# Hyperbolic Brunnian Links

Sheng Bai<EMAIL_ADDRESS>
(Date: 2021 Mar. 28)

###### Abstract.

We present practical methods to show a Brunnian link is hyperbolic. The
methods, belonging to geometric topology, are based on observation, worked by
hand and generally applicable. Typical examples illustrate their efficiency.
We thus determine hyperbolicity for all of the known Brunnian links in
literatures except two series. Especially, we discover 6 infinite series of
hyperbolic Brunnian links.

###### Key words and phrases:

Brunnian links, hyperbolic links, satellite links, essential torus

Figure 1. A hyperbolic Brunnian link: Cirrus in [5].

##  1\. Introduction

Every minimal nontrivial sublink of a link is either a knot or a Brunnian
link. [4] shows that hyperbolic Brunnian links and $(2,2n)-$torus links form
building blocks of Brunnian links in that they generate all Brunnian links by
satellite operations. In this paper, we give practical methods to show a
Brunnian link is hyperbolic.

A classical theorem by Alexander states that every smooth torus bounds at
least one solid torus in $S^{3}$. We say a torus in $S^{3}$ is _knotted_ if it
bounds only one solid torus in $S^{3}$; otherwise, _unknotted_. In [4], A
Brunnian link, other than Hopf link, is called _s-prime_ /_untied_ if there is
no essential unknotted/knotted torus in its complement space. [4] further
states that if a Brunnian link $L$ is s-prime, untied and not a $(2,2n)-$torus
link, then $L$ is a hyperbolic link. So the two main tasks in this paper are
to discriminate whether a Brunnian link is untied and s-prime.

Extending [5], we will cope with these two tasks using geometric topology
methods. [5] provides two intuitive methods, generally effective and performed
by handwork, for detecting Brunnian property of a link if each component is
known to be unknot. In Section 4, we will prove a universal criterion to
detect whether a Brunnian link is untied.

######  Theorem 1.1.

(Untiedness Criterion)

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link and
$k\in\\{1,...,n\\}$. For each $1\leq i\leq k$, let $D_{i}$ be a disk bounded
by $C_{i}$ so that $U=\bigcup_{i=1}^{k}D_{i}\cup L$ is a spanning complex.
Suppose for every $1\leq i\leq k$, $D_{i}$ is (s7). Then $L$ is untied if and
only if there is no incompressible knotted torus in the complement of $U$.

To show that a Brunnian link is s-prime, instead of giving a criterion, we
will first prove a universal intermediate theorem in Section 5, based on which
we will provide four methods to show s-primeness in Section 6.

######  Theorem 1.2.

(Simple Intersection Pattern Theorem)

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link. Let $T$ be an
essential torus in the complement of $L$, splitting $S^{3}$ into two solid
tori $V_{I}$ and $V_{J}$ so that $L\cap V_{I}=L_{I}$ and $L\cap V_{J}=L_{J}$,
where $L_{I}=\cup_{i\in I}C_{i}$ and $L_{J}=\cup_{j\in J}C_{i}$.

Suppose $Q\sqcup R\sqcup S\subset I$ and $J_{0}\subset J$. Let
$D_{q}^{I}$($q\in Q$), $D_{j}^{J}$($j\in J_{0}$), $D_{r}^{E}$($r\in R$) and
$D_{s}^{F}$($s\in S$) be $L_{I}$-interior disks, $L_{J}$-interior disks,
exterior cross disks, and free cross disks respectively, so that the cross
disks are mutually disjoint and each disk is stable. If the union of these
disks and $L$ forms a spanning complex, then after an isotopy of $T$, every
disk intersects $T$ in simple intersection pattern.

We will define _(s7)_ , _stable_ and _spanning complex_ in Section 3, and
_$L_{I}$ -interior disks_, _$L_{J}$ -interior disks_, _exterior cross disks_ ,
_free cross disks_ and _simple intersection pattern_ in Subsection 5.1. In
fact, the two corresponding theorems in the text will be stronger than the two
presented here, but the conditions will be more complicated. We will
illustrate our methods by typical examples.

Then we are interested in how many Brunnian links in history, which, to our
best knowledge, all appear in [6, 8, 20, 1, 3, 2, 15, 12, 10, 18, 16, 5], are
hyperbolic. Our methods work to detect s-primeness for all of them and
untiedness for all except Fountains and Jade-pendants in [5]. Especially, we
list some infinite series of Brunnian links:

  1. (1)

Lamp($n_{1},n_{2},...,n_{2k}$), where $k\in\mathbb{N}_{+}$ and each index is
an odd integer indicating the number of half twists as shown in Fig. 3.

  2. (2)

deBrunner($n$), where $n\leq 5$, as shown in Fig. 3. (c.f. [20])

  3. (3)

W($n$), where $n>1$, as shown in Fig. 4. (c.f. [20])

  4. (4)

Torus($m,n$), where $m,n\in\mathbb{N}_{+}$, as shown in Fig. 5. (c.f. [2, 3])

  5. (5)

Tube($m,n$), where $m>0,n>1$, as shown in Fig. 6.[2, 3, 5]

  6. (6)

Carpet($m,n,p$), where $m,n,p\in\mathbb{N}_{+}$ and $m<n$, as shown in Fig.
7.[2, 3, 5]

######  Theorem 1.3.

The six series of links above are hyperbolic.

Our methods are worked by hand. An alternative to showing a link is hyperbolic
is to use SnapPy. However, a program cannot show hyperbolicity for infinite
series of links. By the theorem of W. Menasco[17], it can be verified that
every alternating Brunnian link is hyperbolic. Nevertheless, these newly
discovered series are not alternating at least seen from the diagrams, and our
methods provide alternative proofs for hyperbolicity of alternating Brunnian
links in literatures.

In the last section, we will generalize our criteria and framework to wider
classes of links.

Figure 2. Lamp($n_{1},n_{2},...,n_{2k}$)

Figure 3. deBrunner(5).

Figure 4. W(5), constructed by H. Brunn in [6]. Figure 5. Torus($m,n$) has
$2m$ rows and each row has $n$ components. Figure 6. Tube(3,4), a tube with 3
rows and 4 columns. Figure 7. Carpet(1,3,4), see [5] for notations.

Acknowledgement: We are grateful to Weibiao Wang who drew most of the figures
in this paper and verified Fig. 1 is hyperbolic.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Credible disks and stable disks
    1. 3.1 Credible disks
    2. 3.2 Stable disks
    3. 3.3 Spanning complex
  4. 4 Criterion for Untiedness
    1. 4.1 Decision Theorem and Example analysis
    2. 4.2 Proof of Theorem 4.1
  5. 5 Simple intersection pattern Theorem
    1. 5.1 Simple intersection pattern
    2. 5.2 Proof of the main Theorem
  6. 6 Criteria for S-primeness
    1. 6.1 Analysis on plausible annuli
    2. 6.2 Components discarded
    3. 6.3 Interior disks only
    4. 6.4 Cross disk with the least intersection points
  7. 7 Conclusions
  8. 8 Appendix: An alternating example

##  2\. Preliminaries

In this paper, all objects and maps are smooth. Without special explanation,
we always consider links in $S^{3}$. All intersections are compact and
transverse. Each arc is assumed to be simple. The notations $C,L,D$ (maybe
with subscripts) are always used to denote an unknot, a link and an embedded
disk respectively. We use $N(\cdot)$ to denote a regular neighborhood.

For a compact set $K$ in a solid torus, $K$ is _geometrically essential_ in it
if any meridian disk intersects $K$. If $V$ is a regular neighborhood of a
knot $K$ in $S^{3}$, the _longitude_ of $V$ is the essential curve in
$\partial V$ that is null-homologous in $S^{3}-intV$, oriented similarly to
$K$; the _meridian_ is the essential curve in $\partial V$ bounding a disk in
$V$ and having link number $+1$ with $K$. (c.f. [7])

######  Definition 2.1.

[4] Let $L$ and $L^{\prime}$ be links in $S^{3}$, and $C\subset L$ and
$C^{\prime}\subset L^{\prime}$ be oriented unknotted components. A
homeomorphism $h:S^{3}-intN(C^{\prime})\longrightarrow N(C)$ maps the oriented
meridian of $N(C^{\prime})$ to the oriented longitude of $N(C)$ and maps the
oriented longitude of $N(C^{\prime})$ to the oriented meridian of $N(C)$,
where $N(C)$ is the regular neighborhood of $C$. Then the link $(L-C)\sqcup
h(L^{\prime}-C^{\prime})$ is the _s-sum_ of $L(C)$ and
$L^{\prime}(C^{\prime})$.

######  Definition 2.2.

[4] Let $L_{0}\sqcup L^{n}$ be a link in $S^{3}$ where
$L^{n}=\sqcup_{i=1}^{n}C^{i}$ is an oriented unlink, and $k_{i}$, $i=1,...,n$,
be nontrivial knot types. Let $h:U_{n}=S^{3}-intN(L^{n})\longrightarrow S^{3}$
be an orientation-preserving embedding so that

$\displaystyle S^{3}-h(U_{n})\cong\sqcup_{i=1}^{n}(S^{3}-N(k_{i})),$

where $h(\partial N(C^{i}))$ corresponds to $\partial N(k_{i})$, and the
oriented meridian of $N(C^{i})$ maps to the oriented null-homologous curve in
$S^{3}-h(U_{n})$ corresponding to the oriented longitude of $N(k_{i})$. Then
$L=h(L_{0})$ is an _s-tie_.

For Brunnian links, we have

######  Theorem 2.3.

[4] If a torus $T$ in $S^{3}$ splits a Brunnian link $L$, then $T$ is
incompressible, and $L$ is decomposed by $T$ as an s-sum of two Brunnian
links.

######  Theorem 2.4.

[4] Every knotted essential torus in the complement of a Brunnian link bounds
the whole link in the solid torus side.

##  3\. Credible disks and stable disks

In this section, we introduce our main tools and some notions for links with
some unknotted components, and confirm these tools are always available.

###  3.1. Credible disks

Our basic tool is credible disk.

######  Definition 3.1.

Let $C_{i}$ be a component of a link $L$ and $D_{i}$ be a disk bounded by
$C_{i}$. $D_{i}$ is _credible_ if there is no circle $C$ in the interior of
$D_{i}$ such that

  * (i)

the disk in $D_{i}$ bounded by $C$ intersects $L$;

  * (ii)

there is a disk $D_{C}$ bounded by $C$ with $D_{C}\subset S^{3}-L$.

To simplify the task of showing a disk is credible, we give a definition used
for the proposition following it.

######  Definition 3.2.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component link and
$I\subset\\{1,...,n\\}$. Take disjoint disks $D_{i}(i\in I)$ bounded by
$C_{i}(i\in I)$ and set $U=\cup_{i\in I}D_{i}\cup L$. A circle $C$ in the
interior of $D_{i}$ is _incredible_ for $U$ if

  * (i)

the disk in $D_{i}$ bounded by $C$ intersects $L$;

  * (ii)

there is a disk $D_{C}$ bounded by $C$ with $intD_{C}\subset S^{3}-U$.

######  Proposition 3.3.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component link and
$I\subset\\{1,...,n\\}$. Take disjoint disks $D_{i}(i\in I)$ bounded by
$C_{i}(i\in I)$ and set $U=\cup_{i\in I}D_{i}\cup L$. Then $D_{i}$ is credible
for each $i\in I$ if and only if there is no incredible circle for $U$.

###### Proof.

Since the“only if” implication is obvious, we need only proof the “if”
implication. Suppose for an $i\in I$, there is a circle $C\subset intD_{i}$
such that the disk in $D_{i}$ bounded by $C$ intersects $L$ and there is a
disk $D_{C}$ bounded by $C$ with $D_{C}\subset S^{3}-L$. We wish to find an
incredible circle for $U$.

