Fractals and Billiard Orbits on Sierpinski...

Fractals and Billiard Orbits on Sierpinski Carpets

Department of Mathematics, Linköping University

Erik Landstedt

LiTH-MAT-EX–2017/05–SE

Credits: 16 hp

Level: G2

Supervisor: Milagros Izquierdo,Department of Mathematics, Linköping University

Examiner: Göran Bergqvist,Department of Mathematics, Linköping University

Linköping: June 2017

Abstract

This Bachelor’s thesis deals with fractals and orbits on Sierpinski carpets. Wepresent the fundamental theory regarding fractals and some illustrative exam-ples together with fractal billiards. In the latter part of the thesis we use el-ementary methods to present an original proof concerning the closure of somebilliard orbits on Sierpinski carpets. A survey of the article Periodic Billiardorbits of self-similar Sierpinski Carpets, see [8], has been done, in which wemake a discussion about one open question regarding reflections on the car-pet. Furthermore, we state and prove some propositions related to this openquestion.

Keywords:Fractals, Geometry, Billiards, Orbits, Surfaces, Translation surfaces, Home-omorphisms.

URL for electronic version:http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva-138562

Landstedt, 2017. iii

http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva-138562

Acknowledgements

Initially, I would like to express my warmest and most sincere gratitude toProfessor Milagros Izquierdo, my supervisor, who has been my guide during thisamazing journey. She showed me the beauty of fractals and of geometry ingeneral and without her enthusiasm and energy this work had never reachedits full potential. I would also like to thank my examiner, Professor GöranBergqvist, for many useful comments and ideas regarding this work.

For many interesting thoughts and ideas I would like to express my deepestthanks to Patrik Petters, my opponent. He has given me many good suggestionsregarding design, pedagogical formulations and, of course, the contents of thethesis. It would be a lie to say that I completed this work on my own. I would liketo illuminate my marvelous classmates Tobias Wängberg, Mikael Böörs, JohanPersson, Ludwig Rahm, Erik Sätterqvist, Rebecca Bauer and André Malm formany helpful discussions regarding mathematics. Thank you all!

Landstedt, 2017. v

Landstedt, 2017. vii

viii

Nomenclature

T Topology.Sn The unit sphere of dimension n.(X, d) Metric space, sometimes written as just X.H(X) The hyperspace of a metric space X.[A,B,C,D] The cross ratio of the distinct points A,B,C and D.dH(x, y) The Hausdorff metric.dhyp(x, y) The hyperbolic metric.C The expanded complex plane, i.e. C = C ∪ ∞.C The Cantor set.A Complex atlas.C Open cover.Hc Hilbert cube.J(x) The Jacobian matrix.M, U , X, E, F Sets.∂U The boundary of the set U .(X, d, µ) Measure space.Γ(E,F ) The family of all paths in X that connect E and F where E,F ⊂ X.L(E) The Lebesgue measure of a set E.fw Weierstrass function.dimH Hausdorff dimension.Cov Lebesgue dimensionind Urysohn-Menger dimension.dimB Minkowski-Bouligand dimension.Sa The Sierpinski carpet with odd parameter a ≥ 3.Sl(Sa) The set of all slopes of nontrivial line segments possible on Sa.σ Slope of nontrivial segment.Aa Set of nontrivial line segments in Sa.θ0 The tangential angle for a point x0.fD A billiard function.Ω(D) A planar billiard.O(x0, θ0) An orbit of a planar billiard.∆k Polygon.S(D) Translation surface.b·c Lower integer part of a fraction.Fn Farey sequence of order n.

Contents

1 Introduction 1

2 Preliminaries: Topology 52.1 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3.1 Topological Surfaces . . . . . . . . . . . . . . . . . . . . . 212.4 Measure spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Fractal Theory 293.1 Metric Spaces and Fractals . . . . . . . . . . . . . . . . . . . . . 293.2 Some Examples of Fractals . . . . . . . . . . . . . . . . . . . . . 30

3.2.1 Weierstrass Fractal Curve . . . . . . . . . . . . . . . . . . 313.2.2 The Koch Curve . . . . . . . . . . . . . . . . . . . . . . . 373.2.3 Julia Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2.4 Heighway’s Dragon . . . . . . . . . . . . . . . . . . . . . . 42

3.3 Fractal Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 433.3.1 Lebesgue Dimension . . . . . . . . . . . . . . . . . . . . . 443.3.2 Urysohn-Menger Dimension . . . . . . . . . . . . . . . . . 453.3.3 Minkowski-Bouligand Dimension . . . . . . . . . . . . . . 483.3.4 Hausdorff Dimension . . . . . . . . . . . . . . . . . . . . . 493.3.5 Homeomorphisms and the Standard Sierpinski Carpet . . 50

4 Fractal Billiards 534.1 Billiards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.2 Unfolding of Billiard Orbits . . . . . . . . . . . . . . . . . . . . . 624.3 Fractal Billiards . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.3.1 Properties of Slopes on the Sierpinski Carpet . . . . . . . 65

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x CONTENTS

5 Results: Periodicity of Orbits on Sierpinski Carpets 77

A Codes Used 103

B Closed Orbits on Sierpinski Carpets 105

List of Figures

2.1 A Hyperbolic Triangle . . . . . . . . . . . . . . . . . . . . . . . . 82.2 The standard Sierpinski carpet . . . . . . . . . . . . . . . . . . . 132.3 The Cantor set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Illustration of stereographic projection . . . . . . . . . . . . . . . 182.5 Illustration of closed intervals on the Cantor set . . . . . . . . . . 202.6 Illustration of holomorphically compatible maps . . . . . . . . . . 232.7 Lebesgue measure . . . . . . . . . . . . . . . . . . . . . . . . . . 252.8 The diameter of a set . . . . . . . . . . . . . . . . . . . . . . . . 272.9 Hausdorff measure . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.1 Weierstrass function . . . . . . . . . . . . . . . . . . . . . . . . . 323.2 The von Koch curve . . . . . . . . . . . . . . . . . . . . . . . . . 373.3 A Julia set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.4 Another Julia set . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.5 The Mandelbrot set . . . . . . . . . . . . . . . . . . . . . . . . . 413.6 A sketch of the Heighway curve . . . . . . . . . . . . . . . . . . . 42

4.1 All possible nontrivial line segments in S3 . . . . . . . . . . . . . 544.2 Mathematical billiard . . . . . . . . . . . . . . . . . . . . . . . . 564.3 Billiard with a closed orbit . . . . . . . . . . . . . . . . . . . . . 584.4 Mathematical billiard with a singular orbit . . . . . . . . . . . . 594.5 A translation surface . . . . . . . . . . . . . . . . . . . . . . . . . 604.6 Mathematical billiard . . . . . . . . . . . . . . . . . . . . . . . . 614.7 Unfoldning of a closed billiard . . . . . . . . . . . . . . . . . . . . 624.8 Illustration of two compatible initial conditions for two orbits . . 64

5.1 Billiard orbit that hits a corner point . . . . . . . . . . . . . . . . 825.2 A billiard orbit not following L1 . . . . . . . . . . . . . . . . . . 835.3 A billiard orbit in a cell of S3 . . . . . . . . . . . . . . . . . . . . 95

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xii LIST OF FIGURES

5.4 Orbits with a new reflection line . . . . . . . . . . . . . . . . . . 955.5 Example of orbit with a nontrivial segment as relfection line . . . 96

B.1 Part of an orbit of a billiard . . . . . . . . . . . . . . . . . . . . . 105B.2 A grid that covers a billiard . . . . . . . . . . . . . . . . . . . . . 106B.3 A closed orbit of a billiard . . . . . . . . . . . . . . . . . . . . . . 106B.4 The grid of a closed orbit of a billiard . . . . . . . . . . . . . . . 107

Chapter 1

Introduction

Analysis and geometry are concepts that span wide across the field of math-ematics, and one of the areas that is overlapped by both of these is fractaltheory. A fractal is a mathematical object that can intuitively be seen as a setconsisting of self-repeating patterns. With this informal definition, we can soonunderstand that it is possible to create many different mathematical objects thatmimic fractals. For example, consider an equilateral triangle in the Euclideanplane. Divide one of its sides into three equal segments. Then, construct anew equilateral triangle having the middle line segment as its base and assumethat it lies in the exterior of the first triangle. Finally, remove the baseline andthe middle segment and repeat the following instructions iteratively for eachnew triangle created. For a more complete description see e.g. [31]. When thenumber of iterations approaches infinity the line segments will converge to ageometric figure that has infinite perimeter, but finite area. The appearance ofthis curve will resemble a snowflake if we start with an equilateral triangle andhas therefore been named the Koch Snowflake after the Swedish mathematicianHelge von Koch, who introduced this fractal in an article in 1904.

Fractals can be represented as geometrical objects and have therefore also adimension which, however, is not easy to define neither topologically nor met-rically. The problem is that fractals have a complex geometrical structure, andthey are defined as sets whose topological dimensions differ from the Euclideanones, see [34]. An object that has fractal structure requires a different approachwhen it comes to dimension measurement. For instance, the topological dimen-sion of the empty set is by definition equal to −1. This gives rise to a dimensionthat does not necessarily have to be an integer. Other types of dimensionalconcepts proved useful, e.g. the Hausdorff dimension, to determine fractals di-mension. Describing a fractal in a geometric manner requires some tools. One

Landstedt, 2017. 1

2 CHAPTER 1. INTRODUCTION

approach is to study the resemblance between fractals and various dynamicalsystems. In M. Lapidus and R. G. Niemeyer’s [21] reasoning, we consider abilliard table that is located in the plane in which the edges of the table aredefined by a polygon in such a way that it makes the set of all points within thepool table a compact set. Now let a billiard ball travel in this compact set inany, known, direction and at a constant speed. Then define a reflection functionthat acts on the edges of the table, which makes the ball reflect in the same wayas light reflects in a mirror. Assume that the ball hits the edge such that theimpact is fully elastic. By allowing many balls to travel on this imaginary pooltable, we have created a dynamical system. This can be generalized mathemat-ically. The pool table just mentioned is a compact set where the boundary iscomposed of an arbitrary polygon. The set is generally called a planar billiardand may be referred as Ω(D), with D as its boundary. If we instead let D be afractal set and regard the billiard balls moving in different directions, interestingchanges between each iteration in the system will occur. For instance, outcomeangles change and the number of collisions will increase. The dynamics of thesystem is transformed in contrast to the previous one.

We may consider that fractal mathematics was illuminated by the Germanmathematician Karl Weierstrass, who gave name to the Weierstrass function.This function is an example of a function curve that is continuous in all pointsallowed in its domain, but lacks any derivative of these points, see e.g. [31]. Itis seen as a fractal curve. Moreover, Thim [31] points out that the Weierstrasscurve, which was discovered in 1872, was not published officially until 1875 by P.du Bois-Reymond. It is worth mentioning that it was only after its publicationin 1875 that these types of features were deemed interesting, as it was thefirst example of a continuous curve without derivatives. This curve can berepresented mathematically with the help of a Fourier series as

fw(x) =

∞∑n=0

an cos(bnπx)

where 0 < a < 1, b = 2k + 1 for some integer k and ab > 1 + 3π/2.The purpose of this thesis is to emphasize fractal mathematics from a primar-

ily geometric perspective, and to examine periodic billiard orbits on Sierpinskicarpets.

As said earlier, this paper seeks to highlight fractals from a geometric pointof view, and to study billiard orbits. The theory regarding fractal billiards isrelatively new, as publications concerning fractals emerged in the second halfof the twentieth century, which means that the availability of materials in theform of literature is limited to this time frame. Names as Niemeyer, Bonkand Gutkin are often related to fractal mathematics. A literature study will

3

be conducted with much material centered around the publications regardingthese mathematicians. This means that all literature used will be criticallyreviewed and chosen with care so that it fits well with the scientific purpose.The literature selected consists mainly of peer scientific articles published inscientific journals. However, other serious literature, like textbooks, will also beused in order to present the theory.

There are in total five chapters including the introduction and two ap-pendixes. Chapter 2 is dedicated to preliminaries and hence presents the math-ematical theory necessary for the coming chapters. Topics regarding metricspaces, topologies, measures and surfaces are discussed. Chapter 3 deals withvarious examples of fractals and their relation to metric spaces. A fractal curveis also described together with some of its properties. In Chapter 4 we make asurvey of the article Periodic Billiard orbits of self-similar Sierpinski Carpets,[8], which means that billiards in general are described and some theorems re-garding the topic is cited with proof. The main focus of this chapter is thepresentation of a main theorem, Theorem 4.3.15, and the complete proof withparts cited from both [8] and [11] to give a complete proof. In the last chapter wediscuss the theory presented in Chapter 4 and we state and prove some propo-sitions building on the survey. Two appendices are written where we presentthe codes we used to generate the fractals in the pictures. Some illustrations ofperiodic orbits are also presented.

Chapter 2

Preliminaries: Topology

This chapter builds up the background theory necessary for the coming sectionson fractals and covers the basic theory of metric and topological spaces, as wellas an introduction to measure theory. The material can be found in [3],[20], [9],[25], [1], [6], [18] and [22].

2.1 Metric SpacesLet us start by establishing the necessary definitions. First of all we introducethe concept of distance and of metric spaces. The material below can be foundin [20], [9], [25] and [1]. Consider Rn and two given points x = (x1, x2, x3, ..., xn)and y = (y1, y2, y3, ..., yn). If we want to study the distance between x and y weneed a good way to express or measure this distance, e.g.

d(x, y) =√

(x1 − y1)2 + (x2 − y2)2 + ...+ (xn − yn)2, (2.1)

which is known as the Euclidean distance between x and y. In measuretheory it is convenient to call d(x, y) a metric, since it measures the distance.In general a metric space is defined as follows:

Definition 2.1.1. A metric space can be seen as a pair (M, d), where M is aset and d is a distance function, d : M×M→ R, such that

(i) ∀x, y ∈M : d(x, y) ≥ 0 and d(x, y) = 0⇔ x = y,

(ii) ∀x, y ∈M : d(x, y) = d(y, x),

(iii) ∀x, y, z ∈M : d(x, y) ≤ d(x, z) + d(z, y).

Landstedt, 2017. 5

6 CHAPTER 2. PRELIMINARIES: TOPOLOGY

The function defined in (2.1) is a distance function and R is a metric space.There are also other examples of metric spaces.

Example 2.1.2. Consider a vector space V on which we define a norm, || · ||.Recall that a norm on a vector space V is a real valued function such that it forall vectors u,v ∈ V satisfies the following conditions:

(i) ||u|| ≥ 0,

(ii) ||u|| = 0 if and only if u = 0,

(iii) ||c · u|| = |c| · ||u|| for all scalars c ∈ C,

(iv) ||u + v|| ≤ ||u||+ ||v||.

Then the norm induces a distance function (metric) as d(x, y) = ||x − y||,where x, y ∈ V. This means that the pair (V, || · ||) is a metric space.

♦

Example 2.1.3. The space of convergent sequences, denoted l p, has the prop-erty that every sequence x = (xi)i∈N in the space converges, i.e.

∞∑i=1

|xi|p <∞. (2.2)

The l p space is a metric space with metric

d(x, y) =( ∞∑i=1

|xi − yi|p)1/p

(2.3)

for p ≥ 1. That (2.3) is indeed a distance function follows from the definition.First of all, we have that ( ∞∑

i=1

|xi − yi|p)1/p

≥ 0

since we only sum powers of |xi − yi|, which is nonnegative, and d(x, y) isa real valued function. Moreover, the sum in (2.3) is only zero, if xi = yi forall i = 1, 2, 3, ..., which means that it is zero only if xi = yi. The symmetricproperty also holds because |xi − yi| = |yi − xi|. Essential is that the series in(2.3) converges, and this can be proved if we consider the Minkowski inequality,which can be found in [20]:( ∞∑

i=1

|xi + yi|p)(1/p)

≤( ∞∑j=1

|xj |p)(1/p)

+( ∞∑k=1

|yk|p)(1/p)

. (2.4)

2.1. METRIC SPACES 7

Here we see that the triangle inequality is satisfied. Therefore, d(x, y) is ametric on the space.

♦

Example 2.1.4. Let us consider an example from hyperbolic geometry.Define

H = z ∈ C∣∣ Im(z) > 0

as the hyperbolic plane. Given x, y ∈ H, let Γ[x, y], for x, y ∈ H, be the set ofall piecewise C1-paths f : [a, b] → H with start point f(a) = x and end pointf(b) = y. We also make the constraint Γ[x, y] 6= ∅. Poincaré defined a metricon H as the function dhyp : H×H→ Rn such that

dhyp(x, y) = inf∫

f

1

Im(z)|dz|

∣∣∣ f ∈ Γ[x, y], (2.5)

see e.g. [2]. The metric dhyp(x, y) is called the hyperbolic metric and one canprove that it satisfies the definition for a metric. A proof of this is presented in[2]. Another equivalent way, with a differential geometrical approach, of definingthe hyperbolic metric is found in [4] and [7] . Let x, y ∈ H be two distinct pointson a given hyperbolic line. A hyperbolic line is either a half circle in H with itscenter on the real axis, or a line that is perpendicular to R, the real line.

In order to make clear to the reader what the distance between the pointsx and y is we will need the cross ratio in C = C ∪ ∞ and we formulate thesenecessities with the help of Beardon in [4]. Initially, Let z1, z2, z3 and z4 be fourdistinct points in C. The cross ratio, denoted [z1, z2, z3, z4], of the four elementsis then given by

[z1, z2, z3, z4] =(z1 − z3)(z2 − z4)

(z1 − z2)(z3 − z4). (2.6)

when z1, z2, z3, z4 ∈ C. If any of the four points are ∞ we get

[∞, z2, z3, z4] =(z2 − z4)

(z3 − z4),

[z1,∞, z3, z4] =(z3 − z1)

(z3 − z4),

[z1, z2,∞, z4] =(z4 − z2)

(z1 − z2),


Figure 2.1: The image illustrates a hyperbolic triangle. Two of its sides areorthogonal to the real line, and the base of the triangle is a semicircle centeredon on R. The triangle in this illustration has the total sum of inner angles equalto zero, since the lines are all parallel.

and

[z1, z2, z3,∞] =(z1 − z3)

(z1 − z2).

Now, let the endpoints of the hyperbolic line seen in Figure 2.1 be denotedby a and b. The hyperbolic distance between x and y becomes

dhyp(x, y) =

log(∣∣[a, x, y, b]∣∣), if x 6= y,

0, if x = y,(2.7)

where [a, x, y, b] is the cross ratio.

♦

Example 2.1.5. Another important example is the Hausdorff metric. It hasthe property that it fulfills the definition of a metric on bounded sets, but noton arbitrary ones.

2.1. METRIC SPACES 9

Definition 2.1.6. Let (X, d) be a metric space and assume that A,B ⊂ X.The subsets A and B are withinHausdorff distance, r > 0, if and only if everypoint of A is within distance r of some point of B, and in the same way, everypoint of B is within distance r of some point of A. The open r-neighborhood ofA is defined as

Dr(A) = y | d(x, y) < r for some x ∈ A (2.8)

and the Hausdorff metric as

dH(A,B) := infrr > 0 |A ⊆ Dr(B) and B ⊆ Dr(A). (2.9)

We will see that dH is a metric on the set consisting of nonempty compactsubsets of (X, d). This set will be denoted H(X) and is called the hyperspaceof X, which will be used later in Chapter 3. To see that dH is a metric on thisspace, we formulate a proposition and use the proof given in [12].

Proposition 2.1.7. The space (H(X), dH) is a metric space.

Proof. The Hausdorff distance satisfies non-negativity and symmetry. We cansee that dH(A,B) ≥ 0 always holds since infrr > 0 |A ⊆ Dr(A) and B ⊆Dr(B) only considers positive values on r. Moreover, dH(A,B) = dH(B,A)holds trivially since infrr > 0 |A ⊆ Dr(B) and B ⊆ Dr(A) = infrr >0 |B ⊆ Dr(A) and A ⊆ Dr(B).

The subsets A and B are both bounded sets, which implies that the value ofdH(A,B) must be finite. Now, assume that dH(A,B) = 0. It is then necessaryto prove that A = B. If dH(A,B) = 0 we have an x ∈ A such that for all ε thereexists an open ε-neighborhood such that x ∈ Nε(B). If x ∈ Nε(B) we musthave d(x, x) = 0, for some x ∈ B. This means that B is compact and thereforealso closed. If B is closed, then x ∈ B and we see that A ⊆ B. In the samefashion, it is possible to show that B ⊆ A. So, then A = B, which is what weneeded to show.

Consider now the case when A = B and let ε > 0. Then we have for all εthat A ⊆ Dε(B), which means dH(A,B) = dH(A,A) = 0.

