Presentations of Groups Acting Discontinuously on Direct Products of Hyperbolic Spaces

Home , Fundamental domain

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This allowed them to obtain presentations for subgroups of finite index in (ZG) for some new class of finite groups G, called groups of Kleinian type. The basic idea canU be explained as follows. If G is a finite group then ZG is an order in the rational group n algebra QG and it is well known that QG = i=1 Mni (Di), where each Di is a division algebra. n If i is an order in Di, for each i, then = i=1 Mni ( i) is an order in QG and the group of unitsO ( ) of is commensurable with O(ZGQ). Recall thatO two subgroups of a given group are said toU beO commensurableO if they have aU commonQ subgroup that is of finite index in both. The n group ( ) is simply Π GLn ( i), the direct product of the groups (Mn ( i)) = GLn ( i). U O i=1 i O U i O i O Moreover ( ( i)) SLn ( i), the direct product of the central units in i and the group consisting U Z O × i O O of the reduced norm one elements in Mni ( i), contains a subgroup of finite index isomorphic O n to a subgroup of finite index of GLni ( i). The group i=1 ( ( i)) may be determined from Dirichlet’s Unit Theorem and it is commensurableO with ( (UZGZ)),O the group of central units of ZG. For a large class of finite groups G one can describeQZ generatorsU of a subgroup of finite index in ( (ZG)) [JOdRVG13] (see also [JdRVG14]). Hence, up to commensurability, the problem of Z U finding generators and relations for (ZG) reduces to finding a presentation of SLn ( i) for every U i O 1 i n. The congruence theorems allow to compute generators up to finite index for SLn ( i) ≤ ≤ i O when ni 3 (without any further restrictions) and also for ni = 2 but then provided Di is neither a totally≥ definite rational quaternion algebra, nor a quadratic imaginary extension of Q and also not Q. The case ni = 1 can also be dealt with in case Di is commutative or when it is a a totally definite quaternion algebra. In case each Mni (Di) is of one of these types and if, moreover, G does not have non-commutative fixed point free epimorphic images, then concrete generators, the so called Bass units and bicyclic units, for a subgroup of finite index in (ZG) have been determined [Seh93, RS91a, RS91b, RS89, JL93]. A finite group G is said to beU of Kleinian type if each non- commutative simple factor Mn (Di) of QG is a quaternion algebra over its centre, i.e. ni 2 and i ≤ the natural image of SLn ( i) in PSL (C) (obtained by extending some embedding of the centre i O 2 of Di in C) is a Kleinian group. A Kleinian group is a subgroup of PSL2(C) which is discrete for the natural topology, or equivalently, its action on the 3-dimensional hyperbolic space via the Poincaré extension of the action by Möbius transformations is discontinuous [Bea95, Theorem 5.3.2]. Poincaré introduced a technique to determine presentations for Kleinian groups via fundamental polyhedra (see e.g. [Bea95, EGM98, Mas88]). An alternative method, due to Swan [Swa71], gives a presentation from a connected open subset containing a fundamental domain. Thus, if G is a finite group of Kleinian type then, in theory, one can obtain a presentation of a group commensurable with (ZG). In practice, it is difficult to execute this procedure because it is usually hard to computeU a fundamental Poincaré polyhedron of a Kleinian group. Groups of Kleinian type have been classified in [JPdR+07] and examples of how to find presentations of (ZG) for some groups of Kleinian type of small order are given in [PdRR05] and [PdR06]. A generalizationU to group rings over commutative orders is given in [OdR07]. Recently, in [JJK+15], an algorithm is given to compute a fundamental domain, and hence generators, for subgroups of finite index in groups that are contained in the unit group of orders in quaternion algebras over quadratic imaginary extensions of Q, in particular for Bianchi groups. This work was in fact a generalization of [CJLdR04]. In the investigations on determining presentations for (ZG) it is now natural to deal with finite groups G that are such that QG has simple componentsU of which the unit group of some order does not necessarily act discontinuously on one hyperbolic space, but does so on a direct product of hyperbolic 2 or 3-spaces. More precisely, one admits simple components A of QG that

2 are quaternion algebras over a number field, say F . Every field homomorphism σ : F C naturally → extends to an isomorphism C A = M (C) and thus to a homomorphism σ : A M (C), ⊗σ(F ) ∼ 2 → 2 via a 1 a. This homomorphism maps SL1(A), the group consisting of reduced norm one 7→ ⊗ 3 elements, into SL2(C). Composing this with the action of SL2(C) on H , by the Poincaré extension 3 of Möbius transformations, yields an action of SL1(A) on H . Note that if σ(F ) R and A is R R H2 ⊆ unramified in σ (i.e. σ(F ) A ∼= M2( )), then σ restricts to an action on . Let σ1,...,σt be representatives (modulo⊗ complex conjugation) of the homomorphisms from F to C. If r is the number of real embeddings of F on which A is unramified and s = t r, then the action of σi on − the i-th component gives an action of SL1(A) on

2 (r) 2 3 (s) 3 Hr,s = H . . . H H . . . H . × × × × × More precisely the action is given by

a (x ,...,xt) = (σ (a) x ,...,σt(a) xt), · 1 1 · 1 · 2 3 where a SL1(A) and x1,...,xr H , xr+1,...,xr+s H . If is an order in A then the ∈ r (t r)∈ ∈ O image of in M2(R) M2(C) − is discrete. This implies that the action of SL1( ) on Hr,s is discontinuousO (see Proposition× 2.1). O This suggests the following program to obtain a presentation of (ZG) (more precisely of a group commensurable with (ZG)) for the finite groups G such thatUQG is a direct product of fields and quaternion algebras.U (Equivalently, the degrees of the irreducible characters of G are all either 1 or 2, or equivalently, by a result of Amitsur in [Ami61], G is either abelian, or has an abelian subgroup of index 2 or G/ (G) is elementary abelian of order 8.) First determine the Wedderburn kZ decomposition QG = Ai. For each i = 1,...,k, let i be an order in Ai and calculate ∼ i=1 O ( ( i)), using for example the Dirichlet Unit Theorem. Next, if Ai is non-commutative and U Z O Q = i (in fact it is enough to consider the components which are not totally definite quaternion O O algebras) then calculate a fundamental domain for the action of SL ( ) on Hr,s and determine 1 O from this a presentation of SL1( ). For this program to be successfulO one needs to solve at least the following problem.

(1) Does Poincaré’s Method remain valid for discontinuous actions on Hr,s? The following problem is one of the first issues to deal with in order to answer the question. (2) Determine methods to calculate fundamental domains effectively. Once this is done, the following problem arises. (3) The sides of the well-known fundamental polyhedra in H2 and H3 are geodesic hyperplanes i.e. lines and circles (respectively, planes and spheres) orthogonal to the border. How should one define the “sides” of some potential fundamental domain in Hr,s? In this paper we show that the three problems can be controlled in case the considered group acts on a direct product of two copies of H2. More precisely, we complete the outlined program for the Hilbert modular group PSL2(R), where R is the ring of integers in Q(√d), with d> 0, and R is 2 2 a principal ideal domain. The group PSL2(R) acts discontinuously on H H , the direct product of 2 copies of hyperbolic 2-space. Hence, this group constitutes the “easiest”× case of the outlined program and as such can be considered a relevant test case. For the construction of a fundamental

3 domain we will make use of some ideas in [Coh65a] and [Coh65b]. The construction is based on the classical construction of a Ford fundamental domain. For more details we refer the reader to [Voi09] or to [Bea95, Section 9.5], where the author calls such domains Generalized Dirichlet domains. The largest part of the paper is devoted to set up the machinery needed to prove an extension of the presentation part of Poincaré’s Polyhedron theorem so that we can extract a presentation from this domain. The assumption that R is a principal ideal domain allows us to get a description of the fundamental domain in terms of finitely many varieties. Note that this restriction is essentially only used in Lemma 3.7F , Theorem 3.8, Lemma 4.12 and Corollary 5.8. The method outlined allows computing a description of the unit group (ZG) for several classes of finite groups G. As a matter of example, we mention that this now easilyU can be done for the dihedral groups D10 and D16 of order 10 and 16 respectively, and the quaternion group Q32 of order 32. The group (ZD ) is commensurable with SL Z[ 1+√5 ] , the group (ZD ) is commen- U 10 2 2 U 16 surable with a direct product of a non-abelian free group and the group SL2(Z[√2]). The group (ZQ ) is commensurable with a direct product of (ZD ) and an abelian group. Corollary 5.8 U 32 U 16 1+√5 √ gives generators of the groups SL2(Z[ 2 ]) and SL2(Z[ 2]) and the algorithm described in Theo- rem 6.24 yields a presentation of these groups. If one does not want to work up to commensurability then the Reidemeister-Schreier method allows to go down to the exact subgroup of the previous groups needed for the computation of the corresponding unit groups. In the case of (ZD16) this exact subgroup is described in [JP93]. U An alternative method for computing a presentation for the group of units of an order in a semi-simple algebra over Q was recently given by Coulangeon, Nebe, Braun and Schönnenbeck in [CNBS]. The authors present a powerful algorithm that is based on a generalisation of Voronoï’s algorithm for computing perfect forms and is combined with Bass-Serre theory. The method differs essentially from the one presented here, as the authors use an action on a Euclidean space. To illustrate their algorithm, several computations are carried out completely in [CNBS]; including orders in division algebras of degree 3. As mentioned above, in this paper we focus on groups that act discontinuously on direct products of two hyperbolic spaces and we extend the presentation part of Poincaré’s Polyhedron Theorem into this context. Developing such new methods are relevant in the bigger scheme of discovering new generic constructions of units that generate large subgroups in the unit group of an order of a rational division algebra and hence solving completely the problem of describing finitely many generic generators for a subgroup of finite index in (ZG) for any finite group (hence without any restriction on the rational group algebra QG). U The paper is organized as follows. In Section 2, we give some background on discontinuous group actions on direct products of copies of H2 and H3 and introduce some notation on the group of our interest, namely PSL2(R) with R the ring of integers of a real quadratic field, and its action on H2 H2. In Section 3, we give a description of a fundamental domain for this action, see Theorem× 3.8. In Section 4, we prove some topological lemmas about theF fundamental domain and they will be used in the two following sections. Finally, in Section 5, we describe the sides of the fundamental domain . In Theorem 5.3, we generalize the generating part of Poincaré’s Polyhedron Theorem to theF case of PSL (R) acting on H2 H2 and in Corollary 5.8, we give an 2 × effective description of the generators of PSL2(R). In Section 6, we describe the edges of and in Theorem 6.24, we generalize the presentation part of Poincaré’s Polyhedron Theorem. NoteF that Theorem 5.3 and Theorem 6.24 could potentially have been deduced from [Mac64a, Corollary of Theorem 2]. However, in order to get an effective set of generators and relations as an application

4 of this theorem one needs all the lemmas in Section 5 and 6 anyway. Moreover, we think that the proofs of Theorem 5.3 and Theorem 6.24, presented in this paper, are more intuitive from a geometric point of view. Furthermore they are algorithmic in nature. For the convenience of the reader we include an appendix on some needed algebraic topological results.

2 Background

The following deﬁnitions are mostly taken from [Bea95, Rat94]. Let C = C and let Hn denote the upper half Poincaré model of the hyperbolic space of dimension n. We∪ consider {∞} the group SL2(C) acting on C by Möbius transformations. We also let SL2(C) actb by orientation-preserving isometries on H3, by the Poincaré extension of Möbius transformations. The hyperbolic plane H2 is invariant under theb action of SL2(R). This deﬁnes an isomorphism between PSL2(C) (respectively 3 2 PSL2(R)) and the group of orientation-preserving isometries of H (respectively H ). 2 (r) 2 3 (s) Consider for non negative integers r and s the metric space Hr,s = H . . . H H . . . 3 (r) (s) × × × × H and the group r,s = SL (R) . . . SL (R) SL (C) . . . SL (C), a direct product of r × G 2 × × 2 × 2 × × 2 copies of SL2(R) and s copies of SL2(C). An element g r,s will be written as an (r + s)-tuple (1) (r+s) ∈ G (g ,...,g ) and elements of Hr,s will be written as (r +s)-tuples, say Z = (Z1,...,Zr+s). The metric ρ on this space is given by

n 2 (i) 2 ρ(X, Y ) = ρ (Xi, Yi) , i=1 X where ρi is the hyperbolic metric on the hyperbolic 2 or 3-space (see [Maa40, Paragraph 2]). The 2 3 action of SL (R) and SL (C) on H and H induces componentwise an action of r,s by isometries 2 2 G on Hr,s: (1) (r+s) g Z = (g Z ,...,g Zr s). · · 1 · + 2 We consider Mn(C) as an Euclidean 2n -dimensional real space. This induces a structure of topological group on r,s. Let G be a subgroup of r,s. One says that G is discrete if it is discrete G G with respect to the induced topology. One says that G acts discontinuously on Hr,s if and only if, for every compact subset K of Hr,s, the set g G : g(K) K = is ﬁnite. { ∈ ∩ 6 ∅} The following proposition is well known and extends the fact that a subgroup of SL2(C) is discrete if and only if its action on H3 is discontinuous.

Proposition 2.1 [Rat94, Theorem 5.3.5.] A subgroup of r,s is discrete if and only if it acts G discontinuously on Hr,s.

