1 Circle maps

Several techniques used for the analysis of unimodal maps have their originin the theory of circle maps. To avoid annoying factors of 2π, we define thecircle S1 as R\Z. We shall sometimes consider the dynamics on R (usuallyfor the purpose of computations) and sometimes on S1 (for conceptual pur-poses). A circle map is a homeomorphism of the circle. Note that circlemaps are invertible, in contrast with unimodal maps. The simplest exampleof a circle map is a rotation. For any angle α ∈ R, we define

Rα(x) = x + α mod 1. (1.1)

There is a fundamental difference between rational and irrational rotations.It is easy to show that if α ∈ Q every orbit is periodic. If α 6∈ Q then no orbitis periodic and every orbit is dense (Exercise 1). This example illustratesthe typical dichotomy between the presence and absence of periodic orbits.We shall mainly focus on homeomorphisms that do not have periodic orbits,for the following reason.

Proposition 1.1. Suppose f is a circle map that has a periodic point. Thenevery orbit is asymptotic to a periodic orbit.

Proof. Suppose fp(x∗) = x∗. Observe that S1\x∗ is an interval. Since fp

is a homeomorphism, it is a monotone map of this interval, thus every orbitfpk, k ≥ 0 asymptotes to a fixed point of fp.

We shall first focus on the topological theory of circle maps. This isthe study of properties of a circle map that remain invariant under home-omorphisms. The fundamental topological invariant for circle maps is therotation number introduced by Poincare. The classical approach to the ro-tation number is introduced in the Exercises. Here we follow a longer routebeginning with the idea of first return times. The connection with the classi-cal theory is that the sequence of first return times constitutes the continuedfraction expansion of the rotation number. This approach is well suited tothe study of a similar invariant for unimodal maps.

In all that follows, we will consider only orientation preserving home-omorphisms. The extension to orientation reserving homeomorphisms aretrivial.

1.1 Lifting circle maps to an interval

We first reduce circle maps to discontinuous maps of the unit interval. Letg : S1 → S1 be a homeomorphism, and g : R → R a lift of g. There is a

1

unique point c ∈ [0, 1] such that g(c) is an integer. We assume that g hasno fixed points, so that c ∈ (0, 1). We define the map f : [0, 1] → [0, 1] by

f(x) =

g(x) mod 1, x ∈ [0, 1]\c,0, x = c.

(1.2)

Note that limx↓c f(x) = f(c), that is f is right-continuous. This is an arbi-trary, but useful convention. Note also that f(0) = f(1) ∈ (0, 1). Figure 1.1illustrates this class of maps. Conversely, it is easy to check that every f ofthis form lifts to a circle map. The point c is called the critical point. Note

f (c)

c f(c )−f(c)

2

Figure 1.1: Lift of a circle map to the class S(I)

that with this definition of S1, there is a unique c associated to every mapg.

In order to define renormalization of circle maps, let us isolate the es-sential features of these maps in slightly greater generality. Let I denote aclosed interval. The class S(I) consists of maps from I → I such that

1. f has a unique point of discontinuity, denoted c, in the interior of I.We call c the critical point of f .

2. f is continuous and strictly increasing in each component of I\c andis right-continuous at c. The left limit f(c−) is the right-endpoint ofI, and the right limit f(c+) is the left-endpoint of I.

3. f maps both boundary points to a single point in the interior of I.

Though c is a point of discontinuity for f , notice that it is a point of conti-nuity for f2. We will use this observation often.

2

1.2 Symbolic dynamics

Let us first introduce the space of symbol sequences.

Definition 1.2. Σ denotes the space of sequences 0, c, 1N with typicalelement x = (x0, x1, . . . , xn, . . .). The shift map σ : Σ → Σ is defined by

(σx)n = xn+1, n ≥ 0. (1.3)

We order Σ lexicographically with the convention 0 < c < 1. That is,x ≺ y if xk < yk where k is the first index where xj 6= yj. The naturaldistance on a space of symbols is usually the Hamming distance . If a, b ∈0, c, 1, the distance d(a, b) is 0 if a = b and 1 otherwise. This distance canbe used to induce a distance between two elements of Σ. We set

d(x, y) =

∞∑

k=0

d(xk, yk)

2k. (1.4)

The metric space (Σ, d) is compact and totally disconnected. σ is a contin-uous function on this metric space.

There is a natural sequence associated to any orbit. The interval I maybe decomposed into

I = I0 ∪ c ∪ I1,

where I0 and I1 denote the subintervals to the left and right of c.

Definition 1.3. The address of a point x ∈ I is the symbol 0 if x ∈ I0, c ifx = c, and 1 if x ∈ I1.

Definition 1.4. Let f ∈ S(I) and x ∈ I. The itinerary of x under the mapf is the sequence of addresses of (x, f(x), . . . , fn(x), . . .) and is denoted

if (x) = (i0(x), i1(x), . . .) , (1.5)

where ik(x) is the address of fk(x).

The following monotonicity lemma is fundamental. Heuristically, it tellsis that the lexicographical ordering on Σ is ‘correct’ for circle maps. Thisshould be contrasted with a similar lemma for unimodal maps where thelexicographical ordering has to be replaced by a different ordering.

Lemma 1.5. Let f ∈ S(I).

(a) If x < y then if (x) if (y).

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(b) If if (x) ≺ if (y) then x < y.

(c) if (fk(x)) = σk(if (x)).

The assertion in (a) cannot be strengthened to strict inequality withoutfurther assumptions. If f has a periodic orbit, then we may have if (x) =if (y) for x, y in an interval.

Proof. (a) Suppose x < y. If if (x) = if (y) there is nothing to prove. There-fore, let k be the first index in which the sequences if (x) and if (y) differ.We must show that ik(x) < ik(y). Since ij(x) = ij(y), 0 ≤ j < k, the pointsf j(x) and f j(y) lie in the same subinterval for 0 ≤ j < k (what happensif ij(x) = c for some j < k?). Since f is increasing on both I0 and I1 wehave inductively f(x) < f(y), f2(x) < f2(y), . . . , fk−1(x) < fk−1(y). Sinceik(x) 6= ik(y), the monotonicity of f now implies ik(x) < ik(y) as desired.

