9
$\begingroup$

Let $N \ge 4$ and let $Y_1(N)$ be the complex manifold $\Gamma_1(N) \backslash \mathcal{H}$, where $\Gamma_1(N) \subset \mathrm{SL}_2(\mathbf{Z})$ is the usual congruence subgroup and $\mathcal{H}$ the upper half-plane.

I know of two models of the modular curve $Y_1(N)$ as a variety over $\mathbf{Q}$: one for which $Y_1(N)$ represents the functor "elliptic curves with a point of order $N$", with the universal object being $(\mathbf{C} / (\mathbf{Z} + \mathbf{Z}\tau), 1/N)$; and another, representing the functor "elliptic curves with an embedding of $\mu_N$", with the universal object being $(\mathbf{C} / (\mathbf{Z} + \mathbf{Z}\tau), \zeta_N \mapsto 1/N)$. The difference between these has come up before, in this question, where it was established that the modular functions whose $q$-expansions are in $\mathbf{Q}[[q]]$ are in the coordinate ring of the second model, but not the first.

My question is this. Modular curves are Shimura varieties for $GL_2$; and Deligne (following Shimura) has defined a "canonical model" for Shimura varieties over the reflex field, which is $\mathbf{Q}$ here. Here "canonical" has a precise, but rather complicated, definition in terms of Galois actions on CM points.

Which open compact subgroups of $\mathrm{GL}_2(\hat{\mathbf{Z}})$ give these two models of $Y_1(N)$ as their Deligne canonical models?

I'm virtually certain that one of the two models for $Y_1(N)$ is the canonical model of level $\{ \begin{pmatrix} * & * \\ 0 & 1 \end{pmatrix} \bmod N\}$, and the other is $\{ \begin{pmatrix} 1 & * \\ 0 & * \end{pmatrix} \bmod N\}$; but I'm really struggling to find a straight answer in the literature as to which is which!

$\endgroup$
8
  • $\begingroup$ Both are quotients of the (disconnected) curve with full level $N$, which parameterizes elliptic curves $E$ with basis $(e_1,e_2)$ for $E[N]$. Giving a point of the quotient by the first compact open amounts to saying that Galois acts on $e_1$ by a (non-trivial) character, while the second amounts to saying that Galois fixes $e_1$. So it appears to me that the first gives you the $\mu_N$ model while the second gives the usual one. $\endgroup$ Apr 23, 2017 at 20:22
  • $\begingroup$ @KeerthiMadapusiPera Both are quotients of Y(N), but to get something out of this one has to know how to identify the canonical model of Y(N) with the moduli space of elliptic curves with full level N structure. There's more than one possible convention for how to do this. (Kato's article in Asterisque 295 gives a model for Y(N) for which the quotient by the first subgroup classifies points of order N; but I'm pretty sure Kato's model is not the canonical model, because the Galois action on the connected components differs by a sign from the one on p109 of Milne's Shimura Varieties notes.) $\endgroup$ Apr 23, 2017 at 20:33
  • $\begingroup$ Seems to me it's a question of whether you're asking for trivializations of the homology or cohomology of the elliptic curve. At least, by Deligne's conventions in his Bourbaki article (which I'm sure has consistent signs), it's homology that's being trivialized, so I think what I said is the correct convention. I must confess though that I'm not completely convinced yet. $\endgroup$ Apr 23, 2017 at 21:08
  • $\begingroup$ I think this is a matter of whether you consider the action of $GL_2$ to be on the left (matrix times column vector) or the right (row vector times matrix). The convention that Shimura takes is to have the matrices act on the right (which is maybe a little bit unusual). With this convention the 2nd would be $Y_{1}(N)$ and the first would be the other one. David Zureick-Brown and I discuss this issue a bit in Section 2 of our preprint here. $\endgroup$ Apr 24, 2017 at 1:11
  • $\begingroup$ Even if you use right actions throughout, there are still two conventions, differing by the automorphism $g \mapsto (\det g)^{-1} g$ of $GL_2(\hat{\mathbf{Z}})$. $\endgroup$ Apr 24, 2017 at 6:19

1 Answer 1

1
$\begingroup$

This was getting too long for a comment I'll post it as an answer.

Though the set of $K$-special points on the canonical model does not depend on the choice of an embedding of $K\hookrightarrow\operatorname{GL}_{2}(\mathbb Q)$, the Galois action does depend on it in general and this dependence is encoded in a choice of involution (very concretely, are you going to embed $K$ through the basis $\{1,z\}$ or through $\{z,1\}$). If I am not mistaken, this corresponds to the general fact that an abelian variety with CM is special on the Siegel modular variety in two different ways depending on a choice of isomorphism of $\operatorname{GL}(H_1(A,\mathbb Q))\simeq\operatorname{GL}(V)$ (composed or not with an involution). Hence, the precise formulation of Shimura's reciprocity law characterizing canonical model, which you left implicit in your question, depends (slightly) on such a choice.

I think these choices of normalization are interchanged by the analytic involution you mention in comments exchanging the two complex curves. If I'm correct, determining which curve is the canonical model for which group depends on this normalization.

Now I have tried myself to write something internally coherent and 1) I found it surprisingly hard and 2) I concluded that if you normalize the fundamental theorem of complex multiplication in the usual way in terms of uniformization (by trivializing the homology of the abelian variety), then the canonical model corresponding to the choice of your second group corresponds to the moduli problem "point of order $N$" and that the canonical model corresponding to the choice of your first group corresponds to the moduli problem "embedding of $\mu_N$". As for how I checked it, I computed the action of an element of the absolute Galois group of $K$ on an Heegner point for the first moduli problem and compared the result with the fundamental theorem of CM. However, in view of point 1), I admit I worry that my computation might have been circular at some point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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