DYNAMICS OF MAPS OF THE CIRCLE

ANTONIO DONALD

Abstract. This paper applies the basics of dynamical systems to circle home-

omorphisms. We lift the circle to the real line and introduce a dynamical in-

variant, the rotation number, to analyze a map’s behavior. We demonstratemany properties of rotation number, such as its continuous dependence on

choice of homeomorphism. We finish by proving the Poincare classification

theorem, which states that all circle maps with irrational rotation number aresemi-conjugate to a rotation.

Contents

1. Introduction 12. Preliminary Definitions — An Introduction to Dynamics 23. Maps of the Circle 33.1. What is a map of the circle? 33.2. Rotations of the Circle 43.3. An example of a lift 53.4. Rotation Number 63.5. Continuity and ordering of circle homeomorphisms 104. Homeomorphisms of the Circle — The Poincare Classification 144.1. Rational Rotation Number 144.2. Irrational Rotation Number 16Acknowledgments 215. bibliography 21References 21

1. Introduction

Dynamics studies the patterns in a map under repeated iteration. In this paper,we will focus on the dynamics of circle homeomorphisms. We begin by studying thesimplest circle homeomorphisms, rotations. We will then discuss how we can userotations to study the dynamics of arbitrary circle homeomorphisms. Our goal isto examine when circle homeomorphisms are dynamically equivalent to rotations.Our main tool in this discussion is an invariant called the rotation number, whichis due to Poincare.

Date: August 2020.

1

2 ANTONIO DONALD

2. Preliminary Definitions — An Introduction to Dynamics

In this section, we introduce some basic definitions. A (discrete) dynamicalsystem is a pair (X, f), where X is a space and f : X → X is a map. In this paper,we restrict our attention to the behavior of homeomorphisms of the circle and thereal line.

Definition 2.1. A map f : X → Y is a homeomorphism if f is continuous, injec-tive, surjective, and f−1 is continuous. If the first r derivatives f, f ′, f ′′, · · · , f (r)exist and are continuous, then f is said to be Cr continuously differentiable. A mapf : X → Y is said to be a Cr diffeomorphism if f is a Cr homeomorphism and f−1

is also Cr.

For a given pair (X, f), as before, dynamics studies the orbits of f .

Definition 2.2. Let f : X → X be a map. The forward orbit of a point x ∈ X,denoted byO+(x), is the sequence {fn(x)}n=∞n=0 . If f is invertible then the backwardsorbit of a point x ∈ X, denoted by O−(x), is the sequence {fn(x)}n=0

n=−∞. Forinvertible f we define the full orbit of a point x ∈ X, denoted by O(x), as thesequence {fn(x)}n=∞n=−∞.

Notation 2.3. By fn(x) we will always mean the nth iterate of a point x. If wewant to refer to the nth derivative of a function f at point x, we will write f (n)(x).

The simplest orbits to understand are those of periodic points and fixed points.

Definition 2.4. Let f : X → X be a map. A point x ∈ X is a fixed point of f iff(x) = x. A point x ∈ X is a periodic point of f with period n, if fn(x) = x. Thesmallest n ∈ N such that fn(x) = x is called the prime period of x.

Remark 2.5. If x is a period n point of f , then x is a fixed point for fn.

Notions of equivalence are useful because it can often be easier to analyze onemap or object than another similar one.

Definition 2.6. Let f : X → X and g : Y → Y be maps. Then, f is semi-conjugate to g if there exists a surjective continuous map h : X → Y such thath◦f = g ◦h. If h is also injective (so h is a bijection), then f and g are topologicallyconjugate.

The definition of topological conjugacy, unlike that of topological semi-conjugacy,is symmetric in f and g. The definition of a topological conjugacy is parallel tothat of a change of basis in Linear Algebra. That is, it can be helpful to thinkof topological conjugacy as changing the space of a map while maintaining itsdynamics. Topological conjugacy is a notion of equivalence because it preservesorbits. On the other hand, for a semi-conjugacy, the dynamics of f are containedin g but not necessarily the other way round. For further reading, Chapter 1.7 in[2] has a good discussion of the material.

Sometimes we want to analyze how much of a space an orbit of a dynamicalsystem covers. To this end, we introduce the following definitions.

Definition 2.7. Let E ⊂ X be a subset of space X. A point p ∈ E is a limit pointof E if every neighborhood of p contains some point q 6= p ∈ E.

Definition 2.8. The closure of D ⊂ X, denoted D, is defined as the union of Dand all it’s limit points. The set D is said to be dense in X if D = X.

DYNAMICS OF MAPS OF THE CIRCLE 3

For example, the set of all rational numbers is dense in the real numbers. Indynamics, we care about the density of orbits.

Definition 2.9. A map f : X → X is said to be topologically transitive if, for anytwo open intervals U, V ⊂ X, there exists k ∈ N such that fk(U) ∩ V 6= ∅.

In this paper, we only consider dynamical systems on the circle or on the realline. For a dynamical system f : X → X in one dimension, topological transitivityis equivalent to having at least one point x ∈ X, such that O+

f (x) is dense in X.

Definition 2.10. A dynamical system f : X → X is called minimal if every orbitof x ∈ X is dense in X.

3. Maps of the Circle

3.1. What is a map of the circle? There are several equivalent ways of viewingthe unit circle S1. One logical way is to think about the unit circle in the complexplane,

S1 = {x ∈ C : |x| = 1}.Euler’s identity, e2πix = cos(2πx) + i sin(2πx), allows us to lift the circle to the realline, which helps simplify constructions for many of the arguments in this paper.If unfamiliar with Euler’s identity observe that it follows by considering the Taylorseries of sin, cos, and the exponential function. It follows that

S1 = {x ∈ C : |x| = 1} = {cos(2πx) + i sin(2πx) : x ∈ R} = {e2πix : x ∈ R} = R/Z.Here, we wrap the whole real line around S1 as shown in Figure 1. With somethought, we can see that we can cover the circle with just the points in an integerinterval. For example, consider the interval [0, 1]. It follows that S1 = {e2πix : x ∈[0, 1]}. This gives rise to what we call additive notation, R/Z (read real numbersmodulo the integers), is an equivalent definition of the circle. We can actually use

0 1-1 2

The lift wraps each integer interval around the circle at θ=0

A . . - . §"

"R

g,

ftp.B.e""

. ]{eZtiZIzeZ}

Figure 1. Lifting S1 to the real line.

Euler’s identity to lift a map f : S1 → S1 of the circle to a map F : R→ R of thereal line. Define the exponential function π : R→ S1, which takes us from the realline to the circle, as π(x) = e2πix. This gives rise to the following definition,

4 ANTONIO DONALD

Definition 3.1. Let f : S1 → S1 be a homeomorphism of the circle. We call afunction F : R→ R a lift of the homeomorphism f if f ◦ π = π ◦ F . See Figure 2.

R F //

π��

R

π��

S1

f// S1

Figure 2. Commutative diagram of the lift of a circle map to thereal line.

In this paper, we only consider orientation-preserving homeomorphisms of thecircle.

Definition 3.2. Let f : S1 → S1 be a homeomorphism of the circle. Then,f is said to be orientation-preserving if we can lift f to a monotone increasinghomeomorphism F : R→ R.

