How can one tell whether a given (finite-dimensional) symplectic manifold is homogeneous (that is, admits a transitive group of symplectomorphisms)?

Note that it is of little help to observe that a homogeneous symplectic manifold must be a covering space of a coadjoint orbit of some Lie group, since this provides no effective test applicable to a given example. Note also that in the Riemannian case, there is a well-known local obstruction to the existence of a transitive group action by sutomorphisms: the covariant derivative of the curvature must vanish. But since all symplectic manifolds of dimension n are locally symplectomorphic, there can be no such obstruction in this case, and any analogous constraint must be simultaneously sensitive to the symplectic form and the topology of the manifold.

nottrue that the condition for a Riemannian metric to be homogeneous is that the covariant derivative of the curvature vanish; that's the condition for the Riemannian metric to be (locally) symmetric, which is amuchstronger condition than homogeneity. For example, any left-invariant metric on a $3$-dimensional Lie group is homogeneous, but the vast majority of them are not symmetric. $\endgroup$ – Robert Bryant Dec 8 '11 at 12:55