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
H^0(X_{\overline{k}}, \mathcal{F}_\rho ) = M^{\pi _1(X_{\overline{k}}, \overline\eta )}
and
H_ c^2(X_{\overline{k}}, \mathcal{F}_\rho ) = M_{\pi _1(X_{\overline{k}}, \overline\eta )}(-1)
Of course if X is not projective, then H^0_ c(X_{\overline{k}}, \mathcal{F}_\rho ) = 0.
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
H_ c^{2-i}(X_{\overline{k}}, \mathcal{F}_\rho ) = H_ i(\pi _1(X_{\overline{k}}, \overline\eta ), M)(-1)
Moreover, there is a derived version, namely
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
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
H^ i(X_{\overline{k}}, \mathcal{F}_\rho ) = H^ i(\pi _1(X_{\overline{k}}, \overline\eta ), M)
Moreover, in this case there is a derived version too.
Proof. (Idea) Show both sides are universal \delta -functors. \square
Remark 64.29.4. By the proposition and Trivial duality then you get
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}
a perfect pairing. If X is projective then this is Poincare duality.
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