Virtually fibred Montesinos links of type $\widetilde{SL_{2}}$

Xiao Guo, Yu Zhang
Department of Mathematics, SUNY at Buffalo E-mail: xiaoguo@buffalo.eduE-mail: yz26@buffalo.edu

Abstract

We find a larger class of virtually fibred classic Montesinos links of type $\widetilde{SL_{2}}$ , extending a result of Agol, Boyer and Zhang.

1 Introduction

A $3$ -manifold is called virtually fibred if it has a finite cover which is a surface bundle over the circle. A link in a connected $3$ -manifold is said to be virtually fibred if its exterior is a virtually fibred $3$ -manifold. Thurston conjectured that all closed hyperbolic $3$ -manifolds and all hyperbolic links in closed $3$ -manifolds are virtually fibred. This conjecture, which has been named as virtually fibred conjecture, is one of the most fundamental and difficult problems in $3$ -manifold topology.

Recall that a link $K$ in $S^{3}$ is called a generalized Montesinos link if the double branched cover $W_{K}$ of $(S^{3},K)$ is a Seifert fibred $3$ -manifold. Such a link $K$ is further said to be of type $\widetilde{SL_{2}}$ if the canonical geometric structure on $W_{K}$ is from the $\widetilde{SL_{2}}$ -geometry. When every component of the branched set in $W_{K}$ is not a fiber of the Seifert fibration of $W_{K}$ , $K$ is called a classic Montesinos link. Recent work of Walsh [Wa] and work of Agol-Boyer-Zhang [ABZ] combined together solved the virtually fibred conjecture for all generalized Montesinos links in $S^{3}$ which are not classic Montesinos links of type $\widetilde{SL_{2}}$ . Agol-Boyer-Zhang [ABZ] also gave an infinite family of virtually fibred classic Montesinos links of type $\widetilde{SL_{2}}$ . In this paper, we extend the latter result of [ABZ] to a larger family of classic Montesinos links of type $\widetilde{SL_{2}}$ . Note that every classic Montesinos link has a cyclic rational tangle decomposition of the form $(q_{1}/p_{1},q_{2}/p_{2},...,q_{n}/p_{n})$ with all $p_{i}\geq 2$ as shown in Figure 1. We prove

Refer to caption — Figure 1: Classic Montesinos link

Theorem 1.1

If a classic Montesinos link $K$ has a cyclic rational tangle decomposition of the form $(\displaystyle{\frac{q_{1}}{p},\frac{q_{2}}{p},\cdots,\frac{q_{n}}{p}})$ with $p\geq 3$ odd, then $K$ is virtually fibred.

This theorem is proved in [ABZ] when $n$ is a multiple of $p$ . Our approach follows closely to that of [ABZ]. At a number of places we need to deal with some new issues that arise. We shall describe our ways of dealing with these issues when we reach these places.

Let $K$ be a link as given in Theorem 1.1. The base orbifold ${\cal B}_{K}$ of the Seifert fibred space $W_{K}$ is a $2$ -sphere with $n$ cone points each has order $p$ . Let $f:W_{K}\rightarrow\mathcal{B}_{K}$ be the Seifert fibration which is invariant under the covering involution $\tau:W_{K}\rightarrow W_{K}$ . The Euler number $e(W_{K})$ of $W_{K}$ and the orbifold Euler characteristic $\chi({\cal B}_{K})$ of ${\cal B}_{K}$ are given by the following formulas:

e(W_{K})=-\frac{\sum_{i=1}^{n}q_{i}}{p},\;\;\;\chi({\cal B}_{K})=2-n+\frac{n}{p}.

Note that $W_{K}$ has the $\widetilde{SL_{2}}$ -geometry precisely when $e(W_{K})\neq 0$ and $\chi({\cal B}_{K})<0$ . Thus $K$ is of type $\widetilde{SL_{2}}$ precisely when

Case (1). $n=3$ , $p\geq 5$ , and $q_{1}+q_{2}+q_{3}\neq 0$ , or

Case (2). $n>3$ and $q_{1}+...+q_{n}\neq 0$ .

We shall split our proof of Theorem 1.1 into these two cases, given in Section 2 and Section 3 respectively. All main ingredients of the proof will occur already in Case (1); the proof of Case (2) will be a quick generalization.

In the remainder of this section, we shall give a few additional notes and notations which will be used throughout this paper. From now on we assume that $K=K(q_{1}/p,...,q_{n}/p)$ is a link satisfying the conditions of Case (1) or (2) listed above.

First note that $K$ is a single component knot if $q_{1}+...+q_{n}$ is odd, and a two component link otherwise. The double branched cover $W_{K}$ has the $\widetilde{SL_{2}}$ -geometry and its base orbifold ${\cal B}_{K}$ is hyperbolic. Let $\widetilde{K}$ be the corresponding branched set in $W_{K}$ and let $K^{*}=f(\widetilde{K})$ , where $f:W_{K}\rightarrow{\cal B}_{K}$ is the Seifert quotient map. By the orbifold theorem we may assume that the covering involution $\tau$ on $W_{K}$ is an isometry and the branch set $\widetilde{K}=Fix(\tau)$ is a geodesic which is orthogonal to the Seifert fibres by Lemma 2.1 of [ABZ]. Also $K^{*}$ is a geodesic in ${\cal B}_{K}$ , and is an equator of ${\cal B}_{K}$ containing all the $n$ cone points which we denote by $c_{1},c_{2},\cdots,c_{n}$ (indexed so that they appear consecutively along $K^{*}$ ). The order of each cone point $c_{i}$ is $p$ . Since $p$ is odd, the restriction map of $f$ to each component of $\widetilde{K}$ is a double covering onto $K^{*}$ .

2 Proof of Theorem 1.1 in Case (1).

In this section we prove Theorem 1.1 in the case $n=3$ , $p\geq 5$ and $q_{1}+q_{2}+q_{3}\neq 0$ . Throughout this section, without further notice, $i,j,k$ as well as numbers expressed by them will be considered as non-negative integers mod $p$ , and $i\neq k$ will also be assumed.

2.1 $K$ is a knot.

Here is an outline of the proof.

Step 1 Take a specific $p^{2}$ -fold orbifold cover $F$ of ${\cal B}_{K}=S^{2}(p,p,p)$ such that $F$ is a smooth surface. There is a corresponding free cover $\Psi:Y\rightarrow W_{K}$ of the same degree such that the base orbifold of the Seifert space $Y$ is $F$ and such that the following diagram commutes

\begin{CD}Y@>{\hat{f}}>{}>F\\ @V{}V{\Psi}V@V{}V{\psi}V\\ W_{K}@>{f}>{}>{\mathcal{B}_{K}}\end{CD}

where $\hat{f}$ is the Seifert quotient map. Note that $Y$ is a locally-trivial circle bundle over $F$ since $F$ is a smooth surface.

Let $L=\Psi^{-1}(\widetilde{K})$ . Then $L$ has exactly $p^{2}$ components, which we denote by $\{L_{i,j}:1\leqslant i,j\leqslant p\}$ . Let $L_{i,j}^{*}=\hat{f}(L_{i,j})$ . Then $\{L_{j,j}^{*}:1\leqslant j\leqslant p\}$ are $p$ mutually disjoint simple closed geodesics in $F$ .

Step 2 Construct a surface semi-bundle structure on $M=Y-\overset{p}{\underset{j=1}{\cup}}\overset{\circ}{N}(L_{j,j})$ , where $N(L_{j,j})$ is a small regular neighborhood of $L_{j,j}$ in $Y$ , $1\leqslant j\leqslant p$ .

Note that $M$ is a graph manifold with non-empty boundary, with vertices $M_{2}^{j}$ and $Y_{1}$ , where $M_{2}^{j}=\hat{f}^{-1}(L_{j,j}^{*})\times[-\epsilon,\epsilon]-\overset{\circ}{N}(L_{j,j})$ , $1\leqslant j\leqslant p$ . $Y_{1}=M-\overset{p}{\underset{j=1}{\cup}}\overset{\circ}{M^{j}_{2}}$ . We construct a surface semi-bundle structure on $M$ by using [WY] and following [ABZ].

Step 3 Isotope all $L_{i,k}$ , $i\neq k$ , such that they are transverse to the surface bundle in $M_{2}^{j}$ , denoted by $\mathcal{F}_{2}^{j}$ . This is one of the key parts of the proof. Let $U_{i,k}=\hat{f}^{-1}(L_{i,k}^{*})$ , the vertical torus over $L_{i,k}^{*}$ . By the construction in Step 2, there are exactly two singular points in the induced foliation by ${\cal F}_{2}^{j}$ on every component of $U_{i,k}\cap M_{2}^{j}$ . We show that $L$ can be oriented such that the arcs $L_{i,k}\cap M_{2}^{j}$ whose images intersect in $F$ , all travel from one of $\hat{f}^{-1}(L_{j,j})^{*}\times\{-\epsilon\}$ and $\hat{f}^{-1}(L_{j,j})^{*}\times\{\epsilon\}$ to the other. This property allows us to arrange these arcs in $U_{i,k}\cap M_{2}^{j}$ such that they always travel from the “ $-$ ” side of the surface bundle in $M_{2}^{j}$ to the “ $+$ ” side.

Step 4 Construct a double cover of $M$ , denoted by $\breve{M}$ , so that $\breve{M}$ has a surface bundle structure. Denote the corresponding double cover of $Y$ as $\breve{Y}$ ,the lift of $Y_{1}$ as $\breve{Y}_{1,1}\cup\breve{Y}_{1,2}$ , and the lift of $L_{i,k}$ as $\breve{L}_{i,k,1}\cup\breve{L}_{i,k,2}$ . Perform certain Dehn twist operations on the surface bundles of $\breve{Y}_{1,1}\cup\breve{Y}_{1,2}$ so that the $\breve{L}_{i,k,s}$ are transverse to the new surface bundles, $s=1,2$ .

This is another key part of the proof. We perform these Dehn twist operations along a union of vertical tori, $\Gamma\subset\breve{Y}_{1,1}\cup\breve{Y}_{1,2}$ . $\Gamma$ contains some boundary parallel tori of $\breve{Y}_{1,1}\cup\breve{Y}_{1,2}$ as in [ABZ], different form [ABZ], these boundary parallel tori intersect some arcs of $\breve{L}_{i,k,s}\cap\breve{Y}_{1,s}$ two times with different direction or do not intersect at all, $s=1,2$ . We call such arcs of $\breve{L}_{i,k,s}\cap\breve{Y}_{1,s}$ “bad” arcs. We construct four specific extra tori as members of $\Gamma$ to deal with the “bad” arcs, $s=1,2$ .

By Steps 1-4, the exterior of the inverse image of $\widetilde{K}$ in $\breve{Y}$ has a surface bundle structure. It is a free cover of the exterior of $\widetilde{K}$ in $W_{K}$ , which in turn is a free double cover of the exterior of $K$ in $S^{3}$ . Thus $K$ is virtually fibred in $S^{3}$ .

Now we fill in the details.

Step 1

Instead of taking the $p$ -fold cyclic cover of $\mathcal{B}_{K}$ as in [ABZ], we construct the $p^{2}$ -fold cover $F$ of ${\cal B}_{K}=S^{2}(p,p,p)$ , by a composition of two $p$ -fold cyclic covers.

Let $\Gamma_{1}=\pi_{1}(S^{2}(p,p,p))$ be the orbifold fundametal group of $S^{2}(p,p,p)$ . It has a presentation

\Gamma_{1}=<x_{1},x_{2},x_{3}\ |\ x_{1}^{p}=x_{2}^{p}=x_{3}^{p}=x_{1}x_{2}x_{3}=1>.

where $x_{r}$ is represented by a small circular loop in $S^{2}(p,p,p)$ centered at $c_{r},r=1,2,3$ . Let $\psi_{1}:F^{\prime}\rightarrow S^{2}(p,p,p)$ be the $p$ -fold cyclic orbifold cover of $S^{2}(p,p,p)$ corresponding to the homomorphism:

h_{1}:\Gamma_{1}\rightarrow\mathbb{Z}/p

where $h_{1}(x_{1})=\overline{1},h_{1}(x_{2})=-\overline{1}$ , and $h_{1}(x_{3})=\overline{0}$ . $F^{\prime}$ is as shown in Figure 2 (while $p=5$ ).

The order of $h_{1}(x_{1})$ and $h_{1}(x_{2})$ are both $p$ , so $\psi_{1}^{-1}(c_{1})=\hat{c}_{1}$ and $\psi_{1}^{-1}(c_{2})=\hat{c}_{2}$ are two points in $F^{\prime}$ (not cone points). The order of $h_{1}(x_{3})$ is $0$ , so $\psi_{1}^{-1}(c_{3})=\{c_{3,1},c_{3,2},\cdots,c_{3,p}\}$ are the only cone points in $F^{\prime}$ , each has order $p$ . We have $F^{\prime}=S^{2}(\underbrace{p,p,\cdots,p)}_{p}$ . $\psi_{1}^{-1}(K^{*})$ is a set of $p$ geodesics $L_{1}^{*},L_{2}^{*},\cdots,L_{p}^{*}$ , such that $L_{i}^{*}$ goes through the cone point $c_{3,i}$ (cf. Figure 2).

Let $\tau_{1}^{*}$ be the deck transformation of $\psi_{1}$ corresponding to $\bar{1}\in\mathbb{Z}_{p}$ . Fix $(\tau_{1}^{*})=\{\hat{c}_{1},\hat{c}_{2}\}$ . We may assume that $\tau_{1}^{*}(L_{i}^{*})=L_{i+1}^{*}$ . Orient $L_{1}^{*}$ and give $L_{i}^{*}$ the induced orientation, $1<i\leqslant p$ . $F^{\prime}$ admits an orientation such that $\tau_{1}^{*}$ is a counterclockwise rotation by ${2\pi}/{p}$ near $\hat{c}_{1}$ and a clockwise rotation by ${2\pi}/{p}$ near $\hat{c}_{2}$ on $F^{\prime}$ , since $h_{1}(x_{1})=\bar{1}$ and $h_{1}(x_{2})=-\overline{1}$ . Note that $L_{i}^{*}$ intersects $L_{j}^{*}$ at $\hat{c}_{1}$ in an angle of ${2\pi(i-j)}/{p}$ , also intersects $L_{j}^{*}$ at $\hat{c}_{2}$ in an angle of ${2\pi(j-i)}/{p}$ . (We always suppose that the counterclockwise direction is the positive direction of angles.)

Let $\Gamma_{2}$ be the orbifold fundamental group of $F^{\prime}$ . It has the presentation

\Gamma_{2}=\pi_{1}(S^{2}(\underbrace{p,p,\cdots,p}_{p}))=<y_{1},y_{2},\cdots,y_{p}\ |\ y_{1}^{p}=y_{2}^{p}=\cdots=y_{p}^{p}=y_{1}y_{2}\cdots y_{p}=1>,

where $y_{j}$ is represented by a small circular loop in $S^{2}(\underbrace{p,p,\cdots,p}_{p})$ centered at $c_{3,j}$ . Let $\psi_{2}:F\rightarrow F^{\prime}$ be the $p$ -fold cyclic orbifold cover of $F^{\prime}$ corresponding to the homomorphism:

h_{2}:\Gamma_{2}\rightarrow\mathbb{Z}/p

where $h_{2}(y_{j})=\overline{1}$ , $1\leqslant j\leqslant p$ .