After a perturbation near $C$, we may assume $D_{C}\cap\cup_{i\in I}D_{i}$ is
a disjoint union of circles. Choose an innermost circle in $D_{C}$, which
bounds a disk, say $D_{0}$, in it. Suppose $C_{0}=\partial D_{0}$ is in
$D_{j}$ for some $j\in I$. If $C_{0}$ encloses some intersection points with
$L$ in $D_{j}$, then $C_{0}$ is an incredible circle for $U$. Otherwise,
$D_{0}$ caps a 3-ball with $D_{j}$. Replacing $D_{0}$ in $D_{C}$ by the disk
in $D_{j}$ bounded by $C_{0}$, a little beyond, we eliminate $C_{0}$. Step by
step, we find an incredible circle for $U$. ∎

Noting that the complement of $U$ is smaller than the component of $L$, this
proposition reduces the task of showing a disk is credible in application. In
fact, the main content of [5] is to show disks are credible, which proposes
two widely applicable methods, including several criteria and a general
procedure.

###  3.2. Stable disks

Sometimes we will use a special kind of credible disks, called stable disks.
Let $C_{i}$ be a component of a link $L$ and $D_{i}$ be a disk bounded by
$C_{i}$. Let $L_{0}$ consist of the components of $L$ which intersects the
interior of $D_{i}$, and $L_{J}$ be a sublink of $L-C_{i}$ containing $L_{0}$.

######  Definition 3.4.

The disk $D_{i}$ is _stable for $L_{J}$_ if $\sharp(D_{i}\cap L_{J})$ is
minimal among all disks bounded by $C_{i}$ whose interior only intersect
$L_{J}$, and $D_{i}$ is _stable_ if it is stable for $L_{0}$.

Clearly a stable disk is credible and the converse is not true in general. To
reduce the task of showing a disk is stable, we introduce the following
definition.

######  Definition 3.5.

A circle $C$ in the interior of $D_{i}$ is _unstable for $L_{J}$_ if $C$
bounds a disk $D_{C1}$ such that

  * (i)

$D_{C1}\cap L=D_{C1}\cap L_{J}$;

  * (ii)

$\sharp(D_{C1}\cap L)<\sharp(D_{C0}\cap L)$, where $D_{C0}$ is the disk in
$D_{i}$ bounded by $C$.

Obviously that $D_{i}$ is stable for $L_{J}$ implies that there is no unstable
circle for $L_{J}$. The following lemma shows the converse is also true.

######  Lemma 3.6.

There is no unstable circle for $L_{J}$ in $D_{i}$ if and only if $D_{i}$ is
stable for $L_{J}$.

###### Proof.

Suppose there is no unstable circle for $L_{J}$ on $D_{i}$. Assume for
contradiction that $D^{\prime}_{i}$ is a disk bounded by $C_{i}$, having less
number of intersection points with $L_{J}$ than $D_{i}$. After perturbation,
$D^{\prime}_{i}$ intersects $D_{i}$ transversely in their interiors. Choose an
innermost circle in $D^{\prime}_{i}$. Using the same argument as in the proof
of Lemma 3.8(2), we get a contradiction. ∎

The task of showing a disk is stable can be further reduced by considering
mutually disjoint disks. Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component link
and $I\subset\\{1,...,n\\}$. Take disjoint disks $D_{i}(i\in I)$ bounded by
$C_{i}(i\in I)$.

######  Definition 3.7.

For any $k\in I$, a circle $C$ in the interior of $D_{k}$ is _unstable
avoiding $\cup_{i\in I}D_{i}$_ if $C$ bounds a disk $D_{C1}$ such that

  * (i)

$D_{C1}\cap\cup_{i\in I}D_{i}=C$;

  * (ii)

$\sharp(D_{C1}\cap L)<\sharp(D_{C0}\cap L)$, where $D_{C0}$ is the disk in
$D_{i}$ bounded by $C$.

######  Lemma 3.8.

(1) Fix $k\in I$. Then there is no unstable circle in $D_{k}$ avoiding
$\cup_{i\in I}D_{i}$ if and only if there is no disk $D$ bounded by $C_{k}$
such that

  * (i)

$D\cap\cup_{i\in I}D_{i}=C_{k}$, and

  * (ii)

$\sharp(D\cap(L-C_{k}))<\sharp(D_{k}\cap(L-C_{k}))$.

(2) For any $k\in I$ there is no unstable circle in $D_{k}$ avoiding
$\cup_{i\in I}D_{i}$ if and only if for any $k\in I$ there is no unstable
circle in $D_{k}$ for $L-\cup_{i\in I}C_{i}$.

###### Proof.

(1). The“only if” implication is trivial. For the “if” implication, suppose
there is an unstable circle $C$ bounding $D_{C0}$ in $D_{k}$ and $D_{C1}$ as
in Definition 3.7. Then the disk $(D_{k}-D_{C0})\cup D_{C1}$ has less
intersection points with $L-C_{k}$, and after perturbation, its interior is
disjoint from $D_{k}$.

(2). Since the“if” implication is obvious, we need only proof the “only if”
implication. Set $L_{J}=L-\cup_{i\in I}C_{i}$. Suppose there is a circle $C$
in the interior of some $D_{k}$, bounding $D_{C0}$ in $D_{k}$ and bounding
$D_{C1}$ as in Definition 3.5. Then $D_{C1}\cap\cup_{i\in I}D_{i}$ is disjoint
union of circles. Let $C_{0}$ be an innermost circle in $D_{C1}$, which bounds
a disk $D_{0}$ in it and bounds a disk $D_{j0}$ in some $D_{j}$. Since $D_{j}$
is stable avoiding $\cup_{i\in I}D_{i}$, $D_{j0}$ has no more number of
intersection points with $L_{J}$ than $D_{0}$. Take the immersed disk
$(D_{C1}-D_{0})\cup D_{j0}$ and push the $D_{j0}$ in it a little downward
$D_{j}$ to eliminate $C_{0}$. We denote the obtained immersed disk by
$E_{C2}$.

Figure 8. Two kinds of surgeries to eliminate double curves.

Generally $E_{C2}$ may not be embedded since $C_{0}$ may not be innermost in
$D_{j}$. Nevertheless, its singularities consists of only double points,
forming disjoint circles. This is an easy case of Dehn’s Lemma([21, 9]), so we
can revise it to get an embedding disk. For each self-intersection circle,
change the immersion map either on two disks or in an annulus region, as
illustrated in Figure 8, and then smooth the corners. We will get an embedding
disk $D_{C2}$ having the same intersection points with $L_{J}$ as $E_{C2}$.

Step by step we eventually get a disk $D_{CN}$, which intersects $\cup_{i\in
I}D_{i}$ in $C$ and has no more number of intersection points with $L_{J}$
than $D_{C1}$. Then $D_{CN}$ has less number of intersection points with
$L_{J}$ than $D_{C0}$. Thus $C$ is unstable avoiding $\cup_{i\in I}D_{i}$, a
contradiction. ∎

In summary, the following proposition simplifies significantly the task of
showing a disk is stable.

######  Proposition 3.9.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component link and
$I\subset\\{1,...,n\\}$. Let $D_{i}(i\in I)$ be disjoint disks bounded by
$C_{i}(i\in I)$. Then every $D_{i}(i\in I)$ is stable for $L-\cup_{i\in
I}C_{i}$ if and only if for any $i\in I$ there is no disk $D$ bounded by
$C_{i}$ such that

  * (i)

$D\cap\cup_{k\in I}D_{k}=C_{i}$;

  * (ii)

$\sharp(D\cap(L-C_{i}))<\sharp(D_{i}\cap(L-C_{i}))$.

###### Proof.

This follows immediately from the previous two lemmas. ∎

Applying this proposition, we present how to prove a disk is stable by two
examples.

######  Example 3.10.

Consider the Milnor link in Fig. 9(1). The components $C_{1},C_{2},C_{3}$
bound mutually disjoint oriented disks $D_{1},D_{2},D_{3}$ as Fig. 9(2), and
the component $C_{0}$ is homotopically nontrivial in
$S^{3}-\cup_{i=1}^{3}C_{i}$, whose fundamental group is freely generated by
$g_{1},g_{2},g_{3}$ as Fig. 9(2). To obtain the representation of a loop in
$\pi_{1}(S^{3}-\cup_{i=1}^{3}C_{i})$, just record the intersections of the
loop and the disks in order with orientations. Notice that $C_{0}$ intersects
$D_{1}$ and $D_{2}$ both in 4 points, and intersects $D_{3}$ in two points.
With the base point $P\in C_{0}$, we see $C_{0}$ represents
$[[g_{1},g_{2}],g_{3}]$ in $\pi_{1}(S^{3}-\cup_{i=1}^{3}C_{i})$. We show
$D_{1},D_{2},D_{3}$ are all stable.

Figure 9. Stable disks for Milnor link.

###### Proof.

Since $[[g_{1},g_{2}],g_{3}]$ is a reduced word, unique when generators of the
free group are fixed, any isotopy of $D_{1}\cup D_{2}\cup D_{3}$ cannot give
less intersection points with $C_{0}$ for each disk. To show $D_{1}$ is
stable, it suffices to show there is no disk bounded by $C_{1}$, whose
interior disjoint from $D_{1}\cup D_{2}\cup D_{3}$, has less intersection
points with $C_{0}$ than $C_{1}$. Suppose such a disk $D^{\prime}_{1}$ exists,
then $S_{1}=D_{1}\cup D^{\prime}_{1}$ is a sphere separating $D_{2}$ and
$D_{3}$. There are two cases.

Case 1: The positive side of $D_{1}$ and $C_{2}$ belong to one side of
$S_{1}$. Consider the element $C_{0}$ represents in
$\pi_{1}(S^{3}-\cup_{i=1}^{3}C_{i})$, that is
$g_{1}g_{2}g_{1}^{-1}g_{2}^{-1}g_{3}g_{2}g_{1}g_{2}^{-1}g_{1}^{-1}g_{3}^{-1}$.
We can assume the base point $P$ is in the negative side of $S_{1}$ and find
an arc from $P$ representing $g_{1}g_{2}g_{1}^{-1}$, which avoids
$D^{\prime}_{1}$. Then the fourth letter $g_{2}^{-1}$ means we have to go to
the positive side of $S_{1}$ to have an arc representing $g_{2}^{-1}$. But the
third letter $g_{1}^{-1}$ means the arc representing the fourth letter
$g_{2}^{-1}$ avoids $D_{1}$, so it must intersect $D^{\prime}_{1}$. In other
words, between the third letter and the fourth letter, $C_{0}$ intersects
$D^{\prime}_{1}$. In the same way, between the fourth letter $g_{2}^{-1}$ and
the fifth letter $g_{3}$, between the fifth letter $g_{3}$ and the sixth
letter $g_{2}$, and between the sixth letter $g_{2}$ and the seventh letter
$g_{1}$, our loop intersects $D^{\prime}_{1}$.

In short, in the representation
$g_{1}g_{2}g_{1}^{-1}g_{2}^{-1}g_{3}g_{2}g_{1}g_{2}^{-1}g_{1}^{-1}g_{3}^{-1}$,
replacing $g_{2}$ and $g_{2}^{-1}$ by $2$, $g_{3}$ and $g_{3}^{-1}$ by $3$,
$g_{1}$ by $1^{-}1^{+}$, and $g_{1}^{-1}$ by $1^{+}1^{-}$, we get a cyclic
sequence

$1^{-}1^{+}21^{+}1^{-}\mid 2\mid 3\mid 2\mid 1^{-}1^{+}21^{+}1^{-}3$.

Since $1^{+},2$ belong to one side, and $1^{-},3$ belong to the other side,
the number of how many adjacent pairs in this cyclic sequence cross through
$S_{1}$ gives the minimum number of $D^{\prime}_{1}\cap C_{0}$, which is 4.

Case 2: The positive side of $D_{1}$ and $C_{3}$ belong to one side of
$S_{1}$. Make replacement in the representation
$g_{1}g_{2}g_{1}^{-1}g_{2}^{-1}g_{3}g_{2}g_{1}g_{2}^{-1}g_{1}^{-1}g_{3}^{-1}$
as before. Now $1^{+},3$ belong to one side, and $1^{-},2$ belong to the other
side. The number of adjacent pairs in this cyclic sequence cross through
$S_{1}$ is 8, as

$1^{-}1^{+}\mid 2\mid 1^{+}1^{-}2\mid 3\mid 21^{-}1^{+}\mid 2\mid
1^{+}1^{-}\mid 3\mid$.

The same method applies to $D_{2},D_{3}$. ∎

Figure 10. Lamp(1,1,1,1,1,1,1,1).

