Let $M$ be a symplectic manifold equipped with a hamiltonian action of a compact Lie group $G$ with moment map $\mu\colon M\to \mathfrak g^\*$. Assume $c\in \mathfrak g^\*$. Then the symplectic reduction theorem tells us that if $G$ acts freely on $\mu^{-1}(c)$ then we can take a quotient of $\mu^{-1}(c)$ by $G$-action and get a symplectic manifold. But what if $G$ doesn't act freely? Do we get a symplectic orbifold in this case? And can we globalize the theorem, say take the quotient $M/G$ with the structure of a not necessarily hausdorff orbifold with a $2$-form $\tilde\omega$ s.t. it becomes the symplectic reduction form $\omega\_{red}$ when restricted on $\mu^{-1}(c)/G$ for each $c\in\mathfrak g^\*$? Is the form $\tilde\omega$ of constant rank? If it is, it defines the distribution of its kernel. Under which conditions is this distribution integrable?



Global symplectic reduction
In the general case, the reduced space $\mu^{-1}(c) / G\_c$ is what is called a stratified symplectic space. This means, that for every orbit type $(H)$ the orbit type subset $\mu^{-1}(c)\_{(H)} / G\_c$ is a smooth symplectic manifold and the symplectic forms on these strata fit together in the sense that there is a compatible Poisson structure on $\mu^{-1}(c) / G\_c$. Details are worked out, for example, in the book "Momentum Maps and Hamiltonian Reduction" by Ortega and Ratiu.


Btw, "moment map" is an erroneous translation of the French notion "application moment". "Momentum map" should be the preferred usage in English, see [Origin of the name ''momentum map''](https://mathoverflow.net/questions/242468/origin-of-the-name-momentum-map) for details about the history of the name.


