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Let $\rho$ be a unitary representation of a torus $G$ on $\mathbb{C}^n$. The action of $\rho$ is Hamiltonian with a moment map $\mu:\mathbb{C}^n \to \mathfrak{g}^*$. Here $\mathfrak{g}^*$ is the dual space of the Lie algebra $\mathfrak{g}$ of $G$. If the weights of $\rho$ span $\mathfrak{g}$ 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: Is the (orbifold) fundamental group of $M$ trivial?


  • If $G$ acts freely on $\mu^{-1}(\alpha)$, then we can show that the above statement is true by using theory of toric varieties.
  • We do not assume here that the action of $G$ is effective. In this case, the fundamental group is isomorphic to the fundamental group of the Borel construction $\mu^{-1}(\alpha) \times_G EG$.
  • A proof which does not require theory of toric varieties is better. If there is such a proof for smooth symplectic quotients, please let me know. It will be a help.

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" §4.2.) 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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I have asked this question on MO. HERE –  H. Shindoh Jan 16 '13 at 11:04
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