The order of $h_{2}(y_{j})$ is $p$ , so $\psi_{2}^{-1}(c_{3,j})=\hat{c}_{3,j}$ is a point, $1\leqslant j\leqslant p$ . Hence $F$ is a smooth closed orientable surface without cone points. For each $r=1,2$ , $\psi_{2}^{-1}(\hat{c}_{r})$ is a set of $p$ points, which we denote by $\hat{c}_{r,1},\hat{c}_{r,2},\cdots,\hat{c}_{r,p}$ . For each $i=1,...,p$ , $\psi_{2}^{-1}(L_{i}^{*})$ is a set of $p$ simple closed geodesics, which we denote by $L_{i,1}^{*},L_{i,2}^{*},\cdots,L_{i,p}^{*}$ .

Denote the deck transformation of $\psi_{2}$ corresponding to $\bar{1}\in\mathbb{Z}/p$ by $\tau_{2}^{*}$ . Fix $(\tau_{2}^{*})=\{\hat{c}_{3,1},\hat{c}_{3,2},\cdots,\hat{c}_{3,p}\}$ . We may assume that $\tau_{2}^{*}(L_{i,j}^{*})=L_{i,j+1}^{*}$ , $\tau_{2}^{*}(\hat{c}_{1,j})=\hat{c}_{1,j+1}$ , and $\tau_{2}^{*}(\hat{c}_{2,j})=\hat{c}_{2,j+1}$ . Then $\psi_{2}^{-1}(L_{i}^{*})=\{L_{i,j}^{*}:1\leqslant j\leqslant p\}$ . Now we fix an orientation for each of $L_{1,1}^{*},L_{2,2}^{*},\cdots,L_{p,p}^{*}$ such that $L_{j,j}^{*}$ goes through $\hat{c}_{3,j}$ , $\hat{c}_{2,j}$ and $\hat{c}_{1,j}$ in order, and give $L_{i,j}^{*}=(\tau_{2}^{*})^{j-i}(L_{i,i}^{*})$ the induced orientation. $F$ admits an orientation such that $\tau_{2}^{*}$ is a counterclockwise rotation by $2\pi/p$ near the fixed points $\hat{c}_{3,i}$ , so $L_{i,1}^{*},L_{i,2}^{*},\cdots,L_{i,p}^{*}$ intersect at $\hat{c}_{3,i}$ , and $L_{i,j}^{*}$ intersects $L_{i,i}^{*}$ at $\hat{c}_{3,i}$ in an angle of $2\pi(j-i)/p$ , $1\leqslant i,j\leqslant p$ .

Since $\hat{c}_{r}=\psi_{2}(\hat{c}_{r,j})$ is not a cone point on $F^{\prime}$ , a small regular neighborhood of $\hat{c}_{r,j}$ on $F$ is a copy of a small regular neighborhood of $\hat{c}_{r}$ on $F^{\prime}$ , $r=1,2$ . Recall that $L_{i}^{*}$ intersects $L_{j}^{*}$ at $\hat{c}_{1}$ in an angle of $2\pi(i-j)/p$ , and at $\hat{c}_{2}$ in an angle of $2\pi(j-i)/p$ . Then $L_{i,j}^{*}$ intersects $L_{j,j}^{*}$ at $\hat{c}_{1,j}$ in an angle of $2\pi(i-j)/p$ , and at $\hat{c}_{2,j}$ in an angle of $2\pi(j-i)/p$ . See the schematic picture, Figure 3, for example $p=5$ . Here we didn’t depict the genus of the surface, so some curves meet in the picture actually do not meet. For convenience, we only draw a part of $L_{j,j}^{*}$ in Figure 3, $1\leqslant j\leqslant p$ .

Summarizing the above discussion, we have the following remark.

Remark 2.1

$\{L_{i,j}^{*}:1\leqslant i,j\leqslant p\}$ only intersects at the points $\{\hat{c}_{r,j},r=1,2,3,1\leqslant j\leqslant p\}$ on $F$ , and $L_{i,j}^{*}$ goes through $\hat{c}_{3,i}$ , $\hat{c}_{2,j}$ and $\hat{c}_{1,j}$ in order following the given orientation of $L_{i,j}^{*}$ , $1\leqslant i,j\leqslant p$ . In particular $L_{j,j}^{*}$ goes through $\hat{c}_{3,j},\hat{c}_{2,j}$ , and $\hat{c}_{1,j}$ , so $\{L_{j,j}^{*},1\leqslant j\leqslant p\}$ are mutually disjoint, and

	$\displaystyle L_{i,j}^{}\ \text{intersects}\ L_{j,j}^{}\ \text{at}\begin{cases}\hat{c}_{1,j}\ \text{in an angle of}\ {2\pi(i-j)}/{p},\\ \hat{c}_{2,j}\ \text{in an angle of}\ -{2\pi(i-j)}/{p},\end{cases}$
	$\displaystyle L_{j,k}^{}\ intersects\ L_{j,j}^{}\ at\ \hat{c}_{3,j}\ \text{in an angle of}\ {-2\pi(j-k)}/{p},$

Note that $i,j,k,i-j,k-j$ are considered as integers mod $p$ .

Let $\psi=\psi_{1}\circ\psi_{2}:F\xrightarrow{\psi_{2}}S^{2}(\underbrace{p,p,\cdots,p}_{p})\xrightarrow{\psi_{1}}S^{2}(p,p,p)$ , which is a $p^{2}$ -fold orbifold covering. We have $\psi^{-1}(c_{r})=\{\hat{c}_{r,1},\hat{c}_{r,2},\cdots,\hat{c}_{r,p}\}$ , $r=1,2,3$ , and $L^{*}=\psi^{-1}(K^{*})=\{L_{i,j}^{*}:1\leqslant i,j\leqslant p\}$ .

Let $\Psi_{1}:Y^{\prime}\rightarrow W_{K}$ be the $p$ -fold cover of $W_{K}$ corresponding to $\psi_{1}$ , $\Psi_{2}:Y\rightarrow Y^{\prime}$ the $p$ -fold cover of $Y^{\prime}$ corresponding to $\psi_{2}$ , and $\Psi=\Psi_{1}\circ\Psi_{2}$ . Then $\Psi:Y\rightarrow W_{K}$ is a $p^{2}$ -fold cover of $W_{K}$ corresponding to $\psi$ . $Y$ has locally-trivial circle bundle Seifert structure with base surface $F$ . Set $L=\Psi^{-1}(\widetilde{K})$ , which is a geodesic link in $Y$ . We have the following commutative diagram:

Note that $L$ has exactly $p^{2}$ components. Let $L=\{L_{i,j}:1\leqslant i,j\leqslant p\}$ , where $L_{i,j}=\hat{f}^{-1}(L_{i,j}^{*}),1\leqslant i,j\leqslant p$ . $\hat{f}|:L_{i,j}\rightarrow L_{i,j}^{*}$ is a 2-fold cover.

Step 2

By Remark 2.1, $\{L_{j,j}^{*},1\leqslant j\leqslant p\}$ are mutually disjoint. Let $F_{2}^{j}=L_{j,j}^{*}\times[-\epsilon,\epsilon]$ , for some small positive number $\epsilon$ , be a neighborhood of $L_{j,j}^{*}$ in $F$ , such that $F_{2}^{j}$ ’s are mutually disjoint. Let $\beta_{1}^{j}=L_{j,j}^{*}\times\{-\epsilon\}$ , $\beta_{2}^{j}=L_{j,j}^{*}\times\{\epsilon\}$ , and $F_{1}=\overline{F-(\underset{j=1}{\overset{p}{\cup}}F_{2}^{j})}$ . We may suppose that $\beta_{1}^{j}$ is on the left side of $L_{j,j}^{*}$ . By Remark 2.1, $L_{i,k}^{*}$ goes through $\hat{c}_{1.k}$ , $\hat{c}_{2,k}$ and $\hat{c}_{3,i}$ , so $L_{i,k}^{*}$ is separated into $2n=6$ arcs by $F_{1},F_{2}^{k}$ , and $F_{2}^{i}$ . We denote the six arcs of $L_{i,k}^{*}$ by $(L_{i,k}^{1})^{*}$ , $(L_{i,k}^{2})^{*}$ , $\cdots$ , $(L_{i,k}^{6})^{*}$ , so that $(L_{i,k}^{2})^{*}\cup(L_{i,k}^{4})^{*}\subset F_{2}^{k}$ , $(L_{i,k}^{6})^{*}\subset F_{2}^{i}$ , $(L_{i,k}^{1})^{*}\cup(L_{i,k}^{3})^{*}\cup(L_{i,k}^{5})^{*}\subset F_{1}$ , and $\hat{c}_{1,k}\in(L_{i,k}^{2})^{*}$ , $\hat{c}_{2,k}\in(L_{i,k}^{4})^{*}$ , and $\hat{c}_{3,i}\in(L_{i,k}^{6})^{*}$ . We also require that as we travel along $L_{i,k}^{*}$ in its orientation, we will pass $(L_{i,k}^{6})^{*}$ , $(L_{i,k}^{5})^{*}$ , $\cdots$ , $(L_{i,k}^{1})^{*}$ consecutively. From the construction, we have Remark 2.2 and Table 1.

Remark 2.2

$\begin{cases}\text{$(L_{i,k}^{1})^{*}$ is the part of $L_{i,k}^{*}$ between $\hat{c}_{3,i}$ and $\hat{c}_{1,k}$ inside $F_{1}$};\\ \text{$(L_{i,k}^{3})^{*}$ is the part of $L_{i,k}^{*}$ between $\hat{c}_{1,k}$ and $\hat{c}_{2,k}$ inside $F_{1}$};\\ \text{$(L_{i,k}^{5})^{*}$ is the part of $L_{i,k}^{*}$ between $\hat{c}_{2,k}$ and $\hat{c}_{3,i}$ inside $F_{1}$}.\end{cases}$

$(L_{i,k}^{1})^{*}$	tail	head	$(L_{i,k}^{3})^{*}$	tail	head	$(L_{i,k}^{5})^{*}$	tail	head
$i-k\leqslant\frac{p-1}{2}$	$\beta_{1}^{k}$	$\beta_{1}^{i}$	$i-k\leqslant\frac{p-1}{2}$	$\beta_{2}^{k}$	$\beta_{2}^{k}$	$i-k\leqslant\frac{p-1}{2}$	$\beta_{2}^{i}$	$\beta_{1}^{k}$
$i-k>\frac{p-1}{2}$	$\beta_{2}^{k}$	$\beta_{2}^{i}$	$i-k>\frac{p-1}{2}$	$\beta_{1}^{k}$	$\beta_{1}^{k}$	$i-k>\frac{p-1}{2}$	$\beta_{1}^{i}$	$\beta_{2}^{k}$

Note: $i-k$ is considered as a nonnegative integer mod $p$

Table 1: The induced orientation on

(L_{i,k}^{2l-1})^{*},1\leqslant l\leqslant 3

We need Table 1 in the following lemma and in Step 3.

Lemma 2.3

$F_{1}$ is connected.

Proof: It suffices to prove that the boundary components of $F_{1}$ , which is the set $\{\beta_{1}^{i},\beta_{2}^{j},1\leqslant i,j\leqslant p\}$ , can be mutually connected to each other by arcs in $F_{1}$ .

From Table 1, the arc $(L_{j+1,j}^{5})^{*}$ connects $\beta_{2}^{j+1}$ and $\beta_{1}^{j}$ , and the arc $(L_{j+1,j}^{1})^{*}$ connects $\beta_{1}^{j}$ and $\beta_{1}^{j+1}$ , since $j+1-j=1\leqslant\displaystyle{\frac{p-1}{2}}$ , where $p\geq 3$ . So $\beta_{2}^{j+1}$ and $\beta_{1}^{j+1}$ can be connected in $F_{1}$ , $1\leqslant{j+1}\leqslant p$ . Also $\beta_{2}^{j+1}$ and $\beta_{1}^{j}$ are connected by the arc $(L_{j+1,j}^{5})^{*}$ . Hence $\{\beta_{1}^{i},\beta_{2}^{j},1\leqslant i,j\leqslant p\}$ can be mutually connected to each other in $F_{1}$ . $\square$

Let $T^{j}=\hat{f}^{-1}(L_{j,j}^{*})\subset Y$ , which is a vertical torus over $L_{j,j}^{*}$ . Then $L_{i,j}$ and $L_{j,i}$ are transverse to $T^{j}$ for all $1\leqslant i\leqslant p$ and $i\neq j$ . The situation near $L_{j,j}^{*}$ in $F$ is described in Figure 4, while $p=5$ .

Since $L_{j,j}^{*}$ is a geodesic, the torus $T^{j}$ is a totally geodesic torus which inherits a Euclidean structure from the $\widetilde{SL_{2}}$ structure on $Y$ . For any two simple closed geodesics $\{a_{j},b_{j}\}\subset T^{j}$ with $H_{1}(T^{j})=<a_{j},b_{j}>$ , $T^{j}$ can be identified to $S^{1}\times S^{1}$ where each $S^{1}\times\{*\}$ is a geodesic isotopic to $a_{j}$ and each $\{*\}\times S^{1}$ is a geodesic isotopic to $b_{j},1\leqslant j\leqslant p$ .

Similar to Proposition 6.1 in [ABZ], we have the following proposition.

Proposition 2.4

The exterior of $\{L_{1,1}\cup L_{2,2}\cup\cdots\cup L_{p,p}\}$ in $Y$ is a surface semi-bundle.

Proof: To prove this, we apply [WY] and compare [ABZ].

Let $Y_{2}^{j}$ be the submanifold of $Y$ lying over $F_{2}^{j}=L_{j,j}^{*}\times[-\epsilon,\epsilon]$ , and $Y_{1}$ be the submanifold of $Y$ lying over $F_{1}$ . Note that $Y_{1}$ is connected since $F_{1}$ is connected by Lemma 2.3. Define

T_{1}^{j}=\hat{f}^{-1}(\beta_{1}^{j}),\ T_{2}^{j}=\hat{f}^{-1}(\beta_{2}^{j}).

Let $Y_{0}$ be the 3-manifold obtained by cutting $Y$ open along $T_{2}^{j}$ , and $F_{0}$ the surface obtained by cutting $F$ open along all $\beta_{2}^{j}$ , $1\leqslant j\leqslant p$ . The restriction of the Seifert fiberation of $Y$ to each of $Y_{0},Y_{1},Y_{2}^{j},j=1,...,p$ , is a trivial circle bundle. We give the circle fibers of $Y_{0}$ a consistent orientation. Choose a horizontal section $B_{0}$ of the bundle $Y_{0}\rightarrow F_{0}$ such that $T^{j}\cap B_{0}$ is a geodesic, $1\leqslant j\leqslant p$ . Let $B_{1}$ , $B_{2}^{j}$ be the restriction of $B_{0}$ in $Y_{1}$ and $Y_{2}^{j}$ respectively. Fix an orientation of $B_{0}$ and let $B_{1},B_{2}^{j}$ have the induced orientation. And let $\partial B_{1}$ and $\partial B_{2}^{j}$ have the induced orientation.

