# Thurston unit ball of a family of $n$-chained links and their fibered face

Juhun Baik  Juhun Baik, Department of Mathematical Sciences, KAIST, 291
Daehak-ro, Yuseong-gu, Daejeon 34141, South Korea<EMAIL_ADDRESS>and
Philippe Tranchida  Philippe Tranchida, Universié Libre de Bruxelles,
Département de Matématiques, C.P.216-Algèbre et Combinatoire, Boulevard du
Triomphe, 1050 Bruxelles, Belgium<EMAIL_ADDRESS>
###### Abstract.

We determine the Thurston unit ball of a family of $n$-chained link, denoted
by $C(n,p)$, where $n$ is the number of link components and $p$ is the number
of twists. When $p$ is strictly positive, we prove that the Thurston unit ball
for $C(n,p)$ is an $n$-dimensional cocube, for arbitrary $n$. Moreover, we
clarify the condition for which $C(n,p)$ is fibered and find at least one
fibered face for any $p$. Finally we provide the Teichmüller polynomial for
the face of Thurston unit ball of $C(n,-2)$ with $n\geq 3$.

###### Key words and phrases:

Thurston norm, fibered faces, Teichmüller polynomial, polytopes.

##  1\. Introduction

Let $M$ be a $3$-dimensional manifold. We will suppose that $M$ has tori
boundaries for simplicity. In one of his many seminal works [20], W. Thurston
introduced a notion of a semi-norm for the second homology vector spaces of
$M$. More precisely, let $[a]\in H_{2}(M,\partial M;\mathbb{Z})$ be an
integral second homology class. Then $[a]$ can be represented by a disjoint
union of properly embedded surfaces $S_{i}$. The Thurston norm of $[a]$ is
then defined to be

$x(a):=\min\\{\sum_{i}\max\\{0,-\chi(S_{i})\\}\\}$

where the minimum is taken over all possible ways to represent $[a]$ as a
disjoint union of properly embedded surfaces. If $M$ is irreducible and
atoroidal, this then extends to a norm on $H_{2}(M,\mathbb{R})$. We sometimes
use $||\cdot||$ to denote the Thurston norm. In the same paper, he proves that
the unit ball with respect to that norm, that we will call Thurston unit ball,
is always a polytope. Even though this concept has had huge theoretical
consequences, it seems that there are very few cases for which we know
precisely what the Thurston unit ball is. An interesting question in that
regard is the following.

######  Question.

Which polytope can appear as a Thurston unit ball of some $3$-manifold?

This question was already posed by Kitayama in [13]. It has been generalized
in terms of groups and their $1$st homology by Friedl, Lück and Tillmann [5].
In [17], Pacheco-Tallaj, Schreve and Vlamis find out the shape of the Thurston
unit ball for tunnel number-one manifolds. We refer [13] for more recent
research of Thurston norm.

In this article we show that the Thurston unit ball can contain highly
symmetric polytopes in arbitrary high dimensions. We will be interested in
determining the Thurston unit ball for a family of complements of links,
denoted by $C(n,p)$. Briefly speaking, $C(n,p)$ is an $n$-chained link with
$p$ positive half-twist on the first component if $p$ is positive or $p$
negative half-twist on the first component if $p$ is negative (see figure 4).
The complements of these links are in some sense generalizations of the magic
manifold, which is the complement of $C(3,0)$. The magic manifold and its
properties are thus good examples to keep in mind.

In a previous article [2], the two authors together with Harry Baik and
Changsub Kim studied the relation between the minimal entropy of pseudo-anosov
maps on a surface $S$ and the action of these maps on $H_{1}(S)$. In order to
do so, the use of the complement of $C(n,-2)$ was crucial.

Here are the main results of this article.

######  Theorem 1.1.

Let $M(n,p)$ be the complement of the link $C(n,p)$ with $n\geq 3$ and
$B(n,p)$ be the Thurston unit ball of $M(n,p)$. Suppose $M(n,p)$ is
hyperbolic. Then

  * •

If $p\geq 1$, $B(n,p)$ is an $n$-dimensional cocube with vertices
$(\pm 1,0,\cdots,0),\cdots,(0,\cdots,0,\pm 1)$. (Corollary 5.2)

  * •

If $p=0$, $B(n,p)$ is the union of an $n$-dimensional cocube and two
simplices. (Theorem 4.1)

Moreover, we also find a fibered face in each case, and determine the
topological type of every fibered surface in that fibered face.

A complete answer for the case of $p<0$ is out of our reach for now. We
nonetheless find a set $V(n,p)$ of points in the Thurston unit ball and
conjecture that their convex hull, denoted by $B(n,p)$ is the whole Thurston
unit ball. This conjecture is partially supported by computational data,
obtained using the program Tnorm and gathered in Appendix B.

######  Conjecture 1.

$B(n,p)$ is equal to the Thurston unit norm ball of $C(n,p)$ when $p<0$.

For the case of $p=-2$, we also managed to compute the Teichmuller polynomial
for every value of $n$.

######  Theorem 1.2.

(Theorem 7.1) The Teichmüller polynomial $P$ for the fibered face
$\mathcal{F}$ is

$P(x_{1},\cdots,x_{n-1},u):=A-\sum_{k=1}^{n}ua_{k}A_{k}$

where $a_{1}=1,a_{2}=x_{1}^{-1},\cdots,a_{n}=(x_{1}\cdots x_{n-1})^{-1}$,
$A:=(a_{1}-u)\cdots(a_{n}-u)$ and $A_{k}=\dfrac{A}{(a_{k}-u)(a_{k-1}-u)}$,
subscript $k\equiv(mod~{}n)+1$.

### Acknowledgement

We would like to thank Harry Hyungryul Baik, Chenxi Wu, Dan Margalit, JungHwan
Park and Livio Liechti for useful conversations. We also thank William Worden
for helping us setting up and using the Tnorm package.

##  2\. Preliminary

We gather here the essential tools that will be used in the rest of the paper.

###  2.1. Murasugi sums

In his papers [7], [8], David Gabai proved theorems related to the fiberedness
of embedded surfaces and about their monodromy map when they are in fact
fibered. He constantly makes use of a geometric operation called “Murasugi
sum”, which is a way to glue surfaces together while preserving some property
of their fibers. We begin with the definition of this Murasugi sum.

######  Definition 2.1 (Murasugi sum, [7]).

The oriented surface $\Sigma\subset S^{3}$ is a _Murasugi sum_ of two
different oriented surface $\Sigma_{1}$ and $\Sigma_{2}$ if

  1. (1)

$\Sigma=\Sigma_{1}\cup\Sigma_{2}$ and $\Sigma_{1}\cap\Sigma_{2}=D$, where $D$
is a $2n$-gon.

  2. (2)

There is a partition of $S^{3}$ into two $3$-balls $B_{1},B_{2}$ satisfying
that

     * •

$\Sigma_{i}\subset B_{i}$ for $i=1,2$.

     * •

$B_{1}\cap B_{2}=S^{2}$ and $\Sigma_{i}\cap S^{2}=D$ for $i=1,2$.

In simple terms, the Murasugi sum is a way to cut-and-paste two surfaces in an
alternating way so that, around the gluing region, it looks like there are
$2n$ legs going up and down alternatively.

Figure 1. Murasugi sum of two surfaces, here $D$ is a hexagon

The power of the Murasugi sum is that it preserves the fiberedness and also
the monodromies. More precisely, Gabai proved the two following theorems.

######  Theorem 2.2 (Gabai, [6]).

Let $S$ be a Murasugi sum of $S_{1}$ and $S_{2}$. Then $S$ is a fibered
surface if and only if both $S_{1}$ and $S_{2}$ are fibered surfaces.

