Lecture Notes on Dynamical Systems Homeomorphisms on the Circle

Lecture Notes on Dynamical Systems

Homeomorphisms on the Circle

Diego A. S. Sanhueza October 30, 2018

Abstract In this lecture we lead very quickly the essential results on the Poincaré’sTheory for maps of the circle. Rotation number is defined and the classification theorem of Poincaréis proved. Further simples facts also are mentioned.

Contents

1 Introduction...... 1 2 Preliminaries...... 2 3 Rotation number...... 3 4 The meaning of ρ(f)...... 5 4.1 Rational rotation number ...... 5 4.2 Irrational rotation number ...... 8 5 Solved problems...... 12

§1. INTRODUCTION

Homeomorphisms of the circle were first considered by Henri Poincaréwho used them to obtain qual- itative results for a class of differential equations on the torus. In [15] he introduced the rotation number and in [16] classified those which have a dense orbit by showing that they are topologically conjugated to a rotation by the angle their rotation number. In [4], Arnaud Denjoy shows the existence of orientation- preserving homeomorphisms without periodic orbit and without dense orbit, and the Poincaré’sresult not assure that are conjugated to a rotation. They are usually called Denjoy C1−diffeomorphisms. Denjoy also proves that such a homeomorphism admits a unique minimal set which is a Cantor set (in contrast to the case when has periodic points. There every minimal set is a periodic orbit). This lead to study of classifying such Cantor sets and in this direction, Nelson Markley [11] characterized those Cantor sets completely in terms of dynamical properties. We mentioned that a Cantor set is always the unique minimal set of some orientation preserving homeomorphism of S1, but it may not be the minimal set of any Denjoy C1−diffeomorphism of S1 (See [14, 17]). Dusa McDuff [12], gives necessary conditions to guarantees when a Cantor set is a minimal set of a Denjoy diffeomorphism, answering partially to an ask posed by Michael Herman [6]. On the other hand, [9] it is showed that complementary intervals of those Cantor sets are permuted freely (without finite cycles) among themselves by the homeomorphism. Continuous map without periodic point were regarded by Joseph Auslander and Y. Katznelson in [2]. If f admits periodic point, the dynamics is very much simple, but in general conjugation to a rotation is not obtained. In [19], M. Zdun gives a necessary and sufficient condition for topological conjugacy of this type of homeomorphisms of the circle. 2 Diego A. S. Sanhueza

On the other hand, H. Furstenberg [5] proved a remarkable theorem in connection with ergodic theory. More precisely, he shows that if f : S1 → S1 is an orientation-preserving homeomorphisms with irrational rotation number and µ is an Borel f−invariant measure, then Fµ ◦ f = Rα ◦ F µ, where α = (Fµ ◦ T )(0) (necessarily irrational). This say that homeomorphisms with irrational rotation number are (metrically) equivalent to an irrational rotation. Moreover, Furstenberg also proved that such homeomorphisms are uniquely ergodic. Diﬀeomorphisms on the circle give rise to a important object of study within of the theory of dynamical systems. We refer to [7, 13] for a more specialized interest. In this short notes, we collect various main results on the Poincar´e’sTheory. The proofs of the principal results are standard and we follow [3, 8] e [18].

§2. PRELIMINARIES

Throughout we consider S1 = R/Z. The quotient map is π : R → R/Z. We denote by π(x) or sometimes x the equivalence class that contains x. We recall that y ∈ π(x) if, and only if, x − y ∈ Z. It is obvious that each class admits a representant in [0, 1).

Deﬁnition 2.1. Let f : S1 → S1 be a continuous function. A lift of f is a continuous function F : R → R which satisﬁes (π ◦ F )(x) = (f ◦ π)(x) (1)

for all x ∈ R.

Suppose that F0 is another lift of f, then by (1), we have that (π◦F )(x) = (f ◦π)(x) = (π◦F0)(x), so F (x) − F0(x) = k ∈ Z for all x ∈ R. By continuity of the lifts, the integral number k = k(F,F0) does not depend on x. In summary, two lift diﬀer by a integer constant, the family of lifts of a given 1 1 continuous function f : S → S is determined knowing a lift: If F0 is a lift of f, then any other lift is in {F0+k : k ∈ Z}. Others important properties of the lifts are summarized in the following proposition.

Proposition 2.2. Let f, g : S1 → S1 be two continuous function and F,G : R1 → R1 lift of f and g, respectively. Then: (i) F ◦ G is a lift of f ◦ G. In particular, F n is a lift of f n. (ii) Id : R1 → R1 is a lift of id : S1 → S1. (iii) If f is a homeomorphism, then F −1 is a lift of f −1.

Proof. (i) π[(F ◦ G)(x)] = (π ◦ F )(G(x)) = f(π(G))(x) = [f ◦ g](π(x)) for all x ∈ R. (ii) Immediate. (iii) By (i) and (ii) id(π(x)) = π(F ◦ F −1)(x) = f(π(F −1(x)) for all x ∈ R. Applying f −1 to this −1 −1 equality, we obtain f (π(x)) = (π ◦ F )(x) for all x ∈ R, but this is just the deﬁnition of lift.

