Finite generation of equivariant cohomology rings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T10:44:53Z http://mathoverflow.net/feeds/question/79086 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79086/finite-generation-of-equivariant-cohomology-rings Finite generation of equivariant cohomology rings Mark Grant 2011-10-25T14:41:21Z 2012-02-28T11:55:55Z <p>Let $G$ be a finite group acting on a topological space $X$. (In my applications, $X$ is the classifying space of a compact Lie group.) Let $H^\ast(-)=H^\ast(-;\mathbb{Z}/2\mathbb{Z})$ denote mod 2 singular cohomology.</p> <p>By the work of Quillen ("The Spectrum of an Equivariant Cohomology Ring: I"), if $H^\ast(X)$ is a finitely generated $\mathbb{Z}/2\mathbb{Z}$-module, then the equivariant cohomology $$H^\ast_G(X) = H^\ast(EG\times_G X)$$ is a finitely generated as a $\mathbb{Z}/2\mathbb{Z}$-algebra.</p> <p>My question is, if we only assume that $H^\ast(X)$ is finitely generated <em>as an algebra</em>, does the conclusion still hold, ie is the equivariant cohomology $H^\ast_G(X)$ necessarily finitely generated as an algebra?</p> <p><strong>More background</strong> </p> <p>The standard tool for computing equivariant cohomology is the Leray-Serre spectral sequence of the fibration $X\to EG\times_G X \to BG$, which has $$H^\ast(BG,H^\ast(X)) \Longrightarrow H^\ast(EG\times_G X)$$ <em>as algebras</em>. </p> <p>Now if $G$ acts trivially on $H^\ast(X)$ then the $E_2$-page can be identified with $$H^\ast(BG)\otimes H^\ast(X)$$ <em>as algebras</em>. By a classical result of Evens, the mod 2 cohomology algebra of a finite group is finitely generated. Since the tensor product of finitely generated algebras is again finitely generated, so is this $E_2$-page, <strike>and hence so is the equivariant cohomology algebra.</strike> <strong>(see the comments made by Algori and Ralph below.)</strong></p> <p>However, in general the coefficients in the $E_2$-page are twisted by the action of $G$ on $H^\ast(X)$ (which in my case happens to be non-trivial), and so I can't see how the argument would go. </p> <p>At the other extreme, if $G$ acts freely on $X$ then we are basically asking if there is a finite covering space $X\to Y$ such that $H^\ast(X)$ is finitely generated as an algebra but $H^\ast(Y)$ is not. Such a beast would give a counter-example.</p> <p>(I'm also rather sure that if the answer to my question was yes, Quillen would have told us so!) </p>