# Brunnian exotic surface links in the 4–ball

Kyle Hayden  Department of Mathematics, Columbia University, USA
<EMAIL_ADDRESS>,  Alexandra Kjuchukova  Department of Mathematics,
University of Notre Dame, USA<EMAIL_ADDRESS>,  Siddhi Krishna  School of
Mathematics, Georgia Institute of Technology, USA<EMAIL_ADDRESS>,
Maggie Miller  Department of Mathematics, Massachusetts Institute of
Technology, USA<EMAIL_ADDRESS>,  Mark Powell  Department of
Mathematical Sciences, Durham University, UK<EMAIL_ADDRESS>and
Nathan Sunukjian  Department of Mathematics and Statistics, Calvin University,
USA<EMAIL_ADDRESS>
###### Abstract.

This paper investigates the exotic phenomena exhibited by links of
disconnected surfaces with boundary that are properly embedded in the 4-ball.
Our main results provide two different constructions of exotic pairs of
surface links that are _Brunnian_ , meaning that all proper sublinks of the
surface are trivial. We then modify these core constructions to vary the
number of components in the exotic links, the genera of the components, and
the number of components that must be removed before the surfaces become
unlinked. Our arguments extend two tools from 3–dimensional knot theory into
the 4–dimensional setting: satellite operations, especially Bing doubling, and
covering links in branched covers.

###### 1991 Mathematics Subject Classification:

57K40, 57K45 (primary); 57N35 (secondary).

KH was partially supported by NSF grant DMS-1803584. SK was partially
supported by NSF grant DMS-1745583. MM was supported by NSF grant DMS-2001675.
MP was partially supported by EPSRC New Investigator grant EP/T028335/1 and
EPSRC New Horizons grant EP/V04821X/1.

##  1\. Introduction

One of the motivating principles of 4-dimensional topology is to understand
exotic phenomena, namely differences between the smooth and topological
categories. This paper is concerned with exotic behavior exhibited by surfaces
in 4-manifolds, which we interpret in the strongest possible sense: we say
that two smooth, properly embedded surfaces $\Sigma$ and $\Sigma^{\prime}$ in
a 4-manifold form an _exotic pair_ if they are topologically ambiently
isotopic rel. boundary but are not smoothly equivalent, meaning that there is
no ambient diffeomorphism carrying $\Sigma$ to $\Sigma^{\prime}$.

Though exotic behavior is well-studied for closed surfaces, even the following
basic question remains open: are there exotic closed, orientable surfaces in
$S^{4}$? In contrast, exotic pairs of orientable surfaces have been
constructed in many 4–manifolds $X$ with $b_{2}(X)>0$; see e.g. the
foundational paper by Fintushel–Stern [FS97]. In a related vein, $S^{4}$ is
known to contain exotic nonorientable surfaces [FKV88]. However, all these
results rely on the topology of the ambient $4$–manifold or nonorientability
of the surfaces in an essential way.

To detect exotic phenomena, it can be helpful to work in the relative setting.
This motivates the study of properly embedded surfaces. The first
constructions of exotic surface knots in $B^{4}$ appeared nearly
simultaneously in papers of Juhász–Miller–Zemke [JMZ20] and Hayden [Hay20],
using significantly different techniques. Hayden constructed, for every $g\geq
0$, exotic pairs of properly embedded genus $g$ surfaces, while
Juhász–Miller–Zemke exhibited an infinite family of properly embedded genus
$g\geq 1$ surfaces, any two of which form an exotic pair. The main objects of
study in the relative setting are surface links.

######  Definition 1.1.

A _surface link_ is a smooth, oriented, properly embedded, 2–dimensional
submanifold $\Sigma\subseteq B^{4}$ whose connected components $\Sigma_{i}$
each have exactly one boundary component in $S^{3}$. When $\Sigma$ is
connected, it is called a _surface knot_. If each component $\Sigma_{i}$ is a
disk, we say that $\Sigma$ is a _disk link_ (or _disk knot_ , in the connected
case). A surface link $\Sigma$ is _unlinked_ if $\partial\Sigma$ is an unlink
in $S^{3}$ and $\Sigma$ is smoothly ambiently isotopic to a Seifert surface
for the unlink.

In this language, Hayden produced exotic pairs of disk and surface knots, and
Juhász–Miller–Zemke found an infinite family of pairwise exotic surface knots.
We extend these results to study exotic linking of surfaces $B^{4}$. To
highlight how linking plays an important role for exotic surfaces, we study a
particularly delicate class of surfaces.

######  Definition 1.2.

A surface link $\Sigma\subseteq B^{4}$ is _Brunnian_ if removing any component
from $\Sigma$ yields a smoothly unlinked surface.111In the literature, one
usually requires that Brunnian links are themselves nontrivial or even non-
split. However in dimension four this criterion is more cumbersome, so we
prefer to allow some trivialities. Namely, according to our definition, an
unlink is Brunnian.

These are higher-dimensional analogues of Brunnian links in $S^{3}$,
introduced over a century ago by Brunn [Bru92]. To study surface links, we
leverage two other techniques originating in 3–dimensional knot theory:
constructing surfaces via satellite operations (especially Bing doubling) and
distinguishing surface links using _covering surfaces_ , a technique inspired
by Cha–Kim [CK08].

###  1.1. 2–component exotic surface links

By adapting Hayden’s [Hay20] examples of exotic disks in $B^{4}$, we produce
an exotic pair of 2–component Brunnian disk links. Constructing these disk
links requires Bing doubling a pair of slice disks, $D_{1}$ and $D_{2}$, in
$B^{4}$. The _Bing double_ of an unknotted disk is shown in Figure 1(a), and
this operation is discussed in more detail in Section 1.2. We will write
$\operatorname{BD}(D)$ to denote the Bing double of a disk $D$. The underlying
disks $D_{1}$ and $D_{2}$ in our construction are bounded by the same slice
knot $K$, pictured in Figure 1(b).

(a)

$K$
---
(b)

Figure 1. (a) The Bing double of a disk consists of two disjoint disks. (b)
The knot $K$, together with two bands, leading to the different slice disks
$D_{1}$ and $D_{2}$, whose Bing doubles appear in Theorem 1.3.

######  Theorem 1.3.

The 4–ball contains an exotic pair of 2–component Brunnian disk links. In
particular, the knot $K$ in Figure 1(b) bounds two different slice disks
$D_{1}$ and $D_{2}$ $($see Figure 3$)$, and their Bing doubles
$\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ form an exotic pair
in $B^{4}$.

We briefly summarize the strategy which distinguishes the diffeomorphism
classes of these links. The surface links $\operatorname{BD}(D_{1})$ and
$\operatorname{BD}(D_{2})$ have the same boundary in $S^{3}$, a link $L$. We
identify a knot $\gamma$ in the exterior of $L$ such that any diffeomorphism
of pairs sending $(B^{4},\operatorname{BD}(D_{1}))$ to
$(B^{4},\operatorname{BD}(D_{2}))$ must preserve the isotopy class of
$\gamma$. By showing that $\gamma$ is slice in the complement
$B^{4}{\smallsetminus}\operatorname{BD}(D_{2})$ but not in
$B^{4}{\smallsetminus}\operatorname{BD}(D_{1})$, we deduce that the disk link
exteriors are not diffeomorphic.

Our next construction produces exotic surface links where one component has
positive genus. These examples rely on rim surgery, first introduced in
[FS97]. This technique takes as input an essential curve $\alpha$ on a surface
$\Sigma$ and a knot $J$, and outputs a new surface $\Sigma(\alpha;J)$. Rim
surgery allows us to construct an infinite family of pairwise exotic
2–component surface links.

######  Theorem 1.4.

Let $K$ be a strongly quasipositive topologically slice knot and let $\Sigma$
denote the surface link of Figure 11. For each integer $n\geq 0$, let
$J_{n}\subseteq S^{3}$ be a knot whose Alexander polynomial has $n$
irreducible factors $($counted with multiplicity$)$. The 2–component Brunnian
surface links $\\{\Sigma(\alpha;J_{n})\\}$, each consisting of a disk and a
genus one surface, are pairwise exotic.

Rim surgery replaces the annulus $\alpha\times I\subseteq\Sigma$ by a knotted
annulus determined by $J$; see Section 4.3 for a precise description. The
surface links of Theorem 1.4 are constructed explicitly in Section 4. A key
ingredient in the construction is a choice of curve $\alpha\subseteq\Sigma$
such that:

  1. (1)

$\alpha$ bounds a locally flat disk with interior in the complement of
$\Sigma$,

  2. (2)

$\alpha$ bounds a smooth disk with interior in the complement of each
component $\Sigma_{i}$, but

  3. (3)

the interior of every smooth disk with boundary $\alpha$ intersects $\Sigma$.

This ensures that rim surgery along $\alpha$ preserves both the topological
isotopy type of the link and the smooth isotopy type of each component, while
potentially changing the smooth isotopy type of the link itself.

To obstruct smooth equivalence, we follow [JMZ20] and use link Floer cobordism
maps. They first showed that the cobordism map induced by a quasipositive
surface is nontrivial, and then applied [JZ18] to show rim surgery can
potentially change the smooth isotopy type of a quasipositive surface. While
the surface $\Sigma$ in Theorem 1.4 is not quasipositive, we may compose it
with another link cobordism and obtain a quasipositive surface. The
composition laws for cobordism maps guarantee that the map associated to
$\Sigma$ is nonvanishing.

###  1.2. Increasing the number and genera of components

With these foundational constructions established, we bootstrap to produce
exotic surface links with any number of components. Inspired by work of
Cha–Kim [CK08] on covering links in $S^{3}$, we develop and apply the notion
of a _covering surface link_ in $B^{4}$.

