## 63.29 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 $X_{\acute{e}tale}$.

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.

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 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 $H_ c^2(X_{\overline{k}}, \mu _ n)=^{\text{can}} \mathbf{Z}/n\mathbf{Z}$. $\square$

Remark 63.29.2. 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 63.29.3. 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 63.29.4. 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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