Let $G$ be a complex reductive group acting linearly on a complex affine variety $X$, and let $K$ be the kernel of the action, i.e. $$K:=\{g\in G:g\cdot x=x\text{ for all }x\in X\}.$$ Is $$X_K:=\{x\in X:\mathrm{Stab}_G(x)=K\}$$ Zariski-open in $X$?
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$\begingroup$ You are correct, I am wrong! Sorry about that. I will delete that comment. $\endgroup$– Jason StarrCommented Aug 8, 2017 at 19:49
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$\begingroup$ By the same argument as in the following case, there is a maximal open subscheme of $X$ over which the stabilizers are finite: mathoverflow.net/questions/277874/… However, as you can see from the case of the standard scalar action of $\mathbb{G}_m$ on $\mathbb{A}^n$, that open subscheme need not be affine . . . $\endgroup$– Jason StarrCommented Aug 8, 2017 at 21:18
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$\begingroup$ I keep thinking about this question, but I do not have an answer. If the open subscheme were affine, then Luna's 'etale slice theorem would solve the problem, as in Friedrich Knop's solution of the earlier question. However, I do not see how to reduced to that case. $\endgroup$– Jason StarrCommented Aug 10, 2017 at 13:40
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The answer is affirmative if $G$ is finite or abelian since in that case subgroups are rigid.
Otherwise, $X_K$ may not even be dense, let alone open. The standard example is due to Luna from his slice paper: Let $G=SL(2,\mathbb C)$ act on the space $X=S^3\mathbb C^2$ of binary cubics. The action is effective so $K=1$. Every non-degenerate cubic $f$ is the product of $3$ distinct linear factors. Then $Stab_G(f)$ contains the cyclic permutation of these factors which shows that $X_K$ is not dense. It is not empty either, since $f=x^2y$ has trivial stabilizer.
In my opinion, the question is ill-posed since $X_K$ may be open for the trivial reason that it is empty. Let, e.g., $G$ be absolutely simple (e.g. $G=SO(2n+1,\mathbb C)$) and $H$ a proper reductive subgroup. Then $X=G/H$ is affine with $K=1$ but $Stab_G(x)\ne K$ for all $x\in X$.
The way to go is to look at the stabilizers of only the closed orbits. Their behavior is much more regular which gives rise to the Luna stratification.
answered Aug 11, 2017 at 8:50
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$\begingroup$ That makes me feel a lot better. In that example, the open subset on which $G$ acts properly discontinuously is non-affine: the complement in $S^3\mathbb{C}^2$ of the image of the Veronese map $v_3:\mathbb{C}^2\to S^3\mathbb{C}^2$. It is also not a union of $G$-invariant open affines. The open has precisely two orbits, one of which is open and one of which is (relatively) closed. So the smallest $G$-invariant open that contains the (relatively) closed orbit, i.e., the orbit of points with trivial stabilizer, contains the entire "proper-discontinuous open" (and thus is not affine). $\endgroup$ Commented Aug 11, 2017 at 9:33
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$\begingroup$ @Jason Starr: Probably you mean the open subset on which $G$ acts with finite stabilizers. The action on this set is in general not properly discontinuous. $\endgroup$ Commented Aug 11, 2017 at 13:07
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$\begingroup$ Yes, I meant the open subset on which $G$ acts with finite stabilizers. $\endgroup$ Commented Aug 11, 2017 at 14:16