Are Calabi-Yau manifolds in dimension >= 3 algebraic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:36:22Z http://mathoverflow.net/feeds/question/30629 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30629/are-calabi-yau-manifolds-in-dimension-3-algebraic Are Calabi-Yau manifolds in dimension >= 3 algebraic? Thomas Koeppe 2010-07-05T15:20:46Z 2010-07-05T15:59:38Z <p>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?</p> http://mathoverflow.net/questions/30629/are-calabi-yau-manifolds-in-dimension-3-algebraic/30634#30634 Answer by Aaron Bergman for Are Calabi-Yau manifolds in dimension >= 3 algebraic? Aaron Bergman 2010-07-05T15:59:38Z 2010-07-05T15:59:38Z <p>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 <a href="http://arxiv.org/abs/math/0108088" rel="nofollow">http://arxiv.org/abs/math/0108088</a>). 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.</p>