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Let $K$ be a complete discrete valuation field of mixed characteristic $(0,p)$ with perfect residue field $k$. Suppose $V$ is an unramified representation with associated continuous homomorphism $\rho:G_K \rightarrow GL(V)$, where $G_K$ is the absolute Galois group of $K$. Then $V$ is a crystalline representation with associated filtered $\varphi$-module $D = (\widehat{K^{nr}} \otimes_{\mathbb{Q}_p} V)^{G_K}$, where $\widehat{K^{nr}}$ is the completion of the maximal unramified extension of $\mathbb{Q}_p$. Note that $D$ is a vector space over the fraction field $K_0$ of the ring of Witt vectors $W(k)$ with coefficients in $k$. It is finite-dimensional with dimension $d = \dim_{\mathbb{Q}_p}V$.

Question: In general, how does the operator $\varphi$ relate with the arithmetic Frobenius $Frob \in G_K/I_K$?

In the case when $K$ is a finite extension of $\mathbb{Q}_p$ (i.e. $k$ is finite of size $q = p^r$), such a relation can be obtained in the following manner: By extending scalars, we have an isomorphism $\alpha: \widehat{K^{nr}} \otimes_{K_0} D \simeq \widehat{K^{nr}} \otimes_{\mathbb{Q}_p} V$. If $\sigma$ denotes the arithmetic Frobenius in $Gal(K_0 / \mathbb{Q}_p)$, then the $\sigma$-linear extension $\sigma \otimes \varphi$ of $\varphi$ to $\widehat{K^{nr}} \otimes_{K_0} D$ satisfies $$ \alpha \circ (\sigma \otimes \varphi) \circ \alpha^{-1} = \sigma \otimes 1_V. $$ Raising both sides to power $r$ and extending $\varphi^r$ linearly to the whole of $\widehat{K^{nr}} \otimes_{K_0} D$, we have $$ \alpha \circ (\sigma^r \otimes 1_D) \circ \varphi^r \circ \alpha^{-1} = \sigma^r \otimes 1_V. \hspace{1in} (1)$$ On the other hand, note that $D$ is the $K_0$-subspace of $\widehat{K^{nr}} \otimes_{\mathbb{Q}_p} V$ fixed by the $\sigma^r$-linear extension of $\rho(Frob)$ to $\widehat{K^{nr}} \otimes_{\mathbb{Q}_p} V$. Extending $\rho(Frob)$ to $\widehat{K^{nr}} \otimes_{\mathbb{Q}_p} V$, we then see that $$ \alpha \circ (\sigma^r \otimes 1_D) \circ \alpha^{-1} = (\sigma^r \otimes 1_V) \circ \rho(Frob). \hspace{1in} (2)$$ Comparing $(1)$ and $(2)$, we obtain the relation $$ \alpha \circ \varphi^r \circ \alpha^{-1} = \rho(Frob^{-1}). $$

I am then interested to know whether this relation generalizes to arbitrary $K$ (with perfect residue field not necessarily finite). I appreciate any ideas you can share.

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Let $k$ be the infinite residue field, then no power of the Frobenius automorphisms fixes each element of $k$, so none is in the Galois group of $k$. Which Galois group element do you refer to? –  Will Sawin Oct 6 '13 at 22:34
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