I have often heard/read a statement (see, e.g., this MathOverflow question) equivalent to the following:
Let $G$ be a connected Lie group and $(M,\omega)$ a connected and simply-connected symplectic manifold admitting a transitive action of $G$ via symplectomorphisms. Then $M$ covers a coadjoint orbit of a one-dimensional central extension of $G$.
but I don't know how to prove this unless I also assume that $G$ is simply connected.
A sketch of the proof is as follows. Let $\xi : \mathfrak{g} \to \mathscr{X}(M)$ be the Lie algebra homomorphism from the Lie algebra of $G$ to the corresponding symplectic vector fields on $M$. Since $M$ is simply-connected, these vector fields are hamiltonian and we can lift $\xi$ to a linear map $\varphi: \mathfrak{g} \to C^\infty(M)$. The obstruction to this map being a Lie algebra homomorphism (with the Lie algebra structure on $C^\infty(M)$ given by the Poisson brackets defined by $\omega$) is a class $[c] \in H^2(\mathfrak{g})$, with representative cocycle $c(X,Y) = \{\varphi_X, \varphi_Y\} - \varphi_{[X,Y]}$. If $[c] = 0$ then we can modify $\varphi$ by constants to make it into a Lie algebra homomorphism, but if $[c] \neq 0$, then it defines a nontrivial one-dimensional central extension $\widehat{\mathfrak{g}}$ of $\mathfrak{g}$ and we can extend $\varphi$ to a Lie algebra homomorphism $\widehat\varphi: \widehat{\mathfrak{g}} \to C^\infty(M)$.
In the former case ($[c]=0$) we have a $G$-equivariant moment map $\mu : M \to \mathfrak{g}^*$ dual to $\varphi$ and whose image is therefore a coadjoint orbit of $G$, but what about in the latter case ($[c]\neq 0$)?
Let $\widehat{G}$ be the connected and simply-connected group with Lie algebra $\widehat{\mathfrak{g}}$. Dual to $\widehat\varphi$ we now have a $\widehat{G}$-equivariant moment map $\widehat \mu : M \to \widehat{\mathfrak{g}}^*$ whose image is then a coadjoint orbit of $\widehat{G}$.
Now, by the Lie correspondence we get a unique Lie group homomorphism $\widehat{G} \to G$ and hence a one-dimensional extension of $G$, but is it necessarily central?
The conditions under which a central extension of the Lie algebra of a given group $G$ integrates to a central extension of $G$ have been studied by Karl-Hermann Neeb (Journal of Lie Theory 6 (1996) 2017-213) refining earlier work of Tuynman and Wiegerinck. These conditions are seen to be satisfied if $G$ is simply connected, as they are equivalent to the fundamental group of $G$ acting trivially on $\widehat{\mathfrak{g}}$.
Now it is of course the case that if $G$ acts transitively on $M$, then so does its universal covering group, so it might seem that without any loss of generality one could always assume that $G$ is simply connected, but I am wondering whether this is essential.
Is the statement at the top of this question correct? Or should it be modified to "... a central extension of the universal covering group of $G$"?
