A uniformizable spherical CR structure on a two-cusped hyperbolic 3-manifold

Let $\langle I_{1}, I_{2}, I_{3}\rangle$ be the complex hyperbolic $(4,4,\infty)$ triangle group. In this paper we give a proof of a conjecture of Schwartz for $\langle I_{1}, I_{2}, I_{3}\rangle$. That is $\langle I_{1}, I_{2}, I_{3}\rangle$ is discrete and faithful if and only if $I_1I_3I_2I_3$ is nonelliptic. When $I_1I_3I_2I_3$ is parabolic, we show that the even subgroup $\langle I_2 I_3, I_2I_1 \rangle$ is the holonomy representation of a uniformizable spherical CR structure on the two-cusped hyperbolic 3-manifold $s782$ in SnapPy notation.


Introduction
Let H 2 C be the complex hyperbolic plane and PU.2; 1/ be its holomorphic isometry group; see Section 2 for more details.It is well known that H 2 C is one of the rank-one symmetric spaces and PU.2; 1/ is a semisimple Lie group.H 2 C can be viewed as the unit ball in C 2 equipped with the Bergman metric.Its ideal boundary @H 2 C is the 3-sphere S 3 .We study the geometry of discrete subgroups of PU.2; 1/.
Let M be a 3-manifold.A spherical CR structure on M is a system of coordinate charts into S 3 such that the transition functions are restrictions of elements of PU.2; 1/.Any spherical CR structure on M determines a pair .; d/, where W 1 .M / !PU.2; 1/ is the holonomy and d W z M !S 3 is the developing map.There is a special spherical CR structure.A uniformizable spherical CR structure on M is a homeomorphism between M and a quotient space = , where is a discrete subgroup of PU.2; 1/ and @H 2

C
is the discontinuity region of .An interesting problem in complex hyperbolic geometry is to find (uniformizable) spherical CR structures on hyperbolic 3-manifolds.
Geometric structures modeled on the boundary of complex hyperbolic space are rather difficult to construct.The first example of a spherical CR structure existing on a cusped hyperbolic 3-manifold was discovered by Schwartz.In [23], Schwartz constructed a uniformizable spherical CR structure on the Whitehead link complement.He also constructed a closed hyperbolic 3-manifold that admits a uniformizable spherical CR structure in [26] at almost the same time.
Let M 8 be the complement of the figure eight knot.In [9], Falbel constructed two different representations 1 and 2 of 1 .M 8 / in PU.2; 1/, and proved that 1 is the holonomy of a spherical CR structure on M 8 .In [11], Falbel and Wang proved that 2 is also the holonomy of a spherical CR structure on M 8 .In [7], Deraux and Falbel constructed a uniformizable spherical CR structure on M 8 whose holonomy is 2 .In [6], Deraux proved that there is a 1-parameter family of spherical CR uniformizations of the figure eight knot complement.This family is in fact a deformation of the uniformization constructed in [7].
Let us return to the Whitehead link complement.It admits a uniformizable spherical CR structure which is different from Schwartz's.In the recent work [22], Parker and Will also constructed a spherical CR uniformization of the Whitehead link complement.By applying spherical CR Dehn surgery theorems to the uniformizations of the Whitehead link complement, one can get infinitely many manifolds which admit uniformizable spherical CR structures.In [28], Schwartz proved a spherical CR Dehn surgery theorem, and applied it to the spherical CR uniformization of the Whitehead link complement constructed in [23] to obtain infinitely many closed hyperbolic 3-manifolds which admit uniformizable spherical CR structures.In [2], Acosta applied the spherical CR Dehn surgery theorem he proved in [1] to the spherical CR uniformization of the Whitehead link complement constructed by Parker and Will in [22] to obtain infinitely many one-cusped hyperbolic 3-manifolds which admit uniformizable spherical CR structures.In particular, the spherical CR uniformization of the complement of the figure eight knot constructed by Deraux and Falbel [7] is contained in this family.
We show that the two-cusped hyperbolic 3-manifold s782 admits a uniformizable spherical CR structure.By studying the action of the even subgroup of a discrete complex hyperbolic triangle group on H 2 C , we prove that the quotient space of its discontinuity region is homeomorphic to s782.That means the holonomy representation of the spherical CR uniformization of s782 is a triangle group.Now let us talk about the complex hyperbolic triangle groups.Let p;q;r be the abstract .p;q; r / reflection triangle group with the presentation where p, q and r are positive integers, or 1 in which case the corresponding relation disappears.A complex hyperbolic .p;q; r / triangle group is a representation of p;q;r in PU.2; 1/, which maps the generators to complex involutions fixing complex lines in H 2 C .The study of complex hyperbolic triangle groups was begun by Goldman and Parker, and in [13] they studied the complex hyperbolic .1;1; 1/ triangle groups.They conjectured that a representation of 1;1;1 into PU.2;1/ is discrete and faithful if and only if the image of 1 2 3 is nonelliptic.The Goldman-Parker conjecture was proved by Schwartz in [24] (and with a better proof in [27]).In particular, the representation with the image of 1 2 3 being parabolic is closely related with the holonomy of the spherical CR uniformization of the Whitehead link complement constructed in [23].In the survey [25], a series of conjectures on complex hyperbolic triangle groups are put forward.
Conjecture 1.1 (Schwartz [25]) Suppose that p Ä q Ä r .Let hI 1 ; I 2 ; I 3 i be a complex hyperbolic .p;q; r / triangle group.Then hI 1 ; I 2 ; I 3 i is a discrete and faithful representation of p;q;r if and only if I 1 I 3 I 2 I 3 and I 1 I 2 I 3 are nonelliptic.Moreover: If 3 Ä p < 10, then hI 1 ; I 2 ; I 3 i is discrete and faithful if and only if I 1 I 3 I 2 I 3 is nonelliptic.
If p > 13, then hI 1 ; I 2 ; I 3 i is discrete and faithful if and only if I 1 I 2 I 3 is nonelliptic.
In a recent work [22], Parker and Will proved Conjecture 1.1 for complex hyperbolic .3;3; 1/ groups.They also showed that, when I 1 I 3 I 2 I 3 is parabolic, the quotient of H 2 C by the group hI 2 I 3 ; I 2 I 1 i is a complex hyperbolic orbifold whose boundary is a spherical CR uniformization of the Whitehead link complement.In [21], Parker, Wang and Xie proved Conjecture 1.1 for complex hyperbolic .3;3; n/ groups with n 4. Furthermore, Acosta [2] showed that when I 1 I 3 I 2 I 3 is parabolic the group hI 2 I 3 ; I 2 I 1 i is the holonomy representation of a uniformizable spherical CR structure on the Dehn surgery of the Whitehead link complement on one cusp of type .1;n 3/.
We give a proof of Conjecture 1.1 for the complex hyperbolic .4;4; 1/ triangle groups and further analyze the group when I 1 I 3 I 2 I 3 is parabolic.Our result is as follows: Theorem 1.2 Let hI 1 ; I 2 ; I 3 i be a complex hyperbolic .4;4; 1/ triangle group.Then hI 1 ; I 2 ; I 3 i is a discrete and faithful representation of 4;4;1 if and only if C by the group hI 2 I 3 ; I 2 I 1 i is a complex hyperbolic orbifold whose boundary is a spherical CR uniformization of the two-cusped hyperbolic 3-manifold s782 in the SnapPy census.
In [29], Wyss-Gallifent studied the complex hyperbolic .4;4; 1/ triangle groups.He discovered several discrete groups with I 1 I 3 I 2 I 3 being regular elliptic of finite order and conjectured that there should be countably infinitely many.It would be very interesting to know what the manifold at infinity is for the group with I 1 I 3 I 2 I 3 being regular elliptic of finite order.
Our method is to construct Ford domains for the triangle groups acting on H 2 C .The space of complex hyperbolic .4;4; 1/ triangle groups hI 1 ; I 2 ; I 3 i is parametrized by the angle Â 2 0; 2 ; see Section 3. Let S D I 2 I 3 , T D I 2 I 1 and D hS; T i.Here S is regular elliptic of order 4, and T is parabolic fixing the point at infinity.For each group in the parameter space, the Ford domain D is the intersection of the closures of the exteriors of the isometric spheres for the elements S, S 1 , S 2 , .S 1 T / 2 and their conjugations by the powers of T .The combination of D is the same except when I 1 I 3 I 2 I 3 is parabolic, in which case there are additional parabolic fixed points.D is preserved by the subgroup hT i and is a fundamental domain for the cosets of hT i in .Its ideal boundary @ 1 D is the complement of a tubular neighborhood of the T -invariant R-circle (or horotube defined in [28]).By intersecting @ 1 D with a fundamental domain for hT i acting on @H 2 C , we obtain a fundamental domain for acting on its discontinuity region; see Section 4.
When I 1 I 3 I 2 I 3 is parabolic, that is Â D 3 , there are four additional parabolic fixed points fixed by T 1 S 2 , S 2 T 1 , S T 1 S and T 1 S T 1 S T , except the point at infinity which is the fixed point of T ; see Section 5.By studying the combinatorial properties of the fundamental domain for acting on its discontinuity region ./,we prove that the quotient ./= is homeomorphic to the two-cusped hyperbolic 3-manifold s782.Motivated by the work of Acosta [2], we guess that there are similar structures on its surgeries.hyperbolic geometry during his visit to Fudan University.We would like to thank the referee for comments which improved a previous version of this paper.Jiang was supported by NSFC 12271148.Wang was supported by NSFC 11701165.Xie was supported by NSFC 11871202 and Hunan Provincial Natural Science Foundation of China 2018JJ3024.

