A uniformizable spherical CR structure on a two-cusped hyperbolic 3-manifold
AA UNIFORMIZABLE SPHERICAL CR STRUCTURE ON ATWO-CUSPED HYPERBOLIC 3-MANIFOLD
YUEPING JIANG, JIEYAN WANG AND BAOHUA XIE
Abstract.
Let (cid:104) I , I , I (cid:105) be the complex hyperbolic (4 , , ∞ ) triangle group.In this paper we give a proof of a conjecture of Schwartz for (cid:104) I , I , I (cid:105) . Thatis (cid:104) I , I , I (cid:105) is discrete and faithful if and only if I I I I is nonelliptic. When I I I I is parabolic, we show that the even subgroup (cid:104) I I , I I (cid:105) is the ho-lonomy representation of a uniformizable spherical CR structure on the two-cusped hyperbolic 3-manifold s
782 in SnapPy notation. Introduction
Let H C be the complex hyperbolic plane, PU(2 ,
1) be its holomorphic isometrygroup. See Section 2 for more details. It is well known that H C is one of the rankone symmetric spaces and PU(2 ,
1) is a semisimple Lie group. H C can be viewedas the unit ball in C equipped with the Bergman metric. Its ideal boundary ∂ H C is the 3-sphere S . The purpose of this paper is to study the geometry of discretesubgroups of PU(2 , M be a 3-manifold. A spherical CR structure on M is a system of coordinatecharts into S , such that the transition functions are restrictions of elements ofPU(2 , M determines a pair ( ρ, d ), where ρ : π ( M ) −→ PU(2 ,
1) is the holonomy and d : (cid:102) M −→ S 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 Ω / Γ, where Γ isa discrete subgroup of PU(2 ,
1) and Ω ⊂ ∂ H C is the discontinuity region of Γ.An interesting problem in complex hyperbolic geometry is to find (uniformizable)spherical CR structures on hyperbolic 3-manifolds.Geometric structures modelled on the boundary of complex hyperbolic space arerather difficult to construct. The first example of a spherical CR structure existingon a cusped hyperbolic 3-manifold was discovered by Schwartz. In the work [24],Schwartz constructed a uniformizable spherical CR structure on the Whiteheadlink complement. He also constructed a closed hyperbolic 3-manifold that admitsa uniformizable spherical CR structure in [27] at almost the same time.Let M be the complement of the figure eight knot. There are several worksshowed that M admits (uniformizable) spherical CR structures. In [9], Falbelconstructed two different representations ρ , ρ of π ( M ) in PU(2 , Date : January 26, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Complex hyperbolic space, spherical CR uniformization, trianglegroups, Ford domain, hyperbolic 3-manifolds.Y. Jiang was supported by NSFC (No.11631010). J. Wang was supported by NSFC(No.11701165). B. Xie was supported by NSFC (No.11871202) and Hunan Provincial NaturalScience Foundation of China (No.2018JJ3024). a r X i v : . [ m a t h . G T ] J a n YUEPING JIANG, JIEYAN WANG AND BAOHUA XIE that ρ is the holonomy of a spherical CR structure on M . In [12], Falbel andWang proved that ρ is also the holonomy of a spherical CR structure on M . In[7], Deraux and Falbel constructed a uniformizable spherical CR structure on M whose holonomy is ρ . In [5], Deraux proved that there is a 1-parameter family ofspherical CR uniformizaitons of the figure eight knot complement. This family isin fact a deformation of the uniformization constructed in [7].Let us back to the Whitehead link complement. It admits a uniformizable spher-ical CR structure which is different from Schwartz’s one. In the recent work [21],Parker and Will also constructed a spherical CR uniformization of the Whiteheadlink complement. By applying spherical CR Dehn surgery theorems to the uni-formizations of the Whitehead link complement, one can get infinity many mani-folds which admit uniformizable spherical CR structures. In [28], Schwartz proveda spherical CR Dehn surgery theorem, and using it to the spherical CR uniformiza-tion of the Whitehead link complement constructed in [24] to obtain infinity manyclosed hyperbolic 3-manifolds which admit uniformizable spherical CR structures.In [2], Acosta used the spherical CR Dehn surgery theorem proved in [1] to thespherical CR uniformization of the Whitehead link complement constructed byParker and Will in [21] to obtain infinity many one-cusped hyperbolic 3-manifoldswhich admit uniformizable spherical CR structures. In particular, the spherical CRuniformization of the complement of the figure eight knot constructed by Derauxand Falbel [7] is contained in this family.There are some hyperbolic 3-manifolds described in the SnapPy census (see[4]) which admit spherical CR structures. In [6], Deraux proved that the cuspedhyperbolic 3-manifold m
009 admits a uniformizable spherical CR structure whoseholonomy representation was constructed by Falbel, Koseleff and Rouillier in [11].In [17], Ma and Xie proved that the one-cusped hyperbolic 3-manifolds m
038 and s
090 admit spherical CR uniformizations.In this paper, we show that the two-cusped hyperbolic 3-manifold s
782 admits auniformizable spherical CR structure. By studying the action of the even subgroupof a discrete complex hyperbolic triangle group on H C , we will prove that thequotient space of its discontinuity region is homeomorphic to s s
782 is a trianglegroup.Now let us talk about the complex hyperbolic triangle groups. Let ∆ p,q,r be theabstract ( p, q, r ) reflection triangle group with the presentation (cid:104) σ , σ , σ | σ = σ = σ = ( σ σ ) p = ( σ σ ) q = ( σ σ ) r = id (cid:105) , where p, q, r are positive integers or ∞ in which case the corresponding relationdisappears. A complex hyperbolic ( p, q, r ) triangle group is a representation of ∆ p,q,r in PU(2 , H C . The study of complex hyperbolic triangle groups was begun by Goldmanand Parker. In [14], Goldman and Parker studied the complex ( ∞ , ∞ , ∞ ) trianglegroups. They conjectured that a representation of ∆ ∞ , ∞ , ∞ into PU(2 ,
1) is discreteand faithful if and only if the image of σ σ σ is nonelliptic. The Goldman-Parkerconjecture was proved by Schwartz in [23] (or a better proof in [26]). In particular,the representation with the image of σ σ σ being parabolic is closely related withthe holonomy of the spherical CR uniformizaiton of the Whitehead link complementconstructed in [24]. In the survey [25], a series of conjectures on complex hyperbolictriangle groups are put forward. Schwartz conjectured that: PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 3
Conjecture 1.1 (Schwartz [25]) . Suppose that p ≤ q ≤ r . Let (cid:104) I , I , I (cid:105) be acomplex hyperbolic ( p, q, r ) triangle group. Then (cid:104) I , I , I (cid:105) is a discrete and faithfulrepresentation of ∆ p,q,r if and only if I I I I and I I I are nonelliptic. Moreover, • If ≤ p < , then (cid:104) I , I , I (cid:105) is discrete and faithful if and only if I I I I is nonelliptic. • If p > , then (cid:104) I , I , I (cid:105) is discrete and faithful if and only if I I I isnonelliptic. In a recent work [21], Parker and Will proved Conjecture 1.1 for complex hy-perbolic (3 , , ∞ ) groups. They also showed that when I I I I is parabolic thequotient of H C by the group (cid:104) I I , I I (cid:105) is a complex hyperbolic orbifold whoseboundary is a spherical CR uniformization of the Whitehead link complement. In[20], Parker, Wang and Xie proved Conjecture 1.1 for complex hyperbolic (3 , , n )groups with n ≥
4. Furthermore, Acosta [2] showed that when I I I I is parabolicthe group (cid:104) I I , I I (cid:105) is the holonomy representation of a uniformizable sphericalCR structure on the Dehn surgery of the Whitehead link complement on one cuspof type (1 , n − , , ∞ ) triangle groups and further analyze the group when I I I I is parabolic.Our result is as follows: Theorem 1.2.
Let (cid:104) I , I , I (cid:105) be a complex hyperbolic (4 , , ∞ ) triangle group.Then (cid:104) I , I , I (cid:105) is a discrete and faithful representation of ∆ , , ∞ if and only if I I I I is nonelliptic. Moreover, when I I I I is parabolic the quotient of H C bythe group (cid:104) I I , I I (cid:105) is a complex hyperbolic orbifold whose boundary is a sphericalCR uniformization of the two-cusped hyperbolic 3-manifold s in the SnapPycensus. In [29], Wyss-Gallifent studied the complex hyperbolic (4 , , ∞ ) triangle groups.He discovered several discrete groups with I I I I being regular elliptic of finiteorder and conjectured that there should be countable infinity many. Thus, it shouldbe very interesting to know what is the manifold at infinity for the group with I I I I being regular elliptic of finite order. Motivated by the work of Acosta [2],we guess that the manifold is the Dehn surgery of the hyperbolic 3-manifold s H C .The space of complex hyperbolic (4 , , ∞ ) triangle groups (cid:104) I , I , I (cid:105) is parameter-ized by the angular θ ∈ [0 , π/
2) (See Section 3). Let S = I I , T = I I andΓ = (cid:104) S, T (cid:105) . Here S is regular elliptic of order 4, and T is parabolic fixing the pointat infinity. For each group in the parameter space, the Ford domain D is the inter-section of the closures of the exteriors of the isometric spheres for the elements S , S − , S , ( S − T ) and their conjugations by the powers of T . The combination of D is the same except when I I I I is parabolic, in which case there are additionalparabolic fixed points. D is preserved by the subgroup (cid:104) T (cid:105) and is a fundamentaldomain for the cosets of (cid:104) T (cid:105) in Γ. Its ideal boundary ∂ ∞ D is the complement of atubular neighborhood of the T -invariant R -circle (or horotube defined in [28]). Byintersecting ∂ ∞ D with a fundamental domain for (cid:104) T (cid:105) acting on ∂ H C , we obtain afundamental domain for Γ acting on its discontinuity region. See section 4.When I I I I is parabolic, that is θ = π/
3, by studying the combinatorialproperties of the fundamental domain for Γ acting on its discontinuity region Ω(Γ),
YUEPING JIANG, JIEYAN WANG AND BAOHUA XIE we prove that the quotient Ω(Γ) / Γ is homeomorphic to the two-cusped hyperbolic3-manifold s T − S , S T − , ST − S and T − ST − ST , except the point at infinity which isthe fixed point of T . See Section 5. Acknowledgements.
We thank Jiming Ma for his help in the proof of Theorem5.23. The third author also would like to thank Jiming Ma for numerous helpfuldiscussions on complex hyperbolic geometry during his visiting at Fudan University.2.
Background
The purpose of this section is to introduce briefly complex hyperbolic geometry.One can refer to Goldman’s book [13] for more details.2.1.
Complex hyperbolic plane.
Let (cid:104) z , w (cid:105) = w ∗ H z be the Hermitian form on C associated to H , where H is the Hermitian matrix H = . Then C is the union of negative cone V − , null cone V and positive cone V + , where V − = (cid:8) z ∈ C − { } : (cid:104) z , z (cid:105) < (cid:9) ,V = (cid:8) z ∈ C − { } : (cid:104) z , z (cid:105) = 0 (cid:9) ,V + = (cid:8) z ∈ C − { } : (cid:104) z , z (cid:105) > (cid:9) . Definition 2.1.
Let P : C − { } → C P be the projectivization map. Thenthe complex hyperbolic plane H C is defined to be P ( V − ) and its boundary ∂ H C isdefined to be P ( V ). This is the Siegel domain model of H C . Let d ( u, v ) be thedistance between two points u, v ∈ H C . Then the Bergman metric on complexhyperbolic plane is given by the distance formulacosh (cid:18) d ( u, v )2 (cid:19) = (cid:104) u , v (cid:105)(cid:104) v , u (cid:105)(cid:104) u , u (cid:105)(cid:104) v , v (cid:105) , where u , v ∈ C are lifts of u, v .There is another model of H C . Definition 2.2.
The ball model of H C is the unit ball in C , which is given by theHermition matrix J = diag (1 , , − ∂ H C is then the 3-dimensionalsphere S ⊂ C . The Cayley transform C is given by C = 1 √ √ − . It satisfies C ∗ HC = J and interchanges the Siegel domain model and the ball modelof H C .Let N = C × R be the Heisenberg group with product[ z, t ] · [ ζ, ν ] = [ z + ζ, t + ν − zζ )] . PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 5
Then, in the Siegel domain model of H C , the boundary of complex hyperbolic plane ∂ H C can be identified to the union N ∪ { q ∞ } , where q ∞ is the point at infinity.The standard lift of q ∞ and q = [ z, t ] ∈ N in C are(2.1) q ∞ = , q = −| z | + it z . The closure of complex hyperbolic plane H C ∪ ∂ H C can be identified to the unionof N × R ≥ and { q ∞ } . Any point q = ( z, t, u ) ∈ N × R ≥ has the standard lift q = −| z | − u + it z . Here ( z, t, u ) is called the horospherical coordinates of H C ∪ ∂ H C . Definition 2.3.
