# Highly twisted plat diagrams

Nir Lazarovich  ,  Yoav Moriah  and  Tali Pinsky  Department of Mathematics
Technion
Haifa, 32000 Israel<EMAIL_ADDRESS>Department of Mathematics
Technion
Haifa, 32000 Israel<EMAIL_ADDRESS>Department of Mathematics
Technion
Haifa, 32000 Israel<EMAIL_ADDRESS>
###### Abstract.

We prove that the knots and links in the infinite set of $3$-highly twisted
$2m$-plats,with $m\geq 2$, are all hyperbolic. This should be compared with a
result of Futer-Purcell for $6$-highly twisted diagrams. While their proof
uses geometric methods our proof is achieved by showing that the complements
of such knots or links are unannular and atoroidal. This is done by using a
new approach involving an Euler characteristic argument.

###### Key words and phrases:

knot diagrams, plats, twist regions, Euler Characteristic

###### 2010 Mathematics Subject Classification:

Primary 57M

NL was supported by the Israel Science Foundation (grant no. 1562/19), and by
the German-Israeli Foundation for Scientific Research and Development.

##  1\. introduction

The prevailing feeling among low dimensional topologists is that “most” links
$\mathcal{L}$ in $S^{3}$ are hyperbolic. That means that the open manifold
$S^{3}\smallsetminus\mathcal{L}$ can be endowed with a complete hyperbolic
metric of sectional curvature $-1$. Being hyperbolic is a property of the
manifold with far reaching consequences. However, proving that a specific link
$\mathcal{L}$ is hyperbolic turns out to be not trivial. This is especially
true if the link $\mathcal{L}$ is “heavy duty”, i.e., has a very large
crossing number. See for example [8].

The question of when can one decide if the complement of a link in $S^{3}$ is
a hyperbolic manifold from its projection diagram has been of interest for a
long time. Just to give three examples: The first result in this direction is
by Hatcher and Thurston who proved that complements of $2$-bridge knots which
have at least two twist regions (they are not torus knots or links) are
hyperbolic, see [3]. The second is Menasco’s result [7] that a non-split prime
alternating link which is not a torus link is hyperbolic. Later Futer and
Purcell proved in [2], among other results, that every link with a $6$-highly
twisted irreducible diagram and which has at least two twist regions is
hyperbolic. Their result is obtained by applying Marc Lackenby’s $6$-surgery
theorem, see [4], to the corresponding fully augmented links. Our main theorem
is:

######  Theorem 1.1.

Let $L$ be a $3$-highly twisted $2m$-plat, $m\geq 2,$ with at least three
twist regions, then $L$ is hyperbolic.

Every link $\mathcal{L}$ in $S^{3}$ has a plat projection [1]. Assume that
$\mathcal{L}$ has a $2m$-plat projection $D(\mathcal{L})=D_{P}(\mathcal{L})$
in some plane $P$, for some $m\geq 2$. Note that every $2m$-plat projection
defines a knot or link projection with $m\geq b(\mathcal{L})$ bridges, and
every $m$-bridge knot or link has a $2m$-plat projection (see [1, p. 24]).
Here, $b(\mathcal{L})$ denotes the bridge number of $\mathcal{L}$ as defined
in the next section. It follows from Theorem 1.1 that not all links have a
$3$-highly twisted plat diagram. The subset of links that do is a “large”
subset in a sense that can be made precise, see the discussion in [6]. In
Theorem 1.1 we weaken the conditions imposed in [2] on $\mathcal{L}$ from
$6$-highly twisted to $3$-highly twisted, at the price of requiring that the
diagram of $\mathcal{L}$ be a plat. It seems that the techniques developed
here using the Euler characteristic, with some additional work, might be
adequate for more general diagrams. Thus, we would like to make the following
conjecture:

######  Conjecture 1.2.

Let $L$ be a link in $S^{3}$ with a link diagram which is prime, twist-
reduced, $3$-highly twisted and has least two twist regions, then the link $L$
is hyperbolic.

##  2\. preliminaries

###  2.1. Bubbles and twist regions.

Given a projection of a link $\mathcal{L}$ onto a plane $P$, surround each
crossing in the projection diagram by a small $3$-ball $B$. Denote the
collection of these $3$-balls by $\mathcal{B}$. Then $\mathcal{L}$ is isotopic
to a link $L$ that is embedded in $P\cup{\partial{\mathcal{B}}}$. For a single
crossing we refer to $\partial B$ as a _bubble_. Note that $P$ divides each
bubble into two hemispheres denoted by $\partial B^{+}$ and $\partial B^{-}$.
Denote the union of all the $\partial B^{\pm}$ by $\mathcal{B}^{\pm}$
respectively. Denote the two disjoint $2$-spheres
$(P\smallsetminus\mathcal{B})\cup\mathcal{B}^{\pm}$ by $P^{\pm}$ respectively.
Each of $P^{\pm}$ bounds a $3$-ball $H^{\pm}$ in $S^{3}\smallsetminus L$.

A _twist region_ $T$ in $L$ is a “cube” $D\times[-\varepsilon,+\varepsilon]$
where $D$ is a maximal disk in $P$ so that and $(T,T\cap L)$ is a trivial
integer 2-tangle. For example, in Figure 1, a box labeled $a_{i,j}$ indicates
a _twist region_ with $a_{i,j}$ crossings, where $a_{i,j}$ can be positive,
negative, or zero. In the example in the figure, $a_{1,1}=a_{2,2}=-3$, and all
other $a_{i,j}=-4$.

$a_{1,1}$$a_{1,2}$$a_{2,1}$$a_{2,2}$$a_{2,3}$$a_{3,1}$$a_{3,2}$$a_{4,1}$$a_{4,2}$$a_{4,3}$$a_{5,1}$$a_{5,2}$
Figure 1. A 6-plat projection of a $3$-bridge knot.

###  2.2. Plats.

Let $b$ be an element in the braid group $\mathcal{B}_{2m}$ on $2m-1$
generators $\\{\sigma_{1},\dots,\sigma_{2m-1}\\}$. We require that $b$ is
written as a concatenation of sub-words as follows:

$b=b_{1}\cdot b_{2}\cdot....\cdot b_{n-1},$

where $n$ is odd, and where $b_{i}$ has the following properties:

  1. (1)

When $i$ is odd, $b_{i}$ is a product of all $\sigma_{j}$ with $j$ even.
Namely:

$b_{i}=\sigma_{2}^{a_{i,1}}\cdot\sigma_{4}^{a_{i,2}}\cdot\dots\cdot\sigma_{2m-2}^{a_{i,m-1}}$

  2. (2)

When $i$ is even, $b_{i}$ is a product of all $\sigma_{j}$ with $j$ odd.
Namely:

$b_{i}=\sigma_{1}^{a_{i,1}}\cdot\sigma_{3}^{a_{i,2}}\cdot\dots\cdot\sigma_{2m-1}^{a_{i,m}}$

Consider the geometric braid on $2m$ strings corresponding to the element $b$.
At the top of $b_{1}$, connect each pair of strands
$\\{1,2\\},\\{3,4\\},\dots\\{2m-1,2m\\}$, (ordered from left to right) by a
small unknotted arc. Similarly connect the same pairs of strands at the bottom
of $b_{n}$.

The obtained knot or link is a $2m$-plat. The number $m$ is the _width_ of the
plat and $n$, is the _length_ of the plat. The braid $b$ will be called the
underlying braid of the plat.

For any knot or link $\mathcal{L}\subset S^{3}$ there is an $m\in\mathbb{N}$
so that $\mathcal{L}$ has a $2m$-plat projection, on some projection plane $P$
as indicated in Figure 1. This follows from the fact that any knot or link can
be presented as a closed braid (proved by Alexander in 1923 see [1, p. 23])
and then strands of the braid closure can be pulled across the braid diagram;
for example see [1, p. 24]. The number $m$ is by no means unique. If
$m=br(\mathcal{L})$, where $br(\mathcal{L})$ is the bridge number of
$\mathcal{L}$ (see Schubert [9]), then the $2m$-plat is a _minimal plat_ for
$\mathcal{L}$.

Given a knot or link in a $2m$-plat projection then the twist regions
corresponding to the $\sigma_{i,j}^{a_{i,j}}$ are called _twist boxes_. The
twist boxes are arranged in a configuration which is “almost” a matrix: There
are $n\in\mathbb{N}$ rows indexed by $i\in\\{1,\dots,n\\}$. For odd $i$’s
there are $m-1$ columns and rows with an even $i$ have $m$ columns. Denote the
crossing number in each twist box by $t_{i,j}=a_{i,j}$.

######  Definition 2.1.

A $2m$-plat will be called _$c$ -highly twisted_ if $\left|t_{i,j}\right|\geq
c$ for some constant $c$, for all $i,j$. Similarly, a knot or link that admits
a $c$-highly twisted plat projection will be called a _$c$ -highly twisted
knot or link_.

###  2.3. Diagram regions

Every knot of link diagram $D(L)$ in a projection plane $P\subset S^{3}$
defines a planar $4$-regular graph. The complementary surfaces of
$P\smallsetminus D(L)$ can be colored black and white so that two surfaces
adjacent to an edge are colored in different colors.

A $2m$-plat diagram $D(L)$ determines a template diagram $\mathcal{T}$ in $P$
where each twist region is replaced by a rectangle. Thinking of these
rectangles as vertices of valency $4$ in the graph determined by
$\mathcal{T}$, one can color the complementary regions of the graph in a black
and white checkerboard manner.

######  Definition 2.2 (Lackenby [5]).

A link diagram $D(L)$ is _prime_ if any simple closed curve in $P$
intersecting $D(L)$ in two points bounds a subdiagram with no crossings.

A link diagram $D(L)$ is _twist-reduced_ if any simple closed curve in $P$
which intersects the edges of $\mathcal{T}$ transversally in four points
composed of two pairs each of which is adjacent to a crossing of $D(L)$,
bounds a subdiagram which is the diagram of an integer $2$-tangle.

######  Remark 2.3.

Let $L$ be a $3$-highly twisted $2m$-plat with $m\geq 3$. Then one can observe
directly that the corresponding diagram $D(L)$ is prime and twist-reduced.

######  Remark 2.4.

An arc in $P$ connecting two regions of different colors must cross an edge of
the graph or go through a rectangle. Hence an arc on $P^{\pm}$ in the
$2m$-plat diagram connecting two regions of different colors must intersect
$L$ or meet a bubble (by which we mean that the arc intersects $\partial B$ in
an arc for some bubble $B$).

######  Definition 2.5.

Given two regions in $A,B\in P\smallsetminus\mathcal{T}$ define the distance
$d(A,B)$ between $A$ and $B$ to be the minimal number of color changes over
all arcs between $A$ and $B$. Similarly, if $a,b$ are points in regions $A,B$
respectively, then define the distance $d(a,b)$ to be the distance $d(A,B)$.

######  Definition 2.6.

The regions in $P\smallsetminus\mathcal{T}$ are composed of quadrilaterals,
triangles, four bigons and a single unbounded region. The bigon regions are
called the _corners_ of the template $\mathcal{T}$.

######  Definition 2.7.

We say that an arc of $L$ connects two regions if its endpoints are contained
in the closure of the two regions respectively (note that the same arc can
connect different pairs of regions).

######  Definition 2.8.

Given a knot or link projection with the over-crossings and under-crossings
indicated, a _bridge_ 111Note that the standard definition of a bridge
requires the arc to be maximal and contain at least one over-crossing. does
not contain any under-crossings, i.e., it is a subarc of $L\cap P^{+}$. Note
that in a plat, a bridge can pass over at most two crossings.

######  Observation 2.9.

If $L$ is 2-highly twisted, and $\alpha$ a bridge which connects two regions
of the diagram $D(L)$ then:

  1. (1)

If $\alpha$ passes over two crossings, then $\alpha$ is the unique bridge
connecting the same regions that passes over two crossings.

  2. (2)

If $\alpha$ passes over two crossings and the two regions that are connected
by $\alpha$ are at distance at least 2, then $\alpha$ is the unique bridge
connecting them.

Note that the only case of an arc $\alpha$ which passes over two crossings and
connects regions of distance less than $2$ occurs on the boundaries of the
corner bigon regions of the diagram. In this case, there is a second arc which
does not pass over any crossings which connects the same regions.

##  3\. Surfaces in link complements.

###  3.1. Normal position.

We are interested in studying compact surfaces $S$ properly embedded in
$S^{3}\smallsetminus\mathcal{N}(L)$. If $\partial S\neq\emptyset$ we extend
$S$ by shrinking the neighborhood $\mathcal{N}(L)$ radially. This determines a
map $i:S\to S^{3}$, whose image we denote by $S$ as well, which is an
embedding on $S\smallsetminus\partial S$ and $i(\partial S)\subseteq L$.

