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Given a smooth affine symplectic variety $V$ with an action of a connected algebraic $G$. If $\mu$ is the moment map, the define the affine quotient to be :

$X = \mu^{-1}(0)// G = \text{Spec}\mathbb{C}[\mu^{-1}(0)]^{G}$

This is an algebraic Variety (may be singular).

The Hamiltonian reduction of $V$ is defined to be

$Y = \mu^{-1}(0)/ G$.

Q : When are these two notions same ?

edited Nov 23 2011 at 2:38

asked Nov 22 2011 at 22:38

J Verma
1,740●1●4●26

	Do you really want $G$ to be compact and algebraic? It seems to me that if so, your action is necessarily trivial: an orbit will be a compact subvariety of $V$ which then must be a point. – Jim Bryan Nov 23 2011 at 1:17
	non necessarily compact, but connected G. I've edited the question accordingly. – J Verma Nov 23 2011 at 2:40

Ben Webster · Accepted Answer · 2011-11-23 04:10:48Z UTC

We can assume that $G$ is affine, since an abelian variety must act trivially on any affine variety. The closed points of $X$ are exactly the closed $G$-orbits on $\mu^{-1}(0)$. On an affine variety, closed orbits can be distinguished by invariant global functions, and if one orbit is in the closure of another they will go to the same point in $X$.

Thus, $X$ and $Y$ are the same if all $G$-orbits on $\mu^{-1}(0)$ are closed.

Hamiltonian Reduction and Affine quotient

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