By viewing $\operatorname{BD}$ as an operator on ordered surface links (see
Section 2 for details), we can perform an iterated Bing doubling of an ordered
surface link. We let $\operatorname{BD}^{k}(\Sigma)$ denote the surface link
obtained after Bing doubling _the first component_ of $\Sigma$ a total of $k$
times. This operation yields the promised generalizations of Theorems 1.3 and
1.4 to any number of connected components.

######  Theorem 1.5.

Let $n\geq 3$.

  1. $(i)$

There exists an exotic pair of $n$–component Brunnian disk links in $B^{4}$.

  2. $(ii)$

There exists an infinite family of pairwise exotic $n$–component Brunnian
surface links that each consist of a single genus one surface and $n-1$ disks.

In particular, let $\Sigma$ and $\Sigma^{\prime}$ be an exotic pair of
Brunnian 2–component surface links constructed as in Theorem 1.3 or Theorem
1.4, with orderings such that the first components are disks. Then
$\operatorname{BD}^{n-2}(\Sigma)$ and
$\operatorname{BD}^{n-2}(\Sigma^{\prime})$ are an exotic pair of $n$–component
surface links.

Theorem 1.5 is established by induction: under (minor) technical conditions,
we show that $\Sigma$ arises as a covering surface link of
$\operatorname{BD}(\Sigma)$, and that a hypothetical diffeomorphism of pairs
sending $\operatorname{BD}(\Sigma)$ to $\operatorname{BD}(\Sigma^{\prime})$
would lift to a diffeomorphism of pairs sending $\Sigma$ to $\Sigma^{\prime}$.
However, we have already proven that the latter cannot exist.

We modify the pairs of surface links from Theorem 1.4 to increase the genera
of the components.

######  Theorem 1.6.

Fix a pair of integers $r\geq 0$ and $s\geq 1$. There exists an infinite
family of Brunnian 2–component surface links in $B^{4}$ consisting of a genus
$r$ surface and a genus $s$ surface, any two of which form an exotic pair.

In particular, let $\Sigma^{r,s}_{0}$ be the surface link constructed in
Figure 18. For each $n\geq 0$, let $J_{n}\subseteq S^{3}$ be a knot whose
Alexander polynomial has $n$ irreducible factors $($counted with
multiplicity$)$. Then the 2–component Brunnian surface links
$\\{\Sigma^{r,s}_{0}(\alpha;J_{n})\\}$, each consisting of a genus $r$ surface
and a genus $s$ surface, are pairwise exotic.

The primary constructive technique requires banding the surface from Theorem
1.4 with another simple Brunnian surface link of controlled genus. An
analogous application of the link Floer argument from Theorem 1.4 obstructs a
diffeomorphism of pairs.

###  1.3. $\boldsymbol{(n,k)}$-Brunnian disk links

Next, we study a natural generalization of Brunnian links.

######  Definition 1.7.

An $n$–component surface link is $(n,k)$–Brunnian if every sublink of fewer
than $k$ components is an unlink, but every sublink of at least $k$ components
is nontrivial.

The notion of $(n,k)$–Brunnian was introduced by Debrunner [Deb61] for links
in $S^{3}$. Debrunner’s original definition requires all sublinks of at least
$k$ components to be nonsplit; in our adaptation to dimension four, we ask
only for these sublinks to be nontrivial. By combining his construction of
$(n,k)$–Brunnian links with our construction of exotic disk links, we prove
the following theorem.

######  Theorem 1.8.

For any integers $n$ and $k$ with $n\geq 2$ and $1\leq k\leq n$, there exists
a pair of $(n,k)$–Brunnian disk links in $B^{4}$ forming an exotic pair.

Heuristically, the index $k$ measures the extent to which the link components
are entangled. At one extreme are $(n,n)$–Brunnian links: these are the
nontrivial $n$–component Brunnian links. The $(n,1)$–Brunnian links lie at the
other extreme: all of their nonempty sublinks are nontrivial. Thus Theorem 1.8
encapsulates a wide spectrum of exotic behavior.

The extreme cases of Theorem 1.8 follow immediately from the above: for $n=k$,
take the disk links from Theorem 1.5, while for $k=1$, consider a split union
of $n$ copies of $D_{1}$ and the same for $D_{2}$ and apply [Hay20]. The
intermediate cases require a new construction.

###  1.4. Exotic closed surfaces

Finally, in Section 8, we show that exotic surface links in $B^{4}$ can be
used to construct exotic links of closed surfaces in larger 4-manifolds. This
is illustrated using the disk links $\operatorname{BD}(D_{1})$ and
$\operatorname{BD}(D_{2})$ from Theorem 1.3. We first consider the 4–manifold
$X$ obtained from $B^{4}$ by attaching 0-framed 2-handles along the link in
$S^{3}$ bounding $\operatorname{BD}^{k}(D_{1})$ and
$\operatorname{BD}^{k}(D_{2})$. Capping off these disk links with the cores of
the 2-handles yields a pair of exotic sphere links in $X$. When $k=1$, we then
show that $X$ further embeds into a closed 4–manifold in which these sphere
links remain exotic. In each case, we distinguish the exotic sphere links by
considering the effect of surgering the ambient 4–manifold along these
spheres.

###  1.5. Open problems

Here are two natural extensions of our results that remain open.

  1. (1)

Increase the genus and the number of components simultaneously.

  2. (2)

Find examples of infinitely many Brunnian disk links that are pairwise exotic.

###  1.6. Organization

In Section 2, we establish foundations on Bing doubling disks. Section 3
constructs the exotic pair of Brunnian 2–component disk links promised in
Theorem 1.3. In Section 4, we recall the necessary Heegaard Floer theory and
construct the pairwise exotic Brunnian surface links of Theorem 1.4. Section 5
extends this construction to produce an infinite family of 2-component
Brunnian links with arbitrary genera. Covering surfaces are introduced in
Section 6; this section also contains the proof of Theorem 1.5. Section 7
constructs exotic pairs of $(n,k)$-Brunnian disk links. Finally, Section 8
produces our examples of exotic closed surfaces. The appendices detail the
computer-assisted calculations used to prove Theorems 1.3 and 1.5.

###  1.7. Conventions

We provide some of our conventions.

  * •

$J$, $K$, and $L$ denote knots or links in $S^{3}$, while $\Sigma$, $S$, and
$D$ denote surfaces in $B^{4}$, with $D$ always a disk.

  * •

We work in the category $\operatorname{DIFF}$ of smooth manifolds with smooth
embeddings and smooth isotopies, and the category $\operatorname{TOP}$ of
topological manifolds with locally flat embeddings and topological ambient
isotopies. We specify whether an isotopy or surface is in
$\operatorname{DIFF}$ or $\operatorname{TOP}$.

  * •

Given a smooth or locally flat proper submanifold $Y\subseteq X$, $\nu Y$ will
denote an open tubular neighborhood of $Y$. In the smooth case, the existence
of tubular neighborhoods is standard differential topology. In the locally
flat case, existence follows from [FQ90, Theorem 9.3] when $X$ is a
4-manifold. A closed tubular neighborhood is denoted by $\overline{\nu}Y$.

###  1.8. Acknowledgements

This work is the product of a research group formed under the auspices of the
American Institute for Mathematics (AIM) in their virtual Research Community
on 4–dimensional topology. We are grateful to AIM, and especially to the
program organizers Miriam Kuzbary, MM, Juanita Pinzón-Caicedo, and Hannah
Schwartz. SK thanks Jen Hom and John Etnyre for helpful conversations. AK and
MP are grateful to the Max Plank Institute for Mathematics in Bonn, where they
were visitors while this paper was written. We used KLO [Swe21], developed by
Frank Swenton, for initial exploration.

##  2\. Background on Bing doubling

The Bing doubling operation replaces one disk component of a surface link with
two disjoint disks in a neighborhood of the original, as illustrated in Figure
1(a).

More formally, the _Bing double_ of a slice disk $D$ in $B^{4}$ for $K$ is
obtained as follows. First define a model Bing double of the core disk
$D^{2}\times\\{0\\}\subseteq D^{2}\times D^{2}$ from four parallel copies of
the disk, joined with two bands, in the manner depicted in Figure 2. The
result is two disjoint slice disks in $D^{2}\times D^{2}\cong B^{4}$ for the
Bing double of $S^{1}\times\\{0\\}\subseteq S^{1}\times
D^{2}\subseteq\partial(D^{2}\times D^{2})\cong S^{3}.$ Now, for a slice disk
$D$ in $B^{4}$, choose an orientation-preserving diffeomorphism of a tubular
neighborhood $\nu D$ with $D^{2}\times D^{2}$ such that $D$ maps to
$D^{2}\times\\{0\\}$ via an orientation-preserving diffeomorphism. Embed the
model Bing double in $B^{4}$ using the inverse of this identification. Call
the resulting pair of disks $\operatorname{BD}(D)\subseteq\nu D$. Their
boundary is the Bing double $\operatorname{BD}(K)\subseteq\nu K$ of the knot
$K$. Since any two orientation-preserving diffeomorphisms of $D^{2}\times
D^{2}$ are isotopic (not necessarily rel. boundary), the isotopy class of
$\operatorname{BD}(D)$ is independent of the choice of identification $\nu
D\cong D^{2}\times D^{2}$.

$K$ at 210 128 $K$ at 620 128 $K$ at 1045 128 $K$ at 1470 128 attach at 338
150 bands at 338 110 resolve at 760 150 isotopy at 1180 150

Figure 2. Constructing $\operatorname{BD}(D)$, the Bing double of a slice disk
$D$, from four copies of $D$ and two bands.