Remark 3.3. Suppose f : S1 → S1 is an orientation-preserving homeomorphism ofthe circle, and let F : R → R be a lift of f . For any θ ∈ S1, there are countablymany x ∈ R such that π(x) = θ; that is, if π(x) = θ, then π(x + 1) = θ. Since fis an orientation-preserving homeomorphism, f(θ) is constant. Therefore, we havethat F (x+ 1) = F (x) + 1 and more generally that F (x+ z) = F (x) + z for z ∈ Z.

The property in Remark 3.3 allows us to bound the difference of lifts of twopoints.

Proposition 3.4. Let f : S1 → S1 be an orientation-preserving homeomorphismof the circle, and let F be a lift of f . If |x − y| < 1, then |F (x) − F (y)| < 1. Itfollows that |Fn(x)− Fn(y)| < 1.

Proof. Since f is orientation-preserving F is monotone increasing. Hence,

|F (x)− F (y)| < |F (x+ 1)− F (x)| = F (x) + 1− F (x) = 1.

�

Remark 3.5. At this point, it is important to note that for each orientation-preserving circle homeomorphism, f , there are countably many lifts F . The liftonly cares about where in an integer interval each point on the circle is placed.Since there are infinitely many of these intervals, we can lift to any one of them.Hence, there are countably infinite lifts differing only by integer values.

3.2. Rotations of the Circle. The simplest example of an orientation-preservingcircle homeomorphism is rotation.

Definition 3.6. Rotation of the circle by α ∈ R is the map Rα(e2πiθ) : S1 → S1,defined by e2πiθ 7→ e2πi(θ+α). In additive notation, rotation is the map Rα(x) :R/Z→ R/Z, defined by x 7→ x+ α.

DYNAMICS OF MAPS OF THE CIRCLE 5

The dynamics of the rotation map are very different when α is rational versuswhen α is irrational. In fact, the dichotomy between rational and irrational rotationwill be a thread throughout this paper.

Suppose that α is rational, that is α = pq for p, q ∈ Z relatively prime. Then,

any point θ ∈ S1 is fixed by Rqα, since Rqα(θ) = θ+ 2π pq q = θ+ 2pπ = θ. Therefore,

every point θ ∈ S1 is a periodic point of Rα with period q. The story is entirelydifferent for irrational α.

Proposition 3.7. If α is irrational, then the map Rα(θ) = θ + 2πα is minimal(see Definition 2.10) in S1.

Proof. Fix α ∈ R\Q and θ ∈ S1. First, we will show that all points in the orbit ofθ are distinct. Then, we will show that the orbit of θ is dense in S1. Suppose thatall the points on the orbit of θ are not distinct, i.e that

Rnα(θ) = θ +m2πα = θ + n2πα = Rnα(θ), and m 6= n.

Then, we must have that (n−m)α ∈ Z. However, since α is irrational, any non-zerointeger multiple of α is also irrational. So, n must equal m, which is a contradiction.Hence, all points in the orbit of θ are distinct.

Fix ε > 0. We need to show that the orbit O(θ) is dense in S1, i.e that for anyarc I ⊂ S1 of length ε, we have that O(θ)∩I 6= ∅. The circle is compact. Therefore,given any infinite sequence of points in O(θ) ⊂ S1, there exists a convergent sub-sequence. Thus, there exist n,m ∈ N such that n > m and |Rnα(θ) −Rmα (θ)| < ε.Rotation is an isometry, or a map that preserves distance. So, since we are rotatingintervals around the circle, the lengths of these intervals must be constant. Thus,we have |Rkα(θ)− θ| < ε.

Let us consider the shortest closed arc between Rkα(θ) and θ. Iteration by therotation map gives us

(θ,Rkα(θ)) 7→ (Rkα(θ),R2kα (θ)) 7→ (R2k

α (θ),R3kα (θ)) 7→ · · · .

We have partitioned the circle into arcs of length ε each containing a point of theorbit of θ. Since ε and θ were picked arbitrarily, O(θ) is dense in S1 and Rα(θ) isminimal in S1. �

3.3. An example of a lift. In this section, we introduce a family of circle mapsto visualize the lift of a circle homeomorphism.

Definition 3.8. The family of circle maps {fω,γ : S1 → S1|ω, γ ∈ R}, which wewill call the standard family, is defined as

(3.9) fω,γ(θ) = θ + 2πω + γ sin(θ).

The map fω,γ is not always a homeomorphism. The choice of ω rotates the mapand does not affect whether fω,γ is a homeo/diffeomorphism. When γ = 0, fω,0 isjust rotation by ω. When 0 ≤ γ < 1, fω,γ is a diffeomorphism. When γ = 1, fω,γ isonly a homeomorphism. Finally, when γ > 1, fω,γ fails to be injective. We will notprove these statements, but the graphs in Figure 3 should make them believable.For a more rigorous treatment, see Chapter 14 in [2].

For fixed ω, γ ∈ R, the collection of maps Fω,γ,k : R→ R, defined by

Fω,γ(x) = x+ ω +γ

2πsin(2πx) + k, where k ∈ Z,

is the collection of lifts of fω,γ .

6 ANTONIO DONALD

(a) F.1(x) = x+√2

2+.1 sin(x) (b) F1(x) = x+

√22

+ sin(x) (c) F2(x) = x+√2

2+2 sin(x)

Figure 3. Lift Fω,γ of fω,γ with ω =√22 and γ = 0.1, 1, and 2

respectively.

Figure 3 shows the lifts of the standard family where ω =√22 and γ is equal to

0.1, 1, and 2 respectively.

3.4. Rotation Number. The central question in this paper is when does an ar-bitrary circle homeomorphism behave like a rotation? To answer this question,we introduce an invariant called rotation number. Given an orientation-preservinghomeomorphism of the circle f , rotation number will tell us which rotation, Rα byα ∈ R, f most behaves like. In fact, we will be able to understand a lot about acircle homeomorphism’s dynamics just by knowing it’s rotation number. Our firststep will be to introduce what is called a lift number. The lift number is a valuedetermined by the choice of lift F of a circle homeomorphism f . We will see thatall lift numbers differ only by integer values, which gives us a natural way to definethe rotation number which will depend only on choice of circle homeomorphism f .

Theorem 3.10. Let f : S1 → S1 be an orientation-preserving homeomorphism of

the circle, and let F : R → R be a lift of f . The limit lim|n|→∞Fn(x)−x

n exists forall x ∈ R, and its value is independent of choice of x ∈ R. We will define the lift

number τ(F ) := lim|n|→∞Fn(x)−x

n . Furthermore, given any two lifts, F1 and F2,their lift numbers differ by an integer value, i.e. τ(F1)− τ(F2) = z ∈ Z.

Proof. Assume that the limit lim|n|→∞Fn(x)−x

n exists for all x ∈ R. We will showthat the lift number, τ(F ), is independent of choice of x ∈ R. Fix x, y ∈ [0, 1).Then, ∣∣∣∣ (Fn(x)− x)− (Fn(y)− y)

n

∣∣∣∣ ≤ |Fn(x)− Fn(y)|+ |x− y|n

≤ 2

n.

Since 2n → 0 as n→∞, the limit does not depend on choice of x ∈ R.

