Symmetric Rigidity for Circle Endomorphisms with Bounded Geometry and Their Dual Maps

City University of New York (CUNY) CUNY Academic Works

All Dissertations, Theses, and Capstone Projects Dissertations, Theses, and Capstone Projects

6-2020

John Adamski The Graduate Center, City University of New York

How does access to this work benefit ou?y Let us know!

More information about this work at: https://academicworks.cuny.edu/gc_etds/3790 Discover additional works at: https://academicworks.cuny.edu

This work is made publicly available by the City University of New York (CUNY). Contact: [email protected] Symmetric rigidity for circle endomorphisms with bounded geometry and their dual maps

John Adamski

A dissertation submitted to the Graduate Faculty in Mathematics in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy, The City University of New York.

2020 ii

c 2020

John Adamski

This manuscript has been read and accepted for the Graduate Faculty in Mathematics in satisfaction of the dissertation requirements for the degree of Doctor of Philosophy.

(required signature)

Date Chair of Examining Committee

(required signature)

Date Executive Oﬃcer

Dr. Yunping Jiang

Dr. Linda Keen

Dr. Frederick Gardiner

Dr. Sudeb Mitra, Dr. Zhe Wang

Supervisory Committee

THE CITY UNIVERSITY OF NEW YORK iv

Abstract

Symmetric rigidity for circle endomorphisms with bounded geometry and their dual maps

John Adamski

Advisor: Dr. Yunping Jiang

Let f be a circle endomorphism of degree d ≥ 2 that generates a sequence of Markov partitions that either has bounded nearby geometry and bounded geometry, or else just has bounded geometry, with respect to normalized Lebesgue measure. We deﬁne the dual symbolic space Σ∗ and the dual circle endomorphism f ∗ = h˜ ◦ f ◦ h−1, which is topologically conjugate to f. We describe some properties of the topological conjugacy h˜. We also describe an algorithm for generating arbitrary circle endomorphisms f with bounded geometry that preserve Lebesgue measure and their corresponding dual circle endomorphisms f ∗ as well as the conjugacy h˜, and implement it using MATLAB.

We use the property of bounded geometry to deﬁne a convergent Mar- v tingale on Σ∗, and apply the study of such Martingales to obtain a rigidity theorem. Suppose f and g are two circle endomorphisms of the same degree d ≥ 2 such that each has bounded geometry and each preserves the normalized Lebesgue probability measure. Suppose that f and g are symmetrically conjugate. That is, g = h ◦ f ◦ h−1 and h is a symmetric circle homeomorphism. We deﬁne a property called locally constant limit of Martingale, and show that if f has this property then f = g. Acknowledgements

Thank you to my advisor, Professor Yunping Jiang, for your steady guidance, abundant wisdom, and inﬁnite patience. For the many discussions on Friday afternoons, when I left your oﬃce with more ideas than I had when I entered,

I will be forever grateful.

Thank you to Professor Zhe Wang and Professor Yunchun Hu for the generosity you showed me with your time and your thoughts. Thank you to

Professor Fred Gardiner and Professor Linda Keen for sharing your passion and your knowledge, taking interest in my work and progress, and for being bottomless sources of encouragement. Thank you to Professor Tao Chen for all of your help, your cheery disposition, and your ﬂexibility with scheduling over the years. Thank you to Professor Sudeb Mitra for putting front and center the joy of methematics and having a good sense of humor, both inside and outside of the classroom.

Thank you to my friends who helped in so many ways big and small,

vi vii providing the right balance of encouragement and distraction: Eli Amzallag,

Dave and Maria Anderson, Jess Bogwicz, Hina Dar, Corey J. Feldman, Ryan and Julia Ford, Nick Hundley, Megan Moskop, Nick and Katie Nacca, Anna

Padilla, Jon and Marlee Patrizio, Shilpa Ray, Dave Schnurman, Leigh Sugar,

Rylan Summit, Warren Tai, and Tara Thomas.

Thank you to my family. And a special thank you to my mom and dad.

I love you both. Contents

1 Quasiconformal and quasisymmetric maps 1

1.1 Conformal maps ...... 1

1.2 Quasi-conformal maps ...... 7

1.3 Quasisymmetric and uniformly quasisymmetric maps . . . . . 11

2 Circle endomorphisms 13

2.1 Circle homeomorphisms and endomorphisms ...... 13

2.2 Symbolic space and topological representation ...... 22

2.3 Bounded geometry and bounded nearby geometry ...... 27

2.4 Measures with bounded geometry and bounded nearby geometry 49

3 Invariant measures 52

3.1 Borel measures ...... 52

3.2 Invariant measures and ergodicity ...... 60

viii CONTENTS ix

3.3 Invariant measures on T with bounded geometry and bounded

nearby geometry ...... 70

3.4 Constructing examples of distribution functions for measures

with bounded geometry preserved by q2 ...... 74

4 The dual circle map 92

4.1 Equivalence of partitions and maps ...... 92

4.2 Dual partitions, dual maps, and dual measures ...... 95

4.3 Properties of the dual homeomorphism ...... 102

4.4 Dual symbolic space ...... 107

5 Martingales 114

5.1 Deﬁnition and convergence theorem ...... 114

5.2 Locally constant limits and ﬁnite martingales ...... 127

5.3 A symmetric rigidity theorem assuming locally constant limit

of martingale ...... 137

6 Appendix 140

6.1 Bernoulli measure ...... 140

6.2 Bounded geometry measure ...... 141

6.3 Dual measure and dual endomorphism ...... 146 CONTENTS x

Bibliography 153 List of Figures

1.1 Conformally mapping a quadrilateral onto a rectangle . . . . . 10

2.1 On the left is the graph of the ternary Cantor function on

[0, 1]. On the right is the graph of F (x) = x + C1/3(x) on

[0, 1]. Both functions map the same sets of measure 0 to sets

of measure 1...... 19

2.2 f −1(1) cuts T into d − 1 closed arcs...... 22

2.3 Gk(x) for d =2...... 32

2.4 One possible way that x − t, x, and x + t may be contained in

partitions ηN−1,f , ηN,f , and ηN+N1,f . The corresponding ﬁgure

for how H(x − t),H(x), and H(x + t) would be contained in

partitions ηN−1,g, ηN,g, and ηN+N1,g is homeomorphic...... 35

2.5 The circle endomorphism Fα shown acting on R/Z. Here α =

.8. The map Fα has bounded geometry but it is not uniformly

quasisymmetric...... 44

xi LIST OF FIGURES xii

2.6 The circle homeomorphism Hα shown acting on R/Z. It is not

−1 quasisymmetric. Fα = Hα ◦ Q2 ◦ Hα ...... 45

2.7 The circle endomorphism F˜ shown acting on R/Z. The map F˜

is uniformly quasisymmetric. Thus is has bounded geometry

and bounded nearby geometry...... 46

2.8 The circle homeomorphism H˜ shown acting on R/Z. It is

˜ ˜ ˜ −1 quasisymmetric. F = H ◦ Q2 ◦ H ...... 47

3.1 For all f ∈ C(X) and for all open intervals I ⊆ [0, 1], the set R of all measures µ ∈ M(X) such that X f dµ ∈ I belongs to

the weak∗ topology on M(X). The weak∗ topology on M(X)

is the smallest topology satisfying this condition...... 53

3.2 On the left are the distribution functions h(x) = µ([0, x]) for

two Bernoulli measures µ. The top Bernoulli measure has

α = .2 and the bottom Bernoulli measure has α = .8. On the

−1 right are the corresponding circle endomorphisms f = h◦q2 ◦h . 78

3.3 On the left are the distribution functions h(x) = µ([0, x]) for

two Bernoulli measures µ. The top Bernoulli measure has

α = .4 and the bottom Bernoulli measure has α = .9. On the

−1 right are the corresponding circle endomorphisms f = h◦q2 ◦h . 79 LIST OF FIGURES xiii

3.4 The interval Iωn ∈ ηn and its subinterval Iωn0 ∈ ηn+1, along

with their preimages belonging to ηn+1 and ηn+2, respectively. 86

3.5 On the left are the distribution functions h(x) = µ([0, x]) for

three arbitrary measures µ that are invariant with respect to

the doubing map q2, and that have bounded geometry with

C = .1. On the right are the corresponding circle endomor-

−1 phisms f = h ◦ q2 ◦ h ...... 91

4.1 The lifts F0, F1, and F2...... 94

4.2 For d = 2 the intervals of ηn are labeled as n-digit binary

numbers, increasing left to right, from 0 to 2n+1 − 1...... 96

∗ 4.3 Shuﬄing intervals of ηn to construct ηn...... 98

∗ 4.4 Topologically conjugate circle endomorphisms qd, f, and f

(the dual circle endomorphism of f)...... 100

4.5 Distribution functions h (left, blue) and dual distribution func-

tions h∗ (left, red), and their corresponding circle endomor-

−1 ∗ ∗ ∗ −1 phisms f = h ◦ q2 ◦ h (right, blue) and f = h ◦ q2 ◦ (h )

(right, red). On the left in black is the dual homeomorphism

h˜ = h∗ ◦ h−1...... 101 LIST OF FIGURES xiv

∞ ˜ 4.6 The increasing sequence {am}m=1 of ﬁxed points of h, con-

structed with n =2...... 105

4.7 Two directed d-nary trees (d=2). The dotted tree shows the

action of σ on left-cylinders of Σ, and the solid tree shows the

action of σ∗ on right-cylinders of Σ∗...... 110

5.1 Finite Martingale of length n0 = 3 and C = 4. The top left

shows the lengths of the intervals belonging to ηn and the top

right shows the values of Xn on intervals belonging to ηn, for

n = 1, 2, 3, 4. The bottom left shows the distribution function

h and the bottom right shows the circle endomorphism f, with

−1 f = h ◦ q2 ◦ h . The circle endomorphism f has bounded

geometry but does not have bounded nearby geometry, thus

it is not uniformly quasisymmetric...... 135 LIST OF FIGURES xv

5.2 Another ﬁnite Martingale of length n0 = 3 and C = 4. The

top left shows the lengths of the intervals belonging to ηn and

the top right shows the values of Xn on intervals belonging to

ηn, for n = 1, 2, 3, 4. The bottom left shows the distribution

function h and the bottom right shows the circle endomor-

−1 phism f, with f = h ◦ q2 ◦ h . The circle endomorphism f

has bounded geometry and bounded nearby geometry, thus it

is uniformly quasisymmetric...... 136 Chapter 1

Quasiconformal and quasisymmetric maps

1.1 Conformal maps

In this section we introduce conformal maps and discuss their existence and importance in the study of functions of a complex variable. A good reference for the following is [5].

Definition 1. Suppose Ω ⊆ C is an open subset of the complex plane and f :Ω → C is a complex-valued function defined on Ω. Given a point z0 ∈ Ω, we say f is differentiable at z0 if the limit

f(z) − f(z ) lim 0 (1.1) z→z0 z − z0 exists. In this case, we call the limit the derivative of f at z0 and denote

0 it by f (z0). If f is diﬀerentiable at every point in Ω, then we say f is holomorphic in Ω.

1 CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 2

If we let

1 i x = (z + z) = Re(z), y = − (z − z) = Im(z), 2 2 u = Re(f(z)), v = Im(f(z)) then we can write

f(z) = f(x, y) = u(x, y) + iv(x, y).

The existence of the limit in equation (1.1) implies that holomorphic functions satisfy the Cauchy-Riemann equations

ux = vy and uy = −vx,

i.e. fx = −ify. (1.2)

Furthermore, we have

∂x 1 ∂x 1 ∂y i ∂y i = , = , = − , = . ∂z 2 ∂z 2 ∂z 2 ∂z 2

Thus,

∂f ∂f ∂x ∂f ∂y 1 = + = (f − if ), and ∂z ∂x ∂z ∂y ∂z 2 x y ∂f ∂f ∂x ∂f ∂y 1 = + = (f + if ). ∂z ∂x ∂z ∂y ∂z 2 x y

Therefore, since they satisfy the Cauchy-Riemann equations (1.2), holomorphic functions satisfy ∂f ∂f = f 0(z), = 0. (1.3) ∂z ∂z CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 3

The fact that a holomorphic function f satisﬁes fz = 0 is one we will revisit shortly. For now, we examine the connection between holomorphic functions and conformal maps.

Deﬁnition 2. Suppose Ω ⊆ C is an open subset of the complex plane and f :Ω → C is a complex-valued function deﬁned on Ω. Given a point z0 ∈ Ω

0 with a deleted neighborhood D (z0; r) ⊂ Ω,

0 D (z0; r) := {z ∈ C : 0 < |z − z0| < r},

in which f(z) 6= f(z0), we say f is conformal at z0, i.e. f preserves angles at z0, if the limit

−iθ iθ lim e Arg[f(z0 + re ) − f(z0)] r→0+ exists and is independent of θ, where

z Arg(z) = . |z|

If f is conformal at every point in Ω then we say f is conformal in Ω.

The following theorem makes clear the connection between holomorphic functions with non-zero derivatives and conformal mappings.

0 Theorem 1. Let f map an open subset Ω ∈ C into the plane. If f (z0) exists

0 at some z0 ∈ Ω and f (z0) 6= 0, then f is conformal at z0. Conversely, if the CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 4

diﬀerential of f exists and is diﬀerent from 0 at z0, and if f preserves angles

0 at z0, then f (z0) exists and is diﬀerent from 0. Here, the diﬀerential of f at

2 2 z0 = (x0, y0) is taken to be the linear transformation L : R → R such that

p 2 2 f(x0 + x, y0 + y) = f(x0, y0) + L(x, y) + x + y (x, y), (1.4)

lim (x, y) = 0. (x,y)→(0,0)

Proof. Without loss of generality, let us assume z0 = f(z0) = 0 throughout the proof. Let f 0(0) = a 6= 0. Then

−iθ iθ −iθ iθ lim e Arg[f(z0 + re ) − f(z0)] = lim e Arg[f(re )] r→0+ r→0+

Conversely, we can write the diﬀerential L(x, y) as az + bz for some a, b ∈

C, not both 0. Thus, equation (1.4) becomes

f(z) = az + bz + |z|(z).

Since f is conformal, the limit

e−iθ(areiθ + bre−iθ + r(reiθ)) lim e−iθ Arg[f(reiθ)] = lim r→0+ r→0+ |e−iθ||areiθ + bre−iθ + r(reiθ)| CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 5

a + be−2iθ = lim r→0+ |a + be−2iθ| exists independent of θ. Since the limit exists, we may ignore any values of θ which cause the denominator to equal 0. Then the limit is equal to

Arg[a + be−2iθ], and the only way for this to not depend on θ is if b = 0.

Since we assume the diﬀerential is non-zero, this implies a 6= 0. Thus

f(z) = az + |z|(z)

f 0(z) = a 6= 0.

Deﬁnition 3. A non-empty, open, connected subset of the complex plane is called a region.

Since holomorphic functions fail to be injective in neighborhoods of criti- cal points (see [2] Theorem 13.7), it follows that given two regions Ω1, Ω2 ⊂ C and a holomorphic bijection f :Ω1 → Ω2, this map f is conformal. The in-

−1 verse map f :Ω2 → Ω1 is also holomorphic, with derivative

1 (f −1)0(z) = 6= 0, f 0(f −1(z)) and so is also conformal. CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 6

Deﬁnition 4. Two regions Ω1, Ω2 ⊂ C are called conformally equivalent if there exists a holomorphic bijection (equivalently, conformal bijection) f :

Ω1 → Ω2. Such a bijection f is called a conformal map.

Let H(Ω) denote the set of all holomorphic functions deﬁned on the region

Ω. Then a conformal map f :Ω1 → Ω2 provides a bijection from H(Ω2) to

H(Ω1) deﬁned by

f(ϕ) = ϕ ◦ f, ϕ ∈ H(Ω2).

Moreover, this map preserves both sums and products and so is a ring iso- morphism from H(Ω2) to H(Ω1). This allows one to answer questions about

H(Ω2) by studying H(Ω1), as long as Ω1 and Ω2 are conformally equivalent.

Theorem 2 (Riemann mapping theorem). Every simply connected region Ω in the complex plane (except the plane itself) is conformally equivalent to the open unit disc ∆.

Proof. See [5] Theorem 14.8.

Thus, conformal maps allow the study of H(∆) to generalize to the study of H(Ω), for any simply connected Ω (C.

Furthermore, certain conformal maps between regions may be extended

(uniquely) to conformal maps between larger regions (supersets). This is the content for the following. CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 7

Theorem 3 (Schwarz reﬂection priciple). Suppose L is a segment of the real axis, Ω+ is a region in the upper half-plane H, and every t ∈ L is the center

+ − of an open disc Dt such that H ∩ Dt lies in Ω . Let Ω be the reﬂection of

Ω+:

Ω− = {z : z ∈ Ω+}.

Suppose f = u + iv is holomorphic in Ω+, and

lim v(zn) = 0 n→∞

+ for every sequence {zn} in Ω which converges to a point of L.

Then there is a function F , holomorphic in Ω+ ∪L∪Ω−, such that F (z) = f(z) in Ω+. This F satisﬁes the relation

F (z) = f(z), z ∈ Ω+ ∪ L ∪ Ω−.

Proof. See [5] Theorem 11.17.

1.2 Quasi-conformal maps

Given two arbitrary rectangular regions R1,R2 ⊂ C, the Riemann mapping theorem shows that there exists a conformal map f : R1 → R2 which maps the interior of one rectangle to the interior of the other. However, if one wishes to continuously extend f so that it also maps the boundary of one CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 8 rectangle to the boundary of the other, with vertices mapped to vertices, this is only possible when the rectangles are similar. The reason for this is based on the following argument. Take two copies of C, tiled by reflections of R1 and R2, respectively. If f was such a map, then by repeated applications of the Schwartz reflection principle, we could extend f to all of C, sending each reflection of R1 to its corresponding reflection of R2. Thus, the extended map F : C → C is one-to-one, and so F (z) = az + b. This implies that R1 and R2 are similar.

For rectangular regions R1 and R2 which are not similar, a natural question to ask is the following. Of all the maps f : R1 → R2 which map vertices to vertices, which is the one that is “closest to conformal”?

In 1928, with this question in mind, Grötzsch introduced the first measure of how “close to conformal” a given sense-preserving, C1 homeomorphism f is. Pointwise, we define

|fz| + |fz| Df = . |fz| − |fz|

Df (z) is called the dilatation of f at the point z. If f is conformal, then its dilatation is identically 1. The supremum of Df over the domain of f measures how “close to conformal” f is.

Deﬁnition 5. When Df is bounded, we say f is quasiconformal. More CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS 9

speciﬁcally, when Df ≤ K, we say f is K-quasiconformal.

Historically, this was the ﬁrst deﬁnition of quasiconformal. Note that it

1 requires the homeomorphism f :Ω1 → Ω2 to be C . Later, the deﬁnition was generalized to apply to a much larger class of homeomorphisms.

Deﬁnition 6. Given a region Ω ⊂ C, a quadrilateral in Ω is a subregion

Q, Q ⊂ Ω, whose boundary is a simple closed curve with two disjoint closed arcs, labelled the b-arcs.

Lemma 1. Given a quadrilateral Q ∈ Ω, there exists a conformal mapping of Q onto the interior of a rectangle R, which extends continuously to the boundary, mapping boundary to boundary and mapping endpoints of the b- arcs to vertices. The rectangle R is unique up to an aﬃne transformation.

Proof. The Riemann mapping theorem provides a conformal map f : Q → H, which we may extend continuously to the boundary. Let [A, B], [C,D] ⊂ R be the images of the b-arcs of Q. Let g be deﬁned by the Schwarz-Christoﬀel mapping Z z 1 g(z) = √ √ √ √ dξ. 0 ξ − A ξ − B ξ − C ξ − D Then the composition F = g◦f maps Q conformally onto a rectangle R with the b-arcs mapped to opposite sides, as desired. If F˜ were another such map from Q to another rectangle R˜, then F˜ ◦ F −1 : R → R˜ would be a conformal CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS10

Figure 1.1: Conformally mapping a quadrilateral onto a rectangle map between rectangles sending vertices to vertices. Hence F˜ ◦ F −1 is an aﬃne transformation, and both rectangles R and R˜ are similar.

The images of the b-arcs form one pair of opposite sides of R with length, say, b. Let the other pair of opposite sides of R have length a.

Deﬁnition 7. The module m of the quadrilateral Q is deﬁned to be m(Q) = a/b. Note that if Q∗ is the same region with complimentary arcs, then m(Q∗) = m(Q)−1.

We can now state a more general deﬁnition of quasiconformal.

Deﬁnition 8. A homeomorphism f :Ω1 → Ω2 is quasiconformal if there exists a constant K such that for every quadrilateral Q ∈ Ω1 we have

K−1m(Q) ≤ m(f(Q)) ≤ Km(Q), CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS11 i.e. modules of quadrilaterals are K-quasi-invariant. More speciﬁcally, such a map is called K-quasiconformal.

It turns out that a K-quasiconformal mapping is H¨older continuous. In fact, if f is a K-quasiconformal mapping of the open unit disc ∆ onto itself, then for any distinct z1, z2 ∈ ∆ we have

1/K |f(z1) − f(z2)| < 16|z1 − z2| .

This is known as Mori’s theorem. A proof can be found in [1, page 30]. As a corollary to this result, every quasiconformal mapping from an open disc onto another open disc can be continuously extended to a homeomorphism between the closed discs. We now shift our attention to these induced homeomorphisms between the boundaries of discs, which we assume to be the unit circle.

1.3 Quasisymmetric and uniformly quasisymmetric maps

What type of homeomorphism h : S1 → S1 of the unit circle is an extension of a quasiconformal map f : ∆ → ∆? Since ∆ and H are conformally equivalent, in order to answer this question more simply, we ask instead, what type of homeomorphism H : R → R (h(−∞) = −∞, h(∞) = ∞) is an extension of a quasiconformal map F : H → H. CHAPTER 1. QUASICONFORMAL AND QUASISYMMETRIC MAPS12

Theorem 4. The boundary values H of a K-quasiconformal mapping F :

H → H satisfy the condition

H(x + t) − H(x) M −1 ≤ ≤ M, ∀x, t ∈ , (1.5) H(x) − H(x − t) R where M > 0 is a constant depending only on K.