(b) This is almost identical to the previous argument. Let k be the firstindex where ik(x) < ik(y). This implies fk(x) < fk(y) and f j(x) and f j(y)are contained in the same subinterval for 0 ≤ j < k. But now we may usethe monotonicity of f to inductively deduce that fk−1(x) < fk−1(y), . . . ,f2(x) < f2(y), f(x) < f(y) and x < y.

(c) This is left as an exercise.

A general theme in symbolic dynamics is to deduce properties of an orbitvia knowledge of if (x). The following lemma is an example of this approach.

Lemma 1.6. If if (x) is periodic with period p, then f has a periodic pointwith period not greater than p.

Proof. We may as well assume c does not appear in if (x). If it did, period-icity of if (x) implies fp(c) = c and we are done.

Let Ir, 0 ≤ r ≤ p − 1, denote the smallest closed interval containingthe points f r(x), fp(f r(x)), f2p(f r(x)), . . .. Then Ir is contained withinI0 or I1 depending on whether ir(x) is 0 or 1. In either case, f is monotoneincreasing on Ir, and maps it homeomorphically to Ir+1. We compose mapsto find fp : I0 → I0 is a homeomorphism. Thus, fp has a fixed point.

The basic invariants we will consider are the symbol sequences

K+(f) = limx↓c

if (x), K−(f) = limx↑c

if (x). (1.6)

Both limits exist by Lemma 1.5. Observe also that

K+(f) = (1, 0) · if (f2(c)), K−(f) = (0, 1) · if (f2(c)),

4

where the symbol · denotes concatenation of two strings. Thus, it will sufficeto consider K+(f).

Definition 1.7. Two maps f, f ∈ S(I) are combinatorially equivalent if theorbit Of (c) of c under f has the same order as the orbit of Of (c). That is,the map h : Of (c) → Of (c) is strictly order preserving.

For any n, the ordering of c, f(c), . . . , fn−1(c) defines a permutation of1, 2, . . . , n. Two maps are combinatorially equivalent if these permutationsare identical for every n.

Proposition 1.8. Suppose f, f ∈ S(I) are maps without periodic orbits.Then f and f are combinatorially equivalent if and only if K+(f) = K+(f).

Proof. The forward implication is trivial. If f and f are combinatoriallyequivalent, then the ordering of Of (c) and Of (c) is identical. Therefore,

if (f2(c)) = if (f2(c), and K+(f) = K+(f).

Let us now assume that K+(f) = K+(f). We must show that order ispreserved, that is for every k, l, fk(c) < f l(c) if and only if fk(c) < f l(c).The proof relies on a strengthened form of Lemma 1.5(a):

Claim: If fk(c) < f l(c), then if (fk(c)) ≺ if (f l(c)).

Proof of the claim. The point here is strict inequality. If the claim werefalse, we would have if (fk(c)) = if (f l(c)). But then Lemma 1.5(c) implies

if (f l(c)) = σl−k(

if (fk(c)))

= if (fk(c)).

Thus, if (fk(c)) is periodic with period l − k and Lemma 1.6 implies theexistence of a periodic orbit, contradicting our assumptions.

We now complete the proof of the Proposition. For any k, l ≥ 2 we have

fk(l) < f l(c) ⇒ if (fk(c)) ≺ if (f l(c))

⇒ if (fk(c)) ≺ if (f l(c)) ⇒ fk(c) < f l(c).

The second implication relies on the hypothesis K+(f) = K+(f) and thelast on Lemma 1.5(b).

The first part of the proof also shows that K+(g) is well-defined for acircle map. Indeed, it is easy to show that any two lifts f, f of g into theunit interval are combinatorially equivalent. Another proof of the converseis outlined in the exercises.

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1.3 First return times and the first return map

We now develop a concrete example of renormalization. Suppose J ⊂ I isa subinterval such that f l(x) ∪ J 6= φ for every x ∈ J and some positiveinteger l(x). We define the first return time

k(x) = minl > 0∣

∣

∣f l(x) ∈ J (1.7)

Maps in the class S(I) satisfy either f(I0) ⊂ I1 or f(I1) ⊂ I0 (why?). Iff(I0) = I1 then f simply permutes the intervals, and f2 has a fixed point.We ignore this case. To be concrete, we shall always denote the interior ofthe ‘smaller’ interval by J0, and the interior of the ‘larger’ interval by J1

in order that f(J0) ⊂ J1. (This may mean that J0 = I1 as in the examplebelow). We also set J = I. We sometimes abuse notation and let the sameletter denote both the open and closed interval with the same endpoints.

The return time k depends on the point x. It is continuous, except atpreimages of c. The following times provide uniform control.

a(f) = minl∣

∣

∣J0 ∩ f l+1(J0) 6= φ (1.8)

Lemma 1.9. Suppose f ∈ S(J) has no fixed points. a(f) is the smallestinteger such that the closure of J0, f(J0), . . ., fa(f)+1(J0) covers the closedinterval J .

Proof. Let cn denote the orbit of c and a′ = supl∣

∣fk(J0) ⊂ J1 for 1 ≤ k ≤l. For k ≤ a′ the closed intervals fk(J0) are [ck+1, ck]. Thus, they areordered and have a common endpoint. Since f does not have a fixed point,a′ < ∞. It follows that a′ = a.

We use a(f) to define the first return map

F(f)(x) = fk(x)(x), (1.9)

whose domain is the interval

J(f) = closure(

J0 ∪ fa+1(J0))

. (1.10)

There are two cases to consider. Our interest is usually in the second case.

Lemma 1.10. (a) If c ∈ fa(f)(J0) then c is a fixed point of fa(f)+1. Inthis case, fa(f)+1(J0) = J0, J(f) = J0, and F(f) is fa(f)+1.

6

(b) If not, J(f) strictly contains J0, the first return map is in S(J(f)),and given by

F(f)(x) =

fa(f)+1(x), x ∈ J0,f(x), x ∈ J1 ∩ J(f).

(1.11)

Proof. This follows directly from the geometry in the proof of Lemma 1.9.

