Question:
Do all connected Lie groups have dense torsion-free subgroups?
Context :
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}) $.
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.
cross-posted from MSE https://math.stackexchange.com/questions/4721843/does-every-connected-lie-group-have-a-dense-torsion-free-subgroup
Best Answer
Let $G$ be a connected Lie group. We know that $G$ has a real analytic structure for which the law is analytic, and we fix it. Also fix a left Haar measure and a smooth positive function of integral 1, thus defining a fully-supported probability $\mu$ on $G$.
Let $F_n$ be the free group on $n$ given generators, and let $R_n(G)$ be the set of words $w\in F_n$ that vanish on $G$ (i.e., $w\in F_n$ such that $w(g_1,\dots,g_n)=1$ for all $g_1,\dots,g_n\in G$). Then $R_n(G)$ is a fully characteristic subgroup of $F_n$ (i.e., stable under all endomorphisms), and in particular is a normal subgroup (this is true for an arbitrary group $G$).
For $w\in F_n$, write $M_w=\{(g_1,\dots,g_n)\in G^n :w(g_1,\dots,g_n)=1\}$. Then $M_w$ is closed in $G^n$, and since the law is real-analytic and $G$ is connected, for $w\notin R_n(G)$, $M_w$ has $\mu$-measure zero. Hence the complement $J_{n,G}$ of $\bigcup_{w\notin R_n(G)}M_w$ in $G^n$ is (G$_\delta$-)dense in $G^n$. For $(g_1,\dots,g_n)\in J_G$, the group generated by $\{g_1,\dots,g_n\}$ is isomorphic to $F_n/R_n(G)$.
[Note: the latter fact is false in the group $\mathrm{O}(2)$, which is not connected, with the word $w(g)=g^2$.]
Next let us check that $F_n/R_n(G)$ is torsion-free. Suppose that $w\in F_n$ and $w^d\in R_n(G)$, $d\ge 1$. The image of the word map $w:G^n\to G$ is connected and contained in the set $T_d(G)=\{g\in G:g^d=1\}$ of $d$-torsion elements in $G$, and contains $1$ (image of $(1,1,\dots,1)$). But in any Lie group, $1$ is isolated in $T_d(G)$. Hence we deduce that $w\in R_n(G)$. This proves torsion-freeness.
It follows that a $\mu$-generic sequence $(g_n)$ generates a torsion-free subgroup. But also, clearly, a $\mu$-generic sequence is dense. Hence a $\mu$-generic sequence generates a dense torsion-free subgroup.
[I think that for $d$ large enough (maybe $d=\dim(G)+1$ always works), a $\mu$-generic $d$-tuple generates a dense subgroup. But this would require more work; I don't immediately see a reference working.]