Hodge-Riemann bilnear form on symplectic manifolds. - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T09:32:15Zhttp://mathoverflow.net/feeds/question/42307http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/42307/hodge-riemann-bilnear-form-on-symplectic-manifoldsHodge-Riemann bilnear form on symplectic manifolds.YCho2010-10-15T16:32:49Z2010-10-15T16:48:43Z
<p>Let $\omega$ a symplectic(may be Kahler) forms on $M^{2n}$. Then we have a symmetric
bilnear two form on $H^2(M,\mathbb{R})$ given by </p>
<p>$ HR_\omega (\alpha,\beta) := < \alpha\beta[\omega]^{n-2}, [M] > $</p>
<p>for $\alpha, \beta \in H^2(M,\mathbb{R})$. (If $n=2$, then this form is just an intersection form.)</p>
<p>Is there any example of $(M^{2n}, \omega, \omega')$ such that $HR_\omega$ and $HR_\omega'$
have different set of eigenvalues? (with repetition) </p>