Going a step further, we define Bing doubling as an operation on ordered
surface links. Let $\Sigma=\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{k}$ be a
surface link, with $\Sigma_{1}\cong D^{2}$. To perform the Bing doubling
operation, and obtain a new surface link $\widehat{\Sigma}$, replace
$\Sigma_{1}$ with $\operatorname{BD}(\Sigma_{1})$, give the two new disks the
labels $\widehat{\Sigma}_{1}$ and $\widehat{\Sigma}_{2}$, and label
$\widehat{\Sigma}_{i}:=\Sigma_{i-1}$ for $3\leq i\leq k+1$. This defines
$\widehat{\Sigma}$ as an ordered surface link obtained from $\Sigma$ by Bing
doubling. For an ordered surface link $\Sigma$, we will also write
$\operatorname{BD}(\Sigma):=\widehat{\Sigma}$, where the doubling operation is
always performed on the first component.

Viewing $\operatorname{BD}$ as an operator on ordered surface links, we may
perform an iterated Bing doubling. We write $\operatorname{BD}^{k}(\Sigma)$
for the surface link obtained after Bing doubling _the first component_ $k$
times. (The ordering of the two components of $\operatorname{BD}(\Sigma_{1})$,
assigned arbitrarily at each stage, does not affect the construction.) This
operation is integral to the proof of Theorem 1.5.

We prove some foundational lemmas about Bing doubling surfaces.

######  Lemma 2.1.

For $\operatorname{CAT}$ either $\operatorname{DIFF}$ or $\operatorname{TOP}$,
let $\Sigma=\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n}$ and
$\Sigma^{\prime}=\Sigma^{\prime}_{1}\sqcup\cdots\sqcup\Sigma^{\prime}_{n}$ be
surface links in $B^{4}$ with $\partial\Sigma=\partial\Sigma^{\prime}\subseteq
S^{3}$. Suppose that $\Sigma$ and $\Sigma^{\prime}$ are
$\operatorname{CAT}$-ambiently isotopic rel. boundary as ordered surface
links. Assume that $\Sigma_{1}\cong D^{2}\cong\Sigma_{1}^{\prime}$ and that
$\partial(\operatorname{BD}(\Sigma_{1}))=\partial(\operatorname{BD}(\Sigma_{1}^{\prime}))$.
Then the surface links
$\operatorname{BD}(\Sigma)=\operatorname{BD}(\Sigma_{1})\sqcup\Sigma_{2}\sqcup\cdots\sqcup\Sigma_{n}$
and
$\operatorname{BD}(\Sigma^{\prime})=\operatorname{BD}(\Sigma_{1}^{\prime})\sqcup\Sigma_{2}^{\prime}\sqcup\cdots\sqcup\Sigma_{n}^{\prime}$,
obtained from Bing doubling the first component of each surface link, are also
$\operatorname{CAT}$-ambiently isotopic rel. boundary.

###### Proof.

Let $F_{t}\colon B^{4}\to B^{4}$, $t\in[0,1]$, be a $\operatorname{CAT}$
ambient isotopy with $F_{t}|_{S^{3}}=\operatorname{Id}_{S^{3}}$ for all $t$,
such that $F_{0}=\operatorname{Id}_{B^{4}}$ and
$F_{1}(\Sigma)=\Sigma^{\prime}$. Let $G\colon D^{2}\times
D^{2}\xrightarrow{\cong}\overline{\nu}\Sigma_{1}$ and $G^{\prime}\colon
D^{2}\times D^{2}\xrightarrow{\cong}\overline{\nu}\Sigma_{1}^{\prime}$ be two
identifications, used for the definitions of $\operatorname{BD}(\Sigma_{1})$
and $\operatorname{BD}(\Sigma^{\prime}_{1})$ respectively.

Every homeomorphism of $D^{2}$ is smoothable, and every orientation preserving
rel. boundary diffeomorphism of $D^{2}$ is isotopic rel. boundary to the
identity by Smale’s theorem [Sma59]. Therefore by the isotopy extension
theorem [EK71] we may and shall assume that the isotopy between $\Sigma$ and
$\Sigma^{\prime}$ sends $\Sigma_{1}$ to $\Sigma_{1}^{\prime}$ respecting fixed
choices of parametrizations $D^{2}\to\Sigma_{1}$ and
$D^{2}\to\Sigma_{1}^{\prime}$. In other words, extending $F_{t}$ with a
further ambient isotopy supported in a neighborhood of $\Sigma_{1}$, we obtain
that the composition

$D^{2}\times\\{0\\}\xrightarrow{G}\Sigma_{1}\xrightarrow{F_{1}}\Sigma_{1}^{\prime}\xrightarrow{(G^{\prime})^{-1}}D^{2}\times\\{0\\}$

is the identity. Similarly, by uniqueness of normal bundles for 2-dimensional
submanifolds of 4-manifolds (which holds for
$\operatorname{CAT}=\operatorname{TOP}$ by [FQ90, Chapter 9]), we may assume
that $F_{1}$ sends a given parametrization of $\overline{\nu}\Sigma_{1}$ as
$D^{2}\times D^{2}$ to a given such parametrization of
$\overline{\nu}\Sigma_{1}^{\prime}$. The disks
$\operatorname{BD}(\Sigma_{1})\subseteq\overline{\nu}\Sigma_{1}$ are therefore
sent to the disks
$\operatorname{BD}(\Sigma_{1}^{\prime})\subseteq\overline{\nu}\Sigma_{1}^{\prime}$
by the new $F_{1}$. ∎

######  Lemma 2.2.

Let $\Sigma=\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n}\subseteq B^{4}$ be a
Brunnian surface link with $\Sigma_{1}\cong D^{2}$. Then
$\widehat{\Sigma}:=\operatorname{BD}(\Sigma_{1})\sqcup\Sigma_{2}\sqcup\cdots\sqcup\Sigma_{n}$
is Brunnian.

###### Proof.

Removing $\Sigma_{i}$ for $i\geq 2$ from $\widehat{\Sigma}$ yields a surface
link smoothly isotopic to an unlinked surface with one component Bing doubled,
by Lemma 2.1 with $\operatorname{CAT}=\operatorname{DIFF}$. This is also
unlinked. On the other hand removing one component of
$\operatorname{BD}(\Sigma_{1})$ renders the second one an unknotted disk,
split from $\Sigma_{2}\sqcup\cdots\sqcup\Sigma_{n}$. Since the latter surface
link is also unlinked by the Brunnian hypothesis, $\widehat{\Sigma}$ is indeed
Brunnian as desired. ∎

##  3\. An exotic pair of disk links

We begin with the pair of disks $D_{1}$ and $D_{2}$ in $B^{4}$ are depicted in
Figure 3; these are obtained using the construction from [Hay20, Section 2]
(as applied to the link $L14n_{40949}$), but are distinct from the examples
used in that paper. The figure shows two handle diagrams for $B^{4}$. To see
this note that, erasing the gray disk from either picture, the remaining
dotted and $0$-labeled components form a Hopf link. Note that these disks are
disjoint from the 1-handle curves and that all intersections between the disks
and the 2-handles’ attaching regions occur in the disks’ interiors. Therefore,
after pushing the disks’ interiors into the 0-handle, we see that the disks
are indeed embedded in $B^{4}$.

By construction, these disks are bounded by the same knot $K$ in $S^{3}$,
which we redraw in the standard diagram for $S^{3}$ in Figure 1(b).

(a)
---
(b)
---
(c)
---
$D_{1}$
---
$D_{2}$
---
$K$
---
$\gamma$
---
$0$
---
$0$
---
$0$
---
$0$
---
Figure 3. Parts (a) and (b) depict disks $D_{1}$ and $D_{2}$ in non-standard
handle diagrams of $B^{4}$. Part (c) depicts the slice knot $K=\partial
D_{1}=\partial D_{2}$ and a distinguished curve $\gamma\subseteq
S^{3}{\smallsetminus}K$, drawn in a non-standard surgery description of
$S^{3}$. $0$
---
$K$
---
Figure 4. Redrawing $K$ in the standard diagram of $S^{3}$.

######  Proposition 3.1.

The disks $D_{1}$ and $D_{2}$ are topologically ambiently isotopic rel.
boundary.

###### Proof.

The disks are smoothly embedded and are bounded by the same knot in $S^{3}$,
so it suffices to show that these disks’ exteriors have
$\pi_{1}\cong\mathbb{Z}$ [CP19, Theorem 1.2]. Handle diagrams for these disk
exteriors are drawn in parts (a-1) and (b-1) of Figure 5; the remaining parts
of Figure 5 manipulate and decorate these diagrams to simplify the $\pi_{1}$
calculation. By tracing the 2-handle curves from parts (a-2) and (b-5)
starting at the labeled arrows, we obtain the following presentations:

$\displaystyle\pi_{1}(B^{4}{\smallsetminus}D_{1})$ $\displaystyle=\langle
x,y\mid x^{-1}xy^{-1}x^{-1}y=1\rangle=\langle x,y\mid
y^{-1}x^{-1}y=1\rangle\cong\mathbb{Z}$
$\displaystyle\pi_{1}(B^{4}{\smallsetminus}D_{2})$ $\displaystyle=\langle
x,y\mid xx^{-1}x^{-1}xyx^{-1}y^{-1}=1\rangle=\langle x,y\mid
yx^{-1}y^{-1}=1\rangle\cong\mathbb{Z}$

It follows that the disks are topologically ambiently isotopic rel. boundary.
∎

$0$
---
$0$
---
$x$
---
$y$
---
$x$
---
$y$
---
$0$
---
$0$
---
$0$
---
$0$
---
$0$
---
(b-1)
---
(b-2)
---
(b-3)
---
(b-4)
---
(b-5)
---
(a-1)
---
(a-2)
---
Figure 5. Redrawing the disk exteriors to compute their fundamental groups.

