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Suppose we have a de Rham Galois representation $G_K\rightarrow GL(V)$ for some $p$-adic field $K$ and some finite dimensional vector space $V$ over $\mathbf{Q}_p$. Then it is a theorem that there is some finite extension $L/K$ such that $D_{dR}^L(V)\cong L\otimes_{L_0}D_{st}^L(V)$ (this is actually a corollary of the theorem that de Rham implies potentially semi-stable).

My question is whether I can still find such an extension $L/K$ without the assumption that the representation is de Rham. In other words, is there always some finite extension $L/K$ such that $dim_L D_{dR}^L(V)=dim_{L_0}D_{st}^L(V)$? I can't imagine that this is true, so what I'm really asking for is a counterexample.

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I suspect that the answer might be "yes": let D be the phi, Gamma module over the Robba ring attached to V. Then I think you can deine a submodule E of D which is the things whose localization at a root of unity of sufficiently high order lands in DdR(V), which is somehow the lrgest de Rham submodule of D. Then applying Berger's results to this submodule E should do the trick. But let's wait and see if Laurent has something to say about this himself! –  David Loeffler Sep 29 '12 at 8:40

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up vote 15 down vote accepted

Actually, I think that Rebecca is right and that the answer is "no". Here's a sketch of the reason why.

Let $V$ be a $p$-adic representation. If $V$ is Hodge-Tate, then $D_{dR}(V) \neq 0$. So it's enough to find a HT representation such that $D_{st}^L (V) = 0$ for any $L$. Although I can't think of an explicit one, in my paper "Représentations potentiellement triangulines de dimension 2", with Gaëtan Chenevier, we prove the following theorem: if $X$ is the universal deformation space of some mod $p$ representation, and if $X_P$ is the subset of $X$ consisting of representations whose Sen polynomial is $P$, then the subset of $X_P$ consisting of potentially trianguline representations is a "thin subset" (i.e. most representations in $X_P$ are not potentially trianguline). It remains to observe that if $D_{st}^L (V) \neq 0$, then $V$ is potentially trianguline.

EDIT : I should have said that $V$ is of dimension 2 here.

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By the way, these "thin subsets" are called "parties fines" in our article. I suspect that my coauthor was playing a joke on me (and the readers) since I later found out that in French, "partie fine" also means "orgy". –  Laurent Berger Sep 29 '12 at 11:56
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This has more spice than Bourbaki's "ensemble flirtant à droite et à gauche" instead of "ensemble filtrant à gauche et à droite". Bravo, Gaëtan ! –  Chandan Singh Dalawat Sep 29 '12 at 16:13
    
Great, thanks! –  Rebecca Bellovin Sep 30 '12 at 19:23

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