######  Example 3.11.

Consider the link in Fig. 10(1). We show the grey disk $D$, bounded by $C_{1}$
in Fig. 10(2) is stable. Let $D^{\prime}$ be a disk bounded by $C_{1}$ whose
interior avoids $D$. It suffices to show $\sharp(D^{\prime}\cap
C_{2})\geq\sharp(D\cap C_{2})$. The disk $D$ cuts $C_{2}$ into 8 arcs
$\alpha_{1},\alpha_{2},...,\alpha_{8}$, successively indexed along $C_{2}$ as
shown in Fig. 10(2). Notice that each $\alpha_{i}$ can only intersect
$D^{\prime}$ in even number of points, since the two endpoints of each
$\alpha_{i}$ are on the same side of $D$. The result will be proved by showing
that for any odd $i$, either $\sharp(D^{\prime}\cap\alpha_{i})\geq 2$ or
$\sharp(D^{\prime}\cap\alpha_{i\pm 1})\geq 2$.

In fact, we can say $lk(\alpha_{i},\alpha_{i\pm 1})=1$ in the sense that
connecting the endpoints of $\alpha_{i}$ by an arc on the positive side of $D$
and connecting the endpoints of $\alpha_{i\pm 1}$ by an arc on the negative
side of $D$ gives two curves with linking number equal to 1. See Fig. 10(3).
If $D^{\prime}$ intersects neither $\alpha_{i}$ nor $\alpha_{i\pm 1}$, it
would cap off $\alpha_{i}$ from $\alpha_{i\pm 1}$ and thus
$lk(\alpha_{i},\alpha_{i\pm 1})=0$, a contradiction.

###  3.3. Spanning complex

We now introduce our main tool in this paper.

######  Definition 3.12.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component link. Fix
$I\subset\\{1,...,n\\}$ and for any $i\in I$, let $D_{i}$ be a credible disk
bounded by $C_{i}$. The union $U=\cup_{i\in I}D_{i}\cup L$ is a _spanning
complex_ for $L$ if the following regularity conditions hold:

  * (Ri)

$U$ has only generic singularities of double or triple points.

  * (Rii)

(No trivial intersection circle.) $\forall i,j\in I$, there is no circle
component $C$ of $D_{i}\cap D_{j}$ such that the disk bounded by $C$ in
$D_{i}$ does not intersect $L$.

  * (Riii)

(No trivial annulus region.) $\forall i,j\in I$, there is no pair of circle
components of $D_{i}\cap D_{j}$ such that they bound annuli in both $D_{i}$
and $D_{j}$ and neither annulus intersects $L$.

We point out that, given the credible disks $D_{i}(i\in I)$, one can always
modify $\cup_{i\in I}D_{i}\cup L$ to meet the above regularity conditions. In
fact, general differential topology guarantees condition (Ri). For condition
(Rii), since $D_{j}$ is also credible, if such $C$ exists, then the disk
bounded in $D_{j}$ by $C$ also does not intersect $L$. Suppose $C$ bounds
$D_{Ci}$ and $D_{Cj}$ in $D_{i}$ and $D_{j}$ respectively. Interchanging
$D_{Ci}$ by $D_{Cj}$ and then smoothing the corners near $C$, we can eliminate
$C$. The trouble is that the new $D_{i}$ and $D_{j}$ may have self-
intersection. Notice that the singularities can only be double points. By
changing the immersion map and smoothing the corners, as in the second
paragraph of the proof of Lemma 3.8(2), we get embedded ones. For condition
(Riii), if such pair of intersection circles exists, say $C_{01}\cup C_{02}$,
which bounds $A_{i}$ in $D_{i}$ and bounds $A_{j}$ in $D_{j}$. Interchanging
$A_{j}$ and $A_{i}$ and then smoothing the corners, we can eliminate
$C_{1}\cup C_{2}$. The new $D_{i}$ and $D_{j}$ may have self-intersection, but
we can get embedded new $D_{i}$ and $D_{j}$ from them as above.

We conclude this section with a definition used in the statements of our
theorems.

######  Definition 3.13.

Let $C_{i}$ be a component of a link $L$ and $D_{i}$ be a credible disk
bounded by $C_{i}$. For a natural number $N$, we say $D_{i}$ is _(s $N$)_ if
either

  * (i)

$\sharp(D_{i}\cap(L-C_{i}))<N$; or

  * (ii)

$D_{i}$ is stable.

##  4\. Criterion for Untiedness

In this section we give a sufficient and necessary criterion to detect
untiedness for Brunnian links and illustrate our method by examples.

###  4.1. Decision Theorem and Example analysis

######  Theorem 4.1.

(Untiedness Criterion)

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link and
$k\in\\{1,...,n\\}$. For each $i\leq k$, let $D_{i}$ be a disk bounded by
$C_{i}$ so that $U=\bigcup_{i=1}^{k}D_{i}\cup L$ is a spanning complex.
Suppose for every $i\leq k$,

$\left\\{\begin{array}[]{ll}D_{i}\ is\ (s7),&if\ n=2\ and\
lk(C_{1},C_{2})=1;\\\ D_{i}\ is\ (s8),&in\ other\ cases.\end{array}\right.$

Then $L$ is untied if and only if there is no incompressible knotted torus in
the complement of $U$.

The proof of this theorem will be given at the end of this section.

Recall that a spanning complex is always available. The “if” implication of
this theorem gives a method to show a Brunnian link is untied. The “only if”
implication indicates this method is theoretically universal. Furthermore, if
a Brunnian link is an s-tie, we can always find a companion torus by choosing
a spanning complex $U$ satisfying the condition in Theorem 4.1, since an
incompressible knotted torus in $S^{3}-U$ is a desired companion.

Let us illustrate our method with an example.

######  Example 4.2.

Figure 11. Brunn’s chain.

Consider Brunn’s chain in Fig. 11(1). Take the disks as shown in Fig. 11(2),
which are all credible as demonstrated in [5]. The union of these disks forms
a spanning complex, say $U$. Since $S^{3}-intN(U)$ is a handlebody, where
$N(U)$ is a regular neighborhood of $U$, by Theorem 4.1, Brunn’s chains are
untied.

From this example, we can see that this method is intuitive and efficient. It
works conveniently to show all the Brunnian links in literatures, except
Fountains and Jade-pendant in [5], are untied.

###  4.2. Proof of Theorem 4.1

The “if” implication:

Suppose that $L$ is an s-tie. Let $T$ be an essential torus in $S^{3}-L$,
bounding a solid torus $V$ in $S^{3}$ containing $L$. We may assume $T$ is an
_outermost_ essential torus. More formally, there is no essential torus
$T_{0}$ in $S^{3}-L$ so that

  * (0i)

$T_{0}$ bounds a solid torus containing $V$,

  * (0ii)

$T_{0}$ is not parallel to $T$.

The intersection $\cup_{i=1}^{k}D_{i}\cap T$ is a union of circles, forming a
graph in $T$. We wish to make $D_{i}$ into $V$ one by one by an isotopy of
$T$. First consider $D_{1}$. We do this in three steps.

Step 1. Eliminate inessential circles in $T$.

Let $C$ be a circle component of $D_{1}\cap T$, bounding a disk $D_{CT}$ in
$T$ and a disk $D_{C1}$ in $D_{1}$. Since $D_{1}$ is credible, $D_{C1}$ does
not intersect $L$. Thus we have the following two observations:

  * (i)

Any circle component of $D_{1}\cap T$ contained in $D_{C1}$ is inessential in
$T$. This is because $T$ is essential.

  * (ii)

For any $i>1$, there is no circle component of $D_{1}\cap D_{i}$ contained in
$D_{C1}$. This is by regularity condition (Rii).

By (i), we may assume $C$ is innermost in $D_{1}$ as a circle component of
$D_{1}\cap T$. Then $D_{C1}\cup D_{CT}$ bounds a 3-ball, say $B$. Pushing
$D_{CT}$ across $B$ a little beyond $D_{C1}$, we eliminate $C$. Do it
inductively. After finitely many steps, $D_{1}\cap T$ contains no inessential
circles in $T$.

To guarantee the validity of making $D_{i}(i>1)$ into $V$ later, we will need
the following two more observations:

  * (iii)

The singularities of $U$ contained in $D_{C1}$ consists of proper arcs. This
follows from that $D_{C1}$ does not intersect $L$ and (ii).

  * (iv)

For any $i>1$, $D_{i}\cap B$ is a proper surface, and each boundary component
of it is either contained in $D_{CT}$ or is an alternating sequence of
properly embedded arcs in $D_{CT}$ and in $D_{C1}$. This follows from (iii).

See Fig. 12 for an example.

Figure 12. The red circles in $D_{CT}$ are intersections with $D_{1}$. The
black graph is intersection with $D_{i}(i>1)$.

Step 2. Eliminate exterior annulus regions in $D_{1}$.

Consider a component of $D_{1}\cap T$ innermost in $D_{1}$. Since it bounds a
disk in $D_{1}$, it must be the meridian of $V$. Thus $D_{1}\cap T$ is a union
of meridians of $V$, cutting $D_{1}$ into regions alternatively in $intV$ and
in $S^{3}-V$. We call a component of $D_{1}-T$ in $intV$ an _interior region_
in $D_{1}$, and call a component of $D_{1}-T$ in $S^{3}-V$ an _exterior
region_ in $D_{1}$. Clearly every exterior region does not intersect $L$.

We claim that if annulus $A$ is an exterior region in $D_{1}$, then $A$ is
$\partial$-parallel to $T$. In fact, if otherwise, then $\partial A$ would cut
$T$ into two annuli $A_{1}$ and $A_{2}$ such that both $A\cup A_{1}$ and
$A\cup A_{2}$ bound knot complements in $S^{3}-intV$. This implies that the
core of $V$ is a connected sum of two knots, contradicting that $T$ is
outermost.

So we can isotope $T$ to make $A$ into $V$. Step by step, we can eliminate all
the exterior annulus regions in $D_{1}$.

Step 3. Verify $D_{1}\subset V$.

Consider $D_{1}$ cut by $D_{1}\cap T$. We have the following three
observations on the regions in $D_{1}$:

  * (v)

The outermost region and all of the innermost regions are interior, in view of
$C_{1}\subset V$ and that every intersection circle is a meridian of $V$.

  * (vi)

If the outermost region is an annulus, it must intersects $L-C_{1}$. In fact,
if not, $C_{1}$ would be parallel to a meridian of $T$. However, by Brunnian
property, $L-C_{1}$ is trivial in $V$. Then $L$ would also be trivial in $V$,
a contradiction.

  * (vii)

For any exterior region $\Omega$, $\partial\Omega$ should be one outer circle
enclosing odd number($\geq 3$) of inner circles in $D_{1}$, and the disk
bounded by each inner circle intersects $L$ in at least $2$ points. In fact,
since $\partial\Omega$ is null-homologic in $S^{3}-intV$, $\partial\Omega$ has
even number of components, and by Step 2, it can not have only 2 components.

We now show by cases that $D_{1}\subset V$.

Case 1: $D_{1}$ is stable. Suppose $D_{1}$ has exterior regions. Then the
components of $D_{1}\cap T$ can not be all innermost in $D_{1}$. By (v), each
innermost region in $D_{1}$ is a meridian disk of $V$. There exist such a disk
region, say $D_{u1}$, and a component of $D_{1}\cap T$ adjacent to $\partial
D_{u1}$ in $T$, say $C_{u+1,1}$, such that

  * (1i)

$\sharp(D_{u1}\cap L)$ is minimal among all the innermost regions,

  * (1ii)

$C_{u+1,1}$ is either not innermost in $D_{1}$, or innermost but not achieving
minimal intersection points with $L$.

Let $A_{T}$ be the annulus in $T$ between $C_{u+1,1}$ and $\partial D_{u1}$.
Then by (vii), the disk $D_{u1}\cup A_{T}$ has less intersection points with
$L$ than the disk in $D_{1}$ bounded by $C_{u+1,1}$. So $C_{u+1,1}$ is an
unstable circle on $D_{1}$, a contradiction.

