Let $G$ be a compact connected Lie group and $\mathfrak g^*$ be dual of Lie algebra $\mathfrak g$. Let $M$ be a compact projective variety and $G$ act on $M$ freely and $M$ is $G$ equivariant, and $\mu:M\to \mathfrak g^*$ be a moment map, then the symplectic quotient $M_\lambda$, $\lambda\in \mathfrak g^*$ is still projective?
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1$\begingroup$ If $M$ is compact then a Hamiltonian action of an non-trivial connected group cannot be free: $\mu$ must have critical points (certainly it may act freely \emph{on a level set} but this is different). $\endgroup$– JeremyCommented Nov 30, 2014 at 17:33
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I believe that the answer is yes.
First note that the complexification $G_{\mathbb{C}}$ of $G$ is reductive and contains $G$ as a maximal compact subgroup. Secondly, $G_{\mathbb{C}}$ acts algebraically on $M$ via an extension of the original $G$-action. Your symplectic quotient is then homeomorphic to the projective Geometric Invariant Theory quotient of $M$ by $G_{\mathbb{C}}$.
A useful reference is Chapter 8 of Kirwan's thesis, Cohomology of Quotients in Symplectic and Algebraic Geometry.
answered Nov 29, 2014 at 14:21
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$\begingroup$ But I think you must add the condition $M^s=M^{ss}$? $\endgroup$– DanielCommented Nov 29, 2014 at 14:50
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$\begingroup$ I don't believe this is necessary. I think you need only assume that the $G$-stabilizers are finite. However, you took care of this by requiring the $G$-action to be free. $\endgroup$ Commented Nov 29, 2014 at 14:54
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$\begingroup$ Here has been written mathoverflow.net/questions/6316/… $\endgroup$– DanielCommented Nov 29, 2014 at 14:58
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$\begingroup$ A reference for the specific fact to which I'm referring is page 84 of cms.zju.edu.cn/UploadFiles/AttachFiles/201096185821505.pdf $\endgroup$ Commented Nov 29, 2014 at 15:01
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