We denote the torus $T_{r}^{j}$ by $T_{1,r}^{j}$ and $T_{2,r}^{j}$ when we think of it as lying in $Y_{1}$ and $Y_{2}^{j}$ respectively, $r=1,2$ . Let $\phi_{k,r}^{j}$ be a fixed circle fiber in the torus $T_{k,r}^{j}$ for each of $r=1,2;k=1,2;1\leqslant j\leqslant p$ . Let $\alpha_{k,r}^{j}=T_{k,r}^{j}\cap B_{0}$ . Then $\alpha_{k,1}^{j}=-\alpha_{k,2}^{j}$ , $\phi_{k,1}^{j}=\phi_{k,2}^{j}$ , and $\{\alpha_{k,r}^{j},\phi_{k,r}^{j}\}$ form a basis of $H_{1}(T_{k,r}^{j})$ , $k=1,2$ ; $r=1,2$ ; $1\leqslant j\leqslant p$ .

By choosing the proper horizontal section $B_{0}$ , we can assume that the Seifert manifold $Y$ is obtained form $Y_{1}$ and $Y_{2}^{j}$ ’s by gluing $T_{1,1}^{j}$ to $T_{2,1}^{j}$ and $T_{1,2}^{j}$ to $T_{2,2}^{j}$ using maps $g_{r}^{j}:T_{2,r}^{j}\rightarrow T_{1,r}^{j}\ (r=1,2)$ determined by the conditions

(g_{1}^{j})_{\ast}(\alpha_{2,1}^{j})=-\alpha_{1,1}^{j}\ \ \ \ \ \ (g_{1}^{j})_{\ast}(\phi_{2,1}^{j})=\phi_{1,1}^{j}

(g_{2}^{j})_{\ast}(\alpha_{2,2}^{j})=-\alpha_{1,2}^{j}+e^{j}\phi_{1,2}^{j}\ \ \ \ \ \ (g_{2}^{j})_{\ast}(\phi_{2,2}^{j})=\phi_{1,2}^{j}

where $e=\sum_{j=1}^{p}e^{j}\in\mathbb{Z}$ is the Euler number of the oriented circle bundle $Y\rightarrow F$ . $Y\rightarrow W_{K}$ is a $p^{2}$ -fold cover so

e=p^{2}e(W_{K})=-p^{2}(\frac{q_{1}}{p}+\frac{q_{2}}{p}+\frac{q_{3}}{p})=-p(q_{1}+q_{2}+q_{3}).

Note that $q_{1}+q_{2}+q_{3}$ is an odd number since $K$ has only one component. For convenience, we may assume $e^{1}=e^{2}=\cdots=e^{p}=\displaystyle{\frac{e}{p}}=-(q_{1}+q_{2}+q_{3})=\tilde{e}$ , so $\tilde{e}$ is odd.

Each $Y_{2}^{j},j=1,...,p$ also has a circle fibration with $L_{j,j}$ as a fiber. As in [ABZ], we call the circle fibers of this circle fibration new fibers and called the fibers of $Y$ original fibers. Give new fibers of $Y_{2}^{j}$ a fixed consistent orientation. We shall denote by $\overline{\phi}^{j}$ a general new fiber in $M_{2}^{j}$ , and $\phi$ a general original fiber of $Y$ .

Let $N^{j}$ be the regular neighborhood of $L_{j,j}$ in $Y_{2}^{j}$ , which is disjoint from other components of $L$ and consists of new fibers of $Y_{2}^{j}$ . Let $M_{2}^{j}=Y_{2}^{j}\setminus\overset{\circ}{N^{j}}$ . Then $M_{2}^{j}$ is a Seifert fibered space whose circle fibers are new fibers of $Y_{2}^{j}$ . Set $T_{3}^{j}=\partial{N^{j}}$ . The exterior of $\{L_{1,1},L_{2,2},\cdots,L_{p,p}\}$ in $Y$ , $M$ , is a graph manifold and has the following JSJ decomposition

M=Y_{1}\cup M_{2}^{1}\cup M_{2}^{2}\cup\cdots\cup M_{2}^{p}

Let $\overline{B}_{2}^{j}$ be the image of one section of the circle fibration of $M_{2}^{j}$ , such that $\overline{B}_{2}^{j}$ intersects $T^{j}$ in a geodesic. Fix an orientation for $\overline{B}_{2}^{j}$ and let $\partial\overline{B}_{2}^{j}$ have the induced orientation. There is another basis of $H_{1}(T_{2,r}^{j})$ , $\{\overline{\alpha}_{2,r}^{j},\overline{\phi}_{2,r}^{j}\}$ , where $\overline{\phi}_{2,r}^{j}$ is a fixed new fiber on $T_{2,r}^{j}$ and $\overline{\alpha}_{2,r}^{j}$ is the component of $\partial\overline{B}_{2}^{j}$ on $T_{2,r}^{j}$ , $r=1,2$ , with their chosen orientations. Let $\overline{\alpha}_{2,3}^{j}=\overline{B}_{2}^{j}\cap T_{3}^{j}$ .

The relation between the old basis $\{\alpha_{2,1}^{j},\phi_{2,1}^{j}\}$ of $H_{1}(T_{2,1}^{j})$ and the new one $\{\overline{\alpha}_{2,1}^{j},\overline{\phi}_{2,1}^{j}\}$ is given by

\overline{\alpha}_{2,1}^{j}=a^{j}\alpha_{2,1}^{j}+b^{j}\phi_{2,1}^{j}\ \ \ \ \ \ \overline{\phi}_{2,1}^{j}=c^{j}\alpha_{2,1}^{j}+d^{j}\phi_{2,1}^{j}

where $a^{j},b^{j},c^{j},d^{j}$ are integers satisfying $a^{j}d^{j}-b^{j}c^{j}=\pm 1$ . We may assume that $a^{j}d^{j}-b^{j}c^{j}=1$ by reversing the orientation of the fibres $\phi$ if necessary. For convenience, we can suppose that $a^{1}=a^{2}=\cdots=a^{p}=a,\ b^{1}=b^{2}=\cdots=b^{p}=b,\ c^{1}=c^{2}=\cdots=c^{p}=c$ , and $d^{1}=d^{2}=\cdots=d^{p}=d$ .

In $M_{2}^{j}$ , we have $\overline{\alpha}_{2,1}^{j}=-\overline{\alpha}_{2,2}^{j}$ and $\overline{\phi}_{2,1}^{j}=\overline{\phi}_{2,2}^{j}$ . Thus

\overline{\alpha}_{2,2}^{j}=a\alpha_{2,2}^{j}-b\phi_{2,2}^{j}\ \ \ \ \ \ \overline{\phi}_{2,2}^{j}=-c\alpha_{2,2}^{j}+d\phi_{2,2}^{j}.

Hence, with respect to the basis $\{\alpha_{1,r}^{j},\phi_{1,r}^{j}\}$ of $H_{1}(T_{1,r}^{j})$ and $\{\overline{\alpha}_{2,r}^{j},\overline{\phi}_{2,r}^{j}\}$ of $H_{1}(T_{2,r}^{j})\ (r=1,2)$ , the gluing maps $g_{1}^{j}:T_{2,1}^{j}\rightarrow T_{1,1}^{j}$ and $g_{2}^{j}:T_{2,2}^{j}\rightarrow T_{1,2}^{j}$ can be expressed as

(g_{1}^{j})_{*}(\overline{\alpha}_{2,1}^{j})=-a\alpha_{1,1}^{j}+b\phi_{1,1}^{j}\ \ \ \ \ \ (g_{1}^{j})_{*}(\overline{\phi}_{2,1}^{j})=-c\alpha_{1,1}^{j}+d\phi_{1,1}^{j}

(g_{2}^{j})_{*}(\overline{\alpha}_{2,2}^{j})=-a\alpha_{1,2}^{j}+(a\widetilde{e}-b)\phi_{1,2}^{j}\ \ \ \ \ \ (g_{2}^{j})_{*}(\overline{\phi}_{2,2}^{j})=c\alpha_{1,2}^{j}+(d-c\widetilde{e})\phi_{1,2}^{j}

The associated matrices are

G_{1}^{j}=(g_{1}^{j})_{*}=\begin{pmatrix}-a&b\\ -c&d\end{pmatrix}\ \ \ \ \ \ G_{2}^{j}=(g_{2}^{j})_{*}=\begin{pmatrix}-a&{a\widetilde{e}-b}\\ c&{d-c\widetilde{e}}\end{pmatrix}

(G_{1}^{j})^{-1}=\begin{pmatrix}-d&b\\ -c&a\end{pmatrix}\ \ \ \ \ \ (G_{2}^{j})^{-1}=\begin{pmatrix}{c\widetilde{}e-d}&{a\widetilde{e}-b}\\ c&a\end{pmatrix}.

The graph of the JSJ-decomposition of $M$ consists of $p+1$ vertices corresponding to $Y_{1},M_{2}^{1},M_{2}^{2},\cdots,M_{2}^{p}$ , and $2p$ edges corresponding to $T_{1}^{j}$ ’s and $T_{2}^{j}$ ’s. See Figure 5.

By [WY], we can get a horizontal surface of $M$ from every non-zero solution of the following equation.

(Y-Z)\left(\begin{array}[]{c}\lambda\\ \bar{\lambda}_{1}\\ \vdots\\ \bar{\lambda}_{p}\end{array}\right)=\left(\begin{array}[]{c}0\\ {\ast}\\ \vdots\\ {\ast}\end{array}\right),\ \lambda,\bar{\lambda}_{1},\cdots,\bar{\lambda}_{p},\ast\in\mathbb{Z}

where $Y$ and $Z$ are $(p+1)\times(p+1)$ matrices defined on page 450 of [WY]. The entries of $Y$ and $Z$ are decided by the gluing matrix. From [WY], we have $y_{1,i}=y_{i,1}=\displaystyle{\frac{2}{c}},2\leqslant i\leqslant{p+1}$ , and other entries of $Y$ are all zeroes. $Z=diag(z_{1},z_{2},\cdots,z_{p+1})$ is a diagonal matrix with $z_{1}=\sum_{i=1}^{p}(\displaystyle{\frac{d}{c}+\frac{c\widetilde{e}-d}{c}})=p\widetilde{e}=e,z_{i}=\displaystyle{\frac{a}{c}-\frac{a}{c}}=0,2\leqslant i\leqslant{p+1}$ . The matrix equation (1.6) of [WY] becomes:

\begin{pmatrix}{-e}&{\displaystyle{\frac{2}{c}}}&{\displaystyle{\frac{2}{c}}}&{\cdots}&{\displaystyle{\frac{2}{c}}}\\ {\displaystyle{\frac{2}{c}}}&0&0&{\cdots}&0\\[8.61108pt] {\displaystyle{\frac{2}{c}}}&0&0&{\cdots}&0\\ {\vdots}&{\vdots}&{\vdots}&{\vdots}&{\vdots}\\ {\displaystyle{\frac{2}{c}}}&0&0&{\cdots}&0\end{pmatrix}_{(p+1)\times(p+1)}\begin{pmatrix}\lambda\\ \overline{\lambda}_{1}\\ \overline{\lambda}_{2}\\ {\vdots}\\ \overline{\lambda}_{p}\end{pmatrix}=\begin{pmatrix}0\\ {\ast}\\ {\ast}\\ {\vdots}\\ {\ast}\end{pmatrix}\Rightarrow\begin{cases}{-e\lambda+\displaystyle{\frac{2}{c}}\overline{\lambda}_{1}+\displaystyle{\frac{2}{c}}\overline{\lambda}_{2}+{\cdots}+\displaystyle{\frac{2}{c}}\overline{\lambda}_{p}=0}\\ {\displaystyle{\frac{2}{c}}\lambda=\ast}\end{cases}

We may assume $\overline{\lambda}_{j}=\overline{\lambda}$ , $1\leqslant j\leqslant p$ . Then

\begin{cases}{-e\lambda+\displaystyle{\frac{2p}{c}}\overline{\lambda}=0}\\ {\displaystyle{\frac{2}{c}}\lambda=\ast}\end{cases}

(1)

We can construct horizontal surfaces $H_{1},H_{2}^{j}$ from the solution of (1) such that the projection of $H_{1}$ to $F_{1}$ has degree $|\lambda|$ and that of $H_{2}^{j}$ to the base of $M_{2}^{j}$ has degree $|\overline{\lambda}|$ ([WY]). Suppose $\lambda\neq 0$ . Then

\frac{\lambda}{\bar{\lambda}}=\frac{2p}{ce}=\frac{2}{c\tilde{e}},\ \ \ \ \frac{\bar{\lambda}}{\lambda}=\frac{ce}{2p}=\frac{c\tilde{e}}{2}.

Suppose that $\partial H_{1}$ on $T_{i}^{j}$ with respect to the basis $\{\alpha_{1,i}^{j},\phi_{1,i}^{j}\}$ is $u_{i}^{j}\alpha_{1,i}^{j}+t_{i}^{j}\phi_{1,i}^{j}\ (i=1,2)$ , and $\partial H_{2}^{j}$ on $T_{i}^{j}$ with respect to the basis $\{\bar{\alpha}_{2,i}^{j},\bar{\phi}_{2,i}^{j}\}$ is $\bar{u}_{i}^{j}\bar{\alpha}_{2,i}^{j}+\bar{t}_{i}^{j}\bar{\phi}_{2,i}^{j}\ (i=1,2,3)$ . Then there are $\epsilon_{1}^{j},\epsilon_{2}^{j}\in\{\pm 1\}$ , so that equation (1.2) of [WY] takes on the form:

\frac{t_{1}^{j}}{u_{1}^{j}}=\frac{\epsilon_{1}^{j}\bar{\lambda}}{-\lambda c}-\frac{d}{c}=\frac{-\epsilon_{1}^{j}c\tilde{e}-2d}{2c},\ \ \ \ \ \ \frac{\bar{t}_{1}^{j}}{\bar{u}_{1}^{j}}=\frac{\epsilon_{1}^{j}\lambda}{-\bar{\lambda}c}+\frac{a}{-c}=\frac{-2\epsilon_{1}^{j}-ac\tilde{e}}{\tilde{e}c^{2}},

\frac{t_{2}^{j}}{u_{2}^{j}}=\frac{\epsilon_{2}^{j}\bar{\lambda}}{\lambda c}-\frac{c\tilde{e}-d}{c}=\frac{2d-(2-\epsilon_{2}^{j})c\tilde{e}}{2c},\ \ \ \ \ \ \frac{\bar{t}_{2}^{j}}{\bar{u}_{2}^{j}}=\frac{\epsilon_{2}^{j}\lambda}{\bar{\lambda}c}+\frac{a}{c}=\frac{2\epsilon_{2}^{j}+ac\tilde{e}}{\tilde{e}c^{2}}.

Without lose of generality, we can assume $\epsilon_{1}^{1}=\epsilon_{1}^{2}=\cdots=\epsilon_{1}^{p}=\epsilon_{1},\epsilon_{2}^{1}=\epsilon_{2}^{2}=\cdots=\epsilon_{2}^{p}=\epsilon_{2}$ .

