Timeline for Why does a group action on a scheme induce a group action on cohomology?
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| Nov 2, 2011 at 14:42 | comment | added | Damian Rössler | @unknown : How do you construct the map $H^i(Y,G)\to H^i(X,F)$ ? Indeed, the pull-back functor $f^*$ does not send injectives into injectives (or other acyclics) in general, so this map cannot be simply constructed from a morphism of complexes... in fact, you need to consider the adjoint map $G\to f_*F$ and use the Leray spectral sequence (there isn't enough space here to write down the details). | |
| Nov 1, 2011 at 16:30 | history | edited | Niels | CC BY-SA 3.0 |
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| Nov 1, 2011 at 13:42 | vote | accept | George P. | ||
| Nov 1, 2011 at 13:22 | comment | added | George P. | I just came across the book "Etale Cohomology Theory" by Lei Fu which seems to explain precisely this stuff with the G-sheaves you mentioned. I will take a look at this now. | |
| Nov 1, 2011 at 13:15 | comment | added | Niels | @Damian Rössler: you are right, for endomorphisms one needs to be somewhat more careful and work with monoids, but I don't think it is fundamentally different. For actions of non-constant schemes, I am not longer sure what the question means (probably one then considers action of the global sections of the group scheme on the cohomology). | |
| Nov 1, 2011 at 13:12 | comment | added | Niels | @ George P. 1 You can associate to each abstract group $G$ a "constant" group scheme $G_X$ on $X$, that represents the functor $Hom_X(\cdot,G\times X)$. In other words the sections of $G_X$ above the open $U\to X$ are $G^{\pi_O(U)}$, where $\pi_O(U)$ is the set of connected components of $U$. 2 The standard reference is Sur quelques points d'algèbre homologique, I Alexander Grothendieck Source: Tohoku Math. J. (2) Volume 9, Number 2 (1957), 119-221. projecteuclid.org/… | |
| Nov 1, 2011 at 12:14 | comment | added | Nicolás | Sheaf cohomology is only functorial on the pairs $(X, F)$ consisting of a scheme $X$ and a sheaf $F$, where a map $(X, F)\to (Y, G)$ is a pair of maps $(f,\phi)$, $f : X \to Y$ and $\phi : f^*G \to F$. Any such map of pairs induces a map on the cohomology $H^i(Y,G)\to H^i(X,F)$ as explained by Niels. | |
| Nov 1, 2011 at 12:00 | comment | added | Damian Rössler | The question of George P. was about general endomorphisms, not just constant group schemes. The functoriality of the cohomology is more difficult to construct for general endomorphisms. | |
| Nov 1, 2011 at 11:48 | comment | added | George P. | Very good, thanks, I'll think about this! Two more questions: 1. What is a constant group? 2. Do you have a good reference for what you just explained? Perhaps something post-SGA to make it shorter :) | |
| Nov 1, 2011 at 10:51 | history | answered | Niels | CC BY-SA 3.0 |