# When does the $\mathfrak g$-invariance of the symplectic form imply $G$-invariance?

Let $G$ be a Lie group with Lie algebra $\mathfrak g$, and let $M$ be a smooth manifold. Suppose $G$ acts on $M$, $G \to \text{Diff}(M)$. This naturally induces an action $\mathfrak g \times M \to M$ by $X \mapsto \exp(X) \cdot m$.

Suppose we have a symplectic form $\omega$ on $M$, then if the $G$-action leaves $\omega$ invariant, this implies that the $\mathfrak g$-action also leaves $\omega$ invariant. My question is about the converse: under what conditions does $\mathfrak g$-invariance of $\omega$ imply $G$-invariance?

Surely it would be sufficient if the exponential map is a surjection. However, in the notes I'm reading it's stated that the $G$-invariance follows when $G$ is connected, and there are examples of connected Lie groups for which the exponential map is not surjective. I'm having trouble proving this claim. How does $G$-invariance follow from $\mathfrak g$-invariance for connected $G$?

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1. The exponential map $\textrm{exp}:\mathfrak g\to G$ is a local diffeomorphism at $0$ and therefore $\textrm{exp}(\mathfrak g)$ is a neighborhood of $\textrm{exp}(0)=e$ in $G$.
2. Obviously the elements of $G$ acting sympctically on $(M,\omega)$ constitute a subgroup $H$. The left multiplications are diffeomorphisms of $G$ onto itself, and $\{g\cdot H:g\in G\}$ is a partition of $G$, therefore: $$H\textrm{ is open in }G\Rightarrow H\textrm{ is closed in }G,$$ $$H\textrm{ is open in }G\Leftrightarrow H\textrm{ is a neighborhood of }e\textrm{ in }G.$$

In conclusion, denoting by $G_0$ the connected component of $e$ in $G$, if $H\supseteq\textrm{exp}(\mathfrak g)$ then by 1. and 2. we get that $H$ is closed-open subset of $G$ containing $e$, therefore $H\supseteq G_0$.

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How does $G$-invariance follow from $\mathfrak{g}$-invariance for connected $G$?

This follows from the following fact (for topological groups):

The identity component $G_0$ of $G$ is generated by an open neighbourhood of the identity.

In other words, take an open neighbourhood $U$ of $1$, then any $g \in G_0$ can be given by $g_1 \dots g_k$ for some $g_1, \dots, g_k \in U$.

The $\mathfrak{g}$-invariance implies "$U$-invariance". So $G_0$-invariance follows.

(I think you can find the proof of the above fact in books on topological groups or Lie groups.)

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