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?
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asked Dec 5, 2013 at 22:02
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$\begingroup$ If you require finite dimensions, this is a rather hyperbolic claim--- consider an arbitrary projective variety. $\endgroup$– user36931Commented Dec 6, 2013 at 2:22
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$\begingroup$ @user: yes, and? $\endgroup$– Qiaochu YuanCommented Dec 6, 2013 at 2:50
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$\begingroup$ 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. $\endgroup$– user36931Commented Dec 6, 2013 at 16:51
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$\begingroup$ Between Kahler manifolds and symplectic quotients (there is obviously some overlap here), you get a lot of the symplectic manifolds people care about. $\endgroup$– user36931Commented Dec 6, 2013 at 16:54
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2 Answers
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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
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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.)
answered Dec 6, 2013 at 2:12
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$\begingroup$ Thanks! The word "reduction" in the last sentence seems to be referring to a construction more general than Hamiltonian reduction though, if I'm not mistaken. $\endgroup$ Commented Dec 6, 2013 at 2:54
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$\begingroup$ @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? $\endgroup$ Commented Dec 6, 2013 at 3:32