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

Question

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.

Answer 1

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.