Math. Proc. Camb. Phil. Soc. (2006), 140, 79 c© 2006 Cambridge Philosophical Society

doi:10.1017/S0305004105008868 Printed in the United Kingdom

79

On hyperbolic 3-manifolds with an infinite number

of fibrations over S1

By HUGH M. HILDEN

Department of Mathematics, University of Hawaii, Honolulu, Hi 96822, U.S.A.

MARIA TERESA LOZANO

Departamento de Matematicas, Universidad de Zaragoza, 50009 Zaragoza, Spain.e-mail: tlozano@unizar.es

and JOSE MARIA MONTESINOS-AMILIBIA

Departamento de Geometrıa y Topologıa, Facultad de Matematicas,

Universidad Complutense, 28040 Madrid, Spain.

(Received 8 October 2004)

Abstract

Arithmetic and non arithmetic hyperbolic 3-manifolds having infinitely many

fibrations over S1 of arbitrary high genus are constructed. A concrete example isstudied in detail.

1. Introduction

Hyperbolic 3-manifolds with a surface bundle structure form a class of interest-

ing 3-manifolds after Thurston’s conjecture that every hyperbolic 3-manifold can

be finitely covered by a such manifold. The first explicit examples were given by

Jorgensen [5]. Others examples can be found in [1, 4, 7].

Tollefson in [9] found some Seifert manifolds having infinitely many different fiber

bundle structures over S1. To our knowledge the analogous result for hyperbolicmanifolds remained open.

In this paper we offer an elementary proof of the following Theorem.

Theorem 1. There are arithmetic and non arithmetic hyperbolic manifolds fibered

over S1 in infinitely many different ways.

The classification of Euclidean 3-orbifolds contains examples with underlying

space S3 and singular set a knot or a link L. For instance L can be the Borromeanrings with isotropy of order 2. Other examples that can be used equally well to obtain

the results in this paper are the figure-eight rational knot 5/3 with isotropy of order3, or the rational link 10/3 with isotropies of orders 2 and 3.We will consider only the Borromean rings case here. This link is more generally

the singular locus of a three parameter family of cone-manifold structures of S3. Theparameters are the angles α, β, γ around the components of L. For α=β = γ =π the

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80 H. Hilden, M. T. Lozano and J. M. Montesinos

cone-manifold is Euclidean; while if the angles are all less than π the cone-manifoldis hyperbolic.

The idea of the proof of the above theorem is the following. If the isotropy of

the components of the hyperbolic orbifold with singular set L are all even, then theorbifold has a finite covering which is a hyperbolic manifoldM . This covering factors

through an 8-fold covering with underlying space the 3-torus S1×S1×S1 and singularset Σ made up of curves parallel to S1 × {0} × {0}, {0} × S1 × {0}, {0} × {0} × S1.There are infinitely many fibrations of S1 × S1 × S1 over S1 with fibers transversalto Σ. These fibrations lift to fibrations of M .

In the last section we study the particular case of the hyperbolic orbifold with

singular set L and isotropy of order 4. The manifold M in this case is acted upon by

the quaternionic group and a number of consequences follow.

Finally, the fact that the list of arithmetic hyperbolic orbifolds with singular set Lis known allows us to claim that there are arithmetic and non arithmetic examples

of manifolds satisfying the theorem.

2. Preliminaries

The following well known Lemma is about factoring regular orbifold coverings and

will be used in several cases.

Lemma 1. Let G be a discontinuous subgroup of isometries of a geometric orbifoldM ,

such that the quotient p:M → M/G is a regular orbifold covering. Assume that there

exists a short exact sequence

0 −→ H −→ G −→ H \ G −→ 0

then there exists a commutative diagram of regular orbifold coverings as

Mq2−→ M/H

p ց ւq1

M/G� (M/H)/(H \ G)

Proof. The normal subgroup H of G acts in M by restriction of the action of G,therefore the action is discontinuous and the quotient q1:M → M/H is a regular

orbifold covering. The map

M/H × H \ G −→ M/H([m], [g]) �−→ [mg]

is a well defined action. In fact, if [g] = [g′] and [m] = [m′], then g′ = h1g for someh1 ∈ H, and m′ = mh2 for some h2 ∈ H. Thus m′g′ = mh2h1g = mgh′ implies that

[m′g′] = [mg]. Therefore the quotient map q2:M/H → (M/H)/(H \ G) is a regularorbifold covering. It is clear that q2 ◦ q1 = p.

