Question about examples of symplectic non-Kahler classes. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:24:16Z http://mathoverflow.net/feeds/question/42424 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42424/question-about-examples-of-symplectic-non-kahler-classes Question about examples of symplectic non-Kahler classes. YCho 2010-10-16T22:13:46Z 2010-10-19T19:54:56Z <p>Let $M$ be an even dimensional smooth manifold.<br> I want to find an example $M$ satisfying the following conditions, </p> <ol> <li>$M$ admits a Kahler structure.</li> <li>$\omega$ is a symplectic form on $M$. </li> <li>There is no Kahler structure $(M,\omega',J)$ such that $[\omega']=[\omega] \in H^2(M;\mathbb{R})$</li> </ol> <p>(I mean, want to find an example $M$ such that "Kahler cone $\neq$ symplectic cone" with non-empty Kahler cone.)</p> <p>Thank you in advance. </p> http://mathoverflow.net/questions/42424/question-about-examples-of-symplectic-non-kahler-classes/42443#42443 Answer by Mike Usher for Question about examples of symplectic non-Kahler classes. Mike Usher 2010-10-16T23:31:39Z 2010-10-16T23:31:39Z <p>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, $b^+>1$).</p> <p>In fact, a paper of Draghici (see the last paper listed on <a href="http://www2.fiu.edu/~draghici/research/resint.html" rel="nofollow">this page</a>) 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 <em>any</em> complex structure, not just the original one.</p> <p>There's a different sort of example in a <a href="http://front.math.ucdavis.edu/0601.5540" rel="nofollow">paper</a> 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).</p> http://mathoverflow.net/questions/42424/question-about-examples-of-symplectic-non-kahler-classes/42830#42830 Answer by Misha Verbitsky for Question about examples of symplectic non-Kahler classes. Misha Verbitsky 2010-10-19T19:54:56Z 2010-10-19T19:54:56Z <p>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.</p>