3 added 15 characters in body; edited title; deleted 14 characters in body; deleted 8 characters in body

# HodgeRiemannHodge-Riemann bilnear form on Symplecticsymplectic manifolds.

Let $\omega$ a symplectic symplectic(may be Kahler) forms on $M^{2n}$. Then we have a symmetric bilnear two form on $H^2(M,\mathbb{R})$ given by

$HR_\omega (\alpha,\beta) := < \alpha\beta[\omega]^{n-2}, [M] >$

for $\alpha, \beta \in H^2(M,\mathbb{R})$. (If $n=2$, then this form is just an intersection form.)

Is there any example of $(M^{2n}, \omega, \omega')$ such that $HR_\omega$ and $HR_\omega'$ have different diagonal decomposition? (up to ordering set of eigenvalues? (with repetition)

2 deleted 4 characters in body; edited body; added 5 characters in body; deleted 5 characters in body

Let $\omega$ a symplectic forms on $M^{2n}$. Then we have a symmetric bilnear two form on $H^2(M,\mathbb{R})$ given by

$HR_\omega (\alpha,\beta)_\omega \alpha,\beta) := < \alpha\beta[\omega]^{n-2}, [M] >$

for $\alpha, \beta \in H^2(M,\mathbb{R})$. (If $n=2$, then this form is just an intersection form.)

Is there any example of $(M^{2n}, \omega, \omega')$ such that $< , >{\omega}$ HR_\omega$and$< , >{\omega'}$HR_\omega'$ have different diagonal decomposition? (up to ordering of eigenvalues)

1

# Hodge Riemann bilnear form on Symplectic manifolds.

Let $\omega$ a symplectic forms on $M^{2n}$. Then we have a symmetric bilnear two form on $H^2(M,\mathbb{R})$ given by

$(\alpha,\beta)_\omega = < \alpha\beta[\omega]^{n-2}, [M] >$

for $\alpha, \beta \in H^2(M,\mathbb{R})$. (If $n=2$, then this form is just an intersection form.)

Is there any example of $(M^{2n}, \omega, \omega')$ such that $< , >{\omega}$ and $< , >{\omega'}$ have different diagonal decomposition? (up to ordering of eigenvalues)