Lemma 64.29.1. 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.
64.29 Cohomology of curves, revisited
Let $k$ be a field, $X$ be geometrically connected, smooth curve over $k$. We have the fundamental short exact sequence
\[ 1 \to \pi _1(X_{\overline{k}}, \overline\eta ) \to \pi _1(X, \overline\eta ) \to \text{Gal}(k^{^{sep}}/k) \to 1 \]
If $\Lambda $ is a finite ring with $\# \Lambda \in k^*$ and $M$ a finite $\Lambda $-module, and we are given
\[ \rho : \pi _1(X, \overline\eta ) \to \text{Aut}_{\Lambda }(M) \]
continuous, then $\mathcal{F}_\rho $ denotes the associated sheaf on $X_{\acute{e}tale}$.
Here the subscript ${}_{\pi _1(X_{\overline{k}}, \overline\eta )}$ indicates co-invariants, and $(-1)$ indicates the Tate twist i.e., $\sigma \in \text{Gal}(k^{^{sep}}/k)$ acts via
\[ \chi _{cycl}(\sigma )^{-1}.\sigma \text{ on RHS} \]
where
\[ \chi _{cycl} : \text{Gal}(k^{^{sep}}/k) \to \prod \nolimits _{l\neq char(k)}\mathbf{Z}_ l^* \]
is the cyclotomic character.
Reformulation (Deligne, Weil II, page 338). For any finite locally constant sheaf $\mathcal{F}$ on $X$ there is a maximal quotient $\mathcal{F}\to \mathcal{F}''$ with $\mathcal{F}''/X_{\overline{k}}$ a constant sheaf, hence
\[ \mathcal{F}'' = (X\to \mathop{\mathrm{Spec}}(k))^{-1}F'' \]
where $F''$ is a sheaf $\mathop{\mathrm{Spec}}(k)$, i.e., a $\text{Gal}(k^{^{sep}}/k)$-module. Then
\[ H_ c^2(X_{\overline{k}}, \mathcal{F})\to H_ c^2(X_{\overline{k}}, \mathcal{F}'')\to F''(-1) \]
is an isomorphism.
Proof of Lemma 64.29.1. 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 $H_ c^2(X_{\overline{k}}, \mu _ n)=^{\text{can}} \mathbf{Z}/n\mathbf{Z}$. $\square$
Remark 64.29.2. Thus we conclude that if $X$ is also projective then we have functorially in the representation $\rho $ the identifications and Of course if $X$ is not projective, then $H^0_ c(X_{\overline{k}}, \mathcal{F}_\rho ) = 0$.
\[ H^0(X_{\overline{k}}, \mathcal{F}_\rho ) = M^{\pi _1(X_{\overline{k}}, \overline\eta )} \]
\[ H_ c^2(X_{\overline{k}}, \mathcal{F}_\rho ) = M_{\pi _1(X_{\overline{k}}, \overline\eta )}(-1) \]
Proposition 64.29.3. Let $X/k$ as before but $X_{\overline{k}}\neq \mathbf{P}^1_{\overline{k}}$ The functors $ (M, \rho )\mapsto H_ c^{2-i}(X_{\overline{k}}, \mathcal{F}_\rho ) $ are the left derived functor of $(M, \rho )\mapsto H_ c^2(X_{\overline{k}}, \mathcal{F}_\rho )$ so Moreover, there is a derived version, namely in $D(\Lambda [[\widehat{\mathbf{Z}}]])$. Similarly, the functors $(M, \rho )\mapsto H^ i(X_{\overline{k}}, \mathcal{F}_\rho )$ are the right derived functor of $(M, \rho )\mapsto M^{\pi _1(X_{\overline{k}}, \overline\eta )}$ so Moreover, in this case there is a derived version too.
\[ H_ c^{2-i}(X_{\overline{k}}, \mathcal{F}_\rho ) = H_ i(\pi _1(X_{\overline{k}}, \overline\eta ), M)(-1) \]
\[ R\Gamma _ c(X_{\overline{k}}, \mathcal{F}_\rho ) = LH_0(\pi _1(X_{\overline{k}}, \overline\eta ), M(-1)) = M(-1) \otimes _{\Lambda [[\pi _1(X_{\overline{k}}, \overline\eta )]]}^\mathbf {L} \Lambda \]
\[ H^ i(X_{\overline{k}}, \mathcal{F}_\rho ) = H^ i(\pi _1(X_{\overline{k}}, \overline\eta ), M) \]
Proof. (Idea) Show both sides are universal $\delta $-functors. $\square$
Remark 64.29.4. By the proposition and Trivial duality then you get a perfect pairing. If $X$ is projective then this is Poincare duality.
\[ H^{2-i}_ c(X_{\overline{k}}, \mathcal{F}_\rho ) \times H^ i(X_{\overline{k}}, \mathcal{F}_\rho ^\wedge (1)) \to \mathbf{Q}/\mathbf{Z} \]
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