Question

Let $G$ be a reductive, affine, algebraic group over $\newcommand{\C}{\mathbb C}\C$. Let $X$ be a $G$-variety. For $x\in X$, we write $$G_x:=\{ g\in G\mid g.x=x\}$$ for its stabilizer and for any subgroup $H\subseteq G$, we write $$X^H:=\{x\in X\mid H.x=x\}$$ for the $H$-invariants of $X$. We say that $x\in X$ is characterized by its stabilizer if $X^{G_x}=\{x\}$. Let $\{V_\lambda\mid \lambda\in\Lambda\}$ be the irreducible $G$-modules.

Given two points $x,y\in X$, then $x\in\overline{G.y}$ implies $\overline{G.x}\subseteq\overline{G.y}$. Hence, $\C[\overline{G.y}]\twoheadrightarrow\C[\overline{G.x}]$ and thus,

$$\DeclareMathOperator{\mult}{mult}\forall \lambda\in\Lambda:\quad \mult\nolimits_\lambda(\C[\overline{G.x}])\le\mult\nolimits_\lambda(\C[\overline{G.y}])$$

Finding $\lambda\in\Lambda$ violating the above is therefore an "obstruction" for the inclusion of orbit closures.

My question now is the following: If $x$ and $y$ are characterized by their respective stabilizers, does the converse hold? I.e., does the above inequality imply that $x\in\overline{G.y}$? I have been trying to come up with a counterexample, but without success so far.

Intuition: If $G$ acts on a variety $Y$ and $y\in Y$ is characterized by its stabilizer, then you can very easily find counterexamples if you give up the condition that both points are characterized by their respective stabilizers: Consider $X:=Y\times\{z_1,z_2\}$ with $G$ acting trivially on $Z=\{z_1,z_2\}$. Now, the points $x_i:=(y,z_i)$ satisfy $x_1\notin\overline{G.x_2}$ and $\C[\overline{G.x_1}]\cong\C[\overline{G.x_2}]$. In the cases of interest to me, however, both points are characterized by their stabilizer and the question arises whether there are counterexamples under this additional condition.

Answer 1

Take the nilpotent cone $\mathcal N$ in $g={\rm Lie}(G)$ and the $G$-orbit of a regular semisimple element, $h$ say. The categorical quotient $g\rightarrow g//G\cong\mathbb{A}^l$, $l={\rm rk}(G),$ is equidimensional and each of its fibres is an trreducible complete intersection and contains a unique open $G$-orbit. Since the orbit $Gh$ is closed it coincides with one of the fibres and the algebra of regular functions $\mathbb{C}[Gh]$ is just a filtered deformation of the graded algebra $\mathbb{C}[\mathcal{N}]=\mathbb{C}[g]/(f_1,\ldots,f_l)$ whose defining ideal is generated by $f_1-\lambda_1,\ldots, f_l-\lambda_l$ for some $\lambda_i\in\mathbb{C}$ (here $f_1,\ldots, f_l$is a set of free homogeneous generators for $\mathbb{C}[g]^G$). Since $G$ is reductive, we are in characteristic $0$ and the action of $G$ on $\mathbb{C}[Gh]$ is rational, we have that $\mathbb{C}[\mathcal{N}]\cong \mathbb{C}[Gh]$ as $G$-modules. So all multiplicities will be the same in both cases. However, $Gh$ is not contained in $\mathcal N$ (and vice versa).

However this example does not answer the question as the stabilisers $G_x$ of regular elements $x\in g$ are not self-normalising (I have completely overlooked the extra condition on $x$ in the first reading, which implies that $N_G(G_x)=G_x$).