Proof. See [1], chapter 4, theorem 1.

Deﬁnition 9. A homeomorphism H : R → R that satisﬁes the condition

(1.5) for some M > 0 is called quasisymmetric or, more speciﬁcally, M- quasisymmetric.

Furthermore, the converse to theorem 4 is also true.

Theorem 5. Every mapping H : R → R which is M-quasisymmetric is extendable to a K-quasiconformal mapping F : C → C for a K that depends only on M.

Proof. See [1], chapter 4, theorem 2. Chapter 2

Circle endomorphisms

2.1 Circle homeomorphisms and endomorphisms

Let T = {z ∈ C | |z| = 1} be the unit circle in the complex plane C. The universal cover of T is the real line R with covering map

2πix π(x) = e : R → T.

Deﬁnition 10. A circle homeomorphism is an orientation preserving homeomorphism h : T → T. Note that every such map h can be lifted to an orientation preserving homeomorphism H : R → R satisfying

π ◦ H(x) = h ◦ π(x) ,H(x + 1) = H(x) + 1, ∀x ∈ R.

We assume that 0 ≤ H(0) < 1. Thus H is unique and there is one-to-one correspondence between h and H. Therefore, we also call such a map H a circle homeomorphism.

13 CHAPTER 2. CIRCLE ENDOMORPHISMS 14

H R R

π π

h T T

Deﬁnition 11. A circle homeomorphism h is called quasisymmetric if its lift H is quasisymmetric, i.e. if there exists a single constant M > 1 such that 1 H(x + t) − H(x) ≤ ≤ M, ∀x ∈ , ∀t > 0. M H(x) − H(x − t) R

Furthermore, h is called symmetric if its lift H is symmetric, i.e. if there exists a bounded function : R → R such that (t) → 0+ as t → 0+ and

1 H(x + t) − H(x) ≤ ≤ 1 + (t) , ∀x ∈ , ∀t > 0. 1 + (t) H(x) − H(x − t) R

In chapter 5 We will make use of the following lemma about quasisymmetric homeomorphisms, proved in [13].

Lemma 2. Suppose H is an M-quasisymmetric homeomorphism of [0, 1] onto [0, 1] with H(0) = 0 and H(1) = 1. Then we have that

|H(x) − x| ≤ M − 1, ∀x ∈ [0, 1].

The bound M − 1 is the sharpest estimation. CHAPTER 2. CIRCLE ENDOMORPHISMS 15

Deﬁnition 12. A circle endomorphism is an orientation preserving covering map f : T → T of (topological) degree d ≥ 2. Every such map has a

ﬁxed point, and we assume it is 1, that is f(1) = 1. Note that every such map f can be lifted to an orientation preserving homeomorphism F : R → R satisfying

π ◦ F (x) = f ◦ π(x) ,F (x + 1) = F (x) + d, ∀x ∈ R.

We assume that F (0) = 0. Thus F is unique and there is a one-to-one correspondence between f and F . Therefore, we also call such a map F a circle endomorphism.

F R R

π π

f T T

A circle endomorphism f is Ck if its lift F is Ck, that is, if the kth derivative F (k) is continuous. Furthermore, f is Ck+α for some 0 < α ≤ 1 if its lift F is Ck+α, that is, if F (k) is α-H¨oldercontinuous, i.e. there exists a single constant C > 0 such that for all x ∈ R,

|F (k)(y) − F (k)(x)| ≤ C|y − x|α. CHAPTER 2. CIRCLE ENDOMORPHISMS 16

Deﬁnition 13. A C1 circle endomorphism is called expanding if there are constants C > 0 and λ > 1 such that

(F n)0(x) ≥ Cλn, n = 1, 2,....

Deﬁnition 14. A circle endomorphism f is called uniformly quasisymmetric if there exists a single constant M > 1 such that

1 F −n(x + t) − F −n(x) ≤ ≤ M, ∀x ∈ , ∀t > 0 , ∀n ≥ 1. M F −n(x) − F −n(x − t) R

Furthermore, f is called uniformly symmetric if there exists a bounded function : R → R such that (t) → 0+ as t → 0+ and

1 F −n(x + t) − F −n(x) ≤ ≤ 1 + (t) , ∀x ∈ , ∀t > 0 ∀n ≥ 1. 1 + (t) F −n(x) − F −n(x − t) R

Lemma 3. If a degree d ≥ 2 circle endomorphism f is expanding and C1+α for some 0 < α ≤ 1 then f is uniformly symmetric.

Proof. By deﬁnition, F 0 is α-H¨oldercontinuous. Thus there exists a constant

C1 > 0 such that

0 0 α |F (x) − F (y)| ≤ C1|x − y| , ∀x, y ∈ R.

Since f is expanding, there exist constants C2 > 0 and λ > 1 such that

n 0 n (F ) (x) ≥ C2λ , ∀x ∈ R , ∀n ≥ 1. CHAPTER 2. CIRCLE ENDOMORPHISMS 17

−k −k For any x, y ∈ R and n ≥ 1, let xk = F (x) and yk = F (y), 0 < k ≤ n.

Thus

−n 0 n 0 n (F ) (x) (F ) (yn) X 0 0 log −n 0 = log n 0 ≤ | log F (xk) − log F (yk)|. (F ) (y) (F ) (xn) k=1

Since f is expanding, the mean value theorem implies

n n X 0 0 1 X 0 0 | log F (xk) − log F (yk)| ≤ |F (xk) − F (yk)|, C2λ k=1 k=1 and since f is C1+α, we have

n n 1 X 0 0 C1 X α |F (xk) − F (yk)| ≤ |xk − yk| . C2λ C2λ k=1 k=1

It follows from the inverse function theorem and the expanding condition for f that

n n α n ! C1 X α C1 X 1 C1 X 1 α |xk − yk| ≤ k |x − y| = 1+α αk |x − y| C2λ C2λ C2λ C λ λ k=1 k=1 2 k=1 C1 1 α ≤ 1+α α |x − y| C2 λ λ − 1

Let C C = 1 . 1+α α C2 λ(λ − 1) In summary, we have the following H¨olderdistortion property.

−n 0 1 (F ) (x) α ≤ ≤ eC|x−y| eC|x−y|α (F −n)0(y) CHAPTER 2. CIRCLE ENDOMORPHISMS 18

Furthermore, let

(t) = eC(2t)α − 1.

Then (t) > 0 is a bounded function such that (t) → 0+ as t → 0+ and such that

1 F −n(x + t) − F −n(x) (F −n)0(ξ) ≤ = ≤ 1 + (t) , ∀x ∈ , ∀t > 0, 1 + (t) F −n(x) − F −n(x − t) (F −n)0(η) R where ξ and η are two numbers provided by the mean value theorem in

(x, x + t) and (x − t, x), respectively, with |ξ − η| < 2t. Thus f is uniformly symmetric.

Remark 1. For degree d ≥ 2 circle endomorphisms, the uniformly symmetric condition is a weaker condition than the C1+α expanding condition for any

0 < α ≤ 1. For example, a uniformly symmetric circle endomorphism could be singular, that is, it could map a set with zero Lebesgue measure to a set with positive Lebesgue measure. It could even be totally singular – that is, it could map a set with zero Lebesgue measure to a set with full Lebesgue measure.

The well-known ternary Cantor function C1/3 : [0, 1] → [0, 1] (whose graph is the “devil’s staircase”, see ﬁgure 1) is a totally singular function, and the map   F (x + 1) − 2 if x < 0 F (x) = x + C1/3(x) if 0 ≤ 0 < 1  F (x − 1) + 2 if 1 ≤ x CHAPTER 2. CIRCLE ENDOMORPHISMS 19 is a singular degree 2 circle endomorphism. However, neither is uniformly symmetric.

1.8 1

1.6

0.8 1.4

1.2 0.6

0.4 0.8

0.6 0.2

0.4

0 0 0.2 0.4 0.6 0.8 1 0.2

0 0 0.5 1

Figure 2.1: On the left is the graph of the ternary Cantor function on [0, 1]. On the right is the graph of F (x) = x+C1/3(x) on [0, 1]. Both functions map the same sets of measure 0 to sets of measure 1.

Deﬁnition 15. Given a C1 circle endomorphism f, deﬁne the function

ω(t) = sup |F 0(x) − F 0(y)|, t > 0. |x−y|≤t

This function ω(t) is called the modulus of continuity of F 0. Then f is called C1 Dini if Z 1 ω(t) dt < ∞. (2.1) 0 t CHAPTER 2. CIRCLE ENDOMORPHISMS 20

Remark 2. For a C1 circle endomorphism f of degree d , we have F (x+1) =

F (x) + d and, more generally,

F n(x + 1) = F n(x) + dn, ∀n ≥ 1.

Thus,

(F n)0(x + 1) = (F n)0(x), ∀n ≥ 1.

It follows from the deﬁnition of ω(t) that

ω(t) = ω(1) < ∞, ∀t ≥ 1.

Lemma 4. A C1 Dini expanding circle endomorphism f is uniformly symmetric.

Proof. The proof begins similarly to lemma 3. Since f is expanding, there exist constants C1 > 0 and λ > 1 such that

n 0 n (F ) (x) ≥ C1λ , ∀x ∈ R , ∀n ≥ 1.

−k −k For any x, y ∈ R and n ≥ 1, let xk = F (x) and yk = F (y), 0 < k ≤ n.

Thus

−n 0 n 0 n (F ) (x) (F ) (yn) X 0 0 log −n 0 = log n 0 ≤ | log F (xk) − log F (yk)|. (F ) (y) (F ) (xn) k=1 Since f is expanding, the mean value theorem implies

n n X 0 0 1 X 0 0 | log F (xk) − log F (yk)| ≤ |F (xk) − F (yk)|, C1λ k=1 k=1 CHAPTER 2. CIRCLE ENDOMORPHISMS 21 and we have

Z ∞ t 1 Z t/C1 ω(u) ω˜(t) ≤ ω x dx = du 0 C1λ ln λ 0 u

It follows from the fact that f is C1 Dini and remark 2, that

ω˜(t) < ∞, ∀0 ≤ t ≤ 1, and

ω˜(t) → 0 as t → 0+.

Letting C = 1/(C1λ), we then have the following Dini distortion property.

(F n)0(x) e−Cω˜(|x−y|) ≤ ≤ eCω˜(|x−y|), ∀x, y ∈ , ∀n ≥ 1. (2.2) (F n)0(y) R

Now deﬁne eCw˜(t) − 1 if 0 < t ≤ 1 (t) = eCw˜(1) − 1 if 1 < t.

Then (t) > 0 is a bounded function such that (t) → 0+ as t → 0+ and such that

1 F −n(x + t) − F −n(x) (F −n)0(ξ) ≤ = ≤ 1 + (t) , ∀x ∈ , ∀t > 0, 1 + (t) F −n(x) − F −n(x − t) (F −n)0(η) R CHAPTER 2. CIRCLE ENDOMORPHISMS 22 where ξ and η are two numbers provided by remark 2 belonging to [0, 1].

Thus f is uniformly symmetric.

2.2 Symbolic space and topological representation

Suppose f is a circle endomorphism of degree d with f(1) = 1. Then the

−1 preimage f (1) consists of d points which partition T into d closed arcs J0,

J1,..., Jd−1. The labeling of these arcs is ordered counter-clockwise, with

J0 and Jd−1 sharing a common endpoint of 1.

Figure 2.2: f −1(1) cuts T into d − 1 closed arcs.

Deﬁne ξ0 to be this partition of T according to f.

ξ0 = {J0,J1,...,Jd−1}

Then ξ0 is a Markov partition. CHAPTER 2. CIRCLE ENDOMORPHISMS 23

Deﬁnition 16. (T, ξ0, f) is a Markov Partition if the following three conditions are satisﬁed.

Sd−1 1. T = i=0 Ji,

2. The restriction of f to the interior of Ji is injective for every 0 ≤ i ≤

d − 1,

3. f(Ji) = T for every 0 ≤ i ≤ d − 1.

Let I0, I1,..., Id−1 be the unique lifts of J0, J1,..., Jd−1 in [0, 1]. Then the following conditions are satisﬁed.

Sd−1 i [0, 1] = i=0 Ii,

ii F (Ii) = [i, i + 1] for every 0 ≤ i ≤ d − 1.

Let

η0 = {I0,I1,...,Id−1}.

Then η0 is a partition of [0, 1].

−n n n Consider the pull-back partition ξn = f ξ0 of ξ0 by f . It contains d intervals and is also a Markov partition of T with respect to f n. Intervals in

ξn can be labeled as follows. Let ωn = i0i1 . . . in−1 be a word of length n of 0’s,

n 1’s,. . . , and (d − 1)’s. There are exactly d distinct words ωn. Each interval CHAPTER 2. CIRCLE ENDOMORPHISMS 24

k J ∈ ξn is labeled with a word ωn, written Jωn , such that f (Jωn ) ∈ Jik for

0 ≤ k ≤ n. Thus

ξn = {Jωn | ωn = i0i1 . . . in−1, ik ∈ {0, 1, . . . , d − 1}, 0 ≤ k ≤ d − 1}.

Let ηn be the corresponding lift of ξn to the interval [0, 1] with the same labeling. That is,

ηn = {Iωn | ωn = i0i1 . . . in−1, ik ∈ {0, 1, . . . , d − 1}, 0 ≤ k ≤ d − 1}, with

k π ◦ F (Iωn ) ∈ Jik , ∀0 ≤ k ≤ n.

Consider the space

∞ Y Σ = {0, 1, . . . , d − 1} n=0

= {ω = i0i1 ... | ik ∈ {0, 1, . . . , d − 1}, k = 0, 1,...}.

Deﬁnition 17. A left-cylinder for a ﬁxed word ωn = i0i1 . . . in−1 of length n is

0 0 0 0 [ωn] = {ω = i0i1 . . . in−1inin+1 ... | in+k ∈ {0, 1, . . . , d − 1}, k = 0, 1,...}.

The set of all left-cylinders forms a topological basis of Σ. We call the topology generated by the set of all left-cylinders the left topology, and we call the set Σ with the left-topology the symbolic space. CHAPTER 2. CIRCLE ENDOMORPHISMS 25

For any ω = i0i1i2 ..., let

σ(ω) = i1i2 ... be the left shift map. Then (Σ, σ) is called a symbolic dynamical system.

For a point ω = i0i1 ... ∈ Σ, let ωn = i0i1 . . . in−1. Then

T ⊃ Jω1 ⊃ Jω2 ⊃ ... ⊃ Jωn ⊃ ....

Since each Jωn is compact and non-empty,

∞ Jω = ∩n=1Jωn 6= ∅.

If every Jω = {xω} contains only one point, then we deﬁne the projection πf from Σ onto T as

πf (ω) = xω.

The projection πf is one-to-one except for on a countable set 00 ..., (d − 1)(d − 1) ..., B = n ≥ 0, ω0 = ∅ ωn(d − 1)00 . . . , ωn0(d − 1)(d − 1) ... on which πf is two-to-one. These are precisely the points in Σ that project

to an endpoint of Jωn for some n. From our construction we have

πf ◦ σ(ω) = f ◦ πf (ω), ω ∈ Σ.

For n ≥ 1, let Σn denote the set of all words ωn of {0, 1, . . . d−1} of length n. For any interval I = [a, b] ∈ [0, 1], let |I| = b − a denote its Lebesgue CHAPTER 2. CIRCLE ENDOMORPHISMS 26 measure. Let

`n,f = max |Iωn |. ωn∈Σn

Deﬁnition 18. Two circle endomorphisms f and g are topologically conjugate if there is an orientation-preserving circle homeomorphism h of T such that

f ◦ h = h ◦ g.

The following result was ﬁrst proved by Shub for C1 expanding circle endomorphisms in [12] using the contraction mapping theorem.

Theorem 6. Let f and g be two circle endomorphisms such that both `n,f and `n,g tend to 0 as n → ∞. Then f and g are topologically conjugate if and only if their topological degrees are the same.

Proof. Topological conjugacy preserves the topological degree. Thus if f and g are topologically conjugate then their topological degrees are the same.

Now suppose f and g have the same topological degree. Then they have the same symbolic space. Since both sets Jω,f = {xω} and Jω,g = {yω} contain only a single point for each ω, we can deﬁne

h(xω) = yω. CHAPTER 2. CIRCLE ENDOMORPHISMS 27

f T T

Πf Πf

σ h Σ Σ h

Πg Πg

g T T

One can check that h is an orientation-preserving homeomorphism with the inverse

−1 h (yω) = xω.

Therefore, for a ﬁxed degree d > 1, there is only one topological model

(Σ, σ) for the dynamics of all circle endomorphisms of degree d with `n → 0 as n → ∞.

2.3 Bounded geometry and bounded nearby geometry

Deﬁnition 19. Given a circle endomorphism f : T → T, the corresponding

∞ sequence {ξn}n=0 of nested partitions of T is said to have bounded nearby geometry if there is a constant M ≥ 1 such that for any n ≥ 0 and any two

0 intervals I,I ∈ ηn which share an endpoint modulo 1 (meaning 0 and 1 are CHAPTER 2. CIRCLE ENDOMORPHISMS 28 identiﬁed), 1 |I0| ≤ ≤ M. M |I|

∞ The sequence {ξn}n=0 of nested partitions of T is said to have bounded geometry if there is a constant C > 0 such that

|L| ≥ C, ∀L ⊂ I,I ∈ η ,L ∈ η , ∀n ≥ 0. |I| n n+1

∞ Since the sequence {ξn}n=0 of nested partitions is determined by the circle endomorphism f, we sometimes say that f has bounded nearby geometry or bounded geometry.

∞ Lemma 5. If a sequence {ξn}n=0 of nested partitions of T has bounded nearby geometry then it has bounded geometry. However, the converse is false.

Proof. Let L ⊂ I with I ∈ ηn and L ∈ ηn+1 for some n ≥ 0. The interval L is just one of d intervals L0,...,Ld−1 (labeled from left to right) from ηn+1 inside of I, each of which satisﬁes

1 |L | ≤ i+1 ≤ M, i = 0, 1, . . . , d − 2, M |Li| where M ≥ 1 is provided by the bounded nearby geometry condition. It follows that for any 0 ≤ i, j, ≤ d − 1,

1 |Li| |i−j| |i−j| ≤ ≤ M . M |Lj| CHAPTER 2. CIRCLE ENDOMORPHISMS 29

Pd−1 Since i=0 |Li| = |I|, there exists some 0 ≤ i0 ≤ d − 1 such that |Li0 | ≥

|I|/(d − 1). Hence

|L | |I| |L | ≥ i0 ≥ , ∀i = 0, 1, . . . d − 1. i M d−1 (d − 1)M d−1

Now set 1 C = > 0. (2.3) (d − 1)M d−1

Since L = Li for some 0 ≤ i ≤ d − 1, we have

|L| ≥ C. |I|

∞ That is, {ξn}n=0 has bounded geometry.

A counterexample to the converse statement is provided by the following sequence of nested partitions for a degree d = 2 circle endomorphism. Let

0 < p < 1/2 and let q = 1 − p. Let |I0| = p and |I1| = q. Then for any n ≥ 1 deﬁne

|Iωn0| = p|Iωn |, |Iωn1| = q|Iωn |.

It is clear that this sequence of nested partitions has bounded geometry, since

|L| ≥ p, ∀L ⊂ I,I ∈ η ,L ∈ η , ∀n ≥ 0. |I| n n+1

0 However, if I = I11...1 and I = I00...0 both belong to ηn, then they share a common endpoint modulo 1, and

|I0| pn = → 0 as n → ∞. |I| q CHAPTER 2. CIRCLE ENDOMORPHISMS 30

∞ Lemma 6. If a sequence {ξn}n=0 of nested partitions has bounded geometry then there exists a constant 0 < τ < 1 such that

|L| ≤ τ, ∀L ⊂ I,I ∈ η ,L ∈ η , ∀n ≥ 0 |I| n n+1

n and `n,f ≤ τ → 0 as n → ∞.

∞ Proof. A sequence {ξn}n=0 of nested partitions with bounded geometry has, by deﬁnition, a constant C > 0 such that

|L| ≥ C, ∀L ⊂ I,I ∈ η ,L ∈ η , ∀n ≥ 0. |I| n n+1

Since there are exactly d subintervals L ∈ ηn+1 contained inside I ∈ ηn, the largest possible |L| is obtained when the other (d − 1) subintervals of I are all as small as possible, size C. Deﬁne

τ = 1 − (d − 1)C < 1. (2.4)

. Since C ≤ 1/d, we have τ ≥ 1/d > 0. Thus,

|L| ≤ τ, ∀L ⊂ I,I ∈ η ,L ∈ η , ∀n ≥ 0. |I| n n+1

Furthermore, it follows that

n |Iωn | ≤ τ , CHAPTER 2. CIRCLE ENDOMORPHISMS 31 and

n `n,f ≤ τ → ∞ as n → ∞.

Theorem 7. If f is a uniformly quasisymmetric circle endomorphism, then

∞ the sequence {ξn}n=0 of nested partitions of T has bounded nearby geometry, and thus bounded geometry.

Proof. Deﬁne

−1 Gk(x) = F (x + k) : [0, 1] → [0, 1], for k = 0, 1, . . . , d − 1.

For any n ≥ 1 and for any word ωn = i0i1 . . . in−1 ∈ Σn, deﬁne

Gωn = Gi0 ◦ Gi1 ◦ ... ◦ Gin−1 .

Note that d−1 −1 [ f (π(x)) = π(Gk(x)). k=0 Then

−n Iωn = Gωn ([0, 1]) = F ([m, m + 1]),

−n −n |Iωn | = |F (m + 1) − F (m)|,

n−1 0 where m = in−1 + in−d + ... + i0d . Suppose Iωn is an interval in ηn having

a common endpoint with Iωn modulo 1. Then

0 −n −n Iωn = F ([m + 1, m + 2]) or F ([m − 1, m]), CHAPTER 2. CIRCLE ENDOMORPHISMS 32

Figure 2.3: Gk(x) for d = 2.

0 −n −n −n −n |Iωn | = |F (m + 2) − F (m + 1)| or |F (m) − F (m − 1)|.