We have $\sum_{j=1}^{p}(\displaystyle{\frac{t_{1}^{j}}{u_{1}^{j}}+\frac{t_{2}^{j}}{u_{2}^{j}}})=0$ , since $H_{1}$ is a horizontal surface. So

p[-\epsilon_{1}c\tilde{e}-2d+2d-(2-\epsilon_{2})c\tilde{e}]=0

-\epsilon_{1}+\epsilon_{2}=2.

From the above equation, $\epsilon_{2}=-\epsilon_{1}=1$ . Then

\displaystyle\frac{t_{1}^{j}}{u_{1}^{j}}=\frac{c\tilde{e}-2d}{2c},

\displaystyle\frac{\bar{t}_{1}^{j}}{\bar{u}_{1}^{j}}=\frac{2-ac\tilde{e}}{\tilde{e}c^{2}};

\displaystyle\frac{t_{2}^{j}}{u_{2}^{j}}=\frac{2d-c\tilde{e}}{2c},

\displaystyle\frac{\bar{t}_{2}^{j}}{\bar{u}_{2}^{j}}=\frac{2+ac\tilde{e}}{\tilde{e}c^{2}}.

Also $\displaystyle{\frac{\overline{t}_{1}^{j}}{\overline{u}_{1}^{j}}+\frac{\overline{t}_{2}^{j}}{\overline{u}_{2}^{j}}+\frac{\overline{t}_{3}^{j}}{\overline{u}_{3}^{j}}}=0$ , so

\frac{\overline{t}_{3}^{j}}{\overline{u}_{3}^{j}}=-(\frac{\overline{t}_{1}^{j}}{\overline{u}_{1}^{j}}+\frac{\overline{t}_{2}^{j}}{\overline{u}_{2}^{j}})=-(\frac{2-ac\tilde{e}+2+ac\tilde{e}}{\tilde{e}c^{2}})=-\frac{4}{\tilde{e}c^{2}}.

(3)

In fact, $L_{j,j}$ double covers $L_{j,j}^{*}$ , so $c=2$ . Then, we can take $\lambda=1,\bar{\lambda}=\tilde{e}$ . As in [ABZ], we take $b=0,a=d=1$ . Since $\tilde{e}$ is odd, we can determine $t_{i}^{j},u_{i}^{j},\overline{t}_{i}^{j}$ and $\overline{u}_{i}^{j}$ , as following

\displaystyle t_{1}^{j}=\frac{\tilde{e}-1}{2},\ u_{1}^{j}=1;\;\;\;\;\;t_{2}^{j}=\frac{1-\tilde{e}}{2},\ u_{2}^{j}=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;

\displaystyle\bar{t}_{1}^{j}=\frac{1-\tilde{e}}{2},\ \bar{u}_{1}^{j}=\tilde{e};\;\;\;\;\;\bar{t}_{2}^{j}=\frac{1+\tilde{e}}{2},\ \bar{u}_{2}^{j}=\tilde{e};\;\;\;\;\;\ \bar{t}_{3}^{j}=-1,\ \bar{u}_{3}^{j}=\tilde{e}.

$\partial H_{1}\cap T_{1,i}^{j}$ has $\displaystyle{|\frac{\lambda}{u_{i}^{j}}|=1}$ component, $i=1,2$ , and $\partial H_{2}^{j}\cap T_{2,i}^{j}$ has $\displaystyle{|\frac{\bar{\lambda}}{\bar{u}_{i}^{j}}|=1}$ component, $i=1,2,3$ . Since $\epsilon_{1}^{j}=-\epsilon_{2}^{j}$ , $H=H_{1}\cup(\underset{j=1}{{\overset{p}{\cup}}}H_{2}^{j})$ is non-orientable, so $M$ is a surface semi-bundle. $\square$

Next, we give the constructions for $H_{1}$ and $H_{2}^{j}$ in detail. The construction of $H_{2}^{j}$ is the same as the construction of $H_{2}$ in [ABZ].

Set $\overline{\alpha}_{0}^{j}=(\overline{B}_{2}^{j}\cap T^{j})$ . $\overline{\phi}_{0}^{j}$ is a fixed new fibre in $T^{j}$ . $\overline{\alpha}_{0}^{j}$ and $\overline{\phi}_{0}^{j}$ are both geodesics. Then $T^{j}$ is identified with $\overline{\alpha}_{0}^{j}\times\overline{\phi}_{0}^{j}$ and $Y_{2}^{j}$ with $\overline{\alpha}_{0}^{j}\times\overline{\phi}_{0}^{j}\times\left[-\epsilon,\epsilon\right]$ . Separate $\overline{\alpha}_{0}^{j}$ into two parts, $\overline{\alpha}_{0}^{j}=I^{j}\cup J^{j}$ , where $I^{j}$ meets $L_{j,j}$ . Take $0<\delta<\epsilon$ , then $M_{2}^{j}$ can be expressed as following

M_{2}^{j}=(\overline{\alpha}_{0}^{j}\times\overline{\phi}_{0}^{j}\times[-\epsilon,\epsilon])\setminus(\overset{\circ}{I^{j}}\times\overline{\phi}_{0}^{j}\times(-\delta,\delta)).

As in [ABZ], we have $H_{2}^{j}=\Theta_{-}^{j}\cup\Theta_{0}^{j}\cup\Theta_{+}^{j}$ . $\Theta_{-}^{j}=\gamma_{1}\times[-\epsilon,-\delta]$ , where $\gamma_{1}$ is a fixed connected simple closed geodesic in $T^{j}\times\{-\epsilon\}$ of slope $\displaystyle{\frac{1}{2\tilde{e}}-\frac{1}{2}}$ with respect to the basis $\{\overline{\alpha}_{0}^{j}\times\{-\epsilon\},\overline{\phi}_{0}^{j}\times\{-\epsilon\}\}$ . $\Theta_{+}^{j}=\gamma_{2}\times[\delta,\epsilon]$ , where $\gamma_{2}$ is a fixed connected simple closed geodesic in $T^{j}\times\{\delta\}$ of slope $\displaystyle{-\frac{1}{2\tilde{e}}-\frac{1}{2}}$ with respect to the basis $\{\overline{\alpha}_{0}^{j}\times\{\delta\},\overline{\phi}_{0}^{j}\times\{\delta\}\}$ , see (2.1). (Note that $\overline{\alpha}_{0}^{j}\times\{\delta\}=-\bar{\alpha}_{2,2}^{j}$ .) $\Theta_{0}^{j}$ is a surface in $J^{j}\times\overline{\phi}_{0}^{j}\times[-\delta,\delta]$ such that for each $t\in(-\delta,\delta),\ \Theta_{0}^{j}\cap T^{j}\times\{t\}$ is a union of $|\tilde{e}|$ evenly spaced geodesic arcs of slope $\displaystyle{-\frac{1}{2}-\frac{t}{2\delta\tilde{e}}}$ . $H_{2}^{j}$ is everywhere transverse to the new fibres. $H_{2}^{j}$ is also transverse to the original fibres except when $t=0$ . Thus we can construct a surface fibration $\mathcal{F}_{2}^{j}$ in $M_{2}^{j}$ by isotoping $H_{2}^{j}$ around the new fibres. Similar as [ABZ], Figure 6 illustrates one fibre of $\mathcal{F}_{2}^{1}$ in $M_{2}^{1}$ , when $K=(\displaystyle{\frac{1}{5},\frac{1}{5},\frac{1}{5}})$ , $\tilde{e}=-3$ and $\delta$ is small enough.

Next we construct $H_{1}$ similarly to [ABZ]. Recall that $B_{1}$ is a horizontal section of $Y_{1}\rightarrow F_{1}$ , and $B_{1}\cap(T_{1,1}^{j}\cup T_{1,2}^{j})$ are geodesics. $F_{1}$ is connected by Lemma 2.3, so $B_{1}$ is connected. For each $j$ , fix a properly embedded arc $\sigma^{j}$ in $B_{1}$ connecting $T_{1,1}^{j}$ and $T_{1,2}^{j}$ such that all $\sigma^{j}$ , $j=1,..,p$ are mutually disjoint (such set of arcs obviously exists). Let $\sigma^{j}\times[-1,1]$ be a regular neighborhood of $\sigma^{j}$ in $B_{1}$ such that $(\sigma^{j}\cap T_{1,1}^{j})\times[-1,1]$ is a subarc of $\alpha_{1,1}^{j}$ which follows the orientation of $\alpha_{1,1}^{j}$ as we pass from -1 to 1. Then $H_{1}$ is obtained by replacing every $\sigma^{j}\times[-1,1]$ in $B_{1}$ by a reimbedding of it which wraps around the $\phi^{j}$ -direction $\displaystyle{(\frac{1-\tilde{e}}{2})}$ times as we pass from -1 to 1 in $\sigma^{j}\times[-1,1]\times\phi^{j}$ . Let ${\cal F}_{1}$ be the corresponding surface fibration of $Y_{1}$ with $H_{1}$ as a surface fiber.

As in [ABZ], we may suppose that $\partial H_{1}=\partial H_{2}^{1}\cup\partial H_{2}^{2}\cup\cdots\cup\partial H_{2}^{p}$ and that $\mathcal{F}_{1}\cup\mathcal{F}_{2}^{1}\cup\cdots\cup\mathcal{F}_{2}^{p}$ forms a semi-surface bundle $\mathcal{F}$ in $M$ , as described in Proposition 2.4

Step 3

Recall that $T^{j}=\hat{f}^{-1}(L_{j,j}^{*})$ is vertical in the Seifert structure of $Y$ . Let $U_{i,k}=\hat{f}^{-1}(L_{i,k}^{*})$ , so $L_{i,k}\subset U_{i,k}$ . By Remark 2.1, we have $U_{i,j}\cap U_{k,j}=\phi_{1,j}\cup\phi_{2,j}$ , $i\neq k$ , and $U_{i,j}\cap U_{i,k}=\phi_{3,i}$ , $j\neq k$ , where $\phi_{r,j}=\hat{f}^{-1}(\hat{c}_{r,j}),r=1,2$ , and $\phi_{3,i}=\hat{f}^{-1}(\hat{c}_{3,i})$ . $L_{j,j}$ is a 2-fold cover of $L_{j,j}^{*}$ , so it intersects each of $\phi_{1,j},\phi_{2,j}$ and $\phi_{3,j}$ twice. Then $U_{i,k}^{0}=U_{i,k}\cap M$ is a 6-punctured torus.

Let $U_{i,k}^{0,2l}=\hat{f}^{-1}((L_{i,k}^{l})^{*})\cap M$ , $1\leqslant i,k\leqslant p,i\neq k,1\leqslant l\leqslant 6$ . $U_{i,k}^{0,2l-1}$ is a vertical annulus, and $U_{i,k}^{0,2l}$ is a vertical annulus with two punctures, $l=1,2,3.$ By the way that we define $(L_{i,k}^{l})^{*}$ in Step 2, $(U_{i,k}^{0,1}\cup U_{i,k}^{0,3}\cup U_{i,k}^{0,5})\subset Y_{1}$ , $(U_{i,k}^{0,2}\cup U_{i,k}^{0,4})\subset M_{2}^{k}$ , $U_{i,k}^{0,6}\subset M_{2}^{i}$ and $U_{i,k}^{0,2l}\cap\phi_{l,k}\neq\emptyset$ , $l=1,2$ , and $U_{i,k}^{0,6}\cap\phi_{3,i}\neq\emptyset$ .

Similarly, $L_{i,k}$ is separated into 12 parts, $L_{i,k}^{1},L_{i,k}^{2},\cdots,L_{i,k}^{12}$ , such that $(L_{i,k}^{l}\cup L_{i,k}^{l+6})\subset U_{i,k}^{0,l},1\leqslant l\leqslant 6$ . We have $\hat{f}(L_{i,k}^{l})=\hat{f}(L_{i,k}^{l+6})=(L_{i,k}^{l})^{*}$ .

Note that $(L_{i,k}^{2l}\cup L_{i,k}^{2l+6})\cap\phi_{l,k}\neq\emptyset$ , $l=1,2$ . and $(L_{i,k}^{6}\cup L_{i,k}^{12})\cap\phi_{3,i}\neq\emptyset$ .

By Remark 2.1,

\text{The angle between $L_{j,j}^{*}$ and $L_{i,j}^{*}$, $i\neq j$, at $\hat{c}_{1,j}$ $\in$}\begin{cases}(0,\pi)&\text{$i-j\leqslant\frac{p-1}{2}$ (mod $p$)},\\ (\pi,2\pi)&\text{otherwise}.\end{cases}

\text{The angle between $L_{j,j}^{*}$ and $L_{i,j}^{*}$, $i\neq j$, at $\hat{c}_{2,j}$ $\in$}\begin{cases}(\pi,2\pi)&\text{$i-j\leqslant\frac{p-1}{2}$ (mod $p$)},\\ (0,\pi)&\text{otherwise}.\end{cases}

\text{The angle between $L_{j,j}^{*}$ and $L_{j,k}^{*}$, $j\neq k$ at $\hat{c}_{3,j}$ $\in$}\begin{cases}(\pi,2\pi)&\text{$j-k\leqslant\frac{p-1}{2}$ (mod $p$)},\\ (0,\pi)&\text{otherwise}.\end{cases}

Now, we reorient $\{L_{i,j}^{*}:1\leqslant i,j\leqslant p\}$ by changing the orientations of $\{L_{i,j}^{*}:1\leqslant i-j\leqslant\displaystyle{\frac{p-1}{2}}\ \text{(mod $p$)},1\leqslant i,j\leqslant p\}$ . We also change the orientations of the corresponding $L_{i,j}$ ’s.

After this reorientation, the angle between $L_{j,j}^{*}$ and $L_{i,j}^{*}$ at $\hat{c}_{1,j}$ $\in(\pi,2\pi)$ ; the angle between $L_{j,j}^{*}$ and $L_{i,j}^{*}$ at $\hat{c}_{2,j}$ $\in(0,\pi)$ ; the angle between $L_{j,j}^{*}$ and $L_{j,k}^{*}$ at $\hat{c}_{3,j}$ $\in(0,\pi)$ , ( $i\neq j,j\neq k$ , here $i$ may equal $k$ ).

If we equip $\{L_{i,k}^{l}:1\leqslant i,k\leqslant p,i\neq k,1\leqslant l\leqslant 12\}$ with the new induced orientation, then

\text{the tails of}\begin{cases}L_{i,k}^{2}\ \text{and}\ L_{i,k}^{8}\ \text{lie in}\ T_{1}^{k},\\ L_{i,k}^{4}\ \text{and}\ L_{i,k}^{10}\ \text{lie in}\ T_{2}^{k},\\ L_{i,k}^{6}\ \text{and}\ L_{i,k}^{12}\ \text{lie in}\ T_{2}^{i}.\end{cases}\text{the heads of}\begin{cases}L_{i,k}^{2}\ \text{and}\ L_{i,k}^{8}\ \text{lie in}\ T_{2}^{k},\\ L_{i,k}^{4}\ \text{and}\ L_{i,k}^{10}\ \text{lie in}\ T_{1}^{k},\\ L_{i,k}^{6}\ \text{and}\ L_{i,k}^{12}\ \text{lie in}\ T_{1}^{i}.\end{cases}

(5)

The induced orientation on other segments of $L_{i,k}$ ’s are described in Table 2, where $r=0$ or $6$ . We get Table 2 from Table 1 by changing the orientation of $\{L_{i,k}^{*}:1\leqslant i,k\leqslant p,1\leqslant i-k\leqslant\frac{p-1}{2}\}$ .

