Jørgensen Lemma

Fabian Cejka Eingereicht bei Prof. Dr. Anna Wienhard an der Universität Heidelberg Betreuer: Prof. Dr. Anna Wienhard, Dr. Gye-Seon Lee

BACHELORARBEIT

im Juni 2016 Declaration

I hereby declare and confirm that this thesis is entirely the result ofmy own original work. Where other sources of information have been used, they have been indicated as such and properly acknowledged. I further declare that this or similar work has not been submitted for credit elsewhere.

Heidelberg, June 13, 2016

i Contents

Declarationi

Abstract iii

1 Introduction1 1.1 Goals and Structure of this thesis...... 1 1.2 Background...... 1 1.3 Classification of Möbius transformations of H ...... 2 1.4 Fuchsian Groups...... 7

2 Jørgensen Inequality 12 2.1 Proof of Jørgensen Inequality...... 12 2.2 A criterion for discreteness...... 15 2.3 Extreme Fuchsian groups...... 16

References 21 Literature...... 21

ii Abstract

In this thesis we will first do some classification of the elements in 푃 푆퐿(2, R). After that we will introduce the notion of Fuchsian groups, i. e. a discrete subgroup of 푃 푆퐿(2, R), and prove two criteria for discreteness in 푃 푆퐿(2, R). One of them is the Jørgensen Inequality, which is the main theorem in this thesis. Finally we look at the special case of equality in the Jørgensen inequality.

iii Chapter 1

Introduction

1.1 Goals and Structure of this thesis

Our first goal in this thesis will be the following Theorem in Chapter 2.1:

Theorem (Jørgensen Inequality). Suppose that 푇, 푆 ∈ PSL(2, R) and < 푇, 푆 > is a non-elementary Fuchsian group. Then

| tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| ≥ 1. (1.1)

The lower bound is best possible.

In order to prove that, we have to distinguish between three types of possible elements in PSL(2, R). In Chapter 1.3 we do this classification of PSL(2, R). After that, in Chapter 1.4 we introduce the notion of Fuchsian groups, i.e. discrete groups in PSL(2, R). We also prove some properties of Fuchsian groups, for example Theorem5 which gives us a complete charac- terisation of elementary Fuchsian groups, which are Fuchsian groups with a finite orbit when acting on the upper half-plane H. As from Chapter 2, we will only consider non-elementary Fuchsian groups, which occur in the Jørgensen Inequality. After that we will be able to prove our next criterion for discreteness in PSL(2, R) in Chapter 2.2:

Theorem. A non-elementary subgroup Γ of PSL(2, R) is discrete if and only if, for each 푇 and 푆 in Γ, the group < 푇, 푆 > is discrete.

Finally, in Chapter 2.3, we find that we have equality in(1.1) if and only if < 푇, 푆 > is a triangle group of order (2, 3, 푝) with 푝 ∈ {7, 8, 9, ..., ∞}.

1.2 Background

We start with the Classification of Möbius transformations of H. For basic introduction of the hyperbolic plane in the upper half-plane H := {푧 ∈

1 1. Introduction 2

C| Im(푧) > 0} with boundary 휕H = R ∪ {∞} and the unit disk model in the unit disk D := {푧 ∈ C||푧| < 1} with boundary 휕D = {푧 ∈ C||푧| = 1} see Walkden’s script [6] or Katok’s book [4]. Definition 1. The set of fractional linear (or Möbius) transforma- tions of H is defined as follows 푎푧 + 푏 PSL(2, R) = {H → H, 푧 → |푎푑 − 푏푐 = 1, 푎, 푏, 푐, 푑 ∈ R}. 푐푧 + 푑 For 푇 ∈ PSL(2, R) we define the trace of T tr(푇 ) = |푎+푑| and set 푇 푟(푇 ) := |푡푟(푡)|. We call 푧 ∈ H˜ = H ∪ 휕H a fixed point if 푇 (푧) = 푧.

1.3 Classification of Möbius transformations of H In what follows the aim is to classify the types of behaviour that Möbius transformations of H exhibit. We will see that there are three different classes of Möbius transformation of H. We start with some Möbius transformation of H called 푇 . Our initial classification of Möbius transformations of H is based on how many fixed points a given Möbius transformation of H has, and whether they lie in H or in 휕H. Clearly the identity map is a Möbius transformation of H which fixes every point. As from now, we will assume that 푇 is not the identity. Let us first consider the case when ∞ ∈ 휕H is a fixed point. As

푎푧 + 푏 푎 + 푏 푇 (푧) = = 푧 푐푧 푑 푑 + 푐 + 푧 1 푎 we note that as 푧 → ∞ we have 푧 → 0 and by that 푇 (∞) = 푐 . Thus ∞ is a fixed point of 푇 if and only if 푇 (∞) = ∞, and this happens if and only if 푐 = 0. Suppose that ∞ is a fixed point of 푇 so that 푐 = 0. What other fixed points 푧0 can 푇 have? Observe that now 푎 푏 푇 (푧 ) = 푧 + . 0 푑 0 푑 푏 Hence 푇 also has a fixed point at 푧0 = 푑−푎 . Note that if 푎 = 푑 then this point may be ∞. Thus if ∞ ∈ 휕H is a fixed point for 푇 then 푇 has at most one other fixed point, and this fixed point also lies on 휕H. Now let us consider the case when ∞ is not a fixed point of 푇 . In this case 푐 ≠ 0. Multiplying

푎푧0 + 푏 푇 (푧0) = = 푧0 푐푧0 + 푑 1. Introduction 3

푑 by 푐푧0 + 푑 (which is non-zero as 푧0 ≠ − 푐 ) we see that 푧0 is a fixed point if and only if 2 푐푧0 + (푑 − 푎)푧0 − 푏 = 0.

This is a quadratic in 푧0 with real coefficients. Hence there are either one or two real solutions, or two complex conjugate solutions. In the latter case, only one solution lies in H ∪ 휕H. Thus, we have proved:

Lemma 1. Let 푇 be a Möbius transformation of H and suppose that 푇 is not the identity. Then 푇 has either: 1. two fixed points in 휕H and none in H. 2. one fixed point in 휕H and none in H. 3. no fixed points in 휕H and one in H. In the first case we call 푇 hyperbolic, in the second parabolic and in the third elliptic.

Corollary 1. Suppose 푇 is a Möbius transformation of H with three or more fixed points. Then 푇 is the identity (and so fixes every point).

Now we go on in our classification by looking at the absolute value Tr of the trace of some Möbius transformation of H called 푇 . Suppose for simplicity that ∞ is not a fixed point (it follows that 푐 ≠ 0). So we know 푧0 is a fixed point of 푇 if and only if √︀ 푎 − 푑 ± (푎 − 푑)2 + 4푏푐 푧0 = . 2푐 Using

푎푑 − 푏푐 = 1, (푎 + 푑)2 = Tr2(푇 ) it is easy to see that

(푎 − 푑)2 + 4푏푐 = Tr2(푇 ) − 4.

