Let $M$ be a manifold. Let a finite group $G$ act on $M$ discretely. Let $F$ be a field.

Suppose the induced action of $G$ on the cohomology algebra $H^*(M,F)$ is known. We want to obtain $H^*(M/G;F)$. Is there any method or procedure to follow? Can I quotient the action directly $H^*(M/G;F)=H^*(M;F)/G$?


If $F$ is a field of characteristic $0$, then $H^k(M/G;F)$ equals the invariants of the action of $G$ on $H^k(M;F)$. For two different proofs of this, see Proposition III.2.4 of Bredon's "Introduction to compact transformation groups" and Proposition 1.1 of my note "The action on homology of finite groups of automorphisms of surfaces and graphs" available on my webpage here. Both of these are really about simplicial complexes and not manifolds, but it is standard that $M$ can be endowed with a triangulation that is preserved by $G$ (e.g. you can lift a triangulation of $M/G$ such that the images of all fixed points of $G$ are vertices).

If the characteristic of $F$ is not $0$, then all bets are off and in general you can't say anything.

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    $\begingroup$ Presumably this can be promoted to a statement about graded algebras? And is it also true if $|G|$ is invertible in $F$? $\endgroup$ – Mark Grant Mar 19 '15 at 7:54
  • $\begingroup$ @MarkGrant : It's definitely true if $|G|$ is invertible in $F$ (with the same proof). As for whether the isomorphisms respect cup products, I rather doubt it, though I don't have an example in mind. If $G$ acts freely (so the projection $M \rightarrow M/G$ is a covering map), then the isomorphism is induced by the transfer map. See the answer to mathoverflow.net/questions/58159/… for an example of where this is not a ring homomorphism (though that example is not over a field of characteristic $0$). $\endgroup$ – Andy Putman Mar 19 '15 at 16:05

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