Two classes of virtually fibered Montesinos links of type $\widetilde{SL_2}$

5.1 MB·English

by  Xiao Guo

Two classes of virtually fibered Montesinos links of
type SL
2
Xiao G(cid:103)uo
∗
0
University at Buffalo, SUNY
1
0
2

n
a
J
Abstract
8
2

We find two new classes of virtually fibered classic Montesinos links of type
]
T
SL .
2
G
.
h (cid:103)
t
a 1 Introduction
m
[

1 A 3-manifold is called fibered if it can be given the structure of a surface bundle over
v
4 the circle. If a 3-manifold can be finitely covered by a fibered 3-manifold, we call it
9
virtually fibered. Thevirtually fibered conjecturestatesthateverycompletehyperbolic
2
5
3-manifold with finite volume is virtually fibered. This conjecture was proposed by
.
1
0 Thurston in 1982 as a question in [Thu]. A link in S3 is called virtually fibered if its
0
exterior is virtually fibered.
1
:
v
A classic Montesinos link has a projection as shown in Figure 1, where a small
i
X
rectangle with q /p stands for a rational tangle, 1 i n. n, q and p are integers,
i i i i
r ≤ ≤
a and we may assume that n 1, p 2, and q and p are relatively prime, 1 i n.
i i i
≥ ≥ ≤ ≤
By assumption, the absolute value of q /p may be greater than one, 1 i n.
i i
≤ ≤
(q /p ,q /p ,...,q /p ) is called a cyclic rational tangle decomposition of the classic
1 1 2 2 n n
Montesinos link which has a projection as shown in Figure 1. Note that a classic
Montesinos link may have different cyclic rational tangle decompositions. Montesinos
link K in S3 has a Seifert fibered 2-fold branched covering space, W . The branch
K
set K is the preimage of K. (W ,K) (S3,K) is induced by the homomorphism
K
→
E–mail: xiaoguo@buffalo.edu
∗ (cid:101) (cid:101)
1
q q q
1 2 n
p1 p2 ··· pn
Figure 1: Classic Montesinos link
◦ ◦
π (S3 N(K)) Z , where the image of meridianal generators in π (S3 N(K)) is
1 2 1
− → −
¯1. K is a Montesinos link of type SL if W has SL geometric structure.
2 K 2
Recently, the virtually fibered conjecture has been solved for classic Montesinos
(cid:103) (cid:103)
links which are not of type SL due to the works of Walsh [Wa] and Agol-Boyer-
2
Zhang[ABZ]. Agol-Boyer-Zhangalsogaveaninfinitefamilyofvirtuallyfiberedclassic
(cid:103)
Montesinos links of type SL , in Sec. 6 of [ABZ]. Those links have cyclic rational
2
tangle decompositions (q /p,q /p,...,q /p) : p 3 odd and n is a multiple of p .
1 2 n
{ ≥ }
(cid:103)
Later, this result is extended by removing the condition that n is a multiple of p
in [GZ]. Note that a classic Montesinos link in the set (q /p,q /p,...,q /p) : p
1 2 n
{ ≥
3 odd is of type SL when q +q + +q = 0, and n = 3,q 5, or n > 3. In
2 1 2 n
} ··· (cid:54) ≥
this paper, we give another two families of virtually fibered classic Montesinos links
(cid:103)
of type SL by extending the techniques used in Sec. 6 of [ABZ].
2
Theorem(cid:103)1.1. If K is a classic Montesinos link with a cyclic rational tangle decom-
position of one of the following forms:
q q q q q q
1 2 k k+1 k+2 n
I. ( , ,..., , , ,..., ) where p,r 3 odd, k is a multiple of p, and
p p p pr pr pr ≥
(n k) is a multiple of p and r;
−
q q q
1 2 n
II. ( , ,..., ) where n 4 even, p = 2m, and m is odd,
p p p ≥
then K is virtually fibered.
Note that K is of type SL if r(q + +q )+q + +q = 0 in Case I, and
2 1 k k+1 n
··· ··· (cid:54)
q + +q = 0 and (p,n) = (2,4) in Case II.
1 n
··· (cid:54) (cid:54)
(cid:103)
In Case I, the rational tangles in the cyclic rational tangle decomposition of K
have different denominators. In Case II, the denominators are same but they are even
numbers. There are some new issues raised by these two new situations. We shall
mention them later.