When 푐 = 0, we must have that ∞ is a fixed point. The other fixed point 푏 is given by 푑−푎 . Hence ∞ is the only fixed point if 푎 = 푑 (in which case we must have that 푎 = 1, 푑 = 1 or 푎 = −1, 푑 = −1 as 푎푑 − 푏푐 = 푎푑 = 1); hence Tr(푇 ) = | ± (1 + 1)| = 2. If 푎 ≠ 푑 then there are two fixed points on 휕H if Tr(푇 ) > 2 and one fixed point in H if Tr(푇 ) < 2. Thus, we have proved:

Lemma 2. Let 푇 be a Möbius transformation of H and suppose that 푇 is not the identity. Then: • T is hyperbolic if and only if Tr(푇 ) > 2. • T is parabolic if and only if Tr(푇 ) = 2. 1. Introduction 4

• T is elliptic if and only if Tr(푇 ) < 2.

Definition 2. Let 푇1, 푇2 be two Möbius transformations of H. We say that 푇1 and 푇2 are conjugate if there exists another Möbius transformation of −1 H called 푆 such that 푇1 = 푆 ∘ 푇2 ∘ 푆.

Remark 1. • Conjugacy between Möbius transformations of H is an equiv- alence relation. • It is easy to see that if 푇1 and 푇2 are conjugate then they have the same number of fixed points, hence they are of the same type of trans- formation (elliptic, parabolic, hyperbolic). • If 푇2 has matrix 퐴2 ∈ 푆퐿(2, R) and 푆 has matrix 퐴 ∈ 푆퐿(2, R) then −1 푇1 has matrix ±퐴 퐴2퐴. • Geometrically, if 푇1 and 푇2 are conjugate then the action of 푇1 on H ∪ 휕H is the same as the action of 푇2 on 푆(H ∪ 휕H). Thus conjugacy reflects a change in coordinates of H ∪ 휕H.

Lemma 3. Let 푇 be a Möbius transformation of H and suppose that 푇 is not the identity. Then the following are equivalent: (i) 푇 is parabolic. (ii) Tr(푇 ) = 2. (iii) 푇 is conjugate to a translation, i.e. 푇 is conjugate to a Möbius trans- formation of H of the form 푧 ↦→ 푧 + 푏 for some 푏 ∈ R. (iv) 푇 is conjugate either to the translation 푧 ↦→ 푧 + 1 or to the translation 푧 ↦→ 푧 − 1. (︂1 1)︂ (︂1 −1)︂ (v) The matrix of 푇 is conjugate to or . 0 1 0 1 Proof. We already know that (푖) and (푖푖) are equivalent. Clearly (푖푣) implies (푖푖푖) and (푖푣) and (푣) are equivalent. 푧 In order to prove that (푖푖푖) implies (푖푣) we have to choose 푆(푧) = 푏 in Definition2 and then 푧 ↦→ 푧 + 푏 is conjugate to 푧 ↦→ 푧 + 1 or 푧 ↦→ 푧 − 1 if 푏 > 0 or 푏 < 0. Notice that 푆 actually√ is an element of 푃 푆퐿(2, R) since 푆 푧 √ 푧√ 푎푑 − 푏푐 √1 · 푏 − ( ) = 푏· 푏 and thus = 푏 0 = 1. Suppose now that (푖푣) holds. Recall that 푧 ↦→ 푧 + 1 has a unique fixed point at ∞. Hence if 푇 is conjugate to 푧 ↦→ 푧 + 1 then 푇 has a unique fixed point in 휕H, and is therefore parabolic. The same argument holds for 푧 ↦→ 푧 − 1. Finally we show that (푖) implies (푖푖푖). Suppose that 푇 is parabolic and has a unique fixed point at 휁 ∈ 휕H. Let 푆 be a Möbius transformation of −1 H that maps 휁 to ∞. Then 푆푇 푆 is a Möbius transformation of H with a unique fixed point at ∞. We claim that 푆푇 푆−1 is a translation. Write

푎푧 + 푏 푆푇 푆−1(푧) = . 푐푧 + 푑 1. Introduction 5

As ∞ is a fixed point of 푆푇 푆−1, we must have that 푐 = 0. Hence 푎 푏 푆푇 푆−1(푧) = 푧 + 푑 푑 −1 푏 −1 and it follows that 푆푇 푆 has a fixed point at 푑−푎 . As 푆푇 푆 has only one fixed point and the fixed point isat ∞ we must have that 푑 = 푎. Thus −1 ′ ′ 푆푇 푆 (푧) = 푧+푏 for some 푏 ∈ R. Hence 푇 is conjugate to a translation.

Lemma 4. Let 푇 be a Möbius transformation of H and suppose that 푇 is not the identity. Then the following are equivalent: (i) 푇 is hyperbolic. (ii) Tr(푇 ) > 2. (iii) 푇 is conjugate to a dilation, i.e. 푇 is conjugate to a Möbius transfor- mation of H of the form 푧 ↦→ 푘푧, for some 푘 > 0. (︂푢 0)︂ (iv) The matrix of 푇 is conjugate to 1 , for some 푢 ∈ R. 0 푢 Proof. We have already seen that (푖) is equivalent to (푖푖). Obviously (푖푖푖) is equivalent to (푖푣) with 푢2 = 푘. Suppose (푖푖푖) holds. Then 푇 is conjugate to a dilation which has 0 and ∞ as fixed points in 휕H, namely 0 and ∞. Hence 푇 also has exactly two fixed points in 휕H. Hence (푖) holds. Finally, we prove that (푖) implies (푖푖푖). We first make the remark that if 푇 fixes both 0 and ∞ then 푇 is a dilation. To see this, write

푎푧 + 푏 푇 (푧) = 푐푧 + 푑 where 푎푑 − 푏푐 = 1. As ∞ is a fixed point of 푇 , we must have that 푐 = 0. 푎푧+푏 푎 Hence 푇 (푧) = 푑 . As 0 is fixed, we must have that 푏 = 0. Hence 푇 (푧) = 푑 푧 so that 푇 is a dilation. Suppose that 푇 is a hyperbolic Möbius transformation of H. Then 푇 has two fixed points in 휕H; denote them by 휁1, 휁2. First suppose that 휁1 = ∞ and 휁2 ∈ R. Let 푆(푧) = 푧 − 휁2. Then the transformation 푆푇 푆−1 is conjugate to 푇 and has fixed points at 0 and ∞. By that above remark 푆푇 푆−1 is a dilation. Now suppose that 휁1 ∈ R and 휁2 ∈ R. We may assume that 휁1 < 휁2. Let 푆 be the transformation 푧 − 휁2 푆(푧) = . 푧 − 휁1