Now take the Bing doubles of the disks $D_{1}$ and $D_{2}$. Note that each of
$\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ are bounded by the
2–component link $\operatorname{BD}(K)$.

######  Corollary 3.2.

The disk links $\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ are
topologically ambiently isotopic rel. boundary.

###### Proof.

This follows from combining Proposition 3.1 and Lemma 2.1 with
$\operatorname{CAT}=\operatorname{TOP}$. ∎

By Lemma 2.2, $\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ are
Brunnian. To prove Theorem 1.3, it therefore remains to establish the
following proposition.

######  Proposition 3.3.

The disk links $\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ are
not smoothly equivalent.

The proof of the proposition uses the following technical lemma.

######  Lemma 3.4.

The knot complement $S^{3}{\smallsetminus}K$ has a hyperbolic structure with
trivial isometry group.

###### Proof.

This is verified using SnapPy [CDGW] and Sage [The19]; see Appendix A for
additional documentation regarding this calculation. ∎

###### Proof of Proposition 3.3.

To distinguish $\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$, we
will examine the simple closed curve $\gamma\subseteq S^{3}{\smallsetminus}K$
shown in Figure 3(c). This curve is redrawn in the exteriors of
$\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ in Figure 6. It is
easy to see that $\gamma$ bounds a smooth disk in the exterior of
$\operatorname{BD}(D_{2})$, since it bounds an obvious disk in the diagram
from Figure 6(b) that only intersects the 2-handle attaching circle (and not
the 1-handle curve).

$\gamma$
---
$\gamma$
---
$0$
---
$0$
---
(a)
---
(b)
---
Figure 6. Handle diagrams for the exteriors of the disk links
$\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ in parts (a) and
(b), respectively.

On the other hand, we claim that $\gamma$ does not bound a smoothly embedded
disk in the exterior of $\operatorname{BD}(D_{1})$. To prove this, we begin by
attaching three additional $-1$-framed 2-handles to the exterior of
$\operatorname{BD}(D_{1})$ as shown in Figure 7, yielding a larger 4-manifold
$W$. Note that, if $\gamma$ were to bound a smoothly embedded disk in the
exterior of $\operatorname{BD}(D_{1})$, then this disk could be capped off
with the core of the $-1$-framed 2-handle attached along $\gamma$ to produce a
smoothly embedded 2-sphere of square $-1$ in $W$. We will show that no such
2-sphere can exist. After simplifying the handle diagram (Figures 7-8), we see
that the resulting 4-manifold $W$ is obtained from $B^{4}$ by attaching a
single $-1$-framed 2-handle along a knot $J\subseteq S^{3}$. In Figure 7(f),
the knot $J$ has been drawn as a Legendrian knot with Thurston-Bennequin
number $tb=0$. (The Thurston-Bennequin number is computed from a Legendrian
knot diagram as the diagram’s writhe less the number of right cusps.) Since
the 2-handle is attached with framing $-1=tb-1$, $W$ admits a Stein structure
[Eli90]. By [LM98], it follows that $W$ cannot contain any smoothly embedded
2-sphere of square $-1$. It follows that $\gamma$ cannot bound a smoothly
embedded disk in the exterior of $\operatorname{BD}(D_{1})$.

This proves that $\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$ are
not smoothly equivalent rel. boundary. To obtain the stronger conclusion
stated in the theorem, we will show that any supposed diffeomorphism
$F\colon(B^{4},\operatorname{BD}(D_{1}))\to(B^{4},\operatorname{BD}(D_{2}))$
must fix $\gamma$ up to isotopy. Observe that, along the boundary, $F$
restricts to a self-diffeomorphism of $(S^{3},\operatorname{BD}(K))$. It
follows that it further restricts to a self-diffeomorphism of
$S^{3}{\smallsetminus}\nu\operatorname{BD}(K)$. By Lemma 3.4, $K$ has a
hyperbolic complement (with trivial isometry group, to be used below). A JSJ
decomposition of $S^{3}{\smallsetminus}\nu\operatorname{BD}(K)$ is therefore
given by

$S^{3}{\smallsetminus}\nu\operatorname{BD}(K)=(S^{3}{\smallsetminus}\nu
K)\cup(S^{1}\\!\\!\times\\!D^{2}{\smallsetminus}\operatorname{BD}(S^{1}\\!\times\\!\\{0\\})).$

The latter piece is diffeomorphic to the complement of the Borromean rings. By
uniqueness of JSJ decompositions of 3-manifolds [JS79, Joh79] (see also
Theorem 6.6) up to isotopy any diffeomorphism of pairs must restrict on
$S^{3}{\smallsetminus}\nu K\subseteq
S^{3}{\smallsetminus}\nu\operatorname{BD}(K)$ to a self-diffeomorphism of
$S^{3}{\smallsetminus}\nu K$. We need to show that every diffeomorphism of
$S^{3}{\smallsetminus}\nu K$ is isotopic to the identity to conclude that
$\gamma$ is fixed up to isotopy. By Mostow rigidity [Mos68], every homotopy
equivalence of a hyperbolic 3-manifold is homotopic to an isometry.
Waldhausen’s [Wal68] work, combined with Hatcher’s proof of the Smale
conjecture [Hat83], shows that for compact, orientable 3-manifolds $N\neq
B^{3}$ with nonempty boundary, homotopic diffeomorphisms are isotopic.
Combining these two statements, it follows that every diffeomorphism of a
hyperbolic knot complement is isotopic to an isometry. Since the isometry
group of $S^{3}{\smallsetminus}\nu K$ is trivial, this completes the proof of
the claim. ∎

$0$
---
(a)
---
(b)
---
(c)
---
(d)
---
(e)
---
(f)
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$0$
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$J$
---
Figure 7. Attaching three 2-handles to the exterior of
$\operatorname{BD}(D_{1})$ and simplifying. All steps except (b) to (c) and
(e) to (f) are isotopies. From (b) to (c), we cancel the red 2-handle with a
1-handle, introducing a full twist into all blue strands passing through the
1-handle. Diagrams (e) and (f) are related as in Figure 8. (d)
---
$-2$
---
(i)
---
(h)
---
(c)
---
(a)
---
(f)
---
(e)
---
(j)
---
(b)
---
(g)
---
$-1$
---
$-1$
---
$-1$
---
$-1$
---
$0$
---
$0$
---
$0$
---
$-1$
---
$-1$
---
$0$
---
$-1$
---
$-1$
---
$-2$
---
$-1$
---
$-1$
---
$n$
---
$n$
---
$n$
---
$n$
---
$n$
---
$n$
---
$n$
---
$n-1$
---
$n$
---
$n$
---
Figure 8. The process of simplifying (e) to (f) in Figure 7. From (a) to (b),
we add a cancelling handle pair. From (b) to (c), we remove a cancelling pair.
From (c) to (d) to (e) is simply isotopy of the diagram. From (e) to (f), we
remove a cancelling pair, and then isotope to obtain (g). From (g) to (h), we
do one handle slice, and then from (h) to (i) we remove a cancelling pair and
further isotope to obtain (j).

##  4\. Infinitely many exotic surface links

In this section, which is structured as follows, we construct the 2–component
surfaces promised in Theorem 1.4. In Section 4.1, we begin with some
background on Heegaard Floer cobordism maps; the obstructions we use arise
from these maps. In Section 4.3 we review the rim surgery construction. In
Section 4.4 we give the construction and show that rim surgery produces exotic
pairs of Brunnian surface links.

###  4.1. Heegaard Floer cobordism maps

In this section, the obstruction to smooth isotopy comes from maps on link
Floer homology. Link Floer homology was introduced by Ozsváth–Szabó [OS04,
OS08], while these cobordism maps were later defined by Juhász [Juh16]. We
refer the reader to e.g. [JMZ20] for a more detailed description, but for our
purposes the following very brief descriptions are sufficient.

######  Definition 4.1.

Given a link $l$ in a closed, oriented, connected 3-manifold $M$, let $\ell$
denote $l$ with two basepoints (labeled $w$ and $z$) chosen per component of
$l$. We call $\ell$ a _multi-pointed link._

There is a functor

$\widehat{\operatorname{{\emph{HFL}}}}\colon\operatorname{{DLink}}^{\times}\to\operatorname{{Vect}}^{\times}$

from the groupoid $\operatorname{{DLink}}^{\times}$ of multi-pointed links in
closed, oriented, connected 3-manifolds, with morphisms as smooth multi-
pointed isotopies, to the groupoid $\operatorname{{Vect}}^{\times}$ of
$\mathbb{F}_{2}$-vector spaces with linear isomorphisms. Given as input a
multi-pointed link $\ell$ in a 3-manifold $M$ we write the output as
$\widehat{\operatorname{{\emph{HFL}}}}(M,\ell)$ and call this the _link Floer
homology_ of $\ell$.