######  Lemma 3.1.

Let $S\subset S^{3}\smallsetminus\mathcal{N}(L)$ be a proper surface with no
meridional boundary components, and let $(T,t)$ be a twist region. Then, up to
isotopy, and $S\cap T$ is a disjoint union of disks $D\subset(T,t)$ of one of
the following three types:

  1. Type 0:

$D$ separates the two strings of $t$.

  2. Type 1:

$\partial D$ decomposes as the union of two arcs $\alpha\cup\beta$ such that
$\alpha\subset t$ and $\beta\subset\partial T$.

  3. Type 2:

$\partial D$ decomposes as the union of four arcs
$\alpha_{1}\cup\beta_{1}\cup\alpha_{2}\cup\beta_{2}$ where $\alpha_{i}\subset
t_{i}$ and $\beta_{i}\subset\partial T$.

Moreover, the isotopy decreases the number of bubbles that $S$ meets, and if
no component of $\partial S$ is a meridian, then we may further assume that
$i|_{\partial L}:\partial S\to L$ is a covering map.

###### Proof.

If no component of $\partial S$ is a meridian, we may assume that up to
isotopy $i|_{\partial L}:\partial S\to L$ is a covering map.

The twist region $(T,t)$ is a trivial 2-tangle. The complement
$T\smallsetminus\mathcal{N}(t)$ can be identified with $P\times[0,1]$ where
$P$ is a twice holed disk. Let $E$ be the disk $\alpha\times[0,1]$ where
$\alpha$ is the simple arc connecting the two holes of $P$. Up to a small
isotopy, we may assume that $S$ intersects $E$ transversely. Since the bubbles
in $T$ are in some neighborhood of $E$, we may assume that $S$ meets a bubble
if it does so in $E$. The intersection $S\cap E$ comprises of simple closed
curves and arcs. All curves and arcs except those connecting
$\alpha\times\\{0\\}$ to $\alpha\times\\{1\\}$ can be eliminated by an isotopy
pushing $S$ off $T$. This isotopy decreases the number of bubbles $S$ meets.
The number of bubbles the resulting surface meets equals the number of such
arcs times the number of twist in the twist box.

Up to isotopy, we may also assume that $S$ intersects
$P\times\\{\tfrac{1}{2}\\}$ transversely. Hence,
$S\cap(P\times\\{\tfrac{1}{2}\\})$ is a collection of simple closed curves and
arcs. By pushing $S$ outwards towards the boundary of the disk $P$, one can
assume that each component of $S\cap(P\times\\{\tfrac{1}{2}\\})$ is of the
following form:

  1. (0)

An arc connecting the boundary of the disk $P$ to itself separating the holes,
and intersecting $\alpha$ once.

  2. (1)

An arc connecting a hole to the boundary of the disk and not intersecting
$\alpha$.

  3. (2)

An arc connecting the two holes and not intersecting $\alpha$.

Thus, $S\cap(P\times[\tfrac{1}{2}-\varepsilon,\tfrac{1}{2}+\varepsilon])$ is a
collection of disks of types (0),(1) or (2) as stated. By an ambient isotopy,
we can stretch the slab
$P\times[\tfrac{1}{2}-\varepsilon,\tfrac{1}{2}+\varepsilon]$ to
$P\times[0,1]=T$. The number of bubbles the resulting surface meets equals the
number of arcs of type (0) times the number of twist in the twist box. The
arcs of type (0) are in one-to-one correspondence with the arcs of $S\cap E$.

The fact that $i:\partial S\to L$ is a covering map was not affected by the
isotopies above. ∎

######  Definition 3.2.

A surface $S\subset S^{3}\smallsetminus\mathcal{N}(L)$ is _in normal position_
if its extension intersects each twist region as specified in Lemma 3.1 and
$i:\partial S\to L$ is a covering map. In particular, $S$ has no meridional
boundary components.

######  Corollary 3.3.

Let $S\subset S^{3}\smallsetminus\mathcal{N}(L)$ be a proper incompressible
surface, and let $(T,t)$ be a twist region. Then, up to isotopy, each
component of the intersection $S\cap T\cap P^{\pm}$ looks as in Figure 2.

\begin{overpic}[height=113.81102pt]{figures/Type0twist.pdf}
\put(5.0,-10.0){Type 0}
\end{overpic}\begin{overpic}[height=113.81102pt]{figures/Type1twist.pdf}
\put(5.0,-10.0){Type 1} \end{overpic}

\begin{overpic}[height=113.81102pt]{figures/Type2twist.pdf}
\put(5.0,-10.0){Type 2} \put(75.0,76.0){$S\cap P^{-}$} \put(75.0,63.0){$S\cap
P^{+}$} \end{overpic}

Figure 2. The possible three types of intersection of $S$ with a twist box.

###  3.2. Curves of intersection.

Let $S\subseteq S^{3}\smallsetminus\mathcal{N}(L)$ be a surface in normal
position. We would like to study the surface $S$ through its curves of
intersection with the planes $P^{\pm}$. However, disks of Type (2) cause some
technical complications. In order to simplify the situation we consider the
surface $\widehat{S}$ obtained by removing those disks from $S$. Explicitly:
Let $\mathcal{T}$ be the union of all twist boxes of the plat $L$. Consider
the collection of disks $\mathcal{D}$ of Type (2) which occur as intersections
$S\cap\mathcal{T}$. We may assume that $\partial\mathcal{D}\subset P\cup L$,
and that the subsurface $\widehat{S}=S\smallsetminus\mathcal{D}$ is
transversal to $P^{\pm}$.

Define $\mathcal{C}^{+}=\partial i^{-1}(\widehat{S}\cap H^{+})$ and
$\mathcal{C}^{-}=\partial i^{-1}(\widehat{S}\cap H^{-})$. Now define
$\mathcal{C}=\mathcal{C}^{+}\cup\mathcal{C}^{-}$. As each of $P^{\pm}$ is a
$2$-sphere, $\widehat{S}\cap H^{\pm}$ is a collection of subsurfaces of
$\widehat{S}$, the boundary of which are simple closed curves $c\subset S$.
For $c\in\mathcal{C}^{+}$, denote by $S_{c}$ the component of $\widehat{S}\cap
H^{+}$ so that $c\subset\partial S_{c}$, and respectively for
$c\in\mathcal{C}^{-}$. Although formally defined as the preimage under $i$, we
think of curves in $\mathcal{C}^{\pm}$ as curves on $P^{\pm}$, as such they
are disjoint outside $L$.

######  Definition 3.4.

For $c\in\mathcal{C}$,

  1. (1)

We say that a curve $c\in\mathcal{C}$ _passes through a bubble_ $B$ if
$c\cap(\partial B\smallsetminus L)\neq\emptyset$.

  2. (2)

An _intersection point_ on $c\in\mathcal{C}$ is an endpoint of the intervals
in $c\smallsetminus L$.

  3. (3)

Denote by $\\#_{b}(c)$ the number of bubbles (with multiplicities) through
which $c$ passes.

  4. (4)

Denote by $\\#_{I}(c)$ the number of intersection points along $c$.

  5. (5)

Define $\\#(c)=\\#_{b}(c)+\\#_{I}(c)$.

  6. (6)

Define $\mathcal{C}_{i,j}=\\{c\in\mathcal{C}\mid\\#_{b}(c)=i,\\#_{I}(c)=j\\}$.

###  3.3. Taut surfaces

######  Definition 3.5.

Given an incompressible surface $S\subset S^{3}\smallsetminus\mathcal{N}(L)$
we define a lexicographic complexity of $S$ as follows:

$\text{Com}(S)=(\left|\mathcal{C}\right|,\,\sum_{c\in\mathcal{C}}\\#_{b}(c),\,\sum_{c\in\mathcal{C}}\\#_{I}(c))$
(3.1)

Recall that a properly embedded surface $S$ in a $3$-manifold $M$ is called
_essential_ if it is either a 2-sphere which does not bound a 3-ball, or it is
incompressible, boundary incompressible and not boundary parallel.

######  Lemma 3.6.

Let $S\subset S^{3}\smallsetminus\mathcal{N}(L)$ be an essential surface in
normal position. Assume that either

  1. (i)

$S$ is an essential 2-sphere, and minimizes complexity among all essential
2-spheres, or

  2. (ii)

$S$ is not a 2-sphere, the link $L$ is not split (i.e., $S^{3}\smallsetminus
L$ is irreducible), and $S$ minimizes complexity in its isotopy class.

Then, for all $c\in\mathcal{C}$ we have:

  1. (1)

$S_{c}\cong\mathbb{D}^{2}$.

  2. (2)

$\\#_{I}(c)$ is even.

  3. (3)

If $\\#_{I}(c)\leq 2$ then $\\#_{b}(c)>0$.

  4. (4)

If $\\#_{I}(c)=0$ then $\\#_{b}(c)$ is even.

  5. (5)

The curve $c$ does not pass twice through the same bubble on the same side of
$L$.

  6. (6)

The curve $c$ does not contain an arc, as depicted in Figure 3, that passes
through exactly one bubble and has two intersection points with an edge of $L$
that emanates from the bubble on both sides of the edge.

  7. (7)

The curve $c$ does not contain an arc, as depicted in Figure 4, that is
contained in a twist box, and passes through exactly one bubble and one
intersection point.

  8. (8)

There is no arc of $c\cap L$ such that a small extension of the arc along $c$
has endpoints in the same region.

###### Proof.

Let $S\subset S^{3}\smallsetminus\mathcal{N}(L)$ be an essential surface
satisfying (i) or (ii). Note that in both cases, compressing along a disk
$D\subset S^{3}\smallsetminus\mathcal{N}(L)$ with $D\cap S=\partial D$ results
either in two essential spheres, or a surface in the same isotopy class of
$S$. Thus, by the assumption on $S$, surfaces obtained by such a compression
cannot have lower complexity.

(1) Since $S$ is essential, each subsurface $S_{c}$ must be planar, as
otherwise it contains a non-trivial compression disk. If $S_{c}$ has more than
one boundary component then compressing along a disk in $H^{+}$ or $H^{-}$
whose boundary separates boundary components of $S_{c}$ will result in a
surface with fewer intersections with $P$ in contradiction to the choice of
$S$.

(2) By definition, $\\#_{I}(c)$ is the number of endpoints of arcs in
$c\smallsetminus L$. Since each arc has two endpoints, $\\#_{I}(c)$ is even.

(3) By (2) $\\#_{I}(c)$ is either two or zero. If $\\#_{I}(c)=0$ and
$\\#_{b}(c)=0$ then $c$ bounds a disk on $P\smallsetminus L$. Compressing $S$
along this disk reduces the number of intersections with $P$.

If $\\#_{I}(c)=2$ and $\\#_{b}(c)=0$ then $c$ bounds a disk $D$ in $P$ such
that $\partial D=\alpha\cup\beta$, where $\alpha$ is an arc in $L$ and $\beta$
is an arc in $P$. Since both $\alpha$ and $\beta$ do not pass through bubbles
$\partial D$ bounds a disk in $P\smallsetminus L$. By choosing an innermost
such $D$ we may assume that $D\subset P\smallsetminus(L\cup S)$ and thus $D$
is a boundary compression disk. Compressing $S$ along $D$ we get an isotopic
surface with less intersections with $P$ which is a contradiction.

(4) Consider the colors of complementary regions of
$P\smallsetminus\mathcal{T}$ which the curve $c$ intersects. If $\\#_{I}(c)=0$
every change of colors, of these regions along $c$, accounts for one bubble
that $c$ meets. Since $c$ is a closed curve the total number of color changes
is even, and correspondingly $\\#_{b}(c)$ is even.

(5) This claim follows directly from Lemma 1(ii) of [7].

(6) In this case there is an ambient isotopy of the surface $S$ which pushes
the disk bounded by the curve $c$ through the bubble thus reducing the number
of bubbles met by $S$ by one. The isotopy is indicated by the arrow in Figure
3. The surface $S$ is assumed to be in normal form, by Definition 3.2. This
contradicts the assumption on the choice of $S$ as minimizing the complexity
as in Definition 3.5.

(7) The proof in this case is the same as in case (6), using the isotopy
described by Figure 4. Note that as a result of the isotopy we see two more
intersection points with $L$ but one less bubble.

(8) If such an arc $\alpha\subseteq c\cap L$ then by pushing $S$ through $P$
in a neighborhood of $\alpha$ we reduce the number of intersection points by
2, in contradiction to the minimal complexity of $S$. ∎

\begin{overpic}[width=99.58464pt]{figures/isotopy1.pdf}
\put(5.0,55.0){$\alpha$} \end{overpic}

Figure 3. The isotopy in Case (6) of Lemma 3.6

\begin{overpic}[width=99.58464pt]{figures/isotopy3.pdf}
\put(60.0,90.0){$\alpha$} \end{overpic}

Figure 4. The isotopy in Case (7) of Lemma 3.6

######  Definition 3.7.

A surface $S$ is _taut_ if it satisfies the conditions specified in Lemma 3.6.