The
Cygan metric d Cyg on ∂ H C − { q ∞ } is defined to be(2.2) d Cyg ( p, q ) = | (cid:104) p , q (cid:105)| / = (cid:12)(cid:12) | z − w | − i ( t − s + 2Im( z ¯ w )) (cid:12)(cid:12) / , where p = [ z, t ] and q = [ w, s ].The Cygan metric satisfies the properties of a distance. The extended Cyganmetric on H C is given by the formula(2.3) d Cyg ( p, q ) = (cid:12)(cid:12) | z − w | + | u − v | − i ( t − s + 2Im( z ¯ w )) (cid:12)(cid:12) / , where p = ( z, t, u ) and q = ( w, s, v ).The formula d Cyg ( p, q ) = | (cid:104) p , q (cid:105)| / remains valid even if one of p and q lies on ∂ H 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 C : complex lines and Lagrangian planes. Definition 2.4.
Let v ⊥ be the orthogonal space of v ∈ V + with respect to theHermitian form. The intersection of the projective line P ( v ⊥ ) with H C is called a complex line . The vector v is its polar vector .The ideal boundary of a complex line on ∂ H C is called a C -circle . In the Heisen-berg group, C -circles are either vertical lines or ellipses whose projections on the z -plane are circles.Let H R = { ( x , x ) ∈ H C : x , x ∈ R } be the set of real points. H R is aLagrangian plane. All the Lagrangian planes are the images of H R by isometriesof H C . The ideal boundary of a Lagrangian plane is called a R -circle . In theHeisenberg group, R -circles are either straight lines or lemniscate curves whoseprojections on the z -plane are figure eight.2.2. Isometries.
Let SU(2 ,
1) be the special unitary matrix preserving the Her-mitian form. Then the projective unitary group PU(2 ,
1) = SU(2 , / { I , ω I , ω I } is the holomorphic isometry group of H C , where ω = ( − i √ / H C is generated by PU(2 ,
1) and complexconjugation.
YUEPING JIANG, JIEYAN WANG AND BAOHUA XIE
Definition 2.5.
Any isometry g ∈ PU(2 ,
1) is loxodromic if it has exactly two fixedpoints on ∂ H C . g is parabolic if it has exactly one fixed point on ∂ H C . g is elliptic if it has at least one fixed point in H C .The types of isometries can be determined by the traces of their matrix realiza-tions, see Theorem 6.2.4 of Goldman [13]. Now suppose that A ∈ SU(2 ,
1) has realtrace. Then A is elliptic if − ≤ tr(A) <
3. Moreover, A is unipotent if tr(A) = 3.In particular, if tr(A) = − , , A is elliptic of order 2, 3, 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 isgiven by the following formula:(2.4) I C ( z ) = − z + 2 (cid:104) z , n (cid:105)(cid:104) n , n (cid:105) n . 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 , Definition 2.7.
A left
Heisenberg translation associated to [ z, t ] ∈ N is given bythe matrix(2.5) T [ z,t ] = − ¯ z −| z | + it z . It is obvious that T [ z,t ] fixes q ∞ and maps [0 , ∈ N to [ z, t ].2.3. Isometric spheres and Ford polyhedron.
Suppose that g = ( g ij ) i,j =1 ∈ PU(2 ,
1) does not fix q ∞ . Then it is obvious that g (cid:54) = 0. We first recall thedefinition of isometric spheres and relevant properties; see for instance[19]. Definition 2.8.
The isometric sphere of g , denoted by I ( g ), is the set(2.6) I ( g ) = { p ∈ H C ∪ ∂ H C : |(cid:104) p , q ∞ (cid:105)| = |(cid:104) p , g − ( q ∞ ) (cid:105)|} . The isometric sphere I ( g ) is the Cygan sphere with center g − ( q ∞ ) = [ g /g , g /g )]and radius r g = (cid:112) / | g | .The exterior of I ( g ) is the set(2.7) { p ∈ H C ∪ ∂ H C : |(cid:104) p , q ∞ (cid:105)| > |(cid:104) p , g − ( q ∞ ) (cid:105)|} . The interior of I ( g ) is the set(2.8) { p ∈ H C ∪ ∂ H C : |(cid:104) p , q ∞ (cid:105)| < |(cid:104) p , g − ( q ∞ ) (cid:105)|} . The isometric spheres are paired as the following.
Lemma 2.9 ( [13], Section 5.4.5) . Let g be an element in PU(2 , which does notfix q ∞ . Then g maps I ( g ) to I ( g − ) , and the exterior of I ( g ) to the interior of I ( g − ) . Besides, for any unipotent transformation h ∈ PU(2 , fixing q ∞ , we have I ( g ) = I ( hg ) . PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 7
Since isometric spheres are Cygan spheres, we now recall some facts about Cyganspheres. Let S [0 , ( r ) be the Cygan sphere with center [0 ,
0] and radius r >
0. Then(2.9) S [0 , ( r ) = (cid:8) ( z, t, u ) ∈ H C ∪ ∂ H C : ( | z | + u ) + t = r (cid:9) . The geographic coordinates for Cygan sphere will play an important role in ourcalculation; see Section 2.5 of [21].
Definition 2.10.
The geographic coordinates ( α, β, w ) of q = q ( α, β, w ) ∈ S [0 , ( r )is given by the lift(2.10) q = q ( α, β, w ) = − r e − iα / rwe i ( − α/ β ) , where α ∈ [ − π/ , π/ β ∈ [0 , π ) and w ∈ [ − (cid:112) cos( α ) , (cid:112) cos( α )]. The ideal bound-ary of S [0 , ( r ) on ∂ H C are the points with w = ± (cid:112) cos( α ).We are interested in the intersection of Cygan spheres. Proposition 2.11 ( [21], Proposition 2.10) . The intersection of two Cygan spheresis connected.Remark . This intersection is often called
Giraud disk .The following property should be useful to describe the intersection of Cyganspheres.
Proposition 2.13 ([21], Proposition 2.12) . Let S [0 , ( r ) be a Cygan sphere withgeographic coordinates ( α, β, w ) . (1) The level sets of α are complex lines, called slices of S [0 , ( r ) . (2) The level sets of β are Lagrangian planes, called meridians of S [0 , ( r ) . (3) The set of points with w = 0 is the spine of S [0 , ( r ) . It is a geodesiccontained in every meridian. A central work in this paper is to construct a polyhedron for a finitely generatedsubgroup of PU(2 , Definition 2.14.
Let G be a discrete subgroup of PU(2 , Ford polyhedron D G for G is the set D G = (cid:8) p ∈ H C ∪ ∂ H C : |(cid:104) p , q ∞ (cid:105)| ≥ |(cid:104) p , g − q ∞ (cid:105)| for all g ∈ G with g ( q ∞ ) (cid:54) = q ∞ (cid:9) . That is to say D G is the intersection of closures of the exteriors of all the isometricspheres for elements of G which do not fix q ∞ . In fact, Ford polyhedron is the limitof Dirichlet polyhedra as the center point goes to q ∞ .3. The parameter space of complex hyperbolic (4 , , ∞ ) trianglegroups In this section, we give a parameter space of the complex hyperbolic (4 , , ∞ )triangle groups.Let θ ∈ [0 , π/ C , C , C be three complex lines in complex hyperbolicspace H C with polar vectors n , n , n , respectively. By conjugating elements inPU(2 , ∂C = { [ z, ∈ N : | z | = √ } . That is thecircle in the z -plane of the Heisenberg group with center the origin and radius √
2. Then, up to the complex conjugation and rotations about the t -axis of the YUEPING JIANG, JIEYAN WANG AND BAOHUA XIE
Heisenberg group, the C -circles ∂C and ∂C can be normalized to be the sets ∂C = { [ − e − iθ , t ] ∈ N : t ∈ R } and ∂C = { [ e iθ , t ] ∈ N : t ∈ R } . That is, ∂C (respectively ∂C ) is the vertical line whose projection on the z -plane of theHeisenberg group is the point − e − iθ (respectively e iθ ). Thus the polar vectors ofthe complex lines can be written as follows n = e iθ , n = − e − iθ , n = . Note that these two C -circles ∂C and ∂C coincide with each other if θ = π/ I , I and I on the complex lines aregiven as I = − e iθ
20 1 2 e − iθ − , I = − − e − iθ
20 1 − e iθ − ,I = − . Proposition 3.1.
Let θ ∈ [0 , π/ and I , I , I be defined as above. Then (cid:104) I , I , I (cid:105) is a complex hyperbolic (4 , , ∞ ) triangle group. Furthermore, the el-ement I I I I is nonelliptic if and only if ≤ θ ≤ π/ .Proof. By computing the products of two involutions, we have I I = e − iθ − − e iθ − − , I I = − − − e − iθ − e iθ ,I I = − θ ) − e − iθ )0 1 4 cos( θ )0 0 1 . It is easy to verify that I I and I I are elliptic of order 4 and I I is unipotent.Thus (cid:104) I , I , I (cid:105) is a complex hyperbolic (4 , , ∞ ) triangle group.Since the trace of I I I I is tr( I I I I ) = 7 + 8 cos(2 θ ), the element I I I I is elliptic if and only if − ≤ tr( I I I I ) = 7 + 8 cos(2 θ ) < , that is π/ < θ ≤ π/
2. Thus I I I I is nonelliptic if and only if 0 ≤ θ ≤ π/ θ = π/
3, the element I I I I is parabolic. (cid:3) If θ = 0, then (cid:104) I , I , I (cid:105) preserves the Lagrangian plane whose ideal boundary isthe x -axis of the Heisenberg group. Thus (cid:104) I , I , I (cid:105) is obviously a discrete subgroupof PU(2 , θ = π/
3, we have the following corollary.
Corollary 3.2.
Let θ = π/ . Then (cid:104) I , I , I (cid:105) is a discrete subgroup of PU(2 , .Proof. Let θ = π/
3. Then I I = − i √ − − − i √ − − , I I = − − − i √ − i √ , PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 9 I I = − − i √
30 1 20 0 1 . Observe that I I and I I are contained in the Eisenstein-Picard modular groupPU(2 , Z [ ω ]), where ω = ( − i √ / , Z [ ω ]) is discrete, (cid:104) I I , I I (cid:105) is also discrete. Therefore, (cid:104) I , I , I (cid:105) is dis-crete. (cid:3) The Ford domain
For θ ∈ [0 , π/ S = I I and T = I I . Then Γ = (cid:104) S, T (cid:105) is a subgroup of (cid:104) I , I , I (cid:105) of index two. In this section, we will mainly prove that Γ is discrete.Our method is as follows: firstly, construct a candidate Ford domain D (see Defi-nition 4.12), then applying the Poincar´e polyhedron theorem to show that D is afundamental domain for the cosets of (cid:104) T (cid:105) in Γ, and as well Γ is discrete. Definition 4.1.
For k ∈ Z , let • I + k be the isometric sphere I ( T k ST − k ) = T k I ( S ), • I − k be the isometric sphere I ( T k S − T − k ) = T k I ( S − ), • I (cid:63)k be the isometric sphere I ( T k S T − k ) = T k I ( S ), • I (cid:5) k be the isometric sphere I ( T k ( S − T ) T − k ) = T k I (( S − T ) ).Note that S and S − T both have order 4, so S = S − , ( S − T ) = ( S − T ) − .The centers and radii of the isometric spheres I + k , I − k , I (cid:63)k and I (cid:5) k are listed in thefollowing table. Isometric sphere Center radius I + k [4 k cos( θ ) , k sin(2 θ )] √ I − k [4 k cos( θ ) + 2 e iθ , √ I (cid:63)k [4 k cos( θ ) + e iθ , k sin(2 θ )] 1 I (cid:5) k [4 k cos( θ ) − e − iθ , k sin(2 θ )] 1 Proposition 4.2.