3. Some coverings

We denote by (S3, Bm,n,p ) the geometric orbifold structure with underlying to-

pological space S3 and singular locus the Borromean rings, B, and such that theisotropy groups of the components of the rings are cyclic of ordersm, n, p, (Figure 1)where m, n, p � 2. This family of orbifolds (Borromean orbifolds) is studied in [3],

where it is shown, among other results, that (S3, Bm,n,p ) is a hyperbolic orbifold for

3 � m, n, p � ∞, and the corresponding orbifold group is arithmetic if and only if

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Manifolds with infinite fibrations 81

a b

c

la lb

lc

Fig. 1. The Borromean link.

(m, n, p) is one of the following unordered triples. (6, 6, 6), (4, 4, 4), (3, 3, 3), (4, 4, 3),(6, 6, 3), (∞,∞,∞), (∞,∞, 3), (∞,∞, 4), (∞, 3, 3), (∞, 4, 3), (∞, 4, 4).The orbifold (S3, B2,2,2) is Euclidean [8].

The orbifold group of the orbifold (S3, Bm,n,p ) has the following presentation

πo1 (S

3, Bm,n,p ) = |a, b, c, ; [a, la ], [b, lb], [ c, lc], am , bn , cp |

where a, b, and c are meridian elements, and la , lb and lc are the canonical longitudesof the components of the Borromean rings. Here the notation [x, y] = xyxy has beenused, so that the first three relators express the commutativity of every meridian

with the corresponding canonical longitude. This group is a quotient of the group

of the Borromean link, π1(S3 \ B) = |a, b, c, ; [a, la ], [b, lb], [ c, lc]|. This presentation

can be easily computed from the link projection in Figure 1.

Next, we restrict our attention to Borromean orbifolds with even indices.

Consider the abelianization homomorphism, Ab, on the orbifold group of the orbi-fold (S3, B2m,2n,2p ). The resulting abelian quotient group has the presentation

Ab(πo1 (S

3, B2m,2n,2p ) ) = |a, b, c; [a, la ], [b, lb], [ c, lc], a2m , b2n , c2p , [a, b], [b, c], [c, a]|

= |a, b, c; a2m , b2n , c2p , [a, b], [b, c], [c, a]| =: C2m × C2n × C2p .

This group is mapped onto the group

C2 × C2 × C2 = |a, b, c; a2, b2, c2, [a, b], [b, c], [c, a]|

in the obvious way, by the map map ν, sending each generator of C2m × C2n × C2p

to the generator in C2 × C2 × C2 with the same name.

These maps give rise to the following short exact sequences

0 −→ K = Ker(Ab) −→ πo1 (S

3, B2m,2n,2p )Ab−→ C2m × C2n × C2p −→ 0

0 −→ H = Ker(ν ◦ Ab) −→ πo1 (S

3, B2m,2n,2p )ν◦Ab−→ C2 × C2 × C2 −→ 0

whereK is a normal subgroup of H, normal subgroup of πo1 (S

3, B2m,2n,2p ):K � H �

πo1 (S

3, B2m,2n,2p ).

Theorem 2. The finite regular orbifold covering of the orbifold (S3, B2m,2n,2p ) associ-

ated toH, is an orbifold whose underlying space is the 3-torus, T 3, and the singular locus

is a finite collection of curves with isotropy groups of order m, n, p,

q : (T 3, Em,n,p ) −→ (S3, B2m,2n,2p )

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82 H. Hilden, M. T. Lozano and J. M. Montesinos

The finite regular orbifold covering of the orbifold (S3, B2m,2n,2p ) associated to K, is amanifold Mm,n,p ,

q2 : Mm,n,p −→ (S3, B2m,2n,2p ).

The following diagram of orbifold coverings is commutative:

Mm,n,p

ցq1

q2 ↓ (T 3, Em,n,p ).ւq

(S3, B2m,2n,2p )

Proof. To construct the regular covering associated to H, the kernel of ρ = ν ◦ Ab

ρ : πo1 (S

3, B2m,2n,2p ) −→ C2 × C2 × C2

a �−→ ab �−→ bc �−→ c

(3·1)

consider the following representation µ of C2 × C2 × C2 into, Σ8, the permutation

group of 8 symbols,

µ : C2 × C2 × C2 −→ Σ8a �−→ (1, 5)(2, 6)(3, 7)(4, 8)b �−→ (1, 8)(2, 7)(3, 6)(4, 5)c �−→ (1, 4)(2, 3)(5, 8)(6, 7).

