# Highly twisted diagrams

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

We prove that the knots and links that admit a $3$-highly twisted irreducible
diagram with more than two twist regions are 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, twist regions, Euler Characteristic

###### 2010 Mathematics Subject Classification:

Primary 57M

NL was supported by the Israel Science Foundation (grant no. 1562/19).

##  1\. introduction

The prevailing feeling among low dimensional topologists is that “most” links
$\mathcal{L}$ in $\mathbb{S}^{3}$ are hyperbolic. That means that the open
manifold $\mathbb{S}^{3}\smallsetminus\mathcal{L}$ can be endowed with a
complete, finite-volume, 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 non trivial. This is especially true if the link $\mathcal{L}$ is “heavy
duty”, i.e., has a very large crossing number. See for example [10].

The question of when can one decide if the complement of a link in
$\mathbb{S}^{3}$ is a hyperbolic manifold from a projection diagram has been
of interest for a long time. 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 [5]. The second is Menasco’s result [9] that a non-split prime alternating
link which is not a torus link is hyperbolic. Later Futer and Purcell proved
in [3], among other results, that every link with a $6$-highly twisted, twist-
reduced diagram which has at least two twist regions is hyperbolic. Their
result is obtained by applying Marc Lackenby’s $6$-surgery theorem, see [6],
to the corresponding fully augmented links. Our main theorem is the following
(for definitions see §2):

######  Theorem A.

Let $D(\mathcal{L})$ be a connected, prime, twist-reduced, $3$-highly twisted
diagram of a link $\mathcal{L}$ with at least two twist regions, then
$\mathcal{L}$ is hyperbolic.

This result is sharp since there are non-hyperbolic links with 2-highly
twisted link diagrams, as Figure 1 shows.

Theorem A weakens the conditions imposed in [3] on $D(\mathcal{L})$ from
$6$-highly twisted to $3$-highly twisted. Two other relevant results in the
spirit of Theorem A are [4] and [2]. They replace the condition that a diagram
be 6-highly twisted by conditions related to its “semi-adequacy” (as defined
there). Theorem A is a generalization of previous work of the authors [7],
where a similar result was proved for knots/links that have 3-highly twisted
plat diagrams.

Clearly not all links have a diagram that satisfies the conditions of Theorem
A. However the subset of links that do is a “large” subset in a sense that can
be made precise, see the discussion in [8].

Figure 1. A non-hyperbolic link with a 2-highly twisted diagram.

As a corollary to Theorem A we obtain a simple method to construct essential
surfaces in complements of links with highly twisted diagrams. This is stated
in Section 7 as Theorem B.

Although there are non-hyperbolic links with 2-highly twisted diagrams, we
believe that the conditions of Theorem A could be weakened so as to include
all alternating links (cf. [9]) and allow _some_ twist regions to have less
than three crossings.

##  2\. preliminaries

###  2.1. Bubbles and twist regions.

Let $\mathcal{L}\subset\mathbb{S}^{3}$ be a link. The projection of a link $L$
in the isotopy class $\mathcal{L}$ onto a plane $P$ together with the crossing
data is a _link diagram_ of $L$ and is denoted by $D(L)$. Let $\varepsilon$ be
sufficiently small so that the closed $\varepsilon$-balls around the crossings
of $D(L)$ are disjoint. Let $\mathcal{B}_{1},\dots,\mathcal{B}_{r}$ be
$\varepsilon$-balls around the $r$ crossings of the diagram. The boundaries
$B_{i}=\partial\mathcal{B}_{i}$, $1\leq i\leq r$, are the _bubbles_ of the
diagram. The link $\mathcal{L}$ is isotopic to a link $L$ which is embedded in
$P\cup\bigcup_{i}B_{i}$. Note that $P$ divides each bubble into two
hemispheres denoted by $B^{+}_{i}$ and $B^{-}_{i}$. Denote the two 2-spheres

$P^{\pm}=(P\smallsetminus\bigcup_{i}\mathcal{B}_{i})\cup\bigcup_{i}B^{\pm}_{i}.$

Each of $P^{+},P^{-}$ bounds a $3$-ball $H^{\pm}$ in
$\mathbb{S}^{3}\smallsetminus L$.

A _twist region_ $T$ is a disk in $P$ which contains a maximal (with respect
to inclusion) chain of bigons in $D(L)$ describing a trivial integer 2-tangle.
See Figure 2 for an example of a diagram with twist regions. We will assume
that a twist region contains the projection of the bubbles around the
crossings in $T$. We will often abuse terminology, and use twist regions to
refer to the regions in $P^{\pm}$ which project to twist regions.
Correspondingly, we treat the bubbles around the crossings of $T$ as being
contained in $T$. A _twist box_ is the tangle $(\mathbf{T},t)$ where
$\mathbf{T}$ is the product $T\times[-2\varepsilon,2\varepsilon]$ where $T$ is
a twist region, and $t$ is the tangle $\mathbf{T}\cap L$ The disk diagram
$D(L)$ uniquely decomposes into disjoint twist regions.

Figure 2. A 3-highly twisted link diagram. The twist regions are the dashed
rectangles.

######  Definition 2.1.

Let $D(L)$ be a link diagram.

  1. (1)

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

  2. (2)

A _twist-reduction subdiagram_ is a subdiagram of $D(L)$ enclosed by a simple
closed curve $\gamma$ in $P$ which intersects the edges of $\gamma$
transversely in four points composed of two pairs each of which is adjacent to
a crossing of $D(L)$ but which is not a chain of bigons describing an integer
2-tangle. The diagram $D(L)$ is _twist-reduced_ if it contains no twist-
reduction subdiagram.

  3. (3)

For $k\in\mathbb{N}$, the diagram is _$k$ -highly twisted_ if every twist
region has at least $k$ crossings.

Note that every diagram can be made twist-reduced by performing flypes on
twist-reduction subdiagrams.

######  Definition 2.2.

A twist region $T$ intersects the link diagram in four points, dividing its
boundary $\partial T$ into four segments. If the twist region has at least two
crossings, then a pair of opposite segments of $\partial T$ can be called the
_length edge_ or _width edge_ of $T$ as in the following figure.

##  3\. Surfaces in link complements.

###  3.1. Normal position.

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

######  Lemma 3.1.

Let $S\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ be a proper surface
with no meridional boundary components, and let $(\mathbf{T},t)$ be a twist
box. Then, up to isotopy, $S\cap\mathbf{T}$ is a disjoint union of disks
$D\subset(\mathbf{T},t)$ 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 $\beta\subset\partial\mathbf{T}$.

  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 $\beta_{i}\subset\partial\mathbf{T}$.

Moreover, the isotopy decreases the number of bubbles that $S$ meets and we
may further assume that $\iota|_{\partial L}:\partial S\to L$ is a covering
map.

###### Proof.

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

The twist box $(\mathbf{T},t)$ is a trivial 2-tangle. The complement
$\mathbf{T}\smallsetminus\mathcal{N}(t)$ 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 $\mathbf{T}$ 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 $\mathbf{T}$. 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 crossings in the corresponding twist region.

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 region. The
arcs of type (0) are in one-to-one correspondence with the arcs of $S\cap E$.
Note that the fact that $\iota:\partial S\to L$ is a covering map was not
affected by the isotopies above. ∎

######  Definition 3.2.

A surface $S\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ is _in normal
position_ if its extension intersects the planes $P^{\pm}$ transversely and
also intersects each twist box as specified in Lemma 3.1 and $\iota|_{\partial
S}:\partial S\to L$ is a covering map. In particular, $S$ has no meridional
boundary components.

######  Lemma 3.3.

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

\begin{overpic}[height=99.58464pt]{Type0twist.pdf} \put(5.0,-10.0){Type 0}
\end{overpic}\begin{overpic}[height=99.58464pt]{Type1twist.pdf}
\put(5.0,-10.0){Type 1} \end{overpic}

\begin{overpic}[height=99.58464pt]{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 3. The possible three types of intersection of $S$ with a twist box.

###  3.2. Curves of intersection.

Let $S\subseteq\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ be a surface in
normal position. We would like to study the surface $S$ through its curves of
intersection with the planes $P^{\pm}$.

Let $\mathcal{T}$ be the union of all twist boxes of $L$. 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 $P^{\pm}$.

Recall the map $\iota:(S,\partial S)\to(\mathbb{S}^{3},L)$. Define
$\mathcal{C}^{+}=\partial\iota^{-1}(\widehat{S}\cap H^{+})$ and
$\mathcal{C}^{-}=\partial\iota^{-1}(\widehat{S}\cap H^{-})$. 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
$c\in\mathcal{C}^{-}$.

We think of curves in $\mathcal{C}^{\pm}$ as curves on $P^{\pm}$, as they are
disjoint outside $L$. Here, and in most of the figures in the remainder of
this paper, curves in $\mathcal{C}^{+}$ are colored blue while curves in
$\mathcal{C}^{-}$ are colored orange.

Assume $S$ is in normal position, and let $\mathbf{T}$ be a twist box. The
curves of intersection $\mathcal{C}$ of $S$ which meet a connected component
of $\iota(S)\cap\mathbf{T}$ must meet the corresponding twist region $T$ in
one of the following three configurations:

In order to annalize the curves $c\in\mathcal{C}$ we need to consider specific
subarcs and points of $c$ which we define next.

######  Definition 3.4.

For a curve $c\in\mathcal{C}$, we define the following arcs and points (they
are illustrated in the figures following the definition):

  1. (1)

A _$\circ$ -joint_ (“interior-joint”) of $c$ is a subarc of $c$ which is a
connected component of $c\cap\iota^{-1}(B)$ for some bubble $B$. The number of
$\circ$-joints of $c$ is denoted $\mathfrak{J}_{\circ}(c)$.

  2. (2)

A _$\partial$ -joint_ (“boundary-joint”) of $c$ is an endpoint of a connected
component of $c\cap\partial S$. The number of $\partial$-joints of $c$ is
denoted $\mathfrak{J}_{\partial}(c)$

  3. (3)

A _joint_ of $c$ is a $\circ$-joint or a $\partial$-joint of $c$. The number
of joints of $c$ is denoted by
$\mathfrak{J}(c)=\mathfrak{J}_{\circ}(c)+\mathfrak{J}_{\partial}(c)$.

  4. (4)

Define
$\mathcal{C}_{i,j}=\\{c\in\mathcal{C}\mid\mathfrak{J}_{\circ}(c)=i,\mathfrak{J}_{\partial}(c)=j\\}$.

  5. (5)

A _bone_ of $c$ is a connected component of $c$ minus its joints. Note that
all bones are arcs which are mapped by $\iota$ to $P$.

  6. (6)

A _$\partial$ -bone_ of $c$ is a bone which is contained in $\partial S$. Note
that the endpoints of $\partial$-bones are $\partial$-joints. All other bones
are _$\circ$ -bones_.

  7. (7)

A _limb_ of $c$ is a subarc $\alpha\subset c$ with endpoints in the interiors
of bones. Two limbs are equal if there is an isotopy of limbs (in $c$) between
them. In particular, their endpoints lie in the interior of the same bones.
The quantities $\mathfrak{J}(\alpha)$, $\mathfrak{J}_{\circ}(\alpha)$ and
$\mathfrak{J}_{\partial}(\alpha)$ are defined as for curves.

  8. (8)

A _turn_ of $c$ is a limb of $c$ that contains exactly one joint, this joint
is an $\circ$-joint and the endpoints of the limb are outside twist regions. A
curve _turns_ at a twist region if it contains a turn in that region.

  9. (9)

A _wiggle_ of $c$ is a limb of $c$ that contains exactly two joints, these
joints are $\circ$-joints through consecutive bubbles of a twist box, and the
endpoints of the limb are outside the twist regions. A curve _wiggles_ through
a twist region if it contains a wiggle in that region.

