# de Rham cohomology with twisted coefficients and group cohomology

I'm trying to follow Goldman's paper 'The symplectic nature of fundamental groups of surfaces'. I'm having trouble understanding the following isomorphism: let's take a principal $G$-bundle $P \to M$ over a closed oriented surface, equipped with a flat connection $A \in \Omega^1(P, \mathfrak{g})$, and let's suppose that the holonomy is given by (the pointwise inverse of) $\phi : \pi_1(M) \to G$ (I'll omit choices of basepoint). Goldman claims in section 1.8 that $H^*(\pi_1(M); \mathfrak{g}_{\text{Ad}_{\phi}}) \simeq H^*(M; \text{ad }P)$, where the left hand side is the group cohomology of $\pi_1(M)$ with values in $\mathfrak{g}$ as a $\pi_1(M)$-module with the $\text{Ad}_{\phi}$ representation, and the right hand side is the de Rham cohomology of $M$ with values in the adjoint bundle $\text{ad }P$ (using as differential the one that comes from the flat connection in $\text{ad }P$ associated to $A$).

Since $M$ is a surface, it's a $K(\pi_1(M), 1)$, so the group cohomology gets identified with the singular cohomology of $M$, which in turn gets in some sense identified with its de Rham cohomology. I'm not sure what are the proper statements in this case, though, since we have coefficients. What exactly are the versions of these theorems for coefficients, and what's the relation between $\text{ad }P$ equipped with the flat connection that comes from $A$ and $\mathfrak{g}$ as a $\pi_1(M)$-module?

• Riemann-Hilbert correspondence tells you that bundles with flat connection are the same as local systems. Cohomology in a local system is (quasi by definition) the same as cohomology of the fundamental group with coefficients in the general fiber of the local system. – user1688 Sep 1 '16 at 19:26
• See the discussion in this paper of Kapovich and Millson math.ucdavis.edu/~kapovich/EPR/KM_1996.pdf which essentially addresses your question. – Andy Sanders Sep 2 '16 at 1:02
• @AntonDeitmar Riemann-Hilbert is a bit overkill here :) . This follows by viewing the universal covering space as a $\pi_1 (M)$-bundle and then applying the representation to the cocycles. This actually works more generally and gives $Rep (G) \cong Rep (Gau (P))$ – user40276 Sep 2 '16 at 2:36