TITLE: Zograf's bound on the index of a modular curve for Shimura curves
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: 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.