Let $p$ be a prime number and $X_0(p)/\mathbf{Q}$ be the classical modular curve for $\Gamma_0(p)$. Let $\tilde{X}_0(p)/\mathbf{Z}$ be the projective arithmetic surface corresponding to the normalization of $\mathbb{P}_{\mathbf{Z}[j]}^1$ inside $\mathbf{Q}(X_0(p))$. Then when one reduces $\tilde{X}_0(p)/\mathbf{Z}$ modulo $p$ one gets two copies of $\mathbb{P}_{\mathbf{F}_p}^1$ with normal crossings at points corresponding to supersingular elliptic curves over $\mathbf{F}_{p^2}$. So let $\mathbf{Z}_{p^2}$ be the unique quadratic unramified extension over $\mathbf{Z}_p$. Having in mind Mumford's dictionary, this begs the following question:

Q: Is $\tilde{X}_0(p)\otimes\mathbf{C}_p$ uniformized by some Schottky group of $PGL_2(\mathbf{Q}_{p^2})$ ?


1 Answer 1


Yes, unless I'm missing some subtlety.

Theorem 4.20 of Mumford's paper reads:

Every stable curve over $S$ with nonsingular generic fiber and $k$-split degenerate closed fiber is isomorphic to $P_{\Gamma}$ for a unique$^\ast$ flat Schottky group $\Gamma \subset PGL(2,K)$.

Here $S = \mathrm{Spec}\ A$ where $A$ is a complete integrally closed noetherian local ring, $K = \mathrm{Frac}\ A$, and $k$ is the residue field of $A$. Saying a stable curve $X$ over $k$ is $k$-split degenerate means (see shortly after Defn 3.2)

  1. The components of the normalization $X$ are all $\mathbb{P}^1_k$'s.

  2. The nodes of $X$ are all $k$ points.

  3. Any node locally looks like $k[x,y]/(xy)$. For example, working over $\mathbb{R}$, a node of the form $\mathbb{R}[x,y]/(x^2+y^2)$ is forbidden.

You say above (and I agree) that conditions (1) and (2) are true in your setting, and condition (3) follows from the fact that your nodes always join two different components, not one component and itself.

$^*$ Mumford must mean "unique up to conjugacy".

  • $\begingroup$ Is it possible to give a finite list of explicit topological generators for $\Gamma$? $\endgroup$ Dec 7, 2013 at 1:48
  • $\begingroup$ In principal, yes, Mumford's proof is constructive. And once you find any generators, arxiv.org/abs/1309.5243 will put them into a convenient normalized form. But I don't know how to get those generators in the first place. $\endgroup$ Dec 7, 2013 at 2:42
  • $\begingroup$ The number of generators is equal to the genus of $X_0(p)$. $\endgroup$ Aug 5, 2016 at 14:36

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