######  Definition 4.2.

Let $M_{0}$ and $M_{1}$ be closed, oriented, connected 3-manifolds. A
_decorated cobordism_ between multi-pointed links $\ell_{0}\subseteq M_{0}$
and $\ell_{1}\subseteq M_{1}$ consists of a compact, oriented surface $\Sigma$
neatly embedded in a compact, connected, oriented 4-manifold $W$ with:

  1. (1)

$\partial W=M_{0}\sqcup-M_{1}$;

  2. (2)

$\partial\Sigma=l_{0}\sqcup-l_{1}$; and

  3. (3)

a collection of arcs $\mathcal{A}$ properly embedded in $\Sigma$ such that:

     * •

$\mathcal{A}$ does not meet any of the $w$ or $z$ basepoints of $\ell_{i}$;

     * •

each component of $\ell_{i}-\\{w$ and $z$ basepoints$\\}$ meets exactly one
endpoint of $\mathcal{A}$;

     * •

the components of $\Sigma{\smallsetminus}\mathcal{A}$ can be sorted into two
subsurfaces $\Sigma_{w}$ and $\Sigma_{z}$ of $\Sigma$ so that all $w$
basepoints are in $\Sigma_{w}$ and all $z$ basepoints are in $\Sigma_{z}$.

We may view a surface $\Sigma$ neatly embedded in a 4-manifold $W$ with
connected boundary as being a cobordism from $(\emptyset,\emptyset)$ to the
boundary $(M,l)$. In this setting, it is particularly easy to decorate
$\Sigma$ after choosing $w$ and $z$ basepoints on each component of $l$: take
$\Sigma_{w}$ to consist of small bigons including each $w$ point and
$\Sigma_{z}=\Sigma{\smallsetminus}\Sigma_{w}$. We call this a _trivial
decoration of $\Sigma$._ If no other decoration is specified, then we will
always assume a surface is decorated trivially.

Juhász [Juh16] showed that decorated link cobordisms give rise to an extension
of the functor $\widehat{\operatorname{{\emph{HFL}}}}$. Let
$\operatorname{{DLink}}$ denote the category of multi-pointed links $(M,\ell)$
with morphisms smooth ambient isotopy classes of decorated cobordisms
$[(W,\Sigma,\mathcal{A})]$. Let $\operatorname{{Vect}}$ denote the category of
$\mathbb{F}_{2}$-vector spaces with linear transformations as morphisms. Since
we can consider a multi-pointed isotopy as a decorated cobordism,
$\operatorname{{DLink}}^{\times}\subseteq\operatorname{{DLink}}$ and
$\operatorname{{Vect}}^{\times}\subseteq\operatorname{{Vect}}$ are
subcategories that contain all the objects but fewer morphisms. Then Juhász
constructed a functor

$\widehat{\operatorname{{\emph{HFL}}}}\colon\operatorname{{DLink}}\to\operatorname{{Vect}}$

with the same value $\widehat{\operatorname{{\emph{HFL}}}}(M,\ell)$ on multi-
pointed links as the functor
$\widehat{\operatorname{{\emph{HFL}}}}\colon\operatorname{{DLink}}^{\times}\to\operatorname{{Vect}}^{\times}$
introduced above; this explains why we use the same notation.

In the first two items of the next remark, we make explicit what is meant by
the above formalism. The final item recalls a key computation of the cobordism
maps from [JMZ20].

######  Remark 4.3.

  1. (1)

Given a decorated link cobordism $(W,\Sigma,\mathcal{A})$ from
$(M_{0},\ell_{0})$ to $(M_{1},\ell_{1})$ there is an induced map

$F_{W,\Sigma}\colon\widehat{\operatorname{{\emph{HFL}}}}(M_{0},\ell_{0})\to\widehat{\operatorname{{\emph{HFL}}}}(M_{1},\ell_{1}).$

We will omit the arcs $\mathcal{A}$ from the notation for cobordism maps. This
map is well-defined up to smooth ambient isotopy of $\Sigma$ rel. boundary.
Given a surface $\Sigma$ whose boundary is a multi-pointed link $\ell$ in a
3-manifold $M$, we may view $\Sigma$ as having trivial decoration and find
that $\Sigma$ naturally induces a map

$F_{W,\Sigma}\colon\widehat{\operatorname{{\emph{HFL}}}}(\emptyset)\to\widehat{\operatorname{{\emph{HFL}}}}(M,\ell).$

The vector space $\widehat{\operatorname{{\emph{HFL}}}}(\emptyset)$ is a copy
of $\mathbb{F}_{2}$, so $F_{W,\Sigma}$ is determined by the image of the
single generator.

  2. (2)

According to [Juh16, Theorem 11.3], link cobordism maps behave well under
composition. That is, if a decorated cobordism $(W,\Sigma)$ from
$(M_{0},\ell_{0})$ to $(M_{1},\ell_{1})$ can be split as the composition of
decorated cobordisms $(W_{0},\Sigma_{0})$ from $(M_{0},\ell_{0})$ to
$(M^{\prime},\ell^{\prime})$ followed by $(W_{1},\Sigma_{1})$ from
$(M^{\prime},\ell^{\prime})$ to $(M_{1},\ell_{1})$, then

$F_{W,\Sigma}=F_{W_{1},\Sigma_{1}}\circ F_{W_{0},\Sigma_{0}}.$

  3. (3)

By [JMZ20, Corollary 8.4], if $\Sigma\subseteq B^{4}$ is obtained by pushing
the interior of a quasipositive Seifert surface for a knot in $S^{3}$ into
$B^{4}$ then $F_{B^{4},\Sigma}$ is nonvanishing. (In fact, $\Omega(\Sigma)=0$;
see Section 4.2)

The composition law of Remark 4.3 (2) is useful as a means of decomposing a
complicated cobordism into simpler ones. Remark 4.3 (3) is useful because in
practice, link cobordism maps are difficult to compute and this gives a large
family of examples where we at least know the induced map is nontrivial. If
the reader is unfamiliar with quasipositive surfaces, then it is useful to
know that, for example, the genus one Seifert surface for any iterated
positive untwisted Whitehead double of a positive knot $K$ (i.e.
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(\cdots(\operatorname{Wh}_{+}(K))\cdots))$)
is quasipositive [Rud01].

###  4.2. The $\boldsymbol{\Omega}$-invariant

The link cobordism map $F_{W,\Sigma}$ is an invariant of $\Sigma$ only up to
isotopy rel. boundary. In order to obstruct two surfaces from being smoothly
equivalent, we need a refined invariant. We make use of the invariant $\Omega$
from [JMZ20, Section 6], defined for surfaces embedded in $B^{4}$.

Given an oriented surface $\Sigma$ neatly embedded in $B^{4}$ with positive
genus and connected boundary, there exists an invariant
$\Omega(\Sigma)\in\mathbb{Z}^{\geq 0}\cup\\{-\infty\\}$ with the property that
if $(B^{4},\Sigma)$ is diffeomorphic to $(B^{4},\Sigma^{\prime})$, then
$\Omega(\Sigma)=\Omega(\Sigma^{\prime})$. The invariant $\Omega(\Sigma)$ is
defined to be $-\infty$ if and only if $F_{B^{4},\Sigma}$ vanishes.

In the paper [JMZ20], the surface $\Sigma$ is assumed to have connected
boundary, so to avoid tedious writing we will use $\Omega$ only in this
setting. The details of the construction of $\Omega$ are outside the scope of
this paper, but here is a heuristic description of $\Omega(\Sigma)$ when
$\Sigma$ has genus $g$.

  * •

View $F_{B^{4},\Sigma}(1)$ as an element of
$\widehat{\operatorname{{\emph{HFL}}}}(S^{3},\ell)\otimes\mathbb{F}[\mathbb{Z}^{2g}]$.

  * •

For each element $a$ of $\mathbb{F}_{2}[\mathbb{Z}^{2g}]$, let $\Omega(a)$
denote the number of irreducible factors of $a$, counted with multiplicity
(here using the fact that $\mathbb{F}_{2}[\mathbb{Z}^{2g}]$ is a UFD).

  * •

Now let $\Omega(\Sigma)=\max\\{\Omega(a)\mid F_{B^{4},\Sigma}(1)=a\cdot y$ for
some $y\\}$.

This is a description and _not_ a true definition. The proper definition
involves deformed knot Floer homology. Please see [JMZ20] for more details.

###  4.3. Rim surgery

We briefly recall the technique of rim surgery. This is a well-known method
for producing potentially-exotic pairs of surfaces in 4-manifolds that was
invented by Fintushel and Stern [FS97] and generalizes the twist-spin
construction of Zeeman [Zee65].

######  Definition 4.4.

Let $X$ be a 4-manifold and let $\Sigma\subseteq X$ be a smoothly embedded
surface in $X$. Let $\alpha\subseteq\Sigma$ be a simple closed curve with
$w_{1}^{\Sigma}(\alpha)=0$, and let $J\subseteq S^{3}$ be a knot.

Choose a framing $\eta$ of the 2-subbundle of the normal bundle of $\alpha$
that is normal to $\Sigma$. Now we can identify $\nu\alpha$ with $B^{3}\times
S^{1}$; choose this identification so that $\alpha=\\{0\\}\times S^{1}$,
$\Sigma\cap\nu\alpha=I\times S^{1}$ for a fixed vertical arc $I\subseteq
B^{3}$, and $\eta$ restricts to the same pair of vectors in $T_{0}B^{3}$ for
each $\\{0\\}\times\theta$. We construct the _rim surgered surface_
$\Sigma(\alpha;J)\subseteq X$ as follows. This surface agrees with $\Sigma$
outside of $\overline{\nu}\alpha$. Note that
$(\overline{\nu}\alpha,\Sigma\cap\overline{\nu}\alpha)=(B^{3}\times
S^{1},I\times S^{1})$, where $I$ is an unknotted arc in $B^{3}$. To obtain
$\Sigma(\alpha;J)$, replace each $(B^{3},I)$ with $(B^{3},\mathring{J})$, i.e.
the tangle obtained from $(S^{3},J)$ by deleting a small ball from $S^{3}$
centered at a point on $J$.

