Consider a short (not necessarily split) exact sequence of groups

$1 \rightarrow N \rightarrow G \rightarrow Q \rightarrow 1$

and suppose we wish to find the cohomology of $G$ with coefficients in a ring $R$. Then, it is known that there is a first quadrant cohomological spectral sequence of algebras converging as an algebra:

$E_2^{p,q} = H^p(Q;H^q(N;R)) \implies H^{p+q}(G;R)$.

Suppose $R$ is replaced by a $Q$-module $M$. (A specific example of interest to me is when $Q = \mathbb Z_2$ and $M = \mathbb Z$ with the inversion action of $Q$.) Then, can we still define a spectral sequence of algebras with an $E_2$-page as given above, so that it converges to $H^{p+q}(G,M)$ as an algebra?

(Knowing the answer for the specific example I mentioned is sufficient.)

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    $\begingroup$ If M isn't a ring, what is the algebra structure on $H^*(G, M)$? I believe it is just a module over the algebra $H^*(G, R)$ where $M$ is an $R$-module (or even an $RG$-module, appropriately defined). $\endgroup$ Nov 26, 2018 at 23:37

1 Answer 1


As Joshua Grochow mentioned in a comment, there is not necessarily an algebra structure on this spectral sequence. (In particular, $H^0(G;M) = M^G$ does not necessarily have a ring structure.) Generally, an equivariant pairing $M \otimes N \to P$ gives rise to a multiplication map on spectral sequences.

In your case of $\Bbb Z$ with the sign action, here is a handy trick that does elaborate on the structure. Let $R$ be the ring $\Bbb Z[x]/(x^2 -1)$, where $G$ acts on $x$ through the quotient $Q$ by sending it to $-x$. Then $R$ is a $G$-equivariant ring, but as a module it decomposes as $\Bbb Z \cdot 1 \oplus \Bbb Z^{sgn} \cdot x$. As a result, the spectral sequence naturally additively decomposes as a direct sum: $$ H^p(Q; H^q(N;R)) \cong H^p(Q;H^q(N;\Bbb Z)) \oplus H^p(Q;H^q(N;\Bbb Z^{sgn})) $$ The multiplication on $R$, however, gives this spectral sequence a multiplication. This both makes the sign-spectral sequence into a module over the trivial-action spectral sequence and gives you a bilinear pairing $$ H^p(Q;H^q(N;\Bbb Z^{sgn})) \otimes H^{p'}(Q;H^{q'}(N;\Bbb Z^{sgn})) \to H^{p+p'}(Q;H^{q+q'}(N;\Bbb Z)) $$ that is compatible with the differentials in the spectral sequence. These converge to a natural module structure $$ H^n(G;\Bbb Z) \otimes H^{n'}(G;\Bbb Z^{sgn}) \to H^{n+n'}(G;\Bbb Z^{sgn}) $$ and a pairing $$ H^n(G;\Bbb Z^{sgn}) \otimes H^{n'}(G;\Bbb Z^{sgn}) \to H^{n+n'}(G;\Bbb Z). $$ (This trick of making a ring out of modules can be used to determine extra structure in a number of other cases.)

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    $\begingroup$ This non-trivial pairing $H^1(C_2; \Bbb Z^{sgn}) \otimes H^1(C_2; \Bbb Z^{sgn}) \to H^2(C_2; \Bbb Z)$ shows up nicely in the homotopy fixed point spectral sequence from $H^s(C_2; KU_t)$ to $KO_{t-s}$ to show that the class in $H^1(C_2; KU_2)$ which detects $\eta$ squares to the class in $H^2(C_2; KU_4)$ which detects $\eta^2$. $\endgroup$ Nov 27, 2018 at 11:43
  • $\begingroup$ A related question: can the above structure be used to establish a "Coefficient Theorem" for $H^n(G)$ which relates cohomology with coefficients in $\mathbb Z$ to coefficients in $\mathbb Z^{sgn}$? $\endgroup$ Nov 27, 2018 at 15:52
  • $\begingroup$ @NarenManjunath I don't think that it's possible, in general, to derive the cohomology with sign coefficients from the cohomology with integer coefficients. Sorry. $\endgroup$ Nov 28, 2018 at 15:39

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