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.