Let k ∈ Z . (1) There is an antiholomophic involution τ such that τ ( I + k ) = I + − k , τ ( I − k ) = I −− k − and τ ( I (cid:63)k ) = I (cid:5)− k . (2) The complex involution I interchanges I (cid:63)k and I (cid:63) − k , interchanges I + k and I −− k , and interchanges I (cid:5) k and I (cid:5)− k +1 .Proof. (1) Let τ : C −→ C be given as follows: τ : z z z (cid:55)−→ ¯ z − ¯ z ¯ z . Then τ is the identity. It is easy to see that τ fixes the polar vector n , andinterchanges the polar vectors n and n . Thus τ conjugates I to itself, I to I and vice versa. Therefore τ conjugates T to T − , S to T − S , S − to S − T , and S to ( T − S ) = ( S − T ) . This implies that τ ( I + k ) = I + − k , τ ( I − k ) = I −− k − and τ ( I (cid:63)k ) = I (cid:5)− k . (2) The statement is easily obtained by the facts I SI = S − and I T I = T − . (cid:3) Before we consider the intersections of two isometric spheres, we would like togive a useful technical lemma. Suppose that q ∈ I +0 . Then by (2.10) the geographiccoordinates of q = q ( α, β, w ) is given by the lift(4.1) q = q ( α, β, w ) = − e − iα √ we i ( − α/ β ) where α ∈ [ − π/ , π/ β ∈ [0 , π ) and w ∈ [ − (cid:112) cos( α ) , (cid:112) cos( α )]. Definition 4.3.
Let ( α, β, w ) be the geographic coordinates of I +0 . We define thefollowing functions. f (cid:63) ( α, β, w ) = 2 w + 1 + cos( α ) − √ w cos( − α/ β − θ ) − √ w cos( α/ β − θ ) ,f − ( α, β, w ) = 2 w + 1 + cos( α ) − √ w cos( α/ β − θ ) − √ w cos( − α/ β − θ ) ,f −− ( α, β, w ) = 2 w + 1 + cos( α ) + √ w cos( α/ β + θ ) + 2 √ w cos( − α/ β + θ ) . Lemma 4.4.
Suppose that θ ∈ [0 , π/ . Let f (cid:63) ( α, β, w ) , f − ( α, β, w ) and f −− ( α, β, w ) be the functions defined in Definition 4.3. Suppose that q ∈ I +0 . Then we have thefollowing properties. (1) q lies on I (cid:63) (resp. in its interior or exterior) if and only if f (cid:63) ( α, β, w ) = 0 (resp. negative or positive); (2) q lies on I − (resp. in its interior or exterior) if and only if f − ( α, β, w ) = 0 (resp. negative or positive); (3) q lies on I −− (resp. in its interior or exterior) if and only if f −− ( α, β, w ) = 0 (resp. negative or positive).Proof. (1) Any point q ∈ I +0 lies on I (cid:63) (resp. in its interior or exterior) if andonly if the Cygan distance between q and the center of I (cid:63) is 1 (resp. less than 1or greater than 1). Using 4.1, the difference between the Cygan distance from q tothe center of I (cid:63) and 1 is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:42) q , − / e iθ (cid:43)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −
1= 4 (cid:12)(cid:12)(cid:12) − e − iα + √ we i ( − α/ β − θ ) − / (cid:12)(cid:12)(cid:12) −
1= 4 (cid:16) w + 1 + cos( α ) − √ w cos( − α/ β − θ ) − √ w cos( α/ β − θ ) (cid:17) = 4 f (cid:63) ( α, β, w ) . Hence, q lies on I (cid:63) (resp. in its interior or exterior) if and only if f (cid:63) ( α, β, w ) = 0(resp. negative or positive). PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 11 (2) Similarly, the difference between the Cygan distance from q to the center of I − and its radius is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:42) q , − e iθ (cid:43)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −
4= 4 (cid:12)(cid:12)(cid:12) − e − iα + 2 √ we i ( − α/ β − θ ) − (cid:12)(cid:12)(cid:12) −
4= 16 (cid:16) w + 1 + cos( α ) − √ w cos( α/ β − θ ) − √ w cos( − α/ β − θ ) (cid:17) = 16 f − ( α, β, w ) . (3) Similarly, the difference between the Cygan distance from q to the center of I −− and its radius is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:42) q , − − e − iθ (cid:43)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −
4= 4 (cid:12)(cid:12)(cid:12) − e − iα − √ we i ( − α/ β + θ ) − (cid:12)(cid:12)(cid:12) −
4= 16 (cid:16) w + 1 + cos( α ) + √ w cos( α/ β + θ ) + 2 √ w cos( − α/ β + θ ) (cid:17) = 16 f −− ( α, β, w ) . (cid:3) Now, we begin to study the intersections of isometric spheres.
Proposition 4.5.
Suppose that θ ∈ [0 , π/ , then each pair of the isometric spheresin { I + k : k ∈ Z } are disjoint in H C ∪ ∂ H C .Proof. It suffices to show that I +0 and I + k are disjoint for | k | ≥
1. Observe that T is a Heisenberg translation associated with [4 cos( θ ) , θ )]. Since the isometricsphere I +0 has center [0 ,
0] and radius √
2, the isometric sphere I + k has center[4 k cos( θ ) , k sin(2 θ )] and radius √
2. According to the Cygan metric given in (2.2),the Cygan distance between the centers of I +0 and I + k is4 (cid:112) | k | cos( θ ) | k cos( θ ) − i sin( θ ) | / ≥ (cid:112) cos( θ ) ≥ √ . Thus the Cygan distance between the centers of I +0 and I + k is bigger than the sumof the radii, except when k = ± θ = π/
3. This implies that I +0 and I + k aredisjoint for all | k | ≥ k = ± θ = π/
3, although the Cygan distance between the centers of I +0 and I + ± is the sum of the radii, we claim that they are still disjoint. Using thesymmetry τ in Proposition 4.2, we only need to show that I +0 ∩ I +1 = ∅ . Supposethat q ∈ I +0 . Using the geographic coordinates of q = q ( α, β, w ) given in (4.1), wecan compute the difference between the Cygan distance of q and the center of I +12 YUEPING JIANG, JIEYAN WANG AND BAOHUA XIE with its radius. That is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:42) q , a − e − iπ/ (cid:43)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −
4= 4 (cid:12)(cid:12)(cid:12) − e − iα + √ we i ( − α/ β − θ ) − e iπ/ (cid:12)(cid:12)(cid:12) −
4= 32 (cid:16) w + √ w (cos( α/ β ) / α/ − β + π/ α + π/
3) + 2 (cid:17) = 32 f ( α, β, w ) . Here f ( α, β, w ) can be seen as a quadratic function of w . Let B = √ α/ β ) / α/ − β + π/ C = cos( α + π/
3) + 2. If B − C <
0, then it is obvious that f ( α, β, w ) > B − C ≥
0, then B ≤ − √ C (it is impossible by numerically computation) or B ≥ √ C . In this case we have B − (cid:112) cos( α ) ≥ B − √ C ≥ α ) ≤ C .This means that the symmetry axes of the f lie on the left side of w = − (cid:112) cos( α ).Besides, one can compute numerically that f ( α, β, − (cid:112) cos( α )) > α and β . So, we have f ( α, β, w ) >
0. This means that every point on I +0 liesoutside of I +1 . Hence I +0 ∩ I +1 = ∅ . (cid:3) By a similar argument, we have the following proposition.
Proposition 4.6.
Suppose that θ ∈ [0 , π/ , then (1) I +0 and I − k are disjoint in H C ∪ ∂ H C , except possibly when k = − , , (2) I +0 and I (cid:63)k are disjoint in H C ∪ ∂ H C , except possibly when k = − , , (3) I +0 and I (cid:5) k are disjoint in H C ∪ ∂ H C , except possibly when k = 0 , .Proof. (1) Since I − has center [2 e iθ ,
0] and radius √
2, the isometric sphere I − k hascenter [4 k cos( θ ) + 2 e iθ ,
0] and radius √
2. The Cygan distance between the centersof I +0 and I − k is (cid:12)(cid:12) k cos( θ ) + 2 e iθ (cid:12)(cid:12) = 2 (cid:113) sin ( θ ) + (2 k + 1) cos ( θ ) , which is bigger than 2 √ k (cid:54) = − , I (cid:63) has center [ e iθ ,
0] and radius 1, the isometric sphere I (cid:63)k has center[4 k cos( θ ) + e iθ , k sin(2 θ )] and radius 1. The Cygan distance between the centersof I +0 and I (cid:63)k is (cid:12)(cid:12) | (4 k + 1) cos( θ ) + i sin( θ ) | − i (4 k sin(2 θ )) (cid:12)(cid:12) / ≥ | k + 1 | cos( θ ) , which is bigger than 1 + √ k (cid:54) = − , I (cid:5) has center [ − e − iθ ,
0] and radius 1, the isometric sphere I (cid:5) k has center[4 k cos( θ ) − e − iθ , k sin(2 θ )] and radius 1. The Cygan distance between the centersof I +0 and I (cid:5) k is (cid:12)(cid:12) | (4 k −
1) cos( θ ) + i sin( θ ) | − i (4 k sin(2 θ )) (cid:12)(cid:12) / ≥ | k − | cos( θ ) , which is bigger than 1 + √ k (cid:54) = 0 , (cid:3) Similarly, we have
Proposition 4.7.
Suppose that θ ∈ [0 , π/ , then PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 13 (1) I (cid:63) and I (cid:63)k are disjoint in H C . Furthermore, when θ = π/ the closure of I (cid:63) and I (cid:63) − (respectively, I (cid:63) ) is tangent at the parabolic fixed point of T − S (respectively, T ( T − S ) T − ) on ∂ H C . (2) I (cid:63) and I (cid:5) k are disjoint in H C ∪ ∂ H C , except possibly when k = 0 , .Proof. (1) I (cid:63)k is a Cygan sphere with center [4 k cos( θ ) + e iθ , k sin(2 θ )] and radius1, thus the distance between the centers of I (cid:63) and I (cid:63)k is d Cyg ([4 k cos( θ ) + e iθ , k sin(2 θ )] , [ e iθ , | k cos( θ ) | ≥ . The equality holds when k = ± θ = π/ θ = π/
3, we have known that I I I I = T − S is unipotent. Since I (cid:63) = I ( S ) = I ( T − S ) , and I (cid:63) − = I ( T − S T ) = I ( S T ) = I ( S − T ) , using Phillips’s theorem (Theorem 6.1 of [22]), the closure of I (cid:63) and I (cid:63) − will betangent at the fixed point of T − S on ∂ H C . Since I (cid:63) = T ( I (cid:63) − ) , I (cid:63) = T ( I (cid:63) ) , the closure of I (cid:63) and I (cid:63) is tangent at the fixed point of T ( T − S ) T − on ∂ H C .(2) The distance between the centers of I (cid:63) and I (cid:5) k is d Cyg ([4 k cos( θ ) − e − iθ , k sin(2 θ )] , [ e iθ , (cid:112) cos( θ ) · | (2 k − cos( θ ) − i sin( θ ) | / ≥ | k − | cos( θ ) , which is bigger than 2, except when k = 0 , (cid:3) Lemma 4.8.
Suppose that θ ∈ [0 , π/ , then I +0 ∩I (cid:63) ∩I −− = ∅ except when θ = π/ ,in which case the triple intersection is the point [ e i π/ , −√ ∈ ∂ H C . Moreover,this point is the parabolic fixed point of T − S .Proof. Suppose that q ∈ I +0 . Using Lemma 4.4, the geographic coordinates ( α, β, w )of q ∈ I +0 ∩ I (cid:63) ∩ I −− should satisfy the following equation(4.2) 2 w + 1 + cos( α ) − √ w cos (cid:16) − α β − θ (cid:17) − √ w cos (cid:16) α β − θ (cid:17) = 0 , (4.3) 2 w + 1 + cos( α ) + √ w cos (cid:16) α β + θ (cid:17) + 2 √ w cos (cid:16) − α β + θ (cid:17) = 0 . Subtracting the two equations (4.2) and (4.3), we have2 √ w cos( β ) (cos( α/ θ ) + 2 cos( α/ − θ )) = 0 . This implies that either w = 0 or β = π/
2, since (cos( α/ θ ) + 2 cos( α/ − θ )) (cid:54) =0 for θ ∈ [0 , π/ w = 0 lie in the meridian with β = π/
2. Therefore, a necessary condition for q ∈ I +0 ∩ I (cid:63) ∩ I −− is that β = π/ β = π/ w + 2 cos ( α/
2) + √ w (sin( α/
2) cos( θ ) − α/
2) sin( θ )) = 0 . Let b ( α, θ ) = sin( α/
2) cos( θ ) − α/
2) sin( θ ). It is easy to see that for every α ,the function θ (cid:55)−→ b ( α, θ ) is decreasing on [0 , π/ w with positive leading coefficient. Thus the equation (4.4) has at least one I − I +0 I (cid:63) I − I +0 I (cid:63) Figure 1.
