visualizing the riemann mapping
TRANSCRIPT
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
1.3 A sequence of finite Riemann mappings under fpzq “ z´i . . . . . . 9 z`i
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
z0 P Ω, then there exists a unique such F with F pz0q “ 0 and F 1pz0q 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
˚
u 1 1? ´? x uy ‹ ˚ ‹
detpDfRpx0, y0qq “ det˝ ‚
“ det 2 2‚ “ 1 ą 0.
vx v1 1
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, ¨ ˛
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
π π
to be fpz iq “ e 4 z with z “ |z|eiargpzq. Then i|fpzq| “ |e 4 ||z| “ |z| and argpfpzqq “ π
arg ei z
arg iπ iarg z i π arg z π p 4 q “ p|z|e e p q p4 q “ argp|z|e p ` qq
4 q “ ` gpzq ar . This shows that f4
is a rigid rotation about the origin by π . The map f preserves angles and orientation 4
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
fR : Ω Ñ R2 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.
π
Example 1.1.5. Let f z ei z 1p q “ 4 “ ? p1 ` i x´y x`yqpx ` yiq “ p ? q ` ip ? q “ u ` vi.
2 2 2
So fRpx, y x´y x`yq “ pp ? q, p ? qq which implies that,
2 2
˚
1 0‹
detpDfRpx0, y0qq “ det ‚ “ ´1 ă 0. 0 ´1
3
Axx¨Axy cos θ1 pxx¨xy pxx¨xy xx? ¨xyp q “ ? “ ? ? “ ? ? “ “ cospθq .
Axx¨Axx pAxy¨Axy lxxl plxyl lxxllxylxx¨xx pxy¨xy
Example 1.1.7. Let f z z2 q z x2 p “ ` “ ´ y2 ` x p2xy ´ yqi “ upx, yq ` vpx, yqi.
So fRpx, y 2 2 q “ px ´ y ` x, 2xy ´ yq which implies that,
So, detpDf x , y 2 2 1 2 p 0 0qq ą 0 if and only if x ` y ă p q . This shows that f preserves 2
orientation for all values inside the circle of radius 1 centered at 0 and does not2
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
serves dot products up to a scalar p (because we have cos θ xx x T ? ?¨y“ ). A A “ pI
¨ xx¨xx ˛
xy ¨xy
˚
x1‹
means that Afx ¨ Afy “ pfx ¨ fy for all fx, fy 2 P R with fx 2“˝ ‚
P R . This isx2
true if and only if A T p fx T q pAfyq “ pfx fy if and only if fxT T pA Aqfy T “ pfx fy if and
only if AT A “ pI which means that if θ1 is the angle between Afx and Afy then
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.
¨ ˛ ¨ ˛
˚
ux uy ‹ ˚
2x ` 1 ´2y ‹ det 2
pDfRpx0, y0qq “ det˝ ‚ “ det ‚ “ ´4y ´ 4x2 ` 1. vx vy 2y 2x ´ 1
4
Theorem 1.1.10. Let Ω Ă C be open. Then f : Ω Ñ C is holomorphic with
f 1pz0q “ 0 for all z0 P Ω if and only if fR : Ω Ñ R2 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.
Proof. Let us recall that if f “ u ` vi then fRpx, yq “ pupx, yq, vpx, yqq and df “
dz
f pz`Δzq´f pzqlimΔzÑ0 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˚
˝ ‹
‚
.
which implies that DfRpx0, y0q preserves
So AT A“ “ “By the Cauchy-Riemann equations, A DfRpx0, y0q vx ux
˛¨˛¨ 2 ` v2 0 1 0
˚
˝ u
˚
˝ ‹
‚ ‹
‚x x
“ 2 ` v2pu qx x
0 u2 ` v2 0 1 x x˛¨
ux ´vx˚
˝ ‹
‚ 2 v2q ą 0. Similarly, detpAq “ detx x “angles at each point of Ω where puvx ux
ˇ
ˇˇ
ˇ
2 2 2 2` v q ě 0 implies that A preserves orientation where pu ` v q ‰ 0. Butpux x x x
2df df u2 ` v2 “ |ux ` ivx|2 “ x x ą 0 since ‰ 0 on Ω.dz dz
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
function fpzq “ z´i which maps the upper half plane onto the unit disk. Calculating z`i
the first complex derivative of f, we get f 1pzq “ pz`
2iiq2 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
unique conformal mapping F of Ω onto D such that F pz0q “ 0 and F 1pz0q 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
D. We do it by choosing f such that |f 1pz0q| 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
function fpzq “ z´i through the theory of circle packings to suggest another proof z`i
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
maps do the same thing on small scales. Ideally fpzq « f 1pz0qpz ´ z0q ` fpz0q near
z0 when f is holomorphic, by linear approximation. So when f 1pz0q ‰ 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.