######  Remark 3.8.

Note that if $S$ is taut then
$\mathcal{C}_{0,0}=\mathcal{C}_{0,2}=\mathcal{C}_{i,2k+1}=\mathcal{C}_{2k+1,0}=\emptyset$
for all $i,k\in\mathbb{N}\cup\\{0\\}$.

######  Remark 3.9.

From now on we assume that the surface $S$ is taut.

##  4\. Euler Characteristic and curves of intersection

###  4.1. Distributing Euler characteristic among curves

For each curve $c\in\mathcal{C}$ we will define the _contribution_ of $c$, and
show that the Euler characteristic of $S$ can be computed by summing up the
contributions of curves $c\in\mathcal{C}$.

######  Definition 4.1.

The _contribution_ $\chi_{+}(c)$ of a curve $c\in\mathcal{C}$ is defined by

$\chi_{+}(c)=\frac{\chi(S_{c})}{|\partial S_{c}|}-\frac{1}{4}\\#(c).$

Note that if $S$ is taut, then $S_{c}\cong D^{2}$, and therefore
$\chi_{+}(c)=1-\frac{1}{4}\\#(c)$.

######  Lemma 4.2.

If $S\subset S^{3}\smallsetminus\mathcal{N}(L)$ is taut then
$\chi(S)=\sum_{c\in\mathcal{C}}\chi_{+}(c)$.

###### Proof.

The union of the collection of all the curves $c\in\mathcal{C}$ on $S$ is an
embedded graph $X$ of $S$. Let $X^{0}$ and $X^{1}$ denote the vertex and the
edge sets of $X$ respectively. The vertices of the graph are the points in $S$
corresponding to the points on $c$ which are on $P\cap\mathcal{B}$ (i.e.,
where $c$ meets a bubble) or $P\cap L$ (i.e., an intersection point of $c$).
The graph $X$ partitions $S$ into disk regions of three types:

  1. (1)

the subsurfaces $S_{c}\subseteq\widehat{S}\cap H^{\pm}$ for
$c\in\mathcal{C}^{\pm}$,

  2. (2)

the regions $R\subseteq\widehat{S}\cap B$ where $B\in\mathcal{B}$ is a 3-ball
bounded by a bubble, or

  3. (3)

regions $D\subset S$ corresponding to Type (2) disks.

In case (3), the regions $D$ are disks whose boundary consists of two arcs on
$L$ and two edges of $X$. By collapsing each such disk $D$ to one of the edges
in $X$ we get a homotopic surface. By abuse of notation, we call it $S$, and
call the corresponding graph $X$. Note that in the new surface, $\partial
S\subset X$. It follows that

$\displaystyle\begin{split}\chi(S)&=\chi(X)+\sum_{S^{\prime}\subseteq S\cap
H^{\pm}}\chi(S^{\prime})+\sum_{R\subseteq S\cap B}\chi(R)\\\
&=|X^{0}|-|X^{1}|+\sum_{S^{\prime}\subseteq S\cap
H^{\pm}}\chi(S^{\prime})+\sum_{R\subseteq S\cap B}\chi(R).\end{split}$ (4.1)

We compute how each $c\in\mathcal{C}$ contributes to each of the summands in
(4.1):

The vertices of $X$. Every curve $c\in\mathcal{C}$ passes through
$2\\#_{b}(c)$ vertices of $X^{0}$ in the interior of $S$ (because it goes in
and out of a bubble). Further, it goes through $\\#_{I}(c)$ vertices of
$X^{0}$ in $\partial S$. Moreover, each of these vertices belongs to two
curves $c\in\mathcal{C}$. Hence,

$|X^{0}|=\sum_{c\in\mathcal{C}}(\\#_{b}(c)+\tfrac{1}{2}\\#_{I}(c))$ (4.2)

The edges of $X$. Every curve $c\in\mathcal{C}$ passes through
$2\\#_{b}(c)+\\#_{I}(c)$ edges in $X^{1}$. Note that every edge in
$X^{1}\cap\text{int}(S)\cap\mathcal{B}$ or in $X^{1}\cap\partial S$ belongs to
exactly one curve in $\mathcal{C}$, while each edge in
$X^{1}\cap\text{int}(S)\smallsetminus\mathcal{B}$ belongs to two curves in
$\mathcal{C}$. Thus, edges in $X^{1}\cap\text{int}(S)\cap\mathcal{B}$ are in
one to one correspondence with the bubbles they meet. Hence,

$|X^{1}\cap\text{int}(S)\cap\mathcal{B}|=\sum_{c\in\mathcal{C}}\\#_{b}(c).$

Similarly each edge in $X^{1}\cap\partial S$ accounts for 2 in $\\#_{I}(c)$.
Hence,

$|X^{1}\cap\partial S|=\sum_{c\in\mathcal{C}}\tfrac{1}{2}\\#_{I}(c).$

Each edge in $X^{1}\cap\text{int}(S)\smallsetminus\mathcal{B}$ accounts for
two vertices in $X^{0}$. So the number of these edges is equal to

$|X^{1}\cap\text{int}(S)\smallsetminus\mathcal{B}|=\tfrac{1}{2}|X^{0}|=\tfrac{1}{2}(\sum_{c\in\mathcal{C}}(\\#_{b}(c)+\tfrac{1}{2}\\#_{I}(c))).$

Adding these contributions together gives

$|X^{1}|=\sum_{c\in\mathcal{C}}(\tfrac{3}{2}\\#_{b}(c)+\tfrac{3}{4}\\#_{I}(c)).$
(4.3)

Regions $S^{\prime}\subset S\cap H^{\pm}$. To every curve $c\in\mathcal{C}$
there is a surface $S_{c}\subseteq S\cap H^{\pm}$, and each such surface is
associated to $|\partial S_{c}|$ curves $c\in\mathcal{C}$. Thus,

$\sum_{S^{\prime}\subseteq S\cap
H^{\pm}}\chi(S^{\prime})=\sum_{c\in\mathcal{C}}\frac{\chi(S_{c})}{|\partial
S_{c}|}.$ (4.4)

Regions $R\subset S\cap B$. Each curve $c\in\mathcal{C}$ passes through the
boundary of $\\#_{b}(c)$ such regions. As each such region has four curves
passing through its boundary, we have

$\sum_{R\subseteq S\cap B}\chi(R)=\sum_{c\in C}\tfrac{1}{4}\\#_{b}(c).$ (4.5)

Summing over all of the above we get,

$\displaystyle\chi(S)$
$\displaystyle=|X^{0}|-|X^{1}|+\sum_{S^{\prime}\subseteq S\cap
H^{\pm}}\chi(S^{\prime})+\sum_{R\subseteq S\cap B}\chi(R)$
$\displaystyle=\sum_{c\in\mathcal{C}}(\tfrac{\chi(S_{c})}{|\partial
S_{c}|}-\tfrac{1}{4}(\\#_{b}(c)+\\#_{I}(c)))$
$\displaystyle=\sum_{c\in\mathcal{C}}\chi_{+}(c).$

∎

###  4.2. Redistributing positive Euler characteristic among curves

In order to prove that $L$ is hyperbolic, we would eventually be interested in
surfaces $S$ with non-negative Euler characteristic, namely spheres, annuli
and tori. It would thus be useful to distribute the Euler characteristic of
$S$ in such a way that each summand is non-positive. This would rule out
2-spheres, and show that each summand must be zero for $S$ to be a torus or an
annulus.

Lemma 4.2 shows that the Euler characteristic of $S$ is the sum of the
contributions of curves in $\mathcal{C}$. However, some curves might have
positive contributions. Our strategy would be to “pass” the contributions of
curves with $\chi_{+}>0$ to “neighbouring” curves with $\chi_{+}<0$. This will
be done by defining $\chi^{\prime}(c)$ in Definition 4.7, and proving in
Lemmas 4.8 and 4.9 respectively that $\chi(S)=\sum\chi^{\prime}(c)$ and
$\chi^{\prime}(c)\leq 0$.

Our search for “neighbouring” curves will use the following definition:

######  Definition 4.3.

  1. (1)

Let $c\in\mathcal{C}$, two subarcs $\alpha_{1},\alpha_{2}$ of $c$ whose
endpoints are not in $\mathcal{B}\cup L$ are _equivalent_ if they are isotopic
in $c$ so that the endpoints of each arc in the isotopy are not in
$\mathcal{B}\cup L$. Note that for equivalent arcs $\alpha_{1},\alpha_{2}$, we
have $\\#_{b}(\alpha_{1})=\\#_{b}(\alpha_{2})$ and
$\\#_{I}(\alpha_{1})=\\#_{I}(\alpha_{2})$. We will use the term _subarc of
$c$_ (or simply _arc_) to indicate its equivalence class. Two arcs
$\alpha,\alpha^{\prime}$ are _disjoint_ if they are subarcs of different
curves or if they have disjoint representatives. Alternatively, two arcs
$\alpha,\alpha^{\prime}$ are non-disjoint, if they are subarcs of the same
curve, and they overlap in a bubble or an intersection point.

  2. (2)

Two curves (or subarcs of curves) $c,c^{\prime}\in\mathcal{C}$ are said to be
_opposite along an arc $\alpha$_, or simply _opposite_ , if $c\neq c^{\prime}$
and $\alpha\subset c\cap c^{\prime}\smallsetminus L$. Note that if
$c,c^{\prime}$ are opposite then one of them is in $P^{+}$ and the other in
$P^{-}$.

######  Definition 4.4.

Denote by $\mathcal{C}_{>0}$ (resp. $\mathcal{C}_{=0},\mathcal{C}_{\leq
0},\mathcal{C}_{<0}$) the collection of all $c\in\mathcal{C}$ such that
$\chi_{+}(c)>0$ (resp. $=0,\leq 0,<0$).

Note that if $S$ is taut, then
$\mathcal{C}_{>0}=\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}$ and
$\mathcal{C}_{=0}=\mathcal{C}_{4,0}\cup\mathcal{C}_{2,2}\cup\mathcal{C}_{0,4}$.
We begin by studying the curves in $\mathcal{C}_{>0}$:

\begin{overpic}[height=99.58464pt]{figures/C20b.pdf}
\put(48.0,90.0){$\kappa^{\prime}$} \put(50.0,78.0){$\kappa^{\prime\prime}$}
\put(-7.0,40.0){$c$} \end{overpic}
\begin{overpic}[height=85.35826pt]{figures/C20a.pdf}
\put(90.0,80.0){$\kappa^{\prime}$} \put(90.0,60.0){$\kappa^{\prime\prime}$}
\put(30.0,55.0){$c$} \end{overpic}

Figure 5. $c\in C_{2,0}$ and the opposite arcs $\kappa^{\prime}$ and
$\kappa^{\prime\prime}$.

Curves in $\mathcal{C}_{2,0}$. Let $c\in\mathcal{C}_{2,0}$. The two possible
configurations for $c$ are shown in Figure 5.

In case (a), opposite to $c$ there are two arcs $\kappa^{\prime}$ and
$\kappa^{\prime\prime}$ shown in Figure 5 passing through two and four bubbles
respectively. Let $\mathcal{K}_{4,0}$ denote the collection of all such arcs
$\kappa^{\prime\prime}$ opposite to some $c\in\mathcal{C}_{2}$. (In this and
following discussion, the subscript $4,0$ indicates that the arcs in
$\mathcal{K}_{4,0}$ pass through four bubbles and zero intersection points.)

In case (b), opposite to $c\in\mathcal{C}_{2,0}$ there are two arcs,
$\kappa^{\prime}$ and $\kappa^{\prime\prime}$ shown in Figure 5. Each of the
arcs $\kappa^{\prime}$ and $\kappa^{\prime\prime}$ passes through three
bubbles. Let $\mathcal{K}_{3,0}$ denote the collection of all such
$\kappa^{\prime},\kappa^{\prime\prime}$ which are opposite to some
$c\in\mathcal{C}_{2,0}$ as described above.

Each arc $\kappa^{\prime}\in\mathcal{K}_{3,0}$ is part of some closed curve
$c^{\prime}\in\mathcal{C}$. We would like to distinguish those
$\kappa^{\prime}$ for which $c^{\prime}\in\mathcal{C}_{4,0}$ as those have
$\chi_{+}(c^{\prime})=0$. We denote the subcollection of all
$\kappa^{\prime}\in\mathcal{K}_{3,0}$ for which
$c^{\prime}\notin\mathcal{C}_{4,0}$ by
$\widehat{\mathcal{K}}_{3,0}\subset\mathcal{K}_{3,0}$.