We will now show that given any orientation-preserving circle homeomorphismf , and any lift F : R→ R of f , the lift number exists. For this proof, we will breakup the argument into two cases: when f has a periodic point and when f does nothave a periodic point.

DYNAMICS OF MAPS OF THE CIRCLE 7

Case 1: Suppose f has a periodic point of period n ∈ N, i.e for some θ ∈ S1,there exists m > 0 such that fm(θ) = 0. Fix x ∈ R such that π(x) = θ. We knowthat Fm(x) = x+ z for some z ∈ Z. It follows from Remark 3.3 that F jm = x+ jzfor j ∈ Z. Immediately, we have that

limj→∞

|F jm(x)− x|jm

= limj→∞

jz

jm=

z

m.

We want to show that

limj→∞

|F jm(x)− x|jm

= limn→∞

Fn(x)− xn

.

We can write any integer n ∈ N as n = jm + r, where r ∈ N is the remainder,0 ≤ r < m. Since r is bounded above, we know that there must exist some M > 0such that, for all x ∈ R, |F r(x)− x| < M . It immediately follows that

|Fn(x)− x− (F jm(x)− x)|n

=|F r(F jm(x))− F jm(x)|

n≤ M

n.

Since Mn → 0 as n→∞, we have shown that

τ(F ) = limn→∞

Fn(x)− xn

= limj→∞

|F jm(x)− x|jm

=k

m.

We have not only shown that τ(F ) exists if f has a periodic point, but also thatτ(F ) must be rational.

Case 2: Suppose that f does not have any periodic points. Fix a lift F of f .Since f has no periodic points, we know that, for all x ∈ R, Fn(x) − x is neveran integer for n 6= 0. Therefore, there must exist some integer z ∈ Z such thatz < Fn(x)− x < z + 1 for all x ∈ R. For simplicity, fix x = 0. It follows that

(3.11) z < Fn(0)− 0 < z + 1.

From (3.4) if we take x = Fn(0), then z < F 2n(0) − fn(0) < z + 1. Furthermore,for m ∈ N, we have that z < Fmn(0)−F (m−1)n(0) < z+ 1. Therefore, we can sumover these m inequalities to get mz < Fmn(0) − 0 < m(z + 1). Dividing by n, wehave that

(3.12)z

n<Fmn(0)− 0

mn<z + 1

n.

We can divide by n in (3.11) and combine with (3.12) to get

(3.13)

∣∣∣∣Fmn(0)− 0

mn− Fn(0)

n

∣∣∣∣ < 1

n.

Repeating the exact same argument with m and n interchanged we get

(3.14)

∣∣∣∣Fmn(0)− 0

mn− Fm(0)

m

∣∣∣∣ < 1

m.

Combining (3.13) and (3.14), it is immediate that∣∣∣∣Fn(0)− 0

n− Fm(0)− 0

m

∣∣∣∣ < 1

n+

1

m.

So, the sequence {Fn(0)−0n }n∈Z ⊂ R is Cauchy and must converge.

Let F1 and F2 be different lifts of f . We want to show that τ(F1)−τ(F2) = z ∈ Z.Since there are countably many lifts F of f that differ only by integer values,

8 ANTONIO DONALD

we know that there exists k ∈ Z such that F1(x) = F2(x) + k. It follows thatFn1 (x) = Fn2 (x) + nk. Hence, τ(F1) = τ(F2) + k. �

We have shown that the lift numbers τ(F ) are well defined and differ only byinteger values, which naturally leads to the following definition:

Definition 3.15. Let f : S1 → S1 be a homeomorphism and let F : R → R be alift of f . We define the rotation number of f as

τ(f) := τ(F ) mod 1 = lim|n|→∞

Fn(x)− xn

mod 1.

One of the useful properties of rotation number is that it is a dynamical invariant.A dynamical invariant is a property that is preserved under topological conjugacy.

Notation 3.16. Let [x] := max{k ∈ Z|k ≤ x}.Theorem 3.17. If f : S1 → S1 and h : S1 → S1 are orientation-preservinghomeomorphisms of the circle, then τ(h−1fh) = τ(f).

Proof. This proof follows that of Proposition 11.1.3 in [3]. Let F and H be liftsof f and h respectively such that F (0), H(0) ∈ [0, 1). We note that (H−1FH)n(x)telescopes and is equal to (H−1FnH)(x). So, by definition of rotation number, weneed to show that

limn→∞

∣∣∣∣ (H−1FnH)(x)− Fn(x)

n

∣∣∣∣ = 0.

Our first step will be to show that H−1 is indeed a lift of h−1. This is true because

π ◦H−1 = h−1 ◦ h ◦ π ◦H−1 = h ◦H−1 ◦ π ◦H ◦H−1 = h−1 ◦ π.Here, the second equality follows because H is a lift of h, so h ◦π = π ◦H. We alsoneed to establish that H−1 ◦ F ◦H is a lift of h−1 ◦ f ◦ h. To do this, we compute

π ◦H−1 ◦ F ◦H = h−1 ◦ π ◦ F ◦H = h−1 ◦ f ◦ π ◦H = h−1 ◦ f ◦ h ◦ π,where in each of the equalities, we use the fact that H−1, F , and H are lifts ofh−1, f , and h respectively.

Since H(0) ∈ [0, 1), for x ∈ [0, 1) we have that −1 < H(x)−x ≤ H(x) < H(1) <2. It follows that |H(x)−x| < 2 for all x ∈ R. A similar process with H−1 gives usthe estimate

(3.18) |H−1(x)− x| < 2 ∀x ∈ R.If |y − x| < 2, then |Fn([y])− Fn([x])| < 3. This gives us the estimate

−3 ≤ [y]− [x]− 1 = Fn([y])− Fn([x] + 1)

≤ Fn(y)− Fn(x)

≤ Fn([y] + 1)− Fn([x])

= [y] + 1− [x] ≤ 3. (3.19)

Combining (3.18) and (3.19) we get

|(H−1FnH)(x)−Fn(x)| ≤ |(H−1FnH)(x)−FnH(x)|+|FnH(x)−Fn(x)| < 2+3 = 5,

where the first inequality is just the triangle inequality. The second inequalityfollows by taking FnH(x) as x in (3.18) and taking H(x) as y in (3.19). We arenow done as

limn→∞

∣∣∣∣ (H−1FnH)(x)− Fn(x)

n

∣∣∣∣ ≤ limn→∞

5

n= 0.

DYNAMICS OF MAPS OF THE CIRCLE 9

�

We now prove a lemma that allows us to work with the rotation number ofiterated circle homeomorphisms, fm.

Lemma 3.20. Suppose f : S1 → S1 is an orientation-preserving homeomorphismof the circle with rotation number τ(f). Then, for m ∈ Z\{0}, τ(fm) = mτ(f).

Proof. Observe that

τ(fm) = limn→∞

(Fm)n(x)− xn

= m limn→∞

Fmn(x)− xmn

= mτ(f) mod 1.

�

Proposition 3.21. Let f : S1 → S1 be an orientation preserving homeomorphismof the circle. Then, f has rational rotation number if and only if it has a periodicpoint.

