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Given some finite-dimensional symplectic manifold $(M,\omega)$, can we answer whether or not it arises as the coadjoint orbit of some finite-dimensional $G$ acting on the dual $\mathfrak{g}^*$ to its Lie algebra, which naturally comes with the Kostant–Kirillov–Souriau Poisson structure?

A result of Patrick Iglesias-Zemmour (Every symplectic manifold is a coadjoint orbit) establishes that the answer is always "yes" in the infinite-dimensional setting where $G$ is the group of symplectomorphisms of $(M,\omega)$.

I am interested in the (more restrictive?) case where $G$ is a finite-dimensional group. Since finite-dimensional Lie algebras are classified, I assumed that we should have an exhaustive list of all $(M,\omega)$ which arise as coadjoint orbits in $\mathfrak{g}^*$.

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    $\begingroup$ This is a neat question! I find the first sentence of the linked article funny: "At the end of the sixties, last century …." Future-proofed against reading in 2071, but not in 2171! $\endgroup$
    – LSpice
    Commented Nov 28, 2021 at 23:45
  • $\begingroup$ I guess higher genus oreintable compact surfaces are never of that type. For the simple reason that they are not even homogeneous manifolds at all, no matter which symplectic you put on them. So you might want to add the assumption that $M$ is at least homogeneous, i.e. admits a transitive Lie group action...? $\endgroup$
    – Stefan Waldmann
    Commented Nov 29, 2021 at 9:15
  • $\begingroup$ Stefan, it's easy to prove that any connected smooth manifold is acted on transitively by its diffeomorphism group. (Connect two points by a path, trivialize in a tubular nbhd of the path, ... ) duetosymmetry, I'm more worried about your statement "finite-dim (real) Lie algebras are classified". If you mean semisimple ones, then yeah, but to give an idea of the horrors, let R act on V=R^2 by speed 1 rotation, and on W=R^2 by speed sqrt(2) rotation, then make the semidirect product R x (V+W). Irrational-flow-on-the-torus issues means that this has coadjoint orbits that aren't locally closed. $\endgroup$
    – Allen Knutson
    Commented Nov 29, 2021 at 15:43
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    $\begingroup$ @Allen. Sure, if you allow for infinite-dim "Lie" groups then of course. But I had the impression that the OP wanted to have coadjoint orbits of finite-dim Lie groups. Then the manifold in question should be at least homogeneous, or do I miss something here? This seems to be a question completely independent of whether also the symplectic structure matches KKS or not. $\endgroup$
    – Stefan Waldmann
    Commented Nov 29, 2021 at 17:13

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