For convenience, we follow the notations used in Sec. 6 of [ABZ]. Let K =
K(q /p ,q /p ,...,q /p ) be the classic Montesinos link with cyclic rational tangle
1 1 2 2 n n
2
decomposition (q /p ,q /p ,...,q /p ). Let be the base orbifold of the Seifert
1 1 2 2 n n K
B
fibered space W . W has the SL geometry structure if
K K 2
n n
q 1
(cid:103)i
e(W ) = = 0; χ( ) = 2 n+ < 0, (1)
K K
− p (cid:54) B − p
i i
i=1 i=1
(cid:88) (cid:88)
where e(W ) is the Euler number of W , and χ( ) is the Euler characteristic of .
k K K K
B B
K is geodesic in W and orthogonal to Seifert fibers of W by Lemma 2.1 of [ABZ].
K K
Let f : W be the Seifert quotient map. By the construction of W , is
K K K K
→ B B
(cid:101)
a 2-sphere with n cone points c ,c ,...,c , and the order of c is p , 1 i n.
1 2 n i i
{ } ≤ ≤
Let K∗ = f(K). Then K∗ is a geodesic equator of containing all the n cone
K
B
points. The number of components of K is also decided by the cyclic rational tangle
(cid:101)
decomposition.
1 if each p is odd and (q + +q ) is odd ,
i 1 n
···
K = 2 if each p is odd and (q + +q ) is even, (2)
| |  i 1 ··· n


# i : p is even otherwise.
i
{ }




We proof Case I of Theorem 1.1 in section 2, and Case II in section 3.
2 Proof of Theorem 1.1 in Case I.
q q q q q q
1 2 k k+1 k+2 n
In this section we prove Theorem 1.1 when K = K( , ,..., , , ,..., )
p p p pr pr pr
with p,r 3 odd, where k is a multiple of p and (n k) is a multiple of p and r. We
≥ −
consider K is of type SL , and k = 0, n = k. Otherwise, K is virtually fibered by
2
(cid:54) (cid:54)
[ABZ]. By(1),K isoftypeSL whene(W ) = (r(q + +q )+q + +q )/pr =
2 K 1 k k+1 n
− ··· ··· (cid:54)
(cid:103)
0. From (2), K has one or two components. We first prove Theorem 1.1 when K is a
(cid:103)
knot in Sec. 2.1. As in Sec. 6.2 of [ABZ], the proof can be extended to the case that
K has two components with several adjustments in Sec. 2.2.
2.1 K is a knot.
We first give the outline of the proof.
At first, we construct a finite cover of W , say Y, so that Y is a locally-trivial
K
circle bundle. Let L be the preimage of K in Y. We prove that the exterior of some
components of L in Y, denoted by M, is a surface semi-bundle. Next, we construct
(cid:101)
3
˘ ˘
M, which is a 2-fold cover of M, such that M is a fibered manifold. This covering
can be extended to Y. Suppose that the corresponding 2-fold cover of Y and L are
˘ ˘ ˘ ˘
Y and L respectively. We can isotope L M and perform Dehn twists to the surface
∩
˘ ˘ ˘
fibers of M such that the components of L M are transverse to the new surface
∩
˘ ˘ ˘
fibers of M. Then the exterior of L in Y has a structure of surface bundle over the
˘ ˘
circle. In addition, the exterior of L in Y is a finite cover of the exterior of K in W ,
K
so it is also a finite cover of the exterior of K in S3. Therefore, K is virtually fibered.
(cid:101)
Different from [ABZ], the tangles in the cyclic rational tangle decomposition of
K have different denominators. We need compose two finite covers to build Y, and
˘
perform additional Dehn twist operations to the surface fibers of M.
Recall that f : W is the Seifert quotient map and K∗ = f(K). is a
K K K
→B B
2-sphere with n cone points. K∗ goes through all cone points of : c ,c ,...,c
K 1 2 n
B { }
(cid:101)
successively. c hasorderpif1 i k,andhasorderpr ifk+1 i n. f : K K∗
i
≤ ≤ ≤ ≤ | →
is a 2-fold cyclic cover because the order of c is odd, 1 i n.
i
≤ ≤
(cid:101)
At first, we construct a proper covering map, ψ : F , such that F is a
K
→ B
smooth orientable closed surface. Since contains cone points of different orders,
K
B
we construct ψ by composing two cyclic covers, ψ and ψ .
1 2
Let Γ be the fundamental group of . Then
1 K
B
Γ = x ,x ,...,x : xp = = xp = xpr = = xpr = x x x = 1 .
