## A proof of simply connectedness of a symplectic quotient

Let $\rho$ be a unitary representation of a torus $G$ on $\mathbb{C}^n$. The action of $\rho$ is Hamiltonian with a moment map $\mu$. If the weights of $\rho$ span the Lie algebra of $G$ and are contained in an open half space, then for a regular value $\alpha$ of $\mu$, the quotient space $M := \mu^{-1}(\alpha)/G$ is a closed symplectic orbifold (whenever the preimage is not empty).

My question 1: Is the orbifold fundamental group of $M$ trivial?

We do not assume here that the action of $G$ on $\mu^{-1}(\alpha)$ is free. If $G$ acts freely on $\mu^{-1}(\alpha)$, then we can show that the above statement is true by using theory of toric varieties.

My question 2: can we prove the simple connectedness without theory of toric varieties if $M$ is smooth?

Namely I am looking for a proof which makes use of moment maps, etc.

Comments on the orbifold fundamental group

• The symplectic quotient is regarded as the action groupoid $G \ltimes \mu^{-1}(\alpha)$ in terms of groupoids, and $\pi_1(G \ltimes \mu^{-1}(\alpha))$ is defined to be the classifying space of the action groupoid. (See Moerdijk "Orbifolds as Groupoids: an Introduction" ยง4.2.) It is known that the classifying space has the same homotopy type as the Borel construction $\mu^{-1}(\alpha) \times_G EG$.
• We have (part of) a long exact sequence: $$\pi_2(BG) \to \pi_1(\mu^{-1}(\alpha)) \to \pi_1(\mu^{-1}(\alpha) \times_G EG) \to 0.$$ I wonder if we can see that the first homomorphism is surjective or the second homomorphism is zero.
• If the quotient is smooth, then we have a very similar exact sequence: $$\pi_1(G) \to \pi_1(\mu^{-1}(\alpha)) \to \pi_1(\mu^{-1}(\alpha)/G) \to 0.$$ I think that a proof of surjectivity of the first homomorphism will be a help.
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 If you form the quotient orbifold as in Geometric Invariant Theory, then the exact sequence of (orbifold) homotopy groups you want is valid. In fact the basic point is very simple. There is an Zariski closed subset of $\mathbb{C}^n$ whose complex codimension is at least $2$ at all points, so that the open complement $U$ is simply connected. Then there is a submersion of orbifolds $U \to [U/G]$ that is a $\mathbb{C}^*$-fiber bundle. Since the fiber is connected, and since $U$ is simply connected, also the orbifold $[U/G]$ is simply connected. – Jason Starr Jan 16 at 12:53 Thank you for your comment. I will try to translate it into symplectic geometry. – H. Shindoh Jan 20 at 15:31