# Are Calabi-Yau manifolds in dimension >= 3 algebraic?

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?

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## 1 Answer

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 http://arxiv.org/abs/math/0108088). 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.

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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