f72
f252
(a) P72 - packing of 49 circles with radius ɛ=1/14
(b) P152 - packing of 225 circles with radius ɛ=1/30
(c) P252 - packing of 625 circles with radius ɛ=1/50
f152
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
the unit disk is the complex analytic function fpzq “ z´i (Figures 1.3q. The process z`i
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
(a) P144 - packing of 144 circles with radius ɛ=1/12
(b) P400 - packing of 400 circles with radius ɛ=1/20
(c) P1024 - packing of 1024 circles with radius ɛ=1/32
f400
f1024
Figure 1.3: A sequence of finite Riemann mappings under fpzq “ z´i . 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
each circle packing Pn of radius size n 1 . 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
a unique such F with F pz0q “ 0 and F 1pz0q 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
F of Ω onto D such that F pz0q “ 0 and F 1pz0q ą 0. See Figure 2.1.
11
Gp0q “ F1pF p0qq “ F1pz0q “ 0, and2
G1 ´1 ´1 1 ´1 1 1 p0q “ F1
1pF p0qq ¨ pF q p0q “ F1
1pz0q ¨ pF q p0q “ F1
1pz0q ¨ ą 0. (2.1)2 2 2 1
´1
F p0q
α ´ z fpzq “ e iθ .
1 ´ αz
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
F1pz0q “ 0, F11pz0q ą 0, and F2pz0q “ 0, F2
1pz0q ą 0. A conformal map from D
onto itself is called an automorphism of D. Then G “ F1 ˝ F2 ´1 : D Ñ D is an
automorphism of the unit disk. Furthermore,
2
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
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 iθG “ F1 ˝ F2 of the unit disk D has the form Gpzq “ e z. Thus, G1pzq “ 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
of D by making |f 1pz0q| 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
g : hpΩq Ñ C by gpzq “ 1 z. Then gphpΩqq contains 0 and is contained in D. SinceR
13
1 F pzq “ .
logpz ´ αq ´ plogpw ´ αq ´ 2πiq
1 F 1pzq “
pz ´ αqplogpz ´ αq ´ plogpw ´ αq ´ 2πiqq2
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]
(2.2)
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,
(2.3)
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
disjoint from fpΩq. Then there exists a sequence pznq8 in Ω such that fpznq Ñ n“1
14
ż 1! fpζq
f 1p0q “ dζ,
2πi pζ ´ 0q2 cR
f w 2πi. But then ef pznq efpwq`2πi p q` Ñ by the continuity of the exponential function.
Now ef pznq elogpzn αq z α while efpw“ “ n ´q`2πi fpwq 2πi logpw´α“ e e “ e q “ 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
F “ tf : 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,
15
ˇ
ˇ
ˇ
ˇ dζ
ˇ
ˇ
ˇ
ˇ
ż 1! fpζq
pζ ´ 0q2
ˇ
ˇ
|f 1p0q| “
2πi ż 2
c
π Rˇ
ˇfpReiθ
R2
1 qď |Rdθ|
2π ż
02π
ˇ
ˇˇ
ˇfpReiθ
R 1 q
“ dθ 2π 0
sup |fpζq|ηPΩ
ď R
1ď R
which implies that
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
psnq8 n“1 Ă S with sn Ñ s in R. Since each sn “ |f 1
np0q| for some fn P F , we
can conclude that there exists a sequence f 8 p q Ă F with s “ |f 1 n n“1 n np0q| for all
n P N by the definition of S. The sequence pfnq8 n“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)
If pfnq8 is a sequence of holomorphic functions on Ω and for all compact n“1
K P Ω there exists M ą 0 such that |fnpzq| ď M for all z P K, then there exists a
subsequence pfnj q8 converging to a holomorphic function f : Ω Ñ C uniformly on j“1
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
a subsequence pfnj q8 j“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.
Proposition 2.2.2. If Ω is a connected open subset of C and pfnq8 n“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
ψαpzq “ 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 logpwq2gpwq “ w “ e .