Since f is uniformly quasisymmetric, there exists a constant C ≥ 1 such that

1 |I0 | ≤ ωn ≤ C. C |Iωn |

∞ Hence the nested sequence {ξn}n=1 has bounded nearby geometry and, by

Lemma 5, bounded geometry.

Remark 3. In fact, the converse to theorem 7 is also true. This will be proved shortly. CHAPTER 2. CIRCLE ENDOMORPHISMS 33

Example 1. The canonical example of a degree d ≥ 2 circle endomorphism is

d qd(z) = z , with lift

Qd(x) = dx

For this example we have

k k + 1d−1 η0 = , d d k=0 and n k k + 1d −1 ηn−1 = n , n . d d k=0

Given ωn = i0i1 . . . in−1 ∈ Σn for any n ≥ 1, let k(ωn) be the integer whose d-nary representation is ωn. That is,

n−1 X n−1−j k(ωn) = ijd , 0 ≤ ij ≤ d − 1. j=0

Then k(ω ) k(ω ) + 1 I = n , n . ωn dn−1 dn−1

Furthermore, for any ω = i0i1 ... ∈ Σ,

∞ X ij x = . ω dj+1 j=0 CHAPTER 2. CIRCLE ENDOMORPHISMS 34

Theorem 8. Any two circle endomorphisms f and g of the same degree d ≥ 2 that both have bounded nearby geometry are topologically conjugate and the conjugacy is a quasisymmetric homeomorphism.

Proof. Suppose

ηn,f = {Iωn,f | ωn ∈ Σn}, ηn,g = {Iωn,g | ωn ∈ Σn}, n = 1, 2,... are two sequences of Markov partitions for f and g, respectively. By theorem

7 and lemma 6, we have a constant 0 < τ < 1 such that for all n = 1, 2,...,

n `n,f = max {Iωn,f } ≤ τ , ωn∈Σn

n `n,g = max {Iωn,g} ≤ τ . ωn∈Σn

Thus `n,f and `n,g both tend to 0 as n → ∞, and theorem 6 implies that f and g are topologically conjugate. Let h be the topological conjugacy,

f ◦ h = h ◦ g.

For any n ≥ 0, ωn ∈ Σn, and j ∈ Z let

Iωn,f + j = {x ∈ R | x − j ∈ Iωn,f , }

Iωn,g + j = {x ∈ R | x − j ∈ Iωn,g.} CHAPTER 2. CIRCLE ENDOMORPHISMS 35

For each n ≥ 0, we now extend the partitions ηn,f and ηn,g of [0, 1] to all of

R by

{Iωn,f + j | j ∈ Z},

{Iωn,g + j | j ∈ Z},

which we still denote as ηn,f and ηn,g, respectively. Since the original partitions had bounded nearby geometry modulo 1, the extended partitions now have bounded geometry everywhere.

Figure 2.4: One possible way that x − t, x, and x + t may be contained in partitions ηN−1,f , ηN,f , and ηN+N1,f . The corresponding ﬁgure for how H(x − t),H(x), and H(x + t) would be contained in partitions ηN−1,g, ηN,g, and ηN+N1,g is homeomorphic.

For all x ∈ R and for all t > 0, let N > 0 be the smallest integer such that ˜ there exists an interval I ∈ ηN,f satisfying I ⊆ [x − t, x + t]. Let I ∈ ηN−1,f CHAPTER 2. CIRCLE ENDOMORPHISMS 36

˜ ˜ ˜ ˜ be the interval such that I ⊂ I. Then [x − t, x + t] ⊆ I ∪ J, where J ∈ ηN−1,f shares an endpoint with I˜.

∞ Let H : R → R be the lift of h. The bounded geometry of {ηn,g}n=0 provides a constant C > 0 such that

|H(I)| ≥ C. |H(I˜)|

If [x − t, x] ⊆ I˜ (in which case, [x, x + t] * I˜), then |H([x − t, x])| ≤ |H(I˜)|, and |H(I)| |H(I)| ≥ ≥ C. |H([x − t, x])| |H(I˜)|

Otherwise [x − t, x] * I˜ (in which case [x, x + t] ⊆ I˜). The bounded nearby

∞ geometry of {ηn,g}n=0 provides a constant M ≥ 1 such that

1 |H(J˜)| ≤ ≤ M. M |H(I˜)|

Thus 1 |H(I˜)| ≤ |H(J˜)| ≤ M|H(I˜)|, M and M + 1 |H(I˜)| ≤ |H(J˜ ∪ I˜)| ≤ (1 + M)|H(I˜)|. M

Since |H([x − t, x])| ≤ |H(J˜ ∪ I˜)|, we have

|H(I)| |H(I)| |H(I)| C ≥ ≥ ≥ . |H([x − t, x])| |H(J˜ ∪ I˜)| (1 + M)|H(I˜)| 1 + M CHAPTER 2. CIRCLE ENDOMORPHISMS 37

Set C1 = C/(1 + M). In summary, we have

|H(I)| |H(I)| ≥ C , and similarly ≥ C . |H([x − t, x])| 1 |H([x, x + t])| 1

We claim that there exists a constant N1 > 0 such that there exist inter-

vals J1,J2 ∈ ηN+N1 contained inside [x − t, x] and [x, x + t], respectively. To

− + justify our claim, let I ,I ∈ ηN be the two intervals which each share one

∞ endpoint with I. Since |I| ≤ 2t, the bounded nearby geometry of {ηn,f }n=0 provides a constant M ≥ 1 such that

|I−|, |I+| ≤ M|I| ≤ 2Mt.

N1−N − + We require N1 to be suﬃciently large so that the 3d subsets of I ∪I ∪I

belonging to ηN+N1,f all have length less than or equal to t/4. Hence, it is suﬃcient for N1 to satisfy

− ln(8M) τ N1 (2Mt) ≤ t/4 ⇔ N ≥ , 1 ln τ

∞ where 0 < τ < 1 is a constant provided the bounded geometry of {ηn,f }n=0 and lemma 6.

We now have

|H(I)| |H(I)| 1 C1 ≤ ≤ ≤ N , |H([x − t, x])| |H(J1)| C 1 and |H(I)| |H(I)| 1 C1 ≤ ≤ ≤ N . |H([x, x + t])| |H(J2)| C 1 CHAPTER 2. CIRCLE ENDOMORPHISMS 38

Therefore |H(I)| N1 C1 |H([x−t,x])| 1/C N ≤ |H(I)| ≤ . 1/C 1 C1 |H([x,x+t])|

N1 Setting M1 = 1/(C1C ) > 1, we have

1 H(x + t) − H(x) ≤ ≤ M1. M1 H(x) − H(x − t)

That is, the topological conjugacy h is a quasisymmetric homeomorphism.

Remark 4. In theorem 8, the same argument shows that the topological conjugacy h−1 is also quasisymmetric.

We can now prove the following.

Theorem 9. A degree d ≥ 2 circle endomorphism f is uniformly quasisym-

∞ metric if and only if its sequence {ξn}n=0 of nested Markov partitions of T has bounded nearby geometry.

∞ Proof. If f is uniformly quasisymmetric, then the sequence {ξn}n=0 of nested

Markov partitions of T has bounded nearby geometry by theorem 7.

To prove the converse, consider the linear degree d circle endomorphism

d qd(z) = z : T → T with lift Qd(x) = dx : R → R. It is uniformly quasisymmetric, and thus yields a sequence of nested partitions with bounded CHAPTER 2. CIRCLE ENDOMORPHISMS 39 geometry. Hence, by theorem 8, there exists a quasisymmetric homeomorphism h : T → T with lift H : R → R such that f = h ◦ q ◦ h−1. We claim that the property of being uniformly quasiymmetric is preserved under conjugation by a quasisymmetric homeomorphism. It follows from the claim that f is uniformly quasisymmetric.

We now justify the claim. The justiﬁcation is very similar to the proof of theorem 8. We must ﬁnd a constant M0 ≥ 1 such that for any x ∈ R,t > 0, and n ≥ 0 we have

1 |H ◦ Q−n ◦ H−1(x + t) − H ◦ Q−n ◦ H−1(x)| ≤ −n −1 −n −1 ≤ M0. M0 |H ◦ Q ◦ H (x) − H ◦ Q ◦ H (x − t)|

Let

−n −1 −n −1 −n −1 x1 = Q (H (x − t)), x2 = Q (H (x)), x3 = Q (H (x + t)).

Since H−1 is quasisymmetric and Q−n(x) = x/dn is linear, there exists a constant M ≥ 1 such that

1 |x − x | ≤ 3 2 ≤ M. M |x2 − x1|

∞ ∞ Let {ηn,f }n=0 and {ηn,qd }n=0 be extended partitions of R with respect to f and qd, respectively, as constructed in the proof of theorem 8. For n ≥ 0, deﬁne

ηn,h = H(ηn,qd ). CHAPTER 2. CIRCLE ENDOMORPHISMS 40

That is, ηn,h is the partition on R into intervals which are the images under

H of the intervals of R belonging to ηn,qd . Since H is quasisymmetric and

n the intervals of ηn,qd all have the same length 1/d , the sequence of partitions

∞ {ηn,qd }n=0 have bounded nearby geometry.

Let N > 0 be the smallest integer such that there exists an interval

˜ I ∈ ηN,qd satisfying I ⊆ [x1, x3]. Let I ∈ ηN−1,f be the interval such that ˜ ˜ ˜ ˜ ˜ I ⊂ I. Then [x1, x3] ⊆ I ∪ J, where J ∈ ηN−1,f shares an endpoint with I.

∞ The bounded geometry of {ηn,h}n=0 provides a constant C > 0 such that

|H(I)| ≥ C. |H(I˜)|

˜ ˜ If [x1, x2] ⊆ I, then |H([x1, x2])| ≤ |H(I)|, and

|H(I)| |H(I)| ≥ ≥ C. |H([x1, x2])| |H(I˜)|

Otherwise [x − t, x] * I˜ (in which case [x, x + t] ⊆ I˜). The bounded nearby

∞ geometry of {ηn,h}n=0 provides a constant M1 ≥ 1 such that

1 |H(J˜)| ≤ ≤ M1. M1 |H(I˜)|

Thus

1 ˜ ˜ ˜ |H(I)| ≤ |H(J)| ≤ M1|H(I)|, M1 and M + 1 |H(I˜)| ≤ |H(J˜ ∪ I˜)| ≤ (1 + M)|H(I˜)|. M CHAPTER 2. CIRCLE ENDOMORPHISMS 41

˜ ˜ Since |H([x1, x2])| ≤ |H(J ∪ I)|, we have

|H(I)| |H(I)| |H(I)| C ≥ ≥ ≥ . |H([x1, x2])| |H(J˜ ∪ I˜)| (1 + M)|H(I˜)| 1 + M

Set C1 = C/(1 + M). In summary, we have

|H(I)| |H(I)| ≥ C1, and similarly ≥ C1. |H([x1, x2])| |H([x2, x3])|

−1 Without loss of generality, assume |x2 − x1| ≤ |x3 − x2|. Since H is uniformly quasisymmetric, there exists a constant M2 ≥ 1 such that

1 1 |x2 − x1| ≥ |x3 − x2| ≥ N+1 . M2 M2d

Thus, setting N1 ≥ 1 such that

1 1 N+N ≤ N+2 , d 1 M2d that is, log M N ≥ 2 + , 1 log d

there exist intervals J1,J2 ∈ ηN+N1,qd contained inside [x1, x2] and [x2, x3], respectively.

We now have

|H(I)| |H(I)| 1 C1 ≤ ≤ ≤ N , |H([x1, x2])| |H(J1)| C 1 and |H(I)| |H(I)| 1 C1 ≤ ≤ ≤ N , |H([x2, x3])| |H(J2)| C 1 CHAPTER 2. CIRCLE ENDOMORPHISMS 42

∞ Where the constant C > 0 is provided by the bounded geometry of {ηn,h}n=0.

Therefore |H(I)| N1 C1 |H([x1,x2])| 1/C N ≤ |H(I)| ≤ . 1/C 1 C1 |H([x2,x3])|

N1 Setting M0 = 1/(C1C ) > 1, we have

1 H(x3) − H(x2) ≤ ≤ M0, M0 H(x2) − H(x1)

1 |H ◦ Q−n ◦ H−1(x + t) − H ◦ Q−n ◦ H−1(x)| ≤ −n −1 −n −1 ≤ M0. M0 |H ◦ Q ◦ H (x) − H ◦ Q ◦ H (x − t)|

−1 That is, the degree d circle endomorphism f = h ◦ qd ◦ h is uniformly quasisymmetric.

We have established that if a circle endomorphsim f is uniformly quasisymmetric, then it has bounded geometry. However, the converse is false.

For example, for any α ∈ (0, 1), α 6= 1/2, the piecewise linear degree 2 circle endomorphism fα (see ﬁgure 2.5) with lift   Fα(x + 1) − 2 if x < 0  x/α if 0 ≤ x < α F (x) = α 1 + (x − α)/(1 − α) if α ≤ x < 1   Fα(x − 1) + 2 if 1 ≤ x has bounded geometry but is not uniformly quasisymmetric.

We can construct the conugacy hα such that

fα ◦ hα = hα ◦ q2. CHAPTER 2. CIRCLE ENDOMORPHISMS 43

A description and geometric construction of its lift Hα : R → R is given in [6], and its graph is shown in figure 2.6. It is strictly increasing, totally singular, and satisfies the Höldercondition

2 |H (x) − H (y)| ≤ |x − y|log2 a, ∀x, y ∈ [0, 1], α α a where a = max{α, 1 − α}. However, hα is not a quasisymmetric homeomorphism.

If we deﬁne F˜ and H˜ , such that  ˜  F (x + 1) − 2 if x < 0   3x/2 if 0 ≤ x < 1/3 F˜(x) = 3x − 1/2 if 1/3 ≤ x < 2/3  3x/2 + 1/2 if 2/3 ≤ x < 1   F˜(x − 1) + 2 if 1 ≤ x ˜ ˜ ˜ ˜ −1 ˜ and H is the circle homeomorphism such that F = H ◦ Q2 ◦ H , then F is uniformly quasisymmetric (thus it has bounded geometry and bounded nearby geometry) and H˜ is quasisymmetric. It is interesting to compare the

˜ ˜ graphs of F and H (ﬁgures 2.7 and 2.8) with those of Fα and Hα (ﬁgures 2.5 and 2.6).

Deﬁnition 20. Let H denote the space of all quasisymmetric homeomorphisms h : T → T with h(0) = 0, and let Fd denote the space of all degree d ≥ 2 uniformly quasisymmetric endomorphisms f : T → T with f(0) = 0.

Deﬁne qd : T → T as

qd(x) = dx mod 1. CHAPTER 2. CIRCLE ENDOMORPHISMS 44

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 2.5: The circle endomorphism Fα shown acting on R/Z. Here α = .8. The map Fα has bounded geometry but it is not uniformly quasisymmetric. CHAPTER 2. CIRCLE ENDOMORPHISMS 45

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 2.6: The circle homeomorphism Hα shown acting on R/Z. It is not −1 quasisymmetric. Fα = Hα ◦ Q2 ◦ Hα . CHAPTER 2. CIRCLE ENDOMORPHISMS 46

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 2.7: The circle endomorphism F˜ shown acting on R/Z. The map F˜ is uniformly quasisymmetric. Thus is has bounded geometry and bounded nearby geometry. CHAPTER 2. CIRCLE ENDOMORPHISMS 47

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 2.8: The circle homeomorphism H˜ shown acting on R/Z. It is qua- ˜ ˜ ˜ −1 sisymmetric. F = H ◦ Q2 ◦ H . CHAPTER 2. CIRCLE ENDOMORPHISMS 48

Deﬁne αd : H → Fd as

−1 αd(h) = h ◦ qd ◦ h .

Theorem 10. The map α is bijective.

Proof. Let d ≥ 2. It follows from theorem 9 and theorem 8 that αd : H → Fd is surjective. Now suppose we have two order preserving quasisymmetric homeomorphisms h1, h2 ∈ H such that αd(h1) = αd(h2). Thus

n −1 n −1 h1 ◦ qd ◦ h1 (x) = h2 ◦ qd ◦ h2 (x), for all x ∈ T and for all n ≥ 1. In particular, we have

n −1 −n n −1 −n h1 ◦ qd ◦ h1 (h1(qd (1))) = h2 ◦ qd ◦ h2 (h1(qd (1)))

n −1 −n ⇒ 1 = qd ◦ h2 (h1(qd (1))).

−1 Therefore, for each n ≥ 1, the homeomorphism h2 ◦ h1 : T → T maps the set k {x | qn(x) = 1} = { | 0 ≤ k ≤ dn − 1} dn

−1 to itself. Furthermore, since h2 ◦ h1 is order preserving, we have

k k h−1 ◦ h ( ) = , ∀0 ≤ k ≤ dn. 2 1 dn dn

That is, k k h ( ) = h ( ), ∀0 ≤ k ≤ dn, ∀n ≥ 1. 1 dn 2 dn CHAPTER 2. CIRCLE ENDOMORPHISMS 49

Since this equality holds on a dense subset of T, it follows that h1 = h2 and thus αd is injective.

2.4 Measures with bounded geometry and bounded nearby geometry

Up to this point, we have treated T and R as metric spaces and used Eu- clidean distance to define lengths of intervals and thus to define bounded nearby geometry and bounded geometry. The open balls defined by this metric form a basis for our topology, and hence they generate the Borel σ- algebra B. Now equipped with a σ-algebra, {T, B(T)} and {R, B(R)} are measurable spaces.

For any probability measure µ on {T, B}, it has a unique lift to {R, B(R)}, which we also denote by µ. We can now generalize the deﬁnitions of bounded nearby geometry and bounded geometry as follows.

Deﬁnition 21. Let {T, B(T), µ} be a probability space. Given a circle en-

∞ domorphism f : T → T with the corresponding sequence {ξn}n=0 of nested partitions of T, the probability measure µ is said to have bounded nearby geometry if there is a constant M ≥ 1 such that for any n ≥ 0 and any two

0 intervals I,I ∈ ηn which share an endpoint modulo 1 (meaning 0 and 1 are CHAPTER 2. CIRCLE ENDOMORPHISMS 50 identiﬁed), 1 µ(I0) ≤ ≤ M. M µ(I)

The measure µ is said to have bounded geometry if there is a constant

C > 0 such that

µ(L) ≥ C, ∀L ⊂ I,I ∈ η ,L ∈ η , ∀n ≥ 0. µ(I) n n+1

Remark 5. When µ = λ, where λ denotes Lebesgue meausre on R and normalized Lebesgue measure on T, the previous deﬁnitions and the new def- initions of bounded nearby geometry and bounded geometry are the same.

Lemma 7. If f : T → T is a degree d ≥ 2 circle endomorphism and

{T, B(T)} is a probability space such that µ has bounded nearby geometry, then it has bounded geometry. However, the converse is false.

Proof. The proof is nearly identical to the proof of lemma 5.

Lemma 8. If f : T → T is a degree d ≥ 2 circle endomorphism and

{T, B(T)} is a probability space such that µ has bounded geometry, then there exists a constant 0 < τ < 1 such that

µ(L) ≤ τ, ∀L ⊂ I,I ∈ η ,L ∈ η , ∀n ≥ 0 µ(I) n n+1 and

max µ(I) ≤ τ n → 0 as n → ∞. I∈ηn,f CHAPTER 2. CIRCLE ENDOMORPHISMS 51

Proof. The proof is nearly identical to the proof of lemma 6.

Lemma 9. If f : T → T is a degree d ≥ 2 circle endomorphism and

{T, B(T)} is a probability space such that µ has bounded geometry, then µ is non-atomic and has full support.

∞ Proof. For any x ∈ T, there exists a sequence of intervals {Jn(x)}n=0 such that Jn(x) ∈ ξn,f , Jn+1(x) ⊂ Jn(x), and x ∈ Jn(x) for all n ≥ 0. By lemma

8, there exists a constant 0 < τ < 1 such that

n µ(Jn(x)) ≤ τ → 0 as n → ∞, and hence µ({x}) = 0. Furthermore, given any open neighborhood U(x) of x, there exists N such that JN (x) ⊆ U(x) and, by the deﬁnition of bounded

N+1 geometry, there exists a constant 0 < C < 1 such that µ(JN (x)) > C .

Thus µ(N(x)) > 0. That is, µ has full support. Chapter 3

Invariant measures

3.1 Borel measures

Throughout this chapter, let X denote a compact metrizable space and let d denote a metric on X. The open balls

Br(x) = {y ∈ X | d(x, y) < r} form a basis for a topology on X, and this is the topology we will always use. Let B(X) denote the smallest σ-algebra containing the open sets of X, i.e. the Borel σ-algebra on X.

Deﬁnition 22. Given a measurable space {X, B(X)}, let M(X) denote the set of all probability measures µ : B → [0, 1]. We call M(X) the set of

Borel measures on (X, B(X)). We turn M(X) into a topological space by endowing it with the weak∗ topology, that is, the smallest topology such

52 CHAPTER 3. INVARIANT MEASURES 53 that the map Z µ → f dµ X is continuous for all continuous functions f ∈ C(X).

Figure 3.1: For all f ∈ C(X) and for all open intervals I ⊆ [0, 1], the set of R ∗ all measures µ ∈ M(X) such that X f dµ ∈ I belongs to the weak topology on M(X). The weak∗ topology on M(X) is the smallest topology satisfying this condition.

∗ Remark 6. In the weak topology, µn → µ in M(X) if and only if for all R R f ∈ C(X), f dµn → f dµ.

Deﬁnition 23. Given a measure space {X, B(X), µ}, the measure µ ∈ M(X) is regular if for all A ∈ B(X) and for all > 0, there exists an open set U and a closed set C satisfying C ⊆ A ⊆ U such that µ(U \ C) < .

Theorem 11. A Borel probability measure µ on a metric space X is regular. CHAPTER 3. INVARIANT MEASURES 54

Proof. Let A be the collection of all subsets A ∈ B(X) such that the regularity conditions hold. We show that A is a σ-algebra that contains the open sets.

First note that X ∈ A since X itself is both open and closed. Now suppose A ∈ A. We show that Ac ∈ A, where Ac denotes the complement of

A, X\A. For any > 0 we have open U and closed C such that C ⊆ A ⊆ U

c c c c c and µ(C \ U) < . Then Ce is open and U is closed with U ⊆ A ⊆ C

c c c and µ(C \ U ) = µ(U \ C) < . Thus, A ∈ A and therefore A is closed under complements.

