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 $X$?



Is the set of points with smallest stabilizer open?
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.