As written, $\Sigma(\alpha;J)$ depends on our choice of framing $\eta$. There
are an integers’ worth of such framings that induce a fixed orientation on the
subbundle. The effect of choosing a different $\eta$ is to twist
$B^{3}\times\theta$ an integer number of times about a fixed axis with
boundary $\partial\mathring{J}$ as $\theta$ goes from $0$ to $2\pi$. (See
Figure 9, second vs. third row.)

One setting in which it is easy to specify a framing $\eta$ is in the case
that $\alpha$ bounds a framed locally flat disk $\Delta$ in the complement of
$\Sigma$. Here the disk $\Delta$ is _framed_ if a section of the normal bundle
of $\alpha$ that lies in $T\Sigma$ extends to a nonvanishing section over all
of $\Delta$. Then we can specify that for each $x\in\alpha$, the first
coordinate of $\eta(x)$ points into $\Delta$. This ensures that $\Delta$
intersects $(\nu\alpha)$ in an annulus of the form (arc)$\times S^{1}$. In
this setting, we say that replacing $\Sigma\cap\overline{\nu}\alpha$ with
$\mathring{J}\times S^{1}$ yields $\Sigma(\alpha;J,0)$. We call this
_0-twisted rim surgery_ (relative to $\Delta$). If we twist the copy of
$\mathring{J}$ a total of $n$ times, as above, then instead we obtain a
surface that we call $\Sigma(\alpha;J,n)$. We call this _$n$ -twisted rim
surgery_. When $\Delta$ has been implicitly specified, we will write
$\Sigma(\alpha;J,n)$ as a well-defined surface without reference to $\Delta$.
In Figure 9, we illustrate a neighborhood of $\alpha$ intersecting $\Sigma$,
$\Sigma(\alpha;T,0)$, and $\Sigma(\alpha;T,1)$ for $T$ the right-handed
trefoil.

$B^{3}\times 0$ at 27 200 $B^{3}\times\pi/2$ at 99 200 $B^{3}\times\pi$ at 170
200 $B^{3}\times 3\pi/2$ at 240 200 $B^{3}\times 2\pi$ at 310 200 $B^{3}\times
0$ at 27 95 $B^{3}\times\pi/2$ at 99 95 $B^{3}\times\pi$ at 170 95
$B^{3}\times 3\pi/2$ at 240 95 $B^{3}\times 2\pi$ at 310 95 $B^{3}\times 0$ at
27 -12 $B^{3}\times\pi/2$ at 99 -12 $B^{3}\times\pi$ at 170 -12 $B^{3}\times
3\pi/2$ at 240 -12 $B^{3}\times 2\pi$ at 310 -12 $\Delta$ at 115 250 $\alpha$
at 20 242 $\Sigma\cap\overline{\nu}\alpha$ at -25 240
$\Sigma(\alpha;T,0)\cap\overline{\nu}\alpha$ at -45 133
$\Sigma(\alpha;T,1)\cap\overline{\nu}\alpha$ at -45 28

Figure 9. Top row: we draw a neighborhood $\overline{\nu}\alpha$ of $\alpha$,
a simple closed curve in $\Sigma$ with annular neighborhood. In blue, we draw
the portion of a framed disk $\Delta$ bounded by $\alpha$ that intersects
$\overline{\nu}\alpha$. We draw $\overline{\nu}\alpha$ as $B^{3}\times S^{1}$,
with $\alpha$ as $0\times S^{1}$. The surface $\Sigma$ intersects
$\overline{\nu}\alpha$ in (vertical arc)$\times S^{1}$ as illustrated. The
parametrization of $\overline{\nu}\alpha$ is chosen so that $\Delta$
intersects it in (arc from $0$ to $\partial B^{3}$)$\times S^{1}$, as
illustrated. Middle row: replacing $\Sigma\cap\overline{\nu}\alpha=I\times
S^{1}$ with $\mathring{T}\times S^{1}$ yields $\Sigma(\alpha;T,0)$. Bottom:
replacing the $I\times\theta$s of $\Sigma\cap\overline{\nu}\alpha$ with copies
of $\mathring{T}$ that rotate once about a vertical axis as $\theta$ runs from
$0$ to $2\pi$ yields $\Sigma(\alpha;T,1)$.

######  Remark 4.5.

Let $\Sigma$ be an unknotted sphere in $S^{4}$ obtained by doubling
$(B^{4},D)$, where $D$ is a boundary-parallel disk in $B^{4}$. Let
$\alpha=\partial D$, so $\alpha$ is a curve in $\Sigma$ that bounds a framed
disk (a parallel copy of $D$) into the complement of $\Sigma$. Then
$\Sigma(\alpha;J,n)$ is the $n$-twist spin of the knot $J$ as constructed by
Zeeman [Zee65]. Zeeman also showed that the 1-twist spin of any knot $J$ is an
unknotted $S^{2}S^{4}$.

The next lemma, proven in [JMZ20], gives a criterion for rim surgery to have
no essential effect.

######  Lemma 4.6.

If $\alpha$ bounds a $\operatorname{CAT}$-embedded framed disk $D$ whose
interior lies in $B^{4}{\smallsetminus}\Sigma$, then $\Sigma(\alpha;J,1)$ and
$\Sigma$ are $\operatorname{CAT}$ ambiently isotopic rel. boundary.

###### Proof.

The proof of this lemma is based on Zeeman’s [Zee65] work on twist-spun knots.
Since $D$ is framed, there is a thickening of $D$ to $D\times I$ meeting
$\Sigma$ in $(\partial D)\times I$; i.e. we can take parallel copies of $D$
that are disjoint from $D$ and have boundary on $\Sigma$. Let $D^{\prime}$ and
$D^{\prime\prime}$ be two parallel copies of $D$, pushed to have boundary
curves on opposite sides of $\alpha$ in $\Sigma$, and compress $\Sigma$ along
both $D^{\prime}$ and $D^{\prime\prime}$ to obtain a surface
$\Sigma^{\prime}$. Then $\Sigma^{\prime}$ has a 2-sphere component $U$
bounding $D\times I$, so $U$ is $\operatorname{CAT}$-unknotted in the
complement of the rest of $\Sigma^{\prime}$.

The surface $\Sigma$ can be obtained from $\Sigma^{\prime}$ by surgery along
two framed arcs $\eta^{\prime}$ and $\eta^{\prime\prime}$ (undoing the
compressions along $D^{\prime}$ and $D^{\prime\prime}$, respectively). See
Figure 10. In addition, the curve $\alpha$ exists in both $\Sigma$ and
$\Sigma^{\prime}$, so rim surgery can be performed on $\alpha$ inside either
surface. By the same logic we then see that $\Sigma(\alpha;J,1)$ can be
obtained from $\Sigma^{\prime}(\alpha;J,1)$ by surgery along $\eta^{\prime}$
and $\eta^{\prime\prime}$. Note that one component of
$\Sigma^{\prime}(\alpha;J,1)$ is the rim-surgered 2-sphere $U(\alpha;J,1)$. As
in Remark 4.5, Zeeman [Zee65] showed that the 1-twist spin $U(\alpha;J,1)$ is
$\operatorname{CAT}$-unknotted for any $J$, so $\Sigma^{\prime}(\alpha;J,1)$
is $\operatorname{CAT}$-isotopic to $\Sigma^{\prime}$. Since $\eta^{\prime}$
and $\eta^{\prime\prime}$ are arcs each meeting $\nu U$ in only one interval,
we may arrange for this isotopy to take $\eta^{\prime}$ and
$\eta^{\prime\prime}$ to $\eta^{\prime}$ and $\eta^{\prime\prime}$,
respectively (this uses homotopy implies isotopy and the fact that $\nu U\cong
S^{2}\times D^{2}$ is simply connected). Therefore, $\Sigma(\alpha;J,1)$ is
$\operatorname{CAT}$-isotopic to the result of surgering $\Sigma^{\prime}$
along $\eta^{\prime}$ and $\eta^{\prime\prime}$, which we noted above is
isotopic to $\Sigma$. Thus $\Sigma(\alpha;J,1)$ is
$\operatorname{CAT}$-isotopic to $\Sigma$ as desired. ∎

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}D}$ at 35
84
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}D^{\prime}}$
at 37.5 48
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}D^{\prime\prime}}$
at 38 121
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\alpha}$ at
-10 83 $U$ at 190 70 $\eta$ at 151 37 $\eta^{\prime}$ at 154 132

Figure 10. Left: a neighborhood of a framed disk $D$ with boundary $\alpha$ on
$\Sigma$. Since $D$ is framed, we may find two parallel copies $D^{\prime}$
and $D^{\prime\prime}$ of $D$ on opposite sides of $D$ (in $\Sigma$) with
boundary on $\Sigma$. Right: compressing $\Sigma$ along $D^{\prime}$ and
$D^{\prime\prime}$ yields a disconnected surface $\Sigma^{\prime}$ with an
unknotted 2-sphere component $U$. Surgering $\Sigma^{\prime}$ along the arcs
$\eta,\eta^{\prime}$ (with framing chosen to yield an surface orientable in
the pictured neighborhood of $D$) yields $\Sigma$ again.

On the other hand, rim surgery does often change the isotopy type of a surface
when outside the setting of Lemma 4.6. One potential obstruction to smooth
isotopy comes from the link cobordism maps of Section 4.1, since if surfaces
$\Sigma_{1}$ and $\Sigma_{2}$ in $X$ are isotopic rel. boundary then
$F_{X,\Sigma_{1}}=F_{X,\Sigma_{2}}$ (where we choose $w,z$ basepoints of
$\partial\Sigma_{1}=\partial\Sigma_{2}$ to agree).

