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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?

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  • $\begingroup$ This spectral sequence is Lindon's. Have a look at MacLane's "Homology" book, Chapter XI, section 10. $\endgroup$
    – Fernando Muro
    Nov 15, 2018 at 23:22
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    $\begingroup$ @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. $\endgroup$
    – Laura
    Nov 15, 2018 at 23:30
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    $\begingroup$ 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. $\endgroup$
    – Fernando Muro
    Nov 15, 2018 at 23:55
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    $\begingroup$ 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. $\endgroup$
    – IJL
    Nov 16, 2018 at 11:50
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    $\begingroup$ 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. $\endgroup$
    – tj_
    Dec 6, 2018 at 11:51

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