Now suppose we have a countable collection of sets A1,A2,... ∈ A. We

S∞ must show that A = n=1 An ∈ A. For any > 0 and any n ≥ 1 there exists an open set U,n and a closed set C,n such that C,n ⊆ An ⊆ U,n and

n µ(U,n \ C,n) < /3 . Deﬁne

∞ ∞ [ ˜ [ U = U,n, C = C,n. n=1 n=1

Note that U is open. Since µ is a probability measure, there exists an integer k ≥ 1 such that k ˜ [ µ(C \ C,n) < /2. n=1 Sk Deﬁne C = n=1 C,n. Note that C is closed. We now have C ⊆ A ⊆ U CHAPTER 3. INVARIANT MEASURES 55 and

˜ ˜ µ(Ue \ C) ≤ µ(U \ C) + µ(C \ C) ∞ X ˜ ≤ µ(U,n \ C,n) + µ(C \ C) n=1 ∞ X ≤ + = . 3n 2 n=1

Thus, A is closed under countable unions. Hence, A is a σ-algebra.

All that remains is to show that A contains the open sets. Given any open set A, let C = Ac be its complement, which is closed. For each n ≥ 1, deﬁne the open set

Un = {x ∈ X | d(C, x) < 1/n}.

Note that these sets are nested,

U1 ⊇ U2 ⊇ ... ⊇ Un ⊇ ..., and ∞ \ C = Un. n=1

Choose k ≥ 1 such that µ(Uk \ C) < . Note that Uk is open. Thus,

C ⊆ C ⊆ Uk with µ(Uk \ C) < and C ∈ A. Since A is a σ-algebra,

A = Cc ∈ A. Therefore A contains the open sets, and so every set of B(X) satisﬁes the regularity condition. That is, µ is regular. CHAPTER 3. INVARIANT MEASURES 56

Corollary 1. Let µ be a Borel probability measure on a metric space X. For any A ∈ B(X), let CA denote the collection of all closed sets contained inside

A and let UA denote the collection of all open sets that contain A. Then

µ(A) = sup µ(C) = inf µ(U). U∈U C∈CA A

Theorem 12. If X is a compact metrizable space then the space M(X) is

∞ metrizable. Moreover, if {fn}n=1 is a dense subset of C(X) (endowed with the supremum norm || · ||) then

∞ R R X | fn dµ1 − fn dµ2| d(µ , µ ) = 1 2 2n||f || n=1 n is a metric on M(X) that induces the weak∗ topology.

∞ Proof. The existence of a dense subset {fn}n=1 of C(X), where X is compact and metrizable, follows from the Stone-Weirstrass theorem [11]. It is straightforward to check that d deﬁnes a metric, and so (M(X), d) is a metric space. For each ﬁxed n0, we have Z Z n0 fn0 µ1 − fn0 dµ2 ≤ 2 ||fn0 ||d(µ1, µ2),

R and so the map µ → fn0 dµ is continuous on (M(X), d). Since the set

∞ {fn}n=1 is dense in C(X), given any f ∈ C(X) and any > 0, there exists

n0 such that ||f − fn0 || < /3. Thus, if

d(µ1, µ2) < , n0 3 · 2 ||fn0 || CHAPTER 3. INVARIANT MEASURES 57 then R R R R R R f µ1 − f dµ2 ≤ f µ1 − fn0 dµ1 + fn0 µ1 − fn0 dµ2 R R + fn0 µ2 − f dµ2 n0 < 3 + 2 ||fn0 ||d(µ1, µ2) + 3 < .

Thus for all f ∈ C(X) the map µ → R f dµ is continuous on (M(X), d).

Therefore, every open set in the weak∗ topology is open in the metric space

(M(X), d). To prove the converse, we show that each open ball

Bµ() = {m ∈ M(X) | d(m, µ) < } in (M(X), d) contains a set of the form

N Z Z \ Uµ(f1, f2, . . . , fN ; δ) = m ∈ M(X) fn dm − fn dµ < δ . n=1

If µ ∈ M(X) and > 0 are given, choose N such that

∞ X 2 < 2n 2 n=N+1 and let N ! X 1 δ = . 2 2n||f || n=1 n CHAPTER 3. INVARIANT MEASURES 58

Now if µ ∈ Uµ(f1, f2, . . . , fn; δ) then

∞ R R X | fn dm − fn dµ| d(m, µ) = 2n||f || n=1 n N R R ∞ R R X | fn dm − fn dµ| X | fn dm − fn dµ| = n + n 2 ||fn|| 2 ||fn|| n=1 n=N+1 N ∞ N ∞ X δ X 2||fn|| X 1 X 2 ≤ n + n = δ n + n 2 ||fn|| 2 ||fn|| 2 ||fn|| 2 n=1 n=N+1 n=1 n=N+1 < + = . 2 2

Thus Uµ(f1, . . . , fn; δ) ⊂ Bµ(), and every open set in the topology induced by the metric d(·, ·) is open in the weak∗ topology.

Theorem 13. If X is a compact metrizable space then M(X) is compact in the weak∗ topology.

Proof. Since M(X) is metrizable, it is enough to show that it is sequentially

∞ compact. Suppose {µn}n=1 is a sequence of measures in M(X). We show that it has a convergent subsequence.

∞ Let {fn}n=1 be a dense subset of C(X). First, consider the sequence of real numbers Z f1 dµn.

This sequence is bounded by ±||f1|| (supremum norm). Thus it has a convergent subsequence and a corresponding subsequence of measures which we CHAPTER 3. INVARIANT MEASURES 59

1 ∞ denote as {µn}n=1. Next, consider the sequence of real numbers Z 1 f2 dµn.

This sequence is bounded by ±||f2||, and so it has a convergent subsequence

2 ∞ and a corresponding subsequence of measures {µn}n=1 (a subsequence of

1 ∞ {µn}n=1). Contunuing in this way, for each j ≥ 1 we have a subsequence

j ∞ {µn}n=1 of measures such that Z j fk dµn converges for all 1 ≤ k ≤ j. Finally, consider the “diagonal” sequence

n ∞ {µn}n=1. For each j ≥ 1, the limit Z n lim fj dµn = Lj (3.1) n→∞ exists. We must show that the limit (3.1) exists for arbitrary f ∈ C(X).

∞ Let {fjl }l=1 be a sequence of the functions f1, f2,... that converges to f in

C(X). Note that Z n |Ljl − Ljl | ≤ lim |fjl − fjl | dµn ≤ ||fjl − fjl ||, 1 2 n→∞ 1 2 1 2

and since fjl → f in C(X), this shows that the sequence Ljl is a Cauchy

sequence in R and so Ljl → L for some L ∈ R. Let > 0 be ﬁxed. Choose l0 suﬃciently large such that

||f − f|| < /3 and |L − L| < /3. jl0 jl0 CHAPTER 3. INVARIANT MEASURES 60

Choose N suﬃciently large such that n ≥ N implies

Z n fj dµ − Lj < /3. l0 n l0

Then for n ≥ N we have

Z Z Z Z n n n n f dµ − L ≤ f dµ − fj dµ + fj dµ − Lj + |Lj − L| n n l0 n l0 n l0 l0 < + + = 3 3 3

Thus, the limit (3.1) exists for arbitrary f ∈ C(X). Ths deﬁnes a map

J : C(X) → R, Z n J(f) = lim f dµn. n→∞

The map J is linear, bounded (J(f) ≤ ||f||), J(1) = 1, and if f ≥ 0, then

J(f) ≥ 0. Thus, it follows from the Riesz representation theorem that there exists a Borel probability measure µ ∈ M(X) such that

Z Z n J(f) = lim f dµn = f dµ, n→∞ for all f ∈ C(X). This shows that M(X) is compact.

3.2 Invariant measures and ergodicity

Deﬁnition 24. Suppose (X, B(X), µ) is a measure space and T : X → X is a continuous function such that for all A ∈ B

µ(T −1(A)) = µ(A), CHAPTER 3. INVARIANT MEASURES 61 where T −1(A) = {x ∈ X | T (x) ∈ A}. In this case we say that T preserves the measure µ, and that µ is an invariant measure with respect to T .

The set of all invariant probability measures with respect to T is denoted

M(X,T ).

Alternatively, any continuous function T : X → X can be used to deﬁne a map T∗ : M(X) → M(X) such that

−1 −1 T∗(µ)(A) = µ ◦ T (A) = µ(T (A)), ∀A ∈ B(X).

The measure T∗(µ) is called the pushforward measure. An invariant measure with respect to T is a ﬁxed point of the map T∗.

Lemma 10. Let T : X → X be a continuous transformation of a compact metric space X. For all f ∈ C(X) we have

Z Z f d(T∗µ) = f ◦ T dµ. (3.2)

Proof. It follows from the deﬁnition of T∗µ that (3.2) holds for characteristic functions f = χA. Then (3.2) holds also for simple functions. For any f ∈ C(X) we let

f + = max{0, f},

f − = max{0, −f}. CHAPTER 3. INVARIANT MEASURES 62

By considering increasing sequences of simple functions converging pointwise to f + and f −, respectively, we see that (3.2) holds for f + and f −, and thus

(3.2) holds for f = f + − f −.

Theorem 14. If T : X → X is a continuous transformation of a compact metric space X, then the set M(X,T ) of invariant measures with respect to

T is non-empty.

∞ Proof. Begin with an arbitrary sequence {σn}n=1 of measures in M(X). (In particular, for any x ∈ X, we may set σn = δx for all n, where δx(A) = 1 if x ∈ A and δx(A) = 0 is x∈ / A.) Using this sequence, we form a new sequence

∞ {µn}n=1 of measures in M(X) such that

n−1 1 X µ = (T k) σ . n n ∗ n k=0

Since M(X) is compact, there exists a subsequence µnj → µ ∈ M(X). CHAPTER 3. INVARIANT MEASURES 63

We claim that µ ∈ M(X,T ). For any f ∈ C(X) we have

Z Z Z Z

f d(T∗µ) − f dµ = f ◦ T dµ − f dµ

Z Z

= lim f ◦ T dµnj − f dµnj j→∞

nj −1 1 Z Z X k k = lim f ◦ T d(T )∗σnj − f d(T )∗σnj j→∞ nj k=0

nj −1 1 Z X k+1 k = lim (f ◦ T − f ◦ T ) dσnj j→∞ nj k=0 1 Z nj = lim (f ◦ T − f) dσnj j→∞ nj 2||f|| ≤ lim = 0. j→∞ nj

Since this holds for all f ∈ C(X), we have T∗µ = µ. That is, µ is invariant with respect to T , and so µ ∈ M(X,T ).

Theorem 15. Given a compact measurable space (X, B(X)) and a continuous function T : X → X, the set M(X,T ) ⊆ M(X) of invariant measures µ is a convex and compact subset of M(X), with respect to the weak∗ topology.

∞ Proof. Suppose {µn}n=1 is a sequence of measures in M(X,T ) and µn → µ in M(X). Then for any f ∈ C(X) we have

Z Z Z Z f dT∗µ = f ◦ T dµ = lim f ◦ T dµn = lim f dµn n→∞ n→∞ Z = f dµ. CHAPTER 3. INVARIANT MEASURES 64

Thus µ ∈ M(X,T ). This shows that M(X,T ) is compact.

Now suppose µ, ν ∈ M(X,T ) and p, q ∈ [0, 1] with p + q = 1. Then for any A ∈ B(X) we have

(pµ + qν)(T −1A) = pµ(T −1A) + qν(T −1A) = pµ(A) + qν(A) = (pµ + qν)(A).

Thus M(X,T ) is convex.

Deﬁnition 25. Let T : X → X be a continuous transformation of a compact metric space, and let µ ∈ M(X,T ) be an invariant probability measure with respect to T . We say µ is ergodic if for any A ∈ B(X), we have

T −1A = A ⇒ either µ(A) = 0 or µ(A) = 1.

Lemma 11. The measure µ ∈ M(X,T ) is ergodic if and only if for any

A ∈ B(X) we have

µ(T −1A 4 A) = 0 ⇒ either µ(A) = 0 or µ(A) = 1.

Proof. Let A ∈ B(X) and µ(T −1A 4 A) = 0. For each integer n ≥ 0 we have

n−1 n−1 [ [ T −nA 4 A ⊆ T −i−1A 4 T −iA = T −i(T −1A 4 A), i=0 i=0 and since T ∈ M(X,T ),

µ(T −nA 4 A) ≤ nµ(T −1A 4 A) = 0. CHAPTER 3. INVARIANT MEASURES 65

Now let

∞ ∞ ∞ ∞ [ −i \ \ [ −i An = T A and A∞ = An = T A. i=n n=0 n=0 i=n We have ∞ ∞ [ −i [ −i A 4 An = A 4 T A ⊂ (A 4 T A), i=n i=n and so ∞ X −i µ(A 4 An) ≤ (A 4 T A) = 0 ∀n ≥ 0. (3.3) i=n

In particular, this shows that µ(An) = µ(A). Since An ⊇ An+1 for all n ≥ 0, it folows that µ(A∞) = µ(A). Also, we have

∞ ∞ ∞ −1 \ [ −i−1 \ T A∞ = T A = An+1 = A∞. n=0 i=n n=0

Since µ ∈ M(X,T ) is ergodic, we must have µ(A∞) = 0 or 1. It follows from

(3.3) that µ(A) = 0 or 1.

The converse is immediate since T −1A = A implies µ(A4T −1A) = 0.

Deﬁnition 26. We say µ ∈ M(X,T ) is an extreme point of the convex set M(X,T ) if whenever there exists a number p ∈ (0, 1) and measures

µ1, µ2 ∈ M(X,T ) such that µ = pµ1 +(1−p)µ2, it must be that µ1 = µ2 = µ.

Theorem 16. If T : X → X is a continuous transformation of a compact metric space X, then µ is an extreme point of the convex set M(X,T ) if and only if µ is ergodic. CHAPTER 3. INVARIANT MEASURES 66

Proof. Assume µ ∈ M(X,T ) is ergodic, and let µ1, µ2 ∈ M(X,T ) and p ∈

(0, 1) such that

µ = pµ1 + (1 − p)µ2.

Fix i ∈ {1, 2}. Note that if A ∈ B(X) and µ(A) = 0 then µi(A) = 0.

That is, µi is absolutely continuous with respect to µ, µi µ. Then the

Radon-Nikodym derivative dµi/dµ exists (see [5]), and for all A ∈ B(X) we have Z dµi µi(A) = dµ. A dµ

Deﬁne dµi Ei = x ∈ X (x) < 1 . dµ

Then Z Z dµi dµi µi(Ei) = dµ + dµ, −1 −1 Ei∩T Ei dµ Ei\T Ei dµ and Z Z −1 dµi dµi µi(T Ei) = dµ + dµ. −1 −1 T Ei∩Ei dµ T Ei\Ei dµ

−1 Since µi ∈ M(X,T ), we have µi(Ei) = µi(T Ei), and so

Z dµ Z dµ i dµ = i dµ. (3.4) −1 −1 Ei\T Ei dµ T Ei\Ei dµ CHAPTER 3. INVARIANT MEASURES 67

Note that

−1 −1 −1 µ(T Ei \ Ei) = µ(T Ei) − µ(T Ei ∩ Ei) (3.5)

−1 = µ(Ei) − µ(T Ei ∩ Ei) (3.6)

−1 = µ(Ei \ T Ei), (3.7)

−1 −1 Since dµi/dµ < 1 on Ei \ T Ei, and dµi ≥ 1 on T Ei \ Ei, equations (3.4) and (3.7) now imply that

−1 −1 µ(T Ei \ Ei) = µ(Ei \ T Ei) = 0, that is,

−1 µ(T Ei 4 Ei) = 0.

Since µ is ergodic, we have µ(Ei) = 0 or 1. If µ(Ei) = 1 (full measure), then

Z dµi µi(X) = dµ < µ(Ei) = 1. Ei dµ

This is a contradiction since µi is a probability measure. And so we must have µi(Ei) = 0.

Similarly, we may deﬁne

dµi Fi = x ∈ X (x) > 1 , dµ and show that µ(Fi) = 0. Hence dµi/dµ = 1 µ-a.e. CHAPTER 3. INVARIANT MEASURES 68

Since the above argument holds for both i = 1 and i = 2, this shows that

µ = µ1 = µ2. Thus, µ is an extreme point of M(X,T ).

Conversely, suppose that µ ∈ M(X,T ) and µ is not ergodic. Then there exists a set E ∈ B(X) such that T −1E = E and 0 < µ(E) < 1. Deﬁne measure µ1 and µ2 by

µ(B ∩ E) A ∩ (X \ E)) µ (A) = , µ (A) = , ∀A ∈ B(X). 1 µ(E) 2 µ(X \ E)

It is easy to check that µ1 and µ2 are in M(X,T ) and that m1 6= µ2. More- over, for any A ∈ B(X) we have

µ(A) = µ(E)µ1(A) + (1 − µ(E))µ2(A).

Thus µ is not an extreme point of M(X,T ).

Corollary 2. Let T : X → X be a continuous transformation of a compact metrizable space X. If the set M(X,T ) of invariant measures with respect to

T contains only one ergodic measure µ, then µ is the only invariant measure with respect to T . Conversely, if µ is the only invariant measure with respect to T , then µ is ergodic.

Proof. This follows immediately from theorem 16 and the fact that M(X,T ) ⊂

M(X) is both closed and bounded. CHAPTER 3. INVARIANT MEASURES 69

Deﬁnition 27. Let T : X → X be a continuous transformation of a compact metrizable space X. If M(X,T ) = {µ} contains just one invariant measure

µ with respect to T , then both T and µ are said to be uniquely ergodic.

Remark 7. The case where is X = T and T is C1 expanding has been studied extensively, and in this context it is common to restrict oneself to the study of absolutely continuous invariant measures (ACIM). That is, invariant Borel probability measures µ on T which are invariant with respect to some C1 expanding transformation T : T → T, such that µ is absolutely continuous with respect to Lebesgue measure λ.

ACIM = {µ ∈ M(T,T ) | µ λ}

The set ACIM is compact and convex, and a measure µ ∈ ACIM is an extreme point if and only if it is ergodic (the proofs are similar to the proofs of theorems 15 and 16). Thus, we may ask which continuous transformations

T : T → T have an ACIM which is uniquely ergodic. It is shown in [13] that if

T is a C1+dini expanding circle endomorphism, then there is an ACIM µ that is uniquely ergodic and, furthermore, µ is smooth. That is, it µ a continuous density function ρ such that

Z µ(A) = ρ dλ. A CHAPTER 3. INVARIANT MEASURES 70

The modulus of continuity ω(t) for the density function ρ may not satisfy the

Dini property (see deﬁnition 2.1). That is, the distribution function for µ is

C1 but may not be C1 +dini. An expanding circle endomorphism that is only only C1 may have more than one ACIM. An example can be found in [14], in which a C1 expanding map T : T → T is explicityly constructed such that there is an ACIM that is not ergodic, and so it is not unique.

3.3 Invariant measures on T with bounded geometry and bounded nearby geometry

In order to simplify things, let us equate the unit circle T with the unit interval [0, 1] with the endpoints 0 and 1 identiﬁed, 0 ∼ 1, endowed with the

Euclidean metric.

Deﬁnition 28. Given a measure space (T, B(T), µ), we may deﬁne a map h : T 7→ T by

h(x) = µ([0, x]).

This function h is called the cumulative distribution function or simply the distribution function for µ.

Theorem 17. If µ has bounded geometry with respect to a degree d circle endomorphism f, then the distribution function h : T 7→ T is a homeomorphism CHAPTER 3. INVARIANT MEASURES 71 with 0 as a ﬁxed point satisfying

µ(A) = λ(h(A)), ∀A ∈ B(T).

∗ −1 That is, µ = h λ = (h )∗λ. The measure µ is invariant with respect to f if and only if Lebesgue measure λ is invariant with respect to g = h ◦ f ◦ h−1.

Proof. Since µ has bounded geometry, lemma 9 implies µ has full support and is non-atomic. Thus, for any x, y ∈ T such that x < y,

h(y) − h(x) = µ([0, y]) − µ([0, x]) = µ([x, y]) > 0 and

lim h(y) − h(x) = lim h(y) − h(x) = 0. x→y− y→x+

That is, h is strictly increasing and continuous. Since µ is a probability measure, we have h(0) = 0 and h(1) = 1, and thus h : T → T is a bijection. Since both h and h−1 map open intervals to open intervals, h is a homeomorphism.

∗ −1 ∗ Consider the measure h λ = (h )∗λ, with h λ(A) = λ(h(A)) for all

A ∈ B(T). From the deﬁnition of h, for all x ∈ T we have

λ(h([0, x])) = λ([h(0), h(x)]) = h(x) = µ([0, x]), and for all x < y ∈ T we have

λ(h([x, y])) = λ([h(x), h(y)]) = h(y) − h(x) = µ([0, y]) − µ([0, x]) = µ([x, y]). CHAPTER 3. INVARIANT MEASURES 72

Since µ and h∗λ agree on the semi-algebra of subintervals of T of the form

[0, x], x ∈ T, and [x, y], x < y ∈ T, it follows that µ = h∗λ and

µ(A) = λ(h(A)), ∀A ∈ B(T).

Now let µ be an invariant measure with respect to f and deﬁne g = h ◦ f ◦ h−1. For any A ∈ B(T), we have

λ(g−1(A)) = λ(h(f −1(h−1(A))))

= µ(f −1(h−1(A))) = µ(h−1(A))

= λ(h(h−1(A))) = λ(A)

Thus λ is invariant with respect to g. Conversely, if λ is invariant with respect to g = h ◦ f ◦ h−1, then for any A in B(T) we have

µ(A) = λ(h(A)) = λ(g−1(h(A))) = λ(h(f −1(h−1(h(A)))))

= λ(h(f −1(A))) = µ(f −1(A))).

Thus µ is invariant with respect to f.

Theorem 18. Given the measure space (T, B(T), µ), the measure µ has bounded nearby geometry with respect to the degree d ≥ 2 circle endomorphism qd : T → T,

qd(x) = dx mod 1, CHAPTER 3. INVARIANT MEASURES 73 if and only if its distribution function h(x) = µ([0, x]) is a quasisymmetric homeomorphism.