Now, since (π◦F )(x+1) = f(π(x+1)) = f(π(x)) = (π◦F )(x) we see that F (x+1)−F (x) ∈ Z and again by continuity of the lift, this diﬀerence is independent of x. Moreover, by the preceding paragraph, also is independent of the choice of lift F . We denote this diﬀerence by deg(f).

Deﬁnition 2.3. The number deg(f) is called the degree of f.

One can prove the identity: deg(f ◦ g) = deg(f) deg(g) and in particular to obtain deg(f n) = [deg(f)]n for any n ≥ 1. Note also that if f is a homeomorphism, then deg(f) = ±1.

Proposition 2.4. The degree of a continuous map is invariant by topological conjugacy.

Proof. It is immediate from preceding paragraph, since deg(f) = deg(h−1◦g◦h) = deg(h−1) deg(g) deg(h) −1 and deg(h) = deg(h ) = ±1.

Deﬁnition 2.5. We say that a homeomorphism f : S1 → S1 is an orientation-preserving map if it admits a lift which is increasing.

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Equivalently, f preserves orientation if, and only if, deg(f) = 1. For instance, rotation maps are orientation-preserving homeomorphisms. It is clear that every lift of a map which preserves orientation is increasing. We denote by Homeo+(S1) the set of all the orientation-preserving homeomorphisms on the circle.

Proposition 2.6. Let f : S1 → S1 be an orientation-preserving homeomorphism and F be a lift of f. Then, (i) F (x + k) = F (x) + k for all x ∈ R, k ∈ Z; (ii) F n(x + k) = F n(x) + k for all x ∈ R, k, n ∈ Z; n n (iii) If Fk = F + k is another lift, then Fk (x) = F (x) + nk for all x ∈ R, k, n ∈ Z; n (iv) The map ϕn(x) = F (x) − x is periodic and in particular it is bounded (and is independent of the lift).

Proof. (i) We note that π(F (x + k)) = f(π(x + k)) = f(π(x)) = π(F (x)), so F (x + k) − F (x) = k0(x) ∈ Z for all x ∈ R. We shall prove that k = k0(x) for all x ∈ R. We suppose ﬁrst that k > 0 and + 1 prove this by induction. Because f ∈ Homeo (S ), the conclusion is clear for k = 1. If k = k0(x) for all x ∈ R, then (k + 1)0(x) = F (x + k + 1) − F (x) = F (x + 1) + k − F (x) = k + 1 for all x. This prove that case k > 0. For if k < 0, put k+ = −k and so F (x + k) − F (x) = −(F (x) − F (x − k+)). We now use induction + + in the expression F (x) − F (x − k ) to obtain k0 (x) = −k. This proves (i). (ii) Use induction. (See exercise 2). (iii) Fix k. We ﬁrst consider n > 0 and shall use induction. For n = 1 there is nothing to prove. n n n+1 n n Suppose Fk (x) = F (x) + nk holds for all x ∈ R, then by (ii), Fk (x) = Fk (Fk(x)) = Fk (Fk(x)) + k = F n+1(x) + nk + k = F n+1(x) + (n + 1)k. The case n < 0 is similar. n n (iv) By (ii), ϕn(x + 1) = F (x + 1) − (x + 1) = F (x) − x = ϕn(x) for all x ∈ R.

§3. ROTATION NUMBER

The most important concept in our study of homeomorphisms of the circle is that of rotation number (Deﬁnition 3.3). The following theorem essentially shows their existence and uniqueness.

Theorem 3.1. Let f : S1 → S1 be an orientation-preserving homeomorphism and let F : R → R be a lift of f. Then, for every x ∈ R, the following limit F n(x) − x ρ(F ) = lim (2) n→∞ n exists and is independent of x.

Proof. We ﬁrst prove that (2) does not depend on x. Suppose that (2) exists for x ∈ R and we write x = [x] + x , where [x] and x denote the integer and fractional part of x, respectively. Then L M L M F n(x) − x F n([x] + x ) − ([x] + x ) lim = lim n→∞ n n→∞ L Mn L M F n x + [x] − ([x] + x ) = lim n→∞ L M n L M F n x − x = lim n→∞ L Mn L M Hence (2) also exists for x and they coincide. For the rest of the proof,L weM can assume that x ∈ [0, 1). Let y ∈ [0, 1), then |x − y| ≤ 1 and |F n(x) − F n(y)| ≤ 1 for all n ∈ N, so |F n(x) − x − (F n(y) − y)| ≤ |F n(x) − F n(y)| + |x − y| ≤ 2.

Dividing by n and letting n → ∞, we obtain that the limit (2) is independent of x ∈ [0, 1). Using the same argument above, by writing every real number in its integer and fractional part, we can conclude that if ρ(F ) exists, it does not depend on x ∈ R.

3 4 Diego A. S. Sanhueza

We now prove the existence of (2). For this purpose it is convenient to divide the proof in two cases.

