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Let $G$ be a group. I have two questions about the homology of $G$:

  1. Consider a finite exact sequence $$0 \rightarrow M_1 \rightarrow \cdots \rightarrow M_m \rightarrow 0$$ of $G$-modules. How are the homology groups $H_k(G;M_i)$ related?

  2. Consider a filtration of $G$-modules $$0 = N_0 \subset N_1 \subset \cdots \subset N_n = N.$$ How can I relate $H_k(G;N)$ to the homology groups $H_k(G;N_{i+1}/N_i)$?

These feel related since the simplest cases of both of them reduce to the long exact sequence in group homology associated to a short exact sequence of coefficients. Maybe there is some kind of spectral sequence?

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  • $\begingroup$ In 1. do you want your chain complex to be an exact sequence? $\endgroup$ Commented Aug 11, 2022 at 7:59
  • $\begingroup$ Filtered complexes give rise to spectral sequences. $\endgroup$
    – Z. M
    Commented Aug 11, 2022 at 8:57
  • $\begingroup$ @Z.M: Yes, but it’s not a filtered complex, just a filtration on the coefficient system. $\endgroup$
    – Laura
    Commented Aug 11, 2022 at 13:11
  • $\begingroup$ @JeremyRickard: Yes, thanks for figuring out what I really wanted! I will fix it. $\endgroup$
    – Laura
    Commented Aug 11, 2022 at 13:12
  • $\begingroup$ Taking the derived $-\otimes^L_{\mathbb Z[G]}\mathbb Z\colon D(\mathbb Z[G])\to D(\mathbb Z)$, you get filtered complexes. $\endgroup$
    – Z. M
    Commented Aug 11, 2022 at 15:27

1 Answer 1

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  1. Think of your exact sequence as a resolution of $M_m$. It's not necessarily a resolution by free $G$-modules, or by projective $G$-modules; it's just a resolution by $G$-modules. You get a spectral sequence whose $E_1$-term is the direct sum of the $H_*(G; M_i)$ for all $i<m$, and which converges to $H_*(G; M_m)$.

    When the $G$-modules $M_i$ are projective for all $i<m$, then the $E_1$-page degenerates on to a single line, and then running the spectral sequence reduces to the usual process for calculating $Tor$: take a projective resolution, apply the tensor product (at this point you have recovered the one nonzero line in the $E_1$-page), then take homology (at this point you have both the $E_2$-page and the final answer).

    This "resolution spectral sequence" is a standard homological tool: see application 5.9.8 in Weibel's homological algebra textbook for its construction.

  2. Having a filtration on the coefficient module $N$ yields a filtration on the bar complex, when you tensor $N$ with the bar resolution of $\mathbb{Z}$ by $G$-modules. The resulting filtered chain complex yields a spectral sequence whose input is homology of $G$ with coefficients in the associated graded $G$-module of your filtration of $N$, and whose output is the homology of $G$ with coefficients in $N$. This was suggested in the comments as well. The resulting spectral sequence is also a standard tool, and although I don't know a textbook that discusses it in exactly this level of generality, any treatment of the spectral sequence of a filtered chain complex will be applicable to this one.

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