Proof. Suppose h is quasisymmetric. For any n ≥ 0, we have

j j + 1 {η } = , | 0 ≤ j ≤ dn+1 − 1 , (3.8) n,qd dn+1 dn+1

n+1 with all d intervals from ηn,qd the same length. Since h is quasisymmetric, there exists a constant M ≥ 1 such that for any n ≥ 0 and any two intervals

0 I,I ∈ ηn which share an endpoint modulo 1 (meaning 0 and 1 are identiﬁed),

1 λ(h(I0)) ≤ ≤ M. M λ(h(I))

For µ = h∗λ, this implies

1 µ(I0) ≤ ≤ M. M µ(I)

Thus µ has bounded nearby geometry with respect to qd.

Now suppose µ has bounded nearby geometry with respect to qd. Then

∞ −1 the sequence of nested partitions {h(ηn,qd )} = {ηn,h◦qd◦h }n=0 has bounded nearby geometry, in the sense of deﬁnition 19 (with respect to Lebesgue

∞ measure λ). From equation 3.8, it is clear that the sequence {ηn,qd }n=0 of nested partitions of T also has bounded nearby geometry with respect to

Lebesge measure λ. Since h ◦ f ◦ h−1 is topologically conjugate to f, it follows from theorem 8 that h is a quasisymmetric homeomorphism. CHAPTER 3. INVARIANT MEASURES 74

We are interested in the following symmetric rigidity problem which is important for our study of geometric Gibbs theory in [13].

Conjecture 1 (Symmetric Rigidity). Suppose f and g are both uniformly quasisymmetric circle endomorphisms of the same degree d ≥ 2 that both preserve the Lebesgue measure λ. Suppose h is a symmetric conjugacy between f and g. Then h must be the identity.

This conjecture was proved in [4] with the additional assumption that either f or g is the standard circle endomorphism qd. In chapter 5 we generalize this result to circle endomorphisms that have bounded geometry and locally constant limits of martingales.

3.4 Constructing examples of distribution functions for measures with bounded geometry preserved by q2

In this section, we describe some algorithms for constructing distribution functions for measures with bounded geometry preserved by q2. By con- jugating q2 by such homeomorphisms, we produce corresponding degree 2 circle endomorphisms that preserve Lebesgue measure. We also construct the corresponding dual measures and dual circle endomorphisms. Different algorithms produce measures/circle endomorphisms with different properties, CHAPTER 3. INVARIANT MEASURES 75 e.g., finite martingales and bounded nearby geometry.

As pointed out in the proof of theorem 17, the distribution function for any measure µ with bounded geometry with respect to q2 : x → 2x (mod 1) is a strictly increasing homeomorphism h : [0, 1] → [0, 1] satisfying a bounded geometry condition. Such a homeomorphism is determined by its values on a dense set, say {k/2n | n ∈ N, 0 ≤ k ≤ 2n}. This can be done according to the following inductive procedure.

Begin with h(0) = 0 and h(1) = 1. Then, assuming h has been deﬁned on

{k/2n | 0 ≤ k ≤ 2n} (3.9) for a ﬁxed n, we deﬁne h on

{(2k + 1)/2n+1 | 0 ≤ k ≤ 2n − 1}, (3.10) the midpoints of the intervals created by the points in (3.9).

This procedure results in a corresponding measure

µ([0, x]) = h(x)

that has bounded geometry with respect to q2 if and only if there exists a constant C ∈ (0, 1/2] such that for all n ≥ 0 and for all 0 ≤ k ≤ 2n − 1 we have 2k+1 k h 2n+1 − h 2n C ≤ k+1 k ≤ 1 − C. (3.11) h 2n − h 2n CHAPTER 3. INVARIANT MEASURES 76

The measure is invariant with respect to q2 if an only if for all n ≥ 0 and for all 0 ≤ k ≤ 2n − 1 we have

k k + 1 k k + 1 2n + k 2n + k + 1 µ , = µ , + µ , , 2n 2n 2n+1 2n+1 2n+1 2n+1 (3.12) i.e.,

k + 1 k k + 1 k k + 1 k h −h = h − h + h − h 2n 2n 2n+1 2n+1 2n+1 2n+1 (3.13)

The simplest construction of a distribution function satisfying both (3.11) and (3.13) is that of a Bernoulli measure. Fix α ∈ (0, 1). Beginning with h(0) = 0 and h(1) = 1, for each n ≥ 0 and for each 0 ≤ k < 2n, set

2k + 1 k k + 1 k h = h + α h − h . (3.14) 2n+1 2n 2n 2n

∞ In other words, if (ηn)n is a sequence of nested Markov partitions of [0, 1]

corresponding to q2 labelled as in section 2.2, then given any interval Iωn ∈ ηn, we have µ(I ) µ(I ) ωn0 = α, ωn1 = 1 − α. (3.15) µ(Iωn ) µ(Iωn )

Note that a Bernoulli measure clearly satisﬁes the bounded geometry condition, and that the Lebesgue measure is a Bernoulli measure with α = 1/2. CHAPTER 3. INVARIANT MEASURES 77

Theorem 19. Let µ be a Bernoulli measure on [0, 1], deﬁned as above for some ﬁxed 0 < α < 1. Then µ is invariant with respect to the doubling map q2.

Proof. Since the collection of intervals A = {Iωn ∈ ηn | n ≥ 0} forms a semi- algebra of sets which generate the σ-algebra of Borel subsets, it is enough to

−1 show that µ(q2 (Iωn )) = µ(Iωn ) for all Iωn ∈ A. That is,

µ(I0ωn ) + µ(I1ωn ) = µ(Iωn ), ∀ωn ∈ Σn, ∀n ≥ 0. (3.16)

We do this by induction.

By deﬁnition, we have µ(I) = 1, µ(I0) = α, and µ(I1) = 1 − α. Thus equation (3.16) holds for n = 1. Now assume (3.16) holds for all ωn ∈ Σn.

Then we have

µ(Iωn0) = αµ(Iωn ) = α µ(I0ωn ) + µ(I1ωn ) = µ(I0ωn0) + µ(I1ωn0), and

µ(Iωn1) = (1 − α)µ(Iωn ) = (1 − α) µ(I0ωn ) + µ(I1ωn ) = µ(I0ωn1) + µ(I1ωn1).

Thus equation (3.16) holds for n + 1.

The algorithm described above for constructing distribution functions for

Bernoulli measures has been implemented in MATLAB. The code is included CHAPTER 3. INVARIANT MEASURES 78 in the appendix, chapter 6 section 6.1. The recursive nature of the algorithm is visible in the fractal geometry of it graph (see ﬁgures 3.4 and 3.4).

1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 3.2: On the left are the distribution functions h(x) = µ([0, x]) for two Bernoulli measures µ. The top Bernoulli measure has α = .2 and the bottom Bernoulli measure has α = .8. On the right are the corresponding −1 circle endomorphisms f = h ◦ q2 ◦ h .

We can generalize the construction of a Bernoulli measure to obtain a CHAPTER 3. INVARIANT MEASURES 79

1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 3.3: On the left are the distribution functions h(x) = µ([0, x]) for two Bernoulli measures µ. The top Bernoulli measure has α = .4 and the bottom Bernoulli measure has α = .9. On the right are the corresponding −1 circle endomorphisms f = h ◦ q2 ◦ h . construction for any invariant measure µ with bounded geometry. This time,

∞ we constuct a sequence of piecewise-linear homeomorphisms {hn}n=0 that converge uniformly to the distribution function h. Then µ = h∗λ. CHAPTER 3. INVARIANT MEASURES 80

First choose a bounded geometry constant 0 < C ≤ 1/2, and deﬁne h0(x) = x to be the identity on I = [0, 1]. Its graph is linear on I and consists

0 of 2 = 1 line segment. Let µ0 be the probability measure on [0, 1] whose distribution function is h0 (µ0 = λ). Now choose a value C ≤ α ≤ 1 − C.

The value α will be used to construct h1.

The homeomorphism h1 will agree with h0 at the endpoints of I (that is, h1(0) = 0 and h1(1) = 1)) and its value at the midpoit of I will be h(1/2) = α.

Between these points, on each closed interval I0 and I1, the graph of h1 is continuous and linear. Its graph over [0, 1] is continuous and piecewise-linear,

1 and consists of 2 = 2 line segments. Let µ1 be the probability measure on

[0, 1] whose distribution function is h1. In particular, we have µ1(I0) = α and µ1(I1) = 1 − α.

Now a number is chosen that will determine α0 and α1 as follows.

α0 = α + µ1(I0) α1 = α − µ1(I1)

Recall that C ≤ α ≤ 1 − C. We require the values α0 and α1 to also lie between these same bounds. That is, we require to satisfy both of the CHAPTER 3. INVARIANT MEASURES 81 following inequalities.

−(α − C)µ1(I0) ≤ ≤ (1 − C − α)µ1(I0)

−(1 − C − α)µ1(I1) ≤ ≤ (α − C)µ1(I1)

Note that may be positive as long as α is strinctly greater than C, or negative as long as α is strictly less than 1 − C.

These values α0 and α1 are now used to construct h2. The homeomorphism h2 will agree with h1 at the endpoints of all intervals belonging to η1

(that is, I0 and I1), and its values at the midpoints of those intervals (1/4 and 3/4) will be determined as follows.

h2(1/4) = h1(0) + α0(h1(1/2) − h1(0)) (3.17)

h2(3/4) = h1(1/2) + α1((h1(1) − h1(1/2)). (3.18)

Now h2 is deﬁned at the endpoints of all intervals in η2. Between these points, on each closed interval belonging to η2, the graph of h2 is continuous and linear. Its graph over [0, 1] is continuous and piecewise-linear, and

2 consists of 2 = 4 line segments. Let µ2 be the probability measure on [0, 1] whose distribution function is h2. The measure µ2 agrees with µ1 on intervals belonging to η1, and equations (3.17) and (3.18) are equivalent to the CHAPTER 3. INVARIANT MEASURES 82 following.

µ2(I00) = α0µ1(I0), µ2(I01) = (1 − α0)µ1(I0),

µ2(I10) = α0µ1(I1), µ2(I11) = (1 − α0)µ1(I1).

It is straightforward to check that µ2 preserves the Lebesgue measure of intervals belonging to η1. We have

−1 µ2(q2 I0) = µ2(I00) + µ2(I10) = α0µ1(I0) + α0µ1(I1)

= α0(µ1(I0) + µ1(I1)) = α0 = µ1(I0) = µ2(I0),

−1 µ2(q2 I1) = µ2(I01) + µ2(I11) = (1 − α0)µ1(I0) + (1 − α0)µ1(I1)

= (1 − α0)(µ1(I0) + µ1(I1)) = 1 − α0 = µ1(I1) = µ2(I1).

In summary, two numbers, α and were chosen and associated with the interval I. The number α was used to deﬁne h1 and µ1, and both α and were used to deﬁne h2 and µ2.

Now we have α0 and α1 associated with the intervals I0 and I1, respec- CHAPTER 3. INVARIANT MEASURES 83

tively. By choosing numbers 0 and 1 satisfying

−(α0 − C)µ2(I00) ≤ 0 ≤ (1 − C − α0)µ2(I00),

−(1 − C − α0)µ2(I10) ≤ 0 ≤ (α0 − C)µ2(I10),

−(α1 − C)µ2(I01) ≤ 1 ≤ (1 − C − α1)µ2(I01),

−(1 − C − α1)µ2(I11) ≤ 1 ≤ (α1 − C)µ2(I11),

to be associated with these same intervals I0 and I1, respectively, we can use the numbers α0, 0, α1, 1 to deﬁne the values α00, α01, α10, α11, the piecewise-linear homeomorphism h3, and the measure µ3 as follows.

0 1 α00 = α0 + , α01 = α1 + , µ2(I00) µ2(I01) 0 1 α10 = α0 − , α11 = α1 − , µ2(I10) µ2(I11)

h3(1/8) = h2(0) + α00(h2(1/4) − h2(0)),

h3(3/8) = h2(1/4) + α01(h2(1/2) − h2(1/4)),

h3(5/8) = h2(1/2) + α10(h2(3/4) − h2(1/2)),

h3(7/8) = h2(3/4) + α11(h2(1) − h2(3/4)), CHAPTER 3. INVARIANT MEASURES 84

µ3(I000) = α00µ2(I00), µ3(I001) = (1 − α00)µ2(I00),

µ3(I010) = α01µ2(I01), µ3(I011) = (1 − α01)µ2(I01),

µ3(I100) = α10µ2(I10), µ3(I101) = (1 − α10)µ2(I10),

µ3(I110) = α11µ2(I11), µ3(I111) = (1 − α11)µ2(I11).

It is straightforward to check that µ3 preserves the Lebesgue measure of CHAPTER 3. INVARIANT MEASURES 85

intervals belonging to η2. We have

−1 µ3(q2 I00) = µ3(I000) + µ3(I100) = α00µ2(I00) + α10µ2(I10)

0 0 = (α0 + )µ2(I00) + (α0 − )µ2(I10) µ2(I00) µ2(I10)

= α0(µ2(I00) + µ2(I10)) = α0µ2(I0) = µ2(I00)

= µ3(I00),

−1 µ3(q2 I01) = µ3(I001) + µ3(I101) = (1 − α00)µ2(I00) + (1 − α10)µ2(I10)

0 0 = (1 − (α0 + ))µ2(I00) + (1 − (α0 − )µ2(I10) µ2(I00) µ2(I10)

= (1 − α0)(µ2(I00) + µ2(I10)) = (1 − α0)µ2(I0) = µ2(I01)

= µ3(I01),

−1 µ3(q2 I10) = µ3(I010) + µ3(I110) = α01µ2(I01) + α11µ2(I11)

1 1 = (α1 + )µ2(I01) + (α1 − )µ2(I11) µ2(I01) µ2(I11)

= α1(µ2(I01) + µ2(I11)) = α1µ2(I1) = µ2(I10)

= µ3(I10),

−1 µ3(q2 I11) = µ3(I011) + µ3(I111) = (1 − α01)µ2(I01) + (1 − α11)µ2(I11)

1 1 = (1 − (α1 + ))µ2(I01) + (1 − (α1 − )µ2(I11) µ2(I01) µ2(I11)

= (1 − α1)(µ2(I01) + µ2(I11)) = (1 − α1)µ2(I1) = µ2(I11)

= µ3(I11).

In general, suppose hn and hn+1 have been deﬁned (and thus µn and µn+1 CHAPTER 3. INVARIANT MEASURES 86

as well), with hn and hn+1 both continuous on [0, 1], and each one linear

on the closed intervals Iωn ∈ ηn and Iωn ∈ ηn, respectively. Suppose µn

and µn+1 are invariant with respect to q2 on the intervals Iωn−1 ∈ ηn−1 and

Iωn ∈ ηn, respectively. Suppose each interval Iωn ∈ ηn has been assigned a

value C ≤ αωn ≤ 1 − C such that

µn+1(Iωn0) µn+1(Iωn1) = αωn , = 1 − αωn . (3.19) µn+1(Iωn ) µn+1(Iωn )

Figure 3.4: The interval Iωn ∈ ηn and its subinterval Iωn0 ∈ ηn+1, along with their preimages belonging to ηn+1 and ηn+2, respectively.

∗ The goal now is to deﬁne hn+2 and µn+2 = (hn+2) λ such that µn+2 agrees

with µn+1 and is q2-invariant on all intervals Iωn+1 ∈ ηn+1. There are many CHAPTER 3. INVARIANT MEASURES 87

n way to do this, but each one is determined by a choice of 2 values ωn , each satisfying

−(aωn − C)µn+1(I0ωn ) ≤ ωn ≤ (1 − C − αωn )µn+1(I1ωn ) (3.20)

−(1 − C − αωn )µn+1(I0ωn ) ≤ ωn ≤ (aωn − C)µn+1(I1ωn ). (3.21)

Then for each ωn we set

ωn ωn α0ωn = αωn + , α1ωn = αωn − . µn+1(I0ωn ) µn+1(I1ωn )

Remark 8. Note that 0 is always an allowable choice for any ωn . If for some

non-negative integer N, 0 is chosen to be the value for all ωn with n ≥ N,

−1 then the resulting circle endomorphism f = h ◦ q2 ◦ h is piecewise-linear and has a ﬁnite martingale deﬁned on the dual symbolic space (see chapter

5).

We then deﬁne hn+2 such that it agrees with hn+1 at the endpoints of all

intervals Iωn+1 ∈ ηn+1 and for any interval Iωn+1 ∈ ηn+1 with left endpoint a, right endpoint b, and midpoint m,

hn+2(m) = h(a) + αωn+1 (h(b) − h(a)). CHAPTER 3. INVARIANT MEASURES 88

Then

µn+2(I0ωn0) = α0ωn µn+1(I0ωn ), µn+2(I0ωn1) = (1 − α0ωn )µn+1(I0ωn ), (3.22)

µn+2(I1ωn0) = α1ωn µn+1(I1ωn ), µn+2(I1ωn1) = (1 − α1ωn )µn+1(I1ωn ). (3.23)

Now hn+2 is deﬁned at the endpoints of all intervals in ηn+2. Between these points, on each closed interval belonging to ηn+2, the graph of hn+2 is continuous and linear. Its graph over [0, 1] is continuous and piecewise-linear, and

n+2 consists of 2 line segments. Let µn+2 be the probability measure on [0, 1] whose distribution function is hn+2. The measure µn+2 agrees with µn+1 on intervals belonging to ηn+1, and equations (3.22) and (3.23) are equivalent to the following.

µn+2(Iωn+10) = αωn+1 µn+1(Iωn+1 ),

µn+2(Iωn+11) = (1 − αωn+1 )µn+1(Iωn+1 ).

It is straightforward to check that µn+2 preserves the Lebesgue measure CHAPTER 3. INVARIANT MEASURES 89

of intervals belonging to ηn+1. We have

−1 µn+2(q2 Iωn0) = µn+2(I0ωn0) + µn+2(I1ωn0)

= α0ωn µn+1(I0ωn ) + α1ωn µn+1(I1ωn ) ωn ωn = αωn + µn+1(I0ωn ) + αωn − µn+1(I1ωn ) µn+1(I0ωn ) µn+1(I1ωn )

= αωn µn+1(I0ωn ) + αωn µn+1(I1ωn )

= αωn µn+1(Iωn ) = µn+1(Iωn0)

= µn+2(Iωn0),

−1 µn+2(q2 Iωn1) = µn+2(I0ωn1) + µn+2(I1ωn1)

= (1 − α0ωn )µn+1(I0ωn ) + (1 − α1ωn )µn+1(I1ωn ) ωn ωn = 1 − αωn + µn+1(I0ωn ) + 1 − αωn − µn+1(I1ωn ) µn+1(I0ωn ) µn+1(I1ωn )

= (1 − αωn )(µn+1(I0ωn ) + µn+1(I1ωn ))

= (1 − αωn )µn+1(Iωn ) = µn+1(Iωn1)

= µn+2(Iωn1).

By construction, it is clear that for any n ≥ 0 we have

n |hn+1 − hn| ≤ (1 − C) ,

and so the sequence hn converges uniformly to a continuous function h. Fur- thermore, h is strictly increasing since it agrees with hn at all endpoints of CHAPTER 3. INVARIANT MEASURES 90

ηn for all n ≥ 0, the union of which are dense in [0, 1], and it is clear from the construction that each hn is strictly increasing. Thus h is a homeomorphism.

Fruthermore, since each αon = µ(Iωn0)/µ(Iωn ) lies between C and 1 − C, the

∞ sequence of nested partitions {ηn}n=1 has bounded geometry with respect to the measure µ = h∗λ.

The algortithm described above has been implememnted in MATLAB with a random number generator supplying the values of (ωn), each inside the maximum allowable range according to the necessary restrictions (3.20) and (3.21). The code is included in the appendix (see chapter 6, section 6.2). CHAPTER 3. INVARIANT MEASURES 91

1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 3.5: On the left are the distribution functions h(x) = µ([0, x]) for three arbitrary measures µ that are invariant with respect to the doubing map q2, and that have bounded geometry with C = .1. On the right are the −1 corresponding circle endomorphisms f = h ◦ q2 ◦ h . Chapter 4

The dual circle map

4.1 Equivalence of partitions and maps

We have seen that a circle endomorphism of degree d ≥ 2 determines a

∞ sequence {ηn}n=0 of nested partitions of T. In fact, the converse is also true.

∞ Theorem 20. Suppose d ≥ 2 and let {ηn}n=0 be a sequence of nested partitions of T such that η0 contains d intervals, two of which share 0 as a common endpoint (modulo 1). Furthermore, for each n ≥ 0 and each interval I ∈ ηn, let I contain exactly d subintervals from ηn+1, and suppose

there exist two constants 0 < C ≤ τ < 1 such that for all Iωn ∈ ηn we have

n+1 n+1 ∞ C ≤ |Iωn | ≤ τ . Then we can use {ηn}n=0 to construct the unique

−n degree d circle endomorphism f such that ηn = f η0, for all n ≥ 0.

Proof. We construct a sequence of piecewise-linear degree d circle endomorphisms fn which converge uniformly to the desired map f. For any n ≥ 0, let

92 CHAPTER 4. THE DUAL CIRCLE MAP 93

n+1 us label the d intervals of the partition ηn with words ωn of length n + 1 composed of the symbols 0, 1,..., (d − 1) exactly as in previous chapters.

That is, ηn consists of the intervals

I00...00,I00...01,...,I(d−1)(d−1)...(d−1),

labeled from left to right. Now for any ωn = i0i1 . . . in, let aωn denote the left

endpoint of Iωn and let a(d−1)(d−1)...d denote the right endpoint of the interval

I(d−1)(d−1)...(d−1), which is the point 1. Now we can begin to define the maps fn by defining their lifts Fn as follows. The map f0 has lift F0 defined such that

F0(a0) = 0,F0(a1) = 1,...F0(ad) = d,

i.e. F0(ai0 ) = i0,

and such that F0 is continuous on [0, 1] and linear on each interval Iω0 ∈ η0.

Next we deﬁne

F1(ai0i1 ) = i0 + ai1 ,

and such that F1 is continuous on [0, 1] and linear on each interval Iω1 ∈ η1.

