MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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).

share|cite|improve this answer
Thank you for your kind answer. Now I understood what I want to know. – Yunhyung Cho Oct 18 '10 at 18:09

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.

share|cite|improve this answer
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 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 '10 at 22:57
You are right - thanks. Sorry for the confusion. – Misha Verbitsky Oct 20 '10 at 18:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.