  10. (10)

Let $\mathcal{B}$ be a 3-ball bounded by a bubble $B$, then the components of
$S\cap\mathcal{B}$ are _saddles_. Those are disks whose boundary is
$\iota(\alpha_{1}^{+}\cup\alpha_{2}^{+}\cup\alpha_{1}^{-}\cup\alpha_{2}^{-})$
where $\alpha_{i}^{\pm}$ are $\circ$-joints of curves in $\mathcal{C}^{\pm}$
respectively. The two $\circ$-joints $\alpha_{1}^{+},\alpha_{2}^{+}$ (and the
two $\circ$-joints $\alpha_{1}^{-},\alpha_{2}^{-}$) are said to be _opposite_.

Figure 4. A saddle of $S$ in a bubble viewed from an angle and from the top.

######  Remark 3.5.

Note that a $\partial$-joint connects a $\partial$-bone and $\circ$-bone,
while a $\circ$-joint connects two $\circ$-bones. The $\partial$-joints and
$\partial$-bones are contained in the boundary of $S$, while $\circ$-joints
and $\circ$-bones are contained in the interior of $S$.

######  Definition 3.6.

Two curves (or limbs of curves) $c,c^{\prime}\in\mathcal{C}$ are _abutting_ if
they share an $\circ$-bone and $c\neq c^{\prime}$. Necessarily, if
$c\in\mathcal{C}^{+}$ then $c^{\prime}\in\mathcal{C}^{-}$ and vice versa.

The following figure shows an example of two abutting curves $c$ and
$c^{\prime}$.

###  3.3. Taut surfaces

######  Definition 3.7.

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

$\text{Com}(S)=(\sum_{c\in\mathcal{C}}\mathfrak{J}_{\circ}(c),\,\sum_{c\in\mathcal{C}}\mathfrak{J}_{\partial}(c),\,\left|\mathcal{C}\right|)$
(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.

######  Definition 3.8.

Let $S\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ be an essential
surface in normal position. The surface $S$ is _taut_ if either

  1. (i)

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

  2. (ii)

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

The next lemma shows that the intersection curves of taut surfaces must have
certain properties.

######  Lemma 3.9.

Assume that the diagram $D(L)$ is connected. Let
$S\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ be a taut surface. Then,
for all $c\in\mathcal{C}$ we have:

  1. (1)

$S_{c}$ is a disk.

  2. (2)

$\mathfrak{J}_{\partial}(c)$ is even.

  3. (3)

If $\mathfrak{J}_{\partial}(c)\leq 2$ then $\mathfrak{J}_{\circ}(c)>0$.

  4. (4)

If $\mathfrak{J}_{\partial}(c)=0$ then $\mathfrak{J}_{\circ}(c)$ is even.

  5. (5)

If a curve $c$ meets a bubble $B$ more than once, then it does so in two
opposite $\circ$-joints.

  6. (6)

If a curve $c$ has two $\partial$-joints on a connected component of
$P^{+}\cap L$ (or $P^{-}\cap L$), then they are the endpoints of a
$\partial$-bone. Moreover, the two $\circ$-bones incident to them are in
different regions of $P\smallsetminus D(L)$.

  7. (7)

The curve $c$ is not a curve in $\mathcal{C}_{1,2}$ as depicted in the
following figure:

###### Proof.

Let $S\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ be an essential
surface satisfying (i) or (ii). Note that in both cases, compressing along a
disk $D\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ 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, $\mathfrak{J}_{\partial}(c)$ is the number of endpoints of
arcs in $c\smallsetminus L$. Since each arc has two endpoints,
$\mathfrak{J}_{\partial}(c)$ is even.

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

If $\mathfrak{J}_{\partial}(c)=2$ and $\mathfrak{J}_{\circ}(c)=0$ then $c$
bounds a disk $D$ in $P$ such that $\partial D=\alpha\cup\beta$, where
$\alpha$ is a $\partial$-bone of $c$ and $\beta$ is an $\circ$-joint of $c$.
However, this is impossible by (2).

(4) The diagram $D(L)$ is a 4-regular graph, and thus it partitions $P$ into
regions which can be given a checkerboard coloring, i.e., can be colored black
and white so that two adjacent regions are colored in different colors.
Consider the colors of complementary regions of $P\smallsetminus\mathcal{T}$
which the curve $c$ intersects. If $\mathfrak{J}_{\partial}(c)=0$ 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 $\mathfrak{J}_{\circ}(c)$ is even.

(5) Without loss of generality assume that $c\in\mathcal{C}^{+}$. The curve
$c$ can meet $B$ in two $\partial$-bones, an $\circ$-joint and a
$\partial$-bone or two $\circ$-joints. If $c$ meets $B$ in two
$\partial$-bones, then the disk $S_{c}$ contains an arc connecting the two
$\partial$-bones. This arc together with an arc on $\partial\mathcal{N}(L)$
bounds (by an innermost argument) a compression disk for $S$. Compressing
along this disk reduces the complexity of $S$.

If $c$ meets $B$ in an $\circ$-joint and a $\partial$-bone, then the disk
$S_{c}$ contains an arc connecting the $\circ$-joint and the $\partial$-bone.
This arc, together with an arc on $B$ bounds a disk. Isotoping $S$ through
this disk reduces the complexity of $S$.

Thus, $c$ meets $B$ in two $\circ$-joints. By Lemma 1(ii) of [9], $c$ does not
have two $\circ$-joints in $B$ on the same side of $L$, i.e., as in the
following figure.

It follows that $c$ has at most two $\circ$-joints in the same bubble, and
they are separated by $L$ in $B^{+}$. The number of components of $S\cap
B^{+}$ separating each of the $\circ$-joints of $c$ from $L$ is the same: Each
component of $S\cap B$ separating an $\circ$-joint of $c$ and $L$ belongs to a
curve $c^{\prime}$ in $\mathcal{C}^{+}$. As curves in $\mathcal{C}^{+}$ do not
intersect, in order to close up, $c^{\prime}$ has to return to $B$ on the
other side of $L$ between $L$ and the other $\circ$-joint of $c$ in $B$. This
is depicted in the following figure.

The intersection of $S$ with the ball bounded by $B$, is a finite collection
of stacked saddles, it follows that the $\circ$-joints of $c$ belong to the
same saddle.

(6) If $c$ has two $\partial$-joints on the same component of $P^{+}\cap L$.
Let $\alpha$ in $P^{+}\cap L$ and $\beta$ in $S_{c}$ be arcs connecting the
two $\partial$-joints. By compressing along the disk bounded by
$\alpha\cup\beta$ (using an innermost such disk) we reduce complexity unless
the arc $\alpha$ is a $\partial$-bone of $c$.

If $c$ contains a $\partial$-bone $\alpha$ such that the two adjacent
$\circ$-joints are in the same region of $P\smallsetminus D(L)$ 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 $S$.

(7) Assume in contradiction that $c$ is such a curve. By (6), any curve
contained in $c$ has to be of a similar form. Assume $c$ is an innermost such
curve. The curve $\bar{c}$ abutting $c$, as depicted in the following figure,
meets the bubble twice, at a $\partial$-bone and an $\circ$-joint, in
contradiction to (5).

∎

######  Remark 3.10.

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 $i,k\in\mathbb{N}\cup\\{0\\}$.

##  4\. Euler Characteristic and curves of intersection

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

###  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

$\chi_{+}(c)=1-\frac{1}{4}\mathfrak{J}(c).$

######  Lemma 4.2.

If $S\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ 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 $S$. The vertices $X^{0}$ of the graph $X$ are the
$\partial$-joints and the endpoints of $\circ$-joints. The edges $X^{1}$ of
the graph $X$ are the bones and the $\circ$-joints. The graph $X$ partitions
$S$ into disk regions of three types:

  1. (1)

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

  2. (2)

saddles $R\subseteq\widehat{S}\cap\mathcal{B}$ where $\mathcal{B}$ 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, $\partial
S\subset X$ and consists of circles comprised of $\partial$-bones and
$\partial$-joints. Moreover, along every $\circ$-bone there are two abutting
curves. 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
$2\mathfrak{J}_{\circ}(c)$ vertices of $X^{0}$ in the interior of $S$ (those
are the endpoints of $\circ$-joints it passes). Furthermore, it goes through
$\mathfrak{J}_{\partial}(c)$ vertices of $X^{0}$ in $\partial S$. Each of
these vertices belongs to two (abutting) curves $c\in\mathcal{C}$. Hence,

$|X^{0}|=\sum_{c\in\mathcal{C}}(\mathfrak{J}_{\circ}(c)+\tfrac{1}{2}\mathfrak{J}_{\partial}(c))$
(4.2)

The edges of $X$. Every curve $c\in\mathcal{C}$ passes through
$2\mathfrak{J}_{\circ}(c)+\mathfrak{J}_{\partial}(c)$ edges in $X^{1}$. Note
that every $\circ$-joint edge and every $\partial$-bone edge belongs to
exactly one curve in $\mathcal{C}$, while each $\circ$-bone edge in belongs to
two curves in $\mathcal{C}$. Each $\circ$-joint edge appears in exactly one
curve $c$ and is counted once in $\mathfrak{J}_{\circ}(c)$. Hence, the number
of $\circ$-joint edges is $\sum_{c\in\mathcal{C}}\mathfrak{J}_{\circ}(c).$
Similarly each $\partial$-bone edge appears in exactly one curve $c$ and is
counted twice in $\mathfrak{J}_{\partial}(c)$. Hence, the number of
$\partial$-bone edges is
$\sum_{c\in\mathcal{C}}\tfrac{1}{2}\mathfrak{J}_{\partial}(c).$ Finally, each
$\circ$-bone edge in $c$ accounts for two vertices in $X^{0}$. So the number
of $\circ$-bone edges is equal to

$\tfrac{1}{2}|X^{0}|=\tfrac{1}{2}(\sum_{c\in\mathcal{C}}(\mathfrak{J}_{\circ}(c)+\tfrac{1}{2}\mathfrak{J}_{\partial}(c))).$

Adding these contributions together gives

$|X^{1}|=\sum_{c\in\mathcal{C}}(\tfrac{3}{2}\mathfrak{J}_{\circ}(c)+\tfrac{3}{4}\mathfrak{J}_{\partial}(c)).$
(4.3)

Regions $S^{\prime}\subset S\cap H^{\pm}$. To every curve $c\in\mathcal{C}$
there is a disk $S_{c}\subseteq S\cap H^{\pm}$. Thus,

$\sum_{S^{\prime}\subseteq S\cap H^{\pm}}\chi(S)=\sum_{c\in\mathcal{C}}1.$
(4.4)

Saddle regions $R\subset S\cap\mathcal{B}$. Each $\circ$-joint of a curve
$c\in\mathcal{C}$ belongs to the boundary such a region. And so each curve
passes through the boundary of $\mathfrak{J}_{\circ}(c)$ such regions. As each
saddle region has four $\circ$-joint in its boundary, we have

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

Summing over all of the above we get,

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

∎

##  5\. Redistribution of Euler characteristic

In this section we redistribute the positive contribution of the Euler
characteristic of curves, $\chi_{+}$, so that after the redistribution each
curve’s contribution is non-positive.

We first characterize the curves of intersection that have a positive
$\chi_{+}$. The characterization is done in the following lemma:

######  Lemma 5.1.

Let $c\in\mathcal{C}$ so that $\chi_{+}(c)>0$ (i.e., $\mathfrak{J}(c)<4$) then
$c\in\mathcal{C}_{2,0}$ or $c\in\mathcal{C}_{1,2}$ and it is one of the
following six forms (up to isotopy).

