Two versions of hamiltonian reduction

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Given a symplectic manifold $X$ with nice $G$ action, equivariant momentmap $\mu$ and invariant $\chi \in \mathfrak{g}^*$ which is a regular value of $\mu$. There are two ways to form the Hamiltonian reduction. What one usually does is take the levelset $X_\chi=\mu^{-1}(\chi)$ and quotient out the group $X_\chi/G$. However one could also quotient out $G$ first and then define $(X/G)_\chi$ to be the set of all points represented by elements of $\mu^{-1}(\chi)$.

I have sometimes heard that these two procedures are equal. Is this true, or more precisely in what situations is it true? Is this written down somewhere? Are there counterexamples one should have in mind?

I am also interested in settings where one replaces spaces by possibly noncommutative Poisson algebras.

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If an orbit contains an element of that level set, then it is completely contained in it by the equivariance of $\mu$ and invariance of $\chi$. – Santiago Canez Sep 13 2010 at 19:43

2 Answers

It depends on whether you think of the symplectic form as giving you a map $TX \to T^*X$ or vice versa. When you restrict to $\mu^{-1}(\chi)$, you only have a "presymplectic form", which gets you a map like the above (and not an isomorphism). If you quotient down to $X/G$, you only have a "Poisson tensor", which gets you a map backwards (that isn't an isomorphism). From your desired generalization it seems like you would prefer the latter picture.

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I seem to recall that, in the algebraic setting (so working with (affine, say) algebraic varieties and their algebraic functions, rather than manifolds and smooth functions), it is necessary to assume that G is reductive for the two processes you propose to yield the same result. On the one hand, one has $(A/J)^\mathfrak{g}$, while on the other, one has $A^\mathfrak{g}/J^\mathfrak{g}$. If it happens that $J$ has an complementary submodule $J^c$ in $A$, then one can write $A=J \oplus J^c$ as a $G$-module, so that $A^\mathfrak{g}=J^\mathfrak{g} \oplus (J^c)^\mathfrak{g}$, and one can finally conclude that $(A/J)^\mathfrak{g} \cong A^\mathfrak{g}/J^\mathfrak{g}$. Otherwise, the definitions can disagree. I learned this argument in a course which didn't produce notes, so I cannot provide a reference....

I didn't think about it, but I think the same sort of thing is going on for $C^\infty$ manifolds: you want a reductive group. I could be mistaken.

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Thanks, I recall the same thing for reductive groups, so it must be true ;-) – Jan Weidner Sep 30 2010 at 9:39
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