1 { 1 2 n 1 ··· k k+1 ··· n 1 2··· n }
x is represented by a circle centered at c and in a small regular neighborhood of c in
i i i
, 1 i n. n k is a multiple of r. There exists a homomorphism h : Γ Z/r
K 1 1
B ≤ ≤ − →
where
¯0 if 1 i k ,
h (x ) = ≤ ≤
1 i
¯1 otherwise.

Let ψ : F(cid:48) be the r-fold cyclic covering map corresponding to h , where F(cid:48)
1 K 1
→B
is the covering space of . F(cid:48) is a 2-dimensional orbifold with underlying surface F(cid:48),
BK 0
whichisaclosedsurfacewithgenusg = (n k 2)(r 1)/2. When1 j k, ψ−1(c )
j
− − − ≤ ≤
is a set of r cone points of order p, since h (x ) = ¯0. Let ψ−1(c ) = c(cid:48) ,c(cid:48) ,...,c(cid:48) ,
1 j j { 1,j 2,j r,j}
1 j k. When k +1 j n, h (x ) = ¯1, and ¯1 has order r in Z/r, so ψ−1(c )
1 j 1 j
≤ ≤ ≤ ≤
is a cone point of order p. Let ψ−1(c ) = c(cid:48), k + 1 j n. In summary, F(cid:48) has
j j ≤ ≤
kr+(n k) cone points of order p.
−
ψ−1(K∗) is a set of r geodesics in F(cid:48). Let ψ−1(K∗) = L∗,...,L∗ . Each L∗
{ 1 r} i
goes through n cone points c(cid:48) ,...,c(cid:48) ,c(cid:48) ,...,c(cid:48) successively, 1 i r. Let τ be
i,1 i,k k+1 n ≤ ≤ 1
4
c!20 c!19 c!18 c!17 c!16 c!2,5 c!2,4 c!2,3 c!2,2 c!2,1
L∗2
c!11 c!12 c!13 c!14 c!15 c!1,5 c!1,4 c!1,3 c!1,2 c!1,1
L∗1
c!3,5 c!3,4 c!3,3 c!3,2 c!3,1
c!10 c!9 c!8 c!7 c!6
L∗3
Figure 2: ψ−1(K∗)
the deck transformation of ψ corresponding to ¯1 Z/r. Fix(τ ) = c(cid:48) ,...,c(cid:48) .
1 ∈ 1 { k+1 n}
Assume that L∗ = τ (L∗), so c(cid:48) = τ (c(cid:48) ), 1 i < r,1 j k. Orient
i+1 1 i i+1,j 1 i,j ≤ ≤ ≤
L∗ and give L∗ the induced orientation, 1 < i r. Fix an orientation on F(cid:48) such
1 i ≤
that τ is a counterclockwise rotation near c(cid:48) by the angle of 2π/rp, k +1 l n.
1 l ≤ ≤
By this construction, L∗ only intersects L∗ at c(cid:48), and the angle from L∗ to L∗ is
i1 i2 l i1 i2
(i i )2π/pr, (we alway assume that counterclockwise is the positive direction.)
2 1
−
k +1 l n,1 i ,i r.
1 2
≤ ≤ ≤ ≤
Figure 2 shows ψ−1(K∗), when K = K(2/5,...,2/5,1/15,...,1/15). Note that
5 15
there are some intersections of L∗’s in Figure 2 which are not in c(cid:48),k +1 l n .
i (cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122){ l (cid:125) ≤ ≤ }
If we draw them on F(cid:48) which is a surface with genus, those intersections will not
appear. Figure 2 is a schematic picture, so are Figure 3, 6, 7, 8, 17, 18, 20, 22 and
23.
Now we want to define ψ . Let Γ be the fundamental group of F(cid:48). Then we have:
2 2
Γ = a ,b ,...,a ,b ,y ,y ,...,y ,...,y ,y ,...,y ,y ,...,y : yp = 1,
2 { 1 1 g g 1,1 1,2 1,k r,1 r,2 r,k k+1 n i,j
g
yp = = yp = 1, [a ,b ] y y y = 1,1 i r,1 j k .
k+1 ··· n i i · i,j · k+1··· n ≤ ≤ ≤ ≤ }
i=1 1≤i≤r,1≤j≤k
(cid:89) (cid:89)
y and y are represented by a small circle on F(cid:48) centered at c(cid:48) and c(cid:48) respectively,
i,j l i,j l
1 i r,1 j k, k +1 l n. a ,b are the generators of π (F(cid:48)), 1 i g.
