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.”

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