As −휁1 +휁2 > 0, this is a Möbius transformation of H. Moreover, as 푆(휁1) = −1 ∞ and 푆(휁2) = 0, we see that 푆푇 푆 has fixed points at 0 and ∞ and is therefore a dilation. Hence 푇 is conjugate to a dilation. 1. Introduction 6

Sometimes it will be more convenient to look at elliptic Möbius trans- formations of H in the unit disk model. We know PSL(2, R) ≃ Aut(H) = −1 푧−푖 푤 Aut(D)푤 where 푤(푧) = 푧+푖 is the map which maps H bijectively to D and 휕H bijectively to 휕D (but 푤 is not a Möbius transformation of H). See Walkden’s Script [6] for details. So the corresponding Möbius transforma- tion of H in the unit disk model of some Möbius transformation of H called 훾 in the upper half plane is given by

푧 ↦→ 푤훾푤−1. (1.2)

By that, it is possible to calculate that a Möbius transformation of H in D is a map of the form

훼푧 + 훽 2 2 푧 ↦→ , 훼, 훽 ∈ C, |훼| − |훽| = 1. 훽푧¯ +훼 ¯ Since 푤 is bijective, one can classify PSL(2, R) in D exactly the same as in H and a transformation 푇 of D is hyperbolic for example if and only if 푇 has two fixed points in 휕D or if and only if Tr(푇 ) > 2. Lemma 5. Let 푇 be a Möbius transformation of H and suppose that 푇 is not the identity. Then the following are equivalent: (i) 푇 is elliptic. (ii) Tr(푇 ) < 2. cos(휃)푧+sin(휃) (iii) 푇 is conjugate in H to a rotation 푧 ↦→ − sin(휃)푧+cos(휃) . (iv) 푇 is conjugate in D to a rotation 푧 ↦→ exp(푖휃)푧. (︂ cos(휃) sin(휃))︂ (v) The matrix of 푇 in is conjugate to . H − sin(휃) cos(휃) (︂푢 0)︂ (vi) The matrix of 푇 in D is conjugate to 1 with |푢| = 1 and 푢 ≠ 1. 0 푢 Proof. We have already seen that (푖) is equivalent to (푖푖). Again it is obvious, that (푖푖푖) and (푣) are equivalent and (푖푣) and (푣푖) are equivalent (푢2 = exp(푖휃)). (푖푖푖) and (푖푣) is equivalent. To see that we need again the fact (1.2). See Theorem 8.19 of [5] for detailed calculation. Suppose that (푖푣) holds. A rotation has a unique fixed point (at the origin). If 푇 is conjugate to a rotation then it must also have a unique fixed point, and so is elliptic. Finally, we prove that (푖) implies (푖푣). Suppose that 푇 is elliptic and has a unique fixed point at 휁 ∈ D. Let 푆 be a Möbius transformation of D that −1 maps 휁 to the origin 0. Then 푆푇 푆 is a Möbius transformation of H that is conjugate to 푇 and has a unique fixed point at 0. Suppose that

훼푧 + 훽 푆푇 푆−1(푧) = 훽푧¯ − 훼¯ 1. Introduction 7 where |훼|2 − |훽|2 > 0. As 0 is a fixed point, we must have that 훽 = 0. Write 훼 in polar form as 훼 = 푟 exp(푖휃). Then

훼 푟 exp(푖휃) 푆푇 푆−1(푧) = 푧 = 푧 = exp(2푖휃)푧 훼¯ 푟 exp(−푖휃) so that 푇 is conjugate to a rotation.

1.4 Fuchsian Groups

Besides being a group, PSL(2, R) is also a topological space in which a 푎푧+푏 4 transformation 푧 ↦→ 푐푧+푑 can be identified with the point (푎, 푏, 푐, 푑) ∈ R . More precisely, as a topological space, 푆퐿(2, R) can be identified with the 4 subset of R

4 푋 = {(푎, 푏, 푐, 푑) ∈ R |푎푑 − 푏푐 = 1}. We define 훿 : 푋 → 푋, (푎, 푏, 푐, 푑) ↦→ (−푎, −푏, −푐, −푑) and topologize

푃 푆퐿(2, R) ≃ 푋/{푖푑, 훿} where 푖푑, 훿 is a cyclic group of order 2 acting on 푋. One can prove that the group multiplication and taking of inverse are actually continuous with respect to the topology on PSL(2, R). A norm on PSL(2, R) is induced from 4 푎푧+푏 R : for 푇 (푧) = 푐푧+푑 in PSL(2, R), we define

2 2 2 2 1 ‖푇 ‖ = (푎 + 푏 + 푐 + 푑 ) 2 .

Hence PSL(2, R) is a topological group with respect to the metric 푑(푇, 푆) := ‖푇 − 푆‖ for 푇, 푆 ∈ PSL(2, R). Definition 3. A set 푆 in a topological space 푋 is called discrete if every point 푥 ∈ 푆 has a neighbourhood 푈 such that 푆 ∩ 푈 = {푥}.

Definition 4. A discrete subgroup of PSL(2, R) is called Fuchsian group.

Examples. (i) The subgroup of integer translations {훾푛(푧) = 푧 +푛|푛 ∈ Z} is a Fuchsian group. For example here, for 푛, 푚 ∈ Z we have

1 (︀ 2 2 2 2)︀ 2 푑(훾푛, 훾푚) = ‖훾푛 − 훾푚‖ = (1 − 1) + (푛 − 푚) + (0 − 0) + (1 − 1) = 푛 − 푚.

The subgroup of all translations {훾푏(푧) = 푧+푏|푏 ∈ R} is not a Fuchsian group, as it is not discrete. (ii) Any finite subgroup of PSL(2, R) is a Fuchsian group. This is because any finite subset of any metric space is discrete. 1. Introduction 8

(iii) As a specific example, let

cos(휃)푧 + sin(휃) 훾휃(푧) = − sin(휃)푧 + cos(휃)

be a rotation around 푖. Let 푞 ∈ N. Then {훾 휋푗 |0 ≤ 푗 ≤ 푞 − 1} is a finite 푞 subgroup of PSL(2, R). (iv) The modular group 푃 푆퐿(2, Z) is Fuchsian. This is the group given by Möbius transformation of Hs of the form 푎푧 + 푏 훾(푧) = , 푎, 푏, 푐, 푑 ∈ Z, 푎푑 − 푏푐 = 1. 푐푧 + 푑

(v) Let 푞 ∈ N. Define 푎푧 + 푏 Γ푞 = {훾(푧) = |푎, 푏, 푐, 푑 ∈ Z, 푎푑 − 푏푐 = 1, 푏, 푐 are divisible by 푞}. 푐푧 + 푑 This is called the level q modular group which is also a Fuchsian group.