Let G be a group of isometries of Hr,s. A fundamental domain of G is a subset of Hr,s whose F boundary has Lebesgue measure 0, such that Hr,s = g G g( ) and if 1 = g G and Z Hr,s then Z,g(Z) is not contained in the interior of . The∈ diﬀerentF sets g( 6 ), for∈ g G, are∈ called tiles given{ by }G and and the set of tiles = gF( )S : g G is calledF the tessellation∈ given by G and . F T { F ∈ } In thisF paper we study a presentation for the special linear group of degree 2 of the ring of integers of real quadratic extensions. So throughout k is a square-free positive integer greater than 1 and 1+ √k 1, if k 1 mod 4; ω = , with k0 = 6≡ (1) k 2, if k 1 mod4. 0 ≡

5 Moreover, K = Q √k and R = Z[ω], the ring of integers of K. For α K, let α′ denote the ∈ 1 √k algebraic conjugate of α. Then ω′ = − and if α = α0+α1ω with α0, α1 Q, then α′ = α0 +α1ω′. k0 ∈ Let N denote the norm map of K over Q and ǫ0 denote the fundamental unit of K, i.e. ǫ0 is the smallest unit ǫ0 of R greater than 1. Then (R) = α R : N(α) = 1 = ǫ0 , by Dirichlet’s Unit Theorem. U { ∈ ± } ±h i We consider SL (R) as a discrete subgroup of SL (R) SL (R) by identifying a matrix A 2 2 × 2 ∈ SL2(R) with the pair (A, A′), where A′ is the result of applying the algebraic conjugate ′ in each a b entry of A. Thus we consider SL (R) acting on H2 H2. More precisely, if γ = SL (K), 2 × c d ∈ 2 Z = (x1 + y1i, x2 + y2i) and γ(Z)=(ˆx1 +ˆy1i, xˆ2 +ˆy2i) then a straightforward calculation yields

2 (ax1 + b)(cx1 + d)+ acy1 y1 xˆ1 = 2 2 2 , yˆ1 = 2 2 2 (2) (cx1 + d) + c y1 (cx1 + d) + c y1 and 2 (a′x2 + b′)(c′x2 + d′)+ a′c′y2 y2 xˆ2 = 2 2 2 , yˆ2 = 2 2 2 . (3) (c′x2 + d′) + c′ y2 (c′x2 + d′) + c′ y2 If Z = (Z ,Z ) H2 H2 then we write 1 2 ∈ ×

Zj = xj + yj i, where xj ,yj R and yj > 0 (j =1, 2). (4) ∈ 2 + 2 Then the four tuples (x1, x2,y1,y2) R (R ) form a system of coordinates of elements of H2 H2. ∈ × × If Z = (x1 + y1i, x2 + y2i) then we use the following notation

2 2 2 2 2 2 cZ + d = ((cx + d) + c y )((c′x + d′) + c′ y ). k k 1 1 2 2 We introduce another system of coordinates (s ,s ,r,h) R2 (R+)2 by setting 1 2 ∈ ×

2 h 2 x = s + s ω, x = s + s ω′, y = , y = rh (5) 1 1 2 2 1 2 1 r 2 or equivalently ω′x ωx x x y s = 1 − 2 , s = 1 − 2 , r = 1 , h = y y . (6) 1 ω ω 2 ω ω y 1 2 ′ − − ′ 2 2 2 2 + 2 So, each element Z of H H is represented by either (x1, x2,y1,y2) R (R ) ,or(s1,s2,r,h) 2 + 2 × 2 + 2 ∈ × ∈ R (R ) , or (x1, x2,r,h) R (R ) . Then the norm, ratio and height of Z are deﬁned respectively× by ∈ ×

2 2 2 2 y1 Z = (x1 + y1)(x2 + y2), r(Z)= r = , h(Z)= h = y1y2. k k y2 Using (2) and (3), we get h(Z) h(γ(Z)) = . (7) cZ + d k k Let Γ = PSL2(R), (8)

6 the Hilbert modular group, and we consider it as a discontinuous group of isometries of H2 H2. Let Γ denote the stabilizer of by the action of Γ (on C). The elements of Γ are represented× ∞ m ∞ ∞ ǫ0 b by the matrices m with m Z and b R. Abusing notation, if γ Γ is represented 0 ǫ− ∈ ∈ b ∈ 0 a b a b by then we simply write γ = . Hence, (2), (3) and (7) imply c d c d m ǫ0 b 4m if γ = m Γ then h(γ(Z)) = h(Z) and r(γ(Z)) = ǫ0 r(Z). (9) 0 ǫ− ∈ ∞ 0 The second rows of the elements of Γ form the following set

= (c, d) R2 : cR + dR = R . S { ∈ } 3 Fundamental Domain

In this section we compute a fundamental domain for the group Γ = PSL2(R) (8) via methods analogue to those used for computing a Ford fundamental domain of a discrete group acting on the hyperbolic 2-space. This part of our work is based on the ideas of H. Cohn in [Coh65b, Coh65a]. We start introducing the following subsets of H2 H2, expressed in the (s ,s ,r,h) coordinates: × 1 2

+, 2 2 1 V ≥ = (s ,s ,r,h) H H : si , i 1 2 ∈ × ≥ 2 , 2 2 1 V − ≥ = (s ,s ,r,h) H H : si , i 1 2 ∈ × ≥−2 for i =1, 2 and

+, 2 2 2 V ≥ = (s ,s ,r,h) H H ; : r ǫ , 3 1 2 ∈ × ≥ 0 , 2 2 2 V3− ≥ = (s1,s2,r,h) H H : r ǫ0− . ∈ × ≥ Moreover, for every c, d with c = 0, let ∈ S 6 2 2 V ≥ = Z H H : cZ + d 1 . c,d { ∈ × k k≥ } , We also deﬁne the sets Vi± ≤ and Vc,d≤ (respectively, Vi± and Vc,d) by replacing by (respectively, =) in the previous deﬁnitions. ≥ ≤ a b For every (c, d) with c = 0 and γ = Γ we have ∈ S 6 c d ∈ 1 cγ(Z)+ a = (10) k− k cZ + d k k and therefore γ Vc,d≥ = V ≤c,a. −

7 We deﬁne

+, , +, , +, , = V1 ≤ V1− ≥ V2 ≤ V2− ≥ V3 ≤ V3− ≥ F∞ ∩ ∩ ∩ ∩ ∩ 2 + 2 1 2 2 = (s ,s ,r,h) R (R ) : s , s ,ǫ− r ǫ ; { 1 2 ∈ × | 1| | 2|≤ 2 0 ≤ ≤ 0} 2 2 = V ≥ = Z H H : cZ + d 1 for all (c, d) ; F0 c,d { ∈ × k k≥ ∈ S} (c,d) \∈S = 0. F F∞ ∩F Lemma 3.1 is a fundamental domain of Γ . F∞ ∞

Proof. We ﬁrst prove that if 1 = γ Γ and Z = (Z1,Z2) = (x1 + y1i, x2 + y2i) = (s1,s2,r,h) then Z,γ(Z) cannot be contained6 in∈ the∞ interior of . Let { } F∞ m ǫ0 b γ = m . 0 ǫ− 0 4m 2 4m 2 By (9), r(γ(Z)) = ǫ0 r. If Z and γ(Z) belong to the interior of Γ then ǫ0− < r,ǫ0 r<ǫ0 and ∞ therefore m = 0. Therefore the transformation γ is simply a translation by the parameter (b,b′) with b R. Now b = b1 + b2ω, with b1,b2 Z. As Z and γ(Z) belong to the interior of , then ∈ 1 ∈ F∞ s1 , s2 , s1 + b1 , s2 + b2 < 2 . Thus b = 0 and hence γ = 1, as desired. | | | | | | | | 2 2 Let Z = (x1 + y1ω, x2 + y2ω′) be an arbitrary point in H H , with xi,yi R. We will show × n ∈ n that there exists γ Γ such that γ(Z) . As ǫ0 > 1, lim ǫ0 = and lim ǫ0 = 0 and ∈ ∞ ∈ F∞ n + ∞ n hence there exists n Z such that → ∞ →−∞ ∈ 4n 2 4n+2 ǫ − r ǫ . 0 ≤ ≤ 0 n ǫ0− 0 2 4n 2 2 2 Let γ = . By(9), ǫ− r(γ(Z)) = ǫ− r ǫ . So we may assume that ǫ− r ǫ . 0 ǫn 0 ≤ 0 ≤ 0 0 ≤ ≤ 0 0 1 b 1 Now consider γ = , with b = b + b ω, b ,b Z and xi + bi . Again by (9), the 0 1 1 2 1 2 ∈ | | ≤ 2 (s1,s2,r,h) coordinates of γ(Z) are (s1 + b1,s2 + b2,r,h) and hence γ(Z) , as desired. 3 + 3 ∈F∞ Moreover the boundary of is included in i=1 Vi i=1 Vi−. This clearly has Lebesgue measure 0. F∞ ∪ S S For a subset C of H2 H2 and µ> 0 let ×

VC,µ = (c, d) R R : cZ + d µ for some Z C . (11) { ∈ × k k≤ ∈ } We say that a subset C of H2 H2 is hyperbolically bounded if it is bounded in the metric of H2 H2 or equivalently if it is bounded× in the Euclidean metric and there is a positive number ǫ such× that y ,y >ǫ for every (x + y i, x + y i) C. 1 2 1 1 2 2 ∈ Lemma 3.2 Let C be a hyperbolically bounded subset of H2 H2 and let µ > 0. Then there are × ﬁnitely many elements (c1, d1),..., (ck, dk) of R R such that VC,µ = (uci, udi) : i =1,...,k,u (R) . × { ∈ U }

8 Proof. As d R : N(d) µ is finite up to units, it is clear that VC,µ has finitely many elements of the{ ∈ form (0, d) with≤ }d R. As C is hyperbolically bounded there is s > 0 such that h(Z) s for every Z C.∈ As the number of elements of R of a given norm is finite ≥ ∈ 2 µ modulo units, there are c ,...,ck R 0 such that if c R 0 with N(c) 2 then 1 ∈ \{ } ∈ \{ } ≤ s c = uci for some i = 1,...,k and some u (R). Moreover, for every i = 1,...,k, the set 2 2 2 2 ∈ U2 2 2 Ci = (x, y) R : (cix1 + x) ci′ y2 + (ci′ x2 + y) ci y1 r for some Z = (x1 + y1i, x2 + y2i) C { ∈ 2 ≤ 2 ∈ } is a bounded subset of R . As (d, d′) : d R is a discrete subset of R , there are ei ,...,eij R { ∈ } 1 i ∈ such that if d R and (d, d′) Ci then d = eili for some 1 li ji. Assume that cZ + d µ ∈ ∈ 2 2 ≤2 ≤ 2 2 2 2 2 k k ≤2 for some Z = (x1 + y1i, x2 + y2i) C. Then N(c) s N(c) h(Z) = c c′ y1y2 ((cx1 + d) + 2 2 2 2 2 ∈ ≤ ≤ c y )((c′x + d′) + c′ y ) = cZ + d µ. Hence c = uci for some i = 1,...,k and u (R). 1 2 2 k k ≤ ∈ U Moreover, as N(u)= uu′ = 1, we have ± 1 2 2 2 1 2 2 2 2 2 2 2 2 2 (cix + u− d) c′ y + (c′ x + (u− d)′) c y = (cx + d) c′ y + (c′x + d′) c y cZ + d µ. 1 i 2 i 2 i 1 1 2 2 1 ≤k k≤ 1 1 1 Therefore (u− d, (u− d)′) Ci. Thus u− d = eil for some 1 li ji. This proves that there ∈ i ≤ ≤ are finitely many elements (c , d ),..., (ck, dk) (R 0 ) R such that VC,µ is contained in 1 1 ∈ \{ } × u(ci, di) : i =1,...,k,u (R) . Note that if u (R) then uciZ + udi = ciZ + di . Hence, the{ result follows. ∈U } ∈U k k k k

Lemma 3.3 = Z H2 H2 : Z has maximal height in its Γ-orbit . F0 { ∈ × } Proof. We ﬁrst claim that for a ﬁxed Z H2 H2, the set cZ + d : (c, d) has a ∈ × {k k ∈ S} minimum. Indeed, let π = Z . Clearly V Z ,π = as it contains (1, 0). By Lemma 3.2, k k { } ∩ S 6 ∅ V Z ,π = u(ci, di) : i = 1,...,k,u (R) for some (c1, d1),..., (ck, dk) . Let 0 = m = { } ∩ S { ∈ U } ∈ S 6 min ciZ + di : i =1,...,k . If cZ + d cZ + d = N(u) ciZ + di m, a contradiction. Hence the claim follows.∈ Consequently, U by (7),k the Γ-orbitk of Z hask an elementk ≥ of maximum height. Consider a Γ-orbit and let Z be an element in this orbit with maximal height. Hence for every g Γ, h(g(Z)) h(Z) and hence by (7), cZ + d 1, for every (c, d) . Thus Z . ∈ ≤ k k≥ ∈ S ∈F0 To prove the other inclusion, let Z 0. Then cZ + d 1, for every (c, d) and hence for every g Γ, h(g(Z)) h(Z). Thus every∈F elementk of reachesk≥ the maximum height∈ S in its orbit. ∈ ≤ F0

Lemma 3.4 [Coh65b, Coh65a] is fundamental domain for Γ. F Proof. We ﬁrst show that contains a point of every Γ-orbit. To prove this, consider an orbit and let Z be an element in thisF orbit with maximal height. By Lemma 3.1, there is W and g Γ such that W = g(Z). By (9), h(Z)= h(W ) and hence we may assume that Z ∈F.∞ Thus, ∈ ∞ ∈F∞ by Lemma 3.3, Z 0 and so Z . Now we will show∈F that if two points∈F of the same orbit are in then they necessarily are on the border of . Suppose that Z,γ(Z) , for some 1 = γ Γ. IfF γ Γ then, by Lemma 3.1, Z F ∈ F 6 ∈ ∈ ∞ a b belongs to the border of and hence it belongs to the border of . Otherwise γ = F∞ F c d with c = 0 and, by (7) we have 6 h(Z) h(Z)= h(γ(Z)) = . cZ + d k k

9 Therefore cZ + d = 1 and thus Z lies on the border of 0 and thus on the border of . k k 3 + F3 F Finally the boundary of is contained in i=1 Vi i=1 Vi− (c,d) Vc,d with c =0. As is countable, the boundary ofF has measure 0. ∪ ∪ ∈S 6 S F S S S Let 1 2 2 B = (s ,s , r) : s , s , ǫ− r ǫ 1 2 | 1| | 2|≤ 2 0 ≤ ≤ 0 Observe that = B R+. F∞ × We can think of as the region above a “floor”, which is given by the sets Vc,d for (c, d) and F ∈ S limited by “six walls”, which are given by the sets Vi± for i =1, 2, 3. We now give an alternative description of on which the “floor” is given by the graph of a F function h0 defined on B. 4 For each c, d R we define the function fc,d : R R by ∈ → 2 2 2 2 2 2 fc,d(x1, x2,y1,y2)= (cx1 + d) + c y1 (c′x2 + d′) + c′ y2 .

2 2 Observe that if Z H H then fc,d(Z)= cZ + d and if d (R) then f ,d(Z) = 1. Thus ∈ × k k ∈U 0 2 2 = Z H H : fc,d(Z) 1, for all (c, d) . (12) F0 { ∈ × ≥ ∈ S} We use the mixed coordinate system described in Section 2 and write

2 2 2 2 2 c 2 2 f (x , x ,r,h)=[(cx + d)(c′x + d′)] + (cx + d) c′ r + (c′x + d′) v + N(c) v . c,d 1 2 1 2 1 2 r If c, d R, (x , x , r) R2 R+ and v R, then we deﬁne ∈ 1 2 ∈ × ∈ 2 2 2 2 2 c 2 2 f (v) = [(cx + d)(c′x + d′)] + (cx + d) c′ r + (c′x + d′) v + N(c) v . (13) c,d,x1,x2,r 1 2 1 2 r and if c = 0 then we set 6

h1(c, d, x1, x2, r)= 2 1 1 d 2 d 2 1 1 d 2 d 2 1 + x + r x + ′ x + r + x + ′ , v 2 1 2 1 2 uN(c) 4 " c − c′ r # − 2 " c c′ r # u t If u (R) then the following statements hold (throughout the paper we will use these without explicit∈ U reference):

fuc,ud = fc,d, (14)

h1(uc,ud,x1, x2, r) = h1(c, d, x1, x2, r), (15) (c, d) (uc,ud) (16) ∈ S ⇔ ∈ S Vuc,ud≥ = Vc,d≥ , (17) Vuc,ud = Vc,d. (18)

10 2 Set = (x1, x2) R : 1 < (cx1 + d)(c′x2 + d′) < 1 . An easy calculation shows that if c = 0 thenC { ∈ − } 6 + Vc,d = (x , x ,r,h (c, d, x , x , r)) : (x , x , r) R . { 1 2 1 1 2 1 2 ∈C× } As is path-connected, Vc,d is path-connected. Moreover, if f = fc,d,x ,x ,r then C 1 2 2 2 2 2 c 2 f ′(h) = (cx + d) c′ r + (c′x + d′) +2N(c) h. 1 2 r and hence f is strictly increasing on [min(0,h1), ). For (s ,s , r) B let ∞ 1 2 ∈ h (s ,s , r) = sup h > 0 : fc,d,x ,x ,r(h )=1, for some (c, d) , with c =0 , 0 1 2 { 1 1 2 1 ∈ S 6 } where we understand that the supremum of the empty set is 0. If (c, d) , c = 0, h1 > 0 and 2 1 ∈ S 6 fc,d,x ,x ,r(h ) = 1 then h 2 1. Hence the supremum deﬁning h (s ,s , r) exists and 1 2 1 1 ≤ N(c) ≤ 0 1 2 h (s ,s , r) 1. (19) 0 1 2 ≤

By (12) and the monotonicity fc,d,x1,x2,r we have = (s ,s ,r,h) R2 (R+)2 : h h (s ,s , r) and (20) F0 { 1 2 ∈ × ≥ 0 1 2 } = (s ,s ,r,h) R2 (R+)2 : (s ,s , r) B and h h (s ,s , r) . (21) F { 1 2 ∈ × 1 2 ∈ ≥ 0 1 2 } From (19) and (21) the following lemmas easily follow.

2 2 1 1 2 2 Lemma 3.5 If Z = (s ,s ,r,h) H H with s , s , ǫ− r ǫ and h 1, then 1 2 ∈ × | 1|≤ 2 | 2|≤ 2 0 ≤ ≤ 0 ≥ Z . Moreover, if the inequalities are strict then Z ◦. ∈F ∈F Lemma 3.6 is path-connected. F In order to have a finite procedure to calculate the fundamental domain of Lemma 3.4 we need to replace in the definition of by a suitable finite set. In our next resultF we obtain this for R a principalS ideal domain (PID,F for short). For that we need the following lemma, which is proved in [Coh65b, Paragraph 5].

2 Lemma 3.7 If R is a PID and (s ,s ,r,h) then h> k0 . 1 2 ∈F0 2k Theorem 3.8 Let k be a square-free integer greater than 1. Let k0 and ω be as in (1), R = Z[ω], Γ = PSL2(R) and the fundamental domain of Γ given in Lemma 3.4. Let 1 be a set of representatives, up to multiplicationF by units in R, of the couples (c, d) R2 satisfyingS the following conditions: ∈ 2k c =0, cR + dR = R, N(c) 2 , 6 | |≤ k0 (22) 2 2 d 2k k0 1+ ω d′ 2k k0 1 ω′ <ǫ0 2 2 + and <ǫ0 2 2 + − . | c | sN(c) k0 − 2k 2 | c′ | sN(c) k0 − 2k 2

≥ Then 1 is ﬁnite and if R is a principal ideal domain then = (c,d) 1 Vc,d. In particular, S F F∞ ∩ ∈S Vc,d = if and only if (c, d) . ∩ F 6 ∅ ∈ S1 T

11 Proof. That is finite is a consequence of the well known fact that the set of elements of R with a S1 given norm is finite modulo multiplication by units and that the image of R by the map x (x, x′) is a discrete additive subgroup of R2, so that it intersects every compact subset in finitely7→ many elements. 2 Assume now that R is a PID. Then, by Lemma 3.7, h (s ,s , r) k0 > 0. Therefore the set 0 1 2 ≥ 2k 2 k0 s ,s ,r = (c, d) : c = 0 and h (c,d,s ,s , r) S 1 2 ∈ S 6 1 1 2 ≥ 2k is not empty and h (s ,s , r) = sup h (c,d,s ,s , r) : (c, d) s ,s ,r . (23) 0 1 2 { 1 1 2 ∈ S 1 2 }

We claim that if (s1,s2, r) B and (c, d) s1,s2,r then (c, d) satisfies the conditions of (22). Indeed, clearly (c, d) satisfies the∈ first two conditions.∈ S It satisfies the third condition since

2 2 2 k0 2 2 N(c) N(c) h fc,d,x ,x ,r(h )=1. 2k ≤ 1 ≤ 1 2 1

To prove that it satisﬁes the last two conditions of (22) let x1 = s1 + s2ω, x2 = s1 + s2ω′ and h = h (c,d,s ,s , r) Recall that s , s 1 and hence 1 1 1 2 | 1| | 2|≤ 2

1 ω 1+ ω 1+ ω′ 1 ω′ − − x and − x − . (24) 2 ≤ 1 ≤ 2 2 ≤ 2 ≤ 2 Furthermore

k2 k2 d 2 d 2 1 N(c)2 0 0 + x + r + x + ′ = 2k 2k 1 c 2 c r ′ ! 2 2 2 2 k0 2 2 2 2 1 k0 N(c) + (cx + d) c′ r + (c′x + d′) c 2k 1 2 r 2k ≤ 2 2 2 2 2 2 1 2 2 N(c) h + (cx + d) c′ r + (c′x + d′) c h + (cx + d) (c′x + d′) = f (h )=1, 1 1 2 r 1 1 2 c,d,x1,x2,r 1 2 2 This, together with ǫ− r ǫ implies 0 ≤ ≤ 0 2 d 1 2k k0 1 x + < 2 2 ǫ 2 k , 1 c r N c k 2k 0 N(c) k1 1 | | √ ( ) 0 − ≤ − ′ k2  d q 2k 0 q 1 x2 + c′ < √r N c 2k2 2k ǫ0 N(c)2k k1. | | ( ) 0 − ≤ 1 − q q Combining this with (24) we also have

2 d 2k k0 1+ω c <ǫ0 N c 2k2 2k + 2 , | | ( ) 0 − ′ k2 ′  d q 2k 0 1 ω c′ <ǫ0 N c 2k2 2k + −2 . | | ( ) 0 − q This proves the claim.  Combining the claim with (15) and (23) we deduce that if (s ,s , r) B then h (s ,s , r) = 1 2 ∈ 0 1 2 sup h (c,d,s ,s , r) : (c, d) s ,s ,r . As is ﬁnite, this implies that h (s ,s , r) = { 1 1 2 ∈ S1 ∩ S 1 2 } S1 0 1 2

12 h1(c,d,s1,s2, r) for some (c0, d0) 1. Then (s1,s2,r,h) if and only if (s1,s2,r,h) V ≥ if ∈ S ∈F ∈ c0,d0 and only if (s1,s2,r,h) Vc,d≥ for every (c, d) . Therefore = c,d Vc,d≥ as desired. ∈ ∈ S F F∞ ∩ ( ) 1 To get the second part of the theorem, notice that V ∂ for (c, d) ∈S , c = 0. Moreover, c,d 0 T similarly as in (20), F ∩ ⊆ F ∈ S 6

∂ = (s ,s ,r,h) R2 (R+)2 : h = h (s ,s , r) . F0 { 1 2 ∈ × 0 1 2 }

Thus by the same reasoning as above, 0 Vc,d = if and only if (c, d) 1 and hence the result follows. F ∩ 6 ∅ ∈ S

Recall that a collection of subsets of a topological space X is said to be locally finite if every point of X has a neighbourhood intersecting only finitely many elements of the collection. As Hr,s is locally compact, a collection of subsets of Hr,s is locally finite if every compact subset of Hr,s intersects only finitely many elements of the collection. A fundamental domain F of a group Γ acting on Hr,s is said to be locally finite if g(F ) : g Γ is locally finite. { ∈ } Lemma 3.9 The fundamental domains of Γ and of Γ are locally finite. F F∞ ∞ Proof. Let C be a compact subset of H2 H2 and let γ Γ such that C γ( ) = . Let × ∈ ∞ ∩ F 6 ∅ Z = (x1, x2,y1,y2) C γ( ). As C is compact, the coordinates of Z are bounded. As ∈ m ∩ F∞ 1 1 ǫ0 b 1 2m γ− Γ , γ− = m for some m Z and b R, and hence γ− (Z) = (ǫ x1 + ∈ ∞ 0 ǫ0− ∈ ∈ 2m 2m 2m b,ǫ− x2 + b′,ǫ y1,ǫ− y2). As y1 and y2 are bounded, there are only finitely many m Z such 1 1 ∈ that γ− (Z) , or equivalently Z γ− ( ), and this for every Z C. Moreover as the first ∈F∞ ∈ F∞ ∈ 2 two coordinates of Z are bounded and (b,b′) : b R is a discrete subset of R for each m only 2m { ∈2m} finitely many b’s in R satisfy that ǫ x1 + b and ǫ− x2 + b′ satisfy the conditions imposed on x1 and x2 for Z to be in . Thus there are only finitely many γ Γ such that C γ( ) = and therefore is locallyF∞ finite. ∈ ∞ ∩ F∞ 6 ∅ Now supposeF∞ C γ( ) = for C a compact subset of H2 H2 and γ Γ. If γ Γ , then we ∩ F 6 ∅ a× b ∈ ∈ ∞ are done by the first part. So we may suppose that γ = with c =0. If Z C γ( ) and c d 6 ∈ ∩ F 1 1 c = 0 then, by (10), cZ+a = cγ− (Z)+ d 1 and therefore cZ + a 1, in other words 6 k− k k k≥ k− k≤ ( c,a) VC,1, where VC,1 is defined as in Lemma 3.2. Using Lemma 3.2 we deduce that ( c,a) − ∈ 1 − a b ua u− b belongs to a finite subset, up to units in R. So suppose γ = and γu = 1 are c d uc u− d such that C γ( ) = and C γu( ) = respectively for some u (R). Then ∩ F 6 ∅ ∩ F 6 ∅ ∈U 1 1 u− γ− γ = , u 0 u∗ where denotes some element of R. Denote the latter matrix by U. Then γ = γuU and U Γ . ∗ 1 1 ∈ ∞ Hence we also have C γuU( ) = or equivalently γu− (C) U( ) = . As γu− (C) is still a compact subset of H2 H2 and∩ U( F) 6 U∅( ), by the first part of∩ theF proof6 ∅ there are only finitely many × F ⊆ F∞ a b units u such that γu satisfies C γu( ) = for every fixed γ = . For every ( c,a) VC, , ∩ F 6 ∅ c d − ∈ 1 a b c,a a b fix a matrix γ c,a = − Γ. Now consider an arbitrary matrix γ = Γ and set − c d c,a ∈ c d ∈ −

13 1 1 1 1 h = γ−c,aγ. Then h Γ and h− (γ−c,a(C)) = if and only if (γ−c,a(C) h( ) = . As 1 − ∈ ∞ − ∩F∞ 6 ∅ − ∩ F∞ 6 ∅ γ−c,a(C) is compact, by the first part of the proof there are only finitely many h Γ that satisfy this− and hence there are also only finitely many γ Γ satisfying C γ( ) = . ∈ ∞ ∈ ∩ F 6 ∅

Corollary 3.10 For every Z H2 H2, there exists λ> 0 such that if B(Z, λ) g( ) = , then Z g( ) for g Γ. ∈ × ∩ F 6 ∅ ∈ F ∈ 2 2 Proof. Let Z H H . Take some λ′ > 0 randomly. As the closed ball B(Z, λ ) is compact, ∈ × ′ B(Z, λ ) g( ) = for only ﬁnitely many g Γ. Hence also B(Z, λ′) g( ) = for only ﬁnitely ′ ∩ F 6 ∅ ∈ ∩ F 6 ∅ many g Γ. Set λ0 = min d(Z,g( )) : g Γ such that B(Z, λ′) g( ) = and Z g( ) , ∈ { F ∈ ∩ F 6 ∅ λ06∈ F } where d(Z,g( )) denotes the Euclidean distance from Z to the the set g( ). Take λ = 2 and the corollary is proven.F F

The following corollary is now easy to prove.

Corollary 3.11 ∂ = g=1 g( ). F 6 F ∩ F S 4 Some topological and geometrical properties of the fundamental domain

In order to determine a presentation of Γ, we need to obtain more information on the fundamental domain of Γ constructed in the previous section. Thus in this section we will study some propertiesF of . First we recall some notions from real algebraic geometry (for more details see [BCR98]).F Recall that a real algebraic set is the set of zeros in Rn of some subset of R [X1,...,Xn]. An algebraic variety is an irreducible algebraic set, i.e. one which is not the union of two proper real algebraic subsets. Moreover, a real semi-algebraic set is a set of the form s ni n x R : fi,j (x) i,j 0 , where fi,j (x) R [X ,...,Xn] and i,j is either = or <, for i=1 j=1 { ∈ ∗ } ∈ 1 ∗ i = 1,...,s and j = 1,...,ni. We will use the notion of dimension and local dimension of semi- S T , algebraic sets as given in [BCR98]. For example the sets Vi± ≥ and Vc,d≥ are real semi-algebraic sets and the sets Vi± and Vc,d are real algebraic sets and in fact, we will prove that they are real algebraic varieties.

The following lemma describes when two sets Vc1,d1 and Vc2,d2 are equal.

Lemma 4.1 Let (c1, d1), (c2, d2) . Then Vc1,d1 = Vc2,d3 if and only if (c2, d2) = (uc1, ud1) for ∈ S 2 2 some u (R). Moreover if c d = c d and N(c ) = N(c ) then Vc ,d Vc ,d = . ∈U 1 2 2 1 1 6 2 1 1 ∩ 2 2 ∅ Proof. One implication has already been given in (18). For the other implication, suppose that

Vc1,d1 = Vc2,d2 . The case c1c2 = 0 follows from the fact that if (c, d) then c = 0 if and only if 2 2 ∈ S Vc,d = H H . So assume c c = 0. We can then rewrite fc,d as × 1 2 6 d 2 d 2 f (x , x ,y ,y )= N(c)2 x + + y2 x + ′ + y2 . c,d 1 2 1 2 1 c 1 2 c 2 " # " ′ # + In particular, if c = 0 then the intersection of Vc,d with A = (x, 0, y, 1) : x R,y R is 6 { ∈d ∈ } formed by the points (x, 0, y, 1) such that (x, y) belongs to the ball with centre c , 0 and radius 14 ′ 1 d 2 d1 d2 1+ ′ . Therefore, if V = V then = , or equivalently, c d = c d , and N(c) c c1,d1 c2,d2 c1 c2 1 2 2 1 q a1 b1 a2 b2 N(c1) = N (c2). As (c1, d1), (c2, d2) , g1 = Γ and g2 = Γ and ∈ S c1 d1 ∈ c2 d2 ∈ 1 1 u− b g2g1− Γ . Thus g2 = g1 for some u (R) and b R and so (c2, d2) = (uc1, ud1). ∈ ∞ 0 u ∈U ∈ The second part follows easily.

Let

+ = Vi , Vi− : 1 i 3 , V∞ { ≤ ≤ } = Vc,d : (c, d) ,c =0 and V { ∈ S 6 } = γ(M) : γ Γ and M . M { ∈ ∈V∪V∞} Observe that

∂ 0 V, ∂ V and ∂ V. (25) F ⊆ F∞ ⊆ F ⊆ V V ∞ V ∞ [∈V ∈V[ ∈V∪V[ , Let V ∞. If V = V ± (respectively, V = Vc,d) then let V ≥ = V ± ≥ (respectively, V ≥ = V ≥ ). ∈V∪V i i c,d Deﬁne V ≤ in a similar way. Clearly, each V ≥ and V ≥ are real semi-algebraic sets, V is the 2 2 intersection with H H of a real algebraic set and it is the boundary of V ≤ and V ≥ and its × intersection. Thus , 0 and and their boundaries are real semi-algebraic sets. aF b∞ F F Let γ = Γ 1 such that if c = 0 then a > 0 and if moreover a = 1 then c d ∈ \{ } b = s + s ω with s ,s Z. Then let 1 2 1 2 ∈

Vc,d≥ , if c = 0; +, 6 V3 ≤, if c = 0 and a< 1;  ,  V3− ≥, if c = 0 and a> 1;  +, Eγ =  V1 ≤, if c =0,a = 1 and b = s1 + s2ω with s1 < 0; (26)  ,  V1− ≥, if c =0,a = 1 and b = s1 + s2ω with s1 > 0; +, V2 ≤, if c =0,a = 1 and b = s2ω with s2 < 0;  ,  V − ≤, if c =0,a = 1 and b = s ω with s > 0.  2 2 2   Let E′ be the set obtained by interchanging in the deﬁnition of Eγ the roles of and and let γ ≤ ≥

Vγ = ∂Eγ = Eγ E′ . ∩ γ The following lemma follows by straightforward calculations.

