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.




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cross-posted from MSE <https://math.stackexchange.com/questions/4721843/does-every-connected-lie-group-have-a-dense-torsion-free-subgroup>



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


