VISUALIZING THE RIEMANN MAPPING

A Thesis

Presented to the

Faculty of

California State Polytechnic University, Pomona

In Partial Fulfillment

Of the Requirements for the Degree

Master of Science

In

Mathematics

By

Vrej Mikaelian

2014 SIGNATURE PAGE

THESIS: VISUALIZING THE RIEMANN MAPPING

AUTHOR: Vrej Mikaelian

DATE SUBMITTED: Spring 2014

Department of Mathematics and Statistics

Dr. John Arlo Caine Thesis Committee Chair Mathematics & Statistics

Dr. Mitsuo Kobayashi Mathematics & Statistics

Dr. John Rock Mathematics & Statistics

ii ACKNOWLEDGMENTS

I would like to express my sincere gratitude to my advisor Dr. John Arlo Caine for his enthusiasm, immense knowledge, and support in completing this work. I could not have imagined having a better advisor and mentor for my thesis. I would also like to thank the rest of my thesis committee: Dr. Kobayahsi and Dr. Rock, for giving me the knowledge and skills to complete my graduate coursework. I am also grateful to my professors: Dr. Michael Green, Dr. Patricia Hale, Dr. Amber

Rosin, and Dr. Robin Todd Wilson for their help to improve while working as a

Graduate Teaching Associate and support in finding a full time job a year before completing my graduate coursework. I also want to thank my beautiful wife, Katya.

She supported me, helped me to generate nice pictures, loved me, and brought up to the world a beautiful little baby daughter for me during my graduate school years. Lastly, and most importantly, I wish to thank my grandmother who raised me, and my grandfather who taught me and planted in me the idea to pursue graduate studies in mathematics during my childhood years. To their memory I dedicate this thesis.

- Vrej Mikaelian

iii ABSTRACT

The Riemann Mapping Theorem states that any proper simply connected open subset of the complex plane is conformally equivalent to the unit disk. This is an amazing result from complex analysis first proved by Riemann in his 1851 PhD thesis. Since then, several other mathematicians have introduced different, or more compact, versions of the proof of the theorem. This project concerns a visual approach to the proof of the Riemann Mapping Theorem via transformations of circle packings. In chapter 1 we explain the connection between transformations of circle packings and conformal maps. In chapter 2 we illustrate the proof of the theorem due to Riesz. In 1985 Thurston proposed a scheme for constructing the Riemann map as a limit of transformations of circle packings. We explain

Thurston’s conjecture and the proof by Rodin-Sullivan in chapter 3.

iv List of Figures

1.1 A Riemann mapping...... 2

1.2 A sequence of finite Riemann mappings under fpzq “ z3 ...... 8

z´i 1.3 A sequence of finite Riemann mappings under fpzq “ z`i ...... 9

2.1 A classical conformal mapping...... 12

3.1 Circle packings of Ω and D...... 21 3.2 Two circle packings and their carriers...... 22

3.3 Angle distortions under f...... 28

3.4 n circles surround the unit disk...... 29

3.5 Eight unequal vs. equal size circles surround the unit disk...... 30

3.6 An n-cycle surrounding the unit disk...... 31

v Contents

Signature Page ...... ii

Acknowledgements ...... iii

Abstract ...... iv

List of Figures ...... v

1 Introduction ...... 1

1.1 The Riemann Mapping Theorem ...... 1

1.1.1 The Geometry of Conformal Maps...... 2

1.2 Thurston’s idea and the method of circle packings ...... 7

1.3 The Rodin-Sullivan Theorem ...... 10

2 The Riemann Mapping Theorem ...... 11

2.1 The Proof of Uniqueness ...... 12

2.2 The Proof of Existence ...... 13

2.2.1 Reduction to the case Ω Ă D ...... 13

2.2.2 Existence of f :Ω Ď D Ñ D ...... 15

2.2.3 f :Ω Ñ D is surjective ...... 17

3 Thurston’s Conjecture and Rodin-Sullivan’s Theorem ...... 20

vi 3.1 Convergence of Domain and Range Carriers...... 24

3.2 The Convergence of the Functions fE as e Ñ 0...... 28

Bibiliography ...... 34

vii Chapter 1

Introduction

1.1 The Riemann Mapping Theorem

The study of mapping properties of holomorphic functions has been of interest to

many mathematicians interested in complex analytic function theory. Let D “ tz P

C : |z ´ 0| ă 1u be the unit disk. Given an open proper subset Ω of the complex

plane C, what conditions on Ω guarantee the existence of a holomorphic bijective

function f :Ω Ñ D with f ´1 : D Ñ Ω being holomorphic as well? Finding a solution to this problem would allow mathematicians to transfer questions about analytic functions on Ω, which may have very little geometric structure, to D which has more useful properties.

In his PhD thesis in 1851, Bernhard Riemann solved this problem.

Theorem 1.1.1. The Riemann Mapping Theorem

For any simply connected proper open subset Ω of C, there exists a holomorphic bijection F of Ω onto D the unit disk with a holomorphic inverse. Furthermore, if

1 z0 P Ω, then there exists a unique such F with F pz0q “ 0 and F pz0q ą 0. (Figure

1 1.1 illustrates the theorem).

Figure 1.1: A Riemann mapping.

Riemann proved his theorem using the Dirichlet principle with an assumption that the boundary of Ω is piecewise smooth. For many years there were contro­ versies about this proof until later when David Hilbert was able to prove that the

Dirichlet principle is valid under the hypotheses that Riemann was working with.

In 1912, Constantin Caratheodory introduced a new proof of the theorem using

Riemann surfaces. Later in 1922, Leopold Fejer and Frigyes Riesz published an even shorter version of the proof using Montel’s theorem whereas in Riemann’s proof the desired mapping was obtained as the solution of an extremal problem.

This proof is given in Chapter 2.

1.1.1 The Geometry of Conformal Maps.

In this subsection we will define conformal mappings and visualize their geometric

structure. Then we will show the beautiful connection between conformal bijections

and bijective holomorphic functions with nonzero complex derivative. Lastly, we will restate the Riemann Mapping Theorem using the theory of conformal maps.

We define a conformal map as a map that preserves angles and orientation at

2 each point of its domain. Here are two examples to get the idea.

Example 1.1.2. Let Ω “ tz P C : Repzq ą 0u and define a mapping f :Ω Ñ C

i π iargpzq i π to be fpzq “ e 4 z with z “ |z|e . Then |fpzq| “ |e 4 ||z| “ |z| and argpfpzqq “

π π π π i 4 i 4 iargp z q ip4 `argp z qq argp e zq “ arg p|z|e e q “ argp|z|e q “ 4 `argpzq . This shows that f π is a rigid rotation about the origin by 4 . The map f preserves angles and orientation which means that it is an example of a conformal map.