######  Lemma 4.7 ([JZ18]).

Let $\Sigma$ be a smooth surface neatly embedded in $B^{4}$. Let $\alpha$ be
an essential, non-separating simple closed curve on $\Sigma$ and $J$ a knot.
Then

$F_{B^{4},\Sigma(\alpha;J)}(1)=\Delta_{J}(z)F_{B^{4},\Sigma}(1).$

In particular, if $F_{B^{4},\Sigma}$ is nonvanishing and $J$ has nontrivial
Alexander polynomial then $\Sigma$ and $\Sigma(\alpha;J)$ are not smoothly
isotopic rel. boundary. More generally, if $F_{B^{4},\Sigma}$ is nonvanishing
and if $J_{1}$ and $J_{2}$ are knots with Alexander polynomials that are
distinct up to multiplication with a monomial $\pm t^{k}$, then the surfaces
$\Sigma(\alpha;J_{1})$ and $\Sigma(\alpha;J_{2})$ are not smoothly isotopic
rel. boundary.

For $\partial\Sigma$ connected and $k\geq 0$ equal to the number of
irreducible factors of the Alexander polynomial of $J$ $($counted with
multiplicity$)$, we conclude $\Omega(\Sigma(\alpha;J))=\Omega(\Sigma)+k$.

###  4.4. Construction of exotic surface links

Now we construct an infinite family of exotic 2–component Brunnian links. Each
link will have a disk component and a genus one component. Fix $K$ to be some
strongly quasipositive topologically slice knot, e.g. the positive untwisted
Whitehead double of the right-handed trefoil. This is topologically slice by
[FQ90, Theorem 11.7B] because the Alexander polynomial of any untwisted
Whitehead double is trivial, and it is strongly quasipositive by [Rud01].

Now let $\Sigma=S_{1}\sqcup S_{2}$ be the surface link pictured in Figure 11.
In this figure, we draw a disk and a genus one surface immersed in $S^{3}$
with ribbon intersections. The interiors of these surface should be pushed
slightly into $B^{4}$ to obtain disjoint surfaces (as usual). Each knot
$\partial S_{i}$ is an unknot, $S_{1}$ is an unknotted disk, and $S_{2}$ is an
unknotted genus one surface.

$S_{2}$ at 20 20
${\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}S_{1}}$
at 0 200

Figure 11. The surface link $\Sigma=S_{1}\sqcup S_{2}$. At the bottom of the
figure, two bands in $S_{2}$ are knotted through $K$. The diagram of $K$
should be taken to have writhe zero.

######  Proposition 4.8.

Let $C$ be the curve on $S_{2}$ pictured in Figure 12. Then $C$ bounds framed
disks $\Delta$ and $\Delta^{\prime}$ into the complement of $S_{2}$, with
$\Delta$ smooth and $\Delta^{\prime}$ locally flat, and with $\Delta^{\prime}$
disjoint from $S_{1}$. Moreover, $\Delta$ and $\Delta^{\prime}$ can be taken
to agree near their boundary.

###### Proof.

In the right half of Figure 12, we illustrate that $C$ bounds a framed locally
flat disk $\Delta^{\prime}$ whose interior is disjoint from both $S_{2}$ and
$S_{1}$. This disk is obtained from the two pictured locally flat disks for
$K$ by gluing them along the arc where their boundaries coincide.

Viewed as a knot in $S^{3}$ on the ribbon surface $S_{2}$, $C$ is an unknot
and the framing induced by $S_{2}$ on $C$ is the 0-framing. We conclude that
$C$ bounds a framed smooth disk $\Delta$ into $B^{4}$ that is disjoint from
$S_{2}$ in its interior. Moreover, since $\Delta^{\prime}$ as drawn induces
the 0-framing on $C$ as a knot in $S^{3}$, we can arrange for $\Delta$ and
$\Delta^{\prime}$ to agree near their boundaries. ∎

Now that we have specified framed disks (that agree near boundary) bounded by
$C$, it makes sense to consider $n$-twisted rim surgery on $C$.

######  Corollary 4.9.

For any knot $J$, $S_{2}(C;J,1)$ is smoothly isotopic rel. boundary to $S_{2}$
and $\Sigma(C;J,1)$ is topologically isotopic rel. boundary to $\Sigma$.

###### Proof.

This follows immediately from Proposition 4.8 and Lemma 4.6. ∎

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}C}$ at 250
220 locally flat disks for $K$ at 715 -20

Figure 12. Left: a curve $C$ on $S_{2}$. Since $K$ has writhe zero as
pictured, $S_{2}$ induces the 0-framing on $C$ (viewed as a surface in
$S^{3}$). Right: because $K$ is topologically slice, $C$ bounds a topological
locally flat disk whose interior is disjoint from both $S_{2}$ and $S_{1}$
(and which is normal to $S_{2}$ near its boundary). The disk is obtained by
gluing the two drawn disks together along an arc. Both drawn disks are behind
$S_{2}$; the left disk has two ribbon intersections with $S_{1}$ when immersed
in $S^{3}$ while the right disk is completely behind $S_{1}$.

In order to apply Lemma 4.7 and obstruct smooth isotopy rel. boundary, we must
show that the map
$F_{B^{4},L}:\widehat{\operatorname{{\emph{HFK}}}}(\emptyset)\to\widehat{\operatorname{{\emph{HFK}}}}(S^{3},\partial\Sigma)$
is nonvanishing. We will achieve this by obtaining a strongly quasipositive
surface from $\Sigma$, even though $\Sigma$ itself is _not_ strongly
quasipositive (since $S_{2}$ is not).

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}b}$ at 60
70

Figure 13. Left: we form a connected surface $S$ from $S_{2}$ and $S_{1}$ via
attaching the band $b$ pictured. Right: We draw the knot $\partial S$ with
four bands contained in $S$ representing index-1 critical points of the radial
height function on $B^{4}$ restricted to $S$. Surgering $\partial S$ along
these bands yields an unlink; disks bounded by this unlink give the minima of
$S$.

######  Proposition 4.10.

Let $S$ be the connected surface obtained from $L$ by attaching $S_{2}$ and
$S_{1}$ via the band depicted in Figure 13 $($left$)$. $($Push the interior of
the resulting surface into $B^{4}$ to obtain a properly embedded surface$)$.
Then $S$ is smoothly isotopic to the standard genus one Seifert surface for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$ $($with interior pushed
slightly into $B^{4})$.

###### Proof.

In Figure 14, we isotope $S$ to see $\partial
S=\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$. The surface $S$ is
embedded in $B^{4}$ with four index-1 and three index-0 points with respect to
the standard radial height function on $B^{4}$. The index-1 points are
flattened and drawn as bands attached to $\partial S$; surgering $\partial S$
along these bands yields a 3–component unlink. Moreover, two of these bands
($b_{1}$ and $b_{2}$, as labeled) lie on the standard genus one Seifert
surface for $\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$. After surgering
$\partial S$ along these bands, we obtain an unknot with two trivial bands
attached. We conclude that the remaining two bands, i.e. index-1 points of
$S$, can be removed via isotopy of $S$. Thus, $S$ is isotopic to the standard
genus one Seifert surface for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$. ∎

slide bands at 640 500 $b_{1}$ at 365 135 $b_{2}$ at 640 105

Figure 14. Left to right, top to bottom: we isotope the diagram for $S$
obtained in Figure 13. At one point, we slide two bands over another band.
Isotopy and slides of bands describe isotopy of $S$. In the final diagram, we
see that $\partial S=\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$ and note
that the bands $b_{1}$ and $b_{2}$ lie on the standard genus one surface for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$.

resolve at 340 195 two at 340 165 bands at 340 135

Figure 15. Starting with the last diagram of Figure 14, we surger $\partial S$
along the two bands lying on the standard surface for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$. We then isotope the result
to obtain the figure on the right: two trivial bands attached to an unknot. We
conclude that the remaining two bands are cancellable, and therefore $S$ is
isotopic to the standard genus one surface for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$.

######  Corollary 4.11.

The map
$F_{B^{4},L}\colon\widehat{\operatorname{{\emph{HFK}}}}(\emptyset)\to\widehat{\operatorname{{\emph{HFK}}}}(S^{3},\partial
L)$ is nonvanishing.

###### Proof.

Let $B$ denote the link cobordism described by the band $b$ (recall Figure
13). Then $B$ is a cobordism from $\partial\Sigma$ to
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$. By Proposition 4.10,
$B\circ\Sigma$ is isotopic to the standard genus one Seifert surface $S$ for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$. By Remark 4.3 (2), we have

$F_{B^{4},S}=F_{S^{3}\times I,B}\circ F_{B^{4},\Sigma}.$

Since $K$ is strongly quasipositive, $S$ is strongly quasipositive [Rud01]. By
Remark 4.3 (3), $F_{B^{4},S}$ is nonvanishing. Therefore, $F_{B^{4},\Sigma}$
is also nonvanishing. ∎

######  Theorem 4.12.

Fix $K$ to be some strongly quasipositive topologically slice knot. Let
$\Sigma_{0}:=S_{1}\sqcup S_{2}$ be the surface link pictured in Figure 11. Let
$C$ be the curve on $S_{2}$ illustrated in Figure 12. For each
$n\in\mathbb{N}$, let $J_{n}$ be a knot whose Alexander polynomial has $n$
irreducible factors counted with multiplicity $($e.g. $J_{n}$ may be an
$n$-fold connected sum of trefoil knots$)$. Let
$\Sigma_{n}:=\Sigma(C;J_{n},1)$. Then each $\Sigma_{n}$ is Brunnian.
Additionally, for $n\neq m$, we find that $\Sigma_{n}$ and $\Sigma_{m}$ form
an exotic pair.

