Is every symplectic manifold a Hamiltonian reduction of a cotangent bundle? Today I heard the claim that in practice, all symplectic manifolds that people care about arise as the Hamiltonian reduction of a cotangent bundle $T^{\ast}(M)$ under the action of a Lie group $G$ ($M$ and $G$ may both be infinite-dimensional in general, I think). For example many moduli spaces of interest arise in this way. Is it literally true that every symplectic manifold arises in this way? Can we moreover arrange for $G$ and $M$ to be finite-dimensional? 
 A: That reminds me of a paper that I believe should answer your question: $\mathbf R^{2n}$ is a universal symplectic manifold for reduction (available here):

The authors show that if a manifold $Q$ is of finite type, that is, $H^k(Q,\mathbf Z)$ is finite-dimensional, then the noncanonical cotangent bundle $(T^∗Q,dθ_Q+τ^∗_QΩ)$ can be obtained by a Marsden-Weinstein reduction of $T^∗\mathbf R^n=\mathbf R^{2n}$ relative to a torus action. Here $Ω$ is a 2-form on $Q$, $τ_Q:T^∗Q→Q$ is the bundle map and $θ_Q$ is the canonical 1-form. Since any symplectic manifold $(Q,Ω)$ is a reduction of $(T^∗Q,dθ_Q+τ^∗_QΩ)$ relative to the zero section, it follows that any symplectic manifold can be obtained by a reduction of $\mathbf R^{2n}$ with the standard symplectic structure. (etc.)

A: Actually if you allow infinite dimension, every symplectic manifold is a coadjoint orbit of its group of symplectomorphisms. That is even more... how to say? Symplectic :-) If you want a reference there is a diffeological version of this theorem here It is also in the Memoir here §10
