Are there any symplectic but not complex Calabi- Yau manifolds in real dimensions 4 and 6?
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1$\begingroup$ 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 $6$". $\endgroup$– Francesco PolizziCommented Jul 30, 2011 at 16:44
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$\begingroup$ Yes, that is what I meant $\endgroup$– ThomCommented Jul 30, 2011 at 16:48
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$\begingroup$ Thomas, I wonder what makes you ask this question? $\endgroup$– Dmitri PanovCommented Jul 30, 2011 at 17:40
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$\begingroup$ No good reasons. Just got curious while reading some recent papers on arxiv on symplectic Calabi-Yau manifolds. $\endgroup$– ThomCommented Jul 30, 2011 at 18:04
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2$\begingroup$ I see, there will be more papers soon, I believe :) $\endgroup$– Dmitri PanovCommented Jul 30, 2011 at 18:09
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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.
answered Jul 30, 2011 at 17:03
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$\begingroup$ Thanks! Do you know if there is an upper bound on the Betti numbers for symplectic Calabi-Yau? $\endgroup$– ThomCommented Jul 30, 2011 at 17:16
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$\begingroup$ 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. $\endgroup$ Commented Jul 30, 2011 at 17:31
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$\begingroup$ 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 $\endgroup$– ThomCommented Jul 30, 2011 at 18:14
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$\begingroup$ Maybe they admit complex structure. not sure about that. Probably this seems very difficult question. $\endgroup$– ThomCommented Jul 30, 2011 at 18:18
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$\begingroup$ Sure, this is a nice article. There is a difference between non-Kahler and non-holomorphic cases (and your question was of course about non-holomorphic). The twistor construction, that I mentioned produces a huge amount on non-Kahler symplectic Calabi Yau six-manifolds -- for example because Kahler manifolds have quite restricted fundamental groups. But we don't know any non-trivial restrictions on complex manifolds... $\endgroup$ Commented Jul 30, 2011 at 18:25