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Let $X$ be a smooth affine scheme over a finite field $k$. Then there exists a smooth affine formal scheme $\mathfrak{X}$ over $W(k)$ with a lift $\sigma$ of the Frobenius. A convergent $F$-isocrystal on $X$ is a vector bundle $\mathcal{V}$ on $\mathcal{X}_{K}$ with an integrable connection and an isomorphism $\sigma^*\mathcal{V}\cong \mathcal{V}$, where we see $\mathcal{X}_K$ as a rigid analytic space.

This is probably naive, but if we have a scheme-theoretic lift to $W(k)$ , that is to say a scheme $\mathcal{X}\to W(k)$ such that $\mathcal{X}\times_{W(k)} k\cong X$, with a lift of Frobenius $F$, is giving a vector bundle $\mathcal{E}$ on $\mathcal{X}_K$ with an integrable connection and an isomorphism $F^*\mathcal{E}\cong \mathcal{E}$ equivalent to the datum of a convergent $F$-isocrystal?

One reason why I'd like this to be true would be that in that case presumably we would have $$H^i_{rig}(X,\mathcal{V})\cong H^i(\mathcal{X}_K,\text{DR}(\mathcal{E}))$$ extending the comparision of crystaline cohomology with the de Rham cohomology of a lift (under favourable conditions). The reason why I hope this to be true is that we have rigid GAGA, but I'm not sure.

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    $\begingroup$ This statement is true, and essentially follows by descent to the affine case. Beware however that in the non-proper (e.g., affine) case, rigid cohomology is something else, involving overconvergent isocrystals. But if the left-hand side just means the cohomology of the convergent isocrystal, I agree with everything. $\endgroup$
    – Peter Scholze
    Commented Mar 26, 2021 at 10:31
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    $\begingroup$ Why should a vector bundle on the lift with integrable connection be considered convergent? Don't we want a distinction between non-convergent crystals and convergent ones such that the Taylor series of solutions to the connection converge on entire residue disks in the convergent case and small disks in the non-convergent one? I'd have thought arbitrary connections on the lift would have small radius of convergence a priori? $\endgroup$
    – Leray Jenkins
    Commented Mar 26, 2021 at 11:37
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    $\begingroup$ The Frobenius structure forces convergence. $\endgroup$
    – Peter Scholze
    Commented Mar 26, 2021 at 12:26
  • $\begingroup$ Ah yes of course, thank you. $\endgroup$
    – Leray Jenkins
    Commented Mar 26, 2021 at 12:34
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    $\begingroup$ Just to use a few more words to slightly elaborate Peter's point above: the claim is that having a Frobenius structure means that the connection is topologically quasi-nilpotent. I think this might be known as "Dwork's trick"; the essence of the argument is in 6.2/6.3 of "Homomorphisms of Barsotti-Tate groups and crystals in positive characteristic" by de Jong. $\endgroup$
    – Raju
    Commented Mar 26, 2021 at 15:46

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