# Global symplectic reduction

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?

• There is a huge literature on this, to which the “Singular Reduction” section of Marsden-Weinstein (2001) is a good entry point. The symplectic reduced space is not $\mu^{-1}(c)/G$ but $\mu^{-1}(c)/G_c$; standard assumptions you have omitted are that $\mu$ be equivariant and $c$ a regular value of $\mu$. The global quotient $M/G$ does not carry a 2-form $\tilde\omega$ but a Poisson structure, of which the $\mu^{-1}(c)/G_c$ are the symplectic leaves, see Weinstein (1983, p. 544). – Francois Ziegler Nov 10 '18 at 13:47

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.