# Twisted cohomology of torus

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?

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It depends on what you consider easy. From my point of view, Poincaré duality with local coefficients and the Kunneth spectral sequence are much more complicated than a direct computation in the case of the torus. – Fernando Muro Apr 15 '12 at 13:48
In most cases I used twisted cohomology was when it was with coefficients in the sheaf of sections ${\mathbb V}$ of a flat vector bundle (associated with a linear representation $\rho$ of $\pi_1$ of the base $B$). Again, in most cases I had, $\rho$ would preserve a nondegenerate bilinear form. Then, provided the base is a closed oriented genus $g$ surface, the computation is: $H^1(B, {\mathbb V})=(2g-2) dim(V) + 2 dim H^0(B, {\mathbb V})$ (Euler characteristic + Poincare duality). – Misha Apr 15 '12 at 16:36
@Fernando Muro: So does this mean there is a Kunneth spectral sequence for twisted coefficients? Wikipedia, the source of all knowledge, only refers to such a spectral sequence with coefficients in a commutative ring... – Earthliŋ Apr 16 '12 at 6:05
@Misha: Are you missing a dim there on the left-hand side? – Earthliŋ Apr 16 '12 at 6:07
Sorry, I should have said universal coefficient spectral sequence, instead of Künneth spectral sequence, which is the homological counterpart. Anyway, for this topic there are better references than Wikipedia, e.g. McCleary's book on spectral sequences. – Fernando Muro Apr 16 '12 at 8:39

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

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

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Right, and in general you may use the Koszul resolution of the augmentation ideal. – Donu Arapura Apr 15 '12 at 15:14
The conditions for the existence of a Koszul resolution are too restrictive to work for any group ring. – Fernando Muro Apr 15 '12 at 15:32
Yes, I should be clearer. I was only referring to the Laurent polynomial ring which was the case of interest to the OP, and to whom the comment was primarily directed. – Donu Arapura Apr 15 '12 at 15:43

Yes, under suitable circumstances there is a Künneth formula for local coefficients. This can be found in Theorems 1.6, 1.7 of the following paper:

R. Greenblatt: Homology with local coefficients and characteristic classes. Homology, Homotopy and Applications, 8(2), 2006, 91-103.

But I agree with Fernando that in case of a torus it is easier to construct a projective resolution. In particular, this way you can handle any twist, while the theorems cited above only adress "product actions".

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