MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

I believe that I once saw a statement that every compact, smooth Calabi-Yau manifold in dimension at least 3 is algebraic, but I can remember neither the reference nor the proof (which would have been quite short) and I might just be confusing this with something else. Is it true?

share|cite|improve this question
up vote 13 down vote accepted

It depends a little bit on your definition of CY. If you're using a good one, it will imply that the Hodge numbers $h^{0,p} = 0$ for $p \neq 0,d$ (see, for example, Prop. 5.3 of Joyce's This implies that $H^2(X) \cong H^{1,1}(X)$. Since the Kaehler cone is an open set in $H^{1,1}(X)$, it contains an rational class, and we can scale that to be an integral class. So, by Kodaira and Chow, we're done.

share|cite|improve this answer
Yes, that's what I was looking for, thanks! (That may even be the original reference I was thinking of.) – Thomas Koeppe Jul 5 '10 at 16:23
Well, in Joyce's paper he uses the vanishing Ricci curvature definition of Calabi-Yau. This is the same as saying that a power of the canonical bundle is trivial. The more typical (I guess) definition of Calabi-Yau is trivial canonical bundle. So if we use the more typical definition, this is still true. – Kevin H. Lin Oct 12 '10 at 21:29
It's not true in either of those cases. Take the example of T^2 cross a non-algebraic K3 surface. Joyce requires that the holonomy group equals SU(N) and isn't a lower rank subgroup. – Aaron Bergman Oct 13 '10 at 2:07
Ah, ok. Thank you for clarifying. – Kevin H. Lin Oct 19 '10 at 7:34

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.