p-adic Poincaré Lemma

Question

suppose $X$ is a proper and smooth rigid analytic variety over $\text{Spa}(k)$, with $k$ a non-archimedean field of characteristic zero.

One has the de Rham complex of analytic differential forms on $X$, $\Omega^{\bullet}_{X/k}$, say on $X_{et}$.

We call $C$ the sheaf $\ker(\mathcal{O}_X\xrightarrow{d}\Omega^1_{X/k})$.

Q1 Is the augmented complex $C\to\Omega^{\bullet}_{X/k}$ exact? Why / why not?

Q2 Is $C$ at all related to the constant étale sheaf $\underline{k}$?

Q3 Is the augmented complex $\underline{k}\to\Omega^{\bullet}_{X/k}$ exact, or does one still need to base change to big period rings from $p$-adic Hodge Theory?

Q4 Most importantly, if the answer to Q3 is no, what is a counterexample?

I would benefit from some references.

Answer 1

In the simplest case where $X$ is smooth and projective, and $k$ is discretely valued, then the answer to Q3 should be no.

EDIT: While waiting for the bus I realized there is a technical error here, which is that the etale topos of the Berkovich space is not the etale topos of the rigid/adic space. Instead it is the partially-proper etale topos of the rigid/adic space, as far as I know this result is due to Huber. However the partially-proper etale and the etale topologies both give the same cohomology for locally constant sheaves, so the argument below still goes through.

We can view this via the formalism of Berkovich spaces where $X$ is necessarily strictly $k$-analytic, and then we can use the change-of-topology spectral sequence for the geometric morphism $\pi$ going from the etale topos to the topos of the underlying Berkovich space $|X|$ of $X$.

$ E_2^{p,q} = H^q(|X|, R^p\pi_* \, \underline{k}) $

This seems bad, but we actually can compute the stalks of $R^p\pi_* \, \underline{k}$ at a point $x \in |X|$ as galois cohomology:

$ (R^p \pi_* \, \underline{k})_x = H^p(\mathrm{Gal}\, \mathcal{H}(x), \underline{k})$

But $k$ is a discrete module and the continuity of the action of $G = \mathrm{Gal}\,\mathcal{H}(x)$ implies that the action must factor through some finite quotient of $G$. However we know (well, I believe in my heart of hearts) that even infinite dimensional rational representations of finite groups are just big direct sums of the $1$-dimensional representations. We can ultimately conclude that the action of $G$ on $\underline{k}$ is conjugate to the trivial action, and finally conclude that $R^p \pi_* \, \underline{k} = 0$ for $p > 0$.

What is the point of this silliness?

Well the upshot is that if $X$ admits a semi-stable model over the DVR, then $H^*(|X|,\underline{k})$ is actually just the $k$-valued singular cohomology of the dual graph of the special fiber of the model. In particular it is zero above degree $n = \mathrm{dim}\, X$ despite the fact that $H^{2n}(X_{\acute{e}t}, \Omega^\bullet_{X/k}) \neq 0$. As a concrete example we could take $\mathbb{P}^1_k$.

Note that this really carefully depends on the discreteness of the valuation. You could perhaps bootstrap this argument up to fields $k$ where we can compare to Berkovich spaces, e.g. where the semi-norm on $k$ is real-valued. I'm not familiar with Fontaine rings, but I imagine they are quite far from being semi-normable.

Answer 2

On Q3, maybe an additional geometric remark. First, to take the non-torsion field $k$ as coefficients, you should use the pro-étale topology, not the étale one, as Scholze does (see Scholze and Bhatt's paper on pro-étale cohomology). This may correct partly the well identified problem with Galois cohomology that appears in the calculation of Jon Brener's answer. Moreover, i think the point is that, even locally for the étale analytic topology (either in Huber's quasi-étale sense or in Berkovich's sense), a smooth rigid space is not a disk at every point, contrary to the complex analytic case, but a rational domain of the affine space, or of a disk (take for example the Gauss point of the circle, whose neighborhoods are annuli). Such a rational domain may be non-contractible in the de Rham sense, i.e., have non-trivial higher de Rham cohomology, so one can't show that the de Rham complex is locally exact, even for the analytic étale cohomology, in higher degrees. This is to be compared with the algebraic situation, where the Zariski topology is not fine enough to make this complex locally exact, and to the complex analytic situation, where manifolds are locally given by polydiscs, that are ``de Rham contractible'' in the above sense, so that one may prove the local Poincaré's lemma as in the case of smooth manifolds (there is a nice mathoverflow answer on this by David Speyer).

However, every point of the rigid analytic space in the classical sense (with residue field finite over the base field) has a fundamental system of étale neighborhoods given by polydiscs, on which the Poincaré lemma is true, so that one may show that the de Rham complex is locally exact at these classical points for the Berkovich étale topology. This is discussed in details in Berkovich's paper ``Integration of one forms on $p$-adic analytic spaces''.

There is also a $p$-adic Poincaré lemma for smooth proper rigid analytic varieties, due to Scholze (corollary 6.3 of ``$p$-adic Hodge theory for rigid analytic varieties''), that uses the pro-étale topology on $X$, and a version of the de Rham complex defined using a sheaf of periods $B^+_{dR}$ as an analog of the constant sheaf. This implies the isomorphism between the (pro-)étale cohomology with coefficients in $B_{dR}$ and de Rham cohomology of $X_{B_{dR}}$, that was previously known in the semi-stable algebraic situation, by Faltings and others (by other methods).

Another Poincaré lemma in the $p$-adic setting based on the $h$-topology and derived de Rham cohomology is due to Beilinson (see also Bhatt's work).

Answer 3

I don't know if this is still of interest to anyone, but all these questions are answered in Scholze's answer to this post:

$p$-adic Hodge Theory for rigid spaces, after P. Scholze

In particular:

Q1. No. Rigid spaces are not locally contractibe. For example look at the torus.

Q2. Let $k$ be a complete algebraically closed extension of $\mathbb{Q}_p$. Then the kernel of the differential is equal to the integral closure of $\mathbb{Q}_p$ in $\mathcal{O}_X$.

Q3. As stated already the answer to this is also no. There would be no point to p-adic Hodge theory otherwise.