≤ ≤ ≤ ≤ ≤ ≤ i i 1 0 ≤ ≤
Because k and n k are both multiple of p, there is a homomorphism h : Γ
2 2
− →
5
Z/p, where
h (a ) = = h (a ) = h (b ) = = h (b ) = ¯0,
2 1 2 g 2 1 2 g
··· ···
h (y ) = h (y ) = = h (y ) = ¯1,1 i r,1 j k.
2 i,j 2 k+1 2 n
··· ≤ ≤ ≤ ≤
Let ψ be the covering of F(cid:48) corresponding to h , and F the covering space. Since the
2 2
orders of h (y ) (h (y )) equals the order of c(cid:48) (c(cid:48) ) are also p, F is a smooth closed
2 l 2 i,j l i,j
orientable surface without cone points, k +1 l n,1 i r,1 j k. Denote
≤ ≤ ≤ ≤ ≤ ≤
cˆ = ψ−1(c(cid:48) ) and cˆ = ψ−1(c(cid:48)), 1 i k,1 j r,k +1 l n.
i,j 2 i,j l 2 l ≤ ≤ ≤ ≤ ≤ ≤
ThepreimageofL∗ isasetofpgeodesics L∗ , ,L∗ , 1 i r. EachL∗ goes
i { i,1 ··· i,p} ≤ ≤ i,j
through n points cˆ ,...,cˆ ,cˆ ,...,cˆ , 1 i r,1 j p. Let τ be the deck
i,1 i,k k+1 n 2
≤ ≤ ≤ ≤
transformation of ψ corresponding to ¯1 Z/p. Fix(τ ) = cˆ ,...,cˆ ,cˆ ,...,cˆ :
2 2 i,1 i,k k+1 n
∈ {
1 i r . Assume that L∗ = τ (L∗ ), 1 i r,1 j < p. Orient L∗ ,
≤ ≤ } i,j+1 2 i,j ≤ ≤ ≤ i,1
such that it passes through cˆ ,...,cˆ ,cˆ ,...,cˆ successively, and give L∗ the
i,1 i,k k+1 n i,j
induced orientation, 1 i r,1 < j p. F admits an orientation such that τ is
2
≤ ≤ ≤
a counterclockwise rotation near the fixed points cˆ and cˆ by 2π/p, 1 i r,1
i,j l
≤ ≤ ≤
j k,k +1 l n. L∗ ,1 i r,1 j p only intersect at those fixed points
≤ ≤ ≤ { i,j ≤ ≤ ≤ ≤ }
of τ , see Figure 3, for K = K(2/5,...,2/5,1/15,...,1/15). The following remark
2
5 15
gives the details.
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
Remark 2.1. L∗ intersects L∗ at cˆ ,...,cˆ ,cˆ ,...,cˆ and the angle from L∗
i,j1 i,j2 i,1 i,k k+1 n i,j1
to L∗ is (j j )2π/p, 1 i r,1 j ,j p. When i = i , L∗ intersects L∗
i,j2 2− 1 ≤ ≤ ≤ 1 2 ≤ 1 (cid:54) 2 i1,j1 i2,j2
at cˆ, and the angle from L∗ to L∗ is (i i )2π/pr+(j j )2π/p, 1 i ,i
l i1,j1 i2,j2 2− 1 2− 1 ≤ 1 2 ≤
r;1 j ,j p,k +1 l n.
1 2
≤ ≤ ≤ ≤
Let ψ = ψ ψ : F ψ2 F(cid:48) ψ1 . ψ is a pr-fold covering map of . Then we
2 1 K K
◦ → → B B
have
cˆ ;1 i r if 1 j k,
ψ−1(c ) = { i,j ≤ ≤ } ≤ ≤
j

cˆ otherwise.
 j
L∗ = ψ−1(K∗) = L∗ : 1 i r;1 j p .