Example 1.1.3. Let Ω “ tz P C : Repzq ą 0u. Define f :Ω Ñ C to be the complex conjugate function defined by fpzq “ z. Then f is the reflection across the real axis. Thus, f is not a conformal map since it preserves angles but reverses the orientation.

Let us now take a look at what it means for a map to preserve orientation. Let f “ u ` vi where u :Ω Ñ R and v :Ω Ñ R, so that f :Ω Ñ C determines

2 fR :Ω Ñ R by fRpx, yq “ pupx, yq, vpx, yqq.

Definition 1.1.4. A map f preserves orientation at px0, y0q if and only if DfRpx0, y0q has positive determinant.

i π 1 x´y x`y Example 1.1.5. Let f z e 4 z ? 1 i x yi ? i ? u vi. p q “ “ 2 p ` qp ` q “ p 2 q ` p 2 q “ ` x y x y So f x, y ?´ , ?` which implies that, Rp q “ pp 2 q p 2 qq

u u ?1 ? 1 x y 2´ 2 detpDfRpx0, y0qq “ det “ det “ 1 ą 0. v v ??1 1 x y 2 2 ˚ ‹ ˚ ‹ ˝ ‚ ˝ ‚ This shows that f preserves orientation at each point as our intuition suggests.

Example 1.1.6. Let fpzq “ z “ x ´ yi. So fRpx, yq “ px, ´yq which implies that,

1 0 det Df x , y det 1 0. p Rp 0 0qq “ ¨ ˛ “ ´ ă 0 ´1 ˚ ‹ ˝ ‚ 3 Example 1.1.7. Let fp zq “ z2 ` z “ x2 ´ y2 ` x ` p2xy ´ yqi “ upx, yq ` vpx, yqi.

2 2 So fRpx, yq “ px ´ y ` x, 2xy ´ yq which implies that,

ux uy 2x ` 1 ´2y det Df x , y det det 4y2 4x2 1. p Rp 0 0qq “ ¨ ˛ “ ¨ ˛ “ ´ ´ ` vx vy 2y 2x ´ 1 ˚ ‹ ˚ ‹ ˝ ‚ ˝ ‚ 2 2 1 2 So, detpDfp x0 , y 0qq ą 0 if and only if x ` y ă p2 q . This shows that f preserves 1 orientation for all values inside the circle of radius 2 centered at 0 and does not preserve orientation on and outside that circle.

Now, to show what it means to preserve angles, we will measure angles between

curves by computing angles between the tangent vectors. Meanwhile, we will apply

the derivative Dfpx0, y0q to see what fR does to the tangent vectors at px0, y0q.

a11 a12 Definition 1.1.8. A two-by-two matrix A “ ¨ ˛ preserves angles if and a21 a22 ˚ ‹ only if there exists p ą 0 such that AT A “ pI. ˝ ‚

This is because A preserves angles (as a linear transformation on R2) if it pre­

xxx¨y T serves dot products up to a scalar p (because we have cos θ “ ? ? ). AA “ pI xx¨xx xy ¨xy

x1 2 2 means that Afx ¨ Afy “ pfx ¨ fy for all fx, fy P R with fx “ ¨ ˛ P R . This is x2 ˚ ‹ true if and only if pAfxqT pAfyq “ pfxT fy if and only if fxT pAA˝T qfy‚ “ pfx T fy if and only if AT A “ pI which means that if θ1 is the angle between Afx and Afy then

Axx¨Axy pxx¨xy pxx¨xy xx¨xy cospθ1 q “ ? ? “ ? ? “ ?? ““ cospθq. Axx¨Axx Axy¨Axy xx¨xx pxy¨xy plxxl plxyl lxxllxyl

Definition 1.1.9. Let Ω be an open subset of R2 . A map f :Ω Ñ R2 is conformal

if and only if f is differentiable on Ω and Dfpx0, y0q preserves angle and orientation at each px0, y0q P Ω.

4 Theorem 1.1.10. Let Ω Ă C be open. Then f :Ω Ñ C is holomorphic with

1 2 f pz0q “ 0 for all z0 P Ω if and only if fR :Ω Ñ R is conformal.

We will prove the forward direction of this theorem. The other direction can be found in Chapter 8 of the book by Stein and Shakarchi r1s.

df Proof. Let us recall that if f “ u ` vi then fRpx, yq “ pupx, yq, vpx, yqq and dz “ f pz`Δzq´f pzq limΔzÑ0 Δz where Δz approaches to zero from any direction. Since f is holomorphic, all of these directions give the same value for the limit. So let us take

Δz “ Δx ` 0i. Then

df fppx ` Δxq ` iyq ´ fpx ` iyq “ lim dz ΔxÑ0 Δx uppx ` Δxq, yq ` ivpx ` Δx, yq ´ pupx, yq ` ivpx, yqq “ lim ΔxÑ0 Δx

“ ux ` ivx.

ux ´vx By the Cauchy-Riemann equations, A Df x , y . So AT A “ Rp 0 0q “ ¨ ˛ “ vx ux ˚ ‹ ˝ ‚ u 2 ` v2 0 1 0 x x 2 2 “ pu x ` vxq which implies that DfRpx0, y0q preserves ¨ 2 2 ˛ ˛¨ 0 ux ` vx 0 1 ˚ ‹ ˚ ‹ ˝ ‚ ˝ ‚ ux ´vx 2 2 angles at each point of Ω where pux `v xq ą 0. Similarly, detpAq “ det ¨ ˛ “ vx ux 2 2 2˚ 2 ‹ pux ` vxq ě 0 implies that A preserves orientation where pux ` ˝vxq ‰ 0. ‚But 2 2 2 df 2 df ux ` v x “ |ux ` ivx| “ dz ą 0 since dz ‰ 0 on Ω. ˇ ˇ ˇ ˇ

With this theorem, we establish the connection between the geometric point of view of conformal maps and the analytic point of view of holomorphic functions with nonzero complex derivative. To see this connection better, let us consider the

5 z´i function fpzq “ z`i which maps the upper half plane onto the unit disk. Calculating 1 2i the first complex derivative of f, we get f pzq “ pz`iq2 which is clearly nonzero everywhere on the upper half plane. This shows that f is a conformal map in the

complex sense. On the other hand, it will be quite difficult to check if f preserves

angles and orientation on the upper half plane by simply computing the real form

of the function f as shown in the early examples of this subsection.

