Complex Hyperbolic Triangle Groups of Type [m,m,0; n_1, n_2, 2]
CCOMPLEX HYPERBOLIC TRIANGLE GROUPSOF TYPE [ m, m, n , n , SAM POVALL
Abstract.
In this paper we study discreteness of complex hyperbolic trianglegroups of type [ m, m, n , n , n , n , m, m,
0. For fixed m , the parameterspace of such groups is of real dimension one. We determine the possible ordersfor n and n and also intervals in the parameter space that correspond todiscrete and non-discrete triangle groups. Introduction
Complex hyperbolic triangle groups are groups of isometries of the complex hy-perbolic plane generated by three complex reflections in complex geodesics. Wewill focus on the case of ultra-parallel groups, that is, the case where the complexgeodesics are pairwise disjoint.Unlike real reflections, complex reflections can be of arbitrary order. Work onhigher order reflections has been discussed by Parker and Paupert [10] and Pra-toussevitch [13]. If an ultra-parallel complex hyperbolic triangle group is generatedby reflections of orders n , n , n in complex geodesics C , C , C with the distancebetween C k − and C k +1 equal to m k for k = 1 , ,
3, then we say that the group isof type [ m , m , m ; n , n , n ].In this paper, we will study discreteness of ultra-parallel complex hyperbolic tri-angle groups of type [ m, m, n , n , n , one of order n , and one of order 2. The fixed point sets of order n and n reflections intersecton the boundary of the complex hyperbolic plane ( m = 0) and the other two dis-tances between fixed point sets coincide ( m = m ). Ultra-parallel triangle groupsof types [ m, m,
0; 2 , ,
2] and [ m, m, m ; 2 , ,
2] have been considered in [15], groupsof type [ m , m ,
0; 2 , ,
2] have been considered in [6] and [5], and groups of type[ m , m ,
0; 3 , ,
2] have been considered in [12].To determine the possible orders ( n and n ) of the complex reflections for thegroups of type [ m, m, n , n ,
2] to be discrete, we use the work of Hersonsky andPaulin [4]. We combine several results which give rise to the following theorem:
Theorem 1.1.
An ultra-parallel complex hyperbolic triangle group of type [ m , m , n , n , n ] can only be discrete if the unordered pair of orders of the Date : January 11, 2021.2010
Mathematics Subject Classification.
Primary 51M10; Secondary 32M15, 22E40, 53C55.
Key words and phrases. complex hyperbolic geometry, triangle groups. a r X i v : . [ m a t h . G T ] J a n SAM POVALL complex reflections ι and ι is one of { , } , { , } , { , } , { , } , { , } , { , } or { , } . The deformation space of groups of type [ m, m, n , n ,
2] for a given m is of realdimension one. A group is determined up to an isometry by the angular invariant α ∈ [0 , π ], see section 2. The main aim is to determine an interval in this one-dimensional deformation space such that for all values of the angular invariant inthis interval the corresponding triangle group is discrete. The main results of thepaper are the following propositions: Proposition 1.2.
A complex hyperbolic triangle group of type [ m, m,
0; 2 , , withangular invariant α is discrete if m ≥ log e (3) and cos( α ) ≤ − . Proposition 1.3.
A complex hyperbolic triangle group of type [ m, m, n, , , for n = { , } , with angular invariant α is discrete if m ≥ log e (cid:16) √ (cid:17) and cos( α ) ≤ − √ . Proposition 1.4.
A complex hyperbolic triangle group of type [ m, m, n, , , for n = { , } , with angular invariant α is discrete if m ≥ log e (cid:16) √ (cid:17) and cos( α ) ≤ − √ . To prove these propositions, we use a version of Klein’s combination theorem,adapted to the configurations in question. Two of the generating reflections sharea fixed point on the boundary of the complex hyperbolic plane. We show that theultra-parallel triangle group satisfies a compression property by carefully studyingthe structure of the stabilizer of this fixed point and of its subgroup of Heisenbergtranslations.On the other hand, we obtain the following non-discreteness results by applyinga theorem from [7]:
Proposition 1.5.
A complex hyperbolic triangle group of type [ m, m,
0; 2 , , withangular invariant α is non-discrete if cos( α ) > − √ (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )24 √ (cid:0) m (cid:1) for some integer q ≥ . Proposition 1.6.
A complex hyperbolic triangle group of type [ m, m,
0; 2 , , withangular invariant α is non-discrete if cos( α ) > −
116 cosh (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )16 cosh (cid:0) m (cid:1) for some integer q ≥ . Proposition 1.7.
A complex hyperbolic triangle group of type [ m, m,
0; 4 , , withangular invariant α is non-discrete if cos( α ) > −
14 cosh (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )4 cosh (cid:0) m (cid:1) OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 3 for some integer q ≥ . Proposition 1.8.
A complex hyperbolic triangle group of type [ m, m,
0; 2 , , withangular invariant α is non-discrete if cos( α ) > − √ (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )4 √ (cid:0) m (cid:1) for some integer q ≥ . Proposition 1.9.
A complex hyperbolic triangle group of type [ m, m,
0; 3 , , withangular invariant α is non-discrete if cos( α ) > − √ (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )2 √ (cid:0) m (cid:1) for some integer q ≥ . Combining these results, we see that there is a gap between the intervals of discrete-ness and non-discreteness. This is illustrated in Figure 1 for the case [ m, m,
0; 2 , , m, α )-space. The light grey box corresponds to discretegroups (Proposition 1.2). The solid and dotted black areas correspond to non-discrete groups (Proposition 1.5). Ultra-parallel complex hyperbolic triangle groups α mπ π π π log e (3) Figure 1.
Discreteness and non-discreteness results in the ( m, α )-space.of type [ m, m, n , n ,
2] with orders ( n , n ) other than (2 ,
2) were first consideredin [11].The paper is organised as follows: In section 2 we discuss the background in-formation on complex hyperbolic and Heisenberg geometry. We then introduce thestandard parametrisation for ultra-parallel [ m , m , n , n , n ]-triangle groups in SAM POVALL section 3. In section 4 we use the compression property to derive a discrete-ness condition for [ m , m , n , n , n ]-groups. Following this, in section 5, weconclude the possible orders of the complex reflections for the groups of type[ m , m , n , n , n ] to be discrete. In section 6 we specialise the standard parametri-sation to the case of ultra-parallel [ m, m, n , n , n and n with the boundary of the complexhyperbolic plane. In section 8 we use the discreteness conditions from section 4 togive a proof of Proposition 1.2, 1.3 for n = 2 ,
4, and 1.4, for n = 2 ,
3. Finallyin section 9 we contrast the discreteness results with non-discreteness results andprove Proposition 1.5, 1.6, 1.7, 1.8 and 1.9.
Remark . We use the following notation: For group elements A and B , theircommutator is [ A, B ] = A − B − AB .2. Background
This section will give a brief introduction to complex hyperbolic geometry, forfurther details see [3, 9].2.1.
Complex hyperbolic plane:
Let C , be the 3-dimensional complex vectorspace equipped with a Hermitian form (cid:104)· , ·(cid:105) of signature (2 , (cid:104) z, w (cid:105) = z ¯ w + z ¯ w − z ¯ w . If z ∈ C , then we know that (cid:104) z, z (cid:105) is real. Thus we can define subsets V − , V and V + of C , as follows V − = { z ∈ C , | (cid:104) z, z (cid:105) < } ,V = { z ∈ C , \{ } | (cid:104) z, z (cid:105) = 0 } ,V + = { z ∈ C , | (cid:104) z, z (cid:105) > } . We say that z ∈ C , is negative , null or positive if z is in V − , V or V + respectively.Define a projection map P on the points of C , with z (cid:54) = 0 as P : z = z z z (cid:55)→ (cid:18) z /z z /z (cid:19) ∈ P ( C , ) . That is, provided z (cid:54) = 0, z = ( z , z , z ) (cid:55)→ [ z ] = [ z : z : z ] = (cid:20) z z : z z : 1 (cid:21) . The projective model of the complex hyperbolic plane is defined to be the collectionof negative lines in C , and its boundary is defined to be the collection of null lines.That is H C = P ( V − ) and ∂H C = P ( V ) . The metric on H C , called the Bergman metric , is given by the distance function ρ defined by the formula cosh (cid:18) ρ ([ z ] , [ w ])2 (cid:19) = (cid:104) z, w (cid:105)(cid:104) w, z (cid:105)(cid:104) z, z (cid:105)(cid:104) w, w (cid:105) , OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 5 where [ z ] and [ w ] are the images of z and w in C , under the projectivisationmap P . The group of holomorphic isometries of H C with respect to the Bergmanmetric can be identified with the projective unitary group PU(2 , Complex geodesics: A complex geodesic is a projectivisation of a 2-dimensionalcomplex subspace of C , . Any complex geodesic is isometric to { [ z : 0 : 1] | z ∈ C } in the projective model. Any positive vector c ∈ V + determines a two-dimensionalcomplex subspace { z ∈ C , | (cid:104) c, z (cid:105) = 0 } . Projecting this subspace we obtain a complex geodesic P (cid:16) { z ∈ C , | (cid:104) c, z (cid:105) = 0 } (cid:17) . Conversely, any complex geodesic is represented by a positive vector c ∈ V + , calleda polar vector of the complex geodesic. A polar vector is unique up to multipli-cation by a complex scalar. We say that the polar vector c is normalised if (cid:104) c, c (cid:105) = 1.Let C and C be complex geodesics with normalised polar vectors c and c re-spectively. We call C and C ultra-parallel if they have no points of intersectionin H C ∪ ∂H C , in which case |(cid:104) c , c (cid:105)| = cosh (cid:18)
12 dist( C , C ) (cid:19) > , where dist( C , C ) is the distance between C and C . We call C and C ideal ifthey have a point of intersection in ∂H C , in which case |(cid:104) c , c (cid:105)| = 1 and dist( C , C ) =0.2.3. Complex reflections:
For a given complex geodesic C , a minimal complexhyperbolic reflection of order n in C is the isometry ι C in PU(2 ,
1) of order n withfixed point set C given by ι ( z ) = − z + (1 − µ ) (cid:104) z, c (cid:105)(cid:104) c, c (cid:105) c, where c is a polar vector of C and µ = exp(2 πi/n ).2.4. Complex hyperbolic triangle groups: A complex hyperbolic triangle is atriple ( C , C , C ) of complex geodesics in H C . A triangle ( C , C , C ) is a complexhyperbolic ultra-parallel [ m , m , m ] -triangle if the complex geodesics are ultra-parallel at distances m k = dist( C k − , C k +1 ) for k = 1 , ,
3. We will allow m k = 0for some or all k . A complex hyperbolic ultra-parallel [ m , m , m ; n , n , n ] -trianglegroup is a subgroup of PU(2 ,
1) generated by complex reflections ι k of order n k in thesides C k of a complex hyperbolic ultra-parallel [ m , m , m ]-triangle ( C , C , C ). SAM POVALL
Angular invariant:
For each fixed triple m , m , m the space of [ m , m , m ]-triangles is of real dimension one. We can describe a parametrisation of the spaceof complex hyperbolic triangles in H C by means of an angular invariant α . Wedefine the angular invariant α of the triangle ( C , C , C ) by α = arg (cid:89) k =1 (cid:104) c k − , c k +1 (cid:105) , where c k is the normalised polar vector of the complex geodesic C k . We use thefollowing proposition, given in [13], which gives criteria for the existence of a trianglegroup in terms of the angular invariant. Proposition 2.1. An [ m , m , m ] -triangle in H C is determined uniquely up toisometry by the three distances between the complex geodesics and the angular in-variant α . For any α ∈ [0 , π ] , an [ m , m , m ] -triangle with angular invariant α exists if and only if cos( α ) < r + r + r − r r r , where r k = cosh( m k / . For m = 0 we have r = 1 and the right hand side of the inequality in Proposi-tion 2.1 is r + r r r (cid:62) , so the condition on α is always satisfied, i.e. for any α ∈ [0 , π ] there exists an[ m , m , m ]-triangle with angular invariant α .2.6. Heisenberg group:
The boundary of the complex hyperbolic space can beidentified with the
Heisenberg space N = C × R ∪ {∞} = { ( ζ, ν ) | ζ ∈ C , ν ∈ R } ∪ {∞} . One homeomorphism taking ∂H C to N is given by the stereographic projection:[ z : z : z ] (cid:55)→ (cid:32) z z + z , Im (cid:18) z − z z + z (cid:19)(cid:33) if z + z (cid:54) = 0 , [0 : z : − z ] (cid:55)→ ∞ . The
Heisenberg group is the Heisenberg space N with the group law( ξ , ν ) ∗ ( ξ , ν ) = ( ξ + ξ , ν + ν + 2 Im( ξ ¯ ξ )) . The centre of N consists of elements of the form (0 , ν ) for ν ∈ R . The Heisenberggroup is not abelian but is 2-step nilpotent. To see this, observe that[( ξ , ν ) , ( ξ , ν )] = ( ξ , ν ) − ∗ ( ξ , ν ) − ∗ ( ξ , ν ) ∗ ( ξ , ν ) = (0 , ξ ¯ ξ )) . Therefore the commutator of any two elements of N lies in the centre.An alternative description of the Heisenberg group N is as the group of uppertriangular matrices x y z | x, y, z ∈ R OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 7 with the operation of matrix multiplication. For any integer k (cid:54) = 0, the subgroup N k generated by the matrices a = , b = and c = k is a uniform lattice in N with the presentation N k = (cid:104) a, b, c | [ b, a ] = c k , [ c, a ] = [ c, b ] = 1 (cid:105) . Moreover, any uniform lattice in N is isomorphic to N k for some integer k (cid:54) = 0, seesection 6.1 in [2].2.7. Chains:
A complex geodesic in H C is homeomorphic to a disc, its intersectionwith the boundary of the complex hyperbolic plane is homeomorphic to a circle.Circles that arise as the boundaries of complex geodesics are called chains .There is a bijection between chains and complex geodesics. We can therefore,without loss of generality, talk about reflections in chains instead of reflections incomplex geodesics.Chains can be represented in the Heisenberg space, for more details see [3]. Chainspassing through ∞ are represented by vertical straight lines defined by ζ = ζ .Such chains are called vertical . The vertical chain C ζ defined by ζ = ζ has apolar vector c ζ = − ¯ ζ ¯ ζ . A chain not containing ∞ is called finite . A finite chain is represented by an ellipsewhose vertical projection C × R → C is a circle in C . The finite chain with centre( ζ , ν ) ∈ N and radius r > ζ r − ζ ¯ ζ + iν − r + ζ ¯ ζ − iν and consists of all points ( ζ, ν ) ∈ N satisfying the equations | ζ − ζ | = r , ν = ν − ζ ¯ ζ ) . Heisenberg isometries:
We consider the space N equipped with the Cyganmetric , ρ (cid:0) ( ζ , ν ) , ( ζ , ν ) (cid:1) = (cid:12)(cid:12)(cid:12) | ζ − ζ | − i ( ν − ν ) − i Im( ζ ¯ ζ ) (cid:12)(cid:12)(cid:12) / . A Heisenberg translation T ( ξ,ν ) by ( ξ, ν ) ∈ N is given by( ζ, ω ) (cid:55)→ ( ζ + ξ, ω + ν + 2 Im( ξ ¯ ζ )) = ( ξ, ν ) ∗ ( ζ, ω )and corresponds to the following element in PU(2 , ξ ξ − ¯ ξ − | ξ | − iν − | ξ | − iν ¯ ξ | ξ | − iν | ξ | − iν . SAM POVALL
A special case is a vertical Heisenberg translation T (0 ,ν ) by (0 , ν ) ∈ N given by( ζ, ω ) (cid:55)→ ( ζ, ω + ν ) . A Heisenberg rotation R µ by µ ∈ C , | µ | = 1 is given by( ζ, ω ) (cid:55)→ ( µ · ζ, ω )and corresponds to the following element in PU(2 , µ . A minimal complex reflection ι C ϕ of order n in a vertical chain C ϕ with polar vector c ϕ = − ¯ ϕ ¯ ϕ is given by( ζ, ω ) (cid:55)→ (cid:16) µζ + (1 − µ ) ϕ, ω − | ϕ | Im(1 − µ ) + 2 Im((1 − µ ) ¯ ϕζ ) (cid:17) and corresponds to the following element in PU(2 , − µ − (1 − µ ) ϕ − (1 − µ ) ϕ − (1 − µ ) ¯ ϕ (1 − µ ) | ϕ | − − µ ) | ϕ | (1 − µ ) ¯ ϕ − (1 − µ ) | ϕ | − (1 − µ ) | ϕ | − , where µ = exp(2 πi/n ). The complex reflection ι C ϕ can be decomposed as a productof a Heisenberg translation and a Heisenberg rotation: ι C ϕ = R µ ◦ T ( ξ,ν ) = T ( µξ,ν ) ◦ R µ , where ξ = (¯ µ − ϕ and ν = − | ϕ | · Im(1 − µ ) = 2 | ϕ | sin(2 π/n ) . Heisenberg translations, Heisenberg rotations and complex reflections are isometrieswith respect to the Cygan metric. The group of all Heisenberg translations isisomorphic to N . The group of all Heisenberg rotations { R µ | µ ∈ C , | µ | = 1 } isisomorphic to U(1). The group of their products N (cid:111) U(1) contains all complexreflections.2.9.
Products of reflections in chains:
What effect does the minimal complexreflection of order n in the vertical chain C ζ have on another vertical chain, C ξ ,which intersects C × { } at ξ ?We calculate − µ − (1 − µ ) ζ − (1 − µ ) ζ − (1 − µ )¯ ζ (1 − µ ) | ζ | − − µ ) | ζ | (1 − µ )¯ ζ − (1 − µ ) | ζ | − (1 − µ ) | ζ | − − ¯ ξ ¯ ξ = − µ − (1 − µ )¯ ζ + ¯ ξ (1 − µ )¯ ζ − ¯ ξ . This vector is a multiple of − µ )¯ µ ¯ ζ − ¯ µ ¯ ξ − (1 − µ )¯ µ ¯ ζ + ¯ µ ¯ ξ = − (cid:0) µξ − ( µ − ζ (cid:1)(cid:0) µξ − ( µ − ζ (cid:1) OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 9 which is the polar vector of the vertical chain that intersects C ×{ } at µξ − ( µ − ζ .This corresponds to rotating ξ around ζ through πn . So if we have a vertical chain C ξ , the minimal complex reflection of order n in another vertical chain C ζ rotates C ξ as a set around C ζ through πn (but not point-wise).2.10. Bisectors and spinal spheres:
Unlike in the real hyperbolic space, thereare no totally geodesic real hypersurfaces in H C . An acceptable substitute are themetric bisectors. Let z , z ∈ H C be two distinct points. The bisector equidistant from z and z is defined as { z ∈ H C | ρ ( z , z ) = ρ ( z , z ) } . The intersection of a bisector with the boundary of H C is a smooth hypersurfacein ∂H C called a spinal sphere , which is diffeomorphic to a sphere. An example isthe bisector C = { [ z : it : 1] ∈ H C | | z | < − t , z ∈ C , t ∈ R } . Its boundary, the unit spinal sphere , can be described as U = { ( ζ, ν ) ∈ N | | ζ | + ν = 1 } . Parametrisation of complex hyperbolic triangle groups of type [ m , m , n , n , n ]For r , r (cid:62) α ∈ (0 , π ), let C , C and C be the complex geodesics withrespective polar vectors c = − r e − iθ r e − iθ , c = r e iθ − r e iθ and c = , where θ = ( π − α ) / ∈ ( − π/ , π/ C , C , C is anultra-parallel [ m , m , α , where r k = cosh( m k / k = 1 , ι k be the minimal complex reflection of order n k in the chain C k for k = 1 , , (cid:104) ι , ι , ι (cid:105) generated by these three complex reflections is an ultra-parallelcomplex hyperbolic triangle groups of type [ m , m , n , n , n ]. Looking at thearrangement of the chains C , C and C in the Heisenberg space N , the finitechain C is the (Euclidean) unit circle in C × { } , whereas C and C are verti-cal lines through the points ϕ = r e iθ and ϕ = − r e − iθ respectively, see Fig-ure 2. For k = 1 ,
2, the reflection ι k rotates any vertical chain as a set through πn k around C k . 4. Compression Property
Let C , C , C be chains in N as in the previous section. Let ι k be the minimalcomplex reflection of order n k in the chain C k for k = 1 , ,
3. We will assume that n = 2. To prove the discreteness of the group (cid:104) ι , ι , ι (cid:105) we will use the followingversion of Klein’s combination theorem discussed in [15]: •• • C C C r e iθ − r e − iθ Figure 2.
Chains C , C and C (figure from [6]). Proposition 4.1.
If there exist subsets U , U and V in N with U ∩ U = ∅ and V (cid:40) U such that ι ( U ) = U and g ( U ) (cid:40) V for all g (cid:54) = Id in (cid:104) ι , ι (cid:105) , then thegroup (cid:104) ι , ι , ι (cid:105) is a discrete subgroup of PU(2 , . Groups with such properties arecalled compressing. Projecting the actions of complex reflections ι and ι to C × { } we obtain ro-tations j and j of C around ϕ = r e iθ and ϕ = − r e − iθ through πn and πn respectively. We will use Proposition 4.1 to prove the following Lemma: Lemma 4.2. If | f (0) | (cid:62) for all f (cid:54) = Id in (cid:104) j , j (cid:105) and | h (0) | (cid:62) for all verticalHeisenberg translations h (cid:54) = Id in (cid:104) ι , ι (cid:105) , then the group (cid:104) ι , ι , ι (cid:105) is discrete.Proof. Consider the unit spinal sphere U = { ( ζ, ν ) ∈ N | | ζ | + ν = 1 } . The complex reflection ι in C is given by ι ([ z : z : z ]) = [ − z : z : − z ] = [ z : − z : z ] . The complex reflection ι preserves the bisector C = { [ z : it : 1] ∈ H C | | z | < − t , z ∈ C , t ∈ R } and hence preserves the unit spinal sphere U which is the boundary of the bisec-tor C . The complex reflection ι interchanges the points [0 : 1 : 1] and [0 : − H C , which correspond to the points (0 ,
0) and ∞ in N . Therefore, ι leaves U invariant and switches the inside of U with the outside.Let U be the part of N \ U outside U , containing ∞ , and let U be the partinside U , containing the origin. Clearly U ∩ U = ∅ and ι ( U ) = U . OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 11
Therefore, if we find a subset V (cid:40) U such that g ( U ) (cid:40) V for all elements g (cid:54) = Idin (cid:104) ι , ι (cid:105) , then we will show that (cid:104) ι , ι , ι (cid:105) is discrete. Let W = { ( ζ, ν ) ∈ N | | ζ | = 1 } be the cylinder consisting of all vertical chains through ζ ∈ C with | ζ | = 1. Let W = { ( ζ, ν ) ∈ N | | ζ | > } and W = { ( ζ, ν ) ∈ N | | ζ | < } be the parts of N \ W outside and inside the cylinder W respectively. We have U ⊂ W and so g ( U ) ⊂ g ( W ) for all g ∈ (cid:104) ι , ι (cid:105) . The set W is a union of verticalchains. We know that elements of (cid:104) ι , ι (cid:105) map vertical chains to vertical chains.There is also a vertical translation on the chain itself. Therefore, we look at boththe intersection of the images of W with C × { } and the vertical displacementof W .Elements of (cid:104) ι , ι (cid:105) move the intersection of W with C × { } by rotations j and j around r e iθ and − r e − iθ through πn and πn respectively. Provided that theinterior of the unit circle is mapped completely off itself under all non-identity ele-ments in (cid:104) j , j (cid:105) , then the same is true for W and hence for U under all elementsin (cid:104) ι , ι (cid:105) that are not vertical Heisenberg translations.A vertical Heisenberg translation will shift W and its images g ( W ) verticallyby the same distance, hence the same is true for U and its images g ( U ).We choose V to be the union of all the images of U under all non-vertical ele-ments of (cid:104) ι , ι (cid:105) . This subset will satisfy the compressing conditions assuming thatthe interior of the unit circle is mapped off itself by any non-identity element in (cid:104) j , j (cid:105) and that the interior of the unit spinal sphere U is mapped off itself by anynon-identity vertical Heisenberg translation in (cid:104) ι , ι (cid:105) . Since the radius of the unitcircle is preserved under rotations, we need to show that the origin is moved thedistance of at least twice the radius of the circle: | f (0) | (cid:62) f ∈ (cid:104) j , j (cid:105) , f (cid:54) = Id . Since vertical translations shift the spinal spheres vertically, we need to show thatthey shift by at least the height of the spinal sphere: | h (0) | (cid:62) h ∈ (cid:104) ι , ι (cid:105) , h (cid:54) = Id . We see that the conditions of this Lemma ensure that the sets U , U and V satisfythe conditions of Proposition 4.1. (cid:3) Orders of Reflection
The group Γ = (cid:104) ι , ι , ι (cid:105) generated by the three complex reflections is an ultra-parallel complex hyperbolic triangle group of type [ m , m , n , n , n ]. We wantto know for what orders of the complex reflections ι and ι is the group Γ discrete.We will use the work of Hersonsky and Paulin [4].Recall the Heisenberg group N endowed with the group law( ζ , ν ) ∗ ( ζ , ν ) = (cid:16) ζ + ζ , ν + ν + 2 Im (cid:0) ζ ¯ ζ (cid:1)(cid:17) introduced in section 2.6. First, we will use Proposition 5 . n = 2) of [4]: Proposition 5.1.
Let Γ be a discrete cocompact subgroup in N . Let π : N → C bethe canonical projection defined by π ( ζ, ν ) = ζ . Then π (Γ) is a cocompact lattice in C .Proof. From the Heisenberg group N there exists a central extension0 → R → N → C → . By this, we have that Ker( π ) = R and N / R = C . Note that Γ ∩ R is a normal subgroup of Γ, since R is in the centre of N . Therefore,the group G = Γ / (Γ ∩ R )which identifies to π (Γ) acts on C . We can see that G acts with bounded quotienton N / R and therefore π (Γ) acts cocompactly on C .We next need to show that G acts discretely on C . For a contradiction, sup-pose not. For a sequence to converge on the plane C , we are able to bound theircorresponding elements in Γ by applying a vertical Heisenberg translation H in Γ,which exists due to Γ being non-abelian. This implies that there is a convergentsubsequence to an element (0 , t ) , t ∈ R . This is the required contradiction since Γis discrete, and hence discrete on C , which completes the proof. (cid:3) We now use two more results from Hersonsky and Paulin [4] (Theorem 5 . . Theorem 5.2.
Let G be a cocompact, discrete, torsion-free subgroup of isometriesof N n − . There exists a universal constant I n such that G contains a cocompactlattice of index less than or equal to I n . Moreover, I n ≤ π ) n ( n − . Proposition 5.3.
We have I = 6 . For n = 2, Theorem 5.2 and Proposition 5.3 tells us that there exists a constant, I , such that the group of Heisenberg translations, which is a cocompact, discrete,torsion free subgroup of the group of Heisenberg isometries, contains a lattice ofindex less than or equal to I = 6. From Proposition 5.1, we know that the canon-ical projection of this isometry group is a cocompact lattice in C .That is, by the classification of Euclidean 2-orbifolds, the group of Heisenbergtranslations contains a lattice subgroup of index m = 1 , , , m = 2 then the canonical projection of the translation subgroup is S (2 , , ,
2) or P (2 , m = 3 , , , − , (2 , , − , or (2 , , − triangle group respectively. As the Heisenbergtranslations are generated by the complex reflections, the orders of the complexreflections have to be contained in one of these groups. This gives rise to Theorem1.1. Remark . The cases when the unordered pair of orders of the complex reflections ι and ι is { , } and { , } were discussed in [6] and [12] respectively. OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 13 Parametrisation of complex hyperbolic triangle groups of type [ m, m, n , n , m , m , n , n , n ]-groups with m = m = m , n = 2 and { n , n } = { , } , { , } , { , } , { , } and { , } . In these cases thesetting described in section 3 is as follows. We consider the following configurationof chains in N : C is the (Euclidean) unit circle in C × { } , whereas C and C are vertical lines through the points ϕ = re iθ and ϕ = − re − iθ respectively, where r = cosh( m/
2) and θ ∈ ( − π/ , π/ C , C , C is an ultra-parallel [ m, m, α = π − θ ∈ (0 , π ).We will consider the ultra-parallel triangle group Γ = (cid:104) ι , ι , ι (cid:105) generated by theminimal complex reflections ι , ι , ι of orders n , n , C , C , C re-spectively.The description of complex reflections in section 2.8 in these cases are as follows:the reflection ι k for k = 1 , ζ, ω ) (cid:55)→ (cid:16) µ k ζ + (1 − µ k ) ϕ k , ω + 2 | ϕ k | Im(1 − µ k ) + 2 Im((1 − µ k ) ¯ ϕ k ζ ) (cid:17) , where µ k = exp(2 πi/n k ), and can be decomposed into a product of a Heisenbergtranslation and a Heisenberg rotation: ι k = R µ k ◦ T ( ξ k ,ν k ) = T ( µ k ξ k ,ν k ) ◦ R µ k , where ξ k = ( ¯ µ k − ϕ k and ν k = − | ϕ k | · Im(1 − µ k ) = 2 | ϕ k | sin(2 π/n k ) . For k = 1 ,
2, the reflection ι k rotates any vertical chain as a set through πn k around C k . 7. Subgroup of Heisenberg Translations
Let Γ = (cid:104) ι , ι , ι (cid:105) be as in section 6. In this section we will consider the structureof the subgroup E = (cid:104) ι , ι (cid:105) in more detail. For the purpose of the followingcalculations, we will use the notation ι a a ...a n = ι a ι a ...ι a n . Proposition 7.1.
Let the minimal complex reflections ι , ι have orders , re-spectively. Let T be the subgroup of all Heisenberg translations in E . Every el-ement of E can be written as a product of a Heisenberg translation and a word w = { Id , ι , ι , ι , ι , ι } . The group T is generated by the elements T = ι and T = ι . Let H = [ T , T ] = ( ι ) . Every element of T is of the form T x T y H n for some x, y, n ∈ Z . The elements T , T , H are Heisenberg translations by ( v , t ) = (4 r √ θ ) · i, √ r cos ( θ )) , ( v , t ) = (2 r cos( θ ) · (3 + i √ , r sin( θ ) cos( θ ) − √ r cos ( θ )) , (0 , ν ) = (0 , r √ ( θ )) , respectively. The subgroup of vertical Heisenberg translations in E is an infinitecyclic group generated by H . The shortest non-trivial vertical translations in E are H and H − . Proof.
We can write every element in E as a word in the generators ι ± and ι ± .Using the relations ι − = ι , ι − = ι and ι − = ι we can rewrite every element asa word in just ι and ι . As µ is a second root of unity, and µ is a third root ofunity, we will obtain Heisenberg translations with words containing an even numberof ι and a multiple of three of ι . Consider the words ι k k k k k of length five.Straightforward computation shows that the elements ι , ι , ι , ι and ι are Heisenberg translations. Consider the group generated by these fiveHeisenberg translations. This group is generated by T = ι and T = ι since ι = T · T − , ι = T · T − and ι = T − . The remaining words of length five which are not Heisenberg translations can beexpressed in terms of T , T and a remainder word of length at most four: ι = T · T − · ι and ι = T − · ι . Let f ∈ E = (cid:104) ι , ι (cid:105) . We can write f as a product of Heisenberg translations interms of T and T and one word of length at most 4. Moreover, using the relations ι = T · T − · ι ,ι = T · T − · ι ,ι = T · T − · ι ,ι = T · T − · ι ,ι = T − · ι , ι = T · T − · ι ,ι = T − · ι ,ι = T · ι ,ι = T · ι , we can rewrite f as a product of an element in (cid:104) T , T (cid:105) and a word w = { Id , ι , ι , ι , ι , ι } . If w = Id, then f is a Heisenberg translation. There-fore T = (cid:104) T , T (cid:105) is the subgroup of all Heisenberg translations in E .Let H = [ T , T ] ∈ T . As a commutator of two Heisenberg translations, the el-ement H is a vertical Heisenberg translation and lies in the centre of N , hence[ H, T ] = [ H, T ] = 1. Direct computation shows that T HT − = ( ι ι ) . On theother hand, T H = HT implies T HT − = HT T − = H , hence H = ( ι ι ) .Using the relations HT = T H , HT = T H and T T = T T H − , every elementof T can be written in the form T x T y H n for some x, y, n ∈ Z . The elements T and T are Heisenberg translations by ( v , t ) and ( v , t ) respectively. The com-mutator H = [ T , T ] is a vertical Heisenberg translation by ν = 4 Im( v ¯ v ). Wedetermine ( v k , t k ) for k = 1 , ν by direct computation. Projection to C maps H to the identity, T k to the Euclidean translation by v k and T x T y H n to the Eu-clidean translation by xv + yv . Hence T x T y H n is a vertical translation if andonly if x = y = 0, i.e. if it is a power of H . Therefore the subgroup of verticalHeisenberg translations in E is generated by H . (cid:3) Proposition 7.2.
Let the minimal complex reflections ι , ι have orders , re-spectively. Let T be the subgroup of all Heisenberg translations in E . Every el-ement of E can be written as a product of a Heisenberg translation and a word w = { Id , ι , ι , ι } . The group T is generated by the elements T = ι and T = ι . OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 15
Let H = [ T , T ] = ( ι ) . Every element of T is of the form T x T y H n for some x, y, n ∈ Z . The elements T , T , H are Heisenberg translations by ( v , t ) = (4 r cos( θ ) · i, r cos ( θ )) , ( v , t ) = (4 r cos( θ ) , r sin( θ ) cos( θ )) , (0 , ν ) = (0 , r cos ( θ )) , respectively. The subgroup of vertical Heisenberg translations in E is an infinitecyclic group generated by H . The shortest non-trivial vertical translations in E are H and H − .Proof. We can write every element in E as a word in the generators ι ± and ι ± .Using the relations ι − = ι , ι − = ι , ι − = ι and ι − = ι we can rewrite everyelement as a word in just ι and ι . As µ is a second root of unity, and µ is afourth root of unity, we will obtain Heisenberg translations with words containing x number of ι and x number of ι with x + x ≡ ι k k k of length three. Straightforward computation shows that the ele-ments ι , ι and ι are Heisenberg translations. Consider the group generatedby these three Heisenberg translations. This group is generated by T = ι and T = ι since ι = T − .The remaining words of length three which are not Heisenberg translations canbe expressed in terms of T , T and a remainder word of length at most two: ι = T · T − · ι and ι = T − · ι . Let f ∈ E = (cid:104) ι , ι (cid:105) . We can write f as a product of Heisenberg translations interms of T and T and one word of length at most 2. Moreover, using the relations ι = T · T − · ι , ι = T − · ι , we can rewrite f as a product of an element in (cid:104) T , T (cid:105) and a word w = { Id , ι , ι , ι } . If w = Id, then f is a Heisenberg translation. Therefore T = (cid:104) T , T (cid:105) is the subgroup of all Heisenberg translations in E .Let H = [ T , T ] ∈ T . As a commutator of two Heisenberg translations, the el-ement H is a vertical Heisenberg translation and lies in the centre of N , hence[ H, T ] = [ H, T ] = 1. Direct computation shows that T HT − = ( ι ι ) . On theother hand, T H = HT implies T HT − = HT T − = H , hence H = ( ι ι ) .Using the relations HT = T H , HT = T H and T T = T T H − , every elementof T can be written in the form T x T y H n for some x, y, n ∈ Z . The elements T and T are Heisenberg translations by ( v , t ) and ( v , t ) respectively. The com-mutator H = [ T , T ] is a vertical Heisenberg translation by ν = 4 Im( v ¯ v ). Wedetermine ( v k , t k ) for k = 1 ,
2, and ν by direct computation. Projection to C maps H to the identity, T k to the Euclidean translation by v k and T x T y H n to the Eu-clidean translation by xv + yv . Hence T x T y H n is a vertical translation if andonly if x = y = 0, i.e. if it is a power of H . Therefore the subgroup of verticalHeisenberg translations in E is generated by H . (cid:3) Proposition 7.3.
Let the minimal complex reflections ι , ι have orders , re-spectively. Let T be the subgroup of all Heisenberg translations in E . Every el-ement of E can be written as a product of a Heisenberg translation and a word w = { Id , ι , ι , ι } . The group T is generated by the elements T = ι , T = ι and H = ι . Let H = [ T , T ] = ( ι ) . Every element of T is of the form T x T y H n for some x, y, n ∈ Z . The elements T , T , H are Heisenberg translations by ( v , t ) = (2 r cos( θ ) · (1 + i ) , r sin( θ ) cos( θ )) , ( v , t ) = ( − r cos( θ ) · (1 − i ) , − r sin( θ ) cos( θ )) , (0 , ν ) = (0 , r cos ( θ )) , respectively. The subgroup of vertical Heisenberg translations in E is an infinitecyclic group generated by H . The shortest non-trivial vertical translations in E are H and H − .Proof. We can write every element in E as a word in the generators ι ± , k for k = 1 ,
2. Using the relations ι − k = ι k , ι − k = ι k and ι − k = ι k we can rewrite everyelement as a word in just ι k for k = 1 ,
2. Consider the words ι k k k k of lengthfour. Using the decomposition ι k = R µ ◦ T ( ξ k ,ν k ) = T ( µξ k ,ν k ) ◦ R µ , we can write ι k k k k = ( R µ ◦ T ( ξ k ,ν k ) ) ◦ ( R µ ◦ T ( ξ k ,ν k ) ) ◦ ( R µ ◦ T ( ξ k ,ν k ) ) ◦ ( R µ ◦ T ( ξ k ,ν k ) )= ( R µ ) ◦ T ( µξ k ,ν k ) ◦ T ( µ ξ k ,ν k ) ◦ T ( µ ξ k ,ν k ) ◦ T ( ξ k ,ν k ) = T ( µξ k ,ν k ) ◦ T ( µ ξ k ,ν k ) ◦ T ( µ ξ k ,ν k ) ◦ T ( ξ k ,ν k ) , hence ι k k k k is a Heisenberg translation. Using the relations above, we can seethat all words ι k k k k of length four can be expressed in terms of T = ι , T = ι and H = ι as ι = T · H − · T − · T − ,ι = T − · T − ,ι = T · H − · T − ,ι = T · T · H − · T − · T − ,ι = T · H · T − ,ι = T · T · H − · T − . ι = H · T − ,ι = T · T ,ι = T − ,ι = T · H,ι = T − , Let f ∈ E = (cid:104) ι , ι (cid:105) . We can write f as a product of Heisenberg translations interms of T , T , H and one word of length at most 3. Moreover, using the relations ι = T · ι ,ι = T · H · T − · ι ,ι = T · ι , ι = H · T − · ι ,ι = T · H − · T − · T − · ι ,ι = T · H · T − · ι , ι = T · ι ,ι = H · T − · ι ,ι = T − · T − · ι , we can rewrite f as a product of an element in (cid:104) T , T , H (cid:105) and a word w = { Id , ι , ι , ι } . If w = Id, then f is a Heisenberg translation. Therefore T = (cid:104) T , T , H (cid:105) is the subgroup of all Heisenberg translations in E .Direct computation shows that ι = ι = H . So H = [ T , T ] ∈ T .The element H is a vertical Heisenberg translation and lies in the centre of N ,hence [ H, T ] = [ H, T ] = 1. Using the relations HT = T H , HT = T H and T T = T T H , every element of T can be written in the form T x T y H n forsome x, y, n ∈ Z . The elements T and T are Heisenberg translations by ( v , t ) OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 17 and ( v , t ) respectively. The vertical Heisenberg translation H is given by (0 , ν )where ν = 4 Im( v ¯ v ). We determine ( v k , t k ) and ν by direct computation. Pro-jection to C maps H to the identity, T k to the Euclidean translation by v k and T x T y H n to the Euclidean translation by xv + yv . Hence T x T y H n is a verticaltranslation if and only if x = y = 0, i.e. if it is a power of H . Therefore the subgroupof vertical Heisenberg translations in E is generated by H . (cid:3) Proposition 7.4.
Let the minimal complex reflections ι , ι have orders , re-spectively. Let T be the subgroup of all Heisenberg translations in E . Every el-ement of E can be written as a product of a Heisenberg translation and a word w = { Id , ι , ι , ι , ι , ι } . The group T is generated by the elements T = ι , T = ι and H = ι . Let H = [ T , T ] = ( ι ) . Every element of T is of the form T x T y H n for some x, y, n ∈ Z . The elements T , T , H are Heisenberg translations by ( v , t ) = (2 r cos( θ ) · (1 + i √ , √ r cos ( θ ) + 8 r sin( θ ) cos( θ )) , ( v , t ) = ( − r cos( θ ) · (1 − i √ , √ r cos ( θ ) − r sin( θ ) cos( θ )) , (0 , ν ) = (0 , √ r cos ( θ )) , respectively. The subgroup of vertical Heisenberg translations in E is an infinitecyclic group generated by H . The shortest non-trivial vertical translations in E are H and H − .Proof. We can write every element in E as a word in the generators ι ± and ι ± l for l = 1 , , , ,
5. Using the relations ι − = ι , and ι − l = ι − l for l = { , , , , } , wecan rewrite every element as a word in just ι k for k = 1 ,
2. As µ is a second rootof unity, and µ is a sixth root of unity, we will obtain Heisenberg translations withwords containing x number or ι and x number of ι where x + x ≡ ι , ι , ι , ι and ι are Heisenberg translations. Consider the group generated by these Heisenbergtranslations. This group is generated by T = ι , T = ι and H = ι since ι = H · T − · T and ι = T − · T · H − . The remaining words of length four which are not Heisenberg translations can beexpressed in terms of T , T , H and a remainder word of length at most two: ι = H · T − · ι ,ι = T · T − · ι , ι = H · T − · T · T − · ι ,ι = T − · ι . Let f ∈ E = (cid:104) ι , ι (cid:105) . We can write f as a product of Heisenberg translations interms of T , T , H and one word of length at most 3. Moreover, using the relations ι = H · T − · ι ,ι = T · T − · ι , ι = H · T − · ι ,ι = T − · T · H − · ι , ι = H · T − · T · T − · ι , we can rewrite f as a product of an element in (cid:104) T , T , H (cid:105) and a word w = { Id , ι , ι , ι , ι , ι } . If w = Id, then f is a Heisenberg translation. There-fore T = (cid:104) T , T , H (cid:105) is the subgroup of all Heisenberg translations in E . Direct computation shows that ι = ι = H . So H = [ T , T ] ∈ T .The element H is a vertical Heisenberg translation and lies in the centre of N ,hence [ H, T ] = [ H, T ] = 1. Using the relations HT = T H , HT = T H and T T = T T H , every element of T can be written in the form T x T y H n forsome x, y, n ∈ Z . The elements T and T are Heisenberg translations by ( v , t )and ( v , t ) respectively. The vertical Heisenberg translation H is given by (0 , ν )where ν = 4 Im( v ¯ v ). We determine ( v k , t k ) for k = 1 ,
2, and ν by direct compu-tation. Projection to C maps H to the identity, T k to the Euclidean translationby v k and T x T y H n to the Euclidean translation by xv + yv . Hence T x T y H n isa vertical translation if and only if x = y = 0, i.e. if it is a power of H . Thereforethe subgroup of vertical Heisenberg translations in E is generated by H . (cid:3) Proposition 7.5.
Let the minimal complex reflections ι , ι have orders , re-spectively. Let T be the subgroup of all Heisenberg translations in E . Every el-ement of E can be written as a product of a Heisenberg translation and a word w = { Id , ι , ι , ι , ι , ι } . The group T is generated by the elements T = ι , T = ι and H = ι . Let H = [ T , T ] = ( ι ) . Every element of T is of the form T x T y H n for some x, y, n ∈ Z . The elements T , T , H are Heisenberg translations by ( v , t ) = ( r cos( θ ) · (3 + i √ , r sin( θ ) cos( θ )) , ( v , t ) = ( − r cos( θ ) · (3 − i √ , − r sin( θ ) cos( θ )) , (0 , ν ) = (0 , √ r cos ( θ )) , respectively. The subgroup of vertical Heisenberg translations in E is an infinitecyclic group generated by H . The shortest non-trivial vertical translations in E are H and H − .Proof. We can write every element in E as a word in the generators ι ± , and ι ± l for l = 1 , , , ,
5. Using the relations ι − = ι , ι − = ι and ι − l = ι − l for l = { , , , , } , we can rewrite every element as a word in just ι k for k = 1 ,
2. As µ is a third root of unity, and µ is a sixth root of unity, we will obtain Heisenbergtranslations with words containing x number of ι and x number of ι where x + x ≡ ι , ι , ι , ι , ι and ι are Heisenberg translations. Consider the group generated by these Heisenbergtranslations. This group is generated by T = ι , T = ι and H = ι since ι = T − · T − , ι = T · T · H and ι = H. The remaining words of length four which are not Heisenberg translations can beexpressed in terms of T , T , H and a remainder word of length at most three: ι = T − · H · ι ,ι = H · ι ,ι = T − · ι . ι = H − · T − · T − · ι ,ι = T · ι , ι = T − · H − · ι ,ι = T · H · T − · ι , OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 19
Let f ∈ E = (cid:104) ι , ι (cid:105) . We can write f as a product of Heisenberg translations interms of T , T , H and one word of length at most 3. Moreover, using the relations ι = T − · ι ,ι = H · T · ι ,ι = H · T · T · H · T − · ι , ι = T − · ι ,ι = H · T − · ι , ι = T − · ι ,ι = T − · T − · T − · ι , we can rewrite f as a product of an element in (cid:104) T , T , H (cid:105) and a word w = { Id , ι , ι , ι , ι , ι } . If w = Id, then f is a Heisenberg translation.Therefore T = (cid:104) T , T , H (cid:105) is the subgroup of all Heisenberg translations in E .Since ι = ι = H , we have that H = [ T , T ] ∈ T . The element H is a verti-cal Heisenberg translation and lies in the centre of N , hence [ H, T ] = [ H, T ] = 1.Using the relations HT = T H , HT = T H and T T = T T H , every elementof T can be written in the form T x T y H n for some x, y, n ∈ Z . The elements T and T are Heisenberg translations by ( v , t ) and ( v , t ) respectively. The verticalHeisenberg translation H is given by (0 , ν ) where ν = 4 Im( v ¯ v ). We determine( v k , t k ) for k = 1 ,
2, and ν by direct computation. Projection to C maps H tothe identity, T k to the Euclidean translation by v k and T x T y H n to the Euclideantranslation by xv + yv . Hence T x T y H n is a vertical translation if and only if x = y = 0, i.e. if it is a power of H . Therefore the subgroup of vertical Heisenbergtranslations in E is generated by H . (cid:3) Remark . For Proposition 7.1 and 7.2, the group T has the presentation T = (cid:104) T , T , H | [ T , T ] = H, [ H, T ] = [ H, T ] = 1 (cid:105) and is isomorphic to the uniform lattice N as defined in section 2.6.For Proposition 7.3 and 7.4, the group T has the presentation T = (cid:104) T , T , H | [ T , T ] = H , [ H, T ] = [ H, T ] = 1 (cid:105) and is isomorphic to the uniform lattice N as defined in section 2.6.For Proposition 7.5, the group T has the presentation T = (cid:104) T , T , H | [ T , T ] = H , [ H, T ] = [ H, T ] = 1 (cid:105) and is isomorphic to the uniform lattice N as defined in section 2.6. Remark . An alternative approach to the understanding of the structure of thesubgroup E = (cid:104) ι , ι (cid:105) is to use the classification of almost-crystallographic groupsby Dekimpe [2]. Definition . An almost-crystallographic group is a uniform discrete subgroup E of G (cid:111) C , where G is a connected, simply connected nilpotent Lie group and C is amaximal compact subgroup of Aut( G ).As a discrete subgroup of N (cid:111) U(1) (see section 2.8), the group E is an almost-crystallographic group with G = N and U(1) ⊂ C ⊂ Aut( N ). The projection of E = (cid:104) ι , ι (cid:105) to C is a wallpaper group Q = (cid:104) j , j (cid:105) , where j k is the rotation of C around ϕ k through 2 π/n k , for k = 1 ,
2, obtained by projecting ι k to C . The case [ m, m,
0; 2 , , Q = (cid:104) j , j (cid:105) is generated by tworotations of orders 2 and 3 respectively, and has a presentation (cid:104) j , j | j = j = ( j ) = 1 (cid:105) . The standard notation for this wallpaper group is p6 ; see for example [1]. In theclassification of three-dimensional almost-crystallographic groups in section 7 . p6 appears in case 16 on page 166. In this case the group E is generated by elements a, b, c, α with relations[ b, a ] = c k , [ c, a ] = [ c, b ] = [ c, α ] = 1 , αa = abαc k , αb = a − αc k , α = c k . We consider the generators ι = α a and ι = α , so that (cid:0) α a (cid:1) = α = 1 . Thehypothesis α = 1 implies that k = 0. The hypothesis (cid:0) α a (cid:1) = 1 can be rewrittenas c − k +2 k +2 k = 1 ⇒ k = 2 ( k + k ) . The translations T and T in Proposition 7.1 are T = ι = α (cid:16) α a (cid:17) α (cid:16) α a (cid:17) α = b − a − c − k +2 k ; T = ι = α aα α (cid:16) α a (cid:17) α = a − b − a − c − k . Their commutator is H = [ T , T ]= (cid:16) b − a − c − k +2 k (cid:17) − · (cid:16) a − b − a − c − k (cid:17) − · b − a − c − k +2 k · a − b − a − c − k = ab abab − a − b − a − = c k . On the other hand, the kernel of the map E = (cid:104) ι , ι (cid:105) → (cid:104) j , j (cid:105) given by ι → j , ι → j is generated by ( ι ) . Calculating ( ι ) we obtain c k + k ) . Us-ing k = 2 ( k + k ) we can rewrite this as ( ι ) = c k . Hence the element H = [ T , T ] = ( ι ) = c k is the shortest vertical Heisenberg translation in E = (cid:104) ι , ι (cid:105) . The case [ m, m,
0; 2 , , Q = (cid:104) j , j (cid:105) is generated by tworotations of orders 2 and 4 respectively, and has a presentation (cid:104) j , j | j = j = ( j ) = 1 (cid:105) . The standard notation for this wallpaper group is p4 . In the classification of three-dimensional almost-crystallographic groups in section 7 . p4 appears in case 10 on page 163. In this case the group E is generated by elements a, b, c, α with relations[ b, a ] = c k , [ c, a ] = [ c, b ] = [ c, α ] = 1 , αa = bαc k , αb = a − αc k , α = c k . We consider the generators ι = α a and ι = α , so that (cid:0) α a (cid:1) = α = 1 . Thehypothesis α = 1 implies that k = 0. The hypothesis (cid:0) α a (cid:1) = 1 can be rewrittenas c k + k = 1 ⇒ k + k = 0 . The translations T and T in Proposition 7.2 are T = ι = α (cid:16) α a (cid:17) α = b − c k ; T = ι = α aα = a − . OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 21
Their commutator is H = [ T , T ]= (cid:16) b − c k (cid:17) − · (cid:16) a − (cid:17) − · b − c k · a − = bab − a − = abc k b − a − = c k . On the other hand, the kernel of the map E = (cid:104) ι , ι (cid:105) → (cid:104) j , j (cid:105) given by ι → j , ι → j is generated by ( ι ) . Calculating ( ι ) we obtain c k . Hence theelement H = [ T , T ] = ( ι ) = c k is the shortest vertical Heisenberg translationin E = (cid:104) ι , ι (cid:105) . The case [ m, m,
