2
$\begingroup$

Let us consider the anticanonical adic tower of modular curves which is described in Peter Scholze's paper "On torsion in the cohomology of locally symmetric varieties", and let us call $\mathcal{X}_{\infty}$ the perfectoid space which corresponds (in a suitable sense) to the projective limit of the anticanonical tower. I'm particularly interested in the case of elliptic curves, hence I assume to work with modular curves instead of more general Shimura Varieties. In the mentioned paper, Scholze says that a geometric point of $\mathcal{X}_{\infty}$, corresponding to a map from $\text{Spa}(C,\mathcal{O}_C)$, with $C$ a complete algebraically closed extension of $\mathbb{Q}_p$, and $\mathcal{O}_C$ its ring of integers, corresponds to an elliptic curve $E$ over $C$ equipped with a trivialization of the Tate module, $T_pE\cong(\mathbb{Z}_p)^2$. Maybe this is really trivial, but I cannot see how to prove it, or even how to figure it out. Is it related to a choice of a splitting of the Hodge-Tate exact sequence for the Tate module? Another related, but maybe more difficult question, is the following, whats gonna happen if we consider the tower of formal models for the modular curves with increasing level? Of course, computing the limit along Frobenius, we get a formal scheme, but is there an analogous description of a "point" as an elliptic curve over a ring plus a trivialization of the Tate module?

$\endgroup$
5
  • 1
    $\begingroup$ When one takes the inverse limit of the anticanonical tower one does get a perfectoid space but this is not the moduli of elliptic curves with a trivialisation of the Tate module. So I assume you are misinterpreting what Scholze says: can you point to the precise statement that you are referring to? $\endgroup$
    – naf
    Commented Jul 8, 2018 at 6:29
  • $\begingroup$ Yes! Look at the proof of Lemma III.3.6. There, after a reduction argument in which he passes to a rank 1 point, he says that such a point corresponds to an abelian variety with a trivialization of the Tate module. You may also have a look to Ana Caraiani's notes for AWS2017, they are titled "Lecture notes on perfectoid Shimura Varieties". In these notes, at page 41, she declares that geometric points of the infinite level modular curve have such a moduli interpretation. But really I don't see the connection. Thanks! $\endgroup$
    – rime
    Commented Jul 8, 2018 at 6:51
  • $\begingroup$ Btw, I agree this is not a moduli interpretation since, I think, it only holds for geometric points, but I still want to understand why is it true at this level and moreover, is there a moduli interpretation for this perfectoid space? I think no. $\endgroup$
    – rime
    Commented Jul 8, 2018 at 6:53
  • 1
    $\begingroup$ I looked at the lemma in Scholze that you mentioned. The space he is considering there, i.e. $\mathcal{X}_{\Gamma(p^{\infty})}$, is not the inverse limit of the anticanonical tower (which is an open in the space denoted by $\mathcal{X}_{\Gamma_0(p^{\infty})}$ in the paper). $\endgroup$
    – naf
    Commented Jul 9, 2018 at 5:22
  • $\begingroup$ Yes, you are right! Here the tower is an inverse limit given essentially by modular curves which, at each step, trivialize the torsion, and the transition morphisms are not given by Frobenius. So, do yo u think it's not possible to get a kind of moduli interpretation for the anticanonical tower? $\endgroup$
    – rime
    Commented Jul 9, 2018 at 6:45

1 Answer 1

4
$\begingroup$

The stack $M(p^n)$ of elliptic curves with a full level $p^n$-structure is canonically defined over $\mathbf{Z}_p[\zeta_{p^n}]$. Its generic fiber is the modular curve $X(p^n)$, and this determines a rigid analytic curve; taking the inverse limit along the inverse system (in the appropriate category) defines maps $X(p^n)^\mathrm{an} \to X(p^{n-1})^\mathrm{an}$ gives a perfectoid space $\mathcal{X}_\infty$. Your question is really about the classical schemes. When $p$ is inverted, i.e., when we work with the generic fiber, the definition of a full level $p^n$-structure $f$ translates into exactly the condition that the map $f:(\mathbf{Z}/p^n)^2 \to E[p^n]$ is an isomorphism (this is classical, and can be found in the first chapter of Katz-Mazur). Now looking at the definition of the connecting maps $X(p^n) \to X(p^{n-1})$, one finds that an element of the inverse limit (denoted $X(p^\infty)$) is exactly an elliptic curve with a trivialization of its Tate module.

$\endgroup$
4
  • $\begingroup$ Thank you very much for your answer. Sorry, but the point I really don't understand is precisely your last assertion=) So, why does it give a trivialization of the Tate module? Are there any obvious fact I don't see? $\endgroup$
    – rime
    Commented Jul 7, 2018 at 21:21
  • 1
    $\begingroup$ My last assertion follows from the definition of the Tate module as the inverse limit of the $E[p^n]$'s, and the definition of the natural maps between the $X(p^n)$'s. $\endgroup$
    – skd
    Commented Jul 7, 2018 at 21:45
  • $\begingroup$ Ok, natural maps are given by Frobenius, which is essentially quotient modulo the canonical subgroup, and as you recalled, the Tate module is the inverse limit of the $E[p^n]$'s. So it seems to me that, at each finite level, this is the same as choosing a splitting of the exact sequence $0\rightarrow H_n\rightarrow E[p^n]\rightarrow H_n^{\check}\rightarrow 0$, given by canonical subgroup of echelon $n$, and its dual. But again, why is it true? And, how can we pass to inverse limit without touching exactness? Sorry, maybe this is stupid and easy. $\endgroup$
    – rime
    Commented Jul 7, 2018 at 22:19
  • $\begingroup$ Essentially I don't see why the definition of Tate module plus the nature of transition maps allows you to produce this moduli interpretation. That's my concern. I mean, you have a projective system of elliptic curves, where each step is a quotient of the previous one by the canonical subgroup. How do you conclude that such a projective system is equivalent to one elliptic curve plus trivialization? $\endgroup$
    – rime
    Commented Jul 8, 2018 at 6:02

You must log in to answer this question.

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