Let
$\kappa^{\prime}\in\mathcal{K}_{3,0}\smallsetminus\widehat{\mathcal{K}}_{3,0}$.
Then $\kappa^{\prime}$ is a subarc of a curve $c^{\prime}$ with
$\\#_{b}(c^{\prime})=4$, which is opposite to some $c\in\mathcal{C}_{2,0}$.
Note that this case can only occur in the “corners” of the plat, as shown in
Figure 9 for the top left “corner”. Let $\widetilde{\kappa}$ be the arc shown
in Figure 9. Let $\widetilde{\mathcal{K}}$ denote the collection of all
$\widetilde{\kappa}$ constructed in this way.

Curves in $\mathcal{C}_{1,2}$. Schematically, a curve $c\in\mathcal{C}_{1,2}$
must be as shown in Figure 6.

######  Remark 4.5.

As $c$ emanates from $c\cap L$ into regions of different colors, $c\cap L$
cannot pass over a crossing which is not the top or bottom in its twist box.

\begin{overpic}[width=99.58464pt]{figures/C20c.pdf} \put(65.0,75.0){$c$}
\put(95.0,23.0){$\kappa$} \end{overpic} Figure 6. $c\in C_{1,2}$ and the
opposite arc $\kappa$.

Similarly, opposite to a curve $c\in\mathcal{C}_{1,2}$ there is an arc
$\kappa$ passing through two bubbles and one intersection point, shown in
Figure 6. Let $\mathcal{K}_{2,1}$ be the collection of all such $\kappa$
opposite to some $c\in\mathcal{C}_{1,2}$.

Define
$\mathcal{K}=\widehat{\mathcal{K}}_{3,0}\cup\mathcal{K}_{4,0}\cup\widetilde{\mathcal{K}}\cup\mathcal{K}_{2,1}$.

######  Lemma 4.6.

Arcs in $\mathcal{K}$ are pairwise disjoint.

###### Proof.

By definition (see Definition 4.3), two arcs are not-disjoint if they overlap
in a bubble or an intersection point.

One can check that the intersection of an arc $\kappa$ with a bubble in a
twist box determines its intersection with the entire twist box, and that
$\kappa$ must pass through either the top or bottom bubble in this twist box.
Because the plat is 3-highly twisted, an arc $\kappa$ cannot pass through both
the top and bottom bubbles of the twist box. The part of the arc leaving the
twist box determines the curve
$c\in\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}\cup\mathcal{C}_{4,0}$ opposite to
it. Thus, two arcs $\kappa_{1},\kappa_{2}$ which overlap in a bubble determine
the same opposite curve $c$, hence they are identical.

Similarly, if two arcs $\kappa_{1},\kappa_{2}$ overlap in an intersection
point, they are opposite to the same curve $c\in\mathcal{C}_{1,2}$, and they
are identical. ∎

Recall that our goal is to redistribute the Euler characteristic among curves
so that each will contribute non-positively. The quantity $\chi^{\prime}$
defined below is the sought for redistribution.

######  Definition 4.7.

Let $c^{\prime}\in\mathcal{C}_{\leq 0}$. Let $n_{3}$ (resp. $\widehat{n}_{3}$;
$n_{4}$; $\widetilde{n}$; $n_{2,1}$) be the number of subarcs
$\kappa\in\mathcal{K}_{3,0}$ (resp. $\widehat{\mathcal{K}}_{3,0}$;
$\mathcal{K}_{4,0}$; $\widetilde{\mathcal{K}}$; $\mathcal{K}_{2,1}$) in
$c^{\prime}$. We associate to $c^{\prime}$ the following quantity

$\chi^{\prime}(c^{\prime})=\chi_{+}(c^{\prime})+\tfrac{1}{4}\widehat{n}_{3}+\tfrac{1}{2}n_{4}+\tfrac{1}{4}\widetilde{n}+\tfrac{1}{4}n_{2,1}.$

The next lemma shows that $\chi^{\prime}$ is a redistribution of the Euler
characteristic of $S$ among curves in $\mathcal{C}_{\leq 0}$.

######  Lemma 4.8.

$\chi(S)=\sum_{c^{\prime}\in\mathcal{C}_{\leq 0}}\chi^{\prime}(c^{\prime}).$

###### Proof.

Since by Lemma 4.2, $\chi(S)=\sum_{c\in\mathcal{C}}\chi_{+}(c)$, it remains to
prove that

$\sum_{c\in\mathcal{C}}\chi_{+}(c)=\sum_{c^{\prime}\in\mathcal{C}_{\leq
0}}\chi^{\prime}(c^{\prime}).$

Subtracting $\sum_{c^{\prime}\in\mathcal{C}_{\leq 0}}\chi_{+}(c^{\prime})$
from both sides and recalling that $\mathcal{C}_{\leq
0}=\mathcal{C}\smallsetminus\mathcal{C}_{>0}$, we have to show

$\sum_{c\in\mathcal{C}_{>0}}\chi_{+}(c)=\sum_{c^{\prime}\in\mathcal{C}_{\leq
0}}(\chi^{\prime}(c^{\prime})-\chi_{+}(c^{\prime})).$

The left hand side is simply
$\tfrac{1}{2}|\mathcal{C}_{2,0}|+\tfrac{1}{4}|\mathcal{C}_{1,2}|$ since
$\mathcal{C}_{>0}=\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}$ and

$\chi_{+}(c)=\begin{cases}\tfrac{1}{2}&\mbox{if }c\in\mathcal{C}_{2,0}\\\
\tfrac{1}{4}&\mbox{if }c\in\mathcal{C}_{1,2}\end{cases}.$

By the definition of $\chi^{\prime}$, the right hand side gives
$\tfrac{1}{4}|\widehat{\mathcal{K}}_{3,0}|+\tfrac{1}{2}|\mathcal{K}_{4,0}|+\tfrac{1}{4}|\widetilde{\mathcal{K}}|+\tfrac{1}{4}|\mathcal{K}_{2,1}|$.
Note that for every arc
$\kappa\in\mathcal{K}_{3,0}\smallsetminus\widehat{\mathcal{K}}_{3,0}$ there
exists a _unique_ arc $\widetilde{\kappa}\in\widetilde{\mathcal{K}}$. Hence we
have
$|\mathcal{K}_{3,0}\smallsetminus\widehat{\mathcal{K}}_{3,0}|=|\widetilde{\mathcal{K}}|$.
Therefore, the sum becomes
$\frac{1}{4}|\mathcal{K}_{3,0}|+\frac{1}{2}|\mathcal{K}_{4,0}|+\tfrac{1}{4}|\mathcal{K}_{2,1}|$.

Since every curve in $\mathcal{C}_{2,0}$ is opposite to either two arcs in
$\mathcal{K}_{3,0}$, or one arc in $\mathcal{K}_{4,0}$ we get
$|\mathcal{C}_{2,0}|=\frac{1}{2}|\mathcal{K}_{3,0}|+|\mathcal{K}_{4,0}|$
which, after dividing by 2, gives

$\tfrac{1}{2}|\mathcal{C}_{2,0}|=\tfrac{1}{4}|\mathcal{K}_{3,0}|+\tfrac{1}{2}|\mathcal{K}_{4,0}|.$
(4.6)

Similarly, every curve in $\mathcal{C}_{1,2}$ is opposite to one arc in
$\mathcal{K}_{2,1}$ and so $|\mathcal{C}_{1,2}|=|\mathcal{K}_{2,1}|$ which,
after dividing by 4, gives

$\tfrac{1}{4}|\mathcal{C}_{1,2}|=\tfrac{1}{4}|\mathcal{K}_{2,1}|$ (4.7)

Adding together (4.6) and (4.7) completes the proof. ∎

The next lemma shows that indeed $\chi^{\prime}$ is non-positive.

######  Lemma 4.9.

$\chi^{\prime}(c^{\prime})\leq 0$ for all $c^{\prime}\in\mathcal{C}_{\leq 0}$.

###### Proof.

Let $c^{\prime}\in\mathcal{C}_{\leq 0}$ and let
$\widehat{n}_{3},n_{4},\widetilde{n},n_{2,1}$ be as in Definition 4.7 of
$\chi^{\prime}(c^{\prime})$. Then we have,

$\displaystyle\begin{split}\chi^{\prime}(c^{\prime})&=\chi_{+}(c^{\prime})+\tfrac{1}{4}\widehat{n}_{3}+\tfrac{1}{2}n_{4}+\tfrac{1}{4}\widetilde{n}+\tfrac{1}{4}n_{2,1}\\\
&\leq(1-\tfrac{1}{4}\\#(c^{\prime}))+\tfrac{1}{4}\widehat{n}_{3}+\tfrac{1}{2}n_{4}+\tfrac{1}{4}\widetilde{n}+\tfrac{1}{4}n_{2,1}\\\
&\leq
1+\sum_{c^{\prime}\supset\kappa\in\widehat{\mathcal{K}}_{3,0}}(\tfrac{1}{4}-\tfrac{1}{4}\\#(\kappa))+\sum_{c^{\prime}\supset\kappa\in\mathcal{K}_{4,0}}(\tfrac{1}{2}-\tfrac{1}{4}\\#(\kappa))\\\
&\quad\quad+\sum_{\begin{subarray}{c}c^{\prime}\supset\kappa\in\widetilde{\mathcal{K}}\end{subarray}}(\tfrac{1}{4}-\tfrac{1}{4}\\#(\kappa))+\sum_{c^{\prime}\supset\kappa\in\mathcal{K}_{2,1}}(\tfrac{1}{4}-\tfrac{1}{4}\\#(\kappa))\\\
&=1-\tfrac{1}{2}(\widehat{n}_{3}+n_{4}+\widetilde{n}+n_{2,1}).\end{split}$
(4.8)

Where the last equality follows since each one of the summands is
$-\tfrac{1}{2}$. We divide into cases depending on the sum
$n=\widehat{n}_{3}+n_{4}+\widetilde{n}+n_{2,1}$.

Case 0. $n=0$. We have $\chi^{\prime}(c^{\prime})=\chi_{+}(c^{\prime})$. But
since $c^{\prime}\notin\mathcal{C}_{>0}$ we have $\chi_{+}(c^{\prime})\leq 0$
and we are done. (Note that this case includes the case
$\\#_{b}(c^{\prime})=4$ and $n_{3}=1$. See Remark 4.10)

Case 1. $n=1$. That is, $c^{\prime}$ contains a single subarc
$\kappa\in\widehat{\mathcal{K}}_{3,0}\cup\mathcal{K}_{4,0}\cup\widetilde{\mathcal{K}}\cup\mathcal{K}_{2,1}$.
We further divide into sub-cases:

_Case 1.a._ $\kappa\in\widehat{\mathcal{K}}_{3,0}$. Clearly
$\\#_{b}(c^{\prime})\geq\\#_{b}(\kappa)=3$. By definition of
$\widehat{\mathcal{K}}_{3,0}$, $\\#_{b}(c^{\prime})\neq 4$. Then, either
$\\#_{b}(c^{\prime})\geq 6$ or $c^{\prime}$ has at least two intersection
points with $K$. Therefore, $\\#(c^{\prime})\geq 5$ which gives

$\chi^{\prime}(c^{\prime})=\chi_{+}(c^{\prime})+\tfrac{1}{4}\leq(1-\tfrac{5}{4})+\tfrac{1}{4}=0.$

_Case 1.b._ $\kappa\in\mathcal{K}_{4,0}$ so $\\#_{b}(\kappa)=4$. The distance
between the endpoints of $\kappa$ is 2 in the dual graph, so the closed curve
$c^{\prime}$ that contains $\kappa$ must contain two additional bubbles or two
additional intersection points. Thus, $\\#(c^{\prime})\geq 6$, which gives

$\chi^{\prime}(c^{\prime})=\chi_{+}(c^{\prime})+\tfrac{1}{2}\leq(1-\tfrac{6}{4})+\tfrac{1}{2}=0.$

_Case 1.c._ $\kappa\in\widetilde{\mathcal{K}}$. Here again
$\\#_{b}(\kappa)=3$. This happens in one of the cases described in Figure 9.
The endpoints of $\widetilde{\kappa}$ are at distance $\geq 2$. Hence, the
closed curve $c^{\prime}$ containing $\kappa$ must have two additional bubbles
or two additional intersection points. It follows that $\\#(c^{\prime})\geq
5$, which gives

$\chi^{\prime}(c^{\prime})=\chi_{+}(c^{\prime})+\tfrac{1}{4}\leq(1-\tfrac{5}{4})+\tfrac{1}{4}=0.$

_Case 1.d._ $\kappa\in\mathcal{K}_{2,1}$ so $\\#_{b}(\kappa)=2$ and
$\\#_{I}(\kappa)=1$. The curve $c^{\prime}$ must contain an additional
intersection point. Let $\alpha$ be the subarc of $c^{\prime}$ between the two
intersection points. Let $\kappa^{*}$ be a small continuation of
$\kappa\cup\alpha$ along $c^{\prime}$.

