Crystalline when restricted to inertial subgroup

Question

$\newcommand{\ur}{\mathrm{ur}}\newcommand{\cris}{\mathrm{cris}}$Let $K$ be a finite extension of $\mathbb{Q}_p$, $G_K=\operatorname{Gal}(\overline{K}/K)$ and $I_K \subset G_K$ its inertial subgroup. Let $V$ be a finite-dimensional representation of $G_K$. Assume that $V$ is crystalline as $G_K$-representation. Is it true that it is crystalline as $I_K$-representation?

By definition, we need to prove that $$\dim_{\widehat{K^{\ur}}}(V \otimes B_{\cris})^{I_K}=\dim_{K_0}(V \otimes B_{\cris})^{G_K}=\dim_{\mathbb{Q}_p} V$$ where $\widehat{K^{\ur}}$ and $K_0$ are the maximal complete unramified extension of $\mathbb{Q}_p$ in $\overline{K}$ and $K$ respectively.

Edit: As remarked by David, $K^{\ur}$ need to be replaced by its completion.

Answer 1

This is purely formal. If $V$ is crystalline, then $V \otimes \mathbf{B}_{\mathrm{cris}}$ has a basis as a $\mathbf{B}_{\mathrm{cris}}$-module in which the action of $G_K$ is trivial. Hence a fortiori it has a basis in which the action of $I_K$ is trivial.

What is much less obvious, but also true, is that the converse holds: if $V$ is crystalline as an $I_K$-representation, then it's actually crystalline as a $G_K$-representation. This is because $(B_{\mathrm{cris}})^{I_K} = \widehat{K^{\mathrm{nr}}}$ contains the periods of all unramified representations. (Incidentally, the assertion in your question that $(B_{\mathrm{cris}})^{I_K} = K^{\mathrm{nr}}$ is incorrect; it is genuinely the completion that you get here.)