0; 4 , , Q = (cid:104) j , j (cid:105) is generated by tworotations of order 4, and has a presentation (cid:104) j , j | j = j = ( j ) = 1 (cid:105) . The standard notation for this wallpaper group is p4 (for more details, see[ m, m,
0; 2 , , ι = αa and ι = α , sothat ( αa ) = α = 1 . The hypothesis α = 1 implies that k = 0. The hypothesis( αa ) = 1 can be rewritten as c − k +2 k +2 k = 1 ⇒ k = 2 ( k + k ) . The translations T and T in Proposition 7.3 are T = ι = αaαaαaα = ba − b − c k ; T = ι = α aαaαa = a − b − ac k − k . Their commutator is H = [ T , T ]= (cid:16) a − b − ac k − k (cid:17) − · (cid:16) ba − b − c k (cid:17) − · a − b − ac k − k · ba − b − c k = a − babab − a − b − aba − b − = c k . On the other hand, the kernel of the map E = (cid:104) ι , ι (cid:105) → (cid:104) j , j (cid:105) given by ι → j , ι → j is generated by ( ι ) . Calculating ( ι ) we obtain c k + k . Us-ing k = 2 ( k + k ) we can rewrite this as ( ι ) = c k . Hence the element H = ( ι ) = c k is the shortest vertical Heisenberg translation in E = (cid:104) ι , ι (cid:105) . The case [ m, m,
0; 2 , , Q = (cid:104) j , j (cid:105) is generated by tworotations of orders 2 and 6 respectively, and has a presentation (cid:104) j , j | j = j = ( j ) = 1 (cid:105) . The standard notation for this wallpaper group is p6 ; (for more details, see[ m, m,
0; 2 , , ι = α a and ι = α , sothat (cid:0) α a (cid:1) = α = 1 . The hypothesis α = 1 implies that k = 0. The hypothesis (cid:0) α a (cid:1) = 1 can be rewritten as c − k +2 k +2 k = 1 ⇒ k = 2 ( k + k ) . The translations T and T in Proposition 7.4 are T = ι = α (cid:16) α a (cid:17) α = b − a − c − k ; T = ι = α (cid:16) α a (cid:17) α = b − c k . Their commutator is H = [ T , T ]= (cid:16) b − c k (cid:17) − · (cid:16) b − a − c − k (cid:17) − · b − c k · b − a − c − k = bab − a − = c k . On the other hand, the kernel of the map E = (cid:104) ι , ι (cid:105) → (cid:104) j , j (cid:105) given by ι → j , ι → j is generated by ( ι ) . Calculating ( ι ) we obtain c k + k . Us-ing k = 2 ( k + k ) we can rewrite this as ( ι ) = c k . Hence the element H = ( ι ) = c k is the shortest vertical Heisenberg translation in E = (cid:104) ι , ι (cid:105) . The case [ m, m,
0; 3 , , Q = (cid:104) j , j (cid:105) is generated by tworotations of orders 3 and 6 respectively, and has a presentation (cid:104) j , j | j = j = ( j ) = 1 (cid:105) . The standard notation for this wallpaper group is p6 ; (for more details, see[ m, m,
0; 2 , , ι = α a and ι = α , sothat (cid:0) α a (cid:1) = α = 1 . The hypothesis α = 1 implies that k = 0. The hypothesis (cid:0) α a (cid:1) = 1 can be rewritten as c − k +3 k +3 k = 1 ⇒ k = 3 ( k + k ) . The translations T and T in Proposition 7.5 are T = ι = α aα aα = a − ; T = ι = α (cid:16) α a (cid:17) α a = b − c − k + k +2 k . Their commutator is H = [ T , T ]= (cid:16) b − c − k + k +2 k (cid:17) − · (cid:16) a − (cid:17) − · b − c − k + k +2 k · a − = bab − a − = c k . On the other hand, the kernel of the map E = (cid:104) ι , ι (cid:105) → (cid:104) j , j (cid:105) given by ι → j , ι → j is generated by ( ι ) . Calculating ( ι ) we obtain c ( k + k ) . Using2 k = 3 ( k + k ) we can rewrite this as ( ι ) = c k . Hence the element H =( ι ) = c k is the shortest vertical Heisenberg translation in E = (cid:104) ι , ι (cid:105) .8. Discreteness Results
Let Γ = (cid:104) ι , ι , ι (cid:105) be as in section 6. We will now use Lemma 4.2 to find conditionsfor the group Γ to be discrete for all possible orders { n , n } of the complex reflec-tions. For each case, we will check that both conditions of Lemma 4.2 are satisfied.That is, (cid:12)(cid:12) h (0) (cid:12)(cid:12) ≥ h (cid:54) = Id in (cid:104) ι , ι (cid:105) and | f (0) | (cid:62) f (cid:54) = Id in (cid:104) j , j (cid:105) . For the purposes of the following calculations,we will denote t = tan( θ ) and t π/ = tan (cid:0) π (cid:1) . OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 23
Proof of Proposition 1.2.
Proof.
Note that m ≥ log e (3) implies that r = cosh (cid:18) m (cid:19) ≥ √ . Any vertical translation in (cid:104) ι , ι (cid:105) is a power of the vertical translation H by(0 , r √ ( θ )). We need the displacement of each vertical translation H n , n (cid:54) = 0, to be at least the height of the spinal sphere, i.e.96 r √ ( θ ) ≥ ⇔ r cos ( θ ) ≥ √ . The hypothesis cos( α ) ≤ − for α ∈ (0 , π ) implies π ≤ α ≤ π and hence | θ | = (cid:12)(cid:12) π − α (cid:12)(cid:12) ≤ π . For cos( θ ) ≥ √ and r ≥ √ we have r cos ( θ ) ≥ > √ , hence the condition | h (0) |≥ h (cid:54) = Id in (cid:104) ι , ι (cid:105) .We can write every element f in (cid:104) j , j (cid:105) as a word in the generators j and j . Fig-ure 3 shows the points f (0) for all words f of length up to 6 in the case r = 1 and θ = 0. The group (cid:104) j , j (cid:105) is the projection to C of the group E = (cid:104) ι , ι (cid:105) . Projecting Figure 3.
Points f (0) for all words f up to length 6. ι and ι to C , we obtain the rotations j and j of C through π and π around ϕ and ϕ respectively. These rotations are given by j k ( z ) = µ k · z + (1 − µ k ) · ϕ k ,for k = { , } where µ = exp( πi ) and µ = exp(2 πi/ E is of the form T x T y H n w for some x, y, n ∈ Z and w ∈ { Id , ι , ι , ι , ι , ι } , where T = ι , T = ι and H = [ T , T ] areHeisenberg translations by ( v , t ), ( v , t ) and (0 , ν ) respectively and v = 4 r √ θ ) · i, v = 2 r cos( θ ) · (3 + i √ . Projection to C maps H to the identity, T k to the Euclidean translation by v k , T x T y H n to the Euclidean translation by xv + yv and w to the rotation el-ements { Id , j , j , j , j , j } respectively. Therefore every element of (cid:104) j , j (cid:105) is a product of a translation by xv + yv for some x, y ∈ Z and a rotation { Id , j , j , j , j , j } . Hence every point in the orbit of 0 under (cid:104) j , j (cid:105) is ofthe form p + xv + yv , where x, y ∈ Z and p ∈ { , j (0) , j (0) , j (0) , j (0) , j (0) } . Using | v | = | v | = 2 Re( v ¯ v ) = 48 r cos ( θ ), we calculate | p + xv + yv | = x | v | + y | v | +2 xy Re( v ¯ v ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | = 48 r cos ( θ ) · ( x + xy + y ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | . We make a coordinate change u = y − x and v = x + y , that is x = ( v − u ) / y = ( u + v ) /
2. Points ( x, y ) ∈ Z are mapped to points ( u, v ) ∈ Z with u ≡ v mod 2. We obtain | p + xv + yv | = 12 r cos ( θ ) · ( u + 3 v − au − bv + a + 3 b )= 12 r cos ( θ ) · (( u − a ) + 3( v − b ) ) , where a = − Re( p (¯ v − ¯ v ))24 r cos ( θ ) = − Re (cid:16) p (3 + i √ (cid:17) r cos( θ ) ,b = − Re( p (¯ v + ¯ v ))72 r cos ( θ ) = − Re (cid:16) p (1 − i √ (cid:17) r cos( θ )and a + 3 b = | p | r cos ( θ ) . Our aim is to show that | p + xv + yv | ≥ r for all ( x, y ) ∈ Z excluding thecase p = 0, x = y = 0 that corresponds to f = Id. This is equivalent to ( u − a ) + 3( v − b ) ≥ sec ( θ )4 for all ( u, v ) ∈ Z with u ≡ v mod 2 excluding the case a = b = u = v = 0. Note that this inequality is always satisfied if | u − a |≥ sec( θ )2 or | v − b |≥ sec( θ )2 √ , so we only need to check that g , p ( u, v ) = ( u − a ) + 3( v − b ) − sec ( θ )4 ≥ u, v ) ∈ Z with u ≡ v mod 2 inside the bounding box (cid:18) a − sec( θ )2 , a + sec( θ )2 (cid:19) × (cid:18) b − sec( θ )2 √ , b + sec( θ )2 √ (cid:19) . In the following table we list the values of a , b and a + 3 b for p ∈ { , j (0) , j (0) , j (0) , j (0) , j (0) } : OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 25 p a b a + 3 b j (0) √ (cid:16) t − √ (cid:17) − √ (cid:16) t + √ (cid:17) (cid:0) t + 1 (cid:1) j (0) − t √ (cid:0) t + 1 (cid:1) j (0) √ (cid:16) t − √ (cid:17) −
16 112 (cid:16) t − t √ (cid:17) j (0) (1 + t √ √ (cid:16) √ − t (cid:17) (cid:0) t + 1 (cid:1) j (0) − √ (cid:16) t + 3 √ (cid:17) − √ (cid:16) t + √ (cid:17) (cid:16) t + 4 t √ (cid:17) Under the assumption | θ | (cid:54) π we have t = tan( θ ) ∈ [ − d, d ] and sec( θ ) ∈ [1 , d ],where d = √ ≈ . a and b andthe size of the bounding box (cid:0) min( a ) − d, max( a ) + d (cid:1) × (cid:18) min( b ) − , max( b ) + 13 (cid:19) . We then calculate g , p ( u, v ) = ( u − a ) + 3( v − b ) − sec ( θ )4= u + 3 v − au − bv + ( a + 3 b ) − (cid:32) t + 14 (cid:33) and check that g , p ( u, v ) ≥ u, v ) ∈ Z with u ≡ v mod 2 inside thebounding box. • p = 0, a = b = 0: The bounding box ( − d, d ) × (cid:0) − , (cid:1) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2, the point ( u, v ) =(0 , f = Id. • p = j (0), a = √ (cid:16) t − √ (cid:17) ∈ (cid:2) − , − (cid:3) , b = − √ (cid:16) t + √ (cid:17) ∈ (cid:2) − , (cid:3) :The bounding box (cid:18) − − d, −
13 + d (cid:19) × (cid:18) − , (cid:19) ⊂ ( − , × ( − , , g , ( u, v ) = u + 3 v + (cid:18) − t √ (cid:19) u + (cid:16) t √ (cid:17) v + ( t + 1)12is non-negative: g , (0 ,
0) = t +112 ≥ > • p = j (0), a = , b = − t √ ∈ (cid:2) − , (cid:3) : The bounding box (cid:18) − d,
12 + d (cid:19) × (cid:18) − , (cid:19) ⊂ ( − , × ( − , , g , ( u, v ) = u + 3 v − u + tv √ g , (0 ,
0) = 0 . • p = j (0), a = √ (cid:16) t − √ (cid:17) ∈ (cid:2) − , − (cid:3) , b = − : The bounding box (cid:18) − − d, −
56 + d (cid:19) × (cid:18) − , (cid:19) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2. • p = j (0), a = (cid:16) t √ (cid:17) ∈ (cid:2) , (cid:3) , b = √ (cid:16) √ − t (cid:17) ∈ (cid:2) , (cid:3) : Thebounding box (cid:18) − d,
12 + d (cid:19) × (cid:18) − , (cid:19) ⊂ ( − , × ( − , , g , ( u, v ) = u + 3 v − u (cid:16) t √ (cid:17) − v (cid:16) − t √ (cid:17) is non-negative: g , (0 ,
0) = 0 . • p = j (0), a = − √ (cid:16) t + 3 √ (cid:17) ∈ (cid:2) − , − (cid:3) , b = − √ (cid:16) t + √ (cid:17) ∈ (cid:2) − , − (cid:3) : The bounding box (cid:18) − − d, −
23 + d (cid:19) × (cid:18) − , (cid:19) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2.In all cases we have shown that g , p ( u, v ) ≥
0, hence | p + xv + yv | ≥ r . Underthe assumption m ≥ log e (3) we have 3 r = 3 cosh (cid:0) m (cid:1) ≥
4. Therefore | f (0) |≥ f (cid:54) = Id in (cid:104) j , j (cid:105) . Hence all conditions of Lemma 4.2 are satisfied and wecan conclude that the group (cid:104) ι , ι , ι (cid:105) is discrete. (cid:3) Proof of Proposition 1.3 for n = 2 . Proof.
