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.



p-adic Poincaré Lemma
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.


