In this question I intentionally omit words like "(non)compact" because I am not sure about the precise setting where this question makes sense.
Let $M$ be a symplectic manifold, $G$ a Lie group, and suppose that $G$ acts on $M$ in a Hamiltonian way with an equivariant moment map $\mu:M\to \mathfrak{g}^*$. For any $\xi\in \mathfrak{g}^*$ we can define the symplectic quotient $M/\!/_\xi G\sim \mu^{-1}(\xi)/G_\xi$, where $G_\xi$ is the stabilizer of $\xi$ in $G$.
If $M$ is Kähler and $\xi$ is invariant, then $M/\!/_\xi G$ is Kähler and can be thought of as $M/G_{\mathbb{C}}$, where some words are required to make sense of the latter.
My understanding is that if $\xi$ is not $G$-invariant, then the above statement doesn't hold. However, I think that at least in some situations it can be rescued by the following construction.
Let $B$ be a Lie subgroup of $G$ such that $\xi|_\mathfrak{b}$ is $B$-invariant. Note that $B$ can be larger than $G_\xi$: infinitesimally we need $[\mathfrak{b},\mathfrak{b}]\in\ker \xi$ while $\mathfrak{g}_\xi$ satisfies a stronger condition $[\mathfrak{g}_\xi,\mathfrak{g}]\in\ker \xi$. We can form the Kähler quotient $M/\!/_\xi B\approx M/B_{\mathbb{C}}$.
Is it ever true that, as symplectic manifolds, $M/\!/_\xi G\simeq M/\!/_\xi B$? For example, I suspect this might be true when $G=\mathrm{SL}_2(\mathbb{R})$ and $B$ is an appropriate Borel subgroup. Any references or keywords to search for are greatly appreciated.