Definition 5. A group 퐺 is called cyclic if

퐺 =< 푔 >= {푔푛|푛 is an integer} for some 푔 ∈ 퐺.

Definition 6. Let 푋 be a metric space, and let 퐺 be a group of homeomor- phisms of 푋. For 푥 ∈ 푋, we call

퐺푥 = {푔(푥)|푔 ∈ 퐺} the G-orbit of the point x.

Definition 7. A subgroup Γ of PSL(2, R) is called elementary if there exists a finite Γ-orbit in H˜ := H ∪ 휕H. Definition 8. Let 퐺 be a group. The centralizer of 푔 ∈ 퐺 is defined by

퐶퐺(푔) = {푔 ∈ 퐺|ℎ푔 = 푔ℎ}.

Lemma 6. If 푆푇 = 푇 푆 then 푆 maps the fixed-point set of 푇 to itself.

Proof. Suppose 푇 (푝) = 푝 for some p. Then

푆(푝) = 푆푇 (푝) = 푇 푆(푝).

So 푇 fixes 푆(푝).

Lemma 7. (i) Any non-trivial discrete subgroup of R is infinite cyclic. (ii) Any discrete subgroup of 푆1 is finite cyclic. 1. Introduction 9

Proof. (i) Let Γ be a discrete subgroup of R. We have 0 ∈ Γ and since Γ is discrete there exists a smallest positive 푥 ∈ Γ. Then {푛푥|푛 ∈ Z} is a subgroup of Γ. Suppose there is a 푦 ∈ Γ, 푦 ≠ 푛푥. We may assume 푦 > 0, otherwise we take −푦 which also belongs to Γ. There exists an integer 푘 ≥ 0 such that 푘푥 < 푦 < (푘 + 1)푥. and 푦 − 푘푥 < 푥, and (푦 − 푘푥) ∈ Γ which contradicts the choice of 푥. (ii) Let Γ now be a discrete subgroup of 푆1. By discreteness there exists 푖휑 푧 = exp 0 ∈ Γ, with the smallest argument 휑0, and for some 푚 ∈ Z, 푚휑0 = 2휋, otherwise we get a contradiction with the choice of 휑0.

Theorem 1. Two non-identity elements of PSL(2, R) commute if and only if they have the same fixed-point set.

Proof. To prove that, let us look at the centralizer of parabolic, elliptic and hyperbolic elements in PSL(2, R). Suppose that 푇 (푧) = 푧 + 1. If 푆 ∈ 퐶PSL(2,R)(푇 ) then 푆(∞) = ∞. Therefore, 푆(푧) = 푎푧 + 푏. 푆푇 = 푇 푆 gives us 푎 = 1. Hence

퐶PSL(2,R)(푇 ) = {푧 ↦→ 푧 + 푘|푘 ∈ R}. The centralizer of an elliptic transformation of the unit disk D fixing 0 (i.e. 푧 ↦→ 푖휙 푧 푧 ↦→ 훼푧+훽 exp( ) ) consists of all transformations of the form 훽푧¯ +¯훼 . fixing 0, i.e. of the form 푧 ↦→ exp(푖휃)푧 (0 ≤ 휃 < 2휋). Let 푇 (푧) = 휆푧 (휆 > 0, 휆 ≠ 1) and 푆 ∈ 퐶PSL(2,R)(푇 ). After some direct calculation we find out that 푆 is given by a diagonal matrix and hence 푆(푧) = 훾푧 (훾 > 0).

Theorem 2. Let Γ be a Fuchsian group. If all non-identity elements of Γ have the same fixed-point set, then Γ is cyclic.

Proof. Suppose all elements of Γ are hyperbolic, so they have two fixed points in R∪{∞}. By choosing a conjugate group we may assume that each 푆 ∈ Γ fixes 0 and ∞. Thus Γ is a discrete subgroup of 퐻 = {푧 → 휆푧|휆 > 0} * * which is isomorphic to (R , ·). As a topological group (R , ·) is isomorphic to R via 푥 ↦→ ln 푥. By Lemma7, Γ is infinite cyclic. If Γ contains a parabolic element, then Γ is an infinite cyclic group containing only parabolic elements. Suppose Γ contains an elliptic element. In D, Γ is a discrete subgroup of orientation-preserving isometries of D. Again by choosing a conjugate group we may assume that all elements of Γ have 0 as a fixed point, and so all elements of Γ are of the form 푧 ↦→ exp푖휑 푧. Thus Γ is isomorphic to a subgroup of 푆1, and it is discrete if and only if the corresponding subgroup of 푆1 is discrete. The rest follows from Lemma7.

Theorem 3. Every Abelian Fuchsian group is cyclic.

Proof. By Theorem1, all non-identity elements in an Abelian Fuchsian group have the same fixed-point set. The theorem follows immediately from Theorem2. 1. Introduction 10

Theorem 4. Let Γ be a Fuchsian subgroup of PSL(2, R) containing besides the identity only elliptic elements. Then all elements of Γ have the same fixed point, and hence Γ is a cyclic group, Abelian and elementary. Proof. We shall prove that all elliptic elements in Γ must have the same fixed point. In the unit disk let us conjugate Γ in such a way that an element 푢 0 푎 푐 id ≠ 푔 ∈ Γ fixes 0, so 푔 = ( 0 푢 ). Let ℎ = ( 푐 푎 ) ∈ Γ, ℎ ≠ 푔. We have tr[푔, ℎ] = tr(푔 ∘ ℎ ∘ 푔−1 ∘ ℎ−1) = 2 + 4|푐|2(Im(푢))2. Since Γ does not contain hyperbolic elements, | tr[푔, ℎ]| ≤ 2. So either Im(푢) = 0 or 푐 = 0. If Im(푢) = 0 푎 0 then 푢 = 푢 and hence 푔 = id, a contradiction. Hence 푐 = 0, and so ℎ = ( 0 푎 ) also fixes 0. Thus all elements of Γ have the same fixed point. By Theorem 2, Γ is a cyclic group, and hence Abelian. The set {0} is a Γ-orbit, and so Γ is elementary.

