The Great Picard Theorem
Dennis Wahlström
Bachelor’s Thesis, 15 Credits Bachelor of Mathematics, 180 Credits
Spring 2018
Abstract
In this essay, we present a proof of the great Picard theorem by showing that a holomorphic function with an essential singularity attains infinitely many complex
values in the vicinity of the singularity.
Sammanfattning
Vi kommer att presentera ett bevis på Picards stora sats genom att visa att en holomorf funktion med en väsentlig singularitet antar oändligt många komplexa värden
Contents
Acknowledgements 3
1. Introduction 5
2. Complex Analysis 7
3. Conformal Mappings and Automorphism Groups 11
3.1. Automorphism Groups 14
3.2. Schwarz Reflection Principle 19
3.3. The Extended Plane and Carathéodory’s Theorem 20
4. A Modular Function 25
5. Covering Maps 33
6. The Great Picard Theorem 41
Acknowledgements
1. Introduction
Complex analysis is one of the mathematical branches, and without any doubt it is an important mathematical field which has contributed to many study applications related to our daily life experience, such as in physics and engineering.
The complex analysis is the analysis of functions of so called complex numbers, which include the square root of negative numbers. These numbers were first encountered in the first century of Anno Domini by Hero of Alexandria, at that time they were considered nonsensical due to the lack of awareness [7]. Even so, it took the works of many mathematicians before the theory of complex variables was commenced in the eighteenth century by Leonhard Euler [17]. The theory of complex variables was heired by the three pioneers Augustin-Louis Cauchy, Bernhard Riemann and Karl Weierstrass with distinct thought on the subject [17]. Cauchy’s view on the theory of complex variables was circular functions and the exponential functions, meanwhile Riemann focused on the geometry and Weierstrass had presented a vision on the convergent power series [17]. These visions have their own significative meaning in the today’s mathematics.
We will focus ourself on one of the succeeders in the theory of complex variables, Charles Émile Picard. In the nineteenth century, Picard presented his thesis in an application of linear complexes where he mainly dealt with geometrical aspect and as the result, it became an important property to the Riccati’s equation [5]. Picard fully became a professor at the age of thirty and during his time as professor, he taught countless students e.g. in rational mechanics, differential and integral calculus, higher algebra and analysis [5]. Moreover, his personalities and teaching skills were acknowledged by many including by one of his PhD student Jacques Hadamard, who had chosen to described his advisor as: ”A notable feature about Picard’s scientific personality was the perfection of his teaching, one of the most, if not the most, marvelous I have experienced. One could say of it, thinking of what Mozart is reported to have said of one of his own works, that ’there was not one word too much, nor one word lacking’.” [5].
A few years after Picard obtained his professorship, he realised that the Casorati-Weierstrass theorem for a holomophic function with essential singularities could be strengthened, which today is known as the great Picard theorem [5, 17].
Theorem 6.5 (The Great Picard Theorem). If f has an essential singularity at z0, then with at most one exception, f attains every complex value infinitely many times in every neighborhood of z0.
The original proof of the great Picard theorem can be found in [15, 16]. It has many gen-eralisations in the area of analytic mappings to an extended plane for essential functions as well as for a punctured disk [11, 14, 18]. Besides these, it also has generalitions to e.g. Montel’s theorem, Schottky’s theorem and Bloch’s theorem [3, 19].
but we encourage to all the readers to explore the other sections filled with interesting proofs.
The layout of this essay is as follows: first a review of complex analysis, and in Section 3 we shall define the automorphism group and construct the extended complex plane. Section 4 contains a construction of a modular function, and in Section 5 we will tie most of our results together preparing for Section 6, which is the great Picard theorem. The main theorems in this thesis are presented in Figure 1.1.
This thesis is based on [20] with the focus on simplifying and clarifying for the readers who has not studied the advanced calculus course as it was one of prerequisites for the book. Moreover, a few definitions were alternatively taken from [9].
Theorem 6.5
(The Great Picard Theorem)
Theorem 6.4
Theorem 6.3 Theorem 6.2
Theorem 6.1
Theorem 2.6, Theorem 4.10, Theorem 5.11, Corollary 5.12, Theorem 5.14, Theorem 5.17, Lemma 5.18, Theorem 5.19
2. Complex Analysis
The goal in this section is to reconstruct the prerequisites from complex analysis in the usage to construct conformal maps and modular functions. We will start off by introducing the concept of complex differentiable.
Proposition 2.1. Suppose that V is an open subset of C, that f : V → C, and z ∈ V . Let u and v be the real and imaginary of f . Then f = u + iv is complex-differentiable at z, if, and only if, it is real-differentiable at z and the real and imaginary parts satisfy the Cauchy-Riemann equations
ux(z) = vy(z), uy(z) = −vx(z)
where x and y are the real part in R, and ux, uy vx and vy are the partial derivative.
Throughout the essay we will denote V as an open set of the set of all complex numbers C. The Proposition 2.1 gives us authority to define the concept of holomorphic function, which is a complex function that is differentiable everywhere in the domain.
Definition 2.2. Suppose that V ⊂ C is open and f : V → C. We say that f is holomorphic in V if f is complex differentiable at every point of V . The class of all
functions holomorphic in V is denoted H(V ).
There are various type of singularities, isolated singularity, removable singularity and
essential singularity. By any means, the singularity is a point in which the mathematical
object is not defined, e.g. differentiability fails. These singularities have their definition as following.
Definition 2.3. Let f be a function in C and z ∈ C, suppose that f has an isolated
singularity at z. The singularity is removable if f can be defined at z such that f becomes holomorphic in a neighborhood of z. The singularity is a pole if limw→zf (w) = ∞.
Finally, an isolated singularity which is neither a pole nor a removable singularity is an
essential singularity.
Let’s introduce an open and a closed disk with z as its center where z ∈ C and a finite radius r > 0:
D(z, r) = {w ∈ C : |z − w| < r} and D(z, r) = {w ∈ C : |z − w| ≤ r}. (2.1) Further through the essay, we will encounter a bounded holomorphic function on the boundary of its domain which can be divided into two different cases, a constant and a nonconstant holomorphic function. However, we will state these in form of two famous theorems that goes by the name of open mapping theorem with bounds and Maximum
Lets start with the open mapping theorem with bounds.
Theorem 2.4. Suppose that f ∈ H(V ) and its second-order complex derivative |f00(z)| ≤
B for all z ∈ V , where B is a bound of V . Suppose that z0∈ V and first-order complex derivative f0(z0) 6= 0. If r > 0 is small enough that D(z0, r) ⊂ V and also r < 2|f
0(z 0)| B then D(f (z0), δ) ⊂ f (D(z0, r)) for δ = r|f0(z0)| − Br2 2 .
Proof. See e.g. in ([17], p. 256-257) or Theorem 5.7 in [20].
Before we proceed with the proof of the Maximum modulus theorem let’s define power series. Furthermore, we shall observe that the power series contains an important property to the analytical functions as well to the Maximum modulus theorem. Hence, the Maximum modulus theorem will be used frequently through essay, where it shows a great deal to handle the boundness of the holomorphic functions, which is what we needed in our proof of the great Picard theorem.
Definition 2.5. A series on the form P∞
j=0aj(z − z0)j is called a power series. The
constants aj are the coefficients of the power series.
Theorem 2.6. Suppose that f ∈ H(V ) and V is connected. Then |f | cannot achieve a (local) maximum in V unless f is constant: If f is nonconstant, then for every a ∈ V and δ > 0 there exists z ∈ V with |f (z)| > |f (a)| and |z − a| < δ.
Proof. Let f ∈ H(V ) and f has a zero order of N at z ∈ V , then there is a function h ∈ H(V ) with h(z) 6= 0 which can be presented together with the power series of f as:
f (z) = f (a) + c(z − a)N+ (z − a)N +1h(z).
From the previous step and select z − a appropriately we are able to get: |f (a) + c(z − a)N| = |f (a)| + |c(z − a)N| > |f (a)|.
This shows that, if the modulus (z − a) is small enough, then the growth (z − a)N is
negligible.