(3·2)

The monodromy map ω1 of the covering is the following composition

π1(S3 \ B)

α−→ πo

1 (S3, B2m,2n,2p )

ρ−→ C2 × C2 × C2

µ−→ Σ8. (3·3)

Each meridian element goes to a product of four disjoint transposition. Therefore,

the above map defines a covering of S3 branched on the Borromean link, and thebranching indices are all equal to 2. This covering q:M → (S3, B) is regular becauseit is associated to a normal subgroup, the kernel of ω1. It can be considered also as anorbifold covering of the Euclidean orbifold (S3, B2,2,2), with monodromy ω2 = µ ◦ ρ2 :πo1 (S

3, B2,2,2)→ Σ8, where ρ ◦ α factors through ρ2 as follows.

π1(S3 \ B)

ρ◦α−→ C2 × C2 × C2

µ−→ Σ8

ց րρ2

πo1 (S

3, B2,2,2).

The Euclidean orbifold structure lifts to a Euclidean orbifold structure in the above

covering space M , with the only possible singularity the link E = q−1(B). But thebranching indices are 2, which is equal to the order of the isotropy groups in B2,2,2.

Therefore,M is a Euclidean manifold. In fact it is possible to construct geometrically

the covering space as follows.

The universal orbifold cover of (S3, B222) is the Euclidean space E3.

u:E3 −→ (S3, B222).

The group of covering transformation of u is the crystallographic group U =

πo1 (S

3, B222) with a cube as fundamental domain. (See Figure 2.) The group U is

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Manifolds with infinite fibrations 83

x

y

z

a

bc

Fig. 2. The fundamental domain C of U .

generated by 180◦ rotations in the axis a : {y = 12} � {z = 1}, b : {z = 1

2} � {x = 1}

and c : {x = 12} � {y = 1}, partially pictured in Figure 2.

Consider the short exact sequence

0 −→ T2iT2−→ U

ρ2−→ C2 × C2 × C2 −→ 0 (3·4)

where T2 is the kernel of ρ2. Observe that the translations by 2 in the directions ofthe three axis x, y, z, belong to T2. Denote them respectively by t2x , t2y , t2z :

ρ2(iT2(t2x) = ρ2(bcbc) = e; ρ2(iT2(t2y ) = ρ2(caca) = e; ρ2(iT2(t2z ) = ρ2(abab) = e.

The subgroup G of isometries of Euclidean space E3 generated by the translations

t2x , t2y , t2z , is included in T2, and a fundamental domain in E3 is a cube with edge

of length 2, composed of 8 cubes like C. The index of T2 in U is also 8, so that

G = T2.

The commutative diagram of orbifold coverings associated to (3·4), (Lemma 1),

E3 u1−→ E3/T2 ≃ T 3

u ց ւ q

E3/U ≃ T 3/(C2 × C2 × C2) ≃ (S3, B2,2,2)

proves that M = T 3.

Therefore we have proven that q:T 3 → (S3, B) is a regular 8 fold covering ofS3 branched over the Borromean link, with monodromy ω1. As the monodromy ω1factors through πo

1 (S3, B2m,2n,2p ) (see 3·3), the orbifold structure (S

3, B2m,2n,2p ) lifts

to an orbifold structure in T 3 with singularity locus E = q−1(B) and isotropy groupsof order m, n, p. We denote this orbifold structure by (T 3, Em,n,p ). Observe that the

link E consists of 12 components, four of them with isotropy group of order m, fourmore with isotropy group of order n, and four with isotropy group of order p.

Let q2:Mm,n,p → (S3, B2m,2n,2p ) be the regular covering associated to the subgroup

K. The orbifold structure (S3, B2m,2n,2p ) lifts to an orbifold structure in Mm,n,p such

that the only possible singular locus is contained in q−12 (B). But the monodromysends the meridian elements a, b and c to cycles or order 2m, 2n and 2p respectively,therefore the points in q−12 (B) are ordinary points. Thus Mm,n,p has a manifold

structure.