So we have the obvious Corollary 2. Any Fuchsian group containing besides the identity only el- liptic elements is a finite cyclic group. We can now state a theorem which describes all elementary Fuchsian groups. Theorem 5. Any elementary Fuchsian group is either cyclic or is conjugate 1 in PSL(2, R) to a group generated by ℎ(푧) = − 푧 and 푔(푧) = 푘푧 for some 푘 > 1. Proof. Case 1. Suppose Γ fixes a single point 푎 ∈ H˜ = H ∪ 휕H. If 푎 ∈ H, then all elements of Γ are elliptic; by Corollary2, Γ is a finite cyclic group. Suppose 푎 ∈ R∪{∞}.Then Γ cannot have elliptic elements. We are going to show that hyperbolic and parabolic elements cannot have a common fixed point. Assume the opposite, and suppose this point is ∞, and 푔(푧) = 휆푧 for some 휆 > 1 and ℎ(푧) = 푧 + 푘 (since 푔 and ℎ have only one common point, 푘 ≠ 0). Then

푔−푛 ∘ ℎ ∘ 푔푛(푧) = 푧 + 휆−푛푘. As 휆 > 1 the sequence ‖푔−푛 ∘ ℎ ∘ 푔푛‖ is bounded and so, {푔−푛 ∘ ℎ ∘ 푔푛} contains a convergent subsequence of distinct terms which contradicts the discreteness of Γ. So, Γ must contain only elements of one type. If Γ only contains only parabolic elements, we know from Theorem2 it is an infinite cyclic group. Now consider the case in which Γ contains only hyperbolic ele- ments. We are going to prove that the second fixed point of these hyperbolic elements must also coincide, and so Γ will fix two points in R∪{∞}. Suppose 2 푎푧+푏 푓(푧) = 휆 푧 where 휆 > 1 so it fixes 0 and ∞ and suppose 푔(푧) = 푐푧+푑 which 1 fixes 0 but not ∞. Then 푏 = 0, 푐 ≠ 0, 푎 ≠ 0 and 푑 = 푎 . Then (︂1 0)︂ [푓, 푔] = 푓 ∘ 푔 ∘ 푓 −1 ∘ 푔−1 = 푡 1 1. Introduction 11

푐 (︀ 1 )︀ with 푡 = 푎 휆2 − 1 . Since 푐 ≠ 0, [푓, 푔] is a parabolic element in Γ, a contra- diction. Case 2. Suppose Γ has an orbit in R ∪ {∞} consisting of two points. An element of Γ either fixes each of them or interchanges them. A parabolic element cannot fix two points. Since each orbit (except for a single point ofa parabolic transformation) is infinite, a parabolic element cannot interchange these points; hence Γ does not contain any parabolic elements. All hyperbolic elements must have the same fixed point set. If Γ contains only hyperbolic elements, then it is cyclic by Theorem3. If it contains only elliptic elements, it is finite cyclic by Corollary2. If Γ contains both hyperbolic and elliptic elements, it must contain an elliptic element of order 2 interchanging the common fixed points of the hyperbolic elements; and then Γ is conjugate to 1 a group generated by 푔(푧) = 푘푧 (푘 > 1) and ℎ(푧) = − 푧 . Case 3. Suppose now Γ has an orbit in H consisting of 푘 = 2 points or an orbit in H˜ consisting of 푘 ≥ 3 points. Γ must contain only elliptic elements, since the parabolic and hyperbolic elements can have only either fixed points at infinite or infinite orbits. So Γ is a finite cyclic group and it 2휋푖 is conjugate to a group generated by 푧 → exp 푘 푧. Chapter 2

Jørgensen Inequality

2.1 Proof of Jørgensen Inequality

Theorem 6. A non-elementary subgroup Γ of PSL(2, R) must contain a hyperbolic element. Proof. Suppose Γ does not contain hyperbolic elements. If Γ contains only elliptic elements (and id), then by Theorem4 it is elementary. Hence Γ contains a parabolic element, say 푓(푧) = 푧 + 1, which fixes ∞. Let 푔(푧) := 푎푧+푏 푛 (푎+푛푐)푧+(푏+푛푑) 푐푧+푑 be an element in Γ. Then 푓 ∘ 푔 = 푐푧+푑 . So we have

tr2(푓 푛 ∘ 푔) = (푎 + 푑 + 푛푐)2.

Since all elements in the group are either elliptic or parabolic, we have 0 ≤ (푎 + 푑 + 푛푐)2 ≤ 4 for all 푛, so 푐 = 0. But then 푔 fixes ∞ as well, so that ∞ is fixed by all elements in Γ; hence Γ is elementary, a contradiction.

Let < 푇, 푆 > be the group generated by Möbius transformations of H ∏︀푟 푙푛 푙푚 called 푇 and 푆. So < 푇, 푆 >= { 푛,푚=1 푇 푆 | 푙푛, 푙푚, 푟 ∈ N}. The Jørgensen Inequality which follows now, states that if a discrete group generated by two elements in PSL(2, R) is non-elementary, then at least one of this elements must differ considerably from the identity.

Theorem 7 (Jørgensen Inequality). Suppose that 푇, 푆 ∈ PSL(2, R) and < 푇, 푆 > is a discrete non-elementary group. Then

| tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| ≥ 1. (2.1)

The lower bound is best possible. Before we are able to prove that, we need three Lemmas:

Lemma 8. Suppose 푇, 푆 ∈ PSL(2, R) and 푇 ≠ id. Define 푆0 = 푆, 푆1 = −1 −1 푆0 ∘ 푇 ∘ 푆0 , ..., 푆푟 ∘ 푇 ∘ 푆푟 , ... . If, for some 푛, 푆푛 = 푇 , then < 푇, 푆 > is elementary and 푆2 = 푇 .

12 2. Jørgensen Inequality 13

Proof. Suppose the case where 푇 has one fixed point 훼, so 푇 is either parabolic or elliptic. 푆푟 has one fixed point as well, since it is conjugate to 푇 . Because of

−1 푆푟+1 ∘ 푆푟(훼) = 푆푟 ∘ 푇 ∘ 푆푟 ∘ 푆푟(훼) = 푆푟(훼) we have that 푆푟+1 fixes 푆푟(훼). So 푆푟+1 fixes the same point as 푆푟. Now by the fact that 푆푛(= 푇 ) fixes 훼 and that each 푆푟 has one fixed point it follows that 푆푟 fixes 훼 for every 푟 ≥ 0. As a consequence we know that all elements in < 푇, 푆 > fix 훼. We conclude that all elements in < 푇, 푆 > are parabolic and by Theorem6. If 푇 is elliptic, all elements in < 푇, 푆 > are elliptic and by Theorem4 < 푇, 푆 > is elementary. Suppose now that 푇 has exactly two fixed points. We may assume then that 푇 (푧) = 푘푧. With he same argument as above 푆1, ..., 푆푛 have exactly two fixed points and for 0 ≤ 푟 ≤ 푛 we have {푆푟(0), 푆푟(∞)} = {0, ∞}. Since 푆푟 (푟 ≥ 1) is conjugate to 푇 , it cannot interchange two points (all orbits of a hyperbolic transformation are infinite with the exception of two fixed points). Thus 푆1, ...푆푛 fix 0 and ∞, and both 푆 = 푆0 and 푇 leave the set {0, ∞} invariant. Therefore < 푇, 푆 > is elementary.

