MathOverflow is a question and answer site for professional mathematicians. 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

If $X$ is a complex Abelian variety of dimension $g$, then

  • The canonical sheaf is trivial
  • $\dim {\rm H}^i(X; \mathcal{O}_X) = \binom{g}{i}$.

When $g =1,2$, then any connected, projective nonsingular $X$ satisfying the above two must be an Abelian variety. Is this true for higher $g$? If not, what other conditions can I add? Or is such a request unreasonable?

Disclaimer: I don't know very much about Abelian varieties, so apologies if this material is standard. A search in the literature turned up some papers about characterizations of Abelian varieties up to birational equivalence, but under weaker assumptions. I really want to know if I'm given a variety $X$, how to tell if it is Abelian or not via some sort of reasonably accessible sheaf-related conditions. I'm most interested in the case $g=3$, but results for other $g$ are welcome also.

share|cite|improve this question
up vote 18 down vote accepted

A result of Kawamata (Kawamata, Yujiro, Characterization of abelian varieties. Compositio Mathematica, 43 no. 2 (1981), p. 253-276) implies that, under your assumptions, $X$ is birational to an abelian variety (in fact you just need the Kodaira dimension of $X$ to be zero and the irregularity to be equal to the dimension of $X$).

Once you know that $X$ is birational to an abelian variety $A$, a Lemma of Deligne implies that if the canonical divisor on $X$ is trivial, then $X$ is in fact an abelian variety. This is not a particularly deep result. First, the rational map $f \colon X \to A$ is actually a morphism (essentially because $A$ cannot contain rational curves). Second, the morphism $f$ induces a morphism $df$ between the cotangent bundles. The determinant of $df$ is a morphism between the canonical bundles of $A$ and $X$, that are both trivial by assumption. Thus the determinant is either identically zero, or it is an isomorphism. Since the morphism $f$ is generically etale, the determinant is not identically zero. But then it is an isomorphism, so that the morphism $f$ is always etale, and we conclude that $X$ is an abelian variety.

EDIT: Ah, as Pete remarked below, I did not answer the question! The answer is "Yes"! Even under weaker assumption: namely it suffices to know that the canonical bundle is trivial and that $\dim (X) = h^1(X,\mathcal{O}_X)$.

share|cite|improve this answer
So, just to be sure...**yes**, right? – Pete L. Clark Aug 8 '10 at 9:56

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.