1.4 Renormalization of rotations

To illustrate these definition, let us compute the first return map when f isa rotation Rα. We assume α is irrational, and α ∈ (0, 1/2). In this case,

J0 = I1 = (1 − α, 1), f(J0) = (0, α), . . . , fk(J0) = ((k − 1)α, kα),

provided kα < 1 − α. Therefore, we obtain

a(f) =

⌊

1

α

⌋

− 1, and J(f) = [a(f)α, 1]. (1.12)

Here byc = maxk∈Zk ≤ y. The corresponding first return map is given by

F(f)(x) =

x + α, a(f)α ≤ x < 1 − α,x + (a(f) + 1)α, 1 − α ≤ x < 1.

(1.13)

If we plot the graph of F(f) we observe that it has the same form as thegraph of Rα. To make this precise, let h : [0, 1] → J(f), denote the affinemap with h(0) = a(f)α and h(1) = 1. The renormalization of Rα is definedby h−1 F(f) h. We use (1.12) and (1.13) to compute

R(Rα) = Rα′ , α′ =1

1 + G(α), α ∈ (0,

1

2), (1.14)

where the Gauss map G : (0, 1] → (0, 1] is defined by

G(x) =1

x−

⌊

1

x

⌋

. (1.15)

Thus, the renormalization of an irrational rotation is another irrational ro-tation. When α ∈ (1/2, 1) we have

J0 = I0 = (0, 1−α), f(J0) = (α, 1), . . . , fk(J0) = (k(α−1)+1, (k−1)(α−1)+1),

7

provided k(α−1)+1 > 1−α. The largest index that satisfies this inequalityis a(f) and after some algebra we find

a(f) =

⌊

1

1 − α

⌋

− 1, J(f) = [0, a(f)(α − 1) + 1].

The return map may be renormalized as before yielding

R(Rα) = Rα′ , α′ =G(1 − α)

1 + G(1 − α), α ∈ (

1

2, 1). (1.16)

Further analysis of this example, and the relation to continued fractions isexplored in the exercises.

1.5 The structure of K+(f)

We now iterate the construction of the first return map to unravel the struc-ture of K+(f). We inductively construct a sequence of first return times an,closest return intervals J (n) and closest return maps ϕn ∈ S(J (n)) as fol-lows. Suppose n ≥ 2 and suppose ak, J (k), and ϕk have been defined for1 ≤ k ≤ n − 1. If ϕ(n−1) has no fixed point, as in Section 1.3 we define

an = a(ϕn−1), J (n) = J(ϕn−1), ϕn = F(ϕn−1). (1.17)

If ϕn−1 has a fixed point we set an = ∞ and the process terminates. Theinitial conditions and first iterate are fixed as follows:

J (0) = J, ϕ0 = f. (1.18)

For n = 1, if J0 is to the right of c we set

a1 = a(f) + 1, J (1) = J(f), ϕ1 = F(f). (1.19)

(There is a slight inconsistency between the definition of a1 and an, n > 1.This notation proves convenient later in (1.24)).On the other hand, if J0 isto the left of c we set

a1 = 1, J (1) = J, ϕ1 = f. (1.20)

Let J(n)0 denote the interior of the left component of J (n)\c if n is odd,

and the right component if n is even. (The normalization in the first step ischosen to ensure this condition). The role of left and right is interchangedat each step by the procedure (1.17). We have

J(n)0 = J

(n−1)1 ∩ J (n), J

(n)1 = J

(n−1)0 ∩ J (n) = J

(n−1)0 . (1.21)

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The first return map is defined as in Lemma 1.10. We have

ϕ1(x) =

f(x), x ∈ J(1)0 ,

fa1 , x ∈ J(1)1 .

(1.22)

For n ≥ 2 we use (1.21) and Lemma 1.10 to obtain

ϕn|J(n)0

= ϕn−1|J(n−1)1

, ϕn|J(n)1

=(

ϕn−1|J(n−1)1

)an

(

ϕn−1|J(n−1)0

)

. (1.23)

To get a feel for this, write out the first few terms. For example, we have

ϕ2|J(2)0

= fa1, ϕ2|J(2)1

= fa1a2+1.

We define a sequence of closest return times by

qn+1 = an+1qn + qn−1, n ≥ 1, (1.24)

q0 = 1, q1 = a1.

The closest return maps ϕn, n ≥ 2 is then given explicitly by

ϕn|J(n)0

= f qn−1, ϕn|J(n)1

= f qn . (1.25)

This process is illustrated in Figures 1.2 and Figure 1.3.

0

J J

J

J

JJ

J

J

q q

q

q

q

c0 1

1

(0) (0)

(3)

(1)

(2)

(3)

(2)

(1)

1

1

3 2

2

1

1

10

0

0

1

Figure 1.2: The geometry of J (n). The points f qk(c) are denoted qk forbrevity. Observe the role of (1.21).

The terminology closest return time or interval is motivated by the fol-lowing lemma.

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n−1 q c

J0

J1

(n) (n)

q qqn−1

n+1 n+ qn

Figure 1.3: Note that qn+1 = qn−1 + an+1qn. The return map is f qn on theleft interval, and f qn−1 on the right.

Lemma 1.11. Suppose j is an index, 0 ≤ j ≤ qn+1. Then f j(c) ∈ J (n) ifand only if j = qn−1 + kqn for some index k, 0 ≤ k ≤ an+1.

Proof. The proof for n = 1 and n = 2 are explained in Figure 1.5. It is clearthat this generalizes.

c0 1

1

J J

c

f f q1

(0)

1(0)0

J(1)

0 J(1)

1

q q+11 1

Figure 1.4: All iterates f j(c) for 1 < j < q1 lie to the left of q1. Observethat c ∈ (f q1(c), f q1+1(c)) by definition. Moreover, the first return map is f

on J(1)0 , and f q1 on J

(1)0 .

Lemma 1.12. The union of ∪qn−1−1k=0 fk(J

(n)0 ) and ∪qn−1

k=0 fk(J(n)1 ) tiles the

interval J . That is, these intervals have disjoint interior, and the closure oftheir union covers J .

