If $p: B \rightarrow C$ is a finite covering space with covering group $G$. Why (rigorously) $H^{*}(B)^{W} \simeq H^{*}(C)$
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1With what coefficients? – Mariano Suárez-Álvarez Jan 18 '13 at 05:13
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It should be G instead of W... – Jacob Jan 18 '13 at 05:30
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Sorry... $G=W$ and real coefficients – ArthurStuart Jan 18 '13 at 05:57
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In general, $G$ is a group acting continuously on a space $X$ and $M$ is an abelian group, there is a spectral sequence going from $E_2=H^\bullet(G,H^\bullet(B,M))$ to $H^\bullet(B/G,M)$; here $H^\bullet(G,\mathord-)$ is group cohomologgy of the discrete group $G$. This was constructed by Grothendieck in his famous Tohôku paper, for example. (The spaces have to be sufficiently nice for their sheaf cohomology to coincide with whatever cohomology you want; paracompact is enough, and that is not exactly a draconian hypothesis :D )
If $p:B\to C$ is a covering with group $G$, then $B/G=C$. If morevoer the group finite and, say, $M=\mathbb Q$ (or any abelian group in which multiplication by the order of $G$ is an isomorphism) then $H^\bullet(B,M)$ is also a abelian group with that property, and $H^p(G,H^\bullet(B,M))=0$ if $p>0$. Therefore the spectral sequence collapses at $E_2$ and we have an isomorphism $H^0(G,H^\bullet(B,M))\cong H^\bullet(C,M)$. Finally, since $H^0(G,\mathord-)$ is canonically isomorphic to $(\mathord-)^G$, we have your isomorphism.
All this technology, though, is not really needed. Suppose for example the spaces in question are manifolds and we compute cohomology using de Rham. Then the group $G$ acts on the de Rham complex $\Omega^\bullet(B)$, and we can consider its fixed subcomplex, $\Omega^\bullet(B)^G$. It is very easy to show that $\Omega^\bullet(B)^G$ is in fact isomorphic to the de Rham complex $\Omega^\bullet(C)$ of the quotient. On the other hand, it is also easy to show that taking invariants and computing cohomology of these vector spaces commute (it is for this that we want the group finite and real vector spaces), so that $H(\Omega^\bullet(B)^G)\cong H(\Omega^\bullet(B))^G$. The same thing will work wit the complex which computes singular cohomology with coefficients in $\mathbb Q$.
Mariano Suárez-Álvarez
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See also http://mathoverflow.net/questions/57071/on-the-cohomology-of-a-finite-covering-map – Mariano Suárez-Álvarez Jan 18 '13 at 05:28
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1Notice that for your iso to hold you need some hypothesis on the coefficients. Take $G=\mathbb Z/2$ acting as usual on $B=S^2$, so that $C=P^2$. If you take coefficients in $\mathbb Z$, then $H^1(B)^G=0$ yet $H^1(C)\neq0$. – Mariano Suárez-Álvarez Jan 18 '13 at 05:32
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Have you also a specral sequence proof that under good hypotesis if $p: E \rightarrow B$ is a bundle with fiber $F$, then $\chi(E)=\chi(B)\chi(F)$? $\chi$ is the Euler Characteristic – ArthurStuart Jan 18 '13 at 06:25
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@ArthurStuart, sure. You can read it in the book by Bott and Tu. – Mariano Suárez-Álvarez Jan 18 '13 at 06:43
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@ Mariano Suárez-Alvarez Have you a reference for your second proof? Because (for me) it isn't very clear... Thanks – ArthurStuart Jan 24 '13 at 19:49
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Why $H^p(G,H^\bullet(B,M))=0$ if $p>0$, why the multiplicantion by order of $G$ must be an isomorphism and why $H^{0}(G,-) \simeq (-)^{G}$ – ArthurStuart Jan 25 '13 at 00:32
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That is not the second proof but the firts one I sketched! In any case, it is a well-know fact about group cohomology that if $G$ is a finite group and $M$ an abelian group on which $G$ acts and such that multiplication by $|G|$ is invertible on $M$, then $H^p(G,M)=0$ for $p>0$. This is done in pretty much every textbook which treats group cohomology. – Mariano Suárez-Álvarez Jan 25 '13 at 00:54