MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question
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? – Donu Arapura Oct 15 '10 at 16:46
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. – Donu Arapura Oct 15 '10 at 16:57
I am sorry for my bad English. I editted the question. I mean the second question you wrote. – Yunhyung Cho Oct 15 '10 at 16:58
No need to apologize, I just wanted to make sure. Let me know if my example makes sense. – Donu Arapura Oct 15 '10 at 17:00
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. – Yunhyung Cho Oct 15 '10 at 17:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.