In this way, for n ≥ 2 deﬁne Fn such that

Fn(ai0i1...in ) = i0 + ai1...in ,

and Fn is continuous on [0, 1] and linear on each interval Iωn ∈ ηn. CHAPTER 4. THE DUAL CIRCLE MAP 94

Figure 4.1: The lifts F0, F1, and F2. CHAPTER 4. THE DUAL CIRCLE MAP 95

By construction, Fn agrees with Fn−k at all endpoints of intervals belonging to ηn−k for 0 ≤ k ≤ n. Moreover, for any N > 0, if m, n > N then

N |Fn(x) − Fm(x)| ≤ τ , ∀x ∈ [0, 1].

This makes the sequence Fn uniformly Cauchy on [0, 1], and so it converges uniformly to a continuous function F . Now we extend F to have domain R by deﬁning F (x+k) = F (x)+kd. This is the lift of the circle endomorphism

−n f which, by construction, satisﬁes ηn = f η0, for all n ≥ 0. Any other circle endomorphism satisfying this property would agree with f at all endpoints of intervals from all partitions ηn, n ≥ 0. Since this is a dense set in [0, 1], this shows that f is unique.

Corollary 3. If {ηn} is a sequence of nested partitions as in theorem 20 with bounded nearby geometry, then the circle endomorphism it determines is uniformly quasisymmetric.

Proof. This follows immediately from theorem 20 and theorem 9.

4.2 Dual partitions, dual maps, and dual measures

Say we have a degree d ≥ 2 uniformly quasisymmetric circle endomorphism f

∞ that preserves Lebesgue measure. This deﬁnes a sequence {ηn}n=0 of nested CHAPTER 4. THE DUAL CIRCLE MAP 96 partitions of [0, 1] with bounded nearby geometry. As before, for each n ≥ 0 we use words consisting of n+1 symbols 0, 1,..., (d−1) to label the intervals of ηn.

η0 = {I0,I1,...,I(d−1)}

η1 = {I00,I01,...,I0(d−1),I10,...,I(d−1)(d−1)}

ηn = {I00 ... 0,...,I(d − 1) ... (d − 1)} | n{z+1 } | n{z+1 } That is, the intervals are labeled from left to right as n-digit (d − 1)-nary numbers increasing by 1 from 0 to dn+1 − 1.

Figure 4.2: For d = 2 the intervals of ηn are labeled as n-digit binary numbers, increasing left to right, from 0 to 2n+1 − 1.

∗ ∞ Now we construct a new sequence of nested partitions {ηn}n=1 by “shuf-

ﬂing” each of the partitions ηn, n ≥ 0. More speciﬁcally, for each n ≥ 0 and CHAPTER 4. THE DUAL CIRCLE MAP 97

for each Iωn ∈ ηn, we reassign a new length to the interval Iωn in such a way

∗ that the sum of the lengths of all intervals belonging to ηn continues to be 1,

∗ ∗ and such that ηn+1 is a reﬁnement of ηn. The construction is performed as follows.

For each n ≥ 0, permute the intervals of ηn such that their labels increase by 1 from 0 to 2n+1 −1 when each is read from right to left. That is, transpose the positions of intervals with labels that are mirror-images of one another – one label read left-to-right is equal to the other label read right-to-left. For example, for d = 2 and n = 5, the labels 00101 and 10100 are mirror-images, and so the intervals I00101,I10100 ∈ η5 are transposed in the construction of

∗ η5. This new partition, together with its new labelling, is the dual partition

∗ ∗ ∗∗ ηn. The dual partition of ηn is ηn, that is ηn = ηn.

∗ Another way to describe the dual partition ηn is by ﬁrst deﬁning a bijection ρ :Σn → Σn such that

ρ(i0i1 . . . in−1) = in−1 . . . i1i0.

∗ Now for each Iωn ∈ ηn change its length to |Iρ(ωn)| and its label to Iωn . The

∗ result is the partition ηn.

Note that since f preserves Lebesgue measure,

|Iωn | = |I0ωn | + |I1ωn | + ... + |I(d−1)ωn |, ∀Iωn ∈ ηn. CHAPTER 4. THE DUAL CIRCLE MAP 98

∗ Figure 4.3: Shuﬄing intervals of ηn to construct ηn.

And since

∗ ∗ ∗ ∗ ∗ ∗ Iωn = I0ωn ∪ I1ωn ∪ ... ∪ I(d−1)ωn , ∀Iωn ∈ ηn,

∗ ∗ we see that ηn+1 is a reﬁnement of ηn for all n ≥ 0. In addition, if the

∞ sequence {ηn}n=0 of partitions satisﬁes the conditions of theorem 20, then so

∞ does the sequence {ηn}n=0 of dual-partitions.

Theorem 21. Given a degree d ≥ 2 uniformly quasisymmetric circle endomorphism f that preserves Lebesgue measure with nested Markov partitions

∞ ∗ {ηn}n=0, there is a unique circle endomorphism f that is topologically con-

∗ ∗ jugate to f with nested Markov partitions {ηn}n=0 (the dual-partitions), and f ∗ also preserves Lebesgue measure.

Proof. The procedure described above shows how to obtain the sequence of

∗ ∞ ∗ dual-partitions {ηn}n=0, and the desired unique circle endomorphism f is CHAPTER 4. THE DUAL CIRCLE MAP 99 provided by theorem 20. Since f and f ∗ have the same degree d, it follows from theorem 6 that f and f ∗ are topologically conjugate.

∗ ∗ By construction, for any n ≥ 0 and for any interval Iωn ∈ ηn, we have

∗ −1 ∗ ∗ ∗ ∗ λ((f ) (Iωn )) = λ(Iωn0) + λ(Iωn1) + ... + λ(Iωn(d−1))

= λ(Iωn0) + λ(Iωn1) + ... + λ(Iωn(d−1))

∗ = λ(Iωn ) = λ(Iωn ).

Since the collection of such intervals forms a semi-algebra which generates the Borel σ-algebra, it follows that f ∗ preserves Lebesgue measure.

Similarly, the degree d ≥ 2 dual circle map f ∗ is also topologically con-

∗ jugate to the map qd : x 7→ dx mod 1. Let h : T → T denote the home-

∗ ∗ ∗ −1 ˜ omorphism such that f = h ◦ qd ◦ (h ) , and let h : T → T denote the homeomorphism such that f ∗ = h˜ ◦ f ◦ h˜−1. Then we have

h˜ = h∗ ◦ h−1.

In summary, the following diagram commutes.

Deﬁnition 29. The homeomorphism h∗ can be used to deﬁne a measure

µ∗. For any interval A ⊂ T of the form [a, b], (a, b], [a, b), or (a, b), deﬁne

µ∗(A) = h∗(b) − h∗(a). That is, µ∗ is the pullback measure of Lebesgue measure via h∗ (µ∗ = (h∗)∗λ). Since the collection of all such intervals forms CHAPTER 4. THE DUAL CIRCLE MAP 100

f T T

h h

qd h˜ T T h˜

h∗ h∗

f ∗ T T

∗ Figure 4.4: Topologically conjugate circle endomorphisms qd, f, and f (the dual circle endomorphism of f).

a semi-algebra that generates the Borel σ-algebra B, µ∗ extends uniquely to a Borel probability measure on T. We call µ∗ the dual circle measure to

µ = h∗λ.

The following ﬁgure shows three pairs of graphs. On the left in blue is the distribution function h for an arbitrary invariant measure with bounded geometry with respect to the doubling map q2. On the left in red is the distribution function h∗, which is also an invariant measure with bounded geometry with respect to the doubling map q2. On the left in black is the dual homeomorphism h˜ = h∗ ◦h−1. On the right are the corresponding degree

2 circle endomorphisms f in blue and f ∗ in red. CHAPTER 4. THE DUAL CIRCLE MAP 101

1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 4.5: Distribution functions h (left, blue) and dual distribution functions h∗ (left, red), and their corresponding circle endomorphisms f = −1 ∗ ∗ ∗ −1 h ◦ q2 ◦ h (right, blue) and f = h ◦ q2 ◦ (h ) (right, red). On the left in black is the dual homeomorphism h˜ = h∗ ◦ h−1. CHAPTER 4. THE DUAL CIRCLE MAP 102

Remark 9. It is interesting to observe how close the homeomorphisms h and h∗ remain to one another. This is mirrored by the observation that the homeomorphism h˜ remains close to the identity. In the following section, we show that h˜ actually has countably many limit points of ﬁxed points (theorem

22).

The endomorphisms f and f ∗ also remain close to one another. Despite their similar appearances, there is an important distinction: the circle endomorphism f has bounded geometry, but the circle endomorphism f ∗ may not have bounded geometry.

The MATLAB code for creating these graphs is included in the appendix

(see chapter 6 section 6.3).

4.3 Properties of the dual homeomorphism

˜ First, for each n ≥ 0, h must send the endpoints of intervals in ηn to the

∗ endpoints of intervals in ηn, preserving their order. In fact, since this set of points becomes dense in T as n → ∞, this is both a necessary and sufficient ˜ ˜ condition for h. For each n = 1, 2 ..., we can define hn : [0, 1] → [0, 1] to be affine on each interval in ηn, sending endpoints of intervals in ηn to endpoints

∗ of intervals in ηn, preserving their order.

For now, let us ﬁx the degree of f (and thus f ∗) to be 2. Since ρ(0) = 0 CHAPTER 4. THE DUAL CIRCLE MAP 103

˜ and ρ(1) = 1, it is clear that h1 is the identity on [0, 1]. Also, since f preserves

Lebesgue measure and Lebesgue measure is additive, we have

|I0| = |I00| + |I10| = |I00| + |I01|

⇒ |I01| = |I10|,

˜ ˜ and so h2 = Id. The same argument shows that h3 = Id as well. ˜ As n → ∞, the number of ﬁxed points of hn also goes to inﬁnty, that is ˜ # Fix(hn) → ∞. In fact, since f preserves Lebesgue measure and Lebesgue measure is additive, we have

|I 0...0 | = |I0 0...0 | + |I1 0...0 | = |I 0...0 0| + |I 0...0 1|

|{z}n−1 |{z}n−1 |{z}n−1 |{z}n−1 |{z}n−1

⇒ |I1 0...0 | = |I 0...0 1|, and

|{z}n−1 |{z}n−1

|I 1...1 | = |I0 1...1 | + |I1 1...1 | = |I 1...1 0| + |I 1...1 1|

|{z}n−1 |{z}n−1 |{z}n−1 |{z}n−1 |{z}n−1

⇒ |I0 1...1 | = |I 1...1 0|.

|{z}n−1 |{z}n−1 That is,

˜ 1 1 1 1 Fix(hn) ⊇ h , h ± , h 1 − 1 ≤ n ≤ N . 2n 2 2n 2n

˜ Thus #F ix(hn) ≥ 4n−3, and the points h(0) = 0 (and h(1) = 1) and h(1/2) are limit points for the set Fix(h˜). CHAPTER 4. THE DUAL CIRCLE MAP 104

There are other limit points for Fix(h˜) as well. For example, the argument above shows that for any n ≥ 1, the endpoints of the interval

I 0...0 1

|{z}n ˜ are ﬁxed by h. Let a1 denote the left endpoint of this interval. Now consider the interval

I 0...0 1 0...0 .

|{z}n |{z}n Note that the label of this interval is a palindrome (i.e. it reads the same forward and backward), that is,

ρ(0 ... 0 1 0 ... 0) = 0 ... 0 1 0 ... 0, | {z } | {z } | {z } | {z } n n n n and so

∗ |I 0...0 1 0...0 | = |I 0...0 1 0...0 |.

|{z}n |{z}n |{z}n |{z}n

Let a2 be the right endpoint of this interval. Since the left endpoint of this ˜ interval a1 is a ﬁxed point of h, this implies that a2 is also a ﬁxed point of h˜. Then we can also say

a2 = left endpoint of I 0...0 1 0...0 1.

|{z}n |{z}n−1 CHAPTER 4. THE DUAL CIRCLE MAP 105

In this way, deﬁne

a3 = left endpoint of I 0...0 1 0...0 1 0...0 1 (4.1)

|{z}n |{z}n−1 |{z}n−1

a4 = left endpoint of I 0...0 1 0...0 1 0...0 1 0...0 1 (4.2)

|{z}n |{z}n−1 |{z}n−1 |{z}n−1 . . (4.3)

am = left endpoint of I 0...0 1 0...0 1 0...0 1... 0...0 1. (4.4) |{z} |{z} |{z} |{z} n n−1 n−1 n−1

| m{z−1 }

∞ ˜ Figure 4.6: The increasing sequence {am}m=1 of ﬁxed points of h, constructed with n = 2.

∞ ˜ The sequence {am}m=1 of ﬁxed points of h is increasing and is bounded above. Thus am → a for some a ∈ I, and a is a limit point of the set CHAPTER 4. THE DUAL CIRCLE MAP 106

Fix(h˜). In fact, when a decimal point is put before the labels of the intervals in (4.4) and the labels are treated as binary numbers, these numbers form a geometric series and we have

1 1 1 1 a = h + + + ... . m 2n+1 22n+1 23n+1 2mn+1

Thus, taking the limit, we have shown the following.

Theorem 22. Suppose f is a degree 2 circle endomorphism with bounded geometry that preserves Lebesgue measure and f ∗ is its dual circle endomorphism. Let h and h∗ be circle homeomorphisms such that

∗ ∗ ∗ −1 f = h ◦ q2 ◦ h−1, f = h ◦ q2 ◦ (h ) , and

h˜ = h∗ ◦ h−1, f ∗ = h˜ ◦ f ◦ h˜−1.

Then for all n, m ≥ 1, there is a ﬁxed point

m ! X 1 x = h ∈ Fix(h˜), n,m 2in+1 i=1 and for all n ≥ 1, there is a limit point of ﬁxed points

∞ ! X 1 1 x = h = h ∈ Fix(h˜). n 2in+1 2(2n − 1) i=1

Switching all 0’s to 1’s and all 1’s to 0’s in the argument above, we arrive at the following. CHAPTER 4. THE DUAL CIRCLE MAP 107

Corollary 4. For all n, m ≥ 1, there is a ﬁxed point

m ! X 1 y = h 1 − ∈ Fix(h˜), n,m 2in+1 i=1 and for all n ≥ 1, there is a limit point of ﬁxed points

∞ ! X 1 1 y = h 1 − = h 1 − ∈ Fix(h˜). n 2in+1 2(2n − 1) i=1 4.4 Dual symbolic space

Consider the space

← Y Σ∗ = {0, 1, . . . , d − 1} n≥0 ∗ = {ω = . . . j2j1j0 | jk ∈ {0, 1, . . . , d − 1}, k = 0, 1,...}.

The set {0, 1, . . . , d − 1} is given the discrete topology and Σ∗ is given the product topology. It is a compact topological space.

∗ Deﬁnition 30. A right-cylinder for a ﬁxed word ωn = jn−1 . . . j1j0 of length n is

∗ ∗ 0 0 0 [ωn] = {ω = . . . jn+1jnjn−1 . . . j1j0 | jn+k ∈ {0, 1, . . . , d − 1}, k = 0, 1,...}.

The set of all right-cylinders forms a topological basis of Σ∗. We call the topology generated by the set of all right-cylinders the right topology, and we call the set Σ∗ with the right-topology the dual symbolic space. CHAPTER 4. THE DUAL CIRCLE MAP 108

∗ For any ω = . . . j2j1j0, let

∗ ∗ σ (ω ) = . . . j2j1 be the right shift map. Then (Σ∗, σ∗) is called a dual symbolic dynamical system.

∗ ∗ ∗ For a point ω = . . . j1j0 ∈ Σ , let ωn = jn−1 . . . j1j0. Then

∗ ∗ ∗ T ⊃ Iω1 ⊃ Iω2 ⊃ ... ⊃ Iωn ⊃ ....

∗ Since each Iωn is compact and non-empty,

∗ ∞ ∗ Iω = ∩n=1Iωn 6= ∅.

∗ ∗ If every Iω = {xω} contains only one point, then we deﬁne the projection πf ∗ from Σ∗ onto T as

∗ πf ∗ (ω ) = xω∗ .

The projection πf ∗ is one-to-one except for on a countable set

... 00,... (d − 1)(d − 1), ∗ B = ∗ ∗ n ≥ 0, ω0 = ∅ ... 00(d − 1)ωn,... (d − 1)(d − 1)0ωn

∗ on which πf ∗ is two-to-one. These are precisely the points in Σ that project

∗ to an endpoint of Iωn for some n. From our construction we have

∗ ∗ ∗ ∗ ∗ ∗ πf ∗ ◦ σ (ω ) = f ◦ πf ∗ (ω ), ω ∈ Σ . CHAPTER 4. THE DUAL CIRCLE MAP 109

Thus (Σ∗, σ∗) is the symbolic dynamical system for the dual circle endomorphism f ∗. We call (Σ∗, σ∗) the dual symbolic dynamical system.

Both the symbolic dynamical system (Σ, σ) and the dual symbolic dynamical system (Σ∗, σ∗) can be visualized as directed d-nary trees, where each vertex is a cylinder and each edge shows the action of σ or σ∗ on that cylinder. We can overlay these two trees on top of one another (see ﬁg- ure 4.7), with the understanding that the left-cylinders [ωn] must be treated

∗ ∗ as right-cylinders [ωn] when following edges that represent the action of σ .

Remark 10. In reference to ﬁgure 4.7, when the vertices represent left- cylinders of Σ, then the dotted arrows point to each cylinder’s “dynamical parent” (image under σ) and the solid arrows point to each cylinder’s “spatial parent” (cylinder one level above which contains it). On the other hand, when the vertices represent right-cylinders of Σ∗, then the solid arrows point to each cylinder’s “dynamical parent” (image under σ∗) and the dotted arrows point to each cylinder’s “spatial parent”. This is the sense in which σ and σ∗ are dual maps of one another.

We now equip the symbolic space Σ∗ with the structure of a probability space. Let N n [ o G = [w∗ ] 1 ≤ N < ∞ ∪ Σ∗. nk k=1 CHAPTER 4. THE DUAL CIRCLE MAP 110

[0] [1]

[00] [01] [10] [11]

[000] [001] [010] [011] [100] [101] [110] [111]

[0000] [0001] [0010] [0011] [0100] [0101] [0110] [0111] [1000] [1001] [1010] [1011] [1100] [1101] [1110] [1111]

......

Figure 4.7: Two directed d-nary trees (d=2). The dotted tree shows the action of σ on left-cylinders of Σ, and the solid tree shows the action of σ∗ on right-cylinders of Σ∗. be the collection of all ﬁnite unions of right-cylinders in Σ∗. This forms an algebra of sets that generates the σ-algebra of Borel sets B (refer to [7] for more).

∗ ∗ ∗ For each n ≥ 1 and for each ω = . . . jnjn−1 . . . j1j0 ∈ Σ , let ωn = jn−1 . . . j1j0 and wn = i0i1 . . . in−2in−1 where i0 = jn−1, i1 = jn−2, ..., in−2 = j1, and in−1 = j0. For n ≥ 0, deﬁne

∗ ∗ P ([wn]) = |Iwn |,P (Σ ) = |I| = 1. (4.5)

This deﬁnes P on G. Now let

N N [ X P ( [w∗ ]) = |I | nk wnk k=1 k=1 CHAPTER 4. THE DUAL CIRCLE MAP 111 where SN [w∗ ] is a disjoint union. Then k=1 nk

∗ P (Σ ) = |I0| + |I1|... + |Id−1| = 1.

∗ For any right cylinder [wn], the collection of right cylinders

∗ {[jwn] | j = 0, . . . , d − 1},

∗ is the collection of all the right sub-cylinders from one level below and [wn] =

d−1 ∗ −1 d−1 ∪j=0[jwn]. Since f (Jwn ) = ∪j=0Jjwn and since f preserves the Lebesgue measure, we have that d−1 X |Iwn | = |Ijwn | j=0 This implies that d−1 ∗ X ∗ P ([wn]) = P ([jwn]). j=0 So P is a probability measure on G. This fact plus the Caratheodary extension theorem implies that P can be extended to a probability measure on the

σ-algebra B.

Thus we have a probability space

(Σ∗, B,P ) associated with f. Furthermore, under the assumption of bounded geometry for f, P is a non-atomic probability measure. CHAPTER 4. THE DUAL CIRCLE MAP 112

We now use the probability measure P to to deﬁne a sequence of functions

∞ ∗ {Xn}n=1 deﬁned on the dual symbolic space Σ . For n ≥ 1, deﬁne

∗ ∗ ∗ P ([σ (wn)]) Xn(w ) = ∗ , (4.6) P ([wn])

∗ ∗ ∗ ∗ ∗ where P ([σ (ω1)]) = P (Σ ) = 1, for all ω1 ∈ Σ1. Then we have a sequence

∗ of positive real functions (Xn)n≥1 on Σ .

Deﬁnition 31. We say that the sequence (Xn)n≥1 for a circle endomorphism f has a locally constant limit X if there exists a positive integer n0 and

n0 ∗ d constants c(ωn0 ) such that

∗ ∗ ∗ X(ω ) = lim Xn(ωn) = c(ωn ) n→∞ 0

∗ ∗ ∗ for ω = . . . jnjn−1 . . . j1j0 ∈ Σ with ωn = jn−1 . . . j1j0.

Remark 11. Even if the sequence (Xn)n≥1 for a circle endomorphism f has a locally constant limit X, it may not imply that Xn is equal to these constants

d locally. For example, for qd(z) = z , d ≥ 2, we have the corresponding

∞ constant sequence {Xqd,n = d}n=1 and thus it has a constant limit Xqd = d.

Now consider an abitrary symmetric circle homeomorphism h and f = h ◦

−1 qd ◦ h . The sequence

∗ ∗ |h(Iqd,σ (ωn))| Xf,n(ω ) = , n = 1, 2,..., |h(Iqd,ωn )| CHAPTER 4. THE DUAL CIRCLE MAP 113

∞ may not be the constant sequence {d}n=1 for all n ≥ 1, but its limit is still the constant d (see the proof of Theorem 25). In the following chapter, we will use martingale theory to prove that under the further assumption that f preserves the Lebesgue measure on T , this cannot happen. Chapter 5

Martingales

In general, a circle endomorhism with bounded geometry, even one that preserves Lebesgue measure, may be totally singular, that is, its derivative may be 0 almost everywhere despite the fact that it is strictly increasing.