In particular, let α, β, θ ∈ R be selected in such way that f (a) = |f (a)|eiα, c = |c|eiβ
and β + N θ = α for a N 6= 0. If z = a + reiθ with r > 0, then
f (a) + c(z − a)N = |f (a)| + |c|rNei(β+N θ) and
|f (a) + c(z − a)N| = |f (a)| + |c|rN.
We know that h is continuous in V and bounded near a. Thus, there is a bound M such that |h(z)| ≤ M for |z − a| ≤ ρ, where ρ > 0 and M < ∞. The triangle inequality shows that, if 0 < r < ρ, then
|f (z)| ≥ |f (a) + c(z − a)N| − |(z − a)N +1h(z)|
Herein, we are free to select δ ∈ (0, ρ) so that M ρ < |c|/2, hence 0 < r < ρ, |f (z)| ≥ |f (a)| + |c|r
N
2 > |f (a)|.
Besides the open mapping theorem and the maximum modulus theorem, we will also be needing the Schwarz lemma, but first lets define the unit disk D as D = D(0, 1), where
D(z, r) is defined as (2.1). The following theorem is the Schwarz lemma.
Theorem 2.7. Suppose that f : D → D is holomorphic and f (0) = 0. Then |f (z)| ≤ |z| for all z ∈ D and |f0(0)| ≤ 1. Furthermore, if |f0(0)| = 1 or |f (z)| = |z| for some
nonzero z ∈ D, then f is a rotation: f (z) = βz for some constant β with |β| = 1. Proof. Since f (0) = 0, we will define a function g ∈ H(D) in such manner:
g(z) =
(f (z)
z , if z 6= 0,
f0(0), if z = 0.
To begin with, we will recall the result from Theorem 2.6 which states that if D is a bounded open set in the complex plane with f ∈ H(D), then there is a bound M ∈ D preventing f to achieve a local maximum in D unless f is constant. It is also indicating that |g| ≤ 1 in D. However, we want to show that |g| ≤ 1 on the boundary of the unit disk, ∂D, problematically, g is undefined on ∂D.
To handle the issue above: suppose that r ∈ (0, 1), the function |g| ≤ 1/r is restricted on ∂D(0, r) and Theorem 2.6 shows that |g| ≤ 1 in D. If |f0(0)| = 1 or |f (z)| = |z| for a nonzero z ∈ D, then |g(z)| = 1 for some z ∈ D since g is constant. Thus, the proof is
then concluded.
Historical note: Carathéodory recognised the strength of the Maximum modulus
the-orem, where he was first to apply it to prove the Schwarz lemma in such manner [1]. We will continue with the definition of compact sets.
Definition 2.8. A compact set in C is a set that is both closed and bounded.
Now with the definition of a compact set we can state the Montel’s theorem for a uniformly bounded family. A family of functions is said to be uniformly bounded if there is a constant in the domain restricting all the functions. The Montel’s theorem is stated as the following.
Theorem 2.9. If F ⊂ H(D) is a uniformly bounded family and (fn) is a sequence in
F , then there exists a subsequence fnj such that fnj → f ∈ H(D) uniformly on compact
subsets of D.
One of the results from a first course in complex analysis is the Riemann mapping
theorem, and it will be strengthened in the connection with the extended complex plane.
Theorem 2.10. Suppose that D ⊂ C is open, connected, nonempty, and has the property that any nonvanishing function holomorphic in D has a holomorphic square root. If D 6= C, then D is conformally equivalent to D.
Proof. See e.g. Theorem 6.4.2 in [4].
Through this essay, we shall encounter functions with poles and singularities, and realise that such particular functions are suitable to be described as a harmonic function. Accordingly, the property of the harmonic function is essential to the proof of the great Picard theorem. The definition of harmonic function is stated as following.
Definition 2.11. A real-valued function φ(x, y) is said to be harmonic in a domain D
if all its second-order partial derivatives are continuous in D and if, at each point of D,
φ satisfies Laplace’s equation.
Hence, the domain D ∈ R2 with a fact that R2 ∈ C, this contributes a connection between holomorphic function and harmonic function stated as in Lemma 2.12. Ex-plicitly, the harmonic function satisfies the Cauchy-Riemann equations, and hence it is considered as a holomorphic function.
Lemma 2.12. Suppose that D ⊂ C is open and u : D → C is twice continuously differentiable, then these hold:
(i) A holomorphic function is harmonic;
(ii) If u is harmonic, then the function f = δu/δz is holomorphic;
(iii) If D is simply connected and u is a real-valued harmonic function in D, then
there exists f ∈ H(D) with u = Re(f ).
3. Conformal Mappings and Automorphism Groups
The goal with this section is to introduce conformal mappings, automorphism groups and the extended complex plane. We will start with the conformal map.
A function f ∈ H(V ) is said to be a conformal map when its derivative f0 has no zero in V . In other words, the function f preserves angles locally as presented in the following proposition:
Proposition 3.1. Suppose that f ∈ H(V ), z0 ∈ V, and f0(z0) 6= 0. Also, suppose that γ0, γ1: [0, 1] → V are curves with nonvanishing (right) derivatives at 0, with γj(0) = z0. Let ˜γj(t) = f (γj(t)). Then arg ˜γ 0 1(0) ˜ γ0 0(0) = arg γ 0 1(0) γ0 0(0) . Proof. Let γ0j(0) = rjeiθj, for j = 0, 1
with rj> 0 and θj ∈ R and let
f0(z0) = reiθ, then ˜ γj0(0) = f0(z0)γ 0 j(0) = rrjei(θj+θ), gives arg ˜γ 0 1(0) ˜ γ00(0) ! = (θ1+ θ) − (θ0+ θ) = θ1− θ0= arg γ10(0) γ00(0) ! .
The angles are preserved.
Remark 3.2. If f ∈ H(V ) and it is a one-to-one function in V with W = f (V ), then the
function f is said to be a conformal equivalence and W is conformally equivalent to V . The conformal map is a type of transformation which is important in complex analysis, and in our concern it is one of the main pieces of introduction to an other type of transformation that allows us to extend the complex plane, that is a linear-fractional
transformation. Before then, we need a few more components.
Definition 3.3. A function f is meromorphic in V , if there exists a relatively closed set S ⊂ V such that every point of S is isolated with f ∈ H(V \ S), and f has a pole or a
removable singularity at every point of S.
If a function is meromorphic in a domain V that is connected, then the function is a holomorphic mapping of V into the extended plane, C∞ (also knows as the Riemann sphere):
C∞= C ∪ {∞}.
To illustrate this we define:
S1= {x ∈ R3: x21+ x 2 2+ x
2 3= 1}.
This is the unit sphere in R3.
and P (z) for z ∈ C by letting P (∞) = (0, 0, 1) with a straight line intersecting (0, 0, 1) and (z, 0) ∈ R3. The illustration of this can be observed in Figure 3.1. The stereographic projection is: P (z) = 2x |z|2+ 1, 2y |z|2+ 1, |z|2− 1 |z|2+ 1 , for z = x + iy ∈ C. x y z (0, 0, 1) (z, 0) 0 P(z)
Figure 3.1 – A sphere on a plane.
The stereographic projection can be used to define a metric on C∞:
d∞(z, w) = kP (z) − P (w)k, for z, w ∈ C∞. (3.1)
where d∞is the default metric on C∞and k · k is the normed vector space. Analogously,
(C∞, d∞) is a compact metric space due to the fact that S1is the unit sphere, a compact
subset of R3.
Herein, C∞ is a complex manifold and we shall not break it down in details, except
our interest is to show that functions from the subsets of C∞ to C∞ are holomorphic
1/∞ = 0, then this shows that there is a function z → 1/z such that C∞ is a bijection
onto itself. Specifically, suppose that V is an open subset of C∞ and f : V → C∞ is a
function. The function f is holomorphic, if the following four functions are continuous
f (z), f (1/z), 1/f (z) and 1/f (1/z). If this is true, then the function is holomorphic in
the open subset and finite on C. We will call f a holomorphic mapping.
Theorem 3.4. If f : C∞→ C∞ is holomorphic, then there exist polynomials P and Q such that
f (z) = P (z)
Q(z), (3.2)
for all z ∈ C∞.
