Let $G$ be a compact Lie group with algebra $\mathfrak{g}$. Let $\beta $ be an element in the dual of the Lie algebra $\mathfrak{g}$. We denote by $G\_\beta$ the stabilizer subgroup of $\beta$ by the cooadjoint action. Let $V$ be symplectic vector space on which $G\_\beta$ acts.


So, we have $V$ is symplectic, $G/G\_\beta$ is symplectic (since it can be identified with a cooadjoint orbit which is symplectic), How can we show that $M:= G \times\_{G\_\beta} V$ is also symplectic ?


Let $[g,v] \in M$, we have $T\_{[g,v]} M = T\_gG \times T\_vV$
How can we define a symplectic form $\omega\_{[g,v]}: T\_{[g,v]} M \times T\_{[g,v]} M \rightarrow \mathbb{C}$ ?



Define a symplectic structure on $G \times_{G_\beta} V$, where $V$ is symplectic
Your tangent space is too big. It has rank $\dim G + \dim V$ but it should have rank $\dim M = \dim G - \dim G\_\beta + \dim V$.


I think that the tangent space at $[g, v]$ is actually $T\_{g} (G/G\_\beta) \times T\_vV$. On the first factor you have a symplectic form $\omega\_\beta$ and on the second factor the symplectic form $\omega\_v$. On the whole tangent space you can then take block-diagonal sum $\omega\_\beta \oplus \omega\_v$.


