Let $G$ be a connected solvable Lie group and let $H$ denote ist commutator subgroup. By definition, every element $g \in H$ can be written as a product of commutators and the minimal number of commutators needed to write $g$ is called the commutator length and denoted ${\rm cl}(g)$.

Since $G$ is amenable, it follows from Bavard duality that the stable commutator length satisfies $${\rm scl}(g) := \lim_{n \to \infty} \frac{{\rm cl}(g^n)}{n}$$ for all $g \in H$. In other words, the commutator length growths at most sublinearly along powers of a fixed element of $H$.

I was wondering what can be said about the actual values of the function ${\rm cl}$. For example, for which $G$ is $cl$ a bounded function on $H$?

  • 2
    $\begingroup$ I think it's always bounded. I'll try to write down a proof at some point if I have time (unless somebody provides a reference). $\endgroup$ – YCor Jan 16 at 1:22
  • $\begingroup$ Thanks for your comment. A proof would still be very much appreciated, but even a few lines on how to approach this problem in principle would be useful. $\endgroup$ – Lyonel Jan 21 at 16:22

Proposition: for every connected Lie group $G$, the commutator length is bounded on the subgroup $[G,G]$.

(1) To start with, for arbitrary groups, the property P that the derived subgroup has bounded commutator length, passes to group quotients. Indeed, say that $G$ satisfies P$_n$, $n\ge 1$, if every product of $n$ commutators is a product of $n-1$ commutators (e.g., P$_1$ means abelian). Then P$_n$ implies P$_{n+1}$, and $G$ satisfies P iff it satisfies P$_n$ for some $n$. Clearly P$_n$ passes to quotients.

(2) Let $G$ be a group and $N$ a normal subgroup such that every element of $N$ is products of boundedly many commutators (in $G$). Then $G$ satisfies P iff $G/N$ satisfies P.

(3) Let $G$ be a simply connected nilpotent Lie group. Then $G$ satisfies P. Indeed, by (2) and induction, it is enough to show that every element of the last term (say $d$-th) $Z$ of the lower central series, has bounded commutator length. Indeed, the case $G$ abelian is void, and if $G$ has nilpotency class $d\ge 2$, the $d$-commutator function $(G/[G,G])^d\to Z$ is multilinear and its image generates linearly $Z$; then it follows that every element of $Z$ is a product of $\dim(Z)$ elements of its image, which are commutators.

I claim that every connected Lie group $G$ satisfies P. By (1), it is enough to assume that $G$ is simply connected. Let $N$ be its derived subgroup; it is closed and nilpotent. By (3), elements of $[N,N]$ have bounded commutator length. Hence by (2) we can suppose that $N$ is abelian.

If the $G/N$-action on $N$ is unipotent, since $G/N$ is abelian it follows that $G$ is nilpotent and we are done by (3). Otherwise, $N$ has an irreducible $G/N$-submodule $V$ that is not unipotent; fix $g\in G/N$ that does not act unipotently on $V$. By irreducibility, 1 is not eigenvalue of $g$ on $V$. Hence $v\mapsto gv-v=[g,v]$ is surjective. Hence $V$ consists of commutators. We conclude by (2). The proof is finished.

(Note: for $G$ not simply connected, $[G,G]$ need not be closed. But this is not a problem and the proposition applies.)


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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