Proof. If f (z) = ∞ for all z ∈ C∞, then (3.2) is true with P = 1 and Q = 0. We know
that C∞is a complex manifold and f is meromorphic, by assuming that f has a pole or
removable singularity at ∞, we need to show that f is a rational function.
If f has an isolated singularity at ∞, then there exists R > 0 such that f is holomorphic in {z ∈ C : |z| > R}. This implies that f has finitely number of poles in C, since zeroes of f are isolated at ∞, then f has also finitely many zeroes in C. Suppose that the zeroes of f in C are {z1, ..., zn} and the poles are {p1, ..., pm}. Let
g(z) = f (z) m Y j=1 (z − pj) n Y k=1 (z − zj) ,
this shows that g is an entire function with a limit at ∞ and has no zeroes in C. Addi-tionally,
lim
z→∞
f (z) czN = 1,
for some N ∈ Z and nonzero c ∈ C. Thus, lim
z→∞
g(z)
czN +m−n = 1,
which requires g to has a pole or removable singularity at ∞, as well as for the function f . If g has a pole at ∞, then it would imply that 1/g is an entire function that approaches to 0 at ∞, and can also be interpreted as 1/g is identically 0. Therefore, g must has a finite limit at ∞ with a condition of N + m − n ≤ 0, which implies that g is a bounded
3.1. Automorphism Groups. We let Aut(V ) denote a group of invertible holomorphic mappings from V to itself (V is an open subset of the complex plane or V ⊆ C∞). Definition 3.5. A binary operation ∗ on a set S is a function mapping S × S into S.
For each (a, b) ∈ S × S, we will denote the element ∗(a, b) of S by (a ∗ b).
In group theory, a group is a set with elements and an operator that satisfies the group axioms. Let X be the set with a binary operator ∗ : X × X → X. The group axioms are:
G0:(Closure), a ∗ b ∈ X for all a, b ∈ X; G1:(Associative), (a ∗ b) ∗ c = a ∗ (b ∗ c);
G2:(Identity), e ∈ X : a ∗ e = e ∗ a = a for all a ∈ X; G3:(Inverse), a−1∈ X : a ∗ a−1= a−1∗ a = e for all a ∈ X.
We are going to define several definitions related to the group theory and not less to the automorphism group for future conventions.
Definition 3.6. Given two sets X and Y . A set isomorphism between X and Y is a
bijection f : X → Y , for some function f .
Definition 3.7. If a subset H of a group G is closed under the binary operation of G
and if H with the induced operation from G is itself a group, then H is a subgroup of G.
Definition 3.8. In group theory, an order of a group is its cardinality. The order of an
element a is the smallest positive integer m such that am = e, then a is said to have a
finite order. If no such m exists, then a is said to have infinite order.
As a result from defining the conformal map and introducing the stereographic pro-jection, we are now able to obtain other type of transformation.
Definition 3.9. A linear-fractional transformation (or Möbius transformation) is a
map-ping in the form of:
φ(z) = az + b cz + d,
where a, b, c, d ∈ C satisfy
ad − bc 6= 0.
The condition ad − bc 6= 0 implies that c and d cannot both vanish, otherwise, the function becomes rational and reduces to a meromorphic function (a function that is holomorphic on a plane except on a set of isolated points) in the complex plane or a holomorphic mapping from C∞ to itself.
Theorem 3.10. Aut(C∞) is the set of all linear-fractional transformations. Proof. Let φ ∈ Aut(C∞) and let ψ be the inverse of φ, then
ψ(z) = dz − b
−cz + a then,
for all z ∈ C∞and φ ∈ Aut(C∞). Since φ : C∞→ C∞is holomorphic, thus, Theorem 3.4
implies that φ is rational,
φ(z) = c n Y k=1 (z − zk) m Y j=1 (z − pj)
where z1, ..., zn are the zeroes of φ ∈ C and p1, ..., pm are the poles in C.
The fact that φ : C∞ → C∞ is one-to-one shows that φ has at most one zero in
C and similarly to ψ, so that n ≤ 1 and m ≤ 1. If n = m = 0, then the function
φ(z) = c is constant, it would also imply that φ 6∈ Aut(C∞). Hence, it leaves us with
three possibilities, φ(z) = cz − z1 z − p1 , for (z16= p1), φ(z) = c 1 z − p1 and φ(z) = c(z − z1).
Since all the cases are linear-fractional transformation in C∞. Hence, Aut(C∞) is the
set of all linear-fractional transformations.
The Schwarz lemma can be applied to the holomorphic mapping which shows that there is a rotation in Aut(D).
Theorem 3.11. Suppose that φ ∈ Aut(D) and φ(0) = 0. Then φ is a rotation: φ(z) = βz
for some β ∈ C with |β| = 1.
Proof. Theorem 2.7 shows that |φ0(0)| ≤ 1, and if we let ψ be the inverse of φ, then ψ
would also satisfy |ψ0(0)| ≤ 1. Furthermore, if we apply a chain rule, then it would give
ψ0(0) = 1/φ0(0) with |φ0(0)| = 1. Hence, this is a rotation according to Theorem 2.7. Our next subject is the Cayley transform and will be using for construction of a map and extend a plane. First, we need to define a few topological properties.
Definition 3.12. A collection T of subsets of X is a topology on X if:
(i) ∅, X ∈ T ;
(ii) if Gα∈ T for α ∈ A, thenSα∈AGα∈ T ;
(iii) if Gi∈ T for i = 1, 2, ..., n, then T n
i=1Gi∈ T .
One of the relations between topological spaces is the homeomorphic function.
Definition 3.13. A function f : X → Y is said to be homeomorphic between the two
topological spaces (X, τx) and (Y, τy), if they follow these properties:
(i) f is a bijection; (ii) f is continuous;
(iii) the inverse function f−1 is continuous.
Now we are ready to introduce the Cayley transform. The Cayley transform is a con-formal mapping from the unit disk onto the upper half-plane, Π+
= {z ∈ C : Im(z) > 0}. The principal behind the Cayley transform is based on the linear-fractional transforma-tion, principally on the topological mapping. Our interest here is to provide a way to map different topological spaces with the focus on the upper half-plane, but first we will consider this question: ”When do we have a complex number in Π+in a signature of the
Cayley transform?” (The question might be obvious to some, since it is necessary we will carefully explain its meaning.) Consider this: For all z ∈ Π+ that |z − i| < |z + i|, since z needs to be closer to +i than −i, thus,
z ∈ Π+↔ z − i z + i < 1.
Let ψC be a conformal map from the upper half-plane onto the unit disk.
ψC(z) =
z − i
z + i. (3.3)
Since there is a conformal mapping from the upper half-plane onto the unit disk, then there exists a function such that ψC−1= φC, where
φC(z) = i
1 + z
1 − z. (3.4)
The function φC is the Cayley transform from a unit disk onto Π+, it follows that ψC is
the inverse of the Cayley transform.
The next step is to use the Cayley transform to classify different elements of Aut(D), by showing that φC: D → Π+ extends to a homeomorphism of D and Π+∞. Define Π+∞
as the closure of the upper half-plane in the extended plane: Π+∞= {z ∈ C : Im(z) ≥ 0} ∪ {∞}
and ∂∞Π+ to be the extended boundary of Π+∞:
∂∞Π+= R∞= R ∪ {∞}.
Note that in some occasion, Aut(D) and Aut(Π+) can be induced isomorphically, if there
is a function φ0∈ Aut(D) such that the element of Aut(Π+) is ψ = φC◦ φ0◦ φ−1C .
∂D, then φ is said to be parabolic. In a similar way as in the parabolic case but with
two fixed point in ∂D, φ is said to be hyperbolic.
Definition 3.14. Let G be a group and a, b ∈ G. A conjugate of b to a is when b = c−1ac
for some c ∈ G.
In the next few theorems, we will construct the element classification of Aut(D). Also, note that for now we will overlook the identity case of non-trivial elements of φ ∈ Aut(D), since, this will be elaborated in Section 5, for now we will refer to ([20], p. 269).
Theorem 3.15. Suppose that φ ∈ Aut(D) is not the identity.
(i) If φ is elliptic, then φ is conjugate to a rotation: There exists a complex number
α 6= 1 with |α| = 1, such that φ is conjugate to ψ, where ψ(z) = αz, for z ∈ D.
