The following may be well-known (or obviously false), but I can't find a counterexample or a reference.

Suppose that $k$ is some perfect field (one can assume algebraically closed, if that makes you happy) and $X/k$ a connected $p$-divisible group. Suppose though that $X=\widehat{G}[p^\infty]$ where $\widehat{G}$ is a $p$-divisible formal group coming from the completion of some $G/k$.

What is the connection between $D(X)$ and $H^1_\mathrm{crys}(G/W(k))$? If $G$ happens to be a super-singular abelian scheme, then we have equality (as follows from Mazur-Messing-Oda), and they are also equal then $G=\mathbf{G}_m$, but I don't know whether to expect equality in general.

More generally, what are properties on $Y/k$ such that $H^1_\mathrm{crys}(Y/W(k))$ has slopes in $[0,1]$? Moreover, if it does, what can you say about the $X/k$ with $D(X)\cong H^1_\mathrm{crys}(Y/W(k))$—must it 'relate' to $X$ in any reasonable way?


EDIT: A helpful comment from a wise elder has suggested that one might be able to say that basically $G$ is semi-abelian and try to prove the result by proving it respects extensions. Have not run through the details, but it seems plausible.

  • $\begingroup$ if you have figured it out, maybe answer your own question so that everyone could benefit from the wisdom of the elder. $\endgroup$
    – user74900
    Apr 8, 2019 at 20:32


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