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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.

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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 '12 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 '12 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 '12 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 '12 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 '12 at 15:23
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1 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 $N_G(G_x)=G_x$).

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Side remark: mathoverflow.net/questions/85560/… - "Are nilpotent orbits degenerations of semisimple ones" –  Alexander Chervov Sep 4 '12 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 '12 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 '12 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 '12 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 '12 at 7:58
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