About the strength of representation-theoretic obstructions for orbit closure problems - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T09:40:43Z http://mathoverflow.net/feeds/question/106256 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106256/about-the-strength-of-representation-theoretic-obstructions-for-orbit-closure-pro About the strength of representation-theoretic obstructions for orbit closure problems Jesko Hüttenhain 2012-09-03T16:18:42Z 2012-09-05T09:49:47Z <p>Let <code>$G$</code> be a reductive, affine, algebraic group over <code>$\newcommand{\C}{\mathbb C}\C$</code>. Let <code>$X$</code> be a <code>$G$</code>-variety. For <code>$x\in X$</code>, we write $$G_x:=\{ g\in G\mid g.x=x\}$$ for its stabilizer and for any subgroup <code>$H\subseteq G$</code>, we write $$X^H:=\{x\in X\mid H.x=x\}$$ for the <code>$H$</code>-invariants of <code>$X$</code>. We say that <code>$x\in X$</code> is characterized by its stabilizer if <code>$X^{G_x}=\{x\}$</code>. Let <code>$\{V_\lambda\mid \lambda\in\Lambda\}$</code> be the irreducible <code>$G$</code>-modules.</p> <p>Given two points <code>$x,y\in X$</code>, then <code>$x\in\overline{G.y}$</code> implies <code>$\overline{G.x}\subseteq\overline{G.y}$</code>. Hence, <code>$\C[\overline{G.y}]\twoheadrightarrow\C[\overline{G.x}]$</code> and thus,</p> <p>$$\DeclareMathOperator{\mult}{mult}\forall \lambda\in\Lambda:\quad \mult\nolimits_\lambda(\C[\overline{G.x}])\le\mult\nolimits_\lambda(\C[\overline{G.y}])$$</p> <p>Finding <code>$\lambda\in\Lambda$</code> violating the above is therefore an "obstruction" for the inclusion of orbit closures. </p> <p>My question now is the following: If <code>$x$</code> and <code>$y$</code> are characterized by their respective stabilizers, does the converse hold? I.e., does the above inequality imply that <code>$x\in\overline{G.y}$</code>? I have been trying to come up with a counterexample, but without success so far.</p> <p><b>Intuition:</b> If <code>$G$</code> acts on a variety <code>$Y$</code> and <code>$y\in Y$</code> 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 <code>$X:=Y\times\{z_1,z_2\}$</code> with <code>$G$</code> acting trivially on <code>$Z=\{z_1,z_2\}$</code>. Now, the points <code>$x_i:=(y,z_i)$</code> satisfy <code>$x_1\notin\overline{G.x_2}$</code> and <code>$\C[\overline{G.x_1}]\cong\C[\overline{G.x_2}]$</code>. In the cases of interest to me, however, <em>both</em> points are characterized by their stabilizer and the question arises whether there are counterexamples under this additional condition. </p> http://mathoverflow.net/questions/106256/about-the-strength-of-representation-theoretic-obstructions-for-orbit-closure-pro/106307#106307 Answer by Alexander Premet for About the strength of representation-theoretic obstructions for orbit closure problems Alexander Premet 2012-09-04T07:37:43Z 2012-09-05T09:32:07Z <p>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).</p> <p>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 $N_G(G_x)=G_x$).</p>