The last step is to show the triangle inequality for dH. For this, assume thatA,B,C ∈ H(X) and that x ∈ A, y ∈ B and z ∈ C. With this constructionwe see that d(x, y) < dH(A,B) + ε and d(y, z) < dH(B,C) + ε holds for someε > 0. This means that we can find an open r-neighborhood of C that containsA for some proper r. If we choose r = (dH(A,B) + dH(B,C) + 2ε) then A iscontained in a neighborhood. In the same way, with the same r, we see that Cis contained in an open r-neighborhood of A. Hence,

dH(A,C) ≤ dH(A,B) + dH(B,C)


and the proof is done.

♦

2.2 Topological SpacesMetric spaces form a subset of a more general class of spaces, namely topologicalspaces. A comprehensive account of topological spaces can be found in [6] and[18]. Topological spaces form the right setting for continuity, i.e. a more generalway of viewing continuous maps.

Definition 2.2.1. Let T be a collection of subsets of a set X. T is then saidto be a topology if

(i) the empty set ∅ and X are elements of T,

(ii) the intersection of any two elements of T is an element of T,

(iii) the union of any subcollection of T is an element of T.

The set (X,T) is then called a topological space. Moreover, a collection B ofsubsets in T is called a basis of T if every element of T is the union of elementsof B. We say also that B generates T.

A subset A ∈ T is called an open set, and a subset S ⊂ X is called closedif S = X \ A = Ac. Let p be a point in (X,T ). A neighborhood of p is asubset B ⊂ X that contains some C ∈ T with p ∈ C.

Example 2.2.2. The so-called trivial topology, T = ∅, X is of course a topol-ogy. So, if we letX = 1, 2, then T = ∅, 1, 2 is a topology forX. Moreover,if we letM = 1 be a subset of X, then T1 = ∅, 1, 1, 2 defines a topologyfor X, which is called the Sierpinski topology.

♦

Example 2.2.3. Let M be a subset of a metric space (X, d), then M is anopen set if it contains a ball around each of its points.

♦

Example 2.2.4. The usual topology on R, (R,T) is defined as the topology Twhere the basis of T consists of all open intervals of the real line.

♦

2.2. TOPOLOGICAL SPACES 11

In the case that our topology space is a metric space we can define opennessand continuity with the help of distances, since we have a metric. For this,let (X, d) be a metric space and consider a point x0 ∈ X. Denote an openball around x0 with B(x0, r) := x ∈ X| d(x, x0) < r, and a closed ball asB(x0, r) := x ∈ X| d(x, x0) ≤ r.

Example 2.2.5. In general, let (X, d) be a metric space. The topology inducedby d is the topology generated by B(x0, r) where x0 ∈ X and r ∈ R+.

♦

When we study fractal billiards later on we will need the concept of densesets. For a metric space X, it is possible to define a dense set and a separablemetric space X in the following way:

Definition 2.2.6. LetM be a subset of a metric space (X, d), thenM is said tobe dense in (X, d) if M = X, where M stands for the closure of M . Moreover,the space X is said to be separable if it has a countable subset which is densein X.

Example 2.2.7. The set of rational numbers Q is a countable and dense setin R. That Q is dense means intuitively that between every two distinct realnumbers, we can always find a rational number. To prove that the rationalsreally are dense in R one can use the Archimedean Property and consider twoarbitrary numbers x, y ∈ R such that x < y and then prove that there must bea number z ∈ Q such that x < z < y. With this said, since Q is dense in R andcountable we can conclude that R is separable.

♦

We will also need some other definitions.

Definition 2.2.8. Assume that (X,T) is a topological space and let S ⊂ X.A point x ∈ X is called a limit point of S if every element of T containing xmeets S in a point other than x. Moreover, the union of S and all of its limitpoints is called the closure of S, denoted S, and another way of characterizingthat a set S ⊂ X is closed is that S = S.

The definition can be interpreted as follows: If we have a limit point x ∈ Xof S, which may or may not lie in S, then we must always, for all neighborhoodsaround this point, be able to find a point that belongs to S but is not x. Asubset S ⊂ X is closed if and only if it contains all of its limits points. Inparticular for metric spaces there is the usual definition. We continue to studysome properties of topological spaces.


Definition 2.2.9. A topological space is called disconnected if there arenonempty subsets A and B such that X = A ∪ B, with A ∩ B = ∅ and withB ∩ A = ∅. A space that is not disconnected is called connected.

Example 2.2.10. Consider the set of real numbers R equipped with a topologywhose basis consists of half open intervals [a, b), where a, b ∈ R. This topologyis called the Sorgenfrey line and it is a disconnected topology. This is the case,since if we consider the two sets A := (−∞, 0) and B := [0,∞), we see thatA and B are both closed and open sets, which means that A ∩ B = ∅ andB ∩ A = ∅. Hence, the Sorgenfrey line is a disconnected topology.

♦

Example 2.2.11. In [5] there is the following algorithm for generating a so-called Sierpinski carpet fractal. The standard Sierpinski carpet is denoted withS3 and can be created by starting with a unit square and splitting it into 9equal squares, then removing the interior of the middle square and repeatingthis recipe with each of the remaining 8 squares. This carpet is an example ofa disconnected topology.

♦

Definition 2.2.12. 1. Let S ⊂ X, (X,T1) and (Y,T2) be topological spaces.A function f : (X,T1) → (Y,T2) is said to be continuous if for everyS ⊂ X we have that f(S) ⊂ f(S).

2. Let (X, d) and (Y, d) be two metric spaces. A mapping F : X → Yis said to be continuous at x0 ∈ X if for all ε > 0 and for all x ∈X there exists a δ > 0 such that if d(x, y) < δ then d(Fx, Fy) < ε.

Equivalently, a continuous map between to topological spaces (X,T1) and(Y,T2) satisfies the condition that for all open subsets U ⊂ Y , f−1(U) is open.These definitions of a continuity are purely topological and do not considerdistances. For metric spaces this definition becomes the usual definition of acontinuous map, see [20].

Definition 2.2.13. Let (X,T) be a topological space and let a and b be pointsin X. A path in X from a to b is a continuous function f : [0, 1]→ X such thatf(0) = a and f(1) = b, where [0, 1] has the usual topology for metric spaces.

An important concept needed in Chapter 3, where we consider dimensionsof fractals, is open coverings of topological spaces. Recall Example 2.2.4 wherewe considered the closed intervals in R with the usual topology. We will startwith an open covering of a topological space.


.

Figure 2.2: The standard Sierpinski carpet is a fractal in the plane that isgenerated by subdividing a unit square into small reproductions of itself. Thecarpet in this figure was created with the help of Geogebra.

Definition 2.2.14. Let A be a collection of open subsets to a topological spaceX such that X =

⋃A. Then A is called an open cover of X.

Definition 2.2.15. A topological space (X,T) is said to be compact if everyopen covering of X includes a finite subcollection that is also a cover of X.

If (X, d) be a metric space, then the space is compact if every sequence xnof points in X has a convergent subsequence.

Example 2.2.16. We have for instance that S1 and S2 are compact topologicalspaces, with the metric induced by the usual metric in R2 and R3.

♦

Example 2.2.17. It follows directly from the definition of a separable spacethat all compact metric spaces are separable.

♦


Example 2.2.18. The topological space (C,T), where T has a basis B(z, r),where for z ∈ C we have B(z, r) := w ∈ C

∣∣ |z − w| < r and B(∞, r) := z ∈C∣∣ |z| > r ∪ ∞. We can conclude that this topological space is compact.

♦

Example 2.2.19. Consider a subset S of the Euclidean space Rn. Then S iscompact if and only if it is closed and bounded. Sometimes this result is namedafter the two mathematicians Eduard Heine and Émile Borel and called TheHeine-Borel theorem, see e.g. [22]. This means for instance that the unit discD := x, y ∈ Rn : x2 + y2 ≤ 1 is compact. The set D is closed since theboundary is included in the set, and it is bounded.

♦

Example 2.2.20. Every closed interval [a, b] ⊂ R is compact, given the usualtopology. To prove it we follow Theorem 5.2. in [6]. The idea of the proof is tocover the interval with an open cover, and then show that this cover is both openand closed. This then leads us to the statement that [0, 1] is a compact space.Start by covering the interval [a, b] with an open cover C and let F ⊂ C be afinite subset of C. Also let S := x ∈ [a, b] : [a, x] ⊂

⋃F. Since we assumed

that C covers [0, 1] we can consider a particular C ∈ C and an x ∈ C. If wecreate a small open interval I = (x− δ, x+ δ) around x such that I ∩ [a, b] ⊂ C,then F ∪ C ⊂ C is finite and we also have that I ∩ [a, b] ⊂ S. This meansthat S is a neighborhood of each x, which means that S is open. On the otherhand, if we study the limit points of S we can prove that the set also has to beclosed. To see this, let y be a limit point of S such that y ∈ C for some C ∈ C.Because y is a limit point we know that if we let I2 = (y − γ, y + γ) such thatI2 ∩ [a, b] ⊂ C, there must be an x ∈ I2 ∩ [a, b]. Moreover, there must be a finitesubset F∗ ⊂ C with [a, x] ⊂ ∪F∗. This means that F∗ ∪ C ⊂ C is finite andthus we can conclude that [a, y] ⊂ ∪F∗ ∪ C. Therefore y ∈ S and the set isclosed. If the subset is both open and closed at the same time on [a, b], which isa connected space, we can conclude that, given the usual topology, every closedinterval [a, b] ⊂ R is in fact compact.

♦

Example 2.2.21. A classical example of a fractal is the Cantor set, which isgenerated as follows. Start with C0 as the closed interval [0, 1] and define C1 as

C1 = C0 \(1

3,

2

3

)=[0,

1

3

]∪[23, 1],


and so on. Generally we can find an expression for Cn, n ≥ 1, by

Cn =

3n−1−1⋃j=0

([3j

3n,

3j + 1

3n

]∪[

3j + 2

3n,

3j + 3

3n

])=⋃Ix1,...,xn

(2.10)

and if we let n→∞ we get the so-called Cantor set:

C := limn→∞

n⋂i=1

Ci. (2.11)

We then get a decreasing sequence Cn. All of these Cn are of compactsets. Therefore the Cantor set is an example of a compact set. They consist ofa finite union of closed intervals.

♦

Figure 2.3: The Cantor set, an example of an inhomogeneous self-similar fractal.The picture was created with the help of Geogebra. In this figure we have n ≤ 3.

Definition 2.2.22. A T2-space, or Hausdorff space, is a topological space inwhich any two distinct points are separated by disjoint neighborhoods.

The definition of a Hausdorff space says that no matter which two points inthe space we choose, we should always be able to find two disjoint neighborhoodsof them in such a way that they separate the points from each other.

The next example illustrates most of the concepts presented above in onemathematical object, namely the Hilbert cube.


Example 2.2.23. (The Hilbert Cube) Let X = Xa : a ∈ A be a collectionof sets given by a set A. The product space of X is then defined as the set ofall functions x : A →

⋃X with x(a) ∈ Xa and we denote it as

∏X. Now, let

us give an example of a compact connected Hausdorff space. We follow Cain in[6].

Definition 2.2.24. Let Xa = x ∈ R : 0 ≤ x ≤ 1 for every a ∈ A. Further-more, let A be countable and infinite. Then the product space

Hc =∏Xa : a ∈ A (2.12)

is called the Hilbert cube.

Since Xa = x ∈ R : 0 ≤ x ≤ 1 = [0, 1] is a compact connected Hausdorffspace we know that the product 2.12 will also have the same properties. Thisholds since products of a collection of sets preserves these properties, see e.g.Theorem 6.18 in [6]. Moreover, the Hilbert cube has metric topologies. We canuse the following theorem to motivate this. The theorem is formulated in [6].

Theorem 2.2.25. Let (Xn, dn) : n ∈ N be a countable infinite collection ofpseudometric1 spaces. Then the pseudometric

d(x, y) =

∞∑n=1

mindn(πn(x), πn(y)), 12−n, (2.13)

where πn :∏

X → Xn, for n = 1, 2, 3, ..., is called the projection map and isgiven by πn(x) = x(n), generates the product topology for∏

Xn : n ∈ N.

Remark 2.2.26. Observe that the pseudometric given by (2.13) in Theorem2.2.25 gives the same topology as:

d(x, y) =

∞∑n=1

1

2n· dn(xn, yn)

1 + dn(xn, yn).

The necessities in Theorem 2.2.25 are satisfied and since each Xn is a com-pleted connected Hausdorff space, we can conclude that the pseudometric is infact a metric on the Hilbert cube.

♦1A pseudometric space is like a metric space in which we allow the distance between two

distinct points to be zero.

2.3. HOMEOMORPHISMS 17

2.3 HomeomorphismsIn this subsection we will focus on functions that are able to deform topologicalspaces in a continuous way. Homeomorphism are mappings that have thatproperty and two topological spaces are viewed as the same space if there existssuch a mapping between them. We use [4], [35], [6], [14], [18] and [30].

Definition 2.3.1. Let X and Y be two metric spaces. A map f : X → Yis said to be a homeomorphism if f is a bijection where both f and f−1

are continuous. An equivalent definition of is that f : X → Y is called ahomeomorphism if it is a bijective, continuous and closed function.

Example 2.3.2. Stereographic projection is a homeomorphism. Let S2 be theunit sphere in R3 as in Example 2.2.16. Let the point N = (0, 0, 1) be the northpole of S. Let z = x+iy with x, y ∈ R. The stereographic projection, ϕ : z → w,of C to S is given by

ϕ(z) = ϕ(x+ iy) =

(2x

|z|2 + 1,

2y

|z|2 + 1,|z|2 − 1

|z|2 + 1

). (2.14)

To prove that the stereographic projection is a homeomorphism one canfollow the line l. We have ϕ : C → S2 and by definition also ϕ(∞) = N . Weuse the same topology as the one given in Example 2.2.18. If we now considerP1 = x + iy in the complex plane C and follow the projection line l we havethat ϕ(x+ iy) satisfies the equation system

x1 = sx

x2 = sy

x3 = 1− sx21 + x22 + x23 = 1

due to the parametrization of l and that

S2 = x1, x2, x3 ∈ R3|x21 + x22 + x23 = 1.

We can now make calculations,

x1 = sx

x2 = sy

x3 = 1− sx21 + x22 + x23 = 1

=⇒ (sx)2+(sy)2+(1−s)2 = 1⇐⇒ s2(x2+y2)−2s = 0,


Figure 2.4: Stereographic projection along the line l from S2 to C.

which has the solutions s = 0

ors = 2

x2+y2+1 .

Observe now that s = 0 means that we are on the North Pole, so we areonly left with

s =2

x2 + y2 + 1.

We also have that

lim√x2+y2→+∞

ϕ(x+ iy) = N

so ϕ(x + iy) is a continuous function over the whole C. Moreover, it is ainjective functions since its inverse is

ϕ−1(x1, x2, x3) =( x1

1− x3,

x21− x3

)=

x11− x3

+x2

1− x3i. (2.15)


Thus we can conclude that the stereographic projection is a homeomorphism.

♦

Example 2.3.3. The real line, R, is homeomorphic to (−π/2, π/2) since wecan use the bijective function f(x) = arctan(x), from its domain, (−∞,∞), toits image, given by (−π/2, π/2). Both f and f−1 are continuous for these x, sowe can conclude that the given intervals are homeomorphic. In fact, no matterwhich two open intervals we consider fo R, it can be proven that they will alwaysbe homeomorphic.

♦

Example 2.3.4. Recall the Cantor set again given in Example 2.2.21. First,introduce 2N as the product space of the topological space 0, 1 with the discretetopology as:

2N =

∞∏i=1

0, 1i. (2.16)

We can now show that the Cantor set, C, is homeomorphic to 2N. This canbe done in different ways, e.g. with the help of ternary expansions, see Willard[35, p. 121]. However, we will use Cantor’s intersection theorem and a binaryrepresentation. First, let πn : 2N → 0, 1n be the projection from 2N to the n:thfactor in the product space 2N. Also let I0 = [0, 1/3] and I1 = [2/3, 1]. Withthe use of binary sequences x1, x2, ..., xn, ..., where xi = 0, 1 for all i = 1, 2, 3...,and closed intervals Ik, as in Example 2.2.21, we are able to represent a pointof the Cantor set in a unique way. Define Ix1,...,xn−1,0 as the left subintervalof the interval Ix1,...,xn−1 and Ix1,...,xn−1,1 as the right subinterval of Ix1,...,xn−1 .We then have Ix1 ⊃ Ix1,x2 ⊃ ... ⊃ Ix1,x2,...,xn,.... So for instance, I0,1 signifiesthe interval [2/9, 3/9] and I1,1 represents the interval [8/9, 1]. Let P ∈ C. Then,with the help of Cantor’s intersection theorem (see [1]), we have that

P =⋂

(xi)i∈N

Ix1,...,xi

for the unique binary sequences x1, x2, ..., xn, .... Now, let ϕ : 2N → C be afunction that takes the sequence (xi) to P as described above. This means thatwe construct ϕ as a bijective function. To show that C is homeomorphic to 2N

we show that ϕ−1 is continuous. Being a product space, T has the basic opensets

π−1n 0, π−1n 1,


which are all sequences with a 0 and a 1 respectively, in position n. Now letU = π−1n 0 and study ϕ−1(U). We then get

(ϕ−1)−1(U) = ϕ(U) =( n−1⋂k=1

Ck

)⋂( ⋃(xi)i∈N

Ix1,...,xn−1,0

)⋂( ∞⋂k=n+1

Ck

).

(2.17)Since Ix1,...,xn−1,0 and Ix1,...,xn−1,1 are both closed intervals we know that

there exists an open subset V ⊆ R such that⋃(xi)i∈N

Ix1,...,xn−1,0 ⊆ V (2.18)

and

V⋂( ⋃

(xi)i∈N

Ix1,...,xn−1,1

)= ∅. (2.19)

Figure 2.5: This figure illustrates how the intervals covers parts of the Cantorset. The red intervals are Ix1,...,xn−1,1 and the black intervals illustrates V .Observe that the intersection is empty.

We now consider the intersection

V⋂C = V

⋂( ∞⋂k=1

Ck)

=

= V⋂[n−1⋂

k=1

Ck⋂[( ⋃

(xi)i∈N

Ix1,...,xn−1,0

)⋃( ⋃(xi)i∈N

Ix1,...,xn−1,1

)]⋂( ∞⋂k=n+1

Ck)]

=

= V⋂[

n−1⋂k=1

Ck⋂( ⋃

(xi)i∈N

Ix1,...,xn−1,0

)⋂( ∞⋂k=n+1

Ck)]

= (ϕ−1(U))−1,

(2.20)


since

V⋂( ⋃

(xi)i∈N

Ix1,...,xn−1,1

)= ∅.

This means that the ϕ−1 is a continuous function and since 2N is a Hausdorffspace and C is compact ϕ−1 is a closed function. Thus we have proved that 2N

is homeomorphic to C.

♦

2.3.1 Topological Surfaces

We will briefly mention topological surfaces and Riemann surfaces. We referto [14], [17], [18], [30] and [36] and recommend them as further reading on thistopic.

Definition 2.3.5. A surface, or closed surface, is a T2 space S such that everypoint in S has a neighborhood, U , which is homeomorphic to an open subset ofthe complex plane.

Example 2.3.6. Consider Figure 2.4 in Example 2.3.2 again. Since we cancover S2 with a family of open sets Ui such that we have a homeomorphismϕi : Ui → Vi, where Vi ∈ C is an open subset of the complex plane, we canconclude that it is a Riemann surface. This tells us that S2 is a surface.

♦

Now, let S(g) denote the 2-sphere with g handles added to it. One questionthat may arise is when the surface S(g) is well defined. We have the followingcondition to conclude whether a closed topological surface is determined or not.We will cite Armstrong in [3] and formulate it as a theorem, but we will firstpresent the basic definitions needed. See also [18].

Definition 2.3.7. S(g) is called the standard orientable surface of genusg.

Example 2.3.8. The torus S1 × S1 is an orientable surface of genus 1. It isalso a topological surface.

♦

Theorem 2.3.9. A closed surface is completely determined if we know its genusand whether or not it is orientable.


In the same spirit, we define a Riemann surface. Intuitively, we can see aRiemann surface as a work of art which consists of parts of the complex planeC that are glued together. A strict definition found in [36], [14] and [18], usescoordinate charts on a surface to get the structure of a Riemann surface.

Definition 2.3.10. A complex chart on a topological surface X is a homeo-morphism ϕ : U → V , where U ⊂ X is an open subset of X and V ⊂ C is anopen subset of the complex plane. Moreover, a complex atlas A on X is theset

A = ϕi : Ui → Vi,

where the elements are pairwise compatible charts such that⋃i Ui = X. A

biholomorphic map is a map between two domains U and V if f : U → V isa holomorphic, bijective function whose inverse is also holomorphic. Moreover,two complex charts, ϕ1 : U1 → V1 and ϕ2 : U2 → V2, are called holomorphi-cally compatible if the map

ϕ2 ϕ−11 : ϕ1(U1 ∩ U2)→ ϕ2(U1 ∩ U2)

is biholomorphic. See Figure 2.6 for an illustration.

All the compatible charts form the maximal atlas. Let us, with the help of[36], define a complex structure on a surface. For this, we need to let a maximalcomplex atlas be the atlas which contains the most compatible charts. Now, wefollow [14] in order to completely define a Riemann surface. First we define abiholomorhpic map with the help of [32].

Suppose that X is a surface. A complex structure on X is a maximalcomplex atlas on X. A Riemann surface is a completely separable connectedHausdorff topological space X together with a complex structure.

Example 2.3.11. Consider the complex plane C. If we take the identity map1d : C→ C as a complex chart we see that the complex plane is also a Riemannsurface with atlas A = id |C→ C.

♦


Figure 2.6: This is an illustration of holomorphically compatible maps ϕ1 andϕ2.

Example 2.3.12. The sphere S2 from Example 2.3.2 is a Riemann Surface.Recall that S2 = C ∪ ∞. To find a satisfying atlas to cover the sphere, weuse the usual topology and study the sphere as a compact Hausdorff topologicalspace. We can now take U1 = C and U2 = (C ∪ ∞) \ 0 and define twohomeomorphisms ϕ1 and ϕ2 such that

ϕ2(z) =

1z , for z ∈ C,0, for z =∞

andϕ1(z) = 1d,

which gives us that S2 = U1 ∪ U2. To see whether these homeomorphismsare holomorphically compatible we need to determine if

ϕ2 ϕ−11 : ϕ1(U1 ∩ U2)→ ϕ2(U1 ∩ U2)

is biholomorphic. In our case we have ϕ2 ϕ−11 (z) = ϕ1 ϕ−12 (z) = 1/z isbiholomorphic on ϕ1(U1 ∩ U2) = ϕ2(U1 ∩ U2) = C \ 0.


♦

2.4 Measure spaces

In this section we present some definitions and properties of measures and mea-sure spaces. The material can be found in [6], [9], [25] and [26].

Let us define a σ-algebra. One approach follows the one of Neely [26]. Startwith a collection A of subsets to a space X. Then A is called a σ-algebra if thecollections satisfy the following properties:

(i) X ⊂ A,

(ii) if E ∈ A, then EC ⊂ A,

(iii) if Ei∞i=1 is a countable sequence of subsets where ∀i Ei ⊂ A, then⋃i=1E

∞i ⊂ A.

Example 2.4.1. If we study the set A = 1, 2, 3, 4 we get that the collectionA = ∅, 1, 2, 2, 3, 1, 2, 3, 4 is a σ-algebra to A since all of the propertiesmentioned above are satisfied. Another example of a σ-algebra to the same setA could be A = ∅, 1, 2, 3, 4. Note that A is the σ-algebra with the smallestcardinality to A. Another way of reasoning makes use of the fact that a setwith n elements has 2n subsets. This means that the largest σ-algebra to analgebra with n elements must contain 2n elements. In our case, we can create aσ-algebra with 24 = 16 elements by choosing all possible subsets of A.

♦

Example 2.4.2. The Borel σ-algebra of a topological space is the smallestσ-algebra that contains the topology.

♦

In order to to talk about the length or size of a set in an algebra A, theconcept of a measure is required. Hence, we will define it with the help of [9].

Definition 2.4.3. A measure is a countably additive set function from analgebra to [0,∞].

Definition 2.4.4. Let µ be a measure defined on a σ-algebra A of subsets toX. Then the triple (X,A, µ) is called a measure space and the sets in X arecalled measurable.

2.4. MEASURE SPACES 25

Example 2.4.5. Consider a subset E ⊂ R. By definition, we know that thelength, l(I), of a bounded open interval I = (a, b) is l(I) = b − a given thatb > a. Lebesgue gave a measure of the "length" of E. He defined the so-calledLebesgue outer measure with the function L(E) : E → [0,∞]. With Ik∞k=1 asa sequence of open intervals, L(E), is defined as

L(E) = inf ∞∑k=1

l(Ik)|E ⊂∞⋃k=1

Ik

. (2.21)

Intuitively this means that we take the smallest covering of E and then sumthe interval lengths. With the help of this outer measure the so-called Lebesgueσ-algebra is defined. The Lebesgue σ-algebra consists of the subsets E ⊂ R suchthat for all A ⊂ R we have L(A) = L(A∩E)+L(A∩EC). The Lebesgue measureof E is then given by L(E) where E is contained in the Lebesgue σ-agebra.

♦

.

Figure 2.7: Illustration of the Lebesgue measure. The interval E = (a, b) iscovered with an open covering.

Example 2.4.6. Let X be a space on which we define the so-called countingmeasure. This measure counts the number of elements in the subsets of X.

♦

Definition 2.4.7. Let A ⊂ R. We say that A has Lebesgue measure zeroif ∀ε > 0 there exists a countable collection of open intervals In such thatM ⊂

⋃∞n=1 In and

∑∞n=1 l(In) ≤ ε.


Example 2.4.8. Every countable set has Lebesgue measure zero since if we letε > 0 and consider an arbitrary countable set M = m1,m2, ... we can, for alln = 1, 2, 3, ..., construct an interval In = (mn − εn,mn + εn) around mn withεn = ε

2n+1 . For clarification, n = 1 gives I1 = (m1− ε4 ,m1 + ε

4 ), which is a coverfor m1. The same is done with the other points, so they are all covered. Thereason we chose εn = ε

2n+1 is because the sum of the length of each interval is

∞∑n=1

εn =

∞∑n=1

ε

2n+1= ε ·

∞∑n=1

1

2n+1=ε

2· 1

1− 1/2= ε.

By definition, we can conclude that every countable set has measure zero.

♦

Example 2.4.9. In [1] Abbott points out that the Cantor set, C = ∩∞n=0Cnseen in Example 2.2.21, has measure zero. In order to show that L(C) = 0 wecan observe that in the construction of C we create Cn with 2n disjoint intervals,where each of them has length 3−n. This means that we get an estimate of theLebesgue measure, namely L(C) ≤ 3−n · 2n. Since

limn→∞

(2

3

)n= 0,

we can conclude that the Lebesgue measure of the Cantor set is 0.

♦

Another measure is the Hausdorff measure, named after the German math-ematician Felix Hausdorff. In order to define this measure we will need somenew definitions.

Definition 2.4.10. The diameter, diam(Y ), of a set Y in a metric space Xis the least upper bound of the distances between pairs of points of the set.

The definition of the diameter of a set can also be expressed as

diam(Y ) = supd(x, y)|x, y ∈ Y

and we also put diam(∅) = 0.

Example 2.4.11. Let Y ⊂ R be the unit square. The diameter of Y is thendiam(Y ) =

√2, given by the Pythagorean theorem.

♦

2.4. MEASURE SPACES 27

Figure 2.8: The diameter, diam(Y ), of the set Y is the maximal distance betweentwo points in the set Y .

Definition 2.4.12. Let X be a metric space and Y ⊂ X be a subset in X.Moreover, let

Hδα = inf

n∑i=1

(diam(Xi))α |X ⊂

∞⋃i=1

Xi, diam(Xi) < δ, (2.22)

then

limδ↓0

Hδα(Y ) (2.23)

is called the Hausdorff (α-dimensional) measure of the set Y .

Observe that⋃∞i=1Xi is the union of all possible covers of Y and that this is

a so-called outer measure. This means thatHδα satisfies the following conditions:

(i) Its domain is a ring that is closed under countable union, i.e. if Y ⊂ X suchthat ∅ ∈ Y and ∀A∀B ⊂ Y [A\B ⊂ Y and A

⋃B ⊂ Y ] we have that

⋃∞i=1Ai ∈

Y whenever Xii∈N ⊂ Y, of X.

(ii) Its range is [0,∞],

(iii) Hδα(∅) = 0 and that for every countable family Xi in our domain ring,

the inequality

Hδα

(⋃i

Xi

)≤∑i

Hδα(Xi) (2.24)


holds.

Since the Hausdorff measure fulfills these conditions it is called an outermeasure. We will use this measure later when we talk about dimensions inChapter 3.

.

Figure 2.9: The diameter, diam(Y ), of the set Y is the maximal distance betweentwo points in the set Y .

Chapter 3

Fractal Theory

We are mostly interested in Sierpinski carpets but will present some generalfractal theory and give illustrative examples. In this chapter we will see therelation between fractals and metric spaces. First of all we consider contractingmaps on a metric space onto itself. We refer to [19], [7], [10], [31] and [12].

3.1 Metric Spaces and FractalsLetH(X) be the hyperspace, seen in Example 2.1.5, for a metric space (X, d) andconsider a family of contracting maps f1, f2, f3, ..., fk inX. A map f : X → Xis a contraction if for all x and y in X there exists a λ ∈ (0, 1) such that

d(f(x), f(y)) ≤ λ · d(x, y).

Now, we shall use these contracting maps f1, f2, f3, ..., fk in X and definea contracting map F : H(X)→ H(X) as

F (K) =

k⋃i=1

fi(K). (3.1)

where K ⊂ X. Observe that F has a unique attractor. We can now define ahomogeneous self-similar fractal set with the help of F (K). The definition canbe found in [19].

Definition 3.1.1. Let K ⊂ X be the unique nonempty compact set such thatF (K) = K, i.e. the attractor. In this case we call K a homogeneous self-similar fractal set and the system of functions f1, f2, f3, ..., fk is called aniterated function system.

Landstedt, 2017. 29

30 CHAPTER 3. FRACTAL THEORY

If, on the other hand, we consider F : H(X)→ H(X) defined as

F (K) =

k⋃i=1

fi(K)⋃Y, (3.2)

where Y ⊂ X is a fixed and compact subset, we can define an inhomogeneousself-similar fractal in the same fashion. Note that F is also a contracting map.

Definition 3.1.2. The nonempty compact set K ⊂ X is called an inhomoge-neous self-similar fractal set if F (K) = K.

Example 3.1.3. Recall the Cantor set, which we introduced earlier in Example2.2.21. This uncountable set is an example of an inhomogeneous self-similarfractal.

♦

Example 3.1.4. In the same way the Sierpinski carpet can be seen as aniterated function system. This can be done with affine transformations. Initially,we use the maps

fi(x) :=

1/3 0 α0 1/3 β0 0 1

x1x21

= AX (3.3)

where fi : R2 → R2 is a contraction map and vi = (α, β) such thatv1 = (0, 0), v2 = (1/3, 0), v3 = (2/3, 0), v4 = (0, 1/3), v5 = (2/3, 1/3), v6 =(0, 2/3), v7 = (1/3, 2/3) and v8 = (2/3, 2/3). With these functions fi for i =0, 1, 2, 3, ... we see that fi8i=1 is an iterative function system with fixed pointas the standard Sierpinski carpet.

♦

Now, let f : R → R be a function which we wish to study the dynamics of.Therefore, we are interested in the behavior of the mapping points under itera-tions of f . With the n:th iteration of the function f we mean the compositionof it n times, for instance: f4 = f f f f .

3.2 Some Examples of FractalsWe have already seen that the Cantor set and the standard Sierpinski carpet aretwo examples of fractals. Intuitively, one can see a fractal curve as a repeatingpattern that goes on and on with no end. So we can see a fractal as a geometrical

3.2. SOME EXAMPLES OF FRACTALS 31

figure that we copy in a slightly smaller scale and put together with the originalshape. If we do this infinitely many times and make the pieces we put togethersmaller and smaller in each iteration, we have created our own fractal. The wordfractal comes from the Latin word fractus, meaning broken, and was coined in1970 by the French mathematician Benoit B. Mandelbrot. In this section wewill present some fractal curves to give an intuition to what such objects looklike. Recall the Sierpinski carpet in Example 2.2.11. It is an example of a fractalset.

3.2.1 Weierstrass Fractal Curve

The Weierstrass function is a so-called fractal curve. It is named after theGerman mathematician Karl Weierstrass who discovered this curve in 1872. Thefunction was presented by P. du Bois-Reymond in [10] in 1875. The function isthere given by

fw(x) =

∞∑n=0

an cos(bnπx)

where 0 < a < 1, b = 2k + 1 for some integer k and ab > 1 + 3π/2.

Bois-Reymond,[10], stated that the function is continuous and proved thatit is not differentiable in any point of its domain. This was done in 1875. Wereconstruct the theorem, with proof, from [10] to see this fact. We also use thehelp of [31] for structure and [23] for a reference to the Weierstrass M -test:

Lemma 3.2.1. Suppose that V is a complete normed vector space, i.e. a normedvector space where each Cauchy sequence converges in V, and let gk : Dgk → Vbe functions such that there are constantsMk with ||gk(x)|| ≤Mk for all x ∈ Dgk

and∑∞k=1Mk converges. Then

∑∞k=1 gk converges uniformly and absolutely.

Theorem 3.2.2. The Weierstrass function

fw(x) =

∞∑n=0

an cos(bnπx)

for 0 < a < 1, b = 2k+1 for some integer k and ab > 1+3π/2 , is continuousand nowhere differentiable for every real value on x.


Figure 3.1: An example of a fractal curve. Weierstrass function for a = 12 and

b = 3 and with the approximation π ≈ 3. The sketch was generated in Maple.

Proof. In order to show that the function fw(x) is nowhere differentiable, welook at the left derivative and compare it with its right derivative, the same wayas Bois-Reymond did. So, we need to prove that

limh→0+

fw(x+ h)− fw(x)

h6= limh→0+

fw(x− h)− fw(x)

−hfor all x in its domain, which is the whole real line.Now, let x0 ∈ R, m ∈ N, and choose αm as an integer such that

−1

2< bmx0 − αm ≤

1

2.

Next, define

xm+1 = bmx0 − αm ⇔ x0 =xm+1 + αm

bm,

ym =αm − 1

bm


and

zm =αm + 1

bm.

We have the inequality ym < x0 < zm since

ym − x0 =αm − 1

bm− xm+1 + αm

bm=−1− xm+1

bm= −1 + xm+1

bm

and

zm − x0 =αm + 1

bm− xm+1 + αm

bm=

1− xm+1

bm,

which means thatym < x0 < zm.

We also have that ym → x0 when m→∞ due to the limit

limm→∞

(−1 + xm+1

bm

)= 0

and for the same reason we get zm → x0 when m→∞ because

limm→∞

1− xm+1

bm= 0.

Now, we get the left derivative

fw(ym)− fw(x0)

ym − x0=

∞∑n=0

(an

cos(bnπym)− cos(bnπx0)

ym − x0

)=

=

m−1∑n=0

((ab)n


bn(ym − x0)

)+

∞∑n=0

(am+n cos(bm+nπym)− cos(bm+nπx0)

ym − x0

).

Let us estimate

m−1∑n=0

((ab)n


bn(ym − x0)

)with the help of cosα− cosβ = −2 sin(α+β2 ) sin(α−β2 ) and the fact that∣∣∣∣ sin(x)

x

∣∣∣∣ ≤ 1.

We get


∣∣∣∣∣m−1∑n=0

((ab)n


bn(ym − x0)

)∣∣∣∣∣ =

=

∣∣∣∣∣m−1∑n=0

(−(ab)n

1

bn(ym − x0)sin

(bnπym + bnπx0

2

)sin

(bnπym − bnπx0

2

))∣∣∣∣∣ =

=

∣∣∣∣∣∣m−1∑n=0

−π(ab)n sin

(bnπ(ym + x0)

2

) sin(bnπ(ym−x0)

2

)bnπ(ym−x0)

2

∣∣∣∣∣∣ ≤≤m−1∑n=0

π(ab)n = π(ab)m − 1

ab− 1≤ π (ab)m

ab− 1.

In analogy we have

∞∑n=0


ym − x0

)=

=

∞∑n=0


ym − x0

)=

=

∞∑n=0

(am+n−(−1)αm − (−1)αm cos(bnπxm+1)

ym − x0

)=

=

∞∑n=0

(am · an−(−1)αm − (−1)αm cos(bnπxm+1)

− 1+xm+1

bm

)=

= (ab)m(−1)αm

∞∑n=0

an1 + cos(bnπxm+1)

1 + xm+1.

We also know that

−1

2< xm+1 ≤

1

2,

which gives us the estimation

∞∑n=0

1 + cos(bnπxm+1)

1 + xm+1≥ 1 + cos(πxm+1)

1 + xm+1≥ 1

1 + 12

=2

3.


These estimations make it possible, for −1 ≤ ε1 ≤ 1, η1 > 1, to

fw(ym)− fw(x0)

ym − x0=

= (−1)αm(ab)mη12

3+(−1)αm(ab)mη1ε1

π

ab− 1(−1)αm(ab)mη1

(2

3+ ε1

π

ab− 1

).

If we study the same quotient but for the right derivative we have

fw(zm)− fw(x0)

zm − x0=

∞∑n=0

(an

cos(bnπzm)− cos(bnπx0)

zm − x0

)=

=

m−1∑n=0

((ab)n

cos(bnπzm)− cos(bnπz0)

bn(zm − x0)

)+

∞∑n=0

(am+n cos(bm+nπzm)− cos(bm+nπx0)

zm − x0

).

Estimation gives of these two partial sums∣∣∣∣∣m−1∑n=0

((ab)n

cos(bnπzm)− cos(bnπx0)

bn(zm − x0)

)∣∣∣∣∣ =

=

∣∣∣∣∣m−1∑n=0

(−(ab)n

1

bn(zm − x0)sin

(bnπzm + bnπx0

2

)sin

(bnπzm − bnπx0

2

))∣∣∣∣∣ =

=

∣∣∣∣∣∣m−1∑n=0

−π(ab)n sin

(bnπ(zm + x0)

2

) sin(bnπ(zm−x0)

2

)bnπ(zm−x0)

2

∣∣∣∣∣∣ ≤≤m−1∑n=0

π(ab)n = π(ab)m − 1

ab− 1≤ π (ab)m

ab− 1

and the second sum can be estimated by the help of

cos(bm+nπzm) = cos(bm+nπαm + 1

bm) = cos(bn(αm + 1)π)

= ((−1)bn

)αm+1 = (−1)αm+1 = −(−1)αm .

Hence,


∞∑n=0

(am+n cos(bm+nπzm)− cos(bm+nπx0)

zm − x0

)=

=

∞∑n=0

(am+n−(−1)αm − (−1)αm cos(bnπxm+1)

zm − x0

)=

=

∞∑n=0

(am · an−(−1)αm − (−1)αm cos(bnπxm+1)

− 1−xm+1

bm

)=

= −(ab)m(−1)αm

∞∑n=0

an1 + cos(bnπxm+1)

1− xm+1

and since xm+1 ∈(− 1

2 ,12

]we get

∞∑n=0

1 + cos(bnπxm+1)

1− xm+1≥ 1 + cos(πxm+1)

1− xm+1≥ 1

1−(− 1

2

) =2

3.

Therefore we can write

fw(zm)− fw(x0)

zm − x0= −(−1)αm(ab)mη1

(2

3+ ε1

π

ab− 1

).

Lastly, we need to prove continuity. We know that for 0 < a < 1 we havethat

∞∑k=0

ak =1

1− a,

because it is a geometrical series, and∣∣ak cos(bkπx)∣∣ ≤ ak

since | cos(bkπx)| ≤ 1 for all x ∈ R.These two properties will directly imply thatthe function is continuous by the Weierstrass M-test. In addition, we see thatthe Weierstrass function can not be differentiable, since its left derivative is notequal to its right derivative, since the sign differs. We have proven that thisholds for all real x.

Remark 3.2.3. The formulation and proof of Theorem 3.2.2 has been citedmostly from the original source [10], but the proof has been presented in a moremodern form by Thim in [31].


3.2.2 The Koch CurveThe Koch curve is named after the Swedish mathematician Helge von Koch andis seen as one of the earliest fractals studied. We follow [12] in order to see howthis type of fractal can be generated.

We start with a triangle ABC with anglesm∠BAC = 2π/3, m∠ABC = π/6and m∠BCA = π/6 and then subdivide it into three smaller triangles wheretwo of these are congruent and isoceles. Let one of these be denoted A′B′C ′

with angles m∠B′C ′A′ = 2π/3, m∠B′A′C ′ = π/6 and m∠A′B′C ′ = π/6 withsides such that A′B′ lies on AB and B′C ′ lies on BC with A′ = A. The thirdtriangle is the equilateral triangle with sides A′C ′B′′ where B′′ is the mirrorpoint of B′ with respect to the axis going through A and the middle point ofthe distance BC. After this is done we have made one iteration. Then we repeatthis step and receive the so-called Koch curve as the limit of iterations as theytend to infinity, see Figure 3.2.

.

Figure 3.2: This iteration will converge to something that is called the Kochcurve.

3.2.3 Julia SetsThe Julia set is named after the French mathematician Gaston Julia who studiedsuch sets in the 1920’s. We will now define the Julia set and we need the conceptof stable points. Keen in [13] gives the following definition:

Definition 3.2.4. We call x a stable point for a map f if there is a neigh-borhood U of x such that the iterates fn form a normal family, i.e. it forms afamily of complex functions fi on a domain Df such that every infinite sub-set of fi contains a subsequence that converges uniformly on every compactsubset of Df .


