Equivariant cohomology of finite group actions and invariant cohomology classes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:04:33Z http://mathoverflow.net/feeds/question/120447 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120447/equivariant-cohomology-of-finite-group-actions-and-invariant-cohomology-classes Equivariant cohomology of finite group actions and invariant cohomology classes Edgar 2013-01-31T19:05:54Z 2013-02-01T12:56:13Z <p>Let $W$ be a finite group acting on a space $X$. In what generality is it true that $H^*_W(X) = H^* (X)^W$? We always have a map $H^*_W(X) \rightarrow H^* (X)^W$, but it is certainly not an isomorphism with $\mathbb{Z}$-coefficients (take $X = E_W$, the total space of a universal bundle). But is it true with $\mathbb{Q}$-coefficients? Perhaps if we invert the order of the group?</p> <p>This is a follow up to my <a href="http://mathoverflow.net/questions/120351/equivariant-cohomology-for-actions-with-finite-stabilizers" rel="nofollow">prior question</a> about equivariant cohomology. Again, I am referring to the notes <a href="http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf" rel="nofollow">http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf</a> on equivariant cohomology by Michel Brion.</p> <p>I am interested in understanding the proof of Proposition 1 on pages 6 and 7. Let $G$ be a compact Lie group, let $T$ be a maximal torus, let $N$ be the normalizer of $T$ in $G$, and let $W = N/T$ denote the Weyl group. In part (i), we have the $W$-bundle $G/T \rightarrow G/N$, from which the author claims $H^* (G/N) = H^* (G/T)^W$ when using $\mathbb{Q}$-coefficients.</p> <p>A few sentences later, a similar statement is made for a more arbitrary $W$ bundle. So it seems like the above statement about $W$-invariants of cohomology is true in some generality. Could someone explain why this is true or give a reference? Or perhaps I am mistaken in interpreting this argument.</p> http://mathoverflow.net/questions/120447/equivariant-cohomology-of-finite-group-actions-and-invariant-cohomology-classes/120458#120458 Answer by Mark Grant for Equivariant cohomology of finite group actions and invariant cohomology classes Mark Grant 2013-01-31T20:30:50Z 2013-02-01T12:56:13Z <p>These results follow from the <em>Cartan-Leray spectral sequence</em>, which for a regular covering map $X\to X/W$ and a commutative ring $k$ of coefficients has $$E_2^{p,q}=H^p(W,H^q(X;k))$$ (cohomology of the group $W$ with coefficients in the $kW$-module $H^\ast(X,k)$) and converges to a graded group associated to $H^\ast(X/W)$. A reference is Ken Brown's "Cohomology of groups", section VII.7.</p> <p>In case the group $W$ is finite, if $|W|$ is invertible in $k$ then $H^p(W;H^q(X;k))=0$ for all $q$ and all $p>0$ (see Brown, Corollary III.10.2). In particular this is true if $k=\mathbb{Q}$. Thus the spectral sequence is concentrated in the $0$ column and therefore collapses, giving $H^\ast(X/W)\cong H^0(W;H^\ast(X;k))$. Since $H^0(W;M)=M^W$ for any group $W$ and any $W$-module $M$, this gives the stated results. So you were exactly right in your first paragraph!</p> <p>More generally, the same collapse happens for the Serre spectral sequence of the fibration $X\to X_W\to BW$, which has $$E_2^{p,q}=H^p(BW;H^q(X))\cong H^p(W;H^q(X)),$$ giving the isomorphism $H^\ast_W(X)\cong H^\ast(X)^W$ you mentioned.</p>