Case 2: $\sharp(D_{1}\cap L)<7$, if $n=2$ and $lk(C_{1},C_{2})=1$;
$\sharp(D_{1}\cap L)<8$, in other cases. Suppose $D_{1}$ has exterior regions.
By (vi) and (vii), $D_{1}$ intersects $L-C_{1}$ in more than 6 points. If the
outermost region intersects another component, say $C_{2}$, in only one point,
then $lk(C_{1},C_{2})=1$ and $n=2$. As shown in Fig 13(1), we have
$\sharp(D_{1}\cap L)\geq 7$. Otherwise, $\sharp(D_{1}\cap L)\geq 8$, as shown
in Fig. 13(2). So under this condition, $D_{1}\subset V$.

Figure 13.

Now we follow the three steps above to make $D_{2}$ into $V$ by an isotopy of
$T$. The key point is that this isotopy of $T$ will keep $D_{1}\cap
T=\emptyset$.

Consider $D_{2}\cap T$. For Step 1, by the same token, choose a circle $C$
innermost in $D_{2}$, which bounds a disk $D_{CT}$ in $T$ and a disk $D_{C2}$
in $D_{2}$, and then push $D_{CT}$ across the 3-ball cobounded with $D_{C2}$
to eliminate $C$. We have two similar observations for $D_{2}$ as clauses
(iii) and (iv) for $D_{1}$. Notice that $D_{CT}$ and $D_{1}$ do not intersect.
Thus $D_{C2}$ also does not intersect $D_{1}$. Therefore the isotopy of
$D_{CT}$, replacing $D_{CT}$ by $D_{C2}$, keeps $D_{1}\cap T=\emptyset$. So we
can eliminate all the inessential circles in $T$ in finitely many steps. For
Step 2, we can eliminate exterior annulus regions in $D_{2}$ just as before
because $D_{1}$ is already contained in $V$. For Step 3, the same argument as
demonstrated before shows that $D_{2}\subset V$.

Performing the same procedure for $D_{i}$ for $i$ from $2$ to $k$, we have
$U\subset V$. Since $T$ is incompressible in $S^{3}-L$, clearly $T$ is
incompressible in $S^{3}-U$.

The “only if” implication:

Suppose $T$ is an incompressible torus in the complement of $U$. Let $V$ be
the solid torus bounded by $T$ containing $U$. We proof by negation that $L$
is an s-tie. If $L$ is untied, then $T$ is compressible in the complement of
$L$ and thus there is a meridian disk $D$ of $V$ not intersecting $L$. Notice
that any meridian disk of $V$ intersects $U$. It follows that for any $i\in
I$, $D\cap D_{i}$ is a disjoint union of circles. The intersection
$\bigcup_{i\in I}D_{i}\cap D$ forms a graph.

Let $C$ be a component of $D_{1}\cap D$ innermost in $D_{1}$, bounding
$D_{C1}$ in $D_{1}$ and bounding $D_{C}$ in $D$. Since $D_{1}$ is credible,
$D_{C1}$ does not intersect $L$. By (Rii), the singularities of $U$ contained
in $D_{C1}$ consists of proper arcs. Replacing $D_{C}$ in $D$ by $D_{C1}$ and
pushing it a little beyond $D_{C1}$, we get a new disk $D$ with less
intersection components with $D_{1}$. After finitely many steps, $D$ is
disjoint from $D_{1}$.

A careful examination shows that the same procedure for $D_{2}$ keeps
$D_{1}\cap D=\emptyset$. Step by step for each $D_{i}$, we eventually get a
meridian disk of $V$ not intersecting $U$, a contradiction.

##  5\. Simple intersection pattern Theorem

To show a Brunnian link is s-prime is to prove no essential torus splits it
into two sublinks. We divide our strategy in two steps. In this section, as
the first step, we give a general intermediate result, stating that a spanning
complex satisfying certain conditions can intersect a splitting torus in a
simple form. In the next section, as the second step, we give criteria to
negate the existence of such torus.

###  5.1. Simple intersection pattern

We first define several disks with respect to a partition of a link.

######  Definition 5.1.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component link. Let $L_{I}=\cup_{i\in
I}C_{i}$ and $L_{J}=\cup_{j\in J}C_{i}$ be proper sublinks of $L$ with
$I\sqcup J=\\{1,...,n\\}$. For any $i\in I$ and $j\in J$, the credible disks
defined as follows are called _$L_{I}$ -interior disk_, _$L_{J}$ -interior
disk_, _exterior cross disk_ and _free cross disk_ respectively, denoted
$D_{i}^{I}$, $D_{j}^{J}$, $D_{i}^{E}$ and $D_{i}^{F}$ respectively.

  * $D_{i}^{I}$

: bounded by $C_{i}$ such that $D_{i}^{I}\cap L_{J}=\emptyset$;

  * $D_{j}^{J}$

: bounded by $C_{j}$ such that $D_{j}^{J}\cap L_{I}=\emptyset$;

  * $D_{i}^{E}$

: bounded by $C_{i}$ such that $D_{i}^{E}\cap(L_{i}-C_{i})=\emptyset$ and
$D_{i}^{F}\cap L_{J}\neq\emptyset$;

  * $D_{i}^{F}$

: bounded by $C_{i}$ such that $D_{i}^{F}\cap(L_{i}-C_{i})\neq\emptyset$ and
$D_{i}^{F}\cap L_{J}\neq\emptyset$.

$L_{I}$-interior disk and $L_{J}$-interior disk are _interior disks_ , and
exterior cross disk and free cross disk are _cross disks_.

Consider a torus $T$ splitting $L$ as $L_{I}$ and $L_{J}$ and the intersection
of $T$ and the disks defined above.

######  Definition 5.2.

An interior disk intersects $T$ in _simple intersection pattern_ if it does
not intersect $T$, and a cross disk intersects $T$ in _simple intersection
pattern_ if each intersection circle is innermost in the disk and is a
meridian of $V_{J}$.

######  Remark 5.3.

Let $\beta_{I}$ and $\beta_{J}$ be the core of $V_{I}$ and $V_{J}$
respectively. Consider a cross disk $D_{i}$ intersecting $T$ in simple
intersection pattern.

(1) Recall that cross disks are credible. It is easy to see that each inner
disk in it is a credible disk for $\beta_{I}\cup L_{J}$, and the outer region
in it corresponds to a credible disk for $L_{I}\cup\beta_{J}$.

(2) If $D_{i}$ is stable, consider $\beta_{I}\cup L_{J}$. It can be verified
that each inner disk in $D_{i}$ is a stable disk bounded by $\beta_{I}$, and
thus all the inner disks have the same number of intersection points with
$L_{J}$.

Now we turn to Brunnian links.

######  Proposition 5.4.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link. Let
$L_{I}=\cup_{i\in I}C_{i}$ and $L_{J}=\cup_{j\in J}C_{i}$ be proper sublinks
of $L$ with $I\sqcup J=\\{1,...,n\\}$.

  * (1)

For any exterior cross disk $D_{i}^{E}$, we have
$\sharp(D_{i}^{E}\cap(L-C_{i}))\geq 4$.

  * (2)

For any free cross disk $D_{i}^{F}$, we have
$\sharp(D_{i}^{F}\cap(L-C_{i}))\geq 6$.

######  Remark 5.5.

If $\sharp(D_{i}^{E}\cap(L-C_{i}))=4$, then $D_{i}^{E}$ is stable for $L_{J}$.

The proofs of this proposition and the following main theorem will be given at
the end of this section.

######  Theorem 5.6.

(Simple Intersection Pattern Theorem)

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link. Let $T$ be an
essential torus in the complement of $L$, splitting $S^{3}$ into two solid
tori $V_{I}$ and $V_{J}$ so that $L\cap V_{I}=L_{I}$ and $L\cap V_{J}=L_{J}$,
where $L_{I}=\cup_{i\in I}C_{i}$ and $L_{J}=\cup_{j\in J}C_{i}$. Suppose
$Q\sqcup R\sqcup S\subset I$ and $J_{0}\subset J$. Let $D_{q}^{I}$($q\in Q$),
$D_{j}^{J}$($j\in J_{0}$), $D_{r}^{E}$($r\in R$) and $D_{s}^{F}$($s\in S$) be
$L_{I}$-interior disks, $L_{J}$-interior disks, exterior cross disks, and free
cross disks respectively, so that the cross disks are mutually disjoint and

$\left\\{\begin{array}[]{ll}D_{q}^{I}\ is\ (s8),&\forall q\in Q;\\\ D_{j}^{J}\
is\ (s8),&\forall j\in J_{0};\\\ D_{r}^{E}\ is\ (s10),&\forall r\in R;\\\
D_{s}^{F}\ is\ stable,\ or\ contains\ a\ longitude\ of\ V_{I}\ with\
\sharp(D_{s}^{F}\cap(L-C_{s}))<8,&\forall s\in S.\end{array}\right.$

If the union of these disks and $L$ forms a spanning complex, then after an
isotopy of $T$, every disk intersects $T$ in simple intersection pattern.

######  Remark 5.7.

We explain the conditions in this theorem in more detail.

(1) The cross disks $D_{r}^{E}$($r\in R$), $D_{s}^{F}$($s\in S$) are required
to be mutually disjoint. As will be seen in the proof, $\cup_{q\in
Q}D_{q}^{I}$ and $\cup_{j\in J_{0}}D_{j}^{J}$ are actually disjoint.

(2) In (iv), the condition “$D_{s}^{F}$ contains a longitude of $V_{I}$” seems
not natural. But actually once $R\neq\emptyset$, $D_{s}^{F}$ has to contain a
longitude of $V_{I}$, since the cross disks are mutually disjoint.

###  5.2. Proof of the main Theorem

We first present a simple lemma on s-sum decomposition of Brunnian links.

######  Lemma 5.8.

[4] Let $T$ be an essential torus in the complement of a Brunnian link $L$,
splitting $L$ into proper sublinks $L_{1}\subset V_{1}$ and $L_{2}\subset
V_{2}$, where $V_{i}$ is a solid torus bounded by $T$, for $i=1,2$.

(1) Any meridian disk in $V_{i}$ intersects $L_{i}$ with at least 2 points,
for $i=1,2$;

(2) If a component of $L_{1}$ bounds a disk whose intersection with $T$
contains a meridian of $V_{2}$, then the intersection contains at least two
meridians of $V_{2}$.

###### Proof of Theorem 5.6.

Let $U$ denote the spanning complex formed by $L$ and all the credible disks.
Then $U\cap T$ is a union of circles, forming a graph in $T$. We will isotope
$T$ to make the $L_{I}$-interior disks, the $L_{J}$-interior disks and the
cross disks into simple intersection pattern in turn. So we divide our proof
in three steps.

Step 1. Make $L_{I}$-interior disks into $V_{I}$ one by one by an isotopy of
$T$.

We shall adopt the same procedure as in the proof of Theorem 4.1. Consider a
disk $D_{q}^{I}$. We make it into $V_{I}$ in three steps.

Step 1.1. Eliminate inessential circles in $T$.

Let $C$ be a circle component of $D_{q}^{I}\cap T$, bounding a disk $D_{CT}$
in $T$ and a disk $D_{Cq}$ in $D_{q}^{I}$. Using the same argument as in the
proof of Theorem 4.1, we may assume $C$ is innermost in $D_{Cq}$. Then
$D_{Cq}\cup D_{CT}$ bounds a 3-ball. Since no sphere splits a Brunnian link,
this 3-ball does not intersect $L$. So we can eliminate $C$ by an isotope of
$T$ as demonstrated before. Do it inductively. After finitely many steps,
$D_{q}^{I}\cap T$ contains no inessential circle in $T$.

Step 1.2. Eliminate annulus regions in $V_{J}$.

Now $D_{q}^{I}\cap T$ is a disjoint union of circles. Consider a circle
component innermost in $D_{q}^{I}$. Since the disk in $D_{q}^{I}$ bounded by
this circle does not intersect $L_{J}$, it is the meridian disk of $V_{I}$.
Thus all the components of $D_{q}^{I}\cap T$ are meridian of $V_{I}$.

The intersection $D_{q}^{I}\cap T$ cuts $D_{q}^{I}$ into regions alternatively
in $V_{I}$ and in $V_{J}$. Let $A$ be an annulus region contained in $V_{J}$.
Since the boundary components of $A$ are longitude of $V_{J}$, it follows that
$A$ must cut $V_{J}$ into two solid tori. Notice that the core of each solid
tori and the core of $V_{I}$ form a Hopf link, which is Brunnian. By Theorem
2.3, one of the two solid tori does not intersect $L_{J}$. Isotope $T$ across
this solid torus to make $A$ into $V_{I}$. Step by step, we can eliminate all
the annulus regions in $V_{J}$.

