Let $G$ be a connected, simply-connected complex semisimple Lie group with Lie algebra $\frak{g}$. Fix a maximal torus $T\subseteq G$, and let $$\frak{g}=\frak{t}\oplus\bigoplus_{\alpha\in\Delta}\frak{g}_{\alpha}$$ be the corresponding decomposition into weight spaces. Given a non-zero root vector $e_{\alpha}\in\frak{g}_{\alpha}$, is there a nice way to describe those root vectors lying in the nilpotent $G$-orbit of $e_{\alpha}$? This is possible in the context of several examples, but is there a more example-independent description?
$\begingroup$
$\endgroup$
1
asked Jul 5, 2013 at 23:02
-
$\begingroup$ To study nilpotent orbits in a semisimple Lie algebra it's best to focus on the individual simple summands. Then your orbit for any long root is the unique minimal nonzero one in the finite partially ordered set of nilpotent orbits. A standard modern reference is section 4.3 in Collingwood-McGovern Nilpotent Orbits in Semisimple Lie Algebras. (Note you can work with any simple algebraic group and its Lie algebra over an alg. closed field of good characteristic.) For short roots you get the unique minimal special orbit: cf. mathoverflow.net/questions/67143 $\endgroup$– Jim HumphreysCommented Jul 8, 2013 at 17:08
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
If $\alpha$ is a long root (e.g. if $\mathfrak g$ is simply-laced), then $e_\alpha$ is in the $G$-orbit of the high weight vectors. Projectively, its orbit looks like $G/P_\Theta$ where $P_\theta$ is the parabolic using those simple roots that are perpendicular to the highest root.
But you asked for the root vectors, not all the vectors. Then the answer is that you get all the root vectors for roots of the same length as $\alpha$.
answered Jul 7, 2013 at 4:32