######  Theorem 2.3 ([7], Cor 1.4).

Suppose that $R$ is a Murasugi sum of $R_{1},R_{2}$ with $\partial
R_{i}=L_{i}$, where $L_{i}$ is a fibered link with monodromy $f_{i}$ fixing
pointwise the boundary $\partial R_{i}$, resp. Then $L=L_{1}\cup L_{2}$ is a
fibered link with fiber $R$ and its monodromy map is $f=f_{2}^{\prime}\circ
f_{1}$ where $f_{i}^{\prime}$ is the map induced on $R$ by inclusion.

Using two theorems it is possible to construct fibered surfaces by gluing
together smaller fibered surfaces while having a nice control on the monodromy
maps. A good starting block for this construction is the Hopf link, which
consists of $2$ circles linked together exactly once. The Hopf band is then a
Seifert surface of the Hopf link. It is thus a fibered surface of
$S^{3}-\\{\text{Hopf link}\\}$.

######  Lemma 2.4 (Monodromy of a Hopf band).

The Hopf band is a fibered surface. Moreover, the monodromy of the positive
(resp. negative) Hopf band is the right-handed (resp. left-handed) Dehn twist
along its core curve.

In fact, in [9] Giroux and Goodman proved that every fibered link in $S^{3}$
is obtained from the unknot, by Murasugi summing or desumming along Hopf
bands. _i.e.,_ The Hopf bands are building blocks to construct any fibered
link in $S^{3}$.

###  2.2. Fibers of alternating knots/links

Suppose a fibered link $L$ in $S^{3}$ is given. In general it is really hard
to detect what is the fiber of $S^{3}-L$. However, if $L$ is alternating,
Seifert showed in [19] how to construct the fiber. We first recall the
definition of an alternating link and then explain the Seifert algorithm. For
more details, we refer [18].

######  Definition 2.5 (Alternating link).

Let $L$ be an oriented link. An alternating diagram for $L$ is a link diagram
such that the crossings alternate under and over as one travels along each
component of the link. A link is alternating if it admits an alternating
diagram.

######  Definition 2.6 (Seifert algorithm).

Let $L$ is an oriented link. The Seifert algorithm can be described as
follows.

  1. (1)

At each crossing, cut at the crossing and paste back in such a way that, near
the crossing, there are $2$ components, as showed in figure 2.

Figure 2. Cut and paste at a crossing in Seifeirt algorithm

  2. (2)

After all these cut-and-paste operations, we are left with a disjoint
collection of oriented simple closed curves. Each curve bounds a disk, unless
two or more curves are nested. In that case, we can consider the innermost to
be lying slightly above the others and repeat this process until there are no
more nested curves. We then assign to each region ”+” sign if the region is on
the left side of the boundary curve, with respect to its orientation, or ”-”
sign otherwise. Note that the result is sometimes called a checkerboard
coloring.

  3. (3)

Finally, we reconnect these discs at each crossing with a twisted strip, the
direction of the twist being determined by the direction of the original
crossing.

The result of this algorithm is a surface $S$ whose (oriented) boundary is
$L$.

The surface obtained from the Seifert algorithm is called the Seifert surface
of $L$. The genus of a link $L$ is defined to be the minimal genus of a
surface in the complement of $L$ whose boundary is $L$. In [8] Gabai proved
that if $L$ is alternating, then the genus of $L$ is equal to the genus of the
Seifert surface of $L$.

######  Theorem 2.7 ([8], Thm 4).

Let $L$ be an oriented link in $S^{3}$. If $S$ is a surface obtained by
applying Seifert’s algorithm to an alternating diagram of $L$, then $S$ is a
surface of minimal genus.

This condition is closely related to the fiberedness.

######  Theorem 2.8 (Theorem 4.1.10 in [10]).

Let $S$ be a Seifert surface for a fibered link $L$. Then the following are
equivalent.

  1. (1)

$S$ attains the minimal Seifert genus.

  2. (2)

$S$ is a fibered surface.

###  2.3. Teichmüller polynomial

The Teichmüller polynomial $\theta_{F}$ for a fibered face $\mathcal{F}\subset
H^{1}(M,\mathbb{R})$ is a polynomial invariant that determines the stretch
factors of all the monodromy of fibers in $\mathcal{F}$. Similarly to the
Alexander polynomial, the Teichmüller polynomial has coefficients in the group
ring $\mathbb{Z}(G)$ where $G=H_{1}(M,\mathbb{Z})/$torsion.

We describe here one way to compute the Teichmüller polynomial. Let
$\varphi\colon S\to S$ be a pseudo-Anosov mapping and let $x=x_{1},\cdots
x_{n-1}$ be a multiplicative basis for

$H=\operatorname{Hom}(H^{1}(M,\mathbb{Z}^{\varphi}),\mathbb{Z})$

, where $H^{1}(M,\mathbb{Z})^{\varphi}$ is the $\varphi$-invariant cohomology.
Remark that we can construct a natural map from $\pi_{1}(S)$ to $H$ by
evaluating cohomology classes on loops. Choose now a lift
$\tilde{\varphi}\colon\tilde{S}\to\tilde{S}$ of $\varphi$ to the cover
$\tilde{S}$ corresponding to $H$ under the previous map.

Let $M=S\times[0,1]/(p,1)\sim(\varphi(p),0)$ be the mapping torus of
$\varphi$. Then we have that

$G=H_{1}(M,\mathbb{Z})/\operatorname{torsion}=H\oplus\mathbb{Z}$

We let $u$ denote the generator of the $\mathbb{Z}$ component of $G$ so that
$G$ is generated by $x_{1},\cdots,x_{n-1}$ and $u$. Let $V$ and $E$ be the
vertices and the edges of an invariant train track $\tau$ on $S$ carrying the
pseudo-Anosov map $\varphi$. The lifts $\tilde{V}$ and $\tilde{E}$ of $V$ and
$E$ to $\tilde{S}$ can respectivily be considered as $\mathbb{Z}(H)$-modules.
Therefore, the lift $\tilde{\varphi}$ asts as matrices $P_{V}(x)$ and
$P_{E}(x)$ on these $\mathbb{Z}(H)$-modules. McMullen showed in [15] that the
Teichmüller polynomial can then be computed in term of these two matrices.

######  Theorem 2.9.

Under the previous notations, the Teichmüller polynomial can be explicitly
computed as follows:

$\theta_{\mathcal{F}}(x,u)=\frac{\det(uI-P_{E}(x))}{\det(uI-P_{V}(x))}$

##  3\. The $n$ chained links and their complements

In [11], Eiko Kin analysed in detail a $3$-manifold, known as the magic
manifold. This manifold has the property that all the faces of its Thurston
unit ball are fibered. She was able to precisely determine all the fibered
faces and, for each integer point in a fibered face, find the topology of the
associated monodromy (i.e determine its genus and the number of boundary
components). In this section we generalize the technique used for the magic
$3$-manifold to study sequences of fibers in more general link complements. We
investigate whether these $n$-chained links are fibered, what are the fibers
and the associated monodromies, and the shape of Thurston unit norm ball of
various $C(n,p)$’s.

######  Definition 3.1 ($n$-chained link).

A $n$-chained link is a link with $n$ components which are linked in a
circular fashion. Some of the components may have self half-twists. One can
always gather all such self half-twists into a single component, with same
clasps shape (see figure 4). Here the word clasp designates a pair of
crossings of two adjacent link components. Remark that there are only $2$
possible shapes of clasps and one can change one to the other by performing a
self-half twist in a suitable direction.

Figure 3. The $2$ different kinds of clasps, we will call the left one as a
$+$ clasp and the right one as a $-$ clasp

We denote such an $n$-chained link by $C(n,p)$ where $n$ is the number of
components and $p$ indicates the number of half twists. Note that $p$ is an
integer, and the sign of $p$ indicates the direction of the half twists. We
choose the positive sign as the direction of twists which makes $C(n,p)$
admits an alternating link diagram.

