Timeline for Hodge-Riemann bilnear form on symplectic manifolds.
Current License: CC BY-SA 2.5
9 events
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Oct 15, 2010 at 17:31 | comment | added | Yunhyung Cho | I have one more question. Is there any example that the numbers of positive eigenvalue of $HR_\omega$ and $HR_\omega'$ are different? (I assume that $\omega$^n and $\omega'^n$ give positive volumes on $M$. ) | |
Oct 15, 2010 at 17:24 | comment | added | Yunhyung Cho | Thank you. I works for $T^6 \equiv (S^1)^6$. I saw some result in toric cases (V.Timorin's paper, 1999, Russ math survey): If $(M, \omega)$ is symplectic toric, then $HR_\omega$ is non-degenerate and the number of positive eigenvalues is one. | |
Oct 15, 2010 at 17:00 | comment | added | Donu Arapura | No need to apologize, I just wanted to make sure. Let me know if my example makes sense. | |
Oct 15, 2010 at 16:58 | comment | added | Yunhyung Cho | I am sorry for my bad English. I editted the question. I mean the second question you wrote. | |
Oct 15, 2010 at 16:57 | comment | added | Donu Arapura | Assuming the latter. Take $M$ to be a torus give as quotient of $\mathbb{R}^{4}$. Take $\omega=dx_1\wedge dx_3+dx_2\wedge dx_4$ and $\omega'= dx_1\wedge dx_3+\lambda dx_2\wedge dx_4$, with $\lambda >1$ real. I haven't checked it, but I suspect that might work. | |
Oct 15, 2010 at 16:48 | history | edited | Yunhyung Cho | CC BY-SA 2.5 |
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Oct 15, 2010 at 16:46 | comment | added | Donu Arapura | Could you clarify what you want exactly? I assume you want to diagonalize the forms corresponding to $\omega$ and $\omega'$. The same basis is unlikely to work for both. Is that what you want? Or do you want to know that the set of eigenvalues (with repetitions) for the two are different? | |
Oct 15, 2010 at 16:38 | history | edited | Yunhyung Cho | CC BY-SA 2.5 |
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Oct 15, 2010 at 16:32 | history | asked | Yunhyung Cho | CC BY-SA 2.5 |