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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?

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Related: mathoverflow.net/questions/47585/… –  Qiaochu Yuan Dec 14 '10 at 7:58
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In particular, the set of pairs of rotations satisfying a given word is Zariski closed, hence measure zero unless it is all of SO(3) x SO(3), so it's enough to exhibit, for each word, a single pair of rotations which does not satisfy it. The discussion at sbseminar.wordpress.com/2007/09/17/… includes several attempts to rule out the second case without writing down a free subgroup, but I don't think they got anywhere. –  Qiaochu Yuan Dec 14 '10 at 8:00
    
Does it help to fix two axes, perhaps for simplicity 90 degrees apart and then look only at rotations very close to the identity? I think working up to first order takes care of words that don't become trivial if you add the relation $ab=ba$. Could working up to second order suffice for these? –  David Feldman Dec 14 '10 at 8:13

7 Answers 7

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...

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Well, it is really $SO(3)\simeq PSU(2)\subset PSL_2(\mathbb{C})$ which is (as a maximal compact subgroup). I shoud have precised that. Sorry. –  BS. Dec 14 '10 at 9:49
    
Thanks for the comment. I think my question was quite silly, so I removed the comment. –  Keivan Karai Dec 14 '10 at 10:27
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Very slick!!!!! –  David Speyer Dec 14 '10 at 13:37

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.

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I doubt that there is a spherical triangle with all angles $2\pi/7$ : they sum to less than $\pi$, right ? –  BS. Dec 15 '10 at 9:05

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.

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A remark: to prove that $O(B;Q(\pi))$ is dense in $O(B;R)$, one may use the fact that $O(B;R)$ is generated by reflections, and that the reflections in $O(B;Q(\pi))$ are dense in the reflections in $O(B;R)$. –  Ian Agol Dec 16 '10 at 2:18

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...)

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Any element of $SO(3)$ is a product of two half-turns (extensions of reflexions inside a plane), which are conjugate, hence it is a commutator (in fact $SO(3)$ is simple as an abstract group). As to the the intersection of the derived series of $F_2$, I don't see a cheap argument yet. –  BS. Dec 14 '10 at 8:56
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The statement about the intersection of the commutators can be proven using certain representations of the free group. For instance, consider the ring $R={\mathbb Z}[x,y]$ of all formal power series in noncommuting variables $x,y$. It is not to hard to show that $1+x$ and $1+y$ generate a free subgroup of the group of the unit elements of $R$. Then a simple induction shows that the lowest order term of a long commutator goes to infinity as the length of the commutator goes to infinity. You can do the same by realizing the free group as a subgroup of $SL(2,Z)$ and use arithmetic. –  Keivan Karai Dec 14 '10 at 9:15
    
Is (something like) this true? Consider the two matrices over ${\Bbb R}[[\epsilon]]$ $$A = \left[ \begin{array}{ccc} \cos(\epsilon) & \sin(\epsilon) & 0 \\ -\sin(\epsilon) & \cos(\epsilon) & 0 \\ 0 & 0 & 1 \end{array} \right] \ {\rm and}\ B = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos(\epsilon) & \sin(\epsilon) \\ 0 & -\sin(\epsilon) & \cos(\epsilon) \end{array} \right] \ .$$ Then if a word $w$ in $A$ and $B$ equals the identity modulo $\epsilon^k$, $w$ belongs the $k$th term of the derived series? –  David Feldman Dec 14 '10 at 19:55
    
>The commutator subgroup of a free group can be freely generated by commutators. Am I missing something? Any subgroup of a free group is free. –  David Feldman Dec 14 '10 at 23:59
    
@David: yes, but can the generators be taken to be commutators themselves, rather than products of commutators? (Also, is the commutator subgroup of a finitely generated free group finitely generated?) –  Qiaochu Yuan Dec 15 '10 at 0:29

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.

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Sorry, I answered what I assumed the question to be. Also, my matrices are not showing up properly. –  John Wiltshire-Gordon Dec 14 '10 at 18:57
    
TeX fixed. Plus some more characters. –  David Speyer Dec 14 '10 at 19:11
    
I'm afraid I don't understand. What does a pair of 3-by-3 matrices over a field with 9 elements do for you in terms of free groups? It seems more likely you want a quadratic unramified extension of the 3-adic field $\mathbb{Q}_3$, so you can identify the matrices with hyperbolic isometries. –  S. Carnahan Dec 14 '10 at 20:04
    
@David: Thanks for fixing my TeX. @Scott: I added some more of what I have in mind. –  John Wiltshire-Gordon Dec 14 '10 at 21:11
    
I can't figure out what you mean by "corresponding matrices." The matrices you write down aren't even invertible, so they don't live in a matrix group. –  David Feldman Dec 14 '10 at 22:53

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

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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).

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Hi Igor! So the question is whether you get the mere existence of free generators from the measure-theoretic (or category-theoretic in the sense of Oxtoby) result, or whether you need at least one explicit set of free generators to get the measure-theoretic argument started (so that you know there's no non-trivial relation that's satsified identically in the group). All answers so far seem to depend on having a more or less explicit construction first. I'm happy to see better motivation for the construction I'd already learned, but that still leaves me curious for a purely soft proof. –  David Feldman Dec 15 '10 at 5:11

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