If $G$ is a finite group then there is the so-called Bockstein spectral sequence $$E_2^n = H^n(G,\mathbb{F}_p) \Rightarrow \begin{cases} \mathbb{F}_p & n =0 \newline 0 & n>0\end{cases}$$ that can be used to compute the integral cohomology out of the mod-$p$ cohomology. In more detail, each non-zero element of $d_r(E_r^{n-1})\subseteq E_r^n$ corresponds to a direct $\mathbb{Z}/p^r$-summand of $H^n(G,\mathbb{Z})$ (Corollary 5.9.12 in Weibel's Homological Algebra book).

Now my question is whether it is possible to compute the integral cohomology ring of $G$ if the mod-$p$ cohomology rings for the primes dividing $|G|$ and the differentials in the associated Bockstein spectral sequences are known ?

  • $\begingroup$ Doesn't the Bockstein end up twisting the cup product, in the sense that $\beta(xy)=\beta(x)y\pm x\beta(y)$? Or something like that. I would guess this makes it difficult to get the integral ring structure. $\endgroup$ Sep 5 '12 at 1:10
  • $\begingroup$ Are you asking whether the cohomology ring can be recovered from just the Bockstein data, or whether two finite groups with (multiplicatively) isomorphic Bockstein spectral sequences must have isomorphic integral cohomology rings? $\endgroup$ Sep 5 '12 at 2:12
  • 1
    $\begingroup$ The BSS is a spectral sequence of algebras (since the Bockstein is a derivation, as mentioned by Chris). So you may be able to glean some information about integral cup products. That said, it may be a better tactic to use the ring homomorphism $H^\ast(G;\mathbb{Z})\to H^\ast(G;\mathbb{F}_p)$ given by reducing coefficients mod $p$. The BSS is, after all, just the SS obtained by wrapping up the long exact coefficent sequence into an exact couple. $\endgroup$
    – Mark Grant
    Sep 5 '12 at 6:30
  • $\begingroup$ @Tyler: My question is about the former. But if it is possible to recover the integral cohomology ring uniquely from the Bockstein data, wouldn't this also answer the 2nd question affirmatively ? $\endgroup$
    – tj_
    Sep 5 '12 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.