# cohomology version of Cartan-Leray spectral sequence that deduces cup product

On page 338, A User's Guide to Spectral Sequences. 2nd Edition, by John McCleary, Theorem 8.9, there is a Cartan-Leray spectral sequence for homology:

If $X$ is a connected pace on which the group $\pi$ acts freely and properly, then there is a spectral sequence, homological type, with $$E^2_{p,q}=H_p(\pi,H_q(X)),$$ converging strongly to $H_*(X/\pi)$.

Is there any cohomology version of Cartan-Leray spectral sequence that can get cup product structure of $H^*(X/\pi)$?

Let $k$ be a field. Suppose $H^*(X;k)$ and $H^*(\pi;k)$ are known. How to get the cup product structure of $H^*(X/\pi;k)$?

• Doesn't it suffice to simply dualize? For the cup product structure, you need more than $H^*(X,k)$ and $H^*(\pi , k)$. You will need $H^*(\pi , H^*(X;k))$ (note that $\pi$ usually acts non-trivially on $H^*(X;k)$). Jan 14, 2015 at 9:20
• There is such a version, and it is compatible with the products (in the usual sense: differentials obey the Leibnitz rule and all isomorphisms involved are multiplicative). Of course, you cannot get the product structure of the limit term, just the corresponding graded ring. Jan 14, 2015 at 9:53
• @AlexDegtyarev: Would you happen to have a reference handy?
– jdc
Aug 31, 2016 at 22:29

The Leray-Serre spectral sequence in cohomology is multiplicative, meaning that there is a multiplication $E_{r}^{pq} \otimes E_r^{p'q'} \to E_r^{p+p',q+q'}$ for each $r$, and the multiplication on the $E_{r+1}$ page is induced by that on the $E_r$ page. Note however that this does not actually give you the cup product on $H^\ast(X/\pi)$ (in your case), only on the bigraded algebra $\bigoplus_{p,q} E_\infty^{p,q} = \bigoplus_{p,q} \mathrm{Gr}_L^p H^{p+q}(X/\pi)$. So you're only getting partial information about the cup product on $H^\ast(X/\pi)$; to get the full information you will need some luck (maybe $E_\infty$ is concentrated along a single row, or something) or trickiness.
In general, knowledge of $H^\ast(X)$ and $H^\ast(\pi)$ does not even give you knowledge of the $E_2$ page $E_2^{pq} = H^p(\pi,H^q(X))$, as remarked in the comments. This is only true if $\pi$ acts trivially on the cohomology of the fibers.