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Are there any symplectic but not complex Calabi- Yau manifolds in real dimensions 4 and 6?

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Since you are looking for non complex manifolds, saying "in complex dimension $2$ and $3$" makes no sense...So I guess you meant "in real dimension $4$ and $6$". –  Francesco Polizzi Jul 30 '11 at 16:44
Yes, that is what I meant –  Thom Jul 30 '11 at 16:48
Thomas, I wonder what makes you ask this question? –  Dmitri Jul 30 '11 at 17:40
No good reasons. Just got curious while reading some recent papers on arxiv on symplectic Calabi-Yau manifolds. –  Thom Jul 30 '11 at 18:04
I see, there will be more papers soon, I believe :) –  Dmitri Jul 30 '11 at 18:09

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up vote 9 down vote accepted

First of all, the notion Symplectic Calabi-Yau is quite new. A few persons who use it (including myself) usually mean by this symplectic manifolds with $c_1=0$, (this is just to make sure that we speak about the same thing)

In real dimension $4$ we know for the moment only two types of symplectic Calabi-Yau manifolds - $K3$ surfaces and $T^2$ bundles over $T^2$. These manifolds have as well the structure of a complex manifold with a non-vanishing holomorphic volume form. It is conjectured that there are no other symplectic Calabi Yau manifolds in dimension $4$.

In real dimension six there are quite a lot of symplectic CY manifolds coming from the twistor construction (you can check here: http://arxiv.org/abs/0802.3648), and some of them do have a complex structure, but this is not known for all of them.

At the same time, probably you know that in dimension $2n\ge 6$ the following question is open:

Question. Is it true that every manifold $M^{2n}$ that has an almost complex structure $J$ has as well a holomorphic structure homothopic to J?

This is an old question and apparently no one has an idea of how to answer it. Now, the answer to your question in dimension $6$ depends on what you mean by a complex Calabi-Yau. This notation is not used in math literature. If by such a manifold you mean a complex manfiold with $c_1=0$, then you would not be able (for the moment) to get any example in dimension $6$ where the answer to your question is no (because the above Question is open). On the other hand, if by complex Calabi-Yau you mean a complex manifold with a non-vanishing holomorphic volume form, then the answer to your question is yes, an example is given in http://arxiv.org/abs/0905.3237. There is a symplectic Calabi Yau 6-manifold in this paper, that has $b_3=0$, hence it can not have a holomorphic volume form of top degree for any complex structure. One can construct further such examples.

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Thanks! Do you know if there is an upper bound on the Betti numbers for symplectic Calabi-Yau? –  Thom Jul 30 '11 at 17:16
No, there is no upper bound on the Betty number of symplectic Calabi-Yaus in dimension higher than $4$. On the other hand there is such a bound in dimension $4$, you can check it here: T. J. Li. Quaternionic vector bundles and Betti numbers of symplectic 4-manifolds with Kodaira dimension zero. Internat. Math. Res. Notices, (2006), 1–28. –  Dmitri Jul 30 '11 at 17:31
Thanks Dmitri for the references! –  Thom Jul 30 '11 at 17:41
I was just checking the following papers on arxiv. They do contain some symplectic, but non Kahler Calabi-Yau 6 manifolds. Here are the links arxiv.org/pdf/1107.2623.pdf arxiv.org/abs/1105.3519 –  Thom Jul 30 '11 at 18:14
Maybe they admit complex structure. not sure about that. Probably this seems very difficult question. –  Thom Jul 30 '11 at 18:18

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