Note that m ≥ log e (3 + 2 √
2) implies that r = cosh (cid:18) m (cid:19) ≥ √ . Any vertical translation in (cid:104) ι , ι (cid:105) is a power of the vertical translation H by(0 , r cos ( θ )). We need the displacement of each vertical translation H n , n (cid:54) = 0,to be at least the height of the spinal sphere, i.e.64 r cos ( θ ) ≥ ⇔ r cos ( θ ) ≥ . The hypothesis cos( α ) ≤ − √ for α ∈ (0 , π ) implies π ≤ α ≤ π and hence | θ | = (cid:12)(cid:12) π − α (cid:12)(cid:12) ≤ π . For cos( θ ) ≥ √ √ and r ≥ √ r cos ( θ ) ≥ √ > , hence the condition | h (0) |≥ h (cid:54) = Id in (cid:104) ι , ι (cid:105) .We can write every element f in (cid:104) j , j (cid:105) as a word in the generators j and j .Figure 4 shows the points f (0) for all words f of length up to 6 in the case r = 1and θ = 0. The group (cid:104) j , j (cid:105) is the projection to C of the group E = (cid:104) ι , ι (cid:105) . Pro-jecting ι and ι to C , we obtain the rotations j and j of C through π and π around OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 27
Figure 4.
Points f (0) for all words f up to length 6. ϕ and ϕ respectively. These rotations are given by j k ( z ) = µ k · z + (1 − µ k ) · ϕ k ,for k = { , } where µ = exp( πi ) and µ = exp( πi/ E is of the form T x T y H n w for some x, y, n ∈ Z and w ∈ { Id , ι , ι , ι } , where T = ι , T = ι and H = [ T , T ] are Heisenbergtranslations by ( v , t ), ( v , t ) and (0 , ν ) respectively and v = 4 r cos( θ ) · i, v = 4 r cos( θ ) . Projection to C maps H to the identity, T k to the Euclidean translation by v k , T x T y H n to the Euclidean translation by xv + yv and w to the rotation elements { Id , j , j , j } respectively. Therefore every element of (cid:104) j , j (cid:105) is a product of atranslation by xv + yv for some x, y ∈ Z and a rotation { Id , j , j , j } . Henceevery point in the orbit of 0 under (cid:104) j , j (cid:105) is of the form p + xv + yv , where x, y ∈ Z and p ∈ { , j (0) , j (0) , j (0) } . Using | v | = | v | = 16 r cos ( θ ) and Re( v ¯ v ) = 0, we calculate | p + xv + yv | = x | v | + y | v | +2 xy Re( v ¯ v ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | = 16 r cos ( θ ) · ( x + y ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | . We make a coordinate change u = y − x and v = x + y , that is x = ( v − u ) / y = ( u + v ) /
2. Points ( x, y ) ∈ Z are mapped to points ( u, v ) ∈ Z with u ≡ v mod 2. We obtain | p + xv + yv | = 8 r cos ( θ ) · ( u + v − au − bv + a + b )= 8 r cos ( θ ) · (( u − a ) + ( v − b ) ) , where a = − Re( p (¯ v − ¯ v ))16 r cos ( θ ) = − Re (cid:0) p (1 + i ) (cid:1) r cos( θ ) ,b = − Re( p (¯ v + ¯ v ))16 r cos ( θ ) = − Re (cid:0) p (1 − i ) (cid:1) r cos( θ )and a + b = | p | r cos ( θ ) . Our aim is to show that | p + xv + yv | ≥ r for all ( x, y ) ∈ Z excluding thecase p = 0, x = y = 0 that corresponds to f = Id. This is equivalent to ( u − a ) + ( v − b ) ≥ sec ( θ )4 for all ( u, v ) ∈ Z with u ≡ v mod 2 excluding the case a = b = u = v = 0. Note that this inequality is always satisfied if | u − a |≥ sec( θ )2 or | v − b |≥ sec( θ )2 , so we only need to check that g , p ( u, v ) = ( u − a ) + ( v − b ) − sec ( θ )4 ≥ u, v ) ∈ Z with u ≡ v mod 2 inside the bounding box (cid:18) a − sec( θ )2 , a + sec( θ )2 (cid:19) × (cid:18) b − sec( θ )2 , b + sec( θ )2 (cid:19) . In the following table we list the values of a , b and a + b for p ∈ { , j (0) , j (0) , j (0) } : p a b a + b j (0) ( t − − ( t + 1) (cid:0) t + 1 (cid:1) j (0) − t (cid:0) t + 1 (cid:1) j (0) ( t + 2) −
12 14 (cid:0) t + 4 t + 5 (cid:1) Under the assumption | θ | (cid:54) π we have t = tan( θ ) ∈ (cid:104) √ − , − √ (cid:105) and sec( θ ) ∈ (cid:20) , √ (cid:16) √ − (cid:17)(cid:21) . In each of the four cases we list the bounds on a and b and thesize of the bounding box(min( a ) − γ, max( a ) + γ ) × (min( b ) − γ, max( b ) + γ ) , where γ = √ − √ . We then calculate g , p ( u, v ) = ( u − a ) + ( v − b ) − sec ( θ )4= u + v − au − bv + ( a + b ) − (cid:32) t + 14 (cid:33) and check that g , p ( u, v ) ≥ u, v ) ∈ Z with u ≡ v mod 2 inside thebounding box. • p = 0, a = b = 0: The bounding box ( − γ, γ ) × ( − γ, γ ) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2, the point ( u, v ) =(0 , f = Id. OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 29 • p = j (0), a = ( t − ∈ (cid:104) − − t π/ , − + t π/ (cid:105) , b = − ( t + 1) ∈ (cid:104) − − t π/ , − + t π/ (cid:105) : The bounding box (cid:18) − − t π/ − γ, −
12 + t π/ γ (cid:19) × (cid:18) − − t π/ − γ, −
12 + t π/ γ (cid:19) ⊂ ( − , × ( − , − , −
1) and (0 , g , ( u, v ) = u + v + (1 − t ) u + (1 + t ) v + ( t + 1)4is non-negative: g , ( − , −
1) = g , (0 ,
0) = t +14 ≥ > • p = j (0), a = , b = − t ∈ (cid:104) − t π/ , t π/ (cid:105) : The bounding box (cid:18) − γ,
12 + γ (cid:19) × (cid:18) − t π/ − γ, t π/ γ (cid:19) ⊂ ( − , × ( − , , g , ( u, v ) = u + v − u + tv is non-negative: g , (0 ,
0) = 0 . • p = j (0), a = ( t + 2) ∈ (cid:104) − t π/ , t π/ (cid:105) , b = − : The boundingbox (cid:18) − t π/ − γ, t π/ γ (cid:19) × (cid:18) − − γ, −
12 + γ (cid:19) ⊂ (0 , × ( − , , − g , ( u, v ) = u + v − ( t + 2) u + v + ( t + 1)is non-negative: g , (1 , −
1) = 0 . In all cases we have shown that g , p ( u, v ) ≥
0, hence | p + xv + yv | ≥ r . Underthe assumption m ≥ log e (3 + 2 √
2) we have 2 r = 2 cosh (cid:0) m (cid:1) ≥
4. Therefore | f (0) |≥ f (cid:54) = Id in (cid:104) j , j (cid:105) . Hence all conditions of Lemma 4.2 are satisfiedand we can conclude that the group (cid:104) ι , ι , ι (cid:105) is discrete. (cid:3) Proof of Proposition 1.3 for n = 4 . Proof.
As in the proof in section 8.2 above (for n = 2), m ≥ log e (3 + 2 √
2) impliesthat r = cosh (cid:18) m (cid:19) ≥ √ . Any vertical translation in (cid:104) ι , ι (cid:105) is a power of the vertical translation H by(0 , r cos ( θ )). We need the displacement of each vertical translation H n , n (cid:54) = 0,to be at least the height of the spinal sphere, i.e.16 r cos ( θ ) ≥ ⇔ r cos ( θ ) ≥ . The hypothesis cos( α ) ≤ − √ for α ∈ (0 , π ) implies | θ | = (cid:12)(cid:12) π − α (cid:12)(cid:12) ≤ π . Forcos( θ ) ≥ √ √ and r ≥ √ r cos ( θ ) ≥ √ > , hence the condition | h (0) |≥ h (cid:54) = Id in (cid:104) ι , ι (cid:105) .We can write every element f in (cid:104) j , j (cid:105) as a word in the generators j and j .Figure 5 shows the points f (0) for all words f of length up to 6 in the case r = 1and θ = 0. The group (cid:104) j , j (cid:105) is the projection to C of the group E = (cid:104) ι , ι (cid:105) . Figure 5.
Points f (0) for all words f up to length 6.Projecting ι and ι to C , we obtain the rotations j and j of C through π around ϕ and ϕ respectively. These rotations are given by j k ( z ) = µ · z + (1 − µ ) · ϕ k ,for k = { , } where µ = exp( πi/ E is of the form T x T y H n ι l for some x, y, n ∈ Z and l ∈ { , , , } , where T = ι , T = ι and H = [ T , T ] are Heisenberg translations by ( v , t ),( v , t ) and (0 , ν ) respectively and v = 2 r cos( θ ) · (1 + i ) , v = − r cos( θ ) · (1 − i ) . Projection to C maps H to the identity, T k to the Euclidean translation by v k , T x T y H n to the Euclidean translation by xv + yv and ι to the rotation j .Therefore every element of (cid:104) j , j (cid:105) is a product of a translation by xv + yv forsome x, y ∈ Z and a rotation j l for some l ∈ { , , , } . Hence every point in theorbit of 0 under (cid:104) j , j (cid:105) is of the form p + xv + yv , where x, y ∈ Z and p ∈ { , j (0) , j (0) , j (0) } . OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 31
Using | v | = | v | = 8 r cos ( θ ) and Re( v ¯ v ) = 0, we calculate | p + xv + yv | = x | v | + y | v | +2 xy Re( v ¯ v ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | = 8 r cos ( θ ) · ( x + y ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | . We make a coordinate change u = y − x and v = x + y , that is x = ( v − u ) / y = ( u + v ) /
2. Points ( x, y ) ∈ Z are mapped to points ( u, v ) ∈ Z with u ≡ v mod 2. We obtain | p + xv + yv | = 4 r cos ( θ ) · ( u + v − au − bv + a + b )= 4 r cos ( θ ) · (( u − a ) + ( v − b ) ) , where a = Re( p (¯ v − ¯ v ))8 r cos ( θ ) = − Re ( p )2 r cos( θ ) ,b = − Re( p (¯ v + ¯ v ))8 r cos ( θ ) = − Im ( p )2 r cos( θ )and a + b = | p | r cos ( θ ) . Our aim is to show that | p + xv + yv | ≥ r for all ( x, y ) ∈ Z excluding thecase p = 0, x = y = 0 that corresponds to f = Id. This is equivalent to ( u − a ) + ( v − b ) ≥ sec ( θ )2 for all ( u, v ) ∈ Z with u ≡ v mod 2 excluding the case a = b = u = v = 0. Note that this inequality is always satisfied if | u − a |≥ sec( θ ) √ or | v − b |≥ sec( θ ) √ , so we only need to check that g , p ( u, v ) = ( u − a ) + ( v − b ) − sec ( θ )2 ≥ u, v ) ∈ Z with u ≡ v mod 2 inside the bounding box (cid:18) a − sec( θ ) √ , a + sec( θ ) √ (cid:19) × (cid:18) b − sec( θ ) √ , b + sec( θ ) √ (cid:19) . In the following table we list the values of a , b and a + b for p ∈ { , j (0) , j (0) , j (0) } : p a b a + b j (0) − ( t + 1) (1 − t ) (cid:0) t + 1 (cid:1) j (0) − − t t + 1 j (0) ( t − − ( t + 1) (cid:0) t + 1 (cid:1) Under the assumption | θ | (cid:54) π we have t = tan( θ ) ∈ (cid:104) √ − , − √ (cid:105) and sec( θ ) ∈ (cid:20) , √ (cid:16) √ − (cid:17)(cid:21) . In each of the four cases we list the bounds on a and b and thesize of the bounding box(min( a ) − γ, max( a ) + γ ) × (min( b ) − γ, max( b ) + γ ) , where γ = √ −
1. We then calculate g , p ( u, v ) = ( u − a ) + ( v − b ) − sec ( θ )2= u + v − au − bv + ( a + b ) − (cid:32) t + 12 (cid:33) and check that g , p ( u, v ) ≥ u, v ) ∈ Z with u ≡ v mod 2 inside thebounding box. • p = 0, a = b = 0: The bounding box ( − γ, γ ) × ( − γ, γ ) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2, the point ( u, v ) =(0 , f = Id. • p = j (0), a = − ( t + 1) ∈ (cid:104) − − t π/ , − + t π/ (cid:105) , b = (1 − t ) ∈ (cid:104) − t π/ , + t π/ (cid:105) : The bounding box (cid:18) − − t π/ − γ, −
12 + t π/ γ (cid:19) × (cid:18) − t π/ − γ,
12 + t π/ γ (cid:19) ⊂ ( − , × ( − , − ,
1) and (0 , g , ( u, v ) = u + v + ( t + 1) u + ( t − v is non-negative: g , ( − ,
1) = g , (0 ,
0) = 0. • p = j (0), a = − b = − t ∈ (cid:104) − t π/ , t π/ (cid:105) : The bounding box( − − γ, − γ ) × (cid:16) − t π/ − γ, t π/ + γ (cid:17) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2. • p = j (0), a = ( t − ∈ (cid:104) − − t π/ , − + t π/ (cid:105) , b = − ( t + 1) ∈ (cid:104) − − t π/ , − + t π/ (cid:105) : The bounding box (cid:18) − − t π/ − γ, −
12 + t π/ γ (cid:19) × (cid:18) − − t π/ − γ, −
12 + t π/ γ (cid:19) ⊂ ( − , × ( − , − , −
1) and (0 , g , ( u, v ) = u + v − ( t − u + ( t + 1) v is non-negative: g , ( − , −
1) = g , (0 ,
0) = 0 . In all cases we have shown that g , p ( u, v ) ≥
0, hence | p + xv + yv | ≥ r . Underthe assumption m ≥ log e (3 + 2 √
2) we have 2 r = 2 cosh (cid:0) m (cid:1) ≥
4. Therefore | f (0) |≥ f (cid:54) = Id in (cid:104) j , j (cid:105) . Hence all conditions of Lemma 4.2 are satisfiedand we can conclude that the group (cid:104) ι , ι , ι (cid:105) is discrete. (cid:3) OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 33
Proof of Proposition 1.4 for n = 2 . Proof.