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84 H. Hilden, M. T. Lozano and J. M. Montesinos

The covering q2:Mm,n,p → (S3, B2m,2n,2p ) factors through q: (T 3, Em,n,p )→(S3, B2m,2n,2p ) producing a commutative diagram of orbifold coverings

Mm,n,p

ցq1

q2 ↓ (T 3, Em,n,p ).ւq

(S3, B2m,2n,2p )

Theorem 3. The orbifold (T 3, Em,n,p ) has an infinite number of T 2-bundle structures

over S1 such that the fibres are transverse to the link Em,n,p .

Proof. The link E = q−1(B) is the projection in T 3 of u−1(B), three families ofparallel straight lines in E3, with the same directions as the three coordinate axis.

Let Ax +By + cZ = 0 be the equation of a plane Π in E3 with characteristic vector

(A, B, C). Suppose that A, B and C are integer numbers such that gcd(A, B, C) = 1.Then the plane Π projects to a torus T 2

Π in T 3 defining a T 2Π-bundle structure over S1.

The plane Π intersects transversely u−1(B), Π ⋔ u−1(B), if and only if the coefficientsA, B and C are all different from 0. Thus, T 2

Π ⋔ E, if and only if the coefficients A,B and C are all integers different from 0 and such that gcd(A, B, C) = 1. Thereforethere exists an infinite number of such T 2

Π-bundle structures over S1 such that thefibres are transverse to the link Em,n,p .

Theorem 4. Letm, n, p � 2 be integers. The hyperbolic manifoldMm,n,p , has infinite

many different Fg -bundle structures over S1, where Mm,n,p is the finite orbifold regular

covering of the hyperbolic orbifold (S3, B2m,2n,2p ) associated to K, and Fg is a connected

genus g orientable surface. Given any integer N there exists an Fg -bundle structure over

S1 in Mm,n,p with g � N .

Proof. The orbifold covering q1:Mm,n,p → (T 3, Em,n,p ) is a covering of T3 branched

over E. Therefore every T 2Π-bundle structure over S1 in T 3, lifts to a Sg (Π)-bundle

structure over S1 inM , where the fibre Sg (Π) is the covering of the torus T 2Π, branched

over the points of the intersection T 2Π � Em,n,p . Observe that the genus of the fibre

g(Π) can be computed by a Euler characteristic argument as a function of thenumbers n1, n2, n3 of points in T 2

Π � Em,n,p of type m, n, p respectively and thenumber of sheets. Therefore, givenN , it is always possible to obtain a number g(Π) >N by choosing an appropriate plane Π. The fiber Sg (Π) might be not connected. But,

since the degree of q1 is mnp, the number of components of Sg (Π) is at most mnp.Since the genus g(Π) is arbitrarily large, as is the genus of each component of Sg (Π).

Therefore we conclude that Mm,n,p can be fibered over S1 with connected fibres ofarbitrarily large genus.

Corollary 1. Any hyperbolic manifoldN which is an orbifold covering of the hyper-

bolic orbifold (S3, B2m,2n,2p ), is finitely covered by a hyperbolic 3-manifold with infinitely

many surface bundle structures.

Proof. Let

p1:N −→ (S3, B2m,2n,2p )

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Manifolds with infinite fibrations 85

be an orbifold covering such that N is in fact a hyperbolic manifold. Then the

branching indices in every component of the preimage of each component of B are

respectively 2m, 2n, or 2p. Thus the covering p1 is a virtually regular covering ([6]),so that the associated regular covering

p2:Nr −→ (S3, B2m,2n,2p )

defines an unbranched covering

p3:Nr −→ N

where p2 = p3 ◦ p1,

Nrp3−→ N

p2 ↓ ւp1

(S3, B2m,2n,2p )

.

Then, there exists a hyperbolic manifold MN which is a finite unbranched cover-

ing of Mm,n,p and Nr , that is, the one corresponding to the intersection of K, thesubgroup of πo

1 (S3, B2m,2n,2p ) associated to

q2:Mm,n,p −→ (S3, B2m,2n,2p ),

with the subgroup associated to p2:Nr → (S3, B2m,2n,2p )

MNq3 ւ ցq4

Mm,n,p ↓ Nrp3−→ N

q2 ց ւp2

(S3, B2m,2n,2p )

.

Every surface bundle structure in Mm,n,p lifts by q3, to a surface bundle structurein MN , which is a finite unbranched covering (p3 ◦ q4) of N .

