So I'm learning a bit about Lie group actions on symplectic manifolds and moment maps, and I came across a question that I'd like an answer to, which I can't work out by myself.
So I'll give some definitions first.
A Lie group action of $G$ on $(M, \omega)$ is said to be symplectic if it acts by symplectomorphisms, i.e. for $g \in G$, if we denote by $g: M \to M$ the map $x \to g \cdot x$, the action is symplectic if every $g$ is a symplectomorphism, that is $g^{*} \omega = \omega$.
A Lie group action of $G$ on $(M, \omega)$ is said to be Hamiltonian if it is symplectic and there exists a moment map $\mu: M \to \frak{g}^*$ (denoting by $(\mu, \xi) \in C^{\infty}(M)$ the function which at $x \in M$ takes the value $\mu(x)(\xi)$) such that:
$d(\mu, \xi) = i_{\xi^{M}} \omega$, where for $\xi \in \frak{g}$ we define the vector field $\xi^M \in \frak{X}$$(M)$ as $$\xi^{M}_{x} = \frac{d}{dt}|_{t=0}(e^{t \xi} \cdot x)$$
$\mu(g \cdot x) = Ad_{g}^{*}\mu(x)$
Now in the text that I am reading it says that if $G$ is connected, we can drop the "symplectic" condition in the defintion of a Hamiltonian action, i.e. if we have an arbitrary action with a function $\mu$ satisfying the two listed properties, we automatically have $g^* \omega = \omega$. However, I can't seem to figure out why this is true.
What I've managed to find is that if $G$ is connected, then for any $g \in G$, there exist some Lie algebra vectors $v_{1}, \dots, v_{k}$ such that $g = \Pi_{i=1}^{k} e^{v_{i}},$ so it suffices to show that $(e^{v})^{*} \omega = \omega$.
I guess I'm supposed to use the first condition of $\mu$ now, but I don't really know how to, since I don't really know anything about the map $(\mu, \xi)$ or about which vectors can be written as $\xi^{M}_{x}$ for some $\xi$.
EDIT: I found out that $$T_{x}(G \cdot x) = \lbrace \xi^{M}_{x} | \xi \in \frak{g} \rbrace,$$ where $G \cdot x$ is the orbit of $x$, so now I'm leaning towards the fact that what I'm trying to prove isn't true for non-transitive actions. Can anyone maybe work out the proof for transitive actions, and maybe even provide a counterxample in the case when the action isn't transitive? Of course, there's still a chance that my hunch is wrong and that the statement is true in the general case.
Let $X$ an element of the Lie Lie algebra and $e_X$ the vector field defined on $M$ by $e_X(x)={d\over{dt}}_{t=0}e^{tX}.x$, we have $L_{e_X}\omega=i_{e_X}d\omega+di_{e_X}\omega=d(d\mu(,X)=0$, this implies that $\phi_t^X=f(exp(tX))$ where $f:G\rightarrow Diff(M)$ is the morphism of Lie groups which defines the action by $g.x=f(g)(x))$. Remark that we have used the fact morphisms between Lie groups commute with the exponential maps.
The flow generated by $e_X$ preserves $\omega$, since a connected Lie group is generated by a neighborhood of the identity, we deduce that $G$ preserves $\omega$.