Cheap, non-constructive, free group generating rotations for Banach-Tarski

Question

Stan Wagon's exposition of Banach-Tarski (for example) includes a beautiful explicit construction of two 2-sphere rotations which generate a free subgroup of the rotation group.

For teaching purposes I'd like the fastest, quick and dirty soft proof.

A further thought or two:

Presumably, given a nontrivial word in the free group on two generators, only a "small" set of pairs of rotations satisfy the word. Small could mean measure zero or nowhere dense. But I don't see an easy proof of this.

Some words such as $aba^{-1}b^{-1}$ have infinitely many solutions even once you choose a rotation to substitute for $a$ (all the rotations with the same axis will work for $b$). But I'd guess that such words are atypical - that usually fixing $a$ would reduce your choices for $b$ to a set of dimension $0$. What is the right theorem?

Answer 1

As already remarked by Quiaochu, the set of $(a,b)$ satisfying any non trivial relation $w\in F_2$ is Zariski closed in $K=SO(3)\times SO(3)$. But this last $K$ is itself Zariski dense in $G=PSL_2(\mathbb{C})\times PSL_2(\mathbb{C})$ : any polynomial relation (here with rational coefficients, but complex ones too), holding identically on $K$, also holds on $G$. [ EDIT (after comment by Keivan) : my $SO(3)$ is really $PSU(2)$, the fixed points of the antiholomorphic involution $g\mapsto {}^t\overline g^{-1}$ of $PSL_2(\mathbb{C})$ ]

But now $\Gamma=PSL_2(\mathbb{Z})$ embeds in $PSL_2(\mathbb{C})$, and it is easy to "see" an $F_2$ subgroup in $\Gamma$. Indeed, $\Gamma$ is the group of orientation preserving isometries of the $3$-regular tree with cyclic orientations at vertices (see e.g. Serre's Trees). For instance the fundamental group of the $\Theta$ graph (with cyclic orientations at both vertices) is an $F_2$ sitting in $\Gamma$, when viewed as acting on the universal cover of the $\Theta$ graph. Maybe it doesn't count as a cheap non constructive proof...

Answer 2

A good way to see this is by thinking of Galois conjugation. Many discrete groups acting on the hyperbolic plane $H^2$ (whose orientation preserving isometry group is $PSL(2,\mathbb R)$ or in hyperbolic 3-space (with isometry group $PSL(2,\mathbb C)$) have Galois conjugates in $PSU(2) = SO(3)$, acting on $S^2$.

It's very easy to geometrically establish certain subgroups of isometries of $H^2$ or $H^3$ are free: in the plane for any 4 disjoint half-planes $U,X,Y,Z$, isometries $A$ sending the complement of $U$ to $X$ and $B$ sending the complement of $Y$ to $Z$ generate a free group. For generic $A$ and $B$, they have enough algebraic independence that they are Galois conjugate to a pair of elements in $PSU(2)$, therefore giving a free subgroup.

Edited to correct error noted in comment: A particular example that is easy to see algebraically before understanding things in greater generality is the $(777)$ triangle group generated by two $2\pi/ 7$ rotations in $H^2$ such that there product is also a $2 \pi/7$ $\left < a, b, c | abc=a^7=b^7=c^7 = 1 \right > $, it has 2 Galois conjugates within $PSL(2,\mathbb C)$, where the rotations become $2/7*2\pi$ and $3/7*2\pi$. The second of these is in $PSU(2) = SO(3)$ and acts on $S^2$: this Galois conjugate comes from a spherical triangle with three angles of $3/7 \pi$: you rotate about the corners of the triangles by twice the angle of the triangle.

The Galois action can be shown in elementary terms for anyone who understands that the different primitive 7th roots of unity are algebraically isomorphic (Galois conjugate): in $SL(2,\mathbb C)$, the trace of any element in a 2-generator group is determined by the traces of $a$, $b$, and $ab$, using the trace relation $T(a*b) + T(a*b^{-1}) = T(a)T(b)$, which is a simple consequence of the fact that a matrix satisfies its characteristic polynomial. Traces of other elements in the group are polynomials in these three traces, so they're determined by the algebraic relationships in this ring. From this it follows it follows that this $(777)$ triangle group acts faithfully on $S^2$ (taking into account that $SO(3) = SU(2)/\pm 1$ and $SU(2) \subset SL(2,\mathbb C)$.)

It's easy to write down free subgroups of any hyperbolic group like $777$: there are simple geometric sufficient criterion by constructing fundamental domains. These turn into free subgroups of $SO(3)$.

