Scaling the parameter of symplectic reduction

Question

Let $(M,\omega)$ be a symplectic manifold and $G$ a compact Lie group acting freely on $M$ by symplectomorphisms with moment map $\mu:M\to\mathfrak{g}^*$. Then, for all fixed point $\xi$ of the coadjoint action of $G$ on $\mathfrak{g}^*$, the Marsden-Weinstein reduction $$M_\xi := \mu^{-1}(\xi)/G$$ is a smooth symplectic manifold.

By playing with some examples, I noticed that the diffeomorphism type of $M_\xi$ doesn't change after scaling $\xi$ by a positive real number. Is this a general phenomenon? Or, at least, is there a known condition or a class of Hamiltonian manifolds which are known to satisfy this?

For example, if $U(1)$ acts on $\mathbb{C}^n$ by $$z\cdot(x_1,\ldots,x_n)=(z^{-1}x_{1},\ldots,z^{-1}x_m,z^{k_1}x_{m+1},\ldots,z^{k_r}x_n,)$$ for some $k_i\ge 0$ and $\xi\in \mathfrak{u}(1)\cong\mathbb{R}$ is $\xi>0$, then the Marsden-Weinstein reduction is always diffeomorphic to the vector bundle $\mathcal{O}_{\mathbb{CP}^{m-1}}(k_1)\oplus\cdots\oplus\mathcal{O}_{\mathbb{CP}^{m-1}}(k_r)$.

Answer 1

The critical points of the moment map correspond precisely to the fixed points of the $G$-action. Therefore if the action is free and $\mu$ is proper, all level sets of $\mu$ will be equivariantly diffeomorphic (by an equivariant analogue of Ehresmann's lemma).

If you drop the freeness assumption, $M$ will need to be noncompact, since in the compact case $\mu^{-1}(K\xi) = \varnothing$ for $K \gg 0$. There are non-free, non-compact examples where $\mu^{-1}(K\xi)/G$ depends on $K$, for example $\mathrm{SU}(2)$-character varieties of punctured surfaces, where $K$ would correspond to the conjugacy class of the holonomy around the punctures.