TITLE: Two versions of hamiltonian reduction
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.