For the other example mentioned, generated by $A$ and $B$, if we lift to $SL(2,\mathbb C)$ and choose the traces of $A$, $B$ and $AB$ to be algebraically independent, then we can map them to 3 transcendentally independent elements in the interval $(-2,2)$ to conjugate the group inside an $SU(2)$. Or, we can do the same thing by choosing appropriate elements within any number field except the rationals and the quadratic imaginary field. (It's also straightforward to get free subgroups of $SO(2, \mathbb Q)$, by using a different geometric picture).

Nearly any subgroup of $SO(3)$ has at least one galois conjugate that has unbounded orbits in $PSL(2,\mathbb C)$. In any such case, you can find a free subgroup geometrically by picking elements that give a good fundamental domain.

Answer 3

Consider the quadratic form $B(x,y,z)=x^2+y^2+\pi z^2$. Then $O(B;\mathbb{Q}(\pi))=\{A\in M_3(\mathbb{Q}(\pi)) | B(Av)=B(v),\forall v\in \mathbb{Q}(\pi)^3\} $ is isomorphic to a subgroup of $O(3)$, by the law of inertia. The Galois automorphism $\sigma:\mathbb{Q}(\pi)\to \mathbb{Q}(\pi)$ sending $\pi$ to $-\pi$ induces an isomorphism $O(B;\mathbb{Q}(\pi)) \cong O(B^{\sigma};\mathbb{Q}(\pi))$, where $B^{\sigma}(x,y,z)=x^2+y^2-\pi z^2$ (just apply $\sigma$ to the defining equation to see this). Again, by the law of inertia, $O(B^{\sigma};\mathbb{Q}(\pi))\leq O(2,1)$, and it is dense since $\mathbb{Q}(\pi)\leq \mathbb{R}$ is dense. But $O(2,1)$ acts by isometries on the hyperbolic plane $\mathbb{H}^2$, and contains many two-generator free subgroups which are stable under perturbations, as one may prove by the ping-pong lemma, for example as Fuchsian Schottky groups. There is of course much flexibility in this construction: $\pi$ could be replaced by any transcendental, or even by any positive real number with a negative Galois conjugate.

Answer 4

Here's an idea. It's easy to see that there is a pair of rotations not satisfying a word $w$ which does not lie in the commutator subgroup $[F_2, F_2]$ of $F_2$, just by picking two rotations about a common axis. Now, here are three things which I think are true but which I don't know how to prove:

• The commutator subgroup of a free group can be freely generated by commutators.
• The intersection $\bigcap G_n$ of the derived series $G_n = [G_{n-1}, G_{n-1}]$ of $F_2$ (where $G_0 = F_2$) is trivial.
• $\text{SO}(3)$ is equal to its own commutator subgroup.

If all of these things are true, it follows that the above argument applies to any $w \in F_2$, by writing $w$ as a word $w_1 ... w_k$ where $w_i$ are commutators of elements in $G_{n-1}$ but $w \not \in G_{n+1}$ for some $n$ and setting the $w_i$ to be rotations about a common axis. (The first assumption is the one in which I have the least confidence...)

Answer 5

These two matrices generate a free group: $$ A=\left( \begin{array}{ccc} \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\ -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array} \right), B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\ 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3} \end{array} \right). $$ To see that they do, note that the entries of $3A$ and $3B$ live in the ring $\mathbb{Z}[\sqrt{2}]$, which admits a surjective homomorphism to the field $\mathbb{F}_3(i)$. This map induces a map on the matrix rings, under which $3A$ and $3B$ become

$$ A' =\left( \begin{array}{ccc} 1 & - i & 0 \\ i & 1 & 0 \\ 0 & 0 & 0 \end{array} \right), B'=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & - i \\ 0 & i & 1 \end{array} \right). $$ If there were some non-trivial reduced group word in $A$ and $B$ that gave the identity, then a similar word in $3A$, $3A^{\top}$, $3B$ , and $3B^{\top}$ would be a multiple of the identity matrix. However, one may check that any monoid word in the generators $A'$, $A'^{\top}$, $B'$, $B'^{\top}$ that evaluates to a multiple of the identity must contain the subword $A'A^{\top}$ or some similar forbidden subword.

Answer 6

I thought the exposition in the Wikipedia article was pretty good-- not hard to follow even without knowing much set theory.

Answer 7

How is this question different from:

Random rotations in SO(3) and free group

( there is a reference to D.B.A. Epstein's classic paper on the subject in my answer there).