Figure 5. The six possibilities for a curve in
$c\in\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}$

Note that the curves $c$ in cases $(i)$ and $(ii)$ are in $\mathcal{C}_{2,0}$.
The curves in cases $(iii),(iv),(v)$ and $(vi)$ belong to $\mathcal{C}_{1,2}$

###### Proof.

It follows from Definition 4.1, Lemma 3.9 and Remark 3.10 that if
$\chi_{+}(c)>0$ then $c\in\mathcal{C}_{2,0}$ or $c\in\mathcal{C}_{1,2}$. Since
the diagram is prime, a curve $c\in\mathcal{C}_{2,0}$ must contain two turns
at different twist boxes. Hence $c$ is as depicted in figures (i) or (ii).

Let $c\in\mathcal{C}_{1,2}$ and let $\alpha$ denote a limb which is a small
extension of the unique $\partial$-bone in $c$. The endpoints of $\alpha$ must
be in regions of $P\smallsetminus D(L)$ of different color since the
complementary limb of $\alpha$ in $c$ is a turn. Since $S$ is taut, Lemma
3.9(7) implies that $\alpha$ must pass over at least one crossing of $D(L)$.
Since the endpoints of $\alpha$ are in regions of different colors, $\alpha$
cannot connect the two regions adjacent to the two length edges of a twist
region. Thus, $\alpha$ enters a twist region $T$ through its length edge, and
exists on $L$. It can meet one or two twist regions. If $\alpha$ meets one
twist region, then it must be as in figures (iii) or (iv). Otherwise, up to
isotopy, it must be as in figures (v) or (vi). ∎

######  Definition 5.2.

Denote by $\mathcal{C}_{>0}$ (resp. $\mathcal{C}_{\leq 0}$) the set of curves
$c\in\mathcal{C}$ such that $\chi_{+}(c)>0$ (resp. $\leq 0$). The lemma above
shows that $\mathcal{C}_{>0}=\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}$. The
_type_ of a curve in $\mathcal{C}_{>0}$ corresponds to the types of curves as
depicted in Figure 5. For example, a curve of $\mathcal{C}_{2,0}$ is of type
(i) or (ii).

Next, we will describe a distinguished set, denoted by $\mathcal{K}$, of limbs
of curves in $\mathcal{C}$ to which we will “reallocte” some of the positive
contribution of curves in $\mathcal{C}_{>0}$. We begin with a definition:

######  Definition 5.3.

An _extremal_ bubble is a first or last bubble of a twist region. A curve
wiggles _extremally_ if it wiggles through an extremal bubble. Assume that a
curve or an arc $\beta$ wiggles through a twist region extremally, then the
$\circ$-bone $\alpha\subset\beta$ which leaves the twist region from the
extremal bubble of the wiggle is called a _core_ of $\beta$.

######  Definition 5.4.

A _vertebra_ is an $\circ$-bone $\tau$ in a curve $c$ connecting two turns of
$c$ in two twist regions $T,T^{\prime}$ so that $\tau$ meets the length edge
of $T$ and the width edge of $T^{\prime}$.

A _rib_ is a closed curve $c\in\mathcal{C}_{4,0}$ which consists of exactly
two turns and an extremal wiggle.

######  Remark 5.5.

Note that the two $\circ$-bones of a curve $c\in\mathcal{C}_{2,0}$ of type (i)
are vertebrae.

######  Lemma 5.6.

Assume that $c_{0}$ is not a rib, and that $c_{0}$ contains a vertebra
$\tau_{0}$. Then, there exists a finite sequence of curves
$c_{0},c_{1},\dots,c_{n}$, limbs $\kappa_{1},\dots,\kappa_{n}$ and bones
$\tau_{0},\dots,\tau_{n-1}$ such that:

  1. (1)

For $1\leq i<n$, $c_{i}$ is a rib and $\tau_{i}$ is the $\circ$-bone
connecting its two turns.

  2. (2)

For $1\leq i\leq n$, $\kappa_{i}$ is a limb of $c_{i}$ with a unique core
$\tau_{i-1}$ and $\mathfrak{J}(\kappa_{i})=3$. In particular, the curve
$c_{i}$ abuts the curve $c_{i-1}$ along the core $\tau_{i-1}$.

  3. (3)

The curve $c_{n}$ has $\chi_{+}(c_{n})<0$.

Moreover, given $\kappa_{n}$ one can uniquely determine the $\circ$-bone
$\tau_{0}$ and hence the curves $c_{i}$, arcs $\kappa_{i}$ and bones
$\tau_{i}$ as above.

######  Definition 5.7.

We will refer to the curves $c_{i}$ (resp. limbs $\kappa_{i}$) in the lemma as
the _layers curves_ (resp. _layer limbs_) of $\tau_{0}$ , and to the curve
$c_{n}$ (resp. arc $\kappa_{n}$) as the _terminal layer curve_ (resp.
_terminal layer limb_) of $\tau_{0}$.

###### Proof of Lemma 5.6.

We produce a sequence of curves, limbs and bones, satisfying the assumptions
above, which terminates at the first curve $c_{n}$ such that
$\chi_{+}(c_{n})<0$.

Let $c_{0}$ and $\tau_{0}$ be as in the statement of the lemma. Let $c_{1}$ be
the curve abutting $c_{0}$ along $\tau_{0}$. The bone $\tau_{0}$ connects an
extremal wiggle and a turn of $c_{1}$, hence it is a core of $c_{1}$. Let
$\kappa_{1}$ be the limb of $c_{1}$ containing $\tau_{0}$ and the adjacent
wiggle and turn. If $\chi_{+}(c_{1})<0$, then stop the process. Otherwise,
$\mathfrak{J}(c_{1})=4$. Hence, the curve $c_{1}$ is a rib, i.e., it consists
of a wiggle and two turns and has a unique core. Let
$\alpha_{1},\alpha_{1}^{\prime}$ be the $\circ$-joint of the two turns of
$c_{1}$, and assume that $\alpha_{1}\subset\kappa_{1}$. Let $\tau_{1}$ be the
$\circ$-bone of $c_{1}$ connecting $\alpha_{1},\alpha_{1}^{\prime}$. The bone
$\tau_{1}$ meets the length edge of the twist region containing $\alpha_{1}$.

Assume first that $\tau_{1}$ is not a vertebra of $c_{1}$, i.e., it meets the
length edge of the twist region containing $\alpha_{1}^{\prime}$. Then, the
curve $c_{2}$ abutting $c_{1}$ along $\tau_{1}$ contains two wiggles in two
different twist regions. It follows that $\chi_{+}(c_{2})<0$ as otherwise
$c_{2}$ bounds a twist reduction subdiagram. Let $\kappa_{2}$ be the limb in
$c_{2}$, abutting $\tau_{1}$, consisting of a wiggle through the bubble of
$\alpha_{1}$, and one more $\circ$-joint at the bubble containing
$\alpha_{1}^{\prime}$. The bone $\tau_{1}$ is the unique core of the limb
$\kappa_{2}$, and the process stops ($n=2$).

If $\tau_{1}$ is a vertebra, then we iterate the process. That is, we consider
the curve $c_{2}$ abutting $c_{1}$ along $\tau_{1}$, and the limb $\kappa_{2}$
of $c_{2}$ containing $\tau_{1}$ and its adjacent wiggle and turn.

Since there are finitely many curves and limbs, the process either terminates
or is periodic. It cannot be periodic because the initial curve $c_{0}$ is not
a rib, but note that all the curves $c_{i}$ for $i<n$ are ribs.

Finally, given $\kappa_{n}$, the curve $c_{n-1}$ is the curve abutting $c_{n}$
along the unique core of $\kappa_{n}$. The curve $c_{n-1}$ has a unique core,
which is also the core of a unique arc $\kappa_{n-1}$ (with
$\mathfrak{J}(\kappa_{n-1})=3$). Repeating this process, we can retrace the
sequence all the way to $\tau_{0}$. ∎

######  Example 5.8.

In the figure below, we see two examples of outputs of the process in Lemma
5.6. Starting with the curve $c_{0}$ which is not a rib, and the vertebra
$\tau_{0}$ of $c_{0}$, we get the curves $c_{0},c_{1},c_{2},c_{3}$, limbs
$\kappa_{1},\kappa_{2},\kappa_{3}$ and $\circ$-bones
$\tau_{0},\tau_{1},\tau_{2}$. The limbs $\kappa_{i}$ are shown in bold the
figure. The $\circ$-bones $\tau_{0},\tau_{1},\tau_{2}$ are the cores of
$\kappa_{1},\kappa_{2},\kappa_{3}$ respectively.

In the left figure, note that the $\circ$-bone $\tau_{2}$ is a vertebra. If
$\chi_{+}(c_{3})<0$ then the process stops at $c_{3}$ and $\kappa_{3}$ is its
terminal limb. Otherwise, $c_{3}$ is again a rib, and the process continues.

In the right figure, the $\circ$-bone connecting the two turns of $c_{2}$ is
not a vertebra, as it meets the length edge of both twist regions. Therefore
the limb $\kappa_{3}$, which is shown in bold in the figure, contains a wiggle
(through the bubble in which both $c_{1}$ and $c_{2}$ turn) and “half a
wiggle” (through the other bubble in which $c_{2}$ turns). As the proof shows,
the curve $c_{3}$ necessarily satisfies $\chi_{+}(c_{3})<0$, and thus it is
the terminal curve of the process.

######  Definition 5.9.

Definition of $\mathcal{K}$.

  1. (1)

Each $c\in\mathcal{C}_{2,0}$ of type (i) has two $\circ$-bones. Each
$\circ$-bone $\tau$ of $c$ is a vertebra and hence by Lemma 5.6 determines a
terminal layer limb $\kappa_{n}$. Let $\mathcal{K}_{3,0}$ be the set of all
terminal layer limbs associated with all $\circ$-bones of curves in
$c\in\mathcal{C}_{2,0}$.

  2. (2)

Each $c\in\mathcal{C}_{2,0}$ of type (ii) determines an arc $\kappa$ which
abuts $c$ and wiggles through the two twist regions. Let $\mathcal{K}_{4,0}$
be the collection of all arcs $\kappa$ obtained in this way.

  3. (3)

Each $c\in\mathcal{C}_{2,1}$, determines an arc $\kappa$ which abuts $c$,
wiggles through the twist region in which $c$ turns, and contains one of the
$\partial$-joints of $c$. Let $\mathcal{K}_{2,1}$ be the collection of all
arcs $\kappa$ obtained in this way.

Finally, let $\mathcal{K}$ be the set
$\mathcal{K}_{3,0}\cup\mathcal{K}_{4,0}\cup\mathcal{K}_{2,1}$.

######  Remark 5.10.

We note that given $\kappa\in\mathcal{K}$ one can uniquely determine the curve
$c\in\mathcal{C}_{>0}$ that determines it. Conversely, to each curve in
$\mathcal{C}_{2,0}$ of type (i) there are two curves of $\mathcal{K}_{3,0}$
corresponding to the two choices of $\circ$-bones of $c$. To each curve in
$\mathcal{C}_{1,2}$ and each curve in $\mathcal{C}_{2,0}$ of type (ii) there
is a unique arc in $\mathcal{K}_{2,1}$ and $\mathcal{K}_{4,0}$ respectively.
In particular,
$\tfrac{1}{2}|\mathcal{K}_{3,0}|+|\mathcal{K}_{4,0}|=|\mathcal{C}_{2,0}|$ and
$|\mathcal{K}_{2,1}|=|\mathcal{C}_{1,2}|$.

######  Lemma 5.11.

The arcs in $\mathcal{K}$ which belong to curves in $\mathcal{C}^{+}$ (resp.
$\mathcal{C}^{-}$) are pairwise disjoint.

