On the cohomology of a finite covering map - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:03:36Z http://mathoverflow.net/feeds/question/57071 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57071/on-the-cohomology-of-a-finite-covering-map On the cohomology of a finite covering map Hugo Chapdelaine 2011-03-02T03:30:35Z 2011-03-03T22:51:16Z <p>So let $X$ be a "nice" topological space and assume that $G$ is a finite group which acts freely on $X$. </p> <p>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?</p> http://mathoverflow.net/questions/57071/on-the-cohomology-of-a-finite-covering-map/57076#57076 Answer by David Sprehn for On the cohomology of a finite covering map David Sprehn 2011-03-02T04:35:27Z 2011-03-03T18:42:07Z <p>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.</p> <p>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 $G=0$.)</p> <p>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 <code>$H^*(X/G)\to H^*(X)$</code> is an isomorphism onto the invariants of $G$, i.e. the subring <code>$H^*(X)^G$</code> 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.</p> http://mathoverflow.net/questions/57071/on-the-cohomology-of-a-finite-covering-map/57080#57080 Answer by Torsten Ekedahl for On the cohomology of a finite covering map Torsten Ekedahl 2011-03-02T06:03:16Z 2011-03-02T06:03:16Z <p>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.</p> <p>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.</p> <p>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.</p>