Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am in the (slow) process of editing my notes on Lie Groups and Quantum Groups (V Serganova, Math 261B, UC Berkeley, Spring 2010. Mostly I can fill in gaps to arguments, but I have found myself completely stuck in one step of one proof. One possibility that would get me unstuck is a positive answer to the following (which may be obviously false or trivial, but I'm not thinking well):

Question: Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb K$, and if necessary you may assume that $\mathbb K = \mathbb C$ and that $\mathfrak g$ is semisimple. Then $\mathfrak g$ acts on itself by the adjoint action, and on polynomial functions $f : \mathfrak g \to \mathbb K$ via derivations. A polynomial $f: \mathfrak g \to \mathbb K\,$ is $\mathfrak g$-invariant if $\mathfrak g \cdot f = 0$. For example, let $\pi: \mathfrak g \to \mathfrak{gl}(V)$ be any finite-dimensional reprensentation. Then $x \mapsto \operatorname{tr}_V \bigl(\pi(x)^n\bigr)$ is $\mathfrak g$-invariant for any $n\in \mathbb N$. Is every $\mathfrak g$-invariant function of this form? Or at least a sum of products of functions on this form?

When $\mathfrak g$ is one of the classical groups $\mathfrak{sl},\mathfrak{so},\mathfrak{sp}$, or the exceptional group $G_2$ the answer is yes, because we did those examples in the aforementioned class notes. But I have no good grasp for the $E$ series, and I don't know if the statement holds for non-semisimples.

What I'm actually trying to prove is a weaker statement, but I figured I'd ask the stronger question, because to me the answer is not obviously "no". The weaker statement:

Claim: Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra over $\mathbb C$. Then every $\mathfrak g$-invariant function is constant on nilpotent elements of $\mathfrak g$. (Recall that $x\in \mathfrak g$ is nilpotent if $\operatorname{ad}(x) = [x,] \in \mathfrak{gl}(\mathfrak g)$ is a nilpotent matrix — some power of it vanishes.)

It's clear that the spectrum of any nilpotent matrix is $\{0\}$, and for a semisimple Lie algebra, any nilpotent element acts nilpotently in all representations. For the classical groups, in the notes we exhibited generators for the rings of $\mathfrak g$-invariant functions as traces of representations, and so we can just check the above claim. But we did not do the $E$ series or $F_4$.

share|improve this question

4 Answers 4

up vote 4 down vote accepted

The answer to the general question is "no":

If $\mathfrak{g}$ is solvable, by Lie's theorem its commutant $\mathfrak{g}^{\prime}=[\mathfrak{g},\mathfrak{g}]$ is represented by strictly upper triangular matrices in a suitable basis in any finite-dimensional module. Hence all "trace generated" polynomials are zero on $\mathfrak{g}^{\prime}$; in other words, they factor through the abelianization $\mathfrak{g}/\mathfrak{g}^{\prime}$ and are generated by linear invariant polynomials. Unless the adjoint action of $G$ with Lie algebra $\mathfrak{g}$ on $\mathfrak{g}^{\prime}$ has a Zariski dense orbit, there are invariant polynomials that cannot be obtained in this way.

The answer to the claim is "yes", this is Kostant's theorem from his celebrated paper:

If $G$ is a complex semisimple group then its nullcone $\mathcal{N}\subset\mathfrak{g}$ is the Zariski closure of a single adjoint orbit consisting of regular nilpotent elements.

Kostant actually proved that the nullcone is the scheme-theoretic complete intersection defined by $rk\;G$ homogeneous positive degree algebra generators of $\mathbb{C}[\mathfrak{g}]^G$ — this is the connection with the Chevalley theorem mentioned by others. But for the present purpose, it is enough to show that regular nilpotents are Zariski open and dense in $\mathcal{N}\cap\mathfrak{n},$ and a good way of doing it was indicated by David Speyer.

share|improve this answer
Great! The section I'm editing is working towards a proof of Kostant's theorem. –  Theo Johnson-Freyd May 21 '10 at 2:38
Kostant's theorem is easily the single most important theorem about semisimple Lie groups/Lie algebras that I didn't properly learn when I first learned the subject. It wasn't in Serre's book or "Seminaire Sophus Lie". I am glad to see that the situation has changed drastically (e.g. Chris and Ginzburg feature a proof). –  Victor Protsak May 21 '10 at 2:52
The nullcone in a semisimple Lie algebra definitely gets too little attention in the structure and classification theory. The problem is partly that its importance doesn't emerge clearly until you get further into representation theory and related algebraic geometry. (And Kostant's early papers were dauntingly long though full of ideas, so they didn't translate easily into textbook treatments.) For a follow-up to standard Lie theory I'd certainly advocate the study of nilpotent orbits and the geometry of the nullcone, though the subject still needs its own comprehensive book. –  Jim Humphreys May 21 '10 at 22:25

The answer to your question is yes for semisimple Lie algebras. This is essentially the content of the Chevalley restriction theorem. See the proof at the beginning of chapter 2 of Gaitsgory's notes.

share|improve this answer
Chevalley's restriction theorem is an important step in describing the center of the enveloping algebra of a semisimple complex Lie algebra (which eventually gets identified with the polynomial algebra of Weyl group invariants) It is explained in quite a few books and lecture notes including Gaitgory's notes and section 23 of my 1972 text, with references there to other sources. The ideas have since been extended to other settings as well, but the theorem itself still requires work. It's well worth consulting multiple sources. –  Jim Humphreys May 21 '10 at 22:16

If the representation is fixed as the fundamental representation, then in the case of $\mathfrak{so}(2n)$, you need Pfaffians as well as traces.

share|improve this answer

Here is a sketch of an alternate proof of the claim; making this rigorous may be harder than the approach you take.

Let $G$ be the lie group corresponding to the lie algebra $\mathfrak{g}$. So $G$ acts on $\mathfrak{g}$. $G$-invariant functions are, as the name suggests, invariant under this action.

If $x$ is nilpotent then we can use the $G$-action to move $x$ into the nilradical $\mathfrak{n}_+$. Let $\psi: \mathbb{C}^* \to T$ be a one parameter subgrop that paris positively with the positive roots. So, for $x$ in $\mathfrak{n}_+$, we have $\lim_{t \to 0} \psi(t) x=0$.

So $0$ is in the closure of $Gx$ and, if $f$ is $G$ invariant, we must have $f(x)=f(0)$. In particular, if $f$ is $G$-invariant and has positive degree, then $f(x)=0$.

share|improve this answer
The one-parameter group construction is a special case of the Hilbert-Mumford criterion: Zariski closure of Gx contains 0 ${\ }\iff$ there is a one-parameter subgroup $T\subset G$ such that Zariski closure of Tx contains 0. –  Victor Protsak May 21 '10 at 1:43
This is essentially the approach I was planning to take. Thanks! –  Theo Johnson-Freyd May 21 '10 at 2:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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