Let G be a lie group, then When a G-Manifold is a Hamiltonian G-manifold and under which condition a manifold is Hamiltonian G manifold for some lie group G
2026-03-30 03:55:05.1774842905
When a G-Manifold is a Hamiltonian G-manifold
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This is a hard question and the answer is not known in general, even for $G = S^1$. The obvious necessary condition for a $G$-action on a compact symplectic manifold $(M, \omega)$ to be Hamiltonian is that the action has fixed points. This is because the Hamiltonian function will have critical points on a compact manifold, and these critical points will correspond to fixed points of the corresponding Hamiltonian action.
A well-known theorem of Ono shows that the existence of fixed points is also a sufficient condition in the case of circle actions on symplectic manifolds of "Lefschetz type."
All Kähler manifolds satisfy the hypothesis of the above theorem, so any symplectic $S^1$-action on a Kähler manifold with fixed points is Hamiltonian. However, McDuff found an example of a $6$-dimensional symplectic manifold (not of Lefschetz type) with a symplectic $S^1$-action possessing a fixed point that is not Hamiltonian. So Ono's theorem has no hope of extending to all symplectic manifolds.
An often studied class of symplectic manifolds are the monotone symplectic manifolds. The situation is better for them:
In the noncompact case, there's an easy general result for exact symplectic manifolds $(M, \omega)$, i.e. where $\omega = -d\theta$ for some $1$-form $\theta$. A symplectic $G$-action on such an exact symplectic manifold is called exact if the action preserves the $1$-form $\theta$.