Background
The purpose of this section is to briefly introduce complex hyperbolic geometry.One can refer to Goldman's book [12] for more details.

Complex hyperbolic plane
Let hz; wi D w H z be the Hermitian form on C 3 associated to H , where H is the Hermitian matrix Then C 3 is the union of the negative cone V , null cone V 0 and positive cone V C , where Definition 2.1 Let P W C 3 f0g !CP 2 be the projectivization map.Then the complex hyperbolic plane H 2 C is defined to be P .V /, and its boundary @H 2 C is defined to be P .V 0 /.This is the Siegel domain model of , which is given by the Hermitian matrix J D diag.1; 1; 1/.In this model, @H 2 C is then the 3-dimensional sphere S 3 C 2 .The Cayley transform C is given by Then, in the Siegel domain model of H 2 C , the boundary of the complex hyperbolic plane @H 2 C can be identified to the union N [ fq 1 g, where q 1 is the point at infinity.The standard lift of q 1 and q D OEz; t 2 N in C 3 are (2-1) 5 : The closure of the complex hyperbolic plane H 2 C [ @H 2 C can be identified to the union of N R 0 and fq 1 g.Any point q D .z;t; u/ 2 N R 0 has the standard lift 5 : Then the Bergman metric on the complex hyperbolic plane is given by the distance formula cosh The formula d Cyg .p;q/ D j2hp; qij 1=2 remains valid even if one of p or q lies on @H 2 C .A Cygan sphere is a sphere for the extended Cygan distance.
There are two kinds of 2-dimensional totally real totally geodesic subspaces of H 2 C : complex lines and Lagrangian planes.Definition 2.4 Let v ?be the orthogonal space of v 2 V C with respect to the Hermitian form.The intersection of the projective line P .v? /with H 2 C is called a complex line.The vector v is its polar vector.
The ideal boundary of a complex line on @H 2 C is called a C-circle.In the Heisenberg group, C-circles are either vertical lines or ellipses whose projections on the z-plane are circles.

Isometries
Let SU.2; 1/ be the special unitary matrix preserving the Hermitian form.Then the projective unitary group PU.2; 1/ D SU.2; 1/=fI; !I; ! 2 I g is the holomorphic isometry group of H The types of isometries can be determined by the traces of their matrix realizations; see Theorem 6.2.4 of Goldman [12].Now suppose that A 2 SU.2; 1/ has real trace.Then A is elliptic if 1 Ä tr.A/ < 3.Moreover, A is unipotent if A is not the identity and tr.A/ D 3. In particular, if tr.A/ D 1; 0; 1, A is elliptic of order 2, 3 or 4, respectively.
There is a special class of elliptic elements of order two: Definition 2.6 The complex involution on complex line C with polar vector n is (2-4) It is obvious that I C is a holomorphic isometry fixing the complex line C .
There is a special class of unipotent elements in PU.2; 1/: Definition 2.7 A left Heisenberg translation associated to OEz; t 2 N is given by It is obvious that T OEz;t fixes q 1 and maps OE0; 0 2 N to OEz; t.

