3
$\begingroup$

In Gross's paper Heegner points on $X_0(N)$, he considers some examples of general modular curve, not only the usual ones $X_0(N),X_1(N),X(N)$. We know these three kinds of modular curves have "simple" moduli interpretation.

Essentially, Gross considers the modular curves of this kind: Let $K$ be an imaginary quadratic field, $p$ be a prime number inert in $K$. Given an embedding $K\to M_2(\mathbb{Q})$, let $R=\mathcal{O}_K+pM_2(\mathbb{Z})$. Let $\Gamma$ be norm 1 element in $R$. Then my question is what is the moduli interpretation of $\Gamma\backslash\mathscr{H}$, in the similar way as the usual ones?

$\endgroup$

1 Answer 1

3
$\begingroup$

By "norm 1 element", you mean "determinant 1 element", right? I guess also that your embedding $K \to M_2(\mathbf{Q})$ is chosen to send $O_K$ into $M_2(\mathbf{Z})$.

Then $\Gamma$ is visibly a subgroup of $\Gamma(1) = SL_2(\mathbf{Z})$ containing the principal congruence subgroup $\Gamma(p)$; and its image in $\Gamma(1) / \Gamma(p)$ is a non-split torus (it's the norm one units of $\mathbf{F}_{p^2}$, embedded as the non-split Cartan subgroup of $SL_2(\mathbf{F}_p)$). So $X(\Gamma)$ is the quotient of $X(p)$ by the non-split Cartan, and it parametrizes -- surprise! -- elliptic curves together with an equivalence class of bases of $E[p]$ modulo the action of the non-split Cartan. In particular, for a field $K$ of char 0, a $K$-point of $X(\Gamma)$ is an elliptic curve over $K$ for which the mod $p$ Galois representation lands in the non-split Cartan.

There is nothing special about the non-split Cartan here -- one can give a similar description of the moduli interpretation of $X(\Gamma)$ for any congruence subgroup $\Gamma$, in terms of elliptic curves with a $\Gamma$-orbit of bases of their $N$-torsion. This is all explained in Deligne-Rapoport.

$\endgroup$
2
  • $\begingroup$ Is there a way to interpret the $\Gamma$-orbits of bases of $N$-torsion? For example, in the usual case, if the non-split Cartan changes to upper triangular, we will get $X_0(N)$, and each orbit correspond to a cyclic subgroup of order $N$. $\endgroup$
    – user58510
    Commented Sep 23, 2014 at 11:51
  • $\begingroup$ The question you ask in your comment is more philosophical than mathematical: the set of orbits is what it is; what you consider to be an "interpretation" of this set is up to you. :-) I don't know of any particularly nice way of thinking about this set. $\endgroup$ Commented Sep 24, 2014 at 3:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .