ABSTRACTTitle of Dissertation: Complex Hyperbolic Triangle GroupsJustin O. Wyss-Gallifent, Doctor of Philosophy, 2000Dissertation directed by: Dr. William GoldmanDr. Richard SchwartzDepartment of MathematicsWe present several approaches to proving discreteness of triangle groups inPU(2; 1) and prove several results. The various methods used are algebraic,geometric, and real-world algorithmic (meant to be implemented on a present-day computer).

Complex Hyperbolic Triangle GroupsbyJustin O. Wyss-GallifentDissertation submitted to the Faculty of the Graduate School of theUniversity of Maryland, College Park in partial ful�llmentof the requirements for the degree ofDoctor of Philosophy2000Advisory Committee:Dr. William Goldman, Chairman/AdvisorDr. Richard Schwartz, AdvisorDr. John MillsonDr. Larry Washington

c Copyright byJustin O. Wyss-Gallifent2000

TABLE OF CONTENTSList of Figures iv1 Introduction 11.1 Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Isosceles Ultra-Ideal Triangle Groups 112.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 First Discreteness Theorem . . . . . . . . . . . . . . . . . . . . . 132.2.1 Discreteness Criterion . . . . . . . . . . . . . . . . . . . . 142.2.2 Discreteness Proof . . . . . . . . . . . . . . . . . . . . . . 162.2.3 Alternate Expressions for Condition D . . . . . . . . . . . 292.2.4 What We Have Proved . . . . . . . . . . . . . . . . . . . . 302.2.5 Ideal Triangle Groups and the Goldman-Parker Results . . 322.3 An Alternate Proof by Construction of Related Fundamental Do-mains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.3.1 Pictures of Fundamental Domains . . . . . . . . . . . . . . 392.3.2 The Quotient Space . . . . . . . . . . . . . . . . . . . . . . 442.4 New Frontiers: Second Discreteness Theorem . . . . . . . . . . . . 512.4.1 Some Preliminaries . . . . . . . . . . . . . . . . . . . . . . 51ii

2.4.2 The Four Spheres . . . . . . . . . . . . . . . . . . . . . . . 542.4.3 A More Detailed Look at the Structure of Chains . . . . . 592.5 Foliated Patches and a Finitely Computable Discreteness Criterion 652.5.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 652.5.2 Computational Issues . . . . . . . . . . . . . . . . . . . . . 652.5.3 The Algorithm . . . . . . . . . . . . . . . . . . . . . . . . 662.5.4 Practical Implementation . . . . . . . . . . . . . . . . . . . 752.6 Return to the Parabolic Ridge: Results . . . . . . . . . . . . . . . 792.7 The General [m; l; k] Case . . . . . . . . . . . . . . . . . . . . . . 803 Let's Go Inside - Some (p; q; r) Triangle Groups 853.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 853.2 The (4; 4;1) case. . . . . . . . . . . . . . . . . . . . . . . . . . . 903.3 An Interesting Attempted Approach . . . . . . . . . . . . . . . . 1023.4 The General (n; n;1) Triangle Group . . . . . . . . . . . . . . . 1034 Some [m;m; 0] Triangle Groups 110References 113

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LIST OF FIGURES2.1 Plot of Condition D . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Projection of Condition D . . . . . . . . . . . . . . . . . . . . . . 232.3 Example of Condition D . . . . . . . . . . . . . . . . . . . . . . . 242.4 [ U 6= ; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.5 R [ S, the Projection of the Fundamental Domain . . . . . . . . . 262.6 Cli�ord Torus and Image for Discrete Ultra-Ideal Group . . . . . 272.7 Cli�ord Torus and Image for Indiscrete Ultra-Ideal Group . . . . 282.8 Type 2 and Type 4 Fundamental Domains in the Unit Disc . . . . 432.9 Type 2 and Type 4 Fundamental Domain in @H2C . . . . . . . . . 452.10 The Parabolic Ridge . . . . . . . . . . . . . . . . . . . . . . . . . 522.11 The Four Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.1 i0i1i2 Ellipticity Wedge . . . . . . . . . . . . . . . . . . . . . . . . 1053.2 Close-Up of i0i1i2 Ellipticity Wedge . . . . . . . . . . . . . . . . 105

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Chapter 1IntroductionThe subject of the discreteness of triangle groups in PU(2; 1) is a beautiful andfar-reaching one. There has been some exploration in this area following researchin the early 1990's by Bill Goldman and John Parker. Their conjecture sum-marizing the discreteness of ideal triangle groups was tackled unsuccessfully butwith some very interesting methods by Hanna Sandler, and then successfully byRichard Schwartz in the late 1990's using an entirely di�erent approach. Sincethen we have, along with Richard Schwartz, done some research into discretenessof the nonideal cases as well as fundamental domain construction and explorationof the structure of the limit sets and domains of discontinuity. This latter explo-ration has led to some fascinating results by Schwartz, while the former has ledto this thesis.In this thesis we push the envelope a little by proving some results pertainingto the cases where the complex geodesics intersect within complex hyperbolicspace, and when they do not intersect at all. These two categories make up thetwo main parts of the thesis, with the latter being more thoroughly touched uponthan the former. We prove some su�cient conditions for discreteness, some for

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indiscreteness, and make several conjectures.I would like to thank Bill Goldman and Richard Schwartz for introducing meto this fascinating area and giving me inspiration when it was sorely lacking.I would also like to thank John Parker and Larry Triplett for several inspiringconversations relating to this research.1.1 Summary of ResultsThe results of this thesis may be broken down into three major areas (bundledinto chapters 2 and 3) and one minor area (chapter 4).First we examine some ultra-ideal triangle groups, where none of the generat-ing geodesics meet. We �nd a two-dimensional family of these groups which arediscrete. This discreteness is proved in two ways, one being through the anal-ysis of a subset of the boundary of H2C which is taken o� itself nontrivially byall elements of the group (Theorem 2.2.0.14), and the other being through theconstruction of explicit fundamental domains for the group actions (section 2.3).Second, by doing extensive computer analysis on the behavior of certain ob-jects in the boundary of H2C , we hypothesize that our discreteness may be ex-tended further (Section 2.4), and we construct an algorithm which will provethe discreteness for speci�c parameter values outside the range proved previously(section 2.5).Thirdly we examine some triangle groups where the generating geodesics in-tersect (Chapter 3). We parametrize a family of these, namely the (n; n;1)

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cases, and we examine various attributes of these groups. In Sections 3.2 and 3.3we look more closely at the (4; 4;1) groups, prove some discreteness results andmake several conjectures. In Section 3.4 we then go on to look at the (n; n;1)cases for large enough n that we can see the connection with the Goldman-ParkerConjecture.Lastly (chapter 4) we give a short, sweeping discreteness criterion for anotherlarge family of ultra-ideal groups.1.2 PreliminariesConsider complex hyperbolic n-space HnC , which is de�ned as the complex projec-tivization of the negative vectors in C n;1 with the Hermitian inner product de�nedby hz; wi = z0 �w0+ :::+ zn �wn� zn+1 �wn+1. We call a vector v 2 C 2;1 positive, null,or negative if its Hermitian norm is positive, zero, or negative respectively. Sincea negative vector z must have zn+1 6= 0, HnC may be identi�ed with the unit ballin C n by normalizing the last co-ordinate and then ignoring it.We will be concerned with the case where n = 2. Call the ball model inthis case B. Projectivizing the null vectors, which must also have nonzero lastco-ordinate, we get @H2C , which may be identi�ed with S3.Let PU(n; 1) be the group of automorphisms of C n;1 which preserve the Hermi-tian norm. The elements of PU(n; 1) fall roughly into three categories as follows.Elliptic elements have at least one �xed point in H2C and may have others onthe boundary. Parabolic elements have a single �xed point which lies in @H2C ,roughly correspond to a rotation around a boundary point. Hyperbolic elements3

have two �xed points which lie in @H2C , and corresponds to rotation around apoint which lies outside H2C [ @H2C .For completeness' sake, we mention that an elliptic element is regular elliptic ifits eigenvalues are all distinct. Parabolic elements are divided into two categories.If an element may be written as element of U(2; 1) with 1 as its only eigenvalue,it is said to be unipotent. If it is not unipotent, it is ellipto-parabolic (alsocalled screw-parabolic). An ellipto-parabolic element preserves a unique complexgeodesic, on which it acts as a parabolic element of PU(1; 1).In [Go], Goldman gives a discriminant function which acts on traces of ma-trices in PU(2; 1) and categorizes the elements accordingly. More precisely, let�f(z) = jzj4 � 8Re(z3) + 18jzj2 � 27Then for A 2 PU(2; 1), denote the cube root of unity by !3, then:1. A is regular elliptic i� �f(Tr(A)) < 0.2. A is hyperbolic i� �f(Tr(A)) > 0.3. A is ellipto-parabolic i� A is not elliptic and Tr(A) 2 �f�1(0)� 3f1; !3; !23g.4. A is a complex re ection i� A is elliptic and Tr(A) 2 �f�1(0)� 3f1; !3; !23g.5. A represents a unipotent automorphism i� Tr(A) 2 3f1; !3; !23g.De�nition 1.2.0.1. We de�ne the Hermitian Cross Product� : C 2;1 � C 2;1 ! C 2;14

byv � w = 266664�1 0 00 �1 00 0 1

377775 (�v � �w) = 266664�v3 �w2 � �v2 �w3�v1 �w3 � �v3 �w1�v1 �w2 � �v2 �w1377775 :The Hermitian cross product plays the same role in the Hermitian vectorspace that the standard cross product plays in Euclidean space. That is, thecross product of two vectors yields a vector perpendicular to the two.De�nition 1.2.0.2. Given two points x and y in H2C [@H2C , we de�ne the com-plex geodesic C � H2C [ @H2C containing these points by lifting x and y to ~xand ~y respectively, and then taking ~C to be the complex span of ~x and ~y. Let Cbe the projectivization of ~C, which is a projective subspace of complex dimension1. C is unique by construction.It is worth noting that by consideration of dimensions, two complex geodesicsare either identical, disjoint, or meet in a single point. It is impossible for twocomplex geodesics to meet in a real line.De�nition 1.2.0.3. The polar vector for C is the unique vector perpendicularto both ~x and ~y and given by ~x�~y. Polar vectors are always positive and thereforeany positive vector corresponds to a complex geodesic.De�nition 1.2.0.4. Two complex geodesics that do not intersect inside H2C ei-ther intersect in the boundary, in which case they are called parallel or asymp-totic, or are disjoint, in which case they are called ultraparallel.5

Given a complex vector space V , there is an underlying real vector spacewhere the vectors are identical, but scalars are taken only from R. Denote thisreal vector space by VR.De�nition 1.2.0.5. Let C1 and C2 be two complex geodesics with polar vectorsc1 and c2 respectively. Denote by spanR(ci) the span of ci under multiplication byreal scalars. Thus spanR(ci) is a real linear subspace of C 2;1R . De�ne the angle](C1; C2) = ](c1; c2) = minc1;c2 f\(c1; c2) : ci 2 spanR(ci)g ;where \ denotes the angle between the two vectors measured normally. SincespanR(ci) is invariant under multiplication by �1, the angle necessarily satis�es0 � ](C1; C2) � �=2.Lemma 1.2.0.6. Suppose c1 and c2 are polar vectors for two complex geodesicsC1 and C2 respectively, and suppose c1 and c2 have been normalized. There arethree possibilities.1. j< c1; c2 >j < 1, in which case j< c1; c2 >j = cos(�), where � is the angle ofintersection between C1 and C2, which meet inside H2C . In this case c1� c2is a negative vector, corresponding to the point of intersection.2. j< c1; c2 >j = 1 in which case C1 and C2 are parallel. In this case, c1 � c2is a null vector, corresponding to the intersection point on the boundary.3. j< c1; c2 >j > 1, in which case j< c1; c2 >j = cosh(�=2), where � is thedistance between C1 and C2. In this case, c1 � c2 is a positive vector, andis the polar vector corresponding to the unique complex geodesic orthogonalto and intersecting C1 and C2. 6

De�nition 1.2.0.7. Given a complex geodesic C, there is a unique involutioni 2 PU(2; 1) which �xes C, called inversion in C. This involution is given inC 2;1 by the standard inversion in a vector subspace. Speci�cally, if p denotes thepolar vector for C, we have V 7! �p?(V )� �p(V ).De�nition 1.2.0.8. Given a complex geodesic C and a unit complex number �,there is an elliptic element which rotates \around" the geodesic C by � radians.This will be called a �-re ection in C.For example, when C has polar vector266664100377775 ;the diagonal matrix (�; 1; 1) is the �-re ection. All others are PU(2; 1)-conjugateto this one.An inversion in a complex geodesic is then simply a �-re ection with � = �1.Clearly, if � is an n-th root of unity, then the �-re ection has order n.Consider that a complex geodesic, in the ball model, is homeomorphic toa disc. Thus its intersection with the boundary is homeomorphic to a circle.We call circles that arise in this manner chains. From two distinct points on achain we can retrieve the complex geodesic, so there is a bijection between chainsand complex geodesics. We can therefore, without loss of generality, talk aboutre ections in chains rather than in complex geodesics.7

Complex geodesics are one of two types of totally geodesic submanifolds ofcomplex hyperbolic space, and the metric on H2C restricts on a complex geodesicto the Poincar�e model of the unit disk. The other type of totally geodesic sub-manifolds are totally real. More concretely, let U be a subspace of C 2;1R . Then Uis said to be totally real if J(U) is orthogonal to U , where orthogonality is deter-mined under the nondegenerate real-valued symmetric bilinear form Re h�; �i andJ : C 2;1R ! C 2;1R such that J � J = �Id. (Such a J is called a complex structure.)The projectivization of U is then a totally real totally geodesic submanifold ofH2C to which the metric restricts yielding the Klein model.The simplest totally real totally geodesic submanifold is the �xed point set ofthe automorphism (z; w) 7! (�z; �w) in the ball model of H2C . All others are imagesof this one under PU(2; 1).The boundaries of these submanifolds are called R-circles. We will not go farinto depth into the structure of these, we will only mention the few types we shallneed. For a more complete description see [Go].De�nition 1.2.0.9. Given three complex geodesics 0, 1 and 2, consider therepresentation � : G = Z2 �Z2 �Z2 ! PU(2; 1), the former having generators g0,g1 and g2, which takes gi to inversion in i. We call the image of � a trianglegroup. Abusing notation, we also say � is a triangle group, and we say that g0,g1 and g2 generate the triangle group �.Naturally various types of triangle groups depend on how the geodesics iintersect one another. We isolate three particular cases for discussion. If the imeet pairwise in @H2C , we call � an ideal triangle group. If the i meet pairwise8

inside H2C (or perhaps some but not all meet in @H2C ), we call � a sub-idealtriangle group. Lastly, if the i do not meet inside H2C (or again some in @H2C )we call � an ultra-ideal triangle group.In all three cases, the question may be asked of when a given triangle group isdiscrete. By discrete we mean that the group is discrete as a subgroup of GL3(C ),the group of 3� 3 matrices over C .The boundary @H2C is homeomorphic to S3, and one of the representationswe choose for this is C � R [ f1g, with points either 1 or (z; r) with z 2 Cand r 2 R, intuitively the �rst co-ordinate corresponds to the xy-plane and thesecond to the z-axis as we are familiar with it. Let H denote this representation,that is, C � R [ f1g. We have the homeomorphism P taking S3 to H given bythe standard stereographic projection:(z1; z2) 7! � z11 + z2 ;�Im�1� z21 + z2��(0;�1) 7! 1There are many stereographic projections based from any point in the bound-ary. The technical de�nition is unnecessary here, but su�ce to say we are usingbut one example based at (0;�1).The most important thing we need to know about the H-representation isthe way chains appear. Speci�cally, the unit circle in C � f0g and vertical lines(with the in�nite point) are all chains. It is straightforward to show that the onlychains through1 are vertical. Other chains are various ellipses (perhaps circles)which project to circles via C � R ! C . See [Go] for more details.9

De�nition 1.2.0.10. For z 2 C , the z-chain is the chain having polar vector266664 1��z�z377775 :The z-chain is the vertical chain in H through the point (z; 0).De�nition 1.2.0.11. For z; r 2 R the (z; r)-chain is the chain having polarvector 266664 01 + r2 + iz1� r2 � iz

377775 :The (z; r)-chain is the circle of radius r centered at the origin in C �fzg � H.De�nition 1.2.0.12. The Cli�ord Torus T isT = ((z; w) 2 B : jzj = jwj =r12) :De�ne the map � : B ! C by (z; w) 7! w. � is the orthogonal projectiononto the complex geodesic having the 0-chain as its boundary. The image of thismap is the unit disk � in C , and in particular the 0-chain is taken to the unitcircle and T is taken to the circle of radiusp1=2. Since � is a complex geodesic,it carries the Poincar�e model of real hyperbolic space.10

Chapter 2Isosceles Ultra-Ideal Triangle Groups2.1 PreliminariesFirst we introduce some notation.De�nition 2.1.0.13. De�ne an [m; l;k] triangle group, for m; l; k � 0 to bethe triangle group whose generating complex geodesics are pairwise distance m; land k apart. We extend this de�nition to say that if one of these is 0, then thecorresponding geodesics are asymptotic.This notation complements the standard notation of a (p; q; r) triangle groupfor which the geodesics intersect or are asymptotic.Inversion in the (z; r)-chain is given by the matrix

iz;r = 266664�1 0 00 1 + z2 + r42r2 �1� 2iz + z2 + r42r20 ��1 + 2iz + z2 + r42r2 1 + z2 + r42r2377775 :

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This map commutes with � (B ! C taking (z; w) 7! w) and takes � to itself.Restricted to �, iz;r is a fractional linear transformation with a single �xed point.That is, it is an elliptic element of PU(1; 1) of order two corresponding to thelower right two-by-two block in the matrix above.The product iz;ri�z;r restricted to � is a hyperbolic element of PU(1; 1),therefore �xes two points on the boundary @�. These points correspond inthe H model to the two �xed points on the vertical chain (invariant under theproduct) through 0, which derive from the attracting and repelling eigenvectorsof the product of the inversions.Let C be the chain with polar vector266664 1�10377775 :In H, C winds itself around the Cli�ord Torus. Inversion in C, denoted ic, takesthe Cli�ord Torus to itself and interchanges the two components of H� T .The collection of ultra-ideal triangles we consider have the property that oneof the remaining two chains is inside the Cli�ord Torus and the other is outside.Unfortunately, there is no particularly nice way to arrange these last two chainsin H. Picking one of the last two complex geodesic and inverting it in C, weproduce a triple of complex geodesics which do lie in a more convenient fashion.Clearly the group generated by this new triple is identical to that produced bythe original. We take this approach in labelling our inversions. When we need todiscuss distances between pairs of sides, though, we will switch back.

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Fix z; r 2 Z. Let i+ = iz;r and i� = i�z;r. Let � be the group generated byic, i+ and i�. Let � be the group generated by i+ and i�. Let C+ and C� bethe chains (and by abuse of notation the geodesics) left invariant by i+ and i�respectively. Let C 0� = ic(C�), so the triple C, C+, C 0+ really form the trianglewe are considering, while the triple C, C+, C� is easier to work with. It followsfrom symmetry that the triangle thus formed is isosceles.2.2 First Discreteness TheoremWe prove the following theorem:Theorem 2.2.0.14. � is discrete given that1. z and r satisfyz2 + 2r4z2 + r8z2 + 2z4 + 2r4z4 + z6 � 8r8 � 0;2. The (z; r)-chain is not contained within the Cli�ord Torus.Call the inequality in (1) Condition D.The cleanest way to express (2) is to note that one foliation of the Cli�ordTorus by (w; s)-chains may be given by the collection of such chains satisfyingw = p6s2 � 1� s4, so that a (z; r)-chain is not inside the Cli�ord Torus i�z2 � 6r2 � 1� r4. Call this condition the Torus Condition.Figure 2.1 illustrates the zeros set of both Condition D and the Torus Con-dition. The picture is one quadrant of a vertical slice through the z-axis of theH-representation. The area of discreteness is the area above both curves.13

1 2 3 4 5z

0.5

1

1.5

2

2.5

r

T

D

Figure 2.1: Plot of Condition D2.2.1 Discreteness CriterionWe use the same discreteness criterion used by Richard Schwartz in [Sc].De�nition 2.2.1.1. Suppose � acts on a set S, and U1; U2; V � S with V ( U1,we say that (U1; U2)V is compressing for � if the following two conditions aremet:1. ic(U1) = U2.2. i(U2) ( V for all non-identity i 2 �.De�nition 2.2.1.2. We say (U1; U2)V is semi-compressing if only the secondof these conditions is met.We say � is compressing if there exists a set S on which � acts, and threesubsets U1, U2, and V ( U1, for which (U1; U2)V is compressing. Note that � isimportant in this de�nition. 14

The following is elementary:Lemma 2.2.1.3. Let � : A ! B be a surjective map of sets on which � acts.Suppose there exist U1, U2, and V ( U1 all subsets of A such that � is semi-compressing with respect to the sets �(U1), �(U2), and �(V ) in B. Then if icinterchanges U1 and U2, then (U1; U2)V is compressing for �.Lemma 2.2.1.4. If � is compressing, then � is a discrete subgroup of PU(2; 1).Proof. Consider an element g 2 �. There are four types of g and each has anaction on either U1 or U2 which is isolated from the identity. Observe that g maybe written as i1ici2:::icin where all the ij are in �. Then notice the four typesand their e�ects in the following. For purposes of clari�cation for the four linesbelow assume the leading and trailing ij's, if they exist, are non-identity.1. i1ici2:::icin(U2) ( V2. i1ici2:::ic(U1) ( V3. ici2:::icin(U2) ( ic(V )4. ici2:::ic(U1) ( ic(V )In all four cases, g is nontrivial and is isolated from the identity. Thus � isdiscrete. 515

2.2.2 Discreteness ProofThis implies discreteness of all the ideal triangle groups up to a certain limit, andmany ultra-ideal groups.Let U1 be the outer component of H � T (the component containing the 0-chain) and let U2 be the inner component. The map c switches U1 and U2. Wewill �nd V such that � is semi-compressing with respect to (�(U1);�(U2))�(V ),proving � is discrete.Let U = �(U2), the inner component of �� �(T ).It su�ces to show that � takes U completely o� itself for the parameters inquestion, and that the images of U under all elements of � are contained in aproper subset of �(U1).Let h = i+i�, and consider h as a hyperbolic element of PU(1; 1) acting on� with the Poincar�e model. Explicitly computing the eigenvectors of h, we �ndthe �xed points of h to be�1� r4 � z21 + r4 + z2�� 2pr4 + z21 + r4 + z2! i:The �rst of these corresponds to the attracting eigenvector and as such shallbe denoted va. The latter of these corresponds to the repelling eigenvector andshall be denoted vr. Denote by the real geodesic connecting the two.De�nition 2.2.2.1. Suppose � 2 PU(2; 1) is hyperbolic with invariant geodesic� not passing through the origin. De�ne the clockwise end of � to be the end onthe clockwise side of the smaller component (in the Euclidean metric) of �� �.16

De�nition 2.2.2.2. We say that two hyperbolic elements in PU(2; 1) are agree-able if they have the same attracting and repelling �xed points.De�nition 2.2.2.3. Suppose � and � are agreeable geodesics. We say � inPU(1; 1) is stronger than � if � moves any given point further than �.Lemma 2.2.2.4. Suppose �1 and �2 are agreeable geodesics, and �i is the at-tracting eigenvalue for �i. Then �1 is stronger than �2 i� �1 > �2.Proof. Under PU(1; 1)-equivalence it su�ces to prove the lemma for the element264 �2 + 12� i�2 � 12��i�2 � 12� �2 + 12� 375where � > 1 (resp. 1=�) is the eigenvalue corresponding to the attracting �xedpoint i (resp. repelling �xed point �i). A simple computation (applying theelement to any point in int(�)) shows that increasing � yields a stronger element.5De�nition 2.2.2.5. Suppose ! is a geodesic in real hyperbolic space, and letd 2 R+ . Let W denote the set of all points whose distance to ! is less than orequal to d. W is bounded by two curves which are said to have distance d from!. Let any curve which arises in this fashion be called parallel to !.De�nition 2.2.2.6. Suppose � is a geodesic in real hyperbolic space. A curveparallel to � is called a hypercycle. 17

Lemma 2.2.2.7. In the Poincar�e model, the hypercycles are all Euclidean circlesthrough the endpoints of the geodesic, except for one, which is the Euclideanstraight line through the two points.Proof. The easiest way to see this is to map the unit disk to the upper half-planemodel of real hyperbolic space by the mapz 7! 1� izz � iwith inverse z 7! zi + 1z + i :In this model, geodesics are either vertical lines or semicircles perpendicularto the real axis. Let p be a point on the geodesic consisting of the complexaxis and let x be the point of distance ln(d) measured along the perpendiculargeodesic to the right of p. That is, x lies on the circle of radius Im(p) centeredat the origin. Since distance is measured along semicircles centered at the originvia the formula dist(x; p) = ln jx� �pj+ jx� pjjx� �pj � jx� pj ;the point x satis�es d = jx� �pj+ jx� pjjx� �pj � jx� pj :Solving this, we �nd that18

Im(x) = � 2dd2 + 1� Im(p):Since jxj2 = Im(p)2, we computeRe(x) =s1� � 2dd2 + 1�2Im(p)so that x lies on the line whose slope is the ratio of these. Speci�cally, the slopeof the line is independent of p. Thus the hypercycle consisting of all such x forall such p consists precisely of the line with this slope. Note that this line passesthrough both the origin and in�nity.Since both of the maps above take circles and lines to circles and lines, thehypercycle is taken back to either a circle or a line in the disk model as claimed.5Lemma 2.2.2.8. Suppose � is the invariant geodesic for a hyperbolic element g.The orbit of a point p under g lies on a hypercycle parallel to �.Proof. We measure distance from p to � along the unique geodesic perpendicularto �. Since g preserves angles and distances, p is taken to a point whose distanceto � is identical. The set of such points is a hypercycle by de�nition. 5In fact, if we take the in�nitesimal generator of a hyperbolic element, theorbit of a point under that generator is precisely a hypercycle.19

Suppose � is a hyperbolic element with attracting and repelling �xed points�a and �r respectively. For any � agreeable with �, �(U) is contained within thetwo unique hypercycles joining �a to �r and tangent to U . It follows that if � isstronger than � and � takes U o� itself, � will also.The attracting eigenvalue for the element h isr4 + 2z(z +pr4 + z2)r4 = 1 + 2z2r4 + 2zr 1r12 + z2r16 :Increasing z and/or decreasing r will increase the attracting eigenvalue, andhence yield a stronger hyperbolic element. It follows that here is a \�rst case" ofwhen U is o� itself, and then it is o� itself for all larger z and smaller r.A straightforward calculation shows that there is a unique hypercycle ~ par-allel to which is perpendicular to �(T ).Let ta be the point in ~ \ �(T ) on the same side of �(T ) as va, and let tr bethe other point.Since elements of PU(1; 1) preserve angles and preserve the property of beinga circle, and since the e�ect of h is to \drag" �(T ) toward va, insisting that h take�(T ) to a tangent circle (the �rst case of U being taken o� itself) is equivalentto the condition that h(tr) = ta.The following is an elementary geometric computation.Lemma 2.2.2.9. Suppose g is a hyperbolic element with conjugate �xed points�� ip1� �220

and invariant geodesic . The unique invariant hypercycle parallel to and per-pendicular to the circle of radius p1=2 intersects that circle at the points23�� ir12 � 49�2:The vectors (in C 2;1) corresponding to ta and tr are therefore given (up toequivalence) by 2666640ab377775 and 2666640�ab

377775where a = 4(1� (r4 + z2)2) + ip18(1 + r4 + z2)4 � 16(1� (r4 + z2)2)2and b = 6(1 + r4 + z2)2:The matrix (in PU(2; 1)) for the element h is:2666641 0 00 � �0 �� ��377775where � = r4(1 + iz) + iz(�i + z)2r421

and � = iz(1 + r4 + z2)r4 :It su�ces to show that �h(tr) = �(ta) for the data we have here. In thenotation above, �h(tr) = �a+ �b��a + ��b;and �(ta) = �ab :In order for these two to be equal, we solve the equation(�a+ �b)b = (��a+ ��b)�a;or equivalently, �nd the solutions of�ab+ �b2 � ��a�a� ��b�a = 0:Cancelling all denominators and simplifying the left-hand side of this equationgives us leaves us solving (1 + r4 + z2)2(1 + 34r4 + r8 + 34z2 + 2r4z2 + z4)(z2 +2r4z2 + r8z2 + 2z4 + 2r4z4 + z6 � 8r8) = 0. Since the �rst two terms are strictlypositive, the condition is that the third term is zero, as we claimed.Figure 2.2 shows the projection via � to � of Condition D and the Cli�ordTorus to �. The region corresponding to Condition D is the region above andbelow the top and bottom curves.22

Φ(Τ)

D

Figure 2.2: Projection of Condition DFigure 2.3 shows an example of Condition D. Included are va, vr, ta, tr,the geodesic connecting va and vr, and the hypercycle � parallel to andperpendicular to �(T ). We also see both �(T ) and its image under the map h.By symmetry, i�i+ satis�es the same criteria. By the parabolicity of h, onceU has left itself, all powers of h merely drag U closer towards va. Similarly, allpowers of i�i+ drag U closer towards vr.All that's left (up to symmetry) is to show that for all n � 0, i�hn(U) isdisjoint from U .Assume for convenience that va is contained in the �rst quadrant. The secondquadrant argument is symmetric. , the invariant geodesic for h, is also leftinvariant by both i+ and i�. Let R be the union of U with the region in the�rst and second quadrants between the two (unique) hypercycles parallel to 23

(Φ(Τ))h

γ

t a

t r vr

va

Φ(Τ)

γ

Figure 2.3: Example of Condition Dand tangent to �(T ). Since these hypercycles are parallel to , they are leftinvariant by h and therefore since U is contained in R, all the images hn(R) arealso contained in R. It su�ces therefore to show that i� takes R o� itself.It su�ces to show that there exists a geodesic 0 through p� which does notintersect R, since then i� leaves this geodesic invariant and switches both sides,so i�(R) \ R = ;.Suppose \ U = ;, then is on the outside of the region R and so we maylet 0 = .On the other hand, if \ U 6= ;, then let � be the geodesic in the �rst andsecond quadrants tangent to �(T ) and symmetric with respect to the x-axis. Notethat inversion in points on � will take R o� itself because �, being to the right of24

p-

γ’

α

γ

Figure 2.4: [ U 6= ; , is therefore to the right of the hypercycles parallel to up until the last casewhen the hypercycle, � and �(T ) are all tangent to one another on the right sideof �(T ). Thus � has all of R on its left-hand side. Let 0 be the unique geodesicpassing through p� with perpendicular bisector passing through the origin. Sincethe angle p� makes with the x-axis is greater than zero, 0 is further clockwisethan � and since jp�j �p1=2, the endpoints of 0 are closer than or equal to �'s,measured along the unit circle with the Euclidean metric. Putting these togethergives us that the most counterclockwise endpoint of 0 is further clockwise thanthat for �. Therefore 0 does not intersect R so i�(R)\R = ;. Figure 2.4 showsa picture of this.25

Figure 2.5: R [ S, the Projection of the Fundamental DomainA similar argument works for i+(i�i+)n, where we denote by S the regionanalogous to R.Thus � takes U completely o� itself. Since we surely missed parts of the �rstand fourth quadrants (or second and third), we are guaranteed that the imagesof U under all elements of � are contained in a proper subset of �(U1), namelyR [ S. Thus we see that the sets (�(U1); U)R[S are semi-compressing for �.In �gure 2.5 we see the region R[S between the bounding hypercycles tangentto �(T ) and parallel to .Now we may de�ne V to be ��1(R[S), which gives us (U1; U2)V compressingfor �. Thus � is discrete. 526

Figure 2.6: Cli�ord Torus and Image for Discrete Ultra-Ideal Group27

Figure 2.7: Cli�ord Torus and Image for Indiscrete Ultra-Ideal Group28

Figure 2.6 shows the Cli�ord torus and its image under h for a discrete case,speci�cally for z = 3 and r = 2. The original torus is the darker, larger of thetwo. Figure 2.7 shows the torus and its image under h for an indiscrete case, forz = 1:5 and r = 2.2.2.3 Alternate Expressions for Condition DThere are two other convenient ways of expressing Condition D. The �rst issimply to project it down into � and express it in terms of the point p+. Doingthis shows that if p+ is given by y = h+ vi, then the borderline for Condition Dis given by1� 4h2 + 6h4 � 4h6 + h8 � 6v2 + 8h2v2 � 14h4v2+ 4h6v2 + 2v4 � 16h2v4 + 6h4v4 � 6v6 + 4h2v6 + v8 = 0and so Condition D is satis�ed if y lies above this curve.Alternately we may convert the above to polar coordinates (�; s). Doing soyields the borderline condition1� 5s2 + 4s4 � 5s6 + s8 + s8 + (s+ s3)2 cos 2� = 0;and so Condition D is satis�ed when (�; s) lies outside this curve.

29

2.2.4 What We Have ProvedFor any pair of ultraparallel complex geodesics, we may compute the distancebetween them. We would like to know what triples of distances we have found tobe discrete. At this point, calculation-wise, we need to switch back to the tripleC, C+ and C 0+ which form a triangle.Denote by vc, v+ and v0+ the normalized polar vectors for C, C+ and C 0+respectively. We havevc = 266664 1�10

377775 ; v+ = 266664 01 + r2 + zi2r1� r2 � zi2r377775 and v0+ = 266664�1� r2 + zi2r0�1 + r2 � zi2r

377775 :Computing distances, we see��< v+; v0+ >�� = (1� r2)2 + z24r2 � 1;so that dist(C+; C 0+) = 2 arccosh�(1� r2)2 + z24r2 � :We also havej< v+; vc >j = ��< v0+; vc >�� =r(1 + r2)2 + z28r2 � 1;so that dist(C+; C) = dist(C 0+; C) = 2 arccoshr(1 + r2)2 + z28r2 :Note that the two inequalities result from the Torus Condition.30

Now then, let d = dist(C+; C). We will show that dist(C+; C 0+) = 2d, thusproving that the triangle groups constructed are of distances (d; d; 2d).Using the two hyperbolic trigonometric identitiescosh(2�) = cosh2 � + sinh2 �and sinh arccosh� = p�2 � 1;we calculate d = dist(C+; C)d = 2 arccoshr(1 + r2)2 + z28r2cosh(d) = cosh(2 arccoshr(1 + r2)2 + z28r2 )cosh(d) = (1 + r2)2 + z28r2 + sinh arccoshr(1 + r2)2 + z28r2 !2cosh(d) = (1 + r2)2 + z28r2 +r(1 + r2)2 + z28r2 � 1cosh(d) = (1� r2)2 + z2 + 4r28r2 + (1� r2)2 + z2 � 4r28r2cosh(d) = (1� r2)2 + z24r2d = arccosh�(1� r2)2 + z24r2 �2d = dist(C+; C 0+)It follows from this calculation and the fact that d may be made as large or(positive) small as possible by appropriate choice of nonnegative z and r that31

for any d we sample at least one [d; d; 2d] ultra-ideal triangle group; In fact, wesample in�nitely many.2.2.5 Ideal Triangle Groups and the Goldman-Parker Re-sultsFor the remainder of this subsection, assume the two chains C+ and C� lie onthe Cli�ord torus. That is, � is an ideal triangle group. Simultaneously solvingthe Curve and Torus Condition gives us the solutions(z; r) = 111q302� 36p70;r 311 �9�p70�!and (z; r) = 111q302 + 36p70;r 311 �9 +p70�!For now let us examine the �rst pair. In searching for a correspondence to[Gp] we again must note that when C+ and C� do lie on the Cli�ord torus, thethree chains C+, C�, and C do not form an ideal triangle, though inversions inthem do yield an ideal triangle group. One triple which yields an ideal triangleis the triple C+, ic(C�), and C.The polar vector for C+ is266664 038� 3p70 + ip302� 36p70�16 + 3p70� ip302� 36p70377775 ;

32

for iC(C�) is 266664�38 + 3p70 + ip302� 36p70016� 3p70� ip302� 36p70377775 ;

and for C is 266664�110377775 :The pairwise points of intersections of the corresponding complex geodesicsare 26666470� 9p70 + ip302� 36p7070� 9p70� ip302� 36p702(�54 + 6p70)

377775 ;266664�16 + 3p70 + ip302� 36p70�16 + 3p70 + ip302� 36p7038� 3p70 + ip302� 36p70377775 ;

and33

266664�16 + 3p70� ip302� 36p70�16 + 3p70� ip302� 36p7038� 3p70� ip302� 36p70377775 :The Cartan angular invariant of these three points is arctanp35, which is thepositive part of the upper bound for discreteness found in [Gp].Following similar computations for the second pair yields the Cartan angularinvariant � arctanp35, which is the negative part of the upper bound found fordiscreteness in [Gp].

2.3 An Alternate Proof by Construction of Re-lated Fundamental DomainsWe emulate the method for constructing a fundamental domain for an index twosubgroup in [Gp] with some modi�cations necessary to cover the greater rangeof groups we are dealing with. This will also result in an alternate proof ofdiscreteness to the one we stated earlier, which may seem quicker at �rst, butrelies upon the following lemmas and de�nitions.The following may all be found in [Gp].De�nition 2.3.0.1. Let x; y 2 HnC and de�ne the half-spaceH(x; y) = fz 2 HnC : �(x; z) < �(y; z)g ;34

which has as its boundary the equidistant hypersurfaceE(x; y) = fz 2 HnC : �(x; z) = �(y; z)g :The construction of Dirichlet fundamental domains for discrete groups extendsin a straightforward fashion to HnC , so we may de�ne, for G acting discretely onHnC and for x 2 HnC , the Dirichlet fundamental domain for G based at x, denotedDG(x), by DG(x) = fz 2 HnC : �(x; z) < �(gx; z), 8g 2 G� Idg= \g2G�IdH(x; gx)The following lemma is a version of the Klein Combination Theorem. Wehave taken this from [Gp].Lemma 2.3.0.2. Let G1, G2 be discrete subgroups of PU(n; 1) with connectedfundamental domains D1 and D2. Let E1 and E2 be the interior of the comple-ment of D1 and D2 respectively. Suppose that E1 \ E2 = ; and D1 \ D2 6= ;.Then G =< G1; G2 > is discrete and D = D1 \D2 is a fundamental domain forG. In particular, if D1 and D2 are Dirichlet polyhedra based at x 2 HnC for G1and G2 then D is the Dirichlet polyhedron based at x for G.Lemma 2.3.0.3. Suppose S is a C -linear subspace of HnC and let �S : HnC ! Sbe orthogonal projection onto S. Suppose that x 2 C and G is a discrete groupof automorphisms of HnC leaving S invariant. Then35

DG(x) = ��1S (DG(x) \ S):Lemma 2.3.0.4. Let S = H1C and let p1; p2 2 S be distinct points. Let �i denoteinversion in pi and � = h�1; �2i. Let x 2 E(p1; p2) � S and let i = E(x; �ix)denote the geodesic bisecting the segment from x to �ix. ThenD�(x) = H(x; �1x) \ H(x; �2x) \ H(x; �1�2x) \ H(x; �2�1x)and has either two or four sides which are segments of the geodesics 1, 2, 12 = E(x; �1�2x), and 21 = E(x; �2�1x). The two-sided case occurs whenthe half-spaces spaces H(x; �1x) and H(x; �2x) are contained in the half-spacesH(x; �1�2x) and H(x; �2�1x) respectively.Consider now the three complex geodesics C, C+ and C�. C+ and C� bothlie in same component of H � T . Let D+ = ic(C+) and D� = ic(C�), so thatD+ and D� also lie in the same component of H � C. Denote by C� (resp.D�) the complex geodesic perpendicular to C+ and C� (resp. D+ and D�).Such geodesics exist and are unique because C+ and C� (resp. D+ and D�) aredisjoint (the chains are unlinked). Conveniently, due to our construction, thepolar vectors for C� and D� are266664100377775 and 266664010

377775respectively, so that they intersect at the origin of the ball.36

Let us focus our attention on C�, by symmetry the situation on D� is anal-ogous. Since C+ and C� are perpendicular to C�, inversion in both of thesepreserves C�. Inversion restricts as a map on C� to inversion in a point, that is,an elliptic element of order two. We looked at the same restriction of inversionin x2.2.2, only this time we will take a di�erent approach from here.Consider the group hi�; i+i acting on C� where each inversion restricts toinversion in a point. By lemma 2.3.0.4 we can construct a Dirichlet domain basedat the origin of the ball and by Lemma 2.3.0.3 we can extend it to a Dirichletdomain for the action of hi�; i+i on H2C . It su�ces by Lemma 2.3.0.2 to showthat this extended Dirichlet domain does not intersect the corresponding one forD�. It su�ces therefore to show that the Dirichlet domain for the action on C�contains the circle of Euclidean radiusp1=2. By symmetry the Dirichlet domainfor the D� situation will also satisfy this criterion, and we will have the necessaryhypothesis for discreteness of �� = hi�; i+; ici�ic; ici+ici by Lemma 2.3.0.2. Since�� is a subgroup of index two in �, this proves � is also discrete.Consider then C� as � like we did before. The intersections of C� and C+(resp. C�) occur at the pointsp+ = �1 + r2 � iz�1� r2 + iz = 1� r4 � z2(r2 + 1)2 + z2 + 2z(r2 + 1)2 + z2 iand its conjugate,p� = �1 + r2 + iz�1� r2 � iz = 1� r4 � z2(r2 + 1)2 + z2 � 2z(r2 + 1)2 + z2 i:Observe that since we are inspecting the Dirichlet domain at the origin, for37

any geodesic in Lemma 2.3.0.4, the point on closest to the origin is the one ofsmallest Euclidean norm. Again referring to Lemma 2.3.0.4, note that � and +have closest point precisely p+ and p� respectively. A straightforward calculationshows that the Euclidean norms of p+ and p� are greater than or equal top1=2.Thus we only need look at �+ and +�. By symmetry we can look only atthe �rst of these. Computationally we �ndi�i+(0) = �z(1 + r4 + z2)(z � i)(r4 + z(z � i))which has Euclidean norm pa, wherea = z2(1 + r4 + z2)2(z2 + 1)(r8 + z2 + 2r4z2 + z4) :Consider that �+ is the geodesic bisecting the segment from 0 to i�i+(0).The point on this geodesic which is closest to 0 is precisely this point of bisectionwhich is located at half the Poincar�e distance from 0 to i�i+(0).The following is a straightforward geometrical computation.Lemma 2.3.0.5. Suppose p0 2 � is located precisely halfway between 0 and apoint p 2 � in the Poincar�e metric. Thenkp0 k = p1 + kp k �p1� kp kp1 + kp k+p1� kp k ;where kz k denotes the Euclidean norm of z.For p0 therefore to have Euclidean norm greater than or equal to p1=2, asimple calculation shows that we must have38

kp k2 � 89 :Assuming p has norm pa, we must havea � 89 :Solving this inequality gives us the solution setz2 + 2r4z2 + r8z2 + 2z4 + 2r4z4 + z6 � 8r2 � 0;which is precisely the same bound found in section 2.2.2.2.3.1 Pictures of Fundamental DomainsThe method used for construction in this section not only proves the discrete-ness of the group ��, but also suggests how to construct fundamental domainsfor these groups. Since H2C is a four dimensional space, explicitly drawings offundamental domains would be di�cult, if not impossible, to construct or puton paper. The next best thing is to intersect the fundamental domains with theboundary @H2C and examine these. In light of this approach we o�er the followingde�nitions, taken from [Go].De�nition 2.3.1.1. Given two points x; y 2 H2C , the spinal sphere Sx;y withvertices x and y is the set

39

@H2C \ E(x; y):De�nition 2.3.1.2. Given a spinal sphere Sx;y, the (unique) complex geodesiccontaining x and y is called the complex spine of S and is denoted SC .De�nition 2.3.1.3. Given a spinal sphere S, the spine of S, denoted Ss, isde�ned to be the set of points in the complex spine which are equidistant from xand y. That is, Ssx;y = E(x; y) \ SC :In other words, this is the perpendicular bisector in SC of the real geodesicjoining x and y.De�nition 2.3.1.4. The points x and y above are called covertices of Sx;y,while the endpoints of the spine Ss are called the vertices of Sx;y.We should mention that the vertices of a spinal sphere are unique. That is,there is a bijection between pairs of distinct points in @H2C (or real geodesicsin H2C ) and spinal spheres. The same is not true of covertices, any two pointswill su�ce as covertices for a given spinal sphere provided that the hypersurfaceE(x; y) used to de�ne the spinal sphere is equidistant from the two points.The following may be found in [Gp].40

Theorem 2.3.1.5. (Mostow)Let E(x; y), SC , and Ss be as above. Let � : H2C ! SC be orthogonal projec-tion. Then E(x; y) = ��1(Ss):De�nition 2.3.1.6. This is called the Mostow Slice Decomposition of E.Using the notation from this section, the group hi�; i+i has fundamental do-main based at the origin bounded by portions of the equidistant hypersurfaces��1( �), ��1( +), ��1( �+), and ��1( +�). The intersection of these hyper-surfaces with @H2C is then composed of pieces of the spinal spheres with �, +, �+, and +� as spines. The vertices of the four spinal spheres are then theendpoints of the four spines, while the pairs (0; i�(0)), (0; i+(0)), (0; i�i+(0)) and(0; i+i�(0)) are covertices for each of the spheres.Let us drop back down to C� again for a moment. Consider p+ = C� \ C+and p� = C� \ C� calculated earlier, where both were taken to be points in C�with the unit disc model. Since these two are conjugate to one another withp+ having positive imaginary part, the following facts may be easily veri�ed byconstruction.We assume p+ is in the �rst quadrant, the second-quadrant argument is sym-metric. The geodesic +� was de�ned as being the geodesic equidistant between0 and its image under i+i�. The point i+i�(0) has larger Euclidean norm thanthe point i+(0), and also has larger imaginary part. It follows that the geodesic41

+ is an arc of a circle with larger Euclidean radius than the geodesic +� andlies (rotationally) closer to the (Euclidean) top of the unit disk C�. Thus it is im-possible for the latter to extend further clockwise than the former. By symmetrythe same is true of the geodesic � with respect to �+.Since p+ and p� are conjugate and i+ is an isometry,�(0; i+(0)) = �(0; i�(0))= �(i+(0); i+i�(0))so that i+(0) 2 +� = E(0; i+i�(0)):It follows that since i+(0) is outside of H(0; i+(0)) by de�nition and by the pre-ceding paragraph, so is the most clockwise end of +�.To summarize, +�'s clockwise end is further counterclockwise than +'s, butis indeed outside of H(0; i+(0)).Since i+ takes + to itself and reverses its orientation, it follows that if i+intersects i+� at a distance d from p+, then another geodesic, namely �, mustintersect it at a distance �d, otherwise i+ would take the segment of + con-tributing to the fundamental domain o� itself, which would mean it would notact properly on the faces. The net e�ect here is that either + intersects both +� and �, or it intersects neither.A similar argument holds for p� in the lower half of the circle.42

+γ

+γ

+γ+γ+γ

+γ

γ

γ

+γ

+γ

+γ

γFigure 2.8: Type 2 and Type 4 Fundamental Domains in the Unit DiscWe therefore have two possibilities for the number of faces in the fundamentaldomain. Either +� is completely outside of H(0; i+(0)), in which case we don'tneed to pay attention to H(0; i+i�(0)) when constructing the fundamental do-main, or +� intersects +. In the former case, we have two disjoint faces, and inthe latter case, four faces that meet in a line. That is, +� intersects +, whichintersects �, which intersects �+.Figure 2.8 shows the two and four-faced cases and the intermediate case (alsotwo-faced) in the middle.This combinatorical argument lifts up to its analogy for the equidistant hy-persurfaces, the E 's. We have two cases, two E 's with no intersection, or four E 'swith \linear" intersection. Call these combinatorical types Type 2 and Type 4respectively.A similar argument yields the fundamental domain for hici�ic; ici+ici whichconsists of ic applied to the hypersurfaces above, and whose intersection with@H2C is ic applied to the spinal spheres above. Intersecting these two yields thefundamental domain F� for ��, the free product of the two. Note that dependingupon the position of the spinal spheres and due to symmetry, F� is either fouror eight-faced.43

The fundamental domain F�@ for the action of the group �� on @H2C , is @H2C \�F�, where �F� is taken to be the extension of F� to the boundary @H2C . In@H2C , we construct the fundamental domain by deleting the bounded region of@H2C �fthe four or eight spinal spheresg from @H2C . That is, F�@ is homeomorphicto S3 with four or eight D3's deleted, where the D3's may overlap.Figure 2.9 shows several pictures of fundamental domains. The top two arethe fundamental domains for the group generated by i� and i+ and by �� in theType 2 case, while the bottom two are the Type 4 cases. For the top two, weclearly see the spheres do not intersect, while for the latter the spheres intersectin two large groups in the manner previously mentioned.2.3.2 The Quotient Space

For simplicity, we shall denote H(0; g(0)) for g 2 � by H(g) and similarlyfor E(0; g(0)). Since the fundamental domain we constructed before is the inter-section of two fundamental domains, one based on C� and one on D�, in thediscussion that follows, any object (such a face, spinal sphere, etc.) which relatesto C� will be called C�'s object, and similarly for D�.To construct the quotient space Q for the action of a discrete � on H2C ,consider the subgroup �� used before to compute the quotient space Q� for this,and then take the quotient of ic. We shall explicitly calculate the face-pairingtransformations on F� (and hence on F�@ ) in order to calculate Q� and Q�@ (thequotient space on the boundary) respectively.44

Figure 2.9: Type 2 and Type 4 Fundamental Domain in @H2C45

To construct the quotient space H2C��� we must examine the face identi�ca-tions of �� on F�. The faces of F� do not intersect for some parameter values,but in some they do. Recall that the F� is either four or eight-sided, so whenwe mention the \sides" or \faces" of F�, we mean whichever ones exist for theparameter values being considered.Denote by S the set fi�; i+; i�i+; i+i�; ici�ic; ici+ic; ici�i+ic; ici+i�icg. Eachg 2 S �xes a real geodesic mentioned in the construction of the fundamentaldomains, so these (in addition to notation before) will be denoted (g) so that,for example, (i+i�) is identical to +� etc. Note that the �rst four of these areC�'s while the second are D�'s.Lemma 2.3.2.1. The faces of F� are mapped via the identi�cationsg�1 : E(g) 7! E(g�1)for g 2 S.Proof. Let x 2 E(g) so that �(x; 0) = �(x; g(0)). Then�(g�1(x); 0) = �(x; g(0))= �(x; 0)= �(g�1(x); g�1(0))so that g�1(x) 2 E(g�1). 5

46

It therefore follows that the quotient space is formed by taking H2C , slic-ing o� four or eight possibly intersecting \caps" (corresponding to the interiorsof the complements of the H's) from the edge of the ball and identifying theexposed edges (corresponding to the E 's) in four self-identi�cations or four self-identi�cations and two pairs. The former case arises when we have:i� : E(i�)$ E(i�)i+ : E(i+)$ E(i+)ici�ic : E(ici�ic)$ E(ici�ic)ici+ic : E(ici+ic)$ E(ici+ic)and the latter when we have, in addition to the above,i+i� : E(i�i+) 7! E(i+i�)ici+i�ic : E(ici�i+ic) 7! E(ici+i�ic)The intersection of this with @H2C is the same as Q�@, @H2C (a copy of S3) withfour or eight 3-discs (each bounded by a spinal sphere) removed in the correctconsistent manner which leaves a connected space remaining, and then identifyingthese spinal spheres in the corresponding manner.Consider next the identi�cations on the faces. Recall that E(g) = ��1( (g))has (g) as its spine. Notice that from by the Mostow slice decomposition,��1( (g)) = [x2 (g)��1(x);47

where each ��1(x) is a complex geodesic and is called a slice at x of the face. Thegeneral picture here is that the face is one-parameter stack of discs put together.Notice also that for g = i� (resp. i+) we have the slice at C�\C� (resp. C�\C+)is precisely C� (resp. C+) and for g = ici�ic (resp. ici+ic) we have the slice atD� \D� (resp. D� \D+) is precisely D� (resp. D+). These can be thought ofas the equators of the face. The two discs at the ends of the spine are degenerate,in that they are simply points.Because the elements in S preserve either C� or D� (depending upon whichthey relate to) and send faces to faces, is follows that the spines are sent to spinesin the correlative manner that the faces are.Two spines contained within the same complex geodesic (known as cospinal)intersect if and only if their corresponding spinal spheres do (This is not necessar-ily true for any two spines.) Thus intersection of faces is characterized completelyby intersections of spines within (without loss of generality due to symmetry) C�.If two spines intersect, then only the parts of the spines which border the equidis-tant hypersurface restricted to C� will be used in the construction of the face ofthe fundamental domain. The meeting point of the spines (of which there canonly be one) will correspond to the unique slices where the E 's intersect.Consider the self-identi�cation arising from i�; the other three work similarly.Recall that +'s orientation is reversed by i�. Correspondingly, E(i+) will undergoan orientation reversing transformation in the sense that a slice at distance d fromthe point p+ will be identi�ed with a slice at distance �d. Another way to saythis is that E(i+) would be identi�ed with its re ection in the slice at p+.Consider now the other intersections. We will look at i�i+ since the other48

three will again arise similarly. Since p+ and p� are conjugate to one another,the (real) geodesic �xed by i�i+ is symmetric with respect to the x-axis. Byconstruction, (i+i�) and (i�i+) are symmetric with respect to the x-axis, andsince i�i+ is parabolic, its e�ect is to re ect (i+i�) in the x-axis, taking it to (i�i+) with reversed orientation.Recall the two combinatoric types we saw in construction of the fundamentaldomain and which we use here. We will discuss < i�i+ > speci�cally becausethe same result holds for < ici�ic; ici+ic >. The face identi�cation arising fromi� and i+ are all that are considered in Type 2, and involve removing caps fromthe open ball that H2C is, and identifying each boundary component E with itselfvia a re ection through its equator. The e�ect of this is easier to visualize in twoor three real dimensions, and is that the newly exposed face simply closes up onitself and seals the open ball back up. Thus in the Type 2 case we are cutting outtwo disks and then sealing up the space. Thus in this case, the quotient spaceQ� is homeomorphic to the original space, that is, the open unit ball in C 2 .In the Type 4 case, the faces intersect in the orderH( +) \ H( +�) 6= ;;H( +�) \ H( �+) 6= ;;H( �+) \ H( �) 6= ;;and the union of these four faces is homeomorphic to a large cap. The identi�-cations on the individual faces combine to an identi�cation which collapses theface and (like in Type II) closes the hole originally created by removing this cap.49

Thus in this case also, Q� is homeomorphic to an open unit ball in C 2 .Passing now to computing Q, we quotient Q� by the element ic. It is clearthat we are simply identifying two halves of an open ball in C 2 , which results inan open hemiball.In looking at the boundary of the quotient space Q@, we are removing eithertwo open balls bounded by spinal spheres and which do not intersect, or removingfour open balls which intersect in a \line" and are bounded by spinal spheres. Inthe former case, the identi�cations map the top half of each spinal sphere to itsbottom half, closing up the hole in @H2C . In the latter case, the boundary of thefour balls that are removed is composed of parts of four spinal spheres, whichclose up all together in an analogous fashion to the equidistant hypersurfaces.In either case, we end up with S3, which, when we quotient out by the �nalinversion ic, gives us Q@ being homeomorphic to half of S3, that is, a hemisphere.It is certainly worth mentioning that the boundary of the quotient space isquite di�erent from the quotient space for the action of the group on the boundary.For the latter, we must �rst note that the action of the group on the boundary isnot properly discontinuous, since we have accumulation points both at attractingpoints of hyperbolic elements and at �xed points of parabolic elements, amongperhaps other places. We must remove all these limit points and look at theremaining space, called the domain of discontinuity. The action of the group onthis space is properly discontinuous. We have not explored this here.50

2.4 New Frontiers: Second Discreteness Theo-remHere is a general statement of our next theorem. After introducing some prelim-inaries, this is more precisely stated as Theorem 2.4.2.1Theorem 2.4.0.2. There is a one-parameter family of [l; l; 2l] triangle groupslying beyond the range proven in Theorem 2.2.0.14 for which several values yielddiscrete groups. We conjecture that they all do.2.4.1 Some PreliminariesIn [Sc], Schwartz proves that in the ideal case, the upper bound for discretenessmay be pushed further than the angular invariant arctanp35 �rst proved in [Gp].In fact, Schwartz shows that the ideal triangle group is discrete if and only if theproduct of the generators is not elliptic. This product is parabolic at the angularinvariant arctanp125=3, and elliptic afterwards. One may ask the followingquestion about the ultra-ideal triangle groups. If we perturb the borderline idealcase by pulling the two chains o� the Cli�ord Torus, precisely as we have in theprevious sections, is there a way to preserve discreteness? In other words, canthe discreteness of the ultra-ideal triangle groups be pushed beyond ConditionD? By pulling the chains in and down along a certain curve, we preserve theparabolicity of the product of the generators (by generators we mean the threethat actually yield the triangle). To compute this curve, note that the trace of51

0.1 0.2 0.3 0.4 0.5z

0.1

0.2

0.3

0.4

0.5

r

T

P

D

Figure 2.10: The Parabolic Ridgethe product i+i�i0 = i+i0(i0i�i0) is� = i(i + z)(r4 + z2 + zi)r4 ;so we simply plug this trace into the discriminant function f . We have avoidedwriting the actual equation for the curve here because it is quite complicated.Su�ce to say it passes through the origin and lies completely below the curve forCondition D. Call this curve the Parabolic Ridge. See �gure 2.10 for a picture ofthe Parabolic Ridge, Condition D, and the edge of the Cli�ord Torus Condition.We will �nd a fundamental domain for the group �� generated by i�, i+, i0i�i0and i0i+i0 >, thereby proving discreteness of the ultra-ideal triangle groups alongthe Parabolic Ridge.Between the Parabolic Ridge and Condition D lies a no-man's land of probablediscreteness. We think that a modi�cation of Schwartz's Dented Torus approach,52

or the more recent work of Parker might give methods with which to settle this.Suppose C is a chain and c 2 C. For x =2 C there is a unique R-circlejoining x and c and passing through C � fcg. To see this, apply an appropriatetransformation so that C is the vertical chain through the origin and c is 1.Then the relevant R-circle is the radial R-circle through x and the vertical axis.For any given pair (C; c) and point x =2 C, de�ne(C; c; x)to be the segment of the aforementioned R-circle which joins x to c but does nottouch C � fcg.De�nition 2.4.1.1. For any set X such that X \ C = ;, de�ne(C; c;X) = [x2X (C; c; x):In both cases we say that (C; c;X) is the cone of X to c, mentioning C onlywhen necessary.If g is an ellipto-parabolic element preserving the pair (C; c), we often write(g;X)instead.De�ne �C : H ! C to be vertical projection. Letting V be the vertical chainthrough the origin, de�ne � : H � V ! S1 � R to be (z; y) 7! (arg(z); y), so �takes H � V to a cylinder.53

2.4.2 The Four SpheresGiven that � = i+i0i� is ellipto-parabolic, we let (P; q) be the chain-point pairsuch that �(q) = q and �(P ) = P . � acts on the complex geodesic associatedto P by a parabolic automorphism of PU(1; 1). Let p denote the polar vectorassociated to P .Let C 0+ = i0(C+) and C 0� = i0(C�). De�ne�#� = (�;C�)and �"� = C�(�#�)for � = +;�. Lastly, de�ne ���� = i0(���)for � = +;� and for � ="; #.Lastly, de�ne �� = �"� [ �#�and ��� = ��"� [ ��#�for � = +;�.54

Call the set (�+;��; ��+; ���) the four spheres. We have yet to prove thatthey are in fact spheres, but the notation gives us some direction.Call q and i�(q) the ends of ��, for � = +;�. Let the ends of the ��'s be theimages of the ends of the �'s under i0. Note the following:1. �+ and �� meet at q.2. One end of ��� is i0i�(q) = i+(q), so ��� meets �+ at i+(q).3. One end of ��+ is i0i+(q) = i�(q), so ��+ meets �� at i�(q).4. ��+ and ��� meet at i0(q).Let C� the equator of �� for � = +;�. De�ne �C� similarly. Note thatinversions in the equators generate the group ��.This may seem confusing at �rst, but we would like the reader to keep inmind �gure 2.11.The picture implies that the points q, i0(q), i+(q) and i�(q) are the onlyplaces where the four spheres meet, and in fact, this is precisely what needs tobe proved. We present our results and conjecture now, and then go on to provethe necessities.Theorem 2.4.2.1. 1. The four �'s are in fact spheres, each of which is leftinvariant by inversion in its equators, which also interchanges the interiorand exterior of the sphere.55

C+

C -

i 0(q)

i 0 (C )-

i 0 (C )+

i i (q)+0i (q)-

=

i i (q)-0i (q)+

=

Σ+

Σ-

Σ--

Σ+

Σ-

Σ- -

Σ+-

Σ+-

q

Figure 2.11: The Four Spheres2. For the values r = 0:3, r = 0:35, r = 0:41, r = 0:4 and r = :39 withappropriate t (along the Parabolic Ridge) the spheres do not intersect exceptat the points mentioned. It follows that the group is discrete for these values.Conjecture 2.4.2.2. For all parameter values, the spheres do not intersect ex-cept at the points mentioned. The group is thus discrete along the entire ParabolicRidge.We prove Theorem 2.4.2.1 with a mix of standard and computational com-puting methods. For the latter, we present an algorithm which will, for twocomputationally signi�cantly disjoint hybrid cones, prove this disjointness in a�nite amount of time. We then run this algorithm on all necessary hybrid conesfor the parameters mentioned. The cases where the cones meet at a point are56

dealt with using standard mathematical proofs.The following may be found in [Sc2].Lemma 2.4.2.3. The �'s are in fact analytically embedded spheres.The fact that inversion in the equator does as is mentioned in Theorem 2.4.2.1follows from construction of the spheres. We have thus proves part 1 of thetheorem.Let R be rotation of � radians around the R-circle C = R�f0g � H. R takesC0 to itself, P to itself, and switches C+ and C�. It follows that R switches i0(C+)and i0(C�). Note also that since R switches C+ and C�, we have i� = R� i+ �R,from whence it follows that R interchanges i+(q) and i�(q) and interchangesi+(P ) and i�(P ). Thus since R preserves the property of being a chain or anR-circle and is an order 2 action on everything used to construct the four spheres,it acts as an element of order 2 on these spheres, interchanging + and �.Consider then that the group F = hi0; Ri �= Z2�Z2 acts on the four spheres.In order to show the requirements of Conjecture 2.4.2.2, we can mod out by theaction of F �rst to simplify matters. More speci�cally, any disjointness necessitynot involving �+ may be written via F as one involving �+. It is therefore onlynecessary that we show three things:Lemma 2.4.2.4. (The Initial Disjointness Lemma)(I) �+ \ �� = fqg.(II) �+ \ ��+ = ;.(III) �+ \ ��� = fi0(q)g.57

It is worth noting that for any pair of spheres meeting at a point, there is anantiholomorphic automorphism (R or i0 �R = R� i0) which interchanges the two,�xing that point.Instead of looking at the � as a whole, we break them into pieces. In otherwords, we look at the disjointness criterion involving the parts " and #. Thus webreak each of I, II and III above into four pieces. Fortunately since i0 switches�#+ and ��#+, any statement involving the latter but not the former may be ig-nored because there is an equivalent (under i0) statement involving the former.Similarly, since R switches �#+ and ��"�, any statement involving the latter butnot the former may be ignored. Lastly, since i0 �R = R � i0 switches �#+ and ��#�,any statement involving the latter but not the former may be ignored.We thus have the following breakdown of the Initial Disjointness Lemma intopieces:Lemma 2.4.2.5. (The Main Disjointness Lemma)(I) �+ \ �� = fqg.(a) �#+ \ �#� = fqg:(b) �#+ \ �"� = ;:(c) �"+ \ �"� = ;:(II) �+ \ ��+ = ;.(a) �#+ \ ��#+ = ;:(b) �#+ \ ��"+ = ;:(c) �"+ \ ��"+ = ;:(III) �+ \ ��� = fi0(q)g.(a) �"+ \ ��"� = fi+(q)g: 58

(b) �#+ \ ��#� = ;:(c) �#+ \ ��"� = ;:A note about a pair of cones: If two chains are coned to the same point thenby the cone construction itself, if they meet other than at the cone point, thenthey meet for an entire R-arc.2.4.3 A More Detailed Look at the Structure of ChainsChains have a fascinatingly symmetric structure which we will review here with-out proof. For a more technical discussion, see [Go]. Every nonvertical chain isuniquely determined by a center point in H and a real radius. The chain struc-ture is preserved under rotation about the line f0g � R � H and under verticaltranslation, so it su�ces to consider chains whose center lies on the positive partof the axis R�f0g � C �R . A chain whose center is at the origin is a circle, andmoving the chain a distance along the axis from the origin yields a chain whichis an ellipse projecting (via C � R ! C � f0g) to a circle, and which has aneccentricity proportional to its distance from the origin. Viewing H from above,the chain is angled up under a clockwise orientation.For any given chain C, we may draw a horizontal radial line segment l whichpasses through the center point and passes through the chain in precisely twopoints. Under �C , l is mapped to the radial line segment which is a diameter of�C (C). Applying � to a chain not intersecting V yields a curve which is thenrotationally symmetric about �(l), which is a point. We take the image of � tobe [0; 2�]� R with the right and left-hand edges identi�ed.59

Lemma 2.4.3.1. If C links V , then �(C) is a curve which has an upper segmentto the left of �(l) and a lower segment to the right of �(l). The upper segmentis concave down while the lower segment, by rotational symmetry, is concave up.Any nontrivial translation of �(C) yields a curve which then intersects �(C) inat most two points.Proof. From [Go] we have a parametrization of a �nite chain in H with center(�0; h0) and radius r0 as the collection of all points (�; h) 2 H satisfyingj� � �0j = r0 h = h0 � 2Im(��0�):Alternately we may convert to polar co-ordinates centered at �0 = x0 + y0i with� = (x0 + r0 cos �) + (y0 + r0 sin �)i. Plugging this reparametrization into theequation for h above yieldsh = h0 � Im((x0 � y0i)((x0 + r0 cos �)� (y0 + r0 sin �)i));which simpli�es to h = h0 � 2r0(x0 sin � � y0 cos �):which is then the function describing h as a polar function of �, centered at �0,under the map �. We then computed2hd�2 = �2r0(y0 cos � � x0 sin �);which equals 0 for de�ned h only when y0 cos � = x0 sin �, which occurs only twiceon [��=2; �=2). The concavity changes only twice, therefore, and an easy checkshows that in one segment it is concave up and in the other, concave down.60

Now then, since C links V , shifting polar co-ordinates to be centered at anypoint inside C other than �0 does not change the pattern of derivative changesin h when taken in polar co-ordinates to that new center. Speci�cally, polarco-ordinates centered at the origin yields the same two changes in concavity. �maps l to one of these points, and results in the rotational symmetry which isthen evident in the formula for h above. 5The following lemma may be proved by straightforward calculation. We givethe avor here.Lemma 2.4.3.2. Suppose we have a chain with center (k; 0) on the positive realaxis and radius l with l < k. Then inversion in this chain followed by inversionin the chain of radius 1 (or really of any value) centered at (0; 0) takes C+ and C�to chains with centers of conjugate �rst co-ordinate, negative second co-ordinate,and identical radius. Call such chains nicely related.Proof. Applying such inversions in the chain with center (0; y) and radius r yieldsa chain with center having �rst co-ordinatek(k2 � l2 � r2 � yi)k4 + l4 � k2(2l2 + r2 + yi) ;second co-ordinate l2(�2k2 + l2)yk8 + l8 � 2k6(2l2 + r2)� 2k2l4(2l2 + r2) + k4(6l4 + 4l2r2 + r4 + y2) ;and radius61

s l4r2k8 + l8 � 2k6(2l2 + r2)� 2k2l4(2l2 + r2) + k4(6l4 + 4l2r2 + r4 + y2) : 5Lemma 2.4.3.3. �#+ and �#+, cannot meet along one or two R-arcs.Proof. We show that one cannot happen, and then that two cannot.Without loss of generality, by symmetry via R, if we have an R-arc joiningp+ 2 C+ and q which passes through p� 2 C�, we also have a R-arc joiningR(p+) 2 C� with R(p�) 2 C+. Thus unless R interchanges p+ and p�, we havetwo R-arcs where they meet. If R does interchange p+ and p�, then this impliesthat each lies on the R-arc connecting the other to q. This is clearly impossible.Suppose they meet along two R-arcs. By the above argument, if we let M 2PU(2; 1) be as in 2.4.1, we have a point on M(C+) which is radially away from apoint on M(C�) and vice versa. This implies, however, that M(C+) and M(C�)are in fact linked, which is not the case. It is easier to see this from the pointof view of �, since under � we get two curves which intersect each other twice,with one intersection being the projection of M(C�) on top of M(C+), and onebeing the opposite. 5Theorem 2.4.3.4. (Ia) is true.Proof. Let M 2 PU(2; 1) be as in 2.4.1. That is, M(P ) = V and M(q) =1. Itsu�ces to show that �(M(C1)) \ �(M(C2)) = ;.62

A straightforward calculation shows that P links both C+ and C� and hascenter on the positive real axis R � f0g � H. We also have q lying on this sameaxis. In fact, q is computed to correspond to the point in H (k; 0) withk =s6r4 �p36r8 � 4(1 + z2)(r8 + z2 + 2r4z2 + z4)2(1 + z2) :We may then choose M to be inversion in a chain linking C+ and C� with centeron the positive real axis (taking P to V and q to 0) followed by inversion in thechain of radius 1 centered at (0; 0) (preserving V and taking 0 to1.) This we cando since a straightforward calculation shows that if we de�ne �M to be inversionin the chain with center (k; 0) with k 2 R and radius l, then the vertical chainthrough 0 is taken to the chain with center (k � l2=(2k); 0) and radius l2=(2k).We can thus pick l and k so that q is taken to the origin and P to vertical. Thende�ne M to be the composition of the diagonal matrix (�1; 1;�1) with �M�1,that is, M = (�1; 1;�1)� �M�1. The former is precisely the inversion in the chainof radius 1 centered at (0; 0) and the latter �nishes the job.By the above lemma, M(C+) and M(C�) are nicely related. By the rota-tionally symmetric structure of chains, M(C+) and M(C�) are then related toeach other by a vertical translation and a rotation. Under �, �(M(C+)) and�(M(C�)) are then related by a translation. It follows then that if �(M(C+))and �(M(C�)) meet, they do so in at most two points. We saw above that �#�and �#+ cannot meet in one or two points. Thus they do not intersect at all, andso the lemma is proved. 5Theorem 2.4.3.5. (IIIa) is true. 63

Proof. First note that R � i0 maps i+(q) to itself, i�(q) to itself, switches q andi0(q), and interchanges the pairs (C+; i0(C�)) and (C�; i0(C+)). More speci�cally,R � i0 sets up a symmetry which switches �"+ and ��"�. We can then use the �rstparagraph of Lemma 2.4.3.3, mutatis mutandi, to show that �"+ and ��"� do notmeet in either one or two R-arc. Notice it is necessary here to note that C+ andi0(C�) do not meet. This is true because by construction C0, C+ and i0(C�)form an ultra-ideal triangle.Observe that �"+ = i+(�#+) and ��"� = i+(i+i0i�(�#�)). It follows that theintersection �"+\ ��"� = i+(�#+\i+i0i�(�#�)). Recall that underM , �#+ and �#� arecreated by adding radial R-arcs toM(C+) andM(C�) respectively, with the latterbeing a rotation and vertical translation of the former. Since i+i0i� is an ellipto-parabolic element �xing (P; q), M � i+i0i� �M�1 is the ellipto-parabolic element�xing (V;1). Such elements are screw-translations, that is, they have a rotationaround V and a vertical shift. Summarizing, i+i0i�(�#�) is a rotation and/or shiftof �#+. Thus �(C+) and �(i+i0i�(C�)) meet in at most two points. Since they donot meet in one or two, they don't meet. Thus �"+\ ��"� = i+(�#+\ i+i0i�(�#�)) =i+(q) as claimed. 5We now have seven disjointness criteria left to prove. We leave this to thenext section with a di�erent approach.

64

2.5 Foliated Patches and a Finitely ComputableDiscreteness CriterionThe methods here were developed by Schwartz, while the application is our own.2.5.1 MotivationSuppose we have two sets A;B � @H2C and we wish to show that A \ B = ;. Itsu�ces of course to show that there exist A0; B0 containing A and B respectivelyand with A0 \ B0 = ;. If A and B are numerically di�cult to handle, we mightbe better o� picking A0 and B0 with A � A0 and B � B0 which are moremanageable. The A and B we will be working with are the hybrid cones in theMain Disjointness Lemma, and this section will be concerned with a computableconstruction of satisfactory A0 and B0.2.5.2 Computational IssuesThe method will be presented �rst in a purely geometric/algorithmic form withno regard for any actual application. We will then move on to the issues ofcoaxing this algorithm to function on a present-day computer. Lastly we willdiscuss some problems that can and do arise in certain situations.

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2.5.3 The AlgorithmThe boundary of Complex Hyperbolic Space represented as H may be given polarco-ordinates provided we ignore the point at in�nity. This parametrization is verynatural, with points (�; r; h), and the obvious map(�; r; h)! (r cos � + ir sin �; h) 2 C � R:Given two radii r1 and r2 with 0 < r1 � r2, two angles �1 and �2 with �1 � �2,and two heights h1 and h2 with h1 � h2, we can look at the set P of points(�; r; h) with �1 � � � �2, r1 � r � r2 and h1 � h � h2. P is a \block" in polarco-ordinates with three very natural foliations: P is foliated by arcs of chainswhich are circles centered on the vertical axis, also by arcs of vertical chains, andlastly by arcs of R-circles which are horizontal lines passing through the verticalaxis.De�nition 2.5.3.1. Given any set of points as above, we abuse notation andde�ne a foliated patch P to be either the set of points itself or the image ofP�1(P ) under an element of PU(2; 1). If no map has been applied, we say thepatch is straightened. Recall that P is stereographic projection, so P�1(P ) is afoliated patch in a \nice" position in @H2C .Given a chain C which links V0 the vertical chain through 0, we can de�ne afoliated patch bounded by C given by all the data above except for the heights.The reason for this is straightforward, since for any two angles we can merelylook at the maximum and minimum height of the chain between those angles.Taking those two heights, we get our patch.66

The vertical projection map is injective on the hybrid cone (V0;1;C), soa foliated patch bounding C actually contains the piece of (V0;1;C) which isbounded by the same angles and radii. It follows that for any hybrid cone instandard position, we may cover � f1g by patches merely by picking a polargrid and taking a patch for each grid member, and then throwing out the patcheswhich don't contain a piece of .De�nition 2.5.3.2. A patching is de�ned to be a collection of patches.Note that a patching of a cone covers the entire cone except the cone point.De�nition 2.5.3.3. For any patch P in H in standard position, de�ne the co-patch P c to be the patch with the same heights and radii as P , but with the twoangles equal to each other and to � + (�1 + �2)=2, where �1 and �2 are the anglesfor P .Note that P c is two-dimensional; it has no angular width. P c will play anintegral role in doing computations on P .De�nition 2.5.3.4. A straightened patch has eight vertices and a center, thelatter with a canonical de�nition. Given a straightened patch P and an elementM 2 PU(2; 1), we can de�ne the center of M(P ) to be M(c) where c is thecenter of P , and we can de�ne the radius of M(P ) to be maximum Euclideandistance (in the ball model) between M(c) and M(v), where v is taken over theeight vertices. Thus each patch, straightened or not, has a center and a radius.For any patch P , denote the former by c(P ) and the latter by r(P ).67

De�nition 2.5.3.5. Say a patch is guaranteed if is contained in the ball ofradius r(P ) centered at c(P ). Say a patching is guaranteed if each patch in thecollection is guaranteed. For M 2 PU(2; 1), say a patch P is M-guaranteed ifM(P ) is guaranteed.De�nition 2.5.3.6. Say two guaranteed patches (resp. two guaranteed patch-ings) are ball disjoint if the balls (resp. sets of balls) containing them aredisjoint.Not all patches are guaranteed. If a patch is quite curved, bits of it couldstick outside of the ball. One of the main parts of the computational part of thealgorithm will be to �nd a guaranteed patching of the hybrid cones in question.In [Sc3], Schwartz goes through the details associated to proving that a patchis guaranteed. We repeat the basic statements here without proof in order togive motivation for our algorithms, pseudocode and code. The reader interestedin speci�cs can read [Sc3]. We have made minor changes to account for our ownspeci�c application.S3 is a Lie group. The group law is given by(z1; z2) � (w1; w2) = (z1w1 � z2 �w2; z1w2 + z2 �w1):Note that the element (1; 0) is the identity here. Using this law, de�ne rightmultiplication by RX(Y ) = Y �X. Let�RX(Y ) = 264 0 1�1 0375 �R�1X (Y ):68

Note that �Rp(p) = (0;�1) for any p 2 S3.For a chain C, suppose the endpoints of an arc A of C are p1 and p2, andq 2 C is o� this arc. De�ne a diameter guarantee for this arc, �(A) to bep2=2 if dist(q; pi) > dist(p1; p2) for either i and 1 otherwise. �(A) satis�es�(A) � �(A), where � denotes the diameter of A; that is, the maximum distancebetween pairs of points on A.For the above arc A, we de�ne a curvature guarantee, K(A) to be 2=s,where s is the maximum distance between pairs in the triple fp1; p2; qg. K(A)satis�es �(A) � K(A), where �(A) is the standard space curvature of A.For an R-circle R, suppose the endpoints of an arc A of R are p1 and p2, andq 2 R is o� this arc. De�ne the diameter guarantee for A as�(A) = �1(�1 + �2)�3 ;where1. �3 is the Euclidean distance between p1 and p2.2. �1 is half the length of the line segment �C � P � �Rq(A). Notice that thislatter expression is a line segment because C, the continuation of the arcA, passes through q, and under this map q passes through 1 in H. Allsuch R-circles are a�ne lines in H, which then project to real lines in C .3. �2 is the smallest � such that �Rq(pi) 2 U�, where U� = f(z; w) 2 S3 :jw + 1j � �g.We de�ne a curvature guarantee for A to be K(A) = p(6=�)2 � 8, where �69

is the Euclidean diameter (in the ball model) of A. In practice, we �nd a lowerbound for � (by testing a few pairs of points) which yields an upper bound forK(A). Since we check curvature by checking K(A) instead, using the biggestK(A) su�ces.De�nition 2.5.3.7. Say an arc A (for us, either an arc in a chain or an R-circle) is straight enough for a ball B with radius r if the endpoints are con-tained in B and both �r < 1 and 2� < 1�r � 1:In [Sc3] Schwartz shows that if an arc is straight enough for a ball, it is containedin the ball.The basic method for showing a patch is guaranteed is as follows: First, showthat the copatch is guaranteed by showing �rst that its top and bottom boundingR-arcs are straight enough, and then that each foliating vertical C -arc is. Thisshows that the copatch is contained within a little ball. Next, show that each ofthe left and right faces of the patch are guaranteed for the patch's ball by thesame method just given. Lastly we verify that all of the circular C -arcs foliatingthe patch between the left and right sides are straight enough, simultaneously,based upon our knowledge of three points on each such arc, those three pointsbeing the intersections of the chain containing the arc with the three at patchespreviously guaranteed.More speci�cally, here is the algorithm. Let P be a patch with copatch P c.Let r and r0 denote the radii and c and c0 denote the centers of the balls B and70

B0 corresponding to P and P c. Denote by d(; ) the Euclidean distance in the ballmodel.1. Use K and � on each (top and bottom) radial R-arc of P c to verify thateach arc is straight enough for B0.2. Suppose A is a vertical C -arc in P c with endpoints p1 and p2. Let q, theother points on A's chain, be (0;�1). Since the endpoints of A are in thearcs checked in step 1, they are contained in B0. Thus d(pi; c0) � r0 andso s � d(q; p1) � d(q; c0)� d(p1; c0) � d(q; c0)� r0, where the second-to-lastinequality is the triangle inequality. Since � � K = 2=s � 2=(d(q; c0)� r0),if 2r0=(d(q; c0)� r0) < 1 then �r0 < 1.3. For the same arbitrary A, the endpoints are within 2r0 of each other (bothare contained in B0), and each is at least d(q; c0)�r0 from q. Put � = p2=2if d(q; c0)�r0 > 2r0, that is, if 3r0 < d(q; c0), and1 otherwise. Since � � �,check if 2� < 1=(�r0)� 1 by checking if this is true for 2�. In other words,�rst check if 3r0 < d(q; c0) and if not, we fail. Otherwise, � = p2=2 so wethen check if p2 < 1=(�r0)� 1, that is, if �r0 < 1=(1 +p2).At this point, steps 1-3 pass if the copatch is guaranteed. Note that ourarbitrary choice of vertical C -arc in steps 2 and 3 was absorbed via estimatesof the endpoints, so the checks hold for any choice of such an arc.4. Do the same three steps for the left and right faces of P for the ball B.5. Let A be any circular C -arc joining the left and right faces. The chain Ccontaining A intersects P c once. Let this intersection point be q, and let p1and p2 be the endpoints of A. q is contained in B0 since P c is guaranteed.71

Both pi are within r of c since each is in a guaranteed face of P . Thus q isat least d(c; c0)� r � r0 from either pi. Like in step 2, s � d(c; c0)� r � r0,so to show �r < 1, it su�ces to show that 2r=(d(c; c0) � r � r0) < 1, since�r � (2=s)r = 2r=(d(c; c0)� r � r0).6. This is like step 3. Since both pi are within 2r of each other, and each isat least d(c; c0)� r � r0 from q, put � = p2=2 if 2r < d(c; c0)� r � r0 and1 otherwise. In other words, �rst check if 2r < d(c; c0)� r � r0 and if not,fail. Otherwise, check if �r < 1=(1 +p2).These last two steps guarantee that the arbitrary A, hence any A, is straightenough for B. Thus P is guaranteed.If the algorithm �nishes with no problems, the patch is guaranteed.One thing to note in the application of this algorithm. When checking thecopatch P c, we check if �r0 < 1 and then later that �r0 < 1=(1 +p2). The �rstcheck is redundant, so in practical application we check only the latter, whichimplies the former. The same thing is true three other times during the algorithm.De�nition 2.5.3.8. Given a straightened patch P bounded by a chain C linkingV0, de�ne a re�ning of P to be the collection of between one and four patchesgiven by breaking P up into four equal pieces (via cuts at halfway between theangles and halfway between the radii) and then recalculating the heights of theindividual pieces. Throw away any patch which does not intersect (V0;1;C).Note that at least one patch will remain.De�nition 2.5.3.9. A re�ning of a patching is the patching yielded by re�ningall its nonguaranteed patches. 72

Frustratingly enough, patchings of cones are still in�nite objects (that is, anin�nite number of patches), and hence by construction inconvenient to computewith. Thus we de�ne the following:De�nition 2.5.3.10. An N-patching of a cone in standard form is a patchingof the cone such that the patches cover the cone out to a radius of greater thanN . For a cone not in standard form, we say an N-patching of the cone is theimage of an N-patching.Since an N -patching is built from a �nite subset of a cone in standard from,it leaves out a piece of the cone around the cone point, so that two N -patchingsbeing disjoint does not prove that the cones are. We will �nd a separate patchingwhich covers the cone point and which is computationally compatible with therest of the patches.Let U be the unit spinal sphere, the spinal sphere with vertices (0;�1) in H.U is not a Euclidean sphere inside H, instead it is the set of points f(z; y) 2 H :jzj4 + y2 = 1g. Let I be inversion in the (�1)-chain. I(U) is contained in thesingle patch with angles 2.5 and 3.75 radians, radii 1 and 3, and heights -5 and5. Call this patch P1. I(P1) is then a patch containing U . Let IR be inversionin the (0; R)-chain. In the ball model, IR � I(P1) is a patch containing (0;�1),corresponding to 1 in H. Because IR takes spinal spheres to spinal spheres,IR(U) is a spinal sphere, but note that in H, IR has switched the interior andexterior, so that IR � I(P1) has boundary inside the spinal sphere IR(U) but theinterior of the patch itself extends outward.IR(U) is a spinal sphere which extends no further than the cylinder of radius73

R2 in H. It follows that for a cone in standard form and M 2 PU(2; 1), we canguaranteeM() with a patching comprised ofM applied to an R2-patching alongwith a subdivision of the cone point patchM(IR�I(P1)) into guaranteed patches.Notice that a priori there is no way of knowing if M(IR � I(P1)) is guaranteed,and in most cases it isn't. Instead we choose a subdivision into smaller patcheswhich are guaranteed. By abuse of notation, call this guaranteed patching P1also. There will be no ambiguity, P1 is simply a guaranteed patching whichcovers 1 2 H.De�ne an N-hat to be the patching IR � I(P1) for R = pN .Another way of looking at this is that for any N -patching of a cone , we cancomplete the patching to the cone point by adding on an N -hat.Suppose 1 and 2 are two cones with chains C1 and C2 in standard form inH, and M1 and M2 are elements of PU(2; 1). We wish to check the disjointnessof M1(1) and M2(2). Our algorithm follows:1. For i = 1; 2, let Ni be the a small number satisfying:(a) The cylinder in H of radius Ni contains the chain Ci.(b) Let Hi be the Ni-hat. The two patchings Mi(Hi) are guaranteed andball disjoint.In practice this is done simply by picking Ni to be slightly bigger thannecessary for the �rst criteria, and then doubling it until it satis�es thesecond.2. Let Pi be a guaranteed Ni-patching of i for i = 1; 2.74

3. Throw out any patch in P1 which is ball disjoint from any patch in P2[H2.4. Throw out any patch in P2 which is ball disjoint from any patch in P1[H1.5. If H1 is not ball disjoint from P2, increase N1 and recompute H1. ExtendP2 out to a guaranteed N1 patching of 1.6. if H2 is not ball disjoint from P1, increase N2 and recompute H2. ExtendP1 out to a guaranteed N2 patching of 2.If the algorithm halts, the cones are disjoint. Only one thing could go wrongwhich would cause the algorithm to run inde�nitely. That would be if there area string of non-guaranteed patches P1; P2; ::: such that Pi is in the re�nement ofPi�1.2.5.4 Practical ImplementationThe major problem with implementing this algorithm on a present-day machine isroundo� error. There are two approaches to dealing with this. The best and mostcomplicated approach would be to build the code around a computational enginewith arbitrary precision. If the sizes of objects become so small that the error inthe calculations interferes (we can test this by testing object size against currentprecision), we merely scrap the current attempt and run the entire algorithm withmore precision. Assuming the two cones are in fact disjoint, there is a minimumdistance between them which provides us with safe knowledge that eventually weshall have enough precision to do the job (assuming we don't run out of machinememory).75

In practicality, though, this is an exceedingly di�cult approach, rife with pro-gramming issues, the major one being that the precision checking would multiplythe running time by a rather large factor. We take the other track, which is tocompute using standard machine precision. Under this approach, if the objectswe are dealing with become to small, the algorithm simply fails to work.We still have to bound our error, though, and make sure it doesn't get outof hand. We use interval arithmetic with slight modi�cations. One problem thatmay still arise is that after numerous calculations, the intervals get larger in size,hence decreasing accuracy. We write the code in a fashion that if the computerruns up against precision issues, the code simply fails to halt. In this manner weguarantee that we will not get an erroneous halting.Our implementation begins with two pairs of data. Each pair is a directedhybrid cone, which is de�ned to be the data (pi;Mi), where pi is the polar vectorfor a chain linking V0, and Mi is an element of PU(2; 1). The implementationwill halt if the cones M1((V0;1; p1)) and M2((V0;1; p2)) are disjoint. It willfail to halt if the cones are either not disjoint, or if the size of the intervals in theinterval arithmetic outgrow the size of the data we are working with.We assume, to minimize error, that the vectors and matrices which are fed tothe program are at least precise to 15 digits. We compute ours with Mathematica,but the choice is up to the user. The code itself is written in Turbo C++, runningthe ANSI C standard. The data types are doubles, which themselves are preciseto 15 digits. Interval arithmetic is implemented in the following manner. Doublesin C are represented by 64 bits. Denoting these bits se1:::e11b1:::b52, let E be thedecimal number between 0 and 2047 represented by the binary e1:::e11, and we76

have a representation of the number(�1)s2E�1023(1 + 52Xj=1 bj2�j):This is a fairly fascinating way to represent a number. Disregarding the sign,the �rst part of the representation can give us every power of two between 2�1023and 21024. For each interval between a power of two and its neighbors, the secondpart (the part in parenthesis) essentially breaks the interval into 252 � 1 furthervalues. In other words, there are 252 � 1 representable numbers strictly between1/4 and 1/2, between 1 and 2, between 2 and 4, and so on.Note that the number 0 is not representable! Instead, there are two approxi-mations of equal precision, namely 2�1023 and �2�1023. In fact, the only naturalnumbers which are exactly representable are those between �253 and 253 (except0). Note also that working with numbers that are smaller in absolute value ismore precise than working with numbers that are larger, however the accuracyis relative to the magnitude.Assigning a value x to a double in C gives that double the value closest tox which the computer can represent. Denote this by ~x. For any number xthat the computer CAN represent, there is a unique predecessor and a uniquesuccessor, which we will represent by x� and x+ respectively. For any number xwhich might not be representable, we take the value to be the interval [~x�; ~x+].Necessary operations may be rede�ned appropriately, being careful, for instance,to give an error when dividing by an interval containing 0. The net result is thatthe result of a calculation to a value x is bounded by the interval resulting fromperforming precisely the same calculation on [~x�; ~x+].77

In summary, we rede�ne our necessary operations as follows:1. [x1; x2] � [y1; y2] = [(x1 + y1)�; (x2 + y2)+] for � = +;�.2. [x1; x2]� [y1; y2] = [(minfxi � yig)�; (maxfxi � yig)+]3. [x1; x2]=[y1; y1] = [x1; x2]� [1=y2; 1=y1] for [y1; y2] not containing 0.4. p[x1; x2] = [(px1)�; (px2)+].5. sin(I) = [(minfsin(x)jx 2 Ig)�; (maxfsin(x)jx 2 Ig)+]. Any other trigfunction may be de�ned from sin.To be consistent with the computer's problematic representation of 0 (theissue is that any calculation resulting in 0 might be represented either way bythe system), we consider both representations as identical. In other words, thesuccessor of either representation is de�ned to be the successor of the higher, andthe predecessor of either is the predecessor of the lower.Whenever dealing with numbers where we have some choice, we choose theinterval to be perfect. That is, reasonably sized natural numbers n 6= 0 areintroduced as the interval [n; n], since C++ can represent n precisely.The angles that bound our initial foliated patches are taken to be representablesums of powers of 2, so they themselves are representable exactly, and dividinga patch in half angle-wise essentially yields no loss in precision. In a worst-casescenario, a patch will have bounding angles [x; x] and [x+; x+] for some x, anddividing this in half will yield two intervals with bounding angles being selectionsfrom the same pair. Any pair of bounding angles [x; x] and [x; x] is simply splitinto two of the same. All we can do then is cycle down into redundancy where the78

code will simply not stop running. Essentially this avoids the algorithm haltingbecause errors have caused it to think it's done.The radii are taken to be perfect intervals, so that when we re�ne a patch,we split not at half the distance between the radii, but at the perfect interval[~x; ~x], where x = (r1 + r2)=2. The issue is not that we split the patch exactlyin quarters, but that we get pieces which are more likely to be guaranteed butwhich still cover the same area when put together.2.6 Return to the Parabolic Ridge: Results1. 20 February 2000: We run our code with r = 0:3 and appropriate z. Thecode takes approximately 690 minutes to run and halts without error. Weconclude that the triangle group is discrete for this r and z.2. 23 February 2000: We run our code, after some optimizations, with r =0:35. The code takes approximately 438 minutes to run and halts withouterror.3. 7 March 2000: We run our code with r = 0:41. The code halts afterapproximately 344 minutes.4. 8 March 2000: We run our code with r = 0:4. The code halts after approx-imately 319 minutes.5. 8 March 2000: We run our code with r = 0:39. The code halts afterapproximately 333 minutes.

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Notice that more or less as r shrinks, the time requirement increases. Thisis a result of the fact that while the two spheres �+ and �� are smaller andless convoluted, the complementary spheres i0(�+) and i0(��) are larger andmore twisted. This results in the need for more re�nement during the guaranteeprocess. On the other hand, more initial re�nement leads to less division duringthe disjointness part of the algorithm. These aspects do not balance out, however,and the time requirement increases as r decreases.At this point we have not entirely proven that the triangle groups are discretealong the entire parabolic ridge, we have only built an algorithm which can provefor us that for any given parameter along the ridge, the group is discrete. In orderto �nish the job, we would need to show that it su�ces to prove discreteness fora �nite number of parameters. This can be done by proving that the topologyfor the entire family may be approximated by a �nite set of parameters. We havenot done this yet.2.7 The General [m; l; k] CaseThe method applied in x2.3 may be extended somewhat to yield a conditionthat is su�cient for some general [m; l; k] triangle groups to be discrete. Themethod in x2.2.2 may be extended by the same method, but the calculations areslightly more di�cult. The advantage of the former is the construction of thefundamental domains and the quotient spaces.The approach we took was to start with chains C+, C and C�, constructD+ = ic(C+) and D� = ic(C�), then look at the perpendicular bisectors C� and80

D�. Since there is an inherent symmetry gained from construction via ic, we needonly look at the fundamental domain for C�. Within this complex geodesic, therewas a symmetry gained from the fact that we were inspecting isosceles ultra-idealor ideal triangle groups.For the general [m; l; k] case, the former symmetry remains, a fact which canbe easily veri�ed computationally, but the latter symmetry vanishes. We can,however, tweak the approach slightly to coax out some results. Given a (z+; r+)-chain and a (z�; r�)-chain with inversions i+ and i� respectively, we still examinethe perpendicular bisector C�, resulting in both our chains reducing to pointsp+ and p� in C�, and having the i's project to inversions in those points inC�. The only di�erence here is that p+ and p� are no longer conjugate, that wasgained only from the isoscelarity. Note that expressions for the p's were explicitlycomputed in x2.3 to be p(z; r) = �1 + r2 � iz�1� r2 + iz :As we saw before, the Euclidean norms of p+ and p� are greater than or equalto p1=2. What changes is that the Euclidean norms of i�i+(0) and i+i�(0) aredi�erent from one another and both less straightforward to compute. Insteadthen we take a more constructive approach.De�nition 2.7.0.1. Suppose is a geodesic. De�ne the angle of , denoted\ , to be the unique angle satisfying ��=2 < \ � �=2 such that rotation of by \ radians puts the image perpendicular to the x-axis.Denote by the geodesic connecting p+ and p�. In the [l; l; 2l]-case, \ = 081

since p+ and p� were complex conjugates, but it is not necessarily so in this case.Theorem 2.7.0.2. Given an [m; l; k] triangle group as above, suppose \ = 0,though p+ and p� may not be complex conjugates, and suppose both p+ and p�each satis�es the curve condition, then the triangle group is discrete.Proof. Suppose without loss of generality that the Euclidean norm of p+ is lessthan that of p�. Since + and � lie entirely outside of the circle of radiusp1=2,it is su�cient to show that +� and �+ do.Let p0� be the conjugate of p+, which lies on , and let i0� be inversion in p0�.Let 0�+ be the geodesic equidistant from 0 and i0�i+(0). It is clear that chainsprojecting to p0� and p+ will yield a discrete group since the two are complexconjugates and satisfy Condition D.To show that �+ is further away from the origin than 0�+, recall one of thecosine rules from hyperbolic geometry. If we have a triangle with sides A, B, andC with opposite angles a, b, and c, thencoshC = coshA coshB � sinhA sinhB cos c:Letting p = p0� with i being inversion in p, consider the triangle with sides(and lengths) as follows: A = �(0; p), B = �(p; i0i+(0)) and C = �(i0i+(0); 0).Now then, c > �=2 and as p moves away from p0� towards p�, c increases but doesnot reach �. It follows then that cos c < 0 at all points along this journey. Bothof A and B increase, and so it follows from the cosine rule above that coshC andhence C increases. Thus in the �nal case when p = p�, the distance from 0 to82

i�i+(0) is greater than that from 0 to i0�i+(0). Hence �+ is further away fromthe origin than 0�+ as claimed.Substituting the triple A = �(0; p+), B = �(p+; i+i(0)) and C = �(i+i(0); 0)in the above proof shows that in the �nal case, the distance from 0 to i+i�(0) isgreater than that from 0 to i+i0�(0). De�ning 0+� in the obvious way, it followsthat +� is further away from the origin than 0+�.It follows that the triangle group is discrete via fundamental domain construc-tion. 5Suppose now that \ 6= 0. What we need only do is rotate our picture of� by \ to see whether Condition D is satis�ed by the two chains. Considerthe version of Condition D given in polar coordinates in x2.2.3. Replacing � by� + \ su�ces for a discreteness condition. We have thus proved:Theorem 2.7.0.3. Suppose we have a triangle group with chains projecting topoints as given in the calculations above, then the group is discrete if the pointsin polar coordinates (�; s) lie outside the curve given by1� 5s2 + 4s4 � 5s6 + s8 + s8 + (s+ s3)2 cos 2(� + \ ) = 0:Let us analyze these fundamental domains. Assuming without loss of gener-ality that the norm of p+ is less than or equal to that of p� and \ = 0, we seethat83

�(0; i+(0)) � �(0; i�(0))= �(i+(0); i+i�(0))so that i+(0) =2 H(0; i+i�(0)):It follows that since i+(0) is outside of H(0; i+(0)) by de�nition, so is the mostclockwise end of +�. Thus +� intersects +. Applying the intersection argu-ment derived from the face identi�cations we used in x2.3.1 gives us that �+intersects �.It follows that the fundamental domains and quotient spaces of these [m; l; k]triangle groups are identical to those for the [l; l; 2l] cases.

84

Chapter 3Let's Go Inside - Some (p; q; r) Triangle Groups3.1 PreliminariesConsider now what happens if the complex geodesics which generate the trianglegroup meet in H2C [ @H2C . Denote by x0, x1, and x2 the points of intersection.Given two complex geodesics that intersect inside the ball, we have seen howto de�ne the angle between them in a nice manner and which extends continuallyto be zero if the geodesics meet at the boundary.De�nition 3.1.0.4. For p; q; r 2 Z[ f1g we may de�ne the (p;q; r)-trianglegroup to be the representation � : � ! PU(2; 1) given by taking a generatorto each of three re ections in complex geodesics meeting at angles �p , �q , and �rrespectively.When p = q = r = 1, we get the (ideal) triangle groups discussed inGoldman-Parker. One thing we immediately notice is that if the images of thethree generators are given by i0, i1 and i2, we have the relations (i0i1)r = (i0i2)q =(i1i2)p = id. 85

This summary of results deals with the (4; 4;1) and (n; n;1) triangle groups.We begin by �nding a parameter which is an extension, of sorts, of the CartanAngular Invariant.De�nition 3.1.0.5. An (n; n;1) triangle group � will be said to be in regularposition if there exists a z 2 C so that � is generated by re ections in the(0; 1)-chain, the z-chain, and the (��z)-chain.Proposition 3.1.0.6. Any (n; n;1) triangle group is PU(2; 1)-equivalent to aunique triangle group in regular position, with jzj = cos �n .Lemma 3.1.0.7. Any (m;n;1) triangle group is PU(2; 1)�equivalent to onegenerated by inversions in the (0; 1)-chain and in two vertical chains.Proof. Let the original generators be i0, i1 and i2, where ii is inversion in ci and(i0i1)m = (i0i2)n = id. Let r be the (unique) point where c1 and c2 intersect andlet C be the complex geodesic containing c0 Let p and q be two points in C whichare endpoints of a real geodesic containing �C(r), where �C : H2C [ @H2C ! C isorthogonal projection. Then A (p; q; r) = 0 by the area equivalence given in [Go].Consider that A ((1; 0; 1); (�1; 0; 1); (0;�1; 1)) = 0 also, and thus by [Go], thereexists an element T (unique up to the choice of p and q) of PU(2; 1) taking p; q;and r to (1; 0; 1); (�1; 0; 1); and (0;�1; 1) respectively. The latter three of thesecorrespond in H to the points in (1; 0), (�1; 0), and 1. Thus T takes c0 to theunit circle and r to1. Since the only chains through in�nity are vertical, c1 andc2 must be mapped by T to vertical chains. 586

Lemma 3.1.0.8. If iz denotes inversion in the z-chain, i0 denotes inversion inthe unit circle, and the order n of i0iz is �nite, then jzj = cos �n .Proof. The inversion i0 is represented by the polar vector p0 = (0; 1; 0) while izis represented by pz = (1;��z; �z). Since p0� pz is negative, we �nd that the anglebetween the two complex geodesics satis�es cos � = j< p0; pz >j = jzj. For � = �nwe have our claim. 5The �rst part of the proposition now follows from the following elementaryfact: The elliptic element R� of PU(2; 1) given by the diagonal matrix (�; 1; 1),with � a unit complex number, rotates H around the 0-chain by an angle ofArg(�) radians. Thus if z1; z2 and w1; w2 are two pairs of complex numbers with](z1; z2) = ](w1; w2), and if each pair, each along with the unit circle, representsa triangle group, then there is a � such that R� takes one to the other.Lastly we prove uniqueness. It su�ces to show that any element T of PU(2; 1)which preserves the property of regular position preserves the angle between thevertical chains. Let C be the complex geodesic corresponding to the (0; 1)-chain.T must leave C invariant and hence commutes with projection to C.Projection from H to C, with the latter considered as the unit disk with thePoincar�e model is given by (z; y) 7! 2z1 + jzj21 7! 087

so that the \unit disk" in H, that is, the set of points f(z; 0) : jzj � 1g, is mappedhomeomorphically onto C in a manner which sends circles centered at the originto circles centered at the origin, preserving the angle between pairs of nonzeropoints.Lastly, note that T also �xes the point at in�nity. It follows that T projectsdown to an action on C which �xes the origin, preserves norm, and acts onangles between points in exactly the same manner as angles between verticalchains. Such an action is an elliptic element of PU(1; 1) �xing the origin and ishence a rotation, preserving angles. We thus have uniqueness. 5In order to facilitate neat computations, we shall use the parameter t = tan �,where 0 � � � �2 is the argument of z. We only concern ourselves with 0 < t � 1,since negative values of t will yield the same results by relabelling of chains. Wehave de�ned tan(�=2) to be 1 for convenience, as once we write down matricesfor the inversions, we see we can simply take the limit as t!1 of these matricesas the matrix for t =1, that is, for � = �=2.The polar vectors corresponding to a triangle group in regular position usingthe parameters t = tan(arg z) and r = jzj are then given by266664010377775 ; 266664pt2 + 1�r + rtir � rti

377775 and 266664pt2 + 1r + rti�r � rti377775 ;while the inversions themselves are given by88

266664�1 0 00 1 00 0 �1377775 ;26666664 1 �2r + 2rtip1 + t2 �2r + 2rtip1 + t2�2r + 2rtip1 + t2 2r2 � 1 2r22r � 2rtip1 + t2 �2r2 �2r2 � 1

37777775 ;and 26666664 1 2r � 2rtip1 + t2 2r � 2rtip1 + t22r + 2rtip1 + t2 2r2 � 1 2r2�2r + 2rtip1 + t2 �2r2 �2r2 � 137777775 :The correspondence with Cartan's Angular Invariant is evident for t =1. Inthis case, the triangle group is ideal, and letting � be the angle between z and��z, we have t = tan � = tan(�2 � �2 ) = tan 12(� � �) = tan Awhere A is the angular invariant of the points where the pairs of geodesics meet.

89

3.2 The (4; 4;1) case.Looking speci�cally at the (4; 4;1) triangle group, we have jzj = p1=2, sowe have parametrized all (4; 4;1) triangle groups by t = tan � where � is theargument of z. Note that for t = 0 the triangle group thus generated is R-fuchsian. That is, it �xes an R-circle, namely the one-point compacti�cationof the horizontal line R � f0g. For t > 0, the representation is never againR-fuchsian, nor is it ever C -fuchsian.Our three generators are denoted i0, i1(t), and i2(t), giving inversions in the(0; 1)-chain, z-chain, and (��z)-chain respectively.We �nd the matrix representation for inversion in the unit circle chain (as anelement of PU(2; 1)) is: 266664�1 0 00 1 00 0 �1

377775The matrix for i1(t) is given by26666664 1 � ip2(t� i)p1 + t2 � ip2(t� i)p1 + t2ip2(t+ i)p1 + t2 0 1p2(1� ti)p1 + t2 �1 �237777775 ;while the matrix for i2(t) is given by

90

26666664 1 p2(1� it)p1 + t2 p2(1� it)p1 + t2p2(1 + it)p1 + t2 0 1�p2(t� i)p1 + t2 �1 �237777775 :Consider the triangle group � with generators i0, i1(t) and i2(t).Recall the discriminant function �f given in Section 1.2. For real z, �f(z) factorsinto �f(z) = (z + 1)(z � 3)3;which yields the same categorization of element types asf(z) = (z + 1)(z � 3):It is clear therefore that an element � is hyperbolic i� Tr(�) > 3 or Tr(�) <�1, regular elliptic i� �1 < Tr(�) < 3, and all other cases when Tr(�) = �1 or3. The following two propositions are almost identical:Proposition 3.2.0.9. Consider the element �t = i0i1i0i2(t). If � is discrete theneither t2 � 3 or f(Tr(�nt )) = 0 for some n.Proof. The eigenvalues of �t are�1; 7� t2 + 4tp3� t2t2 + 1 ; 7� t2 � 4tp3� t2t2 + 1 � :91

The trace of any power of �t is real, so then a straightforward calculation withf above shows that �t is hyperbolic for t2 < 3, unipotent for t2 = 3, and regularelliptic for t2 > 3. Suppose � is discrete and t2 > 3, then �t must have �niteorder, since a regular elliptic element of in�nite order generates a nondiscretesubgroup. Since for t2 > 3 the latter two eigenvalues above are norm 1 nonrealcomplex conjugates of one another, to have �nite order n they must be nth rootsof unity. Thus �nt has all eigenvalues equal to 1 and therefore trace 3, and hencef(Tr(�nt )) = 0. 5Proposition 3.2.0.10. Consider the element �t = i0i1i2i1(t). If � is discretethen either t2 � 7 or f(Tr(�nt )) = 0 for some n.Proof. The eigenvalues of �t are(1; 8 +p(7� t2)(t2 + 9)t2 + 1 ; 8�p(7� t2)(t2 + 9)t2 + 1 ) :The remainder of the proof is precisely the same as the previous proposition,mutatis mutandi. 5It is important to take note that it is not su�cient to show that some elementis regular elliptic to destroy discreteness unless we also show it is of in�nite order,

92

since we are asking that the representation be discrete but not necessarily faithful.A regular elliptic element of �nite order still generates a discrete group.Theorem 3.2.0.11. The triangle groups with t2 = 7; 15 are discrete.Proof. We will be more explicit for the former case, the latter case is similar. LetJ be the matrix which puts the element element i1i2 into Jordan canonical form.We compute J to be 266664 0 1 �12 + 12p�7�1 0 11 0 0377775 :Conjugating each of i0, i1, and i2 by J yields generators

J�1i0J = 266664 �1 0 0�1 +p�7 �1 1�p�7�2 0 1377775 ;

J�1i1J = 266664�1 12 � 12p�7 12 + 12p�70 1 �32 + 12p�70 0 �1377775 ;and

J�1i2J = 266664�1 �12 � 12p�7 10 1 �12 + 12p�70 0 �1377775 :All of the entries of these matrices are in the ring93

(�2 + �2p�7 ����� � + � = even)which is discrete as a subring of C . Thus the group of matrices with elementsin this ring is also discrete, so that the triangle group is conjugate to a discretegroup and hence is discrete.For the t2 = 15 case, the same type of conjugation yields generators all ofwhose matrix entries are in the ring(�2 + �2p�15 ����� � + � = even) ;hence this triangle group is also discrete for the same reason. 5Theorem 3.2.0.12. The triangle group with t2 = 3 is discrete.Proof. This is not so clear as the t2 = 7 and t2 = 15 cases, but it is not di�cult.First conjugate by the appropriate J as in the previous two proofs. We assumethis has been done. Under this conjugation, the group is generated by the threematrices m0 = J�1i0J , m1 = J�1i1J and m2 = J�1i2J . Denote by G the indextwo subgroup generated by the four elements m1, m2, m0m1m0 and m0m2m0.The appropriate matrices arem1 = 266664 �1 0 0�2 + 2p�3 �1 1�p�3�4 0 1

377775 ;94

m2 = 266664�1 12 � 12p�3 12p�30 1 32 + 12p�30 0 �1377775 ;

m0m1m0 = 266664 �3 12 � 12p�3 1 + 12p�3�6 + 2p�3 �1� 2p�3 92 + 12p�3�8 �2� 2p�3 3 + 2p�3377775 ;and

m0m2m0 = 266664 �3 �12 � 12p�3 32�2 + 2p�3 �3 32 � 32p�3�8 �2� 2p�3 5377775It su�ces to show that the group generated by these four elements is discrete.Denote by R the ring composed of elements(�2 + �2p�3 ����� � + � = even) ;and by S the set (�2 + �2p�3 ����� �; � 2 Z) :Denote by Z the ring of elements(� + �p�3 ����� �; � 2 Z) ;and by 2Z and 4Z the sets of multiples of elements in Z by 2 and 4 respectively.The following are either obvious from ring structure or follow from straight-forward calculations: 95

1. ZZ � Z2. RR � R3. 2R � Z4. 2S � Z5. ZR � R6. ZS � S7. Z + Z � Z8. R +R � R9. S + S � S10. Z +R � R11. Z + S � S12. R + S � S13. 4Z � 2Z � Z � R � SDenote by M the set of matrices which have entry (m;n) being an element ofthe set at position (m;n) in the matrix266664 Z R S2Z Z R4Z 2Z Z377775 :Observe that all four generators above and the identity matrix are contained inM and also 96

266664 Z R S2Z Z R4Z 2Z Z377775266664 Z R S2Z Z R4Z 2Z Z

377775= 266664 ZZ + 2ZR + 4ZS ZR + ZR + 2ZS ZS +RR + ZS2ZZ + 2ZZ + 4ZR 2ZR + ZZ + 2ZR 2ZS + ZR + ZR4ZZ + 4ZZ + 4ZZ 4ZR+ 2ZZ + 2ZZ 4ZS + 2ZR+ ZZ

377775� 266664 Z R S2Z Z R4Z 2Z Z

377775 :from whence it follows that M is actually a group. Since M is a discrete set, thegroup generated by the four elements above is discrete as well. 5It is actually worth noting that this last proof also works for the cases t2 = 7and t2 = 15, though we leave the alternate method in for a little variety.Theorem 3.2.0.13. The triangle group with parameter t is indiscrete for t2 > 7and t2 6= 15.The proof of this uses the following interesting proposition. This proof is byLarry Washington.Proposition 3.2.0.14. Suppose � and � are rational multiples of � (so that ei�and ei� are roots of unity) and cos� + cos � = 1, then putting � = 2�c=n withgcd(c; n) = 1, we must have n 2 f1; 2; 3; 4; 6g.97

Lemma 3.2.0.15. Suppose � and � are rational multiples of � (so that ei� andei� are roots of unity) and cos� + cos � = 1. Then cos� is rational.Proof. Assume cos� is irrational. We will �nd a Galois conjugate taking cos�to a negative number.Write � = 2�c=n with gcd(c; n) = 1. Let z = e2�i=n. Apply the inverseof the Galois automorphism z 7! zc. Call this inverse g. This maps (e2�ic=n +e�2�ic=n)=2 = cos(2�c=n) to (e2�i=n + e�2�i=n)=2 = cos 2�=n. We want to �nd aGalois conjugate of z in the 2nd or 3rd quadrant. The conjugates of z are zw withgcd(w; z) = 1. We therefore need to show there exists w with n=4 < w < 3n=4and gcd(w; z) = 1. Bertrand's postulate states that for x > 1 there is always aprime between x and 2x. Therefore for n > 4, there is always a prime p withn=4 < p < n=2. If gcd(p; n) = 1, let w = p and we're done. If not, then n = kp.Since n=4 < n=k < n=2 we must have k = 3. Since p 6= 2 (since this is the n = 6case) n is odd and there is a power of 2, call it w, between n=4 and n=2 withgcd(w; n) = 1.Now then, observe that since the cosine of an angle corresponding to a rootof unity is equal to 1=2 of that root plus its complex conjugate, and since Galoisconjugation takes roots of unity to roots of unity and complex conjugates tocomplex conjugates, it follows that cosines of such angles are taken to cosinesof such angles. This follows from the fact that the cosine may be expressed as(� + ��)=2, where � is the root of unity. Thus applying g followed by z 7! zw tothe equation cos� = 1� cos � yields 0 > cos�0 = 1� cos � 0 � 0, a contradiction.5Proof. (of proposition.)98

