Let $X\subseteq \Bbb C^n$ be a smooth affine variety and $G=K_{\Bbb C}$ a complex reductive group acting linearly on $\Bbb C^n$ preserving $X$ (where $K$ is a maximal compact subgroup of $G$). Suppose that $X$ has a $K$-invariant Kähler metric and that there is a moment map $$\mu:X\longrightarrow {\frak k}^*$$ for the action of $K$ on $X$.

If the Kähler structure on $X$ comes from a $K$-invariant hermitian product on $\Bbb C^n$, then it is well known (by results of Kempf & Ness and Kirwan) that the symplectic reduction of $X$ by $K$ coincides with the GIT quotient of $X$ by $G$, i.e. the inclusion $\mu^{-1}(0)\subseteq X$ induces a homeomorphism $$\mu^{-1}(0)/K\overset{\sim}{\longrightarrow} X//G:=\operatorname{Spec}{\cal O}(X)^G.$$

Question:Is it necessary that the Kähler structure comes from a $K$-invariant hermitian product, or is there a similar result for the more general context of the first paragraph?

**Motivation:** There is a result by Reyer Sjamaar [1] which says that if $X$ is an integral Kähler manifold so that it has a Kodaira embedding $X\subseteq \Bbb P^n$, then the symplectic reduction $\mu^{-1}(0)/K$ coincides with the GIT quotient $X//G$, even though the symplectic form on $X$ may not come from the Fubini-study metric on $\Bbb P^n$. I was wondering if this can be adapted to the affine case.

In our case, there is certainly an analytic version of the GIT quotient $X^{ss}//G$ where $X^{ss}$ is the set of points $x$ where $\overline{G\cdot x}$ intersects $\mu^{-1}(0)$, and two orbits are identified if their closures intersect (see [1]), and we have $\mu^{-1}(0)/K\cong X^{ss}//G$. A simpler question could thus be:

If $X$ is an affine variety, do we have $X^{ss}=X$?

Indeed, for the affine GIT quotient every point is semistable by definition.

*References.*

[1] Sjamaar, R., Holomorphic Slices, Symplectic Reduction and Multiplicities of Representations, Annals of Mathematics, 1995.