Note that m ≥ log e (7 + 4 √
3) implies that r = cosh (cid:18) m (cid:19) ≥ . Any vertical translation in (cid:104) ι , ι (cid:105) is a power of the vertical translation H by(0 , r √ ( θ )). We need the displacement of each vertical translation H n , n (cid:54) = 0, to be at least the height of the spinal sphere, i.e.16 r √ ( θ ) ≥ ⇔ r cos ( θ ) ≥ √ . The hypothesis cos( α ) ≤ − √ for α ∈ (0 , π ) implies | θ | = (cid:12)(cid:12) π − α (cid:12)(cid:12) ≤ π . Forcos( θ ) ≥ √ √ and r ≥ r cos ( θ ) ≥ √ > √ , hence the condition | h (0) |≥ h (cid:54) = Id in (cid:104) ι , ι (cid:105) .We can write every element f in (cid:104) j , j (cid:105) as a word in the generators j and j .Figure 6 shows the points f (0) for all words f of length up to 6 in the case r = 1and θ = 0. The group (cid:104) j , j (cid:105) is the projection to C of the group E = (cid:104) ι , ι (cid:105) . Pro- Figure 6.
Points f (0) for all words f up to length 6.jecting ι and ι to C , we obtain the rotations j and j of C through π and π around ϕ and ϕ respectively. These rotations are given by j k ( z ) = µ k · z + (1 − µ k ) · ϕ k ,for k = { , } where µ = exp( πi ) and µ = exp( πi/ E is of the form T x T y H n w for some x, y, n ∈ Z and w ∈ { Id , ι , ι , ι , ι , ι } , where T = ι , T = ι and H = [ T , T ] areHeisenberg translations by ( v , t ), ( v , t ) and (0 , ν ) respectively and v = 2 r cos( θ ) · (cid:16) i √ (cid:17) , v = − r cos( θ ) · (cid:16) − i √ (cid:17) . Projection to C maps H to the identity, T k to the Euclidean translation by v k , T x T y H n to the Euclidean translation by xv + yv and w to the rotation el-ements { Id , j , j , j , j , j } respectively. Therefore every element of (cid:104) j , j (cid:105) is a product of a translation by xv + yv for some x, y ∈ Z and a rotation { Id , j , j , j , j , j } . Hence every point in the orbit of 0 under (cid:104) j , j (cid:105) is ofthe form p + xv + yv , where x, y ∈ Z and p ∈ { , j (0) , j (0) , j (0) , j (0) , j (0) } . Using | v | = | v | = 2 Re( v ¯ v ) = 16 r cos ( θ ), we calculate | p + xv + yv | = x | v | + y | v | +2 xy Re( v ¯ v ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | = 16 r cos ( θ ) · ( x + xy + y ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | . We make a coordinate change u = y − x and v = x + y , that is x = ( v − u ) / y = ( u + v ) /
2. Points ( x, y ) ∈ Z are mapped to points ( u, v ) ∈ Z with u ≡ v mod 2. We obtain | p + xv + yv | = 4 r cos ( θ ) · ( u + 3 v − au − bv + a + 3 b )= 4 r cos ( θ ) · (( u − a ) + 3( v − b ) ) , where a = − Re( p (¯ v − ¯ v ))8 r cos ( θ ) = Re ( p )2 r cos( θ ) ,b = − Re( p (¯ v + ¯ v ))24 r cos ( θ ) = − Im ( p )2 √ r cos( θ )and a + 3 b = | p | r cos ( θ ) . Our aim is to show that | p + xv + yv | ≥ r for all ( x, y ) ∈ Z excluding thecase p = 0, x = y = 0 that corresponds to f = Id. This is equivalent to ( u − a ) + 3( v − b ) ≥ sec ( θ )4 for all ( u, v ) ∈ Z with u ≡ v mod 2 excluding the case a = b = u = v = 0. Note that this inequality is always satisfied if | u − a |≥ sec( θ )2 or | v − b |≥ sec( θ )2 √ , so we only need to check that g , p ( u, v ) = ( u − a ) + 3( v − b ) − sec ( θ )4 ≥ u, v ) ∈ Z with u ≡ v mod 2 inside the bounding box (cid:18) a − sec( θ )2 , a + sec( θ )2 (cid:19) × (cid:18) b − sec( θ )2 √ , b + sec( θ )2 √ (cid:19) . In the following table we list the values of a , b and a + 3 b for p ∈ { , j (0) , j (0) , j (0) , j (0) , j (0) } : OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 35 p a b a + 3 b j (0) 1 − t √ t + 1 j (0) (cid:16) t √ − (cid:17) − √ (cid:16) t + √ (cid:17) (cid:0) t + 1 (cid:1) j (0) (cid:16) − t √ (cid:17) − (cid:16) t √ (cid:17) (cid:16) t + 4 t √ (cid:17) j (0) ( t √ − − (cid:16) t √ (cid:17) (cid:0) t + 1 (cid:1) j (0) − (cid:16) t √ (cid:17) − √ (cid:16) t + 3 √ (cid:17) (cid:16) t + 4 t √ (cid:17) Under the assumption | θ | (cid:54) π we have t = tan( θ ) ∈ (cid:104) √ − , − √ (cid:105) and sec( θ ) ∈ (cid:20) , √ (cid:16) √ − (cid:17)(cid:21) . In each of the six cases we list the bounds on a and b and thesize of the bounding box(min( a ) − γ , max( a ) + γ ) × (min( b ) − γ , max( b ) + γ )where γ = √ − √ and γ = √ − √ . We then calculate g , p ( u, v ) = ( u − a ) + 3( v − b ) − sec ( θ )4= u + 3 v − au − bv + ( a + 3 b ) − (cid:32) t + 14 (cid:33) and check that g , p ( u, v ) ≥ u, v ) ∈ Z with u ≡ v mod 2 inside thebounding box. • p = 0, a = b = 0: The bounding box ( − γ , γ ) × ( − γ , γ ) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2, the point ( u, v ) =(0 , f = Id. • p = j (0), a = 1, b = − t √ ∈ (cid:104) − t π/ √ , t π/ √ (cid:105) : The bounding box(1 − γ , γ ) × (cid:18) − t π/ √ − γ , t π/ √ γ (cid:19) ⊂ (0 , × ( − , u, v ) ∈ Z with u ≡ v mod 2. • p = j (0), a = (cid:16) t √ − (cid:17) ∈ (cid:20) − − t π/ √ , − + t π/ √ (cid:21) , b = − √ (cid:16) t + √ (cid:17) ∈ (cid:104) − − t π/ √ , − + t π/ √ (cid:105) : The bounding box (cid:32) − − t π/ √ − γ , −
14 + t π/ √
34 + γ (cid:33) × (cid:18) − − t π/ √ − γ , −
14 + t π/ √ γ (cid:19) ⊂ ( − , × ( − , , g , ( u, v ) = u + 3 v + (cid:32) − t √ (cid:33) u + (cid:32) − t √ (cid:33) v is non-negative: g , (0 ,
0) = 0 . • p = j (0), a = (cid:16) − t √ (cid:17) ∈ (cid:20) − t π/ √ , + t π/ √ (cid:21) , b = − (cid:16) t √ (cid:17) ∈ (cid:20) − − t π/ √ , − + t π/ √ (cid:21) : The bounding box (cid:32) − t π/ √ − γ ,
14 + t π/ √
34 + γ (cid:33) × (cid:32) − − t π/ √ − γ , −
34 + t π/ √
34 + γ (cid:33) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2. • p = j (0), a = (cid:16) t √ − (cid:17) ∈ (cid:20) − − t π/ √ , − + t π/ √ (cid:21) , b = − (cid:16) t √ (cid:17) ∈ (cid:20) − − t π/ √ , − + t π/ √ (cid:21) : The bounding box (cid:32) − − t π/ √ − γ , −
34 + t π/ √
34 + γ (cid:33) × (cid:32) − − t π/ √ − γ , −
14 + t π/ √
34 + γ (cid:33) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2. • p = j (0), a = − (cid:16) t √ (cid:17) ∈ (cid:20) − − t π/ √ , − + t π/ √ (cid:21) , b = − √ (cid:16) t + 3 √ (cid:17) ∈ (cid:104) − − t π/ √ , − + t π/ √ (cid:105) : The bounding box (cid:32) − − t π/ √ − γ , −
54 + t π/ √
34 + γ (cid:33) × (cid:18) − − t π/ √ − γ , −
34 + t π/ √ γ (cid:19) ⊂ ( − , × ( − , − , − g , ( u, v ) = u + 3 v − (cid:32) t √ (cid:33) u + (cid:32) t √ (cid:33) v + (cid:16) t √ (cid:17) is non-negative: g , ( − , −
1) = 0 . In all cases we have shown that g , p ( u, v ) ≥
0, hence | p + xv + yv | ≥ r . Under theassumption m ≥ log e (7 + 4 √
3) we have r = cosh (cid:0) m (cid:1) ≥
4. Therefore | f (0) |≥ f (cid:54) = Id in (cid:104) j , j (cid:105) . Hence all conditions of Lemma 4.2 are satisfied and wecan conclude that the group (cid:104) ι , ι , ι (cid:105) is discrete. (cid:3) Proof of Proposition 1.4 for n = 3 . Proof.
As in the proof in section 8.4 above (n=2), m ≥ log e (7 + 4 √
3) implies that r = cosh (cid:18) m (cid:19) ≥ . OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 37
Any vertical translation in (cid:104) ι , ι (cid:105) is a power of the vertical translation H by(0 , r √ ( θ )). We need the displacement of each vertical translation H n , n (cid:54) = 0,to be at least the height of the spinal sphere, i.e.8 r √ ( θ ) ≥ ⇔ r cos ( θ ) ≥ √ . The hypothesis cos( α ) ≤ − √ for α ∈ (0 , π ) implies | θ | = (cid:12)(cid:12) π − α (cid:12)(cid:12) ≤ π . Forcos( θ ) ≥ √ √ and r ≥ r cos ( θ ) ≥ √ > √ , hence the condition | h (0) |≥ h (cid:54) = Id in (cid:104) ι , ι (cid:105) .We can write every element f in (cid:104) j , j (cid:105) as a word in the generators j and j . Fig-ure 7 shows the points f (0) for all words f of length up to 6 in the case r = 1 and θ = 0. The group (cid:104) j , j (cid:105) is the projection to C of the group E = (cid:104) ι , ι (cid:105) . Projecting Figure 7.