1 1 Lemma 4.2 If γ Γ 1 then Vγ , Eγ , γ− ( ) Eγ′ and hence γ− ( ) Vγ . ∈ \{ } ∈V∞ ∪V F ⊆ F ⊆ F ∩ F ⊆ Using (18) it is easy to prove the following lemma.

Lemma 4.3 The set is locally ﬁnite. V∪V∞ We ﬁrst calculate the local dimensions of and ∂ . F F

15 Lemma 4.4 The fundamental domain has local dimension 4 at every point and its boundary ∂ has local dimension 3 at every point. F F Proof. Let Z . We have to prove that for every λ > 0, B(Z, λ) is of dimension 4. This ∈ F ∩F is obvious for Z ◦. So suppose Z ∂ . If Z ∂ 0, then Z V for some V . As ∂ 0 is closed, there is∈λ F> 0 such that B(Z,∈ λF) ∂F 6∈= F. We may assume∈ without loss∈ V of∞ generalityF 0 0 ∩ 0 ∅ that λ0 is smaller than 1 and 2ǫ0. If Z = (s1,s2, r = s3,h) then let Z′ = (s1′ ,s2′ , r′ = s3′ ,h) where for i =1, 2, 3 we have λ0 + si , if Z V ; − √3 ∈ i λ0 s′ = si + , if Z V −; i  √3 ∈ i  si, otherwise.

Then Z′ B(Z, λ0) ◦ 0◦ and thus there exists λ′ > 0 such that B(Z′, λ′) ◦ B(Z, λ0). ∈ ∩F∞ ∩F  ⊆F ∩ Hence B(Z, λ0) is of dimension 4. By the choice of λ0, this is true for any λ < λ0 and of course also for any∩Fλ > λ . Finally suppose Z ∂ ∂ . By(20) and (21) Z is of the form 0 ∈ F ∩ F0 (s1,s2,r,h0(s1,s2, r)) with (s1,s2, r) B. Let λ> 0. By(20) the point Z′ = (s1,s2,r,h0(s1,s2, r)+ λ ∈ ) is in B(Z, λ) ◦. If Z′ ◦, then there exists λ′ > 0 such that B(Z′, λ′) B(Z, λ) ◦ and 2 ∩F0 ∈F ⊆ ∩F hence B(Z, λ) has dimension 4. If not, then Z′ ∂ ◦ and hence by the above there exists ∩F ∈ F∩F0 λ′ > 0 such that B(Z′, λ′) B(Z, λ) and B(Z′, λ′) has dimension 4. Thus also B(Z, λ) has dimension 4. ⊆ ∩F ∩F To prove the second part, take Z ∂ , λ > 0 and set B = B(Z, λ), U1 = B and c c ∈ F 2 2 ∩ F U2 = B ( ∂ ), where denotes the complementary of in H H . Then B, U1 and U2 satisfy the∩ F conditions∪ F of LemmaF A.3 and hence B(Z, λ) ∂ =FU U ×has dimension 3. ∩ F 1 ∩ 2 The next three lemmas give more details on the elements of . V∪V∞ Lemma 4.5 The elements of are non-singular irreducible real algebraic varieties of dimen- V∪V∞ sion 3. Moreover if two diﬀerent varieties M1 and M2 intersect non-trivially, with M1,M2 , then their intersection has local dimension 2 at every point. ∈ V∪V∞

Proof. Applying the general implicit function theorem to the following functions

2 2 2 2 2 2 fc,d(x1, x2,y1,y2) = (cx1 + d) + c y1 (c′x2 + d′) + c′ y2 ,

F (x1, x2,y1,y2) = (fc1,d1 (x1, x2,y1,y2),fc2,d2 (x1, x2,y1,y2)) , for (c, d), (c , d ), (c , d ) with c,c ,c = 0 and Vc ,d = Vc ,d , one gets that Vc,d and Vc ,d 1 1 2 2 ∈ S 1 2 6 1 1 6 2 2 1 1 ∩ Vc2,d2 are ∞-manifolds of dimension 3 and 2 respectively. The fact that the elements of are path-connectedC together with [BCR98, Proposition 3.3.10] yield the desired property. TheV same argument works for the elements of or the combination of an element of and an element of . V∞ V V∞

Lemma 4.6 The elements of are non-singular irreducible real algebraic varieties of dimension 3 and the intersection of two diﬀerentM elements of is of dimension at most 2. M Proof. The ﬁrst part follows easily from Lemma 4.5 and [BCR98, Theorem 2.8.8] and the fact 2 2 that the action of PSL2(R) on H H is a bijective semi-algebraic map. The second part follows trivially from the irreducibility. ×

16 Lemma 4.7 Let M1,M2,M3 be pairwise diﬀerent elements of with M1 M2 M3 = . V∪V∞ ∩ ∩ 6 ∅ Then M1 M2 M3 is a real algebraic set of local dimension 1 at every point and with at most one singular point.∩ ∩

Proof. Suppose that Mi = Vci,di for i = 1, 2, 3. Similar, as in the proof of Lemma 4.5, we apply the implicit function theorem to the function F : R4 R3 deﬁned by →

(x , x ,y ,y ) (fc ,d (x , x ,y ,y ),fc ,d (x , x ,y ,y ),fc ,d (x , x ,y ,y )). 1 2 1 2 7→ 1 1 1 2 1 2 2 2 1 2 1 2 3 3 1 2 1 2 1 So Vc1,d1 Vc2,d2 Vc3,d3 = F − (1, 1, 1). It is suﬃcient to prove that the Jacobian matrix of F is of rank 3∩ except possibly∩ for at most one point. The Jacobian matrix of F has the following form

c1(c1x1 + d1)α1 c2(c2x1 + d2)α2 c3(c3x1 + d3)α3 c1′ (c1′ x2 + d1′ )β1 c2′ (c2′ x2 + d2′ )β2 c3′ (c3′ x2 + d3′ )β3 2  2 2 2  , (27) c1y1α1 c2y1α2 c3y1α3 2 2 2  c′ y2β1 c′ y2β2 c′ y2β3   1 2 3    2 2 2 2 2 2 1 where αi = (ci′ x2 + di′ ) + ci′ y2 and βi = (cix1 + di) + ci y1. Note that βi = αi− because (x , x ,y ,y ) Vc ,d Vc ,d Vc ,d . In particular αi is nonzero for every i = 1,..., 3. The 1 2 1 2 ∈ 1 1 ∩ 2 2 ∩ 3 3 determinant det1 of the submatrix of (27) formed by the ﬁrst three rows is

det1 = (γ1(c1′ x2 + d1′ )+ γ2(c2′ x2 + d2′ )+ γ3(c3′ x2 + d3′ ))y1,

1 with γi = c′ ci ci α− αi αi (ci di ci di ), where the indexes are to be interpreted i +1 +2 i +1 +2 +1 +2 − +2 +1 modulo 3. Note that each γi = 0 because of Corollary 4.1 and the assumption that the three varieties 6 considered are distinct. Similar calculations show that the determinant det2 of the submatrix of (27) formed by the rows 1, 3 and 4 is:

det = ( γ c′ γ c′ γ c′ )y y . 2 − 1 1 − 2 2 − 3 3 1 2 If both determinants are 0, then (because y = 0 and y = 0) 1 6 2 6

γ1c1′ + γ2c2′ + γ3c3′ =0 (γ1d1′ + γ2d2′ + γ3d3′ =0.

Hence the vector (γ1,γ2,γ3) is perpendicular to the vectors (c1′ ,c2′ ,c3′ ) and (d1′ , d2′ , d3′ ) and thus

(γ ,γ ,γ )= t((c′ ,c′ ,c′ ) (d′ , d′ , d′ )) 1 2 3 1 2 3 × 1 2 3 for some t R. So ∈ 1 c1′ c2c3α1− α2α3(c2d3 c3d3) c2′ d3′ c3′ d2′ 1 − − c2′ c3c1α2− α3α1(c3d1 c1d3) = t c3′ d1′ c1′ d3′ .  1 −   −  c′ c c α− α α (c d c d ) c′ d′ c′ d′ 3 1 2 3 1 2 1 2 − 2 1 1 2 − 2 1     Dividing the ﬁrst coordinate by the second and the third, we get that

2 2 α1− = µ2α2− (28) 2 2 α1− = µ3α3− , (29)

17 1 2 for some µ2 and µ3 depending only on ci and di for 1 i 3. Because αi− can be interpreted as ci ≤ ≤ d times the square of the Euclidean distance from the point (x ,y ) to the point ( i , 0), equations 1 1 ci (28) and (29) take the form −

d1 di d (x ,y ), , 0 = λid (x ,y ), , 0 1 1 − c 1 1 − c 1 i with λ = ci 4 µ for i = 2, 3. Hence, modulo a translation (x ,y ) is a solution for (x, y) of the i c1 √ i 1 1 system

x2 + y2 = λ2((x a)2 + y2) 2 − x2 + y2 = λ2((x b)2 + y2), ( 3 − when a and b are different non-zero real numbers. If λi = 1for i = 2or i = 3, then the corresponding equation represents the bisector (which is a line) of (0, 0) and (a, 0) or of (0, 0) and (b, 0). Otherwise, 2 λi a λ2a the equation represents the circle i with centre 1 λ2 , 0 and radius 1 λ2 (because y > 0 and C − − i − 2 a = b, the case λ1 =1= λ2 is impossible). Such two different lines, two different circles or a circle and6 a line intersect in at most one point. So, if there are more than two points in H2 H2 satisfying × the two equations then λi = 1 for i =2, 3 and the equations represent the same circles. In this case 6 6 2 2 λ2a λ3b 1 λ2 = 1 λ2 − 2 − 3  λ2a λ3b 1 λ2 = 1 λ2 .  − 2 − 3

Dividing the first equation by the second we conclude that λ2 = λ3 thus a = b, a contradiction. So we have shown that there exists at most one possible couple (x1,y1) which can be completed to a point (x1, x2,y1,y2) in the intersection and such that det1 = det2 = 0. Therefore, the condition that the rank of the Jacobian matrix in (x1, x2,y1,y2) is less than 3 determines the coordinates x1 and y1. By symmetry, it also determines the coordinates x2 and y2. This means that the intersection Vc ,d Vc ,d Vc ,d is a real algebraic variety of dimension at most 1. 1 1 ∩ 2 2 ∩ 3 3 A similar argument shows that the result remains true if one of several of M1, M2 and M3 are elements of . V∞ In order to determine the border of , we need the following definition. F Definition 4.8 An essential hypersurface of is an element M such that M is of F ∈V∪V∞ ∩F dimension 3. Let e denote the set of essential hypersurfaces of . V F Following the notation of [BCR98], for A a real semi-algebraic set, we denote by A(d) the set of points of A with local dimension d, i.e.

A(d) = Z A : λ> 0, dim(B(Z, λ) A)= d . (30) { ∈ ∀ ∩ } Lemma 4.9 ∂ = (M )= (M )(3), F ∩F ∩F M e M e [∈V [∈V

18 Proof. (M )(3) (M ) ∂ . Let Z ∂ and = M : Z M . Clearly M e M e Z ∈V ∩F ⊆ ∈V ∩F ⊆ F ∈ F V { ∈V∪V∞ ∈ } We have to prove that M F has local dimension 3 at Z for some M Z . By(25) and S ∩ S ∈ V Lemma 4.3, Z is a non-empty ﬁnite set. Thus there is an open ball B = B(Z, λ ) such that V 0 for every M , B M = if and only if M Z . Thus, if 0 < λ λ0, then ∈ V∪V∞ ∩ 6 ∅ ∈ V ≤ ∂ B(Z, λ) = M Z B M and by Lemma 4.5, ∂ B(Z, λ) has dimension 3. Thus F ∩ ∪ ∈V F ∩ ∩ F ∩ B(Z, λ) M has dimension 3 for some M Z . Therefore M has local dimension 3 at Z. F ∩ ∩ ∈V ∩F

Lemma 4.10 If V , then V F has local dimension 3 at every point, i.e. (V )(3) = V . ∈V∞ ∩ ∩F ∩F Proof. This follows by arguments similar to those used in the proof of Lemma 4.4.

The following proposition gives information about essential hypersurfaces and will be important in the next two sections.

Proposition 4.11 Assume Z ∂ and Z is contained in at most two elements M and M ′ of ∈ F . Then M and M ′ are essential and for every λ> 0, the intersections B(Z, λ) M and V∪V∞ ∩ ∩F B(Z, λ) M ′ are of dimension 3. ∩ ∩F Proof. If Z is contained in a single element M of , then the statement is obvious. So suppose V∪V∞ that Z M M ′ for some M,M ′ with M = M ′ and Z is contained in no other element of ∈ ∩ ∈V∪V∞ 6 < > . By Lemma 4.5, M M ′ is of dimension 2. Moreover M = (M M ′ ) (M M ′) (M M ′ ) V∪V∞ ∩ ∩ ∪ ∩ ∪ ∩ and, as by the proof of Lemma 4.5, both varieties M and M ′ are not tangent in Z the ﬁrst and the > third intersections are of dimension 3. Thus B(Z, λ) M M ′ is of dimension 3 and is contained in the boundary of . Hence M is an essential hypersurface∩ ∩ and B(Z, λ) M is of dimension F ∩ ∩F 3. Inverting the role of M and M ′, we get the same result for M ′.

Finally, if R is a PID, we can prove compactness of the intersection of with the elements of . F V

Lemma 4.12 Assume R is PID and let Vc,d, for (c, d) with c =0. Then Vc,d is compact. ∈ S 6 ∩F 1 1 2 2 Proof. Let Z = (s1,s2,r,h) Vc,d . As Vc,d , s1 2 , s2 2 and ǫ0− r ǫ0. ∈ ∩F ∩F ⊆F∞2 | |≤ | | ≤ ≤ ≤ k0 Moreover Vc,d 0 and hence, by Lemma 3.7, h > 2k . As Z ∂ , Lemma 3.5 yields that ∩F ⊆F 2 ∈2 F h 1. Thus Vc,d is hyperbolically bounded and closed in H H and hence it is compact. ≤ ∩F ×

5 Generators of Γ

In the remainder of the paper, denotes the tessellation of H2 H2 given by Γ and , i.e = γ( ) : γ Γ . In order to getT generators for Γ, we have to analyse× the intersections betweenF elementsT { F of .∈ The} next lemma is crucial. T Lemma 5.1 Let T , T and T be three different elements of . Then T T T M M 1 2 3 T 1 ∩ 2 ∩ 3 ⊆ 1 ∩ 2 for M1,M2 two different elements of . In particular the intersection of three different tiles has dimension at most 2. M

19 Proof. For every i = 1, 2, 3, let γi Γ with Ti = γi( ). Let = M : Ti Tj ∈ F N { ∈ M ∩ ⊆ M for some 1 i j 3 . By Lemma 4.2, for every 1 i < j 3, we have Ti Tj γi(Vγ−1γ ). ≤ ≤ ≤ } ≤ ≤ ∩ ⊆ j i Thus = and it is enough to show that has at least two diﬀerent elements. By means of N 6 ∅ N contradiction, assume that = M = γ(V ) for some V and γ Γ. Let M ≥ = γ(V ≥) N { } ∈V∞ ∩V ∈ and M ≤ = γ(V ≤). Then M = γi Vγ−1γ and, by Lemma 4.2, for each 1 i < j 3 either j i ≤ ≤ Ti M ≥ and Tj M ≤ or viceversa. By symmetry one may assume that T M ≥. Then ⊆ ⊆ 1 ⊆ T M ≤ and hence T M ≤ M ≥ = M, a contradiction because dim(T ) = 4 and dim(M)=3. 2 ⊆ 3 ⊆ ∩ 3

Deﬁnition 5.2 For each γ Γ 1 , set ∈ \{ } 1 Sγ = γ− ( ). F ∩ F

A side of with respect to Γ is a set of the form Sγ that has dimension 3. F In that case, we say that γ is a side pairing transformation of Γ with respect to , or simply a pairing F1 transformation. Observe that if γ is a pairing transformation then so is γ− and γ(Sγ ) = Sγ−1 . Hence the pairing transformations “pair” the sides of . More generally, a side of is a side of γ( ) for some γ Γ. Equivalently, the sides of are theF sets of dimension 3 of the formT γ( ) φ( F), with γ, φ ∈Γ. T F ∩ F We now∈ can state the main result of this section.

Theorem 5.3 Let be the fundamental domain of Γ = PSL2(R) described in Theorem 3.4. Then Γ is generated by theF pairing transformations of with respect to Γ. F

Proof. Let L = i I γi( ) : dim( i I γi( )) 2 and consider the set { ∈ F ∈ F ≤ } T T Ω= H2 H2 Y. × \ Y L [∈ This is a set of elements Z H2 H2 that belong either to the interior of a tile of or to a unique side of . Indeed, let∈Z Ω× and suppose Z is not contained in the interior of aF tile of . T ∈ F Then Z γ1( ) γ2( ) for at least two distinct elements γ1 = γ2 Γ. Because of Lemma 5.1, γ ( ) and∈ γ (F )∩ are theF only tiles containing Z. Thus, Z 6 γ ( )∈ γ ( ) and Z γ( ) for 1 F 2 F ∈ 1 F ∩ 2 F 6∈ F γ Γ γ1,γ2 . By the definition of Ω, dim(γ1( ) γ2( )) = 3 and hence γ1( ) γ2( ) is the unique∈ \{ side containing} Z. F ∩ F F ∩ F Note that X = H2 H2 and L satisfy the hypotheses of Lemma A.1 and hence Ω is path- × connected. Let γ Γ and Z ◦. Put W = γ(Z) γ( )◦. There exists a path p in Ω joining Z and W . Let A = ∈h Γ : p ∈Fh( ) = . As p is compact∈ F and as is locally finite by Lemma 3.9, A is finite. We define{ ∈ recursively∩ F a sequence6 ∅} of subsets of A by settingF A = 1 and if i 1 then 0 { } ≥ Ai = h A : h( ) k( ) is a side for some k Ai 1 j\0 because} \ k A B F α(0) = Z and 1 B, so that Z h( ) for each∈ \ h A B. Then α([0,a)) h( ) ◦ S h B and as this∈ union F is closed,∈ α(a) h( 6∈) Fk( ) for some∈h \B and k A B.⊆ As α(∈a) FΩ, ∈ F ∩ F ∈ ∈ \ S ∈

20 h( ) k( ) has dimension 3 and hence it is a side. This contradicts the deﬁnition of B. Hence F ∩ F A = B and in particular γ B. By using the sets Ai, we create a sequence γ0 =1,γ1,...,γk = g ∈ 1 such that for every 1 j, γj 1( ) γj( ) is a side. Hence γj− 1γj ( ) is a side of and thus 1 ≤ − F ∩ F 1 F1 ∩ − F1 F − − − − γj 1γj is one of the proposed generators. As γ = (γ0 γi1 )(γi1 γi2 ) . . . (γik 1γik ), the result follows. − −

Recall that the border of is covered by the essential hypersurfaces of . In fact it is also covered by the sides of . F F F Lemma 5.4 The border of is the union of the sides of . F F Proof. Since the sides of are contained in ∂ , we only have to prove that if Z ∂ , then Z belongs to a side of . AsFZ is in ∂ and isF a locally ﬁnite a fundamental domain,∈ FZ γ( ) F F F ∈ F for only ﬁnitely many γ Γ, say γ =1,...,γk. Choose λ > 0 such that the closed ball B(Z, λ ) ∈ 1 0 0 intersects γi( ) if and only if Z γi( ) (this is possible by Corollary 3.10). Then F ∈ F k Z B(Z, λ )= B(Z, λ ) B(Z, λ ) γi( ) . ∈ 0 0 ∩F ∪ 0 ∩ ∪i=2 F Thus by Lemma A.3, γi( ) is of dimension 3, for at least one 1 i k and hence Z is contained in a side of F. ∩ F ≤ ≤ F In order to give more precise information on the generators of Γ described in Theorem 5.3, we need to analyse the relationship between pairing transformations and essential hypersurfaces. By Lemma 4.9, ∂ is the union of the sets of the form ( V )(3) with V running through the essential hypersurfacesF and by Lemma 5.4, ∂ also is the unionF ∩ of the sides of with respect to Γ. F F Moreover, if γ is a pairing transformation then Sγ Vγ , by Lemma 4.2. Hence Vγ is an essential ⊂ F ∩ hypersurface of and it is the unique essential hypersurface of containing Sγ. Conversely, let V be an essential hypersurfaceF and let F

ΓV = γ : γ is a pairing transformations such that Sγ V . { ⊆ } Clearly, γ ΓV if and only if dim Sγ = 3 and V = Vγ . Moreover, ∈ (3) (V ) Sγ . ∩F ⊆ γ ΓV ∈[ Each pairing transformation belongs to ΓV for some essential hypersurface V . We will show that, in order to generate Γ it is enough to take one element of ΓV for each essential hypersurface. We start dealing with the elements of . Clearly Γ is generated by the following elements: V∞ ∞ 1 1 1 ω ǫ0 0 P1 = , P2 = , P3 = 1 . 0 1 0 1 0 ǫ− 0

Using (19) and (21) it is easy to see that the six sets Vi±, with i =1, 2, 3, are essential hypersurfaces. Moreover, a straightforward calculation shows that if i =1, 2 then 1 ΓV ± = Pi∓ (with opposite signs on both sides). (31) i { } 1 Thus, for i =1, 2, Pi (respectively, Pi− ) is a pairing transformation and its side covers the part of + the boundary given by V (respectively, V − F ). i ∩F i ∩ However to cover V ± with sides, we may need more than one side. F ∩ 3

21 Lemma 5.5 If g ΓV ± then g = QP3∓ (with opposite signs on both sides) for some Q P1, P2 ∈ 3 ∈ h i and the inversion map is a bijection ΓV + ΓV − . Moreover ΓV + and ΓV + are finite. 3 → 3 3 3 Proof. The first statement follows from Lemma 4.2, Lemma 4.5, (26) and some easy computations. 1 ǫ0− b If γ ΓV + then γ = with b R. Moreover, by (2) and (3), γ( ) is a non- ∈ 3 0 ǫ0 ∈ F ∩ F empty subset formed by elements of the form (x1, x2,y1,y2) such that (x1, x2) and (x1, x2) = 2 1 2 (ǫ− x + ǫ− b,ǫ x + ǫ b′). belong to a compact set. As (b,b′) : b R is discrete, we deduce that 0 1 0 0 2 0 { ∈ } b belongs to a finite subset of R. Thus Γ + is finite. Hence Γ − is finite too. V3 V3 c c

We now deal with the essential hypersurfaces of the form Vc,d with (c, d) (and necessarily c = 0). ∈ S 6

Lemma 5.6 Let (c, d) with c =0. Then the second row of every element of ΓV is of the form ∈ S 6 c,d (uc,ud) for some u (R). Equivalently, if the second row of γ Γ is (c, d) then ΓVc,d γΓ . ∈U ∈ ⊆ ∞ 2 Proof. Let γ ΓV and let v R be the second row of γ. Then Sγ Vγ Vc,d, by Lemma 4.2, ∈ c,d ∈ ⊆ ∩ and hence Vγ = Vc,d, by Lemma 4.6. Then v = (uc,ud) for some u (R), by Lemma 4.1. ∈U

Let e denote the set of (c, d) such that Vc,d is an essential hypersurface of . S ∈ S F a b Corollary 5.7 For every (c, d) e choose a,b R with Pc,d = Γ. Then Γ = ∈ S ∈ c d ∈ P , P , P , Pc,d : (c, d) e . h 1 2 3 ∈ S i

Proof. By Theorem 5.3, Γ is generated by V e ΓV and hence it is enough to show that ∪ ∈V P , P , P ,γc,d : (c, d) contains ΓV for each V e. This is a consequence of (31), Lemma 5.5, h 1 2 3 ∈ Si ∈V Lemma 5.6 and the fact that Γ is generated by P1, P2 and P3. ∞ In case R is a PID we can combine Theorem 3.8 and Corollary 5.7 to get the following

Corollary 5.8 Suppose that R is a PID and let 1 be as in Theorem 3.8. For each (c, d) 1 a b S ∈ S choose a,b R such that Pc,d = Γ. Then Γ= P , P , P , Pc,d (c, d) . ∈ c d ∈ h 1 2 3 | ∈ S1i 6 Deﬁning Relations of Γ

In this section we construct the relations associated to the pairing transformations, and hence we obtain a presentation of the group Γ. As in the classical case, there are two types of relations. These will be called the pairing relations and the cycle relations. The pairing relations are quite obvious to establish.

1 Notation 6.1 Given a side S of , let γS denote the unique element of Γ such that S = γ− ( ) F F ∩ S F and let S∗ = γS( ). F ∩ F

22 Observe that if S is a side then γS(S) = S∗. Moreover S γS and γ Sγ deﬁne mutually 7→ 7→ 1 inverse bijections between the sides of and the pairing transformations such that γ− = γS∗ , or F S equivalently Sγ∗ = Sγ−1 . The pairing relation given by S is then simply γSγS∗ = 1. Note that in 2 case S = S∗, we get as pairing relation γS = 1. We will now turn to the cycle relations. For this we need to introduce the following deﬁnition.

Deﬁnition 6.2 A cell C of is a non-empty intersection of tiles of satisfying the following property: for every γ Γ, eitherT C γ( ) or dim(C γ( )) dim(CF) 1. Clearly, the cells of dimension 4 are the tiles.∈ By Lemma⊆ 5.1F, the sides of∩ Fare≤ the cells of− dimension 3. A cell of dimension 2 is called an edge. If a cell or edge is containedT in a tile T , then it is called a cell or an edge of T .

Observe that a cell is always a ﬁnite intersection of tiles. Indeed, consider Z C. As Z is compact, Z is contained in only ﬁnitely many tiles of , and hence so is C.∈ The following{ } proposition generalizes in some sense the notion of relativeF interior of a cell.

Proposition 6.3 Let C be a cell of . Then there exists Z C, such that, for every γ Γ, Z γ( ) if and only if C γ( ). T ∈ ∈ ∈ F ⊆ F

Proof. Let γ1,...,γn be the elements γ Γ such that C γ( ). Suppose, by contradiction, that there do not exist Z C that satisfy the∈ statement of the⊆ proposition.F Then, for every Z C, ∈ ∈ there exists γZ (Γ γi : 1 i n ) such that Z γZ ( ). Put Γ∗ = γZ : Z C . This is a ∈ \{ ≤ ≤ } ∈ F { ∈ } countable set because Γ is countable, and clearly C γ Γ∗ γ( ). Thus ⊆ ∪ ∈ F

C = γ Γ∗ (γ( ) C) . ∪ ∈ F ∩

As C γ( ) for γ Γ∗ and because C is a cell, dim(C γ( )) dim(C) 1 and hence 6⊆ F ∈ ∩ F ≤ − C = γ Γ∗ (γ( ) C) has dimension at most dim(C) 1, a contradiction. ∪ ∈ F ∩ − The next lemma is an obvious consequence of the deﬁnition of an edge.

Lemma 6.4 Let γ Γ. If E is an edge then γ(E) is an edge. In particular, if E is an edge of ∈ 1 F contained in some side Sγ = γ− ( ), with γ Γ, then γ(E) is an edge of . F ∩ F ∈ F In order to prove more results on the edges of , we ﬁrst have to analyse the sides a bit more in detail. F

Lemma 6.5 Let (c, d) with c = 0. If t Γ then there exists (c0, d0) , c0 = 0, such that ∈ S 6 ∈ ∞ ∈ S 6 t(Vc,d)= Vc0,d0 .

ǫ b 2 Proof. Let Z Vc,d. Fix α, β R such that αc + βd = 1. Write t = 1 . Let c0 = ǫ− c R ∈ ∈ 0 ǫ− ∈ 1 2 and d0 = d ǫ− bc R. Then (c0, d0) , because α′c0 + β′d0 = 1 for α′ = ǫβb + ǫ α R and − ∈ 2∈ S 2 ∈ β′ = β R. Moreover, for every Z H H , we have ∈ ∈ × 2 2 1 c t(Z)+ d = ǫ− c(ǫ Z + ǫb)+ d ǫ− bc = cZ + d k 0 0k k − k k k

Hence t(Vc,d)= Vc0,d0 .

Similarly to Lemma 5.1, one may analyse the intersection of three sides.

23 Lemma 6.6 The intersection of three distinct sides of is of dimension at most 1. F

Proof. Let S , S , S be three distinct sides of and let γi = γS . By Lemma 4.2, each Si Vγ 1 2 3 F i ⊆ i and hence Vγ is an essential hypersurface of . If Vγ = Vγ , for every 1 i < j 3, then by i F i 6 j ≤ ≤ Lemma 4.7, S S S is of dimension at most 1. Assume now that Vγ = Vγ = Vγ = Vc,d, for 1 ∩ 2 ∩ 3 1 2 3 some (c, d) , c = 0. Then, by Lemma 5.6 and Lemma 4.2, γ1(S1 S2 S3) = γ1( ) 1 ∈1 S 6 ∩ ∩ F ∩ F ∩ t2− ( ) t3− ( ) V c,a Vt2 Vt3 , for some a R. If Vt2 = Vt3 , then by Lemma 4.7, the last F ∩ F ⊆ − ∩ ∩ ∈ 6 intersection is of dimension at most 1 as desired. Suppose that Vt = Vt = V ±. Using that t = t 2 3 i 2 6 3 it is easy to prove that i = 3 and hence t2,t3 P1, P2 . By Lemma 5.5, t3 = Qt2 for Q P1, P2 . 6∈1 h i ∈ h i Thus t2γ1(S1 S2 S3) t2(V c,a) Q− ( ) t2( ). By Lemma 6.5, t2(V c,a) = Vc′,d′ for ∩ ∩ ⊆ − ∩ F ∩ F ∩ F − some (c′, d′) , c′ = 0 and by Lemma 4.2 and the fact that t2 P1, P2, , t2( ) 3± and 1 ∈ S 6 6∈ h i F ∩F ⊆V Q− ( ) Vi± with i = 1 or 2. Hence Lemma 4.7 allows to conclude. F∞ ∩F∞ ⊆ Next assume that Vγ1 = Vγ2 = Vγ3 = Vi± with i = 1, 2 or 3. By Lemma 4.2 and the fact that the γi’s are diﬀerent we conclude that i = 3 and hence γ1,γ2,γ3 Γ P1, P2 . Then, ∈ ∞ \ h i again by the same argument as above, γ2 = Q2γ1 and γ3 = Q3γ1 for Q2,Q3 P1, P2 . Thus 1 ∈ h i γ (S S S ) γ ( ) Q− ( ) Q ( ) V ± V ± V ±. with i, j 1, 2 . However, as 1 1 ∩ 2 ∩ 3 ⊆ F ∩ 1 F ∩ 2 F ∩ 3 F ⊆ 3 ∩ i ∩ j ∈{ } Q = Q , V ± = V ± and again Lemma 4.7 allows to conclude. 2 6 3 i 6 j So up to permutations of the sides, we are left to deal with the case Vγ1 = Vγ2 = Vγ3 and four possible subcases: 6

1. Vγ = Vc ,d and Vγ = Vc ,d with (c , d ), (c , d ) and c and c diﬀerent from 0; 1 1 1 3 3 3 1 1 3 3 ∈ S 1 3

2. Vγ1 = Vc1,d1 and Vγ3 with (c1, d1) and c1 = 0; ∈V∞ ∈ S 6

3. Vγ1 and Vγ3 = Vc3,d3 for (c3, d3) and c3 = 0. ∈V∞ ∈ S 6

4. Vγ1 , Vγ3 . ∈V∞ We deal with each case separately. In the first case, by Lemma 5.6, γ2 = tγ1, for some 1 = t Γ . Thus γ1(S1 S2 S3) = 1 1 6 ∈ ∞ ∩ ∩ γ1( ) t− ( ) γ1γ3− ( ) Vγ−1 Vt Vγ γ−1 . We claim that Vγ−1 , Vt and Vγ γ−1 are F ∩ F ∩ F ∩ F ⊆ 1 ∩ ∩ 3 1 1 3 1 −1 pairwise different. On the one side Vt while Vγ and hence Vγ1 = Vt. On the other side, ∈V∞ 1 ∈V 6 since Vγ1 = Vc1,d1 , the last row of γ1 is of the form (uc1, ud1) with u (R), by Lemma 4.1. If 1 ∈ U γ3γ1− Γ then the last row of γ3 is of the same form and hence Vc1,d1 = Vc3,d3 contradicting the ∈ ∞ 1 hypothesis. Thus γ1γ3− Γ and therefore Vγ γ−1 = Vt. Finally, if Vγ−1 = Vγ γ−1 then the last ∞ 3 1 1 3 1 1 1 6∈ 6 rows of γ1− and γ3γ1− differs in a unit and hence γ3 Γ . ∈ ∞ In the second case, again by the same reasoning, γ2 = tγ1 for some t Γ and γ3 Γ . Thus 1 1 1 1 1 ∈ ∞ 1 1 ∈ ∞ S1 S2 S3 = γ1− ( ) γ1− t− ( ) γ3− ( ) Vγ1 V γ1− ( ) γ1− t− ( ) with V ∩ ∩ F ∩ F ∩ F1 ∩ F1 1⊆ ∩ ∩ 1 F ∩ F en element of . Observe that γ1 γ1− ( ) γ1− t− ( ) = t− ( ) V ′ for some element V∞ F ∩1 F F ∩ F ⊆ V ′ . Hence S1 S2 S3 Vγ1 V γ1− (V ′). By applying the implicit function theorem, as ∈V∞ ∩ ∩ ⊆ ∩ ∩ in the proof of Lemma 4.5 and distinguishing between the different cases for V and V ′, it is now a 1 matter of a straightforward tedious calculation to prove that Vγ1 V γ1− (V ′) has dimension at most 1. ∩ ∩

In the third case Vγ1 = V3± by (31). and γ2 = Qγ1 P1, P2 P3± for some Q P1, P2 . Then 1 1 1 ∈ h i ∈ h 1 i γ1(S1 S2 S3)= γ1− ( ) Q− ( ) γ1γ3− ( ) Vγ−1 VQ Vγ γ−1 . Since γ3γ1− Γ and 1 3 1 ∞ 1 ∩ ∩ F ∩ F ∩ F ∩ F ⊆ ∩ ∩ 6∈ γ1− P1, P2 we deduce that Vγ−1 , VQ and Vγ γ−1 are diﬀerent and hence the result follows from 6∈ h i 1 3 1 Lemma 4.7.

24 In the fourth case, again Vγ1 = Vγ2 = V3± and γ2 = Qγ1, as above. Moreover S1 S2 S3 = 1 1 1 1 1 1 1 ∩ ∩ γ1− ( ) Q− γ1− ( ) γ3− ( ) Vγ1 Vγ3 γ1− ( ) Q− γ1− ( ) , with Vγ1 and Vγ3 two F ∩ F ∩ F ∩ F ⊆ 1 ∩ ∩ 1 1F ∩ 1 F diﬀerent elements of . As incase2, γ1− ( ) Q− γ1− ( ) = γ1− (V ′) for some V ′ V1±, V2± . ∞ Again the implicit functionV theorem gives theF desired∩ resulFt. ∈{ } We now describe V for V in terms of intersection of tiles. F ∩ ∈V∪V∞

Lemma 6.7 Let V and set ΓV∗ = γ Γ : γ( ) V . Then ∈V∪V∞ { ∈ F ∩ F ⊆ ∩F} V = ( γ( )) . ∩F F ∩ F γ Γ∗ ∈[V Proof. One inclusion is obvious. To prove the other one, take Z V . Suppose V , i.e. a b ∈ ∩F ∈ V V = Vc,d for some (c, d) . Take γ = Γ for some a,b R. By(7), h(γ(Z)) = h(Z) ∈ S c d ∈ ∈ and hence by Lemma 3.3, γ(Z) 0. As is a fundamental domain for Γ , there exists τ Γ such that τγ(Z) . As, by∈F (9), h(τγF(∞Z)) = h(γ(Z)) = h(Z), we have∞τγ(Z) and∈ thus∞ 1 1 ∈ F∞ 1 1 ∈ F Z γ− τ − ( ). By Lemma 4.2, γ− τ − ( ) Vc,d . ∈If FV ∩ a similarF reasoning mayF be ∩ applied. F ⊆ ∩F ∈V∞ Next we show that every edge is contained in precisely two sides. To prove this we will make use of the following lemma.

Lemma 6.8 Let Z and let C denote the intersection of the tiles of containing Z (i.e. ∈ F T C = Z γ( ) γ( )). If dim(C)=2, then the number of sides of containing C is exactly 2. ∈ F F F Proof.T By Lemma 6.6, C is not contained in three different sides. So we only have to show that C is contained in two different sides. Clearly, every intersection of tiles containing Z also contains C. We claim that a similar property holds for the elements M . Indeed, if M then M = γ(V ) for some V and ∈ M 1 ∈ M ∈V∩V∞ some γ Γ. If Z M then γ− (Z) V and hence, by Lemma 6.7, there is γ1 Γ such that 1 ∈ ∈ ⊆ ∈ γ− (Z) γ1( ) V . Therefore Z γ( ) γγ1( ) and hence C γ( ) γγ1( ) γ(V )= M, as desired.∈ F We ∩ claimF ⊆ that a similar property∈ F holds∩ forF the elements ⊆V F ∩ .F Indeed,⊆ if Z V ∈V∪V∞ ∈ then by Lemma 6.7, there is γ1 Γ such that Z γ1( ) V . Hence C γ1( ) V , as desired. ∈ ∈ F ∩ F ⊆ ⊆ F ∩ F ⊆ By Lemma 4.9, Z is contained in at least one essential hypersurface W . Since C has dimension 2, Z cannot be contained in more than two elements of , by Lemma 4.7. Then, the elements of containing Z are essential hypersurfaces of V∪Vby Proposition∞ 4.11. Take λ> 0 such that B(V∪VZ, λ) ∞γ( )= for every γ with Z γ( ) and B(Z,F λ) W = for every essential hypersurface W of ∩notF containing∅ Z. Again by6∈ PropositionF 4.11,∩ if W is∅ an essential hypersurface of containingF Z then B(Z, λ) W is of dimension 3 and B(Z, λ) W ( F ∩ ∩F ∩ ∩F ⊆ 1=γ Γ,Z γ( ) F ∩ γ( ) W ) and thus one of these intersections is of dimension 3. Therefore, if W6 ∈is an∈ essentialF S hypersurfaceF ∩ of containing C then it contains a side containing Z. Thus, if C is contained in exactly two elementsF of then each contains one side containing Z and these two sides have to be different by LemmaV∪V4.5.∞ Hence we may assume that Z is contained in exactly one element W of . Then C is V∪V1 ∞ contained in at least one side S and S W . Let γ ΓS (i.e. S = γ− ( )). Then, by 1 1 ⊆ 1 ∈ 1 1 F ∩ 1 F

25 Lemma 4.2, W = Vγ1 . As dim(C) = 2, dim(Sγ1 ) = 3 and C S1 there is γ2 Γ 1,γ1 with 1 ⊆ F ∩ ∈ \{ } Z γ2− ( ). Thus γ2 = tγ1 with t Γ 1 and if γ1 Γ then t P1, P2 , by Lemma 5.5 and Lemma∈ 5.6F. We consider separately∈ two∞ \{ cases.} ∈ ∞ ∈ h i First assume that W = Vi± with i = 1 or 2 and without loss of generality, we may assume + 1 α β that W = V1 . Then γ1 = P1− by (31) and γ2 = P1 P2 for some integers α and β. Moreover Z = ( 1 ,s ,r,h) and (α + 1 ,s + β,r,h)= γ (Z) . Thus α 0, 1 . However, if α = 0 then 2 2 2 2 2 ∈ F ∈{ − } Vγ = V ± = W . Thus α = 1 and hence β = 1, because γ = γ . Then Z V ±, a contradiction. 2 2 6 − ± 1 6 2 ∈ 2 Assume now that either W or W = V ±. In both cases Vt = Vγ , because in the first ∈ V 3 6 1 case Vγ1 and Vt and in the second case Vt = Vi± with i = 1 or 2. Moreover γ1(C) ∈ V 1 ∈ V∞ 1 1 ⊆ −1 −1 γ1( ) t− ( ) V Vt. As Et, γ1(Sγ1 )= S Et. Let M = γ− (Vt), M ≥ = γ− (Et) γ1 γ1 1 1 F∩ F ∩ F1 ⊆ ∩ F ⊆ ⊆ and M ≤ = γ− (E′). Then, by the previous, Sγ M ≥. By the choice of λ and Lemma 4.5, 1 t 1 ⊆ B(Z, λ) W M ≤ has dimension 3. Thus, by Lemma 5.4, Z S for some side S W and ∩ ∩ F ∩ ∈ ⊆ different from Sγ1 . Hence also C is contained in two different sides.

The following is a consequence of Proposition 6.3 and Lemma 6.8.

Corollary 6.9 If E is an edge of the fundamental domain , then there are precisely two sides that contain E. F

The following lemma is obvious.

Lemma 6.10 Let E1 and E2 be two diﬀerent edges of some tile. Then the intersection E1 E2 is of dimension at most 1. ∩

Finally, in order to be able to describe the relations, we need two more lemmas.

Lemma 6.11 Let γ ,γ Γ. Assume γ ( ) γ ( ) has dimension 2. Then, 1 2 ∈ 1 F ∩ 2 F

1. there exists Z0 γ1( ) γ2( ) such that Z0 γ( ) γ( ) is of dimension 2; and ∈ F ∩ F ∈ F F T 2. for every such Z0, the set Z0 γ( ) γ( ) is an edge contained in γ1( ) γ2( ). ∈ F F F ∩ F Proof. T 1. We first show the existence of a point Z0 γ1( ) γ2( ) such that Z0 γ( ) γ( ) is of ∈ F ∩ F ∈ F F dimension 2. For every Z γ ( ) γ ( ), let ΓZ = γ Γ γ ,γ : Z γ( ) . We claim that ∈ 1 F ∩ 2 F { ∈ \{ 1 2} ∈ FT } ΓZ = for every Z γ1( ) γ2( ). Otherwise there is a λ> 0 such that B(Z, λ) γ( )= for every6 ∅γ Γ γ ,γ ∈. ThenF ∩B(Z,F λ) γ ( ) γ ( ), which is in contradiction with∩ LemmaF ∅A.3. ∈ \{ 1 2} ⊆ 1 F ∪ 2 F γ( ) γ ( ) γ ( ) . As, This proves the claim. Thus γ1( ) γ2( )= Z γ1( ) γ2( ) γ ΓZ 1 2 F ∩ F ∈ F ∩ F ∈ F ∩ F ∩ F by assumption, γ1( ) γ2( ) has dimension 2 and Γ is countable, it follows that there exists F ∩ F S T Z0 γ1( ) γ2( ) with Z0 γ( ) γ( ) of dimension 2. ∈ F ∩ F ∈ F F 2. Since Z0 belongs to only finitely many tiles (by Lemma 3.9), say γ1( ),γ2( ),...,γn( ), we n T n F F F have that Z0 i=1 γi( ) and i=1 γi( ) has dimension 2. We want to prove that this intersection ∈ F F n is an edge. Let γ0 Γ γ1,...,γn . As Z0 γ0( ), it is clear that i=1 γi( ) γ0( ). Hence it T ∈ \{n T } 6∈ F F 6⊆ F remains to prove that i=1 γi( ) intersects γ0( ) in dimension at most 1. Suppose this is not the n F F T n case, i.e. γi( ) is of dimension 2. As in the first part of the proof, there exists Z1 γi( ) i=0 F T ∈ i=0 F such that Z1 γ( ) γ( ) is of dimension 2. Let γ0,...,γm be all elements of Γ (with m n) T ∈ F F m T ≥ such that Z1 γi( ). So i=0 γi( ) is of dimension 2. By Corollary 3.10, let λ1 > 0 and T ∈ F F n B = B(Z , λ ) be such that B γ( ) = if and only if Z γ( ). Then B = (B γi( )) 1 1 T∩ F 6 ∅ 1 ∈ F ∩ i=1 F ∪ S 26 m B i=n+1 γi( ) γ0( ) , where both factors are closed sets of dimension 4 (by Lemma 4.4). ∩ F ∪ F n m Hence by Lemma A.3, (B γi( )) B γi( ) γ ( ) is of dimension 3. Thus S ∩ i=1 F ∩ ∩ i=n+1 F ∪ 0 F there exists 1 j1 n and n +1 j2 m or j2 = 0 such that Z1 γj ( ) γj ( ) and the latter ≤ ≤ S ≤ ≤ S ∈ 1 F ∩ 2 F intersection is of dimension 3. Hence γj ( ) γj ( ) is a side of the tile γj ( ). We now come 1 F ∩ 2 F 1 F back to Z0. Let λ0 > 0 and B′ = B(Z0, λ0) be such that γ( ) = if and only if Z0 γ( ). Thus n ∩ F 6 ∅ ∈ F B′ = (B′ γj1 ( )) i=1,i=j1 (B′ γi( )) and again by Lemma A.3, there exists 1 j3 n and ∩ F ∪ 6 ∩ F ≤ ≤ j3 = j1 such that γj ( ) γj ( ) is of dimension 3. Thus it is a side of the tile γj ( ). Hence 6 1SF ∩ 3 F 1 F Z0 ∂γj1 ( ), such that Z0 γ( ) γ( ) is of dimension 2. Moreover the latter is contained in ∈ F ∈ F F γj ( ) γj ( ), which is a side. Hence by Lemma 6.8, there exists a second side of γj ( ), which 1 F ∩ 3 F T 1 F contains Z0 γ( ) γ( ). Thus this side also contains Z0 and hence there exists 1 j4 n with ∈ F F ≤ ≤ j4 = j1 and j4 = j3 such that γj1 ( ) γj4 ( ) is a side. Thus the tile γj1 ( ) contains three sides, 6 T 6 F ∩ F F m γj1 ( ) γj2 ( ), γj1 ( ) γj3 ( ) and γj1 ( ) γj4 ( ) and their intersection contains i=0 γi( ), F ∩ F F ∩ F F ∩ F n F which is of dimension 2. This contradicts Lemma 6.6. Hence γi( ) is an edge and it contains i=1 F T Z0. T

Lemma 6.12 Let E be an edge. The ﬁnitely many elements γ Γ with E γ( ) can be ordered, ∈ ⊆ F say as γ0,γ1,...,γm = γ0, such that γj 1( ) γj ( ) is a side (containing E) for every 1 j m. Moreover, up to cyclic permutations and− reversingF ∩ F the ordering, there is only one possible≤ ordering≤ with this property.