Figure 4. $C(5,-1)$ and $C(6,3)$.

######  Remark.

Note that any choice of ‘clasp shape’ can be isotoped into the other choice.
If we denote by $\widetilde{C}(n,p)$ the same link as $C(n,p)$ but with a
different choice of a clasp shape, we can perform $-n$ half twists to flip
every clasp into the other shape. Keep in mind that the sign of $p$ depends on
the shape of clasp. Since every self twist is an isotopic move, we get
$\widetilde{C}(n,p)\cong C(n,-p-n)$. But $\widetilde{C}(n,p)$ is isomorphic to
$C(n,p)$ under orientation reversing isomorphism, so we get $C(n,p)\cong
C(n,-p-n)$.

Usually the ‘minimally twisted’ chain link is the one with the maximal
alternating clasp shapes. With our notation, the minimally twisted $n$-chained
link is $C(n,-\lfloor\frac{n}{2}\rfloor{})$.

Let $M(n,p)$ be the complement of a small enough neighborhood
$\mathcal{N}(C(n,p))$ of $C(n,p)$. For example, $M(3,0)$ is the magic
3-manifold. Note that $\partial\mathcal{N}(C(n,p))$ is a disjoint union of $n$
tori. In [16], Neumann and Reid prove that $M(n,p)$ with $n\geq 3$ is
hyperbolic if and only if $\\{|n+p|,|p|\\}\not\subseteq\\{0,1,2\\}$. Moreover
it is fibered, which is proven by Leininger [14]. More precisely, Leininger
shows that, except for $(n,p)=(2,-1)$, $M(n,p)$ is fibered if $n\geq-p\geq 0$
by providing an actual fibers. We will use that construction later.

Now, we focus on the homology of $M(n,p)$. Consider that we draw $C(n,p)$ in
such a way that the top link has the $p$ half-twists, as in the figure 4. We
then denoted the top link component as $L_{1}$, and we enumerate the other
components $L_{2},L_{3},\cdots L_{n}$ in a clockwise fashion.

Each $L_{i}$ bounds a twice punctured disk, the punctures coming from
$L_{i-1}$ and $L_{i+1}$. We will denote this twice punctured disk by $K_{i}$.
Note that $\\{[K_{i}]_{1\leq i\leq n}\\}$ forms a basis of $H_{2}(M,\partial
M)$. Then we have

######  Lemma 3.2 ([2], lemma 4.6).

The fiber $S$ provided from [14] has a coordinate $(1,\cdots,1,-1)$ with
respect to the base $[K_{i}]$. _i.e.,_ $[S]=[K_{1}]+\cdots+[K_{n-1}]-[K_{n}]$.

Note that the fiber $S$ is a genus $1$ surface with $n$ boundaries and so its
Euler characteristic is equal to $n$. Our next goal is to find the fibered
face $\mathcal{F}$ that contains $S$.

##  4\. Thurston unit ball for $C(n,0)$

###  4.1. Thurston unit ball

We start by stating the main theorem of this section, but we will prove it
only when all the needed lemmas will be established. The notation used in the
main theorem will nonetheless be used throughout the whole section.

######  Theorem 4.1.

The Thurston unit ball $B_{n}$ of $C(n,0)$ is the union of:

  1. (1)

The $n$-dimensional cocube with vertices $(\pm
1,0,\cdots,0),\cdots,(0,\cdots,0,\pm 1)$, and

  2. (2)

Two $n$-simplices whose vertices are $(1,0,\cdots,0),\cdots,(0,\cdots,0,1)$,
$\frac{1}{n-2}(1,\cdots,1)$ and its antipodal image.

Hence the fiber $\frac{1}{n}(1,\cdots,1,-1)$ lies in the fibered face
$\mathcal{F}$ whose vertices are
$(1,0,\cdots,0),\cdots,(0,\cdots,1,0),(0,\cdots,0,-1)$ and
$\frac{1}{n-2}(1,\cdots,1)$. Moreover, every face of $B_{n}$ is a fibered
face.

Here the $n$-dimensional cocube is a dual of $[0,1]^{n}$. One can consider it
as a consecutive suspension of a closed interval $[-1,1]$.

At this moment we only note the theorem generalises the case of the magic
$3$-manifold case, whose Thurston unit ball is a parallelogram with vertices
$(\pm 1,0,0)$, $(0,\pm 1,0)$, $(0,0,\pm 1)$ and $(1,1,1),(-1,-1,-1)$. We
observe that

######  Lemma 4.2.

Suppose $\sigma$ is the $n$-dimensional simplex of vertices
$a_{1},\cdots,a_{n}$ in a fibered face $\mathcal{F}$. Then $\sigma$ is a
subset of a fibered face if there exists a point $a\in\sigma$ whose Thurston
norm $-\chi(a)$ is equal to $1$. In this case the linear equation of the
fibered face $\mathcal{F}$ is $\sum_{i=1}^{n}x_{i}/a_{i}=1$.

###### Proof.

The proof is direct consequence of the fact that the Thurston unit ball is a
polytope. ∎

As in the magic manifold case, we can calculate the Euler characteristic of
any primitive points in $\mathcal{F}$.

######  Corollary 4.3.

The convex hull of the points $e_{1},e_{2},\cdots,-e_{n}$ and
$\frac{1}{n-2}(1,1,\cdots,1)$ is a subset of the facet $F$ of $B_{n}$.
Moreover, any primitive point $\alpha:=(\alpha_{1},\cdots,\alpha_{n})$ in the
cone $\mathcal{C}:=\mathbb{R}^{+}\cdot\mathcal{F}$ the Euler characteristic of
the representative of $\alpha$ is $\alpha_{1}+\cdots+\alpha_{n-1}-\alpha_{n}$.

###### Proof.

Set

$a_{i}=\begin{cases}e_{i},&1\leq i\leq n-1\\\ -e_{n},&i=n\end{cases}$

and $a=\frac{1}{n}(1,\cdots,1,-1)$. Since we already observed that $na$ is a
fiber and $-\chi(na)=n$, it means that the linear equation
$x_{1}+\cdots+x_{n-1}-x_{n}=1$ is the equation of a supporting hyperplane for
the fibered face $\mathcal{F}$. Plugging $(\alpha_{1},\cdots,\alpha_{n})$ into
$x_{1}+\cdots+x_{n-1}-x_{n}$, we get the Euler characteristic for $\alpha$. ∎

Now we are ready to prove theorem 4.1.

###### Proof of theorem 4.1.

Note that $C(n,0)$ is circularly symmetric, so the points
$p_{i}:=\frac{1}{n}(1,\cdots,1)-\frac{2}{n}e_{i}$ for all $1\leq i\leq n$ is
also a fiber. Hence by corollary 4.3, the $n$-dimensional parallelograms
$P_{i}$ of vertices $e_{1},\cdots,-e_{i},\cdots,e_{n}$ and
$\frac{1}{n-2}(1,1,\cdots,1)$ are subsets of the boundary of Thurston unit
ball. (each $p_{i}$ is contained in $P_{i}$.) But the union of the $P_{i}$
forms a closed polytope, so it has to contain the boundary of Thurston unit
ball. ∎

###  4.2. Topological type of fibers

Additionally to understanding the Thurston unit ball, we can also get
information about each fibered surface in the fiber face $\mathcal{F}$. To get
the full topological type of representatives of given fibered points, we will
use a slightly generalized version of the boundary formula proven by Kin and
Takasawa, [12].

######  Lemma 4.4 ([2], lemma 4.9).

Suppose $S$ is a minimal representative of
$(a_{1},\cdots,a_{n})\in\mathcal{C}$. Then the number of boundaries of $S$ is
equal to $\sum_{i=1}^{n}\gcd(a_{i-1}+a_{i+1},a_{i})$, where the subscript
follows $(i\mod n)+1$.

