# Commutator length in connected solvable Lie groups

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 $$H$$?

• 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). – YCor Jan 16 at 1:22
• 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. – 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.)