###### Proof.

Assume that the two arcs $\kappa,\kappa^{\prime}\in\mathcal{K}$ meet. Since,
by Remark 5.10, every arc in $\mathcal{K}$ is determined by its core, it
suffices to show that $\kappa$ and $\kappa^{\prime}$ have the same core.

By assumption $\kappa,\kappa^{\prime}$ share a joint, and hence are limbs of
the same curve $c$. If this joint is a $\partial$-joint then
$\kappa,\kappa^{\prime}\in\mathcal{K}_{2,1}$ and their core is the unique
$\circ$-bone incident to their shared $\partial$-joint. If the joint is an
$\circ$-joint that is part of a wiggle of $c$, then, since the diagram is
3-highly twisted, there is a unique core emanating from the extremal bubble of
this wiggle, which is shared by both $\kappa$ and $\kappa^{\prime}$. Finally,
if the joint is an $\circ$-joint that is part of a turn of $c$, then
$\kappa,\kappa^{\prime}\in\mathcal{K}_{3,0}$. The cores of
$\kappa,\kappa^{\prime}$ must be the unique $\circ$-bone which emanates from
the width edge of the twist region in which the turn occurs because it is also
the vertebra of the previous layer curve. ∎

######  Lemma 5.12.

For all $\kappa\in\mathcal{K}$ we have
$\mathfrak{J}(c)\geq\mathfrak{J}(\kappa)+2$ where $c$ is the unique curve in
$\mathcal{C}$ containing $\kappa$.

###### Proof.

Let $\kappa\in\mathcal{K}$ and let $c\in\mathcal{C}$ be the curve containing
$\kappa$. Assume for contradiction that
$\mathfrak{J}(c)<\mathfrak{J}(\kappa)+2$.

If $\kappa\in\mathcal{K}_{4,0}$ then $\mathfrak{J}(c)<\mathfrak{J}(\kappa)+2$
implies that $c\in\mathcal{C}_{4,0}$. If this is the case, then since $c$
wiggles through two twist regions, the projection of $c$ to $P$ gives a twist-
reduction subdiagram in contradiction to the assumption on the diagram.

If $\kappa\in\mathcal{K}_{2,1}$, then $\mathfrak{J}(c)<\mathfrak{J}(\kappa)+2$
implies that $c\in\mathcal{C}_{2,2}$. If this is the case, then $\kappa$ abuts
some curve $c_{0}\in\mathcal{C}_{1,2}$ which could be one of the figures
(iii)-(vi) in Lemma 5.1. For case (iii), there are two possible configurations
for the closed curve $c\in\mathcal{C}_{2,2}$ containing $\kappa$, while for
each of the cases (iv)-(vi) there is only one possible such curve. Thus, all
possible cases for the curve $c$ are shown in the figure below.

In case (iii) the surface is not taut and in all other cases, the curve $c$
bounds a twist-reduction subdiagram.

If $\kappa\in\mathcal{K}_{3,0}$, then $\mathfrak{J}(c)<\mathfrak{J}(\kappa)+2$
implies that $c\in\mathcal{C}_{4,0}$, however this is in contradiction to the
definition of $\mathcal{K}_{3,0}$ given by Lemma 5.6. ∎

######  Remark 5.13.

It follows from the proof of Lemma 5.6 that if $c$ contains a limb
$\kappa\in\mathcal{K}_{3,0}$ such that the core of $\tau$ is not a vertebra of
its abutting curve then $\mathfrak{J}(c)\geq\mathfrak{J}(k)+3$: Indeed, if
$c=c_{n}$, $\kappa=\kappa_{n}$ and $\kappa$ abuts $c_{n-1}$ along a bone which
is not a vertebra of $c_{n-1}$ then $c$ must have an additional $\circ$-joint
which is not contained in $\kappa$.

######  Definition 5.14.

Let $c\in\mathcal{C}$. If $c\in\mathcal{C}_{>0}$ define $\chi^{\prime}(c)=0$.
Otherwise, let $n_{3,0}$ (resp. $n_{4,0}$; $n_{2,1}$) be the number of limbs
$\kappa\in\mathcal{K}_{3,0}$ (resp. $\mathcal{K}_{4,0}$; $\mathcal{K}_{2,1}$)
in $c$. We associate to $c$ the following quantity

$\chi^{\prime}(c)=\chi_{+}(c)+\tfrac{1}{4}n_{3,0}+\tfrac{1}{2}n_{4,0}+\tfrac{1}{4}n_{2,1}.$

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

######  Lemma 5.15.

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

###### Proof.

By Lemma 4.2, $\chi(S)=\sum_{c\in\mathcal{C}}\chi_{+}(c)$. Since
$\chi^{\prime}(c)=0$ for $c\in\mathcal{C}_{>0}$,

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

It thus remains to prove that

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

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

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

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
$\tfrac{1}{4}|\mathcal{K}_{3,0}|+\tfrac{1}{2}|\mathcal{K}_{4,0}|+\tfrac{1}{4}|\mathcal{K}_{2,1}|$.
The proof is now complete by Remark 5.10. ∎

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

######  Lemma 5.16.

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

###### Proof.

If $c\in\mathcal{C}_{>0}$ then $\chi^{\prime}(c)=0$. Let
$c\in\mathcal{C}_{\leq 0}$ and let $n_{3,0},n_{4,0},n_{2,1}$ be as in
Definition 5.14 of $\chi^{\prime}(c)$. By Lemma 5.11, the limbs of
$\mathcal{K}$ in $c$ are disjoint and therefore

$\mathfrak{J}(c)\geq\sum_{c\supset\kappa\in\mathcal{K}}\mathfrak{J}(\kappa).$
(5.1)

Hence,

$\displaystyle\begin{split}\chi^{\prime}(c)&=\chi_{+}(c)+\tfrac{1}{4}n_{3,0}+\tfrac{1}{2}n_{4,0}+\tfrac{1}{4}n_{2,1}=(1-\tfrac{1}{4}\mathfrak{J}(c))+\tfrac{1}{4}n_{3,0}+\tfrac{1}{2}n_{4,0}+\tfrac{1}{4}n_{2,1}\\\
&\leq
1-\sum_{c\supset\kappa\in\mathcal{K}}\tfrac{1}{4}\mathfrak{J}(\kappa)+\tfrac{1}{4}n_{3,0}+\tfrac{1}{2}n_{4,0}+\tfrac{1}{4}n_{2,1}\\\
&=1+\sum_{c\supset\kappa\in\mathcal{K}_{3,0}}(\tfrac{1}{4}-\tfrac{1}{4}\mathfrak{J}(\kappa))+\sum_{c\supset\kappa\in\mathcal{K}_{4,0}}(\tfrac{1}{2}-\tfrac{1}{4}\mathfrak{J}(\kappa))+\sum_{c\supset\kappa\in\mathcal{K}_{2,1}}(\tfrac{1}{4}-\tfrac{1}{4}\mathfrak{J}(\kappa))\\\
&=1+\sum_{c\supset\kappa\in\mathcal{K}_{3,0}}(\tfrac{1}{4}-\tfrac{1}{4}\cdot
3)+\sum_{c\supset\kappa\in\mathcal{K}_{4,0}}(\tfrac{1}{2}-\tfrac{1}{4}\cdot
4)+\sum_{c\supset\kappa\in\mathcal{K}_{2,1}}(\tfrac{1}{4}-\tfrac{1}{4}\cdot
3)\\\ &=1-\tfrac{1}{2}(n_{3,0}+n_{4,0}+n_{2,1}).\end{split}$ (5.2)

Now the argument is divided into cases depending on the sum
$n=n_{3,0}+n_{4,0}+n_{2,1}$.

Case 0. $n=0$. We have $\chi^{\prime}(c)=\chi_{+}(c)$. But since
$c\in\mathcal{C}_{\leq 0}$ we have $\chi_{+}(c)\leq 0$ and we are done.

Case 1. $n=1$. That is, $c$ contains a single subarc
$\kappa\in\mathcal{K}_{3,0}\cup\mathcal{K}_{4,0}\cup\mathcal{K}_{2,1}$. By
Lemma 5.12, $\mathfrak{J}(c)\geq\mathfrak{J}(\kappa)+2$. If
$\kappa\in\mathcal{K}_{4,0}$ we get $\mathfrak{J}(c)\geq 6$ and thus
$\chi^{\prime}(c)=1-\tfrac{1}{4}\mathfrak{J}(c)+\tfrac{1}{2}\leq 0$.
Similarly, if $\kappa\in\mathcal{K}_{3,0}\cup\mathcal{K}_{2,1}$ we get
$\mathfrak{J}(c)\geq 5$ and thus
$\chi^{\prime}(c)=1-\tfrac{1}{4}\mathfrak{J}(c)+\tfrac{1}{4}\leq 0$.

Case 2. $n\geq 2$. In this case we are done by inequality (5.2). ∎

######  Corollary 5.17.

The link $\mathcal{L}$ is non-split nor the unknot.

###### Proof.

Assume, in contradiction, that
$\mathbb{S}^{3}\smallsetminus\mathcal{N}(\mathcal{L})$ has an essential sphere
or a disk bounding a component of $L$. By Lemma 3.9 we may assume that $S$ is
taut. Let $\mathcal{C}$ be its curves of intersection With $P^{\pm}$. By Lemma
5.15,

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

However, this contradicts Lemma 5.16 which states that $\chi^{\prime}(c)\leq
0$ for all $c\in\mathcal{C}$. ∎

######  Lemma 5.18.

Let $S$ be a taut surface with $\chi(S)=0$, then any curve $c\in\mathcal{C}$
is one of the following:

  1. (1)

$c\in\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}$;

  2. (2)

$\mathfrak{J}(c)=4$, i.e.
$c\in\mathcal{C}_{4,0}\cup\mathcal{C}_{2,2}\cup\mathcal{C}_{0,4}$;

  3. (3)

$c$ contains a limb $\kappa\in\mathcal{K}$ and has
$\mathfrak{J}(c)=\mathfrak{J}(\kappa)+2$,

  4. (4)

$c$ is the union of two limbs $\kappa_{1},\kappa_{2}\in\mathcal{K}$.

Moreover, the cores of arcs in $\mathcal{K}_{3,0}$ are vertebrae of their
abutting curves.

###### Proof.

By Lemma 5.15 and 5.16 each curve $c\in\mathcal{C}$ must have
$\chi^{\prime}(c)=0$. It follow from the definition of $\chi^{\prime}$ that
the above are the only cases in which $\chi^{\prime}(c)=0$.

By Remark 5.13 and Case 1 of the proof of Lemma 5.16, if $c$ contains an arc
$\kappa\in\mathcal{K}_{3,0}$ whose core is not a vertebra of the abutting
curve $c^{\prime}$ then $\chi^{\prime}(c)<0$. ∎

In the following table, we summarise the possible sets of curves with
$\chi^{\prime}=0$ and assign them names:

Name of set | $\mathfrak{J}(\cdot)$ | $\mathfrak{J}_{\circ}(\cdot)$ | $\mathfrak{J}_{\partial}(\cdot)$ | The set’s composition/classification
---|---|---|---|---
$\mathcal{C}_{2,0}$ | 2 | 2 | 0 | type (i) or (ii)
$\mathcal{C}_{1,2}$ | 3 | 1 | 2 | type (iii), (iv) or (v)
$\mathcal{C}_{4,0}$ | 4 | 4 | 0 | 4 turns or 1 wiggle and 2 turns or 2 wiggles
$\mathcal{C}_{2,2}$ | 4 | 2 | 2 | 2 $\circ$-joints and 1 $\partial$-bone
$\mathcal{C}_{0,4}$ | 4 | 0 | 4 | 2 $\partial$-bones
$\mathcal{K}_{3,0}+0,2$ | 5 | 3 | 2 | 1 limb of $\mathcal{K}_{0,3}$ \+ $\partial$-bone
$\mathcal{K}_{2,1}+1,1$ | 5 | 3 | 2 | 1 limb of $\mathcal{K}_{2,1}$ \+ 1 $\circ$-joint + 1 $\partial$-joint
$\mathcal{K}_{4,0}+2,0$ | 6 | 6 | 0 | 1 limb of $\mathcal{K}_{0,4}$ \+ 2 $\circ$-joints
$\mathcal{K}_{4,0}+0,2$ | 6 | 4 | 2 | 1 limb of $\mathcal{K}_{0,4}$ \+ $\partial$-bone
$\mathcal{K}_{3,0}+\mathcal{K}_{3,0}$ | 6 | 6 | 0 | 2 limbs of $\mathcal{K}_{3,0}$ = 2 wiggles + 2 turns.
$\mathcal{K}_{2,1}+\mathcal{K}_{2,1}$ | 6 | 4 | 2 | 2 limbs of $\mathcal{K}_{2,1}$ = 2 wiggles + $\partial$-bone.
$\mathcal{K}_{4,0}+\mathcal{K}_{4,0}$ | 8 | 8 | 0 | 2 limbs of $\mathcal{K}_{4,0}$ = 4 wiggles.
Table 1. Classification of all curves with $\chi^{\prime}(c)=0$.