Therefore, its derivative may not be of interest. Instead, we apply the theory of martingales (see [10]) to deﬁne a positive L1 function on the dual symbolic space Σ∗ for a circle endomorphism with bounded geometry that preserves

Lebesgue probability measure (see [7]). Such a positive L1 function can be thought of as the L1 dual derivative. It is deﬁned P − a.e. in Σ∗ and is bounded and bounded away from zero.

5.1 Deﬁnition and convergence theorem

For any ﬁxed n ≥ 1, deﬁne Bn as the σ-algebra generated by all right cylinders

∗ {[ws ], 0 ≤ s ≤ n}. Then Bn is a sub-σ-algebra of B and we have a ﬁltration

114 CHAPTER 5. MARTINGALES 115

of σ-algebras (Bn)n≥1,

Bm ⊆ Bn ⊆ B, ∀0 ≤ m ≤ n.

Deﬁnition 32. Given the probability space (Σ∗, B,P ) and the ﬁltration of

σ-algebras (Bn)n≥1 deﬁned above, a sequence of random variables (Xn)n≥1 is called a martingale if

i E[|Xn|] < ∞ for all n ≥ 0, that is,

Z ∗ |Xn(ω )| dP < ∞, ∀n ≥ 1; Σ∗

ii Xn is Bn-measurable for all n ≥ 1, that is, for each n ≥ 1 and for each

∗ ∗ right cylinder [ωn] ∈ Bn there exists a constant c(ωn) ∈ R such that

∗ ∗ Xn|[ωn]=c(ωn).

iii For all 1 ≤ m ≤ n, Xm = E[Xn|Bm] P − a.e., where E[Xn|Bm] is the

unique (up to a set of measure zero) Bm-measurable function such that

Z Z E[Xn|Bm] dP = Xn dP, ∀A ∈ Bm. A A

∗ ∗ ∗ In particular, for all ω = . . . jm+1jmωm ∈ Σ we have

Z ∗ ∗ Xm(ω )P ([ωm]) = Xn dP. ∗ [ωm] CHAPTER 5. MARTINGALES 116

For a submartingale, part (3) is replaced with the condition that for all

1 ≤ m ≤ n, Xm ≤ E[Xn|Bm] P − a.e.

Theorem 23. Suppose f : T → T is a circle endomorphism that preserves the Lebesgue probability measure m on T . Then the sequence

∗ ∗ ∗ P ([σ (wn)]) Xn(w ) = ∗ , (5.1) P ([wn]) introduced in (4.6), with the ﬁltration of σ-algebras (Bn)n≥1 is a martingale on (Σ∗, B,P ).

Proof. We check that the sequence of positive random variables (Xn)n≥1 satisﬁes (1)-(3) in Deﬁnition 32 as follows:

(1): For all n we have

∗ ∗ X |Iσ (ωn)| ∗ E[Xn] = · P ([ωn]) ∗ ∗ |Iωn | ωn∈Bn

∗ ∗ X |Iσ (ωn)| X ∗ ∗ ∗ = · |Iωn | = |Iσ (ωn)| ≤ d < ∞ ∗ ∗ |Iωn | ∗ ωn∈Bn ωn∈Bn

∗ (2): On every right cylinder [ωn],

∗ ∗ ∗ ∗ ∗ P ([σ (ωn)]) |Iσ (ωn)| Xn(ω ) = = ∗ ∗ P ([ωn]) |Iωn |

∗ is a constant. So {Xn(ω ) ≤ x} is the union of some collection of open sets belonging to Bn. Hence Xn is Bn-measurable. CHAPTER 5. MARTINGALES 117

∗ (3): For all n ≥ 1 and for every right-cylinder [ωn] ∈ Bn we have

Z ∗ ∗ |Iσ (ωn)| ∗ ∗ ∗ Xn dP = · |Iωn | = |Iσ (ωn)| ∗ |I ∗ | [ωn] ωn d−1 d−1 ∗ ∗ X X |Iσ (kωn)| ∗ ∗ ∗ = |Iσ (kωn)| = · |Ikωn | |Ikω∗ | k=0 k=0 n d−1 Z |I ∗ ∗ | Z |I ∗ ∗ | X σ (ωn+1) σ (ωn+1) = dP = dP ∗ |Iω∗ | ∗ |Iω∗ | k=0 [kωn] n+1 [ωn] n+1 Z = Xn+1 dP. ∗ [ωn]

This says that Xn = E(Xn+1|Bn) for all n ≥ 1. Thus (Xn)n≥1 is indeed a martingale.

Deﬁnition 33. A collection H of B-measurable functions X :Σ∗ → R is said to be uniformly integrable if Z lim sup |X| dP = 0. c→∞ X∈H {ω∗ | |X(ω∗)|≥c}

Theorem 24 (Martingale Convergence Theorem). Let (Xn)n≥1 be a martingale on (Σ∗, B,P ) and suppose the collection of B-measurable functions

{Xn}n≥1 is uniformly integrable. Then

∗ ∗ lim Xn(ω ) = X(ω ) exists P − a.e., n→∞

1 1 X ∈ L (P ), and Xn converges to X in L (P ), that is, Z lim |Xn − X| dP = 0. n→∞ Σ∗ CHAPTER 5. MARTINGALES 118

Moreover, Xn = E[X|Bn] P − a.e., that is

Z Z ∗ ∗ ∗ ∗ Xn(ω )P ([ωn]) = Xn dP = X dP, ∀ω ∈ Σ , ∀n ≥ 0. ∗ ∗ [ωn] [ωn]

In order to prove theorem 24, we will need the following deﬁnition and lemma.

Deﬁnition 34. Let (Xn)n≥1 be a submartingale, and ﬁx a, b ∈ R with a < b.

For any ω∗ ∈ Σ∗, conisder the sequence of real numbers

∗ ∗ ∗ ∗ X1(ω ),X2(ω ),X3(ω ),...Xn(ω ). (5.2)

∗ Then Un(ω ) is the number of times the sequence (5.2) first takes a value ≥ b after previously taking a value ≤ a, that is, the number of times the sequence upcrosses the interval [a, b] in its first n values. More precisely, define

A1 = min{k ≥ 1 | Xk ≤ a},

B1 = min{k ≥ A1 | Xk ≥ b},

A2 = min{k ≥ B1 | Xk ≤ a},

B2 = min{k ≥ A2 | Xk ≥ b}, etc., where min ∅ = ∞. Then

Un = max{j | Bj ≤ n}, where max ∅ = 0. CHAPTER 5. MARTINGALES 119

Lemma 12. [Doob’s Upcrossing Inequality] Let (Xn)n≥1 be a martingale as in (4.6), ﬁx a, b ∈ R with a < b, and let Un be the number of upcrossings of

[a, b] in the first n values of the sequence (Xn)n≥1 (as defined in definition

34). Then 1 E[U ] ≤ E[(X − a)+], n b − a n

+ where (Xn − a) = max{Xn − a, 0}.

+ Proof. Let Yn = (Xn −a) = max{Xn −a, 0}. Since (Xn)n≥1 is a martingale, for m ≤ n we have

Z ∗ ∗ Xm(ω )P ([ωm]) = Xn dP, ∗ [ωm] and so Z ∗ ∗ |Xm(ω ) − a|P ([ωm]) ≤ |Xn − a| dP. ∗ [ωm] Since X − a + |X − a| (X − a)+ = m m , m 2

+ + we have (Xm − a) ≤ E[(Xn − a) |Bm] for all m ≤ n. Thus (Yn)n≥0 is a submartingale.

It is straightforward to check that for all n we have

n n X X Yn = Ymin{A1,n} + (Ymin{Bi,n} − Ymin{Ai,n}) + (Ymin{Ai+1,n} − Ymin{Bi,n}). i=1 i=1 (5.3) CHAPTER 5. MARTINGALES 120

Note that Ymin{Ai,n} ≥ 0 and

n X (Ymin{Bi,n} − Ymin{Ai,n}) ≥ Un(b − a). i=1

Thus, n X Yn ≥ Un(b − a) + (Ymin{Ai+1,n} − Ymin{Bi,n}). i=1 Rearranging and and taking the expectation of each side, we obtain

n X E[Un](b − a) ≤ E[Yn] − (E[Ymin{Ai+1,n}] − E[Ymin{Bi,n}]). (5.4) i=1

Since (Yn)n≥0 is a submartingale, and since min{Ai+1, n} ≥ min{Bi, n}, we have

E[Ymin{Bi,n}] ≤ E[E[Ymin{Ai+1,n}|Bmin{Bi,n}]] = E[Ymin{Ai+1,n}].

Hence, the summation in (5.4) is non-negative, and the result follows.

Now we can proceed with the proof of theorem 24.

Proof. Since (Xn)n≥1 is uniformly integrable, for any > 0 there exists a constant c ∈ R such that

E[|Xn|1{|Xn|>c}] ≤ , ∀n ≥ 1.

Therefore, we have

E[|Xn|] = E[|Xn|1{|Xn|>c}] + E[|Xn|1{|Xn|≤c}] ≤ + c, ∀n ≥ 1. (5.5) CHAPTER 5. MARTINGALES 121

In particular, there exists a constant c0 ∈ R such that E[|Xn|] < c0 for all n ≥ 1.

For any a, b ∈ R with a < b, let Un be the number of upcrossings of the interval [a, b] as deﬁned in deﬁntition 34. For every ω∗ ∈ Σ∗, the sequence

{Un}n≥1 is monotonically increasing. Thus U(a, b) = limn→∞ Un exists. It follows from the Monotone Convergence Theorem and theorem 12 that

E[U(a, b)] = lim E[Un] n→∞

1 + ≤ sup E[(Xn − a) ] b − a n≥1 1 c0 + |a| ≤ sup E[|Xn|] + |a| ≤ b − a n≥1 b − a

Thus,

P (U(a, b) < ∞) = 1. (5.6)

Deﬁne

∗ ∗ ∗ ∗ Sa,b = {ω ∈ Σ | lim inf Xn(ω ) ≤ a and lim sup Xn(ω ) ≥ b}. n→∞ n→∞

From (5.6) we have P (Sa,b) = 0. Let

[ S = Sa,b. a < b a, b ∈ Q Then P (S) = 0. However, we have

∗ ∗ ∗ ∗ S = {ω ∈ Σ | lim inf Xn(ω ) < lim sup Xn(ω )}. n→∞ n→∞ CHAPTER 5. MARTINGALES 122

Thus, limn→∞ Xn = X exists P − a.e. In addition, it follows from Fatou’s

Lemma and (5.5) that

E[|X|] = E[lim inf |Xn|] ≤ lim inf E[|Xn|] ≤ c0. (5.7) n→∞ n→∞

1 Thus Xn converges P − a.e. to a ﬁnite limit X ∈ L (P ).

1 To show that Xn converges to X in L (P ), for c ∈ R deﬁne   −c if x < −c, fc(x) = x if − c ≤ x ≤ c  c if c < x.

Since {Xn}n≥0 is uniformly integrable, for any > 0 there exists a constant c such that

E[|Xn − fc(Xn)|] < /3, ∀n ≥ 0, and

E[|fc(X) − X|] < /3.

Since fc is continuous, we have limn→∞ fc(Xn) = fc(X) P − a.e.. Thus, it follows from Lebesgue’s Dominated Convergence Theorem that there exists

N ∈ N such that for all n ≥ N we have

E[|fc(Xn) − fc(X)|] < /3.

Combining the previous three inqualities, we have

E[|Xn − X|] < , ∀n ≥ N, CHAPTER 5. MARTINGALES 123

1 that is, Xn converges to X in L (P ).

Finally, let A ∈ Bn. Since (Xn)n≥0 is a martingale, for m ≤ n we have

Z Z Xm dP = Xn dP. A A

However,

Z Z Z

Xn dP − X dP ≤ |Xn − X| dP, (5.8) A A A which goes to 0 as n goes to ∞. Thus

Z Z Xm dP = X dP, ∀A ∈ Bm A A that is, Xm = E[X|Bm].

When the martingale (Xn)n≥1 converges to the random variable X as in

Theorem 24, we call X the limit of the martingale (Xn)n≥1.

Corollary 5. Suppose f : T → T is a circle endomorphism with bounded geometry that preserves the Lebesque probability measure m on T . Let (Xn)n≥1 as deﬁned in (4.6) be the corresponding martingale on the dual symbolic space

(Σ∗, B,P ). Then there exists a B-measurable function X ∈ L1(P ) such that

i limn→∞ Xn = X(= Xf ) exists P − a.e.,

R ii limn→∞ Σ∗ |Xn − X| dP = 0, CHAPTER 5. MARTINGALES 124

iii Xn = E[X|Bn] P − a.e., that is

Z ∗ ∗ Xn(ω )P ([ωn]) = X dP, ∗ [ωn]

and

iv there is a constant C > 1 such that 1 < X(ω∗) < C for a.e. ω∗ ∈ Σ∗.

Proof. The bounded geometry of f implies that there is a constant 1 < C <

∗ ∗ ∗ ∞ such that 1 < Xn(ω ) ≤ C for all ω ∈ Σ and all Xn belonging to the martingale (Xn)n≥1 deﬁned by (4.6). Thus, the collection of B-measurable functions {Xn}n≥1 is uniformly integrable. Actually, in our case the bounded geometry condition implies that we have a uniformly bounded martingale.

The result then follows directly from Theorem 24.

Remark 12. In our context, we can think of the limit X of the martingale

1 (Xn)n≥1 as the L -dual derivative for f.

Lemma 13. Let f and g be two circle endomorphisms that both have bounded geometry and that both preserve the Lebesque probability measure m on T .

Suppose there exists a symmetric homeomorphism h : T → T such that

∗ ∗ ∗ f ◦ h = h ◦ g. Then for any ω ∈ Σ , the limit limn→∞ Xn,f (ω ) exists if and

∗ ∗ ∗ only if the limit limn→∞ Xn,g(ω ) exists, and for all ω ∈ Σ such that either

∗ ∗ limit exists, we have Xf (ω ) = Xg(ω ). CHAPTER 5. MARTINGALES 125

Proof. Let H : R → R denote the lift of h. Let > 0 and choose N suﬃciently large such that n ≥ N implies that H is M-quasisymmetric on

each Iωn,f ∈ ηn,f , where 1 ≤ M < 1 + . For any Iωn,f ∈ ηn,f , we have

H(Iωn,f ) = Iωn,g ∈ ηn,g. (5.9)

Let Iωn,f = [x0, xd] and let Iωn,g = [y0, yd]. Deﬁne the linear maps lf :

[x0, xd] → [0, 1] and lg :[y0, yd] → [0, 1] as follows.

lf (x) = (x − x0)/(xd − x0)

lg(y) = (y − y0)/(yd − y0)

Then

˜ −1 H := lg ◦ H ◦ lf : [0, 1] → [0, 1] is an M-quasisymmetric homeomorphism. From lemma 2, we have

˜ |H(x) − x| ≤ M − 1 < . (5.10)

n+1 Let x1, . . . , xd−1 ∈ (x0, xd) be the (d − 1) points such that F (xi) ∈ Z.

∗ ∗ For any ω ∈ [ωni], 0 ≤ i ≤ d − 1, we have

∗ ∗ |Iωn,f | 1 Xn+1,f (ω ) = = . (5.11) ∗ |Iωni,f | lf (xi+1) − lf (xi)

n+1 Similarly, let y1, . . . yd−1 ∈ (y0, yd) be the (d − 1) points such that G (yi) ∈

∗ ∗ Z. For any ω ∈ [ωni], 0 ≤ i ≤ d − 1, we have

∗ ∗ |Iωn,g| 1 Xn+1,g(ω ) = = . (5.12) ∗ |Iωni,g| lg(yi+1) − lg(yi) CHAPTER 5. MARTINGALES 126

˜ ∗ ∗ Since H(lf (xi)) = lg(yi), it follows from (5.10) that for all ω ∈ [ωn], n ≥ N we have

∗ ∗ 1 1 |Xn+1,f (ω ) − Xn+1,g(ω )| = − lf (xi+1) − lf (xi) lg(yi+1) − lg(yi) |H˜ (l (x )) − l (x )| + |H˜ (l (x )) − l (x )| ≤ f i+1 f i+1 f i f i (lf (xi+1) − lf (xi))(lg(yi+1) − lg(yi)) 2 ≤ , Cf Cg where Cf and Cg are positive lower bounds for lf (xi+1)−lf (xi) and lg(yi+1)− lg(yi), respectively, provided by the bounded geometry of f and g, independent of n. Thus,

∗ ∗ ∗ ∗ lim |Xn,f (ω ) − Xn,g(ω )| = 0 ∀ω ∈ Σ , (5.13) n→∞ and the result follows from Corollary 5.

Theorem 25. Let f and g be two circle endomorphisms that both have bounded geometry and that both preserve the Lebesque probability measure m on T . Suppose there exists a symmetric homeopmorphism h : T → T such

∗ that f ◦h = h◦g. Then there exists a subset A ⊂ Σ with Pf (A) = Pg(A) = 1

∗ ∗ ∗ such that Xf (ω ) = Xg(ω ) for all ω ∈ A. CHAPTER 5. MARTINGALES 127

Proof. Deﬁne

∗ ∗ ∗ Af = {ω ∈ Σ | lim Xn,f (ω ) exists} n→∞

∗ ∗ ∗ Ag = {ω ∈ Σ | lim Xn,g(ω ) exists} n→∞

A = Af ∩ Ag.

From Corollary 5, we have Pf (Af ) = 1 and Pg(Ag) = 1. From Lemma 13,

∗ ∗ ∗ we have A = Af = Ag, and for any ω ∈ A we have Xf (ω ) = Xg(ω ).

Remark 13. A problem relating to Conjecture 1 is to show that Pf and Pg are equivalent, that is, there is a constant C ≥ 1 such that

−1 C Pf (A) ≤ Pg(A) ≤ CPf (A) for all Borel subsets A in Σ∗. Conjecture 1 is equivalent to C = 1.

5.2 Locally constant limits and ﬁnite martingales

In this section, we discuss those martingales whose limits satisfy Deﬁni- tion 31, that is, whose limits are locally constant.

Deﬁnition 35. A martingale (Xn)n≥1 with a ﬁltration (Bn)n≥1 on the prob-

∗ ability space (Σ , B,P ) is said to be ﬁnite if there is an integer n0 ≥ 1 such CHAPTER 5. MARTINGALES 128

that Xn = Xn0 for all n ≥ n0. We call the smallest such integer n0 the length of (Xn)n≥1.

It is clear that given a uniformly bounded martingale, if it is ﬁnite then the limit X is a locally constant function on Σ∗. The converse is also true.

Theorem 26. Suppose (Xn)n≥1 with the ﬁltration (Bn)n≥1 is a uniformly bounded martingale on the probability space (Σ∗, B,P ). Then it is ﬁnite of length n0 ≥ 1 if and only if its limit X is a locally constant random variable

∗ ∗ ∗ on Σ with X(ω ) = c(ωn0 ).

Proof. We need only to prove the “if” part. Suppose (Xn)n≥1 satisﬁes Def- inition 31. Since it is uniformly bounded, Theorem 24 says that we have

∗ ∗ ∗ a limit X of (Xn)n≥1 on Σ and X(ω ) = c(ωn0 ) for any n ≥ n0 and a.e.

∗ ∗ ∗ ω = . . . jn−1 . . . j1j0 ∈ Σ with ωn = jn−1 . . . j1j0. Using Theorem 24 again, we have that

∗ ∗ ∗ Xn(ω ) = E(X|Bn)(ω ) = c(ωn0 )

∗ ∗ for a.e. ω ∈ Σ . This says that (Xn)n≥n0 is ﬁnite.

A circle endomorphism that has bounded geometry may not be uniformly quasisymmetric. For a circle endomorphism, an equivalent condition to that of uniformly quasisymmetric is the condition of bounded nearby geometry, deﬁned below (refer to [8, 9] for more). CHAPTER 5. MARTINGALES 129

Deﬁnition 36. Suppose f is a circle endomorphism of degree d ≥ 2 and suppose (ξn)n≥1 is the corresponding sequence of Markov partitions on T .

We say f has bounded nearby geometry if there is a constant C ≥ 1 such that 1 m(J ) ≤ ωn ≤ C 0 C m(Jωn )

0 for every pair of arcs Jωn ,Iωn ∈ ξn that share a common endpoint, for any n ≥ 1.

We have the following theorem to characterize the bounded nearby geometry for a circle endomorphism whose corresponding martingale is ﬁnite.

Theorem 27. Suppose f is a circle endomorphism of degree d ≥ 2 that has bounded geometry and preserves the Lebesgue probability measure m on

∗ T . Suppose (Xn)n≥1 is the corresponding martingale on (Σ , B,P ). Suppose

(Xn)n≥1 is ﬁnite of length n0 ≥ 1. Then f has bounded nearby geometry if and only if

∗ ∗ Xn (... 0 ... 0 ) = Xn (... (d − 1) ... (d − 1) ). (5.14) 0 | {z } 0 n0 | {z } n0

Proof. Let (ηn)n≥1 be the corresponding sequence of Markov partitions for f on [0, 1]. By the deﬁnition, |I0 ... 0| ∗ | n{z−1 } Xn(... 0 ... 0 ) = | {z } |I | n 0 ... 0 | {zn } CHAPTER 5. MARTINGALES 130 and |I(d − 1) ... (d − 1)|

∗ | n{z−1 } Xn(... (d − 1) ... (d − 1) ) = . | {z } |I | n (d − 1) ... (d − 1) | {zn }

Since (Xn)n≥1 is ﬁnite of length n0, we have

∗ ∗ Xn(... 0 ... 0 ) = Xn (... 0 ... 0 ) | {z } 0 | {z } n n0 and

∗ ∗ Xn(... (d − 1) ... (d − 1) ) = Xn0 (... (d − 1) ... (d − 1) ) | {z } | {z } n n0 for all n ≥ n0. Hence n X (... (d − 1) ... (d − 1)∗) |I0 ... 0| |I0 ... 0| n0 | {z } | {z }  | {z }  n+n0 n0  n0  = ∗ . |I | |I |  Xn (... 0 ... 0 )  (d − 1) ... (d − 1) (d − 1) ... (d − 1)  0 | {z }  | {z } | {z } n0 n+n0 n0

If f has bounded nearby geometry, then we have a constant C0 ≥ 1 such that |I0 ... 0| | {z } 1 n+n0 ≤ ≤ C0 C0 |I(d − 1) ... (d − 1)| | {z } n+n0 for all n ≥ 1. This implies (5.14).

