Group cohomology versus deRham cohomology with twisted coefficients

Solution 1:

In general, a twisted coefficient system on a manifold $M$ (also called a local system by more algero-geometrically minded people) is given by a representation of $\pi_1(M)$ (the holonomy representation, also called the monodromy representation by more algebrao-geometrically minded people). Conversely, any representation of $\pi_1(M)$ gives a twisted coefficient system on $M$.

(There is no need for $M$ to be a manifold here; any space for which the usual theory of $\pi_1$ and covering spaces goes through would be fine.)

If $V$ is the representation of $\pi_1(M)$, giving rise to the twisted coefficient system $\mathcal L$, then there will be map $H^i(\pi_1(M), V) \to H^i(M,\mathcal L)$. However, these will not be isomorphisms in general unless $M$ is aspherical, i.e. if its universal cover $\tilde{M}$ is contractible, i.e. if $M$ is a $K(\pi,1)$ (for $\pi = \pi_1(M)$). (Here I am recalling the basic topological interpretation of group cohomology, which you can find in many places.)

What happens in general is that there is a spectral sequence (a special case of the Hochschild--Serre spectral sequence) $$H^i(\pi_1(M), H^j(\tilde{M},V) ) \implies H^{i+j}(M,\mathcal L).$$

Note that if $M$ is hyperbolic (or, more generally, negatively curved), then its universal cover is contractible, and so one does get your desired isomorphism.

Added: To get a feeling for what can happen if $\tilde{M}$ is not contractible, you can consider the case when $M = \mathbb RP^2$, so that $\tilde{M}$ is $S^2$ and $\pi_1(M)$ is cyclic of order $2$, acting on $S^2$ by the antipodal map. Take $V$ to be the trivial representation (over $\mathbb Z$, or $\mathbb Z/2\mathbb Z$, say). Then the preceding spectral sequence gives a way to compute the cohomology (with $\mathbb Z$-coefficients, or $\mathbb Z/2\mathbb Z$-coefficients) in terms of the group cohomology of the cyclic group of order two acting on the cohomology of the sphere. (So it acts trivially on $H^0$, and by $-1$ on $H^2$.)

Of course, you could do the analogous computation for $\mathbb R P^3$ as well (which is more directly relevant to your question).