 { i,j ≤ ≤ ≤ ≤ }
Next we construct a covering space of W from ψ as in Sec. 2 of [ABZ]. W is a
K K
Seifert fibered space with basis . There is a covering of W , say Ψ, induced by the
K K
B
Seifert quotient map f : W and associated to ψ. Y is the covering space of
K K
→ B
W corresponding to Ψ. Y has a locally-trivial circle bundle Seifert structure since
K
ˆ
his basis F is a surface. Let f : Y F be the Seifert quotient map. Y inherits the
→
6
cˆ
18
cˆ
19
cˆ
cˆ 17
20
L
∗2,3
L
∗2,2 L∗3,4
L
∗2,4 cˆ16
L
∗2,5
cˆ L∗3,2 cˆ cˆ15
2,1 cˆ 3,1
1,1
L
∗1,1
β
+ β
−
cˆ2,2 cˆ1,2 cˆ3,2 cˆ14
L
∗1,2
cˆ2,3 cˆ1,3 cˆ3,3 cˆ13
L
∗1,3
cˆ2,4 cˆ1,4 L∗1,4 cˆ3,4 cˆ12
L∗1,5 L∗3,3
cˆ
cˆ2,5 1,5 cˆ3,5 cˆ11
L
∗3,1
L
∗2,1 L∗3,5 cˆ10
cˆ
6
cˆ
9
cˆ
7
cˆ
8
Figure 3: L∗ : 1 i r,1 j p
{ i,j ≤ ≤ ≤ ≤ }
7
L L
∗
ˆ
f
Y F
Ψ ψ
f
W
K K
B
K K
∗
!
Figure 4: Construction of Y
SL geometry structure from W . Recall L = Ψ−1(K). We have the commutative
2 K
diagram which is analogous to Diagram (5) in Sec. 2 of [ABZ].
(cid:103) (cid:101)
From the diagram, we can see that L = fˆ−1(L∗). L has exactly pr components
by similar discussion as in Sec. 6.1 of [ABZ]. Let L = L : 1 i r,1 j p ,
i,j
{ ≤ ≤ ≤ ≤ }
where fˆ(L ) = L∗ . fˆ : L L∗ is a two fold cyclic cover, because p and pr are
i,j i,j | i,j → i,j
both odd.
The following proposition can be proved by an analogous proof of Proposition 6.1
in [ABZ].
Proposition 2.2. The exterior of L in Y is a surface semi-bundle.
1,1
Here we do not repeat the proof. We introduce some notations and notes which
we need to use later.
Let T = fˆ−1(L∗ ), which is a vertical torus overlying L∗ . Because L∗ is a
1,1 1,1 1,1
geodesic on F, T is a totally geodesic torus. T inherits a Euclidean structure from
the SL structure of Y. If a,b is a set of two gedesics which intersects at one point,
2
{ }
and H (T) =< a,b >, then T can be identified to S1 S1 where S1 and S1
1
× ×{∗} {∗}×
(cid:103)
are two geodesics which isotopic to a and b respectively.
Let F = L∗ [ (cid:15),(cid:15)], and β = L∗ (cid:15) ,β = L∗ (cid:15) , where (cid:15) is a small
2 1,1 × − − 1,1 ×{− } + 1,1 ×{ }
positive real number. (β and β are shown in Figure 3.) Let
− +
◦
F = F F , Y = fˆ−1(F ), T = fˆ−1(β ), i = 1,2.
1 2 i i ± ±
−
Note that F is connected by the construction of F. Y is a regular (cid:15)-neighborhood
1 2
of T in Y. Define Y = Y Y . The restriction of the Seifert fibration of Y to
0 1 ∪T− 2
8
each of Y ,Y and Y is a trivial circle bundle. Give the circle fibers of Y a consistent
0 1 2 0
orientation. Fix two circle fibers of Y, φ ,φ on T and T respectively.
− + − +
F is the surface obtained by cutting F open along β . Y is a trivial circle bundle
0 + 0
overF . ChooseahorizontalsectionB ofthisstructuresuchthatB T isageodesic.
0 0 0
∩
Define B to be a subset of B in Y , i = 1,2. Orient F and let F , F and their
i 0 i 0 1 2
boundaries have the induced orientation. Also equip B , B , B and their boundaries
0 1 2
the induced orientation.
Recall that Y = T [ (cid:15),(cid:15)]. T is also fibered by geodesics isotopic to L . This
2 1,1
× −
¯
gives us a new fibration of Y with base space F . We call the fiber of Y in this
2 2 2
¯
new fibration structure new fiber, and denoted φ. We call the fiber from the original
¯
fibration of Y the original fiber, denoted φ. Let B be one horizontal section of
2
¯ ¯
Y F such that B T is a geodesic. Let N be a small regular neighborhood of
2 2 2
→ ∩
L in Y , which is disjoint from other components of L and consists of new fibers.