Points f (0) for all words f up to length 6. ι and ι to C , we obtain the rotations j and j of C through π and π around ϕ and ϕ respectively. These rotations are given by j k ( z ) = µ k · z + (1 − µ k ) · ϕ k ,for k = { , } where µ = exp(2 πi/
3) and µ = exp( πi/ E is of the form T x T y H n w for some x, y, n ∈ Z and w ∈ { Id , ι , ι , ι , ι , ι } , where T = ι , T = ι and H = [ T , T ] areHeisenberg translations by ( v , t ), ( v , t ) and (0 , ν ) respectively and v = r cos( θ ) · (cid:16) i √ (cid:17) , v = − r cos( θ ) · (cid:16) − i √ (cid:17) . Projection to C maps H to the identity, T k to the Euclidean translation by v k , T x T y H n to the Euclidean translation by xv + yv and w to the rotation ele-ments { Id , j , j , j , j , j } respectively. Therefore every element of (cid:104) j , j (cid:105) is a product of a translation by xv + yv for some x, y ∈ Z and a rotation { Id , j , j , j , j , j } . Hence every point in the orbit of 0 under (cid:104) j , j (cid:105) isof the form p + xv + yv , where x, y ∈ Z and p ∈ { , j (0) , j (0) , j (0) , j (0) , j (0) } . Using | v | = | v | = − v ¯ v ) = 12 r cos ( θ ), we calculate | p + xv + yv | = x | v | + y | v | +2 xy Re( v ¯ v ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | = 12 r cos ( θ ) · ( x − xy + y ) + 2 x Re( p ¯ v ) + 2 y Re( p ¯ v ) + | p | . We make a coordinate change u = y − x and v = x + y , that is x = ( v − u ) / y = ( u + v ) /
2. Points ( x, y ) ∈ Z are mapped to points ( u, v ) ∈ Z with u ≡ v mod 2. We obtain | p + xv + yv | = 3 r cos ( θ ) · (3 u + v − au − bv + 3 a + b )= 3 r cos ( θ ) · (3( u − a ) + ( v − b ) ) , where a = − Re( p (¯ v − ¯ v ))18 r cos ( θ ) = Re ( p )3 r cos( θ ) ,b = − Re( p (¯ v + ¯ v ))6 r cos ( θ ) = − Im ( p ) r √ θ )and 3 a + b = | p | r cos ( θ ) . Our aim is to show that | p + xv + yv | ≥ r for all ( x, y ) ∈ Z excluding thecase p = 0, x = y = 0 that corresponds to f = Id. This is equivalent to 3( u − a ) + ( v − b ) ≥ sec ( θ )3 for all ( u, v ) ∈ Z with u ≡ v mod 2 excluding the case a = b = u = v = 0. Note that this inequality is always satisfied if | u − a |≥ sec( θ )3 or | v − b |≥ sec( θ ) √ , so we only need to check that g , p ( u, v ) = 3( u − a ) + ( v − b ) − sec ( θ )3 ≥ u, v ) ∈ Z with u ≡ v mod 2 inside the bounding box (cid:18) a − sec( θ )3 , a + sec( θ )3 (cid:19) × (cid:18) b − sec( θ ) √ , b + sec( θ ) √ (cid:19) . In the following table we list the values of a , b and 3 a + b for p ∈ { , j (0) , j (0) , j (0) , j (0) , j (0) } : p a b a + b j (0) √ (cid:16) t − √ (cid:17) − √ (cid:16) t + √ (cid:17) (cid:0) t + 1 (cid:1) j (0) √ (cid:16) t − √ (cid:17) − (cid:16) t √ (cid:17) t + 1 j (0) √ (cid:16) √ − t (cid:17) − √ (cid:16) t + √ (cid:17) (cid:16) t − t √ (cid:17) j (0) − √ ( t + √ − (cid:16) t √ (cid:17) t + 2 t √ j (0) − − t √ (cid:0) t + 1 (cid:1) OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 39
Under the assumption | θ | (cid:54) π we have t = tan( θ ) ∈ (cid:104) √ − , − √ (cid:105) and sec( θ ) ∈ (cid:20) , √ (cid:16) √ − (cid:17)(cid:21) . In each of the six cases we list the bounds on a and b and thesize of the bounding box(min( a ) − γ , max( a ) + γ ) × (min( b ) − γ , max( b ) + γ )where γ = √ ( √ − ) and γ = √ ( √ − ) √ . We then calculate g , p ( u, v ) = 3( u − a ) + ( v − b ) − sec ( θ )3= 3 u + v − au − bv + (3 a + b ) − (cid:32) t + 13 (cid:33) and check that g , p ( u, v ) ≥ u, v ) ∈ Z with u ≡ v mod 2 inside thebounding box. • p = 0, a = b = 0: The bounding box ( − γ , γ ) × ( − γ , γ ) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2, the point ( u, v ) =(0 , f = Id. • p = j (0), a = √ (cid:16) t − √ (cid:17) ∈ (cid:104) − − t π/ √ , − + t π/ √ (cid:105) , b = − √ (cid:16) t + √ (cid:17) ∈ (cid:104) − − t π/ √ , − + t π/ √ (cid:105) : The bounding box (cid:18) − − t π/ √ − γ , −
16 + t π/ √ γ (cid:19) × (cid:18) − − t π/ √ − γ , −
12 + t π/ √ γ (cid:19) ⊂ ( − , × ( − , , g , ( u, v ) = 3 u + v + (1 − t √ u + (cid:18) t √ (cid:19) v is non-negative: g , (0 ,
0) = 0. • p = j (0), a = √ (cid:16) t − √ (cid:17) ∈ (cid:104) − − t π/ √ , − + t π/ √ (cid:105) , b = − (cid:16) t √ (cid:17) ∈ (cid:20) − − t π/ √ , − + t π/ √ (cid:21) : The bounding box (cid:18) − − t π/ √ − γ , −
12 + t π/ √ γ (cid:19) × (cid:32) − − t π/ √ − γ , −
12 + t π/ √
32 + γ (cid:33) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2. • p = j (0), a = √ (cid:16) √ − t (cid:17) ∈ (cid:104) − t π/ √ , + t π/ √ (cid:105) , b = − √ (cid:16) t + √ (cid:17) ∈ (cid:104) − − t π/ √ , − + t π/ √ (cid:105) : The bounding box (cid:18) − t π/ √ − γ ,
56 + t π/ √ γ (cid:19) × (cid:18) − − t π/ √ − γ , −
12 + t π/ √ γ (cid:19) ⊂ (0 , × ( − , contains the point (1 , − g , ( u, v ) = 3 u + v − (5 − t √ u + (cid:18) t √ (cid:19) v + (cid:18) − t √ (cid:19) is non-negative: g , (1 , −
1) = 0. • p = j (0), a = − √ (cid:16) t + √ (cid:17) ∈ (cid:104) − − t π/ √ , − + t π/ √ (cid:105) , b = − (cid:16) t √ (cid:17) ∈ (cid:20) − − t π/ √ , − + t π/ √ (cid:21) : The bounding box (cid:18) − − t π/ √ − γ , −
12 + t π/ √ γ (cid:19) × (cid:32) − − t π/ √ − γ , −
32 + t π/ √
32 + γ (cid:33) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2. • p = j (0), a = − , b = − t √ ∈ (cid:104) − t π/ √ , t π/ √ (cid:105) : The bounding box (cid:18) − − γ , −
23 + γ (cid:19) × (cid:18) − t π/ √ − γ , t π/ √ γ (cid:19) ⊂ ( − , × ( − , u, v ) ∈ Z with u ≡ v mod 2.In all cases we have shown that g , p ( u, v ) ≥
0, hence | p + xv + yv | ≥ r . Under theassumption m ≥ log e (7 + 4 √
3) we have r = cosh (cid:0) m (cid:1) ≥
4. Therefore | f (0) |≥ f (cid:54) = Id in (cid:104) j , j (cid:105) . Hence all conditions of Lemma 4.2 are satisfied and wecan conclude that the group (cid:104) ι , ι , ι (cid:105) is discrete. (cid:3) Non-Discreteness Results
Let Γ = (cid:104) ι , ι , ι (cid:105) be an ultra-parallel [ m, m, n , n , Theorem 9.1.
Let Γ be a discrete subgroup of PU( n, containing a vertical trans-lation g by t > . Let h be any element of Γ not fixing q ∞ , a distinguished point atinfinity, and let r h = (cid:115) | h − h + h − h | be the radius of its isometric sphere. Then either t/r h ≥ or t/r h = 2 cos( π/q ) for some integer q ≥ . For each ultra-parallel [ m, m, n , n , H = (0 , ν ). Let h = ι . The matrixof the element h = ι = ι − is − − . OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 41
The radius of the isometric sphere of h is r h = 1. To see if the element h fixes q ∞ , we first map ∞ from the Heisenberg space to the boundary of the complexhyperbolic 2-space. That is, ∞ (cid:55)→ [0 : 1 : − ∈ ∂H C . We apply h to this point − − − = = [0 : 1 : 1] ∈ ∂H C . Note that h ( ∞ ) (cid:54) = ∞ and so h = ι does not fix q ∞ . Therefore, if the group isdiscrete then t ≥ t = 2 cos( π/q ) for some integer q ≥ . Hence the group is not discrete if(1) t < t (cid:54) = 2 cos( π/q ) for some integer q ≥ . Proof of Proposition 1.5.
Proof.
The vertical Heisenberg translation element H is given as (0 , r √ ( θ )).Substituting this into (1) with t = 96 r √ ( θ ) we have that the group is notdiscrete if 96 r √ ( θ ) < r √ ( θ ) (cid:54) = 2 cos( π/q ) ⇔ cos ( θ ) < r √ ( θ ) (cid:54) = cos( π/q )48 r √ q ≥ ( θ ) = 12 (cid:0) cos(2 θ ) + 1 (cid:1) = 12 (cid:0) − cos( α ) (cid:1) we are able to rewrite the inequalities to conclude that the group Γ is not discreteprovided thatcos( α ) > − √ (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )24 √ (cid:0) m (cid:1) for some integer q ≥ (cid:3) Proof of Proposition 1.6.
Proof.
The vertical Heisenberg translation element H is given as (0 , r cos ( θ )).Substituting this into (1) with t = 64 r cos ( θ ) we have that the group is notdiscrete if 64 r cos ( θ ) < r cos ( θ ) (cid:54) = 2 cos( π/q ) ⇔ cos ( θ ) < r and cos ( θ ) (cid:54) = cos( π/q )32 r for some integer q ≥ Using (2) we are able to rewrite the inequalities to conclude that the group Γis not discrete provided thatcos( α ) > −
116 cosh (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )16 cosh (cid:0) m (cid:1) for some integer q ≥ (cid:3) Proof of Proposition 1.7.
Proof.
The vertical Heisenberg translation element H is given as (0 , r cos ( θ )).Substituting this into (1) with t = 16 r cos ( θ ) we have that the group is notdiscrete if 16 r cos ( θ ) < r cos ( θ ) (cid:54) = 2 cos( π/q ) ⇔ cos ( θ ) < r and cos ( θ ) (cid:54) = cos( π/q )8 r for some integer q ≥ α ) > −
14 cosh (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )4 cosh (cid:0) m (cid:1) for some integer q ≥ (cid:3) Proof of Proposition 1.8.
Proof.
The vertical Heisenberg translation element H is given as (0 , r √ ( θ )).Substituting this into (1) with t = 16 r √ ( θ ) we have that the group is notdiscrete if 16 r √ ( θ ) < r √ ( θ ) (cid:54) = 2 cos( π/q ) ⇔ cos ( θ ) < r √ ( θ ) (cid:54) = cos( π/q )8 r √ q ≥ α ) > − √ (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )4 √ (cid:0) m (cid:1) for some integer q ≥ (cid:3) Proof of Proposition 1.9.
Proof.
The vertical Heisenberg translation element H is given as (0 , r √ ( θ )).Substituting this into (1) with t = 8 r √ ( θ ) we have that the group is notdiscrete if 8 r √ ( θ ) < r √ ( θ ) (cid:54) = 2 cos( π/q ) ⇔ cos ( θ ) < r √ ( θ ) (cid:54) = cos( π/q )4 r √ OMPLEX HYPERBOLIC TRIANGLE GROUPS OF TYPE [ m, m, n , n ,
2] 43 for some integer q ≥ α ) > − √ (cid:0) m (cid:1) and cos( α ) (cid:54) = 1 − cos( π/q )2 √ (cid:0) m (cid:1) for some integer q ≥ (cid:3) Acknowledgements
I would like to thank Anna Pratoussevitch and John Parker for their insightfulcomments and suggestions throughout this work.
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School of Mathematics & Statistics, University of Melbourne, Parkville, VIC, 3052,Australia.
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