(ii) If φ is parabolic, then φ is conjugate to a translation in Π+
: There exists B ∈ R
with B 6= 0 such that the element of Aut(Π+) corresponding to φ is conjugate to ψ, where
ψ(z) = z + B, for z ∈ Π+.
(iii) If φ is hyperbolic, then φ is conjugate to a dilation in Π+ : There exists δ > 0 with δ 6= 1 such that the element of Aut(Π+) corresponding to φ is conjugate to ψ, where
ψ(z) = δz, for z ∈ Π+.
Proof. For part (i), suppose that φ is elliptic with φ(0) = 0, Theorem 3.11 shows that φ
is a rotation.
For part (ii) and (iii), suppose that φ is either parabolic or hyperbolic in Aut(Π+)
and has no fixed point in Π+, and either one or two fixed points on ∂∞Π+. Then,
we can make a change of variables so that φ(∞) = ∞. Since φ can be expressed as a linear-fractional transformation:
φ(z) = az + b cz + d,
for some a, b, c, d ∈ R with ad − bc 6= 0. It may happen that c = 0, then
φ(z) = Az + B,
for some A, B ∈ R with (A, B) 6= (1, 0). If A = 1, then φ is a translation since ∞ is the only fixed point of φ. Suppose now that A 6= 1: let c = B/(1 − A), set ψ(z) = z + c, and define ˜φ = ψ−1◦ φ ◦ ψ. This shows that ˜φ(z) = Az, which implies that 0 is the only
fixed point of ψ, similarly for ∞. In conclusion, φ is a translation.
We will now use the previous theorem and show that there is a bounded nonconstant function for the nontrivial elements of Aut(D).
Theorem 3.16. Suppose that φ ∈ Aut(D).
(i) If φ is either elliptic of finite order, parabolic, or hyperbolic, then there exists a
bounded nonconstant f ∈ H(D) such that f ◦ φ = f,
and in fact such that f (z) = f (w) if, and only if, w = φn(z) for some integer n. (ii) If u is elliptic of infinite order and f ∈ H(D) with f ◦ φ = f , then f is constant.
Proof. We are dividing the proof into three different conditions where we prove each of
condition separately, elliptic, parabolic and hyperbolic, starting with the elliptic. The meaning of finite order and infinite order will be referred back to Definition 3.8. Theorem 3.15 shows that φ is elliptic and rotational with φ(z) = αz and |α| = 1. Since φ has a finite order of N with αN = 1 and αk6= 1 for k = 1, 2, ..., N −1, it follows that f (z) = zN
works for f ◦ φ = f . On the other hand, if φ has infinite order, then the powers of α are dense in the unit circle so that f ∈ H(D) and f ◦ φ = f , then f must be a constant.
Suppose now that φ is parabolic; based on Theorem 3.15 we may use that φ ∈ Aut(Π+)
and the parabolic function can be expressed as:
φ(z) = z + B
where B 6= 0. Furthermore, let
f (z) = e2πiz/|B|.
This shows that f is bounded in Π+, since z = x + iy ∈ Π+,
|f (x + iy)| = eRe(2πi(x+iy)/|B|)= e−2πy/|B|< 1.
If f (z) = f (w), then 2πiw/|B| = 2πiz/|B| + 2πik for some k ∈ Z. Thus, w = z + nB with n = ±k, which is the same as w = φn(z).
Suppose that φ is hyperbolic; the function is (similarly to earlier, φ ∈ Aut(Π+)): φ(z) = δz
with a positive δ 6= 1. We need to define a branch of logarithm in Π+ as log(reit) = ln(r) + it, for r > 0, 0 < t < π, and set,
f (z) = zic= eic log(z).
The idea here is to find c 6= 0 such that f (w) = f (z) is satisfied. Note that
|f (reit)| = eRe(ic(ln(r)+it)) = e−(ct), for r > 0, 0 < t < π,
this shows that f is bounded in Π+ for some c ∈ R. Hence, z and w are both in Π+. We have f (z) = f (w) if, and only if, iclog(z) − iclog(w) = 2πik, for some k ∈ Z with
and
z = e2πk/cw.
In conclusion, there exists c = 2π/ln(δ) that fulfills the relation of f (w) = f (z) if, and only if, z = δkw or z = φk
(w) for k ∈ Z.
3.2. Schwarz Reflection Principle. Schwarz reflection principle is a way to extend a domain of complex functions on a real axis. Suppose that D is an open set of the complex plane, and D is symmetric on the real axis such that ¯z ∈ D whenever z ∈ D.
D+={z ∈ D : Im(z) > 0}, (3.5)
D−={z ∈ D : Im(z) < 0}, (3.6)
D0={z ∈ D : Im(z) = 0}. (3.7)
There are various flavors of the Schwarz reflection principle and for our beneficial toward the great Picard theorem, we will only consider the Schwarz reflection principle for holomorphic functions. However, we need the result from the Schwarz reflection principle for harmonic functions, therefore, we will refer e.g. to [10]. Below is the proof of the Schwarz reflection principle for holomorphic function.
Theorem 3.17. Suppose that D is an open subset of the plane which is symmetric about the real axis. Suppose that f ∈ H(D+) satisfies
lim
n→∞Im(f (zn)) = 0
for every sequence (zn) in D+ which converges to a point of D0. Then, f extends to a
function F ∈ H(D), which satisfies
F (¯z) = F (z), for z ∈ D.
Before proceeding with the proof, we need to be aware of that we are assuming that Im(f ) extends continuously to D+∪ D0. The problem with this is that we cannot define F (z) = f (z) for z ∈ D0 since f is undefined on D0.
Proof. The Schwarz reflection principle for harmonic functions states that there is a
function f = u + iv ∈ D such that v can be extended to a function V in D, and satisfies
V (¯z) = −V (z), for z ∈ D.
Introducing a disk Dt⊂ D with t as its center such that t ∈ D0. Since V is harmonic
in Dt, and Dt is simply connected, then there exists a function ft ∈ H(Dt) such that
Im(ft) = V in Dt. Moreover, we know that Im(f ) = V in D+t, hence we can shift its
position by applying a constant to ftso that f = ftin D+t.
It follows from (3.7) that V = 0 on D0, which indicates that the function fk
t(t) takes
real values on D0
t, and therefore, coefficients produced from its derivatives must also be
real.
We may define F : D → C by, F (z) = f (z), if z ∈ D+, ft(z), if z ∈ Dt, f (¯z), if z ∈ D−.
We know that that ft = f in D+∩ Dt = Dt+, and it follows that ft(z) = f (¯z) for
z ∈ Dt∩ D− = D−t. Then, ft= fsin Dt∩ Ds. The reason we wrote the definition of F
with overlapping cases is that the domain of the function defined in each case is an open set. Thus, in order to show that F is holomorphic we need to show that g ∈ H(D−), where g(z) = f (¯z) (since we already know that f ∈ H(D+) and f
t∈ H(Dt)).
We will use power series to show that g is holomorphic in D−. Suppose that a ∈ D− then ¯a ∈ D+, since f ∈ H(D+) there exists r > 0 and a sequence of complex numbers,
(cn), such that f (z) = ∞ X n=1 cn(z − ¯a)n, where |z − ¯a| < r. It follows that f (¯z) = ∞ X n=1 cn(¯z − ¯a)n, where |¯z − ¯a| < r, so, g(z) = f (¯z) = ∞ X n=1 cn(¯z − ¯a)n= ∞ X n=0 ¯ cn(z − a)n, where |z − a| < r.
Finally, the last step shows that g ∈ H(D(a, r)) for any a ∈ D−, therefore g ∈ H(D−). 3.3. The Extended Plane and Carathéodory’s Theorem. Carathéodory’s theorem is a way to conformally map an open and simply connected subset of the complex plane onto the unit disk, it is sometimes known as an extension of the Riemann mapping theorem. We shall also provide theorems needed in our construction of a modular function that will be introduced in the next section. Herein, we will state a weaker version of the Carathéodory’s theorem which will suffice for us to show that all the boundary points are simple, since we have an explicit region bounded by a simply closed curve. (The original Carathéodory’s theorem has a more general statement which can be seen e.g. in [1].)