$L_{i,k}^{1+r}$	tail	head	$L_{i,k}^{3+r}$	tail	head	$L_{i,k}^{5+r}$	tail	head
${i-k}\leqslant\frac{p-1}{2}$	$T_{1}^{i}$	$T_{1}^{k}$	${i-k}\leqslant\frac{p-1}{2}$	$T_{2}^{k}$	$T_{2}^{k}$	${i-k}\leqslant\frac{p-1}{2}$	$T_{1}^{k}$	$T_{2}^{i}$
${i-k}>\frac{p-1}{2}$	$T_{2}^{k}$	$T_{2}^{i}$	${i-k}>\frac{p-1}{2}$	$T_{1}^{k}$	$T_{1}^{k}$	${i-k}>\frac{p-1}{2}$	$T_{1}^{i}$	$T_{2}^{k}$

Note: ${i-k}$ is considered as a nonnegative integer mod $p$ , $r=0$ or $6$ .

Table 2: The induced orientation on

L_{i,k}^{2l-1},1\leqslant l\leqslant 6

Fix a transverse orientation on the surface bundle $\mathcal{F}_{2}^{j}$ in $M_{2}^{j}$ . By the construction of $H$ in Step 2, there are two singular points in the foliation of $U_{i,k}^{0,2l}$ given by $U_{i,k}^{0,2l}\cap\mathcal{F}_{2}$ if $\delta$ is small enough, $l=1,2,3$ (cf. Figure 7). We can isotope $L_{i,k}^{2l+r}$ in $U_{i,k}^{0,2l}$ such that $L_{i,k}^{2l}$ is transverse to $\overset{p}{\underset{j=1}{\cup}}\mathcal{F}_{2}^{j}$ , and travels from the “ $-$ ” side of $\mathcal{F}_{2}^{j}$ to the “ $+$ ” side ( $1\leqslant l\leqslant 3,\ r=0,6$ ). We only discuss the isotopy in $M_{2}^{1}$ , and others are similar.

By the definition of $U_{i,k}^{0,l}(1\leqslant l\leqslant 6)$ , ( $U_{i,1}^{0,2}\cup U_{i,1}^{0,4}\cup U_{1,i}^{0,6})\subset M_{2}^{1}$ , $2\leqslant i\leqslant p$ . Suppose that $\delta$ is small enough so that the singular foliation on $U_{i,1}^{0,2}$ , $U_{i,1}^{0,4}$ or $U_{1,i}^{0,6}$ induced by $\mathcal{F}_{2}^{1}$ has two singular points. The singular points on $U_{i,1}^{0,2l}\in\phi_{l,1}$ $(l=1,2)$ , and the singular points on $U_{1,i}^{0,6}\in\phi_{3,1}$ , $2\leqslant i\leqslant p$ . Recall that $L_{i,k}^{l}\cup L_{i,k}^{l+6}\in U_{i,k}^{0,l}$ , $1\leqslant l\leqslant 6$ , so $\{L_{i,k}^{2l+r}:1\leqslant i,k\leqslant p,i\neq k,(k,l)=(1,1),(1,2),\text{ or }(i,l)=(1,3),r=0,6\}=L\cap M_{2}^{1}$ .

Let $S=\{(i,k,l,r):1\leqslant i,k\leqslant p,i\neq k,(k,l)=(1,1),(1,2)$ , or $(i,l)=(1,3),r=0,6\}$ . Fix $(i,k,l,r)\in S$ . By Lemma 2.1 in [ABZ], $L_{i,k}^{2l+r}$ is always perpendicular to the original fibers in $U_{i,k}^{0,2l}$ . If $L_{i,k}^{2l+r}$ passes from the “ $+$ ” to the “ $-$ ” side of $\mathcal{F}_{2}^{1}$ ’s leaves along its orientation, we isotope $L_{i,k}^{2l+r}$ passing the singular point smoothly along the original fibres in a small neighborhood of $U_{i,k}\cap M_{2}^{1}$ , and place them as in Figure 7. By (5), $L_{i,1}^{2+r}\subset U_{i,1}^{0,2}$ travels from $T_{2,1}^{1}$ to $T_{2,2}^{1}$ ; $L_{i,1}^{4+r}\subset U_{i,1}^{0,4}$ travel from $T_{2,2}^{1}$ to $T_{2,1}^{1}$ . Then $L_{i,1}^{2}$ and $L_{i,1}^{4}$ are in the different sides of their own singular points, so are $L_{i,1}^{8}$ and $L_{i,1}^{10}$ . Also by (5), $L_{i,1}^{4+r}\subset U_{i,1}^{0,4}$ and $L_{1,i}^{6+r}\subset U_{1,i}^{0,6}$ are both travel from $T_{2,2}^{1}$ to $T_{2,1}^{1}$ , so $U_{1,i}^{0,6}\cap\mathcal{F}_{2}^{1}$ is similar to $U_{i,1}^{0,4}\cap\mathcal{F}_{2}^{1}$ . $U_{i,1}^{0,2}\cap\mathcal{F}_{2}^{1}$ is as shown in Figure 7-1, and $U_{i,1}^{0,4}\cap\mathcal{F}_{2}^{1}$ is as shown in Figure 7-2, in other words, they alternate with the singular points in the original circle fiber direction.

The isotopies are similar to that in [ABZ]. Next, we want to show that the above isotopies of $L_{i,k}^{2l+r}$ won’t block each other. Since $U_{i,1}^{0,2}\cap U_{k,1}^{0,2}=\phi_{1,1}$ , $U_{i,1}^{0,4}\cap U_{k,1}^{0,4}=\phi_{2,1}$ , and $U_{1,i}^{0,6}\cap U_{1,k}^{0,6}=\phi_{3,1}$ , $2\leqslant i,k\leqslant p,i\neq k$ , potential obstructions happen in the set $\{L_{i,1}^{2+r}:2\leqslant i\leqslant p,r=0,6\}$ , $\{L_{i,1}^{4+r}:2\leqslant i\leqslant p,r=0,6\}$ , or $\{L_{1,i}^{6+r}:2\leqslant i\leqslant p,r=0,6\}$ . $L_{i,k}^{2l}$ and $L_{i,k}^{2l+r}$ are in the different halves of $U_{i,k}^{0,2l}$ (cf. Figure 7), so they won’t block each other since the smooth $\phi$ -vertical isotopy can not cross the punctures on $U_{i,k}^{0,2l}$ . Next we check $\{L_{i,1}^{2}:2\leqslant i\leqslant p\}$ , and the discussions for $\{L_{i,1}^{4}:2\leqslant i\leqslant p\}$ and $\{L_{1,i}^{6}:2\leqslant i\leqslant p\}$ are similar. By (5), $L_{i,1}^{2}$ travels from $T_{2,1}^{1}$ to $T_{2,2}^{1}$ for any $2\leqslant i\leqslant p$ . We may assume that we need isotope $L_{i,1}^{2}$ smoothly along $\phi$ above its singular point, as shown in Figure 8-1, $2\leqslant i\leqslant p$ . (There is only half of $U_{i,1}^{0,2}\cap(T^{1}\times[-\delta,\delta])$ in Figure 8-1). Although the singular points of $U_{i,1}^{0,2}$ ’s in $\phi_{1,1}$ are at different positions, and the distribution of the intersection points of $\phi_{1,1}$ and $\{L_{i,1}^{2}:2\leqslant i\leqslant p\}$ may be different for different $(q_{1},q_{2},q_{3})$ , we can isotope $L_{i,1}^{2}$ in the positive direction of $\phi_{1,1}$ above all singular points, so they won’t block each other. Figure 8-2 illustrates the situation near $U_{i,1}^{0,4}\cap T^{1}\times[-\delta,\delta]$ . In Figure 8, $s_{i,1}^{2l}$ denotes the singular point of $L_{i,1}^{2l}$ , for $i=2,3,4,5$ and $l=1,2$ .

By generalizing the above discussion to $M_{2}^{j}$ , $1\leqslant j\leqslant p$ , we have the following proposition which is analogous to Proposition 6.2 in [ABZ].

Proposition 2.5

Suppose $1\leqslant j\leqslant p$ . Fix a transverse orientation on $\mathcal{F}_{2}^{j}$ in $M_{2}^{j}$ . There is a smooth $\phi$ -vertical isotope of $L_{i,j}$ and $L_{j,i}$ $(1\leqslant i\leqslant p,i\neq j)$ in $\overset{p}{\underset{i=1,i\neq j}{\cup}}(U_{i,j}^{0}\cup U_{j,i}^{0})$ , fixed outside a small neighborhood of $(\overset{p}{\underset{i=1,i\neq j}{\cup}}(U_{i,j}^{0}\cup U_{j,i}^{0}))\cap M_{2}^{j})$ , which reposition $L_{i,j}^{l}$ and $L_{j,i}^{k}$ to be transverse to $\mathcal{F}_{2}^{j}$ and to pass from the negative to the positive side of $\mathcal{F}_{2}^{j}$ ’s leaves while traveling along the orientation on $L_{i,j}^{l}$ and $L_{1,j}^{k}$ , where $1\leqslant i\leqslant p,i\neq j;l=2,8,4,10;k=6,12$ .

Proposition 2.5 finishes Step 3.

Step 4

At first we construct a double cover of $M$ similarly to [ABZ]. Set $M_{0}$ to be the 3-manifold obtained by cutting $M$ open along $\{T_{2}^{j}$ , $1\leqslant j\leqslant p\}$ . Take two copies of $M_{0}$ , denoted $\breve{M}_{0,1}$ and $\breve{M}_{0,2}$ . We have $\breve{M}_{0,s}=\breve{Y}_{1,s}\cup\breve{M}_{2,s}^{1}\cup\breve{M}_{2,s}^{2}\cup\cdots\cup\breve{M}_{2,s}^{p}$ , where $\breve{Y}_{1,s}$ is a copy of $Y_{1}$ , and $\breve{M}_{2,s}^{j}$ is a copy of $M_{2}^{j}$ , $1\leqslant j\leqslant p$ and $s=1,2$ . $\breve{M}$ is a 3-manifold obtained by gluing $\breve{M}_{2,1}^{j}$ to $\breve{Y}_{1,2}$ along $\breve{T}_{2,1}^{j}$ , and $\breve{M}_{2,2}^{j}$ to $\breve{Y}_{1,1}$ along $\breve{T}_{2,2}^{j}$ , where $\breve{T}_{2,s}^{j}\subset\breve{M}_{0,s}$ is a copy of $T_{2}^{j}$ , $1\leqslant j\leqslant p$ and $s=1,2$ . The gluing map is the same as the one used to glue $M_{2}^{j}$ back to $Y_{1}$ to get $M$ for every $1\leqslant j\leqslant p$ . This construction is shown in Figure 9. It’s clear that $\breve{M}$ is a free double cover of $M$ . We denote $p_{2}$ to be the covering map. Then

p_{2}^{-1}(Y_{1})=\breve{Y}_{1,1}\cup\breve{Y}_{1,2}

p_{2}^{-1}(M_{2}^{j})=\breve{M}_{2,1}^{j}\cup\breve{M}_{2,2}^{j}

p_{2}^{-1}(T_{r}^{j})=\breve{T}_{r,1}^{j}\cup\breve{T}_{r,2}^{j}\ (r=1,2)

$\breve{M}$ is also a graph manifold and Figure 9 illustrates the graph of the JSJ-decomposition of $\breve{M}$ .

By construction, $\breve{M}$ is fibred. In Step 2, we construct $H=H_{1}\cup(\overset{p}{\underset{j=1}{\cup}}H_{2}^{j})$ as a leaf of the semi-bundle structure of $M$ . Let $\breve{H}=p_{2}^{-1}(H)$ . It’s not hard to see that $\breve{H}$ is a connected orientable surface in $\breve{M}$ . Let

\breve{\mathcal{F}}=\breve{\mathcal{F}}_{1,1}\cup\breve{\mathcal{F}}_{1,2}\cup(\underset{s=1}{\overset{2}{\cup}}\underset{j=1}{\overset{p}{\cup}}\breve{\mathcal{F}}_{2,s}^{j}),

where $\breve{\mathcal{F}}_{1,s}=p_{2}^{-1}(\mathcal{F}_{1})\cap\breve{Y}_{1,s},\breve{\mathcal{F}}_{2,s}^{j}=p_{2}^{-1}(\mathcal{F}_{2}^{j})\cap\breve{M}_{2,s}^{j},s=1,2$ . Then $\breve{F}$ is a surface bundle in $\breve{M}$ , and $\breve{H}$ is one leaf of $\breve{F}$ . Fix a transverse orientation for $\breve{\mathcal{F}}$ and let $\breve{\mathcal{F}}_{1,s}$ and $\breve{\mathcal{F}}_{2,s}^{j}$ have the induced transverse orientation, $s=1,2$ , $1\leqslant j\leqslant p$ .

$p_{2}$ can be extended to a double cover of $Y$ . Denote the covering space by $\breve{Y}$ . $\breve{Y}$ inherits the Seifert fibred structure from $Y$ with fibre $\breve{\phi}$ . Let $\breve{\phi}$ have the inherited orientation from $\phi$ .

\breve{Y}=\breve{Y}_{1,1}\cup\breve{Y}_{1,2}\cup(\underset{s=1}{\overset{2}{\cup}}\underset{j=1}{\overset{p}{\cup}}\breve{Y}_{2,s}^{j})

where $\breve{Y}^{j}_{2,1}\cup\breve{Y}_{2,2}^{j}=p_{2}^{-1}(Y_{2}^{j})$ and $\breve{M}_{2,s}^{j}\subset\breve{Y}_{2,s}^{j}$ , $s=1,2$ , $1\leqslant j\leqslant p$ . By construction, $p_{2}^{-1}(L_{j,j})$ are two copies of $L_{j,j}$ , denoted $\breve{L}_{j,j,1}$ , and $\breve{L}_{j,j,2}$ . $\breve{L}_{j,j,s}\subset{\breve{Y}_{2,s}^{j}}$ , $s=1,2$ . Then

\breve{M}=\breve{Y}\setminus(\underset{s=1}{\overset{2}{\cup}}\underset{j=1}{\overset{p}{\cup}}N(\breve{L}_{j,j,s})).

Equip $\breve{L}_{j,j,s}$ with the inherited orientation from $L_{j,j}$ , $1\leqslant j\leqslant p,s=1,2$ .