Case I: Suppose that F q(x) = x + p for some x ∈ [0, 1), p, q ∈ Z with q ≥ 1. For each n ∈ N there exist k, r ∈ N (depending on n), 0 ≤ r < q such that n = kq + r. Then

F n(x) = F r(F kq(x)) = F r(x + kp) = F r(x) + kp.

r(n) Now, since 0 ≤ r = r(n) < q, the sequence {F (x) − x}n≥1 is bounded, so

F n(x) − x F r(n)(x) + kp − x lim = lim n→∞ n n→∞ n F r(n)(x) − x kp = lim + lim n→∞ n k→∞ kq + r = p/q.

Case II: Suppose that F q(x) 6= x + p for all x ∈ R, p, q ∈ R, q ≥ 1. By continuity of F , we note that F q(x) > x + p or F q(x) < x + p for all x ∈ R, else F q(y) = y + p for some y ∈ R by the intermediate value theorem, contradicting our assumption. In particular, F q(x) − x 6∈ Z. For each n ∈ N, take kn ∈ N satisfying

q kn − 1 < F (x) − x < kn (3)

and observe that (3) is valid for all x ∈ R, and it is the same to say that kn is independent of x. Then for each m ∈ N, by (3), we have

k − 1 < F n(x) − x < k  n n  k − 1 < F n(F n) − F n(x) < k  n n  . . .  f . < . < . m inequalities n (m−2)n (m−2)n  kn − 1 < F (F ) − F (x) < kn  n (m−1)n (m−1)n  kn − 1 < F (F ) − F (x) < kn 

Hence m−1 mn X n kn kn m(kn − 1) < F − x = F (F ) − F (x) < mkn (4) k=0 Dividing by mn we obtain k 1 F mn(x) − x k n − < < n . (5) n n mn n Interchanging the roles of m and n

k 1 F mn(x) − x k m − < < m . (6) m m mn m

Subtracting (6) from (5) we ﬁnd

k k 1 k k 1 n − m − < 0 < n − m + . n m n n m m

1 1 Therefore |kn − km| < m + n and so {kn/n}n≥1 is a Cauchy sequence. By (5), we conclude that the F n(x)−x sequence { n }n≥1 converges. This proves the theorem.

1 1 Proposition 3.2. Let f : S → S be an orientation-preserving homeomorphism. If F1,F2 are two lifts of f, then ρ(F1) − ρ(F2) ∈ Z.

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Proof. Let k = K(F1,F2) such that F2 = F1 + k, then

n n n F2 (x) − x F1 (x) − nk − x F1 (x) − x ρ(F2) = lim = lim = lim + k = ρ(F1) + k. n→∞ n n→∞ n n→∞ n Observe that proposition 3.2 implies that

F n(x) − x ρ(f) = ρ(F ) mod 1 = lim mod 1 n→∞ n is independent of x and F .

Deﬁnition 3.3. The number ρ(f) is called rotation number of f.

Remark 3.4. Suppose that f : S1 → S1 is an orientation-preserving homeomorphism. (i) If F1,F2 = F1 + k are lifts of f, then ρ(F2) − ρ(F1) = k. (ii) There is a unique lift F of f such that ρ(f) = ρ(F ).

Theorem 3.5. The rotation number is invariant by topological (positive) conjugacy.

Proof. Consider two orientation-preserving homeomorphisms f, g : S1 → S1 which are conjugated by a homeomorphism h : S1 → S1, which also is orientation-preserving (positive conjugacy). If G and H are lifts of g and h, respectively, then F = H ◦ G ◦ H−1 is a lift of F . For x ∈ R we have that

F n(x) − x = [H ◦ G ◦ H−1]n(x) − x = (H ◦ Gn ◦ H−1)(x) − x = (H ◦ Gn ◦ H−1)(x) − Gn ◦ H−1(x) + Gn ◦ H−1(x) − H−1(x) + H−1(x) − x | {z } | {z } | {z } an bn cn

Dividing by n and letting n → ∞, we obtain that an/n and cn/n tend to 0, because {an}n≥1 is bounded and {cn}n≥1 is constant. Thus, ρ(F ) = ρ(G) so ρ(f) = ρ(g).

1 Example 1. Let α ∈ S and consider Rα the rotation by the angle α. Then ρ(Rα) = α

In fact, we know that Fα(x) = x + α is a lift of Rα. We compute (3.2) for x = 0 ∈ R

n Fα (0) nα ρ(Fα) = lim = lim = α. n→∞ n n→∞ n

Hence ρ(Rα) = ρ(Fα) mod 1 = α.

§4. THE MEANING OF ρ(f)

The limit (2) says that the behavior of iterates of lifts of a homeomorphism f : S1 → S1 are asymptotically similar to that of the rotation by the angle ρ(f). Hence is very natural to expect that orbits of homeomorphisms of the circle and orbits of rotations have similar properties. On the other hand, the dynamics of rotations is well understand. We know that for rotation by rational angle all its orbits are periodic with equal period, and that rotations by irrational angle are minimal. In this section, we shall see that a given homeomorphism f shares this type of features with Rρ(f). Thus the study of homeomorphisms of the circle will be done in two parts depending on the rationality of its rotational number.

