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 $$X_K:=\{x\in X:\mathrm{Stab}_G(x)=K\}$$ Zariskiopen in $X$?

$\begingroup$ You are correct, I am wrong! Sorry about that. I will delete that comment. $\endgroup$– Jason StarrAug 8, 2017 at 19:49

$\begingroup$ 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 StarrAug 8, 2017 at 21:18

$\begingroup$ 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 StarrAug 10, 2017 at 13:40
1 Answer
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 nondegenerate 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 illposed 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.

$\begingroup$ That makes me feel a lot better. In that example, the open subset on which $G$ acts properly discontinuously is nonaffine: 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 "properdiscontinuous open" (and thus is not affine). $\endgroup$ Aug 11, 2017 at 9:33

$\begingroup$ @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. $\endgroup$ Aug 11, 2017 at 13:07

$\begingroup$ Yes, I meant the open subset on which $G$ acts with finite stabilizers. $\endgroup$ Aug 11, 2017 at 14:16