By the construction of $\breve{M}$ , $p_{2}^{-1}(U_{i,k})$ is a $\breve{\phi}$ -vertical torus, $1\leqslant i,k\leqslant p,i\neq k$ . Let $\breve{U}_{i,k}=p_{2}^{-1}(U_{i,k})$ , and $\breve{U}_{i,k}^{0}=\breve{U}_{i,k}\cap\breve{M}$ . Denote $p_{2}^{-1}(U_{i,k}^{0,l})=\breve{U}_{i,k,1}^{0,l}\cup\breve{U}_{i,k,2}^{0,l}$ , $1\leqslant l\leqslant 6$ , where $(\breve{U}_{i,k,s}^{0,1}\cup\breve{U}_{i,k,s}^{0,3}\cup\breve{U}_{i,k,s}^{0,5})\subset\breve{Y}_{1,s}$ , $(\breve{U}_{i,k,s}^{0,2}\cup\breve{U}_{i,k,s}^{0,4})\subset\breve{M}_{2,s}^{k}$ , and $\breve{U}_{i,k,s}^{0,6}\subset\breve{M}_{2,s}^{i},s=1,2$ . Denote $\breve{L}=p_{2}^{-1}(L)$ . $p_{2}^{-1}(L_{i,k})$ has two components, and two-fold trivially covers its Seifert quotient image, since $L_{i,k}$ is a two-fold cyclic cover of $L^{*}_{i,k}$ . Then $\breve{L}$ has $2p^{2}$ components. Suppose $(\breve{L}_{i,k,s}^{l}\cup\breve{L}_{i,k,s}^{l+6})\subset\breve{U}_{i,k,s}^{0,l}$ to be the lift of $(L_{i,k}^{l}\cup L_{i,k}^{l+6})\subset U_{i,k}^{0,l},1\leqslant l\leqslant 6$ . Equip $\breve{L}_{i,k,s}^{l}$ with the inherited orientation, $1\leqslant l\leqslant 12,s=1,2$ .

Now we fix $s=1,2$ in the following discussion.

By construction, proposition 2.5 holds for every $\breve{M}_{2,s}^{j}$ , $1\leqslant j\leqslant p$ . In particular, we can isotope $\{\breve{L}_{i,j,s}\cup\breve{L}_{j,i,s}:1\leqslant i,j\leqslant p,i\neq j,s=1,2\}$ smoothly along $\breve{\phi}$ such that $\{\breve{L}^{2+r}_{i,j,s}\cup\breve{L}^{4+r}_{i,j,s}\cup\breve{L}^{6+r}_{j,i,s}:1\leqslant i,j\leqslant p,i\neq j,r=0,6,s=1,2\}$ pass from the negative side to the positive side of $\breve{\mathcal{F}}_{2,s}^{j}$ ’s leaves when traveling along their orientation in $\breve{M}_{2,s}^{j}$ . This isotopy is fixed outside a small neighborhood of $\overset{2}{\underset{s=1}{\cup}}(\overset{p}{\underset{i=1,i\neq j}{\cup}}(U_{i,j,s}^{0}\cup U_{j,i,s}^{0}))\cap\breve{M}_{2,s}^{j})$ . Then $\{\breve{L}_{i,k,s}:1\leqslant i,k\leqslant p,i\neq k,s=1,2\}$ are transverse to $\{\breve{\mathcal{F}}_{2,s}^{j},1\leqslant j\leqslant p,s=1,2\}$ . Next, we perform Dehn twist operations a sufficiently large number of times on $\{\breve{\mathcal{F}}_{1,s}:s=1,2\}$ along a set of tori $\Gamma$ , such that the new surface fiber is transverse to $\{\breve{L}_{i,k,s},1\leqslant i,k\leqslant p,i\neq k,s=1,2\}$ everywhere. Unlike in [ABZ], here we need more vertical tori. $\Gamma$ consists $2p+4$ $\breve{\phi}$ -vertical tori, $\{\breve{V}_{s}^{j}:1\leqslant j\leqslant p+2,s=1,2\}$ . $\{\breve{V}_{s}^{j}:1\leqslant j\leqslant p,s=1,2\}$ are $2p$ vertical tori, where $\breve{V}_{s}^{j}$ parallel to $\breve{T}_{1,s}^{j}$ for $1\leqslant j\leqslant p,s=1,2$ . In addition to these boundary parallel tori, we need $4$ extra tori, which we will give the construction later.

$\breve{U}_{i,k,s}^{0,2l-1}$ is a $\breve{\phi}$ -vertical annulus, $1\leqslant l\leqslant 6$ . $\breve{L}_{i,k,s}^{2l-1}\cup\breve{L}_{i,k,s}^{2l+5}\subset\breve{U}_{i,k,s}^{0,2l-1}$ , and transverse to $\breve{\phi}$ by Lemma 2.1 in [ABZ], $1\leqslant l\leqslant 3$ . By the construction of $\mathcal{F}$ in Step 2, we may assume that the oriented arcs $\breve{L}_{i,k,s}^{2l-1}\cup\breve{L}_{i,k,s}^{2l+5}$ are always transverse from the negative to the positive side of $\breve{\mathcal{F}}_{1,s}$ near the boundary of $\breve{U}_{i,k,s}^{0,2l-1}$ , $1\leqslant l\leqslant 3$ . By the construction of $H_{1}$ in Step 2, $\breve{\mathcal{F}}_{1,s}$ is transverse to the original fibre $\breve{\phi}$ . $\breve{\mathcal{F}}_{1,s}\cap\breve{U}_{i,k,s}^{0,2l-1}$ gives another foliation of $\breve{U}_{i,k,s}^{0,2l-1}$ , $1\leqslant l\leqslant 3$ . One leaf of $\breve{\mathcal{F}}_{1,s}\cap\breve{U}_{i,k,s}^{0,2l-1}$ is shown in Figure 10.

Let $\breve{V}_{s}^{j}$ be a $\breve{\phi}$ -vertical torus in a small neighborhood of $\breve{T}_{1,s}^{j}$ in $int(\breve{Y}_{1,s})$ and parallel to $\breve{T}_{1,s}^{j}$ , $1\leqslant j\leqslant p,s=1,2$ . By the above discussion we may assume that $\breve{V}_{s}^{j}$ intersect $\breve{L}_{i,k,s}^{2l-1}\cup\breve{L}_{i,k,s}^{2l+5}$ transversely, $1\leqslant l\leqslant 3$ .

Now we introduce a sign for every intersection point of an arc $\eta$ and a $\breve{\phi}$ -vertical torus $T$ in $\breve{M}$ . Suppose $\eta^{*}$ and $\iota^{*}$ are the images of $\eta$ and $T$ under the Seifert quotient map respectively. We say $\eta$ intersects T positively if the algebraic intersection number $i(\eta^{*},\iota^{*})=+1$ (cf. Figure 11), otherwise we say $\eta$ intersects T negatively. Equip the image of $\breve{V}_{s}^{j}$ under the Seifert quotient map the same orientation as $\breve{L}_{j,j,s}$ , $1\leqslant j\leqslant p$ . It’s easy to see that if the tail of $\breve{L}_{i,k,s}^{2l-1+r}$ is in $\breve{T}_{1,s}^{j}$ , $\breve{L}_{i,k,s}^{2l-1+r}$ intersects $\breve{V}_{s}^{j}$ positively,(cf. Figure 10-1), $1\leqslant l\leqslant 3,r=0,6$ .

The following remark is obtained from Table 2.

Remark 2.6

Any of $\{\breve{L}_{i,k,s}^{1+r},{i-k}\leqslant\displaystyle\frac{p-1}{2},r=0,6,s=1,2\}\cup\{\breve{L}_{i,k,s}^{3+r},{i-k}>\displaystyle\frac{p-1}{2},r=0,6,s=1,2\}$ , intersects $\overset{2}{\underset{s=1}{\cup}}\overset{p}{\underset{j=1}{\cup}}\breve{V}_{s}^{j}$ twice and in different signs, and any of $\{\breve{L}_{i,k,s}^{5+r}:1\leqslant i,k\leqslant p,i\neq k,r=0,6,s=1,2\}$ intersects $\overset{2}{\underset{s=1}{\cup}}\overset{p}{\underset{j=1}{\cup}}\breve{V}_{s}^{j}$ once and positively. Note that $\overset{2}{\underset{s=1}{\cup}}\overset{p}{\underset{j=1}{\cup}}\breve{V}_{s}^{j}$ only intersects the above arcs.

By the above remark, when we perform Dehn twist along $\{\breve{V}_{s}^{j}:1\leqslant j\leqslant p,s=1,2\}$ , the relation between $\{\breve{L}_{i,k,s}^{1+r},{i-k}\leqslant\displaystyle\frac{p-1}{2},r=0,6,s=1,2\}\cup\{\breve{L}_{i,k,s}^{3+r},{i-k}>\displaystyle\frac{p-1}{2},r=0,6,s=1,2\}$ and the new fibres are the same as the relation of those with the old fibres. To make $\{\breve{L}_{i,k,s}^{1+r},{i-k}\leqslant\displaystyle\frac{p-1}{2},r=0,6,s=1,2\}\cup\{\breve{L}_{i,k,s}^{3+r},{i-k}>\displaystyle\frac{p-1}{2},r=0,6,s=1,2\}$ transverse to the new fibres, we need extra Dehn twists. We now construct $\{\breve{V}_{s}^{p+1}\cup\breve{V}_{s}^{p+2}\subset\breve{Y}_{1,s},s=1,2\}$ . $\breve{V}_{s}^{p+r}$ is the lift of $\phi$ -vertical torus $V^{p+r}\subset Y_{1}$ , where $V^{p+r}=\hat{f}^{-1}(l_{r}),r=1,2$ . $l_{1}$ and $l_{2}$ are two simple closed curves in $F_{1}$ and constructed as follows.

Recall that $\beta_{1}^{j}=L_{j,j}^{*}\times\{-\epsilon\}$ , $\beta_{2}^{j}=L_{j,j}^{*}\times\{\epsilon\}$ , $1\leqslant j\leqslant p$ . Let $a_{r}^{j}$ be the intersection of $\beta_{r}^{j}$ and a small disk centered at $\hat{c}_{1,j}$ with radius bigger than $\epsilon$ , $1\leqslant j\leqslant p$ . Push $a_{r}^{j}$ into $F_{1}$ a little bit, as shown in Figure 12. By Remark 2.1, $\{L_{i,j}^{*},1\leqslant i,j\leqslant p,i\neq j\}$ intersect $L_{j,j}^{*}$ at $\hat{c}_{1,j}$ . Then $a_{r}^{j}\cap L_{i,j}^{*}\neq\emptyset$ , $1\leqslant i,j\leqslant p$ , $i\neq j$ , $r=1,2$ . Since $\{L_{i,k}^{*}:1\leqslant i,k\leqslant p,i\neq k\}$ intersects $L_{j,j}^{*}$ only at $\hat{c}_{1,j}$ , $\hat{c}_{2,j}$ and $\hat{c}_{3,j}$ , we can take an arc $b_{r}^{j}$ in $F_{1}$ which is parallel to $\beta_{r}^{j}$ and between $\hat{c}_{2,j}$ and $\hat{c}_{3,j}$ , $1\leqslant j\leqslant p$ , $r=1,2$ (cf. Figure 12).

By Remark 2.1 and Remark 2.2, we can connect

\begin{cases}a_{1}^{j}\ \text{and}\ b_{1}^{j},&\text{ by the arc parallel to $(L_{j+\frac{p+1}{2},j}^{3})^{*}$},\\ b_{1}^{j}\ \text{and}\ a_{1}^{j+\frac{p+1}{2}}&\text{by the arc parallel to $(L_{j,j+\frac{p+1}{2}}^{1})^{*}$};\\ a_{2}^{j}\ \text{and}\ b_{2}^{j},&\text{ by the arc parallel to $(L_{j+\frac{p-1}{2},j}^{3})^{*}$},\\ b_{2}^{j}\ \text{and}\ a_{2}^{j+\frac{p-1}{2}}&\text{by the arc parallel to $(L_{j,j+\frac{p-1}{2}}^{1})^{*}$}.\\ \end{cases}

(6)

$1\leqslant j\leqslant p$ . Note that the indices are in $\mathbb{Z}/p$ . Then we get two simple closed curve $l_{1}$ , and $l_{2}$ in $F_{1}$ , (cf. Figure 12). $l_{1}$ and $l_{2}$ can be oriented as in Figure 12, such that the orientations on $a_{1}^{j}$ and $a_{2}^{j}$ are the same as the orientation on $L_{j,j}^{*}$ .

By the construction of $l_{1}$ and $l_{2}$ , they only intersect $L_{i,k}^{*}$ at the arcs $a_{1}^{j}$ and $a_{2}^{j}$ , $1\leqslant j\leqslant p$ . $\{a_{1}^{j}\cup a_{2}^{j}:1\leqslant j\leqslant p\}$ only intersects the arcs of $L_{i,k}^{*}$ between $\hat{c}_{3,i}$ and $\hat{c}_{1,k}$ , and between $\hat{c}_{1,k}$ and $\hat{c}_{2,k}$ , i.e. $(L_{i,k}^{1})^{*}$ and $(L_{i,k}^{3})^{*}$ by Remark 2.2. $l_{1}\cup l_{2}$ intersects $(L_{i,k}^{1})^{*}$ or $(L_{i,k}^{3})^{*}$ exactly once and does not intersect $(L_{i,k}^{5})^{*}$ , $1\leqslant i,k\leqslant p$ , $i\neq k$ . Moreover $i((L_{i,k}^{1})^{*},l_{r})=i((L_{i,k}^{3})^{*},l_{r})=-1,r=1,2$ , since the angle between $L_{j,j}^{*}$ and $L_{i,j}^{*}$ at $\hat{c}_{1,j}\in(\pi,2\pi)$ after the reorientation in Step 3(cf. Figure 12). Since $\breve{V}_{s}^{p+m}$ and $\breve{L}_{i,k,s}^{2m-1+r}$ are the lifts of $V^{p+m}$ and $L_{i,k}^{2m-1+r}$ respectively, and their images under the Seifert quotient map are $l_{m}$ and $(L_{i,k}^{2m-1})^{*}$ , respectively, $m=1,2,r=0,6$ , we have the following remark.

Remark 2.7

$\overset{2}{\underset{s=1}{\cup}}(\breve{V}_{s}^{p+1}\cup\breve{V}_{s}^{p+2})$ intersects $\breve{L}_{i,k,s}^{1+r}$ and $\breve{L}_{i,k,s}^{3+r}$ exactly once and negatively, and $\overset{2}{\underset{s=1}{\cup}}(\breve{V}_{s}^{p+1}\cup\breve{V}_{s}^{p+2})\cap\breve{L}_{i,k,s}^{5+r}=\emptyset$ , $1\leqslant i,k\leqslant p$ , $i\neq k$ , $r=0,6.$

Note that $l_{1}\cup l_{2}\cap\{L_{j,j}^{*},1\leqslant j\leqslant p\}=\emptyset$ , so we may suppose that the tori in $\Gamma$ are mutually disjoint.

Let $N(\breve{V}_{s}^{j})$ be a small regular neighborhood of $\breve{V}_{s}^{j}$ in $int(\breve{Y}_{1,s})$ consisting of $\breve{\phi}$ -circle fibres, $1\leqslant j\leqslant p+2$ . We may suppose that $N(\breve{V}_{s}^{i})\cap N(\breve{V}_{s}^{j})=\emptyset\ (1\leqslant i,j\leqslant p+2;i\neq j)$ if we take the regular neighborhoods small enough. Define $\partial N(\breve{V}_{s}^{j})=\breve{\partial}_{1,s}^{j}\cup\breve{\partial}_{2,s}^{j},1\leqslant j\leqslant p+2$ . $\breve{\partial}_{1,s}^{j}$ and $\breve{\partial}_{2,s}^{j}$ are two $\breve{\phi}$ -vertical tori, $1\leqslant j\leqslant p+2$ .

