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Let $X$ be a Kähler manifold and $G$ a complex semisimple Lie group acting freely on $X$ by biholomorphisms and such that the Riemannian metric is preserved by a maximal compact subgroup $K$ of $G$. Does the Kähler structure descend to $X/G$ in some way?

I know that if $X\subset\mathbb{P}^n$ is projective with the Kähler structure induced by the Fubini-Study metric on $\mathbb{P}^n$ and if $G$ acts via a representation $G\to\mathrm{GL}(n+1,\mathbb{C})$, then the answer is 'yes' because there is a moment map $\mu:X\to\mathrm{Lie}(K)^*$ for the Kähler form of $X$ such that $\mu^{-1}(0)/K\cong X/G$ and we get a metric and Kähler form on $X/G$ from the symplectic quotient $\mu^{-1}(0)/K$. (I think this is due to Frances Kirwan.)

However, I am not assuming that $X$ is projective (not even compact). Is there still a Kähler structure on $X/G$ and is it also obtained by a symplectic reduction?

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    $\begingroup$ Kirwan (1984, 7.4) actually deals with general (compact?) Kähler $X$. Assuming existence of $\mu$ and finite stabilizers $K_x$ for $x\in\mu^{-1}(0)$, she gets homeomorphism between the symplectic subquotient and $X^{\mathrm{min}}/G$, where $X^{\mathrm{min}}$ is the $G$-saturation of $\mu^{-1}(0)$. $\endgroup$
    – Francois Ziegler
    Commented Oct 31, 2019 at 21:33

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