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 be a smooth rigid space over , a -adic field. Recall that a -local system (which will be denoted as ) is a sheaf on the pro-étale site of which (pro-étale-)locally looks like a finite dimensional -vector space times . When is a point, this recaptures the definition of a -adic Galois representation (which will be denoted as ) of a -adic field. There’s a natural generalization of “de Rham representation” which is called a “de Rham -local system” (okay, let me not recall anymore since I’m feeling lazy. From now on, let’s pretend the readers know the “modern” -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: . In the representation situation (i.e. is a point), one knows that there is an injection and de Rham condition means this is an isomorphism. Now one has that is a filtered coherent sheaf and its fibres will have rank at most that of , those points where the fibres having rank equals to that of 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 which presumably is also injective. If one interprete the former as a sub-vector-bundle inside the latter, if one views 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?