algebraic curves! What, if anything, can we say about it?

At several points in the beginning I think you wrote when you intended to write ? (For example, when you talk about .)

This last bit seems related to a question I asked on MO once about twisted equivariant modular forms; there I asked if anyone knew anything about analogues of modular forms defined on the stack of pairs (an elliptic curve , a -bundle on ) for a finite group.

Thanks, I think I got them all!