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Let $X$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristics $0$. Let $G$ be a connected, linearly reductive, affine algebraic group acting regularly on $X$. Let $L$ be $G$-linearised invertible sheaf on $X$. Now, consider the set of semi-stable points $X^{ss}(L)$. Assume that there is a $G$-invariant, non-empty open set $U\subseteq X^{ss}(L)$ such that the geometric quotient of $U$ exists.

Is it true that $U\subseteq X^{s}(L)$? Or, there is some $G$-invariant open subset $U'\subseteq U$ such that $U'\subseteq X^{s}(L)$?

Note that in the definition of "stable" w.r.t. $L$, Mumford doesn't assume the stabilizer is of dimension $0$. (see page 36 of Mumford's book on GIT)

I'm actually interested in the setting of Quiver representation, related to King's stability conditions.

I feel like the answer would be yes, but having trouble trying to prove it. I'm trying to use the following result from Mumford's book.

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    $\begingroup$ No that is not true. Let $k$ be a field. Let $n\geq 2$ be an integer. Let $G$ be $\mathbb{G}_m = \text{Spec}\ k[t,t^{-1}]$. Let $X$ be affine space $\mathbb{A}^n_k=\text{Spec}\ k[x_1,\dots,x_n]$. Let the action of $G$ on $X$ be the one pulling back each $x_i$ to $tx_i$ on $G\times X$. Let $L$ be the structure sheaf with the linearization for which the global section $1$ is $G$-invariant. Then $X^{ss}(L)$ equals all of $X$, i.e., the distinguished open affine of the invariant section $1$. The GIT quotient is $\text{Spec}(k)$. So $U=X\setminus\{0\}$ is a counterexample. $\endgroup$
    – Jason Starr
    Commented Nov 10, 2022 at 10:30
  • $\begingroup$ Thanks for your answer! $\endgroup$
    – It'sMe
    Commented Nov 10, 2022 at 14:33

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