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Let $X$ be a variety over a $p$-adic field $K$.

Is there a simple or intuitive explanation of why the $G_K$ representation $H^i(X_{ét},\mathbb{Q}_p)$ is Hodge-Tate? More precisely, why do the powers of the cyclotomic character appear and not other characters?

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  • $\begingroup$ For clarity, does "variety" assume smooth and proper as well? $\endgroup$
    – Matt
    Sep 8, 2013 at 20:21
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    $\begingroup$ Just to be precise, what is true is that, after tensoring over $\mathbb{C}_p$, the subspaces on which the (semi-linear!) Galois action is via powers of the cyclotomic character generate the whole space. As for whether there is an 'intuitive' explanation for this phenomenon, I don't know! $\endgroup$
    – Keerthi Madapusi
    Sep 9, 2013 at 3:07
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    $\begingroup$ As Keerthi pointed out, it's after you tensor by Cp that the powers of the cyclo char appear. But a statement like that is true for any Galois repn V by Sen's theory. What is specific to the étale cohomology is that it's integer powers of the cyclo char that appear. This is where you use the geometric input. As to why this is so... $\endgroup$
    – Laurent Berger
    Sep 10, 2013 at 8:02

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You need to assume that your variety is proper and smooth.

The properness implies that the vector space $H^{1}_{et}(X_{\overline{K}}, \mathbb{Q}_p)$ is of finite dimension. And the smoothness implies the following decomposition which is proved by Falting: $(\mathbb{C}_p \otimes H^{i}(X_{\overline{K}}, \mathbb{Q}_p)=\oplus_{j \in \mathbb{Z}} \mathbb{C}_p(-j) \otimes H^{i-j}(X,\Omega^{j}_{X})$

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    $\begingroup$ In fact neither smoothness or properness is necessary for the cohomology to be Hodge-Tate, though the isomorphism of your last line does require both smoothness and properness. $\endgroup$
    – Daniel Litt
    Mar 28, 2018 at 18:38

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