The ideal boundaries of the three spheres I +0 , I −− and I (cid:63) on ∂ H C (On the left is the case when θ = 0 and on the right is θ = π/ b − ( α/ ≥
0, that is b ≥ √ α/
2) (it is impossible since b ≤ b ( α,
0) = sin( α/ b ≤ − √ α/ (cid:112) cos( α ) ≤ cos( α/ b + 2 √ (cid:112) cos( α ) ≤ b + 2 √ α/ ≤
0. This means that the symmetry axes ofthe quadratic function lie on the right hand side of w = (cid:112) cos( α ).Besides, one can compute that b (cid:112) cos( α ) + √ (cid:0) cos( α ) + cos ( α/ (cid:1) ≥ (sin( α/
2) cos( π/ − α/
2) sin( π/ (cid:112) cos( α ) + √ (cid:0) cos( α ) + cos ( α/ (cid:1) = √ (cid:32) (cid:112) cos( α )2 + √
22 sin( α/ (cid:33) + 3 √ (cid:32) √ (cid:112) cos( α ) − √
22 cos( α/ (cid:33) . Then b (cid:112) cos( α ) + √ (cid:0) cos( α ) + cos ( α/ (cid:1) ≥
0. If it is 0, then α = − π/ θ = π/
3. It means that for w ∈ [ − (cid:112) cos( α ) , (cid:112) cos( α )] the equation (4.4) has nosolution except when θ = π/ α = − π/
3, in which case w = (cid:112) cos( α ) = √ / q ∈ I +0 ∩ I (cid:63) ∩ I −− if and only if θ = π/ α = − π/ w = (cid:112) cos( α ) = √ /
2. When θ = π/ T − S is unipotent and its fixed point is the eigenvectorwith eigenvalue 1. One can compute that its fixed point is [ e i π/ , −√ ∈ ∂ H C ,which equals to the point q ( − π/ , π/ , √ / (cid:3) Proposition 4.9.
Suppose that θ ∈ [0 , π/ , then (1) The intersection I (cid:63) ∩ I (cid:5) lie in the interior of I +0 , (2) The intersection I (cid:63) ∩ I −− either is empty or lie in the interior of I +0 .Furthermore, when θ = π/ , there is a unique point on the ideal boundaryof I (cid:63) ∩ I −− on ∂ H C , which is fixed by T − S , lying on the ideal boundaryof I +0 .Proof. (1) Let p = ( z, t, u ) ∈ I (cid:63) ∩ I (cid:5) , then p satisfies the equations (cid:12)(cid:12) | z − e iθ | + u − i (cid:0) t + 2Im( ze − iθ ) (cid:1)(cid:12)(cid:12) = 1 (cid:12)(cid:12) | z + e − iθ | + u − i (cid:0) t + 2Im( − ze iθ ) (cid:1)(cid:12)(cid:12) = 1 . PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 15
Set z = | z | e iφ , by simplifying, we have (cid:12)(cid:12) | z | + 1 + u − | z | cos( φ − θ ) − i ( t + 2 | z | sin( φ − θ )) (cid:12)(cid:12) = 1 (cid:12)(cid:12) | z | + 1 + u + 2 | z | cos( φ + θ ) − i ( t − | z | sin( φ + θ )) (cid:12)(cid:12) = 1 . Now set | z | + 1 + u − | z | cos( φ − θ ) − i ( t + 2 | z | sin( φ − θ )) = e iα (4.5) | z | + 1 + u + 2 | z | cos( φ + θ ) − i ( t − | z | sin( φ + θ )) = e iβ . (4.6)Sincecos α = | z | + 1 + u − | z | cos( φ − θ ) = ( | z | − cos( φ − θ )) + sin ( φ − θ ) + u ≥ β = | z | + 1 + u + 2 | z | cos( φ + θ ) = ( | z | + cos( φ + θ )) + sin ( φ + θ ) + u ≥ , we have − π/ ≤ α ≤ π/ − π/ ≤ β ≤ π/
2. Thus it implies that cos( β/ − α/ ≥
0. By computing the difference of the equations (4.5) and (4.6), we have(4.7) z = e iβ − e iα θ ) = ± sin( β/ − α/ θ ) e i ( ± π/ β/ α/ . Thus φ = ± π/ β/ α/
2. Therefore,( | z | + u ) + t = (cos( α ) − | z | cos( φ − θ )) + (sin( α ) + 2 | z | sin( φ − θ )) = 2 + 4 | z | − α ) − | z | cos( φ − θ ) + 4 | z | cos( φ − θ − α ) ≤ | z | − (cid:0) | z | + 1 − | z | cos( φ − θ ) (cid:1) − | z | cos( φ − θ ) + 4 | z | cos( φ − θ − α )= 2 | z | + 4 | z | cos( φ − θ − α )= 2 | z | + 4 | z | cos( ± π/ − θ + β/ − α/ | z | + 4 | z | ( ± sin( θ ) cos( β/ − α/ ∓ cos( θ ) sin( β/ − α/ ( β/ − α/ ( θ ) + tan( θ ) sin( β − α ) − ( β/ − α/ . Since θ ∈ [0 , π/ sin ( β/ − α/ ( θ ) ≤ ( β/ − α/
2) and tan( θ ) sin( β − α ) ≤√
3. This implies that ( | z | + u ) + t <
4. It means that the intersection I (cid:63) ∩ I (cid:5) lie in the interior of I +0 .(2) Suppose that p = ( z, t, u ) ∈ I (cid:63) ∩ I −− , then p satisfies the equations1 = (cid:12)(cid:12) | z − e iθ | + u − i (cid:0) t + 2Im( ze − iθ ) (cid:1)(cid:12)(cid:12) = (cid:12)(cid:12) | z | + u + 1 − it − ze − iθ (cid:12)(cid:12) (cid:12)(cid:12) | z + 2 e − iθ | + u − i (cid:0) t + 2Im( − ze iθ ) (cid:1)(cid:12)(cid:12) = (cid:12)(cid:12) | z | + u + 4 − it + 4 ze iθ (cid:12)(cid:12) . Now set | z | + u + 1 − it − ze − iθ = e iβ (4.8) | z | + u + 4 − it + 4 ze iθ = 2 e iα . (4.9)By computing the difference of the equations (4.8) and (4.9), we have z = 2 e iα − e iβ − e iθ + 2 e − iθ . (4.10) I (cid:63) I (cid:63) I +0 I +0 I − I − Figure 2.
The ideal boundaries of the three spheres I +0 , I − and I (cid:63) on ∂ H C (On the left is the case when θ = 0 and on the right is θ = π/ u = cos( β ) − (cid:12)(cid:12) ze − iθ − (cid:12)(cid:12) (4.11) t = − sin( β ) − ze − iθ ) . (4.12)Since cos β = u + (cid:12)(cid:12) ze − iθ − (cid:12)(cid:12) ≥ α = u + (cid:12)(cid:12) ze iθ + 2 (cid:12)(cid:12) ≥ , we have − π/ ≤ α ≤ π/ − π/ ≤ β ≤ π/ θ = π/
3. Substituting α = π/ β = 0into the equations (4.10), (4.11) and (4.12), we obtain the point p = (cid:32)(cid:32) √ − (cid:33) + i √ , − √ , −
13 + 8 √ (cid:33) ∈ I (cid:63) ∩ I −− . One can compute that p lies in the interior of I +0 , since (cid:12)(cid:12) | z | + u − it (cid:12)(cid:12) = (cid:12)(cid:12) e iβ − ze − iθ (cid:12)(cid:12) = 4 | z | = 20 / − √ < . We know that the intersection I (cid:63) ∩ I −− is connected from Proposition 2.11. Thus,according to Lemma 4.8, I (cid:63) ∩ I −− lie in the interior of I +0 except the point[ e i π/ , −√ I +0 . See Figure 1.Observe that coordinates of the centers of I (cid:63) and I −− are continues on θ . Thusthe geometric positions of the spheres I (cid:63) and I −− are continuously moved on θ .When θ = 0, since the Cygan distance between the centers of I (cid:63) and I −− is biggerthan the sum of their radii, one can see that I (cid:63) ∩ I −− = ∅ . When θ = π/
3, we haveshown that I (cid:63) ∩ I −− lies in the interior of I +0 . We also have I +0 ∩ I (cid:63) ∩ I −− = ∅ for θ ∈ [0 , π/
3) by Lemma 4.8. Hence, the intersection I (cid:63) ∩ I −− is either empty orcontained in the interior of I +0 . (cid:3) PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 17
Proposition 4.10.
Suppose that θ ∈ [0 , π/ . For k ∈ Z , the three isometricspheres I + k , I − k , I (cid:63)k (respectively, I + k , I − k − , I (cid:5) k ) have the following properties. • The intersections I + k ∩ I − k , I − k ∩ I (cid:63)k , and I (cid:63)k ∩ I + k (respectively, I + k ∩ I − k − , I − k − ∩ I (cid:5) k , I (cid:5) k ∩ I + k ) are discs. • The intersection I + k ∩ I − k ∩ I (cid:63)k (respectively, I + k ∩ I − k − ∩ I (cid:5) k ) is a union oftwo geodesics which are crossed at the fixed point of T k ST − k (respectively T k ( S − T ) T − k ) in H C and whose fours endpoints are on ∂ H C . Moreover,the four rays from the fixed point to the four endpoints are cyclical permutedby T k ST − k (respectively T k ( S − T ) T − k ).Proof. According to Definition 4.1 and Proposition 4.2, it suffices to consider theisometric spheres I +0 , I − , I (cid:63) . See Figure 2.Let q ∈ I +0 . Consider the geographic coordinates ( α, β, w ) of q in (4.1). ByLemma 4.4, if q lies on I +0 ∩ I − , then the α, β, w should satisfy the equation(4.13) 2 w + 1 + cos( α ) − √ w cos( α/ β − θ ) − √ w cos( − α/ β − θ ) = 0 . Similarly, if q lies on I +0 ∩ I (cid:63) , then the α, β, w should satisfy the equation(4.14) 2 w + 1 + cos( α ) − √ w cos( − α/ β − θ ) − √ w cos( α/ β − θ ) = 0Thus the intersection I +0 ∩ I − is the set of solutions of equation (4.13) and theintersection I +0 ∩ I (cid:63) is the set of solutions of equation (4.14). One can easily verifythat the geographic coordinate of the point q (0 , θ, √ / ∈ H C satisfies the equa-tions (4.13) and (4.14), so these intersections I +0 ∩ I − and I +0 ∩ I (cid:63) are topologicaldiscs from Proposition 2.11.The intersection of these two sets gives the triple intersection I +0 ∩ I − ∩ I (cid:63) .Now let us solve the system of the equations (4.13) and (4.14). Let t = β − θ .Subtracting the equations (4.13) and (4.14) and simplifying, we obtain2 w sin( α/
2) sin( t ) = 0 . Thus w = 0 (this is impossible), or α = 0, or t = 0. If t = 0, then setting β = θ inequation (4.13), we get2 w − √ (cid:16) α (cid:17) w + 1 + cos( α ) = 2 (cid:32) w − √
22 cos (cid:16) α (cid:17)(cid:33) (cid:16) w − √ (cid:16) α (cid:17)(cid:17) = 0 . Note that the solutions of the above equation for w should satisfy w ≤ cos( α ).Thus w = √
22 cos (cid:16) α (cid:17) with cos( α ) ≥ . If α = 0, then equation (4.13) becomes to2 w − √ t ) w + 2 = 0 . (4.15)Note that the solutions of equation (4.15) for w should satisfy w ≤ cos( α ) = 1.Thus the solutions of equation (4.15) are w = 3 cos( t ) − (cid:112) ( t ) − √ √ ≤ cos( t ) ≤ , and w = 3 cos( t ) + (cid:112) ( t ) − √ − ≤ cos( t ) ≤ − √ . C C L Figure 3.
