14
$\begingroup$

Given a simple Lie algebra over an algebraically closed field of good characteristic such as $\mathbb{C}$, its subvariety $\mathcal{N}$ of nilpotent elements has dimension $2N$ (where $N$ is the number of positive roots) and is the union of finitely many orbits under the adjoint group $G$. These orbits are partially ordered by $\mathcal{O'} \leq \mathcal{O}$ iff $\mathcal{O'} \subset \overline{\mathcal{O}}$. Some pictures are available here.

Using the Killing-Cartan classification of simple types $A$ - $G$, the orbits have been well studied. One general fact is that all orbits have even dimension, but for detailed results case-by-case work is usually essential. There is a unique dense regular orbit and a unique orbit just beneath it in the partial ordering: the subregular orbit, of dimension $2N-2$. Similarly, there is a unique minimal nonzero orbit (determined by any root vector for a long root), whose dimension was shown by Weiqiang Wang here to be $2 h^\vee -2$ with $h^\vee$ the dual Coxeter number of the root system: one greater than the sum of coefficients of the highest short coroot. Wang's proof uses standard facts about root systems and avoids the classification.

Meanwhile, Lusztig's refinement of Springer's work on Weyl group representations in the context of a desingularization of $\mathcal{N}$ led to a notion of special nilpotent orbit, not readily characterized within the Lie algebra setting. (But it has shown up independently in other active areas of representation theory.) The regular and zero orbits are always special; all Richardson orbits (including the subregular orbit) are special, but there are sometimes other special ones. There is always a unique minimal (nonzero) special orbit; it is determined by any root vector for a short root except in type $G_2$ where it is instead the subregular orbit.

The Lusztig-Spaltenstein duality map on the set of all orbits (generalizing the transpose map for partitions which parametrize orbits in type $A$) has as image the special ones, on which it induces a duality involution which interchanges the regular and zero orbits, etc. It turns out that only in types $A, D, E$ (where all roots have equal length) is the minimal nilpotent orbit special, thus in duality with the subregular orbit. Wang's result recently led me to observe case-by-case:

The minimal special nilpotent orbit has dimension $2h-2$.

Here $h$ is the Coxeter number (order of a Coxeter element in the Weyl group, or one greater than the sum of coefficients of the highest root). The formula seems not to be written down anywhere (?)

Is there a uniform proof of this dimension formula within the Lie algebra framework, avoiding the classification and using as little information as possible from Springer theory or other areas of representation theory?

UPDATE: After observing the dimension formula a week or so ago, I checked around with people close to nilpotent orbits but no one seemed to recognize this from the literature or have a classification-free approach to suggest. Lusztig himself replied briefly that he didn't know any uniform proof, but later thought about it more and pointed out a few hours ago the remarks (b) and (c) at the end of his short 1981 Advances in Mathematics paper "Green polynomials and singularities of unipotent classes" (which I wrote a review of at the time but haven't revisited). So it's clear that he had found a unique orbit in each case of dimension $2h-2$ and had begun to fit these into his work on special orbits and what he later called "special pieces" of the unipotent variety. He was thinking about finite fields of definition, where Kostant had told him a polynomial formula for the number of rational points in the minimal orbit for simply-laced types. Lusztig himself extends the combinatorics to other cases by using the entire special piece. So there are other interesting connections here than what I started with in the parallel study of $\mathcal{N}$. But still no uniform explanation.

$\endgroup$
2
  • 1
    $\begingroup$ It certainly has not escaped anyone's attention that this formula is formally identical to (minus) the Euler characteristic of a genus $h$ surface. Is this just a coincidence? or is there some deeper reason? $\endgroup$ Commented Jun 7, 2011 at 16:03
  • $\begingroup$ Actually, it escaped my attention. It's impossible to know in advance what connections the orbits may have with other matters. Still, one would have to come up with a relevant genus $h$ surface. $\endgroup$ Commented Jun 7, 2011 at 21:25

1 Answer 1

4
$\begingroup$

This isn't an answer to your question, but is a remark which is a bit too long to be a comment. Brylinski and Kostant observed (Nilpotent orbits, normality and Hamiltonian group actions, JAMS 7, 1994) that the minimal special orbit closures in the non-simply-laced cases can be covered by minimal nilpotent orbit closures in simply-laced cases, as follows.

Start with a Lie algebra ${\mathfrak g}$ with simply-laced Dynkin diagram $A_{2n-1}$, $D_n$ or $E_6$, and a graph automorphism (an involution except perhaps in type $D_4$) of ${\mathfrak g}$ with fixed points ${\mathfrak g}_0$. Then there is a ${\mathfrak g}_0$-stable complement to ${\mathfrak g}_0$ in ${\mathfrak g}$, and the ${\mathfrak g}_0$-equivariant projection ${\mathfrak g}\rightarrow{\mathfrak g}_0$ restricts to a finite map $\overline{{\rm O}_{min}({\mathfrak g})}\rightarrow\overline{{\rm O}_{min-sp}({\mathfrak g}_0)}$, which is a quotient by ${\mathfrak S}_2$ or ${\mathfrak S}_3$ and is a universal cover over the minimal special orbit. Note that the Coxeter number of ${\mathfrak g}$ is the same as that of ${\mathfrak g}_0$ (since we exclude the cases $A_{2n}$), so to a certain extent this brings us back to the question you asked (though it certainly isn't the sort of answer you were looking for).

$\endgroup$
1
  • $\begingroup$ Thanks for the extended comment. It adds to my conviction that there are many hidden connections in this subject. But I had some hope at first that one could avoid detailed use of the classification. $\endgroup$ Commented Aug 12, 2014 at 15:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .