Assume one has a finitedimensional vector space $V$ over the complex numbers and a discrete subgroup $G$ of translations on $V$, so that the quotient is a complex manifold $X=V/G$.


My question: can one express the cohomology group $H^1(X,\mathcal{O\_X})$ in terms of the data $V$ and $G$?



Cohomology of a quotient manifold
This answer has been corrected per Torsten's remark.


If $V/G = Y$ is a complex torus, i.e. $G$ has maximal rank, then Hodge symmetry says $H^1(Y,\mathcal O) = \overline{H^0(Y,\Omega)}$. But the cotangent bundle of $Y$ is trivial by translation on the group, so $H^0(Y,\Omega) = H^0(Y,\mathcal O) \otimes (T\_0 Y)^\ast$ (the cotangent space at the identity). Now $T\_0Y \cong V$, so $H^1(Y,\mathcal O) \cong \overline V^\ast$. If $Y=V/G$ is an algebraic torus, i.e. a product $(\mathbf C^\ast)^p$, then $H^1(Y,\mathcal O)$ is trivial. One way to see this is through the exponential sequence and the triviality of the Picard group of $Y$. 


This also takes care of any product of complex tori, algebraic tori and affine spaces. But in general I see no reason that a quotient $V/G$ couldn't be a nontrivial extension of a complex torus by an algebraic torus. (In the algebraic setting this would be a semiabelian variety.) I don't know what happens to the cohomology then.