######  Claim.

The endpoints of $\kappa^{*}$ are at distance greater or equal to 1.

###### Proof.

The arc $\alpha$ cannot contain more than one overpass: Otherwise, $\alpha$
passes over two crossings which occur in different twist boxes. This implies
that the curve $c\in\mathcal{C}_{1,2}$ opposite to $\kappa$ is such that
$c\cap L$ is contained in one of the twist boxes, and connects the regions to
its left and right, which must have the same color. This contradicts the fact
that $c\smallsetminus(c\cap L)$ passes through exactly one bubble.

If $\alpha$ does not contain an overpass the regions containing the endpoints
of $\kappa^{*}$ have different colors, thus the endpoints of $\kappa^{*}$ are
at distance greater or equal to 1.

If $\alpha$ contains one overpass then if the endpoints of $\kappa^{*}$ can be
connected by an arc containing no intersection points or bubbles then the
union of $\kappa^{*}$ and the arc bound a subdiagram of $D(L)$ which
contradicts the assumption that $L$ is twist-reduced. ∎

It follows from the claim that $\\#(c^{\prime})\geq\\#(\kappa^{*})+1=4+1=5$,
which gives

$\chi^{\prime}(c^{\prime})=\chi_{+}(c^{\prime})+\tfrac{1}{4}\leq(1-\tfrac{5}{4})+\tfrac{1}{4}=0.$

\begin{overpic}[height=113.81102pt]{figures/K21a.pdf} \put(61.0,72.0){$c$}
\put(-9.0,55.0){$c^{\prime}$}
\end{overpic}\begin{overpic}[height=113.81102pt]{figures/K21b.pdf}
\put(-6.0,25.0){$c^{\prime}$} \put(70.0,55.0){$c$} \end{overpic}

Figure 7. Two possibilities of a curve containing an arc in
$\mathcal{K}_{2,1}$.

Case 2. $n\geq 2$. In this case we are done by inequality (4.8). ∎

######  Remark 4.10.

Note that, when $n_{3}=1\mbox{ and }c^{\prime}\in\mathcal{C}_{4,0}$, i.e.,
when $c^{\prime}$ contains a subarc in
$\mathcal{K}_{3,0}\smallsetminus\widehat{\mathcal{K}}_{3,0}$, then
$\chi^{\prime}(c^{\prime})=\chi_{+}(c^{\prime})=0$. In this case, the positive
contribution of the unique $c\in\mathcal{C}_{2,0}$ opposite $c^{\prime}$ is
counted not by $\chi^{\prime}(c^{\prime})$ but by $\chi^{\prime}(\tilde{c})$
where $\tilde{c}$ is the curve opposite to $c^{\prime}$ containing the
corresponding arc $\tilde{\kappa}$.

The proof above also gives the following lemma:

######  Lemma 4.11.

The curve $c^{\prime}\in\mathcal{C}_{\leq 0}$ satisfies
$\chi^{\prime}(c^{\prime})=0$ exactly in the following cases:

  1. (0)

$c^{\prime}$ does not contain a subarc in $\mathcal{K}$ and
$\\#(c^{\prime})=4$. (it might have $n_{3}=1$)

  2. (1)

$c^{\prime}$ contains exactly one subarc $\kappa$ of $\mathcal{K}$ and
satisfies either:

    1. (a)

$\\#(c^{\prime})=5$ and $\kappa\in\widehat{\mathcal{K}}_{3,0}$.

    2. (b)

$\\#(c^{\prime})=6$ and $\kappa\in\mathcal{K}_{4,0}$.

    3. (c)

$\\#(c^{\prime})=5$ and $\kappa\in\widetilde{\mathcal{K}}$.

    4. (d)

$\\#(c^{\prime})=5$ and $\kappa\in\mathcal{K}_{2,1}$.

In all cases, $\\#(c^{\prime})=\\#(\kappa)+2$.

  3. (2)

$c^{\prime}$ is the union of exactly two arcs of $\mathcal{K}$.∎

As an immediate corollary to Lemma 4.9 we get the following:

######  Corollary 4.12.

The link $L$ is non-split.

###### Proof.

Assume for contradiction that $S^{3}\smallsetminus L$ contains an essential
sphere. Let $S$ be an essential sphere with least complexity among all
essential spheres. By Lemma 3.6, $S$ is taut. By Lemma 4.8,
$\chi(S)=\sum_{c^{\prime}\in\mathcal{C}_{\leq 0}}\chi^{\prime}(c^{\prime}).$
Thus it follows from Lemma 4.9 that $\chi(S)\leq 0$, which is a contradiction.
∎

##  5\. Analysing the curves

As stated in the introduction our goal is to prove the following:

######  Theorem 5.1.

There are no essential tori or annuli in $S^{3}\smallsetminus\mathcal{N}(L)$.

In what follows, assume that $S$ is a taut essential torus or annulus. By
Lemma 4.8, $\sum_{c^{\prime}\in\mathcal{C}_{\leq
0}}\chi^{\prime}(c^{\prime})=\chi(S)=0$. By Lemma 4.9 each summand
$\chi^{\prime}(c^{\prime})\leq 0$, and thus must be equal to $0$. It follows
that all the curves $c^{\prime}\in\mathcal{C}_{\leq 0}$ are as classified in
Lemma 4.11.

The proof of Theorem 5.1 will proceed by analysing each case of Lemma 4.11
separately, and showing that $S$ must be boundary parallel. Before we give the
proof we first need to define some notation which will be used below.

######  Definition 5.2.

The projection of a twist region in the diagram is a rectangle called a twist
box, we refer to its edges as top/bottom/left/right as obvious from Figure 8.

\begin{overpic}[height=142.26378pt]{figures/box.pdf} \put(16.0,90.0){top}
\put(12.0,-3.0){bottom} \put(-10.0,55.0){left} \put(44.0,55.0){right}
\put(48.0,76.0){cross} \put(48.0,22.0){turn} \end{overpic} Figure 8. A
crossing and turning arc in a twist box

######  Lemma 5.3.

There are no arcs in
$\mathcal{K}_{3,0}\smallsetminus\widehat{\mathcal{K}}_{3,0}$. In particular
$\widetilde{\mathcal{K}}=\emptyset$.

###### Proof.

Assume in contradiction that there exists an arc
$\kappa^{\prime}\in\mathcal{K}_{3,0}\smallsetminus\widehat{\mathcal{K}}_{3,0}$.
By assumption, the curve $c^{\prime}$ containing $\kappa^{\prime}$ is in
$\mathcal{C}_{4,0}$. Let $c\in\mathcal{C}_{2,0}$ be the curve opposite to
$\kappa^{\prime}$, let $\widetilde{\kappa}$ be the corresponding arc in
$\widetilde{\mathcal{K}}$ opposite to $c^{\prime}$, and let $\widetilde{c}$ be
the curve containing $\widetilde{\kappa}$. Without loss of generality, we may
assume that $c\subset P^{+}$. This implies that
$\kappa^{\prime},c^{\prime}\subset P^{-}$ and
$\widetilde{\kappa},\widetilde{c}\subset P^{+}$. The possible configurations
of $c,c^{\prime},\kappa^{\prime},\widetilde{\kappa}$ are depicted in Figure 9.
Note that under the assumptions above, the sign of the twists in the figure
must be as depicted.

\begin{overpic}[height=113.81102pt]{figures/Ktilde1.pdf} \put(40.0,17.0){$c$}
\put(102.0,70.0){$\tilde{\kappa}$} \put(43.0,88.0){$c^{\prime}$} \end{overpic}
\begin{overpic}[height=113.81102pt]{figures/Ktilde2.pdf} \put(40.0,17.0){$c$}
\put(102.0,58.0){$\tilde{\kappa}$} \put(43.0,88.0){$c^{\prime}$} \end{overpic}

\begin{overpic}[height=113.81102pt]{figures/Ktilde4.pdf} \put(23.0,55.0){$c$}
\put(92.0,18.0){$\tilde{\kappa}$} \put(50.0,5.0){$c^{\prime}$} \end{overpic}
\begin{overpic}[height=113.81102pt]{figures/Ktilde3.pdf} \put(3.0,80.0){$c$}
\put(82.0,10.0){$\tilde{\kappa}$} \put(79.0,45.0){$c^{\prime}$} \end{overpic}

Figure 9. The possible cases of
$\widetilde{\kappa}\in\widetilde{\mathcal{K}}$.

The curve $\widetilde{c}$ must satisfy one of the cases of Lemma 4.11. Note
that only Cases (1c) and (2) are applicable.

_Case ( 1c)._ In this case $\\#_{b}(\widetilde{c})=3$ and
$\\#_{I}(\widetilde{c})=2$. The complementary subarc
$\beta=\tilde{c}\smallsetminus\widetilde{\kappa}$ must have two intersection
points and no bubbles. We rule out each of the cases. In Case (a) of Figure 9,
the arc $\beta$ meets $L$ and travels along it in a single arc in $P^{+}$
which cannot go through an underpass. As can be seen from Figure 9, no such
arc exists. In Case (b), the arc $\widetilde{\kappa}$ ends in a twist box, and
so the complementary arc $\beta$ must pass through a bubble. Cases (c) and (d)
are ruled out similarly to Cases (a) and (b) respectively.

_Case ( 2)._ In this case the complementary arc $\beta$ is in $\mathcal{K}$.
The curve $\kappa$ cannot be in $\mathcal{K}_{4,0}$ or $\mathcal{K}_{2,1}$ as
otherwise the curve $c$ would be in $\mathcal{C}_{7,0}$ or
$\mathcal{C}_{5,1}$. However these sets are empty by Lemma 3.6.

In Case (a), the endpoints of $\beta$ are the same as those of
$\widetilde{\kappa}$. If $\beta\in\widetilde{\mathcal{K}}$ then it is one of
the arcs depicted in Figure 9. Hence, it must also be as in Case (a) and must
run parallel to $\widetilde{\kappa}$. But this would imply that
$\widetilde{c}$ passes through the same bubble twice on the same side of $L$,
in contradiction to tautness of $S$. If $\beta\in\mathcal{K}_{3,0}$, then
there is a corresponding subarc $\alpha$ in $L\cap P^{+}$ depicted in Figure
10 that connects the same regions as $\beta$. As Figure 9 is a precise
depiction of the possibilities, one can readily check that there is no such
arc.

\begin{overpic}[height=99.58464pt]{figures/alpha.pdf}
\put(75.0,48.0){$\kappa$} \put(92.0,58.0){$b$} \put(5.0,15.0){$a$}
\put(25.0,56.0){$\alpha$} \end{overpic} Figure 10. The arc $\alpha$ parallel
to $\kappa\in\mathcal{K}_{3,0}$

In Case (b), $\widetilde{\kappa}$ ends in the “middle” of a twist box after
passing through its top bubble. The arc
$\beta=\widetilde{c}\smallsetminus\widetilde{\kappa}$ continuing
$\widetilde{\kappa}$ must pass through the bubble immediately below. The arc
of $\beta$ intersecting this bubble cannot be contained in any arc in
$\mathcal{K}$. Hence, this case is impossible. This finishes the proof of the
claim. ∎

######  Lemma 5.4.

If $c^{\prime}$ contains a subarc $\kappa$ of $\mathcal{K}_{4,0}$, then it is
of the form shown in Figure 11.

\begin{overpic}[height=99.58464pt]{figures/C6a.pdf} \put(47.0,96.0){$a$}
\put(46.0,2.0){$b$} \put(0.0,90.0){$\alpha$} \end{overpic} Figure 11. The
unique form of a curve in $S\cap P^{+}$ containing an arc $\kappa$ in
$\mathcal{K}_{4,0}$.

###### Proof.

Let $\kappa$ be the subarc of $c^{\prime}$ in $\mathcal{K}_{4,0}$, with
endpoints $a$ and $b$. As $c$ must have $\chi^{\prime}(c^{\prime})=0$, it
follows from Lemma 4.11 that the complementary arc of $\kappa$ in $c^{\prime}$
is either in $\mathcal{K}_{4,0}$, or passes through exactly two bubbles, or
through two intersection points with the link $L$.

