Does every connected Lie group have a dense torsion-free subgroup?

Question

For $ SO_2(\mathbb{R}) $ it is possible to find a finite set of irrational rotations that generate a torsion-free dense subgroup. For $ SO_3(\mathbb{R}) $ it is also possible to find a finite set of irrational rotations that generate a torsion-free dense subgroup.

Do all connected Lie groups have dense torsion-free subgroups?

Further Context about $ SO_n(\mathbb{R}) $:

Let $ R_\alpha \in SO_2(\mathbb{R}) $ be a rotation by $ \alpha/2\pi $. If $ \alpha $ is irrational, then $ R_\alpha $ generates a dense torsion free subgroup of $ SO_2(\mathbb{R}) $.

Let $ R_{\alpha,z} \in SO_3(\mathbb{R}) $ be a rotation by $ \alpha/2 \pi $ about the $ z $ axis. Let $ R_{\beta,x} \in SO_3(\mathbb{R}) $ be a rotation by $ \beta/2\pi $ about the $ x $ axis. If $ \alpha, \beta $ are algebraically independent then the two rotations $ R_{\alpha,z}$ and $ R_{\beta,x} $ generate a torsion-free dense subgroup of $ SO_3(\mathbb{R}) $.

It is my guess more generally that: Let $ R_{\alpha_i} \in SO_n(\mathbb{R}) $ for $ i=1, \dots, n-1 $ be rotations where each $ R_{\alpha_i} $ is a block diagonal matrix consisting of all $ 1 $s on the diagonal except a $ 2 \times 2 $ rotation matrix by an angle of $ \alpha_i/2\pi $ between the standard basis vectors $ e_i, e_{i+1} $. If $ \alpha_1, \dots, \alpha_{n-1} $ are algebraically independent then $ \langle R_{\alpha_1}, \dots,R_{\alpha_{n-1}} \rangle $ is a dense torsion-free subgroup of $ SO_n(\mathbb{R}) $.

Is it generally true that every connected Lie group has a dense torsion free subgroup?

It is worth noting that every simply connected solvable Lie group is contractible and thus torsion free. So all simply connected solvable Lie groups trivially have a dense torsion free subgroup. $ \widetilde{SL}_2(\mathbb{R}) $ is a contractible, and thus torsion-free, Lie group. So $ \widetilde{SL}_2(\mathbb{R}) $ also trivially has a dense torsion-free subgroup.

Edit: This question was cross-posted to MO

Answer 1

Yes, every connected Lie group $G$ has a dense torsion-free subgroup.

You can construct this inductively. The key is the following.

Lemma. Let $H \subset G$ be a countable torsion-free subgroup that is not dense ($\overline H \neq G$). Then there exists $g \in G - \overline H$ such that $\langle H, g \rangle$ is torsion-free. In fact, the set of such $g$ is dense in $G$.

To prove the claim using the lemma, start with $H_0 = \{1\}$ and keep adding elements that lie closer and closer to the identity to construct torsion-free subgroups $H_1, H_2, \ldots$. Then either the dimension of $\overline H$ will jump by at least 1 when going from $H_n$ to $H_{n+1}$, or this happens in countably many steps. Either way after reaching at most $\dim(G)$ countable limit ordinals you should have $\overline{H_n} = G$.

Proof of lemma. We use the Baire Category theorem. For that, we show that the following countable many closed sets have empty interior: The set $\overline H$, and for all integers $n,k \geq 1$, all $h_1, \ldots, h_k \in H$ and every word $w : G^{k+1} \to G$, the set $$S = \{g \in G : w(g, h_1,\ldots, h_k)^n = 1 \text{ and } w(g, h_1,\ldots, h_k)\neq 1\} \,.$$ By BCT there then exists an element $g \in G$ not in these sets and such $g$ works.

Clearly $\overline H$ has empty interior because it is a Lie group of lower dimension.

Note that for every $n$, the identity is an isolated point in the set of $n$-torsion elements. When $G$ is linear, this is because torsion elements are semisimple and have eigenvalues equal to $n$th roots of unity. When $G$ is not linear, take a discrete central subgroup $\Gamma \subset G$ such that $G/\Gamma$ is linear, and then it follows similarly. This implies that the set $S$ is indeed closed.

Now if $S$ had nonempty interior, then $w(G,h_1, \ldots, h_k)^n = 1$ by real analyticity of $w$. But $w(1, h_1, \ldots, h_k) = 1$ because $H$ has no torsion, and by connectedness of $G$ and the fact that $1$ is isolated in $n$-torsion, it follows that $w(G,h_1, \ldots, h_k) = 1$, a contradiction to the assumption that $S$ is nonempty. $\square$