# 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?

• If you require finite dimensions, this is a rather hyperbolic claim--- consider an arbitrary projective variety. – user36931 Dec 6 '13 at 2:22
• @user: yes, and? – Qiaochu Yuan Dec 6 '13 at 2:50
• I was just pointing out that I see no reason e.g. the quintic three-fold in CP^4 (which is one of the most studied symplectic manifolds with it's standard Kahler form) is a quotient as above. Of course proving that seems difficult enough, but that seems irrelevant to what you are asking. Certainly it's not naturally presented that way and although the theory of symplectic quotients plays a role in that literature, it's not in this direct fashion. – user36931 Dec 6 '13 at 16:51
• Between Kahler manifolds and symplectic quotients (there is obviously some overlap here), you get a lot of the symplectic manifolds people care about. – user36931 Dec 6 '13 at 16:54

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.)
• @Qiaochu You're right, this is "Cartan" reduction -- the quotient of a submanifold $N$ by the characteristic foliation of $\omega_{|N}$ (assumed constant rank) on it. I wonder (but haven't really thought through) whether there might be a moment map interpretation when $N$ happens to be a level set of some functions? – Francois Ziegler Dec 6 '13 at 3:32