Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.