8
$\begingroup$

"Examples first:"

Consider so(3,C). (Co)Adjoint Orbits can be described by equations x^2+y^2+z^2 = R.

R=0 - is nilpotent cone - algebraic closure of the orbit of nilpotent element. (It is union of two orbits - {0} and {Cone w/o {0}} ).

R$\ne$ 0 are orbits of semi-simple elements. So we have degeneration R->0 - semi-simple orbit degenerates to nilpotent.

Question Is there similar description for the other nilpotent orbits in higher dimensions e.g. for gl(n,c) ? I mean can we write some equations depending on parameters F_t(g)=0, such that for general "t" we get semi-simple orbits, but for specific values we have nilpotent orbit (more precisely their closures)? (Here "t" can be vector and F is vector-valued algebraic function).

Of course this can be done the biggest orbit - for nilpotent cone itself.

Consider matrices "M" which satisfy the condition, that their characterestic polynom is fixed with values eigs $a_i$:

$det(M-x) = (x-a_1)(x-a_2)...(x-a_n)$

For $a_i$ generic - this is semisimple orbit, but if $a_i = 0$ we get nilpotent cone.

Question Reformulated Is it possible to do the same for smaller dimensional orbits ?


As far as I heard nilpotent orbits can be described by the equations on their rank and $M^l=0$, however this does not seems to answer the question.


Part of motivation for asking is related to the following questions:

On an affine analogue of the fact $\mathbb{C}[\mathfrak{g}]^G$ is a polynomial ring

Primitive ideals of the universal enveloping algebras of affine Lie algebras

In particular if the answer would be YES - then probably we can do the same in the "affine case" so answering the question "What replaces the concept of the nilpotent orbit in that case?"

$\endgroup$

3 Answers 3

8
$\begingroup$

The answer is often "yes". Here is the sketch of how to obtain a nilpotent orbit as a degeneration of semisimple orbits in the $GL_n$ case. Let $d$ be a partition of $n$ with $k$ parts and $\overline{\mathcal{O}}_{d'}$ be the closure of the conjugacy class of nilpotent $n\times n$ matrices with Jordan blocks sizes given by the dual partition $d'.$ Denote by $\mathcal{O}_d(t_1,\ldots, t_k)$ the conjugacy class of the block diagonal matrix with scalar diagonal blocks $t_i I_{d_i}.$ Then

$$\lim_{t\to 0}\ \mathcal{O}_d(t_1,\ldots, t_k)=\overline{\mathcal{O}}_{d'}.$$

This is manifested on the level of defining equations using Oshima's approach from

A quantization of conjugacy classes of matrices. Adv. Math. 196 (2005), no. 1, 124–146.

For a general $\mathfrak{g},$ this amounts to the induction of (zero-dimensional) orbits and to the correspondence between semisimple and regular orbits. In particular, every Richardson nilpotent orbit can be obtained as a degeneration in the same way. However, the defining equations are not known to the same degree of explicitness.

On the other hand, if $\mathfrak{g}$ is simple of type other than "A" then the minimal nilpotent orbit is rigid, meaning that it cannot be deformed within the family of adjoint orbits. Existence of rigid orbits makes quantization of orbits a non-trivial task, since a very natural prescription for quantization of semisimple orbits needs to be supplemented by ad hoc quantizations of rigid orbits (several papers of Joseph addressed this question). Rigid orbits have been completely classified: if my memory serves, the answer is in Collingwood-McGovern.

$\endgroup$
4
$\begingroup$

There was a conjecture of Grothendieck, proved by Brieskorn in 1970, which says that:

Let $\mathfrak{g}$ be a simple Lie algebra over $\mathbb{C}$ of ADE type, and $x\in \mathfrak{g}$ a subregular nilpotent element. Let $S\subset \mathfrak{g}$ be a transversal (in $x$) slice to the $G$-orbit of $x$. Then the restriction $(S,x)\to (\mathfrak{t}/W,0)$ of the adjoint quotient is a semi-universal deformation of the corresponding simple ADE singularity.

For other types the situation is a bit trickier, but is discussed in Slodowy, Four lectures on simple groups and singularities.

Also, Lehn, Namikawa and Sorger

http://lanl.arxiv.org/abs/1002.4107

give a classification of the nilpotent orbits for which the restriction of the adjoint quotient to the Slodowy (Kostant) slice gives a universal Poisson deformation of the central fibre.

I may edit this later, as I am not sure if this is the kind of statement you are looking for.

$\endgroup$
1
1
$\begingroup$

The answer to the question in the header is probably "no", but it's hard for me to interpret precisely what is being asked. Classically, the methods of Jacobson and Morozov, Kostant, Slodowy, and others have related nilpotent elements to semisimple elements in subtle ways. There the basic viewpoint is that semisimple elements are easier to study than nilpotent ones, even though the latter form only finitely many orbits under the adjoint group action. But what is being asked here needs a more careful formulation, starting with the notion of "degeneration" of an orbit into another orbit. Should this take a given orbit to one of smaller dimension? (In that case, the regular nilpotent orbit couldn't be the result of such a degeneration.) In general, how does "degeneration" affect the dimension you start with? More detailed examples might clarify the issue better, starting with the 8-dimensional simple Lie algebra $\mathfrak{sl}_3(\mathbb{C})$ amd perhaps others of rank 2, where the nilpotent orbits are easy to describe.

$\endgroup$
4
  • $\begingroup$ @Jim Thank you very much for your answer. Of course I want degeneration preserving dimensions of orbits. However, indeed, thinking of what I asked, I find it might not be good question, since in some sense I can (can I?) achieve degeneration in rather trivial way - just take a curve s(t) in "g" , such that s(t), for t=0 we get some nilpotent element, and for $t\ne 0$ we get semi-simples, SUCH THAT their orbits have the same dimension for all t. I think I can do this, am I right ? (My question required may be more subtle thing, but may be it is not necessary). $\endgroup$ Jan 14, 2012 at 16:13
  • $\begingroup$ @Jim Let me try to describe the answer to the question in situation of rank 2 orbits. As you suggest. Consider gl(3)=Mat(3,3)=C^9. Let us write a system of 9 equations on matrix A: A^2=t^2Id. We see that for t=0, we get a closure of rank 2 nilpotent orbit. And for $t\ne 0$ we get equations defining semisimple orbits (more precisely uninion of several orbits, see below)... $\endgroup$ Jan 14, 2012 at 16:40
  • $\begingroup$ Let us look more carefully what happens for t≠0. Equations $A^2=t^2Id$ easy to solve−they define 4 semisimple orbits diag(t,t,t);diag(−t,−t,−t);diag(t,−t,−t);diag(t,t,−t). Orbits of diag(t,t,t);diag(−t,−t,−t) are just points−notveryinteresting. Orbits diag(t,−t,−t);diag(t,t,−t) have the same dimension as rank2 nilpotent orbit(amIright?). So I almost got what I want−a system of equtions−such that fort=0−closure of rank2 nilp.orbit,and for $t\ne 0$ we have semi-simple orbit(s) of the SAME dimension. Some trouble here is that that in semi-simple case not only one orb $\endgroup$ Jan 14, 2012 at 16:51
  • 2
    $\begingroup$ Alexander, it's not possible to deform e.g. the minimal nilpotent orbit outside of type A while preserving the dimension - it's an interesting calculation for classical Lie algebras. $\endgroup$ Jan 15, 2012 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.