Proof. 1. Let us first verify that the intervals are disjoint. First consider

J(n)0 . The first return map is f qn−1 on this interval, therefore if j 6= k are

10

1

1

J J0 1

(1) (1)

cq1

q1 aq+1

122q+1 q+1

11

Figure 1.5: All iterates fkq1+1(c) for 1 ≤ k < a2 lie to the right of f q2(c)(q2 = a2q1 + 1). Moreover, if the index j > q1 + 1 is not of the form kq1 + 1,

then f j(c) does not lie in J(1)1 (if it did, it would contradict the definition of

f q1 as the first return map).

indices less than qn−1 we must have f j(J(n)0 ) ∩ fk(J

(n)0 ) = φ. The same

argument applies on J(n)1 : we simply note that J

(n)1 = J

(n−1)0 .

2. Next we show that the intervals cover J . For n = 1,

∪q1−1k=1 fk(J

(1)1 ) = [0, c2] ∪ [c2, c3] ∪ . . . [cq1−1, cq1 ].

Moreover,

J(1)1 = [c, 1], J

(1)0 = [cq1 , c].

At the next stage, the same arguments yield that ∪q2

k=0fk(J

(1)0 ) covers J (1).

Therefore, it also covers J(0)0 , and it follows that ∪q1+q2

k=0 fk(J(1)0 ) covers J .

Proceed inductively: if Qn = q1 + q2 + . . . + qn, then ∪Qn

k=0fk(J

(n)0 ) covers J .

3. We are using too many intervals at this stage. The minimal number isobtained as follows. Suppose k is the smallest positive integer such that

f−k(x) ∈ J(n)0 . (The previous step ensure that such k exists). Then also

f−k+qn−1(x) ∈ J(n)0 . Thus, k < qn−1. Similarly, if k is the minimal integer

such that f−k(x) ∈ J(n)1 we find k < qn.

The sequences an and qn impose severe restrictions on the invariantsK+ and K−: an arbitrary string in Σ is not admissible as an itinerary as aconsequence of the following lemma. The notation adopted is as follows. If

11

x is a string, xm denotes the truncation (x0, . . . , xm−1), xm · yn

denotes theconcatenation of the strings xm and y

n, and xp

m denotes the concatenation

xm · xp−1m .

Lemma 1.13. Assume f ∈ S(I) has no periodic orbits. Then for n ≥ 1

K+q2n+2

= K+q2n

·(

K+q2n+1

)a2n+2

, (1.26)

K+q2n+1

= K+q2n−1

·(

K−q2n

)a2n+1 .

Proof. We prove only the first identity. Since q2n+2 = a2n+2 q2n+1 + q2n, itsuffices to show

(

if (cq2n+jq2n+1))

q2n+1= K+

q2n+1, 0 ≤ j ≤ a2n+2 − 1. (1.27)

The point c is contained in the interval J (2n+1) = [cq2n+1 , cq2n]. Since f q2n+1

is the first return map on the interval (c, cq2n), and f has no periodic orbits,

if x ∈ [c, cq2n] then f j(x) does not lie in [c, cq2n

] for 0 ≤ j ≤ q2n+1. That is,the intervals f j([c, cq2n

]) do not contain the point c, so that the itinerary ofevery point in this interval agrees for j ≤ q2n+1. Finally, we note that thepoints cq2n+jq2n+1 ∈ [c, cq2n

], 0 ≤ j ≤ a2n+2 − 1 (by the definition of a2n+2).This proves (1.27).

A simple and powerful consequence of Lemma 1.13 is that the knowledgeof K+ is equivalent to knowledge of an.

Lemma 1.14. Suppose f, f ∈ S(I) have no periodic points. Then K+(f) =K+(f) if and only if an(f) = an(f) for n ≥ 1.

Proof. Suppose an = an. Then qn = qn by (1.24) and Lemma 1.13 impliesK+(f) = K+(f).

Conversely, suppose K+(f) = K+(f). Proposition 1.8 implies f and fare combinatorially equivalent. It is then immediate that an = an.

1.6 The rotation number and Poincare’s theorem

A finite set of positive integers a1, . . . , an determines a continued fraction

[0; a1, . . . , an] =1

a1 +1

a2 + . . .1

an

. (1.28)

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Given an infinite sequence ak, k ≥ 1 we construct the sequence of rationalnumbers

pn

qn= [0; a1, . . . , an]. (1.29)

We adopt the convention that rational numbers are always written in re-duced form, that is pn and qn are relatively prime, thus uniquely determinedby ak, 1 ≤ k ≤ n. The terms pn/qn are called the convergents of the con-tinued fraction. limn→∞ pn/qn is an irrational number in (0, 1) (Exercise 8).The continued fraction expansion of a rational number has finite depth, andwe adopt the obvious convention

[0; a1, . . . , an] = [0; a1, . . . , an,+∞].

Definition 1.15. The rotation number of a circle map g, denoted ρ(g) isdefined to be [0; a1, a2, . . .] if all an < ∞, and [0; a1, . . . , an−1] if an = ∞.

The basic properties of the rotation number are the following.

Proposition 1.16. ρ(g) is a rational number p/q if and only if g has aperiodic orbit with period q.

Proof. g has a periodic point if and only if ϕn has a fixed point for some n.In this case, an < ∞, 1 ≤ k ≤ n and an+1 = ∞. Let p/q = [0; a1, . . . , an],so that f q has a fixed point. The period of the orbit cannot be less than q(else, ϕk would have a fixed point for some k < n.)

Proposition 1.17. The map g 7→ ρ(g) is continuous in the C0 topology.

Proof. For every finite N , the map from g to the first N points of the orbitOg(c) is continuous in the C0 topology. It now follows from the definitionsthat the map g 7→ an(g) is continuous. More precisely, if an(g) < ∞, thereis an ε > 0 such that an(g) = an(g) for ‖g− g‖ < ε. And if an(g) = ∞, thenfor every M > 0, there exists ε > 0 such that an(g) > M if ‖g − g‖ < ε.

Proposition 1.18. ρ(g) is a topological invariant.

Proof. Suppose h is a homeomorphism of the circle, and g = h g h−1.It is easy to check that g and g are combinatorially equivalent, thereforean(g) = an(g).