Lets start with a few results on bounded holomorphic functions.
Lemma 3.18. Suppose that f ∈ H(D) and |f (z)| ≤ 1 for all z ∈ D. Suppose that N is a positive integer, a ∈ R, and let S = {reit: 0 < r < 1, a < t < a + 2π/(2N )}. Suppose
that γ : [0, 1] → D is a continuous map, γ(0) = r0eai and γ(1) = r1e(a+2π/(2N ))i for some rj > 0, and γ(t) ∈ S for 0 < t < 1. If |f (γ(t))| < < 1 for all t ∈ [0, 1], then
Proof. For simplicity, assume that a = 0. Let g ∈ H(D) be a function and define g(z) = f (z)f (¯z) and set h(z) = N −1 Y j=0 ge2πij/Nz.
This shows that |g| < at every point of the set γ∗∪ ˜γ∗, so that |h| < at every point of the set Γ = N −1 [ j=0 e−2πij/N, for γ∗[˜γ∗.
There exists 0 ∈ Ω and ∂Ω ⊂ Γ in a bounded open set Ω ⊂ D. Theorem 2.6 shows that there exists |h(0)| ≤ , thus |f (0)| < 1/(2N ).
Lemma 3.19. Suppose that f ∈ H(D) is bounded. Suppose that 0 < θ < π and let S be the region
S = {reit: 1/2 < r < 1, 0 < t < θ}.
Suppose that the map γn : [0, 1] → D is continuous, γn((0, 1)) ⊂ S, γn(0) > 0 and
γn(1) = rneiθ(rn > 0) for n = 1, 2, .... If L ∈ C and f ◦ γn → L uniformly on [0, 1] as
n → ∞, then f = L everywhere in D.
Proof. Subtracting a constant and dividing by another constant we may assume that L = 0 and |f | < 1 in D. If we apply Lemma 3.18 and define g, h ∈ H(D) with an integer N such that 2π/(2N ) < θ, then there exists |f | < at every point of the set γ∗n. If we replace the curve γn∗ by γ as in Lemma 3.18, then we receive |h| ≤ at every point of
D(0, 1/2) with the fact that > 0 and h must disappear identically on D(0, 1/2). Thus, f is identically zero. (If f is not zero, then it would have only finitely many zeroes in
D(0, 1/2), since h has finitely many zeroes.)
The next subject is the Lindelöf’s theorem and it states that a holomorphic function on the bounded region grows ”moderately” in the unbounded direction. Hereby, we will use Lemma 3.18 and Lemma 3.19 in the proof, where we have shown that there is a restriction to the functions with limits, L = 0, and |f | < 1 for f ∈ H(D). Below is the Lindelöf’s theorem.
Theorem 3.20. Suppose that γ : [0, 1] → D is continuous, γ(t) ∈ D for 0 ≤ t < 1 and γ(1) = 1. Suppose that f ∈ H(D) is bounded. If f (γ(t)) → L as t → 1, then
limr→1−f (r) = L.
Proof. We are going to show that there is a function |f (r)| < 1/6 such that R < r < 1
for an arbitrary > 0, then limr→1−f (r) = 0. We may also assume that L = 0 with the
same argument we used in Lemma 3.19. Let ψr∈ Aut(D),
ψr(z) =
z + r
Furthermore, let C0 = [−i, i] ⊂ D be a curve and Cr = ψr(C0), and there exists tr∈ (0, 1) such that γ(tr) ∈ Crand γ(t) /∈ Crfor tr< t < 1. Also let γr= γ|[tr,1).
Define Γr = ψr−1◦ γr and pr = ψr−1(γr(tr)) = Γr(tr). This shows that pr ∈ [−i, i],
since γr(tr) ∈ Crand Cr= ψr(C0) where C0= [−i, i]. If pr= 0, then f (r) = f (γr(tr)),
so that |f (r)| < < 1/6. Suppose p
r 6= 0, then pr = iy for some y 6= 0, by the
assumption of y > 0.
If F = f ◦ ψr, then |F | < on Γ∗r. The curve Γr lies in the right half-plane with
end-points at 1 and iy (from what we have defined earlier). Thus, this could lead to a problem since 1 lies outside D, and in order to handle this we apply Lemma 3.18 with
S = {ρeit: 0 < ρ < 1, π/6 < t < π/2}.
Hence, |F (0)| < 1/6. But we already know that f (r) = F (0), which conclusively gives |f (r)| < 1/6.
We will tie the previous lemmas into a proposition that will mostly cover all the context of the Carathéodory’s theorem, but first we need a definition on the simple boundary point.
Definition 3.21. Let D be an open subset of the complex plane and ζ ∈ ∂D. We will
say that ζ is a simple boundary point, if (zn) is any sequence of points in D such that
zn → ζ, then there exists a continuous map γ : [0, 1) → D and a sequence (tn) in [0, 1)
such that tn ≤ tn+1, tn→ 1, γ(tn) = zn, and limt→1γ(t) = ζ.
Proposition 3.22. Suppose D is a bounded simply connected open set in the plane, and let φ : D → D be a conformal equivalence.
(i) If ζ is a simple boundary point of D, then there exists L ∈ ∂D such that φ(zn) →
L whenever (zn) is a sequence in D tending to ζ. (In other words, φ can be
extended continuously to the set D ∪ {ζ}, and the extended function satisfies φ(ζ) ∈ ∂D).
(ii) Suppose that ζ1 and ζ2 are two simple boundary points; choose corresponding complex numbers L1 and L2 as in (i). If ζ16= ζ2, then L16= L2.
Proof. First note that if (zn) is any sequence in D tending to a point of the boundary,
then |φ(zn)| → 1: if r < 1 with K as a compact subset of D and K = φ−1(D(0, r)), then
there exists N such that zn∈ K for all n > N . Hence, |φ(z/ n)| > r for every n > N .
Part (i), let ζ be a simple boundary point of D with a sequence (zn) tends to ζ.
Since there is a bounded and closed function in the disk, then there exists L ∈ D and a subsequence (znj) such that φ(znj) → L. To prove this, we need to show that every
sequence (zn) of φ(zn) tends to L.
A simple way to prove this is by proving its contradiction, which would be stated as: the function (φ(zn)) has two subsequences that are pursuing to various limits L1 and L2, and that L1 6= L2, see Figure 3.2. By adjusting the notation we can assume that φ(z2n) → L1and φ(z2n+1) → L2, where Lj∈ ∂D. Moreover, we know that ζ is a simple
bounded point and there exists a continuous curve γ : [0, 1) → D and a sequence (tn) in
[0, 1) such that tn≤ tn+1 where tn→ 1, implies that γ(tn) = zn and limt→1γ(t) = ζ.
Γ(t2n) → L1 and Γ(t2n+1) → L2. For this to occur, a reasonable presumption is that
the curve Γ must cross some sector S infinitely many times as it tends to ∂D, it would either be S1 or S2 and this can be seen in Figure 3.2. When n is large enough, there
exist numbers an and bn lying in between t2n < an < bn < t2n+1, with curves γ(an)
and γ(bn) locating on edges of the S (not at the same edge) and γ((an, bn)) lies in the
interior of S. Let γn= γ|[an,bn].
If we let f = φ−1, then f ∈ H(D) and f is bounded. Then, we have f (Γ(t)) → ζ as
t → 1, so f (Γ(t)) = γ(t), since f ◦ γn → ζ uniformly. Thus, f is constant according to
Lemma 3.19, which is a contradiction to the fact that f is a bijection from D to D. Part (ii), suppose that ζ1 and ζ2are simple boundary points of D and ζ16= ζ2. Part
(i) shows that, φ can be extended to a function continuous on D ∪ {ζ1, ζ2}, and that
|Lj| = 1, if Lj= φ(ζj), where j = 1, 2.
We need to show L1 6= L2; start by employing the continuous curves γj : [0, 1) → D
in such way that γj(t) → ζj as t → 1. Also, let Γj = φ ◦ γj and as before, there exists
an inverse function φ−1such that f = φ−1. Hence, f is bounded in H(D) and the curve Γj lies in D tending to Lj, so f (Γj(t)) → ζj as t → 1. Lastly, Theorem 3.20 shows that
f (rLj) → ζj as r approaches to 1, thus the points ζ16= ζ2, and therefore, L16= L2.
