symplectic action

Definition

Let $G$ be a Lie group that acts on a symplectic manifold $(M,\omega )$ smoothly. We call this action symplectic if for all $g\in G$ action map ${\phi }_g:M\to M$ is a symplectomorphism, i.e.

$${\phi }_g^{\ast }\omega =\omega .$$

Relation to fundamental vector fields

For $G$ acting symplecticly on $M$, every fundamental vector field $X_{\xi }$ for $\xi \in \mathfrak{g}$ is a symplectic vector field.

Proof

The flow of the vector field $X_{\xi }$ is $\exp (-t\xi )\cdot p$ by definition. However, we know that $\exp (-t\xi )\in G$ for every $t\in \mathbb{R}$, and that the action is symplectic, so

$${\phi }_{\exp (-t\xi )}^{\ast }\omega =\omega$$

so the flow is a symplectomorphim for every $t\in \mathbb{R}$.

Therefore, we know that all the generating vector fields of a symplectic action are closed, i.e.

$$d({\iota }_{{\xi }_M}\omega )=0\forall \xi \in \mathfrak{g}.$$