The previous lemma demonstrates the principle that combinatorial equiv-alence is weaker than topological equivalence. The converse is not true,without further assumptions. That is, K+(g) = K+(g) does not imply theexistence of a homeomorphism that conjugates g to g. The best we can hopefor without further assumptions is

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Theorem 1.19 (Poincare). Suppose f ∈ S(I) has irrational rotation num-ber α. There is a continuous, increasing map h of I onto I such that

h f = Rα h. (1.30)

Proof. 1. ρ(f) = ρ(Rα) = α = [0; a1, a2, . . .]. Therefore, by Lemma 1.14,K+(f) = K+(Rα), and f and Rα are combinatorially equivalent by Propo-sition 1.8. Let c and c denote the critical points of f and Rα respectively,and cn and cn their orbits. Then the map h : Of (c) → ORα(c) defined byh(cn) = cn, n ≥ 0 is order-preserving and satisfies (1.30).

2. We now extend h to an increasing map on I by defining

h(x) = suph(cn) |cn ≤ x.

We claim that h is continuous. Suppose not: then there exists x such thath(x−) < h(x+). Since the orbit ORα(c) is dense in I, there exists cn suchthat h(x−) < h(cn) < h(x+). But this implies both x < cn and cn < x(since h is increasing).

Poincare’s theorem provides a semi-conjugacy, but not a conjugacy. Thatis, h while continuous, need not have a continuous inverse. If the closed setF = Of (c) is I then h is strictly increasing. Thus, it is an open mapping,hence a homeomorphism. However, the proof does not preclude the possi-bility that F is a strict subset of I. In this case, I\F is a countable union ofopen intervals, and h is constant on each of these intervals. We now explorethis matter in greater detail.

1.7 Topological conjugacy

Theorem 1.20. Suppose g is a circle map with irrational rotation number.Then there is a unique minimal set K ⊂ S1.

Recall that a set is minimal if it is non-empty, compact, and invariantwith no nonempty, compact, invariant proper subset.

Proof. Pick x ∈ S1. The ω-limit set ω(x) is closed, invariant and minimal.If ω(x) = S1 there is nothing to prove (why?). If ω(x) 6= S1, consider theopen set G = S1\ω(x). Since G is open, it may be written as as a unionof disjoint, open intervals G = ∪j∈ZIj. The proof consists in showing thatα(y), ω(y) ⊂ ω(x) for every y ∈ G. This inclusion follows from the

Claim: The orbit of y visits each Ij only once.

14

Proof of the claim: G is also invariant. Thus, if fk1(y) ∈ Ij1 and fk2(y) ∈Ij2 , then fk2−k1 : Ij1 → Ij2 is a homeomorphism. If j1 = j2 then thishomeomorphism has a fixed point, contradicting the assumption that ρ(f)is irrational.

Theorem 1.21. Suppose g is a circle map with irrational rotation numberand minimal set K 6= S1. Then g is not conjugate to a rotation.

Proof. If g is conjugate to a rotation, then the rotation must be Rα withα = ρ(g). Suppose h : S1 → S1 is a homeomorphism. If D is a dense subsetof S1, then so is h(D). Thus, if g is conjugate to Rα, Og(x) = S1 for everyx ∈ S1. Thus, K = S1.

If the minimal set K has an interior point, then K = S1. Thus, ifK 6= S1, it is a Cantor set (ie. perfect and totally disconnected). This is agenuine obstruction to topological conjugacy:

Proposition 1.22. For every Cantor set K ⊂ S1 and irrational α ∈ (0, 1)there is a circle map with rotation number α and minimal set K.

Proof. This is part of HW 3.

A finer consequence of the proof of Theorem 1.21 is that if ρ(g) is irra-tional and g is conjugate to a rotation, then there is no open interval J suchthat the intervals fk(J)k∈Z are pairwise disjoint.

Definition 1.23. An open interval J is wandering for the map f if:

1. The intervals J , f(J), . . . are pairwise disjoint.

2. The ω-limit set of J is not a single periodic orbit.

The second condition is included to rule out the case when f has periodicorbits. A circle map with periodic points may have an interval J that satisfiescondition (1), but these are not wandering because of (2).

1.8 Rigidity and non-rigidity of circle maps

We may rephrase Theorem 1.21 as the assertion that a circle map withirrational rotation number cannot be conjugate to a rotation if it has a wan-dering interval. Proposition 1.22 shows that the assumption of continuity isnot enough to rule out wandering intervals. Differentiability does not sufficeeither!

15

Theorem 1.24 (Bohl). For every irrational α ∈ (0, 1) there exists a C1

diffeomorphism of the circle with a wandering interval.

Theorem 1.25 (Denjoy). Suppose f is a C1 circle diffeomorphism suchthat log Df has bounded variation. Then f does not have a wandering in-terval.

Remark 1.26. Note that 0 < Df = |Df | for an orientation preservingcircle diffeomorphism. If f is a C2 circle diffeomorphism, it satisfies thehypothesis of the theorem since the variation of log f ′ on an interval [a, b] isestimated by

log f ′(b) − log f ′(a) =

∫ b

a

f ′′

f ′dx ≤

max |f ′′|

min f ′|b − a|.

Remark 1.27. These theorems are loosely called rigidity (and non-rigidity)theorems. Here a circle map is ‘rigid’ if it is topologically equivalent to arotation. The surprising aspect of the theorems is the importance of ‘metric’smoothness assumptions for topological conclusions. The smoothness gapbetween these theorems was closed by Herman: for every ε > 0 and irrationalα ∈ (0, 1) there is a C2−ε diffeomorphism with rotation number α that hasa wandering interval.

Proof of Theorem 1.24. 1. Fix a bi-infinite sequence of positive numbersλnn∈Z such that

∑

k∈Z

λk = 1, lim|k|→∞

λk+1

λk= 1. (1.31)

Let ckk∈Z denote the orbit of the critical point for Rα. For any positiveinteger n, we may choose closed, disjoint intervals Ik,n, |k| ≤ n, with length|Ik,n| = λk that are ordered in the same way as ck, |k| ≤ n. Since S1 is

compact, there exists a subsequence n(0)j such that the intervals I

0,n(0)j

con-

verge to an interval I0 with length λ0 as j → ∞. We may now extract a

subsequence n(1)j of n

(0)j such that I

±1,n(1)j

also converge to intervals denoted

I±1. Since the intervals I−1,n, I0,n and I1,n are ordered in the same way asc−1,c0 and c1, the limiting intervals I−,I+ and I0 respect the same ordering.