##  6\. Atoroidal and unannular

In this section we prove that if the link $L$ has a diagram which satisfies
the conditions of Theorem A, then its complement does not contain essential
annuli or tori (in particular $\chi(S)=0$). Let $S$ denote such a 2-torus or
annulus. After an isotopy, if need be, we may assume that $S$ is taut. We
first need the following technical lemmas.

######  Claim 6.1.

The following three configurations for curves $c\in\mathcal{C}$ are
impossible:

where the bubble marked in pink is non-extremal.

###### Proof.

We argue simultaneously that the three configurations are impossible. In each
of these cases, let $c_{2}$ (marked in orange) be the depicted curve abutting
$c_{1}$.

The limb of $c_{2}$ that is shown in the figure above has three joints. None
of these joints belongs to a limb in $\mathcal{K}$: in all cases, the curve
$c_{1}$ is not in $\mathcal{C}_{2,0}$ or $\mathcal{C}_{1,2}$ nor a rib with a
vertebra. By Lemma 5.18, it follows that $\mathfrak{J}(c_{2})=4$. Thus the
only way that $c_{2}$ can close up is depicted in in the following figures.

Let $c_{3}$ be the depicted curve abutting $c_{1}$. In the two left figures,
one sees, as before, that $\mathfrak{J}(c_{3})=4$. In the figure on the right,
$c_{3}$ is in $\mathcal{K}_{4,0}+{2,0}$ in the notation of Table 1: Indeed,
$c_{3}$ cannot have $\mathfrak{J}(c_{3})=4$, as otherwise it bounds a twist
reduction subdiagram. Thus, $c_{3}$ must contain an arc of $\mathcal{K}$. By
elimination of the possibilities in Table 1, $c_{3}$ must be in
$\mathcal{K}_{4,0}+2,0$. Thus, in all cases, $c_{3}$ can close only after
passing through an additional twist region as follows:

Thus, taking into account all possible configurations of the curves $c_{2}$
and $c_{3}$ determined by the stated configurations of the $c_{1}$ curves, the
diagrams are seen to contained a closed curve depicted by the dashed curves in
the figures. Each of the dashed curves bounds a twist-reduction subdiagram.
This contradicts the assumption that the diagrams are twist reduced which
finishes the proof of the claim. ∎

######  Lemma 6.2.

If a curve $c\in\mathcal{C}$ contains a bone $\tau$ connecting two turns of
$c$ then one of the following holds:

  1. (1)

the curve $c\in\mathcal{C}_{2,0}$,

  2. (2)

the curve $c$ is a rib, or

  3. (3)

the $\circ$-bone $\tau$ meets the width edge of both twist regions.

In particular, $c$ cannot have three consecutive turns.

###### Proof.

Let $\tau$ be a bone connecting two turns of $c$ it is therefore an
$\circ$-bone. Assume in contradiction that $\tau$ and $c$ do not satisfy
neither of (1) - (3) of the lemma. That is, $c$ is not a curve in
$\mathcal{C}_{2,0}$ nor a rib, and $\tau$ does not meet the width edge of both
twist regions. There are two cases to consider depending on whether $\tau$
meets a width edge or not.

If $\tau$ meets a width edge, then $\tau$ is a vertebra, i.e., it meets the
length edge of one twist box and the width of the other. If this occurs set
$c_{0}=c,\tau_{0}=\tau$. Since $c$ is assumed not to be a rib, it follows from
Lemma 5.6 that there exist curves $c_{1},\dots,c_{n}$, limbs
$\kappa_{1},\dots,\kappa_{n}$, and vertebrae $\tau_{1},\dots,\tau_{n-1}$ so
that the terminal layer $c_{n}$ has $\chi_{+}(c_{n})<0$. Moreover, note that
the limb $\kappa_{n}$ is not in $\mathcal{K}_{3,0}$ as otherwise by the
uniqueness property, assured in Lemma 5.6, the “initial” layer curve $c_{0}$
must be a curve in $\mathcal{C}_{2,0}$ of type (i). This implies that
$\kappa_{n}$ does not meet any limb of $\mathcal{K}$. As otherwise, as in the
proof of Lemma 5.11, one can prove that $\kappa_{n}$ and the limb it meets
must be equal. However, by Lemma 5.18, there is no curve, with
$\chi^{\prime}=0$, which has three $\circ$-joints that do not belong to a limb
of $\mathcal{K}$.

If $\tau$ does not meet a width edge, then $\tau$ meets the length edge of
both twist regions. It follows that the curve $c^{\prime}$ abutting $c$ along
$\tau$ has two wiggles which are connected by $\tau$. The curve $c^{\prime}$
cannot be in $\mathcal{C}_{4,0}$ as otherwise it bounds a twist-reduction
subdiagram. By Lemma 5.18, one of the wiggles must meet a limb
$\kappa\in\mathcal{K}$. Since $\tau$ is a core of $c^{\prime}$, it must be the
core $\kappa$. It follows that $\kappa\in\mathcal{K}_{4,0}$ and that
$c\in\mathcal{C}_{2,0}$ is of type (ii), in contradiction to our assumption.

In both cases, whether $\tau$ meets a width edge or not, we arrived to a
contradiction. Hence, $c$ must satisfy one of (1) - (3).

Finally, if $c$ has three consecutive turns then $c$ is not a rib nor a curve
in $\mathcal{C}_{2,0}$. One of the two bones between the turns of $c$ must
meets a length edge and a width edge, contradicting (3). ∎

######  Definition 6.3.

A curve $c\in\mathcal{C}$ is _good_ if it is in $\mathcal{C}_{2,0}$ or one of
the following forms:

Figure 6. The five configurations of good curves which are not in
$\mathcal{C}_{2,0}$

We will say that $c$ is good of type (I) - (V) accordingly. Otherwise, $c$ is
called _bad_.

######  Remark 6.4.

Note that if $S$ is a boundary parallel torus, then its intersection curves
are good. A key observation is that the converse holds. That is, if all the
curves of intersection of $S$ with $P$ are good then $S$ is a boundary
parallel torus. Therefore, our goal in the next claims is to show that the
curves of intersection of a taut surface $S$ with $\chi(S)=0$ are good.

######  Lemma 6.5.

If a curve $c\in\mathcal{C}$ passes through a bubble more than once, then $c$
is good.

###### Proof.

Let $c$ meet the bubble $B_{1}$ more than once. By Lemma 3.9(5) it can do so
only in two opposite $\circ$-joints, $\alpha,\beta$. Therefore, at least one
of those $\circ$-joints, say $\alpha$, is part of a wiggle $\alpha^{\prime}$
of $c$. Hence, $\alpha^{\prime}$ meets an adjacent bubble $B_{2}$. Note that
$c\smallsetminus(\alpha^{\prime}\cup\beta)$ consists of two arcs connecting
$\alpha^{\prime}$ and $\beta$ as depicted in Figure 7. Let $\gamma_{R}$,
$\gamma_{L}$ be the dotted subarcs of $c$ on the right and left of the figure
respectively. The argument is divided into cases according to Table 1:

Figure 7. A curve $c$ that meets a bubble twice.

  1. (1)

The curve $c$ contains three $\circ$-joints, hence it is not in $C_{2,0}$,
$C_{2,2}$ or $C_{0,4}$.

  2. (2)

If $c\in\mathcal{C}_{4,0}$, then the subarc $\gamma_{R}$ of $c$ has no joints
while $\gamma_{L}$ has one $\circ$-joint. Hence $c$ is good of type (I), (II)
or (III).

  3. (3)

Since the three $\circ$-joints of $c$ in $T$ are not part of the same limb in
$\mathcal{K}_{3,0}$ nor $\mathcal{K}_{4,0}$, then $c$ cannot be in
$\mathcal{K}_{3,0}+0,2$ or in $\mathcal{K}_{4,0}+0,2$ (See Table 1).

  4. (4)

The curve $c$ cannot be in $\mathcal{K}_{4,0}+\mathcal{K}_{4,0}$: Otherwise
$\beta$ is part of a wiggle $\beta^{\prime}$ of $c$. The wiggle
$\alpha^{\prime}$ (resp. $\beta^{\prime}$) is part of a limb
$\alpha^{\prime\prime}\in\mathcal{K}_{4,0}$ (resp.
$\beta^{\prime\prime}\in\mathcal{K}_{4,0})$. Since each limb of
$\mathcal{K}_{4,0}$ has a core, the bubble $B_{2}$ must be extremal. Then, the
subarc $\gamma_{R}$ of $c$ contains the other wiggle of
$\alpha^{\prime\prime}$. The closed curve which is the union of $\gamma_{R}$
and an arc on the boundary of the twist region intersects the link diagram
twice, and both sub-diagrams bounded by it are non-trivial. This contradicts
the assumption that the diagram is prime.

  5. (5)

A similar argument shows that $c$ cannot be in
$\mathcal{K}_{2,1}$+$\mathcal{K}_{2,1}$.

  6. (6)

The curve $c$ cannot be in $\mathcal{K}_{2,1}+{1,1}$: Otherwise,
$\alpha^{\prime}$ is the wiggle of some
$\alpha^{\prime\prime}\in\mathcal{K}_{2,1}$ and $\beta$ is a turn. Beside
$\alpha^{\prime}$ and $\beta$, $c$ has a $\partial$-bone on the subarc
$\gamma_{L}$, and no joints on $\gamma_{R}$. Since
$\alpha^{\prime\prime}\in\mathcal{K}_{2,1}$ it abuts some
$c_{0}\in\mathcal{C}_{1,2}$. Hence, $c\cap L$ and $c_{0}\cap L$ share
endpoints. It follows that the union $(c\cap L)\cup(c_{0}\cap L)$ is a
component of $L$ passing over at most two wiggles of the diagram (at $c\cap
L$) and under at most two wiggles (at $c_{0}\cap L$). This contradicts the
assumption that $L$ is 3-highly twisted.

  7. (7)

If the curve $c$ is in $\mathcal{K}_{4,0}+{2,0}$ then $c$ is good of type (V):
If $\beta$ is part of a wiggle $\beta^{\prime}$ of $c$, then at most one of
$\alpha^{\prime}$ and $\beta^{\prime}$ is part of limb in $\mathcal{K}_{4,0}$.
Without loss of generality, assume $\alpha^{\prime}$ is a wiggle of a limb
$\alpha^{\prime\prime}\in\mathcal{K}_{4,0}$. Then, $B_{2}$ is extremal, and
the subarc $\gamma_{R}$ contains the other wiggle of $\alpha^{\prime\prime}$.
The curve which is the union of $\gamma_{R}$ and an arc on the boundary of the
twist region intersects the link diagram twice, in contradiction to the the
assumption that the diagram is prime. If $\beta$ is a turn, then $B_{1}$ is
extremal, and the wiggle $\alpha^{\prime}$ is part of a limb
$\alpha^{\prime\prime}\in\mathcal{K}_{4,0}$. This limb abuts a curve
$c_{0}\in\mathcal{C}_{2,0}$, and it follows that $c$ is good of type (V).

