# A symplectic form on a symplectic vector bundle

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).

• 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. – YHBKJ Mar 16 '17 at 16:08

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