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Suppose $E \to B$ is a symplectic vector bundle, i.e. it possesses a fibrewise linear symplectic form $\omega_F$. Further, suppose $\omega_B$ is a symplectic form on $B$.

Question: is there a symplectic form $\omega$ on $E$, such that (a) on any fibre of $E$, it restricts to $\omega_F$, and (b) on the zero section it restricts to $\omega_B$?

Remark: A closed 2-form can be obtained on $E$ that satisfies the first condition -- this can be done by choosing a connection on the associated principal bundle, and using Weinstein's theorem (Theorem 6.17 in McDuff-Salamon).

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  • $\begingroup$ In Thurston's construction, one usually adds the pullback of a large multiple of $\omega_B$ to $\omega_F$. If you don't scale $\omega_B$, then in general you don't get a symplectic form. $\endgroup$
    – YHBKJ
    Mar 16, 2017 at 16:08

1 Answer 1

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Such a symplectic form does not exist in general. In my paper (theorem 8.2). Gerbes, 2-gerbes and symplectic fibrations, I have shown that the obstruction of the existence of such a symplectic form on $E$ can geometrically be represented by a $2$-gerbes. My paper is available at

http://lanl.arxiv.org/pdf/math/0504274

Tsemo Aristide

Gerbes 2-gerbes and symplectic fibrations.

Rocky Mountain J. Math. Volume 38, Number 3 (2008), 727-777.

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