The hyperbolic manifold Mm,n,p might not be the smallest hyperbolic 3-manifold

covering the orbifold Bm,n,p and having infinitely many surface bundle structures.

We give an example to this fact in the next section.

4. A remarkable arithmetic hyperbolic manifold

Consider the hyperbolic orbifold (S3, B4,4,4). Theorem 4 constructs the hyperbolic

manifold M2,2,2 with infinitely many Fg -bundle over S1 structures, as a coveringorbifold of (S3, B4,4,4). Next, we construct a smaller hyperbolic manifoldM with the

same properties asM2,2,2, and we state some other interesting results. The quaternion

group Q acts on the manifold M with the orbifold (S3, B4,4,4) as quotient.

Lemma 2. The map

ω : πo1 (S

3, B222) −→ Σ4a �−→ (12)(34)

b �−→ (13)(24)

c �−→ (14)(23)

is a transitive homomorphism with image isomorphic to the Klein group

S2,2,2 = |i, j, k ; i2 = j2 = k2 = ijk = e| ⊂ Σ4.

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86 H. Hilden, M. T. Lozano and J. M. Montesinos

x

y

z

a

b

c

Fig. 3. The fundamental domain D of T .

Proof. The map ω is the composition of ρ (3·2), the quotient C2×C2×C2 ։ S2,2,2and the injective homomorphism S2,2,2 Σ4.

Lemma 3. The orbifold covering map defined by the monodromy map ω has the torusT 3 = S1 × S1 × S1 as covering space. This orbifold covering

p : T 3 −→ (S3, B2,2,2)

is a 4 fold regular covering. The group of covering transformations is S2,2,2.

Proof. The map ω is a transitive representation of the fundamental group,

π1(S3\B), of the complement of the Borromean rings B in S3, in the permutation

group Σ4. Then let ω be the monodromy of a 4 fold covering over S3 branched overB. This covering p:M → (S3, B) is regular because the associated subgroup is thekernel of ω, and the group of covering transformations is isomorphic to the mono-dromy group S2,2,2. All branching indices are 2, because the image of every meridianelement is a product of disjoint transpositions.

Recall that (S3, B222) is a Euclidean orbifold, with singular set B and all isotropy

groups are cyclic of order 2. The Euclidean orbifold structure lifts to a Euclidean

orbifold structure in the above covering space M , with the only possible singularity

lying in the link E = p−1(B). But the branching indices are 2, equal to the order ofthe isotropy groups in B. Therefore, M is a Euclidean manifold.

To construct it, consider the short exact sequence

0 −→ TiT−→ U

ω−→ S222 −→ 0

where T , is the kernel of ω. Observe that the translations by 2 in the directions ofthe three axis x, y, z, t2x , t2y , t2z , belong to T :

ω(iT (t2x) = ω(bcbc) = e; ω(iT (t2y ) = ω(caca) = e; ω(iT (t2z ) = ω(abab) = e

but the translation by the vector (1, 1, 1), denoted by tx+y+z also belong to T :

ω(iT (tx+y+z ) = ω(bca) = (13)(24)(14)(23)(12)(34) = e.

The subgroup G of isometries of the Euclidean space E3 generated by the trans-

lations t2x , t2y , tx+y+z has a fundamental domain in E3 composed of four cubes like

C. (See Figure 3.) As G � T , and index(T in U) = index(G in U), then G = T .

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Manifolds with infinite fibrations 87

x

z

a

b

c

(0,0,1)

(2,0,0)

(0,2,0)

(2,0,1)

x

z

a

b

c

(0,0,1)

(2,0,0)

(0,2,0)

(2,0,1)

(2,2,0)

Pasting the tetrahedron in frontCutting along P1 (4,0,0)

Fig. 4. First cutting (along P1) and pasting (in front).

The associated commutative diagram of orbifold coverings (Lemma 1)

E3 u−→ E3/T ≃ T 3

u1 ց ւ p

E3/U ≃ T 3/S222 ≃ (S3, B222)

proves that M = T 3.

In the next lemma we construct a particular T 2-bundle over S1 structure in theabove T 3.

Lemma 4. Let p:T 3 → (S3, B2,2,2) be the 4 fold regular covering defined in Lemma 3.

There exists a presentation of T 3 as a trivial T 2-bundle over S1 such that the fibers aretransversal to E = p−1(B) in exactly 8 points.

