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Let $K$ be a finite extension of the $p$-adic numbers. Suppose that $V$ and $W$ are two (finite dimensional, $p$-adic) continuous representations of $G_K$. Suppose that $V \otimes W$ is crystalline. Is $V$ crystalline up to twist by a character of $G_K$?

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Is the zero-dimensional representation crystalline? If so....(Sorry for this.) – Pete L. Clark Apr 29 '10 at 18:37
Idea - we (i.e., not me) know the combinatorics of how tensoring two filtered phi modules affect the Hodge and Newton polygons, and we know the Hodge and Newton polygons of characters. So I think a proof or counterexample could be constructed by thinking about these pictures. – Hunter Brooks Apr 29 '10 at 18:39
@Hunter: the problem is that V and W might not even be Hodge-Tate! Consider for example a random 1-dimensional non-Hodge-Tate V and let W be its dual! – Kevin Buzzard Apr 29 '10 at 19:40
@FC: presumably you can do the ell-adic case? If V tensor W is unramified, is V a twist of an unramified rep? I am wondering whether you might want to start by looking at Sen operators and twisting so that V and W have integral Hodge-Tate weights at least. – Kevin Buzzard Apr 29 '10 at 19:44
@FC: if you just want to know the answer, it's "yes", and I know this because I asked Berger. He didn't tell me why though. – Kevin Buzzard Apr 30 '10 at 0:05
up vote 17 down vote accepted

I'm indeed pretty sure that the answer is "yes". I'd prefer not to post the idea of the proof here because I asked one of my PhD students to write it down with all the details.

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Just out of curiosity, had you already asked your student to do this before the question came up on MO? – Ben Webster May 1 '10 at 19:55
I asked my student to do this in December of 2009. As far as I can remember, it was Barry Mazur who asked me this question when I was at Harvard (so that was at least 5 or 6 years ago). If I remember correctly, he wanted to know what "Sym^2 V crystalline" implied about V. Between then and now, a couple more people asked me the more general question about V \otimes W (I unfortunately don't remember their names). In both cases, I told them the method which I thought would solve the problem, and didn't hear back from them. – Laurent Berger May 2 '10 at 8:40

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