2 Lemma 9. If 푇 = id for some 푇 ∈ PSL(2, R), 푇 ≠ id then tr(푇 ) = 0 Proof. It is (︂ 푎2 + 푏푐 (푎 + 푑)푏)︂ (︂1 0)︂ 푇 2 = = . (푎 + 푑)푐 푑2 + 푏푐 0 1 So either 푎, 푑 = 0 and 푏, 푐 = 1 or 푎, 푑 = 1 and 푏, 푐 = 0. The second case is not possible since 푇 ≠ id. So tr(푇 ) = 푎 + 푑 = 0.

1 1 (︀ 푎 푏 )︀ Lemma 10. Let 푇 = ( 0 1 ), 푆 = 푐 푑 be two matrices in 푆퐿(2, R). Prove that the Jørgensen inequality for 푇 and 푆 holds if and only if |푐| ≥ 1. Proof.

| tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| = 0 + |2(푎푑 − 푏푐) + 푐2 − 2| = 푐2

So (2.4) is hold if and only if |푐| ≥ 1.

Proof of Theorem7. We know that < 푇, 푆 > is discrete and non-elementary. Now (2.4) holds if 푇 is of order two (because then we know from Lemma9, tr2(푇 ) = 0) so we may assume that if 푇 is not of order two. We define

−1 푆0 = 푆, 푆푛+1 = 푆푛푇 푆푛 . (2.2)

By Lemma8 we know 푆푛 ≠ 푇 for any 푛. It remains only to show that if (2.4) fails, then for some 푛 we have

푆푛 = 푇 (2.3) 2. Jørgensen Inequality 14 and we consider two cases. Case 1. 푇 is parabolic. We first assume that (︂1 1)︂ (︂푎 푏)︂ 푇 = , 푆 = 0 1 푐 푑 where 푐 ≠ 0 (else < 푇, 푆 > was elementary). We are assuming that (2.4) fails and this is by Lemma 10 the assumption that

|푐| < 1.

The relation (2.2) yields

(︂ )︂ (︂ )︂ (︂ )︂ (︂ )︂ (︂ 2 )︂ 푎푛+1 푏푛+1 푎푛 푏푛 1 1 푑푛 −푏푛 1 − 푎푛푐푛 푎푛 = = 2 푐푛+1 푑푛+1 푐푛 푑푛 0 1 −푐푛 푎푛 −푐푛 1 + 푎푛푐푛

2푛 2푛 So by induction 푐푛 = −(−푐) = −푐 for 푛 > 0 and thus 푐푛 → 0 as |푐| < 1. Since we have |푐푛| < 1, by induction we see that 푎푛 ≤ 푛 + |푎0|, so 푎푛푐푛 → 0 and 푎푛+1 → 1. This proves that

푆푛+1 → 푇 which, by discreteness, yields (2.3) for large 푛. (︀ 1 −1 )︀ (︀ 푎 푏 )︀ Actually we have to consider the case 푇 = 0 1 and 푆 = 푐 푑 . But this works completely similar and yields the same result. Case 2. 푇 is hyperbolic or elliptic. Without loss of generality,

(︂푢 0)︂ 푇 = 1 . 0 푢

In the hyperbolic case, 푇 is the matrix for the transformation in H, in the elliptic case, 푇 is the matrix for the transformation in D. 푆 is as in Case 1 and 푏푐 ≠ 0 (else < 푇, 푆 > is elementary). The assumption that (2.4) fails is 1 휇 := | tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| = (1 + |푏푐|)|푢 − |2 < 1. 푢 (︁ )︁ 푆 푎푛 푏푛 푆 푆 ∘ 푇 ∘ 푆−1 Again we write 푛 = 푐푛 푑푛 and obtain from 푛+1 = 푛 푛

(︂ )︂ (︂ 푏푛푐푛 (︀ 1 )︀)︂ 푎푛+1 푏푛+1 푎푛푑푛푢 − 푢 푎푛푏푛 푢 − 푢 = (︀ 1 )︀ 푎푛푑푛 푐푛+1 푑푛+1 푐푛푑푛 푢 − 푢 푢 − 푏푛푐푛푢

1 2 so 푏푛+1푐푛+1 = −푏푛푐푛(1 + 푏푛푐푛)(푢 − 푢 ) . By induction

푛 |푏푛푐푛| ≤ 휇 |푏푐| ≤ |푏푐|. 2. Jørgensen Inequality 15

So 푏푛푐푛 → 0 and 푎푛푑푛 = 1 + 푏푛푐푛 → 1. Also, we obtain 푎푛+1 → 푢 and 1 푑푛+1 → 푢 . We have (︂ )︂ 푏푛+1 1 1 | | = |푎푛 − 푢 | ≤ 휇 2 |푢|. 푏푛 푢

1 푏푛+1 2 푏푛 푏푛 Thus | 푢푛+1 | < 휇 | 푢푛 | for some sufficiently large 푛. So 푢푛 → 0 and similarly 푛 푐푛푢 → 0. So (︂ 푏2푛 )︂ −푛 푛 푎2푛 푢2푛 푇 푆2푛푇 = 2푛 푐2푛푢 푑2푛 Since < 푇, 푆 > is discrete, for large 푛 we have

−푛 푛 푇 푆2푛푇 = 푇 and again we have 푆2푛 = 푇 . Finally, we show that the lower bound (2.4) is best possible. Consider 1 the group generated by 푇 (푧) = 푧 + 1 and 푆(푧) = − 푧 . It is < 푇, 푆 >= PSL(2, Z) (see Katok [4] Chapter 3.2, Example A) which is discrete and −1 −1 2푧+1 non-elementary. We have 푇 ∘ 푆 ∘ 푇 ∘ 푆 (푧) = 푧+1 with trace 3, and hence the equality holds in (2.4).

Remark 2. The Jørgensen Inequality also holds for non-elementary discrete groups in PSL(2, C).

2.2 A criterion for discreteness

In order to prove Theorem8 we need the following two general results.

Lemma 11. If Γ is elementary, for any 푇, 푆 ∈ Γ, < 푇, 푆 > is elementary.

Proof. Conversely, suppose Γ is not elementary. By Theorem6 it contains a hyperbolic element 푇 with fixed points 훼 and 훽. Since Γ is not elementary, there exists 푆 ∈ Γ which does not leave the set {훼, 훽} invariant. Hence < 푇, 푆 > is not elementary.

Lemma 12. Any non-elementary subgroup Γ of PSL(2, R) must contain infinitely many hyperbolic elements, no two of which have a common fixed point.

