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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)

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  • $\begingroup$ 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? $\endgroup$
    – Donu Arapura
    Commented Oct 15, 2010 at 16:46
  • $\begingroup$ 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. $\endgroup$
    – Donu Arapura
    Commented Oct 15, 2010 at 16:57
  • $\begingroup$ I am sorry for my bad English. I editted the question. I mean the second question you wrote. $\endgroup$
    – Yunhyung Cho
    Commented Oct 15, 2010 at 16:58
  • $\begingroup$ No need to apologize, I just wanted to make sure. Let me know if my example makes sense. $\endgroup$
    – Donu Arapura
    Commented Oct 15, 2010 at 17:00
  • $\begingroup$ 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. $\endgroup$
    – Yunhyung Cho
    Commented Oct 15, 2010 at 17:24

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