Denjoy’s theorem on circle diffeomorphisms

In this note I’m just presenting the proof of Denjoy’s theorem in Michael Brin and Garrett Stuck’s Introduction to dynamical systems, Cambridge University Press, 2002.

Let $S^1=ℝ/\mathbb{Z}$. For $\alpha \in ℝ$, define $R_{\alpha }:S^1\to S^1$ by $R_{\alpha }(x)=x+\alpha +\mathbb{Z}$.

We say that a homeomorphism $f:S^1\to S^1$ is orientation preserving if it lifts to an increasing homeomorphism $F:ℝ\to ℝ$: $\pi \circ F=f\circ \pi$.

The rotation number of an orientation preserving homeomorphism $f$ is defined by

$$\rho (f)=\underset{n\to \mathrm{\infty }}{lim}\frac{F^n(x)-x}{n}.$$

One proves that this is independent both of the lift $F$ of $f$ and the point $x\in ℝ$. Some facts about the rotation number: it is an invariant of topological conjugacy, and $\rho (f)$ is rational if and only if $f$ has a periodic point. A periodic point is $x\in S^1$ such that $f^n(x)=x$ for some $n\ge 1$.

There are some lemmas in Chapter 7 that I don’t want to write out. The important theorem that we’re going to use without proof is that if $f:S^1\to S^1$ is an orientation preserving homeomorphism that is topologically transitive with irrational rotation number $\rho (f)$, then $f$ is topologically conjugate to $R_{\rho (f)}$. This reduces our problem to showing that a map is topologically transitive.

We will use the following lemma in the proof of Denjoy’s theorem.

Lemma 1.

Let $f\mathrm{:}{S}^{\mathrm{1}}\mathrm{\to }{S}^{\mathrm{1}}$ be a $C^{\mathrm{1}}$ diffeomorphism and let $J$ be an interval in $S^{\mathrm{1}}$. Let $g\mathrm{=}\log \mathit{}{f}^{\mathrm{\prime }}$. If the interiors of $J\mathrm{,}f\mathit{}\mathrm{(}J\mathrm{)}\mathrm{,}\mathrm{\dots }\mathrm{,}{f}^{n\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}J\mathrm{)}$ are pairwise disjoint, then for any $x\mathrm{,}y\mathrm{\in }J$ and any $n\mathrm{\in }\mathrm{Z}$ we have

$$\mathrm{Var}(g)\ge |\log ({(f^n)}^{\prime }(x))-\log ({(f^n)}^{\prime }(y))|.$$

Proof.

The intervals $[x,y],[f(x),f(y)],\mathrm{\dots },[f^{n-1}(x),f^{n-1}(y)]$ are pairwise disjoint, so they are part of a partition of $[0,1]$. The total variation of $g$ is defined as a supremum over all partitions, so in particular it will be $\ge$ the sum coming from any particular partition or a subset of that partition.

$\mathrm{Var}(g)$ $\ge$ ${\displaystyle \sum _{k=0}^{n-1}}|g(f^k(y))-g(f^k(x))|$

$\ge$ $\left|{\displaystyle \sum _{k=0}^{n-1}}g(f^k(y))-g(f^k(x))\right|$

$=$ $\left|\log {\displaystyle \prod _{k=0}^{n-1}}{f}^{\prime }(f^k(y))-\log {\displaystyle \prod _{k=0}^{n-1}}{f}^{\prime }(f^k(x))\right|$

$=$ $|\log ({(f^n)}^{\prime }(x))-\log ({(f^n)}^{\prime }(y))|.$

∎

Now we can prove Denjoy’s theorem.

Theorem 2.

If $f\mathrm{:}{S}^{\mathrm{1}}\mathrm{\to }{S}^{\mathrm{1}}$ is a $C^{\mathrm{1}}$ diffeomorphism that is orientation preserving, that has irrational rotation number $\rho \mathit{}\mathrm{(}f\mathrm{)}$, and whose derivative $f^{\mathrm{\prime }}\mathrm{:}{S}^{\mathrm{1}}\mathrm{\to }\mathrm{R}$ has bounded variation, then $f$ is topologically conjugate to $R_{\rho \mathit{}\mathrm{(}f\mathrm{)}}$.

Proof.

Suppose by contradiction that $f$ is not topologically transitive. It’s a fact proved in Chapter 7 of Brin and Stuck that this implies that $\omega (x)$ is perfect and nowhere dense, and is independent of the point $x$. (Recall that $\omega (x)={\bigcap }_{n\ge 1}\overline{{\bigcup }_{i\ge n}{f}^i(x)}$.) It follows that there is an interval $I=(a,b)$ in its complement.

The intervals $f^n(I)$, $n\in \mathbb{Z}$, are pairwise disjoint, for otherwise $f$ would have a periodic point. Let $\mu$ be Haar measure on $S^1$. Then

$$\sum _{n\in \mathbb{Z}}\mu (f^n(I))\le 1.$$

Let $x\in S^1$. Suppose for the moment that there are infinitely $n\ge 1$ such that the intervals $(x,f^{-n}(x)),(f(x),f^{1-n}(x)),\mathrm{\dots },(f^n(x),x)$ are pairwise disjoint; we shall prove that this is true later. By applying the lemma we proved with $y=f^{-n}(x)$ we get

$$\mathrm{Var}(g)\ge \left|\log \frac{{(f^n)}^{\prime }(x)}{{(f^n)}^{\prime }(y)}\right|=|\log ({(f^n)}^{\prime }(x){(f^{-n})}^{\prime }(x)|.$$

To see the equality in the above line it helps to write out what ${(f^{-n})}^{\prime }(x)$ is.

Then for infinitely many $n$ we have

$\mu (f^n(I))+\mu (f^{-n}(I))$ $=$ ${\displaystyle {\int }_I}{(f^n)}^{\prime }(x)𝑑x+{\displaystyle {\int }_I}{(f^{-n})}^{\prime }(x)𝑑x$

$=$ ${\displaystyle {\int }_I}({(f^n)}^{\prime }(x)+{(f^{-n})}^{\prime }(x))𝑑x$

$\ge$ ${\displaystyle {\int }_I}\sqrt{{(f^n)}^{\prime }(x){(f^{-n})}^{\prime }(x)}𝑑x$

$=$ ${\displaystyle {\int }_I}\sqrt{\exp \log ({(f^n)}^{\prime }(x){(f^{-n})}^{\prime }(x))}𝑑x$

$\ge$ ${\displaystyle {\int }_I}\sqrt{\exp (-|\log ({(f^n)}^{\prime }(x){(f^{-n})}^{\prime }(x))|)}𝑑x$

$\ge$ ${\displaystyle {\int }_I}\sqrt{\exp (-\mathrm{Var}(g))}𝑑x$

$=$ $\exp \left(-{\displaystyle \frac{1}{2}}\mathrm{Var}(g)\right)\mu (I).$

Since $\mu (I)>0$ this implies that ${\sum }_{n\in \mathbb{Z}}\mu (f^n(I))=\mathrm{\infty }$, a contradiction. Therefore $f$ is topologically transitive, and so it is topologically conjugate to the $R_{\rho (f)}$. ∎

It is indeed necessary that $f^{\prime }$ has bounded variation. Brin and Stuck give an example on p. 161 that they attribute to Denjoy: for any irrational number $\rho \in (0,1)$, there is a nontransitive orientation preserving $C^1$ diffeomorphism of $S^1$ with rotation number $\rho$. The only condition of Denjoy’s theorem that isn’t satisfied here is that $f^{\prime }$ have bounded variation.