Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:51:31Z http://mathoverflow.net/feeds/question/64843 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64843/cohomology-of-the-infinite-loop-space-of-the-affine-grassmanian-as-in-the-genera Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture) Alan H 2011-05-12T21:40:44Z 2011-05-12T23:14:38Z <p>I've been reading Hatcher's survey "A short exposition of the Madsen-Weiss theorem". In it, he outlines a nice proof of the "generalized Mumford conjecture", which asserts that the stable cohomology of the mapping class group is the same as the cohomology of $\Omega^{\infty} AG_{\infty,2}^{+}$. Here $AG_{n,m}$ is the space of affine $m$-planes in $\mathbb{R}^n$ and the "plus" sign indicates the 1-point compactification.</p> <p>Now, the Mumford conjecture I know and love asserts that the (rational) stable cohomology ring of the mapping class group is $\mathbb{Q}[e_1,e_2,\ldots]$, where the $e_i$ are the MMM classes.</p> <p>Can anyone explain to me why the rational cohomology ring of $\Omega^{\infty} AG_{\infty,2}^{+}$ is a polynomial ring with 1 generator in each even dimension?</p> <p>This appears to be explained in Madsen-Tillmann's paper introducing the generalized Mumford conjecture, but that paper is rather formidable and I have been unable to extract an answer to the above question from it (indeed, it doesn't really appear to be talking about the affine Grassmannian at all!).</p> http://mathoverflow.net/questions/64843/cohomology-of-the-infinite-loop-space-of-the-affine-grassmanian-as-in-the-genera/64851#64851 Answer by Neil Strickland for Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture) Neil Strickland 2011-05-12T23:14:38Z 2011-05-12T23:14:38Z <p>Most of this is not special to the case of $AG_{\infty,2}^+$. For any spectrum $X$, we have a Hurewicz map $h:\pi_{\ast}(X)\to H_{\ast}(X)$, which induces a map $h':\mathbb{Q}\otimes\pi_{\ast}(X)\to\mathbb{Q}\otimes H_{\ast}(X)$. Standard calculations show that $h'$ this is an isomorphism when $X=S^n$ for some $n$, and it follows by induction up the skeleta that $h'$ is an isomorphism for all $X$. Next, the homotopy groups $\pi_{\ast}(X)$ are (essentially by definition) the same as the homotopy groups of the space $\Omega^\infty(X)$, so we have an unstable Hurewicz map $h'':\pi_{\ast}(X)=\pi_{\ast}(\Omega^\infty(X))\to H_{\ast}(\Omega^\infty(X))$. By combinining these we get a map $\mathbb{Q}\otimes H_\ast(X)\to \mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$. Next, every infinite loop space is a homotopy-commutative H-space, and this makes $H_\ast(\Omega^\infty(X))$ into a graded-commutative (and graded-cocommutative) Hopf algebra. Now let $A_*$ be the free graded-commutative ring generated by $\mathbb{Q}\otimes H_\ast(X)$, with the Hopf algebra structure for which the generators are primitive. More explicitly, $A_\ast$ is a tensor product of polynomial algebras (one for each even-dimensional generator in $\mathbb{Q}\otimes H_\ast(X)$) and exterior algebras (one for each odd-dimensional generator). It is now quite formal to construct a canonical map $A_\ast\to\mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$ of bicommutative Hopf algebras. One can then show that this map is always an isomorphism. Indeed, it is possible to reduce to the case $X=S^n$ again, and the groups $\mathbb{Q}\otimes H_\ast(\Omega^k S^{n+k})$ can be calculated by repeated use of Serre spectral sequences, and one can then let $k$ tend to infinity. As $\mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$ is free on primitive generators, it is a standard fact that the dual Hopf algebra $\mathbb{Q}\otimes H^\ast(\Omega^\infty(X))$ is also free on primitive generators in the same degrees.</p> <p>For the Madsen-Weiss case, you now just need to know the groups $\mathbb{Q}\otimes H_\ast(AG_{\infty,2}^+)$. This is not hard, because $AG^+_{\infty,2}$ is the Thom space of a virtual vector bundle over $BSO(2)=\mathbb{C}P^\infty$, so we can use the Thom isomorphism theorem.</p>