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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 $H$?

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    $\begingroup$ 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
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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.)

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