If (5.14) holds, then the ratios of all pairs of intervals in ξn with common endpoints belong to a ﬁnite collection of ﬁxed positive numbers for all n ≥ n0.

So f has bounded nearby geometry. CHAPTER 5. MARTINGALES 131

−1 Theorem 28. Suppose f = h ◦ qd ◦ h is a circle endomorphism that preserves the Lebesgue probability measure m and has bounded geometry. Here h is a circle homeomorphism. Suppose (Xf,n)n≥1 is the corresponding martingale. For any k ≥ 1, we deﬁne a new ﬁnite martingale Yn = Xf,min{n,k} of length k. Then we can construct a unique circle endomorphism fk =

−1 hk ◦ qd ◦ hk that preserves the Lebesgue probability measure and has bounded

geometry whose corresponding martingale (Xfk,n = Yn)n≥1 is ﬁnite. Here hk is a circle homeomorphism. Then fk (respectively, hk) tends to f (respectively, h) uniformly on T as k → ∞.

Proof. Let F,Fk,H,Hk : R → R be the lifts of f, fk, h, and hk, respectively.

Since fk is the circle endomorphism generated by the ﬁnite martingale Yn of

length k, the ﬁrst k Markov partitions ξfk,n = ξf,n are identical for n = 1,

..., k. This implies that

max |F (x) − Fk(x)| ≤ τk = max {|Iωk |}, x∈[0,1] ωk∈Σk and

max |H(x) − Hk(x)| ≤ τk = max {|Iωk |}. x∈[0,1] ωk∈Σk

Because of the bounded geometry, τk → 0 as k → ∞. This completes the proof. CHAPTER 5. MARTINGALES 132

Remark 14. In the case that f is a uniformly quasisymmetric circle endomorphism (i.e. f has bounded nearby geometry), it does not necessarily follow that fk is uniformly quasisymmetric (i.e., fk has bounded nearby geometry) for any k ≥ 1. This is because of Theorem 27 and the fact that in general, we do not have

∗ ∗ Xf,k(... 0 ... 0 ) = Xf,k(... (d − 1) ... (d − 1) ). | {z } | {z } k k for a general uniformly quasisymmetric circle endomorphism f.

A trivial example of a circle endomorphism having ﬁnite martingale is

d the map q(z) = z . The corresponding martingale (Xn = d)n≥1 is ﬁnite of length n0 = 1.

Suppose f is a circle endomorphism of degree 2 that preserves the Lebesgue probability measure m on T . Then for any ωn ∈ Σn, n ≥ 0, we have

|Iωn | = |Iωn0| + |Iωn1|, (5.15) and

|Iωn | = |I0ωn | + |I1ωn |. (5.16)

Thus, given the lengths of all intervals belonging to ηn+1 and just the lengths

of intervals belonging to ηn+2 of the form I0ωn0, the lengths of the remaining intervals in ηn+2 are determined by the following equations:

|I0ωn1| = |I0ωn | − |I0ωn0|, (5.17) CHAPTER 5. MARTINGALES 133

|I1ωn0| = |Iωn0| − |I0ωn0|, (5.18)

|I1ωn1| = |Iωn1| − |I0ωn | + |I0ωn0|. (5.19)

In order for the sequence {ηn}n≥1 to have bounded geometry with a bound

C, I0ωn0 has to satisfy the following inequalities:

max{0, |I0ωn | − |Iωn1|} < |I0ωn0| < min{|I0ωn |, |Iωn0|},

1 C − 1 |I | ≤ |I | ≤ |I |, C 0ωn 0ωn0 C 0ωn

1 C − 1 |I | ≤ |I | ≤ |I |. C 1ωn 1ωn0 C 1ωn

This is equivalent to

C − 1 1 |I | − |I | ≤ |I | ≤ |I | − |I |, (5.20) ωn0 C 1ωn 0ωn0 ωn0 C 1ωn which is equivalent to the bounded geometry condition (19).

Equations (5.17)-(5.20) summarize one way in which one can create examples of degree 2 circle endomorphisms preserving Lebesgue measure m on

T with bounded geometry. We give two such examples, both with ﬁnite martingales of length n0 = 3 and with bounded geometry constant C = 4. One has only bounded geometry (not bounded nearby geometry), thus it is not uniformly quasisymmetric. The other has bounded nearby geometry, thus it is uniformly quasisymmetric. CHAPTER 5. MARTINGALES 134

Example 2. The ﬁrst example has only bounded geometry which is illustrated in Figure 5.1 with the corresponding lengths of intervals. Since X2 = 4 = C

4 1 C on I00 and X2 = 3 = 1 = C−1 on I01, to satisfy (5.20), X3 has to be 4 and 1− C 4 3 on the preimages of these intervals, respectively. Since X3(000) 6= X3(111), theorem 27 implies the circle endomorphism deﬁned by these partitions is not uniformly quasisymmetric.

Example 3. The second example has bounded nearby geometry which is il- lustratred in Figure 5.2 with the corresponding lengths of intervals. Since

X3 = 4 on I000 and I111 and Xn = X3 for all n ≥ 3, it has bounded nearby geometry from Theorem 27. CHAPTER 5. MARTINGALES 135

I=[0,1] I=[0,1]

1/3 2/3 X1 =3 X1 =3/2

1/12 1/4 1/4 5/12 X2 =4 X2 =4/3 X2 =8/3 X2 =8/5

1/481/16 1/8 1/8 1/16 3/16 1/8 7/24 X3 =4 4/3 2 2 4 4/3 10/3 10/7

X = 1/192 1/32 1/32 3/80 1/64 3/32 1/32 7/80 4 4 2 4 10/3 4 2 4 10/3

4/3 2 10/7 1/64 1/32 3/32 7/80 3/64 3/32 3/32 49/240 4/3 2 4/3 10/7 4/3 1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 5.1: Finite Martingale of length n0 = 3 and C = 4. The top left shows the lengths of the intervals belonging to ηn and the top right shows the values of Xn on intervals belonging to ηn, for n = 1, 2, 3, 4. The bottom left shows the distribution function h and the bottom right shows the circle −1 endomorphism f, with f = h ◦ q2 ◦ h . The circle endomorphism f has bounded geometry but does not have bounded nearby geometry, thus it is not uniformly quasisymmetric. CHAPTER 5. MARTINGALES 136

I=[0,1] I=[0,1]

4/9 5/9 X1 =9/4 X1 =9/5

1/9 1/3 1/3 2/9 X2 =4 X2 =4/3 X2 =5/3 X2 =5/2

1/361/12 1/6 1/6 1/12 1/4 1/6 1/18 X3 =4 4/3 2 2 4 4/3 4/3 4

X = 1/144 1/24 1/24 1/8 1/48 1/8 1/24 1/24 4 4 2 4 4/3 4 2 4 4/3

4/3 2 4 1/48 1/24 1/8 1/24 1/16 1/8 1/8 1/72 4/3 2 4/3 4 4/3 1 1

0.8 0.8

0.6 0.6

0.4 0.4

0.2 0.2

0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 5.2: Another ﬁnite Martingale of length n0 = 3 and C = 4. The top left shows the lengths of the intervals belonging to ηn and the top right shows the values of Xn on intervals belonging to ηn, for n = 1, 2, 3, 4. The bottom left shows the distribution function h and the bottom right shows −1 the circle endomorphism f, with f = h ◦ q2 ◦ h . The circle endomorphism f has bounded geometry and bounded nearby geometry, thus it is uniformly quasisymmetric. CHAPTER 5. MARTINGALES 137 5.3 A symmetric rigidity theorem assuming locally constant limit of martingale

Theorem 29. Suppose f and g are both circle endomorphisms of the same topological degree d ≥ 2 such that each has bounded geometry and preserves the Lebesgue probability measure m on T . Suppose f has a locally constant limit of martingale and h is a symmetric conjugacy between f and g. Then h must be the identity.

∗ ∗ Proof. Let n0 be the length for the martingale (Xf,n)n≥1. For every ω ∈ Σ ,

∗ ∗ write ω = . . . jnjn−1 . . . j1j0 and ωn = jn−1 . . . j1j0.

Theorem 25 and Deﬁnition 31 imply that

∗ ∗ ∗ Xf (ω ) = Xg(ω ) = c(ωn0 )

n0 ∗ for d constants c(ωn0 ). Theorem 26 implies that

∗ ∗ ∗ Xn,f (ω ) = Xn,g(ω ) = c(ωn0 ) ∀n ≥ n0. (5.21)

Let I be any interval belonging to the partition η . For any wn0 ,f n0,f

wn0+m = wn0 in0 . . . in0+m−1, m ≥ 1, we have

I ⊂ I . wn0+m,f wn0 ,f CHAPTER 5. MARTINGALES 138

From (5.21) we have

|Iw ,f | |F (Iw ,f )| n0+m−1 = n0+m−1 , |I | |F (I )| ωn0+m,f ωn0+m,f and it follows that

|F (I )| |F (I )| |F (I )| wn0+m,f wn0+m−1,f wn0 ,f = = ... = := aw . (5.22) |I | |I | |I | n0 ωn0+m,f ωn0+m−1,f ωn0 ,f The set of endpoints of I for all m ≥ 1 forms a dense subset of I . wn0+m wn0

This implies that

F |I (x) = aω x + bω , ωn0 n0 n0 where {b } are dn0 constants. In other words, the circle endomorphism F ωn0 is a piecewise linear function. Similarly, it follows from (5.21) that for all m ≥ 1 we have

|Iw ,g| |G(Iw ,g)| n0+m−1 = n0+m−1 , |I | |G(I )| ωn0+m,g ωn0+m,g and so

|G(I )| |G(I )| |G(I )| wn0+m,g wn0+m−1,g wn0 ,g 0 = = ... = := aw . |I | |I | |I | n0 ωn0+m,g ωn0+m−1,g ωn0 ,g This implies that

0 0 G|I (x) = a x + b , ωn0 ωn0 ωn0 where {b0 } are dn0 constants. In other words, both F and G are piecewise- ωn0 linear functions. Moreover, from (5.9) and (5.21) we have

|H(Iw )| |Iw | |Iw | n0+m−1,f = n0+m−1,g = n0+m−1,f = c, |H(I )| |I | |I | wn0+m,f wn0+m,g wn0+m,f CHAPTER 5. MARTINGALES 139

∗ ∗ ∗ for some c ∈ {c(ωn0 ) | ωn0 ∈ Σn0 }. This implies that

|H(I )| |H(I )| |H(I )| wn0+m,f wn0+m−1,f wn0,f = = ... = := dw |I | |I | |I | n0 wn0+m,f wn0+m−1,f wn0 ,f

Thus we have

H(x) = d x + e , ∀x ∈ I . wn0 n0 wn0

From the deﬁnition of a symmetric homeomorphism, we see that d = d wn0 and e = e for all w . But H(0) = 0 and H(1) = 1 implies that d = 1 and wn0 n0 e = 0. Thus h is the identity. Chapter 6

Appendix

6.1 Bernoulli measure clf; alpha=.8 %set the value of alpha here

M=20 %set number of Markov partition

%levels in construction -

%2^M points will be plotted subplot(1,2,1) for N=1:M

x=linspace(0,1,2^N+1);

y=zeros(size(x));

y(2^(N-1)+1)=alpha;

y(2^N+1)=1;

140 CHAPTER 6. APPENDIX 141

for n=1:N

for k=1:2^(n-1)

y((2*k-1)*2^(N-n)+1)=

(1-alpha)*y((k-1)*2^(N-n+1)+1)

+alpha*y((k)*2^(N-n+1)+1);

end

plot(x,y); %plot each iteration

hold on; end axis equal; axis([0 1 0 1]); subplot(1,2,2) plot(x,y,’k’); axis equal; axis([0 1 0 1]);

6.2 Bounded geometry measure clf; aub=.9 %alpha upper bound CHAPTER 6. APPENDIX 142 alb=.1 %alpha lower bound

N=20 %number of levels in construction

%2^N points will be plotted x=linspace(0,1,2^N+1); y=zeros(size(x)); y(2^N+1)=1; alpha=zeros(size(x));

%level 1 is set half=2^(N-1)+1; alpha(half)=rand*(aub-alb)+alb; y(half)=alpha(half);

for n=2:N

for k=1:2^(n-2)

stepsize=2^(N-n);

%set coordinates of position and twin

%two preimages of the same point under the doubling map

position=(2*k-1)*stepsize+1; CHAPTER 6. APPENDIX 143

posleft=position-stepsize;

posright=position+stepsize;

twin=position+2^(N-1);

twinleft=twin-stepsize;

twinright=twin+stepsize;

%begin with alpha value determined by the image

image=2*(position-1)+1;

a=alpha(image);

%set initial values

y(position)=(1-a)*y(posleft)+a*y(posright);

y(twin)=(1-a)*y(twinleft)+a*y(twinright);

%epsilon

posuptwindown=[aub*y(posright)+(1-aub)*y(posleft)

-y(position),

y(twin)-(alb*y(twinright)+(1-alb)*y(twinleft))];

maxepsilon=min(posuptwindown); CHAPTER 6. APPENDIX 144

posdowntwinup=[alb*y(posright)+(1-alb)*y(posleft)

-y(position),

y(twin)-(aub*y(twinright)+(1-aub)*y(twinleft))];

minepsilon=max(posdowntwinup);

epsilon=minepsilon+rand*(maxepsilon-minepsilon);

%set final values

y(position)=y(position)+epsilon;

y(twin)=y(twin)-epsilon;

%record alpha values

alpha(position)=(y(position)-y(posleft))/(y(posright)

-y(posleft));

alpha(twin)=(y(twin)-y(twinleft))/(y(twinright)

-y(twinleft));

end CHAPTER 6. APPENDIX 145 end

%plot the homeomorphism subplot(1,2,1); plot(x,y,’k’); axis equal; axis([0 1 0 1]); hold on;

%create endo through conjugation yH=zeros(size(x)); for k=1:2^(N-1)+1

yH(k)=y(2*(k-1)+1); end for k=2^(N-1)+2:2^N+1

yH(k)=y(2*(k-1)-2^N+1)+1; end

%plot endo subplot(1,2,2); CHAPTER 6. APPENDIX 146 plot(y,yH,’k’); hold on; plot(y,yH-1,’k’); axis equal; axis([0 1 0 1]);

6.3 Dual measure and dual endomorphism clf;

aub=.9; %alpha upper bound (Bounded Geometry constant) alb=.1; %alpha lower bound

N=20; %number of levels in construction

%2^N points will be plotted

x=linspace(0,1,2^N+1); y=zeros(size(x)); y(2^N+1)=1; CHAPTER 6. APPENDIX 147 alpha=zeros(size(x));

%level 1 is set half=2^(N-1)+1; alpha(half)=rand*(aub-alb)+alb; y(half)=alpha(half);

for n=2:N

for k=1:2^(n-2) %2^n endpoints in level n

%2^(n-1) NEW endpoints

%and their values are assigned in PAIRS

%hence we do do this in 2^(n-2) steps

stepsize=2^(N-n);

%set coordinates of position and twin

%two preimages of the same point under the doubling map

position=(2*k-1)*stepsize+1;

posleft=position-stepsize; CHAPTER 6. APPENDIX 148

posright=position+stepsize;

twin=position+2^(N-1);

twinleft=twin-stepsize;

twinright=twin+stepsize;

%begin with alpha value determined by the image

image=2*(position-1)+1;

a=alpha(image);

%set initial values

y(position)=(1-a)*y(posleft)+a*y(posright);

y(twin)=(1-a)*y(twinleft)+a*y(twinright);

%epsilon

posuptwindown=[aub*y(posright)+(1-aub)*y(posleft)-y(position)

y(twin)-(alb*y(twinright)+(1-alb)*y(twinleft))];

maxepsilon=min(posuptwindown);%positive number

posdowntwinup=[alb*y(posright)+(1-alb)*y(posleft)-y(position) CHAPTER 6. APPENDIX 149

y(twin)-(aub*y(twinright)+(1-aub)*y(twinleft))];

minepsilon=max(posdowntwinup);%negative number

epsilon=minepsilon+rand*(maxepsilon-minepsilon);

%set final values

y(position)=y(position)+epsilon;

y(twin)=y(twin)-epsilon;

%record alpha values

alpha(position)=(y(position)-y(posleft))/(y(posright)-y(posleft));

alpha(twin)=(y(twin)-y(twinleft))/(y(twinright)-y(twinleft));

end end

%create endo through conjugation yH=zeros(size(x));

HyH=zeros(size(x));

HyHmod=zeros(size(x)); CHAPTER 6. APPENDIX 150 for k=1:2^(N-1)+1

yH(k)=y(2*k-1); end for k=2^(N-1)+2:2^N+1

yH(k)=y(2*(k-1)-2^N+1)+1; end

%Construct dual map dualy=zeros(size(x)); dualy(2^N+1)=1; for k=2:2^N

subset=k-2;

%subsetbinary=de2bi(subset,N);

subsetbinary=zeros(N);

s=subset;

%creat binary array

for i=1:N

remainder=mod(s,2^i);

subsetbinary(i)=min(1,remainder);

s=s-remainder; CHAPTER 6. APPENDIX 151

end

dualsubset=0;

for i=1:N

dualsubset=dualsubset+2^(N-i)*subsetbinary(i);

end

dualsubsetlength=y(dualsubset+2)-y(dualsubset+1);

dualy(k)=dualy(k-1)+dualsubsetlength; end

%create dual endo through conjugation dualyH=zeros(size(x)); for k=1:2^(N-1)+1

dualyH(k)=y(2*k-1); end for k=2^(N-1)+2:2^N+1

dualyH(k)=y(2*(k-1)-2^N+1)+1; end

% blue=h, red=h^* dual, black=h^*(h^{-1})dual homeo subplot(1,2,1); CHAPTER 6. APPENDIX 152 plot(x,y,’b’); hold on; plot(x,dualy,’r’);

%hold on; plot(y,dualy,’k’); axis equal; axis([0 1 0 1]);

% blue=endo, red=dual endo subplot(1,2,2); plot(y,yH,’b’); hold on; plot(y,yH-1,’b’); plot(dualy,dualyH,’r’); plot(dualy,dualyH-1,’r’); axis equal; axis([0 1 0 1]); Bibliography

[1] L. Ahlfors, Lectures on quasiconformal mappings, University Lecture Series 38, 2nd edition. American Mathematical Society, Providence, RI, 2006. [2] J. Bak, D. Newman, Complex Analysis, Springer, 3rd edition. 2010. [3] H. Furstenberg, Disjointness in Ergodic Theory, Minimal Sets, and a Problem in Diophantine Approximation, Mathematical Systems Theory, Springer-Verlag, 1967. [4] Y. Jiang, Symmetric Invariant Measures, Contemporary Mathematics, AMS, Vol. 575, 2012, 211-218. [5] W. Rudin, Real and Complex Analysis, McGraw-Hill, 1974. [6] R. Salem, On Some Singular Monotonic Functions which are Strictly Increasing. Transactions of the American Mathematical Society, Vol. 53, No. 3, 1943, pp. 427-439. [7] Y. Hu, Y. Jiang, and Z. Wang, Martingales for quasisymmetric systems and complex manifold structures. Annales Academiæ Scientiarum Fennicæ Mathematica, Volumen 38, 2013, 1-26. [8] Y. Jiang, Renormalization and Geometry in One-Dimensional and Com- plex Dynamics. Advanced Series in Nonlinear Dynamics, Vol. 10 (1996) World Scientiﬁc Publishing Co. Pte. Ltd., River Edge, NJ, xvi+309 pp. ISBN 981-02-2326-9. [9] Y. Jiang, Lecture Notes in Dynamical Systems and Quasiconformal Mappings: A Course Given in Department of Mathematics at CUNY Graduate Center, Spring Semester of 2009.

153 BIBLIOGRAPHY 154

[10] S. R.S. Varadhan, Probability Theory. Courant Lecture Notes 7, AMS and CIMS, 2001.

[11] H. L. Royden, Real Analysis. Second Edition. The Macmillan Company, New York.

[12] M. Shub, Endomorphisms of Compact Diﬀerentiable Manifolds. Amer- ican Journal of Mathematics, Vol. 91, No. 1 (Jan., 1969), pp. 175-199

[13] Y. Jiang, Geometric Gibbs Theory. Sscience China, Mathematics, to appear in 2020.

[14] A. Quas, Non-ergodicity for expanding maps and g-measures. Ergodic Theory and Dynamical Systems, 16. June 1996.

[15] D. Rudolph, ×2 and ×3 invariant measures and entropy. Ergod. Th. & Dynam. Sys. 10 (1990), 395-406.

[16] P. Walters, An Introduction to Ergodic Theory, Graduate Texts in Math- ematics, 79, Springer-Verlag, New York-Heidelberg-Berlin, 1982.

[17] R. Lyons. On measures simultaneously 2- and 3-invariant. Israel Journal of Mathematics, 61, (1988), 219-224.

[18] R.Mañé, Ergodic Theory and Differentiable Dynamics, Springer-Verlag, Berlin-Heidelberg-New York, 1987.

[19] S. T. Liao, An ergodic property theorem for a diﬀerential system, Sci- entia Sinica 16 (1973), 1–24.

[20] S. T. Liao, On characteristic exponents construction of a new Borel set for the multiplicative ergodic theorem. Acta Sci. Natur. Universitatis Pekinensis, 29 (1993), 277–302.

[21] A. Johsson, Measures on the circle invariant under multiplication by a nonlacunary subsemigroup of the integers. Israel Journal of Mathemat- ics, 77 (1992), 211-240.

[22] Y. Jiang, Lecture Notes in Dynamical Systems and Quasiconformal Mappings: A Course Given in Department of Mathematics at CUNY Graduate Center, Spring Semester of 2009. BIBLIOGRAPHY 155

[23] B. Host, Nombres normaux, entropie, translations. Israel Journal of Mathematics, 91 (1995), 419-428.