Step 1.3. Verify $D_{q}^{I}\subset V_{I}$.

For either the case that $D_{q}^{I}$ is stable or the case
$\sharp(D_{q}^{I}\cap L_{J})<8$, an argument similar to the one used in Step 3
in the proof of Theorem 4.1 shows that $D_{q}^{I}\subset V_{I}$.

Next we isotope $T$ to make another $L_{1}$-interior disk $D_{q^{\prime}}^{I}$
into $V_{I}$ by the same token. We now check that the isotopy of $T$ keeps
$D_{q}^{I}\cap T=\emptyset$. For Step 1.1, similarly as in the proof of
Theorem 4.1, this is due to the regularity condition (Rii) and that
$D_{q}^{I}$ does not intersect any disk in $T$. For Step 1.2, it is guaranteed
by that $D_{q}^{I}$ is already contained in $V_{I}$.

Performing the same procedure for $L_{I}$-interior disks one by one, we have
that $\cup_{q\in Q}D_{q}^{I}\subset V_{I}$.

Step 2. Make $L_{J}$-interior disks into $V_{J}$ one by one by an isotopy of
$T$.

With $\cup_{q\in Q}D_{q}^{I}$ already contained in $V_{I}$, we proceed as in
Step 1 to make $L_{J}$-interior disks into $V_{J}$. Consider a disk
$D_{j}^{J}$. We only need to check that in the similar three steps, the
isotopy of $T$ keeps $D_{q}^{I}\cap T=\emptyset$ for all $q\in Q$.

Step 2.1. Eliminate inessential circles in $T$.

In this step, the disk $D_{Cj}$ contained in $D_{j}^{J}$, corresponding to
$D_{Cq}$ in Step 1.1, does not intersect any $L_{I}$-interior disk by (Rii).
Thus the 3-ball it caps with $T$ does not intersect any $L_{I}$-interior disk.

Step 2.2. Eliminate annulus regions in $V_{I}$.

In such a case, $D_{j}^{J}\cap L_{I}=\emptyset$ implies that $D_{j}^{J}\cap T$
can only be disjoint union of meridians of $V_{J}$. The intersection circles
cut $D_{j}^{J}$ into regions alternatively in $V_{I}$ and in $V_{J}$. Let $A$
be an annulus region contained in $V_{I}$. An argument similar to the one used
in Step 1.2 shows that $A$ caps a solid torus with $T$, say $V_{A}$, which
does not intersect $L_{I}$.

We claim that $A\cap D_{q}^{I}=\emptyset$, and thus $V_{A}\cap
D_{q}^{I}=\emptyset$, for all $q\in Q$. In fact, by (Rii), there is no
component of $D_{q}^{I}\cap A$ inessential in $A$ for any $q\in Q$. Suppose
$C_{A}$ is a component of $D_{q}^{I}\cap A$ essential in $A$ for a $q\in Q$.
Since $C_{A}$ is a core of $V_{I}$, the disk in $D_{q}^{I}$ bounded by $C_{A}$
must intersect $L_{J}$, a contradiction.

An isotopy of $T$ across $V_{A}$ makes $A$ into $V_{J}$.

Step 2.3. Verify $D_{j}^{J}\subset V_{J}$.

This step is a judgement, where everything is fixed.

Perform the same procedure for other $L_{J}$-interior disks one by one. We can
verify that $T$ keeps avoiding $\cup_{q\in Q}D_{q}^{I}$ and the
$L_{J}$-interior disks already contained in $V_{J}$ by the same arguments as
in Step 1 and Step 2 respectively. Now we have $\cup_{j\in
J_{0}}D_{j}^{J}\subset V_{J}$.

Step 3. Make the cross disks into simple intersection pattern.

Recall that all of the cross disks are mutually disjoint, so it is convenient
to consider them together. We proceed as in the proof of Theorem 4.1.

Step 3.1. Eliminate inessential circles in $T$.

The intersection of the cross disks and $T$ consists of mutually disjoint
circles. Let $D_{i}$ be one of the exterior or free cross disks. Let $C$ be a
circle component of $D_{i}\cap T$, bounding a disk $D_{CT}$ in $T$ and a disk
$D_{Ci}$ in $D_{i}$. Using the same argument as in Step 1 in the proof of
Theorem 4.1, we may assume $C$ is innermost in $D_{Ci}$. Then $D_{Ci}\cup
D_{CT}$ bounds a 3-ball. As demonstrated in Step 1.1, this 3-ball does not
intersect $L$, and thus we can eliminate $C$ by an isotope of $T$ .

Step by step for all cross disks, we eliminate all inessential intersection
circles in $T$.

Step 3.2. Eliminate annulus regions $\partial$-parallel to $T$.

It is easy to see that each component of $(\cup_{r\in
R}D_{r}^{E}\cup\cup_{s\in S}D_{s}^{F})\cap T$ is either the longitude of
$V_{I}$ or the meridian of $V_{I}$. Let $D_{i}$ be one of the exterior or free
cross disks. The intersection $D_{i}\cap T$ cuts $D_{i}$ into planar regions
alternatively in $V_{I}$ and in $V_{J}$. In this step, we eliminate all the
proper annulus regions $\partial$-parallel to $T$ without intersecting $L$.
Strictly speaking, let $A$ be a proper annulus region contained in $V_{I}$
which caps a solid torus with $T$, say $V_{A}$, such that $V_{A}\cap
L=\emptyset$. We claim that $A\cap D_{q}^{I}=\emptyset$, and thus $V_{A}\cap
D_{q}^{I}=\emptyset$, for any $q\in Q$. Then an isotopy of $T$ across $V_{A}$
makes $A$ into $V_{J}$.

We now prove the claim. By (Rii), there is no component of $D_{q}^{I}\cap A$
inessential in $A$ for any $q\in Q$. Suppose $C_{A}$ is a component of
$D_{q}^{I}\cap A$ essential in $A$ for a $q\in Q$. Let $P$ be a component of
$D_{q}^{I}\cap V_{A}$ one of whose boundary components is $C_{A}$. Clearly the
boundary components of $P$ are all parallel in $A$. By (Riii), $P$ is not an
annulus. Then $P$ is compressible in $V_{A}$. Let $D_{P1}$ be a compression
disk for $P$ in $V_{A}$. Since $D_{q}^{I}$ is credible, the disk in
$D_{q}^{I}$ bounded by $\partial D_{P1}$, denoted $D_{P0}$, does not intersect
$L$. Without loss of generality, we may reselect $C_{A}$ so that it is
contained in $D_{P0}$. However, since $C_{A}$ is essential in $A$, it is
parallel to either the longitude or the meridian of $V_{I}$. In either case,
any disk bounded by $C_{A}$ intersects $L$, a contradiction.

We can eliminate such annulus regions contained in $V_{J}$ in a similar
manner.

Step 3.3. Verify simple intersection pattern.

Consider all of the cross disks cut by $T$. For both exterior and free cross
disks, the following claim investigates the remaining annulus regions. To make
length short, we arrange the proof of this claim after the end of the proof of
the theorem. Claim: After Step 3.2,

  * (Ai)

any proper annulus region in $V_{I}$ intersects $L_{I}$, and intersects any
component of $L_{I}$ in even number of points;

  * (Aii)

any annulus region in $V_{J}$ intersects $L_{J}$, and intersects any component
of $L_{J}$ in even number of points.

In the remainder of the proof, we discuss exterior cross disks and free cross
disks separately.

Exterior cross disks. For any $r\in R$, the intersection $D_{r}^{E}\cap T$
cuts $D_{r}^{E}$ into regions alternatively in $V_{I}$ and $V_{J}$. Since
$D_{r}^{E}$ is an exterior cross disk, the intersection circles are longitudes
of $V_{I}$. We have the following observations on the regions in $D_{r}^{E}$.

  * (Ei)

Each innermost region is a meridian disk of $V_{J}$. In view of linking
number, each innermost disk intersects any components of $L_{J}$ in even
number of points unless $L_{J}$ has only one component.

  * (Eii)

The outermost region is contained in $V_{I}$. It is not an annulus by Lemma
5.8. Its boundary consists of $C_{s}$ and even number of intersection circles
unless $L_{I}=C_{r}$, in view of linking number.

  * (Eiii)

No region in $V_{I}$ intersects $L-C_{r}$. For each region $\Omega$ in
$V_{I}$, except the outermost one, $\partial\Omega$ is one outer circle
enclosing odd number($\geq 3$) of inner circles in $D_{r}^{E}$. In fact, since
$\partial\Omega$ is null-homologic in $V_{I}$, such a region has even
number($>2$) of boundary components.

Now we show by cases that $D_{r}^{E}$ is of simple intersection pattern.

Case 1: $D_{r}^{E}$ is stable. Let $D_{ur}$ be an innermost region in
$D_{r}^{E}$ so that $\sharp(D_{ur}\cap L_{J})$ is minimal among all the
innermost regions in $D_{r}^{E}$. Let $C_{u+1,r}$ be a component of
$D_{r}^{E}\cap T$ adjacent to $\partial D_{ur}$ in $T$, and $A_{T}$ be the
annulus in $T$ between $\partial D_{ur}$ and $C_{u+1,r}$. Then the disk
$D_{ur}\cup A_{T}$ has the same intersection points with $L$ as $D_{ur}$.
Since $D_{r}^{E}$ is stable, the disk bounded in $D_{r}^{E}$ by $C_{u+1,r}$
must be innermost and have the same number of intersection points as $D_{ur}$.
It follows that all the intersection circles are innermost in $D_{r}^{E}$.

Case 2: $\sharp(D_{r}^{E}\cap(L-C_{r}))<10$. Suppose there is an intersection
circle in $D_{r}^{E}$ which is not innermost. By (Aii), (Ei), (Eii) and
(Eiii), $D_{r}^{E}$ intersects $L-C_{r}$ in at least 10 points, as shown in
Fig. 14.

Figure 14. White regions are in $V_{I}$, and gray regions are in $V_{J}$.

Free cross disks. For any $s\in F$, the intersection $D_{s}^{F}\cap T$ cuts
$D_{s}^{F}$ into regions alternatively in $V_{I}$ and $V_{J}$. The
intersection circles can be either longitudes of $V_{1}$ or meridians of
$V_{1}$. We discuss the two possibilities separately.

First suppose that the circles are longitude of $V_{I}$. We have the following
observations on the regions in $D_{s}^{F}$.

  * (Fi)

Each innermost region is a meridian disk of $V_{J}$, and it intersects any
components of $L_{J}$ in even number of points unless $L_{J}$ has only one
component.

  * (Fii)

The outermost region is contained in $V_{I}$. It is not an annulus by Lemma
5.8. Its boundary consists of $C_{s}$ and even number of intersection circles.

  * (Fiii)

For each region $\Omega$ in $V_{I}$, except the outermost one,
$\partial\Omega$ is one outer circle enclosing odd number of inner circles in
$D_{s}^{F}$.

Now suppose that the intersection circles are meridian of $V_{I}$. We have the
following observations on the regions in $D_{s}^{F}$.

  * (F’i)

The outermost region and all of the innermost regions are contained in
$V_{I}$. Each innermost region is a meridian disk of $V_{I}$.

  * (F’ii)

If the outermost region is an annulus, it intersects $L_{I}-C_{s}$, and thus
it intersects any component of $L_{I}-C_{s}$ in even number of points, in view
of linking number. In fact, if not, $C_{s}$ would be parallel to a meridian of
$V_{I}$. By Brunnian property, $L_{I}-C_{s}$ is trivial in $V_{I}$. This would
implies $L_{I}$ is trivial in $V_{I}$, a contradiction.

  * (F’iii)

For any region $\Omega$ in $V_{J}$, $\partial\Omega$ is one outer circle
enclosing odd number of inner circles in $D_{s}^{F}$, and each disk in
$D_{s}^{F}$ bounded by one of such inner circle intersects $L_{I}$.

We now show by cases that $D_{s}^{F}$ is of simple intersection pattern.