Thus, we can now conclude that a biholomorphism, a holomorphic function with

a holomorphic inverse, refers to the same concept of a bijective conformal mapping

establishing the beautiful connection between analysis and geometry. This puts

us in the position to restate the Riemann Mapping Theorem using the concept of

conformal mappings.

Theorem 1.1.11. The Riemann Mapping Theorem

For any simply connected proper open subset Ω of C and z0 P Ω, there exists a

1 unique conformal mapping F of Ω onto D such that F pz0q “ 0 and F pz0q ą 0.

The proof of the theorem is done in Chapter 2. There, we consider all injective

conformal maps f :Ω Ñ D with fpz0q “ 0 and choose one whose image is all of

1 D. We do it by choosing f such that |f pz0q| is as large as possible because a map

into D which is expanding near z0 as much as possible has the best chance of being onto D. In the next two sections of this chapter, we will look at the behavior of the

z´i function fpzq “ z`i through the theory of circle packings to suggest another proof of the Riemann Mapping Theorem conjectured by Thurston and proved by Rodin and Sullivan. This technique is introduced in Chapter 3.

6 1.2 Thurston’s idea and the method of circle pack­

ings

In his 1985 talk in Purdue, William Paul Thurston introduced a new geometric approach to the Riemann Mapping Theorem. In his conjecture, Thurston proposed the method of constructing a sequence of approximately conformal maps which limit to the Riemann mapping.

The key idea is the following. Consider the linear holomorphic function gpzq “

αpz´z0q`β with α ‰ 0. This is a translation by z0 followed by a rotation by argpαq,

a dilation by |α|, and a translation by β. Note that g1pzq “ α ‰ 0 for all z P C. Such a mapping clearly sends circles to circles. Thurston observed that conformal

1 maps do the same thing on small scales. Ideally fpzq « f pz0qpz ´ z0q ` fpz0q near

1 z0 when f is holomorphic, by linear approximation. So when f pz0q ‰ 0, f will map

small circles near z0 to curves which are close to small circles near fpz0q. Thus, if we pack Ω by small circles, f will map this packing of Ω to an approximate circle packing of fpΩq.

To illustrate this approach, consider Figure 1.2. The mapping f :Ω Ñ Ω1 given

by fpzq “ z3 takes the box Ω “ p0, 1q ˆ p0, 1q onto an ice cream cone shape Ω1 . By a circle packing of Ω we simply mean a connected collection of circles in Ω with

mutually disjoint interiors. Three examples of packings of Ω and their images in

Ω1 under fpzq “ z3 are shown in Figure 1.2. Notice how comparing the domain

packing and the image packing allows to visualize how the distortion of fpzq “ z3 varies with z. Thurston’s idea was that if we choose a circle packing for the box p0, 1q ˆ p0, 1q and cut out a corresponding circle packing of the ice cream cone we form a discrete map from the set of centers of circles to centers of corresponding

7 circles. Thus by constructing circle packings with smaller and smaller circles we can form a sequence of maps that in the limit approach to the classical conformal mapping f :Ω Ñ Ω1 as the size of the circles goes to zero.

f 72

- packing of 49 circles with radius radius with circles 49 of P (a) packing - 2 ɛ=1/14 7

f 152

- packing of 225 circles with radius radius with circles 225 of P (b) packing - 2 ɛ=1/30 15

f 252

- packing of 625 circles with radius radius with circles 625 of P (c) packing - 2 ɛ=1/50 25

Figure 1.2: A sequence of finite Riemann mappings under fpzq “ z3 .

A classical example of a conformal mapping that maps the upper half plane onto

z´i the unit disk is the complex analytic function fpzq “ z`i (Figures 1.3q. The process is the same, we construct a regular hexagonal packing of circles in the complex plane

C and use the boundary δΩ as a cookie cutter to cut out a packing Pn of n circles with each circle having a radius of size 1/n. In both of our examples we have chosen

a rectangular boundary for Ω which may be an arbitrary region of C. We will use

8 f144

- packing of 144 circles with radius radius with P circles (a) 144 of 144packing - ɛ=1/12

f400

- packing of 400 circles with radius radius with P circles (b) 400 of 400packing - ɛ=1/20

f1024

- packing of 1024 circles with radius radius with P (c) circles 1024 of 1024 packing - ɛ=1/32

z´i Figure 1.3: A sequence of finite Riemann mappings under fpzq “ z`i .

those two classical examples to give a more detailed view of Thurston’s conjecture

in Chapter 3.

Now for every domain packing Pn of n circles, we consider its simplicial complex

Kn. Each Kn is formed by connecting the centers of the circles that are tangent to each other with line segments creating a family of triangular faces. Then for each complex Kn of the domain packing Pn we compute the corresponding maximal packing Qn of n circles in the unit disk D (see Figure 1.3q and a transformation of circle packings fn : Qn Ñ Pn. The existence of this corresponding maximal packing in D is provided by the Discrete Uniformization Theorem which we will discuss in Chapter 3. Thus by reducing the size of the circles, we construct the discrete approximately conformal maps f144, f400, and f1024 as you can see in parts (a), (b), and (c) of Figure 1.3 with more and more circles involved in each packing keeping

9 the structure the same. Thurston conjectured that as n Ñ 8 the discrete approxi­ mately conformal maps fn : Qn Ñ Pn converge uniformly on compact subsets of D to the classical conformal mapping F of the Riemann Mapping Theorem.

1.3 The Rodin-Sullivan Theorem

After Thurston conjectured in 1985 that his scheme converges to the Riemann mapping, Burt Rodin and Dennis Sullivan proved Thurston’s conjecture in their famous Rodin-Sullivan Theorem in 1987.

Theorem 1.3.1. The Rodin and Sullivan Theorem

For any simply connected proper open subset Ω of C and n “ 1, 2, ..., let F : D Ñ

Ω be the Riemann mapping and fn : Qn Ñ Pn the discrete conformal mappings for

1 each circle packing Pn of radius size n . Then as n Ñ 8 the mappings fn converge uniformly on compact subsets of D to a conformal equivalence D Ñ Ω.

Since then, several other mathematicians have presented different proofs for the

Rodin-Sullivan Theorem. One approach was to apply more general packings rather

than packings based on the hexagonal combinatorics used by Rodin and Sullivan,

and by giving the convergence of the first and second derivatives of the discrete con­

formal maps between circle packings avoiding the use of the quasiconformal maps.

Another method involves two optimization problems and a quadratic program with

the unconstrained minimization of the circle patterns. The proof in Chapter 3

follows Rodin-Sullivan’s work.

10 Chapter 2

The Riemann Mapping Theorem

In this chapter we will prove the Riemann Mapping Theorem. We will show that there exists a conformal mapping F of Ω onto the unit disk D for any open simply connected proper open subset Ω of C. Furthermore, if z0 P Ω is given then there is

1 a unique such F with F pz0q “ 0 and F pz0q ą 0.

Definition 2.0.1. A one-to-one complex analytic (or holomorphic) function f :

Ω Ñ C is called a conformal mapping.

Definition 2.0.2. Two open sets U1 and U2 are said to be conformally equivalent if there exists a conformal mapping of U1 onto U2.

Theorem 2.0.3. The Riemann Mapping Theorem

Every simply connected proper open subset Ω of C is conformally equivalent to the unit disk D. Moreover, given z0 P Ω, there exists a unique conformal mapping

1 F of Ω onto D such that F pz0q “ 0 and F pz0q ą 0. See Figure 2.1.

11 Figure 2.1: A classical conformal mapping.

2.1 The Proof of Uniqueness

Suppose F1 and F2 are two conformal maps from Ω onto D with the property that

1 1 F1pz0q “ 0, F1pz0q ą 0, and F2pz0q “ 0, F2pz0q ą 0. A conformal map from D

´1 onto itself is called an automorphism of D. Then G “ F1 ˝ F2 : D Ñ D is an automorphism of the unit disk. Furthermore,

´1 Gp0q “ F1pF 2 p0qq “ F1pz0q “ 0, and

1 1 ´1 ´1 1 1 ´1 1 1 1 G p0q “ F1pF 2 p0qq ¨ pF 2 q p0q “ F1pz0q ¨ pF 2 q p0q “ F1pz0q ¨ 1 ą 0. (2.1) F2p0q It is an amazing result from complex analysis that the only automorphisms of the unit disk D are actually linear fractional transformations. See Stein and Shakarchi for the proof of the theorem r1s.

Theorem 2.1.1. If f is an automorphism of the disk, then there exist θ P R and

α P D such that α ´ z fpzq “ e iθ . 1 ´ αz¯ In particular, the only automorphisms of the unit disk that fix the origin (i.e.,

fp0q “ 0) are given by fpzq “ eiθz which represent rotations.

12 Now applying Theorem 2.1.1 to G which fixes 0, we get that the automorphism

´1 iθ 1 iθ G “ F1 ˝ F2 of the unit disk D has the form Gpzq “ e z. Thus, G pzq “ e for all z P D. In particular, G1p0q “ eiθ and from equation (2.1) we see that eiθ “ cosθ ` isinθ ą 0 which implies that θ “ 0. This shows that Gpzq “ z

for all z P D. Therefore, G is the identity map. Thus, we conclude that F1 “ F2, as desired.

2.2 The Proof of Existence

In this section, we will prove the existence part of the Riemann Mapping Theorem in

three different steps. The idea contained in those steps is to consider all conformal maps f :Ω Ñ D with fpz0q “ 0 to form a family of functions. Then choose one from this family such that the image of Ω under this function fills out all of D. We will obtain this function as the limit of a sequence of functions and try to fill in all

1 of D by making |f pz0q| is as large as possible.

2.2.1 Reduction to the case Ω Ă D

In this step we want to show that if Ω a simply connected proper open subset of C then Ω is conformally equivalent to an open subset of the unit disk that contains the origin. First, we suppose Ω is a bounded region, in which case we can build a

function composition of translations, rotations, and shrinks allowing us to map Ω

onto a subset of the unit disk D containing the origin. To see this, let z0 P Ω and define h :Ω Ñ C by hpzq “ z ´ z0. Then hpΩq contains 0 and is bounded so for some R ą 0, |z| ă R for all z P hpΩq. Now, to shrink hpΩq into the unit disk, define

1 g : hpΩq Ñ C by gpzq “ R z. Then gphpΩqq contains 0 and is contained in D. Since

13 h is a translation and g is a shrink, then g ˝ h is conformal equivalence of Ω onto a subset of D containing 0. Next, we consider the case when Ω is an unbounded simply connected proper

open subset of C. Let w P Ω and α R Ω. To prove our claim, consider the function [1] 1 F pzq “ . (2.2) logpz ´ αq ´ plogpw ´ αq ´ 2πiq Since Ω is simply connected, we can always specify the branch cut to define logpz ´

αq. So, we observe that F pzq is holomorphic on Ω since it is a composition of three elementary (holomorphic) functions and is defined on all of Ω. Then, we compute

the derivative,

1 F 1pzq “ (2.3) pz ´ αqplogpz ´ αq ´ plogpw ´ αq ´ 2πiqq2

and notice that F 1pzq is never zero on Ω. Thus, we have shown that the function

F :Ω Ñ C is conformal. Next, we prove that F is injective on Ω by observing that it is a composition of

the function fpzq “ logpz ´ αq followed by the injective functions of a translation

and reciprocal. So we only need to prove that fpzq “ logpz ´ αq is injective on Ω.

So, let z1 P Ω and z2 P Ω be arbitrary and suppose that fpz1q “ logpz1 ´ αq “ logpz2 ´ αq “ fpz2q. Then exponentiating both sides we obtain pz1 ´ αq “ pz2 ´ αq which implies that z1 “ z2. This concludes our proof that F is injective on Ω. Lastly, we prove that F is bounded on Ω by showing that the denominator of F is bounded away from zero, i.e, there exists a disk about logpw ´ αq ` 2πi such that fpzq “ logpz ´ αq does not belong to that disk as z ranges over all points of Ω. Suppose there does not exist a disk about logpw ´ αq ` 2πi which is

8 disjoint from fpΩq. Then there exists a sequence pznqn“1 in Ω such that fpznq Ñ

14 fp w q`2πi. But then ef pznq Ñ efpwq`2πi by the continuity of the exponential function.

f pznq logpzn ´αq fpwq`2πi fpwq 2πi logpw´αq Now e ““ e zn ´ α while e “ e e “ e “ w ´ α

implying that zn Ñ w. This shows that fpznq Ñ fpwq which is a contradiction since fpznq Ñ fpwq`2πi. Thus, we conclude that F pΩq is bounded. So, F is a conformal equivalence of Ω with a subset of some disk about 0. We can then compose with a shrink to obtain a map into D and an automorphism of D to ensure that the image contains zero.

2.2.2 Existence of f :Ω Ď D Ñ D

Since we showed in Section 2.2.1 that a proper open subset Ω of C is conformally equivalent to an open subset of the unit disk containing the origin, we may assume from now on that Ω is an open subset of the unit disk D with 0 P Ω. Now, we consider the family

Ftf “ :Ω Ñ D | f is holomorphic, injective, and fp0q “ 0u of one-to-one, holomorphic functions that map Ω Ă D into the unit disk and fix the origin. Let us also consider the set,

S “ t|f 1p0q| : f P Fu Ă R of the magnitudes of the derivatives of the functions f at zero as a subset of the positive real numbers.

By the Generalized Cauchy Integral Formula applied to a small circle cR about the origin with radius R contained in Ω we compute,

1! fpζq f 1p0q “ dζ, 2πi pζ ´ 0q2 żcR

15 which implies that

1! fpζq |f 1p0q| “ dζ 2πi pζ ´ 0q2 ˇ żcR ˇ ˇ 2π iθ ˇ ˇ1 fpRe q ˇ ď ˇ |Rdθˇ | 2π R2 ż0 ˇ ˇ 1 2π ˇ fpReiθqˇ “ dθ 2π R ż0 ˇ ˇ sup |fpζˇq| ˇ ηPΩ ď R 1 ď R

since every f maps into D so |fpηq| ă 1 for all η P Ω Ă D. This shows that the set S “ t|f 1p0q| : f P Fu is bounded. It is also nonempty since the identity function fpzq “ z belongs to F and |f 1p0q| “ |1| “ 1 P S. Now, since S Ď R is bounded and nonempty the supremum exists by the Completeness Axiom. So, we let s “ sup |f 1p0q| and observe that s ě 1. fPF We now have that s “ suppSq, which means that there exists a sequence

8 1 psnqn“1 Ă S with sn Ñ s in R. Since each sn “ |fnp0q| for some fn P F, we

8 1 can conclude that there exists a sequence pfn qn“1 Ă F with sn “ |fnp0q| for all

8 n P N by the definition of S. The sequence pfnqn“1 is uniformly bounded because each fn has an image in D by assumption so |fnpzq| ď 1 for all z P Ω and n P N.

Theorem 2.2.1. (Montel’s Theorem)

8 If pfnqn“1 is a sequence of holomorphic functions on Ω and for all compact

K P Ω there exists M ą 0 such that |fnpzq| ď M for all z P K, then there exists a

8 subsequence pfnj qj“1 converging to a holomorphic function f :Ω Ñ C uniformly on compact subsets of Ω. See Stein and Shakarchi for the proof. r1s

Now since we showed above that |fnpzq| ď 1 for all z P Ω and n P N, the hypothesis of Montel’s Theorem (Theorem 2.2.1) is satisfied. Thus, there exists

16 8 a subsequence pfnj qj“1 converging uniformly on compact subsets of Ω to a holo­ morphic function f :Ω Ñ C. We will argue that this function f is (almost) the Riemann map. First, we need another technical result.

8 Proposition 2.2.2. If Ω is a connected open subset of C and pfnqn“1 a sequence of injective holomorphic functions on Ω that converges uniformly on every compact subset of Ω to a holomorphic function f, then f is either injective or constant. [1]

Since each fnj fixes 0, f fixes 0 as well. Meanwhile, by Proposition 2.2.2, f is either injective or constant. Suppose f is constant, then f 1p0q “ 0 and |f 1p0q| “ s ě 1 which is a contradiction. Therefore, we conclude that f is injective. Since

|fnj pzq| ă 1 for all z P Ω and for all j P N, we know that |fpzq| ď 1 for all z P Ω by continuity of modulus. Then from the maximum modulus principal, we get that

|fpzq| ă 1 and therefore, f :Ω Ñ D. But since f maps the origin to itself, we conclude that f P F and |f 1p0q| “ s.

2.2.3 f :Ω Ñ D is surjective

In the previous steps we considered the function f :Ω Ñ D and showed that it is injective, holomorphic, and fp0q “ 0. To prove that f is a conformal map from Ω

onto D, we still need to show that f is surjective. Furthermore, we need to show that we can adjust f so that its derivative at zero is greater than zero. So, we suppose there exists α P D such that fpzq ‰ α for all z P Ω and show that this leads to a contradiction. To do so, consider the automorphism (stated in Theorem

2.1.1), α ´ z ψ pzq “ α 1 ´ αz

17 of the unit disk that interchanges 0 and α. Since Ω is a simply connected region of the unit disk, U “ ψαpfpΩqq is also a simply connected region of the unit disk and ? does not contain the origin. Then we can define a square root function, gpwq “ w on U, which by definition is,

? 1 logpwq gpwq “ w “ e2 .

Here, logpwq makes sense as a multiple valued function as long as w ‰ 0 and makes sense as a single valued function when restricted to a simply connected domain not containing 0 (like U “ ψαpfpΩqqq. Now, we consider the following function composition,

F “ ψgpαq ˝ g ˝ ψα ˝ f :Ω Ñ D (2.4)

which is holomorphic and injective as it is a composition of such functions. Note

that F p0q “ 0 from the function composition of F. So, we conclude that F P F and

|F 1p0q| P S. Now, since the square function hpwq “ w2 is the inverse of the square ? root function gpwq “ w, equation (2.4) implies,

´1 ´1 ψα ˝ h ˝ ψg pαq ˝ F “ f (2.5)

´1 ´1 and we set Φ “ ψα ˝ h ˝ ψg pαq ˝ F. Note that Φ maps D into D with Φp0q “ 0. ´1 Also, the composition function Φ is not injective since ψ gpαq is an automorphism of D and h is not injective on D. Thus, we have the equation,

Φ ˝ F “ f (2.6)

and by differentiating both sides of (2.6) and evaluating at zero we get,

1 1 1 1 1 f p0q “ Φ pF p0qqF p0q “ Φ p0qF p0q. (2.7)

18 Next we state the Schwarz Lemma. Then apply it and conclude the proof of the

Riemann Mapping Theorem. See Stein and Shakarchi for the proof of the lemma. r1s

Lemma 2.2.3. (Schwarz Lemma)

Let f : D Ñ D be holomorphic with fp0q “ 0.Then

(i) |fpzq| ď |z| for all z P D.

(ii) If for some z0 ‰ 0 we have |fpz0q| “ |z0| , then f is a rotation.

(iii) |f 1p0q| ď 1, and if equality holds, then f is a rotation. [1]

Now, from the last part of the Schwarz lemma (Lemma 2.2.3q, Φ 1 p0q ă 1. Then after taking absolute values on both sides of equation 2.7 we obtain,ˇ ˇ p q ˇ ˇ

1 1 f p0q ă F p0q ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ which is a contradiction to the maximalityˇ ˇ ˇ of f ˇ1 p0q in S. This proves that f is a surjective map. ˇ ˇ ˇ ˇ Lastly, we know that f 1p0q “ reiθ for some r ą 0 and θ P r0, 2πq. So, f˜ : Ω Ñ D difined by f˜pzq “ e ´iθfpzq has all the same properties as f and f˜1 p0q “ e´ iθf 1 p0q “

e ´iθre iθ “ r ą 0. Thus, f˜ :Ω Ñ D is the Riemann map. This concludes the proof of the Riemann Mapping Theorem.

19 Chapter 3

Thurston’s Conjecture and

Rodin-Sullivan’s Theorem

In chapter 1, we showed that conformal maps can be visualized by tracking the transformations of a packing by small circles. We illustrated two examples of con­ formal mappings on Figures 1.2/1.3 and showed that if we pack Ω by circles of radius e and transform under F :Ω Ñ D, we get an approximate circle packing of D since F is bijective (image circles will not overlap) and conformal (images of small circles are approximately circles). But, if we do not know F explicitly, how can we visualize the conformal equivalence of Ω with D? In this chapter, we will illustrate Thurston’s constructive geometric approach to this problem.

In his 1985 conjecture, Thurston introduced the idea of using transformations between circle packings as discrete approximations of conformal mappings. He conjectured, and Rodin and Sullivan proved in their theorem, that as we continue shrinking the size of the circles in those packings, the discrete conformal maps fE : QE Ď D Ñ PE Ă Ω between the packings will converge uniformly on compact

20 subsets of D to the classical conformal mapping F ´1 . Let us recall from Chapter 1 that a circle packing P of a bounded open set

Ω Ă C is a finite connected collection of circles in Ω with disjoint interiors. A regular hexagonal packing of Ω is a circle packing where each circle except for

boundary circles is tangent to 6 other circles. (See Figure 3.1.)

Figure 3.1: Circle packings of Ω and D.

Given a simply connected proper open set Ω Ď C, let HE be a regular hexagonal

packing of C and let PE “ ΩXHE be a circle packing of Ω by circles of size e ą 0. (See

Figure 3.1.) Then the carrier of PE, denoted carrierpPEq, is the geometric simplicial

complex obtained from PE as follows:

1. The vertices of carrierpPEq are the centers of the circles in the packing PE.

2. Two vertices are connected by a line segment (edge) if the corresponding

circles are tangent.

3. Three vertices define a triangular face in the complex if all three corresponding

circles are mutually tangent. Figure 3.2 illustrates this.

Since PE is the largest portion of the hexagonal packing of C contained in Ω and Ω is simply connected, the simplicial complex of Ω, carrierpPEq is also simply

21 Figure 3.2: Two circle packings and their carriers.

connected and gives a triangulated polygonal approximation of Ω. The circles in the

packing PE corresponding to boundary vertices of carrierpPEq are called boundary circles.

A circle packing QE of D is maximal if each boundary circle of QE is tangent to

the boundary of D and carrierpQEq is simply connected. (See Figure 3.1). Note that a circle packing contains the same data as its carrier. A transformation between circle packings is then the same as a transformation between their carriers. The transformation between carriers is a map between the simplicial complexes which is called a simplicial map. A geometric simplicial map is a continuous piecewise linear map between the carriers of the domain and range packings.

A transformation of circle packings is a bijection between the sets of circles. This is the same as bijection between their centers preserving edge relationships in the

22 carriers. Using barycentric coordinates in the triangular faces of the carriers we can extend such a bijection to a bijection between the geometric simplicial complexes which is piecewise linear and complex linear on each triangular face. In general this map is only piecewise linear unless corresponding triangles are similar.

With the use of Barycentric coordinates, we can express the position of any point located on the triangulated faces of the domain and range carriers and de­

fine the geometric simplical map between the carriers at the continuum of points.

This map is continuous, but not neccessarily conformal since it does not preserve angles unless corresponding triangles are similar. We will discuss this map when we define quasiconfromal maps later in this chapter. Next we state the Discrete

Uniformization Theorem and apply it to state Thurston’s conjecture.

Theorem 3.0.1. (Discrete Uniformization Theorem). If Ω Ă C is a simply con­ nected proper open set and P is a regular hexagonal packing in Ω, then there exists a maximal circle packing Q in D and a bijective simplicial map f : carrierpQq Ñ carrierpP q. Furthermore, Q and f are unique up to automorphisms of D. See Chap­ ters 4, 6 of Stephenson for the proof r3s.

Thurston applied this result to study the Riemann map as follows. Given e ą 0, let HE be the regular hexagonal packing of the complex plane C. Also, let PE “

HE X Ω be the packing of Ω with circles of radius e. Then, applying the Discrete

Uniformization Theorem (Theorem 3.0.1), we obtain a maximal packing QE of D and a bijective simplicial map fE : carrierpQEq Ñ carrierpPEq. Thurston’s conjecture

´1 ´1 was that fE Ñ F as e Ñ 0 where F : D Ñ Ω is a Riemann mapping.

Theorem 3.0.2. (The Rodin and Sullivan Theorem) For any simply connected

proper open subset Ω of C and z0 P Ω. Let F :Ω Ñ D to be the classical Riemann

23 mapping and @e ą 0 let fE : D Ñ Ω be the discrete conformal mappings for each circle packing PE of Ω by circles of radius e. Then as e Ñ 0, the sequence of mappings

´1 fE converge uniformly on compact subsets of D to F , up to an automorphism of

D.

3.1 Convergence of Domain and Range Carriers.

Now, the maps fE carry triangles in the carrier of QE to the corresponding triangles in the carrier of PE. We know that the discrete maps fE are not necessarily conformal since they do not necessarily preserve angles. Our ultimate goal is to show that

the distortion of the angles for these maps get smaller as epsilon gets smaller and

the sequence of quasiconformal maps fE approaches to a conformal map as epsilon goes to zero. But since there is only one conformal mapping of D onto Ω up to automorphisms of D, it must be the classical Riemann mapping discussed in Chapter 2.