By Remark 2.6 $\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})$ intersects each of $\breve{L}_{i,k,s}^{5}$ and $\breve{L}_{i,k,s}^{11}$ in a single arc, $1\leqslant i,k\leqslant p$ , $i\neq k$ , $s=1,2$ . We may assume that the tail of this arc (with the induced orientation) is contained in $\breve{\partial}_{1,s}^{j}$ , since $\{\breve{L}_{i,k,s}^{5}\cup\breve{L}_{i,k,s}^{11}:1\leqslant i,k\leqslant p,i\neq k,s=1,2\}$ intersects $\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})$ in the same sign. On the other hand, $N(\breve{V}_{s}^{p+1})\cup N(\breve{V}_{s}^{p+2})$ intersects each of $\breve{L}_{i,k,s}^{1},\breve{L}_{i,k,s}^{7},\breve{L}_{i,k,s}^{3},\breve{L}_{i,k,s}^{9}$ in a single arc, by Remark 2.7. We also assume that the tail of this arc (with the induced orientation) contained in $\breve{\partial}_{1,s}^{p+r}$ , $r=1,2$ .

Since $\breve{\mathcal{F}}_{1,s}$ is transverse to the old fibres, $\breve{\mathcal{F}}_{1,s}\cap N(\breve{V}_{s}^{j})$ is a foliation of $N(\breve{V}_{s}^{j})$ by annuli, $1\leqslant j\leqslant p+2$ , $s=1,2$ . The Dehn twist operation on $\breve{\mathcal{F}}_{1,s}$ , denoted by $D_{s}^{j}$ wraps these annuli sufficiently large number of times around the $\breve{\phi}$ -fibres in the direction opposite to the transverse orientation of $\breve{\mathcal{F}}_{1,s}$ as we pass from $\breve{\partial}_{1,s}^{j}$ to $\breve{\partial}_{2,s}^{j}$ , $1\leqslant j\leqslant p+2$ . Figure 13 shows one Dehn twist.

We perform Dehn twist operations on $\breve{\mathcal{F}}_{1,s}$ , by two steps. At first, perform $\{D_{s}^{j},1\leqslant j\leqslant p\}$ , then perform $D_{s}^{p+1}$ and $D_{s}^{p+2}$ . Let $\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ be the new surface bundle obtained by Dehn twist operations $\{D_{s}^{j},1\leqslant j\leqslant p\}$ on $\breve{\mathcal{F}}_{1,s}$ . By Remark 2.6, there are two cases for the foliation of $\breve{U}_{i,k,s}^{0,2l-1}$ by $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ , $s=1,2$ .

Case A. $l=1$ and ${i-k}\leqslant\displaystyle{\frac{p-1}{2}}$ , or $l=2$ and ${i-k}>\displaystyle{\frac{p-1}{2}}$ . $1\leqslant i,k\leqslant p,i\neq k$

In this case $\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})\cap\breve{L}_{i,k,s}^{2l-1}$ and $\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})\cap\breve{L}_{i,k,s}^{2l+5}$ are both two arcs. In particular $\breve{L}_{i,k,s}^{1}$ and $\breve{L}_{i,k,s}^{7}$ intersect $\breve{V}_{s}^{i}$ and $\breve{V}_{s}^{k}$ , and $\breve{L}_{i,k,s}^{3}$ and $\breve{L}_{i,k,s}^{9}$ intersect $\breve{V}_{s}^{k}$ twice by Table 2. Then $\breve{L}_{i,k,s}^{2l-1}$ intersect $\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})$ two times, and in different signs by Remark 2.6. If the tail of one arc of $\breve{L}_{i,k,s}^{2l-1}\cap\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})$ lies in $\overset{p}{\underset{j=1}{\cup}}\breve{\partial}_{1,s}^{j}$ , the tail of the other one lies in $\overset{p}{\underset{j=1}{\cup}}\breve{\partial}_{2,s}^{j}$ . The same thing happens for $\breve{L}_{i,k,s}^{2l+5}$ .

$\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is obtained from $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}$ by $D_{s}^{j}$ ’s ( $1\leqslant j\leqslant p$ , $s=1,2$ ) in a regular neighborhood of two $\breve{\phi}$ -circle fibres in $\breve{U}_{i,k,s}^{0,2l-1}$ in the direction opposite to the transverse orientation of $\breve{\mathcal{F}}_{1,s}$ as we pass from $(\overset{p}{\underset{j=1}{\cup}}\breve{\partial}_{1,s}^{j})\cap\breve{U}_{i,k,s}^{0,2l-1}$ to $(\overset{p}{\underset{j=1}{\cup}}\breve{\partial}_{2,s}^{j})\cap\breve{U}_{i,k,s}^{0,2l-1}$ , $1\leqslant l\leqslant 3$ . Figure 10-2 shows $\breve{U}_{i,k,s}^{0,3}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ , and $\breve{U}_{i,k,s}^{0,1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is similar to $\breve{U}_{i,k,s}^{0,3}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ .

Now we perform an isotope of $\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ in a small regular neighborhood of $\breve{U}_{i,k,s}^{0,2l-1}$ in $\breve{Y}_{1,s}$ , say $N(\breve{U}_{i,k,s}^{0,2l-1})$ , such that $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ changes back to $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}$ as shown in Figure 10-1, and $\partial N(\breve{U}_{i,k,s}^{0,2l-1})\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ always keeps to be the same. The isotope is described as following. Push the whole arc $AB$ (like a finger move) along $\breve{\phi}$ in $\breve{U}_{i,k,s}^{0,2l-1}$ in the direction opposite to the Dehn twist and the same times as the Dehn twist, meanwhile keep $\partial N(\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}})$ all the time (cf. Figure 10-2). We still call the surface bundle $\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ after this isotope.

In case A, $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is same as $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}$ .

Case B. $l=3,1\leqslant i,k\leqslant p,i\neq k$ .

In this case, $\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})\cap\breve{L}_{i,k,s}^{2l-1}$ and $\overset{p}{\underset{j=1}{\cup}}N(\breve{V}_{s}^{j})\cap\breve{L}_{i,k,s}^{2l+5}$ are both one arc. The foliation of $\breve{U}_{i,k,s}^{0,5}$ determined by $\breve{U}_{i,k,s}^{0,5}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is obtained from the foliation $\breve{U}_{i,k,s}^{0,5}\cap\breve{\mathcal{F}}_{1,s}$ by $D_{s}^{j}$ ’s $(1\leqslant j\leqslant p)$ in a regular neighborhood of one $\breve{\phi}$ -circle fibre in $\breve{U}_{i,k,s}^{0,5}$ in the direction opposite to the transverse orientation of $\breve{\mathcal{F}}_{1,s}$ as we pass from $(\overset{p}{\underset{j=1}{\cup}}\breve{\partial}_{1,s}^{j})\cap\breve{U}_{i,k,s}^{0,5}$ to $(\overset{p}{\underset{j=1}{\cup}}\breve{\partial}_{2,s}^{j})\cap\breve{U}_{i,k,s}^{0,5}$ (cf. Figure 10-3).

Now we perform the Dehn twist operations $D_{s}^{p+1}$ , $D_{s}^{p+2}$ on $\mathcal{F}^{\prime}_{1,s}$ .

Suppose $\breve{\mathcal{F}}_{1,s}^{{}^{\prime\prime}}$ is the new surface bundle of $\breve{Y}_{1,s}$ obtained by $D_{s}^{p+1}$ and $D_{s}^{p+2}$ . By Remark 2.7, $(N(\breve{V}_{s}^{p+1})\cup N(\breve{V}_{s}^{p+2}))\cap\breve{L}_{i,k,s}^{2l-1}$ and $(N(\breve{V}_{s}^{p+1})\cup N(\breve{V}_{s}^{p+2}))\cap\breve{L}_{i,k,s}^{2l+5}$ are both one arc, $l=1,2$ . Then the situation of $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime\prime}}$ is the same as case B in the first step (cf. Figure 10-3), $l=1,2$ . The foliation of $\breve{U}_{i,k,s}^{0,5}$ determined by $\breve{U}_{i,k,s}^{0,5}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime\prime}}$ is the same as $\breve{U}_{i,k,s}^{0,5}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ because $(N(\breve{V}_{s}^{p+1})\cup N(\breve{V}_{s}^{p+2}))\cap(\breve{U}_{i,k,s}^{0,5})=\emptyset$ .

Finally like [ABZ], we adjust $\breve{\mathcal{F}}_{1,s}^{{}^{\prime\prime}}$ by isotope which is the identity in a small regular neighborhood of $\partial\breve{Y}_{1,s}$ and outside a small regular neighborhood of $\overset{3}{\underset{l=1}{\cup}}(\underset{1\leqslant i,k\leqslant p,i\neq k}{\cup}\breve{U}_{i,k,s}^{0,2l-1})$ such that the interval foliation in each $\breve{U}_{i,k,s}^{0,2l-1}$ becomes transverse to $\breve{L}_{i,k,s}^{2l-1}\cup\breve{L}_{i,k,s}^{2l+5}$ , $1\leqslant l\leqslant 3$ (cf. Figure 10-4). We denote the resulting surface bundle $\breve{\mathcal{F}}_{1,s}^{{}^{\prime\prime\prime}}$ . Since $\breve{\mathcal{F}}_{1,s}^{{}^{\prime\prime\prime}}\cap\partial\breve{Y}_{1,s}$ is same as $\breve{\mathcal{F}}_{1,s}\cap\partial\breve{Y}_{1,s}$ , the resulting surface bundle $\breve{\mathcal{F}}^{{}^{\prime\prime\prime}}$ in $\breve{M}$ is transverse to the link $\overset{2}{\underset{s=1}{\cup}}(\underset{1\leqslant i,k\leqslant p;i\neq k}{\cup}\breve{L}_{i,k,s})$ every where. In other words, the exterior of $\overset{2}{\underset{s=1}{\cup}}(\underset{1\leqslant i,k\leqslant p;i\neq k}{\cup}\breve{L}_{i,k,s})$ in $\breve{M}$ has a surface bundle structure, i.e., $\overset{2}{\underset{s=1}{\cup}}(\underset{1\leqslant i,k\leqslant p;i\neq k}{\cup}\breve{L}_{i,k,s})$ is a fibred link, i.e. $K$ is a virtually fibred Montesinos knot. Now we finish the proof of Theorem 1.1 when $K$ is a classic Montesinos knot, and $n=3$ .

2.2 $\mathbf{K}$ is a link of two components.

The proof in this case is mostly same as in section 2.1, except for the following changes.

In this case, $f|:\widetilde{K}\rightarrow K^{*}$ is a trivial 2-fold cover. Then $L=\Psi^{-1}(\widetilde{K})$ is a geodesic link with $2p^{2}$ components. Let $L=\{L_{i,j}^{r}:1\leqslant i,j\leqslant p,r=1,2\}$ . $\hat{f}(L_{i,j}^{r})=L_{i,j}^{*}$ , $r=1,2$ . Now we take $M$ to be the complement of $\{L_{j,j}^{1}:1\leqslant j\leqslant p\}$ in $Y$ . As in the proof of Proposition 2.4, we have $M=Y_{1}\cup M_{2}^{1}\cup M_{2}^{2}\cup\cdots\cup M_{2}^{p}$ , where $M_{2}^{j}=\hat{f}^{-1}(L_{j,j}^{*}\times[-\epsilon,\epsilon])-\overset{\circ}{N}(L_{j,j}^{1})$ , $1\leqslant j\leqslant p$ . Note that $L_{j,j}^{2}$ is a fibre of the Seifert fibre structure on $M_{2}^{j},1\leqslant j\leqslant p$ .

Since $K$ has two components, and $p$ is odd, we have $q_{1}+q_{2}+q_{3}$ is even.

e=p^{2}e(W_{K})=-p(q_{1}+q_{2}+q_{3}),\ \ \ \ \ \ \tilde{e}=\frac{e}{p}=-(q_{1}+q_{2}+q_{3})

so $|\tilde{e}|$ is even and non-zero. We use the definitions of $a,b,c,d$ in the proof of Proposition 2.4 , and in this case $c=1$ . We can take $a=c=d=1$ and $b=0$ . i.e.

\overline{\alpha}^{j}=\alpha^{j},\ \ \ \ \ \ \overline{\phi}^{j}=\alpha^{j}+\phi^{j},\ \ \ \ \ \ 1\leqslant j\leqslant p.

In the following discussion, we will always assume that $1\leqslant j\leqslant p$ except for special indication.

By a similar analysis in Proposition 2.4, we can see that there are orientable horizontal surfaces $H_{1}$ in $Y_{1}$ , and $H_{2}^{j}$ in $M_{2}^{j}$ which piece together to form the semifibre $H$ where the projection of $H_{1}$ to $F_{1}$ has degree $|\lambda|=1$ and that of $H_{2}^{j}$ to the base of $M_{2}^{j}$ has degree $|\overline{\lambda}|=\displaystyle|\frac{\tilde{e}}{2}|$ .

Further, if we suppose that the slope of $H_{1}$ on $T_{i}^{j}$ is given by $u_{i}^{j}\alpha_{1,i}^{j}+t_{i}^{j}\phi_{1,i}^{j}\ (i=1,2)$ , and that of $H_{2}^{j}$ on $T_{i}^{j}$ is $\overline{u}_{i}^{j}\alpha_{2,i}^{j}+\overline{t}_{i}^{j}\phi_{2,i}^{j}\ (i=1,2,3)$ . Then we get the following result from (2.1) and (3)

\frac{t_{1}^{j}}{u_{1}^{j}}=\frac{\tilde{e}-2}{2},\ \frac{t_{2}^{j}}{u_{2}^{j}}=\frac{2-\tilde{e}}{2};\ \ \ \ \ \ \ \ \frac{\overline{t}_{1}^{j}}{\overline{u}_{1}^{j}}=\frac{2-\tilde{e}}{\tilde{e}},\ \frac{\overline{t}_{2}^{j}}{\overline{u}_{2}^{j}}=\frac{2+\tilde{e}}{\tilde{e}},\ \frac{\overline{t}_{3}^{j}}{\overline{u}_{3}^{j}}=-\frac{4}{\tilde{e}}.