The triple intersection I +0 ∩ I − ∩ I (cid:63) viewed on thevertical axis in the ball model of H C . The blue curve is C and thered one is C . The black point on the curve is the projection of L on the complex line. The left curve is the case when θ = 0 and theone on the lower left is the case when θ = π/ I +0 ∩ I − ∩ I (cid:63) is the union L ∪ C ∪ C , where L = (cid:40) q ( α, t + θ, w ) ∈ I +0 : cos( α ) ≥ , t = 0 , w = √
22 cos (cid:16) α (cid:17)(cid:41) , C = (cid:40) q (0 , t + θ, w ) ∈ I +0 : 2 √ ≤ cos( t ) ≤ , w = 3 cos( t ) − (cid:112) ( t ) − √ (cid:41) , and C = (cid:40) q (0 , t + θ, w ) ∈ I +0 : − ≤ cos( t ) ≤ − √ , w = 3 cos( t ) + (cid:112) ( t ) − √ (cid:41) . Note that L lie in a Lagrangian plane of I +0 , and C ∪ C lie in a complex line of I +0 . It is obvious that C is an arc. One of its endpoints is q (0 , θ, √ / ∈ H C , whichis the fixed point of S . The other one is q (0 , arccos(2 √ / θ, ∈ ∂ H C . Similarly, C is an arc whose endpoints are q (0 , θ, √ /
2) and q (0 , arccos( − √ /
3) + θ, − ∈ ∂ H C . Thus C ∪ C is connected. The endpoints of L are q (arccos(1 / , θ, √ / q ( − arccos(1 / , θ, √ / ∂ H C . It is easy to see that L intersectswith C ∪ C at the point q (0 , θ, √ / ∈ H C .Moreover, C ∪ C is a geodesic. In fact, the complex line containing C ∪ C is C = { ( − , z ) ∈ H C ∪ ∂ H C : | z | ≤ √ } . It is a disc bounded by the circle withcenter being the origin and radius √
2. While C ∪ C lie in the circle with center3 e iθ / / C is mapped to the vertical axis { (0 , z ) ∈ H C ∪ ∂ H C : | z | ≤ } in the ball model of H C . Thus C is isometric to thePoincar´e disc. While C ∪C is mapped by the Cayley transform to an arc containedin the circle with center − e iθ / √ / √ C ∪ C is a geodesic. See Figure 3. PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 19
By the Cayley transform, L is mapped to { ( − tan( α/ i, − e iθ / √ ∈ H C ∪ ∂ H C :cos( α ) ≥ / } . Thus L and C ∪ C are crossed at the point (0 , − e iθ / √ ∈ H C which is the image of q (0 , θ, √ /
2) under the Cayley transform.All the five points are lifted to the vectors in C : q (arccos(1 / , θ, √ /
3) = − (cid:16) − i √ (cid:17)(cid:16) − i √ (cid:17) e iθ , q ( − arccos(1 / , θ, √ /
3) = − (cid:16) + i √ (cid:17)(cid:16) + i √ (cid:17) e iθ , q (0 , arccos(2 √ /
3) + θ,
1) = − (cid:16) + i √ (cid:17) e iθ , q (0 , arccos( − √ /
3) + θ, −
1) = − (cid:16) − i √ (cid:17) e iθ , q (0 , θ, √ /
2) = − e iθ . Recall that S = e − iθ − − e iθ − − . Thus it is easy to check that S ( q (0 , θ, √ / q (0 , θ, √ /
2) and the other fourpoints are cyclical permuted by S as the following − (cid:16) + i √ (cid:17) e iθ S −→ − (cid:16) − i √ (cid:17)(cid:16) − i √ (cid:17) e iθ S −→ − (cid:16) − i √ (cid:17) e iθ S −→ − (cid:16) + i √ (cid:17)(cid:16) + i √ (cid:17) e iθ . Moreover, it is easy to verify that C ∪C = S ( L ), S ( L ) = L and S ( C ) = C .Thus, the four rays from the fixed point to the four endpoints are cyclical permutedby S . (cid:3) By applying powers of T and the symmetries in Proposition 4.2 to Proposition4.6, Proposition 4.7 and Proposition 4.9, all pairwise intersections of the isometricspheres can be summarized in the following result. Corollary 4.11.
Suppose that θ ∈ [0 , π/ . Let S = {I ± k , I (cid:63)k , I (cid:5) k : k ∈ Z } be the setof all the isometrical spheres. Then for all k ∈ Z : (1) I + k is contained in the exterior of all the isometric spheres in S except I − k , I − k − , I (cid:63)k , I (cid:63)k − , I (cid:5) k and I (cid:5) k +1 . Moreover, I + k ∩ I (cid:63)k − (resp. I + k ∩ I (cid:5) k +1 ) iseither empty or contained in the interior of I − k − (resp. I − k ). When θ = π/ , I + k ∩ I (cid:63)k − (resp. I + k ∩ I (cid:5) k +1 ) will be tangent with I − k − (resp. I − k ) on ∂ H C at the parabolic fixed point of T k ( S T ) T − k (resp. T k ( S − T S − ) T − k ). (2) I − k is contained in the exterior of all the isometric spheres in S except I + k , I + k +1 , I (cid:63)k , I (cid:63)k +1 , I (cid:5) k and I (cid:5) k +1 . Moreover, I − k ∩ I (cid:5) k (resp. I − k ∩ I (cid:63)k +1 )is either empty or contained in the interior of I + k (resp. I + k +1 ). When θ = π/ , I − k ∩ I (cid:5) k (resp. I − k ∩ I (cid:63)k +1 ) will be tangent with I + k (resp. I + k +1 ) on ∂ H C at the parabolic fixed point of T k ( ST − S ) T − k (resp. T k ( S T − ) T − k ). (3) I (cid:63)k is contained in the exterior of all the isometric spheres in S except I ± k , I + k +1 , I − k − , I (cid:5) k and I (cid:5) k +1 . Moreover, I (cid:63)k ∩ I (cid:5) k (resp. I (cid:63)k ∩ I (cid:5) k +1 ) is containedin the interior of I + k (resp. I − k ). I (cid:63)k ∩ I + k +1 is described in item (1), and I (cid:63)k ∩ I − k − is described in item (2). When θ = π/ , I (cid:63)k will be tangent with I (cid:63)k +1 (resp. I (cid:63)k − ) on ∂ H C at the parabolic fixed point of T k ( S T − ) T − k (resp. T k ( T − S ) T − k ). (4) I (cid:5) k is contained in the exterior of all the isometric spheres in S except I ± k , I ± k − , I (cid:63)k and I (cid:63)k − . Moreover, I (cid:5) k ∩ I (cid:63)k and I (cid:5) k ∩ I (cid:63)k − are described in item(3). I (cid:5) k ∩ I − k is described in item (2) and I (cid:5) k ∩ I + k − is described in item(1). When θ = π/ , I (cid:5) k will be tangent with I (cid:5) k +1 (resp. I (cid:5) k − ) on ∂ H C atthe parabolic fixed point of T k ( T − S ) T − k (resp. T k ( S T − ) T − k ). Definition 4.12.
Let D be the intersection of the closures of the exteriors of allthe isometric spheres I + k , I − k , I (cid:63)k and I (cid:5) k , for k ∈ Z . Definition 4.13.
For k ∈ Z , let s + k (respectively, s − k , s (cid:63)k , and s (cid:5) k ) be the side of D contained in the isometric sphere I + k (respectively, I − k , I (cid:63)k and I (cid:5) k ). Definition 4.14. A ridge is defined to be the 2-dimensional connected intersectionsof two sides.By Corollary 4.11, the ridges are s + k ∩ s − k , s + k ∩ s (cid:63)k , s + k ∩ s − k − , s + k ∩ s (cid:5) k , s − k ∩ s (cid:63)k and s − k − ∩ s (cid:5) k for k ∈ Z , and the sides and ridges are related as follows: • The side s + k is bounded by the ridges s + k ∩ s − k , s + k ∩ s (cid:63)k , s + k ∩ s − k − and s + k ∩ s (cid:5) k . • The side s − k is bounded by the ridges s + k ∩ s − k , s − k ∩ s (cid:63)k , s − k ∩ s + k +1 and s − ∩ s (cid:5) k +1 . • The side s (cid:63)k is bounded by the ridges s + k ∩ s (cid:63)k and s − k ∩ s (cid:63)k . • The side s (cid:5) k is bounded by the ridges s (cid:5) k ∩ s − k − and s + k ∩ s (cid:5) k . Proposition 4.15.
The ridges s + k ∩ s − k , s + k ∩ s (cid:63)k , s + k ∩ s − k − , s + k ∩ s (cid:5) k , s − k ∩ s (cid:63)k and s − k − ∩ s (cid:5) k for k ∈ Z are all topologically the union of two sectors.Proof. The ridge s + k ∩ s − k is contained in I + k ∩ I − k . According to Proposition 4.10, I + k ∩I − k is topologically a disc and I + k ∩I − k ∩I (cid:63)k is the union of two crossed geodesics.The two crossed geodesics divide the disc into four sectors and one opposite pair ofwhich will lie in the interior of the isometric sphere I (cid:63)k . Thus s + k ∩ s − k is the otheropposite pair of the four sectors in the disc. More precisely, up to the powers of T , let us consider s +0 ∩ s − . Let ∆ be the disc I +0 ∩ I − described in the equation(4.13) and the two crossed geodesics L ∪ C ∪ C are described in Proposition 4.10. PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 21
By Proposition 4.2, the complex involution I preserves ∆ and L ∪ C ∪ C . Recallthat I fixing the complex line C with polar vector n described in Section 3. Onecan compute that the intersection C ∩ ∆ is the curve(4.16) C ∩ ∆ = { q ( α, α/ θ, √ / ∈ I +0 : cos( α ) ≥ / } . Of course C ∩ ∆ intersects with L ∪ C ∪ C at the fixed point of S , and divides∆ into two parts. I fixes C ∩ ∆, and interchanges L and C ∪ C . Thus C ∩ ∆is contained in the union of two opposite sectors. By Lemma 4.4, C ∩ ∆ lie onthe closure of the exterior of I (cid:63) , since f (cid:63) ( α, α/ θ, √ /
2) = 1 − cos( α ) ≥ C ∩ ∆ is the ridge s +0 ∩ s − .Moreover, this ridge is preserved by I . By using the parametrization of the Girauddisk in [8], it can be instructive to draw figure of the Giraud disk I +0 ∩ I − and theirintersection with the isometric spheres I −− , I (cid:63) , I (cid:63) − , I (cid:5) and I (cid:5) . See Figure 4.The other ridges can be described by a similar argument. (cid:3) Figure 4.
The figure shows the ridge s +0 ∩ s − (the shaded regionin the intersection of I +0 ∩I − ) in the plane with spinal coordinatesintroduced in [8]. The triple intersection I +0 ∩ I − ∩ I (cid:63) is the twomutually perpendicular lines. Compare to Fig.16 in [8]. Proposition 4.16. (1)
The side s + k (resp. s − k ) is a topological solid cylinderin H C ∪ ∂ H C . The intersection of ∂s + k (resp. ∂s − k ) with H C is the disjointunion of two topological discs. (2) The side s (cid:63)k (resp. s (cid:5) k ) is a topological solid light cone in H C ∪ ∂ H C . Theintersection of ∂s (cid:63)k (resp. ∂s (cid:5) k ) with H C is the light cone.Proof. (1) The side s + k is contained in the isometric sphere I + k . By Corollary 4.11, s + k intersects possibly with the sides contained in the isometric spheres I − k , I − k − , I (cid:63)k , I (cid:63)k − , I (cid:5) k and I (cid:5) k +1 .Let (cid:52) be the union of the ridges s + k ∩ s − k and s + k ∩ s (cid:63)k , and (cid:52) be the unionof the ridges s + k ∩ s − k − and s + k ∩ s (cid:5) k . By Proposition 4.10, (cid:52) contains the cross I + k ∩ I − k ∩ I (cid:63)k . By Proposition 4.15, (cid:52) is a union of four sectors which are patchedtogether along the cross. Hence, (cid:52) is topologically either a disc or a light cone. By Figure 5.
A schematic view of the side s + k ( s − k ). Figure 6.
