I am trying to gather a list of all known symplectic manifolds which don't have Kahler structure. If you know any please add to the list and give references for it.
Please avoid giving repetitive examples.
Thanks.
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2$\begingroup$ Though one can write down such examples, I think the idea of such a list rather misses the point. There are qualitative differences that can be hard to apply to specific examples. We know, thanks to Gompf, that arbitrary finitely presented groups appear as $\pi_1$ of symplectic 4-manifolds of symplectic Kodaira dimension 1 and also of symplectic Kodaira dimension 2. $\endgroup$– Tim PerutzCommented Sep 21, 2012 at 19:06
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$\begingroup$ We also know that such statements are wildly false for Kaehler surfaces of Kodaira dimension 1 or 2, since e.g. we have only finitely many deformation classes of general type surfaces of fixed $c_1^2$ and $c_2$. Yet it might be hard to decide whether some particular symplectic manifold has a Kaehler structure. $\endgroup$– Tim PerutzCommented Sep 21, 2012 at 19:06
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$\begingroup$ I see; do we know anything about closed simply connected examples? $\endgroup$– Mohammad Farajzadeh-TehraniCommented Sep 21, 2012 at 19:23
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$\begingroup$ Mohammad: once you restrict to the simply connected case I can no longer make any such sweeping comments (and I won't attempt to answer in a comment box). I wonder what happens, though, if you embed Gompf's manifolds symplectically into some high-dimensional $\mathbb{C}P^n$ using Gromov-Tischler and then blow up this submanifold. $\endgroup$– Tim PerutzCommented Sep 21, 2012 at 22:37
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5 Answers
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Honestly, this is like asking for a list of all animals that are not dolphins.
Instructions for making your own:
Step 1 - use the symplectic mapping torus for a symplectomorphism of your favorite Kaehler manifold (acting suitably on homology) to produce something whose Betti numbers violate being Kaehler (one odd degree Betti number is odd)
Step 2 - take the manifold from Step 1 and embed it symplectically in projective space, then blow up. That gets rid of the fundamental group, but raises the dimension a lot.
Step 3 - take Donaldson hypersurfaces to get the dimension down again.
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$\begingroup$ could you add some details or references - how are you blowing up? why are the Donaldson hypersurfaces not Kahler? $\endgroup$– Ian AgolCommented Sep 22, 2012 at 1:24
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$\begingroup$ Agol: in step 1, the symplectic mapping torus $X$ is $S^1$ times the usual one. A general theorem of Gromov-Tischler embeds the resulting (integral) symplectic $2N$-manifold symplectically into $\mathbb{C}P^{2N+1}$. You can blow up along such a submanifold much as you would in Kaehler geometry (see e.g Voisin's book). If $b_1(X)$ is odd, we get a symplectic $2N$-manifold with odd $b_3$. Donaldson hypersurfaces obey the Lefschetz hyperplane theorem, so you can then cut down to 8 dimensions preserving the odd $b_3$. $\endgroup$ Commented Sep 22, 2012 at 15:45
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$\begingroup$ It seems interesting to ask what happens if you take e.g. some Noether-violating simply connected symplectic 4-manifold and build a symplectic 8-manifold by this procedure; can it ever be Kaehler? $\endgroup$ Commented Sep 22, 2012 at 15:49
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There are lots of simply-connected four-dimensional examples in Gompf's symplectic sum paper paper which are shown to be non-Kahler by virtue of violating the Noether inequality (see Theorem 6.2). There are also higher-dimensional examples in the last section of Gompf's paper, including infinitely many simply-connected non-Kahler ones in any even dimension at least 8. Some of these are obtained directly by symplectic sum, and others by taking four-dimensional examples and embedding them in CP^n and blowing up, as suggested by Tim and eigenbunny--in these cases Gompf uses the Hard Lefschetz theorem to prove that the result isn't Kahler.
The standard symplectic surgery operations in four-dimensions (symplectic sum, Luttinger surgery, rational blowdown...) should generally be expected to produce non-Kahler manifolds more often than not, though it's not always feasible to see that the result isn't Kahler--the standard ways of doing so are by the parity of $b_1$ or by the Noether inequality. In particular this paper of Fintushel-Park-Stern gives another large collection of simply connected examples violating the Noether inequality.
One can also show that a symplectic manifold is not Kahler by showing that it is not formal (in the sense of rational homotopy theory). Often nonformal examples also don't satisfy hard Lefschetz (so could instead just be shown to be non-Kahler by that criterion), but there is a nonformal example of Cavalcanti which does satisfy hard Lefschetz.
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$\begingroup$ Might I ask one related question: In dimension $4$, do we know some compact symplectic manifold with $b_1$ even but violating the Hard Lefschetz condition？ $\endgroup$– DLINCommented Nov 30, 2023 at 2:48
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I would recommend the Tralle-Oprea book, Symplectic manifolds with no Kähler structure.
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$\begingroup$ I did not know there is a book on it, thanks for sharing. $\endgroup$ Commented Sep 23, 2012 at 3:19
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The original example of a torus bundle over a torus which is symplectic but admits no Kahler structure is due to Thurston.
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Edited: Corrected after Eugene Lerman's comment of Sep 22, 2012, which also implies this is not an answer to the question (edits are in italic).
In [Invent. Math. 131 (1998), no. 2, 311–319] Chris Woodward gives examples of multiplicity-free compact Hamiltonian manifolds that are not compatibly Kähler. This proved that Delzant's result that all compact multiplicity-free torus actions are compatibly Kähler [Bul. Soc. math. France 116, 315–339 (1988)] does not extend to the nonabelian case. Woodward uses a result from Susan Tolman's paper "Examples of non Kähler Hamiltonian torus actions" [Invent. Math. 131 (1998), no. 2, 299–310].
So, back-to-back papers with examples of symplectic manifolds with a lot of symmetry that are not compatibly Kähler.
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$\begingroup$ Wait, Tolman only proves that there are no {\em invariant} Kaehler structure on her examples, not that the examples are not Kaehler. I don't remember Woodward's paper well, but I think it is the same kind of result. The example may admit a Kaehler structure, just not invariant. $\endgroup$ Commented Sep 22, 2012 at 12:40
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$\begingroup$ @Eugene Lerman: you are right. Thanks for pointing this out. I have edited my "answer" accordingly. $\endgroup$– user9154Commented Sep 22, 2012 at 18:07