Let $G$ be a Lie group and let $M \rightarrow B$ be a $G$-equivariant vector bundle with typical fiber $E$.


Suppose that $B$ and $E$ are both symplectic manifolds with a $G$-Hamiltonian action. Can we conclude from this that $M$ is also a symplectic manifold with a Hamiltonian action?


I have been told that given a $G$-equivariant vector bundle of the form $G\times\_K V \rightarrow G/K$, where $V$ is a $K$-vector space of even dimension and $K$ is the stabilizer of some point $x \in V$, then since $G/K$ (which is identified with a coadjoint orbit) and $V$ have $G$-Hamiltonian actions then $G\times\_K V$ also has a symplectic structure with $G$-Hamiltonian action. Could you please explain why this is true?



Symplectic structure on a vector bundle
Let $f:M\rightarrow B$ be the symplectic fibration. A fibration whose typical fibre $F$ is a symplectic manifold, and the coordinate change are elements of the group of symplectomorphisms of $(F,\omega\_F)$. For the first question, a natural idea is to have the symplectic form $\omega\_M$ of $M$ compatible with the symplectic form of the fibres, that is the fibers are symplectic submanifolds of $M$. Thurston has shown that if there exists a form $[c]\in H^2(M,\mathbb{R})$ whose restriction to $F$ is $[\omega\_F]\in H^2(F,\mathbb{R})$, for every $n>>>0$, there exists a compatible form in the class of $c+nf^\*[\omega\_B]$.


Since The bundle is a vector bundle, $H^2(F,\mathbb{R})=0$, thus the form $c$ exists.


W. Thurston, Some simple examples of symplectic manifolds, Proc. Amer.
math. Soc. 55 (1976), 467-468.