A schematic view of the side s (cid:63)k ( s (cid:5) k ).a straight computation, the ideal boundary of (cid:52) on H C is a simple closed curveon the ideal boundary of I + k . See Figure 2. Thus (cid:52) is a topological disc. By asimilar argument, (cid:52) is a topological disc.Since I + k ∩ I (cid:63)k ∩ I − k − = ∅ except when θ = π/ ∂ H C , (cid:52) and (cid:52) are disjoint except when θ = π/ ∂ H C . See Figure 9 and Figure 10. Note that isometric spheresare topological balls and their pairwise intersections are connected. So, s + k is atopological solid cylinder. See Figure 5. s − k can be described by a similar argument.(2) The side s (cid:63)k is contained in the isometric sphere I (cid:63)k . According to Corollary4.11, s (cid:63)k only intersects with s + k and s − k . Let (cid:52) be the union of s + k ∩ s (cid:63)k and s − k ∩ s (cid:63)k .By Proposition 4.10 and Proposition 4.15, (cid:52) is a union of four sectors which arepatched together along the cross I + k ∩ I − k ∩ I (cid:63)k . By a computation for the case k = 0, one can see that the ideal boundary of (cid:52) is a union of two disjoint simpleclosed curves in the ideal boundary of I (cid:63)k . See Figure 2. Thus (cid:52) is a light cone.Hence, s (cid:63)k is topologically a solid light cone. See Figure 6. s (cid:5) k can be described bya similar argument. (cid:3) By applying a Poincar´e polyhedron theorem in H C as stated for example in [21],[8] or [18] (see [3] for a version in the hyperbolic plane), we have our main result asfollows. Theorem 4.17.
Suppose that θ ∈ [0 , π/ . Let D be as in Definition 4.12. Then D is a fundamental domain for the cosets of (cid:104) T (cid:105) in Γ . Moreover, Γ is discrete andhas the presentation Γ = (cid:104)
S, T | S = ( T − S ) = id (cid:105) . PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 23 s +0 ∩ s − s +0 ∩ s − s +0 ∩ s (cid:63) s − ∩ s (cid:63) Figure 7.
A schematic view of the ridges s +0 ∩ s − , s +0 ∩ s (cid:63) and s − ∩ s (cid:63) . The thick cross is the triple intersection I +0 ∩ I − ∩ I (cid:63) ,that is the union L ∪ C ∪ C described in Proposition 4.10. Proof.
The sides of D are s + k s − k , s (cid:63)k and s (cid:5) k . The ridges of D are s + k ∩ s − k , s + k ∩ s (cid:63)k , s + k ∩ s − k − , s + k ∩ s (cid:5) k , s − k ∩ s (cid:63)k and s − k − ∩ s (cid:5) k . To obtain the side-pairing maps andridge cycles, by applying powers of T , it suffices to consider the case where k = 0. The side-pairing maps: s +0 is contained in the isometric sphere I ( S ) and s − in the isometric sphere I ( S − ). The ridge s +0 ∩ s − is contained in the disc I ( S ) ∩ I ( S − ), which is defined by the triple equality |(cid:104) z , q ∞ (cid:105)| = |(cid:104) z , S − ( q ∞ ) (cid:105)| = |(cid:104) z , S ( q ∞ ) (cid:105)| . And the ridge s − ∩ s (cid:63) is contained in the disc I ( S − ) ∩ I ( S ), which is defined bythe triple equality |(cid:104) z , q ∞ (cid:105)| = |(cid:104) z , S ( q ∞ ) (cid:105)| = |(cid:104) z , S − ( q ∞ ) (cid:105)| . Since S maps q ∞ to S ( q ∞ ), S − ( q ∞ ) to q ∞ and S ( q ∞ ) to S ( q ∞ ) = S − ( q ∞ ), S maps the disc I ( S ) ∩ I ( S − ) to the disc I ( S − ) ∩ I ( S ). Note that I ( S ) ∩I ( S − ) ∩ I ( S ) is the union of two crossed geodesics (see Proposition 4.10) whosefour rays are cyclical permutated by S . Since the ridge s +0 ∩ s − lies in the closureof the exterior of the isometric sphere I ( S ), according to the equation (4.16), thepoint q ( π/ , π/ θ, √ /
2) is contained in s +0 ∩ s − . One can easily verify that S ( q ( π/ , π/ θ, √ / I ( S ). Thus S ( q ( π/ , π/ θ, √ / s − ∩ s (cid:63) , which lie in the closure of the exterior of the isometricsphere I ( S ).Hence S maps the ridge s +0 ∩ s − to the ridge s − ∩ s (cid:63) . Similarly, S maps s +0 ∩ s (cid:63) to s +0 ∩ s − . See Figure 7. Since S maps I +0 ∩ I −− ∩ I (cid:5) to I − ∩ I +1 ∩ I (cid:5) , a similarargument shows that S maps s +0 ∩ s −− to s − ∩ s (cid:5) and s +0 ∩ s (cid:5) to s − ∩ s +1 .By a similar argument, s (cid:63) (respectively, s (cid:5) ) is mapped to itself by the ellipticelement of order two S (respectively, ( T − S ) = ( S − T ) ), which sends s +0 ∩ s (cid:63) to s − ∩ s (cid:63) (respectively, s (cid:5) ∩ s −− to s +0 ∩ s (cid:5) ) and vice-versa.Hence, the side-pairing maps are: T k ST − k : s + k −→ s − k T k S T − k : s (cid:63)k −→ s (cid:63)k T k ( T − S ) T − k : s (cid:5) k −→ s (cid:5) k . The cycle transformations:
According to the side-pairing maps, the ridgecycles are: ( s + k ∩ s − k , s + k , s − k ) T k ST − k −−−−−→ ( s (cid:63)k ∩ s − k , s (cid:63)k , s − k ) T k S T − k −−−−−−→ ( s + k ∩ s (cid:63)k , s + k , s (cid:63)k ) T k ST − k −−−−−→ ( s + k ∩ s − k , s + k , s − k ) , and ( s + k ∩ s − k − , s + k , s − k − ) T k ( T − S ) T − k −−−−−−−−−→ ( s (cid:5) k ∩ s − k − , s (cid:5) k , s − k − ) T k ( T − S ) T − k −−−−−−−−−−→ ( s + k ∩ s (cid:5) k , s + k , s (cid:5) k ) T k ( T − S ) T − k −−−−−−−−−→ ( s + k ∩ s − k − , s + k , s − k − ) . Thus the cycle transformations are T k ST − k · T k S T − k · T k ST − k = T k S T − k , and T k ( T − S ) T − k · T k ( T − S ) T − k · T k ( T − S ) T − k = T k ( T − S ) T − k , which are equal to the identity map, since we have S = id and ( T − S ) = id . The local tessellation : There are exactly two copies of D along each side,since the sides are contained in isometrical spheres and the side-pairing maps sendthe exteriors to the interiors. Thus there is nothing to verify for the points in theinterior of every side.According to the ridge cycles and cycle transformations, there are exactly threecopies of D along each ridge. • s + k ∩ s − k , s (cid:63)k ∩ s − k and s + k ∩ s (cid:63)k : These three ridges are in one cycle. Thus weonly need to consider the ridge s + k ∩ s − k . Since the cycle transformation of s + k ∩ s − k is T k ST − k · T k S T − k · T k ST − k = id, the three copies of D along s + k ∩ s − k are D , T k S − T − k ( D ) and T k ST − k ( D ).We know that s + k ∩ s − k is contained in I ( T k ST − k ) ∩ I ( T k S − T − k ), whichis defined by the triple equality |(cid:104) z , q ∞ (cid:105)| = |(cid:104) z , T k S − T − k ( q ∞ ) (cid:105)| = |(cid:104) z , T k ST − k ( q ∞ ) (cid:105)| . For z in the neighborhoods of s + k ∩ s − k in D , |(cid:104) z , q ∞ (cid:105)| is the smallest ofthe three quantities in the above triple equality.For z in the neighborhoods of s (cid:63)k ∩ s − k in D , |(cid:104) z , q ∞ (cid:105)| is no more than |(cid:104) z , T k ST − k ( q ∞ ) (cid:105)| and |(cid:104) z , T k S − T − k ( q ∞ ) (cid:105)| . Applying T k S − T − k givesneighborhood of s + k ∩ s − k in T k S − T − k ( D ) where |(cid:104) z , T k S − T − k ( q ∞ ) (cid:105)| isthe smallest of the three quantities in the above triple equality.For z in the neighborhoods of s + k ∩ s (cid:63)k in D , |(cid:104) z , q ∞ (cid:105)| is no more than |(cid:104) z , T k S − T − k ( q ∞ ) (cid:105)| and |(cid:104) z , T k S − T − k ( q ∞ ) (cid:105)| . Applying T k ST − k givesneighborhood of s + k ∩ s − k in T k ST − k ( D ) where |(cid:104) z , T k ST − k ( q ∞ ) (cid:105)| is thesmallest of the three quantities in the above triple equality.Thus the union of D , T k S − T − k ( D ) and T k ST − k ( D ) forms a regularneighborhood of each point in s + k ∩ s − k . • s + k ∩ s − k − , s (cid:5) k ∩ s − k − and s + k ∩ s (cid:5) k : We only need to consider the ridge s + k ∩ s − k − . Since the cycle transformation of s + k ∩ s − k − is T k ( T − S ) T − k · T k ( T − S ) T − k · T k ( T − S ) T − k = id, PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 25 the union of D , T k ( T − S ) − T − k ( D ) and T k ( T − S ) T − k ( D ) forms a regularneighborhood of each point in s + k ∩ s − k − , by a similar argument as in thefirst item. Consistent system of horoballs : When θ = π , there are accidental idealvertices on D . The sides s (cid:63)k and s (cid:63)k +1 will be asymptotic on ∂ H C at the fixedpoint of the parabolic element T k ( S T − ) T − k , and the sides s (cid:5) k and s (cid:5) k +1 will beasymptotic on ∂ H C at the fixed point of the parabolic element T k ( ST − S ) T − k . Toshow that there will be consistent system of horoballs, it suffices to show that allthe cycle transformations fixing a given cusp is non-loxodromic.Let p be the fixed point of T − S and q be the fixed point of ( T − S ) T (thecoordinates of p and q are given in Definition 5.2, or see Figure 9). Then allthe accidental ideal vertices { T k ( p ) } and { T k ( q ) } are related by the side-pairingmaps as the following: (cid:47) (cid:47) T − ( q ) (cid:15) (cid:15) (cid:47) (cid:47) q T − ST (cid:15) (cid:15) ( S − T ) (cid:47) (cid:47) T ( q ) S (cid:15) (cid:15) (cid:47) (cid:47) T ( q ) (cid:15) (cid:15) (cid:47) (cid:47) (cid:47) (cid:47) T − ( p ) (cid:59) (cid:59) (cid:47) (cid:47) p S (cid:61) (cid:61) S (cid:47) (cid:47) T ( p ) (cid:58) (cid:58) (cid:47) (cid:47) T ( p ) (cid:47) (cid:47) Thus, up to powers of T , all the cycle transformations are S · S · S − = id and T − ST · ( S − T ) − · S = ( T − S ) = ( I I I I ) , which is parabolic. This means that p is fixed by the parabolic element ( T − S ) .Therefore, D is a fundamental domain for the cosets of (cid:104) T (cid:105) in Γ. The side-pairingmaps and T will generate the group Γ. The reflection relations are ( T k S T − k ) = id and ( T k ( T − S ) T − k ) = id . The cycle relations are T k S T − k = id and T k ( T − S ) T − k = id . Thus Γ is discrete and has the presentationΓ = (cid:104) S, T | S = ( T − S ) = id (cid:105) . (cid:3) Since Γ is a subgroup of (cid:104) I , I , I (cid:105) of index 2, as a corollary, we have Corollary 4.18.
Let (cid:104) I , I , I (cid:105) be a complex hyperbolic (4 , , ∞ ) triangle group asin Proposition 3.1. Then (cid:104) I , I , I (cid:105) is discrete and faithful if and only if I I I I is non-elliptic. This answers Conjecture 1.1 on the complex hyperbolic (4 , , ∞ ) triangle group.5. The manifold at infinity
In this section, we study the group Γ in the case when θ = π/
3. That is thegroup Γ = (cid:104)
S, T (cid:105) = (cid:104) I I , I I (cid:105) with T − S = I I I I being parabolic.In this case, the Ford domain D has additional ideal vertices on ∂ H C , which areparabolic fixed points corresponding to the conjugators of T − S . By intersectinga fundamental domain for (cid:104) T (cid:105) acting on ∂ H C with the ideal boundary of D , weobtain a fundamental domain for Γ acting on its discontinuity region Ω(Γ).Topologically, this fundamental domain is the unknotted cylinder cross a ray(see Proposition 5.14). By cutting and gluing we obtain two polyhedra P + and P − (see Proposition 5.19). Gluing P − to P + by S − , we obtain a polyhedron P . By studying the combinatorial properties of P , we show that the quotient Ω(Γ) / Γ ishomeomorphic to the two-cusped hyperbolic 3-manifold s Remark . We will use ˜ s + k (respectively, ˜ s − k , ˜ s (cid:63)k , and ˜ s (cid:5) k ) to denote the idealboundary of the side of D contained in the ideal boundary of the isometric sphere I + k (respectively, I − k , I (cid:63)k and I (cid:5) k ). Definition 5.2.
In the Heisenberg coordinate, we define the points q = (cid:104) ( − i √ / , −√ (cid:105) ,q = (cid:104) (1 + i √ / , √ (cid:105) ,p = (cid:104) ( − i √ / , −√ (cid:105) ,p = (cid:104) (3 + i √ / , √ (cid:105) ,p = (cid:20) (cid:16) √ i (4 √ − √ (cid:17) , (cid:21) ,p = (cid:20) (cid:16) − √ i (4 √ √ (cid:17) , (cid:21) ,p = (cid:34) (cid:16) − √ i (2 √ √ (cid:17) , − √ (cid:35) ,p = (cid:34) (cid:16) √ i (2 √ − √ (cid:17) , √ (cid:35) ,p = (cid:34) (cid:16) − √ i (2 √ √ (cid:17) , √ (cid:35) ,p = (cid:34) (cid:16) − − √ i (2 √ − √ (cid:17) , − √ (cid:35) ,p = (cid:20) (cid:16) − √ i (4 √ √ (cid:17) , (cid:21) ,p = (cid:20) (cid:16) − − √ i (4 √ − √ (cid:17) , (cid:21) , and the other points p = T ( p ) , p = T ( p ) , p = T ( p ) , p = T ( p ) . By Proposition 4.10 and Corollary 4.11, we have the following.
Proposition 5.3.
The points defined in Definition 5.2 have the following proper-ties. • p , p , p , p are the four points on the ideal boundary of I +0 ∩I − ∩I (cid:63) , whichare described in Proposition 4.10; • p , p , p , p are the four points on the ideal boundary of I +0 ∩ I −− ∩ I (cid:5) ; • p , p , p , p are the four points on the ideal boundary of I +1 ∩ I − ∩ I (cid:5) ; • p (resp. p ) is the parabolic fixed point of T − S (resp. S T − ), whichis the intersection of four isometric spheres I +0 ∩ I −− ∩ I (cid:63) ∩ I (cid:63) − (resp. I +1 ∩ I − ∩ I (cid:63) ∩ I (cid:63) ); PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 27 p p p p p p p p p q αβ Figure 8.
Intersections of the isometric spheres I − , I −− , I ∗ , I ∗− , I (cid:5) , I (cid:5) with I +0 in ∂ H C , viewed in geographical coordinates.Here α ∈ [ − π/ , π/
2] in the vertical coordinate and β ∈ [0 , π ] inthe horizontal one. The region exterior to the six Jordan closedcurves has two connect components. One of them is the topologicaloctagon with vertices p , p , p , p , q , p , p , p . The other oneis a topological quadrilateral with vertices p , p , q , p . • q (resp. q ) is the parabolic fixed point of ST − S (resp. T − ST − ST ),which is the intersection of the four isometric spheres I +0 ∩ I − ∩ I (cid:5) ∩ I (cid:5) (resp. I + − ∩ I −− ∩ I (cid:5) ∩ I (cid:5)− ).Proof. As described in Proposition 4.10, all of the triple intersections I +0 ∩ I − ∩ I (cid:63) , I +0 ∩ I −− ∩ I (cid:5) and I +1 ∩ I − ∩ I (cid:5) have exactly four points lying on ∂ H C . Whenwrite the standard lifts of p , p , p , p , one can see that they are the four points inthe proof of Proposition 4.10. Thus the first item is proved.By Proposition 4.2, the four points of I +0 ∩I −− ∩I (cid:5) are the images of p , p , p , p under the antiholomorphic involution τ , which are p , p , p , p .The second item and the fact that I +1 ∩ I − ∩ I (cid:5) = T ( I +0 ∩ I −− ∩ I (cid:5) ) imply thethird item.We will only prove the statement for p in the last two items. The others canbe described by similar arguments. By Lemma 4.8, p is the parabolic fixed pointof T − S and is the triple intersection I +0 ∩ I −− ∩ I (cid:63) . By Corollary 4.11 (3), I (cid:63) istangent with I (cid:63) − at p . This completes the proof. (cid:3) Now we study the combinatorial properties of the sides. See Figure 8 and Figure9.
Proposition 5.4.
The interior of the side ˜ s +0 has two connected components. • One of them is an octagon, denoted by O +0 , whose vertices are p , p , p , p , q , p , p and p . • The other one is a quadrilateral, denoted by Q +0 , whose vertices are p , p , q and p .Proof. By Proposition 4.16, when θ < π/
3, the side ˜ s +0 is topologically an annulusbounded by two disjoint simple closed curves which are the union of the idealboundaries of the ridges s +0 ∩ s −− and s +0 ∩ s (cid:5) , respectively s +0 ∩ s − and s +0 ∩ s (cid:63) .When θ = π/
3, these two curves intersect at two points, which divide ˜ s +0 into twoparts. That is to say the interior of the side ˜ s +0 has two connected components.By Proposition 5.3, the ideal boundary of the ridge s +0 ∩ s −− (resp. s +0 ∩ s (cid:5) )is a union of two disjoint Jordan arcs [ p , p ] and [ p , p ] (resp. [ p , p ] and[ p , p ]), the ideal boundary of the ridge s +0 ∩ s − (resp. s +0 ∩ s (cid:63) ) is a union of twodisjoint Jordan arcs [ p , p ] and [ p , p ] (resp. [ p , p ] and [ p , p ]). Since p is theintersection of four isometric spheres I +0 ∩ I −− ∩ I (cid:63) ∩ I (cid:63) − , it lies on [ p , p ] and[ p , p ]. Similarly, q lies on [ p , p ] and [ p , p ].By Proposition 4.2, the antiholomorphic involution τ preserves ˜ s +0 and inter-changes its boundaries. It is easy to check that τ interchanges p and q , p and p , p and p , p and p , p and p . Thus one part of ˜ s +0 is a quadrilateral withvertices p , p , q and p , denoted by O +0 . The other one is an octagon with ver-tices p , p , p , p , q , p , p and p , denoted by Q +0 . Both of them are preservedby τ . (cid:3) According to the symmetry I in Proposition 4.2, which interchanges I +0 and I − , we have the following. Proposition 5.5.
The interior of the side ˜ s − has two connected components. • One of them is an octagon, denoted by O − , whose vertices are p , p , p , p , p , p , p and q . • The other is a quadrilateral, denoted by Q − , whose vertices are p , p , q and p .Proof. Note that side s − is bounded by the ridges s − ∩ s +1 , s − ∩ s (cid:5) , s − ∩ s +0 and s − ∩ s (cid:63) . By Proposition 4.2, the side s − is isometric to s +0 under the complexinvolution I . Thus its ideal boundary ˜ s − will be also isometric to ˜ s +0 . This impliesthat ˜ s − has the same combinatorial properties as ˜ s +0 . One can check that I : ( q , p , p , p , p , p , p , p ) ↔ ( q , p , p , p , p , p , p , p ) . Thus one part of ˜ s − is an octagon, denoted by O − , whose vertices are p , p , p , p , p , p , p and q . The other one is a quadrilateral, denoted by Q − , whosevertices are p , p , q and p . See Figure 9. (cid:3) Remark . q lies on the C -circle associated to I , that is the ideal boundary ofthe complex line fixed by I . One can also observe that p is fixed by I . Proposition 5.7.
The interior of side ˜ s (cid:63) is a union of two disjoint triangles,denoted by ( T ) (cid:63) and ( T ) (cid:63) , whose vertices are p , p , p and respectively, p , p , p . PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 29 Tq p p p p q p p p p p p p v p p p q p p p q p p p O +0 O − Q +0 Q − ( T ) (cid:63) ( T ) (cid:63) ( T ) (cid:5) ( T ) (cid:5) O + − O −− Q + − v − Q −− ( T ) (cid:63) − ( T ) (cid:63) − ( T ) (cid:5)− ( T ) (cid:5)− c − c Figure 9.
A combinatorial picture of ∂U . The top and bottomcurves are identified. O ± (resp. O ±− ) is divided by c (resp. c − )into a quadrilateral Q (cid:48)± (resp. Q (cid:48)±− ) and a heptagon H ± (resp. H ±− ). v is the intersection of c with the arc [ p , p ]. v − is theintersection of c − with the arc [ T − ( p ) , T − ( p )]. Proof.
By Proposition 4.16, the side ˜ s (cid:63) is the union of two disjoint discs, which arebounded by the ideal boundary of the ridges s +0 ∩ s (cid:63) and s − ∩ s (cid:63) .As stated in Proposition 5.3, the ideal boundary of I +0 ∩ I − ∩ I (cid:63) contains thefour points p , p , p , p . Thus ˜ s (cid:63) is the union of two disjoint bigons whose verticesare p , p , and respectively p , p . Proposition 5.3 also tells us that p and p lieon different component of the boundaries of the two bigons.Therefore, both of the components are triangles, denoted by ( T ) (cid:63) and ( T ) (cid:63) ,whose vertices are p , p , p and respectively, p , p , p . (cid:3) According to the symmetry τ in Proposition 4.2, the side ˜ s (cid:5) has the same topo-logical properties as the side ˜ s (cid:63) . Thus by a similar argument, we have the following. Proposition 5.8.
The interior of side ˜ s (cid:5) is a union of two disjoint triangles,denoted by ( T ) (cid:5) and ( T ) (cid:5) , whose vectors are q , p , p and respectively, q , p , p . Let U be the ideal boundary of D on ∂ H C . Then the union of all the sides { ˜ s + k } , { ˜ s − k } , { ˜ s (cid:63)k } and { ˜ s (cid:5) k } for k ∈ Z form the boundary of U . Proposition 5.9.
Let L = { [ x + i √ / , √ x ] ∈ N : x ∈ R } , then L is a T -invariant R -circle. Furthermore, L is contained in the complement of D .Proof. It is obvious that L is a R -circle, since it is the image of the x -axis of N bya Heisenberg translation along the y -axis. For any point [ x + i √ / , √ x ] ∈ L , wehave T ([ x + i √ / , √ x ]) = [( x + 2) + i √ / , √ x + 2)] which lies in L . Thus L is a T -invariant R -circle. See Figure 10.Note that T acts on L as a translation through 2. To show L is contained in thecomplement of D , it suffices to show that a segment with length 2 is contained in theinterior of some isometric spheres. By considering their Cygan distance between apoint in L and the center of a isometric sphere, one can compute that the segment { [ x + i √ / , √ x ] : − / ≤ x ≤ / } lie in the interior of I +0 and the segment { [ x + i √ / , √ x ] : 1 / ≤ x ≤ / } lie in the interior of I − . (cid:3) I (cid:5)− L I −− I − I (cid:63) I (cid:63) − I +0 I + − I +1 Figure 10.
A realistic picture of the ideal boundaries of theisometric spheres: I +0 , I +1 and I + − (red); I − and I −− (blue); I (cid:63) and I (cid:63) − (pink); I (cid:5) , I (cid:5)− and I (cid:5) (yellow). The line L is the T -invariant R -circle. Definition 5.10.
Let Σ − = { [ − / iy, t ] ∈ N : y, t ∈ R } and Σ = { [1 / iy, t ] ∈N : y, t ∈ R } be two planes in the Heisenberg group.In fact, the vertical planes Σ − and Σ are boundaries of fans in the sense of[16]. Let D T be the region between Σ − and Σ , that is D T = { [ x + iy, t ] ∈ N : − / ≤ x ≤ / } . It is obvious that Σ = T (Σ − ). Thus D T is a fundamental domain for (cid:104) T (cid:105) actingon ∂ H C . Proposition 5.11.
The intersections of Σ and Σ − with the isometric spheres I ± k , I (cid:63)k and I (cid:5) k are empty, except the following: • Each one of Σ ∩ I ± and Σ ∩ I (cid:63) is a circle and Σ ∩ I (cid:5) = Σ ∩ I (cid:5) = { q } . • Each one of Σ − ∩I ±− and Σ − ∩I (cid:63) − is a circle and Σ − ∩I (cid:5)− = Σ − ∩I (cid:5) = { q } .Proof. Since the isometric spheres are strictly convex, their intersections with aplane is either a topological circle, or a point or empty. Note that Σ = T (Σ − ).Thus it suffices to consider the intersections of Σ with the isometric spheres. By astrait computation, each one of Σ ∩ I ± and Σ ∩ I (cid:63) is a circle. (See Figure 11). (cid:3) Lemma 5.12.
