4
$\begingroup$

Let ${\mathbb g}$ be a simple complex finite dimensional Lie algebra, $X\subseteq{\mathbb g}$ a nilpotent orbit. Did anyone study maximal vector subspaces of the closure $\overline{X}$?

In particular, I'd like to know the maximal dimension of a vector subspace $V\subseteq\overline{X}$. Any references?

PS It seems clear what it is in type $A$ but I am more interested in exceptional types.

PPS If $X$ is even, there is a reasonable candidate for a maximal $V$: the positive part of the associated grading. It is clearly a maximal Lie subalgebra in $\overline{X}$ but I am not sure why it must a maximal vector subspace.

Improve this question
$\endgroup$
3
  • $\begingroup$ @PeterMichor they are "cones" but in the complex sense, so closed under multiplication by any complex number. Think about Jordan normal form; for any classical group, that determines the nilpotent. $\endgroup$
    – Ben Webster ♦
    Commented May 17, 2015 at 10:50
  • $\begingroup$ What does "$X$ is even" mean? It's always even-dimensional, so I presume that's not what you mean. $\endgroup$
    – Allen Knutson
    Commented May 18, 2015 at 11:17
  • 1
    $\begingroup$ @AllenKnutson If $(x,h,y)$ is an $SL_2$-triple where $x$ is the given nilpotent orbit, then $h$ defines a grading on the Lie algebra. The orbit is even if the grading is even. One can chose a dominant $h$, and then the weights of simple roots are in $\{0,1,2\}$, this gives a labelling of the Dynkin diagram which is characteristic of the orbit. Even means there are no 1's. $\endgroup$
    – Daniel Juteau
    Commented May 18, 2015 at 16:28

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .