LAMINATIONS: A TOPOLOGICAL APPROACH

by

DEBRA L. MIMBS

JOHN MAYER, COMMITTEE CO-CHAIRLEX OVERSTEEGEN, COMMITTEE CO-CHAIR

IMACULADA ABANALEXANDER BLOKH

VO THANH LIEMKYLE SIEGRIST

A DISSERTATION

Submitted to the faculty of the University of Alabama at Birmingham,in partial fulfillment of the requirements of the degree of

Doctor of Philosophy

BIRMINGHAM, ALABAMA

2010

LAMINATIONS: A TOPOLOGICAL APPROACH

DEBRA L. MIMBS

PH.D. IN APPLIED MATHEMATICS

ABSTRACT

Given a topological space Z and a function, f : Z → Z, one may examine the

sequence of iterates of f , i.e. {fn(z)}n∈N where N = {0, 1, 2, ...} for z ∈ Z, called the

orbit of z. Then one may classify the points of Z based upon their behavior under

iterates of f . Specifically, let P : C→ C, where C denotes the complex plane, be a

polynomial of degree at least two. We denote by F (P ) the Fatou set, which is the

maximal open set on which the iterates of P form a normal family in the sense of

Montel. Further, we let J(P ) = C \ F (P ) denote the Julia set, the set on which the

dynamics is chaotic.

It is well known that J(P ) is a nonempty, perfect, compact set, which is either

connected or has uncountably many components. We consider the case where J(P ) is

connected. As J(P ) is where complex, chaotic behavior occurs, and the dynamics on

F (P ) are well understood, we are interested in studying the behavior of P on J(P ).

Typically, Julia sets exhibit very complex behavior. Thus, we desire to simplify our

study of Julia sets by using a less complicated model. One method of modeling Julia

sets is to use a lamination, which is a closed collection of chords in the unit disc, D,

any two of which intersect at most at an endpoint on the boundary of D. By requiring

that the lamination be sibling d-invariant, one achieves a space whose dynamics are

easier to study than a Julia set, while the dynamics on the two spaces are related.

Keywords: complex dynamics, laminations, julia set, fatou set

ii

DEDICATION

To my family, who has been a vital source of encouragement over the past five

years. And especially to my husband, whose unwavering support saw me through this

challenge. Thank you all. I love you.

iii

ACKNOWLEDGEMENTS

It is difficult to acknowledge everyone who has helped me over the past five years

on one page, but I will attempt to do so. I must thank all the professors here at UAB

who have taught me over the past five years. I have learned so much from you all. You

have made my time here both challenging and enjoyable. I am extremely indebted to

my advisors, Dr. John Mayer and Dr. Lex Oversteegen, for their support and tireless

efforts to help me learn and write down what I have learned. Thank you both.

I must also thank my fellow “classmates.” We have suffered through texts which use

the word “obviously” entirely too much, road trips to conferences, and the tediousness

of writing and rewriting until we get it right. Thank you for being the sounding board

for my thesis for an entire semester and for your encouragement and friendship. You

have made my time at UAB most enjoyable.

I would be remiss if I did not thank our administrative assistants, Susan and

Cheryl, for putting up with me. Thank you for all the help with everything and for

our wonderful conversations. I will miss you both.

And, finally, I thank my friends and family. Getting a Ph.D. is definitely a family

effort, as you all may attest now. Thank you for understanding the times I missed

vacations and important events and for encouraging me when I was able to be there.

I have a loving family and am so grateful. I love you all.

Shannon, without you, this would not have been possible. Thank you for believing

in me more than I did in myself. Thank you for praying for me and keeping me sane

over the past five years. I am so blessed to call you my husband.

iv

TABLE OF CONTENTS

ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii

DEDICATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi

CHAPTER 1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . 1

CHAPTER 2. INVARIANT LAMINATIONS . . . . . . . . . . . . . . . . . 51. Laminations Due to Thurston . . . . . . . . . . . . . . . . . . . . . . . 62. Laminations: An Alternative Definition . . . . . . . . . . . . . . . . . . 83. Equivalence of Sibling d-Invariance and Thurston d-Invariance . . . . . 29

CHAPTER 3. EQUIVALENCE LAMINATIONS . . . . . . . . . . . . . . . 361. Equivalence Relations on Laminations - Definitions . . . . . . . . . . . 362. Finest Equivalence Relation, ≈ . . . . . . . . . . . . . . . . . . . . . . . 373. Properties of the Equivalence relation ≈ . . . . . . . . . . . . . . . . . . 424. Omega Continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

CHAPTER 4. CONSTRUCTING LAMINATIONS . . . . . . . . . . . . . . 501. Properties of Leaves and Gaps . . . . . . . . . . . . . . . . . . . . . . . 502. A Method of Constructing Laminations . . . . . . . . . . . . . . . . . . 513. An Example: The Douady Rabbit . . . . . . . . . . . . . . . . . . . . . 554. Wandering Triangles and Central Strips . . . . . . . . . . . . . . . . . . 59

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

v

LIST OF FIGURES

1.1 Julia set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Triangle corresponding to three rays landing . . . . . . . . . . . . . . . . . 3

1.3 Another triangle corresponding to three rays landing . . . . . . . . . . . . 4

2.1 Choice of siblings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Sibling “arcs” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Siblings and critical leaves . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4 Placements of critical leaves . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5 Factorization of σ∗d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.6 Illustration of the result of Lemma 2.21 . . . . . . . . . . . . . . . . . . . . 26

2.7 Thurston lamination containing a leaf with no sibling leaves . . . . . . . . 30

3.1 Two distinct but equivalent laminations . . . . . . . . . . . . . . . . . . . 41

4.1 Constructing a sibling d-invariant lamination . . . . . . . . . . . . . . . . . 53

4.2 Forward invariant triangle, T . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.3 Pullbacks of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.4 Ambiguity in Pullbacks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.5 Adding a critical leaf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.6 Gaps formed in pulling-back . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.7 Topological Julia Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.8 Douady Rabbit Julia Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

vi

CHAPTER 1

INTRODUCTION

Given a topological space Z and a function, f : Z → Z, one can examine the

sequence of iterates of f , i.e {fn(z)}n∈N where N = {0, 1, 2, ...} for z ∈ Z, called the

orbit of z. Then one may classify the points of Z based upon their behavior under

iterates of f . It is particularly interesting to know if the behavior of these points is

somehow chaotic.

In order to limit our study, we will suppose the function given above is a polynomial,

i.e. let P : C → C be a polynomial of degree at least two. However, the following

definitions and concepts may be applied to any function. Basic facts stated for Fatou

sets and Julia sets may be found in John Milnor’s book entitled Dynamics in One

Complex Variable, [9].

Definition 0.1. We denote by F (P ) the Fatou set, which is the maximal open set

on which the iterates of P form a normal family in the sense of Montel. Further we

let J(P ) = C \ F (P ) denote the Julia set, the set on which the dynamics is chaotic.

Geometrically, the Fatou set may be described as the set of points where orbits

sufficiently close behave similarly. The Julia set of a polynomial is the set of points

where orbits, even if sufficiently close, have unpredictable behavior. It is well known

that J(P ) is a nonempty, perfect, compact, nowhere dense set, which is either connected

or has uncountably many components, and F (P ) is an open and dense set. We consider

the case where J(P ) is connected. As J(P ) is the set where complex, chaotic behavior

occurs, and the dynamics on F (P ) have been studied and are well understood, we are

interested in studying the behavior of P on J(P ).

One of the simplest polynomials to study is P : C→ C defined by P (z) = z2. In

this case, F (P ) consists of two components, the open unit disc and the complement

1

of the closed unit disc. Notice that any two points in the open unit disc exhibit the

same behavior under iteration of P , i.e. they tend to 0, and any two points in the

complement of the closed unit disc tend to infinity. Here, J(P ) is the unit circle.

Parameterizing the unit circle by [0,1), where we identify t ∈ S, 0 ≤ t < 1 with the

point e2πit, we see several different types of behavior occur under iteration of P . For

example, there is a fixed point at 0. There are periodic points, e.g. 13

and 23. There

are pre-fixed and pre-periodic points, e.g. 12

and 16, respectively. There are even points

with infinite orbits.

Polynomials are not typically as simple as the one mentioned above. However, for

any polynomial with connected Julia set, P (z) = a0 + a1x + · · · + anxn, there is a

Bottcher uniformization conjugating P to the polynomial F (z) = zn on the unbounded

component of C \ J(P ). According to Douady and Hubbard, every rational ray will

land on the Julia set of the polynomial, J(P ) (See Figure 1.1). These rays correspond

to angles in the unit disc, where the unit disc, D = R/Z, is parameterized by [0, 1)

(See Figure 1.2). Where there is a landing point of m > 1 rays, we may draw a polygon

connecting the angles at which the rays land. Thus, the polygon will have the same

number of sides as the number of rays landing at the corresponding point. In this

manner, we obtain a picture of the disc with chords, called leaves such that any two

leaves meet at most at a point on the unit circle. If we take the closure of this set

of leaves, we arrive at a lamination. This is a natural way to arrive at a lamination

starting from an arbitrary Julia set of a polynomial.

Figure 1.1 is the Douady Rabbit Julia set, i.e. the Julia set corresponding to the

function P (z) = z2 − 0.123 + 0.745i. The Julia set is the boundary of the black figure,

called the filled-in Julia set. The Julia set is the boundary between two types of

behavior. Within the black figure is a period three attractive orbit, and outside of the

black figure is the basin of attraction for infinity. The rational rays at angles 17, 2

7, and

47

are shown. They all land at a fixed point in the Julia set. Thus, these three angles

must form a period three orbit. At this point, the action of the polynomial looks like

2

Figure 1.1. Julia setNotice that there are three rays landing at the specified point.

0!

T1/7

2/7

4/7

Figure 1.2. Triangle corresponding to three rays landingThe three points on the circle that create the vertices of the triangle correspond tothe angles of the rational rays which land at the point in the Julia set in the previous

picture.

rotation with expansion. This is reflected in Figure 1.2 in the triangle whose vertices

rotate under z → z2.

Additionally, the 0 ray lands at the extreme right point of the Julia set in Figure

1.1. This is the other fixed point in J(P ). The 12

ray lands at the extreme left point

in J(P ).

In Figure 1.3, we see another triangle, T0. This triangle corresponds to the other

main juncture point in the Julia set (Figure 1.1). Since there are three rays landing,

3

0!

T

T0

1/72/7

4/7

1/14

9/14 11/14

Figure 1.3. Another triangle corresponding to three rays landingNotice that the triangle T0 maps to the triangle T , which corresponds in Figure 1.1 to

the juncture point mapping to the fixed point.

in the lamination, there is a corresponding triangle. Notice that the vertices of T0

map to the vertices of T under application of angle doubling, which corresponds to

the behavior z → z2. T0 mapping to T corresponds to the juncture point in Figure

1.1 mapping to the fixed point.

Since every rational ray lands [4], we have a dense set of angles where we know

whether to draw leaves or polygons. These are unlinked, i.e. any two leaves touch

at most at an endpoint. Closing this set of leaves yields a collection which remains

unlinked. In fact, if a Julia set of any polynomial is locally connected, all rays land,

and the lamination constructed in this manner is already closed. Hence the topological

Julia set, which is the quotient space of the lamination found by identifying all points

in the lamination which are on the same leaf, is homeomorphic to the Julia set. In cases

where J(P ) is not locally connected, the lamination still provides information about

J(P ), even though the quotient space of the lamination is no longer homeomorphic to

J(P ). For information on laminations in the non-locally connected case, see [1], [7]

and [8]. Our purpose in this paper is not to study how laminations relate to Julia sets.

Rather, we desire to study laminations in the abstract, in hopes that information may

be obtained which will further the study of Julia sets.

4

CHAPTER 2

INVARIANT LAMINATIONS

Let S ⊂ C be the unit circle (S = R/Z) and σd : S→ S be defined by σd(z) = dz

mod 1, where d ≥ 2. A prelamination L is a collection of chords in the unit disk (D),

called leaves, such that any two leaves of L meet at most in a point on the circle. If,

in addition,⋃L is a closed subset of D, then we will call L a lamination. Hence, one

may obtain a lamination by closing a prelamination. If ` ∈ L and ` ∩ S = {a, b} then

we write ` = ab. If x ∈ S is a point then we say that xx is a degenerate leaf. We will

use the term “leaf” to refer to a non-degenerate leaf in the lamination, and we will

specify when a leaf may be degenerate, i.e. a point in S.

Given a leaf ` = ab ∈ L we denote by σd(`) the chord connecting the points σd(a)

and σd(b). If σd(a) = σd(b) then we call ` a critical leaf and σd(a) a critical value. Let

L∗ = S ∪⋃L and let σ∗d : L∗ → D be the linear extension of σd over all the leaves

in L. It is not difficult to check that σ∗d is continuous. Additionally, σd is locally

one-to-one on S, and σ∗d is one-to-one on any given non-critical leaf. Note that if L is a

lamination, then L∗ is a nonempty, compact, connected metric space, i.e a continuum.

Definition 0.2 (Convex Hull). Given a set A, we define the convex hull of A,

denoted CH(A), as the intersection of all closed half planes containing the set A.

Note that CH(A) has the following properties:

(1) CH(A) is the smallest convex set containing A.

(2) For any two points x, y ∈ CH(A), the straight line segment connecting x to

y is in A.

Definition 0.3 (Gap). A gap G of a lamination L is the closure of a component

of D \ L∗.5

For each gap G we denote by σd(G) the convex hull of σd(G∩S). The set of vertices

of a gap G is the intersection of G with S, denoted by ∂(G) = G∩S. Also, for a gap, G,

we denote its topological boundary by Bd(G). Notice that the topological boundary

of G may consist of both leaves and points on S, so that Bd(G) ∩ S = G ∩ S = ∂(G).

Every leaf ` in L is either approximated by a sequence of leaves on both sides or is

on the boundary of at least one gap. If ` is not on the boundary of a gap, then ` is a

two-sided limit leaf. If ` is on the boundary of exactly one gap, then it is a one-sided

limit leaf. Finally, if ` is on the boundary of exactly two distinct gaps, then ` is an

isolated leaf.

Definition 0.4 (Gap-Leaf). A gap-leaf of a lamination L is either a two-sided

limit leaf, a degenerate leaf which is separated from other points in S by a sequence of

leaves of L, or a gap of L. By abuse of notation, we will often denote a gap-leaf by

G (even though it may be a leaf). Moreover, a gap-leaf G is all-critical if σd(G) is a

single point.

1. Laminations Due to Thurston

Definition 1.1 (Covering Map). Let E and B be topological spaces, and let

p : E → B be a continuous and surjective function. The open set U of B is said to

be evenly covered by p if the inverse image p−1(U) can be written as the union of

pairwise disjoint open sets, called slices, {Vα}α∈J in E such that, for each α ∈ J , the

map p|Vα : Vα → U is a homeomorphism. The function p is called a covering map if

every point of B has a neighborhood that is evenly covered by p.

Note that if p : E → B is a covering map then the following topologically well-

known facts hold:

(1) the subspace p−1(b) of E, where b ∈ B has the discrete topology;

(2) p is open;

(3) p is a local homeomorphism.

(4) on every connected component of B, the cardinality of slices is the same;

6

(5) If E is compact and B is connected, then p−1(b) is finite and constant for

every b ∈ B.

Definition 1.2 (Monotone Map). Let X and Y be topological spaces and f : X →

Y be a continuous map, then f is said to be monotone if f−1(y) is connected for each

y ∈ Y

Definition 1.3 (Thurston Invariant Lamination). A lamination L is Thurston

d-invariant if it satisfies the following conditions:

(1) Forward d-invariance: for any leaf pq ∈ L, either σd(p) = σd(q), or σd(p)σd(q) ∈

L.

(2) Backward invariance: for any leaf pq ∈ L, there exists a collection of d

disjoint leaves in L (this collection of leaves may not be unique), each joining

a pre-image of p to a pre-image of q (That is, ∀pi ∈ σ−1d (p) where 1 ≤ i ≤ d,

∃ leaf piqi whose image is pq, and likewise for q.)

(3) Gap invariance: For any gap G, the convex hull of the image of G is either

(a) a gap,

(b) a leaf, or

(c) a single point.

In the case of a gap, Bd(G) must map as the composition of a monotone map and a

covering map to the boundary of the image gap, where the covering map has positive

orientation. i.e the image of a point moving clockwise around G must move clockwise

around the image of G.

This is the original definition of a lamination given by Thurston (see [9]), except

Thurston stated that in the case of a gap, the boundary of G must map as a covering

map. This is not what he meant, as can been seen by examining a gap which contains

a critical leaf. This critical leaf maps to a point, so σ∗d|Bd(G) is not an open map and

hence not a covering map. Additionally, gap invariance can be cumbersome to prove in

7

many situations. Thus, we have elected to use an alternative definition of an invariant

lamination.

2. Laminations: An Alternative Definition

Definition 2.1 (Sibling Invariant Lamination). 1 A lamination is sibling d-

invariant if:

(1) for each ` ∈ L either σd(`) ∈ L or σd(`) is a point in S,

(2) for each ` ∈ L there exists `′ ∈ L such that σd(`′) = `,

(3) for each ` ∈ L such that σd(`) is a non-degenerate leaf, there exist d disjoint

leaves `1, ..., `d in L such that ` = `1 and σd(`i) = σd(`) for all i.

We will call the leaves `, `2, ..., `d defined in (3), siblings of `. The collection of

siblings for a given leaf is not necessarily unique. Also, the endpoints of the sibling

leaves are found by rigid rotation; each may be found by rotating the endpoints of the

given leaf by 1d. However, the sibling leaves themselves are not necessarily found by

rigid rotation, as there may be multiple ways to connect the endpoints to create a

different set of siblings. Note that if L is sibling d-invariant, then σ∗d(L∗) = L∗.

Suppose v is a critical value and C is a non-degenerate component of (σ∗d)−1(v)

and let Q be the convex hull of C, i.e. the smallest convex set containing C, denoted

Q = CH(C). If |Q ∩ S| ≥ 3, then Q is called an all-critical polygon, its image

σ∗d(Q) = σ∗d(C) = {v} is a single point. Note that any gap G of the lamination

contained in an all-critical polygon satisfies σd(G) = {v} is a single point. If ` = ab

is a leaf in the lamination, then we may make a similar definition. Let C be a non-

degenerate component of (σ∗d)−1(ab), and let Q be the convex hull of C. If |Q∩S| ≥ 3,

then Q is called a collapsing polygon. In particular, collapsing quadrilaterals play a

significant role in later material. Note that any gap contained in a collapsing polygon

is either all-critical or also maps to the same leaf `.

1The alternative definition and many of the ideas in this section are due to Lex Oversteegen.

8

Figure 2.1. Choice of siblings

Here is an example of two different choices of siblings under σ3. The given leaf is 17

27.

Its siblings may be either the two solid leaves or the two dashed leaves.

Theorem 2.2. If a lamination satifies the definition of a sibling d-invariant

lamination (Definition 2.1) then it also satisfies the definition of a Thurston d-invariant

lamination (Definition 1.3).

The part of this theorem that takes a bit of work to prove is that sibling d-invariant

laminations are gap invariant in the sense of the definition of Thurston d-invariant

laminations. This is formally stated in the following theorm.

Theorem 2.3. Suppose that G is a gap of a sibling d-invariant lamination L.

Then either

(1) σd(G) is a point in S or a leaf of L, or

(2) There is a gap H of L such that σd(G) = H, and the map σ∗d : Bd(G) →

Bd(H) is the positively oriented composition of a monotone map m : Bd(G)→

S, where S is a simple closed curve, and a covering map g : S → Bd(H).

9

We will need the following lemmas and proposition in the proof of Theorem 2.3.

However, some of the following lemmas are useful in other situations. Unless otherwise

stated, we will always assume that the lamination L is sibling d-invariant.

Given a leaf ab, call any point of a′ ∈ S such that σd(a′) = a an α-point, and define

β-points similarly. By an αβ-leaf, we mean a leaf that maps to an ab-leaf. The word

“chord” is used in lieu of “leaf” in reference to a chord of the disc which may not be

a leaf in the lamination. Also, when referring to an arc, we may use the notation

[a, b] to mean the closed arc from a to b and (a, b) to mean the open arc from a to b.

The direction of the arc, whether traveling from a to b clockwise or counterclockwise,

will be clear from the context. We will subscript these arcs by the ambient space for

clarity. Hence, [a, b]S refers to the closed arc from a to b on the circle. Since σd is

locally one-to-one on S, the following proposition follows immediately:

Proposition 2.4. For any chord αβ such that a = σd(α) 6= σd(β) = b, the circular

arcs (α, β)S and (β, α)S each have the same number of α-points and β-points inside

them, i.e the number of α-points in (α, β)S is equal to the number of β-points in

(α, β)S, and similarly for (β, α)S.

The next lemma is not difficult to prove; however, it is an important result. In

the quadratic case, laminations are rotational by 12. This is not typically true for

laminations of higher degree. (See the figure below.) In some cases, when looking at a

lamination one can see small intervals whose siblings are found by rigid rotation, but

as soon as leaves get close to a critical value in the lamination, this falls apart. This

is, in large part, why the proofs in this section are not trivial. However, the endpoints

of leaves are always found by rigid rotation. Looking at the figure, one can see that

the endpoints of the leaves of siblings are 1d

apart, forming “patches” of endpoints

which are found by rigid rotation. However, the leaves themselves may run into other

“patches.” The following lemma states that the leaves must connect in the same order.

So, while leaves may enter different “patches,” they do so in a manner which preserves

order.

10

Figure 2.2. Sibling “arcs”This is an example of siblings “arcs” under σ3. Notice that while the arcs connect

points in different “patches” and are not found by rigid rotation, the manner in whichthe endpoints connect preserves order.

Lemma 2.5. Suppose that < is the positive (counterclockwise) order on the unit

circle S and x1 < a1 < b1 < x2 < · · · < xn < an < bn < x1 are 3n points in S, where

σd(xi) 6= σd(aj) 6= σd(bk) 6= σd(xi) for i,j,k ∈ {1, ..., n}. Suppose for each i there exists

a chord Ai and n(i) ∈ {1, ..., n} (Bi and m(i) ∈ {1, ..., n}) in the closed unit disk with

one endpoint xi and the other endpoint a distinct an(i) (bm(i) respectively) such that

Ci = Ai ∪Bi are all pairwise disjoint arcs. Then for each i, xi < an(i) < bm(i).

Proof. Suppose there exists a chord Bi connecting xi to some bm(i) such that

xi < bm(i) < an(i). By Proposition 2.4, there exists the same number of b-points as

x-points on each side of Bi. But, counting the number of a points between xi and

bm(i) yields one more than the number of x-points. Hence, there exists ak for some k

such that ak is not connected to xj for some j, a contradiction. �

The next lemma states that under certain circumstances, siblings touching opposite

endpoints of a critical leaf (or finite concatenation of critical leaves) must be in the

same circular order. Geometrically, this means that under certain circumstances,

we can guarantee siblings are located on opposite sides of a critical leaf (or finite

11

concatenation thereof). Notice that while we are restricted in this section to studying

sibling d-invariant laminations, the following lemma is applicable to both sibling

d-invariant laminations and Thurston d-invariant laminations, as the proof does not

make use of the lamination being sibling d-invariant.

Lemma 2.6. Suppose that A ⊂ L∗ is an arc with endpoints a1 and a2 so that A

is a (finite) concatenation of critical leaves. Suppose there exists a leaf a1b1 ∈ L so

that a1 < a2 < b1 (a1 < b1 < a2) and σd(a1) 6= σd(b1). Then the sibling a2b2 of a1b1

satisfies a1 < b2 < a2 (a2 < b2 < a1, respectively). Moreover, if a1a2 is any chord

of the circle such that σd(a1) = σd(a2) and no leaf ` of L which maps onto σd(a1b1)

crosses the chord a1a2, then the same conclusion holds.

Figure 2.3. Siblings and critical leavesSiblings must be on opposite sides of finite concatenation of critical leaves. Here, thecritical leafs under σd are denoted by a dotted line, while siblings are solid. Note that

there may be more critical leaves than are drawn.

Proof. Suppose, without loss of generality, that a1 < a2 < b1. Between a1 and

a2 there are m other α-points on the arc A, where m ≤ d − 2. Denote by am the

one closest to a2. Since there are d disjoint siblings of a1b1 and because the α-points

alternate on S with β-points, there must be m − 1 β-points within arc (a1, am)S to

12

connect with the m α-points. Hence it must be the case that the sibling ambm is such

that a1 < bm < am so that every β-point is connected to an α-point.

Examine the other siblings between am and a2. (These are necessarily not on the

arc A.) Suppose there are n siblings. Between am and a2, there must be n−1 β-points

so that α-points and β-points alternate. Since am is already connected to a β-point,

a2 must connect to one of these n− 1 β-points. Hence, b2 < a2. �

Definition 2.7. Given a leaf ` = xy, we define the corresponding open leaf to be

`◦ = ` \ {x, y}.

In the following lemma, we start with a leaf ` = ab ∈ L, and examine a component

of {(σ∗d)−1(`)\∪i`◦i }, where `◦i are open critical leaves. One may think of this component

in the following manner: Begin at an α-point and walk along leaves in the lamination

which are preimages of ` without traversing a critical leaf. These are leaves which

connect α and β-points. There are cases when critical leaves will not impact the

number of components when pulling back. However, there are cases when removing

these open critical leaves will impact the number of components. See the following

example.

Lemma 2.8. Suppose that ` = ab ∈ L. Let `◦i be the finite collection of open critical

leaves in the lamination. Define C to be a component of {(σ∗d)−1(`) \ ∪i`◦i }. Let G be

the convex hull of C ∩ S. Then every chord xy in the boundary of G is a leaf in the

lamination L and σd(xy) = ab. In particular, the boundary of G is a simple closed

curve (or a single leaf), S, consisting of leaves in L and, in the circular order, G ∩ S

alternates between points which map to a and those which map to b. Hence G is a

collapsing polygon.

Proof. Suppose that C is a component of {(σ∗d)−1(ab) \ ∪i`◦i }, G is the convex

hull of G′ = C ∩ S, and suppose that x, y are adjacent points in the boundary of

G ∩ S. By assumption, C is connected and C ∩ S is finite. Thus, there exists an arc

A ⊂ C joining x to y. This arc may be chosen as a finite concatenation of leaves such

13

Figure 2.4. Placements of critical leavesIn each picture, the critical leaf is denoted by a dashed line. Notice that in the firstexample, removing the open critical leaf does not disconnect the polygon, while in thesecond example, removing the open critical leaf disconnects a previously connected

set, resulting in two components.

that A ∩ S is minimal. If A is a leaf from x to y, then this leaf is in L and we are

finished. Hence, we may assume that |A ∩ S| ≥ 3. Let z1 be the first point of A ∩ S

after x along A. Note that z1 6= y.

We claim σd(x) 6= σd(y), for if this were not the case, xy would be a critical chord.

No αβ-leaf could cross xy or G would contain the point on S between x and y which

is the endpoint of this αβ-leaf. Then G would become larger than previously defined.

By Lemma 2.6, siblings must be located on opposite sides of a critical chord. Thus,

the sibling of xz1 touching y must have an endpoint between x and y. Then, G would

become larger than previously defined. Hence, without loss of generality, we assume

that x is an α-point and y is a β-point.

Since x is an α-point, z1 must be a β-point by definition of A. By Proposition 2.4,

the number of α-points equals the number of β-points in the arc (x, z1)S. Thus, there

must be a point between z1 and y which is an α-point. Call this point z2. Hence,

|A ∩ S| ≥ 4, z2 6= y, and xz1 and z1z2 are leaves in the lamination. Assume, without

loss of generality, that y < z2 < z1 < x in the circular order.

14

Now, examine the chord z1y. No leaf in the lamination which maps onto σd(ab)

crosses z1y, or either A would not be minimal or G would be larger than previously

defined. Examine the leaf z1z2. It is touching a critical chord, and hence by Lemma

2.6 must have a sibling on the opposite side of the critical leaf. Since z1z2 is an αβ-leaf,

there must be an αβ-leaf connecting y to an α-point in the lamination. Suppose that

there is an α-point, α′, such that x < α′ < y and α′y is a leaf in the lamination. Then

G must be larger than previously defined. Hence there is no such α-point. Thus, the

sibling must be xy. �

From Figure 2.1, one can see that there is ambiguity in the choice of siblings for a

given leaf. Siblings are not necessarily unique. A natural question at this point is:

When are siblings unique? The following lemma states that if a leaf is sufficiently

small and its image is sufficiently far enough away from any critical value, then the

siblings of a given leaf must connect in the “short” way so that siblings are found by

rigid rotation.

Denote by Ri the rigid rotation of the disk D by i/d.

Lemma 2.9. For each ε > 0 there exists δ > 0 such that for any leaf ` ∈ L with

diam (`) < δ and for any critical value v, d(v, σd(`)) > ε. If `′ ∈ L is any leaf with

σ∗d(`) = σ∗d(`′) then there exists i such that `′ = Ri(`). In particular, Ri(`) ∈ L for all

i = 0, 1, ..., d− 1, and these are all the siblings of `.

Proof. Suppose not. Then there exists a sequence `n of leaves converging to a

point x, such that σd(x) is not a critical value, and there exist leaves `′n ∈ L such that

σ∗d(`′n) = σ∗d(`n) and the `′n are not a rigid rotation of the `n, i.e. diam (`′n) > 1

d+1.

Without loss of generality, limn→∞ `′n = `∞ ∈ L, where `∞ is non-degenerate. By

continuity, σ∗d(`∞) = σd(x). This contradicts the fact that σd(x) is not a critical

value. �

Lemma 2.10. Suppose that `c ∈ L is a critical leaf. Then either `c is isolated (i.e.,

`c is the intersection of two adjacent gaps), or `c is on the boundary of an all-critical

15

polygon, C, which is a finite union of all-critical gap-leaves, and every leaf in the

boundary of C is a critical leaf `′c which is a limit of leaves in L \ `′c. Moreover, if

C = `′c, then `′c is a two-sided limit leaf.

Proof. Suppose that `c is not isolated. Then there exists a sequence `i ∈ L \ `c

such that lim `i = `c. If `c is a two-sided limit leaf, we are done. Otherwise, note that

σd(`i) is a sequence of leaves which separate the point σd(`c) from the rest of the circle

and `c is a leaf in the boundary of a gap G.

We claim that σd(G) is a point. Suppose σd(G) is not a point. It is easy to see

that there cannot exist a leaf ` ⊂ G whose image is non-degenerate and one endpoint

of which is σd(`c) (in particular, G cannot be finite). If so, σd(`i) would cross σd(`), a

contradiction.

In general, we argue as follows: Suppose that `c = a∞b∞ = lim `i = lim aibi

with lim ai = a and lim bi = b. Consider all siblings of `i. We may assume that

a1 < a2 < · · · < a∞ < b∞ < · · · < b2 < b1. There are sibling leaves `1i of `i one of

whose endpoints converge to a∞ and are located on the opposite side of `c. The other

endpoints of the leaves, b1i , cannot be converging to b∞, or `c would be a two-sided

limit leaf. Thus, b1i must be converging to a point, b1∞, such that b∞ < b1∞ < a∞.

Additionally, these leaves, `1i , do not converge to a point, as their endpoints must

converge to the endpoints of siblings of `c, which must be 1d

apart. Then lim `1i = `1c is

a critical leaf with endpoints a∞ and b1∞.

Continuing this way in the clockwise direction around S, we will eventually return

to b∞ and obtain the desired finite union of all critical gap-leaves G, every boundary

leaf of which is a limit leaf. �

Lemma 2.11. Let L be a lamination. L does not contain a wedge of leaves, i.e.,

there does not exist an interval I and a point a ∈ S such that for all x ∈ I there is a

leaf `x ∈ L such that `x = ax.

16

Proof. Suppose every x ∈ I is connected to a by a leaf in L. Since σ∗d is an

expanding map, there is n ∈ N such that σnd (I) covers all of S. Hence, by condition

(1) of Definition 2.1, each point of S must be connected to σnd (a) through a leaf.

Then there will not be d disjoint siblings for each leaf, contradicting condition (3) of

Definition 2.1. �

While it is possible to have a lamination which consists solely of leaves, i.e. no

gaps exist in the lamination, such a lamination is not very interesting. In this case, a

picture of the lamination would be a solid disc, as D would be foliated completely by

leaves. We will call such a lamination a gap-free lamination. The following lemmas

and propositions discuss some of the properties of a gap-free lamination.

Lemma 2.12. Let L be a gap-free lamination, then all leaves in L are disjoint.

Proof. Suppose not. Then there are leaves `1, `2 ∈ L such that `1 ∩ `2 = {a},

i.e they have an endpoint in common. Let `1 = ab1 and `2 = ab2 and without loss

of generality let b1 < b2. Examine the arc (b1, b2)S. By Lemma 2.11, there is not an

interval I ⊂ (b1, b2)S where every point x ∈ I is connected to a through a leaf. Choose

x ∈ (b1, b2)S such that there is no leaf in the lamination connecting x to a. Choose x1

and x2 ∈ S to be the closest points to x such that `xi = axi ∈ L with x1 > x > x2.

Then there is a wedge W such that W ⊂ D is bounded by two leaves ax1 and ax2, and

no other points in (x1, x2)S are connected by a leaf to a. Examine the chord `x = ax,

which is not in the lamination. Choose w ∈ `x such that w is on the same side of the

chord x1x2 as a. Then w is in a component of D \ L∗, which must necessarily be a

gap. But, this contradicts L being gap-free. Hence, every two leaves of a gap-free

lamination are disjoint. �

Later we will prove a stronger result. We will show that in a gap-free lamination,

all leaves are parallel.

Proposition 2.13. Let L be a gap-free lamination and Q be the quotient space of

L∗ found by identifying all points which are on the same leaf. Then Q is an arc.

17

Proof. By Lemma 2.12, all leaves in L are disjoint; hence, every non-degenerate

leaf in the lamination cuts D into exactly two pieces. The image of every non-degenerate

leaf under the quotient map q splits Q into exactly two pieces.

We first claim that the set of leaves in L is linearly ordered. Suppose not. Then

there are three leaves `i = aibi ∈ L, i = 1, 2, 3, such that all three are not comparable.

Hence, there are three closed arcs on the circle, A,B,C which are the closure of

S \ (∪`i) such that each arc has endpoints in dinstinct leaves. Let W be the set of

leaves connecting points in A to points in C. Note that W 6= ∅, as one of the `i must

have an endpoint in A and one in C. Without loss of generality, suppose this leaf is

`1. Then there is a leaf `′ = a′1b′1 ∈ W which connects a point in A to a point in C

and separates the set W \ a′1b′1 from the leaf a′1b′1. It is easy to see that this leaf must

be on the boundary of a gap, a contradiction.

We claim that there are exactly two leaves in L which are degenerate. To see

this, we first construct two sequences of leaves `i, `′i ∈ L whose diameter shrinks to 0.

Because the set of leaves is linearly ordered, there is a largest leaf, i.e. a diameter,

`0 and on each side of that diameter, the leaves must shrink in size. Define the two

sides of the disc separated by the diameter to be D and D′. Define each leaf of the

sequence recursively: Let d0 = |`0| = 1. Define di = |`i−1|2

. There are two leaves of this

given length in L, one in D and the other in D′. Choose the leaf in D to be equal to

`i+1. Define the sequence `′i similarly with leaves in D′. Taking the limit of these two

sequences yields two leaves with diameter 0, i.e. degenerate leaves x1, x2 ∈ L.

To see that there are exactly two such degenerate leaves, suppose there is a

third, say, x3. Choose three sequences of leaves, `x3i , `

x2i and `x3

i such that `xji cuts

xj (j = 1, 2, 3) off from the rest of the circle and each `xji . Then, any set of three

leaves, one taken from each sequence, is not comparable, contradicting the leaves

being linearly ordered.

18

Finally, upon collapsing leaves to points, q(x1) and q(x2) are the only points in Q

which are not cut points. Hence, Q is a continuum with exactly two points which are

not cut points, i.e. an arc [N, 6.17]. �

Notice that, given any lamination L, the extended map σ∗d : L∗ → L∗ is not

necessarily an open map. Take a polygon or leaf which collapses to a point or a

polygon which collapses to a leaf. Both of these actions result in an open set being

mapped to a set which is not open. In fact, σ∗d is open if and only if there are no

leaves collapsing to points nor polygons which collapse to leaves. However, even in

the case where σ∗d is not open, it is still a nice map. It is confluent.

Definition 2.14 (Confluent). Let X and Y be topological spaces. A continuous

map f : X → Y is confluent provided for each continuum K ⊂ Y and each component

C of f−1(K), f(C) = K.

Clearly, all monotone maps are confluent, and it is well-known that open maps

between continua are confluent. In particular, all compositions of open and monotone

maps between continua are confluent. In the case where L is a gap-free lamination, it

is easy to show that σ∗d is confluent, as it is a composition of a monotone map and an

open map. However, this is not generally the case.

Lemma 2.15. Suppose L is a gap-free lamination. Then σ∗d : L∗ → L∗ is confluent.

Proof. Away from any critical leaf, σ∗d is a homeomorphism and, hence, an open

map. So, we only need to examine around critical leaves. Let g : L∗ → L∗/ ≈, where

≈ is the equivalence relation defined by: x ≈ y iff x, y ∈ ` = ab and σd(a) = σd(b).

Since L is gap free, critical leaves do not touch by Lemma 2.12 and are linearly ordered,

as seen in the proof of Lemma 2.13. Hence L∗/ ≈ is a space which looks like m discs,

touching at the images under g of the critical leaves. This is a “line” of discs, each of

which touches at most two others and at most one other disc at the same point. This

is a monotone map, as if y ∈ L∗/ ≈ is a point where the discs connect, then g−1(y) is

a critical leaf and hence connected.

19

Now, define h : L∗/ ≈→ L∗ to be the map which does the rest of the collapsing

which σ∗d does, i.e. h identifies the m disks in L∗/ ≈, mapping each disc home-

omorphically onto L∗. Since h is a homeomorphism on each disc, it is an open

map.

Finally, the composition g ◦ h = σ∗d is a composition of a monotone map and an

open map. Hence σ∗d is confluent. �

Figure 2.5. Factorization of σ∗dThis illustrates the proof of Lemma 2.15. The critical leaves are marked by c1 and c2.They are drawn as parallel, and they are, but the proof of this has been postponed

until a later section.

In the much more interesting case where L is not gap-free, the proof that σ∗d is

confluent is not nearly as straightforward, as σ∗d cannot be factored into monotone

and open maps. The problem lies in collapsing polygons. Collapsing a polygon to

a leaf is not a monotone action, as the pre-image of a point on an open leaf which

is the image of a collapsing polygon contains a point on each boundary leaf of the

polygon. In order to prove confluence in the more general case, we must know some

more information about gaps.

20

Lemma 2.16. Suppose a lamination L contains some gap G. Then gaps are dense

in D.

Proof. Suppose L contains a gap G and gaps are not dense in D. Then there

exists an open subset U ⊂ D containing no gaps. Hence every point in U must be

in a leaf. Thus, there are two disjoint intervals I1 and I2 ⊂ S such that each point

of I1 is connected to a point in I2 by a leaf and vice versa. It cannot be the case

that Ii = {a} for i = 1, 2 by Lemma 2.11. Since the map σd is an expanding map

and leaves map to leaves under σ∗d, repeated iteration of σ∗d yields all of D foliated by

leaves, a contradiction to G ⊂ L. �

Lemma 2.17. Suppose L is not a gap-free lamination. Let K ⊂ L∗ be a continuum,

and ` = ab a leaf in L. Suppose K ∩ ` 6= ∅ and K \ ` 6= ∅. Then K must contain a or

b, and K ∩ ` is either connected or consists of exactly two components, one containing

a and the other b.

Proof. Since L is not a gap-free lamination and by Lemma 2.16, gaps are dense

in D. If K∩` = {a} or K∩` = {b}, the result is trivial. So, we may assume K∩`◦ 6= ∅.

Choose an open set U ⊂ D such that Bd(U) ∩K 6= ∅. Let L be the component of

U ∩K containing C. By the boundary bumping theorem, L∩Bd(U) 6= ∅. So, there is

a point z ∈ L ∩ Bd(U) ∩K. Since gaps are dense in the disc, between z and `, there

is a gap, G, separating z from U , which is a contradiction. Moreover, K ∩ ` cannot

contain more than two components, or a component would have to be contained in `◦,

and we may repeat the above argument. �

Theorem 2.18. The map σ∗d : L∗ → L∗ is confluent.

Proof. 2 Since Lemma 2.15 shows that σ∗d is confluent when L is a gap-free

lamination, we assume that L contains gaps. Suppose K ⊂ `0, for some `0 ∈ L.

There exist at least d disjoint preimages of `, each of which maps onto `0; hence σ∗d is

confluent.

2Thanks to Ed Tymchatyn for suggesting this method of proof, thereby shortening a previous proof.

21

Now, suppose K 6⊂ `0 for any `0 ∈ L. Suppose, by way of contradiction, that σ∗d is

not confluent. Let K ⊂ L∗ be a continuum. There is a component C of (σ∗d)−1(K)

such that σ∗d(C) 6= K. Since σ∗d(C) ⊂ K, there is a y ∈ K such that C ∩ (σ∗d)−1(y) = ∅.

By the cut wire theorem [N, 5.2], and since C is component of (σ∗d)−1(K), there are

disjoint closed sets U and V such that (σ∗d)−1(K) = U ∪ V and (σ∗d)

−1(y) ⊂ U and

C ⊂ V . Note that U and V are also open in (σ∗d)−1(K).

Now, σ∗d is clearly a closed map. Hence, σ∗d(V ) is closed. Note also that σ∗d(V ) 6= K,

as y 6∈ σ∗d(V ). All that is left to prove is that σ∗d(V ) is open in K, and then σ∗d(V ) will

be a clopen proper subset of a continuum, K, which is absurd.

To show σ∗d(V ) is open in K, we show K \ σ∗d(V ) is closed. Given w ∈ K and a

sequence wi → w such that wi ∈ K \ σ∗d(V ) for all i, assume by way of contradiction

that w ∈ σ∗d(V ). If wi ∈ S, then by continuity, w ∈ S. Since w ∈ σ∗d(V ) there is an

x ∈ V such that σ∗d(x) = w. If x is in the interior of D, then it is on a critical leaf,

every point of which maps to w; hence, we may take the endpoint of the critical leaf to

be x instead. Since σ∗d is an open map on S, (σ∗d)−1(wi) contains a sequence of points,

xi ∈ S such that xi → x. Since V is an open neighborhood of x in (σ∗d)−1(K), it must

necessarily contain points of xi. Hence, σ∗d(xi) = wi ∈ σ∗d(V ) for i sufficiently large, a

contradiction, as these points were chosen to be in K \ σ∗d(V ). Thus, we assume that

wi 6∈ S.

Then, there are leaves `wi ∈ L∗ such that wi ∈ `wi and `wi = aibi → `w = ab with

w ∈ `w, such that ai → a and bi → b. (Note that it may be that a = b = w so that `w

is a degenerate leaf.) Since K is a continuum and wi ∈ K for all i, by Lemma 2.17,

K must contain an endpoint, ai or bi, for all i. Without loss of generality, suppose

for infinitely many i, the arc [wi, ai]K ⊂ K. Then, since [wi, ai]K → [w, a] and K is a

continuum, [w, a] ⊂ K. There is a point w′ ∈ V such that σ∗d(w′) = w, and there is a

leaf `′ such that w′ ∈ `′ = a′b′ and σ∗d(`′) = `w.

Since σd is a covering map on S, there are points a′i converging to a′, and b′i and

leaves `′i = a′ib′i such that there are points w′i ∈ `′i with σ∗d(w

′i) = wi. It must be that

22

b′i → b′′, where σd(b′′) = b, and `′i → `′′ where `′′ = a′b′′. Then w′i → w′′ ∈ `′′ and, by

continuity σ∗d(w′′) = w. Let C ′ be the component of (σ∗d)

−1(K) containing w′ ∈ V .

Then the arc [w′, a′] ⊂ C ′ since C ′ is connected and contains w′. Since the arc [w′′, a′]

contains a′ and C ′ is connected, [w′′, a′] ⊂ C ′, and hence is a subset of V. Then since

w′′ ∈ V and V is an open neighborhood of w′′, for sufficiently large i, w′i ∈ V , meaning

that σ∗d(w′i) = wi ∈ σ∗d(V ), a contradiction. �

Recall Lemma 2.9, which states that a sufficiently small leaf whose image is far

enough away from critical values has siblings found by rigid rotation. We wish to

be able to track more than just siblings of leaves found by rigid rotation. We would

like to be able to track entire arcs in L∗ and discover when an entire arc has disjoint

siblings. The following lemma gives a criterion for such a situation. The picture is

complicated by that fact that the arcs may be made up of leaves, the siblings of which

may not be found by rigid rotation. In this case, the disjoint sibling arcs may not be

found by rigid rotation but will be made up of subarcs. Additionally, the given arc

may not be monotone (i.e. the points on successive endpoints on leaves in the arc may

not walk along the circle in a monotone fashion), adding a another complication to

finding its disjoint siblings.

Lemma 2.19. Let A ⊂ L∗ be a continuum. Let C be a component of (σ∗d)−1(σ∗d(A))

such that C contains no collapsing polygons and no critical leaves. Then σ∗d|C is locally

one-to-one.

Proof. Suppose, by way of contradiction, that A is a continuum, C is a component

of (σ∗d)−1(σ∗d(A)) with no critical leaves and no collapsing polygons, and σ∗d is not

one-to-one on C. Then there is a point x ∈ C and two sequences of points x1i , x

2i ∈ C

such that σd(x1i ) = σd(x

2i ) for all i. Suppose first that each xji , j = 1, 2 are on distinct

leaves. Then x1ix

1i+1 has as a sibling x2

ix2i+1 for each i, but this cannot be, as siblings

cannot be arbitrarily close. Similarly, if the x1i sequence is all on one leaf, ` but the

x2i sequence is on distinct leaves, there are siblings of ` arbitrarily close to `. Thus,

23

there must be two leaves protruding from x, `1 and `2, such that xji ∈ `j for each i

and j = 1, 2.

Define C ′ to be the component of {(σ∗d)−1(σd(`1)) \ ∪i`◦i } containing `1. Notice C ′

also contains `2 and C, as C contains no critical leaves. Let G = CH(C ∩ S), then by

Lemma 2.8, G is a collapsing polygon. Hence, C contains a collapsing polygon, as

well, a contradiction. �

Lemma 2.20. Suppose that A ⊂ L∗ is an arc with endpoints in S such that

diam (A ∩ S) < 1d+1

and σ∗d(A) contains no critical value. Then there exist d disjoint

arcs A1, ..., Ad such that A = A1 and σd(Ai) = σd(A) for all i. In other words, the

arcs A1, A2, ..., Ad are siblings of A = A1.

Moreover, suppose A∩S ⊂ [a, b]S, where a and b are endpoints of A, diam ([a, b]S) <

1d+1

, and σ∗d(A) \ σd(a) contains no critical value. Then A has d disjoint siblings arcs,

A1, A2, ..., Ad.

Proof. Note first that σ∗d(A) is an arc, since points on S map locally one-to-one,

so points stay in order. Let C be a component of (σ∗d)−1(σ∗d(A)). Since σ∗d is confluent

by Theorem 2.18, σ∗d(C) = σ∗d(A).

Since σ∗d(A) contains no critical value, C contains no critical leaves. If C contains

no collapsing polygons, then by Lemma 2.19 σ∗d|C is a local homeomorphism. By [6],

σ∗d|C is a homeomorphism. Thus, no preimages of σ∗d(A) can meet, as σ∗d is one-to-one

on each component, and meeting preimages would force “folding” at the point where

they meet, so that σ∗d would not be one-to-one at that point, a contradiction. Hence,

the d disjoint components of (σ∗d)−1(A) are the d siblings of A.

Suppose C contains a collapsing polygon, P alternating between α-points and

β-points. Then P must have 2n sides. Recall the diameter of A is small. Hence,

within P there are n siblings of A. This may be repeated finitely many times to

count the siblings of A within collapsing polygons. The rest of the siblings are located

outside of collapsing polygons and are found by the previous case.

24

Suppose next thatA∩S ⊂ [a, b]S where a and b are endpoints ofA and diam ([a, b]S) <

1/(d+ 1). (Note that this means that A cannot contain a critical leaf.) Additionally,

suppose σ∗d(A) contains, at most, the critical value σd(a). If A is a finite union of

leaves, we are done, as each leaf must have d disjoint siblings. So, suppose A is not

a finite union of arcs. If there exists a leaf ax ⊂ A, then the subarc B of A with

endpoints x and b has disjoint siblings B1, B2, ..., Bd by the first part of the lemma.

In each Bi, let xi denote the point in Bi such that σd(xi) = σd(x). The leaf ax has d

disjoint siblings aixi by definition. Then Ai = Bi ∪ aixi are the required sibling arcs.

Hence, we may assume that a is a limit point of A∩ S. Since A∩ S ⊂ [a, b]S, A∩ S

is located on exactly one side of a (i.e., the points in A ∩ S converge to a from one

side in the circle S). Choose a monotone sequence xi ∈ A ∩ S such that limxi = a.

By the first part of the proof, each of the arcs Bi = [xi, b] has disjoint siblings Bji ,

j = 1, 2, ..., d, and we may assume that Bjn ⊂ Bj

n+1 for all j. Let Aj =⋃nB

jn. We

claim that each Aj is an arc. If this is not the case, then the ray⋃nB

jn must sin( 1

x)

to a critical leaf `c joining two points a′, a′′ ∈ σ−1d (σd(a)). It then follows from the

monotonicity of the sequence xi that then the siblings Bjn cannot be pairwise disjoint,

a contradiction. Hence, A, A2, . . ., Ad are arcs. They are pairwise disjoint, as the

Bji were disjoint at each step and the diameter of each was small, and the proof is

complete. �

Lemma 2.21. Suppose that A, B, C ⊂ L∗ are arcs (disjoint except possibly at

endpoints) with endpoints in S such that A ∪B ∪C = J ⊂ Bd(G) is an arc contained

in the boundary of a gap G of L∗; C = [c1, c2] is a finite concatenation of critical

leaves or a single point c1 = c2 with c1 ∈ A and c2 ∈ B; σ∗d(J) is an arc and for

each X ∈ {A,B} either X is a single non-critical leaf, or diam (X) < 12(d+1)

. Let

a (b) denote the endpoint of A (B, respectively) distinct from c1 (c2, respectively).

Let x ∈ C ∩ S. If σd(x) < σd(a) < σd(b) and σ∗d(J) contains no critical value except

possibly σd(x), then x < a < b.

25

Figure 2.6. Illustration of the result of Lemma 2.21

Proof. Since σd(C ∩ S) = σd(x) is the only possible critical value in σ∗d(J),

[a, c1]∪ [c2, b] = A∪B contains no critical leaves. Let Y = {σd(x), σd(a), σd(b)} where

σd(x) < σd(a) < σd(b), and let X = σ−1d (Y ). Then X = {x1 < a1 < b1 < x2 < ... <

bd < x1} where x1 < a1 < b1 < x2 < ... < bd < x1, and assume that x1 = x.

Since A and B are in the boundary of a gap, A ∩ S ⊂ [a, c1]S or A ∩ S ⊂ [c1, a]S;

hence the hypotheses of Lemma 2.20 are satisfied. By Lemma 2.20 each of the arcs

A = [a, c1] and B = [c2, b] has d disjoint siblings. There must exist a sibling of B with

endpoint c1. Pairing up all the siblings of A and B around the points xi ∈ σ−1d (σd(x))

yields disjoint arcs, Ci = Ai ∪ Bi with the endpoints ai and bi. Since all these arcs

satisfy the assumptions of Lemma 2.5, for all of them xi < ai < bi. Since J is contained

in the boundary of a gap, then a1 = a.

Now, beginning at b, there is a closest x-point in J , which is connected to an

a-point. If that a-point is a itself, we are finished, and x < a < b as required. If not,

continue around J to the next x-point, and repeat the process finitely many times.

At some point, one of the a-points must be a, and since J is contained in Bd(G), it

must be that x < a < b, or the A and B arcs would not be contained in Bd(G). �

26

Proof of Theorem 2.3: Suppose that G is a gap of L and H = CH(σ∗d(G)).

We may assume that |σd(G∩ S)| > 1, or the image of G is a point and we are finished.

Let g ◦m = σ∗d|Bd(G) be the monotone light factorization of σ∗d|Bd(G) such that g is

finite to one and for all y ∈ m(Bd(G)), m−1(y) is a point or a maximal arc in Bd(G)

which is a concatenation of critical leaves.

Suppose first that g is not locally one-to-one. Since L is sibling d-invariant and σd is

locally one-to-one, there must exist two leaves x1c1 and xncn and a concatenated chain

of critical leaves c1c2, . . ., cn−1cn inBd(G) such that σd(x1) = σd(xn) 6= σd(c1) = σd(ci)

for all i = 2, ..., n. (Notice that it cannot be the case that there is a null sequence

converging to either c1 or cn, as then g would be locally one-to-one at m(ci).) If

x1 = xn, σd(G) = σd(x1)σd(c1) is a leaf, and we are finished.

Otherwise, proceed as follows. By lemma 2.6, x1c1 and xncn are not siblings.

Hence there exists a sibling xncn+1 of x1c1. Since the chord xnx1 runs inside a gap,

no leaf of the lamination may cross it. (It could be that there is a leaf xnx1, but then

the image of the gap is a leaf, and we are finished.) By Lemma 2.6 x1 < xn < cn+1.

Let C be a maximal arc (or in the degenerate case just the point cn+1) in Bd(G)

which is a concatenated union of critical leaves. Let cn+1 and cm be the endpoints

of C. Then there must exist a sibling cmxm of cnxn; moreover, it must be the case

that cn < cm < xm, by the same argument as before. Continuing this process finitely

many times, we obtain a simple closed curve S such that σ∗d(S ∩ S) = {σd(x1), σd(c1)}

and Bd(G) ⊂ CH(S). This implies that σ∗d(G) = σd(x1)σd(c1) (a leaf), and we are

finished.

Hence, we may assume that g is locally one-to-one on m(Bd(G)) and |σd(G∩S)| ≥ 3.

For each w ∈ m(Bd(G)) ∩ S, one of three things may happen on m(Bd(G)). Either

there are two sequences of leaves, one on each side of w, each converging to w, or there

is a leaf protruding from w to one side with the other side of w being a limit of leaves,

or there are two leaves protruding from w. Also, m−1(w) is either a point in G with

the same behavior as w or a concatenated string of critical leaves, with the endpoints

27

of the string combined having the same behavior as w, i.e. the behavior on each side

of w occurs at one endpoint of the string. Choose for each x ∈ Bd(G) ∩ S such that x

is not on a concatenated string of critical leaves an arc Ax = [a−x , a+x ]D ⊂ Bd(G) with

x ∈ Ax \ {a−x , a+x }, such that a±x ∈ S and either diam ([a−x , x]) < 1

2(d+1)or xa−x ∈ L,

and similarly for a+x , and |σd({x, a−x , a+

x })| = 3. If x is on a concatenated string of

critical leaves, then label the endpoints of that string x− and x+. Then choose an

arc Ax = [a−x , a+x ]D ⊂ Bd(G) with x ∈ Ax \ {a−x , a+

x } and a−x < x− < x+ < a+x < a−x

such that a±x ∈ S and either diam ([a−x , x−]) < 1

2(d+1)or x−a−x ∈ L, and similarly for

a+x , and |σd({x, a−x , a+

x })| = 3. Additionally, the arc Ax may be divided into two arcs,

A−x and A+x with endpoints a±x in S such that A−x ∪ A+

x = Ax, A−x ∩ A+

x = {x}.

Since g is locally one-to-one, there exists an arc [p′, q′] = B ⊂ m(Bd(G)) such

that g(p′) = g(q′) = s, g(B) = S is a simple closed curve, and for all y ∈ S \ {s},

|g−1(y) ∩ B| = 1, while g−1(s) ∩ B = {p′, q′}. In Bd(G), there are points p and q,

which are the endpoints of a finite concatenation of critical leaves or are disjoint from

critical leaves, such that m(p) = p′ and m(q) = q′. Choose q to be the endpoint

of m−1(q′) such that when “walking” around the circle away from m−1(q′) toward

x ∈ m−1(B \ {p′, q′}) it is not necessary to cross m−1(p′) in order to get to x.

We claim that S = g(m(Bd(G))). If g(m(Bd(G))) 6= S there exists x0, x1 ∈

S ∩ Bd(G) such that x1 > q > p, p < x0 < q, σd(x0) = σd(x1) and σd([q, x1]D) ⊂ S.

There also exists x1 < · · · < x3 < x2 < p in Bd(G) such that limxi = x1 and

σd(xi) /∈ S. Since g is locally one-to-one, g|m(Axi )is one-to-one, and [g(m((A+

x1)) \

g(m(x1))] ⊂ T (S). Then, for some i ∈ {0, 1} and z ∈ {+,−}, g(m(Azxi)) “separates”

g(m(A+x(i+1) mod 2

)) from g(m(A−x(i+1) mod 2)), forming a “triod” of arcs, T , with endpoints

g(m(azxi)) and g(m(a±x(i+1) mod 2)). This may not be a true triod, as two of the legs may

have many points in common. Without loss of generality, suppose g(m(A+x(i+1) mod 2

))

is the one which is not contained on the simple closed curve S. Choose a point

u ∈ g(m(A+x(i+1) mod 2

)) such that u /∈ S and a point v ∈ g(m(A+x(i+1) mod 2

)) ∩ S

such that u < v < g(m(a+x(i+1) mod 2

)) and there are no other points on the arc

28

g(m(A+x(i+1) mod 2

)) between u and g(m(a+x(i+1) mod 2

)) which are also on S. Then, the arc

[u, v] on g(m(A+x(i+1) mod 2

)) intersects S only at v. Also, the arcs [v, g(m(a+x(i+1) mod 2

))]

and [σd(x0), v] divide the arc g(m(A+x(i+1) mod 2

)) into two pieces at v. Hence at v, there

is a triod, T = [u, v] ∪ [v, g(m(a+x(i+1) mod 2

))] ∪ [σd(x0), v], in H, with σd(x0) < u <

g(m(a+x(i+1) mod 2

)) on S.

Examine the component of (σ∗d)−1(T ) containing x0. Since σ∗d is confluent (by

Theorem 2.18), this component must map onto T . Hence, there is a point t ∈

Bd(G) such that σd(t) = v. And, at t, there is a triod of arcs with endpoints

x0, d, a+x(i+1) mod 2

mapping to σd(x0), u, σd(a+x(i+1) mod 2

), respectively. Therefore, the

hypotheses of Lemma 2.21 are satisfied, and if σd(x0) < u < σd(a+x(i+1) mod 2

), we

may conclude that x0 < d < a+x(i+1) mod 2

. Thus, there exists an arc from w to d

cutting through the gap G, a contradiction. If it is not the case that σd(x0) <

u < σd(a+x(i+1) mod 2

), then we may repeat the above process using the component of

(σ∗d)−1(T ) containing x0, and we will obtain a similar contradiction. Therefore, we

have shown that g(m(Bd(G))) = σd(Bd(G)) = S and σ∗d|Bd(G) is positively oriented.

Suppose that S is not the boundary of a gap H of L. Then there exists a leaf

` = s1s2 of L which meets the interior of T (S). Then by the previous argument

concerning triod arcs, we obtain a contradiction. Hence, σ∗d|Bd(G) is a positively

oriented map, and the image of Bd(G) is the boundary of some other gap in L. �

3. Equivalence of Sibling d-Invariance and Thurston d-Invariance

A natural question at this point is, “Is Thurston d-invariance equivalent to sibling

d-invariance?” The answer is, “No.” However, in some sense, the laminations where

these two definitions are not equivalent are unusual laminations. Let us consider an

example which illustrates that these definitions are distinct.

In this example, we will consider the map to be σ2. Refer to Figure 2.6. Here, G

is an infinite rotational gap. The solid lines are leaves in the lamination, and that

drawn in Figure 2.6 is a forward invariant set. The rest of the lamination is not

29

pictured. However, it is formed by pulling back the picture drawn. G is an infinite

rotational gap, where a given leaf maps one leaf to the left under application of σ2. T

is a collapsing triangle. The entirety of T maps to the critical leaf c = 012.

Figure 2.7. Thurston lamination containing a leaf with no sibling leavesThe dashed lines illustrate where the sibling leaves of leaf m should be. However,

these leaves are not picked up in the limit. Hence, they do not exist in the lamination.Thus, leaf m has no siblings.

The dashed chord is not in the lamination, which is where the problem arises. This

chord would need to be a leaf in the lamination in order for m to have two disjoint

siblings. Examine m. It maps to the critical leaf, c and hence any siblings of m would

have to do the same. Notice that the only other leaves in the lamination which map

to c are not disjoint from m, and hence cannot be siblings of m, as siblings must be

disjoint. Therefore, the lamination formed from this forward invariant set is not a

sibling d-invariant lamination.

However, this lamination formed from pulling-back this forward invariant set is a

Thurston d-invariant lamination. Forward invariance and backward invariance hold

by construction. The only concern is that this lamination be gap invariant. The gap

30

G maps to itself with a positive orientation. The triangle, T , maps to a leaf. When

pulling back, a copy of the sequence of leaves in the gap G will be formed between

the circle and leaf m. This gap will map to G is an order-preserving fashion.

This lamination is a simple example to show that the two definitions of d-invariant

laminations are, indeed, distinct. However, this lamination is not a lamination that

we would actually see arise from a Julia set.

Definition 3.1. Given a lamination, L, we define L≈ to be the lamination con-

taining all chords which are located on the boundary of a convex hull of an equivalence

class in L under the equivalence relation ≈.

Definition 3.2 (Invariant Equivalence Relation). Given a closed equivalence

relation ≈ on S, we define ≈ to be d-invariant under the map σ∗d if the following hold:

(1) CH([x]) ∩ CH([y]) = ∅ for x 6≈ y.

(2) For every x ∈ S there exists a y ∈ S such that σ([x]) = [y].

(3) L≈ is sibling d-invariant.

We will leave the proof of the following equivalence of item three in Definition 4.4

to the reader, although one may use tools implemented throughout this chapter for

the proof.

Proposition 3.3. The following are equivalent:

(1) L≈ is sibling d-invariant.

(2) σd|[x] : [x]→ [y] for some y ∈ S is a positively oriented covering map.

(3) σ∗d|Bd(CH([x])) : Bd(CH([x])) → Bd(CH([y])) for some y ∈ S is a positively

oriented covering map.

Kiwi [7] and Blokh and Levin [2] introduced invariant equivalence relations (as

defined in Definition 4.4 and with item three replaced by condition two of Lemma 3.3)

as a replacement for Thurston d-invariant laminations. Notice that the equivalence

class [x] is either 0-dimensional or the entire circle. Thus, the positively oriented

31

covering map of item two in Lemma 3.3 is on a 0-dimensional set, while that in item

three is on the boundary of a convex set.

Definition 3.4 (q-lamination). A lamination is said to be a q-lamination if the

only leaves in the lamination are on the boundary of the convex hull of equivalence

classes of an invariant equivalence relation.

A property which is closely related to q-laminations is that of cleanness, which

was originally defined by Thurston.

Definition 3.5. A lamination is said to be clean if for any point z ∈ S there are

no more than two leaves which contain z.

In some sense q-laminations are the smallest ones of all the ones which are

“equivalent.” One important property of q-laminations is the following:

Proposition 3.6. A q-lamination is clean.

Proof. Suppose not, then from some point z ∈ S, there are three leaves `1, `2,

and `3 containing z. Since they all contain z, each `i is in the same equivalence class.

But then, only two of the three leaves can be on the boundary of the convex hull of

this equivalence class, a contradiction to the definition of q-lamination. �

It is important to note that laminations which are not in the closure of the set of

q-laminations are not dynamically relevant to Julia sets, as there is no polynomial

which corresponds to them. Hence, it would make sense to limit our study to closures of

q-laminations. We will show that being a sibling lamination is a closed property. This

is important for the following reason. In the degree two case, the set of polynomials

whose Julia sets are locally connected is dense in the set of all polynomials with

connected Julia sets. Also, locally connected Julia sets correspond directly to q-

laminations. Hence, considering the laminations which are limits of q-laminations will

give a complete picture of laminations which correspond to connected Julia sets.

32

In light of this discussion, the following lemma gives a criterion for when Thurston

d-invariant laminations and sibling d-invariant laminations are equivalent. As this is

based on q-laminations, the two definitions are practically equivalent, which we will

prove shortly. First, we wish to recall a lemma from a previous section, Lemma 2.6,

which stated that under certain conditions siblings must connect on opposite sides of

a finite concatenation of critical leaves. We proved this lemma under the hypothesis

of the lamination being sibling d-invariant. However, this hypothesis was not used in

the proof, as was pointed out before the statement of the lemma. Hence, it applies to

both sibling d-invariant laminations and Thurston d-invariant laminations. We will

need this lemma in the proof of equivalence of the definitions. Hence, we restate it

next.

Lemma 3.7. Suppose that A ⊂ L∗ is an arc with endpoints a1 and a2 so that A is

a (finite) concatenation of critical leaves. Suppose there exists a leaf a1b1 ∈ L which

has d disjoint siblings and such that a1 < a2 < b1 (a1 < b1 < a2) and σd(a1) 6= σd(b1).

Then the sibling a2b2 of a1b1 satisfies a1 < b2 < a2 (a2 < b2 < a1, respectively).

Moreover, if a1a2 is any chord of the circle such that σd(a1) = σd(a2) and no leaf ` of

L which maps onto σd(a1b1) crosses the chord a1a2, then the same conclusion holds.

Lemma 3.8. If L is a q-lamination which is Thurston d-invariant, then L is sibling

d-invariant.

Proof. Let L be a q-lamination which is Thurston d-invariant, and let ` = αβ

and σd(`) = ab be leaves (nondegenerate) in L. We must show ` has d disjoint siblings,

as the other two parts of the definition of sibling d-invariance follows directly from

Thurston invariance. Since L is Thurston d-invariant, there are leaves αiβi, 1 < i ≤ d

such that σd(αiβi) = ab. Assume α = αi while β = βj , where i 6= j, or we are finished,

as the collection {αiβi} would be the d disjoint siblings. Then αiβi ∪ αiβj ∪ αjβj

is contained in the boundary of one equivalence class, [αi], as the lamination is a

q-lamination, and [αi] corresponds to a nondegenerate gap, G.

33

We claim first that G collapses to the leaf ab. If not, then σd(G) = H is a gap,

and σd|Bd(G) : Bd(G)→ Bd(H) is not positively oriented, a contradiction.

We also claim that Bd(G) does not contain a critical leaf. Suppose it does

contain a critical leaf or a finite concatenation thereof. Without loss of generality,

suppose the critical leaf has α endpoints. Label this concatenation of critical leaves

A = αkαk+1 ∪ ... ∪ αn−1αn. Then there is a leaf, αkβk, and a leaf, αnβn. Without loss

of generality, βk < αk < αn < βn < βk. No leaf in the lamination may cross the chord

αkαn, as all leaves mentioned so far are on the boundary of the convex hull of [αi].

Hence the previous lemma applies, and there is a sibling, αnβm of αkβk such that

αk < βm < αn. But, now we have three leaves protruding from the point αn, namely

αnβm, αnαn−1, and αnβn, contradicting Lemma 3.6.

By the previous two claims, we see that G must be a collapsing polygon, which

has 2n sides, as it alternates between α-points and β-points. Hence, given the leaf `,

we may label the leaves in G as odd or even. We call ` itself an even leaf and make

the definition that a leaf in G is even iff in the positive order around G it is an even

number of leaves away from `. Then, all even leaves in G, together with all leaves αiβi

disjoint from G, will constitute the necessary disjoint siblings for `. �

Theorem 3.9. Let Q be the closure of the set of Thurston d-invariant q-laminations.

Then every member of Q is sibling d-invariant. Hence, for elements of Q, Thurston

d-invariance is equivalent to sibling d-invariance.

Proof. Let L = limLi, where Li are sibling d-invariant laminations by Lemma

3.8, as they are Thurston d-invariant q-laminations. Suppose αβ ∈ L, σ∗d(αβ) = ab is

a nondegenerate leaf. We must show that αβ has d disjoint siblings.

Since L = limLi, αβ = limαiβi, where αiβi ∈ Li. Without loss of generality,

suppose that αi → α and βi → β monotonically in the circular order. Since Li are

sibling laminations, we may assume that each αiβi has d disjoint siblings. (This is

certainly the case from some point on. Take a subsequence leaving off the first part of

the sequence.) We will label the siblings of αiβi with superscripts, i.e. αki βki , where

34

1 ≤ k ≤ d. Then, without loss of generality, limαki βki = αkβk. In this manner, we

have arrived at d leaves which are in the limit lamination, L. By continuity of σ∗d, it

must be the case that σ∗d(αkβk) = σ∗d(αβ) for each k, and it must be that for some k,

αkβk = αβ.

We claim that αkβk are all disjoint. By way of contradictions, suppose that they

are not disjoint, then αkβk ∩ αsβs for some k and some s. These leaves either cross in

the interior of the disc or meet at an endpoint. Suppose that they cross in the interior

of the disc, then for the leaves to converge some αki βki must cross some αsiβ

si , but this

would contradict αki βki , α

siβ

si ∈ L. Now, suppose the leaves meet at an endpoint, i.e.

without loss of generality αk = αs and αk < βk < βs. Without loss of generality,

suppose that αki < αk for all i. Then, αsi < αs for all i. Hence, d(βki , βk) ≥ d for all i

and for k 6= s. This contradicts the βki → βk. �

Now, we may state the following main result. One direction is the previous theorem,

theorem 3.9, and the other direction is theorem 2.2.

Theorem 3.10. For a lamination, L, which is a limit of q-laminations, the

following are equivalent:

• L is a sibling d-invariant lamination.

• L is a Thurston d-invariant lamination.

35

CHAPTER 3

EQUIVALENCE LAMINATIONS

Let ≈ be an equivalence relation on S. We desire to study equivalence relations

on sibling d-invariant laminations and relate the properties of those relations to the

laminations themselves. Unless otherwise stated, d-invariant laminations are assumed

to be sibling d-invariant. We have already begun this process with the gap-free

lamination. In Proposition 2.13, we showed that the quotient space of the gap-free

lamination is homeomorphic to an arc. We begin with some definitions.

1. Equivalence Relations on Laminations - Definitions

Definition 1.1. Given an equivalence relation ≈, we say that ≈ is trivial if

each equivalence class consists of exactly one point and degenerate if ≈ has only one

equivalence class.

Definition 1.2. An equivalence relation ≈ is unlinked if for any two distinct

equivalence classes C1 and C2, their convex hulls are disjoint, i.e CH(C1)∩CH(C2) =

∅.

Proposition 1.3. An equivalence relation ≈ is unlinked if and only if when

a1 ≈ a2 ∈ S and a3 ≈ a4 ∈ S but a2 6≈ a3, then there exist intervals I1 and I2 in S

with endpoints a1,a2 and a3,a4, respectively, and these intervals are disjoint.

Proof. Since the convex hull of unlinked equivalence classes on S are disjoint

intervals, the proposition follows immediately from the definition of unlinked. �

An equivalence relation with closed classes need not be a closed equivalence relation.

In our case, we will deal with equivalence relations with closed classes. However, we

often need to know that the relation itself is closed.

36

Definition 1.4. An equivalence relation ≈ on S is closed if the graph G≈ =

{(x, y) ∈ S× S|x ≈ y} of ≈ is a closed set in S× S.

Proposition 1.5. An equivalence relation ≈ is closed if and only if every equiva-

lence class is closed and the equivalence classes are upper semi-continuous (i.e., for

every xi and yi, if xi ≈ yi for every i, limxi = x∞, and lim yi = y∞, then x∞ ≈ y∞).

Definition 1.6. Given a sibling d-invariant lamination L and an equivalence

relation ≈ we say ≈ respects L provided for every leaf ` ∈ L with endpoints a and b,

a ≈ b.

2. Finest Equivalence Relation, ≈

Given a sibling d-invariant lamination L, there are many possible equivalence

relations which respect L. Even when restricting the possible equivalence relations to

those which are closed and unlinked, we obtain many possible such relations. However,

among the closed and unlinked equivalence relations, there does exist a finest one

which respects L. It happens that this finest one among closed relations is the same

as the finest equivalence relation with closed classes which respects L.

Proposition 2.1. For a lamination L, let R be the set of all closed and unlinked

equivalence relations which respect L. There exists a finest relation ≈∈ R.

Proof. We know R 6= ∅, as it contains the equivalence relation which identifies

everything in the circle. Choose a relation r ∈ R, and ` ∈ L. Denote the equivalence

class of ` by [`]r. Define a new relation r∗ by [`]r∗ =⋂r∈R[`]r. This is well-defined, as

such intersections are either disjoint or coincide, and since each r ∈ R respects L, so

will r∗.

Moreover, r∗ is unlinked. Suppose r∗ is linked. Then by Definition 1.2, there exist

two distinct r∗-classes, A and B such that CH(A) ∩ CH(B) 6= ∅. For any r ∈ R,

A and B are contained in two disjoint r-classes, i.e A ⊂ [`A]r and B ⊂ [`B]r. Since

CH(A)∩CH(B) 6= ∅, CH([`A]r)∩CH([`B]r) 6= ∅, which contradicts r being unlinked.

37

Since r∗ was created by intersecting all other closed, unlinked equivalence relations,

r∗ is the finest such equivalence relation and is unique. �

From this point, when we refer to ≈, we mean the finest unlinked closed equivalence

relation which respects L found in Proposition 2.1.

Definition 2.2. A separation of a topological space, X, is a pair A, B of disjoint,

nonempty, open subsets of X whose union is X. X is connected if there does not

exist a separation of X.

Notice that this definition could have been stated in terms of A and B being closed

subsets of X.

Suppose, now, that L is a lamination. Let E be the set of all equivalence relations

on S whose equivalence classes are closed and respect L. Notice that E 6= ∅, as

the degenerate equivalence relation where all of S is one equivalence class is in E .

Notice that a relation chosen from E must have classes which are closed, but it is

not necessary for it to be a closed equivalence relation like ≈. Hence, while ≈ is the

finest equivalence relation among those which are closed, unlinked and respect L, the

relation, E, defined in the next lemma will be the finest equivalence relation among

those which are unlinked and respect L.

Lemma 2.3. Let E ∈ E be the equivalence relation where the classes are defined by

[`]E =⋂e∈E [`]e, where ` ∈ L. Suppose an equivalence class [x]E of E is not connected,

with separation A, B ⊂ [x]E, then there exists a leaf `0 ∈ L such that `0 has one

endpoint in A and the other endpoint in B.

Proof. Assume [x]E = A ∪ B, where A, B are two non-empty, disjoint, closed

subsets of [x]E, and suppose that no leaf ` ∈ L has an endpoint in both A and B.

Define a new equivalence relation, E∗ to be a refinement of E by choosing a point

x ∈ A and a point y ∈ B. By hypothesis, there is no leaf joining the two. Let

[x]E∗ = A and [y]E∗ = B, which contradicts the definition of E, as it was defined to

be the finest equivalence relation with closed classes. �

38

Corollary 2.4. Let ≈ be the finest closed equivalence relation of S which respects

L. Then Lemma 2.3 holds for ≈.

Proof. This follows immediately from Lemma 2.3, since ≈ has closed equivalence

classes and is the finest such relation. �

Proposition 2.5. Given a lamination L in the set of all equivalence relations

whose classes are closed and respect L, the finest such equivalence relation is ≈.

Proof. Given a lamination L and E ∈ E as in 2.3. By construction, E is the

finest equivalence relation in E . We must show that E is unlinked and closed.

To show E is unlinked, assume by way of contradictions that E is linked. Then

there are points x, y ∈ S with x 6≈ y such that [x]E and [y]E are linked. By Proposition

1.3, we can find an arc Ax ⊂ S with both of its endpoints being elements of [x]E,

and such that Ax ∩ [y]E 6= ∅ and (S \ Ax) ∩ [y]E 6= ∅. Since the endpoints of Ax are

elements of [x]E, these intersections define two nonempty disjoint closed subsets of

[y]E whose union is [y]E. Therefore, by Lemma 2.3, there is a leaf `0 ∈ L with one

endpoint y0 ∈ Ax∩ [y]E and the other y1 ∈ (S\Ax)∩ [y]E. Notice that these endpoints

are disjoint from [x]E and y0 ∈ Ax while y1 ∈ S \ Ax. This implies that, for the two

components C0 and C1 of S \ {y0, y1}, C0 ∩ [x]E and C1 ∩ [x]E defines two nonempty,

disjoint, closed subsets of [x]E. By Lemma 2.3, there is a leaf `1 ∈ L with one endpoint

x0 ∈ C0∩ [x]E and the other endpoint x1 ∈ C1∩ [x]E. Now we have found two leaves `0

and `1 in L that intersect in a point not on the circle. This contradicts the definition

of a lamination; hence E must be an unlinked equivalence relation.

Clearly all the equivalences classes in E are closed. So by Proposition 1.5, we

only need to show that the equivalence classes of E form an upper semi-continuous

decomposition. For a point x ∈ S, let Cx denote the convex hull of [x]E. Then

since the equivalence relation is unlinked, Cx ∩ Cz = ∅ for every z 6∈ [x]E. Now, let

{pi}∞i=1 be a sequence of points in S converging to some point p ∈ S. To show upper

semi-continuity, we prove that for every z ∈ S such that for every open neighborhood

39

U of z, [pi]E ∩ U 6= ∅ for infinitely many i, then z ∈ [p]E. Let Ap denote the set of

all such points z. We intend to show Ap ⊂ [p]E. If Ap = {p}, then we are finished,

as p ∈ [p]E. Suppose there exists q ∈ Ap, q 6= p. Then without loss of generality, we

may assume that there exists a sequence of points {qi}∞i=1, with qi ∈ [pi]E for each i,

that converges to q. Since the equivalent classes are closed, we can also assume that

pi 6∈ [pj]E for every j 6= i. Observe that, since Cpi is an infinite sequence of pairwise

disjoint closed subset of the disk, the areas of Cpi must converge to zero. Let {zi}∞i=1

be a sequence of points with zi ∈ [pi]E, converging to some z. It z 6∈ {p, q}, then the

areas of Cpi could not converge to zero (by definition of convex hull). This implies

that, for the assumed sequence of [pi]E, Ap = {p, q}.

Now since L is closed, we need only show that there exists a sequence of leaves

{`i}∞i=1 from L that converges to the leaf pq. For each i, let Di be the smallest arc

in S that contains {p, pi}, and likewise let Bi be the arc containing {q, qi}. By the

convergence of pi and qi, we can find integers N1 < N2 < N3 such that DN3 ⊂ DN2 ⊂

DN1 and BN3 ⊂ BN2 ⊂ BN1 with DN1 ∩ BN1 = ∅. By the construction of these

intervals, and since the convex hulls CpN1, CpN2

, CpN3are pairwise disjoint, this implies

that [pN2 ]E ⊂ (BN1 \BN3)∪(DN1 \DN3). In particular, [pN2 ]E is contained in the union

of two pairwise disjoint arcs K and D, both having nonempty intersection with [pN2 ]E.

Recall that [pN2 ]E is the set if orbits equivalent to pN2 under the finest equivalence

relation E. So by Lemma 2.3, there is a leaf `1 ∈ L such that `1 ∩K ∩ [pN2 ]E 6= ∅ and

`1 ∩D ∩ [pN2 ]E 6= ∅. By induction, this procedure may be repeated for infinitely many

i, which constructs a sequence of leaves `i ∈ L converging to pq. �

In this section so far, we have discovered the equivalence relation ≈. We have left

unresolved the issue of ≈ being sibling d-invariant. However, in section four of this

chapter, we will introduce the idea of omega continua. This concept may be used to

show that ≈ is indeed sibling d-invariant. However, there is another issue at hand here.

After applying the quotient map relating to the relation ≈, we obtain a topological

Julia set. However, there are cases where two distinct laminations yield the same

40

topological Julia set. Hence, we may talk about the equivalence classes of laminations

themselves. We define two laminations to be equivalent when their topological Julia

sets are the same. For an example, refer to Figure 3.1.

Hence, for any given lamination L in an equivalence class of laminations [L] there is

one corresponding topological Julia set. An interesting question at this point is given

a topological Julia set, if one tries to undo the process of the quotient map to arrive

at a lamination, which lamination will one develop? Since there are many members in

one equivalence class, the one developed will be in some sense the “smallest” one. We

have already defined such a lamination in a previous section, but we will redefine that

now.

Definition 2.6 (q-lamination). A lamination is said to be a q-lamination if the

only leaves in the lamination are on the boundary of the convex hull of equivalence

classes, where the invariant equivalence relation identifies any two points which are

located on the same leaf.

Figure 3.1. Two distinct but equivalent laminationsThe above pictures are the beginnings of two laminations. They are distinct, as thefirst has a critical leaf, c, while the second does not have that leaf. However, they areequivalent, as they share the same topological Julia set. Notice that the second is a

q-lamination while the first is not.

When going from a topological Julia set back to a lamination, one cannot distinguish

leaves which are not on the boundary of a convex hull of an equivalence class of points

in that lamination. Thus, the only leaves that are picked up in the reverse process are

those which are on the boundary of a convex hull of an equivalence class of points.

41

Now, one may better understand why the concept of q-laminations is important and

in some sense all we really need to understand a complete picture of laminations.

3. Properties of the Equivalence relation ≈

Definition 3.1. Given a lamination L, we know L∗ = L ∪ S. For any subset

A ⊂ S, similarly defiine A∗ = A ∪ {` ∈ L : ` ∩ A 6= ∅}.

We know that L∗ is a continuum. The next corollary shows that every ≈-class

defines a subcontinuum of L∗.

Corollary 3.2. Let A ⊂ S be an ≈-class for a given lamination L. Then A∗ is

a continuum.

Proof. First note that since A is an ≈-class, we have that A∗ ∩ S = A. Since

every ≈-class is compact, this implies that A∗ is compact. Now, we will show that A∗

is connected. Suppose, by way of contradiction, that A∗ is not connected. Then we

can find two disjoint and closed sets K,D ⊂ D such that K ∩ A∗ 6= ∅ 6= D ∩ A∗ and

A∗ ⊂ K ∪D. Observe that leaves are connected sets. This implies, since K and D

form a separation of A∗, that every leaf in A∗ is contained in either K or D. Also, by

definition of A∗, every leaf in L that intersects A is contained in A∗. This means that

K ∩A 6= ∅ and D∩A 6= ∅. Hence K ∩A and D∩A are two disjoint, nonempty, closed

subsets of S whose union is A. By Corollary 2.3, there is `0 ∈ L with one endpoint

in K ∩A ⊂ K and the other in D ∩A ⊂ D. Since K and D are disjoint, this leaf is

not contained in K nor in D. But, `0 ∩ A 6= ∅, and so we have that `0 ⊂ A∗. This

contradicts the leaf `0 ∈ L being connected. �

Returning for a moment to an example which we have seen before, if we assume

that a sibling d-invariant lamination is gap-free, we proved in Proposition 2.13 that

the quotient space is an arc. We used Lemma 2.12 which stated that all leaves in

a gap-free lamination are disjoint. We now return to showing that these leaves are

42

actually parallel. The proof of this statement is due to R. Ptacek. First, we will need

two propositions.

Proposition 3.3. Let L be a sibling d-invariant gap-free lamination and ` = ab ∈

L be a leaf in the lamination. Then each of the circular arcs (a, b) and (b,a) contain

exactly one degenerate point of L.

Proof. Refer to the proof of Proposition 2.13. The paragraph proving there are

two degenerate leaves proves this statement. �

Proposition 3.4. Let L be a sibling d-invariant gap-free lamination with degen-

erate leaves x 6= y. Then the degenerate leaves satisfy exactly one of the following:

• σd(x) = x, σd(y) = y (A pair of fixed points.)

• σd(x) = x = σd(y) (A fixed point and an immediate preimage of that point.)

• σd(x) = y, σd(y) = x (The points form a period two orbit.)

Proof. If x is degenerate, then σnd (x) is also degenerate by backward invariance

of L. Since there are exactly two degenerate leaves of L, the full orbit of x may contain

at most two points. The result follows. �

Lemma 3.5. Let L be a sibling d-invariant gap-free lamination. Then the leaves of

L are all parallel.

Proof. Let x 6= y be the degenerate leaves of L. We may assume that x = 0 since

rigid rotation of L by a fixed point preserves d-invariance and parallel leaves. Since a

sibling d-invariant lamination is also sibling dk-invariant, without loss of generality we

may assume that x is fixed under d′ = d2. Then σ−1d′ (0) = {0, 1

d′, 2d′, ..., d

′−1d′}. Since all

(except possible one) preimage of 0 is nondegenerate, we must decide how to connect

the preimages with leaves. Note that any leaf that has a preimage of 0 as one end

point must be a critical leaf. The analysis now slightly diverges depending on whether

d′ is even or odd.

43

Suppose d′ is even, then there are d′ − 1 (an odd number) or preimages of 0 other

than 0 itself. Since by Lemma 2.12 all leaves in a gap-free lamination are disjoint,

it is impossible that each of the d′ − 1 preimages of 0 is an endpoint of a critical

leaf. Therefore, y ∈ σ−1d′ (0) \ {0}. We claim that y = 1

2. Let T = σ−1

d′ (0) ∩ (0, 12)

and B = σ−1d′ (0) ∩ (1

2, 0). By way of contradiction, suppose that y ∈ T . Then, 1

2is

a nondegenerate point. Let ` be the critical leaf containing 12

as one endpoint and

z be the other endpoint of `. If x ∈ T , then we have accounted for two points of T

and none from B, so there must be a leaf connecting two points of B. But this is

impossible since this would mean both degenerate leaves lie on the same side of `. On

the other hance, z ∈ B directly implies that both degenerate leaves would lie on the

same side of `. Hence, y 6∈ T . A similar argument shows that y 6∈ B. Hence, we may

conclude that y = 12. Finally, we have to determine how to position critical leaves that

connect the nondegenerate preimages of 0. We cannot join two points of T by a leaf;

otherwise both degenerate leaves would lie below the leaf. Similarly, we cannot join

two points of B. The only remaining option is to join each point of T to a point of B.

Let tb ∈ L with t ∈ T and b ∈ B. It must be that |T ∩ (b, t)| = |B ∩ (b, t)|; otherwise

we would be forced to connect two points of T (or B) or have a third degenerate leaf,

both contradictions to a gap-free lamination. Then, the only way to connect is to

draw vertical leaves between kd′

and d′−kd′

. So, we may conclude that L has a vertical

leaf coming out of every preimage of 0. However, L is sibling dn-invariant, and since d

is odd, dn is also odd. So, the previous argument shows that we have vertical leaves

coming out of every point of σ−1d′n(0) for every n ∈ N. As n grows, this collection of

leaves becomes increasingly dense, disallowing any non-vertical leaves. So, L consists

of all parallel leaves.

Now, suppose that d′ is odd. Then there are an even number of preimages of 0

other than 0 itself. If one of them were degenerate, we would be left with an odd

number of points to connect via critical leaves. This is impossible since leaves must

be disjoint. So no other preimage of 0 may be degenerate. Now, we find ourselves in

44

the first case, where we must join points of T to points of B by leaves, and the only

choice is the vertical connection. Once again, taking powers of the map yields a dense

collection of vertical leaves, completing the argument. �

4. Omega Continua

The concept of omega continua is due to Oversteegen, and many proofs in this

section are due to Childers.

We are ready to give a constructive definition of the sibling d-invariant lamination

which, as we prove later, coincides with ≈. This is an alternative method for arriving

at an invariant equivalence relation which respects a given sibling d-invariant lami-

nation, rather than using the finest equivalence relation as in the preceding sections.

Additionally, using omega continua is the easiest way to prove that ≈ is invariant.

Definition 4.1 (ω-continuum). Call a continuum K ⊂ L∗ which meets S in a

countable set an ω-continuum. Given a sibling d-invariant lamination L, let ≈L be

the equivalence relation in S induced by L as follows: x ≈L y iff there exists an

ω-continuum K ⊂ L∗ containing x and y.

Clearly, ≈L above is an equivalence relation which respects L. For x ∈ S we

denote by [x] the≈L-class of x. Additionally, by L≈ we denote the sibling d-invariant

lamination containing all chords which are located on the boundary of a convex hull

of an equivalence class in L under the equivalence relation ≈.

Lemma 4.2. If K ⊂ L∗ is a continuum then the following claims hold.

(1) If ` ∈ L is a leaf with K ∩ ` 6= ∅ and K does not contain an endpoint of `

then K ⊂ `. In particular, if K ∩ S 6= ∅ then K contains an endpoint of `

and if K meets two distinct leaves then it meets S.

(2) If G is a gap and x, y ∈ S∩G∩K then either (x, y)∩G ⊂ K or (y, x)∩G ⊂ K.

Proof. (1) Given a leaf ` with endpoints a, b, choose small disks U, V centered

at a, b. Set W = D \ (U ∪ V ). Since gaps are dense, arbitrarily close to ` ∩W from

45

either side there are “in-gap” curves Q, T connecting points of U ∩ D with points of

V ∩ D and disjoint from L∗. Hence ` ∩W is a component of L∗ ∩W . If a continuum

K ⊂ L∗ is non-disjoint from ` and does not contain an endpoint of ` then U, V can be

chosen so small that K ⊂ W and so by the above K ⊂ `.

(2) Suppose that u ∈ (x, y)∩G \K and v ∈ (y, x)∩G \K. Connect points u and

v with an arc T inside G. Then T separates x from y in D and is disjoint from K, a

contradiction. �

Theorem 4.3. If L is a sibling d-invariant lamination, so is L≈, and ≈L=≈, the

finest equivalence relation compatible with L.

Proof. We show first that equivalence classes are closed. We may assume that

there exists a sequence {xi} in [x1] such that x1 < x2 < · · · < x∞ with limxi = x∞

and x1x∞ 6∈ L (otherwise trivially x∞ ∈ [x1]), where < denotes the induced circular

order on S = R/Z. We show that x∞ ∈ [x1]. Since xi ∈ [x1] for every i then there

exists an ω-continuum Ki containing both x1 and xi. Also, let L1,∞ be the collection

of all leaves ` = pq ∈ L with p and q in distinct components of S \ {x1, x∞}. Because

L is unlinked, L1,∞ has a linear order defined by ` < `′ provided that ` separates the

(open) disk between `′ and x1 (the leaves ` and `′ may meet on the unit circle in which

case we can only talk about separation in the open unit disk).

First assume that L1,∞ = ∅. Then x1, x∞ belong to the same gap G. By Lemma 4.2,

Ki contains either [x1, xi]∩G or [xi, x1]∩G for any i. If [xi, x1]∩G ⊂ Ki then x∞ ∈ Ki

and hence x1 ≈L x∞ as desired. If [x1, xi] ∩ G ⊂ Ki for any i then [x1, x∞] ∩ G is

countable and the part Q of ∂G which extends, in the positive direction, from x1 to

x∞, is an ω-continuum containing x1 and x∞. Hence again x1 ≈L x∞ as desired.

Next, assume that L1,∞ 6= ∅. Let M = sup{L1,∞} = pq (p and q may coincide, in

which case M = {x∞}). Since L∗ is closed, M ⊂ L∗. Let us show that {p, q} ⊂ [x1].

Indeed, choose a sequence (or a finite set) of leaves `0 < `1 < · · · , lim `i = M and

points xn(i), i = 1, 2, . . . such that the component Li of D \ [`i−1 ∪ `i], whose boundary

46

contains `i and `i+1, separates x1 and xn(i). Then (Li ∩Kn(i))∪ `i−1 ∪ `i = Ri is an ω-

continuum which extends from `i−1 to `i. Also, let R0 be the closure of the intersection

of Kn(1) with the component of D \ `0 containing x1 ∪ `0. Then R = M ∪ (∪∞i=1Ri) is

an ω-continuum, and hence p, q ∈ [x1]. If x∞ ∈M then we are done. If x∞ 6∈M then

it follows from the definition of L1,∞ that there exists a gap G such that ∂G contains

M and x∞. Let p ∈ (x1, x∞). Then the argument from the previous paragraph applies

to p (playing the role of x1) and x∞ and so p ≈L x∞. Together with x1 ≈L p this

implies x1 ≈L x∞ as desired.

Let us show that distinct ≈L-classes are disjoint. Suppose that there exists x, y ∈ S

such that x 6≈L y and [x] ∩ [y] 6= ∅. Then there is a point w ∈ [x] ∩ [y], and there is

an ω-continuum, Kx, containing both x and w and an ω-continuum, Ky, containing

both y and w. But, then Kx ∪ Ky is an ω-continuum containing both x and y, a

contradiction.

Suppose next that xi ≈L yi and (xi, yi)→ (x∞, y∞) in S× S. We must show that

x∞ ≈L y∞. Assume that x∞ 6= y∞; since classes are closed we may also assume that

xi 6= x∞, yi 6= y∞ for any i. Let Ki be ω-continua containing xi and yi. We may

assume that limKi = K∞ ⊂ L∗ exists (in the sense of Hausdorff metric). Since classes

are closed and pairwise unlinked and x∞ 6= y∞, we may also assume that all Ki are

pairwise disjoint and xi, yi are such that the chord xiyi is disjoint from the chord

x∞y∞ for any i which implies that the convex hull of each [xi] is disjoint from x∞y∞.

Each Ki is contained in the convex hull of [xi]. Since the convex hulls of the [xi]’s are

disjoint, then K∞ must be a leaf in L. So x∞ ≈L y∞, and ≈L is a lamination.

Let us show that ≈L and ≈ are the same. Since ≈L is compatible with L, ≈ is

finer than ≈L. We show that ≈L is finer than any lamination compatible with L.

Let ≈ be a lamination compatible with L. Then for any two points x, y ∈ S with

x ≈L y we have to prove that x ≈ y. Since x ≈L y then there exists an ω-continuum

K ⊂ L∗. To proceed we first extend the equivalence ≈ onto D by declaring two points

u, v ∈ D equivalent if and only if for some class A we have u, v ∈ CH(A). Clearly the

47

new equivalence ≈ is an extension of ≈. Set Z = D/ ≈ and let π : D → Z be the

corresponding quotient map. Let us show that π(K) = π(K ∩ S). Indeed, if x ∈ K \ S

then x belongs to the appropriate leaf ` and by Lemma 4.2 and endpoint a of ` belongs

to K. Since π(a) = π(x) we see that π(x) ∈ π(K ∩ S) and so π(K) = π(K ∩ S). Since

K is an ω-continuum then K ∩ S is countable, hence π(K) = π(K ∩ S) is at most

countable and therefore a point. Thus, π(x) = π(y) and x ≈ y as desired. �

At this point, it remains to be shown that the equivalence relation ≈ is an invariant

equivalence relation, the definition of which we restate below. We prove in Lemma 4.5

below that conditions (1) and (2) of the definition hold. The proof that condition (3),

and consequently, Theorem 4.6, hold will appear in a subsequent paper.

Definition 4.4 (Invariant Equivalence Relation). Given a closed equivalence

relation ≈ on S, we define ≈ to be d-invariant under the map σ∗d if the following hold:

(1) CH([x]) ∩ CH([y]) = ∅ for x 6≈ y.

(2) For every x ∈ S there exists a y ∈ S such that σ([x]) = [y].

(3) L≈ is sibling d-invariant.

Lemma 4.5. Given a sibling d-invariant lamination L, the finest closed equivalence

relation ≈which respects L satisfies the following conditions:

(1) CH([x]) ∩ CH([y]) = ∅ for x 6≈ y.

(2) For every x ∈ S there exists a y ∈ S such that σ([x]) = [y].

Proof. We have already proven (1) in the proof of Theorem 4.3.

Now, we prove (2) as follows: Let [x] be an equivalence class under the equivalence

relation ≈. Let y ∈ [x]. We must show that σd(x) ≈ σd(y). By Theorem 4.3, there

exists an ω-continuum, K ⊂ L∗, containing x and y. As σ∗d is a continuous function,

σ∗d(K) ⊂ L∗ is also a continuum, and its intersection with S must be countable.

Hence, σ∗d(K) is an ω-continuum contained in L∗, and σd(x) ≈ σd(y) by Theorem 4.3.

Therefore, σ∗d([x]) ⊂ [σ∗d(x)].

48

Let z ∈ [σ∗d(x)]. Since σ∗d([x]) ⊂ [σ∗d(x)], σ∗d(x) ≈ z. By Theorem 4.3, there is an

ω-continuum, M ⊂ L∗, containing z and σ∗d(x). By Theorem 2.18, σ∗d is a confluent

map. Let Kx be the component of (σ∗d)−1(M) containing x. Since σ∗d is continuous,

Kx is closed and hence a continuum. And, Kx ⊂ L∗, as L∗ is sibling d-invariant. Since

σ∗d is confluent, there is w ∈ Kx such that σ∗d(w) = z. �

Theorem 4.6. Given a sibling d-invariant lamination L the finest closed equiva-

lence relation which respects L, i.e. ≈, is d-invariant.

We must note that the preceding theorem is also true for Thurston d-invariant

laminations.

49

CHAPTER 4

Constructing Laminations

In this section, whenever possible, our statements and proofs will be for laminations

of degree d. However, there may be times when it is more convenient to have statements

and proofs specifically for σ2.

1. Properties of Leaves and Gaps

Proposition 1.1. Given any leaf ` ∈ L, the number of leaves mapping to ` under

σd is finite. Moreover, this is true if ` is a degenerate leaf.

Proof. Suppose {pαqα}α∈A is the set of leaves mapping to ` = pq under σd,

labeled such that σd(pα) = p and σd(qα) = q for all α ∈ A. Since σd maps d-to-1, A is

a finite set. Thus there are finitely many chords connecting a pα to a qα. �

Definition 1.2 (length of leaf). Given a leaf ` of a lamination L, we define the

length of ` to be the distance between its endpoints, where S is parameterized by [0, 1).

Hence, the maximum length of a leaf is 12. In the case of σ2, a leaf of length 1

2

is a critical leaf. Notice also that if S is a set of chords in D all with equal length

such that the chords intersect in only possibly an endpoint, then |S| is finite, as if all

chords are of length x, only finitely many points of distance equal x will fit within the

parameterization of S.

Proposition 1.3. Let L be a lamination under σd. L has a finite number of

critical leaves and can contain at most d− 1 disjoint critical leaves.

Proof. Let a be a critical value. Since σd is d-to-1 on S, there are αi ∈ S for

1 ≤ i ≤ d such that σd(αi) = a for each i. A critical leaf must connect αi to αj for

50

i 6= j and cannot cross other chords. Hence, there must be a finite number of critical

leaves. The only way for there to be d critical leaves is for each critical leaf to touch

two more. Hence, there are at most d− 1 disjoint critical leaves. �

Corollary 1.4. Let L be a sibling d-invariant lamination under σd, then the sum

of the number of critical leaves and the number of critical gaps of L is finite.

Proof. By Proposition 1.3, the number of critical leaves is finite. Suppose there

were infinitely many critical gaps, Gα for α ∈ A. Each Gα contains (either in its

interior or its boundary) a critical chord pαgα with both of its endpoints mapping

to the same point. Since each Gα can only meet another gap in a boundary chord

or a boundary point, we have infinitely many chords of D of length 1d, which is a

contradiction as there can only be finitely many such chords by Proposition 1.3. �

Definition 1.5. Given a leaf ` in a lamination L, if there exists an infinite

sequence of distinct leaves converging onto `, then we call ` a limit leaf. If ` is not a

limit leaf in L, then we call ` an isolated leaf.

Notice there are two kinds of limit leaves: a two-sided limit leaf, which cannot be

on the boundary of any gap, and a one-sided limit leaf which must be on the boundary

of some unique gap.

We will call a leaf which is the preimage of another leaf under σd a preimage leaf.

2. A Method of Constructing Laminations

At this point, a natural questions is: How does one construct a sibling d-invariant

lamination? One simple method is to begin with a forward invariant prelamination and

take preimages of the leaves in that set to construct a sibling d-invariant lamination.

In order to do this, one first needs some definitions.

Definition 2.1 (Degree of a Gap). Let L be a sibling d-invariant lamination and

G a gap of L. The degree of G is defined to be the degree of the map from G to its

image gap, provided its image is a gap. If its image is a leaf `, then the degree of G is

51

half the number of leaves on ∂G mapping to `. If its image is a point, then the degree

is zero. If the degree of G is not one, then G is said to be a critical gap, and a leaf `

is said to be a critical leaf if the image of the leaf is a single point.

We wish to discover when two different laminations are in some sense the same.

Suppose we have two laminations L and M and a map f : L → M. Suppose also

that f |S : S→ S is a positively oriented covering map.

Definition 2.2 (Corresponding Laminations). Laminations L and M correspond

by f if f satisfies the conditions of Definition 2.1 or, equivalently, if f satisfies the

conditions of Definition 1.3.

We will remark that extending the map f into the interior of the disk works in

the case of corresponding laminations. The definitions of degree of a gap, critical gap,

and critical leaf also extend to corresponding laminations.

Proposition 2.3 (Sibling d-invariant Lamination). Let L be a lamination without

critical gaps which satisfies the following conditions:

• L is forward invariant by the map σd, and

• there are d− 1 critical leaves in L.

Then there is a sibling d-invariant lamination, the σd-saturation, S(L) of L, which

contains L.

Proof. Let L = L0 be a forward invariant lamination under σd with no critical

gaps, and let `c1 , `c2 , ..., `cd−1 be the set of critical leaves. The d − 1 critical leaves

divide D into d connected pieces, D1, ..., Dd. Notice that the boundary of each of the

Di is not necessarily a connected subset of S; however, the sum of the length of the

pieces of the boundary of each Di is 1d. By collapsing the critical leaves, `ci , to points,

ci, we obtain a union of d disks, Di, each of which is connected to at least one more

disk by the point which is the collapsed critical leaf.

52

Figure 4.1. Constructing a sibling d-invariant laminationThis is an illustration of the proof of Proposition 2.3. Critical leaves are dotted, and

critical values are denoted by a bold dot.

Suppose first that L = L0 contains no critical value. Then, we define Li0 to be the

collection of leaves in Di formed by placing a homeomorphic copy of L0 into Di with

each ci being identified with the critical value of its corresponding leaf, i.e. σ∗d(`ci), and

then “undoing” shrinking the critical leaves to a point. Hence, we have homeomorphic

copies of L0 in each Di, i.e. for each ` ∈ L0 and for each i = 1, ..., d, there is an `i ∈ Di

such that σ∗d(`i) = `. Thus, we have defined the inverse of σ∗d along its d branches, so

that the new set L1 = L0 ∪ L0 is forward invariant. For any leaf ` ∈ L0, there is a

leaf `′ in L0 ⊂ L1 (in fact d such leaves) such that σ∗d(`′) = `. Additionally, suppose

` ∈ L1. If ` ∈ L0, then by definition of L0, ` has d disjoint siblings, one in each Di. If

` ∈ L0, (One only needs to consider when ` is not a critical leaf, as a critical leaf maps

to a point.), then ` ∈ Di0 for some i0 = 1, ..., d. Hence, when placing a homeomorphic

copy of L0 into Di0 , ` = `i0 . Hence, ` has d disjoint siblings, namely the copies of `

placed into each Di. Hence L1 corresponds to L0 by σ∗d.

Suppose now that L = L0 contains a critical value, σd(`ci), for some critical leaf `ci .

And suppose that ` ∈ L0 is a leaf containing the critical value, σd(`ci). Then, there are

two discs, Dj and Dj+1 such that their intersection, ci, is the collapsed critical leaf `ci .

Place a homeomorphic copy of L0 within each Di, obtaining d copies of ` labeled `i.

53

(For i such that ci does not touch any leaves in these homeomorphic copies, stretching

ci back to a leaf is the same process as above and results in homeomorphic copies of

L0 in those portions on D.) However, we must make a careful construction of leaves

in L0 to get d disjoint siblings in the case that ci touches leaves of the homeomorphic

copies of L0. Now, within Dj, let C be the component of L0 containing `i. Stretch

the ci back out to a critical leaf, `ci , and examine CH(C ∪ `ci). Take the leaves on the

boundary of this convex hull to be in L0. Note that this process must be duplicated

for Dj+1 and for each other critical value in L0. Compare L0 to what we now have in

L1. It may be the case that L0 contains a leaf, `0, in either Dj or Dj+1 which is in

the interior of CH(C ∪ `ci) for some ci. If so, include that leaf in L0. Also, rotate the

leaf 180◦ around the critical leaf `ci to obtain one of its sibling leaves. Include this

leaf in L0 as well. This completes the construction of L1 = L0 ∪ L0.

Now, we show that L1 corresponds to L0 by σ∗d. The first two parts of Definition

2.1 are the same as in the easier part. We need to show that each leaf has d disjoint

siblings. This is why we had to be careful in our construction of L0 in the previous

paragraph. Suppose ` ∈ L1. If ` does not intersect a critical leaf, then ` has d disjoint

siblings by the previous case. So, suppose that ` intersects a critical leaf, then either `

is in Dj on the boundary of CH(C ∪ `ci) for some ci or ` cuts through CH(C ∪ `ci).

If ` is on the boundary of CH(C ∪ `ci), then there is also a symmetric convex hull in

either Dj−1 or Dj+1 with a disjoint sibling of ` on its boundary. If ` cuts through a

convex hull, we have included the sibling found by rotating ` 180◦ around the critical

leaf `ci , and these are disjoint. All other siblings are found in the other Di, and are,

hence, disjoint from `. Thus, L1 corresponds to L0 by σ∗d.

Now, we proceed by induction, defining Ln such that Ln+1 corresponds to Ln

by σ∗d for each n, with the above process. Note that Ln ⊂ Ln+1 for each n. Define

L∞ = ∪Ln. Then we claim that L∞ is the desired sibling d-invariant lamination. If

` ∈ L∞, then conditions (1), (2), and (3) of Definition 2.1 hold as they held at each

step of the inductive process. We only need to check these when ` ∈ L∞ \ L∞. So,

54

we assume ` = lim `i, where `i ∈ L. Then each `i has d disjoint siblings, `ji , where

j = 1, ..., d. Now, either ` = ab is such that σd(a) = σd(b), in which case ` maps to

a point in L∞, or σd(a) 6= σd(b). Examine σd(`). Since `i are forward invariant and

`i → `, σd(`) = lim `i. Hence σd(`) is a leaf in L∞. Hence, property (1) holds for L∞

Similarly, for each `i, there is a leaf `0i Such that σd(`0i ) = `i. Since, `i → `, there must

be a leaf `0 such that `0i → `0. By continuity, σd(`0) = `. Hence property (2) holds for

L∞. Also, since for each `i there are d disjoint siblings, the limit of these d disjoint

sequences are d disjoint leaves which are siblings. The limit leaves must be disjoint,

or else there would be leaves of the sequences crossing in the interior of the disc. �

3. An Example: The Douady Rabbit

The goal of this section is to provide a concrete example for constructing a

lamination from a forward invariant set. Further, we will take this lamination and

demonstrate how the equivalence relation forms a topological Julia set which is

homeomorphic to a particular Julia set. Due to simplicity, we will restrict this example

to one under σ2. Let us begin with a forward invariant set (Figure 4.2).

0!

T1/7

2/7

4/7

Figure 4.2. Forward invariant triangle, T

Now, we will make several pullbacks of T (Figure 4.3). Since T is forward invariant,

one of its pullbacks is itself. The other is labeled by T0. The pullbacks of T0 are

labeled T00 and T01, and so forth.

55

0!

T

T0

T01

T00

T011

T010

Figure 4.3. Pullbacks of T

Notice that the pullbacks so far have not entered the central region of the disc

formed by T and T0. We call this region the Central Strip. However, at some point

pullbacks will enter the central strip. This is where an ambiguity arises, as there are

two different ways of connecting endpoints to form disjoint siblings (Figure 4.4).

0!

T

T0

T01

T00

T0110

T0111

T011

T010

0!

T

T0

T01

T00

T0110

T0111

T011

T010

Figure 4.4. Ambiguity in Pullbacks

This ambiguity is why in Proposition 2.3 there is the requirement of d− 1 critical

leaves. In this example, we need one critical leaf to resolve this ambiguity. Figure 4.5

exhibits one such critical leaf. This critical leaf need not be a leaf in L. It is enough

to make the rule that in pulling-back, no leaf may cross this critical chord. Thus, we

see that making this restriction forces the left picture in Figure 4.4.

56

0!

T

T0

T01

T00

T0110

T0111

T011

T010

c

Figure 4.5. Adding a critical leafNotice that any critical leaf placed in this region will resolve this ambiguity.

In pulling back leaves, gaps are essentially pulled back to other gaps as well.

Triangles pull back to triangles, and Cantor gaps (gaps whose intersection with S is a

Cantor set) pull back to Cantor gaps. These are labeled in Figure 4.6 with the letter

G and using the same number convention as with the triangles.

0!

T

T0

T01

T00

T0110

T0111

T011

T010

G

G11

G10

G1

G0

Figure 4.6. Gaps formed in pulling-back

Now, we may form the topological Julia set by taking the quotient space of the

disc by identifying all points on the same leaf as described previously (Figure 4.7).

57

Notice that leaves not on a gap become ordinary points; triangles become cut points

of order 3; and Cantor gaps become Fatou domains.

T0

T01

T00

T0111

T011

T010

G

G11

G10

G1

G0

T

T0110

Figure 4.7. Topological Julia Set

The topological Julia set that is formed from this process is homeomorphic to a

real Julia set. In this case, it is homeomorphic to the well-known Douady Rabbit Julia

set (Figure 4.8).

Figure 4.8. Douady Rabbit Julia Set

58

4. Wandering Triangles and Central Strips

Let V = a, b, c ⊂ S. The convex hull CH(V ) forms a triangle inscribed in S.

The idea of this section is to discover if wandering triangles exist for σd. However,

we introduce first the concept of the central strip.

Definition 4.1. For a leaf ` = ab, the length of ` is defined to be the length of

the shorter of the subtended arcs of the circle. Under the map σ2 = 2t (mod 1), the

length of σ2(`) is given by the function

(1) τ2(x) =

2x, if 0 ≤ x < 1

4

1− 2x, if 14≤ x ≤ 1

2

Lemma 4.2 (Thurston’s Central Strip Lemma for σ2). Let σ2(t) = 2t (mod 1) and

`1 be a leaf such that |`1| > 13. Define C`1 as the gap bounded by `1 and its sibling, `2.

The first iterate of σ2(`1) that intersects C`1 connects two components of S ∩ Bd(C`1).

Proof. Let `1 be a leaf such that |`1| > 13. Note that |`1| = |`2|. Define the

two components of S ∩ Bd(C`1) to be holes. The length of each hole, h, is given by

h = 1−2|`1|2

. From the length function τ2, |σ2(`1)| = 1 − 2|`1|. Since |`1| > 13

it is

easy to verify that, |`1| > σ2(`1) > h. Thus the first iterate of `1 does not map into

C`1 . Iterates of `1 will grow in length until some, k, at which point |σk2(`1)| > 13.

Then, either |σk2(`1)| > |`1| (meaning we map into C`1 and the theorem holds) or,

|σk2(`1)| < |`1|. In the later case, τ2 gives us that 13> |σk+1

2 (`1)| > |σ2(`1)| > h, so

σk+12 (`1)∩C`1 = ∅. This implies that, σi2(`1) cannot map into C`1 unless there is some

j < i such that |σj2(`1)| ≥ |`1|, forcing it to connect two components of S∩Bd(C`1). �

The following generalization of Thurston’s Central Strip Lemma is due to Jeffrey

Houghton. First, a definition.

59

Definition 4.3. For a leaf ` = ab, under the map σ3 = 3t (mod 1), the length of

σ3(`) is given by the function

(2) τ3(`) =

3x, if 0 ≤ x < 1

6

1− 3x, if 16≤ x < 1

3

3x− 1, if 13≤ x ≤ 1

2

Lemma 4.4 (Central Strip Lemma for σ3). Let σ3(t) = 3t (mod 1) and |`1| ≥

|`2| ≥ |`3| be three sibling leaves. Suppose |`2| > 14. Define C1 as the component of

D \⋃`i such that Bd(C1)∩ (`i) 6= ∅, i = 1, 2, and define the components of S∩Bd(C1)

as holes, h. Then, the first iterate, i, such that |σi3(`2)| < |h| implies that |σi−13 (`2)| is

closer to 13

than |`2|.

Proof. The length of each hole h is given by |h| = 1−3|`2|3

. From the length function

τ3, |σ3(`2)| = 1− 3|`2|. Since 14< |`2| < 1

3, it is easy to verify that 1

4> |σ3(`1)| > |h|.

Subsequent iterates of `2 will grow in length until |σk3(`1)| > 14

for some k > 1. Then

by τ3(`), if σk3(`2) is closer to 13

than |`2|, then |σk+13 | < |h|. Otherwise, |σk+1

3 | > |h|.

Thus, σi3(`2) is bounded below by h until σi3(`2) gets sufficiently close to the critical

length of 13, at which time the next iterate has length less than |h|. �

If the critical leaves of the lamination are known, then this lemma can be stated

in such a way that iterates of `2 must map into one of two strips with holes of length

|h|. One strip will be C1,which will always contain a critical chord. The other strip

will not be fixed, but will be rigid in size and will contain the other critical leaf.

Lemma 4.5 (Gap Lemma for σ2). Let G be a periodic gap under the map σ2 that

is not critical. Then G is finite and contains at most 2 orbits of leaves.

Lemma 4.6 (Gap Lemma for σ3). Let G be a periodic gap under the map σ3. Then

G has at most 3 orbits of leaves. If G has 3 orbits of leaves, then G is a triangle which

returns by the identity.

60

Proof. By way of contradiction, suppose there are four distinct orbits of leaves,

then each orbit has a leaf of length closest to 13. Label these a, b, c, and d such that a

is the closest and d the farthest.

Consider the central gaps Ca, Cb, Cc, and Cd. By the central strip lemma, iterates

of these four leaves cannot enter their respective strips or a narrower one without

violating the “closest to 13” property. Thus, since Ca is the narrowest, none of the

iterates of G may lie inside Ca.

Ca contains some critical chord `. If Cb also contains `, then the iterate of G that

contains a lies inside Cb. This means c and d must map inside their central strips, a

contradiction. So, Cb must contain a different critical chord, m.

The strip Cc must contain either ` or m because there is not enough room in the

circle for a third critical leaf that will not cause the strips to cross each other. Without

loss of generality, let ` ∈ Cc. As above, there is some iterate of G, call it Gk containing

the leaf a. This iterate can contain a leaf from the orbit of b without violating the

central strip lemma, and it can contain one from the orbit of c, but it must be c itself.

There cannot, however, be a leaf from the orbit of d. Thus, we cannot have the fourth

orbit on the leaves of G.

Finally, note that in the case of three orbits, when Gk returns to itself, c must be

fixed. This means the entire gap must be fixed, which means that all leaves are in

separate orbits. Hence, there are only three leaves. Thus, for three orbits of leaves, G

must be a triangle. �

Note that while the previous lemma is a generalization of the one preceding it,

this may be generalized to a statement about σd, which may be attributed to Kiwi [7].

Lemma 4.7 (Gap lemma for σd). Let G be a periodic gap under the map σd. Then

G has at most d orbits of leaves. If G has d orbits of leaves, then G is a d-gon.

Definition 4.8 (Wandering Triangle). We will call a triangle wandering if

(1) T (σnd (V )) ∩ T (σmd (V )) = ∅ whenever n 6= m, and

61

(2) σnd (V ) does not collapse, i.e. it remains three distinct points in S for every

n ∈ N.

It may seem that a wandering triangle should be possible for any lamination.

However, it turns out that these cannot occur in a sibling 2-invariant lamination.

There are generalizations of this idea to wandering gaps of any finite size. The following

theorem is due to Thurston.

Theorem 4.9 (No Wandering Triangles in σ2). For sibling 2-invariant lamination,

L, there exists no wandering triangles.

Proof. Suppose there exists a wandering gap. Without loss of generality, let it

be a triangle W with sides a, b, and c. Since W is wandering, it must get arbitrarily

small, which means leaves must get arbitrarily close to 12

in length. Let n be an iterate

of W such that corresponding iterates of the sides have been greater than 13

in length

for some iterate less than n. Take the longest side of all the Wi’s (assume it is aj)

and consider its central strip Caj . The next time W maps into Caj , at time m, side a

will be greater than 13

in length. One side of this iterate of W will be shorter than

13. Assume it is bm. Now, for 1 < i < m, let bk be the longest. When we consider

Cbk , we note that Cam ⊂ Cbk . However, bm ∈ Cbk , which contradicts the central strip

lemma. �

The following theorem is due to Kiwi [7], but the proof is an alternate proof to

that he gave.

Theorem 4.10 (No Wandering Quadrilaterals in σ3). For a sibling 3-invariant

lamination, L, there exists no wandering gaps with more than 3 sides.

Proof. Suppose there exists a wandering gap with more than 3 sides. Without

loss of generality, let it be a quadrilateral W with sides a, b, c, and d. Since W is

wandering, leaves must get arbitrarily small, which means leaves have to get arbitrarily

close to 13

in length. Let n be an iterate of W such that corresponding iterates of the

62

sides have been closer to 13

than 14

in length for some iterate less than n. Take the

side of all the Wi’s closest to 13; assume it is aj. Consider its central strip Caj . Caj

contains some critical element of the circle. There is a second critical element. Take

the side of all the Wi’s that is closest to this second critical element; assume it is bk.

Consider its central strip Cbk . Since sides of W approach 13

and there is at most these

two critical elements, there exists an iterate Wm that maps into either Caj or Cbk . Let

cs and dt be the closest approach of sides c and d to 13, for a ≤ s, t ≤ m. Caj and Cbk

are both narrower than Ccs or Cdt , so by the central strip lemma, cm and dm are both

closer in length to 13

than cs and dt, respectively. Also by the central strip lemma,

bm must map closer in length to 13

than bk (either through direct application of the

lemma, or due to Caj being narrower than Cbk). Thus, we have 3 sides of Wm greater

than 14

in length, which only happens when the critical elements are inside Wm. This

prevents sides of W from approaching 13, giving us a contradiction. �

The previous theorem is important, in that a long-term goal is to classify all gaps

in a lamination. This has been done by Thurston for gaps in laminations which

are Thurston 2-invariant. However, it remains incomplete for laminations which are

Thurston 3-invariant, not to mention higher degrees and sibling d-invariant laminations.

Before we can classify what types of gaps exist in a sibling d-invariant lamination it is

helpful to know what cannot happen in a sibling d-invariant lamination.

63

References

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