  8. (8)

Finally, if the curve $c$ is in $\mathcal{K}_{3,0}+\mathcal{K}_{3,0}$ then $c$
is good of type (IV): The wiggle $\alpha^{\prime}$ is a wiggle of some limb
$\alpha^{\prime\prime}\in\mathcal{K}_{3,0}$. If $\beta$ is part of a wiggle
$\beta^{\prime}$ of $c$, then $\beta^{\prime}$ is a wiggle of some other limb
$\beta^{\prime\prime}\in\mathcal{K}_{3,0}$. It follows that each of
$\gamma_{R},\gamma_{L}$ contains exactly one $\circ$-joint, which is
impossible. If $\beta$ is a turn, then it is the turn of some limb
$\beta^{\prime\prime}$ in $\mathcal{K}_{3,0}$. As the core of
$\beta^{\prime\prime}$ meets the width of the twist region, its wiggle must be
on $\gamma_{L}$. Similarly, the turn of $\alpha^{\prime\prime}$ must be on
$\gamma_{L}$ as well. This implies that $\gamma_{R}$ does not contain any
joints. If the turn of $\alpha^{\prime\prime}$ and the wiggle of
$\beta^{\prime\prime}$ are in two different twist regions then the curve
abutting (both of) their cores contains three turns. However, this curve is a
non-terminal layer curve (in the sense of Lemma 5.6), and those contain at
most two turns. Thus, the turn of $\alpha^{\prime\prime}$ and the wiggle of
$\beta^{\prime\prime}$ are in the same twist region $T^{\prime}$. Each of
$\alpha^{\prime\prime}$ and $\beta^{\prime\prime}$ meets an extremal bubble of
$T^{\prime}$. Then only option for them to close up is if they meet the same
extremal bubble of $T^{\prime}$. It follows that $c$ is good of type (IV).

∎

######  Proposition 6.6.

All curves in $\mathcal{C}$ are good or in $\mathcal{C}_{0,4}$.

In the proof of the proposition, we will assume in contradiction that such a
curve exists. The proof will follow from the next four lemmas.

######  Lemma 6.7.

Assume that there are bad curves which are not in $\mathcal{C}_{0,4}$. Let $c$
be an innermost bad curve in $P^{+}$ which is not in $\mathcal{C}_{0,4}$. Let
$D$ be the disk bounded by $c$. Then the curve $c$ does not turn or wiggle
through a twist region $T$ which has a bubble contained in $D$.

###### Proof.

Assume $c$ turns at $T$: Let $B$ be the extreme bubble in a twist region $T$
through which $c$ turns. Since the diagram is 3-highly twisted $T$ contains at
least two more bubbles let $B^{\prime}$ be the bubble adjacent to $B$ in $T$.
By assumption $B^{\prime}$ is contained in the disk $D$. Consider the curve
$c^{\prime}$ whose $\circ$-joint is opposite to the $\circ$-joint of $c$ in
$B$. Since $c$ is innermost, the curve $c^{\prime}$ is good. It must be of
good type (I) as in the next configuration.

The curve $\bar{c}$ abutting both $c$ and $c^{\prime}$, passes through the
bubble $B^{\prime}$ twice. Hence, by Lemma 6.5, $\bar{c}$ is good. By the
definitions, it cannot be in $\mathcal{C}_{2,0}$ nor good of type (I). If
$\bar{c}$ is good of type (II) or (III) it has a turn at a bubble
$B^{\prime\prime}$, then $c$ passes twice through $B^{\prime\prime}$ which by
Lemma 6.5 contradicts the assumption that $c$ is bad. If $\bar{c}$ is good of
type (IV) or (V) then it meets an extremal bubble $B^{\prime\prime}$. The
curve $c$ turns at $B^{\prime\prime}$ and hence belongs to $C_{2,0}$ which
again contradicts the assumption that $c$ is bad.

Assume $c$ wiggles through $T$: The curve $c$ wiggles through the bubbles
$B_{1},B_{2}$. Let $B_{0}$ be a bubble of $T\cap D$ so that
$B_{0},B_{1},B_{2}$ are consecutive, as in the following figure.

Consider the curve $c^{\prime}$ whose $\circ$-joint is opposite to the
$\circ$-joint of $c$ in $B_{1}$. The curve $c^{\prime}$ is contained in $D$
and wiggles through $T$ passing through the bubbles $B_{0},B_{1}$.

By assumption $c^{\prime}$ must be good, and so it wiggles through $T$ and
then returns to $T$, passing through $B_{0},B_{1},B_{0}$ in that order. By
Lemma 3.9(5), the two $\circ$-joints of $c^{\prime}$ in $B_{0}$ are opposite
sides of the same saddle. Next, consider the curve $\bar{c}$ abutting
$c^{\prime}$ along the two $\circ$-bones of $c^{\prime}$ connecting $B_{0}$
and $B_{1}$. The curve $\bar{c}$ passes through $B_{1}$ twice. Hence, it is
good by Lemma 6.5. It must be of type (I) and in addition passes through
$B_{2}$. It follows that $c$ abuts $\bar{c}$ along the two $\circ$-bones of
$\bar{c}$ connecting $B_{1}$ and $B_{2}$. Hence, $c$ passes through $B_{2}$
twice, and by Lemma 6.5 $c$ is good, contradicting the assumption. ∎

######  Lemma 6.8.

Assume that there are bad curves which are not in $\mathcal{C}_{0,4}$. Let $c$
be an innermost bad curve in $P^{+}$ which is not in $\mathcal{C}_{0,4}$. Let
$D$ be the disk bounded by $c$. Then the curve $c$ does not wiggles through a
twist region.

###### Proof.

Assume that $c$ wiggles through a twist region $T$. By Lemma 6.7, the disk $D$
does not contain a bubble of $T$. The curve $c$ wiggles extremely through the
twist region by passing through two bubbles $B_{0},B_{1}$, where $B_{0}$ is
extremal. By Lemma 6.5, $c$ meets the bubble $B_{0}$ once.

The curve $c^{\prime}$ turning at $B_{0}$ is good by choice of $c$. Therefore,
$c^{\prime}\in\mathcal{C}_{4,0}$ is good of type (II) or (III) or
$c^{\prime}\in\mathcal{C}_{2,0}$ of type (i) or (ii) (as in Lemma 5.1):

Figure 8. The four configurations of $c^{\prime}$

Let $T^{\prime}\neq T$ be the other twist region which $c^{\prime}$ meets, let
$B_{0}^{\prime}$ denote the extremal bubble in $T^{\prime}$ through which
$c^{\prime}$ passes, and let $B_{1}^{\prime}$ be its adjacent bubble.

Case 1. If $c^{\prime}\in\mathcal{C}_{4,0}$, (i.e., as depicted in the left
two sub-figures in Figure 8), then consider the curve $\bar{c}$ abutting $c$
and $c^{\prime}$ (shown in orange below). None of the bubbles of $\bar{c}$
belongs to an arc of $\mathcal{K}$. Therefore, $\bar{c}\in\mathcal{C}_{4,0}$,
and it follows that it must close up as shown in the dotted curves below:

As $c$ must follow the dotted bone of $\bar{c}$, we see that $c$ passes
through the bubbles $B_{1}$ twice. By Lemma 6.5. This contradicts the
assumption that $c$ is bad.

Case 2. Let $c^{\prime}\in\mathcal{C}_{2,0}$, (i.e., as depicted in the right
two sub-figures of Figure 8) and assume that $c$ _does not_ meet
$B_{0}^{\prime}$. Consider, now, the curve $c^{\prime\prime}$ whose
$\circ$-joint in $B_{0}^{\prime}$ is opposite to that of $c^{\prime}$. By the
choice of $c$, the curve $c^{\prime\prime}$ is good. It must be good of type
(I). And the configuration is as depicted here:

Considering the curve $\bar{c}$ abutting all of
$c,c^{\prime},c^{\prime\prime}$, we see that it must be good by Lemma 6.5 of
type (IV) and (V) respectively. As in Case 1, it follows that $c$ meets
$B_{1}$ twice contradicting the assumption that $c$ is bad.

Case 3. Assume that $c^{\prime}\in\mathcal{C}_{2,0}$ of type (i) and $c$ _does
meet_ $B_{0}^{\prime}$ (as in the third, counted from the left, sub-figure of
Figure 8). In this situation are three sub-cases to consider:

  1. (1)

$c$ wiggles through $T^{\prime}$ at the bubbles $B_{0}^{\prime}$ and
$B_{1}^{\prime}$.

  2. (2)

$c$ turns at $B_{0}^{\prime}$.

  3. (3)

$c\cap L$ meets the bubble $B_{0}^{\prime}$.

Sub-case 3.1. The curve $c$ wiggles through $T^{\prime}$ and through the
bubbles $B_{0}^{\prime}$ and $B_{1}^{\prime}$. After passing through $B_{1}$,
the curve $c$ must exit $T$ at its right length edge and enter $T^{\prime}$ on
its right length edge before meeting $B_{0}^{\prime}$. Otherwise the curve $c$
would be forced to pass through a bubble twice in contradiction to Lemma 6.5.
Thus, the curve $c$ is as in the following figure:

If $c\in\mathcal{C}_{4,0}$ then we get a twist-reduction subdiagram, in
contradiction to the assumption. Thus, by Lemma 5.18, at least one of the
wiggles of $c$, via $B_{0},B_{1}$ or via $B_{0}^{\prime},B_{1}^{\prime}$, must
be part of an arc $\kappa\in\mathcal{K}$. Since the curves
$\bar{c}_{1},\bar{c}_{2}$ abutting $c^{\prime}$ are not in $C_{2,0}$ nor in
$\mathcal{C}_{1,2}$ it is clear that
$\kappa\notin\mathcal{K}_{4,0}\cup\mathcal{K}_{2,1}$. Hence
$\kappa\in\mathcal{K}_{3,0}$. Since $c$ has two wiggles, it must be in
$\mathcal{K}_{3,0}+\mathcal{K}_{3,0}$ (as in Table 1), and each of the dotted
subarcs (in the figure above) must contain a turn.

Let $\kappa\in\mathcal{K}_{3,0}$ be the limb that wiggles through
$B_{0}^{\prime},B_{1}^{\prime}$. Then, $c$ is the terminal curve of some
vertebra $\tau$ in a curve in $\mathcal{C}_{2,0}$ (in the sense of Lemma 5.6).
By retracing backwards, we see that the sequence of curves terminating in $c$
starts with $c^{\prime}\in\mathcal{C}_{2,0}$, then produces
$\bar{c}_{2}\in\mathcal{C}_{4,0}$, and then finally produce the curve $c$ (and
the limb $\kappa$). In particular, $\bar{c}_{2}$ is a rib, and the
$\circ$-bone connecting its two turns is a vertebra. The following figure
shows an example of such a configuration of curves. As one can see, there are
subarcs of $\bar{c}_{2}$ and $c$ that together bound a twist-reduction
subdiagram which is a contradiction.

Sub-case 3.2 Assume $c$ turns at the bubble $B_{0}^{\prime}$. As explained at
the beginning of the previous sub-case 3.1, the curve $c$ must be as in the
following figure:

Figure 9. The curve $c^{\prime}$ is of type (i) and $c$ turns at
$B_{0}^{\prime}$.