Case 1: $D_{s}^{F}$ is stable. Suppose that the intersection circles are
meridian of $V_{I}$. By (F’i), the components of $D_{s}^{F}\cap T$ can not be
all innermost in $D_{s}^{F}$. Notice that each innermost region is a meridian
disk of $V_{I}$. There exist an innermost region in $D_{s}^{F}$, say $D_{us}$,
and a component of $D_{1}\cap T$ adjacent to $\partial D_{us}$ in $T$, say
$C_{u+1,s}$, such that

  * (1i)

$\sharp(D_{us}\cap L_{I})$ is minimal among all the innermost regions,

  * (1ii)

$C_{u+1,s}$ is either not innermost in $D_{s}^{F}$, or innermost but not
achieving minimal intersection points with $L_{I}$.

Let $A_{T}$ be the annulus in $T$ between $C_{u+1,s}$ and $\partial D_{us}$.
Then by (F’iii) and (Aii), the disk $D_{us}\cup A_{T}$ has less intersection
points with $L$ than the disk in $D_{s}^{F}$ bounded by $C_{u+1,s}$. This
implies that $C_{u+1,s}$ is an unstable circle on $D_{s}^{F}$, a
contradiction.

Hence $D_{s}^{F}\cap T$ consists of longitudes of $V_{I}$. The remainder of
the argument is analogous to that in Case 1 for exterior cross disks and is
left to the reader.

Case 2: $\sharp(D_{s}^{F}\cap(L-C_{s}))<8$ and $D_{s}^{F}$ contains a
longitude of $V_{I}$. All of the intersection circles are longitude of
$V_{I}$. Suppose there is a circle in $D_{s}^{F}$ which is not innermost. By
Lemma 5.8(1), every meridian disk of $V_{J}$ intersects $L_{J}$ in at least 2
points. Then by (Fi), (Fii), (Fiii), (Ai) and (Aii), $D_{s}^{F}$ intersects
$L-C_{s}$ in at least 8 points, as shown in Fig. 15.

Figure 15. White regions are in $V_{I}$, and gray regions are in $V_{J}$.

∎

###### Proof of the claim.

We only prove (Ai), and (Aii) follows in a similar manner by interchanging
$V_{I}$ and $V_{J}$ in the proof below.

Let $A$ be a proper annulus region in $V_{I}$. Assume for contrary that $A\cap
L_{I}=\emptyset$. First suppose that all the intersection circles are
longitude of $V_{I}$. Then $A$ must cut $V_{I}$ into two solid tori. Since the
core of each solid tori and the core of $V_{J}$ form a Hopf link, by Theorem
2.3, one of the two solid tori does not intersect $L_{I}$. However, we have
eliminated such $A$ in Step 3.2, a contradiction.

Now suppose that all the intersection circles are meridian of $V_{I}$. If $A$
is $\partial$-parallel in $V_{I}$, then it cuts off a compression solid torus
$V_{A}$ in $V_{I}$. If $V_{A}$ contains some components of $L_{I}$, by
Proposition 3.2.1 in [4], the proper sublink in the closure of $V_{I}-V_{A}$
would not be geometrically essential. A compression disk in this solid torus,
after isotopy, would be a meridian disk in $V_{I}$. However, again by
Proposition 3.2.1 in [4], $L_{I}$ is geometrically essential in $V_{I}$, a
contradiction. So $V_{A}$ does not intersect $L$. Recall that we have
eliminated such $A$ in Step 3.2. Hence $A$ cannot be $\partial$-parallel in
$V_{I}$. Then $\partial A$ cuts $T$ into two annuli such that $A$ and one of
the annuli cobound a knotted solid torus $V_{AI}$ in $V_{I}$. By Theorem 2.4,
the sublink in $V_{AI}$ is not geometrically essential. A compression disk in
$V_{AI}$, after isotopy, would be a meridian disk in $V_{I}$, again a
contradiction.

We have shown that $A$ intersects $L_{I}$. Since $A$ separates $V_{I}$, it
intersects any component of $L_{I}$ in even number of points. ∎

###### Proof of Proposition 5.4.

Let $T$ be an essential torus in the complement of $L$, splitting $S^{3}$ into
two solid tori $V_{I}$ and $V_{J}$ so that $L\cap V_{I}=L_{I}$ and $L\cap
V_{J}=L_{J}$.

(1) Applying Theorem 5.6, we may assume $D_{i}^{E}$ intersects $T$ in simple
intersection pattern. By Lemma 5.8, any meridian disk of $V_{I}$ has more than
one intersection points, and the outermost region has more than two boundary
components. So if $\sharp(D_{i}^{E}\cap(L-C_{i}))\leq 4$ , the only
possibility is as shown in Fig. 16(1).

(2) First suppose all the intersection circles in $D_{i}^{F}$ are longitude of
$V_{I}$. Applying Theorem 5.6, we may assume $D_{i}^{F}$ is of simple
intersection pattern. By Lemma 5.8(1) and (Fii) in the proof of Theorem 5.6,
there are at least two inner disks each intersecting $L_{J}$ in at least 2
points. In view of linking number, the outer region intersects $L_{I}-C_{i}$
in even number of points. So $D_{i}^{F}$ with least number of intersection is
as shown in Fig. 16(2).

Now suppose that all the intersection circles in $D_{i}^{F}$ are meridian of
$V_{I}$. By Lemma 5.8(1), and (Aii) and (Fii) in the proof of Theorem 5.6, if
$\sharp(D_{i}^{F}\cap(L-C_{i}))\leq 6$, the intersections in $D_{i}^{F}$ can
only be as shown in Fig. 16(3).

Figure 16. White regions are in $V_{I}$, and gray regions are in $V_{J}$.

∎

##  6\. Criteria for S-primeness

Based on the theorem in the previous section, we present four methods to prove
a Brunnian link is s-prime and illustrate them by typical examples. In this
section, we call the following paragraph _Standard premise_ :

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link. Let $T$ be an
essential torus in the complement of $L$, splitting $S^{3}$ into two solid
tori $V_{I}$ and $V_{J}$, such that $L\cap V_{I}=L_{I}=\cup_{j\in J}C_{i}$ and
$L\cap V_{J}=L_{J}=\cup_{j\in J}C_{i}$. Suppose $Q\sqcup R\sqcup S\subset I$
and $J_{0}\subset J$. Let $D_{q}^{I}$($q\in Q$), $D_{j}^{J}$($j\in J_{0}$),
$D_{r}^{E}$($r\in R$) and $D_{s}^{F}$($s\in S$) be $L_{I}$-interior disks,
$L_{J}$-interior disks, exterior cross disks, and free cross disks
respectively, such that

  * •

the cross disks are mutually disjoint;

  * •

the union of these disks and $L$ forms a spanning complex $U$;

  * •

every disk intersects $T$ in simple intersection pattern.

###  6.1. Analysis on plausible annuli

To show a hypothetical torus cannot exist, we may estimate the shape of the
torus. We thus introduce some notations.

Under Standard premise, the components of the intersection of $T$ and the
cross disks are _plausible circles_ , denoted
$\tilde{C}_{1},\tilde{C}_{2},...,\tilde{C}_{t}$, successively indexed along
$T$. For any $k=1,2,...,t$, the annulus in $T$ between $\tilde{C}_{k}$ and
$\tilde{C}_{k+1}$ is a _plausible annulus_ , denoted $\tilde{A}_{k}$; the disk
bounded by $\tilde{C}_{k}$ in a cross disk is a _plausible disk_ , denoted
$\tilde{D}_{k}$; the connected component of $V_{J}-(\cup_{r\in
R}D_{r}^{E}\cup_{s\in S}D_{r}^{F})$ containing $\tilde{A}_{k}$ is a _plausible
cylinder_ , denoted $\tilde{B}_{k}$.

######  Proposition 6.1.

Under Standard premise, let $\alpha$ be an arc component of $L_{J}-\cup_{r\in
r}D_{r}^{E}\cup\cup_{s\in s}D_{s}^{F}$.

(1) $\alpha$ connects two consecutive plausible disks on the corresponding
sides of cross disks unless two endpoints of $\alpha$ are on one side of a
cross disk.

(2) If $\alpha$ connects two plausible disks $\tilde{D}_{k}$ and
$\tilde{D}_{k+1}$ in one cross disk $D_{i}$ , and $\alpha$ is _unknotted_ , i.
e. forming unknot with an arc connecting $\partial\alpha$ in $D_{i}$, then
$\alpha$ is the core of $\tilde{B}_{k}$.

Here, a _core_ of a plausible cylinder is a trivial arc in the plausible
cylinder connecting its two bases.

###### Proof.

Since (1) is obvious, we need only prove (2). There are two cases depending on
whether $\tilde{D}_{k}$ and $\tilde{D}_{k+1}$ are on the same side of $D_{i}$.
First suppose they are on the same side of $D_{i}$. Let $\beta_{\alpha}$ be an
arc in $D_{i}$ connecting $\partial\alpha$ so that $\beta_{\alpha}$ intersects
both $\tilde{C}_{k}$ and $\tilde{C}_{k+1}$ in exactly one point. Let
$D_{\alpha}$ be a disk bounded by $\alpha\cup\beta_{\alpha}$. Consider
$D_{\alpha}$ and $D_{i}$. A standard innermost circle/outermost arc argument
shows that we may assume $D_{\alpha}\cap D_{i}=\beta_{\alpha}$. Then consider
$D_{\alpha}$ and $\tilde{A}_{k}$. By the choice of $\beta_{\alpha}$,
$D_{\alpha}\cap\tilde{A}_{k}$ consists of an arc connecting $\tilde{C}_{k}$
and $\tilde{C}_{k+1}$ and some trivial circles in $\tilde{A}_{k}$. We can
eliminate all the circle components by a standard innermost circle argument.
Notice that the remained arc component is parallel to the core of
$\tilde{B}_{k}$. Thus $\alpha$ is the core of $\tilde{B}_{k}$. See Fig. 17(1).

Figure 17. The arc $\alpha$ is the core of $\tilde{B}_{k}$.

Now suppose two plausible disks are on different sides of $D_{i}$. The proof
is almost identical, the major change being the substitution of
$\beta_{\alpha}$ into a shape “T” graph for $D_{\alpha}\cap D_{i}$. See Fig.
17(2). ∎

######  Example 6.2.

Recall the link in Fig. 10(1). Using the notations in Example 3.11, we prove
it is s-prime. Suppose there is an essential torus $T$ splitting the link.
Take the stable exterior cross disk $D_{1}$ bounded by $C_{1}$ as shown in
Fig. 10(2). Since $lk(C_{1},C_{2})=0$ and $C_{1}$ and $C_{2}$ are symmetric,
we may assume $lk(C_{1},\beta_{J})=0$. Thus there is a plausible cylinder, say
$\tilde{B}_{k}$, whose bases are on the same side of $D_{1}$. Without lose of
generality, assume its bases are on the positive side of $D_{1}$. Clearly,
$\tilde{B}_{k}$ intersects $C_{2}$ in some odd labeled arcs.

Let $\alpha_{i}$ be a component of $\tilde{B}_{k}\cap C_{2}$. We claim
$\alpha_{i}$ is the core of $\tilde{B}_{k}$. In fact, if not,
$\partial\alpha_{i}$ would be in one plausible disk, say $\tilde{D}_{k}$. Then
once delete all the other odd labeled arcs, $\tilde{A}_{k}\cup\tilde{D}_{k+1}$
would be a disk capping $\alpha_{i}$ with $D_{1}$. This implies
$lk(\alpha_{i},\alpha_{i+1})=0$, a contradiction.

Hence all the components of $\tilde{B}_{k}\cap C_{2}$ are core of
$\tilde{B}_{k}$. By Lemma 5.8(1), $\tilde{B}_{k}\cap C_{2}$ has more than one
component. Consider two of the components. Then they are parallel to each
other in $\tilde{B}_{k}$. Thus for any even labeled arc $\alpha_{j}$, this two
arcs have the same linking number with $\alpha_{j}$. This is impossible as we
know $lk(\alpha_{i},\alpha_{i\pm 1})=1$ and otherwise the linking number is
$0$.

###  6.2. Components discarded

This method is powerful for large links.

######  Proposition 6.3.

Under Standard premise with $R\cup S\neq\emptyset$, if delete some connected
components of $U$ contained in $V_{I}$, there is no connected component of $U$
contained in $V_{J}$ split out by a sphere.