Proof. Recall that there are only finitely many γ Γ with E γ( ). Let γ0 be such an element. Then E is an edge of γ ( ) and hence, by Corollary∈ 6.9, there⊆ existsF two sides, say γ ( ) γ 0 F 0 F ∩ 1F and γ0( ) γm 1( ) of γ0( ) containing E. Now E also is an edge of γ1( ) and γ0( ) γ1( ) F ∩ − F F F F ∩ F is one of the two sides of γ1( ) containing E. Hence there exists a third tile, say γ2( ), such that γ ( ) γ ( ) and γ ( F) γ ( ) are the two different sides of γ ( ) containingF E. So 1 F ∩ 0 F 1 F ∩ 2 F 1 F γ2 γ0,γ1 . One may continue this process and have a sequence γ 1,γ0,γ1,γ2,... of elements 6∈ { } − of Γ such that γi 1( ) γi( ) and γi( ) γi+1( ) are the two sides of γi( ) containing E for − F ∩ F F ∩ F F every i 0. In particular every three consecutive elements of the list of γi’s are different. As ≥ there are only finitely many tiles containing E, after finitely many steps we obtain γj Γ with ∈ γi = γj and 0 i < j. Let i be minimal with this property. We claim that i = 0. Indeed, if ≤ 0

Based on the previous lemma, we give the following new deﬁnition.

Deﬁnition 6.13 Let E be an edge. We call an ordering (γ0,...,γm 1,γ0) of the elements γ Γ such that E γ( ) as in Lemma 6.12 an edge loop of E. − ∈ ⊆ F

27 We now come to a description of the relations of Γ. We first define what is called a cycle. To do so, we fix some notations. Definition 6.14 Let E be an edge of and S a side of containing E. Then define recursively F F the sequence (E1,S1, E2,S2,...) as follows:

1. E1 = E and S1 = S,

2. Ei+1 = γSi (Ei),

3. Si is the only side of different from S∗ that contains Ei . +1 F i +1 Note that parts 2 and 3 are justified by Lemma 6.4 and Corollary 6.9. For every i, clearly Ei Si ⊆ and Ei+1 Si∗, as γSi (Si) = Si∗. Moreover, each pair (Ei,Si) determines the pairs (Ei 1,Si 1) ⊆ − − and (Ei+1,Si+1). This is clear for the subsequent pair (Ei+1,Si+1) but also for the previous one (Ei 1,Si 1) because if i 2, then Si∗ 1 is the only side of containing Ei and different from Si − − ≥ − F 1 ∗ and Ei 1 = γS− − (Ei)= γS − (Ei). Thus each pair (Ei,Si) determines the sequence (E1,S1,...). − i 1 i 1 Now we relate the sequence (E1,S1, E2,S2,... ) with an edge loop of E. Let L = γ Γ : 1 F { ∈ E γ( ) . Clearly 1,γS− L and as S1 is a side of , there is a unique edge loop of E ⊆ F } 1 ∈ F F of the form (γ0 = 1,γ1 = γS1 ,γ2,...,γm 1,γ0). As S1∗ = Sγ1 and S2 are the two sides of 1 1 − 1 1 F containing E2, S1 = γ1− (S1)∗ and γ1− (S2) are the two sides of γ1− ( ) containing γ1− (E2) = E. 1 1 F Therefore γ− (S )= γ− ( ) γ ( ) and hence S = γ γ ( ). A similar argument shows that 1 2 1 F ∩ 2 F 2 F ∩ 1 2 F Si = γ γ ...γi( ) for every i. In particular, the sequence (E ,S , E ,S ,... ) is periodic, i.e. F ∩ 1 2 F 1 1 2 2 there is n> 0 such that (Ei+n,Si+n) = (Ei,Si) for every i. If n is minimal with this property, then (Ei,Si) = (Ej ,Sj ) for 1 i < j n. We call (E1,S1,...En,Sn) the cycle determined by (E1,S1). Hence this6 proves the following≤ lemma.≤ Lemma 6.15 Let E be an edge of and S a side of containing the edge E . The cycle starting 1 F 1 F 1 with (E1,S1,...) is a finite cycle.

Note that S and S∗ are the two diﬀerent sides of containing E and thus there are two cycles 1 n F 1 starting with the edge E1, namely (E1,S1,...En,Sn) and (E1,Sn∗, En,Sn∗ 1,...,E2,S1∗). It is now also clear that if E is an edge and S is a side containing E, then− all the cycles containing E are cyclic permutations of the cycle starting with (E,S) and the cycles obtained by replacing in those the sides by their paired sides and reversing the order. In particular, if Ei = Ej with 1 i < j n, then Si = Sj and hence ≤ ≤ 6 (Ei,Si, Ei+1,Si+1,...,En,Sn, E1,S1,...,Ei 1,Si 1) − − = (Ej ,Sj∗ 1, Ej 1,Sj∗ 2,...,S2∗, E1,Sn∗, En,Sn∗ 1,...,Ej+1,Sj∗). − − − −

Lemma 6.16 If (E ,S , E ,S ,...,En,Sn) is a cycle of then γS γS − ,...,γS has finite order. 1 1 2 2 F n n 1 1 k Proof. Let γ = γSn γSn−1 ,...,γS1 . Clearly γ(E1)= E1 and thus γ (E1)= E1 for all non-negative k integers k. Hence E1 γ ( ) and because every edge is contained in only finitely many tiles, γ has finite order. ⊆ F

Because of Theorem 5.3, we thus obtain a natural group epimorphism

ϕ : ∆ Γ : [γS] γS (32) → 7→ where ∆ is the group given by the following presentation

28 Generators: a generator [γS] for each side S of • F m Relations: [γS] [γS ] = 1 if the sides S and S∗ are paired and ([γS ] . . . [γS ]) = 1 if • ∗ n 1 (E1,S1, E2,S2,...,En,Sn) is a cycle and m is the order of γSn ...γS1 . Our next aim is to show that ϕ is also injective and thus we obtain a presentation for Γ. To that end we introduce the following deﬁnition.

Definition 6.17 A loop of tiles is a finite list of tiles (h ( ),h ( ),...,hn( )) with hi Γ such 0 F 1 F F ∈ that h0( )= hn( ) and hi 1( ) hi( ) is of dimension 3, for every i =1,...n. F F − F ∩ F 1 The last condition means that hi− hi 1( ) is of dimension 3, which is equivalent to γi = 1 F ∩ − F hi− 1hi being a side-paring transformation. Moreover we get the relation γ1γ2 ...γn = 1 which is called− a loop relation. In Theorem 6.24 we will show that these relations form a complete set of relations of Γ. It is easy to see that the pairing and cycle relations are determined by loop relations. Indeed, let S and S∗ be two paired sides of . Then ( ,γs( ),γS γS∗ ( )= ) is a loop of tiles which gives F F F F F as loop relation the pairing relation. Let (E1,S1, E2,S2,...,En,Sn) be a cycle. We have seen that m there exists a positive integer m such that (γSn γSn−1 ...γS1 ) = 1. Set h0 = 1 and hi = hi 1γSj − where j i mod (n) and i 1,...,mn . Consider (h ( ),...,hmn( )). Clearly hmn = 1 and ≡ ∈ { } 0 F F hence hmn( )= = h0( ). Also, for every i 1,...,mn , hi 1( ) hi( )= hi 1( γSj ( )), F F F ∈{ } − F ∩ F − F ∩ F where γSj is a side-paring transformation and hence γSj ( ) and thus also hi 1( ) hi( ) is F ∩ F − F ∩ F of dimension 3. So, by definition, (h0( ),...,hmn( )) is a loop of tiles and the associated loop relation is the cycle relation. F F Consider the union of intersections of varieties of such that these intersections have dimension at most 1. Let Ω denote the complement of this setM in H2 H2. Note that by Lemma A.1, Ω is path-connected. × In the remainder we fix α to be a continuous map

α : [0, 1] Ω, → such that α(0) g( )◦ and α(1) g′( )◦, ∈ F ∈ F for some g,g′ Γ and such that α is made up of a finite number of line segments, which are parametrized by∈ a polynomials of degree at most one. Moreover, for each line segment forming α, we suppose that at least one of its end-points does not belong to any element in . Note that such M a map exists. Indeed, let α denote the set consisting of the elements of that have non-empty M M is dense in Ω. Hence, intersection with (the image of) α. Then it is easy to see that Ω M α by Lemma A.4, such a map α indeed exists. \ ∈M S Lemma 6.18 The set t [0, 1] : α(t) ∂g( ) for some g Γ is finite. { ∈ ∈ F ∈ } Proof. We know from Lemma 3.9 that the compact set α([0, 1]) only intersects finitely many tiles and thus also only finitely many sides. By Lemma 4.2, for every side S of some tile g( ), g( ) S is contained in a precise variety M . As the elements of are real semi-algebraicF varieties,F ∩ which are given by polynomials, a line∈ M segment is either containedM in such a variety or it intersects it in finitely many points. The path α consists of finitely many line segments, such that at least one of the end-points of these line segments does not belong to any element in . If such a line M

29 segment l is parametrized by a polynomial of first degree then there are only finitely many t [0, 1] such that l(t) M for some M . Moreover, if a line segment l is parametrized by a polynomial∈ of degree 0, i.e.∈ the image of l is∈M just a point, then by definition this point is not contained in any element of and l M = for every M . Thus α(t) belongs to a side for only finitely many t. Hence theM lemma∩ follows∅ by Lemma 5.4∈M.

Lemma 6.19 Let α be a continuous map [0, 1] Ω such that → 1. α(0) g( )◦ and α(1) g′( )◦ for some g,g′ Γ, ∈ F ∈ F ∈ 2. α is made up of a ﬁnite number of line segments, 3. for each of these line segments at least one of its two end-points does not belong to any element in . M Then there exists a unique ordered list = (a ,g ,a ,g ,a ...,gn,an), where gi Γ, 0 = a < L 0 1 1 2 2 ∈ 0 a <... 0, such that α((ai,ai + ǫ)) gi( )= . 0 ∩ F ∅ We call the partition of α. L

Proof. As α(0) g( )◦, we set g = g and a = 0. By Lemma 6.18, there are only ﬁnitely many ∈ F 1 0 t [0, 1] such that α(t) ∂g( ) for some g Γ, say t

Remark 6.20 As, by assumption, α(0) g( )◦ and α(1) g′( )◦ for some g,g′ Γ, the ﬁrst ∈ F ∈ F ∈ element g1 of the partition of α equals g and the last element gn equals g′.

The remainder of the paper is based on ideas from a recent proof of the presentation part of the classical Poincaré theorem [JKDR15]. Let g,h Γ. Let C be a cell of of dimension m 2 and ∈ T ≥ that is contained in g( ) h( ). We deﬁne κC(g,h) ∆ as follows. F ∩ F ∈ If m = 4 then κC (g,h)=1. • 1 If m = 3 then κC (g,h) = [g− h]. •

30 If m = 2 then C is an edge contained in g( ) h( ) and thus, by Lemma 6.12, g and h • belong to an edge loop of C. Up to a cyclic permutation,F ∩ F we can write the edge loop of C as (g = k0,...,kt = h, kt+1,...,km = g) (or the equivalent edge loop (g = km, km 1,...,kt = − h, kt 1,...,k1, k0 = g)) and we set − 1 1 1 1 1 κC (g,h) = [k0− k1][k1− k2] [kt− 1kt] = [km− km 1] [kt−+1kt]. · · · − − · · ·

Observe that κC (g,g) = 1 in the three cases. Lemma 6.21 Let g,h Γ and let C be a cell of of dimension m 2 and that is contained in g( ) h( ). The following∈ properties hold. T ≥ F ∩ F 1 1. κC(g,h)= κC(h,g)− .

2. If D is cell of contained in C and of dimension at least 2 then κD(g,h)= κC(g,h). T n 3. If g1,...,gn Γ and C i gi( ) then κC (g1,gn)= κC(g1,g2)κC (g2,g3) κC(gn 1,gn). ∈ ⊆ =1 F · · · − 1 1 Proof. 1. If m = 4, then g = hTand there is nothing to prove. If m = 3 then g− h and h− g are pairing transformations and hence by the pairing relations of the group ∆, κC(g,h)κC (h,g) = 1. Finally, if m = 2, then we can write the edge loop of C as (g = k0,...,kt = h, kt+1,...,km = g) and thus

1 1 1 κC(g,h) = [k0− k1][k1− k2] [kt− 1kt], · · · − 1 1 1 κC(h,g) = [kt− kt+1][kt−+1kt+2] [km− 1km]. · · · − It is now easy to see that κC(g,h)κC (h,g) = 1 and hence the result follows. 2. If C = D then there is nothing to prove. So assume that C = D. If C is a side then D is an 6 1 edge and g and h are two consecutive elements of the edge loop of D. Then κD(g,h) = [g− h] = κC (g,h). Otherwise, C is a tile and hence g = h. Thus κD(g,h)=1= κC (g,h). 3. By induction it is enough to prove the statement for n = 3. So assume n = 3. If either g1 = g2 or g2 = g3 then the desired equality is obvious. So assume that g1 = g2 and g2 = g3. If C is an edge then, up to a cyclic permutation, possibly reversing the order6 and making use6 of Lemma 6.12, the edge loop of C is of the form (g1 = k0,...,g2 = kt,...,g3 = kl,...,km = g1). Then,

1 1 1 κC (g1,g3) = [k0− k1][k1− k2] [kl− 1kl] · · · − 1 1 1 1 1 = ([k0− k1][k1− k2] [kt− 1kt]) ([kt− kt+1] [kl− 1kl]) · · · − · · · − = κC (g1,g2)κC (g2,g3)

1 1 1 Otherwise, S = g− (C) is a side of , gS = g− g , gS′ = g− g and g = g . Then, 1 F 1 2 2 1 1 3 κC (g1,g3)=1=[gS][gS′ ]= κC (g1,g2)κC (g2,g3).

