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 $\mathcal{O}_{\mathbb{CP}^{m-1}}(k_1)\oplus\cdots\oplus\mathcal{O}_{\mathbb{CP}^{m-1}}(k_r)$.


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.