Definition 3.2.5. Let Ωr denote the set of all stable points of r and let r be arational function. The Julia set is the complement of Ωr and is often denotedas J .

An interesting question about the Julia set is whether it is empty or not. Thefollowing proposition, with proof, is formulated in [13, p. 60] and illuminatesthis mystery.

Proposition 3.2.6. The Julia set, J, is nonempty.

Proof. Assume, on the contrary, that the set is empty. Consider a point x ∈ Jand a subset U of X that contains an open set that includes x. Let fn bea subsequence which converges to an analytic function ρ : U → C. In orderto prove the proposition, we follow [13] and use analytical continuation, whichtells us that ρ can be continued to the whole of C in an analytical way. Thismeans that ρ is a rational function of finite degree. Moreover, we know thatthe function ρ is the limit of rational functions whose degrees tend to ∞, whichgives us a contradiction to the assumption that the Julia set is empty. Thiscompletes the proof.

In order to generate the Julia set we use Newton-Raphsons method and ifwe follow [12] we can introduce a function ϕ(z) = z2 + c, where c ∈ C. Withf(z) :=

√(z − c) as the inverse1 of ϕ we have the following iterative function

for the Julia set:

Jn+1 = f0[Jn] ∪ f1[Jn]. (3.4)

Then, when we let n→∞ we get the Julia set J .To understand how this iteration is done we study some examples.

Example 3.2.7. We start with the function ϕ(z) = z2 + 14 but only consider

real values on z in the beginning. If we start with the value z0 = 1 we get thefollowing scheme of iterations.

1Every complex number, that is not zero, has two square roots but it will not matter whichone we choose, so the branch that gives us the positive one will do.


Iterations of f(zk) = z2k − 14 for real values on zk

Values of zk Iterative valuesz0 = 1 f(1) = 1− 1

4 = 0.75z1 = 0.75 f(f(1)) = 0.3125z2 = 0.3125 f(f(f(1))) = −0.15234375z3 = −0.15234375 f(f(f(f(1)))) = −0.2267913818z4 = −0.2267913818 f(f(f(f(f(1))))) = −0.1985656691z5 = −0.1985656691 f(f(f(f(f(f(1)))))) = −0.2105716750z6 = −0.2105716750 f(f(f(f(f(f(f(1))))))) = −0.2056595697z7 = −0.2056595697 f(f(f(f(f(f(f(f(1))))))) = −0.2077041414z8 = −0.2077041414 f(f(f(f(f(f(f(f(f(1)))))))) = −0.2068589896z9 = −0.2068589896 f(f(f(f(f(f(f(f(f(f(1))))))))) = −0.2072093584

It seems like the iteration converges to approximately −0.2. Imagine now thatwe start at z0 = a+b · i where a, b ∈ R, and then mark the points of convergencefor many values on a and b, for instance in a given interval. We will then endup with a Julia set.

♦

Example 3.2.8. Let f(z) = z2 be a function for which we want to generate theJulia set for. In order to see what this fractal look like we pursue the followingargument. First, we are squaring complex numbers, z = a + bi = reiθ wherea, b, r, θ ∈ R, which means that if r ≤ 1 the iteration will converge. If wemark all points of convergence, we will end up with the unit disk, D := x, y ∈R |x2 + y2 ≤ 1.

♦

Example 3.2.9. Consider now instead f(z) = z2 − 1. If we plot all the pointsthat converge for different complex arguments, we will end up with a set that isillustrated in Figure 3.3. In order to plot this in a proper way we use the Maplecode written in [33].

♦

It is not always the case that the Julia set of a function is connected. Theset we see in figure 3.3 was an example of a connected fractal but if we considerthe function f(z) = z2 − 0.8 · i we get a so-called disconnected Julia set. SeeFigure 3.4. Observe that the shape of the Julia set differs depending on whatvalue on the parameter c is chosen. For a plot of f(z) = z2−0.8 · i we use Maplecode from [33].


.

Figure 3.3: The Julia set for the complex valued function f(z) = z2 − 1. Thepicture was created in Maple with the help of code from [33].

We have seen plots that illustrate different Julia sets for the function

f(z) = z2 + c

for values on the constant c. Recall that different values on c give differentfractal sets. A way to classify Julia sets is done with the help of connectedness.This is the main idea behind the Mandelbrot set. We follow [13] for a definition.

Definition 3.2.10. The Mandelbrot set consists of the set of all parametervalues c for which the Julia set of f(z) = z2 + c is connected.

Remark 3.2.11. This iteration starts with z0 = 0, i.e. in the origin and fromthe definition it follows that if the sequence of iterations for f(z) = z2 + c isbounded, then c is in the Mandelbrot set. Moreover, if the sequence is notbounded, then c is not included in the Mandelbrot set.


.

Figure 3.4: This is an illustration of the Julia set for f(z) = z2 − 0.8 · i. It wasgenerated in Maple.

Figure 3.5: The Mandelbrot set is a fractal consisting of all possible Julia sets.


3.2.4 Heighway’s DragonThe Heighway dragon is another fractal set. It is generated thus: Start witha vertical line segment P0 of length 1 and construct another segment P1 byreplacing P0 with two line segments, both of length 1/

√2, attached at a right

angle.2 For Pn we replace each segment of Pn−1 with two line segments in thesame way as we did for P1. The limit of Pn as n→∞ is called the Heighwaydragon curve.

Figure 3.6: An approximation of the Heighway dragon curve. It is a fractal setin the plane as the approximations tend to infinity.

The Heigway dragon can be approximated from Pn for some n ∈ N. We alsohave a theorem that says that the approximation Pn is a bounded set for allvalues on n. The following theorem and proof idea is taken from Edgar in [12].

Theorem 3.2.12. All of the approximations Pn remain in some bounded regionof the plane.

Proof. Consider an arbitrary point on P0. The distance from that point to anyof the endpoints on P0 is at most 1, since the length of P0 is 1 length unit. In

2It is possible to attach two line segments to one another in two ways so that they form aright angle. It does not matter which one we choose as long as we stay consistent with theother iterations.

3.3. FRACTAL DIMENSION 43

the same fashion, the maximal distance between a point of P0 and P1 is 1/2.From the construction of Pn we see that an arbitrary point of Pk−1 has distanceat most(1/

√2)k+1 to Pk for k = 1, 2, ..., n. The maximal distance of any point

on Pk is then given by

1 +

k∑i=1

( 1√2

)i+1

and since the Heighway dragon is defined as the limit of Pn when n → ∞we study the limit

1 + limk→∞

k∑i=1

( 1√2

)i+1

<∞.

The series is finite so the proof is done.

Another interesting property of the dragon curve is that it is a space-fillingcurve. Intuitively, a space-filling curve can be seen as a continuous curve whoserange covers a finite part of the plane. The definition of a space filling curvewas first introduced by Peano. Therefore, such curves are often called Peanocurves.

Definition 3.2.13. A continuous curve f : [0, 1] → Rn with n ≥ 2 is called aspace-filling curve if and only if f([0, 1]) contains a ball Br(x).

We will now present a theorem that claims that the Heighway dragon is acurve that fills a finite region of the plane. The proof of this statement can beread in [12, p. 74].

Theorem 3.2.14. The Heighway dragon is a space-filling curve.

3.3 Fractal Dimension

From our intuition we know that points have dimension 0, curves have dimension1, surfaces have dimension 2 and solids have dimension 3. In this section wewill study some dimensions of fractals. We shall see that the dimensions notnecessarily must be an integer. This chapter will mainly be a survey of [12]Chapter 3 and Chapter 6, and we recommend it as further reading of thissubject. Initially we will start with zero-dimensional spaces and start therebywith two essential definitions.


Definition 3.3.1. Let X and Y be two sets. The set Y is then called subor-dinate to X if and only if there for every subset F ⊂ Y exists a subset E ⊂ Xsuch that F ⊆ E. Moreover, if (X, d) is a metric space and C is an open coverof (X, d) then a refinement of C is an open cover B of X that is subordinateto C.

We will now continue with the definition of a zero-dimensional metric space.Recall that a metric space is called clopen if (and only if) it is both open andclosed.

Definition 3.3.2. A set in the metric space (X, d) is called zero-dimensionalif and only if every finite open cover of X has a finite refinement that is formedby clopen sets.

Example 3.3.3. The Cantor set, presented in Example 2.2.21, is zero-dimensional.We will give a proof idea of why this is indeed the case and we will use the helpof an open cover A of C to do it. If A is an open cover of the Cantor set, weknow that there for any point x ∈ C there must exist an open set A ∈ A suchthat x ∈ A. As C is a subspace of R we know that there must exist a ball Br(x)with a positive radius r such that Br(x) ∩ C ⊆ A. Observe that the subintervalI of Cn that contains x must have length 3−n and the fact I ∩ C ⊆ A meansthat the complement of C in the interval I is a finite union of closed intervalsdue to the construction of the Cantor set. Hence I ∩ C is clopen in C and thismeans that we can construct a finite number of finite subcovers of the Cantorset. These subcovers are all disjoint, which induces a clopen partition of C. Thisshows that the Cantor set is a zero-dimensional metric space.

3.3.1 Lebesgue Dimension

The focus will now be on the Lebesgue dimension of metric spaces. To describeit we need the following. We cite again [12].

Definition 3.3.4. Assume that n ≥ −1 is an integer. The order of a family Aof sets is ≤ n if and only if any n+ 2 of the sets have empty intersection. Also,if n ≥ 0, then A is said to have order n if and only if it has order ≤ n but not≤ n− 1.

Definition 3.3.5. Let n ≥ −1 be an integer and let X be a metric space. Wesay that X has Lebesgue dimension, or covering dimension, Cov(X) ≤ nif and only if every finite open cover of X has a refinement with order less orequal to n. Also, the Lebesgue dimension is equal to n if and only if it is lessor equal to n but not less or equal to n− 1.


An equivalent definition is using that if C is an open cover of a metric spaceX, then its so-called mesh is supA∈C diamA and the following theorem. RecallDefinition 2.4.10 for a discussion of the concept of the diameter of a set.

Theorem 3.3.6 ([12]). Let X be a compact metric space and let n ≥ −1 be aninteger. Then CovX ≤ n if and only if for every ε > 0 there is an open coverof X with order ≤ n and mesh ≤ ε.

Example 3.3.7. In this example we will construct the proof from the idea ofProposition 3.2.18 in [12, p. 97] to see that the standard Sierpinski carpet hasLebesgue dimension 1. Theorem 3.3.6 will also be used. From the constructionof S3 let Sk3 be the k:th approximation of the carpet, i.e. the k:th iteration. Thismeans that Sk3 is made of 8k filled squares. Each of these subsquares have sidelength 3−k. The minimal distance between two subsquares that does not sharean edge in Sk3 is 3−k so with 0 < r < 3−k/2 we get r-neighborhoods of thesesubsquares that are open sets, i.e. a cover of sk3 of order 1. This means that theLebesgue dimension of the Sierpinski carpet is 1 since Theorem 3.3.6 gives thatCov(S3) ≤ 1 and since the carpet contains line segments we have Cov(S3) ≥ 1.That the Lebesgue dimension of a real interval is 1 can be motivated in thefollowing way. Assume that a < b where a, b ∈ R. We need to motivate whyCov([a, b]) ≤ 1. The inequality holds since for a, arbitrarily, chosen ε > 0 wehave that (k − 1

n,k + 1

n

) ∣∣∣∣∣ k ∈ Z

(3.5)

is an open cover of the interval [a, b] with mesh ≤ ε if we choose n ∈ Nlarge enough so that 1/n ≤ ε/2. We also se that the collection in (3.5) hasorder 1. This proves that Cov([a, b]) = 1. In addition one can conclude thatCov(S3) = 1.

Example 3.3.8. One can also prove that the entire real line R has Lebesguedimension 1. This can be done by considering the real line as the union ofcompact symmetrical intervals [−n, n] for n ∈ N. We know that Cov([−n, n]) =1 from Example 3.3.7. Since this holds for all natural integers n one can concludethat Cov(R) ≤ 1. For a proof of the latter statement we suggest [12, p.95].

3.3.2 Urysohn-Menger Dimension

In the same way as the Lebesgue dimension is an integer n ≥ −1 the Urysohn-Menger dimension, sometimes called the weak inductive dimension, can be as-signed to any metric space X with notation ind(X).


Definition 3.3.9. Let k be a nonnegative integer. Then ind(X) ≤ k if and onlyif there is a base for the open sets of X consisting of sets U with ind(∂U) ≤ k−1.We say that ind(X) = k if ind(x) ≤ k but ind(X) k−1. To make the definitioncomplete we demand that ind(∅) = −1 and if ind(X) ≤ k does not hold for allintegers k, then ind(X) =∞.

Next we will formulate and prove two very important theorems which canbe used to determine the Urysohn-Menger dimension of a metric space. We citethe theorem, Theorem 3.4.2., from [12, p. 104] and follow the proof presented.

Theorem 3.3.10. Let X and Y be metric spaces. If X and Y are homeomor-phic, then ind(X) = ind(Y ).

Proof. Note that the definition of the Urysohn-Menger dimension is inductive.Hence we will use induction to prove the theorem. The base case is whenind(X) = −1 and then X is empty. By the assumption that X and Y arehomeomorphic we see that in this case Y is also empty. This means ind(Y ) =−1. The base case is correct. Now assume that the theorem holds for spacesX that fulfill ind(X) ≤ k. Then consider X with ind(X) = k + 1. We leth : X → Y be a homeomorphism and let h[B] : B ∈ B be a base for theopen sets of X. The base B for the opens sets of X has the property thatind(∂B) ≤ k. This means that for B ∈ B we have that h[∂B] = ∂h[B]. Now usethe induction hypothesis on ∂h[B]., i.e. that ind(∂h[B]) = ind(∂B) ≤ k. Thismeans that there exists a base for the open sets of X consisting of sets withboundary of dimension less or equal to k. Hence, ind(X) ≤ k+ 1. On the otherhand, if ind(X) ≤ k, then the induction hypothesis would give ind(X) ≤ k andthis is a false statement. Therefore we conclude that ind(x) = k + 1. Finally,this means that ind(X) = ind(Y ). There is however a final case to check. It iswhen ind(Y ) = ∞. In this case we see that the inequality ind(Y ) ≤ k is falsefor all integers k and so ind(Y ) =∞ and we are done with the proof.

The other theorem uses the Cantor set. Recall from Example 2.3.4 thatthe Cantor set C is homeomorphic to the product space 2N. The Cantor sethas Urysohn-Menger dimension zero, which is a special case of the followingtheorem. We recall once again that we cite [12]. First, however, we need thefollowing lemma. The proof can be found in [12, p. 105]

Lemma 3.3.11. Let X be a metric space and let Y ⊆ X. Then ind(Y ) ≤ind(X).

Theorem 3.3.12. Let X be a nonempty separable metric space. then ind(X) =0 if and only if X is homeomorphic to a subset of 2N.


The proof found in [12] uses the fact that ind(2N) = 0, which we will notprove.

Proof. Initially, assume that X is homeomorphic to Y ⊆ 2N. We use Theorem3.3.10 and conclude that ind(X) = ind(Y ). Lemma 3.3.11 tells us that ind(X) ≤ind(2N) and since ind(2N) = 0 we see that ind(X) ≤ 0. This implies thatind(X) = 0, because S 6= ∅.

The other direction can be shown in the following way. Suppose that ind(X) =0. Then there is a base B1 for the open sets of X that consists of clopen sets. Wewill now use the Linderlöf property, i.e. that if X is a metric space and there isa countable base for the open sets of X, then every open cover of X has a count-able subcover. The property yields that there exists a countable base B ⊆ B1.Use the representation B = U1, U2, ... where Ui(1) = Ui and Ui(0) = S \ Ui.First of all, we notice that all Ui:s are clopen sets and if α = e1e2 · · · ek ∈ 2N

then we have

U(α) =

k⋂i=1

Ui(ei). (3.6)

Next, let h : X → 2N be a map such that for a given x ∈ X the i:th letter ofh(x) is either 0 or 1 depending on if x belongs to Ui(0) or Ui(1). The function his injective. To see this we must show that if x 6= y then h(x) 6= h(y). Therefore,assume that x 6= y. We then see that S \ y is an open set that contains xand hence there must be an i such that x ∈ UI ⊆ S \ y, i.e. h(x) 6= h(y),so h is injective. This is important because the inverse of h is needed. Leth−1 : h[X]→ X, where h[x] denotes the image set, i.e. h[X] = h(x) |x ∈ X.We must also prove that h is continuous. This is seen due to the fact that thatthe image sets [α] give a base for the open sets of 2N. Recall the definition of acontinuous function in Definition 1. We also get that

h[Ui] =⋃

α∈0,1i−1

(h[X] ∩ [α1]

), (3.7)

is open in h[X] for all i, so h−1 is continuous. This proves that h is ahomeomorphism. Hence, X is homeomorphic to 2N and the proof is done.

Example 3.3.13. The theorem directly gives us that ind(C) = 0. Sometimesthe Cantor set is called a universal zero-dimensional space due to Theorem3.3.12.


3.3.3 Minkowski-Bouligand DimensionThe Minkowski-Bouligand Dimension, sometimes also called the box dimension,is a dimension which not necessarily need to be integer values. For this theorywe follow [27] and start with a definition.

Definition 3.3.14. Let ε > 0 and Nε(X) be the smallest ε-neighborhood thatcovers X. Then the Minkowski-Bouligand dimension is given as

dimB(X) = lim supε↓0

lnNε(X)

ln 1ε

. (3.8)

Example 3.3.15. Recall Example 2.2.21 and the Cantor set. The Cantor sethas Minkowski-Bouligand dimension ln 2

ln 3 . To motivate this pretension we usethe calculations of [27]. With εn = 1

3 it will be possible to cover the Cantor setas we have seen before in Example 2.3.4 with length 1

3n . This means that thesmallest εn-neighborhood can be estimated with 2n−1 ≤ Nεn(C) ≤ 2n since weneed 2n intervals to cover C and that Nεn(C) is the smallest εn-neighborhood.This means that we for instance have the following inequality chain:

ln(2n−1)

ln 3n+1=n− 1

n+ 1· ln 2

ln 3≤ lnNεn(C)

1εn+1

≤ lnNε(C)ln 1

ε

≤ lnNεn(C)ln 1

εn

≤ n− 1

n· ln 2

ln 3

since εn+1 ≤ ε ≤ εn and Nεn(C) ≤ Nε(C) ≤ Nεn+1(C) for all natural numbers

n ≥ 1. Next, we study the limits and use the squeeze theorem. We get

limn→∞

(n− 1

n+ 1· ln 2

ln 3

)≤ limn→∞

(lim supε↓0

lnNε(C)ln 1

ε

)≤ limn→∞

(n+ 1

n· ln 2

ln 3

),

which means that dimB(C) = ln 2ln 3 . See [27] for a wider discussion.

Example 3.3.16. The standard Sierpinski carpet S3 presented in Example2.2.11 has dimB(S3) = ln 8

ln 3 . To see this we must see that we can cover eachapproximation s3 with 8n boxes where each box has side length 3−n, hencewe also get εn = 1

3n . Now we continue in Pollicotts, [27], reasoning. Notethat this cover is also the smallest cover possible and therefore we may chooseεn−1 ≤ ε ≤ εn together with Nεn(S3) ≤ Nε(S3) ≤ Nεn+1

(S3) so that we get

ln 8n

ln 3n+1≤

lnNεn+1(S3)

ln 1εn

≤ lnNε(S3)

ln 1ε

≤ lnNεn(S3)

ln 1εn

≤ ln 8n+1

ln 11/3n

=ln 8n+1

ln 3n

and the squeeze theorem gives in the same way as in Example 3.3.15 thatdimB(S3) = ln 8

ln 3


3.3.4 Hausdorff DimensionThis type of viewing dimension is very important for fractals since it is a gener-alization of the normal topological or euclidean dimension. Recall the definitionof the Hausdorff measure presented in Definition 2.4.12. Let us start with thedefinition.

Definition 3.3.17. Let F be a set. The Hausdorff dimension of the set Fis Fs0(F ) , where s0 ∈ [0,∞] is the unique critical value such that

(i) Hs(F ) =∞ for all s < s0,

(ii) Hs(F ) = 0 for all s > s0.

Henceforth we will write s0 = dimH(F ).

To get an intuitive feeling for the Hausdorff dimension we present this propo-sition with proof, found in [12, p. 180].

Proposition 3.3.18. The Hausdorff dimension of R is 1, i.e. dimH(R) = 1.

Proof. We shall, with the help of Edgar in [12], prove that dimH(R) = 1. Forthis we need a lemma about the Hausdorff measure. We will formulate it asfollows and remind the reader of the Lebesgue measure in Example 2.4.5.

Lemma 3.3.19. In the metric space R, the one-dimensional Hausdorff measureH1 coincides with the Lebesgue measure L.

Consider the unit interval [0, 1]. The lemma then gives us that H1([0, 1]) =L([0, 1]). Since [0, 1] ⊆ R we know that dimH([0, 1]) ≤ dimH(R) since there forall Borel sets A,B is true that if A ⊆ B, then dimH(A) ≤ dimH(B) holds. Thisproperty follows directly from the definition. Moreover, if s > 1 the Hs([0, 1]) =0 and we then get that

Hs(R) ≤ Hs( ∞⋃k=−∞

[k, k + 1])

=

∞∑k=−∞

Hs([k, k + 1]) = 0,

since the interval [k, k+1] is isomorphic to [0, 1] for all integers k and that theHausdorff measure is an outer measure. Consequently we see that dimH(R) ≤ sfor all s > 1, so with the definition of Hausdorff dimension in mind we canconclude that dimH(R) = 1.

We will now focus on how to determine the Hausdorff dimension of somefractals sets. We therefore need the following theorem, which can be found in[27, p. 21] together with a proof.


Theorem 3.3.20. Let X be a metric space. If there exists a probability measureµ such that we can find constants C > 0 and d > 0 such that for all ε > 0 andfor all x ∈ X we have that µ(B(x, ε)) ≤ Cεd, then dimH(X) ≥ d.

Example 3.3.21. In this example we shall prove that dimH(C) = ln 2ln 3 and

we use the reasoning presented in [27]. The approximation Cn of the Cantorset consists of 2n subintervals. Let µ be the probability measure that fulfillsµ(Cn) = 2n

3n for all integers n ≥ 1. Observe that we have 2n subintervals, sayIn, that covers each interval on the n:th level in Cn. We have that µ(In) = 1

3n

for all n ≥ 1, which gives us

1

3·(2

3

)n+1 ≤ µ(B(x, ε)) ≤ 1

3·(2

3

)n,

and with 13n+1 < ε < 1

3n we get that µ(B(x, ε)) ≤ 13ε

ln(2)ln(3) . With the help of

the theorem just presented we have proven that dimH(C) = ln 2ln 3 .

3.3.5 Homeomorphisms and the Standard Sierpinski Car-pet

This part will describe the connection between homeomorphisms and the Sier-pinski carpet. We refer mostly to Bonk, [5].

Definition 3.3.22. A metric space X is called a carpet if it is homeomorphicto the standard Sierpinski carpet S3.

Recall for instance the standard Sierpinski carpet in Example 2.2.11. Weneed a good way to determine whether a metric space is a carpet or not. Thereis a theorem that is useful to do this. It requires three new definitions, whichall can be found in [5].

Definition 3.3.23. A set is called topologically planar if it is homeomorphicto a subset of C.

Example 3.3.24. Every open subset around a point on a Riemann surface willbe topologically planar. This follows from the definition of a Riemann surfaceand is illustrated in Figure 2.6.

Definition 3.3.25. A local cut point p of X is a point that has a connectedneighborhood U such that U \ p is not connected.

Example 3.3.26. Any point of the Cantor set C will be a local cut point

Another essential concept is a locally connected continuum. Let us writedown a definition.


Definition 3.3.27. A locally connected continuum is a compact metricspace that is locally connected, i.e. a Hausdorff space that is compact andconnected.

Theorem 3.3.28 ([5]). Let X be a metric space. Then X is homeomorphic tothe standard Sierpinski carpet if and only if X is a locally connected continuum,is topologically planar, has Lebesgue dimension 1, and has no local cut points.

Chapter 4

Fractal Billiards

This is the main chapter of this thesis and is a survey of J. P. Chen and R.G. Niemeyers paper Periodic Billiard Orbits of self-similar Sierpinski Carpetspublished in the Journal of Mathematical Analysis and Applications year 2014(volume 416 and pages: 1350-1373).1 We will use other sources to describe thematerial, see e.g. [16] and [15].

Recall the iterative function system in Example 3.1.4 where we said that theSierpinski carpet could be generated with the help of fi8i=1 as affine transfor-mations. In this paper we find a more general consideration. We can generalizethe system with the help of a nonnegative odd parameter a ≥ 3. More generalfunctions, φi : R2 → R2, are given by

φi(x) =1

a(x+ (ui, vi)), (4.1)

with ui, vi ∈ N∪0 where it is essential that the inequalities 0 ≤ ui ≤ a−1and 0 ≤ vi ≤ a − 1 hold for all i = 0, 1, 2, ... such that (ui, vi) 6= (a−12 , a−12 ).With these maps the Sierpinski carpet is given by

⋃a2−1i=1 φi. A remark on this

is that this approach is still built on affine transformations in the same way asExample 3.1.4 was, but the mappings φi are written in another form.

In the same fashion as we saw how the inverse map of a multicurve, f−1(γj),may be peripheral, otherwise it is homotopic, to one of the curves γj on S2, wewant to be able to study parts of approximations of the Sierpinski triangle. Inorder to do this we need the concept of so-called peripheral squares.

Definition 4.0.1. A peripheral square of the Sierpinski carpet is the bound-ary of an open square that has been removed in the construction of the carpet.

1If nothing else is said through the chapter we cite Chen and Niemeyers article.

Landstedt, 2017. 53

54 CHAPTER 4. FRACTAL BILLIARDS

Remark 4.0.2. Traditionally, one does not see the unit square as a peripheralsquare.

Example 4.0.3. Recall the Sierpinski carpet in Figure 2.2. Then, the boundaryof every white square in the illustration are examples of peripheral squares ofthe Sierpinski carpet.

♦

If we consider the Sierpinski carpet S3 again, we call, following Chen andNiemeyer [8], a line segment that is contained in S3 for a nontrivial line segmentof S3 if the segment, which is a straight line, has nonzero length.

Figure 4.1: These are examples of line segments in S3. Observe that the redlines, which are associated with the set B3, never intersect or touch any periph-eral squares while the blue ones, associated to A3, touch the vertices of someperipheral squares.

Notation 4.0.4. The set of all slopes of the line segments in the Sierpinskicarpet is denoted with Sl(S3) and each element in Sl(S3) takes values in theinterval [0, 1].

With this definition in mind, the following theorem for slopes on the Sier-pinski carpet S3 is given.

55

Theorem 4.0.5. We have that Sl(S3) = A3 ∪B3, where

A3 =pq

∣∣∣ p+q ≤ 3, 0 ≤ p ≤ q ≤ 2, with p, q ∈ N∪0 such that p+q is odd,

(4.2)and

B3 =pq

∣∣∣ p+ q ≤ 2, 0 ≤ p ≤ q ≤ 1, with p, q ∈ N such that p and q are odd.

(4.3)

Remark 4.0.6. Chen and Niemeyer give a more general theorem, originallycited from E. Durand-Cartagena and E. T. Tyson in [11]. If we let Sa denotethe Sierpinski carpet that is generated from the iterative functions given in 4.1.If a = ( 1

a ,1a ,

1a ...) is a constant sequence, we have that Sl(Sa) is the union of

Aa =pq

∣∣∣ p+q ≤ a, 0 ≤ p ≤ q ≤ a−1, with p, q ∈ N∪0 such that p+q is odd,

(4.4)and

Ba =pq

∣∣∣ p+q ≤ a−1, 0 ≤ p ≤ q ≤ a−2, with p, q ∈ N such that p and q are odd.

(4.5)Chen and Niemeyer made a refinement of this result since they found somethings that needed to be corrected. We will discuss this more thoroughly laterin this chapter.

Example 4.0.7. For the Sierpinski carpet seen in Figure 2.2 we have thatA3 = 0, 12 and B3 = 1 since, for A3, we see that the only p that can satisfythe condition that p+q ≤ 3 at the same time as p+q is odd and 0 ≤ p ≤ q ≤ 2is for p = 0, p = 1 and q = 2 . The restrictions for p and q in B3 give that p = 1and q = 1. Observe that we only look at slopes in the interval [0, 1] because theother slopes are given by symmetries in the carpet.

We can also formulate another theorem that connects line segments to pe-ripheral squares of the Sierpinski triangle.

Theorem 4.0.8. Let the slope σ1 ∈ A3 and the slope σ2 ∈ B3 . Then eachline segment in the Sierpinski carpet S3 with slope σ1 touches vertices of periph-eral squares and each line segment with slope σ2 is disjoint from all peripheralsquares.


Corollary 4.0.9. For every σ ∈ A3 ∪ B3 there are maximal line segments inS3 with slope σ.

Remark 4.0.10. In both the theorem and the corollary one can substitute S3,A3 and B3 with Sa, Aa and Bb and the results still hold.

4.1 BilliardsRecall the billiard table presented in the Introduction as an example of oneapproach to the relation between fractals and dynamical systems. We considereda billiard table in the plane with a boundary shaped as a polygon such that thetable forms a compact set. As mentioned earlier, the set is generally called aplanar billiard.

Definition 4.1.1. A planar billiard, Ω(D), with boundary D is a path-connected region in the plane with sufficiently piecewise smooth boundary.

On this planar billiard it is possible to define reflections and we define thereflections in Ω(D) be given by the Law of Reflection, i.e. the angle of reflectionequals the angle of incidence. We need a starting point for the so-called billiardflow and denotes it as (x0, θ0) where the point x0 ∈ D and θ0 is the tangentialangle for x0. If we let fD be a map such that the next point x1 ∈ ∂D is givenby fD(x0, θ0) = (x1, θ1), where θ1 is an inward pointing vector based at x1.

.

Figure 4.2: This billiard table has a rectangle as boundary and this is an exampleof a mathematical billiard. Please note that the boundary must not necessarilybe a polygon. The boundary could be an ellipse or another continuous curve inthe plane.

4.1. BILLIARDS 57

When we have a billiard function fD it is reasonable for us to study the iter-ations fkD for k = 1, 2, 3, ... of these maps. Observe that the angles θ0, θ1, θ2, ...θkare determined from the Law of Reflection. When we study the iteration chain,we will also get iterative points x1, x2, x3, ..., xk, which can intuitively be seenas the points on the billiard table where the billiard ball hits the boundary, D,of the table. We make the same type of restrictions that the authors made in [8]as we define an equivalence relation, ∼, on the set Ω(D)× S1. The equivalencerelation is identified by the angle between the outward pointing vector based atxk and the inward pointing vector based at the same point. In [21] this equiv-alence relation is thoroughly described and we use the same description. With(x, θ), (y, γ) ∈ Ω(D)× S1 we say that (x, θ) ∼ (y, γ) if either

(i) θ = γ and x = y is not a vertex of the boundary D,

(ii) x = y is a point of the boundary to a polygonal boundary, D, that is nota vertex and θ = ri(γ) where ri denotes the reflection function for eachside si of the boundary.

Note that this means that each equivalence class has exactly two elements,namely a point and an angle (given by two vectors). Now, let us see what the setΩ(D)× S1 looks like. The Cartesian product between D and S1 gives us a solidtorus. We will, like Chen and Niemeyer, only consider the space Ω(D)× S1/ ∼,which is the space of points and reflection angles. This is a comfortable spaceto work with. First, the topological product Ω(D) × S1 gives us all possibledirections around every point in Ω(D), since the circle S1 is a representation ofthe direction vectors for a billiard ball. We use the quotient group Ω(D)×S1/ ∼in order to always (except in singularities) have a direction vector to follow.

Around every point in the table we have many possible directions for the ballto follow. These directions are represented with S1, i.e. a circle representation.The usefulness of the quotient group is seen when we study a boundary point.Since we mod out with ∼ every possible direction that points out of Ω(D) dueto the reflected according to the Reflection law, which means that we alwayshave a direction vector.

Definition 4.1.2. Let fD be the billiard map describing the discrete billiardflow in the phase space (Ω(D) × S1)/ ∼. An orbit O(x0, θ0) of Ω(D) is thengiven by the expression f iD(x0, θ0)∞i=0.

Definition 4.1.3. We call an orbit O(x0, θ0) of a billiard Ω(D) closed ifO(x0, θ0) consists of finitely many elements.

Definition 4.1.4. If there is a least integer m ≥ 1 such that fm(x0, θ0) =(x0, θ0) for a closed orbit O(x0, θ0), we call the orbit periodic.


So, we see that an orbit is periodic if it is closed and comes back to thestarting point x0 for some m ∈ N and angle θ0.

Figure 4.3: In this billiard we can see an orbit with θ0 = π6 .

Definition 4.1.5. If a path of a billiard flow intersects a point of D for whichreflection cannot be determined in a well-defined manner, we call the orbitsingular.

Example 4.1.6. Consider a boundary that has the shape of a rectangle. If thebilliard ball enters one of the corners of the boundary, the billiard ball trajectoryterminates at that point since it is impossible to determine how the ball willcontinue to move. See an example of this in Figure 5.2.

Remark 4.1.7. Chen and Niemeyer point out, in [8], that if the orbit O(x0, θ0)is singular, there will exist a constant k ∈ N such that a previous base point,x−k, associated with k of f−k(x0, θ0), which is the inverse map, i.e. it sendsthe ball backwards, will not have a derivative. This means that we can notdetermine any future paths for the billiard ball since we do not know in whichdirection it should go. In this case, the orbit is not said to be closed. However,it is possible to define a proper way to continue the trajectory. For instance, onecould reflect the direction vector, which leads to a corner point, in the bisectorof that point and reverse the direction of the vector.

We have already mentioned that the boundary of a billiard, D, can havemany different shapes. We can choose this boundary to be polygons and thereare a special type of polygons called rational polygons, which we will define now.

4.1. BILLIARDS 59

Figure 4.4: The illustration shows an example of a singular orbit. Note thatthe orbit terminates at the upper left corner of the billiard table since it is notpossible to determine any well-defined trajectories for the billiard ball.

Definition 4.1.8. A rational polygon is a polygon where each interior anglevi is of the form pi

qiπ where pi, qi ∈ Z such that qi 6= 0. Moreover, a polygonal

billiard is a billiard table with a rational polygon as boundary.

Example 4.1.9. Most of the polygons we are familiar with are rational poly-gons. For instance, an equilateral triangle is an example of a rational polygon.An example of a polygonal that is not rational is a triangle with one angle

√2π

since the number√

2 can not be written as the quotient between to integers,i.e. the number is irrational.

♦

Definition 4.1.10. If the path traversed by a billiard ball is dense in Ω(D),regardless of whether it is forward or backward in time, then the orbit O(x0, θ0)is said to be dense in Ω(D).

Chen and Niemeyer combine the results from E. Gutkin and C. Judge in [16]and [15] to state the following theorem. However, we need one more definition forthe theorem, which deals with translation curves. LetM be a closed topologicalsurface with genus g ≥ 1. According to H. Masur, [24], a translation surface is asurface that contains a finite set of points, denoted Σ = x1, x2, ..xk, and withan open cover of M \ Σ by sets Uα together with charts ϕα : Uα → R2. Thecharts ϕα : Uα → R2 fulfill that for every α and β with Uα ∩ Uβ we have that

ϕα ϕ−1β (v) = v + c. (4.6)

So, v is a vector in R2 and c is a translation vector. For more information, see[24]. We follow the same author in [24] to formulate the definition.


Definition 4.1.11. A translation surface is a finite number of (Euclidean)polygons ∆1,∆2, ...,∆n such that the boundary of each triangle is orientatedin such a way that the triangle lies to the left of the boundaries and for every1 ≤ j ≤ n and for all orientated sides, denoted sj , of ∆j there exist a k suchthat 1 ≤ k ≤ n together with an orientated side sk of ∆k such that sj and sk areparallel and have the same length. The sides are glued together in the oppositeorientation by a parallel translation.

Remark 4.1.12. The part of the definition that describes how the sides areglued together means that if we walk on one of the edges that are glued together,we will have one triangle to the left and another to the right.

Example 4.1.13. Start with an equilateral triangle with corners in the points(−1, 0), (1, 0) and (0, 1) as Ω(D). It is then possible for us to create a translationsurface in the following way. We copy the triangle and form a set of polygons∆1,∆2, ...,∆n where all the ∆k, k = 1, 2, ..., n are exact copies of the originaltriangle. We now place them in the plane in such a way as seen in Figure 4.5.

Figure 4.5: This illustrates the beginning of a translation surface that has beenconstructed from the triangle mentioned in Example 4.1.13.

In [8] we also see a theorem, which relies on integrable billiards.

4.1. BILLIARDS 61

Definition 4.1.14. Let Ω(D) be a billiard table. If Ω(D) tiles the plane it iscalled an integrable billiard table.

We then cite the following theorem from [8] which refered to [15]. The origi-nal theorem in [15] requires branch coverings, but Theorem 4.1.15 is equivalent.

Figure 4.6: This billiard table has a rectangle as boundary and this is an exampleof a mathematical billiard. Please note that the boundary must not necessarilybe a polygon. The boundary could be an ellipse or another continuous curve inthe plane.

Theorem 4.1.15 ([8], [15]). Let Ω(D) be a rational billiard table that is tiled byan integrable billiard table Ω(P ). Then an orbit on Ω(D) is closed if and only ifthe orbit has an initial direction that is rational. In addition, an orbit on S(D)is dense if and only if the orbit has an initial direction that is irrational.

We will now, briefly, discuss the genus of a translation surface that has beenobtained from a rational polygonal billiard. We follow [8] in order to formulatethe following theorem.

Theorem 4.1.16. Let the boundary of the billiard Ω(D) be a rational polygonwith k sides and internal angles of the form pi

qiπ for i = 1, 2, ..., k and let N =

lcmq1, q2, ..., qk. The genus, denoted g, of a translation surface constructed fromΩ(D) is then given by

g = 1 +N

2

(k − 2−

k∑i=1

1

ni

). (4.7)


4.2 Unfolding of Billiard OrbitsLet Ω(D) be a polygonal billiard and let O(x0, θ0) be an orbit of the billiard withstart point x0 and start angle θ0. Assume that there is a base point x1 that lieson the orbit in which a reflection is made. This means that we have an ingoingtrajectory to an edge of the polygonal billiard. Imagine now that, instead ofletting the billiard ball reflect in the side, we unfold the table so that we followthe original trajectory in a straight line. See Figure 4.8 for an illustration. Wethen see that the period length of the orbit O(x0, θ0) is the same as the numberof copies of the billiard table that are made while unfolding.

.

Figure 4.7: We have here unfolded the closed orbit seen in the first billiardΩ(x0, θ0). In the end we obtained a straight line, which means that the periodlength of this orbit is 4 since we made 4 copies of the original billiard.

We see that the way of unfolding a billiard is built on translation surfaces.Chen and Niemeyer [8] claim that there is an equivalence between the dynamicalflow on the billiard, and the corresponding geodesic flow on the translationsurface.

4.3. FRACTAL BILLIARDS 63

The Sierpinski carpet is an example of a self-similar fractal and because ofthis, people often talk about cells of the carpet. We use the same formulationof the definition of a cell as Chen and Niemeyer did in [8].

Definition 4.2.1. Let a0 = 2k0 + 1 with k0 ∈ N. Consider a partition of theunit square Q = S0 into a20 squares of side-length a−10 . A subsquare of thepartition is called a cell of S0 and is denoted by C0,a0 . Let Sa be Sierpinskicarpet. Consider a partion of the prefractal approximation Sa,n into subsquareswith side-length (a · a1 · · · an)−1. A subsquare of the partition of Sa,n is calleda cell of Sa,n and is denoted by Cn,a·a1···an and has side-length (a · a1 · · · an)−1.

In the same fashion we will talk about orbits of fractal cells. Therefore adefinition is in order. The definition is needed for the study of unfolding fractalbilliards. We see that the orbits are repeated in the cells, which makes it easierfor us to study cells instead of entire fractals.

Definition 4.2.2. Let Ck,a denote the cell of the k:th prefractal approximationSa,k.Now, consider the boundary of a cell Ck,ak of Ω(Sa,k) as a barrier to the billiardflow. Then an orbit determined by reflecting in the boundary of the cell andwith an initial condition contained in the cell is called an orbit of the cell Ck,aof Ω(Sa,k).

4.3 Fractal BilliardsIn this section we study billiards on fractals. Mostly we will consider Sierpinskicarpets but we will cite the definitions necessary to built up a more generaltheory. Initially we use On(x0,n, α0,n) two denote the orbit of the n:th approxi-mation of the Sierpinski fractal Sa, with starting point in (x0,n, α0,n). Moreover,we make the assumption that the direction vector corresponding to the angleθ0,n is pointing inwards for all approximations n.

Definition 4.3.1. Let n and m be two nonnegative integers such that n > mwith (x0,n, θ0,n) ∈ (Ω(Sa,n)×S1)/ ∼ and (x0,m, θ0,m) ∈ (Ω(Sa,m)×S1)/ ∼ as ini-tial conditions of the two orbits On(x0,n, θ0,n) and Om(x0,m, θ0,m). We call theconditions On(x0,n, θ0,n) and Om(x0,m, θ0,m) compatible initial conditionsif θ0,n = θ0,m and if x0,n and x0,m lie on a segment determined from θ0,n or θ0,mthat intersects the boundary of the Sierpinski carpet, ∂Sa,n, only at x0,n. More-over, we denote that On(x0,n, θ0,n) and Om(x0,m, θ0,m) are compatible initialconditions with (x0,n, θ0) and (x0,m, θ0) respectively.