###### Proof.

By Theorem 2.3, the link $L_{J}\cup\beta_{I}$ is Brunnian. Thus no sphere
splits $L_{J}\cup\beta_{I}$. By construction of $U$, no sphere splits
$\beta_{I}\cup(U\cap V_{J})$. Notice that a cross disk in simple intersection
pattern contains a longitude of $V_{I}$, which is parallel to $\beta_{I}$. It
follows from the construction of $U$ that in the complex $U$ with some
connected components in $V_{I}$ discarded, no connected component in $V_{J}$
can be split out by a sphere. ∎

This proposition provides a method to rule out essential torus splitting
$L_{I}$ and $L_{J}$, as the hypothetical torus $T$ cannot exist if discarding
some components of $U$ in $V_{I}$ makes some component of $U$ in $V_{J}$ split
out.

######  Example 6.4.

Figure 18. deBrunner(5).

deBrunner(5) is s-prime. Denote the link in Fig 18 by $L$. In view of the
symmetry of $L$, it suffices to show there is no essential torus splitting
$C_{1}$ and $C_{2}$. Assume for contrary that $T$ is a such torus. Take a
cross disk $D_{2}$ bounded by $C_{2}$ as shown in Fig 18. Depending on whether
$T$ splits $C_{2}$ and $C_{3}$, the following discussion is complete.

  * (i)

$V_{I}$ contains only $C_{2}$;

  * (ii)

$V_{I}$ contains exactly $C_{2}\cup C_{3}$;

  * (iii)

$V_{I}$ contains more than $C_{2}\cup C_{3}$ and $V_{J}$ contains more than
$C_{1}$;

  * (iv)

$V_{J}$ contains only $C_{1}$;

  * (v)

$V_{J}$ contains exactly $C_{1}\cup C_{3}$;

  * (vi)

$V_{J}$ contains more than $C_{1}\cup C_{3}$ and $V_{I}$ contains more than
$C_{2}$.

None above is possible. The main point of our argument is that in $L\cup
D_{2}$, after deleting a connected component other than $C_{1}\cup D_{2}\cup
C_{3}$, then all the other components split out.

For (ii) and (iv), they are impossible by Proposition 5.4(2). Clause (i) is
equivalent to (iv) by the symmetry of $L$, thus also impossible. For (iii) and
(iv), they are impossible by Proposition 6.3. In fact, in $L\cup D$, once
delete a circle component in $V_{I}$, each circle component in $V_{J}$ splits
out. Finally, (v) is contained in (iii) after a rotation of $L$.

Proposition 6.2.1 in [4], the uniqueness of s-sum decomposition for Brunnian
links, can simplify the task of showing s-primeness. Specifically, as any two
tori splitting $L$ are disjoint after isotopy, if a torus $T$ splits $L$ into
$L_{I}$ and $L_{J}$ and a torus $T^{\prime}$ splits $L$ into $L_{I^{\prime}}$
and $L_{J^{\prime}}$, then one of $L_{I}\cap L_{I^{\prime}}$, $L_{I}\cap
L_{J^{\prime}}$, $L_{J}\cap L_{I^{\prime}}$ and $L_{J}\cap L_{J^{\prime}}$ is
empty.

######  Example 6.5.

Figure 19. Torus($m,n$) is s-prime.

For any $m>1,n>2$, Torus($m,n$) is s-prime. Up to symmetry, there is only one
kind of component, as $C_{ij}$ in Fig 19. It suffices to show that no
essential torus splits any one of $C_{ij}$, $C_{i-1,j}$ and $C_{i+1,j}$ from
the other two. Take the cross disk $D_{ij}$. By Proposition 5.4(2), no
essential torus splits $C_{i+1,j}$ and $C_{ij}\cup C_{i-1,j}$ or splits
$C_{i-1,j}$ and $C_{ij}\cup C_{i+1,j}$.

Suppose there is an essential torus $T$ splitting $C_{ij}$ and $C_{i-1,j}\cup
C_{i+1,j}$. In $L\cup D_{ij}$, once delete a circle component other than
$C_{i+1,j}$, all other circle components split out. Thus by Proposition 6.3,
$T$ either splits $C_{i-1,j}\cup C_{i+1,j}$ from all the other components, or
splits $C_{ij}\cup C_{i+1,j}$ from all the other components. For the first
case, by the symmetry of $L$, there would be a torus splits $C_{i-3,j}\cup
C_{i-1,j}$ from all the other components. by Proposition 6.2.1 in [4], this is
impossible. Similarly, the second case is impossible.

######  Definition 6.6.

Let $C$ be a component of a Brunnian link $L$. If there is an essential torus
in $S^{3}-L$ splitting $C$ and $L-C$, then $C$ is _multiple_ ; otherwise,
_simple_.

Figure 20. Carpet(1,3,4) is s-prime.

######  Example 6.7.

Carpet(1,3,4) is s-prime. Denote this link by $L$. Up to symmetry, there are 6
kinds of components. We will take cross disks in Fig. 20 one by one to show
$L$ is s-prime.

On account of the symmetry of this link, the following discussion is complete.

  * (i)

No essential torus splits one of $C_{11}$, $C_{12}$ and $C_{21}$ from the
other two;

  * (ii)

No essential torus splits one of $C_{12}$, $C_{13}$ and $C_{22}$ from the
other two;

  * (iii)

No essential torus splits $C_{31}$ and $C_{12}\cup C_{21}$;

  * (iv)

No essential torus splits $C^{\prime}_{21}$ and $C_{13}\cup C_{22}$;

  * (v)

Every component is simple.

For (iii), take cross disk $D_{21}$ and use Proposition 5.4(2). For (i), take
cross disk $D_{11}$. By Proposition 5.4(2), any essential torus neither splits
$C_{12}$ and $C_{11}\cup C_{21}$, nor splits $C_{21}$ and $C_{11}\cup C_{12}$.
Suppose there is an essential torus $T$ splitting $C_{11}$ and $C_{12}\cup
C_{21}$. In $L\cup D_{11}$, once delete a circle component, all other circle
components split out. By Proposition 6.3, either $C_{11}$ is multiple or $T$
splits $C_{12}\cup C_{21}$ from all the other components. By (iii), it only
remains to show $C_{11}$ is simple.

For (iv), take cross disk $D_{22}$ and use Proposition 5.4(2). For (ii), take
cross disk $D_{12}$. By Proposition 5.4(2), any essential torus neither splits
$C_{13}$ and $C_{12}\cup C_{22}$, nor splits $C_{22}$ and $C_{12}\cup C_{13}$.
As demonstrated before, by Proposition 6.3, it remains to show $C_{12}$ is
simple, which is proved in (i).

For (v), we have shown that $C_{12}$, $C_{21}$ are simple when proving (i),
$C_{13}$ and $C_{22}$ are simple when proving (ii), and $C_{31}$ is simple by
(iii). For the remained case that $C_{11}$ is simple, take cross disk $D_{13}$
intersecting $C_{11}$ in 4 points. This is Case (1.0.0) in Lemma 6.10(1).
Delete the ear $E$. Then using the approach in Example 5.3 in [5], we can show
the asserted incredible circle in Proposition 6.11(1) does not exist.

###  6.3. Interior disks only

If there is no cross disk, we can give a sufficient and necessary criterion to
detect s-primeness, parallel to Theorem 4.1.

######  Theorem 6.8.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link. Let $I$ and $J$
be disjoint nonempty subset of $\\{1,...,n\\}$, and let $D_{i}^{I}$($i\in I$)
and $D_{j}^{J}$($j\in J$) be $L_{I}$-interior disks and $L_{J}$-interior disks
respectively. Suppose each interior disk is (s8) and

$U_{I}=L_{I}\bigcup_{i\in I}D_{i}^{I}$

$U_{J}=L_{J}\bigcup_{j\in J}D_{j}^{J}$

are disjoint spanning complex. Then there is no essential torus in the
complement of $L$ splitting $L_{I}$ and $L_{J}$ if and only if any torus
splitting $U_{I}$ and $U_{J}$, if exists, is $\partial$-parallel to a
component of $L$.

###### Proof.

We first prove the “if” implication. Suppose that $T$ is an essential torus in
the complement of $L$ splitting $L_{I}$ and $L_{J}$. Then by Theorem 5.6,
$(U_{I}\cup U_{J})\cap T=\emptyset$. Clearly $T$ is incompressible in the
complement of $U_{I}\cup U_{J}$. Thus by the condition, $T$ is
$\partial$-parallel to a component of $L$, contradicting that $T$ is essential
in the complement of $L$.

Now we prove the “only if” implication. Suppose that $T$ is torus in the
complement of $U_{I}\cup U_{J}$ splitting $U_{I}$ and $U_{J}$ and not
$\partial$-parallel to any component of $L$. Then $T$ splits $L_{I}$ and
$L_{J}$. By Theorem 2.3, $T$ is incompressible in the complement of $L$. This
completes the proof. ∎

For Brunnian links with at least 3 components, the “if” implication of this
theorem gives a method to prove s-primeness, while the “only if” implication
indicates this method is theoretically universal. However, this method does
not work for 2-component links, and will become complicated when the number of
components is large.

######  Example 6.9.

Figure 21. W(5).

For the link W($5$) in Fig. 21(1), we show no torus splits $C_{1}\cup C_{2}$
and $C_{0}\cup C_{3}\cup C_{4}$. Take interior disks $D_{3}^{I}$ and
$D_{4}^{I}$ as shown in Fig. 21(1). By Theorem 6.8, we only need to prove no
torus splits $C_{1}\cup C_{2}$ and $C_{0}\cup D_{3}^{I}\cup D_{4}^{I}$. The
regular neighborhood of the later complex is isotopic to the regular
neighborhood of the graph $U_{I}$ in Fig. 21(2). The result follows from the
claim that no torus splits the red link $C_{01}\cup C_{02}\cup C_{03}$ and
$C_{1}\cup C_{2}$, which will be proved in the last paragraph of this paper by
a generalized version of Theorem 6.8.

###  6.4. Cross disk with the least intersection points

Recall Proposition 5.4. From its proof we see that if a cross disk $D_{i}$
intersects $V_{J}$ in its meridian disks, then $\sharp(D_{i}\cap L_{J})\geq
4$. In this subsection, we analyze cross disks intersecting $L_{J}$ in the
least number of points in detail.

Let $L=\cup_{i=1}^{n}C_{i}$ be an $n$-component Brunnian link. Let $T$ be an
essential torus in the complement of $L$, splitting $S^{3}$ into two solid
tori $V_{I}$ and $V_{J}$, such that $L\cap V_{I}=L_{I}=\cup_{j\in J}C_{i}$ and
$L\cap V_{J}=L_{J}=\cup_{j\in J}C_{i}$. Let $D_{i}$ be a cross disk bounded by
$C_{i}$ intersecting $V_{J}$ in its meridian disks such that $\sharp(D_{i}\cap
L_{J})=4$. We give the following 6 cases on the intersection relationship of
$D_{i}$, $T$ and $L_{j}$. See Fig. 22 for illustration.

If $D_{i}$ intersects only one component of $L_{j}$:

$\begin{array}[]{llllll}Case&lk(C_{i},\beta_{J})&lk(C_{j},\beta_{I}),\ \forall
j&Case&lk(C_{i},\beta_{J})&lk(C_{j},\beta_{I}),\ \forall j\\\
(1.0.0):&0&0&(1.2.0):&2\ \ (L_{I}=C_{i})&0\\\ (1.0.2):&0&2\ \
(L_{J}=C_{j})&(1.2.2):&2\ \ (L_{I}=C_{i})&2\ (L_{J}=C_{j})\end{array}$

If $D_{i}$ intersects more than one component of $L_{j}$:

$\begin{array}[]{llllll}Case&lk(C_{i},\beta_{J})&lk(C_{j},\beta_{I}),\ \forall
j&Case&lk(C_{i},\beta_{J})&lk(C_{j},\beta_{I}),\ \forall j\\\
(2.0.0):&0&0&(2.2.0):&2\ \ (L_{I}=C_{i})&0\end{array}$

Figure 22. A cross disk with the least intersection points with $L_{j}$.

######  Lemma 6.10.

(1) If $D_{i}$ is an exterior cross disk, one of the 6 cases above happens.

(2) If $D_{i}$ is a free cross disk, one of Case (1.0.0), (1.0.2) and (2.0.0)
happens.