Isometric spheres and Ford polyhedron
Suppose that g D .g ij / 3 i;j D1 2 PU.2; 1/ does not fix q 1 .Then it is obvious that g 31 ¤ 0. We first recall the definition of isometric spheres and relevant properties; see for instance [20].
Definition 2.8 The isometric sphere of g, denoted by I.g/, is the set The isometric sphere I.g/ is the Cygan sphere with center and radius r g D p 2=jg 31 j.
The interior of I.g/ is the set The exterior of I.g/ is the set The isometric spheres are paired as follows: Lemma 2.9 [12, Section 5.4.5]Let g be an element in PU.2; 1/ which does not fix q 1 .Then g maps I.g/ to I.g 1 / and the exterior of I.g/ to the interior of I.g 1 /.
Since isometric spheres are Cygan spheres, we now recall some facts about Cygan spheres.Let S OE0;0 .r/ be the Cygan sphere with center OE0; 0 and radius r > 0. Then (2-9) S OE0;0 .r/ D f.z; t; u/ 2 The geographical coordinates on the Cygan sphere will play an important role in our calculation; see Section 2.5 of [22].
Definition 2.10 The point q D q.˛; ˇ; w/ 2 S OE0;0 .r/ with geographical coordinates .˛;ˇ; w/ is the point whose lift to Remark 2.12 This intersection is often called a Giraud disk.
The following property should be useful to describe the intersection of Cygan spheres; see Proposition 2.12 of [22] or Example 5.1.8 of [12].
(1) The level sets of ˛are complex lines, called slices of S OE0;0 .r/.
(3) The set of points with w D 0 is the spine of S OE0;0 .r/.It is a geodesic contained in every meridian.
A central part of this paper is constructing a polyhedron for a finitely generated subgroup of PU.2; 1/.Definition 2.14 Let G be a discrete subgroup of PU.2; 1/.The Ford polyhedron D G for G is the set for all g 2 G with g.q 1 / ¤ q 1 g: That is to say D G is the intersection of closures of the exteriors of all the isometric spheres for elements of G which do not fix q 1 .In fact, the Ford polyhedron is the limit of Dirichlet polyhedra as the center point goes to q 1 .
3 The parameter space of complex hyperbolic .4;4; 1/ triangle groups In this section, we give a parameter space of the complex hyperbolic .4;4; 1/ triangle groups.
Let Â 2 0; 2 .Let I 1 , I 2 and I 3 be the complex involutions on the complex lines C 1 , C 2 and C 3 in complex hyperbolic space H 2 C with polar vectors n 1 , n 2 and n 3 , respectively.By conjugating elements in PU.2; 1/, one can normalize so that @C 3 D fOEz; 0 2 N W jzj D p 2g, @C 1 D fOEz 1 ; t 2 N W t 2 Rg and @C 2 D fOEz 2 ; t 2 N W t 2 Rg.That is, @C 3 is the circle in the z-plane of the Heisenberg group with center the origin and radius p 2, and @C 1 (resp.@C 2 ) is the vertical line whose projection on the z-plane of the Heisenberg group is the point z 1 (resp.z 2 ).Thus the polar vectors of the complex lines can be written as Since tr.I 1 I 3 / D tr.I 2 I 3 / D 1, we have jz 1 j D jz 2 j D 1.Then, up to rotation about the t-axis of the Heisenberg group, the C-circles @C 1 and @C 2 can be normalized to be the sets @C 1 D fOE e iÂ ; t 2 N W t 2 Rg and @C 2 D fOEe iÂ ; t 2 N W t 2 Rg.
Note that the C-circles @C 1 and @C 2 coincide with each other if Â D 2 .According to (2-4), the complex involutions I 1 , I 2 and I 3 on the complex lines are given as 0; 2 , and I 1 , I 2 and I 3 be defined as above.Then hI 1 ; I 2 ; I 3 i is a complex hyperbolic .4;4; 1/ triangle group.Furthermore, the element Proof By computing the products of two involutions and It is easy to verify that I 2 I 3 and I 3 I 1 are elliptic of order 4 and I 2 I 1 is unipotent.Thus hI 1 ; I 2 ; I 3 i is a complex hyperbolic .4;4; 1/ triangle group.
Since the trace of .In both cases, the group hI 1 ; I 2 ; I 3 i is arithmetic.Thus, we have the following proposition: Proposition 3.2 (1) If Â D 0, then the group hI 1 ; I 2 ; I 3 i is discrete and preserves a Lagrangian plane.

The Ford domain
For Â 2 0; 3 , let S D I 2 I 3 and T D I 2 I 1 .Then D hS; T i is a subgroup of hI 1 ; I 2 ; I 3 i of index two.In this section, we will mainly prove that is discrete.Our method is to construct a candidate Ford domain D (see Definition 4.12), then apply the Poincaré polyhedron theorem to show that D is a fundamental domain for the cosets of hT i in , and also that is discrete.I k be the isometric sphere I.T k .S 1 T / 2 T k / D T k I..S 1 T / 2 / and c k be its center.

The isometric spheres
Note that S and S 1 T both have order 4, so S 2 D S 2 and .S 1 T / 2 D .S 1 T / 2 .The centers and radii of the isometric spheres I C k , I k , I ?k and I k are listed in the following table : isometric sphere center radius (1) There is an antiholomorphic involution such that .Proof (1) Let W C 3 !C 3 be given by Then 2 is the identity.It is easy to see that fixes the polar vector n 3 , and interchanges the polar vectors n 1 and n 2 .Thus conjugates I 3 to itself, I 1 to I 2 and vice versa.So conjugates The statement is easily obtained by the facts Before we consider the intersections of two isometric spheres, we would like to give a useful technical lemma.Suppose that q 2 I C 0 .Then by (2-10) the lift of q D q.˛; ˇ; w/ is given by (4-1) q D q.˛; ˇ; w/ D Lemma 4.4 Suppose that Â 2 0; 3 .Let f ?0 .˛;ˇ; w/, f 0 .˛;ˇ; w/ and f 1 .˛;ˇ; w/ be the functions in Definition 4.3.Suppose that q 2 I C 0 .Then (1) q lies on I ?0 (resp.in its interior or exterior) if and only if f ?0 .˛;ˇ; w/ D 0 (resp.is negative or positive), (2) q lies on I 0 (resp.in its interior or exterior) if and only if f 0 .˛;ˇ; w/ D 0 (resp.is negative or positive), (3) q lies on I 1 (resp.in its interior or exterior) if and only if f 1 .˛;ˇ; w/ D 0 (resp.is negative or positive).
Proof (1) Any point q 2 I C 0 lies on I ?0 (resp.in its interior or exterior) if and only if the Cygan distance between q and the center of I ?0 is 1 (resp.less than 1 or greater than 1).Using (4-1), the difference between the Cygan distance from q to the center of D 4f ?0 .˛;ˇ; w/: Hence, q lies on I ?0 (resp.in its interior or exterior) if and only if f ?0 .˛;ˇ; w/ D 0 (resp.negative or positive).
The rest of the proof runs as before.

The intersection of isometric spheres
Proposition 4.5 Suppose that Â 2 0; 3 .Then each pair of isometric spheres in Proof It suffices to show that I C 0 and I C k are disjoint for jkj 1. Observe that T is a Heisenberg translation associated with OE4 cos.Â /; 8 sin.2Â /.According to the Cygan metric given in (2-2), the Cygan distance between the centers of I C 0 and Thus the Cygan distance between the centers of I C 0 and I C k is bigger than the sum of the radii, except when k D ˙1 and Â D 3 .This implies that I C 0 and I C k are disjoint for all jkj 2.
When k D ˙1 and Â D 3 , although the Cygan distance between the centers of I C 0 and ˙1 is the sum of the radii, we claim that they are still disjoint.Using the symmetry in Proposition 4.2, we only need to show that Using the geographical coordinates of q D q.˛; ˇ; w/ given in (4-1), we can compute the difference between the Cygan distance of q and the center of I C 1 with its radius.That is, Here f .˛;ˇ; w/ can be seen as a quadratic function of w.Let Proof Suppose that q 2 I C 0 .Using Lemma 4.4, the geographical coordinates .˛;ˇ; w/ of q 2 We know that the points with w D 0 lie in the meridian with ˇD 2 .Therefore, a necessary condition for q 2 I C 0 \ I ?0 \ I 1 is that ˇD 2 .Substituting ˇD 2 into (4-2) and simplifying, It is easy to see that, for every ˛, the function Â 7 !b.˛; Â / is decreasing on 0; 3 .
The left-hand side of (4-4) can be seen as a quadratic function of w with positive leading coefficient.Thus (4-4) has at least one solution only if .When Â D 3 , T 1 S 2 is unipotent and its fixed point is the eigenvector with eigenvalue 1.One can compute that its fixed point is OEe i2 =3 ; p 3 2 @H 2 C , which equals the point q 3 ; 2 ; Proposition 4.9 Suppose that Â 2 0; 3 .Then: (1) The intersection I ?0 \ I 0 lies in the interior of I C 0 .
(2) The intersection I ?0 \ I 1 either is empty or lies in the interior of I C 0 .Further, when Â D 3 , there is a unique point on the ideal boundary of I ?0 \ I 1 on @H 2 C , which is fixed by T 1 S 2 , lying on the ideal boundary of I C 0 . Proof Since Â 2 0; 3 , we have that sin 2 1 2 ˇ 1 2 ˛ =.2 cos 2 .Â // Ä 2 sin 2 1 2 ˇ 1 2 ˛ and tan.Â/ sin.ˇ ˛/ Ä p 3. This implies that .jzj 2 C u/ 2 C t 2 < 4, so the intersection (2) Suppose that p D .z;t; u/ 2 I ?0 \ I 1 .Then p satisfies the equations Now set By computing the difference of (4-8) and (4-9), we have  One can compute that p 0 lies in the interior of We know that I ?0 \ I 1 is connected from Proposition 2.11.Thus, according to Lemma 4.8, I ?0 \ I 1 lies in the interior of I C 0 except the point OEe i2 =3 ; p 3, which lies on the ideal boundary of I C 0 ; see Figure 1.
Observe that coordinates of the centers of I ?0 and I 1 are continuous on Â .Thus the geometric positions of the spheres I ?0 and I 1 depend continuously on Â .When Â D 0, since the Cygan distance between the centers of I ?0 and I 1 is bigger than the sum of their radii, one can see that I ?0 \ I 1 D ∅.When Â D 3 , we have shown that 0 \ I 1 lies in the interior of I C 0 .We also have The intersections 0 \I 0 is the set of solutions of (4-13) and I C 0 \I ?0 is the set of solutions of (4-14).One can easily verify that the geographical coordinates of the point q 0; Â; Note that the solutions of the above equation for w should satisfy w 2 Ä cos.˛/.Thus So the triple intersection where Note that L 1 lies in a Lagrangian plane of I C 0 , and C 1 [ C 2 lies in a complex line of I C 0 .It is obvious that C 1 is an arc.One of its endpoints is q 0; Â; , which is the fixed point of S. The other one is q 0; arccos 2 Similarly, C 2 is an arc with endpoints q 0; Â; and q 0; arccos 2 and q arccos 1 3 ; Â; . This is a disc bounded by the circle with center the origin and radius p 2, while C 1 [ C 2 lies in the circle with center 3 2 e iÂ and radius 1 2 , which is orthogonal to the boundary of the complex line.The Cayley transform given in Definition 2.2 maps C to the vertical axis f.0; Algebraic & Geometric Topology, Volume 23 (2023)

The Cayley transform maps
, which is the image of q 0; Â; under the Cayley transform.
It is easy to check that S q 0; Â; 2 , and that the other four points are cyclically permuted by S: Moreover, it is easy to verify that C 1 [C 2 D S.L 1 /, S 2 .L 1 / D L 1 and S 2 .C 1 / D C 2 .Thus, the four rays from the fixed point to the four endpoints are cyclically permuted by S.
By applying powers of T and the symmetries in Proposition 4.2 to Propositions 4.6, 4.7 and 4.9, all pairwise intersections of the isometric spheres can be summarized: (2) I k is contained in the exterior of all the isometric spheres in S except I C k , I C kC1 , k , I ?kC1 , I k and I kC1 .Moreover, (3) I ?k is contained in the exterior of all the isometric spheres in S except I k , I C kC1 , I k 1 , I k and I kC1 .Moreover, , and I ?k \ I k 1 is described in (2).When Â D 3 , I ?
k will be tangent with I ?kC1 (resp.I ?k 1 ) on @H 2 C at the parabolic fixed point of T k .S 2 T 1 /T k (resp.T k .T 1 S 2 /T k ).(4) I k is contained in the exterior of all the isometric spheres in S except I k , k and I ?k 1 .Moreover, I k \ I ?k and I k \ I ?k 1 are described in (3).I k \ I k is described in (2) and I k \ I C k 1 is described in (1).When Â D 3 , I k will be tangent with I kC1 (resp.I k 1 ) on @H 2 C at the parabolic fixed point of