Let us now study some facts about the range and domain of these discrete maps and their behavior for different e. In the packing PE, circles have constant radius e and for each value of e, we have a different packing that has hexagonal structure.

From Figure 1.3 we can see that as e Ñ 0 carriers of those packings exhaust the entire region Ω. On the other hand, in the domain of fE, we have the maximal

packing QE with boundary circles tangent to the boundary of the unit disk. We

want to show that carrier(QE) exhausts all of D but, it is sufficient to show that the radius of the boundary circles go to zero since they are tangent to the boundary of D by maximality of QE and carrier(QE) is simply connected. The Length-Area

Lemma will help us prove that the radii of the boundary circles of QE will tend to

24 zero.

Lemma 3.1.1. (The Length-Area Lemma) Let Q be a maximal packing of D and let c be a circle in Q. Also, let S1,S2,...,Sm be m disjoint chains of circles in Q, such that each of the m chains either separate c from the boundary of D or from the origin and a boundary point of D. If n1, n2, . . . , nk are the combinatorial lengths

of the chains in S1,S2,...,Sm, then

1 ´1 ´1 ´1 ´ 2 rpcq ď pn1 ` n 2 ` ¨ ¨ ¨ ` n k q ,

where rpcq is the radius of circle c.

Proof. Suppose rij are the radii of circles of the chains Si, 1 ď j ď ni. Let us consider the sum,

2 ni 2 rij “ pri1 ` ri2 ` ¨ ¨ ¨ ` rini q . ˜j“1 ¸ ÿ By the Schwarz inequality,

2 ni 2 2 rij “ pri1 ` ri2 ` ¨ ¨ ¨ ` rini q “ p1 ¨ ri1 ` 1 ¨ ri2 ` ¨ ¨ ¨ ` 1 ¨ rini q ˜j“1 ¸ ř ni ď p1 ` 1 ` ¨ ¨ ¨ ` 1qpr2 ` r2 ` ¨ ¨ ¨ ` r2 q “ n r2 . i1 i2 ini i ij j“1 ni times ř So we have, looooooooomooooooooon

2 ni ni 2 rij ď ni rij . (3.1) ˜j“1 ¸ j“1 ÿ ÿ Let li be the geometric length of the chain Si. So,

ni l ni l “ 2 r ñi “ r . (3.2) i ij 2 ij j“1 j“1 ÿ ÿ

25 Now, (3.1) and (3.2) imply that,

l 2 ni i ď n r2 2 i ij j“1 ˆ ˙ ÿ which shows that, ni 2 ´1 2 li ni ď 4 rij j“1 ÿ and, ni 2 2 ´1 2 rij ď 1 ñ li ni ď 4 rij ď 4. j“1,ni i“1 j“1,ni i“ÿ1,m ÿ i“ÿ1,m

Let l “ mintl1, l2, . . . , lmu. So,

m m m 2 ´1 2 ´1 2 ´1 l ni “ l n i ď li ni ď 4 i“1 i“1 i“1 ÿ ÿ ÿ and we have that,

m m ´1 2 ´1 2 ´1 l ni ď 4 ñ l ď 4 ni . (3.3) i“1 ˜i“1 ¸ ÿ ÿ Since the circles in the unit disk D as well as in any chain of circles Si have mutually disjoint interiors, l ě dpcq, where dpcq is the diameter of the circle c. Thus, by (3.3)

l ě dpcq “ 2rpcq and ´1 l2 m rpcq2 ď ď n´1 4 i ˜i“1 ¸ ÿ which implies 1 m ´ 2 ´1 rpcq ď ni . ˜i“1 ¸ ÿ

26 To apply the Length-Area Lemma, let us fix an e and let v1 be the vertex

of carrierpPEq corresponding to the origin of the unit disk under fE. Let w be a

boundary vertex of carrierpPEq which is i`1 generations away from v1. From Figure

3.1 we can see that w has been separated from v1 by at least i pairwise disjoint chains

S1,S2, ...Si of circles with each chain starting and ending with a boundary circle in

Ω. Enumerating the generations beginning from w, the combinatorial lengths nj of each Sj and the number i of chains satisfy the following inequalitie,

nj ď 6j. (3.4)

th Here nj is the number of circles in j chain, and nj ď 6j because the packing PE is hexagonal meaning that each circle inside is surrounded by six other circles in the

packing.

At the same time, let h1, h2, ..., hi represent the corresponding chains in QE. Let cw be the circle in QE corresponding to w in PE. Each chain starts and ends with a

boundary circle since boundary circles of PE correspond to boundary circles of QE

under fE. They separate cw from the rest of the unit circle as well as the origin. Ssee Figure 3.1. Now from the inequality (3.4) and the Length-Area Lemma we

have that,

2 1 pradiuspcwqq ď p1{n1 ` 1{n2 ` ... ` 1{niq 1 1 ď “ . p1{6 ` 1{12 ` ... ` 1{6iq p1{6 ` 1{12 ` ... ` 1{6iq The denominator on the right side of our inequality above is one sixth of the ith

partial sum of the divergent harmonic series. This means that the whole fraction approaches to zero and therefore, the radiuspcwq goes to zero as e Ñ 0. This

proves that the radii of the boundary circles of QE converge to zero as e Ñ 0. The

27 convergence is uniform over the set of boundary circles. Thus, the carriers of QE

end up exhausting the entire unit disk D in the limit as e Ñ 0.

3.2 The Convergence of the Functions fe as e Ñ 0.

From 3.1, we know that carrierpPEq converges to Ω and carrierpQEq converges to D as e Ñ 0. Now we need to show that the piecewise complex linear continuous maps

fE : carrierpQEq Ñ carrierpPEq converge to a function f : D Ñ Ω and that f is a conformal equivalence.

By construction, the triangular faces of carrierpPEq are equilateral, but those of carrierpQEq are not necessarily equilateral. The map fE: carrierpQEq Ñ carrierpPEq would be conformal if all triangles is carrierpQEq were equilateral. We now introduce a measure of the angle distortion at each point and aim to show that this measure tends to 1 as e Ñ 0. Suppose Ω Ď C has nonempty interior and that f :Ω Ñ C is

continuous. Given z0 P Ω and δ ą 0 sufficiently small, set w0 “ fpz0q and define

Lδ and lδ as the following,

Figure 3.3: Angle distortions under f.

Lδ “ maxt|fpzq ´ w0| : |z ´ z0| “ δu, lδ “ mint|fpzq ´ w0| : |z ´ z0| “ δu.

