# Convergence of the Lyndon-Hochschild-Serre spectral sequence as an algebra

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.)

• 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). Nov 26, 2018 at 23:37

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.)
• 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 $\eta^2$. Nov 27, 2018 at 11:43
• 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 $\mathbb Z^{sgn}$? Nov 27, 2018 at 15:52