Skip to main content
MathOverflow

Return to Question

added 15 characters in body; edited title; deleted 14 characters in body; deleted 8 characters in body
Source Link
Yunhyung Cho
Yunhyung Cho
  • 1k
  • 7
  • 16

Hodge Riemann-Riemann bilnear form on Symplecticsymplectic manifolds.

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

$ 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 decompositionset of eigenvalues? (up to ordering of eigenvalueswith repetition)

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

$ 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 of eigenvalues)

Hodge-Riemann bilnear form on symplectic manifolds.

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

$ 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 set of eigenvalues? (with repetition)

deleted 4 characters in body; edited body; added 5 characters in body; deleted 5 characters in body
Source Link
Yunhyung Cho
Yunhyung Cho
  • 1k
  • 7
  • 16

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] > $$ 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 $< , >_{\omega}$$HR_\omega$ and $HR_\omega'$ $< , >_{\omega'}$ havehave different diagonal decomposition? (up to ordering of eigenvalues)

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)

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) := < \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 of eigenvalues)

Source Link
Yunhyung Cho
Yunhyung Cho
  • 1k
  • 7
  • 16

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)