Let $G$ be a finite group and let us consider a chain complex $C^{\bullet}$ of $\mathbb C[G]$-modules. In other words, we have a chain complex
$$\ldots \to C^{i-1} \xrightarrow{d_{i-1}} C^i \xrightarrow{d_i} C^{i+1} \to \ldots$$
where each $C^i$ is a complex representation of $G$ and the differentials $d_i$ are $G$-equivariant linear morphisms.
I'd like to find some representation theoretic condition, if any such exists, that would be equivalent to the chain complex $C^{\bullet}$ being quasi-isomorphic to $H^{\bullet}(C^{\bullet})$, which is the chain complex
$$\ldots \to H^{i-1}(C^{\bullet}) \xrightarrow{0} H^{i}(C^{\bullet}) \xrightarrow{0} H^{i+1}(C^{\bullet}) \to \ldots$$
where all differentials are zero, and $H^i(C^{\bullet}) := \mathrm{Ker}(d_i)/\mathrm{Im}(d_{i-1})$. Such a quasi-isomorphism is the datum of a family of $G$-equivariant morphisms $h_i:C^i \to H^i(C^{\bullet})$ such that $h_{i+1}\circ d_i = 0$ for all $i\in \mathbb Z$, and such that the induced maps $H^i(h_i):H^i(C^{\bullet}) \to H^i(C^{\bullet})$ are isomorphisms.
For instance, if all the $C^i$'s are multiplicity-free representations of $G$ (in particular finite dimensional), then they decompose canonically into a sum of irreducible representations. The maps $h_i$ can be defined as the projections onto the irreducible components which occur in the subquotient $H^i(C^{\bullet})$.
A little bit more generally, the same argument seems to apply if all irreducible representations arising in $H^i(C^{\bullet})$ occur with multiplicity one in $C^i$.
However, these are of course far from being necessary conditions. For instance if all the differentials $d_i$'s were $0$ from the beginning, then the conclusion is trivial no matter what the $C^i$'s are.
In fact, I also struggle to find out a counter example where the conclusion would fails, ie. for a certain group $G$, write down a chain complex $C^{\bullet}$ which is not quasi-isomorphic to $H^{\bullet}(C^{\bullet})$. For this reason, I fail to grasp what is the obstruction to having such a quasi-isomorphism. Any help or comment would be appreciated!
Best Answer
It may help to think about this problem in a more general setting, as this is really a special case of a basic technique in homological algebra.
Suppose that $R$ is a ring and that $C$ is a complex of $R$-modules. (Later, we will take $R = \mathbb{C}[G]$.) Associated to $C$ are the short exact sequences $$ 0 \to Z^n \to C^n \to B^{n+1} \to 0 $$ $$ 0 \to B^{n} \to Z^{n} \to H^{n} \to 0 $$ where $Z^n$ are the cycles, $B^n$ are the boundaries and $H^n$ is the homology. Suppose that these sequences are split for every $n$. Then, after choosing splittings we have decompositions $$ C^n = \widetilde{B}^n \oplus B^n \oplus \widetilde{Z}^n, $$ where $\widetilde{B}^n \cong H^n$ is a complement of $B^n$, and $\widetilde{Z}^n \cong B^{n+1}$ is a complement of $Z^n$.
The subcomplex $A$ defined by $A^n = B^n\oplus \widetilde{Z}^n$ is contractible since, by definition, the differential restricts to an isomorphism $\widetilde{Z}^n \cong B^{n+1}$. Therefore, we get a decomposition of chain complexes $C = \widetilde{B}\oplus A$, where $\widetilde{B} \cong H(C)$ and $A$ is contractible. In particular, $C$ is homotopy equivalent to its homology (seen as a complex with zero differentials).
If $R$ is a semisimple ring, then every short exact sequence of $R$-modules is split. By the above, every complex of $R$-modules is homotopy equivalent to its homology. For the problem at hand, we can apply this to the ring $R= \mathbb{C}[G]$, which is semisimple by Maschke's Theorem.