There is no expansive homeomorphism on $S^1$

Question

I want to prove that there is no expansive homeomorphism on $S^1$.

$f: X \to X $ is expansive if \begin{align} \exists\varepsilon>0 ,\quad \sup d(f^n(x),f^n(y)\leq \varepsilon \Rightarrow x=y, \quad n\in \mathbb{N} \cup \{0\} \end{align}

This is how I solved this question:

Suppose not, if $f$ is an expansive homeomorphism on $S^1$ then so is $f^2$ let $f^2=g$ then $g^n$ converges to $1$ as $n \to \infty$ take two points $x$ and $y$ very close to $1$ for example in $[1-\varepsilon, 1)$ so $d(g^n(x),g^n(y))<\varepsilon$ for all $n$. This contradicts expansivity.

I consider $[0,1]$ in my proof.

Are things ok with my proof?

Answer 1

No, your proof does not work at all, starting with the claim that the sequence $(g^n)$ converges to $1$.

The fact that there are no expansive self-homeomorphisms of the circle is nonelementary and was first proven in a combination of work by Bryant on one hand and by Jakobsen and Utz on the other:

1. Bryant proved in his 1954 PhD thesis, published in two installments:

Bryant, B. F., Expansive self-homeomorphisms of a compact metric space, Am. Math. Mon. 69, 386-391 (1962). ZBL0107.16502.

Bryant, B. F., On expansive homeomorphisms, Pac. J. Math. 10, 1163-1167 (1960). ZBL0101.15504.

that there are no expansive self-homeomorphisms of $[0,1]$.

2. In the follow-up work (which appeared earlier)

Jakobsen, J. F.; Utz, W. R., The nonexistence of expansive homeomorphisms on a closed 2-cell, Pac. J. Math. 10, 1319-1321 (1960). ZBL0144.22302.

Jakobsen and Utz used Bryant's result to prove nonexistence of expansive self-homeomorphisms of the circle (this reduction is easier than Bryant's work).

Lastly: Bryant's work was generalized later on by various authors, for instance, in

Williams, R. K., A note on expansive mappings, Proc. Am. Math. Soc. 22, 145-147 (1969). ZBL0177.25604.

where Williams gives a new proof of nonexistence of expansive self-maps of the unit interval (without the assumption that the map is a homeomorphism).

Answer 2

A more personal proof (maybe the same as one reference given by Moishe)

If you consider a homeo of the interval, your proof is almost correct. Obviously if $f$ is expansive so is $f^2$. Therefore you may assume in this case $f$ is increasing. Every point is attracting to a fixed point (but $1$ is not necessarily attracting as you claimed it). This implies $f$ can not be expansive.

Now if $f$ is a homeomorphism of the circle with rational rotation number (equivalently $f$ has a periodic point) you are reduced to the case of interval maps. If the rotation number is irrational and $f$ has no wandering domain, it is conjugated to a rotation by Denjoy theory. But $f$ being assumed expansive it can not have wandering domains. Therefore we are reduced to rotations which are clearly not expansive.

Answer 3

I would like to point out that even though the OP calls the property in question expansivity the kind of expansivity he is talking about is stronger than the kind of expansivity the other two answers seem to focus on. As such, there is an elementary proof (that is, the proof does not require any technology apart from basic metric space theory and some combinatorics). Still, it is slightly more involved than the argument the OP presents, as Moishe Kohan pointed out said argument is not quite sufficient.

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First let me give some definitions. Let $(X,d)$ be a compact metric space, $\epsilon^\ast\in\mathbb{R}_{>0}$, and $\mathcal{F}\subseteq C^0(X,X)$ be a collection of continuous self-maps of it. $\mathcal{F}$ is called $\epsilon^\ast$-expansive if

$$\forall x,y\in X: \sup_{f\in\mathcal{F}} d(f(x),f(y))\leq \epsilon^\ast \implies x=y.$$

Note that if $\mathcal{G}\subseteq C^0(X,X)$ is another family of continuous self-maps and $\mathcal{F}\subseteq \mathcal{G}$, then the $\epsilon^\ast$-expansivity of $\mathcal{F}$ would imply the $\epsilon^\ast$-expansivity of $\mathcal{G}$, so that expansivity gets stronger if the family is made smaller.

If $f\in C^0(X,X)$, then $f$ is called one-sided $\epsilon^\ast$-expansive (alternatively: positively $\epsilon^\ast$-expansive or forward $\epsilon^\ast$-expansive) if the family $\{f^n\,|\, n\in\mathbb{Z}_{\geq0}\}$ is $\epsilon^\ast$-expansive.