Since $\tilde{e}$ is even, the values of the coefficients $u_{i}^{j},t_{i}^{j},\overline{u}_{i}^{j},\overline{t}_{i}^{j}$ are given as following

	$\displaystyle t_{1}^{j}=\frac{\tilde{e}}{2}-1,\ \ u_{1}^{j}=1;\ \ \ \ \ \ t_{2}^{j}=1-\frac{\tilde{e}}{2},\ \ u_{2}^{j}=1;$
	$\displaystyle\overline{t}_{1}^{j}=1-\frac{\tilde{e}}{2},\ \ \overline{u}_{1}^{j}=\frac{\tilde{e}}{2};\ \ \ \ \ \overline{t}_{2}^{j}=1+\frac{\tilde{e}}{2},\ \ \overline{u}_{2}^{j}=\frac{\tilde{e}}{2};$
	$\displaystyle(\overline{t}_{3}^{j},\overline{u}_{3}^{j})=\begin{cases}(-2,\displaystyle{\frac{\tilde{e}}{2}})&\text{if $\tilde{e}=4k+2$},\\[8.61108pt] (-1,\displaystyle{\frac{\tilde{e}}{4}})&\text{if $\tilde{e}=4k$},\end{cases}k\in\mathbb{Z}.$

As in 2.1 $H_{2}^{j}$ is a surface which interpolates between the slope $\displaystyle{\frac{2-\tilde{e}}{\tilde{e}}}$ on $T_{1}^{j}$ and $\displaystyle{\frac{2+\tilde{e}}{-\tilde{e}}}$ on $T_{2}^{j}$ . $\partial H_{1}\cap T_{1,r}^{j}$ has $\displaystyle{|\frac{\lambda}{u_{r}^{j}}|=|\frac{1}{1}|=1}$ component $r=1,2$ , and $\partial H_{2}^{j}\cap T_{2,r}^{j}$ has $\displaystyle{|\frac{\overline{\lambda}}{\overline{u}_{r}^{j}}|=|\frac{\tilde{e}/2}{\tilde{e}/2}|=1}$ component $r=1,2,3$ . Still $\epsilon_{1}=-\epsilon_{2}$ implies that $M$ is a surface semi-bundle. Denote the associated fibring in $M_{2}^{j}$ by $\mathcal{F}_{2}^{j}$ . $\mathcal{F}_{2}^{j}$ is transverse to all the new fibres, and in particular to $L_{j,j}^{2}$ .

Now $L_{j,j}^{1}$ only intersects the original fibre once, so $U_{i,j}^{0,2l}$ is a once-punctured annulus, and there is only one singular point on $U_{i,j}^{0,2l},1\leqslant i,j\leqslant p,i\neq j,l=1,2,3$ .

The intersection of $U_{i,1}^{0,2}$ with $\mathcal{F}_{2}^{1}$ is depicted in Figure 14-1, $U_{i,1}^{0,4}\cap\mathcal{F}_{2}^{1}$ is depicted in Figure 14-2, and $U_{1,i}^{0,6}\cap\mathcal{F}_{2}^{1}$ is similar to $U_{i,1}^{0,4}\cap\mathcal{F}_{2}^{1},2\leqslant i\leqslant p$ . The previous argument can be proceed without significant change to produce the desired surface bundle structure on $\breve{M}_{2,s}^{j}$ which is transverse to $\breve{L}^{\prime}=\underset{s=1}{\overset{2}{\cup}}(\underset{r=1}{\overset{2}{\cup}}\underset{1\leqslant i,j\leqslant p;i\neq j}{\cup}\breve{L}_{i,j,s}^{r}\cup(\overset{p}{\underset{j=1}{\cup}}\breve{L}_{j,j,s}^{2}))$ . By similar Dehn twist operation on $\breve{\mathcal{F}}_{1,s}$ , we can get a surface bundle in $\breve{Y}_{1,s}$ , say $\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ such that $\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is transverse to $\breve{L}^{\prime},s=1,2$ . The proof of Theorem 1.1 is finished in Case (1).

3 Proof of Theorem 1.1 in Case (2).

In this section, we consider classic Montesinos link $K({q_{1}}/{p},{q_{2}}/{p},\cdots,{q_{n}}/{p})$ , where $n>3$ , $p$ odd and $p\geq 3$ . We do not consider the case $e(W_{K})=0$ , otherwise $K$ is virtually fibred by Theorem 1.2 [ABZ].

We still consider $K$ is a knot first, and prove Theorem 1.1 by the same four steps as in Sec 2.1.

The Seifert structure is $f:W_{K}\rightarrow S^{2}(\underbrace{p,p,\cdots,p}_{n})$ , and $\Gamma_{1}=\pi_{1}(S^{2}(\underbrace{p,p,\cdots,p}_{n}))$ has a presentation

\Gamma_{1}=<x_{1},x_{2},\cdots,x_{n}\ |\ x_{1}^{p}=\cdots=x_{n}^{p}=x_{1}x_{2}\cdots x_{n}=1>.

Define $h_{1}$ similarly as in Sec. 2.1.

h_{1}:\Gamma_{1}\rightarrow\mathbb{Z}/p

where $h_{1}(x_{1})=\overline{1},h_{1}(x_{2})=-\overline{1}$ , and $h_{1}(x_{i})=\overline{0}$ , for $3\leqslant i\leqslant n$ . Let $\psi_{1}:F^{\prime}\rightarrow S^{2}(\underbrace{p,p,\cdots,p}_{n})$ be the $p$ -fold cyclic orbifold cover of $S^{2}(\underbrace{p,p,\cdots,p}_{n})$ corresponding to $h_{1}$ . Then $F^{\prime}=S^{2}(\underbrace{p,p,\cdots,p}_{(n-2)p})$ and

\Gamma_{2}=\pi_{1}(F^{\prime})=<y_{1},y_{2},\cdots,y_{(n-2)p}\ |\ y_{1}^{p}=y_{2}^{p}=\cdots=y_{(n-2)p}^{p}=y_{1}y_{2}\cdots y_{(n-2)p}=1>.

Define

h_{2}:\Gamma_{2}\rightarrow\mathbb{Z}/p

where $h_{2}(y_{i})=\overline{1}$ for $1\leqslant i\leqslant(n-2)p$ . Suppose that $\psi_{2}:F\rightarrow F^{\prime}$ is the $p$ -fold cyclic orbifold cover of $F^{\prime}$ corresponding to $h_{2}$ , and $\psi=\psi_{1}\circ\psi_{2}$ . Now let $\Psi:Y\rightarrow W_{K}$ be the associated $p^{2}$ -fold cover. $L=\Psi^{-1}(\widetilde{K})$ is still a geodesic link in $Y$ with exactly $p^{2}$ components. We use the same notations as in section 2. $L=\{L_{i,j},1\leqslant i,j\leqslant p\}$ , $\hat{f}(L_{i,j})=L_{i,j}^{*}$ , $1\leqslant i,j\leqslant p$ , and $\psi^{-1}(c_{i})=\hat{c}_{i,j}$ , $1\leqslant i\leqslant n,1\leqslant j\leqslant p$ .

Let $M=Y\setminus(\overset{p}{\underset{j=1}{\cup}}\overset{\circ}{N}(L_{j,j}))$ . By the same process from Step 2 to Step 4 as described in section 2.1, we can find a double cover of $M$ , say $\breve{M}$ and a surface bundle $\breve{\mathcal{F}}^{{}^{\prime\prime}}$ in $\breve{M}$ such that it is transverse to the lift of $L_{i,j}$ , $1\leqslant i,j\leqslant p,i\neq j$ .

The only thing different here is the construction of $\Gamma$ . In Step 4, we perform Dehn twist operations on $\breve{\mathcal{F}}_{1,s}$ along a set of $\breve{\phi}$ -vertical tori $\Gamma$ such that the new surface fibre is transverse to $\breve{L}_{i,k,s}^{2l-1}$ everywhere, $1\leqslant l\leqslant 2n$ , $s=1,2$ , $1\leqslant i,k\leqslant p$ , $i\neq k$ . $\Gamma=\{\breve{V}_{s}^{1},\breve{V}_{s}^{2}:s=1,2\}$ contains four tori not $2p+4$ as in Case (1). They are the lifts of two $\phi$ -vertical tori, $V^{r}=\hat{f}^{-1}(l_{r})$ , $r=1,2$ . $l_{1}$ and $l_{2}$ are two simple closed curves on $F$ , which intersect all $(L_{i,k}^{2l-1})^{*}$ ’s once or twice when $n\geq 5$ in the same signs, $1\leqslant i,k\leqslant p,i\neq k$ , and $1\leqslant l\leqslant n$ .

Note that we always suppose that $1\leqslant i,j,k\leqslant p$ , $i\neq k$

Recall that $F_{2}^{j}=L_{j,j}^{*}\times[-\epsilon,\epsilon]$ , and $F_{1}=F-\overset{p}{\underset{j=1}{\cup}}{\overset{\circ}{F_{2}^{j}}}$ . As before, $L_{i,k}^{*}$ is separated into $2n$ parts by $\overset{p}{\underset{j=1}{\cup}}F_{2}^{j}$ and $F_{1}$ . Table 1 is still true in this case. In addition, if $4\leqslant l\leqslant n$ ,

the tail of $(L_{i,k}^{2l-1})^{*}$ in $\begin{cases}\beta_{2}^{i}&\displaystyle{i-k}\leqslant\frac{p-1}{2},\\[8.61108pt] \beta_{1}^{i}&\displaystyle{i-k}>\frac{p-1}{2},\end{cases}$

the head of $(L_{i,k}^{2l-1})^{*}$ in $\begin{cases}\beta_{1}^{i}&\displaystyle{i-k}\leqslant\frac{p-1}{2},\\[8.61108pt] \beta_{2}^{i}&\displaystyle{i-k}>\frac{p-1}{2}.\end{cases}$

As in Step 3 in Section 2, we can reorient $\{L_{i,k},1\leqslant i,k\leqslant p,i\neq k\}$ such that

the tail of $\begin{cases}\text{$L_{i,k}^{2}$ and $L_{i,k}^{2+2n}$ lie in $T_{1}^{k}$},\\ \text{$L_{i,k}^{4}$ and $L_{i,k}^{4+2n}$ lie in $T_{2}^{k}$},\\ \text{$L_{i,k}^{2l}$ and $L_{i,k}^{2l+2n}$ lie in $T_{2}^{i}$, $3\leqslant l\leqslant n$}.\end{cases}$

Table 2 is also true in this case and the tail of $L_{i,k}^{2l-1+r}$ is in $T_{1}^{i}$ , and the head of $L_{i,k}^{2l-1+r}$ is in $T_{2}^{i}$ , $4\leqslant l\leqslant n$ , $r=0,2n$ .

The construction of $l_{1}$ and $l_{2}$ is similar as before.

$a_{r}^{j}$ is still the intersection of $\beta_{r}^{j}$ and a small disk centered at $\hat{c}_{1,j}$ with radius bigger than $\epsilon$ , $1\leqslant j\leqslant p$ , $r=1,2$ . Push $a_{r}^{j}$ into $F_{1}$ a little bit. $b_{r}^{j}$ is different. We take $b_{r}^{j}$ in $F_{1}$ which is parallel to $\beta_{r}^{j}$ and between $\hat{c}_{2,j}$ and $\hat{c}_{n,j}$ , $1\leqslant j\leqslant p$ , $r=1,2$ . Then $\overset{p}{\underset{j=1}{\cup}}(a_{1}^{j}\cup a_{2}^{j})$ intersects $(L_{i,k}^{2l-1})^{*}$ once, where $l=1,2$ , and $\overset{p}{\underset{j=1}{\cup}}(b_{1}^{j}\cup b_{2}^{j})$ , intersects $(L_{i,k}^{5})^{*}$ and $(L_{i,k}^{2n-1})^{*}$ both once and intersect $(L_{i,k}^{2l-1})^{*}$ twice where $3<l<n$ , $1\leqslant i,k\leqslant p$ , $i\neq k$ .

We still can connect $a_{r}^{j}$ ’s and $b_{r}^{j}$ ’s by the same arcs as in (6), $1\leqslant j\leqslant p$ , $r=1,2$ . See Figure 15 in the case $n=4$ , and Figure 16 in the case $n=5$ . When $n>5$ , $l_{1}$ and $l_{2}$ are similar to the ones in the case $n=5$ .

$\breve{V}_{s}^{1}=p_{2}^{-1}(V^{1})\cap\breve{Y}_{1,s}$ , $\breve{V}_{s}^{2}=p_{2}^{-1}(V^{2})\cap\breve{Y}_{1,s}$ , $s=1,2$ . Recall that $p_{2}:\breve{M}\rightarrow M$ is the double cover of $M$ . We have Remark 3.1 by the construction of $l_{1}$ and $l_{2}$ .

Remark 3.1

$\breve{L}_{i,k,s}^{2l-1+r}$ ’s intersect $\{\breve{V}_{s}^{1},\breve{V}_{s}^{2}:s=1,2\}$ negatively. In addition, $\breve{L}_{i,k,s}^{2l-1+r}$ intersects $\{\breve{V}_{s}^{1},\breve{V}_{s}^{2}:s=1,2\}$ exactly once when $l=1,2,3,n$ , and twice when $3<l<n$ , where $1\leqslant i,k\leqslant p$ , $i\neq k$ , $r=0,2n$ .

By Remark 3.1 we can define $\partial(N(\breve{V}_{s}^{r}))=\breve{\partial}_{1,s}^{r}\cup\breve{\partial}_{2,s}^{r}$ such that the tail of $\breve{L}_{i,k,s}^{2l-1}$ and $\breve{L}_{i,k,s}^{2l-1+2n}$ on $\breve{\partial}_{1,s}^{r}$ , $r=1,2$ , $s=1,2$ , $1\leqslant i,k\leqslant p$ , $i\neq k$ , $1\leqslant l\leqslant n$ .

Now we perform a Dehn twist operation $D_{s}^{r}$ which wraps these annuli $m$ times around the $\breve{\phi}$ -fibres in the direction opposite to the transverse orientation of $\breve{\mathcal{F}}_{1,s}$ as we pass from $\breve{\partial}_{1,s}^{r}$ to $\breve{\partial}_{2,s}^{r}$ , $r=1,2$ , $s=1,2$ . By Remark 3.1 $\breve{L}_{i,k,s}^{2l-1}$ intersects $(\breve{V}_{s}^{1}\cup\breve{V}_{s}^{2})$ once, $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is similar as in Figure 10-3, when $l=1,2,3,n$ . When $3<l<n$ , $\breve{L}_{i,k,s}^{2l-1}$ intersects $(\breve{V}_{s}^{1}\cup\breve{V}_{s}^{2})$ twice and in the same sign. $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is shown in Figure 17-1. In this case, $\breve{L}_{i,k,s}^{2l-1}$ goes from $\breve{T}_{1,s}^{i}$ to $\breve{T}_{2,s}^{i}$ . Now $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ is wrapped $2m$ times around the $\breve{\phi}$ -fibres in the direction opposite to the transverse orientation of $\breve{\mathcal{F}}_{1,s}$ as we pass from $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{T}_{1,s}^{i}$ to $\breve{U}_{i,k,s}^{0,2l-1}\cap\breve{T}_{2,s}^{i}$ , $1\leqslant i,k\leqslant p$ , $i\neq k$ , $3<l<n$ . We adjust $\breve{\mathcal{F}}_{1,s}^{{}^{\prime}}$ by isotope, and denote the resulting surface bundle $\breve{\mathcal{F}}^{{}^{\prime\prime}}$ in $\breve{M}$ such that it is transverse to $\breve{L}_{i,k,s}^{2l-1}$ as shown in Figure 17-2.

Now we finish the proof of Theorem 1.1 in Case (2) when $K$ is a knot. If $K$ has two components, Theorem 1.1 can be proved by combining Sec. 2.2 with the argument in this section.

References

[ABZ] Ian Agol, Steven Boyer, and Xingru Zhang, Virtually fibred Montesinos Links, Journal of Topology, to appear.
[Wa] G. Walsh, Great circle links and virtually fibred knots, Topology, 44 (2005) No. 5, 947-958.
[WY] S. Wang and F. Yu, Graph manifolds with non-empty boundary are covered by surface bundles, Math. Proc. Camb. Phil. Soc. 122 (1997) 447-455.

Virtually fibred Montesinos links of type S​L2~\widetilde{SL_{2}}