S3 S1 S2 S4 L1 L2
Figure 3.2 – Circle with four sections and two boundary points
Lemma 3.23. Suppose that D is a bounded open set in the complex plane, φ : D → C is continuous, and for every ζ ∈ ∂D there exists a function φζ continuous on D ∪ {ζ}
which agrees with φ on D. Then φ extends to a function continuous onD. Proof. We start by extending φ to a function ψ : D → C such that
ψ(ζ) = φζ(ζ), for ζ ∈ ∂D.
with > 0, then |z − ζ| < δ. Let w ∈ ∂D, then ψ(z) = limw→zψ(w), and this follows
that |ψ(z) − ψ(ζ)| ≤ .
Finally, we are now able to state our anticipated theorem, the Carathéodory’s theorem. To demonstrate this, we are going to use that a function can be continuously extended on D and that there is a sequence tending to a constant on the plane, which has been shown in Lemma 3.23 and Proposition 3.22.
Theorem 3.24. Suppose D is a bounded simply connected domain in the complex plane and every boundary point of D is simple. Let φ : D → D be a conformal equivalence. Then φ extends to a homeomorphism of D and D.
Proof. We have seen that φ can be continuously extended to D ∪ {ζ} for every ζ ∈ ∂D.
Also that φ extends continuously to the closure of D from Lemma 3.23 and Proposition 3.22 (i).
For the extended function, φ(D) must be a compact subset of D containing D such that φ(D) = D. Proposition 3.22 (ii) shows that φ is a bijection of D, so that there exists an inverse function φ−1: D → D. The only thing that remains is to show that the inverse of φ is continuous on D. Since D is compact and there is a continuous bijection
φ : D → D, then there exists an open set O ⊂ D such that (φ−1)−1(O) = φ(O) ⊂ D.
4. A Modular Function
The goal with this section is to construct a subgroup Γ(2) of the modular group Γ, which is a group consisting of all the linear-fractional transformations. Furthermore, we will demonstrate that the subgroup Γ(2) is freely generated by elements τ and σ, and see that this is the implication from Lemma 4.4.
A modular function is a meromorphic function in an upper half-plane Π+, that is
invariant under certain actions of the modular group Γ. Let us recall the linear-fractional transformations, φ from Definition 3.9,
φ(z) = az + b
cz + d (4.1)
where a, b, c and d are integers and ad − bc 6= 0.
The modular group, Γ is a subgroup of all the transformations of Aut(Π+) correspond
to the group SL2(Z) (special linear group, SL2(Z) is the group of 2×2 matrices belonging
to a so called Lie group. We will leave details of the Lie groups e.g. to [6]). The subgroup, Γ(2) is consisting of all φ with requirements of a and d being odd, b and c being even. If so, this constructs
a b c d
∈ SL2(Z). (4.2)
Hence, the condition required for Γ(2), a b c d ≡1 0 0 1 mod 2.
This shows that, Γ(2) is a subgroup of Γ, because every matrix of Γ(2) is the identity matrix with mod 2. Now, let σ and τ be defined by
σ(z) = z
−2z + 1 (4.3)
and
τ (z) = z + 2. (4.4)
Lemma 4.1. The group Γ(2) is generated by σ and τ .
Proof. It is clear that σ and τ are elements from Γ(2). Let G be the subgroup of Γ(2)
that generated by σ and τ . Meaning, we need to show that Γ(2) = G. Define χ : Γ(2) → Z+ by
χa b c d
= |a| + |c|.
Suppose that φ ∈ Γ(2). Hence, we need to show that φ ∈ G. Define
G ◦ φ = {ψ ◦ φ : ψ ∈ G}.
Since, χ can only attain positive integers, then there must exists
φ0= a b c d ∈ G ◦ φ, with χ(φ0) minimal.
Suppose that c 6= 0. If |a| > |c|, then this would lead to a contradiction which is that there exists n ∈ Z such that −|c| ≤ a + 2nc < |c|. This implies that −|c| < a + 2nc < |c| since, a is odd and c is even, hence,
χ(τnφ0) = |a + 2nc| + |c| < |c| + |c| < |a| + |c| = χ(φ0),
which contradicts our choice of φ0. If we now let, |a| ≤ |c|, then 0 < |a| < |c| since a
is odd, and with a similar to the previous argument that there exists n ∈ Z such that
χ(σnφ
0) < χ(φ), once again our choice of φ0 is unsatisfied.
So this only works if c = 0. Since, ad − bc = 1 implies that a = d = ±1, so that
φ0= τn for some n ∈ Z. In particular, φ0∈ G. Hence, φ ∈ G.
From Lemma 4.1, one can see that σ and τ are conjugate in Γ. We define j ∈ Γ by
j(z) =−1
z . (4.5)
So that
σ = j−1◦ τ ◦ j.
Now, we define new σ and τ that only cover the upper half-plane:
T = {z ∈ Π+: −1 ≤ Re(z) < 1} (4.6) and S = z ∈ Π+: z +1 2 > 1 2, z −1 2 ≥ 1 2 . (4.7)
T −1 1 Figure 4.1 S −1 1 Figure 4.2
Lemma 4.2. The following statements are true:
(i) the sets τn
(T ), (n ∈ Z) are disjoint, and their union is Π+;
(ii) the sets σn
(S), (n ∈ Z) are disjoint, and their union is Π+;
(iii) if n ∈ Z and n 6= 0, then τn(T ) ⊂ S; (iv) if n ∈ Z and n 6= 0, then σn(S) ⊂ T .
Proof. The statements (i) and (ii) can be drawn from Figure 4.1 and Figure 4.2, which
indicate that T is a fundamental domain of the subgroup Γ(2) generated by τ , and similarly for S generated by σ. Part (iii) and (iv) follow from part (i) and (ii), since
σ = j−1◦ τ ◦ j and S = j(T ), where j(z) = −1/z takes lines and circles to lines and
circles with the preservation of angles.
Let G be a group generated by a, b ∈ G. Then, a nontrivial reduced word of a and b is expressed as: gn1 1 g n2 2 ...g nk k (4.8)
where k ≥ 1, each nj is a nonzero integer, and each gj is either a or b. For k = 0, we
would have the identity element of G.
Remark 4.3. A group G is freely generated by a and b, meaning that G is generated by a and b, and no nontrivial reduced word in a and b is the identity element of G.
Lemma 4.4 is known as the Ping-Pong lemma:
Lemma 4.4. Suppose that G is a group of bijections of a set X (with composition as the group operation), and G is generated by a, b ∈ G. Suppose that A, B ⊂ X satisfy the following:
(i) the sets an
(A), (n ∈ Z) are disjoint, and their union is X; (ii) the sets bn(B), (n ∈ Z) are disjoint, and their union is X; (iii) if n ∈ Z and n 6= 0, then an(A) ⊂ B;
(iv) if n ∈ Z and n 6= 0, then bn(B) ⊂ A.
Let E = A ∩ B, and assume that E 6= ∅. If w is a nontrivial reduced word in a and b, then
E ∩ w(E) = ∅.
Proof. Assume that, w starts and ends with a power of a as in (4.8), and as the following w = an1bn2...ank. If we apply (iii) and (iv) repeatably, then this shows that
w(A) = an1bn2...bnk−1ank(A) ⊂ an1bn2...bnk−1(B) ⊂ ... ⊂ an1(A).
Part (i) shows that A ∩ w(A) = ∅, which implies that (A ∩ B) ∩ w(A ∩ B) = ∅. Similarly, we can use this to prove part (ii), if we let w ends with a power of b so that w =
an1bn2...bnk. This gives: w(B) ⊂ an1(A) and A∩w(B) = ∅, thus (A∩B)∩w(A∩B) = ∅.
The results show that the sets an(A) and bn(B) are disjoint.
We are one step closer to be able to construct a fundamental domain of the modular group, Γ(2) with the aids from Lemma 4.2 and Lemma 4.4. The fundamental domain we want to construct contains the following properties:
Remark 4.5. We want to construct a set Q ⊂ Π+ such that:
(i) the sets φ(Q) for (φ ∈ Γ(2)) are disjoint; (ii) Π+=S
φ∈Γ(2)φ(Q).