By hypothesis, ei� is a primitive n-th root of unity. Since ei� is a root ofx2 � (2 cos�)x+ 1, which is a polynomial over Q since cos� 2 Q by the lemma,the �eld extension containing all n-th roots of unity has degree at most 2 over Q .Such a �eld extension has degree �(n), where � is the Euler phi-function. Thuswe must have �(n) � 2, so that n 2 f1; 2; 3; 4; 6g. 5The theorem now follows from the following observation. For t2 > 7, both �tand �t are both regular elliptic. �t has an eigenvalue a in the 2nd quadrant and�t has one, call it b, in the 1st. These were computed earlier to bea = 7� t2 + 4tp3� t2t2 + 1and b = 8 +p(7� t2)(t2 + 9)t2 + 1 :Assume that both are roots of unity, and speci�cally that a = 2�c=n withgcd(c; n) = 1. Notice that Re(a)+1=Re(b). Let � be the angle between thenegative x-axis and a and let � be the angle between the positive x-axis andb. Then cos � = Re(b) = Re(a) + 1 = � cos� + 1 so that cos� + cos � = 1.Thus by the proposition n 2 f1; 2; 3; 4; 6g. The corresponding values of t2 arethen fNo Soln; 3; 13=3; 7; 15g. Since t2 > 7 and t2 6= 15, we have a contradiction.Thus both cannot simultaneously be roots of unity, so one must have in�niteorder. Thus the group is indiscrete. 599

Conjecture 3.2.0.16. The triangle group with parameter t is discrete for t2 < 3and discrete for in�nitely many t satisfying 3 � t2 � 7.The basic rationale behind this conjecture is that it seems that in the region3 � t2 � 7, only � is regular elliptic with in�nitely many parameters for whichit has �nite order, and for t2 < 3, there are no elliptic elements of in�nite order,and perhaps no elliptic elements other than the generators.We present some preliminary computer data that supports this conjecture.We �nd the of roots of the discriminant function for various powers of theelement �, and then start testing these values of t. By testing, we mean we havetaken all words of up to a certain length, and started raising them to powers.What we have found is that in all cases for which the root R satis�es 3 � R2 � 7,all words become non-regular elliptic (and non ellipto-parabolic) by a certainpower. That is, they become either the identity or unipotent. Both of these aregood for discreteness.Here are the roots for powers up to 6:Power Roots1 �p32 �p3;�p73 �p3;�q133 ;�p154 �p3;�p7;�q 9715+8p2 ;�q 9715�8p25 �p3;�q 20923+8p5 ;�q 20923�8p5 ;�q 18135+8p5 ;�q 18135�8p56 �p3;�q133 ;�p15;�p7;�q 19331+16p3 ;�q 19331�16p3100

Next we have a table of results for each of the above roots. We check allwords of up to length listed in the \Length" column, and raised each element toall powers less than or equal to\Power". For the cases where a speci�c element isnot listed, then all the elements were either not regular elliptic to begin with, orbecame non-regular elliptic by the time they were raised to that power. Wherean element is listed, we have found that element to not satisfy this criterion; thatis, the element is regular elliptic for all powers less than or equal to the powerlisted. It is important to note that the existence of such an element does notdestroy the chance of discreteness, as some power of that element might be theidentity, which would save the day. This does not seem to happen, however, as weshall illustrate shortly. The �nal column lists the power n such that the element�t is the identity or, as in the �rst line, simply notes that it is unipotent to beginwith.

101

t2 value � Length Power n so �nt = id3 3 10 16 n.a. (unipotent)7 7 10 16 4133 4.333333 10 16 615 15 10 16 69715 + 8p2 3.686292 10 16 89715� 8p2 27.31371 6 i0i1i2i1 is r.e.20923 + 8p5 5.111456 10 16 1020923� 8p5 40.88854 6 i0i1i2i1 is r.e.18135 + 8p5 3.422291 10 16 1018135� 8p5 10.57771 6 i0i1i2i1 is r.e.19331 + 16p3 3.287187 10 16 1219331� 16p3 58.71281 6 i0i1i2i1 is r.e.3.3 An Interesting Attempted ApproachLet p be the vector polar to the geodesic which is perpendicular to c0 and i1(c2).Note that p exists only for t2 < 7, where i0i1i2i1 is hyperbolic so c0 and i1(c2) aredisjoint.Conjecture 3.3.0.17. For 3 < t2 < 7, We have < w(p); p >� (t2 + 1)=(7� t2)for all w 2 �= < i0i1i0i2 >.Note that this will yield discreteness for all the isolated triangle groups wehave conjectured to be as such. This follows from the fact that for t2 > 3,102

(t2+1)=(7� t2) > 1, so for such t, the geodesic given by p is taken o� itself for allw, and there is a lower bound on the distance away that it may be. Thus everynontrivial element is away from the element by a nontrivial amount independentof the element. Since i0i1i0i2 has �nite order in these cases, the discreteness of�= < i0i1i0i2 > proves the discretenss of �.This conjecture is supported by overwhelming computer evidence. The lowerbound was found for various values of t for all word lengths up to 30.3.4 The General (n; n;1) Triangle GroupIn [Gp] and [Sc], it was shown that in the ideal case, � is indiscrete i� i0i1i2is elliptic i� t2 � 125=3. It seems sensible then that for some N perhaps verylarge, the behavior of the (n; n;1) triangle group for n � N is in uenced by thiselement. We clarify precisely the manner in which this occurs with the followingproposition.Proposition 3.4.0.18. There is a \wedge" of parameter pairs (n; t) for whichthe (n; n;1) triangle group with parameter t is indiscrete as a result of i0i1i2being elliptic of in�nite order.In the proof of this proposition we shall clarify what we mean by a \wedge".We compute the trace of i0i1i2(r; t), to bei(1 + 16r2 + it)t� i = �1 + 16r2 + t2t2 + 1 + 16r2tt2 + 1 i:103

Plugging this in f(x) and setting equal to 0 yields solutionst2 =s2 + 11r2 � 80r4 + 64r6 + r2(8r2 � 7)3=2p8r2 + 12(r2 � 1)and t2 =s2 + 11r2 � 80r4 + 64r6 � r2(8r2 � 7)3=2p8r2 + 12(r2 � 1) :Since t � 0 is real, t only exists if 8r2 � 7 in which case the two resultingvalues of t are identical. It follows that the solution set (r; t) is the edge alongwhich the element i0i1i2 may go into and out of regular ellipticity.The �rst of these solution sets shall be called the outer curve while the secondshall be called the inner curve. The outer curve intersects the n = 1 (r = 1)case at t = 1 while the inner curve intersects at t = p125=3. A simple test ofan inside and outside element gives us i0i1i2 regular elliptic inside this wedge andhyperbolic outside.Notice that cos(�=8) < p7=8 < cos(�=9), so the �rst triangle group whichhas a regular elliptic i0i1i2 is the (9; 9;1) case. For lower n, the element is alwayshyperbolic.Figures 3.1 and 3.2 show pictures of the wedge and a close up, along withcircles of appropriate distance for the n = 8, n = 9, and n = 10 triangle groups.We only need now show that i0i1i2 has in�nite order on the triangle-groupradii inside the wedge. To do this, we appeal to Richard Schwartz. The followingis a (very slight) modi�cation of a proof found in [Sc].104

Figure 3.1: i0i1i2 Ellipticity WedgeUpper

Lower

n=8

n=9n=10

Figure 3.2: Close-Up of i0i1i2 Ellipticity Wedge105

Lemma 3.4.0.19. Suppose T the trace of i0i1i2(n; t) satis�es(T + 9)( �T + 9) � 64; T 6= �1; (3.1)then i0i1i2 has in�nite order.Proof. Suppose i0i1i2 has �nite order, then denoting by !n the n-th root of unity,we have T = !pn + !qn + !rn; p+ q + r = 0: (3.2)Where n is taken as small as possible.For k relatively prime to n, let �k : Q [!n ] ! Q [!n ] be the Galois automor-phism de�ned by �k(!n) = !kn. Equation (3.1) is de�ned in Q [!n ], so it remainstrue if we replace T by �k(T ). In particular, (3.1) gives us that Re(�k(T )) < �1.Restricting k to f1; :::; n� 1g and summing we haveXk �k(T ) < ��(n)where � denotes the Euler phi function. Hence�����Xk �k(T )����� > �(n): (3.3)For m = p,q, or r, let (m;n) denote the GCD of m and n. Let dm = n=(m;n).Note that !mn is a primitive dm-th root of unity, and the sum of all such rootsis one of �1, 0, or 1. The map Z=n! Z=dm induces a map (Z=n)� ! (Z=dm)�which is onto, with multiplicity �(n)=�(dm). This gives the bound106

�����Xk �k(!mn )����� � �(n)�(dm) : (3.4)Combining equations (3.1),(3.2),(3.3),(3.4), we get1�(dp) + 1�(dq) + 1�(dr) > 1:There is no positive integer k such that �(k) = 3, and �(k) = 2 only fork 2 f2; 3; 4; 6g. We conclude that all three roots are at most 12-th roots of unity.A rapid computer search rules out this possibility. 5It is only necessary therefore to show (3.1). We compute T to be9 + it+ 8 cos(2�=n)�1� itso that (T + 9)( �T + 9) = 32(2 + t2 + t2 cos(4�=n))t2 + 1which is � 64 provided that n � 8. Since the triangle group radii intersect thewedge only for n � 9, this criteria is satis�ed, and i0i1i2 has in�nite order in thewedge. It follows that the representation is indiscrete there. 5107

One may ask whether or not the behavior of the pair i0i1i0i2 and i0i1i2i1 mightalso occur for the general case.We have shown that every (n; n;1) triangle group is represented by the unitcircle chain in H and two vertical chains through points z and ��z where jzj =cos �n . De�ning i1 to be inversion in the (rei tan�1 t)-chain and i2 to be inversion inthe (rei(��tan�1 t))-chain, with i0 as before, and putting �r;t = i0i1i0i2, we computethe trace of �r;t to be3(t2 + 1) + 16r4(t2 + 1)� 16r2(t2 � 1)t2 + 1 :Thus, since this is real and using our discriminant function, we �nd �r;t isregular elliptic i� t <r1 + r21� r2and hyperbolic i� t >r1 + r21� r2 :An interesting thing to note is the simplicity of the solution set fort =r1 + r21� r2 ;it is the lemniscate r = p� cos 2� (recall t = tan �). We must be careful in notingthat if r is not cos �n for some n, the elements i0i1 and i0i2 are elliptic of in�niteorder, obstructing discreteness. Thus for all the following, we assume r to satisfythis criterion.108

The �rst thought is whether the embedding is discrete for values of t less thanr1 + r21� r2 ;though this is shown to be false by noting that from the calculations related tothe i0i1i2 ellipticity wedge, for n � 14, the element i0i1i2 becomes regular ellipticof in�nite order for values of t less than the above.The obvious conjecture is thenConjecture 3.4.0.20. For n � 13, the embedding is discrete fort �r1 + r21� r2 ;and for n � 14, the embedding is discrete for values of t less than or equal to thecritical value where i0i1i2 goes regular elliptic.In making an analogy to the (4; 4;1) case, we o�er the additional followingconjecture:Conjecture 3.4.0.21. For n � 13, there are in�nitely many t's in the range0 � t �r1 + r21� r2such that the embedding is discrete.

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Chapter 4Some [m;m; 0] Triangle GroupsGlancing at the arrangement of the chains for the (n; n;1) triangle groups, thevertical chains link the (0; 1)-chain. An obvious question would be: What sortsof triangle groups do we get if we place the vertical chains outside the (0; 1)-chain? Since they will then not link this chain, the triangle group we get isultraideal, with the vertical chains at distance 0. If the vertical chains are placedsymmetrically (like the (n; n;1) cases), we get [m;m; 0] groups. If not, we get[m; l; 0] groups.The proof of the 1-1 correspondance between the value t = tan � and the(n; n;1) triangle group given in proposition 3.1.0.6 extends in the canonicalfashion to the [m;m; 0] groups. In the [m;m; 0] case, if the distance between avertical chain and the (0; 1)-chain is to be m, we must choose z such that this isso. For ultraparallel geodesics, we use the notation of lemma 3.1.0.8 and the factthat jzj = j < p0; pz > j = cosh(m=2) so that m = 2 arccoshjzj.Here is a quick proof that most of these [m;m; 0] triangle groups are discrete.We use the compressing criteria described in de�nition 2.2.1.1 and the fact thata compressing triangle group is discrete.110

Let U be the unit spinal sphere described earlier. That is, the spinal spherewith vertices (0;�1) in H. Note that U is not a Euclidean sphere inside H,instead it is the set of points f(z; y) 2 H : jzj4 + y2 = 1g. It thus contains the(0; 1)-chain and in fact, inversion in the (0; 1) chain preserves the spinal sphereand switches its inside and outside.Let i0 denote inversion in the (0; 1) chain, let i1 and i2 denote inversion inthe z-chain and the ��z chains respectively. Let � be the group generated by allthree inversions, and let � be the group generated by i1 and i2.Let U1 be the part of H � U outside U and U2 be the part inside. If we can�nd a V ( U1 such that i(U2) ( V for all non-identity i 2 �, we have proventhat � is compressing and hence discrete.Inversion in a vertical chain acts on all the other vertical chains simply bya rotation of � radians. Some vertical translation of any individual chain willgenerally occur, but the chain is setwise sent to the vertical chain rotated �radians around the chain we are inverting in.Let Y be the set of vertical chains through all w with jwj = 1. Let Y2 be thecomponent of H� Y containing the origin, and let Y1 the other component. Wehave U2 � Y2 and so i(U2) � i(Y2) for all non-identity i 2 �.Now then, we can look at the images i(Y2) merely by looking at the intersectionof these images with C � f0g � H. Since these intersections are all circles, �simply moves the unit circle around in C by rotations of � radians around z and��z. Provided that the unit circle is mapped o� itself by all non-identity i 2 �,the same is true for Y2 and hence for U2. We can then let V be the union of111

all the images of U2. We are sure that V 6= U1 since V is missing all images ofY2 � U2.Since the radii of the circle is preserved, it su�ces to show that the origin ismoved to a distance of at least 2 by all non-identity i 2 �. Clearly we must havejzj � 1 so that ji1(0)j = 2. In general, rotating a point � 2 C in another point� 2 C yields the point 2� � �. Consider applying i1 then i2 then i1 and so on tothe origin 0. i1 takes 0 to 2z, so then i2 takes 2z to 2(��z) � 2z = �4Re(z). i1gets us to 2z + 4Re(z), and then another i2 gets us to �8Re(z). The pattern isclear, we have (i2i1)n(0) = �4nRe(z) and i1(i2i1)n(0) = 2z + 4nRe(z).In order for j � 4nRe(z)j � 2 for all n, we must have Re(z) � 1=2. If thiscriteria is met, then automatically since 4nRe(z) � 2 we have j2z + 4nzj � 2 aslong as n > 0. Provided jzj � 1, we have it always.It follows then that �, the [m;m; 0] triangle group, is discrete for generatorshaving Re(z) � 1=2. A little arithmetic restates this in a more analogous fashionto the (n; n;1) groups: Given an m, the [m;m; 0] ultra-ideal group is discretefor t �q4 cosh2(m=2)� 1.

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References[Gp] W.M. Goldman and J. Parker, Complex Hyperbolic Ideal Triangle Groups,J. reine angew. Math 425, 1992.[Go] W.M. Goldman, Complex Hyperbolic Geometry, Oxford University Press,1999.[Ka] S. Katok, Fuchsian Groups, University of Chicago Press, 1992.[Kr] B. Kernighan and D. Ritchie, The C Programming Language, PTR PrenticeHall, 1988.[Sc] R. Schwartz, Ideal Triangle Groups, Dented Tori, and Numerical Analysis,Annals of Mathematics, 2000.[Sc2] R. Schwartz, The Modular Group, The Whitehead Link, and Complex Hy-perbolic Geometry, Preprint.[Sc3] R. Schwartz, Proving Disjointness with Foliated Patches, Preprint, 1999.[Tc] Turbo C++ User's Guide, Borland International, 1992.[St] S. Wolfram, Mathematica: A System for Doing Mathematics by Computer,Addison Wesley, 1991.

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