That trick where you embed the free group into a Lie group September 17, 2007Posted by David Speyer in Number theory, representation theory.
The Banach-Tarski theorem states that a three dimensional ball can be chopped into finitely many pieces, which can then be rotated and translated to form two balls of the same volume as the first one. In the course of proving this theorem, one needs the following lemma:
The free group on two generators has an embedding into .
In other words, there are two rotation matrices, and , so that the only way for a product of ‘s, ‘s, ‘s and ‘s to be the identity is if that product is formally forced to be the identity by cancelling elements with their inverses. To see how you prove the Banach-Tarski theorem from here, read any number of excellent expositions, such as this one (PDF). In this post, I’m going to explain how you might find such an and a . The proof I give isn’t the shorest, but I think it is the best motivated and it will lead into a follow up post where I talk about the Grand Picard Theorem, groups acting on trees, Schottky groups and Mumford curves.
The first thing one might think of when trying to prove this theorem is to remember places we’ve seen the free group on two generators before. We’ll write for the free group on two generators. The canny reader might recall that is isomorphic to . Here is the subgroup of consisting of matrices where and are odd while and are even. So is a subgroup of . The group even acts on the sphere by Mobius transformations: takes to . Using this, we can prove a Banach-Tarski theorem for the sphere — if we allow ourselves to change pieces by a Mobius transformation. Unfortunately, Mobius trasformations are not area preserving, so no one will be impressed. Hmmm. Back to the drawing board.
How different is from anyway? Not very! The Lie group is isomorphic to ; the group of determinant one transformations of three-space preserving the quadratic form . We are off from by just a sign. If only were a real number, we’d be done.
Well, isn’t in . But there are plenty of fields which do contain . For example, the field of 5-adic numbers which Scott told us about. (Why does contain a square root of -1? Well, in the 5-adic absolute value, . Now, start with and repeatedly apply the recurrence . Basic estimates show that is a Cauchy sequence, and its limit must be a square root of -1. This is a simple case of Hensel’s lemma.) So, here is the idea. We’ll try to embed into , which is isomorphic to the subgroup of preserving , which in turn is isomorphic to . If we get lucky, our matrices and in will have rational entries, so we can also consider them as elements of .
At this point, it will help a great deal to remember how we prove that embeds into . I
started out by referring to the fact that , but it will actually be easier to show that is isomorphic to the subgroup of generated by
We’ll denote the upper half plane by . A fundamental domain for the action of on is the region bounded by the four semicircles with diameters (-3,3), (-3,-1), (-1,1) and (1,3). Let , , and be the regions of which are, respectively, outside the semicircle with diameter (-3,3), inside the semicircle with diameter (-3,-1), inside the semicircle with diameter (-1,1) and inside the semicircle with diameter (1,3). So (up to issues on the boundary, which won’t matter), is the disjoint union of , , , and .
The key computation is to check the following equations:
Now, we show that is the free group. Here is the point. Suppose that is a reduced word in . This means that each is one of , , , and we do not have a generator and its inverse next to each other. Pick a point in the interior of . We will show that . More specifically, we will show that, if is (respectively) , , , then is in , , or respectively. The proof is just induction on , using the equations above. Since isn’t in any of , , or , this shows that . So is not the identity and we are done.
Now, let’s try to fit the free group on two generators into and copy the above proof. Before, the action of on the Riemman sphere was extremely important. How do we see in terms of the group ? Take the hypersurface in and take the quotient under scaling by . The resulting surface is isomorphic to , the action of on corresponds to the action of on and the action of corresponds to the action of . (There is also a copy of contained in , which gives us the ordinary rotational symmetry group of the sphere, but this is transverse to and hence not helpful.)
Let’s try to mimic 5-adically the steps that worked in the complex world. Consider the hypersurface in , and mod out by the action of ; once again, I’ll call the result . While it is not strictly necessary for our argument, I find it extremely helpful to visualize as a toplogical space with the quotient topology. In this case, is isomorphic to , the space we get by attaching two copies of to each other, gluing to . I fix the convention that denotes the square root of -1 in such that . We’ll take
Why did I choose these matrices? I wanted matrices whose action on had an attracting fixed point and a repelling fized point, just as with my previous choice that worked in did. Both and are diagonalizable with eigenvalues . This means that the corresponding matrices in have eigenvalues . Since, 5-adically, and , this corresponds to an action with one repelling and one attracting fixed point. People who are used to Mobius transformations will call this a hyperbolic action on .
The above explains why we might think that and would work. Let’s actually show that they do. Let and be the attracting and repelling fixed points of on . Explicitly, and . Similarly, define and to be the attracting and repelling fixed points of . We’ll now take , , and to be discs around , , and . More specifically, let be a point of . By scaling the homogenous coordinates , we can guarantee that , and are all in , and not all in . Let denote the reduction of modulo 5; this is an element of , modulo scaling. Then we define to be in if . (Explicitly, and we can give similar expressions for the other .) Now, check that the relations still hold and mimic the above proof to show that the group generated inside by and is free. But, since the entries in and were rational, this also shows that the free group embeds into and into .
There are ways to write this proof purely in terms of the rational numbers and divisibility by 5, but I find them unnatural. Seeing an action on the 5-adic priojective line makes it clear to me what is happening.
In a comment on Tim Gowers’ blog, Terry Tao suggests that embedding the free group into has a number of applications. I’m afraid that I don’t know any of them other than Banach-Tarski, although I imagine you could construct useful examples in dynamical systems by using this action. However, I know lots of reasons why you might want to embed the free group into or . This will be the subject of my follow up post.
In closing, here is an exercise for you. Many papers on the Banach-Tarski theorem state that the group generated by
is free. Explain how to adapt the above proof to show this.