The Ford domain
k and s k 1 \ s k for k 2 Z are all topologically the union of two sectors.
According to Proposition 4.10, k is the union of two crossed geodesics.The two crossed geodesics divide the disc into four sectors, one opposite pair of which will lie in the interior of the isometric sphere I ?k .Thus s C k \s k is the other opposite pair of the four sectors in the disc.More precisely, up to the powers of T , let us consider s C 0 \ s 0 .Let be the disc I C 0 \ I 0 described in (4-13) and the two crossed geodesics L 1 [ C 1 [ C 2 be as described in Proposition 4.10.By Proposition 4.2, the complex involution I 2 preserves and L 1 [ C 1 [ C 2 .Recall that I 2 fixes the complex line C 2 with polar vector n 2 described in Section 3. One can compute that the intersection 2 / D 1 cos.˛/0. Therefore, the union of two opposite sectors containing C 2 \ is the ridge s C 0 \ s 0 .Moreover, this ridge is preserved by I 2 .By using the parametrization of the Giraud disk in [8], we can draw the Giraud disk I C 0 \ I 0 and the intersection with the isometric spheres I 1 , I ?0 , I ? 1 , I 0 and I 1 ; see Figure 4.
The other ridges can be described by a similar argument.
is the disjoint union of two topological discs.(2) The side Proof  k is topologically a solid light cone; see Figure 6.A similar argument describes s k .
By applying a Poincaré polyhedron theorem in H 2 C as stated for example in [22], [8] or [19] (see [3] for a version in the hyperbolic plane), we have our main result: Theorem 4.17 Suppose that Â 2 0; 3 .Let D be as in Definition 4.12.Then D is a fundamental domain for the cosets of hT i in .Moreover, is discrete and has the presentation D hS; T j S 4 D .T 1 S/ 4 D idi: To obtain the side-pairing maps and ridge cycles, by applying powers of T , it suffices to consider the case where k D 0.
Hence S maps the ridge s C 0 \ s 0 to the ridge s 0 \ s ?0 .Similarly, S maps s C 0 \ s ?
Hence, the side-pairing maps are The cycle transformations According to the side-pairing maps, the ridge cycles are Thus the cycle transformations are which are equal to the identity map, since S 4 D id and .T 1 S/ 4 D id.
The local tessellation There are exactly two copies of D along each side, since the sides are contained in isometric spheres and the side-pairing maps send the exteriors to the interiors.Thus there is nothing to verify for the points in the interior of every side.
According to the ridge cycles and cycle transformations, there are exactly three copies of D along each ridge.
k These three ridges are in one cycle.Thus, we only need to consider the ridge s is the smallest of the three quantities in the above triple equality.
For z in the neighborhoods of s where jhz; T k S 1 T k .q 1 /ij is the smallest of the three quantities in the above triple equality.
For z in the neighborhoods of s C k \s ?k in D, jhz; where jhz; T k S T k .q 1 /ij is the smallest of the three quantities in the above triple equality.
Thus the union of D, T k S 1 T k .D/ and T k S T k .D/ forms a regular neighborhood of each point in the union of D, T k .T 1 S / 1 T k .D/ and T k .T 1 S/T k .D/ forms a regular neighborhood of each point in s C k \ s k 1 by a similar argument as in the previous case.
Consistent system of horoballs When Â D 3 , there are accidental ideal vertices on D.
The sides s ?k and s ?kC1 will be asymptotic on @H 2 C at the fixed point of the parabolic element T k .S 2 T 1 /T k , and the sides s k and s kC1 will be asymptotic on @H 2 C at the fixed point of the parabolic element T k .S T 1 S/T k .To show that there is a consistent system of horoballs it suffices to show that all the cycle transformations fixing a given cusp are nonloxodromic.Let p 2 be the fixed point of T 1 S 2 and q 2 be the fixed point of .T 1 S/ 2 T (the coordinates of p 2 and q 2 are given in Definition 5.1, or see Figure 9).Then all the accidental ideal vertices fT k .p 2 /g and fT k .q 2 /g are related by the side-pairing maps as follows: Thus, up to powers of T , all the cycle transformations are S S S 2 D id and which is parabolic.This means that p 2 is fixed by the parabolic element .T 1 S 2 / 2 .
Algebraic & Geometric Topology, Volume 23 (2023) Therefore, D is a fundamental domain for the cosets of hT i in .The side-pairing maps and T will generate the group .The reflection relations are The cycle relations are Thus is discrete and has the presentation Since is a subgroup of hI 1 ; I 2 ; I 3 i of index 2: Corollary 4.18 Let hI 1 ; I 2 ; I 3 i be a complex hyperbolic .4;4; 1/ triangle group as in Proposition 3.1.Then hI 1 ; I 2 ; I 3 i is discrete and faithful if and only if This answers Conjecture 1.1 on the complex hyperbolic .4;4; 1/ triangle group.

The manifold at infinity
In this section, we study the group in the case when Â D 3 .That is, the group D hS; T i D hI 2 I 3 ; I 2 I 1 i with T 1 S 2 D I 1 I 3 I 2 I 3 being parabolic.
In this case, the Ford domain D has additional ideal vertices on @H 2 C , which are parabolic fixed points corresponding to the conjugators of T 1 S 2 .By intersecting a fundamental domain for hT i acting on @H 2 C with the ideal boundary of D, we obtain a fundamental domain for acting on its discontinuity region ./.
Topologically, this fundamental domain is the product of an unknotted cylinder and a ray; see Proposition 5.13.By cutting and gluing we obtain two polyhedra P C and P ; see Proposition 5.18.Gluing P to P C by S 1 , we obtain a polyhedron P. By studying the combinatorial properties of P, we show that the quotient ./= is homeomorphic to the two-cusped hyperbolic 3-manifold s782.
Let U be the ideal boundary of    1 \I 0 \I 1 .p 2 (resp.p 3 ) is the parabolic fixed point of T 1 S 2 (resp.S 2 T 1 ), which is the intersection I C 0 \ I 1 \ I ?0 \ I ? 1 (resp.I C 1 \ I 0 \ I ?0 \ I ? 1 ).q 3 (resp.q 2 ) is the parabolic fixed point of S T 1 S (resp.T 1 S T 1 S T ), which is the intersection of the four isometric spheres I C 0 \ I 0 \ I 0 \ I 1 (resp.
.  with I C 0 in @H 2 C , viewed in geographical coordinates.Here ˛2 2 ; 2 in the vertical coordinate and ˇ2 OE0; in the horizontal one.The region exterior to the six Jordan closed curves has two connect components.One of them is the topological octagon with vertices p 2 , p 6 , p 4 , p 7 , q 3 , p 8 , p 11 and p 9 .The other one is a topological quadrilateral with vertices p 2 , p 5 , q 3 and p 10 .
We will only prove the statement for p 2 in the last two items; the others follow by similar arguments.By Lemma 4.8, p 2 is the parabolic fixed point of T 1 S 2 and is the triple intersection I C 0 \I 1 \I ?0 .By Corollary 4.11(3), I ?0 is tangent with I ? 1 at p 2 .

The sides of U
Now we study the combinatorial properties of the sides; see Figures 8 and 9.
Proposition 5. Proof Note that side s 0 is bounded by the ridges s 0 \s C 1 , s 0 \s 1 , s 0 \s C 0 and s 0 \s ?0 .By Proposition 4.2, the side s 0 is isometric to s C 0 under the complex involution I 2 .Thus its ideal boundary Q s 0 will be also isometric to Q s C 0 .This implies that Q s 0 has the same combinatorial properties as Q s C 0 .One can check that I 2 W .q 3 ; p 5 ; p 2 ; p 10 ; p 8 ; p 11 ; p 9 ; p 6 / $ .q 3 ; p 7 ; p 3 ; p 12 ; p 14 ; p 13 ; p 15 ; p 4 /: Thus one part of Q s 0 is an octagon, denoted by O 0 , whose vertices are p 5 , p 6 , p 4 , p 3 , p 15 , p 13 , p 14 and q 3 .The other is a quadrilateral, denoted by Q 0 , whose vertices are p 3 , p 7 , q 3 and p 12 ; see Figure 9.
Remark 5.5 The vertex q 3 lies on the C-circle associated to I 2 , that is, the ideal boundary of the complex line fixed by I 2 .One can also observe that p 2 is fixed by I 1 .Proof By Proposition 4.16, the side Q s ?0 is the union of two disjoint discs, which are bounded by the ideal boundary of the ridges s C 0 \ s ?0 and s 0 \ s ?0 .As stated in Proposition 5.2, the ideal boundary of I C 0 \ I 0 \ I ?0 contains the four points p 4 , p 5 , p 6 and p 7 .Thus Q s ?0 is the union of two disjoint bigons, one with vertices p 5 and p 6 , and the other with p 4 and p 7 .Proposition 5.2 also tells us that p 2 and p 3 lie on different components of the boundaries of the two bigons.Therefore, both of the components are triangles, denoted by .T 1 / ?0 and .T 2 / ?0 , whose vertices are p 2 , p 5 and p 6 and p 3 , p 4 and p 7 , respectively.
According to the symmetry in Proposition 4.2, the side Q s 0 has the same topological properties as the side Q s ?0 .Thus by a similar argument: Proposition 5.7 The interior of side Q s 0 is a union of two disjoint triangles, denoted by .T 1 / 0 and .T 2 / 0 , whose vectors are q 2 , p 9 and p 11 , and q 3 , p 8 and p 10 , respectively.Note that T acts on L as a translation through 2. To show L is contained in the complement of D, it suffices to show that a segment with length 2 is contained in the interior of some isometric spheres.By considering the Cygan distance between a point in L and the center of an isometric sphere, one can compute that the segments ˚

A fundamental domain for the subgroup hT i
x Ci p 3 2 ; lie in the interiors of I C 0 and I 0 , respectively.
« be two planes in the Heisenberg group.
In fact, the vertical planes † 1 and † 0 are boundaries of fans in the sense of [14].Let D T be the region between † 1 and † 0 , that is Proof Since the isometric spheres are strictly convex, their intersections with a plane are either a topological circle, a point or empty.Note that † 0 D T .† 1 /.Thus it suffices to consider the intersections of † 0 with the isometric spheres.By a strait computation, each one of † 0 \ I 0 and † 0 \ I ?0 is a circle; see Figure 11.
Proof It suffices to consider † 0 .The C-circle associated to I 2 , that is, the ideal boundary of the complex line fixed by It is obvious that † 0 contains q 3 , which is the tangent point of I 0 and I 1 .The intersections † 0 \ I C 0 , † 0 \ I 0 and † 0 \ I ?0 are circles by Proposition 5.10.One can compute that the intersection † 0 \ I C 0 \ I 0 contains q 3 , and v 0 D 1 2 C i Remark 5.15 U is the complement of a tubular neighborhood of the T -invariant R-circle L, that is, a horotube for T .(See [28] for the definition of a horotube.) 5.5 The manifold Definition 5.16 Suppose that the cylinder S 1 OE0; 1 has a cell structure with finite faces fF i g.A canonical subdivision on S 1 OE0; 1 R 0 is a finite union of 3-dimensional pieces f y F i g where y F i D F i R 0 .
Proposition 5.17 There is a canonical subdivision on U \ D T .
Let ./be the discontinuity region of acting on @H 2 C .Then U \ D T is obviously a fundamental domain for .By cutting and gluing, we can obtain the following fundamental domain for acting on ./:Proposition 5.18 Let P C be the union of y H C 0 , .y T 1 / 0 , .y T 2 / 0 , T .y 1 /, T .y Q 1 / and T .. y T 2 / ? 1 /.Let P be the union of y Q C 0 , .y T 1 / ?0 , y Q 0 0 and T .y H 1 /.Then P C [ P is a fundamental domain for acting on ./.Moreover, P C (resp.P ) is combinatorially an eleven-pyramid (resp.nine-pyramid ) with cone vertex q 1 and base Proof Since † 0 D T .† 1 / and c 0 D T .c 1 /, U \ D T and T .U \ D T / can be glued together along c 0 R 0 .Note that U \ D T is a fundamental domain for acting on ./and has a subdivision as described in Proposition 5.17.Therefore P C [ P is also a fundamental domain.
Proposition 5.21 Let be the discontinuity region of acting on H 2 C .Then the fundamental group of = has a presentation hu; v; w j w 1 vu 1 v 1 wu D v 2 wuw 3 u D idi: Proof Let x i for i D 1; 2; 3; 4; 5; 6; 7; 8 be the corresponding gluing maps of P given in Lemma 5.20.These are the generators of the fundamental group of = .
By considering the edge cycles of P under the gluing maps, we have the relations Therefore M will be the connected sum of s782 and a closed 3-manifold with trivial fundamental group by the prime decompositions of 3-manifolds [15].The solution of the Poincaré conjecture implies that the closed 3-manifold is the 3-sphere.Thus M is homeomorphic to s782.

C i p 3 2
all entries of I 1 , I 2 and I 3 are in the ring of integers Z, and if Â D 3 all entries of I 1 , I 2 and I 3 are in the ring of Eisenstein integers Z 1

Definition 4 . 1
For k 2 Z, let I C k be the isometric sphere I.T k S T k / D T k I.S / and c C k be its center, I k be the isometric sphere I.T k S 1 T k / D T k I.S 1 / and c k be its center, I ?k be the isometric sphere I.T k S 2 T k / D T k I.S 2 / and c ? k be its center, Algebraic & Geometric Topology, Volume 23 (2023)

)
The complex involution I 2 interchanges I ?k and I ?k , interchanges I C k and I k , and interchanges I k and I ˘ kC1 .

Figure 1 : 8 p 3 /
Figure 1: The ideal boundaries of the three spheres I C 0 , I 1 and I ?0 on @H 2 C .On the left is the case when Â D 0 and on the right is Â D 3 .

0; 3 Proposition 4 . 10
by Lemma 4.8.Hence, the intersection I ?0 \ I 1 is either empty or contained in the interior of I C 0 .Suppose that Â 2 0; 3 .For k 2 Z, the three isometric spheres I C k , I k and I ?k (resp.I C k , I k 1 and I k ) have the following properties:

2 CFigure 2 :p 2w cos 1 2 ˛C ˇ Â 2 p 2w cos 1 2
Figure 2: The ideal boundaries of the three spheres I C 0 , I 0 and I ?0 on @H 2 C .(On the left is the case when Â D 0 and on the right is Â D 3 .)andwhose fours endpoints are on @H 2 C .Moreover, the four rays from the fixed point to the four endpoints are cyclically permuted by T k S T k (resp.T k .S 1 T /T k ).

