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 $A$ be an abelian variety over a field $k$ and let $K_A$ be its canonical divisor. Then I'm almost certain that $K_A$ is trivial, but I can't seem to prove it, nor find a counter example, nor find any reference on abelian varieties that even mentions canonical divisors.

Does anyone know a reference saying (or an idea for a short proof) that the canonical divisor on an abelian variety is trivial?

I'd prefer not to assume anything about $k$, but a result with $\operatorname{char}(k) = 0$ would be better than nothing.

share|cite|improve this question
up vote 9 down vote accepted

The tangent bundle of any group variety is trivial: take a basis of the tangent space at any one point and translate it around using the group law.

Therefore the cotangent bundle of an abelian variety, being the dual of a trivial vector bundle, is trivial. This implies -- but is in general much stronger than! -- that the top exterior power of the cotangent bundle is trivial.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.