Proof. We choose a hyperbolic element 푇 in Γ with fixed points {훼, 훽} and an element 푆 in Γ which does not leave {훼, 훽} fixed. We can find such 푆 like in the proof of Lemma 11. Suppose first that the sets {훼, 훽} and −1 {푆(훼), 푆(훽)} do not intersect. In this case, the elements 푇 and 푇1 = 푆푇 푆 both are hyperbolic and have no common fixed point푆 ( (훼) and 푆(훽) are the 푛 −푛 fixed points of 푇1). The sequence {푇 푇1푇 } consists of hyperbolic elements with fixed points 푇 푛푆(훼) and 푇 푛푆(훽) which are pairwise different. 2. Jørgensen Inequality 16

If the set {훼, 훽} and {푆(훼), 푆(훽)} have one point of intersection, say 훼, then we wish to show that 푃 = [푇, 푇1] is parabolic with 훼 as the only fixed point. So we conjugate Γ so that the fixed points of 푇 are 0 and ∞, and the fixed point it shares with 푇1 is ∞. Then (︂푢 0)︂ (︂푎 푏)︂ 푇 = 1 , 푇1 = 0 푢 푐 푑 where 푢 > 1 and 푎, 푏, 푐, 푑 ∈ R. If 푇1 is to fix ∞, then 푐 = 0, and since 0 is (︀ 1 푏 )︀ not a fixed point, 푏 ≠ 0. This means 푇1 has the form 푇1 = 0 1 . We can then compute (︂1 푏(푢2 − 1))︂ 푃 = [푇, 푇 ] = 푇 푇 푇 −1푇 −1 = , 1 1 1 0 1 so that its transform is of the form 푇1(푧) = 푧 + 푡 for non-zero 푡, and other than ∞, every point is fixed. Since {훼} cannot be Γ-invariant there exists 푈 ∈ Γ not fixing 훼. So 푄 = 푈푃 푈 −1 is parabolic and does not fix 훼. Therefore 푄 and 푇 have no common fixed points. Then for large n, the elements 푇 and 푄푛푇 푄−푛 are hyperbolic and have no common fixed point, and the problem is reduced to the first case.

Theorem 8. A non-elementary subgroup Γ of PSL(2, R) is discrete if and only if, for each 푇 and 푆 in Γ, the group < 푇, 푆 > is discrete. Proof. If Γ is discrete, then every subgroup of it is also discrete. Suppose now that every subgroup < 푇, 푆 > in Γ is discrete, but Γ itself is not. So we find a sequence of distinct transformations in Γ, 푇1, 푇2, ..., such 2 that 푇푖 ≠ id and lim푛→∞ 푇푛 = id. From Lemma9 we know that 푇 = id implies tr(푇 ) = 0 and as tr is a continuous function on PSL(2, R), we may choose a subsequence that contains no elements of order 2. For any 푆 ∈ Γ we have 2 −1 −1 | tr (푇푛) − 4| + | tr(푇푛푆푇푛 푆 ) − 2| → 0 and so by Theorem7, there is some 푁 := 푁(푆) ∈ N such that for 푛 ≥ 푁, the group < 푇푛, 푆 > is elementary. By Lemma 12 we know that Γ contains two hyperbolic elements 푆1 and 푆2 with no common fixed points. So for 푛 ≥ max(푁(푆1), 푁(푆2)) both < 푇푛, 푆1 > and < 푇푛, 푆2 > are elementary, discrete and according to Theorem5, they leave the (distinct) fixed point pair of 푆1 and that of 푆2 invariant. Since 푇푛 is not elliptic of order 2, it cannot interchange a pair of points, so 푇푛 must fix four distinct points which implies 푇푛 = id which is a contradiction.

2.3 Extreme Fuchsian groups

In this section we will work with the notion of an Extreme Fuchsian group and a triangle group. We will write down the definitions first: 2. Jørgensen Inequality 17

Definition 9. Suppose that 푇, 푆 ∈ PSL(2, R) and < 푇, 푆 > is a discrete non-elementary group. Then < 푇, 푆 > is called extreme if | tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| = 1, (2.4) so if there is equality in the Jørgensen Inequality. Definition 10. A Fuchsian triangle group of signature (푛, 푚, 푝) with 푛, 푚, 푝 ∈ N ∪ {∞} is a subgroup of PSL(2, R) generated by three elliptic 1 1 1 elements 훿1, 훿2, 훿3 whith orders 푛, 푚, 푝 respectively and 푛 + 푚 + 푝 < 1. Our last result in this thesis Theorem9 was proved by Jørgensen himself in [3]. We need two lemmas and the following identity for 푇, 푆 ∈ PSL(2, R) tr2(푇 ) + tr2(푆) + tr2(푇 푆) = tr(푇 푆푇 −1푆−1) + tr(푇 ) tr(푆) tr(푇 푆) + 2 (2.5) which can be proven by calculation. Lemma 13. Suppose that 푇, 푆 ∈ PSL(2, R) and < 푇, 푆 > is a discrete non-elementary group and | tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| = 1. (2.6) −1 Then < 푇, 푆1 > is a discrete non-elementary group where 푆1 = 푆푇 푆 and 2 −1 −1 | tr (푇 ) − 4| + | tr(푇 푆1푇 푆1 ) − 2| = 1. (2.7)

Proof. The group generated by 푇 and 푆1 is discrete because it is a subgroup of the discrete group generated by 푆 and 푇 . Suppose 푇 is parabolic. Then the group generated by 푇 and 푆1 were elementary if and only if the fixed point of 푇 where fixed by 푆1 and hence by 푆. As the group generated by 푆 and 푇 is non-elementary, this is not so. If tr(푇 푆푇 −1푆−1) where equal to 2, then 푇 and 푆 would have a common fixed point. This is not so, the group generated by 푇 and 푆 being non-elementary. Thus | tr(푇 ) − 2| is strictly less than 1, by (2.6), and hence the order of 푇 exceeds 6. Therefore assuming now 푇 is elliptic, if the group generated by 푇 and 푆1 were elementary, then either 푆1 would keep the fixed points of 푇 fixed or 푆1 would interchange the fixed points of 푇 . In the former case we deduce that also 푆 would either keep the fixed points of 푇 fixed or 푆 would interchange the fixed points of 푇 . In the latter case, 푆 would have to interchange the fixed points of 푇 . None of the two cases can thus occur since the group generated by 푇 and 푆 is non-elementary. We have proved that the group < 푇, 푆1 > is non-elementary. −1 −1 To verify (2.6), we apply (2.5) to 푇 and 푆1 . Since the 푇 and 푆1 are conjugate we have tr(푇 ) = tr(푆1). After some elementary calculations we −1 −1 −1 −1 find out that tr(푇 푆1 푇 푆1) = tr(푇 푆1푇 푆1 ) and −1 −1 [︀ −1 −1 ]︀ [︀ −1 −1 2 ]︀ tr(푇 푆1푇 푆1 ) − 2 = tr(푇 푆푇 푆 ) − 2 tr(푇 푆푇 푆 ) − tr (푇 ) + 2 . (2.8) 2. Jørgensen Inequality 18

Because of (2.6) and the triangle inequality, we get from (2.8)

−1 −1 −1 −1 | tr(푇 푆1푇 푆1 ) − 2| ≤ | tr(푇 푆푇 푆 ) − 2|. (2.9) Applying (2.6) once more, we get from (2.9)

2 −1 −1 | tr (푇 ) − 4| + | tr(푇 푆1푇 푆1 ) − 2| ≤ 1.

