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?


1 Answer 1


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.

  • $\begingroup$ 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
    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. $\endgroup$ Sep 24, 2014 at 3:07

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.