The distance between $a$ and $b$ is 2: It is clearly smaller or equal to $2$.
It must be even as the regions have the same color and if it is $0$ the
diagram would not be twist-reduced. So, by Lemma 2.9 the only subarc
$\alpha\subset L\cap P^{+}$ connecting the region containing $a$ to the region
containing $b$ is the bridge adjacent to $\kappa$. See Figure 11. The
complement of $\kappa$ in $c^{\prime}$ cannot be an arc of $\mathcal{K}_{4,0}$
since any arc of $\mathcal{K}_{4,0}$ connecting these regions must follow
$\alpha$ from the same side, contradicting the assumption that $S$ is taut.
The complement of $\kappa$ in $c^{\prime}$ cannot contain exactly two
intersection points, as the arc spanning them must be $\alpha$, which again
results in a contradiction to the tautness of $S$. Thus, the complement of
$\kappa$ in $c^{\prime}$ must contain two bubbles. The only possible such arc
is an arc on the other side of $\alpha$ adjacent to $\kappa$, as in Figure 11.
∎

######  Lemma 5.5.

If $c^{\prime}$ contains a subarc $\kappa\in\mathcal{K}_{3,0}$ then it is of
the form shown in Figure 12.

\begin{overpic}[width=99.58464pt]{figures/C6.pdf} \put(4.0,20.0){$a$}
\put(17.0,63.0){$\alpha$} \put(5.0,35.0){$c^{\prime}$} \put(78.0,67.0){$b$}
\end{overpic} Figure 12. The unique form of a curve $c^{\prime}$ containing an
arc of $\mathcal{K}_{3,0}$.

###### Proof.

Let $c^{\prime}$ be a curve containing a subarc $\kappa\in\mathcal{K}_{3,0}$.
Let $a,b$ be the endpoints of $\kappa$ as in Figure 10. As
$\chi^{\prime}(c^{\prime})=0$, it follows from Lemma 4.9 that the
complementary subarc $\beta=c^{\prime}\smallsetminus\kappa$ is either in
$\mathcal{K}_{3,0}$ (in which case $c^{\prime}\in\mathcal{C}_{6,0}$), or
contains exactly two intersections and no bubbles (in which case
$c^{\prime}\in\mathcal{C}_{3,2}$).

If $\beta\in\mathcal{K}_{3,0}$, let $\alpha$ be the the subarc of $L\cap
P^{+}$ connecting the regions containing $a$ and $b$, adjacent to $\kappa$, as
in Figure 10. Similarly, let $\alpha^{\prime}$ be the subarc of $L\cap P^{+}$
connecting the regions containing $a$ and $b$, adjacent to $\beta$. As each of
$\alpha,\alpha^{\prime}$ passes over two crossings and the uniqueness implied
by Lemma 2.9, $\alpha=\alpha^{\prime}$. Since $S$ is taut, $\kappa$ and
$\beta$ must be two different sides of $\alpha$, resulting in the
configuration depicted in Figure 12.

Next suppose $\beta$ has two intersection points and no bubbles. Note that
$\kappa$ must pass through two bubbles in one twist box and through a single
bubble in another. Let $T$ be the second twist box. The arc $\beta$ starts
from $b$ and first meets $L$ at some intersection point $p$. The two endpoints
of $\beta$ must belong to regions of different colors since its complement in
$c^{\prime}$ passes through three bubbles. In particular $\beta$ cannot
connect the two regions to the right and left of the twist box $T$. It follows
that the point $p$ cannot be any of the points depicted by small empty squares
in Figure 13(a), as otherwise the arc $c^{\prime}\cap L\subseteq\beta$ would
be one of the arcs in $P^{+}\cap L\cap T$ connecting its right and left sides.
The point $p$ also cannot be the point depicted by a small empty circle as
otherwise $S$ is not taut.

\begin{overpic}[width=113.81102pt]{figures/C32_a.pdf} \put(70.0,47.0){$b$}
\put(32.0,80.0){$T$} \put(52.0,80.0){$\vdots$} \put(40.0,18.0){$c^{\prime}$}
\end{overpic} \begin{overpic}[width=113.81102pt]{figures/C32_b.pdf}
\put(34.0,52.0){$q$} \put(52.0,80.0){$\vdots$} \put(70.0,47.0){$b$}
\put(72.0,24.0){$p$} \put(40.0,57.0){$\tilde{c}$}
\put(40.0,20.0){$c^{\prime}$} \end{overpic}

Figure 13. A possible configuration of $c^{\prime}\in\mathcal{C}_{3,2}$ on
$P^{+}$ that has an arc of $\mathcal{K}_{3,0}$: (a) The side $P^{+}$, the
twist box $T$ and the forbidden positions for the point $p$. (b) The side
$P^{-}$, the opposite curve $\tilde{c}$, and one of the possible positions for
the point $p$.

Assume that $c^{\prime}$ is a curve on $P^{+}$. Consider the curve $\tilde{c}$
opposite to $c^{\prime}$ at $b$ on $P^{-}$. See Figure 13(b). The curve
$\tilde{c}$ must pass through two bubbles in $T$ and through the intersection
point $p$. Denote a point on $\tilde{c}$ just beyond this second bubble by $q$
(see Figure 13(b)). The points $b$ and $q$ are on opposite sides of the twist
box $T$. The curve $\tilde{c}$ does not contain any $\eta$ in $\mathcal{K}$:
If it does, then $\eta$ will not pass through the two bubbles or intersection
point specified above; thus, $\\#(\tilde{c})\geq\\#(\eta)+3$ in contradiction
to Lemma 4.11.

Therefore, as $\tilde{c}$ passes through at least two bubbles and two
intersection points and contains no arc of $\mathcal{K}$, by Lemma 4.11 it
must be $\tilde{c}\in\mathcal{C}_{2,2}$. Thus, $\tilde{c}\cap L$ is an arc of
$L\cap P^{-}$ connecting the regions containing $b,q$ and ending at the point
$p$. Since $\tilde{c}\cap L$ connects the two regions to the left and right of
a twist box it must be an arc that passes through the twist box. Since
$\tilde{c}$ is opposite to $c^{\prime}$, the point $p$ cannot be any of the
points depicted by small empty squares and circles in Figure 13(a). This
leaves only one possible such arc, depicted by a black dotted line in Figure
13(b). However, if $\tilde{c}$ contains such an arc, this would imply that $S$
is not taut in contradiction to the assumption on $S$. ∎

######  Lemma 5.6.

There is no curve $c^{\prime}\in\mathcal{C}$ containing a subarc $\kappa$ in
$\mathcal{K}_{2,1}$

Note that, in general, if $c\in\mathcal{C}_{1,2}$ then the arc $\alpha=c\cap
L$ is a bridge. If $\alpha$ contains no crossings then it corresponds to
Figure 3, which would imply that $S$ is not taut. Thus, $\alpha$ passes
through either one or two crossings.

###### Proof.

Assume, for contradiction, that $c^{\prime}\in\mathcal{C}$ is a curve
containing a subarc $\kappa\in\mathcal{K}_{2,1}$. By Lemma 4.11 there are two
cases: Either $c^{\prime}$ is a union of two arcs in $\mathcal{K}_{2,1}$, or
$c^{\prime}\in\mathcal{C}_{3,2}$ and it contains one arc in
$\mathcal{K}_{2,1}$.

Assume first that $c^{\prime}$ is the union of exactly two arcs
$\kappa_{1},\kappa_{2}\in\mathcal{K}_{2,1}$. It follows that
$c^{\prime}\in\mathcal{C}_{4,2}$, and can be viewed as the union of three arcs
$\alpha^{\prime},\beta_{1},\beta_{2}$, where $\alpha^{\prime}=c^{\prime}\cap
L$ and $\beta_{1},\beta_{2}$ are subarcs of $\kappa_{1},\kappa_{2}$
respectively each passing through two bubbles. Let $c_{i}$, $i=1,2$, be the
curve in $C_{1,2}$ opposite to $\kappa_{i}$, and let $\alpha_{i}=c_{i}\cap L$.

Assume first that $\alpha^{\prime}$ does not pass over any crossing. Thus,
$\beta_{1},\beta_{2}$ emanate from $\alpha^{\prime}$ into adjacent regions
with opposite colors. Each of them passes through two bubbles, and hence they
cannot meet to form a closed curve.

Next assume that $\alpha^{\prime}$ passes over two crossings. Then,
$\alpha_{1}$ (and $\alpha_{2}$) passes over a crossing which is not the top or
bottom in its twist box, contradicting Remark 4.5. See Figure 14.

\begin{overpic}[width=99.58464pt]{figures/3arcs.pdf}
\put(50.0,45.0){$\alpha^{\prime}$} \put(90.0,100.0){$\alpha_{1}$}
\put(40.0,82.0){$c_{1}$} \put(-4.0,2.0){$\alpha_{2}$} \put(50.0,11.0){$c_{2}$}
\end{overpic} Figure 14. The arc
$\alpha_{1}\cup\alpha^{\prime}\cup\alpha_{2}$.

Finally, assume $\alpha^{\prime}$ goes through a single crossing. By Remark
4.5, $\alpha_{1}$ (and $\alpha_{2}$) must pass over the top or bottom
crossings of a twist box. The only possible configuration is if
$\alpha^{\prime}$ passes over the middle crossing of a twist box with three
crossings. The fact that $c_{1}$ and $c_{2}$ pass through bubbles at the top
or bottom of two other twist boxes, determines the orientations of the two
adjacent twist boxes as depicted in Figure 15. However, as can be readily
checked, a plat diagram $L$ cannot contain a subdiagram as in the figure.

\begin{overpic}[width=199.16928pt]{figures/C12a.pdf}
\put(27.0,40.0){$\alpha^{\prime}$} \put(53.0,56.0){$\alpha_{1}$}
\put(39.0,27.0){$\alpha_{2}$} \put(46.0,72.0){$c_{1}$}
\put(46.0,10.0){$c_{2}$} \end{overpic} Figure 15.

Now, assume that $c^{\prime}\in\mathcal{C}_{3,2}$ and contains exactly one arc
$\kappa\in\mathcal{K}_{2,1}$. Let $c\in\mathcal{C}_{1,2}$ be the curve
opposite to $\kappa$.

Assume first that $\alpha=c\cap L$ passes through two crossings. Since
$c\smallsetminus\alpha$ passes through exactly one bubble, the regions
containing the endpoints of $\alpha$ are at distance 1. Such a bridge $\alpha$
exists only in the corners of the diagram, in one of the configurations
depicted in Figure 16. The corresponding curve $c^{\prime}$ in each of the
possible configurations, contains an arc $\eta$ as in the figure, which
prolongs $\kappa$. The endpoints of $\eta$ are at distance greater than one,
and hence $c^{\prime}\smallsetminus\eta$ cannot pass through only one bubble,
in contradiction to the assumption.

\begin{overpic}[height=113.81102pt]{figures/c21corners1.pdf}
\put(5.0,80.0){$c$} \put(102.0,46.0){$\eta$}
\end{overpic}\begin{overpic}[height=113.81102pt]{figures/c21corners2.pdf}
\put(5.0,80.0){$c$} \put(96.0,14.0){$\eta$} \end{overpic}

Figure 16. The two possibilities for a curve $c\in\mathcal{C}_{1,2}$ with a
bridge passing through two crossings.

Next assume that $\alpha$ passes through one crossing. Then let $\eta$ be
small prolongation of $\kappa\cup(c^{\prime}\cap L)$ in $c^{\prime}$. The arc
$c^{\prime}\cap L$ contains zero, one or two crossings.

If it contains two crossings then the arc $c\cap L$ passes over a crossing
which is not the top or bottom of its twist box, in contradiction to Remark
4.5 similar to Figure 14.

If $c^{\prime}\cap L$ contains one crossing. The curve $c$ bounds a disk $D$
on the plane, containing a segment of $L$ which connects two twist boxes. The
five possibilities for such a curve $c$ are depicted in Figure 17. The subarc
$\eta$ of $c^{\prime}$ is the arc depicted in blue in the figure. In cases (a)
and (b), the endpoints of $\eta$ are in regions with the same color,
contradicting the assumption that $c\smallsetminus\eta$ passes through a
single bubble. In case (d) and (e), the arc $\kappa$ cannot be continued to an
arc $\eta$ such that $\eta\cap L$ has one crossing because of a conflict in
orientation, hence (d) and (e) do not occur. We are left with case (c). In
this case $\eta$ can be closed to form $c^{\prime}$ as shown in figure.
However, the arc $\gamma$ opposite to $c^{\prime}$, at the bottom of the
figure, passes through two bubbles and two intersection points, none of which
belong to an arc of $\mathcal{K}_{2,1}$. By Lemma 4.11, the curve containing
$\gamma$ must close up with no further bubbles or intersection points, which
is impossible.

\begin{overpic}[width=71.13188pt]{figures/K21_1.pdf} \put(40.0,50.0){$c$}
\put(45.0,3.0){$\eta$} \end{overpic}
\begin{overpic}[width=71.13188pt]{figures/K21_3.pdf} \put(70.0,52.0){$c$}
\put(0.0,10.0){$\eta$} \end{overpic}
\begin{overpic}[width=71.13188pt]{figures/K21_4.pdf} \put(30.0,5.0){$\gamma$}
\put(30.0,82.0){$c$} \put(8.0,40.0){$c^{\prime}$} \end{overpic}
\begin{overpic}[width=71.13188pt]{figures/K21_5.pdf} \put(35.0,8.0){$c$}
\put(73.0,60.0){$\kappa$} \end{overpic}

\begin{overpic}[width=71.13188pt]{figures/K21_6.pdf} \put(30.0,100.0){$c$}
\put(80.0,64.0){$\kappa$} \end{overpic} Figure 17. The five possibilities for
a curve $c\in\mathcal{C}_{1,2}$ (in red) and the subarc $\eta$ of the curve
$c^{\prime}$ (in blue) opposite to $c$ such that $c\cap L$ passes under one
crossings. Note that case (a) can occur also on the top boundary of the plat.

