Why is it not (that) crazy.

Not-fun-fact: de Rham was born in the French-speaking area of switzerland, so I’ve probably called his name wrong for my whole life (should I change it?). Damn it!

This starts as a discussion with a friend. We were discussing a surprising result of Liu–Zhu about rigidity of local systems on a smooth rigid space being de Rham.

Let $X$ be a smooth rigid space over $K$, a $p$-adic field. Recall that a $\mathbb{Q}_p$-local system (which will be denoted as $\mathbb{L}$) is a sheaf on the pro-étale site of $X$ which (pro-étale-)locally looks like a finite dimensional $\mathbb{Q}_p$-vector space times $X$. When $X$ is a point, this recaptures the definition of a $p$-adic Galois representation (which will be denoted as $V$) of a $p$-adic field. There’s a natural generalization of “de Rham representation” which is called a “de Rham $\mathbb{Q}_p$-local system” (okay, let me not recall anymore since I’m feeling lazy. From now on, let’s pretend the readers know the “modern” $p$-adic Hodge theory after Scholze. Previous statement is true if there’s no reader at all.). There is an amazing/surprising theorem of Liu–Zhu saying that if such a local system is “de Rham” at one point, then it is in fact de Rham everywhere. Both my friend and I knew this statement, and we had a discussion today for why it is not that crazy.

Recall that one has a morphism between sites: $\nu \colon X_{pro\acute{e}t} \to X_{\acute{e}t}$. In the representation situation (i.e. $X$ is a point), one knows that there is an injection $(V \otimes B_{dR})^{Gal} \otimes B_{dR} \to (V \otimes B_{dR})$ and de Rham condition means this is an isomorphism. Now one has that $\nu_{*} (\mathbb{L} \otimes \mathcal{O}\mathbb{B}_{dR})$ is a filtered coherent sheaf and its fibres will have rank at most that of $\mathbb{L}$, those points where the fibres having rank equals to that of $\mathbb{L}$ are presumably the “de Rham locus”. Then by semi-continuity, we convince ourselves that “being de Rham” is a (Zariski)-closed condition.

On the other hand, we have $\nu^* \nu_* (\mathbb{L} \otimes \mathcal{O}\mathbb{B}_{dR}) \otimes_{\nu^* \mathcal{O}_X}\mathcal{O}\mathbb{B}_{dR} \to \mathbb{L} \otimes \mathcal{O}\mathbb{B}_{dR}$ which presumably is also injective. If one interprete the former as a sub-vector-bundle inside the latter, if one views $\mathcal{O}\mathbb{B}_{dR}$ as some sort of structure sheaf, then the semi-continuity would tell us the locus where this is an isomorphism is an open condition. Therefore we also convince ourselves that “being de Rham” is a (pro-étale)-open condition.

Of course we didn’t read the proof (and perhaps we should), so what I said here perhaps is completely a BS.

BTW, anyone knows how to type accents in LaTeX in wordpress?

2 thoughts on “Why is it not (that) crazy.”

1. mathnewb says:

Something like $\text{\'etale}$?

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1. mathnewb says:

\text{\’etale}

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