Proof. We have already shown that if f has a periodic point then f has rationalrotation number in Theorem (3.10)

We now show the other direction. Suppose τ(f) = pq ∈ Q, where p and q are

relatively prime. Lemma 3.20 tells us that τ(fq) = qτ(f) = p mod 1 = 0. Hence, itsuffices to show that if τ(f) = 0, then f has a fixed point.

Suppose that f does not have a fixed point. Choose a lift F of f such that F (0) ∈[0, 1). Then, for all x ∈ R, F (x)− x /∈ Z. If not, π(x) would be a fixed point for f .Since F (0) ∈ [0, 1), then 0 < F (x)−x < 1. Furthermore, F (x)−x is continuous onthe compact interval [0, 1], so it must attain a minimum and a maximum on [0, 1].Hence, there exists some δ > 0 such that 0 < δ ≤ F (x) − x ≤ 1 − δ < 1, which,by periodicity, holds for any choice of x ∈ R. In particular, we can take x = F i(0),where 0 ≤ i ≤ n− 1. It follows that

nδ ≤n−1∑i=0

F i+1(0)− F i(0) = Fn(0) ≤ n(1− δ).

The sum telescopes, so we have that δ ≤ Fn(0)n ≤ 1 − δ. So, as n → ∞, τ(f)

does not approach 0. But, we assumed τ(f) = 0. Hence, we have arrived at acontradiction, which means that f must have a fixed point. �

Proposition 3.21 immediately implies that if f : S1 → S1 is an orientation-preserving homeomorphism, then τ(f) is irrational if and only if f has no periodicpoints. In addition, if τ(f) is rational, then in fact all periodic points have the sameperiod.

Proposition 3.22. Let f : S1 → S1 be a homeomorphism with rational rotationnumber τ(f) = p

q ∈ Q, p, q ∈ Z relatively prime. Then, all periodic points are of

period q.

Proof. Choose x ∈ R such that π(x) ∈ S1 is a periodic point of f . We need toshow that there exists some lift F of f , for which F q(x) = x + p. Since π(x) isperiodic, for any lift F we have that F i(x) = x + j where i, j ∈ Z. Take k ∈ Z.Then, because t(f) = p

q we have

k +p

q= τ(f) = lim

n→∞

F in(x)− xin

= limn→∞

jn

in=j

i,

10 ANTONIO DONALD

where the second equality follows from the definition of rotation number and thethird equality holds by Lemma 3.20.

Fix a lift F of f such that k = 0. We have that pq = j

i . It follows that there

exists m ∈ N such that j = mp and i = mq. We want to show that F q(x) −p = x by showing that F q(x) − p ≥ x and that F q(x) − p ≤ x. We will showboth inequalities by contradiction. First, assume that F q(x) − p > x. Since fis orientation-preserving, then F is monotone increasing. Thus, it follows thatF 2q(x) − 2p = F q(F q(x) − p) − p ≥ F q(x) − p > x. Furthermore, by inductionit follows that Fmq(x) −mp(x) > x for all m ∈ N. But, we know that Fmq(x) −mp(x) = F i(x)− j = x, so we have a contradiction! The other case, where F q(x)−p < x, follows similarly. Hence, we know that F q(x)−p ≤ x and that F q(x)−p ≥ x.Therefore, F q(x)− p = x. �

3.5. Continuity and ordering of circle homeomorphisms. The goal of thissubsection is to show that the rotation number depends continuously on the choiceof the homeomorphism f . To this end, we will define an ordering of orientation-preserving circle homeomorphisms. Then, given a one-parameter family of orientation-preserving homeomorphisms {fω}, we will see that for fp/q ⊂ {fω} with rationalrotation number, there is an interval around fp/q where all other circle homeomor-phisms have the same rotation number. On the other hand, there is never such aninterval for fir ⊂ {fω} with irrational rotation number.

To show that the rotation number depends continuously on choice of orientation-preserving homeomorphisms, we must specify a distance for Cr(X,X) maps.

Definition 3.23. Let f : X → X and g : X → X be two Cr maps defined on X.Then, we define the Cr-distance between f and g as

dr(f, g) := supx∈X{|f(x)− g(x)|, |f ′(x)− g′(x)|, · · · , |f (r)(x)− g(r)(x)|}.

Proposition 3.24. Let f : S1 → S1 be an orientation-preserving homeomorphismof the circle. For every ε > 0, there exists δ > 0 such that if g : S1 → S1

is an orientation-preserving homeomorphism of the circle and d0(f, g) < δ, then|τ(f)− τ(g)| < ε.

Proof. This proof follows that of Corollary 14.7 in [2]. Fix ε > 0, and choose n ∈ Nsuch that 2

n < ε. Let F be a lift of f . Then, r− 1 < Fn(0) < r+ 1 for some r ∈ Z.Choose δ > 0, and a lift G of g such that r− 1 < Gn(0) < r+ 1. From Lemma 3.20we have that m(r−1) < Fmn(0) < m(r+1) and that m(r−1) < Gmn(0) < m(r+1).Since the length of the interval (r − 1, r + 1) is 2, we get that for all m ∈ N,∣∣∣∣Fnm(0)

nm− Gnm(0)

nm

∣∣∣∣ < 2

n< ε.

Since, by definition, limm→∞Fnm(0)nm = τ(F ) and limm→∞

Gnm(0)nm = τ(G), it follows

that |τ(f)− τ(g)| < ε. �

We will now present an ordering of orientation-preserving homeomorphisms ofthe circle, adapted from Chapter 14B of [4]. This is not the only such ordering (seeProposition 11.1.8 of [3]).

In Section 2, we presented topological conjugacy between maps as a way to studythe orbit structure of a simpler map to understand the orbit structure of a morecomplicated map. In that vein, we introduce symbolic dynamics here to create

DYNAMICS OF MAPS OF THE CIRCLE 11

a topological conjugacy between the circle and the sequence space. Representingthe orbit of a point via a sequence of 0s and 1s gives us a methodical method forcomparing orientation-preserving circle homeomorphisms.

Definition 3.25. We define the sequence space Σ2 := {s = (s0s1s2 · · · ) : sj =0 or 1} of all infinite sequences of 0s and 1s. We define a distance function fors = (s0s1s2 · · · ), t = (t0t1t2 · · · ) ∈ Σ2 by

d[s, t] =

∞∑i=0

|si − ti|2i

.

In fact, Σ2 paired with d[s, t] is a metric space. For an in depth discussion of theconstruction of the sequence space see Chapter 6 in [2].

Let f : R/Z → R/Z be an orientation-preserving homeomorphism of the circle,and choose F : R → R to be the unique lift of f such that 0 ≤ F (0) < 1. ByRemark 3.3, F (1) = F (0)+1 > 1. Since F is continuous on [0, 1] and by Remark 3.3we know that F (1) = F (0) + 1 > 1. The intermediate value theorem therefore tellsus that there must be some point ξ−1 ∈ [0, 1] such that F (ξ−1) = 1. Then, wedefine I0 := [0, ξ−1), which by monotonicity of F is the half closed, half openinterval consisting of all x ∈ [0, 1] such that F (x) < 1. Similarly, let I1 = [ξ−1, 1],which by monotonicity of F is the closed interval containing all points x ∈ [0, 1]such that F (x) ≥ 1.

