The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03VM. Beware of the difference between the letter 'O' and the digit '0'.