§§4.1. RATIONAL ROTATION NUMBER

Theorem 4.1. Let f : S1 → S1 be an orientation-preserving homeomorphism. Then, the rotation number of f is rational if, and only if, f has a periodic point.

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Proof. (⇒) Let ρ(f) = p/q, with (p, q) = 1. Let F : R → R be the unique lift of f such that p < F q(0) < p + q It exists since F q(I) is an interval of length q. To prove the existence of a periodic point, it is suﬃcient to ﬁnd a point x0 ∈ R and r ∈ Z such that q F (x0) = x + r, since it will be projected in a periodic point for f. Suppose that there does not exist such a point, then the intermediate value theorem guarantees

q x + p < F (x) < x + p + q for all x ∈ R, or equivalently, q p < F (x) − x < p + q for all x ∈ R. q Since ϕq(x) = F (x) − x is bounded (Theorem 2.6(iv)), there is ε > 0 such that

q p + ε < F (x) − x < p + q − ε for all x ∈ R. So p + ε < F q(x) − x < p + q − ε p + ε < F q(F q(x)) − F q(x) < p + q − ε ...... p + ε < F q(F q(m−1)(x)) − F q(m−1)(x) < p + q − ε Summing this expressions, we obtain m(p + ε) < F qm(x) − x < m(p + q − ε) Dividing by mq and letting m → ∞, we get p p p + q − ε + ε < = ρ(f) < . q q q

Absurd. Thus there exists x ∈ R such that we can take x0 = x + p or x0 = x + p + q. (⇐) The converse is contained in proof of Theorem 3.1, Case I. Theorem 4.2. Let f : S1 → S1 be an orientation-preserving homeomorphism with rational rotation number ρ(f) = p/q ∈ Q, (p, q) = 1. Then, every periodic point has period equal to q. Proof. Let ρ(f) = p/q, with (p, q) = 1. Let x be a periodic point for f with period q0, that is

0 0 π(F q (x)) = f q (x) = x

0 0 hence F q (x) − x ∈ Z, we put p0 = F q (x) − x. By proof of theorem 3.1, Claim I, ρ(f) = p0/q0. Because of (p, q) = 1, p0 = d0p and q0 = d0q, where d0 = (p0, q0). Note that, in particular, q0 ≥ q. We shall prove that F q(x) = x + p. Without loss of generality suppose that F q(x) > x + p. Applying F q to this last inequality, and taking in account the monotonicity of F q, we have F 2q(x) > F q(x + p) = F q(x) + p > x + 2p Continuing this form, we can conclude that

0 F d q(x) > x + d0p

0 but q0 = d0q and p0 = d0p, so F q (x) > x + p0, contradicting the fact that this is an equality. Thus F q(x) = x + p. Now f q(x) = π(F q(x)) = π(x + p) = π(x) = x. 0 0 This implies that q ≥ q . Therefore q = q and each periodic orbit has period equal to q.

1 1 Each point x ∈ S induces an order ≤x in S by considering the natural order on the interval [x, x + 1), where x is any ﬁxed point in π−1(x) (See Figure 1). We called x a distinguished point for the 1 order ≤x. Observe that in the order ≤x, the distinguished point x is the smallest element in S .

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Figure 1: The order ≤x depend on the distinguished point x.

Theorem 4.3. Let f : S1 → S1 be an orientation-preserving homeomorphism with rational rotation 1 number ρ(f) = p/q ∈ Q, (p, q) = 1. If x ∈ S is a periodic point of f, then the order of Of (x) = q−1 {x, f(x), ..., f (x)} is the same order as that of ORρ(f) (0) = {0, ρ(f), ..., (q − 1)ρ(f)}, where Rρ(f) is the rotation by the angle ρ(f).

Proof. Let x ∈ S1 be a periodic point of f. We pick j ∈ {0, 1, ..., q − 1} being the unique number such that f j(x) is immediately after x in counterclockwise direction in T (see ﬁgure 2). Then f 2j is 2j m j m 2j immediately after f . If not, there is f (x), with m > j such that f (x) ≤f j (x) f (x) ≤f j (x) f (x) m−j j and so x ≤x f (x) ≤x f (x), contradicting our choice of j. Therefore, in the ordering ≤x, the orbit Of (x) has the following order:

j 2j (q−1)j x ≤x f (x) ≤x f (x) ≤x · · · ≤x f (x).

We now determine the order of the orbit ORρ(f) (x) under ≤0. It is obvious that x is a periodic point j j for f with period q and the orbit Of j (x) goes around of circle just one time, so ρ(f ) = 1/q (remember j ∼ the meaning of p), but we know that ρ(f ) = jp/q mod 1, so jp = 1 mod q. Thus 0 ≤0 jp/q ≤0 jk 2jp/q ≤0 · · · ≤0 (q − 1)jp/q, but jkp/q = Rρ(f)(0). This proves that the orders of 0f (x) and ORρ(f) (0) are the same.

Deﬁnition 4.4. Let f : S1 → S1 be a continuous function. A point x ∈ S1 is say to be homoclinic to y ∈ S1 if lim d(f n(x), f n(y)) = 0 |n|→∞

1 1 The point x ∈ S is called heteroclinic to the points y1, y2 ∈ S if

n n −n −n lim d(f (x), f (y1)) = lim d(f (x), f (y2)) = 0 n→∞ n→∞

Theorem 4.5. Let f : S1 → S1 be an orientation-preserving homeomorphism with rational rotation number ρ(f) = p/q ∈ Q, (p, q) = 1. Then nonperiodic points are heteroclinic under f q to two periodic points. More exactly, (a) If f has a unique fixed point, then every point is heteroclinic (in this case homoclinic) to the fixed point. (b) If f has more than one fixed point, then every point is heteroclinic to two different fixed points. (c) If f has a unique periodic point with period greater than 1, then every nonperiodic point is heteroclinic under f q to two different points on the periodic orbit.

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Figure 2: Choice of j and order of the orbits Of (x) and ORρ(f) (0).

(d) If f has more than one periodic point, then every nonperiodic point is heteroclinic under f q to two diﬀerent periodic points.

Proof. (a) We can suppose that the fixed point is 0, so there is a lift F : R → R with F (k) = k for all integer k. By restricting this lift to the interval [0, 1], we study the dynamics of the F on the interval. The dynamics of f is projected by the dynamics of F : [0, 1] → [0, 1] (see figure 3). The part (a) it follows. (b) Similarly, the dynamics of f is reflected from the map F : [0, 1] → [0, 1] as in figure 4. Here the orbit of each nonperiodic point is heteroclinic to two different fixed points. (c) Consider f n instead f and (a). n (d) Consider f instead f and (b).

Figure 3: Dinamics on the interval with a unique ﬁxed point.

§§4.2. IRRATIONAL ROTATION NUMBER

Continuing our analysis of rotation number, we now shall study the case ρ(f) irrational and a ﬁrst immediate observation is that ρ(f) is irrational if, and only if, f has no periodic points, as indicated in Theorem 4.1. 1 By an interval I = [a, b] in T , we mean the set {x ∈ S : a ≤a x ≤a b}.

Lemma 4.6. Suppose that ρ(f) ∈ R \ Q. Then, for any x ∈ S1 and m, n ∈ Z, m > n, every orbit of f intersects the interval I = [f m(x), f n(x)].

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Figure 4: Dynamics of a homeomorphism with more than one ﬁxed point.

1 S∞ −k 1 Proof. It is suﬃcient to prove that S = k=0 f (I). Fix x ∈ S and m, n ∈ N, m > n. Observe that ∞ ∞ [ [ f −k(m−n)(I) = f −k(m−n)[f m(x), f n(x)] k=1 k=1 ∞ [ = [f −(k−1)m+kn(x), f (k+1)n−km(x)] k=1 = [f n(x), f 2n−m(x)] ∪ [f 2n−m(x), f 3n−2m(x)] ∪ [f 3n−2m(x), f 4n−3m(x)] ∪ · · · = [f n(x), lim f (kj +1)n−kj m(x)] j→∞

S∞ −k(m−n) 1 for some increasing subsequence {kj}j≥1 in N (by compactness). If k=1 f (I) 6= S , then z = (k+1)n−km limk→∞ f (x) exists. By continuity, f m−n(z) = f m−n( lim f (k+1)n−km(x)) = lim f kn−(k−1)m(x) = z. k→∞ k→∞

That is, z is a periodic point of f, contradicting the irrationality of ρ(f).

We recall that a set X is nowhere dense if int(X) = ∅. A set X is said to be perfect is each point in X is an accumulation point of X. A set is called a Cantor set if it is perfect and nowhere dense.

Proposition 4.7. Let f ∈ Homeo+(S1) with irrational rotation number. Then: (i) ω(x) = ω(y) for any x, y ∈ S1; (ii) ω(x) is either the whole circle S1 or a Cantor set. 1 n Proof. (i) Let x, y ∈ S . Let x0 ∈ ω(x) and {nk}k≥1 in N such that limk→∞ f k (x) = x0. By m n n above lemma, for each k ∈ N we can choose mk ∈ N such that f k (y) ∈ [f k−1 (x), f k (x)]. This implies m that limk→∞ f k (y) = x0. Hence ω(x) ⊆ ω(y). By symmetry, ω(y) ⊆ ω(x) and hence equality. (ii) Let z ∈ ω(x). We shall prove that z is an accumulation point of ω(x). For z ∈ ω(x) = ω(z), then n n there exists nk such that f k (z) converges toward z. But ω(x) is invariant by f, so f (z) ∈ ω(x) for all n ∈ Z. Hence z is an accumulation point of ω(x). It remains to show that either ω(x) = S1 or ω(x) is nowhere dense. We suppose that ω(x) 6= S1, then ∂ω(x) 6= ∅ and closed. It is also f−invariant because if y ∈ ∂ω(x), then f(y) ∈ ∂ω(f(x)) = ∂ω(x). Now, if z ∈ ∂ω(x), by f−invariance, ω(x) = ω(z) ⊆ ∂ω(x). This implies that ω(x) = ∂ω(x), thus intω(x) = ∅.

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Remark 4.8. It is easy to see that if ω(x) = S1 if, and only if, O(x) is dense in S1. By (i), every orbit is dense or no orbit is dense.

Lemma 4.9. Suppose that ρ(f) ∈ R \ (Q) and consider F : R → R the lift of f such that ρ(f) = ρ(F ). Then for each x ∈ R,

n1 n2 n1ρ(f) + m1 < n2ρ(f) + m2 if and only if F (x) + m1 < F (x) + m2

for any n1, n2, m1, m2 ∈ Z.

n1 n2 Proof. Suppose F (x) + m1 < F (x) + m2 or, equivalently

n1−n2 F (x) < x + m2 − m1

n1−n2 This inequality is valid to every x ∈ R, if not F (x0) = x0 + m2 − m1 for some x0 and by Theorem n1−n2 n1−n2 4.1 ρ(f) would be rational. In particular, F (0) < m2 − m1. Applying F to this inequality, (k − 1)−times, we obtain

n1−n2 F (0) < m2 − m1 2(n1−n2) n1−n2 n1−n2 F (0) < F (m2 − m1) = F (0) + (m2 − m1) < 2(m2 − m1) ......

k(n1−n2) F (0) < k(m2 − m1)

If n1 − n2 > 0, then F k(n1−n2)(0) m − m ρ(f) = lim < 2 1 k→∞ k(n1 − n2) n1 − n2 Therefore, n1ρ(f) + m1 < n2ρ(f) + m2

If n1 − n2 < 0, then F k(n1−n2)(0) m − m −ρ(f −1) = lim > 2 1 k→∞ k(n1 − n2) n1 − n2 Thus, n1ρ(f) + m1 < n2ρ(f) + m2. The converse follows from same inequalities. Deﬁnition 4.10. A homeomorphism f : S1 → S1 is transitive if it has a dense orbit. If each orbit of f is dense, then it is called minimal.

Before continuing we remember some properties of the real number which we shall need. If α is an irrational number, then the set {nα + m : n, m ∈ Z} is dense in R (see [10]). Other important fact is wether A, B ⊆ R are bounded, then sup{a + b : a ∈ A, b ∈ B} = sup A + sup B (see [1]). Ultimately, every surjective monotone real function is continuous. We are now in a position to state and prove one of the most fundamental result of this notes.

Theorem 4.11 (Poincar´eClassiﬁcation Theorem[16], 1985). Let f : S1 → S1 be an orientation- preserving homeomorphism with ρ(f) ∈ R \ Q. Then: (a) If f is transitive, then f is topologically conjugate to Rρ(f). (b) If f is intransitive, then Rρ(f) is a factor map of f.

1 1 Proof. We are looking for a map h : S → S such that h ◦ f = Rρ(f) ◦ h. Equivalently, we need ﬁnd a map H : R → R such that H ◦ F = Fρ(f) ◦ H, which is equivalent to ﬁnd a map H such that n n (H ◦ F )(x + m) = (Fρ(f) ◦ H)(x) + m = nρ(f) + m + H(x). Let F : R → R be a lift of f such that ρ(f) = ρ(F ) and x0 ∈ R. Put

n A = {F (x0) + m : n, m ∈ Z} and B = {nρ(f) + m : n, m ∈ Z}.

10 Homeomorphisms on the circle 11

n1 n2 Observe that A has no repeated elements, otherwise if equality F (x0) + m1 = F (x0) + m2 holds, n1 n2 n1−n2 (n1−n2) then F (x0 + m1) = F (x0 + m2), or equivalently F (x0) − x0 = m2 − m1. Applying F k(n1−n2) (k − 1)−times to this last expression, we have F (x0) − x0 = k(m2 − m1). Dividing by k(n1 − n2) and letting k → ∞, we obtain m2 − m1 ρ(f) = ∈ Q. n1 − n2

n1 Absurd and this proves our assertion. Hence the map H1 : A → B, F (x0) + m1 7→ n1ρ(f) + m1 is well-deﬁned, bijective and increasing. We extend H1 to a mapping H with domain R by setting

n H(x) = sup{nρ(f) + m : F (x0) + m < x}.

n Observe that, by the density of B, H can be deﬁned by H(x) = inf{nρ(f) + m : F (x0) + m > x}, else n there will be an element H(x) < n0ρ(f) + m0 < inf{nρ(f) + m : F (x0) + m > x} for some x ∈ R, by n0 lemma 4.9 x = F (x0) + m0, and so H(x) = n0ρ(f) + m0. Contradicting the above strict inequality.

Claim I. H is surjective: Let y ∈ R, since B is dense in R, there is a sequence {bk = nkρ(f)+mk}k≥1 n in B such that bk ↑ y. Now note that {F k (x0) + mk : k ∈ N} is bounded for any element of the form n n F (x0) + m, where nρ(f) + m > y. Let x = sup{F k (x0) + mk : k ∈ N}, then H(x) = sup{nρ(f) + m : n n n F (x0) + m < sup{F k (x0) + mk : k ∈ N}} ≤ y, but also H(x) = inf{nρ(f) + m : F (x0) + m > n sup{F k (x0) + mk : k ∈ N}} ≥ y, thus H(x) = y and our claim is proved.

Claim II. H is continuous: By a well-know result in real analysis (see paragraph previous to state- ment) it is suﬃce to prove that H is increasing. But it is immediate from deﬁnition of H and in fact x < y implies H(x) ≤ H(y).

Now, for any x ∈ R, we have

n H(x + 1) = sup{nρ(f) + m : F (x0) + m < x + 1} n = sup{nρ(f) + m : F (x0) + (m − 1) < x} n = sup{nρ(f) + (m + 1) : F (x0) + m < x} n = sup{nρ(f) + m : F (x0) + m < x} + 1 = H(x) + 1

Thus, H induces a continuous surjective function h : S1 → S1. On the other hand, for any x ∈ R

n H(F (x)) = sup{nρ(f) + m : F (x0) + m < F (x)} n−1 = sup{nρ(f) + m : F (x0) + m < x} n = sup{(n + 1)ρ(f) + m : F (x0) + m < x} n = sup{nρ(f) + m : F (x0) + m < x} + ρ(f) = H(x) + ρ(f)

1 Denoting by Fρ(f) a lift of the rotation Rρ(f) and applying the projection π : R → S = R/Z on the above equality, we obtain

(h ◦ f)(π(x)) = π(H ◦ F ) = π(ρ(f) + H(x))

= π(Fρ(f) ◦ H)(x) = (Rρ(f) ◦ h)(π(x)) for all x ∈ R. Therefore, h ◦ f = Rρ(f) ◦ h.

11 12 Diego A. S. Sanhueza

(a) Let x ∈ S1 → S1 be a transitive point for f. Then A is dense in R, so H(x) < H(y) in Claim II, and hence h is injective. (b) If no point in S1 is transitive for f, then A is not dense, so there is an interval I ⊆ R \ A where H is constant, so h is not injective and Rρ(f) is a factor map of f. Remark 4.12. (a) h can be chosen to be orientation-preserving. (b) If f ∈ Homeo+(S1) has irrational rotation number, if f is transitive, then it is minimal. It follows from Remark 4.8 (or directly from the conjugation with a rotation by an irrational angle). (c) The homeomorphism h depends on the choice of the point x0. Thus we have construct a lot of (semi-)conjugacy.

§5. SOLVED PROBLEMS

Problem 1. Let f : S1 → S1 be an orientation-preserving homeomorphism. Suppose that f has a ﬁxed point. Prove that ρ(f) = 0

Problem 2. Suppose that f : S1 → S1 is a homeomorphism which reverses orientation. Prove that: (1) f has exactly two ﬁxed points. (2) ρ(f 2) = 0.

Problem 3. Suppose that F : R → R is a lift of an orientation-preserving homeomorphism f : S1 → S1. Let {xn}n≥1 be a sequence in R. Prove that F n(x ) − x lim n n (7) n→∞ n exists.

Problem 4. Let f : S1 → S1 be a orientation-preserving homeomorphism. Suppose that F : R → R is a lift of f and prove that F n(x) − x F n(x) − x τ +(f, F ) = lim max and τ −(f, F ) = lim min n→∞ x∈R n n→∞ x∈R n

Problem 5. Let f : S1 → S1 be a orientation-preserving homeomorphism. Show that there is a unique lift F of f such that ρ(f) = ρ(F ).

Problem 6. Show that ρ(f ◦ g) = [ρ(f) + ρ(g)] mod 1.

Problem 7. Show that the rotation number depends continuously on the map in the C0−topology.

Exercise 1. Prove that if f : S1 → S1 is an orientation-preserving homeomorphism, there exists a unique lift F : R → R which satisﬁes ρ(f) = ρ(F ). Exercise 2. Prove proposition 2.6(ii).

Exercise 3. Suppose that f : S1 → S1 is an orientation-preserving homeomorphism with rational rotation number ρ(f) = p/q, (p, q) = 1. Prove that if π(x) is a periodic point for f, then there is a unique lift F such that F q(x) = x + p.

Exercise 4. Prove that two rotations Rα and Rβ are conjugated if and only if |α| = |β|.

Exercise 5. Show that if f : S1 → S1 is deﬁned by f(x) = 2x, then ρ(F ) could not exist.

Exercise 6. Let f : S1 → S1 deﬁned by f(x) = x + π + 0.41 sin 2πx. Show that L M F n(0.638938) − 0.638938 F n(0.328056) − 0.328056 lim 6= lim n→∞ n n→∞ n for any lift F of f.

12 Homeomorphisms on the circle 13

Exercise 7. Consider the following map on [0, 1]. 1 f (x) = x + sin(kπx). k (k + 1)π 1 (a) Prove that Fk induces a homeomorphism f on S que preserves orientation. (b) Find all ﬁxed points of fk and compute #P er(fk) for all k ≥ 1. (c) Drawing the graph of fk for all k ≥ 1

1 1 Exercise 8. Consider F1(x) = x + 2 sin(2πx) and F2(x) = x + 4π sin(2πx). Decide whether F1 and F2 are lifts of circle homeomorphisms. If so, decide whether that homeomorphism is orientation-preserving. If it is, determine the rotation number.

Exercise 9. Show that if f : S1 → S1 is an orientation-preserving homeomorphism, then can be to exist two periodic point with diﬀerent periodic.

Exercise 10. Let f : S1 → S1 deﬁned by

1 f(x) = (ax + b sin(2πx) + c) mod 1 x ∈ S a, b, c ∈ R. (a) Determine what conditions on the parameters a, b and c must be satisﬁed for f to be an orientation- preserving circle homeomorphism. (b) Compute the rotation number for a = 1, b = 1/8, c = 1/2. (c) Show that the map in (b) is not conjugated to Rρ(f).

+ 1 n 1 Exercise 11. Let f ∈ Homeo (S ) and suppose that z = limn→∞ f (x) exists for some x ∈ S . Prove that z ∈ P er(f).

Exercise 12. Let f ∈ Homeo+(S1). Find Ω(f).

Exercise 13. By using Proposition 4.7, prove that if Rα is a rotation by irrational angle, then every orbit is dense in T .

Exercise 14. (a) Exhibit an example of an orientation-preserving homeomorphism such that every orbit is periodic. + 1 (b) Let f ∈ Homeo (S ) as in (a). Prove that it is conjugated to Rρ(f).

References

[1] Apostol, T., Mathematical analysis, second ed., Addison-Wesley Publishing Co., Reading, Mass.- London-Don Mills, Ont., 1974. [2] Auslander, J. and Katznelson, Y., Continuous maps of the circle without periodic points, Israel J. Math. 32 (1979), no. 4, 375–381. [3] Brin, M. and Stuck, G., Introduction to dynamical systems, Camb. Univ. Press, Cambridge, 2002. [4] Denjoy, A., Sur les courbes definies par les équationsdifférentielles àla surface du tore., J. Math. Pures Appl. (9) 11 (1932), 333–375. [5] H. Furstenberg, Strict ergodicity and transformation of the torus, Amer. J. Math. 83 (1961), 573–601. [6] Herman, M., Sur la conjugaison différentiable des difféomorphismesdu cercle àdes rotations, Inst. Hautes Etudes´ Sci. Publ. Math. (1979), no. 49, 5–233. [7] , Une méthode pour minorer les exposants de Lyapounov et quelques exemples montrant le caractère local d’un théorèmed’Arnol0d et de Moser sur le tore de dimension 2, Comment. Math. Helv. 58 (1983), no. 3, 453–502. [8] Katok, A. and Hasselblatt, B., Introduction to the modern theory of dynamical systems, Cambridge University Press, Cambridge, 1995.

13 14 Diego A. S. Sanhueza

[9] Katznelson, Y., Sigma-finite invariant measures for smooth mappings of the circle, J. Analyse Math. 31 (1977), 1–18. [10] Lima, E., Curso de análise.Vol. 1, IMPA, Rio de Janeiro, 1976. [11] Markley, N., Homeomorphisms of the circle without periodic points, Proc. London Math. Soc. (3) 20 (1970), 688–698. [12] McDuff, D., C1-minimal subsets of the circle, Ann. Inst. Fourier (Grenoble) 31 (1981), no. 1, vii, 177–193. [13] Navas, A., Grupos de difeomorfismos del c´ırculo, SBM, Rio de Janeiro, 2007. [14] Norton, A., Denjoy minimal sets are far from affine, Ergodic Theory Dynam. Systems 22 (2002), no. 6, 1803–1812. [15] Poincaré, H., Mémoire sur les courbes définies par une équation différentielle (I), Journal de MathématiquesPures et Appliquées 7 (1881), 375–422 (fre). [16] , Sur les courbes définiespar les équationsdifférentielles (III), Journal de Mathématiques Pures et Appliquées 1 (1885), 167–244 (fre).

[17] Portela, A., New examples of cantor sets in S1 that are not C1−minimal, Bull. of the Brazilian Math. Soc. 38 (2007), no. 4, 623–633. [18] Walsh, J., The dynamics of circle homeomorphisms: a hands-on introduction, Math. Mag. 72 (1999), no. 1, 3–13. [19] Zdun, M.-C., On conjugacy of homeomorphisms of the circle possessing periodic points, J. of Math. Analysis and Applications 330 (2007), no. 1, 51–65.

Diego A. S. Sanhueza, IM, UFRJ, Rio de Janeiro, Brasil E-mail address, [email protected]