Now we focus on the monodromy map of various fibers $S$, provided from [14].

In [14], the fiber is a sum of $n+1$ Hopf bands. By Gabai’s celebrated
theorems ([7]) the monodromy of our given fiber $S$ is equal to the
composition of Dehn twists. For example, the minimal representative $S$ of
$(1,\cdots,1,-1)$ in $M(n,0)$ admits a monodromy of $n$ vertical Dehn twist,
which the direction of last $2$ only differs from the others, followed by $1$
horizontal Dehn twist.

##  5\. Thurston unit ball for $C(n,p)$ with $p>0$

Since the link $L=C(n,p)$ with $p>0$ is admits an alternating link diagram as
we defined, its Seifert surface $S$ is the minimal genus surface for $L$.

We describe a simple algorithm to compute the genus of $C(n,p)$ using the
Seifert algorithm.

######  Theorem 5.1.

Let $L=C(n,p)$ with $p\geq 1$. Given arbitrary signs on $x=(\pm 1,\cdots,\pm
1)\in H_{2}(M(n,p),\partial\mathcal{N}(L))$, the Thurston norm of $x$ is $n$.

###### Proof.

We will show this by performing the Seifert algorithm explicitly. Assume first
that we arbitrarily fix the signs of each component of $x$. Note that the sign
determines the orientation of each component of links. Let $L$ be the link
with orientations corresponding to $x$. Imagine the link $L$ is drawn in a
circular way such that the twisted one lies at the top (such as in 4). Label
the twisted link as $L_{1}$ and continue the labeling clock-wise.

We begin applying the algorithm, starting with the crossing involving $L_{1}$.
We get $p-1$ circle obtained at the half twists by Seifert algorithm, and two
(unfinished) arcs on both sides of the circle. Now, focus on the right arc and
the next link $L_{2}$.

If signs of $L_{1},L_{2}$ agrees, then the Seifert algorithm produces one
circle and the arc is still not closed. Otherwise, Seifert algorithm makes the
arc to be tied and ends up with a circle, and another arc will created on the
right side of $L_{2}$.

Inductively doing this, we get $n+p$ circles which does not depend on signs of
links. As the number of linking in the diagram is equal to $2n+p$, we conclude
the genus of $S$ is

$\text{Genus of }S=\dfrac{2+(2n+p)-(n+p)-n}{2}=1$

By Gabai’s theorem, it is the minimal genus surface of given link (with
specific orientations we’ve fixed at the beginning). Therefore, $S$ is a
minimal representative of the given $x$ and $||x||=n$. ∎

The above lemma implies that the Thurston unit ball of $M\cong M(n,p)$ is an
$n$ dimensional cocube.

######  Corollary 5.2.

Thurston unit ball of $M(n,p)$ with $p\geq 1$ is an $n$ dimensional cocube
with vertices $(\pm 1,0,\cdots,0),\cdots,(0,\cdots,0,\pm 1)$.

###### Proof.

Let $e_{i}$ be a canonical basis of $Z^{n}\cong H_{2}(M(n,p),\partial
M(n,p))$. Since $e_{i}$ is represented by a $2$ punctured disk, it lies on the
Thurston unit ball. By the above theorem, we know that $(\pm 1/n,\cdots,\pm
1/n)$ is also on the unit ball. For each $(\pm 1/n,\cdots,\pm 1/n)$, it is a
convex combination of the canonical basis (with suitable signs). Therefore we
conclude that the convex hull of $\\{\pm e_{i}\\}_{1\leq i\leq n}$ is exactly
the Thurston unit ball. ∎

###  5.1. Detecting fibered faces

Together with theorem 4.1 and corollary 5.2, we know the shape of Thurston
unit ball of $C(n,p)$ with $p\geq 0$. Next, we want to investigate which faces
are fibered.

We denote the surface $S(n,p)_{x}$ to be the surface obtained from the process
in the theorem 5.1 at $C(n,p)$ with given orientation $x=(\pm 1,\cdots,\pm
1)$. It has genus equal to $1$ and $n$ punctures.

Choose one clasp. If the orientation of the two links at that clasp coincides,
then at the clasp there are new disks and one strip. The $2$ half twists will
connect them. i.e., the Hopf band is Murasugi summed at that place.

If the orientation of two links do not coincide, then choose one of link and
twist half so that their orientations coincide locally at the clasp. In this
case, there is a Hopf band Murasugi summed with one half twists, as in the
Figure 5.

Figure 5. (Murasugi) desum the Hopf bands in the case of different
orientations. Observe that the twist of result is compatible to the positive
half twist with respect to the clasp shape.

In the language of $H_{2}(M(n,p),\partial M(n,p))$, the sign change of the
given orientation $x$ can be interpreted as a number of half twists after we
(Murasugi) desum each vertical Hopf band. Also, we introduce a useful lemma in
[1].

######  Lemma 5.3 (Example 3.1 in [1]).

Suppose $L$ is a $(2,2n)$-torus link with the orientation so that the Seifert
surface is a full-twisted annulus. Then $L$ is fibered if and only if $|n|=1$.
Each case is a positive/negative Hopf band.

Remark that the $(2,2n)$-torus link is fibered if the orientation of the two
link components is parallel. However in our case it cannot happen since their
orientations are inherited by one link component so that they must be
opposite.

We summarize all the observations in the following theorems, which split into
$2$ cases. Recall that $p$ is the number of half-twists on $L_{1}$.

######  Theorem 5.4 ($p$ is even).

Let $p$ is nonnegative even integer. For a given orientation $x=(\pm 1,\pm
1,\cdots,\pm 1)$, denote by $s$ the number of sign changes. _i.e.,_
$s=\sum_{i=1}^{n}\delta_{-1,x_{i}x_{i+1}}$.

Then $S(n,p)_{x}$ is fibered if and only if $(p,s)=(0,2),(2,0)$.

###### Proof.

As in the observation, we will desum whenever there arise a Murasugi sum of
Hopf bands. Consequently, it remains the twisted band whose boundary is a
$(2,p+s)$-torus knot. By the lemma 5.3, it is fibered if and only if $p+s=2$,
it finishes the proof. ∎

######  Theorem 5.5 ($p$ is odd).

Let $p$ is non negative odd integer and $x,s$ is defined same as in the above.
Then $S(n,p)_{x}$ is fibered if and only if $(p,s)=(1,0)$.

###### Proof.

Only difference from the case $p$ is even is the last desumming process. Since
there are odd number of half twists, the leftmost and rightmost part of the
top link $L_{1}$ do not coincides. Hence after the desumming process, the
remaining part is a twisted band whose boundary is a $(2,p+s+1)$-torus knot.
Again by the lemma 5.3, it is fibered if and only if $p+s+1=2$ and
$(p,s)=(1,0)$ is the only solution. ∎

Note that $x=\frac{1}{n}(\pm 1,\cdots,\pm 1)$ is the barycenter of $\pm
e_{i}$’s, the fiberedness with such orientation implies the fiberedness of
faces. Therefore, together with the theroem in [14] we have the following
corollary.

######  Corollary 5.6 (Fiberedness of $C(n,p)$).

The link $C(n,p)$ is fibered if and only if $-n-2\leq p\leq 2$. Moreover,
every face of the Thurston unit ball of $C(n,0)$ is a fibered face. In
contrast there are only $2$ fibered faces of $C(n,1)$ and $C(n,2)$, which
contain $(1,1,\cdots,1)$ and its antipodal point.

###### Proof.

For $p=0$, by theorem 5.4 $S(n,0)_{x}$ is fibered if and only if $x$ has only
one $-1$ entry and the others are all $1$ or its antipodal points. By
corollary 4.3, each $x_{i}$ is in the distinct fibered face, hence every face
of Thurston unit ball for $C(n,0)$ is fibered.

Suppose $p=1$ or $p=2$. By theorem 5.4 and 5.5, $S(n,p)_{x}$ is fibered if and
only if $x=(1,\cdots,1)$ or $(-1,\cdots,-1)$. By corollary 5.2, there are only
$2$ faces whose supporting planes are $\sum_{i=1}^{n}x_{i}=\pm 1$. ∎

##  6\. Thurston unit ball for $C(n,p)$ with $p<0$

In [14] Leininger proved that $C(n,-p)$ is fibered for $0\leq p\leq n$. In
this section we investigate what is the shape of the Thurston unit ball for
the complements of $n$-chained links with negative twists. Suppose we have a
$n$-chain link $C(n,-p)$ and that we have labeled each link components as
before. Note that we can untwist all the negative twists. After resolving the
negative twists on $L_{1}$, our link became a chain link with no twists.
However the shape of clasps may have changed during this process. We re-assign
the orientations of each $L_{i}$’s in a circular way after resolving all the
twists on $L_{1}$, and, for each $i=1,\cdots,n$, we define $e_{i}\in
H_{2}(M(c,p),\partial M(c,p);\mathbb{Z})$ to be the twice punctured disk
bounded and oriented by $L_{i}$. See the figure 8.

Each link component has $2$ clasps, which may now be $+$ clasp or $-$ clasp.
Since the $-$ clasp only arise whenever one negative twist is resolved, the
number of $-$ clasps is equal to $|p|$ after resolving all twists. We define
the shape vector of $C(n,p)$.

######  Definition 6.1.

Suppose $n\geq 4$ and $-\lfloor n/2\rfloor\leq p<0$. The shape vector for
$C(n,p)$ is an $n$-tuple and each entry is either $+$ or $-$. The $i$’th entry
records the shape of clasp between the link components $L_{i}$ and $L_{i+1}$.
For each $L_{i}$, we will say that $L_{i}$ has clasp shape $(\alpha,\beta)$
with $\alpha,\beta\in\\{+,-\\}$ if the clasp between $L_{i_{1}}$ and $L_{i}$
is an $\alpha$ shape clasp and the clasp between $L_{i}$ and $L_{i+1}$ is a
$\beta$ shape clasp.

Note that the number of $-$ entries in any shape vector is equal to $|p|$.

Suppose $L_{i}$ has $-$ shape with $L_{i-1}$ and $+$ shape with $L_{i+1}$.
Here are $2$ isotopic operations we can perform on each $L_{i}$.

Figure 6. Flip Figure 7. Full twist.

  1. (1)

A flip: we flip $L_{i}$ so that the $+$ clasp changes to a $-$ clasp and vice
versa. Hence a flip exchanges the $(i-1)$’th entry and $i$’th entry of the
shape vector.

  2. (2)

A full twists: There is a twice punctured disc $D$ bounded by $L_{i}$ whose
second homology class is $e_{i}$. Cut $M(n,p)$ at $D$. At the slice, there are
$2$ copies of $D$, say $D_{1},D_{2}$. Now re-glue $D_{1}$ and $D_{2}$ after
twisting either $D_{1}$ or $D_{2}$ by $360$ degrees.

Now we have the following proposition.

######  Proposition 6.2.

Let $n\geq 4$. Suppose $L_{i}$ admits $(+,-)$ or $(-,+)$ clasp shape. Then
$(\underbrace{1,\cdots,1}_{n})-e_{i}\in H_{2}(M(c,p),\partial
M(c,p);\mathbb{Z})$ admits a $(n-1)$-punctured sphere as representative, and
its Thurston norm is thus equal to $n-3$.

###### Proof.

After performing a full twist on $L_{i}$, the $2$ consecutive link components
$L_{i-1}$ and $L_{i+1}$ form a clasp (its shape depends on the direction of
the full twist). If we forget about $L_{i}$, then the other link components
became a chain link of $n-1$ components. Apply Seifert algorithm with all
positive orientations, the Seifert surface $S$ is a $n-1$ punctured sphere.
Since the surface $S$ does not admit $L_{i}$ as its boundary component, $S$ is
an embedded surface in $M(n,p)$. Since it has no genus, this is the minimal
Thurston norm representative of the homology class. ∎

Start with $C(n,-1)$. After untwisting once, we obtain an $n$-chained link
with shape vector $(-,+,\cdots,+)$. By proposition 6.2,
$\frac{1}{n-3}(1,0,1,\cdots,1)$ is a point of Thurston norm equal to $1$. If
we flip $L_{2}$, then the shape vector changes to $(+,-,+,\cdots,+)$. Now we
can perform a full twists on $L_{3}$ and then, using the same method as in the
proof of proposition 6.2, we can deduce that the point $(1,-1,0,1,\cdots,1)$
also is on the unit sphere. Note that we have a $-1$ on the second entry this
time. Simply put, as the $0$ coordinate moves one step on the right, it also
introduces a minus sign. Hence, by this process, we get a total of $2n$ points
on the unit sphere, namely

$(1,0,1,\cdots,1),(1,-1,0,1,\cdots,1),\cdots,(1,-1,\cdots,-1,0)\text{ and
}(0,-1,\cdots,-1)$

and their antipodal points.

Now consider $C(n,-2)$. The shape vector contains two $-$ entries. We can
perform a full twist on the two link components $L_{i}$ and $L_{j}$ whose
clasps on both sides are different, unless $L_{i}$ and $L_{j}$ are consecutive
link components. In this case, we get $2$ zero entries in the new points and
hence it represents a sphere with $n-2$ punctures. Therefore it has Thusrton
norm $n-4$. See the figure 8 and 9 also.

Figure 8. $C(5,-2)$. Note that the orientation of each link component is re-
assigned in a circular way. Starting from the top link component, we label the
components $x_{1},x_{2},\cdots,x_{5}$, clockwise. Figure 9. After two full
twists, one on $L_{1}$ and one on $L_{3}$, we get the above link. Note that
$(0,1,0,1,1)$ is represented by $S_{0,3}$, which is obtained by oriented sum
of the $3$ disks bounded by $L_{2},L_{4},L_{5}$.

Therefore, we get the following corollary.

######  Corollary 6.3.

By following the process described above, we obtain a set $V:=V(n,p)$ of
points that lie on the unit sphere of Thurston norm. Every points in $V$ is
obtained by flipping the link components and taking full twists. The flip
operation slides the $0$ entry to the next coordinate. All points
$(x_{1},\cdots,x_{n})$ in $V$ have the following properties.

  1. (1)

There is an integer $N$ such that $N\times x_{i}\in\\{-1,0,1\\}$. Such $N$
corresponds to its Thurston norm.

  2. (2)

No two consecutive entries are equal to $0$.

Let $B:=B(n,p)$ be the convex hull of $V\cup\\{\pm e_{i}\\}$. Then $B$ is
contained in the unit Thurston norm ball.

We describe the topological property of various $B(n,p)$’s.

######  Proposition 6.4.

Let $C(n,p)$ be a negative twisted $n$-chained link. Choose any $1\leq i\leq
n$ and collect all points in $V(n,p)\cup\\{\pm e_{i}\\}$ with $x_{i}=0$. Then
the convex hull of such points forms an $(n-1)$-dimensional polytope and is
contained in the union of $B(n-1,p+1)$ and $B(n-1,p)$.

###### Proof.

After flipping some of link components, we can suppose $L_{i}$ has a clasp
shape $(-,+)$. Perform a full twist on $L_{i}$ and forget $L_{i}$ for a
moment. Then the remaining link components form a new link, isomorphic to
$C(n,p+1)$ or $C(n,p)$, depending on the direction of the twist. (More
precisely, if the full twist yields a negative shape clasp between $L_{i-1}$
and $L_{i+1}$, $C(n,p+1)$ is produced and vise versa.) For any points in
$V(n,p)$ or $V(n,p+1)$, if we plug a $0$ at the $i$’th tuple it becomes a
point which lies on the unit Thurston norm ball. Furthermore, such point is
obtained by doing only one full twist, hence it is in $V(n,p)$.

∎

Some faces of the polytope $B$ are actually faces of the Thurston unit ball.
We introduce another isotopic operation of for link components which have the
same clasp shape on both sides.

######  Definition 6.5 (Squeezing).

Suppose $L_{i}$ has a clasp shape $(+,+)$ or $(-,-)$. Perform a half twists on
both sides so that each clasp alters its shape. We will call this operation as
_squeezing the link_ $L_{i}$.

Figure 10. The link component in the middle has a $(-,-)$ shape. After
squeezing, the shape of the clasp changes to $(+,+)$ with a squeezed link
component.

######  Theorem 6.6.

Let $n\geq 4$ and $C(n,p)$ be a negative twisted $n$-chained link. There is a
fibered face which contains the fiber obtained by squeezing one of the link
components.

###### Proof.

We will proceed by induction. In this proof, every full twist will be
performed such that the clasp has a $+$ shape after the operation.

  1. (1)

$p=-1$. Choose any point $q$ in $V$ and any link component $L_{i}$ which has a
clasp shape $(+,+)$. There is exactly one $0$ entry in $q$. Let $k$ be its
index. By proposition 6.4, the slice of the unit Thurston norm ball of
$C(n,p)$ at $x_{k}=0$ must contain the union of $B(n-1,0)$ and $B(n-1,-1)$.
Choose one face in $B(n-1,0)$. Since its shape vector is all $+$ (or $-$), any
link component of $L_{k}$ has $(+,+)$ shape (or $(-,-)$). We choose $L_{i}$
except $i=k-1,k,k+1$ and squeeze it. Taking the inverse orientation of
$L_{i}$,
$(1,\cdots,\underbrace{-1}_{i\text{th}},\cdots,\underbrace{0}_{k\text{th}},\cdots,1)$
is represented by one horizontal Hopf band Murasugi summed by $n-1$ vertical
Hopf bands. _i.e.,_
$x=\frac{1}{n-1}(1,\cdots,\underbrace{-1}_{i\text{th}},\cdots,\underbrace{0}_{k\text{th}},\cdots,1)$
is in the unit sphere of $C(n,-1)$.
Now the convex sum $\mathbf{x}:=\frac{n-1}{n}\times x+\frac{1}{n}\times e_{k}$
is $\frac{1}{n}(1,\cdots,\underbrace{-1}_{i\text{th}},\cdots,1)$. This is
still a fiber, since we choose $i$ carefully so that the squeezing still works
even if we undo the full twist. Note that since this point is in the convex
hull of $n+1$ vertices, the face containing $\mathbf{x}$ is fibered.

  2. (2)

$\lfloor n/2\rfloor\leq p\leq-2$. By induction, we already have squeezing
fibers on the face of $C(n-1,p+1)$. See the figure 11. So it remains to show
that such squeezing still works after we undo the full twists. But since
$|p+1|$ is strictly smaller than $\lfloor n/2\rfloor$, there always exists a
link component of shape $(+,+)$ or $(-,-)$. Hence by undoing full twists
except near the link component, we get the fibered face which contains a
squeezing fiber.

∎

Figure 11. The vertical axis is $x_{k}$ coordinates, orthogonal to
$\mathbb{R}^{n-1}$. The convex hull of $V(n-1,p+1)$ lies at the bottom and
taking the cone with the apex $x_{k}=1$. The blue dot is a fiber obtained by
squeezing.

The theorem implies that most of the faces in $B(n,p)$ are actually fibered
faces of the Thurston unit norm ball. We end this section with a question and
some remarks.

######  Question 1.

Is $B(n,p)$ equal to the unit Thurston norm ball of $C(n,p)$ when $p<0$?

As far as we know, the answer is positive. We thanks William Worden and the
program Tnorm[21] which helped us to calculate and verify the theorems. We
provide the table of all the vertices of the Thurson unit normal ball ,
calculated by Tnorm, for various $C(n,p)$’s up to $n\leq 6$ in the appendix B.

As we already mentioned, $C(n,p)$ is fibered for $0\leq-p\leq n$. Assigning
proper signs, Leininger’s fibered surface has coordinates $(1,\cdots,1)$ and
has genus $1$ and $n$ punctures. Since each vector $e_{i}$ of the canonical
basis represents a twice punctured disk, we can deduce that there is a fibered
face $\mathcal{F}$ which contains the standard $(n-1)$ simplex spanned by the
$\\{e_{i}\\}$’s. Furthermore, using lemma 4.2 and similar methods as in the
proof of corollary 4.3, we can get that the Euler characteristic of any
primitive points of $(x_{1},\cdots,x_{n})$ with all positive entries is equal
to $\sum_{i=1}^{n}x_{i}$. We provide some calculations for the $p<0$ case in
the appendix B. In the remaining section, we will cover the special case of
$C(n,-2)$, in which case more explicit calculations can be made.

##  7\. Teichmüller Polynomial for one fibered face of $C(n,-2)$

In this section we compute explicitly the Teichmüller polynomials for one
fibered face of $C(n,-2)$. Let $M_{n}$ be the exterior complement of the link
$C(n,-2)$. We denote the surface obtained from performing the Seifert
algorithm to $C(n,-2)$ by $S_{n}$. We will sometimes omit the subscript $n$ if
it is not important in the context. Since $M_{n}$ is the complement of
$C(n,-2)$, the second homology $H_{2}=H_{2}(M_{n},\partial M_{n};\mathbb{Z})$
is a free abelian group of rank $n$, with a canonical basis given by the
meridians of the link components. With that in mind, we remark that $S_{n}$ is
a surface of genus one with $n$ boundaries and its coordinates in $H_{2}$ are
$(1,1,\cdots,1)$. Since $S_{n}$ is a Murasugi sum of one horizontal Hopf bands
with $n$ vertical Hopf bands, it is a fiber. By Gabai’s theorem, the monodromy
$\varphi_{n}$ of this fibering is the composition of the Dehn twists along the
cores of the Hopf bands.

Figure 12. The Seifert surface for $C(n,-2)$, the horizontal band is a
positive Hopf band and each vertical band is a negative Hopf band. Here we
omit the full twists which are supposed to be at each band, as they have no
role in the remainder of the calculations.

Thus, if $S_{n}$ is placed as suggested in figure 12, the monodromy
$\varphi_{n}$ is the composition of the $n$ vertical multi-twists directed
downward followed by the left Dehn twist along the core of horizontal band.
Following the methods in [3] and [15], we now compute the Teichmüller
polynomial corresponding to the fibered face of the Thurston unit ball which
contains the fiber $(1,\cdots,1)$.

