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 $H^*(X/\pi;k)$?

  • $\begingroup$ 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 $H^*(X;k)$). $\endgroup$
    – user43326
    Jan 14, 2015 at 9:20
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 14, 2015 at 9:53
  • 1
    $\begingroup$ @AlexDegtyarev: Would you happen to have a reference handy? $\endgroup$
    – jdc
    Aug 31, 2016 at 22:29

1 Answer 1


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.


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