Lemma 63.29.1. There is a canonical isomorphism

$H_ c^2(X_{\overline{k}}, \mathcal{F}_\rho )=(M)_{\pi _1(X_{\overline{k}}, \overline\eta )}(-1)$

as $\text{Gal}(k^{^{sep}}/k)$-modules.

Proof of Lemma 63.29.1. Let $Y\to ^{\varphi }X$ be the finite étale Galois covering corresponding to $\mathop{\mathrm{Ker}}(\rho ) \subset \pi _1(X, \overline\eta )$. So

$\text{Aut}(Y/X)=Ind(\rho )$

is Galois group. Then $\varphi ^*\mathcal{F}_\rho =\underline M_ Y$ and

$\varphi _*\varphi ^*\mathcal{F}_\rho \to \mathcal{F}_\rho$

which gives

\begin{align*} & H_ c^2(X_{\overline{k}}, \varphi _*\varphi ^*\mathcal{F}_\rho ) \to H_ c^2(X_{\overline{k}}, \mathcal{F}_\rho )\\ & =H_ c^2(Y_{\overline{k}}, \varphi ^*\mathcal{F}_\rho )\\ & =H_ c^2(Y_{\overline{k}}, \underline M) = \oplus _{\text{irred. comp. of } \atop Y_{\overline{k}}}M \end{align*}

$\mathop{\mathrm{Im}}(\rho ) \to H_ c^2(Y_{\overline{k}}, \underline M) = \oplus _{\text{irred. comp. of } \atop Y_{\overline{k}}} M \to _{\mathop{\mathrm{Im}}(\rho ) \text{equivalent}} H_ c^2(X_{\overline{k}}, \mathcal{F}_{\rho }) \to ^{\text{trivial } \mathop{\mathrm{Im}}(\rho ) \atop \text{action}}$

irreducible curve $C/\overline{k}$, $H_ c^2(C, \underline M)=M$.

Since

${\text{set of irreducible } \atop \text{components of }Y_ k} = \frac{Im(\rho )}{Im(\rho |_{\pi _1(X_{\overline{k}}, \overline\eta )})}$

We conclude that $H_ c^2(X_{\overline{k}}, \mathcal{F}_\rho )$ is a quotient of $M_{\pi _1(X_{\overline{k}}, \overline\eta )}$. On the other hand, there is a surjection

$\mathcal{F}_\rho \to \mathcal{F}'' = {\text{ sheaf on } X\text{ associated to } \atop (M)_{\pi _1(X_{\overline{k}}, \overline\eta )}\leftarrow \pi _1(X, \overline\eta )}$

$H_ c^2(X_{\overline{k}}, \mathcal{F}_\rho )\to M_{\pi _1(X_{\overline{k}}, \overline\eta )}$

The twist in Galois action comes from the fact that $H_ c^2(X_{\overline{k}}, \mu _ n)=^{\text{can}} \mathbf{Z}/n\mathbf{Z}$. $\square$

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