Let $\bar{c}$ be the curve abutting $c$’ as in the figure above. The argument
proceeds by dividing into cases according to Table 1:

  1. (1)

The curve $c$ contains three $\circ$-joints, hence it is not in $C_{2,0}$,
$C_{2,2}$ or $C_{0,4}$.

  2. (2)

The curve $c$ is _not_ in $\mathcal{C}_{4,0}$: Otherwise, $c$ has to be a rib.
The bone $\tau$ connecting the two turns cannot meet both width edges of the
corresponding twist regions by Claim 6.1. Hence it is a vertebra, e.g., as in
the following figure:

As $\bar{c}$ has at least five $\circ$-joints, by Table 1 $\bar{c}$ is in
$\mathcal{K}_{3,0}+\mathcal{K}_{3,0}$, and its two cores are those of the
limbs in $\mathcal{K}_{3,0}$. The core of $c$ ending in $B_{0}$ meets the
length edge of both twist regions it connects. This is impossible for a core
of a limb in $\mathcal{K}_{3,0}$.

  3. (3)

The curve $c$ cannot be in $\mathcal{K}_{4,0}+0,2$,
$\mathcal{K}_{2,1}$+$\mathcal{K}_{2,1}$, or
$\mathcal{K}_{4,0}+\mathcal{K}_{4,0}$ (See Table 1) since $c$ contains a turn.

  4. (4)

The curve $c$ cannot be in $\mathcal{K}_{3,0}+0,2$: Otherwise, the three
$\circ$-joints of $c$ are part of the same limb in $\mathcal{K}_{3,0}$. The
core of such a limb connects a width edge to a length edge. This is not the
case here.

  5. (5)

The curve $c$ cannot be in $\mathcal{K}_{2,1}+{1,1}$ or in
$\mathcal{K}_{4,0}+{2,0}$: Otherwise, the wiggle of $c$ is part of a limb in
$\mathcal{K}_{2,1}$ or $\mathcal{K}_{4,0}$ respectively. It follows that the
curve $\bar{c}$ abutting $c$ and $c^{\prime}$ in Figure 9 must be in
$\mathcal{C}_{1,2}$ or $\mathcal{C}_{2,0}$ respectively, which is clearly not
the case.

  6. (6)

Finally, the curve $c$ cannot be in $\mathcal{K}_{3,0}+\mathcal{K}_{3,0}$:
Otherwise, it follows that $\gamma_{R}$ must contain a single wiggle. However,
in this case one can close $\gamma_{R}$ with an arc along $c^{\prime}$ to
obtain a curve in $P$ intersecting the diagram in two points contradicting the
assumption that the diagram is prime.

Sub-case 3.3 If $c\cap L$ meets the bubble $B_{0}^{\prime}$, then it is as
depicted below:

If $c\in\mathcal{C}_{2,2}$ then the left dotted line passes through no bubbles
or intersection points, and we get a contradiction to the parity. Therefore,
by Lemma 5.18, $c$ must contain some arc $\kappa\in\mathcal{K}$. Clearly,
$\kappa$ must contain the subarc of $c$ wiggling through $T$ via
$B_{0},B_{1}$. The curve $\bar{c}_{1}$ is not in
$\mathcal{C}_{1,2}\cup\mathcal{C}_{2,0}$ and therefore
$\kappa\notin\mathcal{K}_{2,1}\cup\mathcal{K}_{4,0}$. If follows
$\kappa\in\mathcal{K}_{3,0}$ and is the result of the process $c^{\prime}$,
then $\bar{c}_{1}$, then $\kappa$, which is discussed in the proof of Lemma
5.6. If this is the case then, as in the end of sub-case 3.1, a subarc of
$\bar{c}_{1}$ and a subarc of $c$ bound a twist-reduction subdiagram as in the
example shown in the following figure:

Case 4. Assume that $c^{\prime}\in\mathcal{C}_{2,0}$ of type (ii) and $c$
_does meet_ $B_{0}^{\prime}$. (as depicted in the rightmost sub-figure of
Figure 8). Let $\bar{c}_{1}$ be the curve abutting $c$ as in the following
figure.

The wiggle of $c$ passing via $B_{0},B_{1}$ is not part of any arc $\kappa$ in
$\mathcal{K}$: If it were, then the curve $\bar{c}_{1}$ will either be in
$\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}$ or will be the non-terminal layer
curve of a process terminating in $\kappa$. Clearly, $\bar{c}_{1}$ is not in
$\mathcal{C}_{2,0}\cup\mathcal{C}_{1,2}$. It is also not a non-terminal layer
of a process defining $\mathcal{K}_{3,0}$ as in Lemma 5.6, since at any step
of a process every curve is a rib, i.e., it has two turns and a wiggle, and
the next step of the process abuts the $\circ$-bone connecting its the two
turns, however here $c$ does not abut the $\circ$-bone of $\bar{c}_{1}$
connecting its two turns.

As in the previous sub-case there are three further sub-sub-cases to consider:

  1. (1)

$c$ wiggles through $T^{\prime}$ at the bubbles $B_{0}^{\prime}$ and
$B_{1}^{\prime}$.

  2. (2)

$c$ turns at $B_{0}^{\prime}$.

  3. (3)

$c\cap L$ meets the bubble $B_{0}^{\prime}$.

Sub-case 4.1 If $c$ wiggles through the bubbles
$B_{0}^{\prime},B_{1}^{\prime}$, the exact same argument as above shows that
the subarc of $c$ passing through $B_{0}^{\prime},B_{1}^{\prime}$ is not a
subarc of any arc in $\mathcal{K}$. It follows that $c\in\mathcal{C}_{4,0}$
and bounds a twist-reduction subdiagram which is a contradiction.

Sub-case 4.2 If $c$ turns at the bubble $B_{0}^{\prime}$. The curve $c$ must
be as in the following figure:

Figure 10. The curve $c^{\prime}$ is of type (ii) and $c$ turns at
$B_{0}^{\prime}$

The argument is further divided into cases according to Table 1:

  1. (1)

The curve $c$ contains three $\circ$-joints, hence it is not in $C_{2,0}$,
$C_{2,2}$ or $C_{0,4}$.

  2. (2)

The curve $c$ is not in $\mathcal{C}_{4,0}$: Otherwise, $c$ has to be a rib.
The bone $\tau$ connecting the two turns cannot meet both width edges of the
corresponding twist regions by Claim 6.1. Hence it is a vertebra, as in the
following figure.

It follows that the curve $\bar{c}$ abutting $c$ and $c^{\prime}$ has at least
three consecutive turns in contradiction to Lemma 6.2.

  3. (3)

The curve $c$ cannot be in $\mathcal{K}_{4,0}+0,2$,
$\mathcal{K}_{2,1}$+$\mathcal{K}_{2,1}$ or
$\mathcal{K}_{4,0}+\mathcal{K}_{4,0}$ (See Table 1) since $c$ contains a turn.

  4. (4)

The curve $c$ cannot be in $\mathcal{K}_{3,0}+0,2$: Otherwise the three
$\circ$-joints of $c$ are part of the same limb in $\mathcal{K}_{3,0}$.
However this would imply that the three $\circ$-joints of $c$ are consecutive,
which is impossible by the checkerboard coloring of the diagram.

  5. (5)

The curve $c$ cannot be in $\mathcal{K}_{2,1}+{1,1}$ or in
$\mathcal{K}_{4,0}+{2,0}$: Otherwise, the wiggle of $c$ is part of a limb in
$\mathcal{K}_{2,1}$ or $\mathcal{K}_{4,0}$ respectively. It follows that the
curve $\bar{c}$ abutting $c$ and $c^{\prime}$ in Figure 10 must be in
$\mathcal{C}_{1,2}$ or $\mathcal{C}_{2,0}$ respectively, which is clearly not
the case.

  6. (6)

Finally, the curve $c$ cannot be in $\mathcal{K}_{3,0}+\mathcal{K}_{3,0}$:
Otherwise, it follows that the curve $\bar{c}$ abutting $c$ has three
consecutive turns. This is impossible by Lemma 6.2.

Sub-case 4.3 If $c\cap L$ meets the bubble $B_{0}^{\prime}$, then by Table 1
$c$ is either in $\mathcal{C}_{2,2}$, or $\mathcal{K}_{2,1}+1,1$ or
$\mathcal{K}_{2,1}+\mathcal{K}_{2,1}$. It cannot be in $\mathcal{K}_{2,1}+1,1$
or $\mathcal{K}_{2,1}+\mathcal{K}_{2,1}$ since otherwise the dotted subarc
$\gamma_{R}$ has a turn or a wiggle respectively, which would contradict
primeness. Thus, $c\in\mathcal{C}_{2,2}$.

If $c\cap L$ passes over one crossing of $L$, then $c$ bounds a twist-
reduction subdiagram as in the following figure:

Otherwise, we are in the following configuration:

Now consider how the curve $\bar{c}_{1}$ abutting $c^{\prime}$ on its left can
close up. It must be as depicted here:

Thus, $\bar{c}_{1}$ must be as one of the configurations that were ruled out
in Claim 6.1. ∎

######  Lemma 6.9.

Assume that there are bad curves which are not in $\mathcal{C}_{0,4}$. Let $c$
be an innermost bad curve in $P^{+}$ which is not in $\mathcal{C}_{0,4}$. Let
$D$ be the disk bounded by $c$. Then $c\notin\mathcal{C}_{4,0}$.

###### Proof.

By lemma 6.8, $c$ only turns. However, this is impossible by Lemma 6.2. ∎

######  Lemma 6.10.

Assume that there are bad curves which are not in $\mathcal{C}_{0,4}$. Let $c$
be an innermost bad curve in $P^{+}$ which is not in $\mathcal{C}_{0,4}$, and
let $D$ be the disk bounded by $c$. Then $c\notin\mathcal{C}_{2,2}$.

###### Proof.

Assume $c\in\mathcal{C}_{2,2}$, Since $c\smallsetminus(c\cap L)$ passes
through two bubbles, the endpoints of (a small continuation of) $c\cap L$ have
the same color in the checkerboard coloring of $P\smallsetminus D(L)$. Thus it
is one of the following:

The complement $c\smallsetminus(c\cap L)$ contains two turns, in twist regions
which do not contain bubbles in $D$. Thus, the possible configurations are:

In each of the cases, consider the curve $\bar{c}$ abutting $c$ (a subarc of
which is shown in orange).

Case $(a1)$ is ruled out by Claim 6.1. In cases $(a2),(b1),(b2)$, it is clear
that the curve $\bar{c}\in\mathcal{C}_{4,0}\cup\mathcal{C}_{2,2}$ and bounds a
twist-reduction subdiagram, which is a contradiction. Cases $(a3),(b3)$ are
impossible by Lemma 6.2. ∎

###### Proof of Proposition 6.6.

Assume in contradiction that there are bad curves which are not in
$\mathcal{C}_{0,4}$. Let $c$ be an innermost bad curve in $P^{+}$ which is not
in $\mathcal{C}_{0,4}$.

By the last two lemmas, $c$ is not in $\mathcal{C}_{4,0}$ nor in
$\mathcal{C}_{2,2}$. Thus, by Lemma 5.18, $c$ must contain an arc of
$\mathcal{K}$. Since every arc in $\mathcal{K}$ wiggles through some twist
region, we get a contradiction to Lemma 6.8. This contradiction finishes the
proof of Proposition 6.6. ∎

######  Corollary 6.11.

If the diagram of $L$ is 3-highly twisted, connected, prime, and twist-reduced
then $\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ is atoroidal. In particular,
$L$ is prime.