We define the discontinuous lift map F : [0, 1]→ [0, 1) as

F(x) = F (x)− sj(x) where

{sj(x) = 0 if x ∈ I0sj(x) = 1 if x ∈ I1.

It follows that F is just the normal lift map for x ∈ I0, but for x ∈ I1, F shiftsF (x) from [1, 2) to [0, 1).

1

0 1

Figure 4. F for F (x) = x+ 13 + 1

2π sin(2πx).

Definition 3.26. Let ξ−1 7→ 0 7→ ξ1 7→ ξ2 7→ · · · be the orbit of ξ−1 underF . We define the itinerary of ξ−1 to be the sequence S(F ) = s1s2s3 · · · , wheresk = 0 if ξk ∈ I0, and sj = 1 if ξk ∈ I1. Notice that the itinerary of ξ−1 beginswith s1(ξ1), because for any map s−1 = 1 and s0 = 0, by design.

12 ANTONIO DONALD

Proposition 3.27. If f is an orientation-preserving homeomorphism, and F isthe unique lift of f such that 0 ≤ F (0) < 1, then

τ(F ) = limm→∞

s1 + · · ·+ smm

.

Proof. First, we will prove by induction that Fm(0) =∑mi=1 sm+ξm. Our induction

hypothesis holds for m = 1 as F (0) = s1 + ξ1 = 0 + ξ1 = F (0) (since we choseF with F (0) ∈ [0, 1)). Now we prove the induction step. Suppose that Fm(0) =s1 + · · ·+ sm + ξm for some m ∈ N. Then,

Fm+1(0) = F (s1+· · ·+sm+ξm) = s1+· · ·+sm+F (ξm) = s1+· · · sm+sm+1+ξm+1.

Now, it follows that s1 + · · ·+ sm ≤ Fm(0) < s1 + · · ·+ sm + 1. Then,

(3.28) limm→∞

s1 + · · ·+ smm

≤ τ(F ) ≤ limm→∞

s1 + · · ·+ sm + 1

m.

Since limm→∞s1+···+sm

m = limm→∞s1+···+sm+1

m , the squeeze theorem implies τ(f) =

limm→∞s1+···+sm

m . �

We now have the machinery to introduce an ordering of the orientation-preservinghomeomorphisms!

Definition 3.29. Let F and G be lifts of orientation preserving homeomorphismsf and g of the circle such that 0 ≤ F (0), G(0) < 1. We say S(F ) < S(G) if thereexists m ∈ N such that sk(F ) = sk(G) for k < m, and sm(F ) < sm(G) .

We now have an ordering of circle homeomorphisms! We will show that thisordering carries over for rotation numbers of circle homeomorphisms.

Proposition 3.30. Let f and g be orientation preserving homeomorphisms of thecircle. Let F and G be lifts of f and g, respectively, such that 0 ≤ F (0), G(0) < 1.If S(F ) < S(G), then 0 ≤ τ(F ) ≤ τ(G) < 1. It follows that τ(f) ≤ τ(g). Theinequalities become strict when τ(F ) or τ(G) is irrational.

Proof. Assume that S(F ) < S(G). It follows, by definition, that s1(F ) + · · · +sm(F ) + 1 = s1(G) + · · ·+ sm(G), since sm(F ) + 1 = 0 + 1 = sm(G). Then, directlyfrom (3.28), we get,

τ(F ) ≤ s1(G) + · · ·+ sm(G)

m≤ τ(G).

Suppose without loss of generality that τ(F ) is irrational. Then the left inequality

is strict because s1(G)+···+sm(G)m ∈ Q.

�

Proposition 3.31. Suppose f : S1 → S1 is an orientation preserving homeomor-phism with rational rotation number τ(f) = p

q ∈ Q. Then there exists a neigh-

borhood If around f such that all orientation-preserving circle homeomorphismsfi ∈ If have rotation number τ(fi) = p

q .

Proof. See Proposition 11.1.10 in [3]. �

We now give an example to illustrate the results of this section.

DYNAMICS OF MAPS OF THE CIRCLE 13

Example 3.32. Let’s return to the standard family fω,ε(θ) = θ + 2πω + ε sin(θ),where ω, ε ∈ R. Consider the lift Fω,ε(x) = x + ω + ε

2π sin(2πx). Fix ε ∈ (0, 1] sothat f is a homeomorphism. We can order the lift, as before, by choice of ω; noticethat if ω1 > ω2, then Fω1

(x) > Fω2(x) for all x ∈ R. Therefore it follows that

Fnω1> Fnω2

, so τ(fω1) ≥ τ(fω2

). We have already shown that at irrational values ofrotation number this inequality must be strict. Now, suppose that τ(fω0

) = pq ∈ Q.

For this family, we can show directly that there exists a neighborhood η about ω0

such that τ(fω) = pq for ω ∈ η.

Figure 5. Lift of fω,1(θ) = θ + 2πω + sin(θ), where ω1 =√22 in

blue and ω2 = 1 in purple.

We know that there exists θ ∈ S1 such that fq(θ) = θ. So, choose x0 ∈ [0, 1)such that π(x0) = θ. Now, we break the proof up into cases.

First, suppose that (F qω0)′(x0) 6= 1. Then, the graph of F q intersects the line

y = x+p at (x0, x0+p). We can apply the implicit function theorem to the functionG(x, ω) = (x + p) − (F qω(x0)), since G(x0, ω0) = 0 and ∂G

∂ω 6= 0. This immediatelytells us that there exists a neighborhood η of ω0 such that F qω(x0) = x0 + p for allω ∈ η. So, the maps fω with ω ∈ η have rotation number τ(fω)) = p

q .

For our second case, suppose that (F qω0)′(x0) = 1. Notice that F qω0

is analytic(can be represented by a convergent power series). Thus, there exists j ∈ N suchthat (F qω0

)(n) 6= 0; otherwise, F qω0would be identically equal to x + p. If j is odd,

then F qω0(x0) cannot be a maximum or minimum since (F qω0

)(j)(x0) 6= 0. Therefore,sufficiently nearby perturbations must also cross the line y = x+p. If j is even, thenF qω0

is either concave up or concave down at x0 (not an inflection point). Hence,the one sided neighborhoods, where ω < ω0 and ω > ω0 respectively, must crossy = x+ p. So, they must also have rotation number τ(fω)) = p

q . For more details,

see Example 14.10 in [2].So, we have shown that for this family of functions, rotation number depends

monotonically and continuously on ω. At rational values of τ(f), nearby circlehomeomorphisms have the same rotation number. Whereas at irrational valuesof τ(f), there are no constant intervals of τ(f). Hence, the graph of τ(fω) is a

14 ANTONIO DONALD

Cantor function (see Proposition 11.1.11 in [3]). Figure 6 is an example of a Cantorfunction.

ω

τ(f)

DAD

Figure 6. Devil’s staircase of rotation number τ(fω,ε) forfω,ε(θ) = θ + 2πω + ε sin(θ).

