# Kunneth formula of Cartesian product modulo orders of coordinates

Let $X$ be a topological space and $F$ a field. Let the $n$-th permutation group $\Sigma_n$ act on $$\prod_n X$$ by $$\sigma(x_1,\cdots,x_n)=(x_{\sigma(1)},\cdots,x_{\sigma(n)}), \sigma\in \Sigma_n.$$ Then we have a quotient space $$(\prod_n X)/\Sigma_n.$$ By Kunneth formula, $$H^*(\prod_nX; F)=\bigotimes_nH^* ( X; F).$$

Question: does $$H^*((\prod_n X)/\Sigma_n; F)=(\bigotimes_nH^* ( X; F))/\Sigma_n=\text{Symmetric tensor product }_nH^* ( X; F)$$ or not? For example, how to compute $$H^*((\prod_n S^m)/\Sigma_n;\mathbb{Z}_2)$$ explicitly?

• Cohomology is contravariant, so you expect to be taking the invariants under the $\Sigma_n$ action, not the coinvariants. This works for free actions by finite groups and over a field of characteristic zero, but the action of $\Sigma_n$ on $X^n$ is almost never free. – Qiaochu Yuan Sep 15 '15 at 14:20
• Thanks! Then how to compute? – QSR Sep 15 '15 at 14:40
• @QiaochuYuan Actually, there is no need to assume the action is free, as discussed here: mathoverflow.net/questions/18898/… For Z/2 coefficients, the problem is subtler because 2 divides the order of the symmetric groups. – Dan Ramras Sep 15 '15 at 20:08

## 1 Answer

Firstly, your space $(\prod_n X)/\Sigma_n$ is better known as the $n$-fold symmetric product of $X$, and denoted $SP^n(X)$. Milgram (building on work of Steenrod, Cartan, Dold, Nakoaka and others) has shown how to compute $H_\ast(SP^n(X);k)$ for any field $k$ and any connected CW complex $X$ of finite type.

One of the key points is the Dold-Thom Theorem, which states that $SP(M(G,m))\simeq K(G,m)$. (Here $M(G,m)$ denotes a Moore space whose only non-trivial reduced homology is a $G$ in dimension $m$, and $K(G,m)$ the corresponding Eilenberg-Mac Lane space. $SP(X)$ stands for the infinite symmetric product, which is the direct limit of the inclusions $SP^n(X)\hookrightarrow SP^{n+1}(X)$). In particular, since $S^m=M(\mathbb{Z},m)$, the limit of the spaces $SP^n(S^m)$ as $n$ increases is a $K(\mathbb{Z},m)$. This indicates that, even for spheres, the $\mathbb{Z}/2$ (co)homology of these spaces can be rather complicated, although calculable due to classic results of Cartan and Serre.