Is it true that two random (w.r.t. Haar measure) rotations in $SO(3)$ generate a free group?
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2 Answers
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Yes. Here's what should be a proof: the set of pairs of elements satisfying any particular relation is Zariski closed, hence has measure zero (to show that it is not $SO(3) \times SO(3)$ it suffices to know that at least one subgroup generated by two elements is free), and there are countably many relations.
answered Nov 28, 2010 at 13:22
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$\begingroup$ Right, the only non-trivial part of showing that there exists at least one pair of rotations, which generate a free group, was proved first by Felix Hausdorff. $\endgroup$ Commented Nov 28, 2010 at 13:26
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4$\begingroup$ Two such elements are given in this blog post: sbseminar.wordpress.com/2007/09/17/… $\endgroup$– S. Carnahan ♦Commented Nov 28, 2010 at 13:26
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3$\begingroup$ @Andreas a conceptual (though not terribly elementary) way to see that $SO(3)$ contains a free group is by the following argument. First note that it is enough to show that $SU(2)$ contains one. To see that $SU(2)$ has one, by a Baire category argument, it is enough to see that for each word in $F_2$ there is a pair in $SU(2)$ which does not satisfy it. As this is an algebraic question, it is enough to show it for the complexification, $SL_2(\mathbb{C})$. But this is obvious, as it contains $SL_2(\mathbb{Z})$, which contains a free group... $\endgroup$ Commented Mar 30, 2016 at 20:52
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See also: Epstein - Almost all subgroups of a Lie group are free (MR).
answered Nov 28, 2010 at 17:35