10
$\begingroup$

Let $R$ be a ring such that $p^nR=0$ for some integer $n$, and $G$ be a $p$-divisible group over $R$.

We think of a $p$-divisible groups as an fppf sheaf $G\colon \mathrm{Alg}^{op}_{R}\to \mathbf{Gps}$ such that

$1) \ G=\mathrm{colim} \ G[p^n]$,

$2) \ [p]\colon G \to G$ is surjective,

$3) \ G[p]$ is a finite, locally-free group scheme.

Question: Is $G$ represented by an $R$-formal scheme (with a finitely generated ideal of definition)?

The answer is positive if $G$ is 'etale or connected. More generally, the answer is positive if $G$ is isogenous to an extension of an 'etale $p$-divisible group by a connected $p$-divisible group (Lemma 3.3.1). In particular, this always holds if $R$ is a field.

However, Scholze and Weinstein write Section 3 of their paper as though there are examples of non-representable $p$-divisible groups but never provide an example of such a group. So, I guess the answer to the question above should be negative in general. But it will be nice to see a particular counter-example.

We can try to take $Y=\mathrm{Spec} \ R$ to be the affine modular curve over $\bar{\mathbf{F}}_p$ (with some full level structure) and $G$ the $p$-divisible group of the universal elliptic curve over $Y$. Then, presumably, $G$ should not be representable. But I don't know a rigorous way to prove it.

$\endgroup$

1 Answer 1

10
$\begingroup$

Your supposed example works indeed. More generally, I think whenever the étale part is not of locally constant height one will run into problems.

Here's a proof that the $p$-divisible group $G$ of the universal elliptic curve $E$ in characteristic $p$ (with auxiliary level structure) is not representable by a formal scheme. Assume it was, and look at an open affine subset $U\subset G$ containing a supersingular point (whose preimage in $G$ is still topologically just a point). Then $U$ necessarily contains the whole preimage of the generic point of the base $Y=\mathrm{Spec}\, R$, as this whole preimage specializes to the given supersingular point (by properness of all $G[p^n]$). But this preimage is not quasicompact, because the étale part of $G$ over the ordinary locus gives a decomposition into countably many connected components. On the other hand, $U$ being affine, it had to be quasicompact, giving a contradiction.

$\endgroup$

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.