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I have in mind the following question for some time. Is there an example of a compact symplectic manifold with a Hamiltonian S^1 action with isolated fixed points, that does not admit a compatible S^1 invariant Kahler strucutre? One would say, of course there should be such an example. But I have not seen any...

Added. Apperently this is an open problem.

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  • $\begingroup$ wiki'd, per policy for known open problems. $\endgroup$ – Scott Morrison Nov 16 '09 at 21:56
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Lerman has constructed such an example see http://arxiv.org/abs/dg-ga/9601012v1 , well it has an isolated fixed point but also some fixed connected submanifolds, so maybe not exactly what you want.

The construction has been later revisited by Kim http://www.mathjournals.org/mrl/2002-009-004/2002-009-004-002.pdf

(note to editors: I guess the creation of 'torus-action' tag migh be useful.)

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  • $\begingroup$ Isn't it toric-action? $\endgroup$ – Ilya Nikokoshev Oct 24 '09 at 18:17
  • $\begingroup$ Actually, may be circle-action or localization? $\endgroup$ – Ilya Nikokoshev Oct 24 '09 at 18:18
  • $\begingroup$ It's definitely 'torus-action', see e.g. this famous book by Audin books.google.fr/books?id=uKXO9BDsk_kC That said we might also need 'toric-variety' or 'toric-geometry' at some point, but that's something else. $\endgroup$ – Thomas Sauvaget Oct 24 '09 at 18:22
  • $\begingroup$ Thanks a lot for both refferences! But still I want an action such that all fixed points are isolated. Though indeed it is nice to know that there are non-Kahler examples with at least one isolated point. $\endgroup$ – Dmitri Panov Oct 24 '09 at 18:28
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Sue Tolman has shown that there are non-Kahler 6-dimensional manifolds admitting Hamiltonian 2-torus actions all of whose fixed points are isolated ( http://arxiv.org/abs/dg-ga/9511007 ) . This might be a good first place to look to see if one of the components of the action satisfy the criteria you want.

Interestingly, Karshon has shown ( in dg-ga/9510004 [sorry, I can only post one link as a new user]) that if a 4-manifold admits a Hamiltonian circle action, it must be Kahler.

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  • $\begingroup$ I now this article. But surprisingly, as far as I understand, you can not just take a subgroup S^1 in this example to get something that does not admit a compatible Kahler metric. At least this is what I understood from Dusa McDuff. Apperently you do need a T^2 in this example... $\endgroup$ – Dmitri Panov Oct 24 '09 at 18:49
  • $\begingroup$ Surprising and interesting! Perhaps the next thing to do is do a literature search for papers that cite Tolman's paper. $\endgroup$ – C. Lee Oct 24 '09 at 18:56

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