Thus, suppose $c^{\prime}\cap L$ contains no crossings. In this case, the
endpoints of $\eta$, shown in blue in Figure 18, are in the same two regions
as the endpoints of the maximal bridge (dashed) passing above the curve $c$.
As $c^{\prime}$ contains one additional bubble the endpoints of the bridge are
at distance 1 and thus the bridge is at a corner of the plat. In each corner,
there are two possible such bridges, and on each of these the curve $c$ can be
oriented in two ways. Thus, altogether there are four possibilities that are
depicted in Figure 19.

\begin{overpic}[width=113.81102pt]{figures/C12_bridge.pdf}
\put(40.0,60.0){$c^{\prime}$} \end{overpic} Figure 18. The bridge of a curve
$c\in C_{1,2}$ such that $c\cap L$ has one crossing.

\begin{overpic}[width=93.89418pt]{figures/C12corners1.pdf}
\put(42.0,14.0){$c$} \put(-7.0,45.0){$c^{\prime}$} \put(32.0,63.0){$\tau$}
\end{overpic} \begin{overpic}[width=93.89418pt]{figures/C12corners2.pdf}
\put(4.0,27.0){$c$} \put(45.0,10.0){$c^{\prime}$} \put(92.0,15.0){$\tau$}
\end{overpic} \begin{overpic}[width=93.89418pt]{figures/C12corners3.pdf}
\put(30.0,46.0){$c$} \put(72.0,42.0){$c^{\prime}$} \put(40.0,2.0){$\tau$}
\end{overpic} \begin{overpic}[width=93.89418pt]{figures/C12corners4.pdf}
\put(30.0,47.0){$c$} \put(75.0,75.0){$c^{\prime}$} \put(90.0,16.0){$\tau$}
\end{overpic}

Figure 19. The four possibilities for a curve $c\in\mathcal{C}_{1,2}$ with a
bridge passing through zero crossings.

In each of these possibilities, there is an arc $\tau$ opposite to
$c^{\prime}$, so that $\tau$ has three or four bubbles and its endpoints are
at distance at least two. Thus, the closed curve containing $\tau$ cannot be
in $\mathcal{K}$ and thus must have $\chi^{\prime}<0$. ∎

######  Definition 5.7.

If a curve enters and exists a twist box from the left or right side edges
then we say that it $\it{crosses}$ the twist box. Otherwise, if it enters from
a side edge and exists from a top/bottom edge or vice versa we say that it
$\it{turns}$ at the twist box.

######  Lemma 5.8.

There are no curves in $\mathcal{C}_{2,2}$.

###### Proof.

For any curve $c$ in $\mathcal{C}_{2,2}$. Consider the subarcs $\alpha=c\cap
L$ and $\beta=c\smallsetminus\alpha$.

The two endpoints of a small extension of $\alpha$ on $c$ must be contained in
regions of the same color as the rest of $c$ contains exactly two bubbles. As
$S$ is taut, these points cannot be in the same region. Thus the arc $\alpha$
passes over at least one crossing of the projection diagram.

######  Claim 5.9.

If $\mathcal{C}_{2,2}$ is nonempty then there exists a curve
$c\in\mathcal{C}_{2,2}$ such that $\alpha$ passes over one crossing, and
$\beta$ crosses one twist box.

###### Proof of claim.

Case 1. Assume first, that $\alpha$ passes over two crossings. Following
$\beta$ from one of its endpoints, $p,q$, we meet a twist box either from its
side or its top/bottom.

If, following $\beta$ from either endpoint, we meet a twist box from its
top/bottom, then $\beta$ meets two different twist boxes. Consider the curve
$c^{\prime}$ opposite to $c$ sharing the arc of $\beta$ between the bubbles.
The curve $c^{\prime}$ crosses the same two twist boxes. The diagram is twist-
reduced and thus $c^{\prime}\notin\mathcal{C}_{4,0}$ as otherwise it would
bound a reducing subdiagram. Therefore, we must have
$\chi^{\prime}(c^{\prime})<0$, a contradiction.

Thus, following $\beta$ from one of its endpoints, say $p$, $\beta$ meets a
twist box $T$ from the side. We claim that one of the two curves opposite to
$c$ containing $p$ or $q$ must cross $T$: Let $c^{\prime}$ be a curve opposite
to $c$ which contains $p$. If $c^{\prime}$ crosses $T$, we are done. Thus
assume that $c^{\prime}$ turns at $T$ and exists through, say, its _bottom_.
This implies that $\beta$ must cross $T$ and pass through its two bottom
bubbles. Therefore, the curve $c^{\prime\prime}$ opposite to $c$ containing
$q$ crosses $T$ and passes through its second and third bubbles (counted from
the bottom).

Let $c^{\prime}$ be a curve opposite to $c$ containing an endpoint of $\beta$
and crossing $T$, as in the previous paragraph. As $c^{\prime}$ passes through
at least two bubbles and has at least two intersection points, it must be in
$\mathcal{C}_{2,2}$ by Lemma 4.11 and Lemma 5.6. Since $c\cap L$ passes over
two crossings, the arcs $c\cap L$ and $c^{\prime}\cap L$ are as the arcs
$\alpha^{\prime}$ and $\alpha_{1}$ in Figure 14. In particular,
$c^{\prime}\cap L$ passes over one crossing and therefore, $c^{\prime}$ is the
required curve.

Case 2. Now assume that $\alpha$ passes over one crossing. Repeating the
argument above we conclude that there is a curve
$c^{\prime}\in\mathcal{C}_{2,2}$ opposite to $c$ that crosses a twist box. If
$\alpha^{\prime}=c^{\prime}\cap L$ passes over two crossings, we are back to
Case 1. Otherwise, it passes over one crossing and we are done. ∎

Let $c$ be a curve as in the claim. A small pushout of $c$ bounds a subdiagram
of $L$ as in the definition of a twist-reduced diagram (Definition 2.2). Since
$L$ is twist-reduced, the subdiagram must be contained in a twist box $T$.
Thus, both $\alpha$ and $\beta$ are contained in $T$. Since $S$ passes through
all bubbles in $T$, and every other bridge of $T$, there exists an innermost
curve such that both $\alpha$ and $\beta$ pass through the same bubble.
However, this implies that $S$ is not taut in contradiction. ∎

######  Lemma 5.10.

Every curve $c^{\prime}\in\mathcal{C}_{4,0}$ is of one of the forms depicted
in Figure 20.

\begin{overpic}[height=85.35826pt]{figures/C40a.pdf} \put(65.0,40.0){$c$}
\end{overpic}\begin{overpic}[height=113.81102pt]{figures/C40b.pdf}
\put(40.0,30.0){$c$} \end{overpic}

\begin{overpic}[height=113.81102pt]{figures/C40c.pdf} \put(40.0,35.0){$c$}
\end{overpic}

Figure 20. The possible cases of a curve in $\mathcal{C}_{4,0}$

###### Proof.

Assume by contradiction that $c^{\prime}\in\mathcal{C}_{4,0}$ and it is not of
the form in Figure 20. The curve $c^{\prime}$ passes through four bubbles
which are divided between at most four twist boxes. Assume that $c^{\prime}$
only turns at twist boxes. Consider one of the turns of $c^{\prime}$ at a
twist box $T$. Such a curve has a curve $c^{\prime\prime}$ opposite to it
which crosses the twist box. The curve $c^{\prime\prime}$ must also be in
$\mathcal{C}_{4,0}$ as it cannot be of any of the types formerly investigated:
$\mathcal{C}_{2,2}$ is empty and furthermore $c^{\prime\prime}$ cannot have an
arc in $\mathcal{K}$ since $c^{\prime\prime}$ is opposite to an arc in
$\mathcal{C}_{4,0}$ and not to $\mathcal{C}_{>0}$.

In addition, $c^{\prime\prime}$ cannot be of one of the forms in Figure 20:
Otherwise one of the arcs $c^{\prime}\cap c^{\prime\prime}$ continues in
$c^{\prime}$ to cross a twist box, in contradiction to the assumption that
$c^{\prime}$ only turns. By replacing $c^{\prime}$ by $c^{\prime\prime}$, if
need be, we assume from now on that

  1. (i)

the curve $c^{\prime}$ is a curve in $\mathcal{C}_{4,0}$ which crosses a twist
box, and

  2. (ii)

$c^{\prime}$ is not of one of the forms in Figure 20.

Let $c^{\prime}$ be an innermost such curve. Let $D$ be the innermost disc
bounded by $c^{\prime}$. Consider the intersection of $D$ with the twist boxes
meeting $c^{\prime}$. Since the diagram is twist-reduced, the curve
$c^{\prime}$ cannot cross two different twist boxes, this leaves us with three
cases: Case 0. The curve $c^{\prime}$ passes twice through the same bubble.
Then it is of the desired form, in contradiction to the assumption. Case 1.
The intersection of one of these twist boxes with $D$ contains a (projection
of a) bubble that is not met by $c^{\prime}$. Let $B,B^{\prime}$ be adjacent
bubbles of a twist box so that $B$ is contained in $D$ and $c^{\prime}$ passes
through $B^{\prime}$, see Figure 21. There must be a curve $c^{\prime\prime}$,
on the same side of $P$ as $c^{\prime}$, passing through the bubbles $B$ and
$B^{\prime}$, as follows from Figure 2. Since $c^{\prime}$ is innermost, and
the curve $c^{\prime\prime}$ crosses the twist box. By the previous lemmas,
namely Lemmas 4.11, 5.3, 5.5, 5.4, and 5.6, it is in one of the forms shown in
Figures 11, 12 or 20. In all of the cases, $c^{\prime\prime}$ contains an arc
outside the twist box, connecting $B$ and $B^{\prime}$. There are two cases to
consider, either $c^{\prime}$ crosses the twist box, or turns at the twist
box, see Figure 21. If $c^{\prime}$ crosses, the curve $c^{*}$ opposite to
$c^{\prime}$ (and $c^{\prime\prime}$), shown in Figure 21, passes through four
bubbles, it must therefore close up without passing through any additional
bubbles, which would imply that $c^{\prime}$ passes through the same bubble
$B^{\prime}$ twice, in which case we are done by case 0. If $c^{\prime}$
turns, then the curve $c^{*}$, passes three times through two bubbles in the
twist box, and in order to close up, must turn at another twist box at a
bubble $B^{*}$, it follows that $c^{\prime}$ passes through the bubble $B^{*}$
twice. In both cases, we are done by Case 0.

\begin{overpic}[height=113.81102pt]{figures/C40proof1.pdf}
\put(69.0,7.0){$c^{*}$} \put(10.0,7.0){$c^{\prime}$}
\put(63.0,84.0){$c^{\prime\prime}$} \put(5.0,36.0){$D$}
\end{overpic}\begin{overpic}[height=113.81102pt]{figures/C40proof2.pdf}
\put(69.0,12.0){$c^{*}$} \put(69.0,3.0){$c^{\prime}$}
\put(63.0,87.0){$c^{\prime\prime}$} \put(5.0,75.0){$D$} \end{overpic}

Figure 21. The curve $c^{\prime}$ crosses the twist box on the left, and turns
on the right. The minimal disk $D$ whose boundary is $c^{\prime}$ is shown in
gray.

