I think I could write down a projective resolution, tensor with the twisted coefficients and find the first cohomology of the standard torus.


BUT, I was wondering if there is an easier way to understand the first (co)homology groups. (Just checking, by Poincaré duality they should be isomorphic, right?)


Is there something like a Künneth formula for twisted coefficients?



Twisted cohomology of torus
Let me offer you a projective resolution of $\mathbb{Z}$ as a module over $\pi\_1(T)=\mathbb{Z}^2$, obtained from the cellular chain complex of $\mathbb{R}^2$, the universal cover of $T$. The cell structure on $\mathbb{R}^2$ is induced by the usual cell structure on $T$ with one vertex, two edges and one $2$-cell.


The group ring is the ring of Laurent polynomials in two variables $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The resolution is


$$0\rightarrow R\stackrel{({y-1},{1-x})^t}\longrightarrow R^2\stackrel{(x-1,y-1)}\longrightarrow R\rightarrow \mathbb{Z}\rightarrow 0$$