Figure 3 :
Figure 3: The triple intersection I C 0 \ I 0 \ I ?0 viewed on the vertical axis in the ball model of H 2 C .The blue curve is C 1 and the red one is C 2 .The black point on the curve is the projection of L 1 on the complex line.The left curve is the case when Â D 0 and the one on the lower left is the case when Â D 3 .model of H 2 C .Thus C is isometric to the Poincaré disc.Then C 1 [ C 2 is mapped by the Cayley transform to an arc contained in the circle with center 3e iÂ =.2 p 2/ and radius 1=.2 p 2/, which is orthogonal to the unit circle.Hence C 1 [ C 2 is a geodesic; see Figure 3.

Corollary 4 . 11
Suppose that Â 2 0; 3 .Let S D fI k ; I ?k ; I k W k 2 Zg be the set of all the isometric spheres.Then for all k 2 Z: (1) I C k is contained in the exterior of all the isometric spheres in S except I k , I k 1 , I ?k , I ?k 1 , I k and I kC1 .Moreover, I C k \I ?k 1 (resp.I C k \I kC1 ) is either empty or contained in the interior of I k 1 (resp.I k ).When Â D 3 , I C k \ I ?k 1 (resp.

Definition 4 .
12 Let D be the intersection of the closures of the exteriors of the isometric spheres I C k , I k , I ?k and I k , for k 2 Z. Definition 4.13 For k 2 Z, let s C k (resp.s k , s ?k and s k ) be the side of D contained in the isometric sphere I C k (resp.I k , I ?k and I k ).Definition 4.14 A ridge is defined to be the 2-dimensional connected intersections of two sides.By Corollary 4.11, the ridges are s

Figure 4 :
Figure 4: The ridge s C 0 \ s 0 (the shaded region in the intersection of I C 0 \ I 0 ) in the plane with spinal coordinates introduced in [8].The triple intersection I C 0 \I 0 \I ?0 is the two mutually perpendicular lines.Compare to [8, Figure 16].

Figure 5 :
Figure 5: A schematic view of the side s C k (s k ).Propositions 4.10 and 4.15, 4 3 is a union of four sectors which are patched together along the cross I C k \ I k \ I ?k .By a computation for the case k D 0, one can see that the ideal boundary of 4 3 is a union of two disjoint simple closed curves in the ideal boundary of I ?k ; see Figure 2. Thus 4 3 is a light cone.Hence, s ?k is topologically a solid light cone; see Figure6.A similar argument describes s k .

Figure 6 :
Figure 6: A schematic view of the side s ?k (s k ).
and Q s k ) be the ideal boundary of the side of D contained in the ideal boundary of the isometric sphere I C k (resp.I k , I ?k and I k ).Then the union of all the sides fQ s C k g, fQ s k g, fQ s ?k g and fQ s k g for k 2 Z form the boundary of U .

5. 1
The vertices of U Definition 5.1 In the Heisenberg coordinates we define the points

p 8 ,
p 9 , p 10 and p 11 are the four points on the ideal boundary of I C 0 \ I 1 \ I 0 .p 12 , p 13 , p 14 and p 15 are the four points on the ideal boundary of I C As described in Proposition 4.10, all of the triple intersections I C 0 \ I 0 \ I ?0 , I C 0 \I 1 \I 0 and I C 1 \I 0 \I 1 have exactly four points lying on @H 2 C .When writing the standard lifts of p 4 , p 5 , p 6 and p 7 , one can see that they are the four points in the proof of Proposition 4.10.Thus the first item is proved.By Proposition 4.2, the four points of I C 0 \ I 1 \ I 0 are the images of p 4 , p 5 , p 6 and p 7 under the antiholomorphic involution , which are p 11 , p 10 , p 8 and p 9 .The second item and the fact that I C 1 \ I 0 \ I 1 D T .I C 0 \ I 1 \ I 0 / imply the third item.

1 Figure 10 :C 2 /
Figure 10: A realistic picture of the ideal boundaries of the isometric spheres.I C 0 , I C 1 and C 1 (red), I 0 and I 1 (blue), I ?0 and I ? 1 (pink), I 0 , I ˘ 1 and I 1 (yellow).The line L is the T -invariant R-circle.

Figure 11 :
Figure 11: The intersection of † 0 with I C 0 , I 0 and I ?0 viewed in † 0 .c 0 D c C 0 [ c 0 is a simple closed curve, where c C 0 and c 0 are the solid line parts of † 0 \ I C 0 and † 0 \ I 0 , respectively.Proposition 5.10 The intersections of † 0 and † 1 with the isometric spheres I k , I ?k and I k are empty, except: † 0 \ I 0 and † 0 \ I ?0 are circles, and † 0 \ I 0 D † 0 \ I 1 D fq 3 g.† 1 \ I ˙ 1 and † 1 \ I ? 1 are circles, and † 1 \ I ˘ 1 D † 1 \ I 0 D fq 2 g.

11 .
These two points divide the circles on I C 0 and I 0 into two arcs.Let c C 0 Algebraic & Geometric Topology, Volume 23 (2023)

Figure 12 :Definition 5 . 19
Figure 12: A schematic view of the fundamental domain of on ./.The red vertices are the parabolic fixed points.The yellow polygon is O 0 [ Q C 0 [ .T 1 / ?0 .