4. Homeomorphisms of the Circle — The Poincare Classification

In this section we answer the big question: when does a circle homeomorphismbehave like a rotation? First, we will look at orientation-preserving circle homeo-morphisms f : S1 → S1 with rational rotation numbers τ(f) = p

q ∈ Q. We will

show that the orbits of periodic points of f have the same order (see Proposition 4.2)in the unit interval as the orbit of 0 under R p

q. We also see that circle homeomor-

phisms with rational rotation number are very rarely even semi-conjugate to ratio-nal rotation R p

q. Since all orbits of R p

qare periodic and have the same period, q,

any sort of conjugacy will result in a map with only periodic points. However, mostorientation-preserving circle homeomorphisms that are not rotation do not havethis sort of orbit structure. Due to the difference in behavior of irrational rotationof the circle (described in Proposition 3.7), orientation-preserving circle homeomor-phisms with irrational rotation number can be conjugate to an irrational rotation,which is the main result of this paper.

Theorem 4.1 (Poincare Classification Theorem). Let f : S1 → S1 be an orientation-preserving homeomorphism with irrational rotation number τ(f) ∈ R\Q.

(1) If f is topologically transitive in S1 then f is conjugate to rotation by τ(f),Rτ(f).

(2) If f is not topologically transitive in S1 then f is semi-conjugate to rotationby τ(f), Rτ(f).

4.1. Rational Rotation Number. We begin this section by discussing the orderof periodic orbits of orientation-preserving circle homeomorphisms with rationalrotation number.

Proposition 4.2. Let f : R/Z → R/Z be an orientation-preserving circle homeo-morphism with rational rotation number τ(f) = p

q ∈ Q, with p, q relatively prime.

We know that there exists x ∈ S1 such that fq(x) = x. Let m,n ∈ N∪0 be such that0 ≤ m,n ≤ q−1. The orbit of x under f has the same order in the unit interval as

DYNAMICS OF MAPS OF THE CIRCLE 15

the orbit of 0 under R pq

. By order, we mean that if mpq mod 1 ≤ npq mod 1, then

fm(x) ≤ fn(x).

Proof. For a complete proof refer to Proposition 11.2.1 in [3]. �

Definition 4.3. Let (X, d) be a metric space, and let f : X → X be a homeomor-phism. A point x ∈ X is said to be homoclinic to a point y ∈ X if

lim|n|→∞

d(fn(x), fn(y)) = 0.

A point x ∈ X is said to be heteroclinic to points y1 and y2 ∈ X if

limn→∞

d(fn(x), fn(y1)) = 0 and limn→∞

d(fn(x), fn(y2)) = 0.

See Figure 7.

Homoclinic Orbit Heteroclinic Orbit

Goes both backwards and forwards to the same periodic point.

Goes backwards to a different period point than forwards.

HEAFigure 7. Example of Homoclinic and Heteroclinic Orbits.

Definition 4.4. Let f : X → X be a map. Suppose that p ∈ X is a periodic pointof f with period n. A point x ∈ X is forward asymptotic to p if limm→∞ fmn(x) =p. Similarly, if f is invertible, a point y ∈ X is backwards asymptotic to p iflimm→−∞ fmn(y) = p.

Definition 4.5. Let f : X → X be a map. A point y ∈ X is said to be

(1) an ω-limit point for x ∈ X if there exists a sequence {nk}∞k=0 going to +∞such that fnk(x)→ y as k →∞.

(2) an α-limit point for x ∈ X if there exists a sequence {nk}∞k=0 going to −∞such that fnk(x)→ y as k → −∞.

Lemma 4.6. Let J = [a, b] ⊂ R be a closed interval. Suppose that f : J → J isan orientation-preserving homeomorphism. Then, all x ∈ J either are fixed pointsor are positively and negatively asymptotic to adjacent fixed points. That is, forx0 ∈ J not fixed, x0 is negatively asymptotic to the maximum of Fix(f) ∩ [a, x0],and positively asymptotic to the minimum of Fix(f)∩ [x0, b] (note that we can takea maximum and a minimum as the set of fixed points for f cannot be dense in J).

16 ANTONIO DONALD

Proof. Since f is orientation-preserving, we know that it is monotone increasing.Fix x ∈ J . Without loss of generality, suppose f(x) > x (the case where f(x) < xis similar). Then, f2(x) = f(f(x)) > f(x), and inductively, fn(x) = f(fn−1(x)) >fn−1(x). So, {fn(x)}n∈N is monotone increasing and bounded above by b. There-fore, the sequence {fn(x)} converges to some y = supn∈N{fn(x)} ∈ J . Further-more, by continuity of f , we get that

f(y) = f( limn→∞

fn(x)) = limn→∞

fn+1(x) = y.

So, x ∈ J is forward asymptotic to fixed point y ∈ J , and the ω−limit set of {fn(x)}is y. Since f is a bijection, we can apply the same argument to f−1 to show thatx is backwards asymptotic to some fixed z ∈ J . �

When we apply this lemma to the circle, it shows that for an orientation-preserving circle homeomorphism with rational rotation number, non-periodic or-bits are asymptotic to periodic orbits. The following proposition adapted fromProposition 11.2.2 in [3] fully explains the details.

Proposition 4.7. Let f : S1 → S1 be an orientation-preserving homeomorphismwith rational rotation number p

q ∈ Q. Then, there are two possible outcomes for the

non-periodic orbits of f .

(1) Suppose that f has exactly one periodic orbit. Then every point not in theperiodic orbit is heteroclinic under fq to two adjacent points on the periodicorbit. These points are distinct if τ(f) 6= 0, i.e periodic points do not haveperiod one.

(2) Suppose f has more than one periodic orbit. Then, each non-periodic pointis heteroclinic under fq to two points on separate periodic orbits.

Proof. Let θ ∈ S1 be fixed by fq, i.e. fq(θ) = θ. Let F be a lift of f . Then, fix somex ∈ R such that π(x) = θ. We know that F q(x)− p = x and F q(x+ 1)− p = x+ 1.If we restrict F q to the closed interval [x, x+ 1], then (1) follows immediately fromLemma 4.6.

Now, suppose that f has more than one periodic orbit. Lemma 4.6 tells us thatany non-periodic y ∈ S1 is asymptotic to adjacent periodic points in S1. Hence,all we need to show is that these two periodic orbits are always distinct. Supposenot, i.e that there exists some interval I = [a, b] ⊂ R such that a and b are adjacentzeros of F q(x)− x− p, and a, b are on the same periodic orbit. Therefore, we havethat π(a) = x ∈ S1 and π(b) = y ∈ S1, so fk(x) = y for 0 < l ≤ q − 1. Byassumption, there are no periodic points in the interval π(a, b) ⊂ S1. This means

that⋃q−1n=0 f

nkπ(a, b) contains no periodic points. We can see that⋃q−1n=0 f

nkπ(a, b)covers S1\{fnπ(a)}, which is the whole circle minus the periodic orbit of a, b. But,this means there is only one periodic orbit, which is a contradiction! Hence, wecannot have two adjacent zeros of F q(x)−x−p from the same orbit. So, Lemma 4.6proves (2). �

4.2. Irrational Rotation Number. We will now build up machinery to provethe Poincare Classification Theorem.

Definition 4.8. Let (X, d) be a metric space. A set E ⊂ X is a Cantor set if it istotally disconnected, perfect subset of X. In other words, E contains no intervals,is closed, and every point of E is a limit point of E.

DYNAMICS OF MAPS OF THE CIRCLE 17

Note that our presentation of the Poincare classification follows that of [3] fromProposition 11.2.4 onwards.

