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So let $X$ be a "nice" topological space and assume that $G$ is a finite group which acts freely on $X$.

Q: Is there a simple relationship between the cohomology groups $H^i(G,\mathbf{Z}), H^i(X,\mathbf{Z})$ and $H^i(X/G,\mathbf{Z})$? Does the Leray spectral sequence simplifies in this special case?

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If $G$ is finite then $H^\ast(G;\mathbb{Z})$ is concentrated in degree zero. The quick fact is that the $G$-invariant cohomology of $X$ is isomorphic to $H^\ast(X/G)$. I'm still unsure what special kind of relationship you're looking for! –  Somnath Basu Mar 2 '11 at 4:20
    
Careful, hypotheses are needed here (to show that $H^*(X/G)$ is the $G$ invariants of $H^*(X)$). Example: take the covering space $S^2\to\mathbb{R}P^2$ with integer coefficients. Then $H^2(X)=0$ but $H^2(X/G)\neq0$. –  David Sprehn Mar 2 '11 at 4:44
    
(added) The thing that goes wrong in that example is that 2, the order of $G$, is not a unit in the coefficient ring. –  David Sprehn Mar 2 '11 at 4:46
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What? $H^*(G)$ need NOT be concentrated in degree zero.... consider periodic cohomology for instance. –  Chris Gerig Mar 2 '11 at 4:46
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Theorem VII.7.9 of Brown's $\textit{Cohomology of Groups}$: If $X$ is a free $G$-complex then we have the Cartan-Leray spectral sequence $E^2=H^p(G,H^qX)$ which abuts to $H^*(X/G)$. I do not think there are any simplifications from this. –  Chris Gerig Mar 2 '11 at 4:52
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2 Answers 2

There is a precise relation at the level of complexes: $C^\ast(X,\mathbb Z)$ is a $G$-complex and as such it is perfect (that is quasi-isomorphic to a finite complex consisting of projective modules) and furthermore $C^\ast(X/G,\mathbb Z)$ is quasi-isomorphic to the derived functor value $R\Gamma(G,C^\ast(X,\mathbb Z)$. The latter is mostly used through its consequent spectral sequence $H^\ast(G,H^\ast(X,\mathbb Z))\implies H^\ast(X/G,\mathbb Z)$ but (as with all spectral sequences) it contains some ambiguity which (somehow) has to be resolved.

This is mainly going from knowledge of the cohomology of $X$ to that of $X/G$. Going the other direction is more difficult as the $G$-cohomology kills a lot of information (if one works with rational coefficients instead, it just picks out the trivial representations). However, the fact that $C^\ast(X,\mathbb Z)$ is perfect helps out even though it can still be difficult to say something.

As an example of the relevance of perfectness consider the case when $G$ is cyclic (of order $n$, say) acting by fixed point free orientation preserving maps on the $k$-sphere. Then the cohomology of $X$ is the trivial representation in degrees $0$ and $k$. Such a complex is classified (this is essentially the Yoneda Ext-description) by an element $\alpha$ in $H^{k+1}(G,\mathbb Z)$. For $C^\ast(X,\mathbb Z)$ to be perfect we must have that $\alpha$ must have order exactly $n$. This excludes $k$ even as the order then is always $1$ and for $k$ odd $\alpha$ must be a generator of $H^{k+1}(G,\mathbb Z)=\mathbb Z/n$. It is now easy to compute $R\Gamma(G,C^\ast(X,\mathbb Z)$ (and its additive structure is actually independent of $\alpha$) but we have also obtained a (necessarily) non-trivial invariant of the action. When $k=3$ this is a well-known invariant of lens spaces.

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The niceness condition you want is on the action, not on the space $X$. Specifically, you want to have that $X\to X/G$ is a principle $G$-bundle, so that we have a Serre spectral sequence for $G\to X\to X/G$. Of course, since you're assuming that $G$ is a finite discrete group, the singular cohomology of $G$ is free, and only in degree 0. In fact, the requirement of being a "principle $G$-bundle" is the same as $X\to X/G$ being a covering space.

The problem: though the spectral sequence looks simple, and collapses immediately at the $E_2$ page, it's not really very useful, since all the interesting data is hidden in the local coefficient system (which is absolutely not trivial unless $G=0$.)

However, we can perhaps get the relationship you want in a much easier way if you're willing to modify the coefficient ring a bit. In particular, the answer is much simpler if you use a ring in which the order of $G$ is a unit. In that case, it's not hard to show directly (using covering space theory) that $H^*(X/G)\to H^*(X)$ is an isomorphism onto the invariants of $G$, i.e. the subring $H^*(X)^G$ of classes which are invariant under the action of $G$. This is an exercise in Milnor's Characteristic Classes, and I believe some form of it appears in Hatcher as well.

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Also: I have a vague memory that there's another sequence you can use, in case you want the more general case (I haven't checked this!): If memory serves, there is a different fibration, $X\to X/G\to BG$. The latter map is the classifying map of the principle bundle $X\to X/G$. But studying this would give a relationship involving the group cohomology of $G$ (maybe that's what you meant in the question though....) –  David Sprehn Mar 2 '11 at 4:39
    
Thats the principal bundle $X\rightarrow X_G\rightarrow BG$ where $X_G$ is the Borel construction –  Chris Gerig Mar 2 '11 at 4:52
    
Right, perfect! Of course, in this case (free action) the Borel construction is homotopy-equivalent to the orbit space $X/G$. A small quibble: it's not a principle bundle (the fiber is $X$, not $G$) but merely a fiber bundle with structure group $G$. Of course, it's still a fibration so we get a Serre spectral sequence as desired. –  David Sprehn Mar 2 '11 at 5:11
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A couple of comments/corrections. First the spectral sequence of $X \to X_{hG} \to BG$ need not collapse for all coefficients, even when the action is free so that the Borel construction on $X$ (which I wrote $X_{hG}$) is just $X/G$. Think of $\mathbb{Z}/2$ acting on $S^1$ by the antipodal map, which induces the identity on cohomology; the quotient is still $S^1$. On the $E_2$ page you have two copies of the cohomology of the group, which is very big, while the target $S^1$ has a cohomology in degrees 0 and 1 only. Plenty of differentials all over the place. (continued) –  Pierre Mar 2 '11 at 6:24
    
Second, the last statement, about coefficients for which the order of the group is invertible, follows immediately from the spectral sequence: in this case it has only one column, which is $H^0(G, H^*(X)) = (H^*(X))^G$. –  Pierre Mar 2 '11 at 6:36
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