Proof. We are going to modify the fundamental domainD in Figure 3 by a cutting

and pasting method such that we finally obtain a parallelepiped where the three pairs

of opposite faces are identified by translation, and such that one of these pairs cuts

the rotation axis p−1(B) transversely. The plane P1 : x+y+2z−2 = 0 intersectsD in

a triangle with vertices (2, 0, 0), (0, 2, 0), (0, 0, 1). CutD along this triangle removing

the tetrahedron ∆1 with vertices (2, 0, 0), (0, 2, 0), (0, 0, 1), (0, 0, 0). The translationt2x sends this tetrahedron to the tetrahedron with vertices (4, 0, 0), (2, 2, 0), (2, 0, 1),(2, 0, 0), which intersects D in the triangle (2, 2, 0), (2, 0, 1), (2, 0, 0) contained in thefront face. Paste t2x(∆1) to this triangle. See Figure 4.

The plane P2 : x + y + 2z − 4 = 0 intersects D in a triangle with vertices (2, 0, 1),(0, 2, 1), (2, 2, 0), cutting off a tetrahedron ∆2 with vertices (2, 0, 1), (0, 2, 1), (2, 2, 0),(2, 2, 1). Paste t−2x(∆2) to the triangle (0, 0, 1), (0, 2, 0), (0, 2, 1). The result is depic-ted in Figure 5 and Figure 6. The front face is a torus intersecting E = p−1(B)transversely in exactly 8 points. These points are labelled in Figure 6 with a, b or ccorresponding to the component of B to which they are projected.

Corollary 2. The orbifold (T 3, Em,n,p ), (m, n, p � 2), with underlying space T 3,

and singular locus E = p−1(B), where the order of the cyclic isotropy group in everycomponent of E, is m, n or p according to the component of B to which it projects, is ahyperbolic orbifold which is a (T 2, (2 points)m , (2 points)n , (4 points)p )-bundle over S1

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88 H. Hilden, M. T. Lozano and J. M. Montesinos

Cutting along P2x

y

z

a

b

c(0,0,1)

(2,0,0)

(0,2,0)(2,0,1)

(4,0,0)

(0,2,1)

(2,2,0)

Pasting the tetrahedron in backx

y

z

a

b

c(0,0,1)

(2,0,0)

(0,2,0)(2,0,1)

(4,0,0)

(0,2,1)

(2,2,0)

(–2,2,1)

Fig. 5. Second cutting (along P2) and pasting (in back).

x

y

z

a

b

c

(0,0,1)

(2,0,0)

(0,2,0)(2,0,1)(2,0,1)

(4,0,0)

(0,2,1)

(2,2,0)

(–2,2,1)

a

b

c

c

c

c

c

c

a

a

a

b

b

The fundamental domain The torus in front

Fig. 6. The fundamental domain and the torus section.

structure. The hyperbolic orbifold (T 3, Em,n,p ) is arithmetic if and only if m = n = p =2, 3.

Proof. Consider the hyperbolic orbifold structure (S3, B2m,2n,2p ), n � 2. Then the

covering p:T 3 → S3 branched over the Borromean rings with branching indices 2,induces a hyperbolic orbifold structure in T 3 with singular locus E = p−1(B) andisotropy groups cyclic of order 2m/2 = m, 2n/2 = n, 2p/2 = p. The arithmeticcondition restricting the integers m = n = p to the values 2 and 3 follows from the

list of hyperbolic orbifolds (S3, B2m,2n,2p ) that are arithmetic ([3]).

Lemma 5. The map

µ : π1(S3 \ B) −→ Σ8a �−→ (1, 2, 5, 6)(3, 4, 7, 8)b �−→ (1, 3, 5, 7)(2, 8, 6, 4)c �−→ (1, 8, 5, 4)(2, 7, 6, 3)

is a transitive homomorphism with image isomorphic to the quaternion group

S2,2,2 = |i, j, k; i2 = j2 = k2 = ijk|.

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Manifolds with infinite fibrations 89

Proof. The map µ is the composition of the quotient homomorphisms α and β

π1(S3 \ B)

α−→ πo

1 (S3, B4,4,4)

β−→ S2,2,2

a �−→ a �−→ ib �−→ b �−→ jc �−→ c �−→ k

and the injective homomorphism γ

γ : S2,2,2 −→ Σ8i �−→ (1, 2, 5, 6)(3, 4, 7, 8)j �−→ (1, 3, 5, 7)(2, 8, 6, 4)k �−→ (1, 8, 5, 4)(2, 7, 6, 3).