Again, the modular function we want is a meromorphic function in Π+. Part (i) follows
One might wonder what was the connection between the Ping-Pong lemma and our modular group of the upper half-plane. The definitions below might be able to clarify that inquiry.
ping : Π+→ T and pong : Π+ → S.
The unique element of T is ping(z) = τn
(z) for some n ∈ Z, similarly for pong(z) but for element S, pong(z) = σn
(z) for some n ∈ Z. The reason to define them in such way is that we want to find φ ∈ Γ(2) such that φ(z) ∈ Q for given z ∈ Π+. Moreover, if there
is such φ, then there is a unique sequence that must converge at a point of Q, hence it is more reasonable to define sets z1= ping(z) and z2= pong(z1).
The purpose here is to make ping and pong move closer to i, since i is the upper half-plane. We will also see that ping and pong are constants and invariants under j defined as in (4.5), otherwise, they will approach to the extended upper half-plane, ∂∞Π+. Thus,
we need to define a certain function: Define ρ : Π+→ [0, ∞) by ρ(z) = z − i z + i . (4.9)
If we have Φ : Π+ → D as the Cayley transform, then this would give ρ(z) = |Φ(z)|.
As we have stated in Section 3, that such function has limits 0 ≤ ρ(z) < 1 for all z ∈ Π+.
Furthermore, the set of all z with ρ(z) ≤ r is a compact subset of Π+for any r ∈ (0, 1). The function ρ as in (4.9) satisfies the following properties:
Lemma 4.6. For ρ as in (4.9), j as in (4.5) and for every z ∈ Π+, we have
(i) ρ(j(z)) = ρ(z); (ii) ρ(ping(z)) ≤ ρ(z); (iii) ρ(pong(z)) ≤ ρ(z).
Proof. Part (i) is straightforward, since j(z) = −1
z and with some algebra it can be
shown that (i) is satisfied:
ρ(j(z)) = −1 z − i −1 z + i = (1 + zi) (1 − zi) = i i (1 + zi) (1 − zi) = z − i z + i = ρ(z)
For part (ii), we are using the fact that z = x + iy and assuming z /∈ T , where T is defined as (4.6), thus,
(ρ(z))2= 1 − 4y
x2+ (y + 1)2.
If z ∈ T , then we would have ping(z) = z since our z /∈ T , and ping(z) = x0+ iy for some x0 ∈ Π+. This leads to |x| ≥ 1 and |x0| ≤ 1, which concludes (ii). Similarly for part
(iii), which follows from part (i) and (ii), together with the fact that σ = j−1◦ τ ◦ j. One of the main concerns here is to ensure that sequences of the holomorphic function in a disk D are bounded, which is significant to Remark 4.5 when we apply the Ping-Pong lemma to construct the fundamental domain. Now, it is a suitable moment to look at the compactness of our holomorphic function in Π+and ∂∞Π+. With the reason of that
Lemma 4.7. Suppose that fn : D → D is holomorphic. If z ∈ D and |fn(z)| → 1, then
|fn(w)| → 1 for every w ∈ D.
Proof. We will prove a contradiction of the statement. It follows from Theorem 2.9, that
there exists a subsequence (fnj) of (fn) such that f converges uniformly on compact
sets to f ∈ H(D). Suppose that there is w ∈ D with the subsequence (fnj) such that
fnj(w) → α with |α| < 1. It implies that |f | ≤ 1 in D. Since |f (z)| = 1, and f is a
constant according to Theorem 2.6, which contradicts the fact that |f (w)| < 1. From Lemma 4.7 follows by two corollaries.
Corollary 4.8. Suppose that fn : Π+ → Π+ is holomorphic, z ∈ Π+ and fn(z) →
∂∞Π+. Then fn(w) → ∂∞Π+ for all w ∈ Π+.
Proof. If (zn) ⊂ Π+, then we will say that zn → ∂∞Π+. If for every compact set K ⊂ Π+,
then there exists N so that zn∈ Π+\ K for all n > N . Moreover, if we let Φ : Π+→ D
be the Cayley transform that leads to zn → ∂∞Π+ with the only requirement, which is
|Φ(zn)| → 1. The proof is concluded according to Lemma 4.7.
A short notice, that if we have ρ as in (4.9) and z ∈ Π+with a, b, c and d as elements
in a group of SL2(Z), then ρ ai + b ci + d 2 = a 2+ b2+ c2+ d2− 2 a2+ b2+ c2+ d2+ 2. (4.10) Corollary 4.9. If (φn) is a sequence of distinct elements of Γ (that is, φn 6= φm for
n 6= m), then φn(z) → ∂∞Π+ for all z ∈ Π+.
Proof. As we have shown in (4.10) that ρ(φn(i)) → 1, since for a given R < ∞ there are
only finitely many elements (a, b, c, d) ∈ Z4with a2+ b2+ c2+ d2< R. This implies that
|Φ(φn(i))| → 1, so that φn(i) → ∂∞Π+. Hence, φn(z) → ∂∞Π+for all z ∈ Π+according
to Corollary 4.8.
For future convention we will use this opportunity to tie up a few theorems and lemmas in this section into one single theorem, for constructing the normal subgroup, Γ(2). First, we need to show that the group Γ(2) is in Γ.
Theorem 4.10. The group Γ(2) is freely generated by τ and σ. Further, Γ(2) is a normal subgroup of Γ, every nontrivial element of Γ(2) is either parabolic or hyperbolic, and if Q = S ∩ T , then
(i) the sets of φ(Q) for (φ ∈ Γ(2)) are disjoint; (ii) S
φ∈Γ(2)φ(Q) = Π
+.
Proof. To prove (i), we will use Lemma 4.2 and Lemma 4.4, these show that Γ(2) is
freely generated by σ and τ . Thus, (i) holds. In part (ii), we need to show for z ∈ Π+
there exists φ ∈ Γ(2) such that φ(z) ∈ Q. Assume that z ∈ Π+\ Q.
Since, Π+= S ∪ T we must either have z ∈ S \ T or z ∈ T \ S. Also, by the fact that
etc., and there exists φn ∈ Γ(2) with zn= φn(z). If there exists n with zn∈ Q, then we
are finished with the proof. Suppose not, then we would have zn+16= zn for all n ≥ 2.
Now, the definitions of ping and pong show that for every m and n with m > n, there exists w, a nontrivial word in σ and τ such that φm = w ◦ φn. This implies that
φm6= φn for m > n. Corollary 4.9 states that there is a sequence zn→ ∂∞Π+, which is
the contradiction to Lemma 4.6 since, ρ(zn) ≤ ρ(z) < 1 for all n, so that the sequences
(zn) are all contained in some fixed compact subset of Π+.
Suppose that ψ ∈ Γ(2), z ∈ Π+ and ψ(z) = z. Choose φ ∈ Γ(2) so that z ∈ φ(Q). We
also have z = ψ(z) ∈ ψ ◦ φ(Q) and (i) implies that φ = ψ ◦ φ. Hence, ψ is the identity. Furthermore, suppose that
A =a b c d ∈ Γ(2), then A = I + 2M
for some matrix M with integer entries. Suppose that B ∈ SL2(Z), then BAB−1= I + 2BM B−1,
5. Covering Maps
We have introduced the conformal map in Section 3, which is a function that preserves all angles locally. Now, we are going to give more topological properties and introduce another kind of projection, namely covering projection or covering maps. At the end of this section, finally, the readers should be able to see the connection between the fundamental domain that we just proved in the previous section and the covering maps. Lets start by defining covering maps.
Definition 5.1. Let X and Y be topological spaces such that there is a map p : X → Y
that is continuous, then p is said to be a covering map, if each point of z in Y has a connected open neighborhood N such that
p−1(N ) = [
α∈A
˜
Nα,
where each ˜N is an open subset of X and ˜N is pairwise disjoint. Each of the point α ∈ A is mapped homeomorphically from ˜N onto N by p. For any open neighborhood N satisfies the condition above, N is then called elementary neighborhood.
Besides the definition of the covering maps we also need several definitions from topol-ogy.
