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 \mathrm{Homeo}^+(\mathbb{S}^1) denotes the set of orientation-preserving homeomorphisms of the circle, endowed with the distance

d(f,g)= \max \left( \| f-g \|_{\infty}, \| f^{-1}-g^{-1} \|_{\infty} \right), \ \forall f,g \in \mathrm{Homeo}^+(\mathbb{S}^1)

where \| \cdot \|_{\infty} = \sup\limits_{ \mathbb{S}^1 } \| \cdot \|, we prove

Theorem: The set of pairs (f,g) \in \mathrm{Homeo}^+(\mathbb{S}^1) \times \mathrm{Homeo}^+(\mathbb{S}^1) generating a free subgroup of rank two in \mathrm{Homeo}^+(\mathbb{S}^1) is a dense G_{\delta} in \mathrm{Homeo}^+(\mathbb{S}^1) \times \mathrm{Homeo}^+(\mathbb{S}^1).

Proof. With the distance given above, \mathrm{Homeo}^+(\mathbb{S}^1) is a complete metric space, so Baire category theorem applies.

If w \in \mathbb{F}_2 is a reduced word, let X_w denote the set of

(f,g,x) \in \mathrm{Homeo}^+( \mathbb{S}^1) \times \mathrm{Homeo}^+( \mathbb{S}^1) \times \mathbb{S}^1

satisfying w(f,g)(x) =x. Let us show that each X_w is a closed set whose interior is empty; it will allow us to deduce that the set of  (f,g) \in \mathrm{Homeo}^+(\mathbb{S}^1) \times \mathrm{Homeo}^+(\mathbb{S}^1) satisfying w(f,g)= \mathrm{Id} is a closed set whose interior is nonempty: indeed, if w(f,g)= \mathrm{Id} and x \in \mathbb{S}^1, then (f,g,x) \in X_w, so there exist f_0, g_0 and x_0 arbitrarily close to f, g and x respectively, so that (f_0,g_0,x_0) \notin X_w ie. w(f_0,g_0)(x_0) \neq x_0, hence w(f_0,g_0) \neq \mathrm{Id}; consequently, according to Baire category theorem, the set

\displaystyle \{ (f,g) \mid \langle f,g \rangle \not\simeq \mathbb{F}_2 \} = \bigcup\limits_{w \in \mathbb{F}_2} \{ (f,g) \mid w(f,g)= \mathrm{Id} \}

has empty interior and its complement  is a dense G_{\delta}, completing the proof.

By contradiction, let w be a word of minimal length k such that X_w has nonempty interior. In particular, there exists a nonempty open set U \subset X_w. For all words w_1,w_2 \in \mathbb{F}_2, let

F(w_1,w_2)= \{ (f,g,x) \mid w_1(f,g)w_2(f,g)(x)=w_2(f,g)(x) \}

denote the image of X_{w_1} by the homeomorphism  (f,g,x) \mapsto (f,g,w_2(f,g)^{-1}(x)); in particular, F(w_1,w_2) is closed set whose interior is empty if \mathrm{\ell g}(w_1),\mathrm{\ell g}(w_2)<k.

Because \bigcup\limits_{\mathrm{\ell g}(w_1),\mathrm{\ell g}(w_2)<k} F(w_1,w_2) is itself a closed set whose interior is empty, there exists (f,g,x) \in U such that (f,g,x) \notin F(w_1,w_2) for all words satisfying \mathrm{lg}(w_1), \mathrm{lg}(w_2)<k.

Let w(f,g)=h_1 \circ \cdots \circ h_k where h_i \in \{ f,f^{-1},g,g^{-1}\} (1 \leq i \leq k) and set

x_1=h_1(x), ~ \dots, ~ x_{k-1}=h_{k-1}(x_{k-2}), ~ x_k=h_k(x_{k-1}).

First, notice that x_k=w(f,g)(x)=x since (f,g,x) \in X_w and that x_i \neq x_j if i \neq j. Indeed, if it was not the case, there would exist a proper subword w_1(f,g) of w(f,g) fixing x_i, while x_i=w_2(f,g)(x) for some word w_2, hence (f,g,x) \in F(w_1,w_2), a contradiction.

To conclude, let \tilde{f} and \tilde{g} be two homeomorphisms arbitrarily close to f and g respectively, and satisfying x_1= \tilde{h}_1(x_1), ~ \dots, ~ x_{k-1}= \tilde{h}_{k-1}(x_{k-2}) but x_k \neq \tilde{h}_k(x_{k-1}).