We proceed in this manner, obtaining subsequences n(N)j such that I

k,n(N)j

converges to an interval Ik with length λk for |k| ≤ N . Since the subse-

quences are nested, ie. n(m)j ⊂ n

(m)j , Ik is independent of N for all |k| < N

and Ik is ordered in the same way as ck.

16

2. Let A = ∪∞k=0I

ok . By construction, A is open and has full measure (thus, it

is dense). We use bump functions to construct a C∞ diffeomorphism fk fromIk → Ik+1 such that f ′

k = 1 at the endpoints of Ik and ‖f ′k −1‖ ≤ 2λk+1/λk.

We the define f : A → A and h : A → S1 by

f(x) = fk(x), h(x) = ck, x ∈ Ik.

It is easy to check that hf = Rα h for x ∈ A. Since A and ck are densein S1, h and f extend to continuous maps of the circle. Thus, ρ(f) = α.Since fk is a diffeomorphism, so is f restricted to A with f−1(x) = f−1

k−1(x),x ∈ Ik. The inverse also satisfies h f−1 = R−α h.

3. We must show that f is a C1 diffeomorphism. It will suffice to show thatf is differentiable at each y ∈ S1\A with derivative 1, that is

limz→y

f(z) − f(y)

z − y= 1. (1.32)

We will prove that the derivative from above

limz↓y

|f ([y, z])|

|[y, z]|= 1 (1.33)

A similar argument for the lower derivative establishes (1.32). Finally, thisestimate also establishes that f−1 has derivative 1 on A.

4. If y is a left endpoint of In then (1.33) follows from the construction offn. Thus, we consider the case when [y, z]∩Ik 6= φ for infinitely many k. LetN1 = n |In ∩ [y, z] 6= φ and N2 = n |In ⊂ [y, z]. Notice that N2 ⊂ N1

and #N1\N2 ≤ 2. It follows from these definitions that

∑

n∈N2

λn ≤ |[y, z]| ≤∑

n∈N1

λn.

Similarly, since f(In) = In+1 we find,

∑

n∈N2

λn+1 ≤ |f ([y, z])| ≤∑

n∈N1

λn+1.

Therefore,∑

n∈N2λn+1

∑

n∈N1λn

≤|f ([y, z])|

|[y, z]|≤

∑

n∈N1λn+1

∑

n∈N2λn

As z → y, we must have m = minn|n ∈ Ni → ∞. Thus, the secondcondition in (1.31) implies (1.33).

17

1.9 Denjoy’s theorem

Definition 1.28. The distortion of a C1 map g on an interval J is

Dist(g, J) = supx,y∈J

log|Dg(y)|

|Dg(x)|. (1.34)

The introduction of the logarithm is natural for iterated maps.

Lemma 1.29. Dist(fn, J) ≤∑n−1

k=0 Dist(f, fk(J)).

Proof. By the chain rule,

Dfn(x) = Df(fn−1(x))Df(fn−2(x)) . . . Df(x).

Therefore,

log|Dfn(y)|

|Dfn(x)|= log

n−1∏

k=0

∣

∣Df(fk(y))∣

∣

|Df(fk(x))|

=

n−1∑

k=0

log

∣

∣Df(fk(y))∣

∣

|Df(fk(x))|≤

n−1∑

k=0

Dist(f, fk(J)).

As a corollary, if Df(x) 6= 0 in J and log |Df(x)| is Lipschitz withconstant C we have

Dist(fn, J) ≤ Cn−1∑

k=0

∣

∣

∣fk(J)

∣

∣

∣. (1.35)

Proof of Theorem 1.25. 1. Suppose J is a wandering interval. Then theintervals fk(J), k ∈ Z are disjoint and have finite total length

∑

k∈Z

∣

∣

∣fk(J)

∣

∣

∣≤ 1. (1.36)

2. The main geometric observation is that for every x ∈ S1 the intervalsfk([x, f−qn(x)]) are disjoint for 0 ≤ k ≤ qn−1. This follows from Lemma 1.12when x is c. It would then also hold for every x if f were simply Rρ(f). Butthe ordering of points by f is equivalent to Rρ(f), thus it also holds for f .

3. We let T be the smallest subinterval containing both J and f−qn(J) thatis contained in S1\f qn(J). Then fk(T ) are disjoint for 0 ≤ k ≤ qn−1. Thisagain follows by considering Rρ(f) as in step (2).

18

4. Let V = var(log |Df |, S1) denote the total variation of log |Df on S1.The distortion of f qn on T is uniformly controlled by V as follows. ByDefinition 1.28 and Lemma 1.29, we have

Dist(f qn , T ) ≤ var(

log |Df |,∪qn−1k=0 fk(T )

)

≤ var(

log |Df |, S1)

= V.

(1.37)

5. By the mean value theorem, there exists x ∈ J and y ∈ f−qn(J) suchthat

|Df qn(x)||J | = |f qn(J)|, |Df qn(x)||f−qn(J)| = |J |

Now both x, y ∈ T , therefore

log|Df qn(y)|

|Df qn(y)|≤ Dist(f qn , T ) ≤ V.

Therefore, in contradiction to step (1) we have

|J ||J |

|f qn(J)||f−qn(J)|≤ eV . (1.38)

1.10 Smooth conjugacy and KAM theory

Denjoy’s theorem settles the question of topological conjugacy, and opensthe door to questions of smooth conjugacy. If a circle map g with ρ(g)irrational is smooth enough to rule out wandering intervals, is it true thatthe conjugacy h is smooth too? For example, if g is (real) analytic, is italso true that h is analytic? The answers to this question are surprisinglydelicate. In order to understand the basic obstruction let us proceed formallyat first.

