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.


1 Answer 1


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 $\dim (X) = h^1(X,\mathcal{O}_X)$.

  • $\begingroup$ So, just to be sure...**yes**, right? $\endgroup$ Aug 8, 2010 at 9:56
  • $\begingroup$ I guess this is a very late comment, but I will try it. In the above, does one actually need the field to be the complex numbers, or could it be any field of characteristic zero? $\endgroup$
    – Stefano
    Aug 15, 2019 at 5:10
  • $\begingroup$ @Stefano Yes, the criterion I asked for is not enough for surfaces in characteristics 2 and 3, but one can use etale cohomology to further separate these exotic examples from abelian surfaces. Since this answer helped me with a paper, I'll post it here: arxiv.org/abs/1203.2575 (the reference for what I said is in Remark 3.2). $\endgroup$
    – Steven Sam
    Aug 16, 2019 at 4:03
  • $\begingroup$ @StevenSam Thank you! It is exactly what I was looking for! $\endgroup$
    – Stefano
    Aug 16, 2019 at 15:42

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