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,
ψα ´1
˝ h ˝ ψ g´
p
1 αq ˝ F “ f (2.5)
and we set Φ “ ψ ´1 ˝ h ˝ ψ g´
p
1 αq ˝ F. Note that Φ maps D into D with Φp0q “ 0.α
Also, the composition function Φ is not injective since ψ ´1 is an automorphism of gpαq
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, ˇ
ˇ 1 Φ p0q
ˇ
ˇă 1. Then
after taking absolute values on both sides of equation p2.7q we obtain,
ˇ
ˇ
ˇ1 f p0q
ˇ
ˇ
ˇă ˇ
ˇ
ˇ1
F p0q ˇ
ˇ
ˇ
which is a contradiction to the maximality of ˇ
ˇ 1 f p0q
ˇ
ˇ in S. This proves that f is a
surjective map.
Lastly, we know that f 1p0q “ reiθ for some r ą 0 and θ P r0, 2πq. So, f : Ω Ñ D
difined by fpzq “ e ´iθfpzq has all the same properties as f and f1 p0q “ e ´iθf
1 p0q “
´iθ iθe re “ 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
was that fE Ñ F ´1 as e Ñ 0 where F ´1 : 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
fE converge uniformly on compact subsets of D to F ´1 , 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
´1 ´1 ´1 ´ 1 rpcq ď pn ` n ` ¨ ¨ ¨ ` n q 2 ,1 2 k
ÿ
niÿ
nl i
ili “ 2 rij ñ “ rij .
2 j“1 j“1
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
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,
˜
ÿ
2 ni
rij “ pri1 ` ri2 ` ¨ ¨ ¨ ` rini q2 .
j“1
¸
By the Schwarz inequality,
˜ ¸2 ř
ni rij “ pri1 ` r 2 2
i2 ` ¨ ¨ ¨ ` rini q “ p1 ¨ ri1 ` 1 ¨ ri2 ` ¨ ¨ ¨ ` 1 ¨ rini qj“1
ř
ni
ď plo
1ooooooo
2 2 ` 1 omo
` ¨ooooooo
¨ ¨ ` 1on
qpr ` i1 r2 2i2 ` ¨ ¨ ¨ ` rin q “
i ni rij .
j“1 ni times
So we have,
˜ ¸
ÿ
2 niÿ
ni
r 2ij ď ni rij .
j“1 j“1
(3.1)
Let li be the geometric length of the chain Si. So,
(3.2)
25
ˆ ˙2 ÿ
nl ii 2
ď n i r2 ij
j“1
˜ ¸´1 l2
ÿ
m2
ď ´1 rpcq ď n4 i
i“1
˜ ¸
ÿ
´ 1 m 2
rpc ď n´ 1q i .
i“1
Now, (3.1) and (3.2) imply that,
which shows that, ÿ
ni
2 ´1 2 li ni ď 4 rij j“1
and, ÿ ÿ
niÿ
r 2 2 ´1 2ij ď 1 ñ li ni ď 4 rij ď 4.
j“1,ni i“1 j“1,ni i“1,m i“1,m
Let l “ mintl1, l2, . . . , lmu. So,
ÿ
mÿ
mÿ
m l2 2 n´ 1
“ i l n ´1 l2 ´1
i ď i ni ď 4i“1 i“1 i“1
and we have that,
˜ ¸
ÿ ÿ
´1 m m´1 l2 ni ď 4 ´1
ñ l2 ď 4 ni .
i“1 i“1
(3.3)
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
which implies
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)
Here nj is the number of circles in jth 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, 1
pradiuspcwqq2
ď p1{n1 ` 1{n2 ` ... ` 1{niq
1 1ď “ .
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
LδDf 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
decreasing sequence e1, e2, e3... with ei Ñ 0 as i Ñ 8 such that pfEi q8 converges i“1
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,
ˇ
ˇ
ˇ
ˇradiuspciq
1 ´ radiuspcq
ˇ
ˇ
ˇ
ˇď Sn,
where tSnu is a sequence decreasing to zero as 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 1Thus 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 tsnEi u and conclude from the Hexagonal Packing Lemma that the triple edges
of the triangular faces of carr(QEi ) are p1 ` snEi q-bound. Meanwhile, from Lemma
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, nEi Ñ 0 as ei Ñ 8 which implies that snEi Ñ 0 and therefore,
p1 ` snEi q Ñ 1 meaning that the triangular faces of the carr(QEi ) become infinites
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