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Gaussler
Gaussler
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Given a Hamiltonian action of a compact Lie group $G$ on a symplectic manifold $(M,\omega)$, we may choose a moment map $\mu \colon M\to \mathfrak{g}^* $ and obtain the symplectic reduction $M/\!\!/G = \mu^{-1}(0)/G$. This construction clearly depends on the choice of moment map, and it is known to vary quite drastically depending on this choice. However, I wonder if it is still unique up to some sort of (very?) weak equivalence in the symplectic category?

Given a Hamiltonian action of a compact Lie group $G$ on a symplectic manifold $(M,\omega)$, we may choose a moment map $\mu \colon M\to \mathfrak{g}^* $ and obtain the symplectic reduction $M/\!\!/G = \mu^{-1}(0)/G$. This construction clearly depends on the choice of moment map, and it is known to vary quite drastically depending on this choice. However, I wonder if it is still unique up to some sort of (very?) weak equivalence in the symplectic category?

Given a Hamiltonian action of a compact Lie group $G$ on a symplectic manifold $(M,\omega)$, we may choose a moment map $\mu \colon M\to \mathfrak{g}^* $ and obtain the symplectic reduction $M/\!\!/G = \mu^{-1}(0)/G$. This construction clearly depends on the choice of moment map. However, I wonder if it is still unique up to some sort of (very?) weak equivalence in the symplectic category?

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Gaussler
Gaussler
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Is the symplectic quotient $\mu^{-1}(0)/G$ unique up to something?

Given a Hamiltonian action of a compact Lie group $G$ on a symplectic manifold $(M,\omega)$, we may choose a moment map $\mu \colon M\to \mathfrak{g}^* $ and obtain the symplectic reduction $M/\!\!/G = \mu^{-1}(0)/G$. This construction clearly depends on the choice of moment map, and it is known to vary quite drastically depending on this choice. However, I wonder if it is still unique up to some sort of (very?) weak equivalence in the symplectic category?