1,1 2
◦
Let M = Y N, and M = Y M . Then M is the exterior of L in Y. ∂M =
2 2− 1∪T− 2 1,1 2
T T T , where T = ∂N, T = T ,T = T . M is a graph manifold
2,− 2,+ 2,0 2,0 2,− − 2,+ +
∪ ∪
with boundary T and characteristic tori T and T . Let B¯0 = B¯ M , which is an
2,0 − + 2 2∩ 2
annulus with one puncture. Orient B¯0 and give ∂B¯0 the induced orietation. Equip
2 2
the new fibers of M a fixed orientation.
2
By Prop.6.1 in [ABZ], we can construct essential horizontal surfaces H Y and
1 1
⊂
H M .
2 2
⊂
Let e be the Euler number of the oriented circle bundle of Y F. Since Y is a
→
pr-fold cover of W ,
K
n
q 1 1
i
e = pr e(W ) = pr( ) = pr( (q + +q ) (q + +q ))
K 1 k k+1 n
· − p −p ··· − pr ···
i
i=1
(cid:88)
= (r(q + q )+q + +q )
1 k k+1 n
− ··· ···
by (1). Since r is odd and K only has one component (c.f. (2)), e is odd which is the
same as in the proof of Prop. 6.1 of [ABZ]. Then by choosing the horizontal sections
¯
B and B properly, H and H will have the same boundary slope as shown in Table
0 2 1 2
3 in Sec. 6.1 of [ABZ]. We describe the construction of H and H in the following
1 2
for later use. (c.f. the discussion after the proof of Prop. 6.1 of [ABZ].)
¯ ¯
Set α¯ = B T which is a geodesic. Take a fixed new fiber in T, denoted φ .
0 2 0
∩
¯ ¯
Then T = α¯ φ and Y = α¯ φ [ (cid:15),(cid:15)]. M can be expressed as following
0 0 2 0 0 2
× × × −
◦
¯ ¯
M = (α¯ φ [ (cid:15),(cid:15)]) (I φ ( δ,δ)),
2 0 0 0
× × − − × × −
9
φ¯
0
original fiber
singular point
one half of fˆ 1(L )
− ∗i,j
one of the e pieces of H
2
| |
!
−
δ
o− α¯0
δ
!
Figure 5: H
2
where I = α¯ N, and δ is a positive real number which is less than (cid:15). Let γ be
0 1
∩ 1 1
a fixed simple closed geodesic in T (cid:15) of slope with respect to the basis
×{− } 2e − 2
1 1
¯
α¯ (cid:15) ,φ (cid:15) , γ a fixed simple closed geodesic in T δ of slope
0 0 2
{ ×{− } ×{− }} ×{ } −2e − 2
¯
with respect to the basis α¯ δ ,φ δ . As in [ABZ], H = Θ Θ Θ .
0 0 2 − 0 +
{ × { } × { }} ◦ ∪ ∪
¯
Θ = γ [ (cid:15), δ], and Θ = γ [δ,(cid:15)]. Θ is a surface in (α¯ I) φ [ δ,δ]
− 1 + 2 0 0 0
× − − × − 1× t× −
such that Θ (T t ) is a union of e geodesic arcs of slope where
0
∩ × { } | | −2 − 2δe
t ( δ,δ). These e geodesic arcs are evenly distributed in (T t ) M . H is
2 2
∈ − | | ×{ } ∩
transverse to all the new fibers, and also transverse to the original fibers except when
t = 0. Figure 5 illustrates one of the e pieces of H (c. f. Figure 6 in [ABZ]). The
2
| |
surface fibration in M is generated by isotoping H around the new fibers.
2 2 2
F
Now see the construction of H , which is the same as in [ABZ]. It’s not hard to
1
see that there exists a properly embedded arc, say σ, in B connecting T and T .
1 − +
Let σ [ 1,1] be a regular neighborhood of σ in B . Suppose we pass (σ T ) 1
1 −
× − ∩ ×{− }
to (σ T ) 1 along the orientation of B T . Wrap σ [ 1,1] in B around the
− 1 − 1
∩ ×1{ }e ∩ × −
φ-direction ( − ) times as we pass from 1 to 1 in σ [ 1,1] φ. The resulting
2 − × − ×
surface is H . Let be the corresponding surface fibration of Y with H as a
1 1 1 1
F
horizontal surface.
As in [ABZ], we may suppose that ∂H = ∂H . Let H = H H . H is a non-
1 2 1 2
∪
oriented horizontal surface in M. Then forms a semi-surface bundle in M,
1 2
F ∪F F
as described in Proposition 2.2.
Next we want to reorient L : 1 i r,1 j p,(i,j) = (1,1) such that
i,j
{ ≤ ≤ ≤ ≤ (cid:54) }
10