Lemma 4.9. Let f : S1 → S1 be an orientation preserving homeomorphism withirrational rotation number τ ∈ R\Q. Let m and n be integers such that m < n. Forany point θ ∈ S1, define an interval I = [fm(θ), fn(θ)]. Note that I is non-trivialsince f cannot have periodic points and it does not matter which of the two possiblearcs for I is chosen. Then, for any x ∈ S1, the positive and negative orbits of xboth intersect I, i.e {fn(x)}∞n=0 ∩ I 6= ∅ and {fn(x)}−∞n=0 ∩ I 6= ∅.

Proof. Fix θ ∈ S1. Without loss of generality, consider the forward orbit {fn(x)}∞n=0.Our strategy will be to show that the backwards iterates of I cover S1, so allpoints eventually end up in I under forward iteration. Define the interval Ik :=f−k(n−m)(I). Notice that for all k ∈ N, Ik and Ik−1 share an endpoint.

Suppose that S1 is not covered by backwards iterates of I, i.e that that S1 *∪k∈NIk. Then, the sequence of endpoints converge to some point ξ ∈ S1. So,

ξ = limk→∞

f−k(m−m)fm(θ) = limk→∞

f−k+1(n−m)fm(θ)

= limk→∞

f−k+1(n−m)fm(θ)

= limk→∞

fn−mf (−k)(n−m)fm(θ)

= f (n−m) limk→∞

f−k(m−m)fm(θ)

= f (n−m)(ξ),

where the first equality holds because the end points of Ik and Ik−1 are the same.The next two equalities hold by expanding parenthesis. The final equality holds bydefinition of ξ. This is a contradiction as ξ would be a periodic point of f , but fhas irrational rotation number so it cannot have any periodic points. Hence, thebackwards iterates of I cover the whole circle, and we are done. �

Proposition 4.10. Let f : S1 → S1 be an orientation-preserving homeomorphismwith irrational rotation number τ(f) ∈ R\Q. Then the ω-limit set ω(x) is indepen-dent of x for any x ∈ S1. We will denote ω := ω(x).

Proof. We need to show that for all x, y ∈ S1, ω(x) = ω(y). Fix x, y ∈ S1 andz ∈ ω(x) ⊂ S1. Then, there exists a sequence {ln} ⊂ N such that f ln(x) → z asln →∞. Define an interval I = [fm(x), fn(x)]. By Lemma 4.9, there exists km ∈ Nsuch fkm(y) ∈ Im = [f lm , f lm+1 ]. Then, we can take the limit limm→∞ fkm(y) =limm→∞[f lm , f lm+1 ] = z. Hence, if z ∈ ω(x), then z ∈ ω(y). So, ω(y) ⊂ ω(x) andsimilarly, ω(x) ⊂ ω(y), and we have that ω(x) = ω(y). �

Proposition 4.11. Let f : S1 → S1 be an orientation-preserving homeomorphismwith irrational rotation number τ(f) ∈ R\Q. The set ω from Proposition 4.10 iseither the whole of S1 or a Cantor set in S1.

Proof. The key to this proof is the observation that ω is the only minimal, closed,nonempty, f -invariant proper subset of S1. We will now show this observation.Since the orbit of any point x ∈ S1 is well-defined, ω must be nonempty. Becauseω ⊂ S1 is the ω-limit set of any point x ∈ S1, every limit point of x ∈ S1 must becontained in ω, so ω is closed.

18 ANTONIO DONALD

Suppose that ω is not the only f -invariant set of this type, i.e. suppose that A ⊂S1 is non-empty, closed, and f -invariant. But, then for x ∈ A, {fn(x)}n∈Z ⊂ A.So, ω(x) ⊂ A. This means that A must be the union of all its ω-limit sets. Hence,A = ω. Therefore, the only closed invariant subsets of ω are the empty-set and ωitself.

Let ∂ω denote the boundary of ω in S1. We know from analysis that ∂ω mustbe a closed invariant subset of ω. So, we have two possible cases, either ∂ω = ∅ or∂ω = ω. If ∂ω = ∅, then ω = S1.

If, on the other hand, ∂ω = ω, then we want to show that ω is a Cantor set.To do this, we need to show that ω is nowhere dense and perfect. Since ∂ω = ω,ω is nowhere dense. So, all that remains is to show that ω is perfect. We havealready shown that ω is closed. Fix x ∈ ω. We need to show that x is a limit pointof ω. From Proposition 4.10 we know that ω = ω(x). So, there exists a sequence{ln} ⊂ Z, such that f ln(x) → x as n → ∞. Since ω is f -invariant and x ∈ ω,f ln ∈ ω for all n ∈ N. Hence, x is a limit point of ω. �

Lemma 4.12. Let f : S1 → S1 be an orientation-preserving homeomorphism withirrational rotation number τ(f) ∈ R\Q (we will denote τ(f) as τ). Let F : R→ R bea lift of f . Let n1, n2,m1,m2 be (not necessarily distinct) integers. Finally, let x bea point in R. Then, n1τ+m1 < n2τ+m2 if and only if Fn1(x)+m1 < Fn2(x)+m2.

Proof. This proof follows that of Proposition 11.2.4 in [3]. Fix n1, n2,m1,m2 ∈Z. We will show the if direction first, i.e if Fn1(x) + m1 < Fn2(x) + m2, thenn1τ +m1 < n2τ +m2. Our first step is to show that the inequality Fn1(x) +m1 <Fn2(x) + m2 is independent of choice of x ∈ R. To this end, define p : R → R asp(a) = Fn1(a)+m1−(Fn2(a)+m2). We claim that p never changes sign. Given theproperties of the lift, we know that p is continuous. Hence, if it did change sign thenthere would exist some z ∈ R such that p(z) = 0, i.e Fn1(z)− Fn2(z) = m2 −m1.Since m2 and m1 are integers, m2 −m1 ∈ Z, which implies that π(z) is a periodicpoint for f . This is a contradiction, since f has irrational rotation number and thuscannot have periodic points. Therefore, Fn1(x)+m1 < Fn2(x)+m2 is independentof choice of x ∈ R.

Now, fix x = 0 and define y = Fn2(0). Since F is increasing, we have thatFn1−n2(y)− y < m1 −m2. We already showed that this inequality is independentof the choice of y ∈ R, so we can set y = 0. It then directly follows that Fn1−n2(0) <m2−m1. Replacing 0 with Fn1−n2(0), we get that Fn1−n2(Fn1−n2(0))−Fn1−n2(0) <m2 −m1. So, F 2(n1−n2)(0) < (m2 −m1) + Fn1−n2(0) < 2(m2 −m1). Inductively,it follows that Fn(n1−n2) < n(m2 −m1). Hence,

τ(f) = limn→∞

Fn(n1−n2)(0)− 0

n(n1 − n2)< limn→∞

n(m1 −m2)

n(n1 − n2)=m2 −m1

n1 − n2,

where the inequality is strict since τ(f) is irrational. It directly follows that n1τ +m1 < n2τ +m2.

