Is every G-invariant function on a Lie algebra a trace? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:53:23Z http://mathoverflow.net/feeds/question/25439 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25439/is-every-g-invariant-function-on-a-lie-algebra-a-trace Is every G-invariant function on a Lie algebra a trace? Theo Johnson-Freyd 2010-05-21T00:06:41Z 2010-05-21T05:17:05Z <p>I am in the (slow) process of editing my <a href="http://math.berkeley.edu/~theojf/QuantumGroups10.pdf" rel="nofollow">notes on Lie Groups and Quantum Groups (V Serganova, Math 261B, UC Berkeley, Spring 2010</a>. 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):</p> <blockquote> <p><strong>Question:</strong> 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 <code>$f: \mathfrak g \to \mathbb K\,$</code> is <em>$\mathfrak g$-invariant</em> if $\mathfrak g \cdot f = 0$. For example, let $\pi: \mathfrak g \to \mathfrak{gl}(V)$ be any finite-dimensional reprensentation. Then <code>$x \mapsto \operatorname{tr}_V \bigl(\pi(x)^n\bigr)$</code> 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?</p> </blockquote> <p>When $\mathfrak g$ is one of the classical groups $\mathfrak{sl},\mathfrak{so},\mathfrak{sp}$, or the exceptional group <code>$G_2$</code> 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.</p> <p>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:</p> <blockquote> <p><strong>Claim:</strong> 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 <em>nilpotent</em> if <code>$\operatorname{ad}(x) = [x,] \in \mathfrak{gl}(\mathfrak g)$</code> is a nilpotent matrix &mdash; some power of it vanishes.)</p> </blockquote> <p>It's clear that the spectrum of any nilpotent matrix is <code>$\{0\}$</code>, 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 $F_4$.</p> http://mathoverflow.net/questions/25439/is-every-g-invariant-function-on-a-lie-algebra-a-trace/25441#25441 Answer by Evan Jenkins for Is every G-invariant function on a Lie algebra a trace? Evan Jenkins 2010-05-21T00:13:59Z 2010-05-21T00:13:59Z <p>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 <a href="http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf" rel="nofollow">Gaitsgory's notes</a>.</p> http://mathoverflow.net/questions/25439/is-every-g-invariant-function-on-a-lie-algebra-a-trace/25447#25447 Answer by David Speyer for Is every G-invariant function on a Lie algebra a trace? David Speyer 2010-05-21T01:17:09Z 2010-05-21T01:17:09Z <p>Here is a sketch of an alternate proof of the claim; making this rigorous may be harder than the approach you take.</p> <p>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.</p> <p>If $x$ is nilpotent then we can use the $G$-action to move $x$ into the nilradical <code>$\mathfrak{n}_+$</code>. Let <code>$\psi: \mathbb{C}^* \to T$</code> be a one parameter subgrop that paris positively with the positive roots. So, for $x$ in <code>$\mathfrak{n}_+$</code>, we have <code>$\lim_{t \to 0} \psi(t) x=0$</code>.</p> <p>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 $f(x)=0$.</p> http://mathoverflow.net/questions/25439/is-every-g-invariant-function-on-a-lie-algebra-a-trace/25450#25450 Answer by Victor Protsak for Is every G-invariant function on a Lie algebra a trace? Victor Protsak 2010-05-21T02:14:10Z 2010-05-21T05:17:05Z <p>The answer to the general question is "no": </p> <blockquote> 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 <em>linear</em> 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. </blockquote> <p>The answer to the claim is "yes", this is Kostant's theorem from his celebrated paper: </p> <blockquote> 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. </blockquote> <p>Kostant actually proved that the nullcone is the scheme-theoretic complete intersection defined by $rk\;G$ homogeneous positive degree algebra generators of $\mathbb{C}[\mathfrak{g}]^G$ &mdash; 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.</p> http://mathoverflow.net/questions/25439/is-every-g-invariant-function-on-a-lie-algebra-a-trace/25455#25455 Answer by Q.Q.J. for Is every G-invariant function on a Lie algebra a trace? Q.Q.J. 2010-05-21T03:50:17Z 2010-05-21T03:50:17Z <p>If the representation is fixed as the fundamental representation, then in the case of $\mathfrak{so}(2n)$, you need Pfaffians as well as traces. </p>