28 Then, a measure of distortion to the circle |z ´ z0| “ δ under the map f is given by the ratio of Lδ and lδ. (See Figure 3.3). Thus, at the point z0 we define the dilatation of f (distortion under f) by

Lδ Df pz0q “ lim sup . δÑ0 lδ

For the maps fE : carrierpQEq Ñ carrierpPEq the dilatation is 1 at each point interior to a triangular face of carrierpQEq because fE is complex linear on the interior of each triangle and therefore conformal there. But on the edges and vertices, the dilatation will typically be greater than 1. The dilatation varies continuously over carrierpQEq and has a maximum value κ “ maxtDf pz0q : z0 P carrierpQEqu. We then

say that fE is κ´quasiconformal. Next we discuss the Ring Lemma and apply it to explain the intuition for why

the maps fE are quasiconformal.

Lemma 3.2.1. (The Ring Lemma) For n ě 3, there is a constant rn depending only on n such that if n circles surround the unit disk (i.e., they form a cycle all tangent to the unit disk; see Figure 3.4) then each circle has radius at least rn.

Figure 3.4: n circles surround the unit disk.

Proof. To show the existence of such rn, we will argue that there exists a lower

bound for the radii of each of the n circles, and take rn to be smaller than the

29 smallest such radius. Here is a sketch of the idea. If n “ 3, then we can use elementary geometry to work out the relationship between the radii of the three

circles and observe that you can not make one of the circles arbitrarily small without

making another arbitrarily large and, in fact, the radius of the growing circle tends to 8 as the radius of the shrinking circle reaches a fixed positive size.

Let n ą 3 be fixed. If all n circles are the same size, we are done because if the radius gets smaller those n circles will not form a cycle (see Figure 3.5). Now, if those n circles are not all the same size, we can argue geometrically that there is a uniform lower bound for the largest of those n circles (i.e., the largest circle has a radius larger than the radius when all circles are same size). Figure 3.5 illustrates this situation.

Figure 3.5: Eight unequal vs. equal size circles surround the unit disk.

Let c0 be the unit disk and c1 the largest circle of the n-cycle as shown on Figure

3.6. Consider the circle c2. As we can see in the picture, the smaller the circle c2

gets, c3 and other circles get small as well. This means that to keep the cycle complete, other circles should get larger which will eventually force c1 no longer to be the largest circle in our n-cycle. This is because eventually all but 3 of the n

circles will be dragged into the valley between c0 and c1, reducing to the n “ 3 case.

30 Thus, we must have a lower bound on the radius of circle c2. This same argument shows that all the other circles in the n-cycle must have a lower bound for their

radii. Let rn be smaller than the radii of all of those circles in the cycle and we are done.

Figure 3.6: An n-cycle surrounding the unit disk.

We conclude from the Ring Lemma that the ratio of the geometric lengths of the radii of nearby circles in a ring of QE is bounded by a constant r6 which

means that the measure of how different triangles in the carrierpQEq are from being

equilateral is bounded. Since PE and therefore QE are hexagonal packings, lδ which

is proportional to r6, can not be arbitrarily small and therefore DfE pz0q can not be arbitrarily large. This shows that each fE map is quasiconformal. As spelled out in Stephenson r3s, discrete versions of Montel’s Theorem and the Schwarz Lemma for quasiconformal maps allow us to conclude that there exists a

8 decreasing sequence e1, e2, e3... with ei Ñ 0 as i Ñ 8 such that pfEi qi“1 converges

uniformly on compact subsets of D to a limiting bijection f : D Ñ Ω which is quasi-conformal.

Next we prove the Hexagonal Packing Lemma and show that the limiting func­

tion f discussed above is 1´quasiconformal and therefore conformal.

31 Lemma 3.2.2. (The Hexagonal Packing Lemma) Let c be a circle in a circle pack­

ing P. If c is surrounded by circles which form n generations of a hexagonal packing,

then the circles ci immediately around c have to be almost the same size as the circle c and this difference in size with c can be limited by an inequality of the form,

radiuspc q 1 ´ i ď S , radiuspcq n ˇ ˇ ˇ ˇ ˇ ˇ where tSnu is a sequence decreasingˇ to zero asˇ n tends to infinity. In other words, the deeper a circle is in a hexagonal packing the more regular the packing must be nearby.

8 1 Thus far, we have shown that we can always find a sequence teiui“1 of e s decreas­ ing to zero such that the corresponding simplicial maps tfEi u converge uniformly

on compact sets to a quasiconformal bijective map f : D Ñ Ω. Our final step will be to show that the above map f is not only quasiconformal

but also conformal. So, let ei be an element of the sequence teiu and let us find

a positive integer nEi so that for any triangular face of the carr(QEi ), the circles

forming the triangular face are a minimum nEi combinatorial generations inscribed in the packing QEi of the unit disk D. Then, we can construct a sequence of num­ bers ts u and conclude from the Hexagonal Packing Lemma that the triple edges nEi of the triangular faces of carr(Q ) are p1 ` s q-bound. Meanwhile, from Lemma Ei nEi

A.9(a) Stephenson, we conclude that there is a bound kEi for the distortion of the

angles of fEi on any triangular face of the carr(QEi ) which implies that fEi is kEi ­ quasiconformal. So, n Ñ 0 as e Ñ 8 which implies that s Ñ 0 and therefore, Ei i nEi p1 ` s q Ñ 1 meaning that the triangular faces of the carr(Q ) become infinites­ nEi Ei imal equilateral triangles in the limit. Next, from Lemma A.9(b), Stephenson, we

conclude that kEi Ñ 1 meaning that the limiting function f is 1-quasiconformal.

32 When κ is equal to one, angles are not being distorted. Thus, we can conclude that

the 1-quasiconformal map f is also conformal. Meanwhile, from the Riemann Map­

ping Theorem proved in Chapter 2, we know that there is only one such conformal

mapping from Ω onto D up to an automorphism of D. Therefore, f must be the same map as the Riemann map F ´1 up to an automorphism of D. This concludes the proof of the Rodin-Sullivan Theorem.

33 Bibliography

[1] E. M. Stein and R. Shakarchi, Complex Analysis, Princeton University Press,

2003.

[2] J. Bak and D. J. Newman, Complex Analysis, third edition, Springer, 2010.

[3] K. Stephenson Introduction to Circle Packing, Cambridge University Press,

2005.

[4] B. Rodin and D. Sullivan, The convergence of circle packings to the Riemann

mapping, Journal of differential geometry, 1987, Pages: 349-360.

34