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Let $K$ be a field complete with respect to some discrete valuation, with perfect residue field of characteristic $p$. Let $\mathbb{C}_p$ be the completion of an algebraic closure of $K$, and set $G_K := \text{Gal}(\mathbb{C}_p/K)$.

Let $V$ be a $\mathbb{C}_p$ vector space. Is it true in general that $\text{Hom}_K(V^{G_K}, K) \simeq \text{Hom}(V, \mathbb{C}_p)^{G_K}$?

Now let $T_p(X)$ be the Tate module of an abelian variety $X/K$. Then it is true that

$$\text{Hom}_K((T_p(X) \otimes_{\mathbb{Z}_p} \mathbb{C}_p)^{G_K}, K) \simeq \text{Hom}_{\mathbb{Z}_p}(T_p(X), \mathbb{C}_p)^{G_K}.$$

I'm looking for a proof of this which is somewhat elementary: I'm okay with using vanishing theorems for the Galois cohomology of $\mathbb{C}_p(i)$, but preferably not much more than that. Does anybody have such a simple way to see this?

Thanks!

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  • $\begingroup$ Concerning your first question: a $G_K$-equivariant map from V to $C_p$ exists, for example, whenever $V$ is of Hodge-Tate type with one weight equal to $0$. This happens for lots of repns for which $V^{G_K} = 0$. $\endgroup$ Dec 17, 2012 at 8:41
  • $\begingroup$ I'm afraid that I'm not familiar enough with $C_p$-representations to see why such a thing exists. If $V$ is Hodge-Tate with a weight of $0$, wouldn't there be a corresponding copy of $K$ in the invariants? $\endgroup$
    – Tony
    Jan 15, 2013 at 16:54

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