As explained in section 2.3, we first need to compute
$H=\operatorname{Hom}{(H^{1}(S,\mathbb{Z})^{\varphi},\mathbb{Z})}$ and then
understand how the lift $\tilde{\varphi}_{n}$ of $\varphi_{n}$ acts on the
cover $\tilde{S}_{n}$ of $S_{n}$ which has $H$ as a deck transform group. In
this case, as noted in [3], the group $H$ is simply equal to the $\varphi_{n}$
invariant first homology $H_{1}(S_{n}:\mathbb{Z})^{\varphi_{n}}$ . We choose
$c_{0},\cdots,c_{n}$ as a basis for $H_{1}(S_{n};\mathbb{Z})$, where $c_{0}$
is the curve corresponding to the core of the horizontal band and
$c_{1},\cdots,c_{n}$ are the curves corresponding to the cores of the vertical
bands, $c_{1}$ being the leftmost one and $c_{n}$ the rightmost one. Then
$H_{1}(S_{n}:\mathbb{Z})^{\varphi_{n}}$ is the subspace of
$H_{1}(S_{n};\mathbb{Z})$ generated by the column vectors of

$B_{n}=\begin{bmatrix}0&0&\cdots&0\\\ 1&1&\cdots&1\\\ -1&0&\cdots&0\\\
0&-1&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\ 0&0&\cdots&-1\\\
\end{bmatrix}$

We now need to figure out what the cover $\tilde{S}_{n}$ is and how
$\tilde{\varphi}_{n}$ acts on it. Once again, the details are all given in
[3], so instead of repeating them here, we will give some graphical
explanation for the simplest no trivial example, that is $M_{3}=C(3,-2)$. In
this case, the cover $\tilde{S}_{n}$ is explicitly drawn in figure 14.

Figure 13. $S_{3}$, the fiber associated to the link $C(3,-2)$ Figure 14.
Galois covering $\tilde{S}_{3}$ of $S_{3}$ with deck transform $H$

Let $T$ be the matrix representing the $H$-module action of
$\tilde{\varphi}_{n}$ on $\tilde{S}_{n}$. The monodromy $\varphi_{n}$ being
the composition of one horizontal Dehn twist and $n$ horizontal ones, we can
decompose the matrix $T$ into $T_{V}$ and $T_{H}$. These matrices represent
the action of the lifts of the vertical multi-twist and the horizontal Dehn
twist on $\tilde{S}$. Note that the entries of these matrices are in
$\mathbb{Z}[G]$, where $G$ is a deck transformation of $\widetilde{S}$,
isomorphic to $\mathbb{Z}^{n-1}$.

Using our conventions, $T_{V},T_{H}$ are $2n\times 2n$ matrices whose entries
are as below;

$T_{V}=\begin{bmatrix}(x_{1}\cdots x_{n-1})^{-1}&0&\cdots&0&0&0&\cdots&0\\\
0&1&\cdots&0&0&0&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&(x_{1}\cdots x_{n-2})^{-1}&0&0&\cdots&0\\\
1&0&\cdots&0&1&0&\cdots&0\\\ 0&x_{1}^{-1}&\cdots&0&0&x_{1}^{-1}&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&(x_{1}\cdots x_{n-1})^{-1}&0&0&\cdots&(x_{1}\cdots x_{n-1})^{-1}\\\
\end{bmatrix}$

$T_{H}=\begin{bmatrix}1&0&\cdots&0&1&1&\cdots&1\\\
0&1&\cdots&0&1&1&\cdots&1\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&1&1&1&\cdots&1\\\ 0&0&\cdots&0&1&0&\cdots&0\\\
0&0&\cdots&0&0&1&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&0&0&0&\cdots&1\\\ \end{bmatrix}$

We cut these matrices by 4 block matrices of $n\times n$ to abbreviate it as

$T_{V}=\begin{bmatrix}D_{s}&0\\\ D&D\end{bmatrix}\qquad
T_{H}=\begin{bmatrix}I&\mathbf{1}\\\ 0&I\end{bmatrix}$

where $D_{s}$ is an $n\times n$ matrix whose diagonal entries are same as $D$
shifted to the right by one and $\mathbf{1}$ is a matrix filled with $1$.

By [15], the Teichmüller polynomial can be obtained by

$P(x_{1},\cdots,x_{n-1},u):=\dfrac{\det(T_{V}T_{H}-uI)}{\det(D-uI)}$

The remaining calculation are showed in Appendix A. Let $a_{k}$ be the $k$th
diagonal entry of $D$. _e.g.,_
$a_{1}=1,a_{2}=x_{1}^{-1},\cdots,a_{n}=(x_{1}\cdots x_{n-1})^{-1}$. Then we
have the following formula.

######  Theorem 7.1 (Teichmüller polynomial).

The Teichmüller polynomial $P$ for the fibered face $\mathcal{F}$ is

$P(x_{1},\cdots,x_{n-1},u):=A-\sum_{k=1}^{n}ua_{k}A_{k}$

where $A:=(a_{1}-u)\cdots(a_{n}-u)$ and
$A_{k}=\dfrac{A}{(a_{k}-u)(a_{k-1}-u)}$, subscript $k\equiv(mod~{}n)+1$.

The manifold $M_{n}$ can be viewed at the same time as a link complement and
has a fibration.

Both point of view lead to natural coordinates on $H_{2}=H_{2}(M_{n},\partial
M_{n};\mathbb{Z})$.

It is sometimes more convenient to use the coordinates coming from the llink
complements point of view for the Teichmüller polynomials. For example, if we
wanted to use the Teichmüller polynomial to compute the stretch factor of the
monodromy of the fiber which has coordinates $(1,1,\cdots,1)$ in the basis
given by the link components.

The Teichmüller polynomials we computed are using the basis coming from the
fibration point of view. We thus need to find the explicit change of
coordinates for going from one basis to the other.

Let us fix the notation clearly. The basis $Y$ given by the link complements
will be denoted as $y_{1},\cdots,y_{n}$, with $y_{1}$ corresponding to the
link complement with the self twist. If the monodromy for the fibration of
$M_{n}$ is denoted by $\varphi_{n}$, the corresponding basis $X$ will be
$u,x_{1},\cdots,x_{n-1}$ where the $x_{i}$ form a basis for the $\varphi_{n}$
invariant cohomology and $u$ corresponds to the suspension flow. We also let
$a_{0},\cdots,a_{n-1}$ be the canonical basis for $H^{1}(S_{n},\mathbb{Z})$.
By the computation above, we already know that $x_{i}=a_{1}-a_{i+1}$.
Moreover, as suggested by figure 15, we see that $a_{i}=y_{i}-y_{i+1}$, where
the indices are taken modulo $n$ as always. Finally, the basis element $u$
corresponding to the suspension flow is simply mapped to $y_{1}$.

Figure 15. The surface $S_{n}$ with the geometric representation of the
$a_{i}$’s. On the bottom, we see how the $a_{i}$’s are related to the link
components, since $a_{i}$, $y_{i}$ and $y_{i+1}$ always bound a disk in
$M_{n}$

To sum it up, the change of coordinates is given by

$u\to y_{1}$
---
$x_{1}\to y_{1}-y_{3}$
$x_{2}\to y_{1}-y_{2}+y_{3}-y_{4}$
$\vdots$
$x_{n-2}\to y_{1}-y_{2}-y_{n-1}+y_{n}$
$x_{n-1}\to-y_{2}+y_{n}$

Hence the image of the fiber whose coordinates in the basis $X$ are
$(1,1,\cdots,1)$ has $(0,0,\cdots,0,1)$ as coordinates in the basis $Y$. The
specialization of the Teichmüller polynomial to the point $p=(0,0,\cdots,0,1)$
is then given by

$(1-u)^{n}-nu(1-u)^{n-2}=(1-u)^{n-2}(1-(n+2)u+u^{2})$

A simple calculation shows that the largest root of this polynomial is
$\frac{n+2+\sqrt{n^{2}+4n}}{2}$.

##  Appendix A Proof of Theorem 7.1

In this appendix we finish the calculations of the Teichmüller polynomial of
section 7.

We need to compute the determinant of block matrices, so we make use of the
following lemma.

######  Lemma A.1.

Let $M=\begin{bmatrix}A&B\\\ C&D\end{bmatrix}$ a block matrix, where $A,D$ are
square matrices. Then the determinant of $M$ is
$\det(A)\det(A-BD^{-1}C)=\det(D)\det(D-CA^{-1}B)$, Assume further that if $C$
and $D$ commutes, then $\det(M)=\det(AD-BC)$.

