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Suppose we have a spectral sequence of algebras and know that it degenerates at some $E_r$, take for example the cohomology Leray Serre spectral sequence associated to some fibration $F\hookrightarrow E\rightarrow B$. Suppose we are working over a field so there is no extension problem and so $H^n(E)\cong\oplus_{r+s=n} E_{\infty}^{r,s}$. Under what conditions can we read off algebra structure of $H^*(E)$ from the $E_{\infty}$ page? (e.g. one very special case is when the $E_{\infty}$ page is free as an algebra.)

I think it is not possible in general, since we can have two different algebras with some filtrations such that the associated graded objects are isomorphic, but I would like to know if there are certain conditions under which one can compute the algebra structure.

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I would also like to know an answer to this question. I guess you are already familiar with McCleary's "A User's Guide to Spectral Sequences", section 1.5? –  Mark Grant Jul 6 '12 at 9:55
@Grant Yes of course –  George Jul 6 '12 at 10:21
I guess you want something that goes beyond stuff about simple systems of generators, e.g. 5.13 in McCleary's book? –  Dan Ramras Jul 6 '12 at 22:40
@ Ramras I want to have something fairly general than that for simple system of generators. –  George Jul 7 '12 at 5:56

1 Answer 1

I want to highlight a case that is similar to the free case but is used in several cohomology computations in literature: Let $C=(C^i)_{i\ge 0}$ be a cochain complex with filtration $$C^i=F^0C^i \supseteq \cdots \supseteq F^iC^i \supseteq F^{i+1}C^i=0.$$

Assume that $E_\infty^{\ast,\ast}=E_\infty^{\ast,0}\otimes_k E_\infty^{0,\ast}$ and $E_\infty^{0,\ast}=k[x_1,...,x_n]$ is a polynomial algebra with homogeneous $x_i$. Then

$\qquad\qquad H^\ast(C) = E_\infty^{\ast,0}\otimes_k k[X_1,...,X_n] \cong E_\infty$

The proof is straightforward. As an application let $$B\mathbb{Z}/2 \to E \to B$$ be a fibration with path-connected base such that $\Pi_1(B)$ acts trivially on $H^\ast(B\mathbb{Z}/2;\mathbb{F}_2)$. Write ${H^\ast(B\mathbb{Z}/2;\mathbb{F}_2) =\mathbb{F}_2[z]}$ and let $h$ be minimal with

$d_{2^{h}+1}(z^{2^h})=0$. Assume moreover that $y_i := Sq^{2^i}\cdots Sq^1(d_2z)$ $(i=0,...,h-1)$ is a regular sequence in $H^\ast(B;\mathbb{F}_2)$. Then

$\qquad H^\ast(E;\mathbb{F}_2)=\mathbb{F}_2[x] \otimes H^\ast(B;\mathbb{F}_2)/(y_1,...,y_{h-1}),\quad \deg(x)=2^h$.

This was used by Quillen in his computation of the cohomology of the extra-special 2-groups and the Spinor groups.

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I should add that the requirement on the filtration makes $E_r$ a first quadrant spectral sequence with $E_\infty^{\ast,0} \le H^\ast(C)$. –  Ralph Jul 8 '12 at 20:00
Thank you Ralph, Is there something similar in the case when the "fiber" is an affine ring instead of a polynomial algebra? –  George Jul 9 '12 at 7:53
I don't think so. I haven't a spectral sequence example at hand, but as a filtered ring you can take $R=k[x]/(x^4)$ and filter it by the powers of $I=(x^2)$. Then $$gr(R):=\bigoplus_{i\ge 0}I^i/I^{i+1}=k[u]/(u^2) \otimes k[v]/(v^2).$$ Hence one of $u,v$ can't be lifted. –  Ralph Jul 10 '12 at 17:01

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