2
$\begingroup$

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 $PGL_2(\mathbf{Q}_{p^2})$ ?

$\endgroup$

1 Answer 1

1
$\begingroup$

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".

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.