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I've been reading Voight's paper on Shimura curves and it prompted the following question; see http://www.cems.uvm.edu/~voight/articles/shimura-clay-proceedings-071707.pdf for which notes I'm talking about

Let $F$ be a totally real number field, $B$ a quaternion algebra which splits at exactly one real place, and $\mathcal{O}\subset B$ a maximal order. Let $\Gamma^B(1)$ be the associated discrete arithmetic subgroup of $\mathrm{PSL}_2(\mathbf{R})$. This group is the analogue of $\Gamma(1)=\mathrm{SL}_2(\mathbf{Z})$. Let $X^B(1) = \Gamma^B(1)\backslash \mathbf{H}$.

Question 1. Let $\Gamma \subset \Gamma^B(1)$ be a finite index subgroup. Do I understand correctly that $X^B(\Gamma) = \Gamma \backslash \mathbf{H}$ is a compact Riemann surface, and that the inclusion $\Gamma \subset \Gamma^B(1)$ induces a finite morphism $X^B(\Gamma) \to X^B(1)$? What are the branch points of this morphism?

Zograf showed that, if $\Gamma \subset \mathrm{SL}_2(\mathbf{Z})$ is a congruence subgroup, the index of $\Gamma$ in $\mathrm{SL}_2(\mathbf{Z})$ is bounded by $128(g+1)$, where $g$ is the genus of the compactification of the Riemann surface $\Gamma \backslash \mathbf{H}$. (This is how he proved that there are only finitely many congruence subgroups of $\mathrm{SL}_2(\mathbf{Z})$ of given genus.)

Question 2. Is there a similar theorem for Shimura curves? That is, assume that $\Gamma \subset \Gamma^B(1)$ is a congruence subgroup (does this term make sense? what should it mean?). Is the index of $\Gamma$ in $\Gamma^B(1)$ bounded by the genus of $X^B(\Gamma)$? If yes, can we make this bound explicit?

I'm not sure what I mean by a congruence subgroup of $\Gamma^B(1)$ yet, but I want the curve $X_\Gamma(B)$ to be a Shimura curve.

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2 Answers 2

up vote 6 down vote accepted

Here are the answers to some of your questions. If $P$ is a (non-zero) prime ideal of the integers of $F$ then $B$ will either be split or ramified at $P$, depending on whether $B\otimes_F F_P$ is isomorphic to $M_2(F_P)$ or not. Now $B$ will only be ramified at a finite set of primes, and if $P$ is a prime which splits $B$ then $\mathcal{O}/P^n$ will be isomorphic to $M_2(R/P^n)$ with $R$ the integers of $F$. More generally if $N$ is an ideal of $R$ which is coprime to all the ramified primes, $\mathcal{O}/N$ will be isomorphic to $M_2(R/N)$. In particular you can define congruence subgroups as pre-images of upper-triangular matrices mod $N$ just as in the classical case. They all have finite index.

The quotient you talk about is a Riemann surface, and is compact except for the special case $F=\mathbb{Q}$ and $B=M_2(\mathbb{Q})$, when it is the usual non-compact modular curve. The inclusion from your finite index subgroup into $\Gamma^B(1)$ will induce a finite flat covering of your Riemann surface as you say. The ramification is controlled, just as in the classical case, by the amount of torsion you lose when passing to the finite cover. Any point in the upper half plane will have a finite stabiliser in $\Gamma/Z$ with $Z$ the scalars and this finite group is always cyclic. Vaguely, you can compute the ramification index in the covering by noting how the stabilizer changes -- classically this is like noting that elliptic points have stabilisers of size greater than 1 in $PSL(2,\mathbb{Z})$ but may have stabilizers of size 1 in a finite index subgroup of this. This is easy to understand -- it's just a local calculation -- you might want to think about how to put a Riemann surface structure on the quotient at a point where the stabilizer is non-trivial (it locally acts as rotations).

About the analogue of Zograf's theorem for Shimura curves: I didn't even know that theorem for modular curves so I can't possibly answer. But you might want to think about what Riemann-Hurwitz gives you. Often the base genus of these things is at least 2, even for full level, and Zograf's theorem looks very much like an application of Riemann-Hurwitz in some sense, but complicated by the fact that one has to control elliptic points.

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Do I understand correctly that by base curve you mean the curve $X^B(1)$? It is a bit counterintuitive for me that often this is of genus $\geq 2$ I'm more used to the modular curve $X(1)$ of genus zero. Could we characterize when the curve $X^B(1)$ has genus zero? If I'm not mistaken, this happens when we look at certain "triangle groups" $\Delta(a,b,c)$. If we suppose that $X_B(1)$ is of genus zero, is it easy to determine when the number of branch points of $X^B(\Gamma) \to X^B(1)$ is less than or equal to three? Finally, the RH formula gives $g\leq [\Gamma:\Gamma(1)]$ in general. It won't –  Ariyan Javanpeykar Feb 14 '12 at 21:12
    
It won't give an upper bound. One needs to combine that the smallest positive eigenvalue $\lambda_1$ is bounded from below by $3/16$ (Selberg) with Zograf's general upper bound for $\lambda_1$ to obtain an upper bound for the index. –  Ariyan Javanpeykar Feb 14 '12 at 21:15
    
The genus of $X^B(1)$ can be huge -- even if $F=\mathbf{Q}$! If $B$ is non-split, then the Jacquet-Langlands correspondence implies that the holomorphic differentials on $X^B(1)$ are, as a Hecke module, isomorphic to the space of weight 2 cuspidal newforms of level $\Gamma_0(D)$, with $D$ the discriminant of $B$. This amazing result, whose proof is hugely non-constructive, gives you a ridiculously easy way of computing the genus of these abstract objects: just get a computer algebra package to compute the dimension of some space of classical newforms! –  Kevin Buzzard Feb 16 '12 at 20:38
    
Riemann-Hurwitz can sometimes be used to get exact formulae. If for example you shrink $\Gamma$ until there's no torsion at all that isn't in the centre, then after this point (if you're in the compact case) all the covers will be unramified everywhere! So now there's a very tight link between the genus of the cover and the degree of the cover. More generally there is a Riemann-Hurwitz for algebraic stacks, or $\mathbf{Q}$-varieties, or whatever language you like best, which again will give exact formulae; the maps of stacks are unramified everywhere even at ell points. –  Kevin Buzzard Feb 16 '12 at 20:40

You could have a look at the paper of Long and Reid. Even though they are considering genus 0 surfaces, they cite Zograf's result in the general case (see Theorem 4.4). Actually the estimate goes back to Yang-Yau, and in fact is a generalization of a much older result of Hersch and others. Anyway, Long-Reid cite the necessary eigenvalue estimate for congruence subgroups Theorem 4.3, which is attributed to Vigneras, but based on results of Gelbart-Jacquet and Jacquet-Langlands (see also Burger-Sarnak). The translation between eigenvalue estimates for congruence groups and automorphic forms is confusing and daunting. Sarnak told me to look at Gelfand-Graev-Piatetskii-Shapiro for the translation. Coupling Theorems 4.3 and 4.4 gives the desired index estimate. One may obtain the same finiteness result of Zograf in the general case of Shimura curves of bounded genus this way.

As for branching, it is more natural to think of the quotient as an orbifold, and the cover as a cover in the category of orbifolds (although note that the Yang-Yau/Zograf technique uses conformal maps to the sphere of the underlying Riemann surface, ignoring torsion). To determine the branched locus, you need to make a comparison of the torsion in $\Gamma^B(1)$ and $\Gamma$. There are only finitely many conjugacy classes of torsion subgroups in an arithmetic group, so in principle this can be done.

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