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We know this important fact from A.A.Kirillov that :

Every homogeneous symplectic $G$-manifold is locally isomorphic to an orbit in the coadjoint representation of the group $G$ or a central extension of it.

But when can we precisely say that a homogeneous symplectic $G$-manifold is locally isomorphic to an orbit in the coadjoint representation of the group $G$

or

A homogeneous symplectic $G$-manifold is central extension of it?

PS:Definition of coadjoint orbit;

Let $G$ be a Lie Group and $\mathfrak{g}$ be its lie algebra,and also $\mathfrak{g^*}$ be the dual of Lie algebra, the coadjoint orbit is as follows

$\mathfrak{G}=\{Ad^*(g)F, g\in G\}$ where $F\in\mathfrak{g^*}$.

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Certainly, if $G$ has a central extension, $G$ will act on the coadjoint orbits of the central extension. So I think your question might be "When can we be sure that our group has no central extensions?"

The central extensions of the Lie algebra are given by elements of $H^2({\mathfrak g})$. So you want that to vanish, which it does for e.g. compact semisimple groups.

The two cautionary examples to keep in mind are ${\mathbb R}^{2n}$ acting on $T^* {\mathbb R}^n$ by translation, which leads one to the Heisenberg group central extension of ${\mathbb R}^{2n}$, and the free loop group $LK$ acting on the based loop space $\Omega K \cong LK/K$, which leads one to the affine Lie group central extension of $LK$. There, $H^2(L{\mathfrak K}) = H^2(LK) = \pi_2(LK)^* = \pi_3(K)^* = H^3(K)$ which is one-dimensional for $K$ simple and simply-connected.

There is an (in my opinion wrongheaded) definition of moment map that includes the composite ${\mathcal O} \hookrightarrow \widetilde{\mathfrak g}^* \twoheadrightarrow {\mathfrak g}^*$ where the first map is the inclusion of the coadjoint orbit, and the latter uses a noncanonical splitting of the central extension $\tilde{\mathfrak g}$. You can recognize this in the literature when people speak of "non-equivariant moment maps". Horrible. If a central extension is implicit I'd rather it be explicit.

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