[To do that, let \mathbb{S}^1= \mathbb{R}/ \mathbb{Z} 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, (\tilde{f}, \tilde{g},x) \in U \subset X_w because U is open, but w(\tilde{f},\tilde{g})(x)=\tilde{h}_k(x_{k-1}) \neq x_k=x by construction, a contradiction. \square

We deduce from our theorem that \mathrm{Homeo}^+(\mathbb{S}^1) 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 G be a group acting on a set S such that there exist x,y \in G and X,Y \subset S satisfying

  • X and Y are non-empty and disjoint,
  • x^n \cdot X \subset Y and y^n \cdot Y \subset X for all n \in \mathbb{Z} \backslash \{ 0\}.

Then \langle x,y \rangle is a free group of basis \{x,y \}.

Proof. Clearly, the second condition implies that x and y has infinite order. Moreover, any reduced word w over \{ x^{\pm 1}, y^{\pm 1} \} not in \langle x \rangle \cup \langle y \rangle can be written (up to conjugacy) as

w= y^{m_0} \cdot x^{n_1} \cdot y^{m_1} \cdots x^{n_r} \cdot y^{m_r} with r \geq 1 and n_i,m_i \in \mathbb{S} \backslash \{ 0 \}.

Let a \in Y. If w=1 in G, then a = w \cdot a. But a \in Y, hence y^{m_r} \cdot a \in X, hence x^{n_r} \cdot y^{m_r} \cdot a \in Y, and so on; finally, w \cdot a \in X. We deduce that a = w \cdot a \in X \cap Y = \emptyset, a contradiction. Therefore, w \neq 1 in G and \langle x,y \rangle is a free subgroup of basis \{x,y \}. \square

The first example of free subgroup of \mathrm{Homeo}^+(\mathbb{S}^1) comes from the natural action of GL(2, \mathbb{Z}) on the set of lines passing through the origin, that is on \mathbb{R}P^1. An explicit homeomorphism between \mathbb{R}P^1 and \mathbb{S}^1 is illustrated by Figure 1, where the two blue points are identified.

Figure1

Figure 1.

Now we wan to apply ping-pong lemma to A = \left( \begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix} \right) and B= \left( \begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix} \right). The actions of A and B on the circle are illustrated by Figures 2 and 3 respectively, through the previous homeomorphism.

Figure 2.

Figure 2.

Figure 3.

Figure 3.

Therefore, ping-pong lemma applies with X and Y as indicated:

Figure4

Therefore, \{ A, B \} is a free basis and generates a free subgroup of GL(2,\mathbb{Z}), and a fortiori of \mathrm{Homeo}^+(\mathbb{S}^1).

Our second example comes from the homeomorphism \mathbb{R} / \mathbb{Z} \simeq \mathbb{S}^1 and is based on piecewise linear homeomorphisms:

Definition: Let f : \mathbb{R} \to \mathbb{R} be a homeomorphism satisfying f(x+1)=f(x)+1 for all x \in \mathbb{R}. If there exists a sequence of reals \{ x_i \mid i \in \mathbb{Z} \} satisfying \lim\limits_{i \to \pm \infty} x_i = \pm \infty and such that f is affine on each [x_i,x_{i+1}], then the induced homeomorphism \tilde{f} : \mathbb{R} / \mathbb{Z} \to \mathbb{R} / \mathbb{Z} is said piecewise linear.

The set of piecewise linear homeomorphisms of the circle is denoted by PL_+(\mathbb{S}^1).

Let f and g be two piecewise linear homeomorphisms of the circle whose graphs are respectively

graphs

Moreover, the graphs of f^{-1} and g^{-1} are respectively

inversegraphs

Easily, we deduce that for all n \in \mathbb{Z} \backslash \{ -1,0,1 \},

f^n \left( \left( \frac{1}{3}, \frac{2}{3} \right) \right) \subset \left( 0, \frac{1}{3} \right) \cup \left( \frac{2}{3}, 1 \right) and g^n \left( \left( 0,\frac{1}{3} \right) \cup \left( \frac{2}{3},1 \right) \right) \subset \left( \frac{1}{3}, \frac{2}{3} \right).

Therefore, ping-pong lemma applies, and \langle f^2,g^2 \rangle is a free subgroup of rank two of the group PL_+(\mathbb{S}^1) \subset \mathrm{Homeo}^+(\mathbb{S}^1).

More information about the group of homeomorphisms of the circle can be found in Etienne Ghys’ document, Groups acting on the circle.

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