Hodge-Riemann bilnear form on symplectic manifolds. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:32:15Z http://mathoverflow.net/feeds/question/42307 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42307/hodge-riemann-bilnear-form-on-symplectic-manifolds Hodge-Riemann bilnear form on symplectic manifolds. YCho 2010-10-15T16:32:49Z 2010-10-15T16:48:43Z <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>$HR_\omega (\alpha,\beta) := &lt; \alpha\beta[\omega]^{n-2}, [M] >$</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>