The plane Σ (respectively Σ − ) is preserved by I (respectively T − I T ). The intersection Σ ∩ ∂U (respectively Σ − ∩ ∂U ) is a simple closed curve c (respectively c − ) in the union ˜ s +0 ∪ ˜ s − (respectively ˜ s + − ∪ ˜ s −− ), which containsthe points q and v = [1 / i √ / , −√ (respectively q and v − = T − ( v ) ). PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 31 Σ ∩ I +0 Σ ∩ I − Σ ∩ I (cid:63) c +0 c − q v Figure 11.
The intersection of Σ with I +0 , I − and I (cid:63) . Viewedin Σ . c = c +0 ∪ c − is a simple closed curve, where c +0 and c − arethe thicker parts of Σ ∩ I +0 and Σ ∩ I − , respectively. Proof.
It suffices to consider Σ . The C -circle associated to I , that is the idealboundary of the complex line fixed by I , is { [1 / i √ / , t ] ∈ N : t ∈ R } , whichis contained in Σ . Thus Σ is preserved by I .It is obvious that Σ contains q , which is the tangent point of I (cid:5) and I (cid:5) .The intersections Σ ∩ I +0 , Σ ∩ I − and Σ ∩ I (cid:63) are circles by Proposition 5.11.One can compute that the intersection Σ ∩ I +0 ∩ I − contain two points q and v = [1 / i √ / , −√ I +0 and I − into two arcs. Let c +0 be the arc with endpoints q and v on I +0 lyingin the exterior of I − and c − be the one on I − lying in the exterior of I +0 . Then c = c +0 ∪ c − is a simple closed curve. Observe that Σ ∩ I (cid:63) lie in the union of theinteriors of I +0 and I − . Thus Σ does not intersect ˜ s (cid:63) . By Proposition 5.11, c +0 lieon ˜ s +0 and c − lie on ˜ s − . Therefore, the intersection Σ ∩ ∂U is c , which is a simpleclosed curve containing q and v . (cid:3) Let U c be closure of the complement of U in N . Proposition 5.13.
The closure of the intersection U c ∩ D T is a solid tube home-omorphic to a 3-ball.Proof. It suffices to show that the boundary of U c ∩ D T is a 2-sphere. Now let usconsider the cell structure of U c ∩ D T . See Figure 9. According to Lemma 5.12,the intersection of U c with Σ (resp. Σ − ) is a topological disc with two vertices q and v (resp. q and v − )and two edges c ± (resp. c ±− ). Moreover, c (resp. c − )divides O ± (resp. O ±− )into a quadrilateral Q (cid:48)± (resp. Q (cid:48)±− ) and a heptagon H ± (resp. H ±− ).Since p , p and T − ( p ) are contained in D T , one can see that D T contains Q (cid:48)− , Q (cid:48) + − , H +0 and H −− . Besides, D T contains Q +0 , Q −− , ( T ) (cid:5) , ( T ) (cid:5) , ( T ) (cid:63) and( T ) (cid:63) − . Thus the boundary of U c ∩ D T consists of 12 faces, 23 edges and 13 vertices.See the region between c and c − in Figure 9. Therefore the Euler characteristicof the boundary of U c ∩ D T is 2. So the boundary of U c ∩ D T is a 2-sphere. (cid:3) Proposition 5.9 and Proposition 5.13 imply the following result.
Proposition 5.14. U ∩ D T is an unknotted cylinder cross a ray homeomorphic to S × [0 , × R ≥ .Proof. As stated in Proposition 5.9, U c contains the line L . Thus U c ∩ D T is atubular neighborhood of L ∩ D T . It cannot be knotted. Therefore ∂U ∩ D T is ununknotted cylinder homeomorphic to S × [0 , U ∩ Σ is c cross a ray and U ∩ Σ − is c − cross a ray. Both of them are homeomorphic to S × R ≥ . Hence U ∩ D T is an unknotted cylinder cross a ray homeomorphic to S × [0 , × R ≥ . (cid:3) Applying the powers of T , Proposition 5.14 immediately implies the followingcorollary. Corollary 5.15. U is an unknotted cylinder cross a ray homeomorphic to S × R × R ≥ .Remark . 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 horotube). Definition 5.17.
Suppose that the cylinder S × [0 ,
1] has a combinatorial cellstructure with finite faces { F i } . A canonical subdivision on S × [0 , × R ≥ is afinite union of 3-dimensional pieces { (cid:98) F i } where (cid:98) F i = F i × R ≥ . Proposition 5.18.
There is a canonical subdivision on U ∩ D T .Proof. As described in the proof of Proposition 5.13, the combinatorial cell struc-ture of ∂U ∩ D T has 10 faces Q (cid:48)− , Q (cid:48) + − , H +0 , H −− , Q +0 , Q −− , ( T ) (cid:5) , ( T ) (cid:5) , ( T ) (cid:63) and ( T ) (cid:63) − . By Proposition 5.14, U ∩ D T is the union of 3-dimensional pieces (cid:100) Q (cid:48)− , (cid:91) Q (cid:48) + − , (cid:100) H +0 , (cid:100) H −− , (cid:100) Q +0 , (cid:100) Q −− , (cid:91) ( T ) (cid:5) , (cid:91) ( T ) (cid:5) , (cid:91) ( T ) (cid:63) and (cid:92) ( T ) (cid:63) − . Combinatorially, these3-dimensional pieces are the cone from q ∞ to the faces of ∂U ∩ D T . (cid:3) Let Ω(Γ) be the discontinuity region of Γ acting on ∂ H C . Then U ∩ D T is obviousa fundamental domain for Γ. By cutting and gluing, we can obtain the followingfundamental domain for Γ acting on Ω(Γ). Proposition 5.19.
Let P + be the union of (cid:100) H +0 , (cid:91) ( T ) (cid:5) , (cid:91) ( T ) (cid:5) , T ( (cid:91) Q (cid:48) + − ) , T ( (cid:100) Q −− ) and T ( (cid:92) ( T ) (cid:63) − ) . Let P − be the union of (cid:100) Q +0 , (cid:91) ( T ) (cid:63) , (cid:100) Q (cid:48)− and T ( (cid:100) H −− ) . Then P + ∪P − is a fundamental domain for Γ acting on Ω(Γ) . Moreover, P + (resp. P − ) iscombinatorially an eleven pyramid (resp. nine pyramid) with cone vertex q ∞ andbase O +0 ∪ Q − ∪ ( T ) (cid:5) ∪ ( T ) (cid:5) ∪ ( T ) (cid:63) (resp. O − ∪ Q +0 ∪ ( T ) (cid:63) ).Proof. Since Σ = T (Σ − ) and c = T ( c − ), U ∩ D T and T ( U ∩ D T ) can be gluedtogether along c × R ≥ . Note that U ∩ D T is a fundamental domain for Γ actingon Ω(Γ) and has a subdivision as described in Proposition 5.18. Therefore P + ∪ P − is also a fundamental domain.As described in Proposition 5.13, c (resp. c − ) divides O ± (resp. O ±− )intoa quadrilateral Q (cid:48)± (resp. Q (cid:48)±− ) and a heptagon H ± (resp. H ±− ). Note that O ± = T ( O ±− ) and ( T ) (cid:63) = T (( T ) (cid:63) − ). Thus the base of P + (resp. P − ) is O +0 ∪Q − ∪ ( T ) (cid:5) ∪ ( T ) (cid:5) ∪ ( T ) (cid:63) (resp. O − ∪ Q +0 ∪ ( T ) (cid:63) ) which is combinatorially ahendecagon (resp. an enneagon). See Figure 9 and Figure 12. (cid:3) Definition 5.20.
Let p (cid:48) = S − ( p ), p = S − ( q ∞ ) = [0 ,
0] and p (cid:48) = S − ( p ). PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 33 p p q p p p p p p q p p p p q p p p Figure 12.
A schematic view of the fundamental domain of Γ onΩ(Γ). The vertices colored with red are the parabolic fixed points.The polygon colored with yellow is O − ∪ Q +0 ∪ ( T ) (cid:63) . Lemma 5.21.
Let P be the union P + ∪ S − ( P − ) . Then P is combinatorially apolyhedron with 8 triangular faces, 4 square faces, 2 pentagonal faces and 2 hexag-onal faces. The faces of P are paired as follows: T : ( q ∞ , p , p , q ) (cid:55)−→ ( q ∞ , p , p , q ) ,S − T : ( q ∞ , q , p , p , p ) (cid:55)−→ ( p , p , p , p , p ) , ( S − T ) : ( q , p , p ) (cid:55)−→ ( q , p , p ) ,S − : ( q ∞ , p , q ) (cid:55)−→ ( p , p (cid:48) , p ) ,S − T − S : ( p , p , q ) (cid:55)−→ ( p , p (cid:48) , p (cid:48) ) ,S − : ( q , p , p , p ) (cid:55)−→ ( p (cid:48) , p (cid:48) , p , p ) ,S − : ( q ∞ , p , p , p (cid:48) , p , p ) (cid:55)−→ ( p , p (cid:48) , p , p , p , q ) ,S − : ( p , p (cid:48) , p ) (cid:55)−→ ( p , p , p ) . p p q p p p p p p p q ´p ´p p p Figure 13.
The combinatorial picture of P . The vertices of P colored with red are the parabolic fixed points. Proof.
The bases of P + and P − are paired as follows: S − : O − −→ O +0 ( p , p , p , p , q , q , p , p ) (cid:55)−→ ( q , p , p , p , p , p , p , p ) ,S : Q +0 −→ Q − ( p , p , q , p ) (cid:55)−→ ( q , p , p , p ) ,S : ( T ) (cid:63) −→ ( T ) (cid:63) ( p , p , p ) (cid:55)−→ ( p , p , p ) , ( S − T ) : ( T ) (cid:5) −→ ( T ) (cid:5) ( q , p , p ) (cid:55)−→ ( q , p , p ) . Thus S − ( P − ) and P + are glued along O +0 . According to Lemma 2.9, S − ( P − ) liein the interior of I +0 , since P − lie in the exterior of I +0 . Moreover, p = S − ( q ∞ ) =[0 ,
0] is the center of the isometric sphere I +0 . See Figure 13. (cid:3) Proposition 5.22.
Let Ω be the discontinuity region of Γ acting on H C . Then thefundamental group of Ω / Γ has a presentation (cid:104) u, v, w | w − vu − v − wu = v wuw − u = id (cid:105) . PHERICAL CR STRUCTURE ON A TWO-CUSPED HYPERBOLIC 3-MANIFOLD 35
Proof.
Let x i , i = 1 , , , , , , , P given inLemma 5.21. These are the generators of the fundamental group of Ω / Γ.By considering the edge cycles of P under the gluing maps, we have the relationsas follows.(1) x − · x · x · x = id, (2) x − · x · x = id, (3) x · x − · x − · x · x = id, (4) x − · x − · x · x = id, (5) x · x · x = id, (6) x − · x · x = id, (7) x · x · x = id, (8) x · x · x = id, (9) x − · x = id. For example, the edge cycle of [ q ∞ , p ] is[ q ∞ , p ] x −→ [ q ∞ , p ] x −→ [ p , q ] x −→ [ p , p (cid:48) ] x − −−→ [ q ∞ , p ] . Thus x − · x · x · x = id. This is the relation in (1). The others can be given by a similar argument.Simplifying the relations and setting u = x , v = x , w = x , we obtain thepresentation of the fundamental group of Ω / Γ. (cid:3) Now, we are ready to show the following theorem.
Theorem 5.23.
Let Ω be the discontinuity region of Γ acting on H C . Then thequotient space Ω / Γ is homeomorphic to the two-cusped hyperbolic 3-manifold s in the SnapPy census.Proof. Let M = Ω / Γ. According to Proposition 5.22, the fundamental group of M has a presentation π ( M ) = (cid:104) u, v, w | w − vu − v − wu = v wuw − u = id (cid:105) . The manifold s
782 is a two-cusped hyperbolic 3-manifold with finite volume. Itsfundamental group provided by SnapPy has a presentation π ( s ) = (cid:104) a, b, c | a cb c = abca − b − c − = id (cid:105) . Using
Magma , we get an isomorphism Ψ : π ( M ) −→ π ( s u ) = c − b − , Ψ( v ) = b − , Ψ( w ) = a. Therefore M will be the connect sum of s
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