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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 $G$?

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  • $\begingroup$ 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)? $\endgroup$ – YCor Apr 16 '20 at 19:11
  • $\begingroup$ @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 $w_1=ab$, $w_2=ba$. $\endgroup$ – Dmitri Scheglov Apr 16 '20 at 19:48
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    $\begingroup$ $SO(3)$ 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 $\{a,b:[(ab)^{60},(ba)^{60}]\neq 1\}$). $\endgroup$ – YCor Apr 16 '20 at 20:01
  • $\begingroup$ @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 $F_2$. $\endgroup$ – Dmitri Scheglov Apr 16 '20 at 20:07
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    $\begingroup$ 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. $\endgroup$ – YCor Apr 16 '20 at 23:19
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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.

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  • $\begingroup$ 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.. $\endgroup$ – Dmitri Scheglov Apr 20 '20 at 21:34
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    $\begingroup$ @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. $\endgroup$ – Ian Agol Apr 21 '20 at 16:28
  • $\begingroup$ let me think about it.. $\endgroup$ – Dmitri Scheglov Apr 21 '20 at 16:45
  • $\begingroup$ 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. $\endgroup$ – Ian Agol Apr 21 '20 at 16:58
  • $\begingroup$ I will try to see.. May be it makes sense to see first how it works at least for {ab, ba}.. $\endgroup$ – Dmitri Scheglov Apr 21 '20 at 17:20

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