We present here an unusual application of Baire category theorem between topology, group theory and dynamical systems. Roughly speaking, we prove that almost all pairs of homeomorphisms of the circle are unrelated; in particular, it leads to natural occurrences of free groups.
More precisely, if denotes the set of orientation-preserving homeomorphisms of the circle, endowed with the distance
where , we prove
Theorem: The set of pairs generating a free subgroup of rank two in is a dense in .
Proof. With the distance given above, is a complete metric space, so Baire category theorem applies.
If is a reduced word, let denote the set of
satisfying . Let us show that each is a closed set whose interior is empty; it will allow us to deduce that the set of satisfying is a closed set whose interior is nonempty: indeed, if and , then , so there exist , and arbitrarily close to , and respectively, so that ie. , hence ; consequently, according to Baire category theorem, the set
has empty interior and its complement is a dense , completing the proof.
By contradiction, let be a word of minimal length such that has nonempty interior. In particular, there exists a nonempty open set . For all words , let
denote the image of by the homeomorphism ; in particular, is closed set whose interior is empty if .
Because is itself a closed set whose interior is empty, there exists such that for all words satisfying .
Let where () and set
First, notice that since and that if . Indeed, if it was not the case, there would exist a proper subword of fixing , while for some word , hence , a contradiction.
To conclude, let and be two homeomorphisms arbitrarily close to and respectively, and satisfying but .
[To do that, let and show that each homeomorphism of the circle can be approximated by a piecewise linear homeomorphism (using a classical argument based on uniform continuity); then we deduce easily that a homeomorphism of the circle can be approximated by a homeomorphism agreeing on a finite set of points and disagreeing on another finite set of points.]
Thus, because is open, but by construction, a contradiction.
We deduce from our theorem that has a lot of free subgroups, but the given proof does not provide any explicit example. In order to exhibit free subgroups, we use ping-pong lemma:
Lemma: Let be a group acting on a set such that there exist and satisfying
- and are non-empty and disjoint,
- and for all .
Then is a free group of basis .
Proof. Clearly, the second condition implies that and has infinite order. Moreover, any reduced word over not in can be written (up to conjugacy) as
with and .
Let . If in , then . But , hence , hence , and so on; finally, . We deduce that , a contradiction. Therefore, in and is a free subgroup of basis .
The first example of free subgroup of comes from the natural action of on the set of lines passing through the origin, that is on . An explicit homeomorphism between and is illustrated by Figure 1, where the two blue points are identified.
Now we wan to apply ping-pong lemma to and . The actions of and on the circle are illustrated by Figures 2 and 3 respectively, through the previous homeomorphism.
Therefore, ping-pong lemma applies with and as indicated:
Therefore, is a free basis and generates a free subgroup of , and a fortiori of .
Our second example comes from the homeomorphism and is based on piecewise linear homeomorphisms:
Definition: Let be a homeomorphism satisfying for all . If there exists a sequence of reals satisfying and such that is affine on each , then the induced homeomorphism is said piecewise linear.
The set of piecewise linear homeomorphisms of the circle is denoted by .
Let and be two piecewise linear homeomorphisms of the circle whose graphs are respectively
Moreover, the graphs of and are respectively
Easily, we deduce that for all ,
Therefore, ping-pong lemma applies, and is a free subgroup of rank two of the group .
More information about the group of homeomorphisms of the circle can be found in Etienne Ghys’ document, Groups acting on the circle.