Decomposing the homology of a finite-index subgroup into isotypic components

Question

$\newcommand\C{\mathbb{C}}$Let $\Gamma$ be a discrete group and let $M$ be a $\C[\Gamma]$-module. Let $G \lhd \Gamma$ be a finite-index normal subgroup with quotient $Q = \Gamma/G$. The conjugation action of $\Gamma$ on $G$ descends to an action of $Q$ on $H_r(G;M)$. In other words, $H_r(G;M)$ is a representation of the finite group $Q$ (possibly infinite-dimensional, but that shouldn't matter for my question and I'm happy to assume that it is finite-dimensional if that makes things easier). Let $V_1,\dotsc,V_k$ be the irreducible representations of $Q$, and for $1 \leq i \leq k$ let $H_r(G;M)_{V_i}$ be the $V_i$-isotypic component of $H_r(G;M)$.

Question: Is there some way of interpreting $H_r(G;M)_{V_i}$ in terms of the homology of $\Gamma$ with respect to some other coefficient system?

Here are a few comments about why I am asking this:

1. If $V_1 \cong \C$ is the trivial representation, then it is well-known that $H_r(G;M)_{V_1} \cong H_r(\Gamma;M)$. For instance, you can prove this using the Hochschild–Serre spectral sequence.

2. In a paper I'm reading, the authors assert casually and without proof (as if it should be obvious) that if $Q$ is the cyclic group of order $2$ and $V_2 \cong \C_{-1}$ is the nontrivial representation where the generator of $Q$ acts by $-1$, then $H_r(G;M)_{V_2} \cong H_r(\Gamma;V_2 \otimes M)$.

Answer 1

I claim $H_*(G,M)_L \cong H_*(\Gamma, M \otimes L^*) \otimes L$ as $Q$-representations.

For a simple $Q$-module $L$, let $(-)_{L}$ be the functor assigning a representation of $Q$ to the isotypic component for $L$. The functor $(-)_L$ is exact, so $H_*(G,M)_L$ is the left derived functor of $H_0(G,-)_L$.

If $V$ is a $Q$-module, then the isotypic component $V_L$ can be realized as the $Q$-representation $H_0(Q,(V \otimes L^*)) \otimes L$. Thus $$H_0(G,M)_L = H_0(Q, H_0(G,M)\otimes L^*))\otimes L = H_0(\Gamma, M \otimes L^*) \otimes L $$ as a representation of $Q$. Now derive both sides in $M$ to obtain the result.