Theorem 5. The 8-fold coverM of S3 branched over the Borromean rings, with mono-dromy µ is an arithmetic hyperbolic manifold with an F5-bundle structure over S1.

Proof. Let q:M → S3 be the 8-fold regular cover of S3 branched over the Bor-romean rings, defined by µ. The hyperbolic orbifold structure in (S3, B4,4,4), lifts to

an orbifold hyperbolic structure in M with q−1(B) as the only possible singular set.The order of an isotropy group for a point of q−1(B), is the quotient of the orderof the isotropy group in the image (a point in B), by the branching index (four inthis covering, because the monodromy maps a meridian element to a product of two

cycles of order 4). Therefore q−1(B) is a set of ordinary points and, being not singular,the orbifold M is a hyperbolic manifold.

The group of covering transformations is isomorphic to the monodromy group

S2,2,2.

There exists a short exact sequence

0 −→ C2 = |A;A2|f

−→ S2,2,2g

−→ S2,2,2 −→ 0

i �−→ ij �−→ jk �−→ k

A �−→ i2 �−→ e.

The associated commutative diagram of orbifold coverings (Lemma 1)

Mq2−→2:1

M/C2 ≃ (T3, E2,2,2)

qց 8:1 4:1

ւp

(S3, B4,4,4) ≃ (T3, E2)/S2,2,2

shows that M is a two fold covering of T 3 branched over E.

The (T 2, (8 points)2)-bundle over S1 structure on the hyperbolic orbifold (T 3, E2,2,2)

(Corollary 2, n = 2), lifts to an Fg -bundle over S1 structure on M , where the fibre

Fg is the double covering of T 2, fibre of T 3, branched on the 8 points of intersection

with E. Then the fibre Fg is connected. The computation of the Euler characteristic

of branched coverings of 2-dimensional orbifolds gives:

2χ(T 2, (8 points)2) = χ(Fg ) =⇒ 2

(0−

8

2

)= 2− 2g =⇒ g = 5.

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90 H. Hilden, M. T. Lozano and J. M. Montesinos

x

y

z

a

b

c

Fig. 7. Part of the dodecahedron.

x

y

z

a

b

cA

B

C

Fig. 8. The splitting complex of the Borromean link.

Corollary 3. There exists a hyperbolic manifold which is an F5-bundle over S1,such that the quaternion group acts on it as a subgroup of isometries, giving the orbifold

(S3, B4,4,4) as quotient.

It can be proven, by similar arguments to those in the proof of Theorem 4, that

the manifold M has in fact an infinite number of Fg -bundle structures over S1. Weconjecture that g = 5 is the minimal genus of the fibre in such structures.

Theorem 6. There is a immersed surface F inM , globally invariant under the action

of the quaternion group, cutting the manifold in eight regular hyperbolic dodecahedra

which are permuted by the action of the quaternion group. The surface F consists of six

components which are genus four non orientable surfaces.

Proof. The universal orbifold cover of (S3, B4,4,4) is H3

r : H3 −→ (S3, B4,4,4)

and the group of covering transformations of r, is the group U studied in [2].The fundamental domain of U is a regular dodecahedron D where all the dihedral

angles are 90◦. Figure 7 shows the intersection of D, centered at the origin, with thepositive octant, in the Klein model of hyperbolic space H3. The whole dodecahedron

D is obtained by reflection of the part depicted in Figure 7, through the coordinate

planes. The labelled lines a, b and c go, in the quotient orbifold (S3, B4,4,4), to the

Borromean rings. The faces of the dodecahedron project on the splitting complex

for B made up of three discs having a single common point. See Figure 8, where

the Borromean rings and the splitting complex are obtained by reflection in the

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Manifolds with infinite fibrations 91

x

y

z

a

b

c

A

A

P1

P2

b

P1

P2

m1

m2

la A

Fig. 9. The disc A.

coordinated planes of the partial figure depicted there. Denote these discs byA, B, C,where ∂A = a, ∂B = b and ∂C = c.The theory of coverings implies that the manifold M is constructed by pasting

together eight hyperbolic dodecahedrons like D, according to the monodromy mapµ, This cell decomposition C of M has n0 vertices, n1 edges, n2 pentagonal faces andeight dodecahedral 3-cells. These numbers are easily computed whereas the dihedral

angles of D are 90◦, therefore each edge belongs to four 3-cells, each vertex belongs

to eight 3-cells,

n0 = 20× 8/8 = 20, n1 = 30× 8/4 = 60, n2 = 12× 8/2 = 48.