Similarly if $f\in \operatorname{Homeo}(X)$, then $f$ is called one-sided $\epsilon^\ast$-expansive if the family $\{f^n\,|\, n\in\mathbb{Z}_{\geq0}\}$ is $\epsilon^\ast$-expansive. In the case of homeomorphisms one can also consider the backward iterates, and $f$ is called two-sided $\epsilon^\ast$-expansive (or simply $\epsilon^\ast$-expansive, for reasons that will be clear in a bit) if the family $\{f^n\,|\, n\in\mathbb{Z}\}$ is $\epsilon^\ast$-expansive. Since the smaller the family the stronger the expansivity, one-sided expansivity of a homeomorphism is stronger than two-sided expansivity.

Any one of the above objects is called expansive if it's $\epsilon^\ast$-expansive for some $\epsilon^\ast\in\mathbb{R}_{>0}$.

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For $f$ a homeomorphism of the compact metric space $(X,d)$, one can introduce new distances on $X$ (In ergodic theory circles these are often called Bowen metrics nowadays). I will be using them because to me they make the quantifier logistics easier to handle. Namely, for any subset $S\subseteq \mathbb{R}$ let us put

$$d_S=d_S^f:X\times X\to \mathbb{R}_{\geq0},\quad (x,y)\mapsto \sup_{n\in S\cap \mathbb{Z}} d(f^n(x),f^n(y)).$$

Thus e.g. $d_{[0,n]}^f(x,y)$ gives the largest distance that can be achieved between the orbits of $x$ and $y$ when $f$ acts on them for $n$ time units (the points in the respective orbits ought to be synchronized in time, so that we only compare $f^k(x)$ and $f^k(y)$).

Using these new metrics note that one-sided $\epsilon^\ast$-expansivity of $f$ is equivalent to:

$$\forall x,y\in X: d_{\mathbb{Z}_{\geq0}}^f(x,y)\leq\epsilon^\ast\implies x=y.$$

There is an alternative characterization of $\epsilon^\ast$-expansivity (I won't prove it, but the ideas in the next section are adaptations of the proof of it.):

Prop.: Let $(X,d)$ be compact metric, $\epsilon^\ast\in\mathbb{R}_{>0}$.

1. Let $f:X\to X$ be continuous. Then $f$ is one-sided $\epsilon^\ast$-expansive iff

$$\forall \epsilon\in\mathbb{R}_{>0},\exists M\in\mathbb{Z}_{>0},\forall x,y\in X: d_{[0,M]}^f(x,y)\leq \epsilon^\ast \implies d(x,y)<\epsilon.$$

2. Let $f: X\to X$ be a homeomorphism. Then $f$ is two-sided $\epsilon^\ast$-expansive iff

$$\forall \epsilon\in\mathbb{R}_{>0},\exists M\in\mathbb{Z}_{>0},\forall x,y\in X: d_{[-M,M]}^f(x,y)\leq \epsilon^\ast \implies d(x,y)<\epsilon.$$

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We claim that no infinite compact metric space can carry a one-sided expansive homeomorphism. According to Gottschalk & Hedlund's Topological Dynamics (pp.85-86, Thm.10.30) this was first proved by Schwartzman in his 1952 thesis. Over the years it seems this result has been optimized; I'll be following Coven & Keane's paper "Every compact metric space that supports a positively expansive homeomorphism is finite". Also see Aoki & Hiraide's Topological Theory of Dynamical Systems - Recent Advances (p.45, Thm.2.2.12) for a more high-brow proof.

The proof uses the following lemma. Roughly speaking it says that in the case of an injective one-sided $\epsilon^\ast$-expansive map, if the orbits of any two points are $\epsilon^\ast$-indistinguishable for a sufficiently long time duration in the future, then any two such indistinguishable points will be $\epsilon^\ast$-indistinguishable also before said duration in the future.

Lemma: Let $(X,d)$ be compact metric, $\epsilon^\ast\in\mathbb{R}_{>0}$, $f:X\to X$ be continuous. If $f$ is injective and one-sided $\epsilon^\ast$-expansive, then

$$\exists M^\ast\in\mathbb{Z}_{>0}, \forall x,y\in X, \forall p\in\mathbb{Z}_{\geq0}: d_{[p+1,p+M^\ast]}^f(x,y)\leq \epsilon^\ast \implies d_{[0,p]}^f(x,y)\leq \epsilon^\ast$$

Pf.: First let us prove the statement for $p=0$, that is, we claim first that

$$\exists M^\ast\in\mathbb{Z}_{>0}, \forall x,y\in X: d_{[1,M^\ast]}^f(x,y)\leq \epsilon^\ast \implies d(x,y)\leq \epsilon^\ast.$$

Suppose otherwise. Then there are two sequences $x_\bullet,y_\bullet:\mathbb{Z}_{\geq0}\to X$ such that

$$\forall n\in\mathbb{Z}_{\geq0}: d_{[1,n]}^f(x_n,y_n)\leq \epsilon^\ast \quad\text{ and }\quad d(x_n,y_n)> \epsilon^\ast.$$