10 Figure 13 :
Figure 13: The combinatorial picture of P. The red vertices of P are the parabolic fixed points.

x 1 7 x 5 2 x 4 x 1 Dx 6 1 4x 5 x 2 1 8x 7 ; p 2 x 1 ! 1 ; p 3 x 7 ! 1 ; q 3 x 5 ! 1 ; p 0 2 x 1 7 ! 1 7x 5
x 7 x 1 D id; x 1 id; x 2 x x 1 D id; x 2 x 3 x 2 D id;x D id;x 7 x 8 x 6 D id;x 8 x 7 x 6 D id;x D id:For example, the edge cycle of OEq 1 ; p 2 isOEq 1 OEq OEp OEp OEq 1 ; p 2 : Thus x x 7 x 1 D id:This is the first relation.The others can be given by a similar argument.Simplifying the relations and setting u D x 1 , v D x 2 and w D x 7 , we obtain the presentation of the fundamental group of = .Now we are ready to show the following theorem:Theorem 5.22 Let be the discontinuity region of acting on H 2 C .Then the quotient space = is homeomorphic to the two-cusped hyperbolic 3-manifold s782 in the SnapPy census.Proof Let M D = .According to Proposition 5.21, the fundamental group of M has a presentation1 .M / D hu; v; w j w 1 vu 1 v 1 wu D v 2 wuw 3 u D idi:The manifold s782 is a two-cusped hyperbolic 3-manifold with finite volume.Its fundamental group, provided by SnapPy, has a presentation 1 .s782 / D ha; b; c j a 2 cb 4 c D abca 1 b 1 c 1 D idi: Using Magma, we get an isomorphism ‰ W 1 .M / ! 1 .s782/given by ‰.u/ D c 1 b 1 ; ‰.v/ D b 1 and ‰.w/ D a: and interchanges the Siegel domain model and the ball model of H 2 C .Let N D C R be the Heisenberg group with product OEz; t OE ; D OEz C ; t C 2 Im.N z /: Rg be the set of real points.H 2 R is a Lagrangian plane.All the Lagrangian planes are the images of H 2 R by isometries of H 2 C .The ideal boundary of a Lagrangian plane is called an R-circle.In the Heisenberg group, R-circles are either straight lines or lemniscate curves whose projections on the z-plane are the figure eight.
? 0 and I ?k are disjoint in H 2 C .Furthermore, when Â D 3 the closures of I ? ) are tangent at the parabolic fixed point of T 1 S 2 (respectively, T .T 1 S 2 /T 1 ) on @H 2 Suppose that Â 2 0; 3 .Then I C 0 then it is obvious that f .˛;ˇ;w/ > 0. If B 2 4C 0, then B Ä 2 p C (which is impossible by numerically computation) or k are disjoint in H 2 C [ @H 2 C , except possibly when k D 0; 1. Proposition 4.7 Suppose that Â 2 0; 3 .Then: Algebraic & Geometric Topology, Volume 23 (2023)(1) I This means that the symmetry axes of the quadratic function lie on the right-hand side of If it is 0, then ˛D 3 and Â D 3 .
and the sides and ridges are related as follows:The side s C k is bounded by the ridgess C k \ s k , s C k \ s ?k , s C k \ s k 1 and s C k \ s k .The side s k is bounded by the ridges s C k \s k , s k \s ?k , s k \s C kC1 and s 0 \s kC1 .The side s ?k is bounded by the ridges s C k \ s ?k and s k \ s ?k .The side s k is bounded by the ridges s k \ s k 1 and s C k \ s k .
the fixed point of S, and divides into two parts.I 2 fixes C 2 \ and interchanges L 1 and C 1 [ C 2 .Thus C 2 \ is contained in the union of two opposite sectors.By Lemma 4.4, C 2 \ lies on the closure of the exterior of I ? p 1) The side s C k is contained in the isometric sphere I C k .By Corollary 4.11, s C k might intersect with the sides contained in the isometric spheres I k , I k 1 , I ?, and 4 2 be the union of the ridges s C k \s k 1 and s C k \s k .By Proposition 4.10, 4 1 contains the cross I C k \I k \I ?k .By Proposition 4.15, 4 1 is a union of four sectors which are patched together along the cross.Hence, 4 1 is topologically either a disc or a light cone.By a straightforward computation, the ideal boundary of 4 1 on H 2 C is a simple closed curve on the ideal boundary of I C k ; see Figure 2. Thus 4 1 is a topological disc.By a similar argument, 4 2 is a topological disc.If Â ¤ 3 then I C k \ I ?k \ I k 1 D ∅, so 4 1 and 4 2 are disjoint, and if Â D 3 they intersect at two points on @H 2 C ; see Figures 9 and 10.Note that isometric spheres are topological balls and their pairwise intersections are connected.So, s C k is a topological solid cylinder; see Figure 5.A similar argument describes s k .
k , I ?k 1 , I k and I kC1 .Let 4 1 be the union of the ridges s C k \ s k and s C k \ s ?k (2) The side s ?k is contained in the isometric sphere I ?k .According to Corollary 4.11, s ?k only intersects with s C k and s k .Let 4 3 be the union of s C k \ s ?k and s k \ s ?k .By T .p 9 /; p 13 D T .p 8 /; p 14 D T .p11 /; p 15 D T .p 10 /: By Proposition 4.10 and Corollary 4.11, we have the following: Proposition 5.2 The points in Definition 5.1 satisfy: p 4 , p 5 , p 6 and p 7 are the four points on the ideal boundary of I C 0 \ I 0 \ I ?0 , which is described in Proposition 4.10.

3
The interior of the side Q s C 0 has connected components an octagon, denoted by O C 0 , with vertices p 2 , p 6 , p 4 , p 7 , q 3 , p 8 , p 11 and p 9 , a quadrilateral , denoted by Q C 0 , with vertices p 2 , p 5 , q 3 and p 10 .Proof By Proposition 4.16, when Â < 3 , the side Q s C 0 is topologically an annulus bounded by two disjoint simple closed curves which are the union of the ideal boundaries of the ridges s C 0 \ s 1 and s C 0 \ s 0 , respectively s C 0 \ s 0 and s C By Proposition 5.2, the ideal boundary of the ridges C 0 \ s 1 (resp.sC 0 \ s 0 ) isa union of two disjoint Jordan arcs OEp 9 ; p 10 and OEp 8 ; p 11 (resp.OEp 10 ; p 8 and OEp 11 ; p 9 ), and the ideal boundary of the ridge s C 0 \ s 0 (resp.s C 0 \ s ?0 ) is a union of two disjoint Jordan arcs OEp 5 ; p 7 and OEp 4 ; p 6 (resp.OEp 7 ; p 4 and OEp 6 ; p 5 ).Since p 2 is the intersection of four isometric spheres I C 0 \ I 1 \ I ?0 \ I ? 1 , it lies on OEp 9 ; p 10 and OEp 6 ; p 5 .Similarly, q 3 lies on OEp 5 ; p 7 and OEp 10 ; p 8 .It is easy to check that interchanges p 2 and q 3 , p 5 and p 10 , p 4 and p 11 , p 6 and p 8 , and p 7 and p 9 .Thus one part of Q s C 0 is a quadrilateral with vertices p 2 , p 5 , q 3 and p 10 , denoted by O C 0 .The other is an octagon with vertices p 2 , p 6 , p 4 , p 7 , q 3 , p 8 , p 11 and p 9 , denoted by Q C 0 .Both of them are preserved by .The interior of the side Q s 0 has connected components an octagon, denoted by O 0 , with vertices p 5 , p 6 , p 4 , p 3 , p 15 , p 13 , p 14 and q 3 , a quadrilateral , denoted by Q 0 , with vertices p 3 , p 7 , q 3 and p 12 .
The top and bottom curves are identified.O 0 (resp.O ˙ 1 ) is divided by c 0 (resp.c 1 ) into a quadrilateral Q 0 0 (resp.Q 0 ˙ 1 ) and a heptagon H 0 (resp.H ˙ 1 ), v 0 is the intersection of c 0 with the arc OEp 4 ; p 6 and v 1 is the intersection of c 1 with the arc OET 1 .p 4 /; T 1 .p6 /.The interior of side Q s ?0 is a union of two disjoint triangles, denoted by .T 1 / ?0 and .T 2 / ?0 , whose vertices are p 2 , p 5 and p 6 and p 3 , p 4 and p 7 , respectively.