Definition 5.2. In a topological context, a homotopy is a continuous deformation of
one continuous function to another continuous function.
Definition 5.3. A topological space is simply connected, if any two paths between two
points are homotopic to each other, i.e. there exists a homotopy between them.
Definition 5.4. Let X, Y and ˜X be topological spaces, and let g : X → ˜X be a covering
map. A lift of a function f : X → Y is a function h : ˜X → Y such that h ◦ g = f .
Now that we have defined the covering maps, we need to return to Section 4 and tie up one last resource for our modular function. So far, we have shown that the normal subgroup of Γ is generated by τ and σ precisely as in Theorem 4.10. Moreover, we have constructed a fundamental domain of Γ(2) such that Q ∈ Π+. The next theorem is one
of the main pieces for the great Picard theorem.
Theorem 5.5. There exists a holomorphic covering map λ for Π+
onto C \ {0, 1}, with the following properties:
(i) λ ◦ φ = λ for all φ ∈ Γ(2); (ii) λ is one-to-one on Q;
(iii) if (zj) is a sequence in Π+ with Im(zj) → ∞, then λ(zj) → ∞.
Proof. In this, we will start off with the conformal maps case and then afterward, we will
show that λ is a holomorphic covering map. For part (i) and (ii), we will start our proof by defining Q+ as an interior of the right half of Q where Q is a fundamental domain of
At this point, we have proved various theorems, and a few of them will be recalled with the significative interest to this proof. Theorem 2.10 shows that there is a conformal map h from Q+onto Π+, and together with Theorem 3.24, it shows that h extends to a homeomorphism of the extended closure of Q+, that is, Q+∪ {∞}, onto Π+
∞. Moreover,
if there is a bounded simply connected open set and every point of it is simple, for a positively oriented triple of distinct points e.g. (0, 1, ∞), there is a conformal equivalence of h such that h(0) = 0, h(1) = 1 and h(∞) = ∞ (proof of this can be seen e.g. in ([20], p. 310)).
Let us also recall Theorem 3.17, which indicates that there is a reflection on a real axis such that:
h(x + iy) = h(−x + iy)
for h(iy) ∈ R and y > 0. Notice that, x + iy signifies the upper right of the circle and −x + iy the upper left of the circle, which are symmetric according to Theorem 3.17. Since there is a reflection h−1({1}) = {1, −1}, h−1({0}) = {0}, h−1({∞}) = {∞} and (Q ∪ {∞}) \ Π+= {−1, 0, 1, ∞}, which in fact gives h(Q ∩ Π+
) = C \ {0, 1}, so
h(Q) = C \ {0, 1}.
The conformal map, h is one-to-one on the closure of Q+, additionally, there exists a
reflection so that h is one-to-one on Q, and containing sequences of zn in Q such that if
Im(zn) → ∞, then h(zn) → ∞. Define a function ˜h : Q+→ C by
˜
h(z) = 1 h(j(z)),
where j(z) is a function in the modular group Γ defined as in (4.5). Since, j defines the conformal map from Q+ onto the interior of the left half of Q, thus, h ◦ j defines
a conformal map from Q+ onto the lower half-plane, then this indicates that ˜h is a
conformal map from Q+onto Π+.
This shows that ˜h extends continuously to the extended closure, Q+∪ {∞}, and one
can verify that by: ˜h(1) = 1 = h(1), ˜h(0) = 0 = h(0) and ˜h(∞) = ∞ = h(∞) (which
is similar as when we argued for h). Then, ˜h = h in Q+, and by the continuity and
uniqueness of analytic continuation (can be seen e.g. in ([20], p. 298-299)) it follows that
h(j(z)) = 1
h(z), for z ∈ Q.
We now define λ : Π+→ C by
λ(z) = h(φ−1(z)) where φ ∈ Γ(2) and z ∈ φ(Q).
Combining Theorem 4.10 with the properties of h shown above yields that λ(Π+) = λ(Q) = C \ {0, 1}, λ ◦ φ = φ for all φ ∈ Γ(2) and that λ is one-to-one. This concludes
the proof of part (i) and (ii) for the conformal map case. (So far we have only proved for the conformal map case and we need to show that λ is a covering map)
Part (iii), if Im(z) > 1, then there exists k ∈ Z such that τk(z) ∈ Q. Let (zn) be
sequences in Q such that ˜zn = τkzn and ˜zn ∈ Q where Im(zn) > 1 . Thus, this leads to
Primarily, we need to show that λ is a holomorphic covering map. To begin with, let us show that λ is holomorphic in (Q ∪ τ−1(Q))◦. Suppose that −1 − x + iy ∈ τ−1(Q) and −1 − x > −3. Since τ (−1 − x + iy) = 1 − x + iy ∈ Q, we see that
λ(−1 − x + iy) = λ(1 − x + iy) = h(1 − x + iy) = h(−1 + x + iy).
Once again, Theorem 3.17 implies that λ is holomorphic in (Q ∪ τ−1(Q))◦ at any point in z with Im(z) = −1. Similarly, λ is also holomorphic in (Q ∪ σ−1(Q))◦, since Γ(2) is freely generated by τ and σ according to Lemma 4.1.
Notice that, Γ(2) is a normal subgroup of Γ, and there is a function λ ◦ j that is invariant under Γ(2). If φ ∈ Γ(2), then there exists ψ ∈ Γ(2) such that j ◦ φ = ψ ◦ j,
λ ◦ j ◦ φ = λ ◦ ψ ◦ j = λ ◦ j.
But we already know that λ ◦ j is invariant in Q, since both sides are invariant under Γ(2). It follows that λ ◦ j = 1/λ in all of Π+. Thus,
j(Q ∪ τ−1(Q)) = j(Q) ∪ j(τ−1(Q)) = j(Q) ∪ σ−1(j(Q)) = Q ∪ σ−1(Q)
shows that λ ◦ j and λ are holomorphic in (Q ∪ σ−1(Q))◦, since λ ◦ j = 1/λ. In addition,
λ is holomorphic in Ω = (Q ∪ τ−1(Q))◦∪ (Q ∪ σ−1(Q))◦ and with the fact that λ is
invariant under Γ(2) shows the holomorphism of λ inS
φ∈Γ(2)φ(Ω).
Let us define
Q \ T◦= {−1 + iy : y > 0},
this gives that Q \ T◦⊂ (Q ∪ τ−1(Q))◦. Again, if we apply j(z) = −1/z, then Q \ S◦⊂
(Q ∪ σ−1Q)◦ and Q◦ = T◦∩ S◦, this shows that
Q \ Q◦= (Q \ T◦) ∪ (Q \ S◦) ⊂ (Q ∪ τ−1(Q))◦∪ (Q ∪ σ−1(Q))◦= Ω. Since we have Q◦ ⊂ Ω, it follows that Q ⊂ Ω, which would give Π+ =S
φ∈Γ(2)φ(Q) ⊂
S
φ∈Γ(2)φ(Ω), so that λ is holomorpic in Π
+.
Now, we define an open set N such that N = λ(Q◦) with λ holomorphic inS
φ∈Γ(2)φ(Ω),
then
λ−1(N ) = [
φ∈Γ(2)
φ(Q◦). (5.1)
This indicates that λ−1 is the union collection of disjoint open sets Oφ, such that λ
restricted to any Oφ and homeomorphic onto N . Conclusively, N has an elementary
neighborhood of each of its points. We have handled most points of C \ {0, 1}, and the next step here is to show that every points of λ(Q) = C \ {0, 1} has an elementary neighborhood.
Suppose there is a g ∈ Γ with Q0 = g(Q) and φ0∈ Γ(2) is not the identity, then there
exists ψ ∈ Γ(2) such that φ0◦ g = g ◦ ψ. It holds that φ0(Q0) = φ0(g(Q)) = g(ψ(Q)).
Since g is a bijection and Q ∩ ψ(Q) = ∅, it follows that Q0∩ φ0(Q
0
) = ∅. Hence, φ(Q0) and φ ∈ Γ(2) are pairwise disjoint and together with (5.1), it shows that N0 = λ((Q0)◦), thus, N0 is elementary for λ.
Let define functions in Γ, such that g1 ∈ Γ by g1(z) = z − 1 and Q1 = g(Q). Also