###### Proof.

Let $S\subset\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ be an incompressible
taut torus. Let $\mathcal{C}$ be the curves of intersection of $S$ with
$P^{\pm}$. Since $S$ has no boundary, $\mathcal{C}_{0,4}=\emptyset$. By
Proposition 6.6 all curves in $\mathcal{C}$ are good. Since each of the curves
$c\in\mathcal{C}$ bounds a disk in either $B^{+}$ or $B^{-}$ then after adding
these disks along the curves in $\mathcal{C}$ it is readily seen that the
torus $S$ is boundary parallel.

If $L$ was a composite knot, then the swallow-follow torus would be an
essential torus in $\mathbb{S}^{3}\smallsetminus L$. ∎

######  Proposition 6.12.

If the diagram of $L$ is 3-highly twisted, connected, prime and twist-reduced
then $\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ is unannular.

###### Proof.

Let $S$ be an essential annulus. Since $L$ is prime by Corollary 6.11, the
annulus can be assumed not to have meridional boundary components. Thus, we
may assume that $S$ is taut. Let $\mathcal{C}$ be its curves of intersection
with $P$. By Proposition 6.6 all curves in $\mathcal{C}$ are either good or in
$\mathcal{C}_{0,4}$. Since $S$ has boundary components, not all curves in
$\mathcal{C}$ are good, and there is at least one curve
$c\in\mathcal{C}_{0,4}$. We show that this is impossible.

If there exists a curve $c\in\mathcal{C}_{0,4}$ then the curve $\bar{c}$
abutting $c$ has two intersection points which are not connected by an arc of
$\bar{c}\cap L$ therefore $\bar{c}\in\mathcal{C}_{0,4}$. Repeating this
argument shows that all the curves are in $\mathcal{C}_{0,4}$, i.e.,
$\mathcal{C}=\mathcal{C}_{0,4}$.

Case 1. There exists a curve $c\in\mathcal{C}_{0,4}$ which passes twice at the
same twist region.

Denote the two connected components of $c\cap L$ by $\alpha$ and $\beta$. Let
$n$ be the number of crossings of $T$ in-between $\alpha$ and $\beta$. We
further divide the proof into sub-cases depending on $n$.

Sub-case 1.0: $n=0$. By Lemma 3.9(5), $\alpha$ and $\beta$ do not meet the
same bubble of $T$. Since we assume $n=0$, they must meet adjacent bubbles of
$T$. The annulus $S$ must spiral 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 Sub-case 1.0. Thus, we may assume that $l_{1}\neq l_{2}$. Consider the
disk $\Delta$ as depicted in Figure 11(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 11(b). Let $U$ be the torus that meets
$\Delta$. Let $U_{-}$ be the component of $\mathbb{S}^{3}\smallsetminus U$
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
$\Delta$ does intersect $l_{2}$ once. It is also incompressible in $U_{+}$, as
if 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
$S$. By Corollary 6.11, $U$ must be boundary parallel to either
$\partial\mathcal{N}(l_{2})$ or $\partial\mathcal{N}(l_{1})$.

\begin{overpic}[height=99.58464pt]{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=113.81102pt]{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 11. (a) The disk $\Delta$. (b) A cross section of the twist box, the
annulus $S$ and the tori $U,V$

If $U$ is parallel to $\partial\mathcal{N}(l_{2})$, then since $l_{1}$ is
parallel to a curve in $U$ crossing $\Delta$ once there exists an annulus
$A\subset\mathbb{S}^{3}\smallsetminus L$ whose boundary is $l_{1}\cup l_{2}$.
This 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 5.17. 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 $\mathbb{S}^{3}$, 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 $S$, which is
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. None of the curves $c\in\mathcal{C}_{0,4}$ passes twice at the same
twist region. Let $c_{1}$ be a curve in $\mathcal{C}_{0,4}$, then $c_{1}\cap
L$ is the disjoint union of two arcs $\alpha,\beta$. Since all of the curves
are in $\mathcal{C}_{0,4}$ and at least one curve passes over a non-extremal
bubble, we may assume, by changing $c_{1}$ that one of the components, say
$\alpha_{1}$, passes over a non-extremal bubble in a twist region. The
component $\beta$ cannot pass over one bubble, as in this case, either $c_{1}$
passes twice in the same twist region, or defines a twist-reduction
subdiagram. Thus, $\beta$ must be the arc connecting two twist regions,
passing over their two extremal bubbles, and the situation is as depicted in
the following figure.

This case was ruled out in Claim 6.1. ∎

######  Remark 6.13.

The Sub-case 1.1 $n=1$ in the proof of Proposition 6.12 follows also from the
following well known general statement:

Non-split, annular atoroidal links in $\mathbb{S}^{3}$ are either torus knots
or a link consisting of a torus knot on the “standard torus” $\mathbb{T}^{2}$
in $\mathbb{S}^{3}$ and one or both of the core curves of the solid tori
components of $\mathbb{S}^{3}\smallsetminus\mathcal{N}(\mathbb{T}^{2})$.

For completeness we include a proof.

###### Proof.

Let $L\subset\mathbb{S}^{3}$ be a non-split and an atoroidal link in
$\mathbb{S}^{3}$ containing an essential annulus $A$. If $L$ is a knot then
boundary $A$ cuts $\partial\mathcal{N}(L)$ into two annuli $A_{1}$ and
$A_{2}$. The surfaces $A\cup A_{1}$ and $A\cup A_{2}$ are tori which bound
solid tori $V_{1}$ and $V_{2}$ as $L$ is atoroidal. The two solid tori are
glued to each other along $A$ and since the result together with a regular
neighborhood of $L$ is $\mathbb{S}^{3}$, then by Seifert (see [11]), their
complement is a regular neighborhood of a torus knot.

If $L$ is a non-hyperbolic, non-split link whose exterior is atoroidal then by
[12] it is a Seifert link, i.e. its exterior is a Seifert fiber space. Links
in Seifert spaces were classified by Burde and Murasugi in [1]. They are
either a connected sum of Hopf links or consist of a union of Seifert fibers
in some Seifert fibration of $\mathbb{S}^{3}$. Atoroidal such links can
contain at most three fibers. Hence the link $L$ is a torus knot $K$ on an
unknotted solid torus $\mathbb{T}$ and the Hopf link which is the core curves
of the complementing solid tori. ∎

###### Proof of Theorem A.

By Thurston [12], it suffices to prove that
$\mathbb{S}^{3}\smallsetminus\mathcal{N}(L)$ has incompressible boundary, and
is irreducible, atoroidal and unannular. By Corollary 5.17 it has
incompressible boundary and is irreducible, by Corollary 6.11 it is atoroidal,
and by Proposition 6.12 it is unannuluar. ∎

##  7\. Essential holed spheres in highly twisted link complements

In this section we use Theorem A to show that certain holed spheres in the
complement of a highly twisted links are essential. We begin with three
definitions:

######  Definition 7.1.

Let $D(T)$ be a projection of a tangle $(B,T)$ onto a disk $\Delta\subset P$
where $\partial\Delta$ is the simple closed curve $\gamma\subset P$ and so
that the end points of the strings of $T$ denoted by $\partial T$ are
contained in $\gamma$. We call $D(T)$ a _tangle diagram_. Let $\bar{D}(T)$
denote the reflection of $D(T)$ along $P$ (i.e. the diagram with the same
projection, but with reverse crossing data).

######  Definition 7.2.

The diagram $D(T)$ is _relatively prime_ (resp._relatively twist reduced_ ,
_relatively $k$-highly twisted_) if the link diagram obtained by gluing $D(T)$
and $\bar{D}(T)$ along their boundary is prime (resp. twist reduced,
$k$-highly twisted).

######  Definition 7.3.

A surface $S$ in $(\mathbb{S}^{3},\mathcal{L})$ is _pairwise-incompressible_
if every disk $D$ in $\mathbb{S}^{3}$ with $\partial D=D\cap S$, and which
intersects $\mathcal{L}$ transversely in a single point, is isotopic to a disk
in $S$ by an isotopy preserving $\mathcal{L}$ set-wise.

The surface $S$ is _acylindrical_ if the complement of $S$ in
$\mathbb{S}^{3}\smallsetminus\mathcal{N}(\mathcal{L})$ contains no essential
annuli whose boundary is on $S\cup\partial\mathcal{N}(\mathcal{L})$.

######  Theorem B.

Let $D(L)$ be $3$-highly twisted diagram and let $\gamma$ be a simple closed
curve in $D(L)$ intersecting $D(L)$ transversely. Assume that both tangle
diagrams bounded by $\gamma$ are connected, relatively prime, relatively
twist-reduced, relatively $k$-highly twisted and contain at least two twist
regions each. Let $\Sigma$ be the sphere in $\mathbb{S}^{3}$ which intersects
$P$ transversely in $\gamma$, and does not intersect $\mathcal{L}$ outside
$\gamma$. Then the punctured sphere
$\Sigma^{\prime}=\Sigma\smallsetminus\mathcal{N}(\mathcal{L})$ is
incompressible, boundary-incompressible, pairwise-incompressible and
acylindrical in $\mathbb{S}^{3}\smallsetminus\mathcal{N}(\mathcal{L})$.

###### Proof of Theorem B.

Assume in contradiction that
$\Sigma^{\prime}=\Sigma\smallsetminus\mathcal{N}(\mathcal{L})$ is either
incompressible, boundary-incompressible, pairwise-incompressible or
acylindrical in $\mathbb{S}^{3}\smallsetminus\mathcal{N}(\mathcal{L})$.

Let $B_{1},B_{2}$ be the two 3-balls that $\Sigma$ bounds in $\mathbb{S}^{3}$,
and let $E_{1}=P\cap B_{1},\;E_{2}=P\cap B_{2}$ be the corresponding two disks
in $P$ bounded by $\gamma$. The induced tangle diagrams on $E_{1}$ and $E_{2}$
are assumed to be relatively prime, relatively twist-reduced and relatively
$3$-highly twisted. After doubling each of $E_{1}$ and $E_{2}$, as in
Definition 7.2, we get two link diagrams $D(L_{1}),D(L_{2})$ which are
connected, prime, twist-reduced and $3$-highly twisted. By Theorem A, the
associated links $\mathcal{L}_{1},\mathcal{L}_{2}$ are hyperbolic.

  1. (1)

If $\Sigma^{\prime}$ is compressible then the doubling of an innermost
compressing disk $\Delta\subset B_{i}$ will give rise to an essential 2-sphere
in $\mathbb{S}^{3}\smallsetminus\mathcal{N}(\mathcal{L}_{i})$.

  2. (2)

If $\Sigma^{\prime}$ is boundary compressible with boundary compression disk
$\Delta\subset B_{i}$, then the doubling of $\Delta$ along
$\Sigma^{\prime}\cap\Delta$ results in a disk which is bounded by a component
of $\mathcal{L}_{i}$. Hence either $\mathcal{L}_{i}$ is the unknot or has a
split unknot component.

  3. (3)

If $\Sigma^{\prime}$ is pairwise boundary compressible, then the doubling of
the essential disk $\Delta\subset B_{i}$ intersecting $\mathcal{L}$ once is a
2-sphere intersecting the boundary of $\mathcal{N}(\mathcal{L}_{i})$ in two
meridians. Thus $\mathcal{L}_{i}$ is not prime.

  4. (4)

If $\Sigma^{\prime}$ contains an essential annulus $A\subset B_{i}$ (i.e.,
$\Sigma^{\prime}$ is cylindrical), then $\mathcal{L}_{i}$ is toroidal if both
boundaries of $A$ are on $\Sigma^{\prime}$, or annular if one of these
boundaries is on $\Sigma^{\prime}$ and the other on
$\partial\mathcal{N}(\mathcal{L})$.

In all these cases we get that one of the links
$\mathcal{L}_{1},\mathcal{L}_{2}$ is not hyperbolic and thus a contradiction.
∎

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