In this section we will work on the covering space R. We consider g :R → R such that g(x+1) = g(x)+1. An equivalent definition of the rotationnumber is ρ(g) = limn→∞

∑n−1k=0 gk(x)/n. We shall consider the smoothness

of the conjugacy for small perturbations of rotations. That is, we considermaps of the form

g = x + α + u(x) (1.39)

where u(x + 1) = u(x). Suppose ρ(g) = α, and consider the conjugacy

g h = h Rα. (1.40)

19

When u ≡ 0, h is the identity. Therefore, we substitute h(x) = x + v(x) in(1.40) to obtain

v(x + α) − v(x) = u(x + v(x)). (1.41)

For small u we expect that the first step in the solution of this functionalequation is to solve the linearized problem

v(x + α) − v(x) = u(x). (1.42)

The linearized problem is called the homological equation. A formal solutionis easily obtained using Fourier series. We substitute the Fourier expansions

u(x) =∑

k∈Z

e2πikxuk, v(x) =∑

k∈Z

e2πikxvk, (1.43)

in (1.42) to obtain

vk =uk

e2πikα − 1, k 6= 0. (1.44)

In addition, we must assume u0 = 0. Elegant as this formal solution maybe, it comes with a subtle convergence problem. The denominators e2πikα−1 never vanish because α is irrational. Nevertheless, if we consider theconvergents pn/qn of α and set k = qn we find the small divisors

∣

∣

∣e2πikα − 1

∣

∣

∣=

∣

∣e2πiqnα − e2πipn∣

∣ ≤ 2πqn

∣

∣

∣

∣

α −pn

qn

∣

∣

∣

∣

≤2π

qn.

Worse yet, there are irrational numbers (the Liouville numbers) which arearbitrarily badly approximated by rational numbers, that is for every posi-tive integer m there are integers pm, qm such that |pm/qm − α| ≤ 1/qm. Wecan no longer prove results such as Theorems 1.24 and 1.25 that hold for allirrational numbers. Further assumptions on α are required. The fundamen-tal assumption is that α is badly approximated by rational numbers in thefollowing sense:

Definition 1.30. An irrational number α is Diophantine of class (K,σ) if

∣

∣

∣

∣

α −p

q

∣

∣

∣

∣

≥K

|q|2+σ(1.45)

for all integers p, q and some fixed K,σ > 0.

Not all irrational numbers satisfy Diophantine conditions. However theexceptions form a set of measure zero (but of first category).

20

We shall now construct a solution to the conjugacy equation q1.41, bas-ing our analysis on the homological equation(1.42). A decay assumption onuk is clearly necessary. This is essentially an assumption on the smooth-ness of u. We assume that u is analytic. More precisely, let Sρ denote thehorizontal strip x + iy ||y| < ρ. For u defined on Sρ we define the norm‖u‖ρ = supz∈Sρ

|u(z)|. An analytic function is univalent if it is one to one.A univalent map g : Ω → C defines a conformal map from Ω to g(Ω). Thefundamental result on conformal conjugacy is

Theorem 1.31 (Arnol’d). Let α be Diophantine of class (K,σ) and ρ > 0.There exists ε(K,σ, ρ) > 0 such that if g is a circle map with rotation numberα that extends to a univalent function on Sρ and u = g − Rα satisfies‖u‖ρ < ε, then g is conformally conjugate to Rα on the strip Sρ/2.

This is the first of a circle of results called the KAM theory (for Kol-mogorov, Arnol’d, and Moser). The unifying aspects of KAM theorems are(a) loss of smoothness in the linear problem because of small divisors, (b)Diophantine assumptions, and (c) accelerated convergence as in Newton’smethod. The proof of Theorem 1.31 demonstrates all these aspects. Wefirst study the linear problem, then the iteration scheme.

At an abstract level, Theorem 1.31 has the character of an implicit func-tion theorem. Roughly, solvability of a linear problem at a point impliessolvability of a nonlinear problem in a neighborhood of the point (look upthe statement and proof in a book on advanced calculus). The catch here isthe loss of smoothness in the linear problem. This affects the ‘implicit func-tion theorem’ as we obtain a solution to the nonlinear problem in a worsespace.

Lemma 1.32. Assume u is a 1-periodic function analytic in Sρ and con-tinuous in Sρ. Then its Fourier coefficients satisfy

|uk| ≤ e−2πρ|k|‖u‖ρ, k ∈ Z. (1.46)

Proof. The Fourier coefficients are defined by

uk =

∫ 1

0e−2πikxu(x) dx.

Assume k > 0. We shift the contour of integration down by −iρ and useCauchy’s theorem to obtain

uk = e−2πkρ

∫ 1

0e−2πikxu(x − iρ) dx.

21

The Diophantine condition is used in the following lower estimate.

Lemma 1.33. Assume α is of type (K,σ). Then for every integer k 6= 0we have

∣

∣

∣e2πikα − 1

∣

∣

∣≥

4K

|k|1+σ. (1.47)

Proof. Suppose θ ∈ (0, π). Then (draw a picture!)

θ ≥ |eiθ − 1| ≥2θ

π.

Now apply the Diophantine condition (1.45).

We now apply this estimate to the solution of the linear equation (1.42).In all that follows, we assume that u is analytic in Sρ. We do not assumethat u has mean zero, and must consider the modified homological equation

v(x + α) − v(x) = u(x) − u0. (1.48)

Lemma 1.34. Assume (1.48) holds and α satisfies the Diophantine condi-tion (1.45). There exists a constant C(K,σ) such that

‖v‖ρ−δ ≤ C‖u‖ρ δ−(2+σ), 0 < δ < ρ. (1.49)

Proof. Since u0 = 0 we may solve the homological equation (1.44). We fixz ∈ Sρ−δ and estimate |v(z)| using Lemmas 1.32 and 1.33 to obtain

|v(z)| ≤∑

k∈Z

|vk(z)||e2πikz | ≤‖u‖ρ

4K

∑

k∈Z

|k|1+σe−2π|k|δ.

Let S(δ) denote the sum∑∞

k=1 k1+σe−2πkδ. S(δ) is a continuous and de-creasing function of δ with the following asymptotics.

limδ→0

δ2+σS(δ) =

∫ ∞

0x1+σe−2πx dx.

As δ → ∞, S(δ) decays exponentially with any rate less than 2π. Thus,there is c(σ) such that S(δ) ≤ cδ−(2+σ), δ > 0 and C = c/2K.