The matrix $T_{V}T_{H}-uI$ can be expressed by the block matrix as below,

$T_{V}T_{H}-uI=\begin{bmatrix}D_{s}-uI&D_{s}\cdot\mathbf{1}\\\
D&D\cdot\mathbf{1}+D-uI\end{bmatrix}$

By performing some row reductions, we can simplify the matrix:

$\begin{bmatrix}I&0\\\
-DD_{s}^{-1}&I\end{bmatrix}\times\begin{bmatrix}D_{s}-uI&D_{s}\cdot\mathbf{1}\\\
D&D\cdot\mathbf{1}+D-uI\end{bmatrix}=\begin{bmatrix}D_{s}-uI&D_{s}\cdot\mathbf{1}\\\
uDD_{s}^{-1}&D-uI\end{bmatrix}$

Such operation does not affect the determinant and now the bottom two block
matrices are both diagonals, so they commutes. Hence we can apply lemma A.1 to
compute the determinant

$\displaystyle\det(T_{V}T_{H}-uI)$
$\displaystyle=\det((D_{s}-uI)(D-uI)-uD_{s}\cdot\mathbf{1}\cdot DD_{s}^{-1})$
$\displaystyle=\det(D_{s}((D_{s}-uI)(D-uI)-u\mathbf{1}\cdot D)D_{s}^{-1})$
$\displaystyle=\det((D_{s}-uI)(D-uI)-u\mathbf{1}\cdot D)$

Let $B_{k}:=(a_{k-1}-u)(a_{k}-u)$ with the indices taken module $n$. Then the
matrix $(D_{s}-uI)(D-uI)-u\mathbf{1}\cdot D$ is

$\begin{bmatrix}B_{1}-ua_{1}&-ua_{1}&\cdots&-ua_{1}\\\
-ua_{2}&B_{2}-ua_{2}&\cdots&-ua_{2}\\\ \vdots&\vdots&\ddots&\vdots\\\
-ua_{n}&-ua_{n}&\cdots&B_{n}-ua_{n}\end{bmatrix}$

Note that the determinant formula for $n\times n$ matrix $[m_{ij}]$ is

$\sum_{\sigma\in S_{n}}sgn(\sigma)m_{1,\sigma(1)}\cdots m_{n,\sigma(n)}$

We sort each summand $m_{1,\sigma(1)}\cdots m_{n,\sigma(n)}$ by how many
diagonal entries they contain. There is the only term with $n$ diagonal
entries, $(B_{1}-ua_{1})\cdots(B_{n}-ua_{n})$. Consider this as a polynomial
of $u$, say $f(u)$. There are no terms containing $n-1$ diagonal entries. Now,
consider the terms containing $n-2$ diagonal entries. If the term misses $i$
and $j$th diagonal entries, then it has to be
$-u^{2}a_{i}a_{j}(B_{1}-ua_{1})\cdots(B_{n}-ua_{n})/(B_{i}-ua_{i})(B_{j}-ua_{j})$.
Note that the sign is negative, since the permutation $\sigma$ is $(i,j)$
which has a negative sign. Since this term offsets every term of $u^{2}$ in
the diagonal product, after summing these terms $f(u)$ has no $u^{2}$ terms.
Iterating this process, $f(u)$ only has constant and linear terms and $u^{n}$
term with respect to $u$.

There are a lot of $u^{n}$ terms from the determinant formula, but it offsets
each others since it is just the determinant of $u\mathbf{1}\cdot D$. Hence we
conclude that the determinant of given matrix is

$P^{\prime}(u)=B_{1}\cdots B_{n}-u(\sum_{k=1}^{n}B_{1}\cdots B_{n}/B_{k})$

The formula follows from dividing $P^{\prime}$ by $\det(D-uI)$.

##  Appendix B Some calculations of $C(n,p)$ with $p<0$

In this section we give our calculations and tables of vertices for some
$C(n,p),p<0$’s.

We thanks to William Worden and his paper [4] and the program called ‘Tnorm’
which helped us verify our theorem. Tnorm is able to compute the vertices of
the Thurston unit ball of given links complements. In the tables in this
section, we list the vertices, except for vertices of the form $\pm e_{i}$’s,
together with the topological type of their representatives. So the left
columns of the tables are the coordinates of the vertices and the right
columns are the corresponding surfaces representing them in the second
homology groups.

$C(4,-1)$
---
$\pm(1,0,1,1)$ | $S_{0,3}$
$\pm(1,-1,0,1)$ | $S_{0,3}$
$\pm(1,-1,-1,0)$ | $S_{0,3}$
$\pm(0,-1,-1,-1)$ | $S_{0,3}$

$C(5,-1)$ | $C(5,-2)$
---|---
$\pm(1/2,0,1/2,1/2,1/2)$ | $\frac{1}{2}S_{0,4}$ | $\pm(0,1,0,1,1)$ | $S_{0,3}$
$\pm(1/2,-1/2,0,1/2,1/2)$ | $\frac{1}{2}S_{0,4}$ | $\pm(0,1,-1,0,1)$ | $S_{0,3}$
$\pm(1/2,-1/2,-1/2,0,1/2)$ | $\frac{1}{2}S_{0,4}$ | $\pm(1,0,-1,0,1)$ | $S_{0,3}$
$\pm(1/2,-1/2,-1/2,-1/2,0)$ | $\frac{1}{2}S_{0,4}$ | $\pm(1,0,-1,-1,0)$ | $S_{0,3}$
$\pm(0,-1/2,-1/2,-1/2,-1/2)$ | $\frac{1}{2}S_{0,4}$ | $\pm(1,-1,0,-1,0)$ | $S_{0,3}$

$C(6,-1)$
---
$\pm(1/3,0,1/3,1/3,1/3,1/3)$ | $\frac{1}{3}S_{0,5}$
$\pm(1/3,-1/3,0,1/3,1/3,1/3)$ | $\frac{1}{3}S_{0,5}$
$\pm(1/3,-1/3,-1/3,0,1/3,1/3)$ | $\frac{1}{3}S_{0,5}$
$\pm(1/3,-1/3,-1/3,-1/3,0,1/3)$ | $\frac{1}{3}S_{0,5}$
$\pm(1/3,-1/3,-1/3,-1/3,-1/3,0)$ | $\frac{1}{3}S_{0,5}$
$\pm(0,-1/3,-1/3,-1/3,-1/3,-1/3)$ | $\frac{1}{3}S_{0,5}$

$C(6,-2)$
---
$\pm(0,1/2,0,1/2,1/2,1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,0,-1/2,0,1/2,1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,-1/2,0,-1/2,0,1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,-1/2,1/2,0,-1/2,0)$ | $\frac{1}{2}S_{0,4}$
$\pm(0,1/2,-1/2,-1/2,0,1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,0,-1/2,-1/2,-1/2,0)$ | $\frac{1}{2}S_{0,4}$
$\pm(0,1/2,-1/2,0,1/2,1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,0,-1/2,-1/2,0,1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,-1/2,0,-1/2,-1/2,0)$ | $\frac{1}{2}S_{0,4}$

$C(6,-3)$
---
$\pm(0,1/2,1/2,0,1/2,1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,0,-1/2,1/2,0,-1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,-1/2,0,1/2,-1/2,0)$ | $\frac{1}{2}S_{0,4}$
$\pm(0,-1/2,1/2,0,1/2,-1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(-1/2,0,1/2,1/2,0,-1/2)$ | $\frac{1}{2}S_{0,4}$
$\pm(1/2,1/2,0,-1/2,-1/2,0)$ | $\frac{1}{2}S_{0,4}$
$\pm(0,1,0,-1,0,1)$ | $S_{0,3}$
$\pm(-1,0,1,0,1,0)$ | $S_{0,3}$

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