We are left with the following case:

Case 2. The disc $D$ does not contain a bubble in a twist box that
$c^{\prime}$ meets, and $c^{\prime}$ does not pass twice through the same
bubble. Therefore, $c^{\prime}$ must cross once some twist box, which we
denote by $U$, and turn at two other twist boxes which we denote by $V$ and
$W$ (see Figure 22. Let $B,B^{\prime}$ be the two adjacent bubbles in $U$
through which $c^{\prime}$ crosses. As we are not in Case 1, the innermost
disk bounded by $c^{\prime}$ contains no bubbles of $U$, $V$ or $W$. Thus, one
of the bubbles $B,B^{\prime}$, say $B$, is extremal in the twist box $U$,
meaning its the first bubble in $U$ counted from the top or bottom. Let $\tau$
be the subarc of $c^{\prime}$ exiting $B^{\prime}$, and connecting $U$ to one
of $V,W$, say $V$. If $\tau$ enters $V$ from the side, then opposite to $\tau$
there is a curve $c^{*}$ crossing the twist boxes $U$ and $V$, which we have
shown can not exist because $L$ is twist-reduced. Thus, $\tau$ must enter $V$
from the bottom or the top. A similar argument shows that the subarc $\eta$ of
$c^{\prime}$ connecting $V$ to $W$, must enter $W$ from the top or the bottom.
Thus, the subarc $\mu$ of $c^{\prime}$ connecting $U$ and $W$, emerges from
the side of $U$ and ends in the side $W$ as depicted in Figure 22.

\begin{overpic}[height=142.26378pt]{figures/C40case2.pdf} \put(18.0,50.0){$W$}
\put(50.0,50.0){$U$} \put(80.0,50.0){$V$} \put(38.0,42.0){$B^{\prime}$}
\put(58.0,11.0){$B$} \put(8.0,11.0){$B^{*}$} \put(70.0,7.0){$D$}
\put(66.0,33.0){$\tau$} \put(94.0,11.0){$\eta$} \put(33.0,26.0){$\mu$}
\put(58.0,22.0){$c^{**}$} \put(3.0,27.0){$c^{*}$} \put(37.0,11.0){$c^{***}$}
\put(33.0,17.5){$\vdots$} \put(33.0,12.0){$\vdots$} \end{overpic} Figure 22.
The only possible configuration in Case 2

Let $c^{*}$ be the curve opposite to $c^{\prime}$ along $\mu$. The curve
$c^{*}$ passes through three bubbles, one in $U$ and two in $W$, and must
close after an additional turn. By the checkerboard coloring, the additional
turn must be at the bubble $B^{*}$ at the bottom of $W$, as shown in Figure
22. Opposite to $c^{*}$ at $B^{*}$, there is another curve $c^{**}$ which
passes through $B^{*}$ parallel to $c^{\prime}$. By the previous lemmas, as in
case 1, $c^{**}$ in $\mathcal{C}_{4,0}$. If it is in $\mathcal{C}_{4,0}$, one
can check that in order to close up $c^{**}$ must cross $U$ parallel to
$c^{\prime}$. Iterating this argument gives an infinite collection of nested
curves intersecting $B$ and $B^{*}$, which is a contradiction. This finishes
the proof of the lemma. ∎

##  6\. No tori and no annuli

In this section we prove Theorem 5.1 and Theorem 1.1

######  Lemma 6.1.

If $S$ is a taut torus or annulus and if all the curves in $\mathcal{C}_{\leq
0}$ are of the form depicted in Figures 11, 12 and 20 then $S$ is a boundary
parallel torus.

###### Proof.

Let $S$ intersect $P^{\pm}$ only in curves as in Figures 11, 12 and 20. The
surface $S$ is obtained by capping each curve in $P^{\pm}$ by the disk it
bounds in $H^{\pm}$ respectively. Thus, $S$ is a torus following one of the
components of $L$. That is, $S$ is boundary parallel. ∎

######  Corollary 6.2.

The link complement $S^{3}\smallsetminus\mathcal{N}(L$) does not contain
essential tori. In particular, the link $L$ is prime.

###### Proof.

Let $S$ be an essential torus in $S^{3}\smallsetminus\mathcal{N}(L)$. By Lemma
3.6, we may assume that $S$ is taut. By Lemmas 5.3, 5.4, 5.5, and 5.10, all
the curves in $\mathcal{C}$ are of the form depicted in Figures 11, 12 and 20.
Hence, by Lemma 6.1, the torus $S$ is boundary parallel which contradicts the
choice of $S$. ∎

######  Proposition 6.3.

The link complement $S^{3}\smallsetminus\mathcal{N}(L$) does not contain
essential annuli.

###### Proof.

Let $S$ be an incompressible and boundary incompressible annulus. Since by
Corollary 6.2 $L$ is prime, we may assume that $\partial S$ has no meridional
component. By Lemma 3.6, we may assume that $S$ is taut.

Assume first that $S$ passes through all twist boxes in disks of Type (0) or
(2) only, as in Lemma 3.1. Therefore, as $S$ is an annulus, there must be a
twist box $T$ such that $S$ meets $T$ in a Type (2) disk. When $S$ emerges
from $T$ it intersects $P$ in a curve $c\in\mathcal{C}$. By Lemmas 5.6 and 5.8
such a curve must be in $\mathcal{C}_{0,4}$. The curve $c$ cannot meet a twist
box as otherwise this twist box will meet $S$ in a disk of Type (1). Consider
the disk $D$ in $P$ bounded by $c$. Either $\text{int}(D)\cap L=\emptyset$ in
which case $S$ is not taut, or the diagram $D(L)$ is not prime, a
contradiction.

Therefore, there is a twist box $T$ which intersects $S$ in a disk of Type
(1). Since $\partial S$ contains a string of $T$, there is a curve $c$ in
$\mathcal{C}$ which passes through the “middle” of the twist box $T$ – i.e.,
the intersection $c\cap L$ contains a bridge which does not meet the top and
bottom of $T$. Let $\alpha$ be a small continuation of this bridge in $c$.
Lemmas 5.6 and 5.8 rule out curves containing $\mathcal{K}_{2,1}$ and curves
in $\mathcal{C}_{2,2}$. Thus, $c$ must be in $\mathcal{C}_{0,4}$, and must
contain another bridge. Let $\beta$ be a small continuation of that bridge in
$c$. The arcs $\alpha$ and $\beta$ belong to different components of $\partial
S$: Otherwise, the arc on $c$ connecting $\alpha$ and $\beta$ together with
$\partial S$ bounds a disk in $S$. An innermost curve $c\in\mathcal{C}$ in
this disk must have only two intersection points with $L$, which is a
contradiction to Lemma 3.6. The endpoints of $\alpha$ in $c$ belong to regions
of the same color, hence the same holds for $\beta$. Note that a bridge in a
plat passes at most two crossings. We divide the proof into cases depending on
the number of crossings $\beta$ passes over.

Case 0: If $\beta$ does not pass over any crossing. Then, as the endpoints of
$\beta$ belong to regions of the same color, they must belong to the same
region, contradicting the assumption that $S$ is taut.

Case 1: If $\beta$ passes over one crossing. A small pushout of the curve $c$
bounds a subdiagram of $L$ containing the two crossings points over which
$\alpha$ and $\beta$ pass. Since we assumed that $L$ is twist-reduced these
two crossing points must belong to the same twist box $T$. Let $n$ be the
number of bridges of $T$ in-between $\alpha$ and $\beta$. We further divide
into sub-cases according to $n$. Sub-case 1.0: $n=0$. First, if $\alpha$ and
$\beta$ meet the same bridge of $L$ then $c$ bounds a boundary compression
disk for $S$, which is a contradiction. So, assume that $\alpha$ and $\beta$
meet adjacent bridges of $S$. The annulus $S$ must thus spiral in-between the
strands of $L\cap T$. Thus we obtain a disk of Type 2 (as in Lemma 3.1) and
hence, by the definition of $\mathcal{C}^{+}$ and $\mathcal{C}^{-}$ as in the
beginning of §3.2 this curve does not appear in $\mathcal{C}$.

Sub-case 1.1: $n=1$. The tangle $L\cap T$ has two components
$\lambda_{1},\lambda_{2}$. Let $l_{1},l_{2}$ be the corresponding components
of $L$ (possibly $l_{1}=l_{2}$). Because $n=1$ the arcs $\alpha$ and $\beta$
meet the same string of $L\cap T$, say $\lambda_{1}$. Hence the two boundary
components of $S$ are contained in the same component $l_{1}$ of $L$. If
$l_{1}=l_{2}$, then there exists a curve $c^{\prime}\in\mathcal{C}_{0,4}$
which meets the bridge in-between $\alpha$ and $\beta$, and this curve must be
as in Case 1.0. Thus, we may assume that $l_{1}\neq l_{2}$. Consider the disk
$\Delta$ as depicted in Figure 23(a). Its interior intersects $L$ in a single
point in $l_{2}$, and its boundary is the union of an arc on $S$ and an arc on
$l_{1}$. The manifold $\mathcal{N}(S)\cup\mathcal{N}(l_{1})$ has two torus
boundary components $U$ and $V$. See Figure 23(b). Let $U$ be the torus that
meets $\Delta$. Let $U_{-}$ be the component of $S^{3}\smallsetminus U$
containing $l_{2}$, and let $U_{+}$ be the other component. The torus $U$ is
incompressible in $U_{-}$, as such a compression must be on $\Delta$ and
$\Delta$ meets $l_{2}$ once. It is also incompressible in $U_{+}$, as if such
a compression disk exists then since it cannot intersect $l_{1}$, it gives a
compression of the annulus $S$, in contradiction to the incompressibility of
$S$.

\begin{overpic}[height=113.81102pt]{figures/C04proofDelta.pdf}
\put(17.0,40.0){$\Delta$} \put(30.0,-3.0){$l_{2}$} \put(50.0,-3.0){$l_{1}$}
\put(38.0,30.0){\Tiny$\beta$} \put(38.0,80.0){\Tiny$\alpha$} \end{overpic}
\begin{overpic}[height=142.26378pt]{figures/C04proofDelta2.pdf}
\put(60.0,88.0){$l_{1}$} \put(73.0,75.0){$V$} \put(85.0,65.0){$S$}
\put(90.0,50.0){$U$} \put(42.0,30.0){$l_{2}$} \put(45.0,45.0){$U^{-}$}
\end{overpic}

Figure 23. (a) The disk $\Delta$. (b) A cross section of the twist box, the
annulus $S$ and the tori $U,V$

By Corollary 6.2, $U$ must be boundary parallel to either
$\partial\mathcal{N}(l_{2})$ or $\partial\mathcal{N}(l_{1})$. If $U$ is
parallel to $\partial\mathcal{N}(l_{2})$, then since $l_{1}$ is parallel to a
curve in $U$ crossing $\Delta$ once, then there exists an annulus $A\subset
S^{3}\smallsetminus L$ whose boundary is $l_{1}\cup l_{2}$. The annulus $A$ is
incompressible, since otherwise $l_{1}\cup l_{2}$ would be a 2-component
unlink that is not linked with $L$, i.e., $L$ is split, contradicting
Corollary 4.12. The annulus $A$ is trivially boundary-incompressible because
the boundary components of $A$ are on two different components of $L$. If we
run the argument for $A$ instead of $S$, Case 1.1 cannot occur because the
boundary components of $A$ are on two different components of $L$.

If $U$ is parallel to $\partial\mathcal{N}(l_{1})$, the intersection
$\Delta\cap U$ is a curve on $U$ which meets the meridian of
$\mathcal{N}(l_{1})$ exactly once: as if it meets it more than once, then the
union $\mathcal{N}(\Delta)\cup\mathcal{N}(l_{1})$ determines a once-punctured
non-trivial lens space contained in $S^{3}$, which is impossible. Thus,
$\partial\Delta$ which is parallel to $\Delta\cap U$ is also parallel to
$l_{1}$. Therefore, the arcs $\partial\Delta\smallsetminus l_{1}\subset S$ and
$l_{1}\smallsetminus\partial\Delta\subset L$ bound a disk. Since the arc
$\partial\Delta\smallsetminus l_{1}$ connects different components of $S$ it
is an essential arc, and the disk is a boundary compression for $S$, a
contradiction.

Sub-case 1.2: $n\geq 2$. As the boundary of the annulus $S$ must pass through
every other bridge in $T$, there must be another curve of $\mathcal{C}_{0,4}$
in between $\alpha$ and $\beta$. By choosing an innermost such curve we are
back in one of the previous cases.

Case 2: If $\beta$ passes over two crossings, then since the endpoints of
$\beta$ have the same color they are at distance 2. Therefore, we are in Case
(2) of Observation 2.9. By the uniqueness of such arcs, $\alpha=\beta$.
However, since $\alpha$ is a bridge in the “middle” of a twist box, it has
only one crossing, a contradiction. ∎

Theorem 5.1 immediately follows from Corollary 6.2 and Proposition 6.3.

###### Proof of Theorem 1.1.

If $m=2$ then $L$ is a $2$-bridge knot/link which is not a torus knot/link,
since there are at least two twist regions, hence it is atoroidal and
unannular by [3]. If $m\geq 3$ then the theorem follows directly from Theorem
5.1. In both case apply Thurston’s result (see [10]) that a link complement
which contains no essential annuli or tori is hyperbolic. ∎

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