Definition 4.3.2. If two orbits Om(x0,m, θ0,m) and On(x0,n, θ0,n) have com-patible initial conditions we call the orbits compatible.


Figure 4.8: This is an example of two compatible initial conditions for two orbitson two billiards.

It is also possible to study sequences of initial conditions. Such sequencesalso have the property of being compatible. What this means is given by thefollowing definition.

Definition 4.3.3. Consider a sequence (x0,i, θ0,i)∞i=N of initial conditionsfor some N ∈ N. The sequence is called a sequence of compatible initialconditions if we for all n > m ≥ N have that (x0,n, θ0,n) and (x0,m, θ0,m) arecompatible initial conditions. We then write the sequence as (x0,i, θ0)∞i=N .

The same can be done with orbits, which is shown in the following definition.

Definition 4.3.4. Let (x0,i, θ0)∞i=N be a sequence of compatible initial condi-tions. Then the sequence of orbits, related to the compatible initial conditions,is called a sequence of compatible orbits and is denoted O(x0,n, θ0)∞n=N .

We do not present the proof of Theorem 4.2 in the article, but we presentthe theorem itself:

Theorem 4.3.5. Let θ ∈ Sl(Sa).

1. If θ ∈ Aa-2, then the sequence of compatible orbits O((0, 0), θ)∞n=0 is asequence of compatible periodic orbits.

2. If θ ∈ Ba and r < 2an is an odd positive integer, then the sequence ofcompatible orbits O(( r

2an , 0), θ)∞n=0 is a sequence of compatible periodicorbits.


4.3.1 Properties of Slopes on the Sierpinski Carpet

A refinement of Theorem 4.1 in [11] concerning the sets of slopes, A and Bdefined earlier, was made in [8]. We cite the original theorem that E. Duran-Cartagena and J. Tyson formulated. Recall the generalization of Theorem 4.0.5.We will now present it in its original form from [11].

Theorem 4.3.6. ([11], [8]) Let a = 1a ,1a ,

1a , ...) be a constant sequence. Then

the set of slopes Sl(Sa) is the union of the follow two sets:

Aa =pq


and

Ba =pq


Moreover, if α ∈ Aa, then each nontrivial line segment in Sa with slope αtouches vertices of peripheral squares, while if α ∈ Ba, then each nontrivial linesegment in Sa with slope α is disjoint from all peripheral squares. For eachα ∈ Aa ∪ Ba, there exists a maximal nontrivial line segment in Sa with slopeα. Finally, if b < a then any maximal nontrivial line segment in Sb is alsocontained in Sa. In particular, Sl(Sb) ⊂ Sl(Sa).

Remark 4.3.7. In section 3 of the article there is a discussion about the proof ofthis theorem, originally given in [11] as Theorem 4.1. The first part of the proofrelies on the fact that nontrivial line segments with slope α ∈ A necessarilyavoid peripheral squares. The authors of [11] claim that such line segmentsmust intersect corners of the peripheral squares. However, this is not alwaysthe case. Chen and Niemeyer present examples of non-trivial line segmentsthat do not intersect any corners of the peripheral squares. We list two of thecounterexamples they gave:

(i) Consider S7 and a nontrivial segment line that starts in ( 12 , 0) with slope

α = 23 ∈ A.

(ii) Consider again S7 and a nontrivial segment line that starts in (0, 0) withslope α = 2

3 ∈ A.

The refinement of Theorem 4.3.6 will be presented later together with aproof. In order to do so we need the lemmas below. All of them are given in [8]So first, let Aa-2 be the set


Aa-2 =pq

∣∣∣ p+q ≤ a−2, 0 ≤ p ≤ q, with p, q ∈ N∪0 such that p+q is odd

and let us formulate the following lemma.

Lemma 4.3.8. Let Sa be a self-similar Sierpinski carpet. Then

Aa/Aa-2 =pq

∣∣∣ 1 ≤ p < q, p+ q = a and p and q are relatively prime.

Proof. If α ∈ Aa it is possible to write α as α = pq where p and q have the same

properties as in 5, so the integers p and q satisfy in particular that

p+ q ≤ a and 0 ≤ p < q ≤ a− 1

and in the same way we see that if α ∈ Aa-2 then we have the constraints

p+ q ≤ a− 2 and 0 ≤ p < q.

Since the set difference Aa/Aa-2 will contain quotients of p and q such thateither the conditions given in Aa or in Aa-2, we see that for α ∈ Aa/Aa-2 wehave αpq such that either

p+ q ≤ a and 0 ≤ p < q ≤ a− 1

or

p+ q ≤ a− 2 and 0 ≤ p < q and q ≥ 3.

hold. Let us see what these two conditions give us. The first constraint,p+ q ≤ a and 0 ≤ p < q ≤ a− 1, gives that p+ q = a and therefore we see thatthe sum p+q is odd, since a is an odd integer. If we study the second constraint,p + q ≤ a − 2 and 0 ≤ p < q and q ≥ 3., we see that p = a − 1 and q = a − 2is the only possible values on p and q. Even here, the sum will be odd since ifwe consider p+ q = a− 1 + a− 2 = 2a− 3 = 2a− 2− 1 = 2(a− 1)− 1, whichis an odd number. Because p + q must be odd at the same time as p + q ≤ awe see that the only possible values are p = 2 and q = 1 or p = 1 and q = 2.The disjunction of the two constraints will contain all integers p and q such thatp+ q = a, 1 ≤ p < q and gcd (p, q) = 1. This proves the lemma.

These constraints imply that the sum p + q must be odd and that p and qare relatively prime.


Another important lemma in order to present the refinement of Theorem4.2 in [11] is Lemma 4.3.10. In order to prove it we need a result from numbertheory, which we learned in the first year maths, namely:

Lemma 4.3.9. Consider the Diophantine equation ax + by = k where k is aconstant and a, b ∈ N. If (x0, y0) is an integer solution to the given equationand if gcd(a, b)|k, then all integer solutions are given by

x = x0 +m · bgcd (a,b)

y = y0 −m · agcd (a,b) ,

(4.8)

where m ∈ Z.

For a proof of this lemma we recommend [28]. We will need another lemmafor the main theorem. We use the formulation and proof formulated in [8, p.16].

Lemma 4.3.10. Let Sa be a self-similar Sierpinski carpet. If α ∈ Ba, n ≥ 1and r and odd positive integer with r ≤ 2an, then the line y = α(x− r

2an ) avoidspoints of the plane of the form

(ual, vam

), where u, v, l,m ∈ Z.

Proof. We will prove this using the principle of contradiction. See Sa as a(fractal) billiard and assume that α ∈ Ba, n ≥ 1 and that r is a positive oddinteger with r ≤ 2an. Moreover, assume that there exist constants u, v, l,m ∈ Zsuch that the line y = α(x− r

2an ) passes through the point(ual, vam

). This means

that we have

v

am= α

( ual− r

2an

)(4.9)

if we substitute x and y. Now, multiply both sides with 2al+m+n we get equiv-alently

2van+l = αam(

2uan − ral)

(4.10)

Use that α ∈ Ba implies that α = pq , where p and q are both positive odd

integers, to rewrite (4.10) to

2qvan+1 = pam(2uan − ral). (4.11)

The equality (4.11) can not possibly hold since the left hand side is evenand the right hand side odd, since r is odd. This gives a contradiction and thelemma is proved.


It is also crucial for us to be able to determine whether a nontrivial linesegment lies in the Sierpinski carpet or not. Therefore we use the followinglemmas, both formulated in [11].

Lemma 4.3.11. If there is a nontrivial line segment of slope α emanating froma point (c, 0), c ∈ [0, 1], and contained in the carpet Sa, i.e. if the set

LSa = (x, y) ∈ Sa∣∣ y = α(x− c) (4.12)

contains a line segment containing (c, 0), then the line emanating from theorigin, i.e. c = 0, does not intersect any member of the collection

aZ2 +Qx,y := (ak, al) +Qx,y∣∣ (k, l) ∈ Z2, (4.13)

where

Qa = (a, y) ∈ R2∣∣ − 1 < x < 0, 0 < y < 1.

The next lemma will require the concept of a Farey sequence, named afterJohn Farey, who coined this term in 1816. We cite the following from S. Das etal., [29], in order to give the definition.

Definition 4.3.12. A Farey sequence Fn of order n is a sequence of irre-ducible, proper and positive fractions that have denominators less than or equalto n, and are arranged in increasing order of their values.

Lemma 4.3.13. Farey sequences have the following properties

(i) Fn ⊂ Fn+1 If p1/q1 < p2/q2 are consecutive in Fn and separated inFn+1, then the fraction p1+p2

q+q2lies in between p1/q1 and p2/q2 and no

other elements of Fn+1 lie between p1/q1 and p2/q2.

(ii) If p1/p2 and p2/q2 are consecutive in any Fn, then they satisfy the uni-modular relation: p1 · q2 = p2 · q−1.

(iii) If pq = p1+p2

q1+q2, then we have that p2q − q2p = pq1 − qp1 = 1 and p and q

are coprime.

Lemma 4.3.14. Assume that p1/q1 and p2/q2 are consecutive fractions inSl(Sa), both in reduced form. Then p1/q1 and p2/q2 are separated in Sl(Sa+2)if and only if p1 + q1 + p2 + q2 ≤ a+ 2.

Let us now formulate the refinement of Theorem 4.3.6.


Theorem 4.3.15. Let a = 1a ,1a ,

1a , ...) be a constant sequence. Then the set

of slopes Sl(Sa) is the union of the following two sets:

Aa =pq


Ba =pq


Moreover, if α ∈ Aa/Aa-2, then each nontrivial line segment in Sa with slopeα beginning from (0, 0) touches vertices of peripheral squares, while if α ∈ Aa-2,then each nontrivial line segment in Sa with slope α beginning from (0, 0) isdisjoint from all peripheral squares. If α ∈ Ba, then each nontrivial line segmentin Sa with slope α beginning at ( r2a , 0), n ≥ 1 and r < 2an being a positive oddinteger, is disjoint from all peripheral squares with side-length < 1

an .In addition, for each α ∈ Aa ∪ Ba, there exist maximal line segments in Sa

with slope α. Finally, ifb < a, then any maximal nontrivial line segment in Sbis also contained in Sa. In particular, Sl(Sb) ⊂ Sl(Sa).

The proof in [8] focuses on the cases where α ∈ Aa/Aa-2 since the proofgiven in [11] is valid when α ∈ Ba. We present a complete proof using both [8]and [11] for the different cases.

Proof. ([8], [11]) Assume that α ∈ Aa ∪ Ba. First of all we will consider thecase when α ∈ Ba. The proof idea is to show that the lines y = α(x− c) do notintersect any peripheral squares.Therefore, let L be defined as the set of pointsgiven by:

L :=

(x, y) ∈ R2∣∣ y = α(x− 1

2).

The claim here is that L never meets any peripheral squares of Sa. Assumeon the contrary that the claim is false, i.e. that there exists a point (x, y) ∈ Lthat also is contained in a peripheral square. There are also odd non-negativeintegers p and q such that p+ q ≤ a− 1, 0 ≤ p ≤ q ≤ a− 2 since α ∈ Ba. Thisgives us the following equivalence:

y = α(x− 1

2) ⇐⇒ p

2+ qy = px.

The fact that we can express p2 as

p

2=(p− 1

2,a− 1

2,a− 1

2,a− 1

2, ...)a

in base a since we for |a| > 1 have


p− 1

2+

1

2=(p− 1

2,a− 1

2,a− 1

2,a− 1

2, ...)a

=p− 1

2a0+

∞∑k=1

a− 1

2ak=p− 1

2+

1

2=p

2.

We know that the point (x, y) is inside a peripheral square. Moreover, wecan express x and y in base a as x = (0, x1, x2, ...)a and y = (0, y1, y2, ...)a if wehave

x =

∞∑k=1

xkak

and y =

∞∑k=1

ykak. (4.14)

This means that (x, y) /∈ Sa and thus we have the inequalities

(0.x1, x2, ..., xn−1,

a− 1

2, 0, 0, ...

)a< x <

(0.x1, x2, ..., xn−1,

a+ 1

2, 0, 0, ...

)a

(4.15)and

(0.y1, y2, ..., yn−1,

a− 1

2, 0, 0, ...

)a< y <

(0.y1, y2, ..., yn−1,

a+ 1

2, 0, 0, ...

)a.

(4.16)It is possible to see that the inequalities hold geometrically. Observe that the

lowest corner to the right in a peripheral square of Sa is ( a+12an+1 ,

a−12an+1 )+

(uan ,

van

)for some u, v ∈ N such that u 6= v and 0 ≤ u, v ≤ an. The coordinates for thebottom left corner are in the same fashion ( a−1

2an+1 ,a−1

2an+1 ) +(uan ,

van

). Since the

x-coordinate of the point (x, y) lies inside the peripheral square, the inequalityfor x follows from the estimation

(0.x1, x2, ..., xn−1,

a− 1

2, 0, 0, ...

)a

=

n−1∑k=1

xkak

+a− 1

2an< x =

∞∑k=1

xkak

<xkak

+a+ 1

2an=

=(

0.x1, x2, ..., xn−1,a+ 1

2, 0, 0, ...

)a. (4.17)

The inequality for y can be derived in the same way. If we now multiply(4.15) with p and (4.16) with q we get

(x0, x1, x2, ..., xn−1,

(a− 1)p

2, 0, 0, ...

)a< px <

(x0, x1, x2, ..., xn−1,

(a+ 1)p

2, 0, 0, ...

)a

(4.18)


and

(y0, y1, y2, ..., yn−1,

(a− 1)q

2, 0, 0, ...

)a< qy <

(y0, y1, y2, ..., yn−1,

(a+ 1)q

2, 0, 0, ...

)a,

(4.19)where xj and yj for i = 0, 1, 2, , , are the values obtained on the position

when xj and yj are multiplied by p and q respectively. Now we add p2 to (4.19)

and get

(p− 1

2+y0,

a− 1

2+y1,

a− 1

2+y2, ...,

a− 1

2+yn−1,

(a− 1)(q + 1)

2,a− 1

2,a− 1

2, ...)a<

<p

2+ qy <

<(p− 1

2+y0,

a− 1

2+y1,

a− 1

2+y2, ...,

a− 1

2+yn−1,

q(a+ 1) + (a− 1)

2,a− 1

2,a− 1

2, ...)a.

(4.20)We can now equivalently write (4.18) as(

x0, x1, x2, ...,p− 1

2+ xn−1,

a− p2

, 0, 0, ...)a< px <

<(x0, x1, x2, ...,

p− 1

2+ xn−1,

a+ p

2, 0, 0, ...

)a, (4.21)

since for instance(x0, x1, x2, ..., xn−1,

(a− 1)p

2, 0, 0, ...

)a

=

n−1∑k=0

xkak

+(a− 1)p

2an=

n−2∑k=0

xkak

+xn−1an−1

+p

2an−1− p

2an=

=

n−2∑k=0

xkak

+1

an

(xn−1+

p− 1

2an−1

)+a− p2an

=(x0, x1, x2, ...,

p− 1

2+xn−1,

a− p2

, 0, 0, ...)a.

In the same way we are able to rewrite (4.20) as

(p− 1

2+y0,

a− 1

2+y1, ...,

a+ q − 2

2+yn−1, a−

(q + 1)

2,a− 1

2,a− 1

2, ...)a<p

2+qy <

<(p− 1

2+ y0,

a− 1

2+ y1, ...,

a+ q

2+ yn−1,

(q − 1)

2,a− 1

2,a− 1

2, ...)a. (4.22)


To make this believable we make the following calculations, starting from(4.20).

(p− 1

2+y0,

a− 1

2+y1,

a− 1

2+y2, ...,

a− 1

2+yn−1,

(a− 1)(q + 1)

2,a− 1

2,a− 1

2, ...)a

=

=p− 1

2+ y0 +

n−1∑k=1

(a− 1

2ak+ykak

)+

(a− 1)(q + 1)

2an+

∞∑k=n+1

a− 1

ak=

=p− 1

2+y0+

n−2∑k=1

(a− 1

2ak+ykak

)+a− 1

2an−1+yn−1an−1

+q

2an−1+a− q − 1

2an+

∞∑k=n+1

a− 1

ak=

=p− 1

2+y0+

n−2∑k=1

(a− 1

2ak+ykak

)+

1

an−1

(a+ q − 2

2+1+yn−1

)+−q − 1

2an+

∞∑k=n+1

a− 1

ak=

=p− 1

2+y0+

n−2∑k=1

(a− 1

2ak+ykak

)+

1

an−1

(a+ q − 2

2+yn−1

)+

2a− q − 1

2an+

∞∑k=n+1

a− 1

ak=

(p− 1

2+ y0,

a− 1

2+ y1, ...,

a+ q − 2

2+ yn−1, a−

(q + 1)

2,a− 1

2,a− 1

2, ...)a.

We now use the conditions on p and q, which are given from the fact thatα ∈ Ba, to obtain

a+ p

2≤ a− q − 1

2and

q − 1

2≤ a− p

2. (4.23)

Now, let us make a conclusion. We combine the inequalities (4.21), (4.22)and (4.23) in order to see that there can not be a point (x, y) that satisfies all ofthem. This is true since the first (n − 1) elements in the boundary expansionsin (4.21) are equal, i.e. the coefficients for xk, k = 1, 2, ..., n − 1 are equal.However, this is not the case for (4.22), where the (n − 1):th coefficient of theright hand side is one greater than the (n−1)th coefficient of the left hand side.Moreover, the inequalities (4.23) imply that there is no x and y such that (4.15)and (4.16) hold. The geometric interpretation of this is that the gap between xand y makes it impossible for the pair (x, y) to be inside a peripheral square.


We also need to show that for n ≥ 1, r < 2an and α ∈ Ba all nontrivialline segments in Sa, beginning at ( r2a , 0), are disjoint from all peripheral squareswith side-length strictly less than 1

an . To show this, let n ≥ 0 and r < 2an bean odd positive integer. The idea is to study a nontrivial line segment startingfrom ( 1

, 0) in Sa and then scale it so the line segment fits a cell of the carpet.The scaling factor is 1

ak, k ≥ n, since we want a nontrivial line segment that fits

the cell. This means that there is a non-positive constant c ≤ ak such that if wetranslate the line segment by c

akwe move the segment to a new cell with base

midpoint ( r2an , 0). The segment that was translated still has slope α. If we fill

every possible cell of Σa with translated nontrivial line segments we get a tilingof the plane such that the line y = α(c− r

2an ) never touches the corners of anyperipheral square. This follows from Lemma 4.3.10.

Next, assume that α ∈ Aa\Aa-1 and consider an arbitrary line segment pass-ing through the origin with slope α. In the reasoning of Chen and Niemeyer, [8],we show that this line segment intersects the lower right corner of a peripheralsquare of Sa. If the line segment intersects the corner, the equation

1

p

(v +

a− 1

2a

)=

1

q

(u+

a+ 1

2a

)(4.24)

must hold where u 6= v are non-negative integers and 0 ≤ u, v ≤ an. Recallthat the lower right corner has coordinates Sa is ( a+1

2an+1 ,a−1

2an+1 ) +(uan ,

van

)for

the constants u and v.We will now follow the footsteps of Chen and Niemeyer and use Lemma 4.3.8

to conclude that p+ q = a for all α ∈ Aa \Aa-1. Hence,

1

p

(v +

a− 1

2a

)=

1

q

(u+

a+ 1

2a

)⇐⇒ 2qv +

q(a− 1)

a= 2pu+

p(a+ 1)

a

and if we use that p+ q = a, which gives 2p = 2a− 2q and therefore we get

2qv+q(a− 1)

a= 2pu+

p(a+ 1)

a⇐⇒ (2a− 2q)u− 2qv =

q(a− 1)

a− p(a+ 1)

a,

which gives us equivalently

2(a− q)u− 2qv = 2q − a− 1. (4.25)

See (4.25) as a straight line in the uv-plane and define a lattice in that planegiven by the set

(u, v)∣∣u, v ∈ Z ∩ [0, an − 1],


where we demand that u 6= v. The lattice is then the Cartesian product ofall integer values in the interval [0, an−1]. The purpose of the lattice is to pointout integer points in order for us to show that the line (4.25) passes trough alower right corner of any peripheral square. Since both u and v are integers,we are only interested in intersection points that are integers. First, notice thatif 2q − a − 1 = 0 or p = q − 1, we get (0, 0) as an intersection point. If theright hand side is not zero, we move along the line (4.25) and search for integervalues on u and v. To find these points, let consider different integer values ofu, namely u ∈ (u, v)

∣∣u, v ∈ Z ∩ [0, an − 1], in increasing order starting at (0, 0)and then check if v is an integer. The v-coordinates can be expressed as:

v =(a+ 1

2q− 1)

+ u(a− q

q

)=

2au+ a+ 1

2q− u− 1 =

1

q

(a+ 1

2+ au

)− u− 1.

(4.26)This means that if v is an integer, then q

∣∣∣(a+12 + au

), which gives us the

following Diophantine equation:

au+ qr = −a+ 1

2, (4.27)

where r is an integer. We want to solve this equation with the help of Lemma4.3.9 and since p

q ∈ Aa \ Aa-2 we know, from Lemma 4.3.8, that p and q arerelatively prime, i.e. gcd(p, q) = gcd(a, q) = 1. Lemma 4.3.9 now yields possiblesolutions (us, rs) ∈ 0, 1, ..., q−1×Z to (4.27) and we then get the point (us, vs)with

vs =2au+ a+ 1

2q− u− 1 =

1

q

(a+ 1

2+ aus

)− us − 1.

This means that the line intersects the lowest right corner of any peripheralsquare.

Furthermore, let us study the case when α ∈ Aa-2. Make the assumptionthat a nontrivial line segment, emerging from (0, 0) with slope α, intersects aperipheral square with lower right coordinates ( a+1

2an+1 ,a−1

2an+1 ) +(uan ,

van

). Since

α ∈ Aa-2 we get the same equation as (4.24). Simplification2 now provides uswith the following, equivalent, equation:

2vq − 2up− p+ q =p+ q

a. (4.28)

2Observe that α ∈ Aa-2, which means that p + q ≤ α. We do not have an equality as wehad in the case for α ∈ Ba.


Observe now that this yields a contradiction. The left hand side is an integer,while the right hand side is not. This is the case since p + q ≤ α − 2 and thatα − 2 < α. The conclusion is therefore that the line segment can not intersectthe lower corner of any peripheral square.

The next part of the proof is the case when α 6∈ Aa ∪ Bb. We then wantto prove that there cannot exist any nontrivial line segments in Sa. To do thiswe use Lemma 4.3.11 and show that the line L, passing through (0, 0), of slopeα intersects the planar tiling aZ2 + Qx,y. Note that L might cross the inferiorright vertex of any peripheral square and those that do therefore have slopel/k. However, if L crosses the superior left vertex of any peripheral square, thenL has slope (al + 1)/(ak − 1). Thus, another line L′, with another slope β,intersects a square from aZ2 +Qx,y if and only if

l

k< β <

al + 1

ak − 1,

where 0 ≤ l < k and gcd(k, l) = 1.Due to the fact that α is a nontrivial line segment, we can always find two

slopes p1q1, p2q2 ∈ Sl(Sa) such that

p1q1

< α <p2q2. (4.29)

Letαn =

p1 + np2q1 + nq2

for integers n ≥ 0 be the iterative mediants3 of αn. Then we have

αn =p1 + np2q1 + nq2

< β <a(p1 + np2) + 1

a(q1 + nq2)− 1

for n ≥ 0. Moreover, if(p1q1,p2q2

)⊂∞⋃n=0

(p1 + np2q1 + nq2

,a(p1 + np2) + 1

a(q1 + nq2)− 1

), (4.30)

we can conclude that the line L must intersect at least one peripheral squaregiven from the tiling. In order to show (4.30) we need to show that the doubleinequality

p1 + (n− 1)p2q1 + (n− 1)q2

<p1 + np2q1 + nq2

<a(p1 + (n− 1)p2) + 1

a(q1 + (n− 1)q2)− 1(4.31)

3A mediant of two fractions, ab

and cd, is the ratio between the sum of the numerators and

the sum of the denominators of the fractions, i.e. a+bc+d

.


holds for all n ≥ 0. The first inequality follows from (4.29) and the rightinequality can be written as

a(q1p2 − p1q2) < n(p2 + q2) + p1 + q1

and Lemma (iii) tells us that q1p2−p1q2 = 1, which reduces the problem to justshowing that

a < p2 + q2 + p1 + q1

and we know that p1/q1 and p2/q2 are consecutive fractions, both contained inSl(Sa). Lemma 4.3.14 now tells us that the inequality (4.3.1) holds.

Finally, the conclusions of all of the cases discussed above prove the theorem.

We have given a survey of the paper of Chen and Niemeyer. A note regardingTheorem 4.1.15 is that the proof requires advanced homological methods andbranched coverings, which are beyond the scope of this thesis. We provide thenext chapter with an elementary proof of the theorem given a billiard flow onthe Sierpinski carpet when the flow starts at the origin.

Chapter 5

Results: Periodicity of Orbitson Sierpinski Carpets

In this chapter we give an original and elementary proof concerning the period-icity of a billiard orbit on Sierpinski carpets.

We know from [8] that the possible slopes for the nontrivial line segmentsare given by

Sl(Sa) = Aa ∪Ba, where

Aa =pq

∣∣∣ p+q ≤ a, 0 ≤ p ≤ q ≤ a−1, with p, q ∈ N∪0 such that p+q is odd

and

Ba =pq


The reflections on Sierpinski carpets are well-defined except in the cornerpoints. Hence, we will define the reflection in a corner point as a reflection inthe bisector given at that point. We will only try to answer this question forthose starting points that give sequences of periodic orbits. From part 1. ofTheorem 4.3.5 we know that O((0, 0), θ)∞n=0 consists of periodic orbits, whichmeans that the orbits are closed. This statement relies on Theorem 4.1.15 andproof of that theorem can be found in [15], where orbifolds and orbifold-coveringsare used. These concepts are beyond the scope of this thesis. However, in thecase with Sierpinski carpets we have a square as boundary to the billiard flow.We will prove that this flow is periodic. The proof idea is to follow the billiardflow, initially starting at (0, 0), and prove that we must hit another corner point.

Landstedt, 2017. 77

78CHAPTER 5. RESULTS: PERIODICITY OF ORBITS ON SIERPINSKI CARPETS

If we can prove this, then the rest follows by symmetry since we define reflectionat the corner point as the reflection in the bisector between two adjacent squaresides. Note that the square under study has corner points; (0, 0), (1, 0), (0, 1)and (1, 1). We do not consider any peripheral squares because the allowed slopesfor the Sierpinski carpets are given in Theorem 4.3.15. Nonetheless, we will onlyassume rational slopes since all the slopes given in that theorem are rational.We will present the results as theorems (and one lemma) but we will first makea definition.

Definition 5.0.1. We call the number of reflection points given by a periodicorbit on a Sierpinski carpet the order of the orbit. We will denote this withnr(p, q) where p, q ∈ Q.

Lemma 5.0.2. Assume that θ = pq ∈ Aa-2 where

Aa-2 =pq

∣∣∣ p+q ≤ a−2, 0 ≤ p ≤ q ≤ a−3, with p, q ∈ N∪0 such that p+q is odd.

(5.1)If pq ≤ 1 and q

p ∈ Z then x qp−1 is a corner point. Moreover, if qp is odd, then

x qp−1 = (1, 1) and if q

p is even, then x qp−1 = (0, 1).

Proof. That the nontrivial line segments does not intersect any peripheral squaresfollows from the construction of Aa-2, see [8]. Let ABC be the triangle spannedby the points A = (0, 0), B = (1, 0) and C = (p, q). Then the slope of thehypotenuse is p/q. When we reflect ABC in the nontrivial line segment goingthrough (1, p/q) we get a new triangle A′B′C ′ where A′ = (0, 2pq ), B′ = (1, 2pq )

and C ′ = C. Since the height of ABC is p/q and the height of Sa is one we seethat since

1

p/q=q

p

is an integer, the billiard must hit a corner after qp−1 reflections. If qp is odd,

then the line segment starting from (0, 0) will have positive slope, which meansthat it will also end with positive slope since we reflect it in an even number oftrivial line segments, i.e. we hit (1, 1). In the same way we see that if qp is even,we must have an odd number of trivial reflection lines and therefore end up in(0, 1).

Consider the case when qp is even. The case when q

p is odd is proved inthe same way. The definition of reflection at a corner singularity says thatwe should reflect the line segment in the bisector of the corner. Let An bethe n-th representation of the point A after n reflections in the trivial line

79

segments τ1, ..., τn−1. Then we see that |CnBn| = p/q and the next step is toreflect triangle AnBnCn in the line y = −x + 1. Call the resulting triangleDEF , where the image of An = D, which is invariant, Bn = E = (0, 0) andCn = F = (0, p/q). Observe that we now can apply the same result to concludethat we must hit Bn = (1, 1), due to symmetry. If we then reflect in the bisectory = x, we get in the same way another sequence of flipping triangles until wehit the corner (0, 1). One last iteration sequence of reflections give that we stopat (0, 0) where we began. This shows that if pq ≤ 1 and q

p ∈ Z then the billiardorbit is closed.

Theorem 5.0.3. Consider the slopes θ of nontrivial line segments in the Sier-pinski carpet Sa given in Aa-2. Then the orbit starting at a corner point withθ0 = p/q is closed and the order of the orbit is

nr(p, q) =

2p+ 2q − 2, if q −⌊(p−1)qp

⌋− 1 is even,

4p+ 4q − 4, if q −⌊(p−1)qp

⌋− 1 is odd.

Proof. The case when qp is an integer is covered by Lemma 5.0.2. We can assume

that pq < 1 since the case when p

q > 1 will be symmetric. We know that theorbit will intersect the segment stop = (x, y) ∈ R2 | 0 < x < 1 and y = 1. Wewant to show that the reflection sequence will hit a corner of Sa. First, we seethat if the lower integer part of q

p is even, then the last reflection point beforethe billiard ball hits Stop is on x = 0. If the lower integer part of qp is odd, thenthe last reflection point is on x = 1. Hence we will study these cases separately.Let y0 = p

qx be the line intersecting the origin, representing the first nontrivialline segment. Then the first reflection point, say P1, is given by

P1 =(

1,p

q

). (5.2)

Since we reflect y0 in the line y = pq in order to obtain the next nontrivial

line segment, we can construct a line y1 = −pqx+ 2pq . This is the line y0 reflected

in y = pq . We get that

Pn =

(

0, n·pq

), if n is even

(1, n·pq

), if n is odd.

(5.3)


First we will study the case when n is even. In order to find the reflectionpoint on x = 0 that sends yn to intersection with Stop we study

minn∈N

∣∣∣pq· n− 1

∣∣∣. (5.4)

We see that n = b qpc, which means that the point we seek is

Pb qp c =(

0,b qpcpq

).

Observe that this is the case when n is an even integer, which we will assumealways is the case for now. We know that the slope of the line yb qp c is p/q. Hence,

yb qp c =p

qx+b qpcpq

.

This implies that next point is

Pb qp c+1 =

(1− b

qp cpq

pq

, 1

). (5.5)

We know here that

1− bqp cpq

pq

6= 1 (5.6)

since p/q /∈ Z. In the same fashion as before we get

yb qp c+1 = −pqx+ 1 +

(pq

)(1− bqp cpq

pq

).

The intersection with the line x = 1 is

Pb qp c+2 =

(1,−p

q+ 2−

b qpcpq

). (5.7)

Assume that

1− bqp cpq

pq

<1

2

81

and consider the distance of d1 from Pb pq c and Pb qp c+1. Now, draw all parallellines to the line yb qp c+1 with distance (In a arctan(p/q) angle) m · d1 for m ∈N \ 0. Then the point Pb qp c+3 is the same as one of the points(

0,1−

( b qp cpq

)m · dy

), (5.8)

which are the intersections of the parallel lines to yb qp c+1, with distance d1,and the y-axis. Observe that d1y is the distance between the intersection points,i.e.

d1y =

√d21

(2− 2 · cos

(2 arctan(

p

q)))

=

=

√√√√2 ·

[(1− b

qp cpq

pq

)2

+

(1−b qpcpq

)2]·[1− cos

(2 arctan(

p

q))]

(5.9)

due to the cosine theorem. Do the same thing as before and draw all parallellines to yb pq c+2 with distance m · dy with m ∈ N \ 0. This construction willnow build a sort of grid and since our starting line goes trough the origin, weknow that if this grid intersect a corner point we are done. This happens e.g.if d1y divides

b qpcpq

. (5.10)

For instance, with p = 3 and q = 7 we get

d1y =

√√√√[(1− b 73c ·37

37

)2

+

(1−b 73c · 3

7

)2]·[1− 2 · cos

(2 arctan(

3

7))]

=

√4

49=

2

7

and since

6/7

2/7= 3

we know that it must hit a corner point. See the Figure 5.1.


Figure 5.1: In this illustration we see that with p = 3 and q = 7 we end up in acorner point.

If

1− bqp cpq

pq

≥ 1

2

we get the same result due to symmetry. Now, assume that d1y does notdivide (5.10). We then know that the grid created by the first distance d1 willnot cover the orbit of the billiard since we start in a corner point. Hence wemust study many different grids generated by the different trajectories. First,let G1 be the grid generated of d1. Now, we let d2 be the distance between aline in the first lattice and the first line that is not in the lattice given from thereflection sequence.

The first lattice failed to intersect a corner point, see for instance Figure 5.2for an illustration over the situation.

Let G′1 be the grid that contains more than 1 (the origin) corner pointconstructed with distance d1, then G1, which was assumed not to have morethan the origin as a corner point, must be a translation of L′1. Let ymax denotethe line which has the closest distance to one of the corner points such that ymax

83

Figure 5.2: This is a part of a billiard orbit that have θ0 = 29/100. As seen inthis figure, the orbit does not follow G1.

intersects the Sierpinski carpet in two points. Then we know that the reflectionof ymax in either x = 0 or x = 1 will construct L2. This is the case since thedistance d1 changes to a smaller distance d2. If d1 = d2 we would have a closedorbit. We need to find an expression for d2. We know that the lines in L1 canbe expressed as

YG1=

(x, y) ∈ R2, a ∈ Z∣∣ y = ±p

qx+ a · dy

(5.11)

and henceforth we define d2 as the length of the nontrivial line segment inYG1

⋂[0, 1]× [0, 1]

that has shortest length. Next, we form the grid G2 in

the same way as we did for G1. Observe that G1 ⊂ G2. The grid G2 can onlyhit a corner point if

1

d2y∈ N. (5.12)

We will now follow the billiard flow in the grid generated of G1, G2, ..., Gn.If we do that, then we get an equivalent description of the correlation betweenthe distances d1, d2, d3, ...dn and the closure of the orbit. First, we study thepoint Pb qp c+2. Since the slope of the lines are either p/q or −p/q we get, in the


same way as for G1, that the point in which the line ymax reflects in order tointersect with Sf is yn where

n =

⌊ −pq + 2− b

qp cpq

pq

⌋=⌊− 1− bq

pc+ 2

q

p

⌋= −1 + b2q

pc − bq

pc, (5.13)

since n > 0. This means that

Pn =

(0, −pq + 2− b

qp cpq −

⌊−1−b qp c+2 q

p

⌋·p

q

)=(

0, −pq + 2− bqp cpq − n·p

q

), if n is even(

1, −pq + 2− bqp cpq −

⌊−1−b qp c+2 q

p

⌋·p

q

)=(

1, −pq + 2− bqp cpq − n·p

q

), if n is odd.

(5.14)Consider the even case and verify that

Pn+1 =

(q

p

(−pq

+ 2−b qpcpq− n · p

q

), 0

)=

(− 1 +

2q

p− bq

pc − n, 0

). (5.15)

and that

Pn+2 =

(1,

2p

q− 2 +

b qpcpq

+pn

q

). (5.16)

If we continue we get that

n2 =

⌊1−

(2pq − 2 +

b qp cpq + pn

q

)pq

⌋=

⌊1−

(2pq − 2 +

b qp cpq +

p

⌊−pq

+2−b qpcp

qpq

⌋q

)pq

⌋=

=

⌊3q

p−2−bq

pc−⌊−1−bq

pc+2

q

p

⌋⌋=⌊−1+

3q

p−b2q

pc⌋

= −1+b3qpc−b2q

pc

(5.17)

85

and

Pn2=

(0, 2pq − 2 +

b qp cpq + pn

q +p(−1+b 3qp c−b

2qp c)

q

), if n2 is even(

1, 2pq − 2 +b qp cpq + pn


2qp c)

q

), if n2 is odd.

(5.18)

Hence, we get in the even case

Pn2+1 =

(1−

(2pq − 2 +

b qp cpq + pn


2qp c)

q

)pq

, 1

). (5.19)

Now, let us do this generally for the even case. We have, due to symmetryof the unit square and the reflection sequence, that

nn = −1 + b (n+ 1)q

pc − bn · q

pc. (5.20)

To motivate this statement we make the following reasoning. Assume thatnk >

12 and even for some k = 1, 2, 3, ..., n. The other cases follows by symmetry.

Initially we get with [Pnk−1+2]/∈Z as the current non integer coordinate for thepoint Pnk−1+2 that

ynk=p

qx+ [Pnk−1+2]/∈Z +

pnkq

and hence

Pnk+1 =

(1−

([Pnk−1+2]/∈Z + pnk

q

)pq

, 1

),

which implies that

ynk+1 =−pqx+ 1 +

p

q

(1−

([Pnk−1+2]/∈Z + pnk

q

)pq

)and

Pnk+2 =

(1,−pq

+ 1 +p

q

(1−

([Pnk−1+2]/∈Z + pnk

q

)pq

)).


This means that

nk+1 =

⌊ −pq + 1 + p

q

(1−([Pnk−1+2]/∈Z+

pnkq

)pq

)pq

⌋=

⌊−1+

q

p+

(1−

([Pnk−1+2]/∈Z + pnk

q

)pq

)⌋

=

⌊− 1 +

q

p+q

p− q

p[Pnk−1+2]/∈Z − nk

⌋=

=

⌊2q

p− 1 +

q

p[Pnk−1+2]/∈Z − nk

⌋=

=

⌊2q

p− 1 +

q

p[Pnk−1+2]/∈Z − nk

⌋.

Now, nk is given from nk−1 in the following way. We have two cases, either

ynk−1=−pqx+ [Pnk−2+2]/∈Z −

pnk−1q

(5.21)

or

ynk−1=p

qx+ [Pnk−2+2]/∈Z −

pnk−1q− p

q. (5.22)

We continue with (5.21) and get

Pnk−1+1 =

(q

p

([Pnk−2+2]/∈Z −

pnk−1q

), 0

)=

(q

p[Pnk−2+2]/∈Z − nk−1, 0

)

and

ynk−1+1 =p

qx− p

q·(q

p[Pnk−2+2]/∈Z − nk−1

),

which gives

Pnk−1+2 =

(1,p

q−(

[Pnk−2+2]/∈Z −pnk−1q

))(5.23)

and thus

87

nk =

⌊1−

(pq −

([Pnk−2+2]/∈Z − pnk−1

q

))pq

⌋=

⌊q

p−1+

q

p[Pnk−2+2]/∈Z− nk−1

⌋,

The same holds for (5.22). Hence we get

nk+1 =

⌊− 1 +

q

p+

(1−

([Pnk−1+2]/∈Z + pnk

q

)pq

)⌋=

=

⌊−1+

q

p+

(1−

(pq −

([Pnk−2+2]/∈Z − pnk−1

q

)+

p

(⌊qp−1+

qp [Pnk−2+2]/∈Z−nk−1

⌋)q

)pq

)⌋=

=

⌊−1+

q

p+q

p−1+

q

p[Pnk−2+2]/∈Z− nk−1−

⌊q

p−1+

q

p[Pnk−2+2]/∈Z− nk−1

⌋⌋=

= −1 +

⌊2q

p+q

p[Pnk−2+2]/∈Z −

⌊q

p+q

p[Pnk−2+2]/∈Z

⌋⌋=

= −1 +

⌊2q

p+q

p

(−pq

+ 1 +p

q

(1−

([Pnk−3+2]/∈Z + pnk−2

q

)pq

))−

−

⌊q

p+q

p

(−pq

+ 1 +p

q

(1−

([Pnk−3+2]/∈Z + pnk−2

q

)pq

))⌋⌋=

= −1+

⌊2q

p−1+

q

p+q

p−qp

[Pnk−3+2]/∈Z−nk−2−

⌊q

p−1+

q

p+q

p−qp

[Pnk−3+2]/∈Z−nk−2

⌋⌋=

= −1 +

⌊4q

p− q

p[Pnk−3+2]/∈Z −

⌊3q

p− q

p[Pnk−3+2]/∈Z

⌋⌋= · · · =


= −1+

⌊(k + 2)q

p− qp·0−

⌊(k + 1)q

p− qp·0

⌋⌋= −1+

⌊(k + 2)q

p

⌋−

⌊(k + 1)q

p

⌋,

since [Pnk+2]/∈Z = 0 when k < 0 since we start at the origin. This shows theprevious claim.

Note that we have restrictions on nk because the inequality

p · nkq

< 1, (5.24)

must hold for all k = 1, 2, 3, .... With this in mind we have

Pnk=

(0, [Pnk−1+2]/∈Z +

pnkq

)=

=

(0,p

q−(


)+pnkq

)=

=

(0,p

q− [Pnk−2+2]/∈Z +

pnk−1q

+pnkq

)=

=

(0,p

q−

(−pq

+ 1 +p

q

(1−

([Pnk−3+2]/∈Z + pnk−1

q

)pq

))+pnk−1q

+pnkq

)=

=

(0,p

q+p

q− 1−

(1−

([Pnk−3+2]/∈Z +

pnk−2q

))+pnk−1q

+pnkq

)=

=

(0,p

q+p

q− 1− 1 + [Pnk−3+2]/∈Z +

pnk−2q

+pnk−1q

+pnkq

)=

=

(0,p

q+p

q−1−1+

p

q−[Pnk−4+2]/∈Z+

pnk−3q

+pnk−3q

+pnk−2q

+pnk−1q

+pnkq

)= · · ·

=

(0,p · kq− 1− k

2+p

q

k∑n=0

nn

)

89

since [Pn0+2]/∈Z = −pq + 2− b

qp cpq .

Consider the point sequence [Pnk]/∈Nnk=1. We must now determine whether

this sequence is a limit point to either 1 or 0, or if there exists an N ∈ N suchthat

[PnN]/∈Z =

1

0⇐⇒

[Pnk−1+2

]/∈Z + pnk

q = 1 if k is even,[Pnk−1+2

]/∈Z + pnk

q = 0, if k is odd.

We get for even k:s that

[Pnk]/∈N =

p · kq− 1− k

2+p

q

k∑n=0

nn =

p · (k + 1)

q− 1− k

2+p

q

k∑n=0

(− 1 +

⌊(n+ 1)q

p

⌋−

⌊n · qp

⌋)=

=p · kq− 1− k

2− kp

q+p

q

k∑n=0

(⌊(n+ 1)q

p

⌋−

⌊n · qp

⌋)=

= −1− k

2+p

q

⌊(k + 1)q

p

⌋.

In order to study this, let pq = α and consider the function

ϕ(α, k) = −1− k

2+ α

⌊ (k + 1)

α

⌋. (5.25)

So, we get

Pnk−1+2 =p

q−(


)=

=p

q−

((−pq

+ 1 +p

q

(1−

([Pnk−3+2]/∈Z + pnk−2

q

)pq

))− pnk−1

q

)=

=p

q+p

q− 1− 1 + [Pnk−3+2]/∈Z +

pnk−2q

+pnk−1q

=

=p

q+p

q− 1− 1 +

(pq−(


))+pnk−2q

+pnk−1q

=


=p

q+p

q+p

q− 1− 1− [Pnk−4+2]/∈Z +

pnk−3q

+pnk−2q

+pnk−1q

= · · ·

=k · pq− k +

k−1∑n=0

pnnq.

since we must end at k = 0. This means that

k · pq− k +

k−1∑n=0

pnnq

=k · pq− k +

p

q

k−1∑n=0

(− 1 +

⌊(n+ 1)q

p

⌋−

⌊n · qp

⌋)=

pk

q− k − (k − 1)p

q+p

q

⌊(k + 1) · q

p

⌋= −k +

p

q+p

q

⌊k · qp

⌋.

Now, we must determine if [Pnk−1+2]/∈Z + pnk

q = 1. we get

[Pnk−1+2]/∈Z +pnkq

= −k+p

q+p

q

⌊k · qp

⌋+p

q

(− 1 +

⌊(k + 1)q

p

⌋−

⌊k · qp

⌋)=

= −k +p

q

⌊(k + 1)q

p

⌋= 1

so if we let k + 1 = p we get

−(p− 1) +p

qbqc = 1.

This proves that the orbit is closed if all the nks are even. Observe that thecases when k are odd follows from symmetry. However, we must also prove thatit holds when nk is odd and when the sequence [Pnk

]/∈Nnk=1 is a composition ofboth even and odd expressions. When all nk:s are odd, we can use the symmetryof the situation and conclude that the sequence must converge. This can be seenif we reflect the Sierpinski carpet in y = 1/2. For the combined cases we write

[Pnk−1+2]/∈Z +pnkq

= Px0,x1,...,xk, (5.26)

where we let the indicies x0, x1, ..., xk be a binary sequence where

91

xi =

1, if ni is odd,0, if ni is even.

Observe that the only difference we get is for the points Pnk+1 and Pnk+2

that will differ since if xi = 0 for some i = 2, 4, ..., then we have(0, [Pni ]/∈Z

)and we get

y =p

qx− 1− i

2+p

q

⌊(i+ 1)q

p

⌋and hence, since i is even in this case,

Pni+1 =

(q

p

(1−(−1− i

2+p

q

⌊(i+ 1)q

p

⌋)), 1

)=

(2q

p+q

p

i

2−

⌊(i+ 1)q

p

⌋, 1

)

and then we get

Pni+2 =

(1,−p

q+ 1 +

p

q

(2q

p+q

p

i

2−

⌊(i+ 1)q

p

⌋))and hence

Pni+1 =

(xi+1,−

p

q+1+

p

q

(2q

p+q

p

i

2−

⌊(i+ 1)q

p

⌋)−pq

(−1+b (i+ 1)q

pc−b i · q

pc

))=

=

(xi+1, 1 +

p

q

(2q

p+q

p

i

2−

⌊(i+ 1)q

p

⌋)− p

qb (i+ 1)q

pc+

p

qb i · qpc

))=

=

(xi+1, 3 +

i

2− 2p

q

⌊(i+ 1)q

p

⌋+p

qb i · qpc

)).

If we now starts on (1, [Pni

]/∈Z

)


we get

y = −pqx− 1− i

2+p

q

⌊(i+ 1)q

p

⌋+p

q

and then

Pni+1 =

(q

p

(−2− i

2+p

q

⌊(i+ 1)q

p

⌋+p

q

), 1

)=

(−2q

p−qp· i2

+

⌊(i+ 1)q

p

⌋+1, 1

).

Thus,

Pni+2 =

(0, 1− p

q

(− 2q

p− q

p· i

2+

⌊(i+ 1)q

p

⌋+ 1

))=

=

(0, 3 +

i

2− p

q

⌊(i+ 1)q

p

⌋− p

q

)).

This implies that

Pni+1=

(xi+1, 3 +

i

2− p

q

⌊(i+ 1)q

p

⌋− p

q− p

q

(− 1 + b (i+ 1)q

pc − b i · q

pc

))=

=

(xi+1, 3 +

i

2− p

q

⌊(i+ 1)q

p

⌋− p

qb (i+ 1)q

pc+

p

qb i · qpc

))=

=

(xi+1, 3 +

i

2− 2p

q

⌊(i+ 1)q

p

⌋+p

qb i · qpc

)).

We get the same point regardless if xi = 0 or xi = 1. This means thatthe sequence P0,0,0...,01... = P0,0,0... and this means that all different binarysequences give the same points. When nk is odd follows by symmetry (reflectthe situation in y = 1/2). This proves that if we begin at a corner point theorbit is always closed. This happens for np−1. The question is now how manyreflections we must do to receive this. Well, the number of reflections for eachk is nk plus the reflection in either y = 0 or y = 1. Hence, the total number ofreflections needed for the closed orbit is

93

2 ·

(∑p−1k=1(nk + 1) + (p− 1)

), if np−1 is even,

4 ·

(∑p−1k=1(nk + 1) + (p− 1)

), if np−1 is odd,

since p − 1 of the reflections happens on y = 0 or y = 1. With nr(p, q) asthe number of reflections we get the following equivalences;

nr(p, q) =

2 ·

(∑p−1k=1

(⌊(n+1)qp

⌋−

⌊n·qp

⌋)+ (p− 1)

), if nk is even,

4 ·

(∑p−1k=1

(⌊(n+1)qp

⌋−

⌊n·qp

⌋)+ (p− 1)

), if nk is odd,

⇐⇒

nr(p, q) =

2p+ 2q − 2, if q −⌊(p−1)qp

⌋− 1 is even,

4p+ 4q − 4, if q −⌊(p−1)qp

⌋− 1 is odd.

Observe that if pq ≥

12 the same result holds due to symmetry since when

we reflect the ingoing trajectory towards the origin in the bisector we get thereflection line that has slope q/p. Hence, the argument works for this case.

Next we will discuss an open question presented in [8]. The question in mindis the following:

Question 5.0.4. If one can construct a well defined notion of reflection forΩ(Sa), is it possible to show that, for a fixed direction θ0, the billiard flow in thedirection θ0 is either closed or dense in Ω(Sa)?

A notion of reflection on a Sierpinski carpet could be a definition of a reflec-tion on each peripheral square. In this case all possible line segments are allowedand the natural question to ask is then whether the billiard flow is closed andperiodic or dense. We will formulate this as a lemma.

Proposition 5.0.5. If each peripheral square on the Sierpinski carpet Sa isequipped with the same reflection property as the boundary of the Sierpinskicarpet, then the billiard flow is periodic if θ0 ∈ Q and dense if θ0 /∈ Q.

Proof. We consider the boundary of Sa with left bottom corner at (0, 0), i.e.a unit square with center (1/2, 1/2). Form Theorem 4.1.16 we know that on a


rational polygon all orbits are closed if and only if the initial direction is rational,and if the initial direction is irrational then the orbit is dense. Now, since thepolygon given by the boundary of Sa is rational the claim in the propositionis true. Now, add the first peripheral square, P1, according to the generatingalgorithm for Sa. Consider the tessellation given by this peripheral square andassume that we start from a point (a, 0) where 0 ≤ a ≤ 1 and with a rationalangle. Then the starting point lies on the boundary of one of the tessellationsquares, and we know from Theorem 4.1.16 that the orbit on this square isclosed and periodic. Call this square Q1 and the periodic orbit O1. We start at(a, 0) and follow O1 until we hit a side of Q1 that is not shared with P1. Then,we get a new start point for O2 with a new initial angle. However, this initialangle will be rational since the billiard flow on O2 will be the reflection of O1 inone of the sides of Q1. Hence, we know that O2 is closed. Thus, all of the orbitsO1, O2, ..., On will be closed. Observe that the last orbit in this sequence willbe reflected back to O1. This means, since the flow of O1, ..., On is the same,that the entire flow is closed. If we make the same reasoning with an arbitrarycell of Sa we get that they also are periodic. Finally, we can conclude that thebilliard flow is closed. The same reasoning can be done in order to show that ifwe start with an irrational angle, then the entire flow is dense. See figure 5.3.So, we have control over the billiard flow due to the unfolding property of eachOj . This means that the flow we seek will be the union of all the other orbits,

n⋃i=1

Oi, (5.27)

or the flow will end before in a subset S ⊂⋃ni=1Oi. Either way, the billiard is

closed.

95

.

Figure 5.3: This illustrates the idea behind the proof of Proposition 5.0.5.

There are other possible notions of reflections for the Sierpinski carpet. Firstwe will consider different lines of reflection added to the carpet. For instance,consider the line y = x as a new reflection line. In this case we see that therecannot be a flow that avoids the peripheral squares of the carpet. See Figure 5.4.In the case in Figure 5.4 the reflection line intersects the peripheral squares. Thereflection line has to be a nontrivial line in order to avoid peripheral squares.However, even if we choose to define a reflection on a nontrivial line segment wecan see that it is impossible in general. See Figure 5.5.

.

Figure 5.4: In this figure we see that there cannot be any nontrivial line segmentsthat avoids all peripheral squares for this carpet with this reflection line.


.

Figure 5.5: We can see that even when the reflection line is given by a nontrivialline segment the flow does not avoid peripheral squares.

So, instead of focusing on peripheral squares we expand the idea of lettingthem be a part of the reflections. We then have a stronger result to present.

Proposition 5.0.6. If each peripheral square on the Sierpinski carpet Sa isequipped with the same reflection property as the boundary of the Sierpinskicarpet and if we add a reflection line with rational slope relative to the x-axis,also equipped with the reflection property, then the billiard flow is periodic ifθ0 ∈ Q and dense if θ0 /∈ Q.

Proof. We will use the same idea as in the proof of Proposition 5.0.5, but in thiscase we need to be more careful with the cells we study. Initially, consider anapproximation of Sa and a line l intersecting Sa in two points. Let C1, ..., Cnbe cells in this approximation such that Ci ∩ l = ∅ ∀i = 1, ..., n and let I1, ...Ikbe cells such that Ij ∩ l 6= ∅ ∀j = 1, ..., k. From the proof of Proposition 5.0.5we have that the billiard orbits on Ci are closed. We must show that the samething is true for Ij . First, assume that l has positive slope and split the cell intofour quadrants. When l has negative slope (or slope 0) this follows in the sameway due to symmetry and we observe that l does not intersect the peripheralsquare and has a slope less than 1/2, then the orbit of the cell is closed if weapply Theorem 4.1.16. If l intersects Sa in the first quadrant of the cell, thenthere must exist a parallel line l′ to l that passes through the top right cornerof the peripheral square P1. Let L1 be a parallel line to the x-axis and L2 bea line parallel to the y-axis, so that they intersect in the middle of the cell.Moreover, let C and D as the intersection points of l with the cell and let L′1and L′2 be the parallel lines to L1 and 2 that passes through C and D. Now,denote the intersection point of l′ and L′1 as A, and denote the intersection pointbetween l′ and the reflection line of l in L2 as B. We see that A,B,C and D

97

are points in a parallelogram with rational angles. This follows since l||l′. If L′1has a y-coordinate that is grater than L1 we get two triangles ∆1 and ∆2 thatare given by l, L2 and P1 (peripheral square) and they are both uniform sincethey have an equal angle. If this is not the case we get a trapezoid which canbe dealt with in analogous ways. Now, these triangles and the parallelogramtogether with the triangle spanned between C,D and the left upper corner pointof the cell all consist of rational polygons. The union of these and P1 will forma polygon with right angles. Hence the rest of the cell can be dealt with inthe same fashion as we did in the proof of Proposition 5.0.5. Now, let us useTheorem 4.1.16 on these rational polygons. We get the same case as before,but this time with different shapes on the rational polygons. However, we knowthat the ingoing billiard flow to this cell must be closed since the reflection linel has rational angle. This means that the billiard flows in

k⋃j=1

Ij (5.28)

is closed, and since we know the same thing for Ci, we can conclude that anybilliard flow with rational starting angle will be closed. Secondly we considera reflection line that passes trough this peripheral square. In the same way asbefore we can form rational polygons. The only difference here will be that theparallelogram is replaced by a third triangle ∆3 and the same reasoning as inthe first case yields the result. This completes the proof.

In addition, we have presented some propositions regarding the periodicity ofbilliard orbits on the Sierpinski carpet. One can here ask if there are other waysof defining reflections on the Sierpinski carpet. Instead of considering reflectinglines we could perhaps consider reflection curves or other more general concepts.

Bibliography

[1] S. Abbott. Understanding Analysis. Springer science+Busisness Media,Inc, 2001.

[2] J. W. Anderson. Hyperbolic Geometry. Springer-Verlag London, 2005.

[3] M. A. Armstrong. Hyperbolic Geometry. Springer-Verlag New York, 1983.

[4] A. F Beardon. Algebra and Geometry. Cambridge University Press, 2005.

[5] M. Bonk. Quasiconformal geometry of fractals. European MathematicalSociety, pages 21–66, 2006.

[6] G. Cain. Introduction to General Topology. Addison-Wesley PublishingCompany, inc, 1994.

[7] J. Cederberg. A Course in Modern Geometries. Springer Science-BusinessMedia, inc, 2001.

[8] J. P. Chen and R. G. Niemeyer. Periodic Billiard orbits of self-similarSierpinski Carpets. J. Math. Anal. Appl., (416):1350–1373, 2014.

[9] J.L Doob. Measure Theory. Springer Verlag, 1993.

[10] P. du Bois-Reymond. Versuch einer Classification der willkürlichen Func-tionen reeller Argumente nach ihren Aenderungen in den kleinsten Inter-vallen. Journal für die reine und angewandte Mathematik Berlin, pages21–66, 1875.

[11] E. Durand-Cartagena and J. T. Tyson. Rectifiable Curves in SierpinskiCarpets. Indiana university. Math. J., 60(416):285–310, 2011.

[12] G. Edgar. Measure, Topology and Fractal Geometry. Springer Sci-ence+Business Media, LLC, 2008.

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100 BIBLIOGRAPHY

[13] D. Feldman. Chaos and Fractals. CPI Group (UK) Ltd, Croydon, 2012.

[14] O. Forster. Lectures on Riemann Surfaces. Springer-Verlag New York inc,1981.

[15] E. Gutkin. Billiards on Almost Integrable Polyhedral Surfaces. Erg. Th.and Dyn. Syst., 4:569–584, 1984.

[16] E. Gutkin and C. Judge. The Geometry and Arithmetic of Flat Surfaceswith Applications. Math. surveys., 3(47):5–80, 1992.

[17] A. Hatcher. Algebraic Geometry. United Kingdom, Cambridge, 2001.

[18] G. Jones and D. Singerman. Complex Functions. Cambridge UniversityPress, 1987.

[19] A. Kirillov. A Tale of Two Fractals. Springer Science+business Media newYork, 2013.

[20] E. Kreyszig. Introductory Functional Analysis with Applications. Wiley,2013.

[21] M. Lapidus and Robert G. Niemeyer. The Current State of Fractal Bil-liards. Applied Mathematics II: Fractals in Applied Mathematics, Con-temporary Mathematics, Amer. Math. Soc., Providence, RI, (601):251–288,2013.

[22] B. Rabern M. Macauley and L. Rabern. A Novel Proof of the Heine-BorelTheorem. arXiv:0808.0844 [math.HO], 2008.

[23] J. E. Marsden and M. J. Hoffman. Elementary Classical Analysis. Freemanand Company, 1974.

[24] H. Masur. Ergodic Theory of Translation Surfaces. Handbook of Dynamicalsystems, 1B(601):527–546, 2006.

[25] G. H. Meisters. Lebesgue Measure on the real line.https://www.math.unl.edu/ gmeisters1/papers/Measure/measure.pdf,1997.

[26] M. J. Neely. On Probability Axioms and Sigma Algebras.http://ee.usc.edu/stochastic-nets/docs/probability-axioms-sigma-algebras.pdf, 2012.

[27] M. Pollicott. Lectures on Fractals and Dimension Theory.https://homepages.warwick.ac.uk/ masdbl/dimension-total.pdf, 52, 2009.

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[28] K. Rosen. Elementary Number Theory. Pearson education, Inc, 2011.

[29] S. Pratihar S. Das, K. Halder and P. Bhowmick. Properties of Farey Se-quence and their Applications to Digital Image Processing. J. Math. Anal.Appl., (416):1350–1373, 2014.

[30] R. E. Schwartz. Mostly Surfaces. American Mathematical society, 2010.

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[32] S. Tillmann. Riemann mapping theorem and Riemann sur-faces. http://www.maths.usyd.edu.au/u/tillmann/2007-complex/ComAna-Lectures.pdf, 2007.

[33] A. F. Walz. Category: Escape-time Fractal. http://ftp.informatik.rwth-aachen.de/maple/mfrjulia.htm, 1999.

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Appendix A

Codes Used

This appendix will be used to present the codes which have been used to generatethe plots in this work.

For the Weierstrass function illustrated in Figure 3.1 we used the followingMaple code:

> with(plots);> Weierstrass := -> add(cos(3*3^k*x)/2^k, k = 0 .. 15):> plot(Weierstrass, -Pi .. Pi, numpoints = 1000, color = "Red");

For the Julia set in Figure 3.3 the following code was used. We give referenceto [33] for the code.

> with(plots);> julia := proc (c, x, y)> local z, m;> z := evalf(x+I*y);> for m from 0 to 30 while abs(z) < 3 do> z := z^2+c end do;> m> end:> J := proc (d) global phonyvar;> phonyvar := d; (x, y) -> julia(phonyvar, x, y)> end:> plot3d(0, -2 .. 2, -1.3 .. 1.3, style = patchnogrid,orientation = [-90, 0], grid = [250, 250], axes = box,scaling = constrained, color = J(-1), shading = zgrayscale);

Landstedt, 2017. 103

104 APPENDIX A. CODES USED

For the Mandelbrot set sketched in Figure 3.5 we use the following Maplecode cited from [33].

> with(plots);> mandelbrot := proc (x, y) local c, z, m;> c := evalf(x+I*y);> z := c;> for m from 0 to 25 while abs(z) < 2 do> z := z^2+c end do; m end proc;> plot3d(0, -2 .. .7, -1.2 .. 1.2,orientation = [-90, 0], grid = [250, 250], style = patchnogrid,scaling = constrained, color = mandelbrot);

Appendix B

Closed Orbits on SierpinskiCarpets

Figure B.1: This is a part of the orbit of a billiard with p = 7 and q = 19 on aSierpinski carpet. We start at the origin and end at (1, 1) where we reflect inthe bisector.

Landstedt, 2017. 105

106 APPENDIX B. CLOSED ORBITS ON SIERPINSKI CARPETS

Figure B.2: This the grid G1 for the billiard with p = 7 and q = 19.

Figure B.3: This the closed orbit for the billiard with p = 7 and q = 19.

107

Figure B.4: The union of the two grids G1 and G2 for the billiard with p = 7and q = 19.

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