To show the only if direction, we first apply the exact same argument to Fn1(0)+m1 > Fn2(0) + m2, to imply that n1τ + m1 > n2τ + m2. We can never haveequality in Fn1(0) +m1 > Fn2(0) +m2, because f has no periodic points. We canalso never have equality in n1τ +m1 > n2τ +m2, because τ is irrational. Supposethat the only if direction did not follow, i.e that n1τ + m1 < n2τ + m2 impliesFn1(0)+m1 > Fn2(0)+m2. Immediately, we have that Fn1(0)+m1 > Fn2(0)+m2

DYNAMICS OF MAPS OF THE CIRCLE 19

implies n1τ + m1 > n2τ + m2, which is a clear contradiction. Therefore, the onlyif direction must follow. �

Finally, we have reached the point where we can prove the Poincare classificationtheorem (Theorem 4.1).

Proof of the Poincare classification theorem. We are going to construct a continu-ous monotone semi-conjugacy between any circle homeomorphism with irrationalrotation number and its corresponding rotation.

Pick a lift F : R → R of f . We will denote τ(f) by τ . Fix θ ∈ S1. Choosex ∈ R such that π(x) = θ. Define the set B = {Fn(x) + m|m,n ∈ Z}, which isthe complete lift of the orbit {fn(π(x))}n=∞n=−∞. Define the map Rτ : R → R byRτ (x) = x+τ . The notation Rτ is a bit of an abuse of notation, which is deliberateas we want to think of the map as additive rotation on the real line. Our strategyis to construct a monotone continuous map H : B → R that is a semi-conjugacybetween F |B and Rτ on R. We will then extend this map to the whole real line tocreate a semi-conjugacy between F and Rτ . Finally, we will show that H descendsto a semi-conjugacy between f : S1 → S1 and Rτ .

Let us define H : B → R by Fn(x) + m 7→ nτ + m. Immediately, Lemma 4.12tells us that for n1, n2,m1,m2 ∈ Z, if Fn1(x)+m1 < Fn2(x)+m2, then n1τ+m1 <n2τ +m2. So, H(Fn1(x)+m1) = n1τ +m1 < n2τ +m2 = H(Fn2(x)+m2). Hence,H is monotone.

It follows immediately from Example 3.7 that H(B) is dense in R. So, on B ⊂ Rwe have that H ◦ F = Rτ ◦H, since for any (Fn(x) +m) ∈ B

H ◦ F (Fn(x) +m) = H ◦ (Fn+1(x) +m)

= (n+ 1)τ +m

= Rτ ◦ (nτ +m)

= Rτ ◦H(Fn(x) +m).

Now, in order to extend H to R, we will first extend H continuously to the closureof B, denoted B. Fix y ∈ B\B. By definition, there exists a sequence {xn}n∈N ⊂ Bsuch that y = limn→∞ xn. We want to be able to define H(y) = limn→∞H(xn).To do this, we need to show that limn→∞H(xn) exists and doesn’t depend onthe choice of our sequence xn. Since H is monotone, we know that the liminf andlimsup, estimating y from below and above, must exist and be independent of choiceof sequence. Now, suppose that these limits did not agree. Then, R\H(B) wouldhave to contain an interval, as H is continuous on B. But, we have already shownthat H(B) is dense in R, so this cannot be the case. We have thus shown that theleft and right limits agree, and that we can extend H : B → R as a continuousmonotone map.

Finally, we want to extend H to the whole real line. Notice that B projects,under π, to the closure of the orbit of π(x), which is exactly ω(π(x)) = ω. So, Bis the lift of ω. From Proposition 4.11, we know that ω is either the whole of S1

or a Cantor set in S1. We will break our argument up into two cases. In the casethat f is transitive, B = R, so we have already extended H to R. Furthermore, weshowed that H must be injective, and thus a conjugacy.

Otherwise, B is a Cantor set. We know that H : B → R is continuous, monotoneand surjective, since H is monotone and continuous on B, and H(B) is dense in R.Hence, to extend H to R we need to define H on intervals R\B. We can simply set

20 ANTONIO DONALD

H as constant and equal to the endpoints on these intervals. This is possible sinceH is continuous on Cantor set B, which is just a set of points. Hence, if on theintervals between these points, H is constant, then H must still be continuous onthe extension. Note that since H is constant on intervals, it fails to be injective andis only a semi-conjugacy. We have now constructed a semi-conjugacy H : R → Rsuch that H ◦ F = Rτ ◦H (see Figure 8).

R F //

H��

R

H��

RRτ

// R

Figure 8. Commutative diagram of semi-conjugacy between liftF and Rτ = x+ τ .

We now want to show that H : R→ R descends to a semi-conjugacy h : R/Z→R/Z. For z ∈ B we have

H(z + 1) = H(Fn(z + 1) +m) = H(Fn(z) +m+ 1) = nτ +m+ 1 = H(z) + 1.

Therefore, our conjugacy is periodic at integer intervals, and we have shown thatf : S1 → S1 is semi-conjugate to Rτ (θ) = θ + 2πτ by the monotone continuousmap h : S1 → S1 (see Figure 9).

R/Zf //

h

��

R/Z

h

��R/Z

Rτ(f)

// R/Z

Figure 9. Commutative Diagram of the Semi-Conjugacy betweenf and Rτ(f).

�

DYNAMICS OF MAPS OF THE CIRCLE 21

Table 1 is based on the table on page 399 of [3] and summarizes the results weproved in this section. Let f : S1 → S1 be an orientation-preserving

homeomorphism of the circle. We have proved the followingτ(f) = p

q ∈ Q τ(f) ∈ R\QPoints in periodic orbit. Case one: f is transitive.All points in periodic orbit haveprime period q. Orbit is ordered inthe same way as orbit R p

q.

The map f is topologically transi-tive, so orbit is dense in S1 and or-dered in the same way as Rτ(f).

Points outside periodic orbit. Case two: f is not transitive.Case 1: f has one periodic orbit,so points are heteroclinic under fq

to two points on that one periodicorbit.

The map f is not topologically tran-sitive, so the orbit is dense in a Can-tor set.

Case 2: f has more than one peri-odic orbit, so points are heteroclinicunder fq to two points on separateorbits.

Points outside Cantor set are homo-clinic to the Cantor set.

Table 1. Poincare Classification Table.

Acknowledgments

I would like to thank my mentor, Meg Doucette, for the incredible amount ofguidance and support she has given me throughout the whole REU program. Shemade an otherwise dreary summer quite enjoyable, so I am very grateful. I wouldlike to thank Peter May for getting the program up and running given all thedifficulties that 2020 has brought forward. Finally, I would like to thank DaniilRudenko for all the work organizing the apprentice program.

5. bibliography

References

[1] de Melo, W. and van Strien, S. One-Dimensional Dynamics. Springer. 1993.

[2] Devaney, R. An Introduction to Chaotic Dynamical Systems Second Edition. Addison-WesleyPublishing Company. 1989.

[3] Katok, A. and Hasselblatt, B., Introduction to the modern theory of dynamical systems,Cambridge University Press, Cambridge, 1995.

[4] Milnor, J. Introductory Dynamics Lectures Notes. http://www.math.stonybrook.edu/ jack/DYNOTES/

[5] Rudin W. Principles of Mathematical Analysis, Third Edition. McGraw-Hill Inc. 1976.[6] Desmos Graphing Calculator. https://www.desmos.com/calculator 2020