Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture)

2011-05-12T21:40:44Z

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.

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.

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?

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!).

Answer by Neil Strickland for Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture)

2011-05-12T23:14:38Z

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.

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.

Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture) - MathOverflow

Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture)

Answer by Neil Strickland for Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture)