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).

Because $\bigcup\limits_{\mathrm{\ell g}(w_1),\mathrm{\ell g}(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).

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.

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 3.

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

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

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

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.