###### Proof.

On account of linking number and Brunnian property, these are all the allowed
cases. ∎

In the following proposition, we call an arc component of $L_{J}-D_{i}$ an
_ear_ if its endpoints are on the same side of $D_{i}$.

######  Proposition 6.11.

Under Standard premise, suppose that there is only one cross disk, denoted
$D_{i}$, and $\sharp(D_{i}\cap L_{J})=4$.

(1) In Case (1.0.0) (resp. (1.2.0)), there are 2 ears on one side (resp.
different sides) of $D_{i}$ such that after deleting each ear $E$, there is an
incredible circle for $U-E$ on the same (resp. the other) side of $D_{i}$.

(2) In Case (2.0.0) (resp. (2.2.0)), once deleting an ear $E$, there is an
incredible circle for $U-E$ on the same (resp. the other) side of $D_{i}$.

###### Proof.

We only give proof for Case (1.0.0), and other cases are similar. Using labels
indicated in Fig. 22, after deleting $E$, the disk
$\tilde{D_{1}}\cup\tilde{A_{2}}$, after a perturbation, shows that
$\tilde{C_{2}}$ is an incredible circle for $U-E$. ∎

In the following example, we use this proposition in Step 2, and then use
Proposition 6.1 in the first paragraph of Step 3. Finally we reduce the
problem of showing s-primeness into showing two knotoids are not isotopic.

######  Example 6.12.

Figure 23. W(5) and spanning complex.

The link W($5$) is s-prime. Suppose $T$ is an essential splitting torus.

Step 1. Take the disk $D_{4}$ bounded by $C_{4}$ as depicted in Fig 23. By
Proposition 5.4, $C_{0}\cup C_{4}$, also $C_{0}\cup D_{4}$, are at the same
side of $T$.

The red arc $\gamma$ in $D_{4}$ and the left arc in $C_{0}$ form a circle, say
$C_{0l}$, and $C_{0l}\cup C_{1}\cup C_{2}\cup C_{3}$ is just W($4$). Notice
that W($4$) is a Milnor link, an s-sum of two Borromean rings, as there is a
torus splits $C_{0l}\cup C_{3}$ and $C_{1}\cup C_{2}$. Thus either $T$ splits
$C_{0}\cup D_{4}$ and $C_{1}\cup C_{2}\cup C_{3}$, or splits $C_{0}\cup
D_{4}\cup C_{3}$ and $C_{1}\cup C_{2}$. As we have ruled out the second
possibility in Example 6.9, $T$ splits $C_{0}\cup C_{4}$ and $C_{1}\cup
C_{2}\cup C_{3}$.

Step 2. Take $L_{J}$-interior disk $D_{4}^{J}$ and exterior cross disk
$D_{3}^{E}$ as depicted in Fig 23. Then $\sharp(D_{3}^{E}\cap C_{0})=4$ and
this is Case (1.0.0) in Lemma 6.10. We claim $p_{1l}\cup p_{1r}$ are in one
plausible disk and $p_{6l}\cup p_{16r}$ are in another plausible disk.

In fact, otherwise once delete $\gamma_{l}$, by Proposition 6.11(1),
$\gamma_{r}$ would be capped by a disk whose boundary is the asserted
incredible circle. Let $\beta_{r}$ be an arbitrary arc in the plausible disk
connecting $\partial\gamma_{r}$. Then $\gamma_{r}\cup\beta_{r}$ is a circle
capped by that disk. However, $\gamma_{r}\cup\beta_{r}$ is homotopically
nontrivial in the complement of $C_{1}\cup C_{2}$, a contradiction.

Step 3. Take $L_{J}$-interior disk $D_{4}^{J}$ and exterior cross disk
$D_{1}^{E}$, $D_{2}^{E}$ and $D_{3}^{E}$ as depicted in Fig. 23. Since
$p_{1l}$ and $p_{1r}$ are in one plausible disk, by Proposition 6.1(1),
$\alpha_{1l}$ and $\alpha_{1r}$ are in one plausible cylinder and $p_{2l}$ and
$p_{2r}$ are in one plausible disk. Notice that by Example 3.10 the disks are
all stable. It follows that $p_{il}\cup p_{ir}$ is the intersection of a
plausible disk and $C_{0}$ for any $i=1,2,...9$, and $\alpha_{il}$ and
$\alpha_{ir}$ are in one plausible cylinder, for any $i=1,2,...8$.

Consider $D_{1}^{E}\cup D_{3}^{E}\cup\alpha_{1l}\cup\alpha_{1r}$, then
$\alpha_{1l}$ and $\alpha_{1r}$ are parallel. A standard innermost circle
argument shows that $\alpha_{1l}$ and $\alpha_{1r}$ are parallel as well in
the plausible cylinder. The shape in other plausible cylinders are similar.
Consequently, there is a disk $D_{0}$ bounded by $C_{0}$ which is contained in
$V_{J}$, intersects each plausible disk in a red arc and intersects
$D_{4}^{J}$ in two green arcs as shown in Fig. 24(1).

We see that $D_{1}^{E}$, $D_{2}^{E}$, $D_{3}^{E}$ and $L$ are all symmetric
about the middle sphere $\Pi$, and we may assume $D_{4}^{J}\subset\Pi$.
Clearly $D_{0}\cap\Pi$ consists of circles and an arc connecting
$\partial\gamma$. We can eliminate all the circle components by a standard
innermost circle argument so that $D_{0}\cap\Pi$ is an arc, say
$\gamma_{\Pi}$, and then isotope $D_{0}$ so that the shape on $D_{0}$ of
$\gamma_{\Pi}$ and the red arcs in Fig. 24(1) are as shown in Fig. 24(2).

Figure 24. (1) $D_{4}^{J}\cup D_{0}$ in $V_{J}$. (2) The green arc is
$\gamma_{\Pi}$.

Step 4. The left half of $D_{0}$ cut by $\gamma_{\Pi}$ indicates that $C_{0l}$
can be projected to $\Pi$ as a planar curve $\gamma_{\cup}\gamma_{\Pi}$, that
is, parallel to $\gamma_{\cup}\gamma_{\Pi}$ and intersecting the three cross
disks in the left halves of red arcs. To complete the proof, it remains to
show this is impossible. In fact, on account of how $C_{0l}$ pass through the
three cross disks, $\gamma_{\cup}\gamma_{\Pi}$ can only be as shown in Fig
25(1). On the other hand, $C_{0l}$, projected to $\Pi$, is as shown in Fig
25(2). Our problem reduces to prove these two knotoids are not isotopic. In
fact, connecting $a_{1},b_{2},b_{3}$ form two Brunnian links. One can verify
that the left one is an s-sum of Whitehead link and Borromean ring, while the
right one is s-prime, which can be proven by Proposition 6.11(1).

Figure 25. The thick arcs are the projection of left halves of $C_{1}$,
$C_{2}$ and $C_{3}$.

##  7\. Conclusions

1\. Series of Hyperbolic Brunnian links.

Based on the examples, we are now confident that Theorem 1.3 can be proved.

###### Proof of Theorem 1.3.

We have pointed out that all these links are untied in Subsection 4.1. We now
prove they are s-prime.

(1) The proof in Example 6.2 is without loss of generality to show the links
in this series are all s-prime.

(2) The proof in Example 6.4 is without loss of generality for $n\geq 5$. For
$n=3$, the proof is just simpler since it suffices to prove every component is
simple. For deBrunner($2$), use Case (1.2.0) in Proposition 6.11(1). For
deBrunner($4$), the case (ii) in the proof of Example 6.4 need to be modified
by using Proposition 6.11(1) and then the approaches in [5] to show there is
no incredible circle.

(3) A proof of s-primeness will follow Example 6.12 by induction on $n$. We
know W($5$) is s-prime. For any $n>5$, W($n-1$) is s-prime, an argument
similar to the one used in the first paragraph in Example 6.12 shows that the
only possibility is that $T$ splits $C_{0}\cup D_{n}$ from all the other
components of the link. The remainder of the proof is identical, except that
in the last paragraph we distinguish two knotoids in a simpler manner as
follows. For each thick arc, connecting the red arcs from below as shown in
Fig. 25, we get two links. The left one will be a Milnor link, which is an
s-sum[4], while the right one will be W($n-1$), which is s-prime by induction.

(4) For any $m>1,n>2$, we have shown that Torus($m,n$) is s-prime in Example
6.5. For $m>1,n=2$, the proof is analogues to that when $m>1,n>2$ and is left
to the reader. For $n=1$, the link Torus($m,1$) is isotopic to deBrunner($m$).
For $m=1,n>2$, Torus($1,n$) is Brunn’s chain (See [6, 5] or recall Example
4.2). The proof of that Brunn’s chains are s-prime is quite similar to that
for deBrunner($n$) and so is omitted.

(5) By suitable modification to the proof of Torus($m,n$) and Carpet ($m,n,p$)
are s-prime, we can show Tube($m,n$) is s-prime for $m>0,n>1$. We leave the
details to the reader.

(6) When $n=1$, Carpet($m,n,p$) is the $p$-component Brunn’s chain. When
$n>1$, the proof in Example 6.7 is without loss of generality to show
Carpet($m,n,p$) is s-prime. ∎

2\. More Hyperbolic Brunnian links.

By our methods, we discriminate s-primeness for all the Brunnian links in
literatures. Recall Subsection 4.1 and the geometric classification theorem
for Brunnian links in [4]. We thus get much more hyperbolic Brunnian links.
Some complicated series include Brunnian solids (Fig. 16 in [2]), Snakes,
Cirrus and Wheels[5]. We only do not know whether Fountains and Jade-
pendant[5] are hyperbolic.

3\. Generalizations.

While we have considered only Brunnian links in this paper, the methods can be
extended to links with some unknotted components. Replacing the conditions of
disks all by “stable”, Theorem 4.1 detects whether there is a knotted torus
containing the whole link in the solid torus side, Theorem 6.8 detects whether
there is an unknotted essential torus splitting the link, and Theorem 5.6
holds as well. The proofs are almost identical, the major change being using
“stable” condition instead of Brunnian property in Step 1.2 and Step 2.2 in
the proof of Theorem 5.6. As an application, we can use the modified theorems
to prove the links in [14] Section 5 (Fig. 2) are hyperbolic.

Our methods can be further generalized by using surfaces instead of disks. For
instance, to show no torus splits $C_{01}\cup C_{02}\cup C_{03}$ and
$C_{1}\cup C_{2}$ in Fig 26, we may take the annulus $A$ bounded by
$C_{01}\cup C_{02}$, intersecting $C_{03}$ in 4 points. Firstly, we can prove
there is no incredible circle on $A$. Then if such a torus exists, say $T$,
which bounds a solid torus $V$ containing $C_{01}\cup C_{02}\cup C_{03}$, we
can prove $A\subset V$ after isotopy. In fact, we can eliminate inessential
circle in $T$ as demonstrated before. To eliminate annulus regions in $A$
outside $V$, the main point of our argument is that for any $i=1,2,3$,
$C_{0i}\cup C_{1}\cup C_{2}$ is a Borromean ring, which is atoroidal, and thus
$C_{01}$ and $C_{02}$ can only be the core of $V$, and such annulus regions in
$A$ can only be $\partial$-parallel to $T$. Finally, it is easy to see no
torus splitting $C_{1}\cup C_{2}$ and $A\cup C_{03}$ by considering a
deformation retraction of $A\cup C_{03}$.

Figure 26. No torus splits $C_{01}\cup C_{02}\cup C_{03}$ and $C_{1}\cup
C_{2}$.

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##  8\. Appendix: An alternating example

We provide an alternative proof that a series of alternating Brunnian links
are s-prime to illustrate the method in Subsection 6.3.

######  Example 8.1.

Figure 27. A link in one of Tait series in [12].

Fig. 27(1) shows a link in a Tait series. In view of symmetry, we only need to
prove no essential torus splits $C_{3}$ from $C_{1}$ and $C_{2}$. Take the
disk $D_{1}$ as shown in Fig. 27(2), which is stable by an argument similar to
the one used in Example 3.11. Then a deformation retraction of $D_{1}\cup
C_{2}$ is as shown in Fig. 27(3). In view of the complement space of the red
part, we see that the only splitting torus is $\partial$-parallel to $C_{3}$.