I am in the (slow) process of editing my [notes on Lie Groups and Quantum Groups (V Serganova, Math 261B, UC Berkeley, Spring 2010](http://math.berkeley.edu/~theojf/QuantumGroups10.pdf). 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 $F\_4$.



Is every G-invariant function on a Lie algebra a trace?
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.