Let Z Ω, g,h Γ and Z C g( ) h( ) for some cell C. Then, by the deﬁnition of Ω, the dimension∈ of C ∈is at least 2∈ and⊆ we deﬁneF ∩ F

κZ (g,h)= κC(g,h).

31 This is well defined. Indeed, suppose D is another cell containing Z and contained in g( ) h( ) F ∩ F with κC (g,h) = κD(g,h). Then g = h and C = D. Hence neither C nor D is a tile. Both are not edges, because6 otherwise Z C D6 , where the6 latter is an intersection of tiles and it has dimension 1, which contradicts the definition∈ ∩ of Ω. Thus either C or D is a side and contains the other. 1 Therefore, g( ) h( ) is a side and hence, by part 2 of Lemma 6.21, κC (g,h) = [g− h]= κD(g,h), F ∩ F a contradiction. This proves that indeed κZ (g,h) is well defined. By parts 1 and 3 of Lemma 6.21 1 n we have κZ (g,h)= κZ (h,g)− and if Z gi( ) with g ,...,gn Γ then ∈ ∩i=1 F 1 ∈

κZ (g1,gn)= κZ (g1,g2) κZ (gn 1,gn). · · · − Deﬁnition 6.22 Let α be a continuous map [0, 1] Ω as in Lemma 6.19 and let = (a ,g ,a ,g , → L 0 1 1 2 a2 ...,gn,an) be the partition of α. We deﬁne

Φ( )= κα(a )(g1,g2) κα(a )(g2,g3) κα(a − )(gn 1,gn), L 1 2 · · · n 1 − if n =1. If n =1, we set Φ( )=1. 6 L

Observe that if i 1,...,n then α(ai) gi( ) gi ( ). By the definition of Ω, gi( ) ∈ { } ∈ F ∩ +1 F F ∩ gi ( ) has dimension 2 or 3. If it has dimension 3, then gi( ) gi ( ) is a side containing α(ai). +1 F F ∩ +1 F If gi( ) gi+1( ) has dimension 2, then g, α(ai) g( ) g( ) has dimension 2 by the definition F ∩ F ∈ F F of Ω and thus, by Lemma 6.11, gi( ) gi ( ) contains an edge that contains α(ai). Hence F ∩ T+1 F α(ai) C gi( ) gi+1( ), for some cell C and thus κα(ai)(gi,gi+1) is well defined. Let∈Z, W⊆ ΩF and∩ let PFbe the set of all the continuous maps α : [0, 1] Ω with α(0) = Z and α(1) = W , and∈ such that α verifies the conditions of Lemma 6.19. The set→P may be considered as a metric space with the metric determined by the infinite norm: α = max α(t) : t [0, 1] . We define the map k k∞ {| | ∈ } Φ : P ∆ → by Φ(α)=Φ( ), where is the partition of α. This is well defined as by Lemma 6.19 the partition of α exists andL is unique.L The next lemma will be a crucial part in the proof of the injectivity of the map ϕ defined in (32).

Lemma 6.23 If both Z and W belong to the interior of some tile then Φ : P ∆ is constant. → Proof. We claim that it is suﬃcient to show that Φ is locally constant. Indeed, assume this is the case and let α, β P. By Lemma A.2, α and β are homotopic in Ω and by Lemma A.5, there is a homotopy H(t, ∈) in P from α to β. Let c denote the supremum of the s [0, 1] for which Φ(H(s, )) = Φ(α).− Since, by assumption, Φ is constant in a neighbourhood of H∈ (x, ), it easily follows that− c = 1 and thus Φ(α)=Φ(β). − To prove that Φ is locally constant, let α, β P and let 1 = (a0,g1,a1,...,gn,an) and 2 be the partition of α and β respectively. Moreover we∈ denote by dL( , ) the Euclidean distance.L Let 0 < − − { d < d <... 0 such that for every i 0, 1,...,m +1 and every g Γ, if B(α(di), 2δ1)) g( ) = then ∈{ } di di−∈1 ∩ F 6 ∅ α(di) g( ). Since α is continuous there is ǫ < min − : i 1,...,m +1 such that, for ∈ F 2 ∈{ } every i 0, 1,...,m+1 , d(α(t), α(di)) <δ1 for everynt with t di <ǫ. For everyoi 1,...,m , ∈{ } | − | ∈{ }

32 let d′ = di ǫ and d′′ = di + ǫ. We also set d′ = 1 and d′′ = 0. Observe that d′′ d′ for i − i m+1 0 i ≤ i+1 every i 0,...,m +1 . Each α([di′′ 1, di′ ]) is compact and it is contained in ki( )◦. Again using that ∈{is locally ﬁnite} we obtain a− positive number δ such that for every i F0, 1,...,m +1 , F 2 ∈ { } d(α(t),g( )) >δ2 for every t [di′′ 1, di′ ] and every g Γ with g = ki. Let δ = min δ1,δ2 . F ∈ − ∈ 6 { } We will prove that if β B ∞ (α, δ), then Φ(α)=Φ(β). So assume β P with α β <δ. ∈ k k ∈ k − k∞ Then d(α(t),β(t)) <δ for every t [0, 1]. In particular, as d(β(t), α(di)) < 2δ , ∈ 1

if t (d′ , d′′) and β(t) g( ) then α(di) g( ). (33) ∈ i i ∈ F ∈ F

Moreover, since d(α(t),β(t)) <δ2,

if t [di′′ 1, di′ ] and β(t) g( ) then g = ki. (34) ∈ − ∈ F The interval [0, 1] may be written as

[0, d′ ] [d′ , d′′] [d′′, d′ ] . . . [d′ , d′′ ] [d′′ , dm ] . 1 ∪ 1 1 ∪ 1 2 ∪ m m ∪ m +1 Based on this information, we construct and prove that Φ( )=Φ( ). By (34), the elements L2 L1 L2 k ,...km appear in that order in . Between those elements may appear other elements h Γ. 1 +1 L2 ∈ For each 1 i m + 1, there are two possibilities: ki and ki are equal or they are different. ≤ ≤ +1 If ki = ki+1 and ki and ki+1 are two consecutive elements in 2, then β( di′′ 1, di′+1 ) ki( ) = L − ⊆ F ki ( ). According to the definition of the partition of β, ki and ki are represented by just +1 F +1 one element in 2. Hence we may suppose, without loss of generality, that if ki and ki+1 are two consecutive elementsL of , then they are different. Thus L2

= (0, k ,b , , k ,b , ,...,km,bm, m, km , 1), L2 1 1 ∗1 2 2 ∗2 ∗ +1 where d′ < bi < d′′ and i represents a, possibly empty, sequence (hi ,bi ,hi ,gi ...,hin ,bin ) i i ∗ 1 1 2 2 i i with hij Γ and d′

β( di′′ 1, di′ ) ki( ), − ⊆ F β( d′′, d′ ) ki ( ), i i+1 ⊆ +1 F β(bi) ki( ) ki ( ). ∈ F ∩ +1 F

By (33) and the fact that bi [d′ , d′′], for every g Γ such that β(bi) g( ) we have that ∈ i i ∈ ∈ F α(di) = α(aj ) g( ). Hence, by Lemma 6.11, g( ) is a cell, which is contained in ∈ F g, α(aj ) g( ) F every cell containing β(b ). Thus both α(a ) and β(b ) are∈ F included in a same cell C contained in i j T i gj ( ) gj ( ). Thus, F ∩ +1 F

κα(aj )(gj ,gj+1)= κC (gj ,gj+1)= κC (ki, ki+1)= κβ(bi)(ki, ki+1).

Suppose now that i is not empty. We will analyse the subsequence ∗

(ki,bi,hi1,bi1,hi2,bi2 ...,hini ,bini , ki+1,bi+1).

By lemma 6.11, α(di) C, β(bi) Ci and β(bij ) Cij for C, Ci and Cij cells for 1 j ni. ∈ ∈ ∈ ≤ ≤ As bi (d′ , d′′) and also bij (d′ , d′′) for every 1 j ni, by (33) we have that α(di) ∈ i i ∈ i i ≤ ≤ ∈

33 ni ni ki( ) ki ( ) hij ( ). Hence C Ci Cij . If ki = ki , then ki = gl, ki = gl F ∩ +1 F ∩ j=1 F ⊆ ∩ i=1 6 +1 +1 +1 and di = al for some 1 l n and by parts 2 and 3 of Lemma 6.21 T ≤ ≤ T

κα(al)(gl,gl+1) = κC (ki, ki+1)

= κC (ki,hi1)κC (hi1,hi2) ...κC (hini ki+1) (k ,h )κ (h ,h ) ...κ (h k ) = κCi i i1 Ci1 i1 i2 Cini ini i+1 = κ (k ,h )κ (h ,h ) ...κ (h k ) β(bi) i i1 β(bi1) i1 i2 β(bini ) ini i+1

If ki = ki+1, then ki = ki+1 = gl for some 1 l n. Then in Φ(α), there is no κ-term corresponding to the subsequence above and by parts 2 and≤ 3≤of Lemma 6.21, we have κ (k ,h )κ (h ,h ) ...κ (h k ) = κ (k ,h )κ (h ,h ) ...κ (h k ) β(bi) i i1 β(bi1) i1 i2 β(bini ) ini i+1 Ci i i1 Ci1 i1 i2 Cini ini i+1

= κC (ki,hi1)κC (hi1,hi2) ...κC (hini ki+1)

= κC (ki, ki+1)

= κC (ki, ki) = 1. This shows that Φ( )=Φ( ). L1 L2 We are now ready to prove that ϕ : ∆ Γ (see (32)) is injective. Proving the injectivity is → equivalent to proving that if g ...gn = 1 with each gi Γ and gi( ) a side, then [g ] . . . [gn]=1 1 ∈ F ∩ F 1 in ∆. So, suppose that g ...gn = 1 with each gi Γ and gi( ) a side. For every i =0, 1,...,n, 1 ∈ F ∩ F put hi = g ...gi and put h = 1. Choose Z ◦ and, for every i = 1,...,n, choose a path 1 0 ∈ F from hi 1(Z) to hi(Z) which is only contained in hi 1( ) hi( ) and which is made up of a ﬁnite number− of line segments, whose end-points do− notF belon∪ gF to any element in . This is possible by Lemma 3.6 and Lemma A.4. Let α be the path obtained by juxtaposingM those paths. So α(0) = Z = α(1) and α is contained in the space P. Let = (a =0,h ,a ,...,hn,an = 1) L1 0 0 1 +1 be the partition of α. Let β be the constant path β(t)= Z for every t [0, 1] and let 2 = (0, 1, 1) be the partition of β. Clearly β is also contained in P. Now we have that∈ L

1 1 1 [g1] . . . [gn] = [h0− h1][h1− h2] . . . [hn− 1hn] − = κa1 (h0,h1)κa2 (h1,h2) ...κan (hn 1,hn) − = Φ( ) L1 = Φ(α).

By Lemma 6.23, Φ(α)=Φ(β). However Φ(β)=Φ( 2)= κ1(1, 1) = 1 and hence [g1] . . . [gn]=1. To sum up we have proven the following theorem.L Theorem 6.24 Let be the fundamental domain of Γ as defined above. Then the following is a presentation of Γ: F Generators: the pairing transformations of , • F Relations: the pairing relations and the cycle relations. • It is well known that Γ is finitely presented. However it is not clear whether the presentation given in Theorem 6.24 is finite. The following proposition implies that at least the presentation is finite in case R is a PID.

34 Lemma 6.25 If R is a PID then every tile of intersects only finitely many other tiles. In particular the fundamental domain has finitelyF many cells and thus finitely many sides and edges. F Proof. It is enough to prove that intersects only finitely many other tiles. Moreover, if 1 = γ Γ F 6 ∈ then γ( ) V e V and e is finite by Theorem 3.8. Thus it is enough to prove that F ∩ F ⊆ ∪ ∈V F ∩ V if V then XV = γ Γ : V γ( ) = is finite. If V then V F ∈ V∩V∞ { ∈ ∩ F ∩ F 6 ∅} ∈ V ∩ is compact, by Lemma 4.12. Since is locally finite (Lemma 3.9), XV is finite. So assume that V . By(19), (21) and LemmaF 3.7, there are positive real numbers a 0. Assume that Z1 = γ(Z2) γ( ) ∞ 2 2 ∈ F∞ ∩ F∞ with γ Γ . Let Z1′ and Z2′ be the elements of H H obtained by replacing the last coordinate ∈ ∞ × of Z1 and Z2 by h0. Then Z1′ = γ(Z2′ ) (B h0 ) γ( ). Thus (B h0 ) γ( ) = . ∈ ×{ } ∩F∞ ∩ F∞ ×{ } ∩ F∞ 6 ∅ As B h0 is compact and is a locally finite fundamental domain of Γ , only finitely many elements×{ satisfies} the last condition.F∞ This finishes the proof. ∞

A Appendix: Topological Lemmas

For completeness’ sake we state some results on real algebraic topology that have been used in the previous sections. The proofs of these are probably well known to the specialists. We would like to thank Florian Eisele for making us aware of these and for providing us proofs. Some of these results can be stated in greater generality for arbitrary diﬀerential manifolds, but their proofs need more sophisticated algebraic topology methods.

n Lemma A.1 Let X be a path-connected open subset of R . Assume that if Z0 and Z1 are distinct points in X, then there exists ǫ > 0 such that the open cylinder with axis [Z0,Z1] and radius ǫ is contained X. If L is a locally ﬁnite collection of semi-algebraic varieties of dimension at most n 2 in X then X L is path-connected. − \ n Lemma A.2 Let n 3 and let X be a connected open subset of R . Assume π1(X)=1. If Ti i is a locally ﬁnite family≥ of algebraic varieties in X of co-dimension at least 3 then { }

π X Ti =1. 1 \ i ! [ n Lemma A.3 Let B be a Euclidean closed ball in R . If B = U1 U2, where U1 and U2 are closed semi-algebraic sets, with U U and U U , then dim(U U∪) n 1. 1 6⊆ 2 2 6⊆ 1 1 ∩ 2 ≥ − Lemma A.4 Let X be a connected open subset of Rn. If Z and W are points in X then there exists a path α : [0, 1] X with α(0) = Z and α(1) = W and such that it is built from of a ﬁnite number of line segments→ that are parametrized by polynomials of degree at most 1. Moreover, if Y is a subset of X with dense complement in X and such that Z, W Y then we can choose the end-points of the line segments in X Y. 6∈ \

35 n Lemma A.5 Let X be a connected open subset of R . Let γ and γ′ be two homotopic paths in X starting and ending at x0 which are built from line segments that are parametrized by polynomials of degree at most 1. Assume we are additionally given a subset Y X with dense complement in X and which satisﬁes the property that a line which is not wholly contained⊂ in Y has ﬁnite intersection with it. If x Y then there is an integer N and a homotopy 0 6∈ H : [0, 1] [0, 1] X × −→ from γ to γ′ such that each loop H(t, ) for t [0, 1] starts and ends in x and there exist 0 = − ∈ 0 τ < τ < ...< τN = 1 such that H(t, ) is a line segment for each, t (0, 1), which 1 2 − |[τi,τi+1] ∈ is parametrized by polynomials of degree at most 1 and at least one of the two points H(t, τi) and H(t, τi+1) lies outside Y.

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Department of Mathematics, Vrije Universiteit Brussel, Pleinlaan 2, 1050 Brussel, Belgium emails: [email protected] and [email protected] Departamento de Matemáticas, Universidad de Murcia, 30100 Murcia, Spain email: [email protected]