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.
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1$\begingroup$ Are you familiar with the fact that $F_2$ admits such an embedding, and also with the Tits alternative? $\endgroup$– Qiaochu YuanOct 7, 2020 at 0:43
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$\begingroup$ 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. $\endgroup$– muchadoaboutNoetherOct 7, 2020 at 1:55
1 Answer
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.
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1$\begingroup$ 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. $\endgroup$ Oct 7, 2020 at 7:41
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1$\begingroup$ 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/… $\endgroup$ Oct 7, 2020 at 19:19
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1$\begingroup$ 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). $\endgroup$– YCorOct 9, 2020 at 22:40
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$\begingroup$ @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. $\endgroup$ Oct 9, 2020 at 22:47
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