The solution to the nonlinear equation (1.48) is built out of a sequenceof linear approximations. Having solved for v, we construct a circle homeo-morphism h = x + v(x) and a new circle map g1 = x+ u1 via the conjugacyg1 = h−1gh. We then solve the homological equation (1.48). This process

22

is then iterated. The difficulty we face is the unavoidable loss of smoothnessin the solution of the linear problem: (1.49) diverges as δ → 0. It is crucialto obtain accelerated convergence so that the deterioting linear estimates donot sabotage the scheme. The margin of victory is the quadratic factor ‖u‖2

ρ

below. This is akin to Newton’s method for solving an equation. Supposef : R → R and we are trying to find a zero x∗ of f via Newton’s iterationscheme xn+1 = xn − f(xn)/f ′(xn). For xn near x∗ the scheme convergesrapidly, because |xn+1 − x∗| ≤ C|xn − x∗|

2.

Lemma 1.35. There are positive constants c,C and δ∗ such that if

‖u‖ρ ≤ cδ3+σ , 0 < δ < δ∗, (1.50)

then‖u1‖ρ−δ ≤ C‖u‖2

ρ δ−(3+σ). (1.51)

Proof. 1. We first verify that h = x + v defines an analytic diffeomorphismon a strip Sρ−δ for suitable 0 < δ < ρ. Let C be as in Lemma 1.34 andsuppose η and ‖u‖ρ are chosen so small that

‖u‖ρ ≤η3+σ

2C≤

η

2. (1.52)

Then we use Lemma 1.34 and Cauchy’s formula to obtain

‖v‖ρ−η ≤η

2, ‖v′‖ρ−2η ≤

1

2. (1.53)

Since h(z) = z + v(z), for z1, z2 ∈ Sρ−2η we then have

1

2|z1 − z2| ≤ |h(z1) − h(z2)| ≤

3

2|z1 − z2|.

Thus, h is one-one on Sρ−2η with a derivative uniformly bounded away from0. The implicit function theorem guarantees the existence of an analyticinverse on the image h(Sρ−2η). In particular, h−1 is defined on Sρ−3η.

Let δ = 5η < ρ/2. Then h(Sρ−δ) ⊂ Sρ−4η and gh(Sρ−δ) ⊂ Sρ−4η+‖u‖ρ⊂

Sρ−3η. Thus, g1 = h−1 g h is analytic on Sρ−δ.

2. We now estimate u1. Since h g1 = g h, we have

u1(z) = v(z) − v(z + α + u1(z)) + u(z + v(z)). (1.54)

Since v solves the homological equation (1.48), we have

u1(z) = (u(z + v(z)) − u(z)) −(

v(z + α + u1(z)) − v(z + α))

+ u0. (1.55)

23

We apply Lemma 1.34 and Cauchy’s theorem to the first term to obtain

‖u(z + v(z)) − u(z)‖ρ−δ ≤ ‖u′‖ρ−δ‖v‖ρ−δ ≤‖u‖ρ

δ‖v‖ρ−δ ≤ C‖u‖2

ρ δ−(3+σ).

The second term admits the estimate

‖v(z + α + u1(z)) − v(z + α)‖ρ−δ ≤ ‖v′‖ρ−δ‖u1‖ρ−δ ≤

‖v‖ρ−δ/2

δ/2‖u1‖ρ−δ .

We combine these estimates and Lemma 1.34 to find(

1 − C‖u‖ρδ−(3+σ)

)

‖u1‖ρ−δ ≤ C‖u‖2ρ δ−(3+σ) + |u0|.

3. In order to estimate u0 we use the fact that g1 and g have the samerotation number. It follows that u1(x) = 0 for some x. Set x = z in (1.55)to obtain

|u0| = |u(z + v(z)) − u(z)| ≤ ‖u(z + v(z)) − u(z)‖ρ−δ ≤ C‖u‖2ρδ

−(3+σ).

If we choose c > 0 sufficiently small that (1.50) holds, we obtain (1.51)

Lemmas 1.34 and Lemma 1.35 constitute the main estimates for theiterative scheme. We construct mappings un defined on the strip Sρn whereρn = ρn−1 − δn−1 and ρ0 = ρ, δ0 = δ. We fix β ∈ (1, 2) and choose

δn = δβn−1, δ0 < min(δ∗, 1) (1.56)

In this case, by further reducing δ0 if necessary, we have

δn = δβn

, and∞∑

n=0

δn <ρ

2. (1.57)

Therefore, ρn > ρ/2, n ≥ 0 and the limiting map will be defined on Sρ/2.Lemma 1.35 yields the estimate

‖un‖ρn ≤ C‖un−1‖2ρn−1

δ−(3+σ)n−1 . (1.58)

We solve this recurrence to obtain

‖un‖ρn ≤ C1+2+...+2n

(‖u‖ρ)2n

δ−(3+σ)rn , (1.59)

where the exponent rn is

rn = 2n

(

1 +β

2+ . . . +

βn

2n

)

≤2n

1 − β/2.

24

Therefore, we have

‖un‖ρn ≤(

C2‖u‖ρδ−(3+σ)/(1−β/2)

)2n

.

If we choose ‖u‖ρ ≤ δκ for κ large enough, we obtain a constant denotedγ > 0 and estimates of the form

‖un‖ρn ≤ δ−γ2n

,

along with the estimate

‖vn‖ρn ≤ δn = δβn

.

Finally, we consider the analytic diffeomorphim Hn = h0 h1 . . . hn−1

on the strip Sρ/2.By the chain rule, the derivative

(Hn)′(z) =n−1∏

k=0

(hk)′(zk)

for some points zk ∈ Sρ/2. Since hk(z) = z + vk(z) this yields the upper andlower bounds

n−1∏

k=0

(1 − δk) ≤ ‖Hn‖ρ/2 ≤

n−1∏

k=0

(1 + δk).

The choice of δk ensures both infinite products converge. This bound ensuresthat Hn converges. Indeed,

∣

∣Hn(z) − Hn+1(z)∣

∣ = |Hn(z) − Hn(hn(z))| ≤ ‖(Hn)′‖ρ/2‖vn‖ρ/2.

This completes the proof of Arnol’d’s theorem.

25