Let $e_1,...,e_4\in\mathbb{C}^2$ whose coordinates are all algebraically independent. Let $\Lambda$ be the lattice spanned by these vectors. Why is $\mathbb{C}^2/\Lambda$ not an abelian variety?
Asked
Active
Viewed 533 times
5
-
If $\Pi$ is your period matrix with respect to, say, the canonical basis, then if $\mathbb{C}^2/\Lambda$ were an abelian variety with polarization $H$ who's real part has matrix $J$, then you would have that $\Pi J\Pi^t=0$ (see Birkenhake-Lange). This would produce some algebraic relations. – rfauffar Mar 30 '12 at 13:24