# Example of a $p$-divisible group that is not representable by a formal scheme

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.

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.