But here equality must hold good since otherwise the Jørgensen inequality would be violated.

Lemma 14. Suppose that 푇, 푆 ∈ PSL(2, R) and < 푇, 푆 > is a discrete non-elementary group and

| tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| = 1.

Then 푇 is elliptic of order at least 7 or 푇 is parabolic. Furthermore, if 푇 is elliptic, then tr(푇 푆푇 푆−1) = 1. Proof. Suppose that 푇 is not parabolic. Then we have tr(푇 ) ≠ 2. Consider −1 푆1 = 푆푇 푆 . As a corollary to the proof of Lemma 13, we have

| tr(푇 푆푇 −1푆−1) − tr2(푇 ) + 2| = | tr(푇 푆푇 −1푆−1) − 2| + | tr2(푇 ) − 4|. this because (2.9) was seen to hold good with equality. Consequently, the ratio between tr(푇 푆푇 −1푆−1) and 4 − tr2(푇 ) must be a positive real num- ber. Notice here that tr(푇 푆푇 −1푆−1) ≠ 2 since 푇 and 푆 generate a non- elementary group. −1 Repeating the argument, now with 푆2 = 푆1푇 푆1 instead of 푇1, we de- −1 −1 2 duce that tr(푇 푆1푇 푆1 ) − 2 must be a positive multiple of 4 − tr (푇 ). By (2.8) we know that 4 − tr2(푇 ) is a positive real number. Because of (2.6) it is less than 1. Hence, 푇 is elliptic of order at least 7. Furthermore, we see that tr(푇 푆푇 −1푆−1) − 2 is a positive real number. Thus (2.6) may be written as

tr(푇 푆푇 −1푆−1) − tr2(푇 ) + 2 = 1 and since tr(푇 푆푇 −1푆−1) + tr(푇 푆푇 푆−1) = tr2(푇 ) (2.10) also the last assertion in Lemma 14 is proved.

Theorem 9. Suppose that 푇, 푆 ∈ PSL(2, R) and < 푇, 푆 > is a discrete non-elementary group. Then

| tr2(푇 ) − 4| + | tr(푇 푆푇 −1푆−1) − 2| = 1 if and only if < 푇, 푆 > is a triangle group of signature (2, 3, 푞) where 푞 ∈ {7, 8, 9, ..., ∞}. 2. Jørgensen Inequality 19

Proof. Consider the case in which 푇 is parabolic, then we get from (2.5)

tr(푇 푆푇 −1푆−1) − 2 = (tr(푆) ± tr(푇 푆))2. (2.11)

Together (2.6) and (2.11) give tr(푇 푆푇 −1푆−1) = 3. Using (2.10), we have tr(푇 푆푇 푆−1) = 1. So by Lemma 14, we have tr(푇 푆푇 푆−1) = 1 in any case, so 푇 푆푇 푆−1 is elliptic of order 3. As seen from the general identities (2.5) and (2.10), it means that

1 = tr2(푆) + tr2(푇 푆) − tr(푇 ) tr(푆) tr(푇 푆) (2.12) or, what is easily seen to be the same

1 = tr2(푆) − tr(푇 푆) tr(푆−1푇 ). (2.13)

−1 Consider the subgroup < 푇, 푆1 > of < 푇, 푆 > where 푆1 = 푆푇 푆 . By Lemma 13, we know, that

−1 −1 | tr(푇 ) − 2| + | tr(푇 푆1푇 푆1 | = 1.

* We may as well take 푇 and 푆 := 푇 푆1 (elliptic of order 3) as generators. Substituting 푆* for 푆 in (2.13), we obtain

* 0 = tr(푇 푆 tr(푆1)

* and since tr(푆1) = tr(푇 ) ≠ 1, we see that 푇 푆 is elliptic of order 2. 1 Example. Consider the group generated by 푇 (푧) = 푧+1 and 푆(푧) = − 푧 as in the proof of the Jørgensen Inequality. We already know < 푇, 푆 >= PSL(2, Z) is an extreme Fuchsian group. One can prove that PSL(2, Z) is a triangle group of signature (2, 3, ∞). Since 푇 is either elliptic of order at least 7 or parabolic, we have proved that < 푇, 푆1 > is one of the triangle groups spoken of in Theorems9. This is because being elliptic of order at least 7 means rotation of angle at least 2휋 7 . But such groups are maximal, that is, such groups cannot be subgroups of strictly larger Fuchsian groups (see Greenberg [1]). Thus < 푇, 푆1 >=< 푇, 푆 >.

Finally the question arises whether the condition as stated in the Jør- gensen Lemma is a consequence of stronger inequalities such as

| tr2(푇 ) + tr(푇 푆푇 −1푆−1) − 6| ≥ 1 or | tr2(푇 ) − tr(푇 푆푇 −1푆−1) − 2| ≥ 1 2. Jørgensen Inequality 20

by means of "unnecessary" use of the triangle-inequality. The answer to each of these questions is in the negative. For that look at (︃ )︃ (︃ )︃ √3 √1 √3 − √1 푇 = √2 4 2 , 푆 = √2 4 2 . √1 − √1 2 2 2 2 2 2

One can prove that < 푇, 푆√ > is indeed a Fuchsian group (see Jørgensen [2]). Then, we have tr2(푇 ) = (2 2)2 = 8, tr(푇 푆푇 −1푆−1) = −2 and thus

tr2(푇 ) + tr(푇 푆푇 −1푆−1) − 6 = 0. References

Literature

[1] Leon Greenberg. “Maximal Fuchsian Groups”. In: Bulletin of the Amer- ican Mathematical Society 4 (1963), pp. 569–573 (cit. on p. 19). [2] Troels Jørgensen. “Comments On A Discreteness Condition For Sub- groups Of 푆퐿(2, C)”. In: Canadian Journal of Mathematics XXXI.1 (1979), pp. 87–92 (cit. on p. 20). [3] Troels Jørgensen and Maire Kiikka. “Some Extreme Discrete Groups”. In: Annales Acedemiae Scientiarum Fennicae, Series A. I. Mathematica 1 (1975), pp. 245–248 (cit. on p. 17). [4] Svetlana Katok. Fuchsian Groups. Chicago: The University of Chicago Press, 1992 (cit. on pp.2, 15). [5] Rubí E. Rodríguez, Irwin Kra, and Jane P. Gilman. Complex Analysis. 2nd ed. New York: Springer-Verlag, 2013 (cit. on p.6). [6] Charles Walkden. Hyperbolic Geometry. 2015. url: http://www.maths. manchester.ac.uk/∼cwalkden/hyperbolic-geometry/hyperbolic_geometry. pdf (cit. on pp.2,6).

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