# Dense subgroup of a compact connected Lie group generated by two words

Let $$G$$ be a compact connected Lie group and $$w_1$$, $$w_2$$ be two positive words in alphabet $$\{a, b\}$$ which are not the powers of some another word $$w$$. Positive means that $$a^{-1}$$ and $$b^{-1}$$ can not be used.

For example the pair $$w_1=ab$$ and $$w_2=ba$$ is allowed. And the pair $$w_1=ababab$$ and $$w_2=abab$$ is not allowed. The pair $$w_1=aba^{-1}b^{-1}$$, $$w_2=a$$ is also not allowed, as $$w_1$$ is not positive.

Question: is it true that for a typical ( with respect to the Haar measure on $$G$$) pair $$(a, b)\in G^2$$ the subgroup generated by $$w_1(a,b)$$ and $$w_2(a,b)$$ is dense in $$G$$?

• For which pairs do you know this to hold (apart from pairs generating $F_2$)? or for which $G$ (other than the easy abelian cases)? – YCor Apr 16 '20 at 19:11
• @YCor, I know only these two easy cases. But I guess for pairs like $w_1=a^2b$, $w_2=b^5$ it should be eassily reducible to these easy cases, even though formally these words do not generate $F_2$. What is of interest is some $\textit{non-trivial pairs}$ whatever "nontrivial" means, like, for example $w_1=ab$, $w_2=ba$. – Dmitri Scheglov Apr 16 '20 at 19:48
• $SO(3)$ should be easy anyway for $(ab,ba)$, because the list of proper closed subgroups is short. Namely, if $\langle ab,ba\rangle$ is not dense, then $(ab)^{60}$ and $(ba)^{60}$ commute (indeed either $ab,ba$ preserve a common axis hence their squares commute, or $ab,ba$ generate a tetrahedral/cubic/icosahedral group, whose order divides $60$). Just exhibiting a pair $(a,b)$ for which this fails shows that $(ab,ba)$ is generically topologically generating (actually, for all $a,b$ in the Zariski-open subset $\{a,b:[(ab)^{60},(ba)^{60}]\neq 1\}$). – YCor Apr 16 '20 at 20:01
• @YCor, sure, by 'non-trivial pair' I mean a pair for which the statement ( if true) does not immediately follow by some obvious general reasons, like for $G$ abelian or for a pair generating $F_2$. – Dmitri Scheglov Apr 16 '20 at 20:07
• I guess that in a compact connected semisimple Lie group $G$, there are finitely many maximal closed proper subgroups up to conjugation, and that the set $\Xi_G\subset G^2$ of pairs $(a,b)$ belonging to one of those is Zariski-closed. If so, one has the alternative, for given $w_1,w_2$: either the induced map $G^2\to G^2$, mapping $(a,b)$ to $(w_1(a,b),w_2(a,b))$ maps into $\Xi_G$, or the inverse image of $\Xi_G$ is a proper Zariski-closed subset. If so (this is maybe known), the question whether the set of pairs has full measure is the same as asking whether it's non-empty. – YCor Apr 16 '20 at 23:19

See the result of Gerstenhaber-Rothaus, which says that if the abelianization of the word map has full rank, then the map $$G\times G\to G\times G$$ has non-zero degree. This is a necessary condition, as one can see if $$G$$ is abelian or has an abelian quotient (e.g. $$U(n)$$). So this won't apply to $$\{ab,ba\}$$.

Once the map is non-zero degree, the pushforward of the Haar measure on $$G\times G$$ should be absolutely continuous with respect to Haar measure. This is because the map is also algebraic, and hence the preimage of points are smaller dimension, so the preimage of a set of measure $$0$$ will be measure $$0$$.

A theorem of Weyl implies that a compact subgroup of an algebraic group over $$\mathbb{R}$$ is an algebraic subgroup. Now we follow the argument in Barnea-Larsen, section 3.

Barnea, Y.; Larsen, M., Random generation in semisimple algebraic groups over local fields., J. Algebra 271, No. 1, 1-10 (2004). ZBL1049.20028.

Let's assume that $$G$$ is semisimple; I think that the general case can be reduced to this case. Since $$G$$ is compact, we may complexify to get a semisimple algebraic group $$G^{\mathbb{C}}$$ over $$\mathbb{C}$$. By Lemma 3.2, there is a countable set $$\{X_0,X_1,\ldots\}$$ of proper closed subvarieties such that if $$\gamma\in G^{\mathbb{C}}- \cup_i X_i(\mathbb{C})$$, then the Zariski closure of $$\gamma$$ is a maximal torus. Passing to $$G=G^{\mathbb{R}}$$, the real subgroup, we see that the same is true for $$G$$. Hence with probability $$1$$, any element $$\gamma\in G$$ will have closure a maximal torus.

Proposition 3.3 states that there is a proper closed subvariety $$X \subset G^{\mathbb{C}}\times G^{\mathbb{C}}$$ so that for any proper algebraic subgroup $$H$$ containing a maximal torus, $$H\times H \subset X$$.

Now choose a random pair of elements $$(\gamma_1,\gamma_2)\in G\times G$$ with respect to a measure absolutely continuous with respect to Haar measure. Then with probability $$1$$, $$\overline{\langle\gamma_i\rangle}$$ is a maximal torus, since $$\cup_i{X_i(\mathbb{R})}$$ has measure $$0$$. Then if $$\langle \gamma_1,\gamma_2\rangle$$ is not dense in $$G$$, then $$\overline{\langle \gamma_1,\gamma_2\rangle}=H < G$$, where $$H$$ is closed and contains a maximal rank torus. So $$(\gamma_1,\gamma_2)\in X$$, again occurring with probability $$0$$.

I think this gives an outline of a proof under these assumptions.

• Thank you so much for your informative answer. The problem though is that I have very little control over the possible words. They appear from some dynamics on higher genus surfaces ( and I do not fix the genus) and the conditions formulated in the original question are the only ones which I can currently prove, this is why I formulated it this way. However your answer made me think if I can establish full rank of abelinization. I did not think about it and that might be possible after extra-work.. however not guaranteed.. – Dmitri Scheglov Apr 20 '20 at 21:34
• @DmitriScheglov: Part of the argument goes through in your case, I think: with probability 1, $\gamma_i$ will be Zariski dense in a maximal torus by the Gerstenhaber-Rothaus argument (because $\gamma_i$ has non-zero exponent sum, so we can pair it with another element giving something of non-zero degree to $G\times G$, then project to the first coordinate). So then you need the map to avoid the set $X$ again. This seems quite plausible to me under your assumption. – Ian Agol Apr 21 '20 at 16:28
• let me think about it.. – Dmitri Scheglov Apr 21 '20 at 16:45
• You might be able to compute the derivative of the map at the identity (which should be a map between Lie algebras) to show that the image of the map doesn't lie in the subgroups H, and hence not lying in the subvariety X. – Ian Agol Apr 21 '20 at 16:58
• I will try to see.. May be it makes sense to see first how it works at least for {ab, ba}.. – Dmitri Scheglov Apr 21 '20 at 17:20