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.



Are D_dR and D_st "potentially comparable"?
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.