By compactness up to subsequences we have $\lim_{n\to\infty} x_n=x^\ast\in X$, $\lim_{n\to\infty} y_n=y^\ast\in X$, and since the terms are at least $\epsilon^\ast$ apart we have that $x^\ast\neq y^\ast$. Let $k\in\mathbb{Z}_{\geq1}$. Then for $n\in\mathbb{Z}_{\geq k}$: $d(f^k(x_n),f^k(y_n))\leq d_{[1,n]}^f(x_n,y_n)\leq\epsilon^\ast$, thus taking the limit as $n\to\infty$ we have $d(f^k(x^\ast), f^k(y^\ast))\leq \epsilon^\ast$. In other words

$$d_{\mathbb{Z}_{\geq1}}^f(x^\ast,y^\ast)=d_{\mathbb{Z}_{\geq0}}^f(f(x^\ast),f(y^\ast))\leq \epsilon^\ast.$$

By the one-sided $\epsilon^\ast$-expansivity of $f$, $f(x^\ast)=f(y^\ast)$, but $f$ is assumed to be injective, a contradiction. Thus there is an $M^\ast$ as claimed.

Next we claim that this $M^\ast$ works uniformly for $x,y$ and $p$ as in the statement. Suppose $p\in\mathbb{Z}_{>0}$ and fix $x,y\in X$ with $d_{[p+1, p+M^\ast]}^f(x,y)=d_{[1,M^\ast]}^f(f^p(x),f^p(y))\leq \epsilon^\ast$. Applying the argument for the $p=0$ case to $f^p(x)$ and $f^p(y)$, we get $d(f^p(x),f^p(y))\leq\epsilon^\ast$, whence we also have

$$d(f^p(x),f^p(y))\leq d_{[p,(p-1)+M^\ast]}^f(x,y)\leq d_{[p, p+M^\ast]}^f(x,y)\leq \epsilon^\ast.$$

Again, by the argument for the $p=0$ case applied to $f^{p-1}(x)$ and $f^{p-1}(y)$ we get

$$d(f^{p-1}(x),f^{p-1}(y))\leq d_{[p-1,(p-2)+M^\ast]}^f(x,y)\leq d_{[p-1, (p-1)+M^\ast]}^f(x,y)\leq \epsilon^\ast.$$

Iterating this process we have the result.

Theorem: Let $(X,d)$ be compact metric, $f:X\to X$ be continuous. If $f$ is injective and one-sided expansive, then $X$ is finite.

Pf.: Say $f$ is $\epsilon^\ast$-expansive for $\epsilon^\ast\in\mathbb{R}_{>0}$ and take $M^\ast\in\mathbb{Z}_{>0}$ be as in the previous lemma. For $z\in X$, put

$$U_z:=\left\{y\in X\,\left|\, d_{[1,M^\ast]}^f(z,y)<\dfrac{\epsilon^\ast}{2}\right.\right\}.$$

By compactness for some finite set $\{z_1,...,z_r\}\subseteq X$, $X=\bigcup_{s=1}^r U_{z_s}$. We claim that $\#(X)\leq r$. Suppose otherwise and fix a subset $S\subseteq X$ with $\#(S)=r+1$. For $k\in\mathbb{Z}_{\geq0}$, $f^k(S)$ has exactly $r+1$ elements, thus $S$ has at least two distinct elements that are mapped to the same $U_{z_s}$ by $f^k$. This way we may choose two sequences $x_\bullet,y_\bullet:\mathbb{Z}_{\geq0}\to S$ such that

$$\forall k\in\mathbb{Z}_{\geq0}: x_k\neq y_k, \,\,d_{[1,M^\ast]}^f(f^k(x_k),f^k(y_k))=d_{[k+1,k+M^\ast]}^f(x_k,y_k)\leq \epsilon^\ast.$$

By compactness up to subsequences we have $\lim_{n\to\infty} x_n=x^\ast\in S$, $\lim_{n\to\infty} y_n=y^\ast\in S$, and since $S$ is finite for some $K\in\mathbb{Z}_{\geq0}$, $x_k=x^\ast$ and $y_k=y^\ast$ for any $k\in\mathbb{Z}_{\geq K}$. In particular $x^\ast\neq y^\ast$ and $d_{[k+1,k+M^\ast]}^f(x^\ast,y^\ast)\leq \epsilon^\ast$ for any $k\in\mathbb{Z}_{\geq K}$. Applying the lemma with $p=K$ we also have $d_{[0,K]}^f(x^\ast,y^\ast)\leq \epsilon^\ast$, so that $d_{\mathbb{Z}_{\geq0}}^f(x^\ast,y^\ast)\leq \epsilon^\ast$, a contradiction to the $\epsilon^\ast$-expansivity of $f$.

It's clear that the theorem implies that the circle does not carry any one-sided expansive homeomorphism.