A manifold $M$ is said to be homogeneous if there exists a transitive action $G ↷ M$. Let $(M,ω)$ be a homogeneous symplectic manifold, i.e., there exists a transitive and symplectic action $G ↷ M$. Prove that if $M$ is simply connected and $H^2(g) = 0$, then $M$ is symplectomorphic to an adjoint orbit.
Here $g$ is the lie algera of $G$ (the action group) and $H^*(g)$ is the Chevalley-Eilenberg cohomology of the lie algebra $g$.
I would appreciate it if someone could help me understand and prove this statement. Specifically, I would like to know the steps and reasoning behind the proof. Could you please provide a detailed explanation or point me to any relevant resources that can help me grasp this concept?
Thank you in advance for your assistance!
The point is that under your hypothesis, there exist a moment map $J=M\to g^*$, and this map is equivariant. This is called strongly hamiltonian in the literature.
Let $\bf X$ the space of hamiltonian vector fields on $M$. The map $C^{\infty }(M)\to \bf X$ is the map which sen a function $f$ to the corresponding hamiltonian vector field is a Lie algebra homomorphism (for the Poisson bracket) with kernel the constant fonction : hence it is a central extension, and define a certain class $\omega \in H^2(\bf X)$. This map is surjective as $X$ is simply connected.
The fact that the action is strongly hamiltonian means that the map $g\to \bf X$ lifts to a Lie algebra homomorphism $g\to C^{\infty}(M)$
The obstruction to lift this Lie algebra homomorphism is a certain element in $H^2(g)=H^1(g,g^*)$, which is zero in your case.