# Kunneth formula for de Rham cohomology twisted by flat vector bundle

It is well known that for two manifolds $M$ and $N$ (let's say they are compact), the Kunneth formula says that $H(M\times N)=H(M)\otimes H(N)$, where $H$ denotes de Rham cohomology with complex coefficients. We also know that one can twist de Rham cohomology by a complex flat vector bundle $E\to M$ with a flat connection $\nabla^E$. Then the differential of the chain complex $\Omega(M; E)$ is coupled with $\nabla^E$. Denote by $H(M, E)$ the corresponding de Rham cohomology.

Let $E\to M$ and $F\to N$ be complex flat vector bundles with flat connections $\nabla^E$ and $\nabla^F$. Write $p_M:M\times N\to M$ be the standard projection map and similarly for $p_N$. Then $p_M^*E\otimes p_N^*F\to M\times N$ is a complex flat vector bundle with flat connection $p_M^*\nabla^E\otimes p_N^*\nabla^F$. I wonder there should be a Kunneth formula for de Rham cohomology twisted by complex flat vector bundles, that is,

$$H(M\times N, p_M^*E\otimes p_N^*F)\cong H(M, E)\otimes H(N, F).$$

Is this assertion obviously true (or obviously wrong)? If the former case is true, is there a reference for a proof of this?

• @abx: I am intimidated by the algebra in the paper. Denoting by $\mathcal F_{\rho}$ the rank 1 local system with monodromy $\rho: \pi_1(X) \to \mathbb C^*$, does theorem 1.6 (or 1.7) imply that $H^k(X\times Y, \mathcal F_{(\rho_1, \rho_2)}) = \sum_{p+q=k} H^p(X, \mathcal F_{\rho_1}) \otimes H^q(X, \mathcal F_{\rho_2})$ if the spaces $X,Y$ are sufficiently well behaved? – klirk Aug 27 '19 at 20:33