# Hochschild-Serre spectral sequence via explicit filtration

Let

$$1 \longrightarrow K \longrightarrow G \longrightarrow Q \longrightarrow 1$$

be a short exact sequence of groups and let $$M$$ be a $$\mathbb{Z}[G]$$-module. The Hochschild--Serre spectral sequence in homology is of the form

$$E^2_{pq} = H_p(Q;H_q(K;M)) \Longrightarrow H_{p+q}(G;M).$$

All the sources I've consulted construct this via a Grothendieck spectral sequence. For something I am doing, what would be great is if there is a free resolution $$R_{\bullet} \rightarrow \mathbb{Z}$$ of the trivial $$\mathbb{Z}[G]$$-module $$\mathbb{Z}$$ and a filtration $$\mathcal{F}_{\bullet} R_{\bullet}$$ of $$R_{\bullet}$$ such that for all $$\mathbb{Z}[G]$$-modules $$M$$, the above spectral sequence is the one associated to the filtration

$$\mathcal{G}_k(R_{\bullet} \otimes M) = (\mathcal{F}_k R_{\bullet}) \otimes M$$ of the chain complex $$R_{\bullet} \otimes M$$ computing $$H_p(G;M)$$. It would be even better if this filtration resolution was reasonably explicit, though I could get by with something non-explicit if I had to. Does anyone know where I could find this?

• This spectral sequence is Lindon's. Have a look at MacLane's "Homology" book, Chapter XI, section 10. – Fernando Muro Nov 15 '18 at 23:22
• @FernandoMuro: It's spelled Lyndon. He had the main idea, but in the form I have stated it the correct attribution is Hochschild-Serre. The reference you give contains the usual proof, which is not of the form I asked for. – Laura Nov 15 '18 at 23:30
• Laura, the proof in MacLane's book is based on the fact that the tensor product of bar constructions $B(Q)\otimes B(G)$ is a free resolution of the trivial $G$-module, and the spectral sequence is recovered filtering by the degree of the first tensor factor. – Fernando Muro Nov 15 '18 at 23:55
• I like the account in Len Evens' book `The cohomology of groups'. It sets up the Lyndon-Hochschild-Serre spectral sequence using the tensor product of a free $\mathbb{Z}Q$-resolution for $\mathbb{Z}$ and a free $\mathbb{Z}G$-resolution for $\mathbb{Z}$: similar to the proof that Fernando mentions but using arbitrary resolutions. I find that double complexes are easier to handle than the more general filtered complexes. – IJL Nov 16 '18 at 11:50
• You can take projective resolutions $X \to \mathbb{Z}$ over $G$ and $Y \to \mathbb{Z}$ over $Q$. Then $R := X \otimes_{\mathbb{Z}}Y$ is a projective resolution of $\mathbb{Z}$ over $G$ with diagonal $G$-action. $R$ and $R \otimes_G M$ are filtered in the usual way and satisfy $F_k(R \otimes_G M) =F_kR \otimes_G M$ and $R\otimes_G M$ yields the LHS-SS. – tj_ Dec 6 '18 at 11:51