## About the strength of representation-theoretic obstructions for orbit closure problems

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-

I do not fully grasp the role of the condition "$x\in X$ is characterized by its stabilizer", but I want to point out that it fails for any linear action of $G$ on a non-zero vector space $V$, because the stabilizer of any non-zero vector fixes all its scalar multiples. This difficulty persists for all $G$-invariant cones in $V$. Of course, it could be avoided by passing to the projectivization $\mathbb{P}(V).$    – Victor Protsak Sep 4 at 21:10
Inclusions of the closures of nilpotent orbits have been studied in various contexts. If my memory serves, for quiver representations, Bobinski and Zwara proved a complete representation-theoretic characterization of the inclusion of the closures of nilpotent orbits, but it involved finer invariants than just the G-multiplicities. – Victor Protsak Sep 4 at 21:21
Concerning your first comment, yes, this is usually dealt with by passing to the projectivization, or by defining "characterized by its stabilizer" different, i.e. requiring $X^{G_x}=\mathbb{C}x$.   I have added a paragraph to somewhat explain the origin behind this condition. Thanks also for the reference, I will look into it – Jesko Hüttenhain Sep 5 at 8:04
With the condition   $X^{G_x}=\mathbb{C}x$  imposed on $X$  my counterexample below works perfectly for  $g=sl_2(\mathbb{C})$   and the adjoint action of  $G=SL_2(\mathbb{C})$. In this case $l=1$ and  $f_1$ is just the determinant of a $2\times 2$ matrix.   One cannot pass to projectivisations either as the regular functons on the orbit closures would then reduce to scalars (for  $G$ connected) and all multiplicities would be zero. – Alexander Premet Sep 5 at 11:45
So you mean $X=\mathfrak{sl}_2(\C)$ with $G=\mathrm{SL}_2(\C)$ acting on $X$ by conjugation (i.e. $g.x = gxg^{-1}$) and the points are the semisimple element $h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and the nilpotent $x=\begin{pmatrix}0&0\\1&0\end{pmatrix}$. Is that correct? I am confused because my calculations yield $\overline{G.x}=X$, while $G.h=\C h$. – Jesko Hüttenhain Sep 5 at 15:23

## 1 Answer

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 $N_G(G_x)=G_x$).

-

Side remark:  mathoverflow.net/questions/85560/… - "Are nilpotent orbits degenerations of semisimple ones" – Alexander Chervov Sep 4 at 12:17
Thanks a lot already, but I am having some difficulties understanding the counterexample, mostly for lacking terminology: I assume $G$ acts on $g$ by $\mathrm{Ad}$. What is a filtered deformation? Why does it follow from $Gh$ being closed that  $\mathbb C[Gh]$ is a filtered deformation of $\mathbb C[\mathcal N]$? It would probably take a while to clear up all these questions (and their possible follow-ups). Maybe you could, instead, apply your (very general) counterexample to the case $G=Gl_n(\mathbb C)$ and explain what $g$, $\mathcal N$, etc. are, in this case?  – Jesko Hüttenhain Sep 4 at 15:45
It is quite problematic that I do not know the definition of a filtered deformation. I really think that making this more concrete would help immensely. – Jesko Hüttenhain Sep 4 at 17:46
I confess that I do not fully grasp the role of the condition "$x$ is characterized by its stabilizer", but it's not fulfilled in the present case. The stabilizer of a regular semisimple element $h$ is its centralizer $H=Z(h)$, which is a Cartan subgroup, so it is commutative and stabilizes each element of $H.$ Similarly, the stabilizer of a regular nilpotent element $e$ is its centralizer $Z(e)$, which is commutative and stabilizes each of its elements. Thus in both cases, $X^{G_x}$ has dimension $\ell.$   – Victor Protsak Sep 4 at 20:58
Hm. In this case, one can construct easier counterexamples, see my edit. The crux is for both points to be characterized by their respective stabilizers. – Jesko Hüttenhain Sep 5 at 7:58
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 66, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9359521865844727, "perplexity_flag": "head"}