The fact that each edge belong to four 3-cells, thus to four pentagonal faces, suggests

that each edge e be considered as the intersection of two mutually orthogonal planes,each one made up of two pentagonal faces. Therefore the 2-skeleton is an immersed

surface F cuttingM in eight regular hyperbolic dodecahedra which are permuted by

the action of the quaternion group. The surface F is the preimage by q of the threediscs A, B, C. We can consider a tessellation T in F by pentagons, induced by the

faces of the dodecahedra, hence by C. Every pentagon in T has a unique edge marked

with a, b or c, and the opposite vertex marked with c, a or b respectively. We orientedthe marked edges according to the orientation of the components of B to which

they project. We label the eight regular hyperbolic dodecahedra in M according to

the monodromy. Each pentagon belongs to two dodecahedra. We label the pentagon

with the label of the dodecahedron below, following the positive meridian around the

marked edge. This is done in order to use the monodromy to construct F . All thepentagons with an edge marked by a, hence with the opposite vertex marked by c,project to half of the disc A.Next we construct q−1(A). The discA intersects the Borromean rings in the bound-

ary ∂A = a and two points P1, P2 in the component c. Let m1 and m2 be meridians

of c around P1 and P2 respectively, and let la be the canonical longitude of a (seeFigure 9).

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92 H. Hilden, M. T. Lozano and J. M. Montesinos

P1

P2

m1

m2

la

A

Q Q

P1P1

Q

Q

la

m1 m2

Fig. 10. The splitting of the disc A.

Q

P2P1

Q

Q

la

m1 m2

Q

P2P1

Q

Q

la

m1 m2

1

5

4

8

α

α

β

β

γ

γδ

δ

ε

ε

φ

φ

η

η

�

�

κ

κ

λλ

Fig. 11. The surface A1.

The image by the monodromy map of these elements are

µ(m1) = µ(c) = (1, 8, 5, 4)(2, 7, 6, 3)

µ(m2) = µ(bcb−1) = (1, 3, 5, 7)(2, 8, 6, 4)(1, 8, 5, 4)(2, 7, 6, 3)(1, 7, 5, 3)(2, 4, 6, 8)

= (1, 4, 5, 8)(2, 3, 6, 7)

µ(la ) = µ(m2m−11 ) = (1, 4, 5, 8)(2, 3, 6, 7)(1, 4, 5, 8)(2, 3, 6, 7)

= (1, 5)(2, 6)(3, 7)(4, 8).

Therefore the preimage of A consists of two components. The first one, A1, is made

up of the copies 1, 4, 5 and 8, and the other one, A2, of copies 2, 3, 6 and 7. Lets

construct A1. We get a splitting complex for the disc A by cutting along the two linesdepicted in Figure 10. Observe that it is composed of two regular pentagons.

Take four copies of this splitting complex numbered 1, 4, 5, and 8 and paste them

according to monodromy. Observe that 1 is pasted to 5 along a, and that 4 is pastedto 8 along a. See Figure 11. Now, observe that, because µ(la ) sends 1 to 5, the surfaceis non orientable. Denote it by Nk .

The same study can be done with the discs B and C. Hence the surface F consists

of six similar components Nk . To obtain the genus of one component, an Euler char-

acteristic argument can be used. This can be done directly from the representation

given in Figure 11, or by computing the Euler characteristic of the surface F from

the abstract decomposition T of F by regular right-angles hyperbolic pentagons (the

splitting of the disc A consists of two pentagons). Observe that each edge in the cell

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Manifolds with infinite fibrations 93

decomposition C ofM corresponds to two edges in the abstract decomposition T of F(each edge inM is the orthogonal selfintersection of F ), each vertex in C correspondsto three vertices in T,

χ(F ) = 3n0 − 2n1 + n2 = 60− 120 + 48 = −12

χ(F ) = 6χ(Nk ) = 6(2− k) = 12− 6k ⇒ 12− 6k = −12 ⇒ k = 4.

Acknowledgements. This research was partially supported by BMF-2002-04137-

C02-01.

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