# What finitely-generated amenable groups arise as subgroups of compact Lie groups?

I am looking for examples of (edit: amenable) finitely-generated subgroups of any compact Lie group which are infinite and not virtually abelian. An example with polynomial growth would be especially nice.

• Are you familiar with the fact that $F_2$ admits such an embedding, and also with the Tits alternative? Oct 7, 2020 at 0:43
• I was not so much, but I'm looking that up now, thanks! I know of the Tits alternative, so I guess that's helpful here because it says any such example is either something like an embedding of a free group or is virtually solvable. I probably should have said that I'd like something at least amenable for my example, so I want something in the latter category that isn't virtually Z^n. Oct 7, 2020 at 1:55

There are no such groups.

By the Tits alternative a finitely generated amenable subgroup of a compact Lie group (which embeds into some $$U(n)$$ and hence into some $$GL_n(\mathbb{C})$$ by the Peter-Weyl theorem) is virtually solvable. (And by Gromov's theorem if we require polynomial growth then it must even be virtually nilpotent, but it turns out we won't need this.) Let $$F$$ be its solvable subgroup of finite index. The closure $$\overline{F}$$ is then a compact solvable Lie group; in particular it has finitely many connected components, and the connected component $$G = \overline{F}_0$$ is a compact connected solvable Lie group.

Proposition: A compact connected solvable Lie group $$G$$ is abelian (hence a torus).

Proof. Consider the adjoint representation of $$G$$ on $$\mathfrak{g}$$. By Lie's theorem $$\mathfrak{g}/Z(\mathfrak{g})$$ is upper triangular in $$\mathfrak{gl}(\mathfrak{g})$$ with respect to some basis of $$\mathfrak{g}$$, so exponentiating, $$G/Z(G)$$ is also upper triangular. The maximal compact subgroup of the upper triangular matrices $$U_n(\mathbb{C})$$ is given by the diagonal subgroup $$U(1)^n$$, so $$G/Z(G)$$ is abelian, hence $$G$$ is nilpotent.

Now by Engel's theorem $$\mathfrak{g}/Z(\mathfrak{g})$$ is strictly upper triangular in $$\mathfrak{gl}(\mathfrak{g})$$ with respect to some basis, so exponentiating, $$G/Z(G)$$ is upper triangular with $$1$$s on the diagonal. The maximal compact subgroup of this group is trivial, so $$G = Z(G)$$. $$\Box$$

Corollary: A solvable subgroup of a compact Lie group is virtually abelian. Hence a finitely generated amenable (or polynomial growth) subgroup of a compact Lie group is virtually abelian.

• If you drop amenability and polynomial growth there are interesting examples like the free groups or free products $C_n \ast C_m$ but they will, by the Tits alternative, all contain $F_2$. I don't know an example off the top of my head that isn't virtually free. Oct 7, 2020 at 7:41
• Also, it's a bit less obvious than I thought that the closure of a solvable subgroup remains solvable, but it is true and you can see a proof here: math.stackexchange.com/questions/86128/… Oct 7, 2020 at 19:19
• It's quite obvious if one has in mind that $G$ is solvable iff for some $n$, every $n$-commutator is trivial (where $n$-commutator means element of the form $c_n(x)$ for some $2^n$-tuple $x$, and $c_n(x)=[c_{n-1}(y),c_{n-1}(z)]$ for $x=yz$, $y,z$ $2^{n-1}$-tuples).
– YCor
Oct 9, 2020 at 22:40
• @YCor: yes, but it takes a tiny argument I haven't seen anywhere to argue that this is equivalent to the usual definition in terms of iterated commutator subgroups. Oct 9, 2020 at 22:47
• Yes, that's correct.
– YCor
Oct 9, 2020 at 22:48