## Question about examples of symplectic non-Kahler classes.

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Let $M$ be an even dimensional smooth manifold.
I want to find an example $M$ satisfying the following conditions,

1. $M$ admits a Kahler structure.
2. $\omega$ is a symplectic form on $M$.
3. There is no Kahler structure $(M,\omega',J)$ such that $[\omega']=[\omega] \in H^2(M;\mathbb{R})$

(I mean, want to find an example $M$ such that "Kahler cone $\neq$ symplectic cone" with non-empty Kahler cone.)

Thank you in advance.

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## 2 Answers

One sort of example arises from the fact that if one starts with a Kahler form $\omega$ (which represents a class of type (1,1) in the Hodge decomposition by definition of a Kahler form), then if $\phi$ is the real part of any closed form of Hodge type (2,0), $\omega+\phi$ will still be a symplectic form (it tames the complex structure $J$), but won't any longer be Kahler, at least if one regards the complex structure as being fixed--in principle there could be another complex structure with respect to which the form is Kahler. Thus you get examples this way on any Kahler manifold with $H^{2,0}\neq 0$. In the case of Kahler surfaces (symplectic $4$-manifolds) this is equivalent to the geometric genus being nonzero (or, in language more familiar to topologists, $b^+>1$).

In fact, a paper of Draghici (see the last paper listed on this page) shows essentially that, on a minimal Kahler surface of general type, if one starts at $\omega$ and goes out sufficiently far on the ray in the direction of $\phi$, then one eventually gets to classes that aren't represented by Kahler forms with respect to any complex structure, not just the original one.

There's a different sort of example in a paper of T.-J. Li and myself: we observe that if the Kahler surface $(M,\omega,J)$ contains any smooth J-complex curve (real 2D surface) $C$ of negative self-intersection other than a sphere of square $-1$, then one can obtain symplectic forms in the class $[\omega_t]=[\omega]+tPD[C]$ (where PD means Poincare dual) for a range of values of $t$ including some large enough that $[\omega_t]$ evaluates negatively on $C$. So the resulting symplectic form $\omega_t$ can't even be tamed by $J$. Again, in general $\omega_t$ might in principle be Kahler after deforming $J$ to some different complex strucutre, but Section 4.1 of that paper gives an example where this is carried out on a rigid surface (i.e. one admitting no deformations of the complex structure).

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Thank you for your kind answer. Now I understood what I want to know.      – YCho Oct 18 2010 at 18:09

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This is probably the simplest example. Take the Fubini-Study form $\omega$ on $CP^2$. Then $-\omega$ is symplectic, but never Kaehler, because by Yau's theorem $CP^2$ admits a unique (standard) complex structure.

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If J is the standard complex structure on CP2, isn't -J a complex structure with respect to which $-\omega$ is Kahler?  Yau's theorem in the form that I know it (see pnas.org/content/74/5/1798.full.pdf) says that any complex surface homotopy equivalent to CP2 is biholomorphic to it.  In the case of the complex structure -J, the biholomorphism is just $[z_0:z_1:z_2]\mapsto [\bar{z_0}:\bar{z_1}:\bar{z_2}]$. – Mike Usher Oct 19 2010 at 22:57
You are right - thanks. Sorry for the confusion. – Misha Verbitsky Oct 20 2010 at 18:02
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