Denjoy’s theorem on circle diffeomorphisms

Jordan Bell
April 3, 2014

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 S1=/. For α, define Rα:S1S1 by Rα(x)=x+α+.

We say that a homeomorphism f:S1S1 is orientation preserving if it lifts to an increasing homeomorphism F:: πF=fπ.

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


One proves that this is independent both of the lift F of f and the point x. Some facts about the rotation number: it is an invariant of topological conjugacy, and ρ(f) is rational if and only if f has a periodic point. A periodic point is xS1 such that fn(x)=x for some n1.

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:S1S1 is an orientation preserving homeomorphism that is topologically transitive with irrational rotation number ρ(f), then f is topologically conjugate to Rρ(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:S1S1 be a C1 diffeomorphism and let J be an interval in S1. Let g=logf. If the interiors of J,f(J),,fn-1(J) are pairwise disjoint, then for any x,yJ and any nZ we have


The intervals [x,y],[f(x),f(y)],,[fn-1(x),fn-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 the sum coming from any particular partition or a subset of that partition.

Var(g) k=0n-1|g(fk(y))-g(fk(x))|
= |logk=0n-1f(fk(y))-logk=0n-1f(fk(x))|
= |log((fn)(x))-log((fn)(y))|.

Now we can prove Denjoy’s theorem.

Theorem 2.

If f:S1S1 is a C1 diffeomorphism that is orientation preserving, that has irrational rotation number ρ(f), and whose derivative f:S1R has bounded variation, then f is topologically conjugate to Rρ(f).


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 ω(x) is perfect and nowhere dense, and is independent of the point x. (Recall that ω(x)=n1infi(x)¯.) It follows that there is an interval I=(a,b) in its complement.

The intervals fn(I), n, are pairwise disjoint, for otherwise f would have a periodic point. Let μ be Haar measure on S1. Then


Let xS1. Suppose for the moment that there are infinitely n1 such that the intervals (x,f-n(x)),(f(x),f1-n(x)),,(fn(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


To see the equality in the above line it helps to write out what (f-n)(x) is.

Then for infinitely many n we have

μ(fn(I))+μ(f-n(I)) = I(fn)(x)𝑑x+I(f-n)(x)𝑑x
= I((fn)(x)+(f-n)(x))𝑑x
= Iexplog((fn)(x)(f-n)(x))𝑑x
= exp(-12Var(g))μ(I).

Since μ(I)>0 this implies that nμ(fn(I))=, a contradiction. Therefore f is topologically transitive, and so it is topologically conjugate to the Rρ(f). ∎

It is indeed necessary that f has bounded variation. Brin and Stuck give an example on p. 161 that they attribute to Denjoy: for any irrational number ρ(0,1), there is a nontransitive orientation preserving C1 diffeomorphism of S1 with rotation number ρ. The only condition of Denjoy’s theorem that isn’t satisfied here is that f have bounded variation.