As stated above, the positive untwisted Whitehead double of the right-handed
trefoil is one possible choice for $K$.

######  Remark 4.13.

In Theorem 4.12, if we ask only that for $n\neq m$, $J_{n}$ and $J_{m}$ have
inequivalent Alexander polynomials, up to multiplication by a unit in
$\mathbb{Z}[t,t^{-1}]$, rather than that their Alexander polynomials have
different numbers of irreducible factors, then we may still conclude that
$\Sigma_{n}$ and $\Sigma_{m}$ are not smoothly isotopic rel. boundary by
applying Lemma 4.7 and Corollary 4.11.

###### Proof.

Since $\Sigma_{0}$ is Brunnian, it follows from Corollary 4.9 that each
$\Sigma_{n}$ is Brunnian. Moreover, Corollary 4.9 implies $\Sigma_{n}$ is
topologically isotopic rel. boundary to $\Sigma_{0}$ for all $n$.

Assume $n<m$. Let $S$ denote the genus one Seifert surface for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))$ with interior pushed
slightly into $B^{4}$. By Proposition 4.10, there is a band $b$ so that
$\Sigma_{n}\cup b$ is isotopic rel. boundary to $S(C;J_{n},1)$. Applying
Remark 4.3 (3) and Lemma 4.7 shows that $\Omega(S(C;J_{n},1))=n$. Suppose
there is a diffeomorphism $f\colon(B^{4},\Sigma_{n})\to(B^{4},\Sigma_{m})$. We
cannot assume that $f$ restricts to the identity on $\partial B^{4}$, so this
does not yield an equivalence from $S(C;J_{n},1)$ to $S(C;J_{m},1)$. Instead,
we obtain an equivalence from $(B^{4},S(C;J_{n},1))$ to
$(B^{4},f(b)\cup\Sigma_{m})$, where $f(b)$ is some other band attached to
$\partial\Sigma_{m}$. Note that $f(b)\cup\Sigma_{m}$ is obtained from
$f(b)\cup\Sigma_{0}$ by $1$-twist rim surgery along $C$ using $J_{m}$.
Applying Lemma 4.7, we have either

$\displaystyle\Omega(f(b)\cup\Sigma_{m})$ $\displaystyle=-\infty\text{ if
$F_{B^{4},f(b)\cup\Sigma_{0}}$ vanishes, or}$
$\displaystyle\Omega(f(b)\cup\Sigma_{m})$ $\displaystyle\geq m\text{ if
$F_{B^{4},f(b)\cup\Sigma_{0}}$ does not vanish}.$

In both cases, $\Omega(f(b)\cup\Sigma_{m})\neq\Omega(S(C;J_{n},1))=n$,
yielding a contradiction. We conclude that there is no diffeomorphism from
$(B^{4},\Sigma_{n})$ to $(B^{4},\Sigma_{m})$. ∎

##  5\. Surfaces of higher genus

Now we extend the construction of Section 4.4 to produce higher genus
surfaces, proving Theorem 1.6. From now on, let $\Sigma_{0}=S_{1}\cup S_{2}$
be as in Theorem 4.12 (i.e. the surface link in Figure 11, with $K$ a
topologically slice and strongly quasipositive knot). Recall that $S_{1}$ is a
disk and $S_{2}$ is genus one. In shorthand, we write that $\Sigma_{0}$ has
genus $(0,1)$.

In order to use $\Sigma_{0}$ to produce exotic surfaces of higher genus, it
will be useful to understand a certain high genus Brunnian link. Fix a
positive integer $g$ and let $R^{g}=R^{g}_{1}\sqcup R^{g}_{2}$ be the surface
link in $B^{4}$ illustrated in Figure 16. We also draw a band $b^{g}_{R}$
between the two components of $R^{g}$. The link $R^{g}$ is Brunnian of genus
$(0,g)$.

$R^{g}_{1}$ at 60 10 $R^{g}_{2}$ at -10 100 $b^{g}_{R}$ at 5 40 $\cdots$ at
249 110 genus $g$ at 249 95 $\cdots$ at 249 15

Figure 16. A 2–component surface link $R^{g}=R_{1}^{g}\sqcup R_{2}^{g}$ and a
band $b^{g}_{R}$ between the components of $R^{g}$. Note that $R_{1}^{g}$ is
an unknotted disk and $R^{g}_{2}$ is an unknotted genus $g$ surface. Viewed as
a ribbon surface in $S^{3}$, there are $g$ arcs of intersection between the
pictured ribbon surfaces.

######  Proposition 5.1.

The surface $\widetilde{R}^{g}$ obtained from gluing $R^{g}_{1}$ and
$R^{g}_{2}$ together along $b^{g}_{R}$ is isotopic to a positive Seifert
surface $($pushed slightly into $B^{4})$.

###### Proof.

In Figure 17, we illustrate an isotopy from $\widetilde{R}^{g}$ to a positive
Seifert surface (slightly pushed into $B^{4}$) for a connected sum of $g$
right-handed trefoil knots. ∎

$\cdots$ at 250 690 $\cdots$ at 250 495 $\cdots$ at 250 300 $\cdots$ at 250
105

Figure 17. We draw the boundary of $\widetilde{R}^{g}$ with bands indicating
index-1 critical points into $B^{4}$. From top to bottom, we isotope
$\widetilde{R}^{g}$. To get from the third to the fourth picture, we also
slide the bottommost band over all of the circled bands. In the end, we obtain
a positive genus $g$ Seifert surface (slightly pushed into $B^{4}$) for a
connected sum of $g$ right-handed trefoil knots.

Now we use $R^{g}$ to increase the genus of $\Sigma_{0}$. For any integers
$r,s$ with $r\geq 0$ and $s\geq 1$, let $\Sigma^{r,s}_{0}$ be obtained as
follows (see Figure 18 (top)).

  1. 1.

Start with a copy of $\Sigma_{0}$.

  2. 2.

If $r>0$, then band a copy of $R^{r}$ to $\Sigma_{0}$. The bands are indicated
in Figure 18 (top); note that the disk component of $R^{r}$ is banded to
$S_{2}$ while the positive genus component of $R^{r}$ is banded to $S_{1}$.

  3. 3.

If $s>1$, then band a copy of $R^{s-1}$ to $\Sigma_{0}$. The bands are again
indicated in Figure 18 (top); note that the disk component of $R^{s-1}$ is
banded to $S_{1}$ while the positive genus component of $R^{s-1}$ is banded to
$S_{2}$.

  4. 4.

Call the result genus $(r,s)$ surface $\Sigma^{r,s}_{0}$.

$R^{s-1}$ at 500 535 $R^{r}$ at 500 380 $C$ at 15 70 $b$ at 70 145

Figure 18. Top: The 2–component surface link $\Sigma^{r,s}_{0}$ (here we draw
as if $r>0$ and $s>1$). Bottom: a band $b$ between the components of
$\Sigma^{r,s}_{0}$ and a curve $C$ lying on $\Sigma^{r,s}_{0}$.

In Figure 18 (bottom), we illustrate a certain band $b$ between the two
components of $\Sigma^{r,s}_{0}$ and a curve $C$ (the same as in Theorem 4.12)
that lies on $\Sigma^{r,s}_{0}$. Just as in Proposition 4.8, we see that $C$
bounds a smooth framed disk into the complement of the component on which it
lies, and a locally flat framed disk into the complement of
$\Sigma^{r,s}_{0}$. The following proposition illustrates why we chose the
specific bands in Figure 18 (top).

######  Proposition 5.2.

The surface obtained by gluing the components of $\Sigma^{r,s}_{0}$ together
along $b$ is isotopic to a quasipositive Seifert surface $($pushed slightly
into $B^{4})$.

###### Proof.

In Figure 19, we isotope the result of gluing $\Sigma^{r,s}_{0}$ along $b$ to
obtain a copy of $\Sigma_{0}\cup b$ trivially boundary-summed with $R^{r}\cup
b^{r}_{R}$ and $R^{s-1}\cup b^{s-1}_{R}$. (If $r=0$ or $s=1$ then ignore the
corresponding $R$.) From Propositions 4.10 and 5.1, we find that this is a
quasipositive Seifert surface (pushed into $B^{4}$) for
$\operatorname{Wh}_{+}(\operatorname{Wh}_{+}(K))\\#(\\#_{r+s-1}($right-handed
trefoils$))$. ∎

Figure 19. We glue the components of $\Sigma^{r,s}_{0}$ together along the
band $b$. From top to bottom, we perform isotopy via band slides. We draw
arrows to preemptively indicate band slides. Note at the end the resemblance
to Figures 14 and 17.

######  Theorem 5.3.

Fix integers $r,s$ with $r\geq 0$ and $s\geq 1$. For each $n\in\mathbb{N}$,
let $J_{n}$ be a knot in $S^{3}$ whose Alexander polynomial has $n$
irreducible factors counted with multiplicity. Let
$\Sigma^{r,s}_{n}=\Sigma^{r,s}_{0}(C;J_{n},1)$. Then for all $m>n\geq 0$, we
have that:

  * •

$\Sigma^{r,s}_{n}$ is a Brunnian genus $(r,s)$ link,

  * •

$\Sigma^{r,s}_{n}$ and $\Sigma^{r,s}_{m}$ are topologically isotopic rel.
boundary,

  * •

$(B^{4},\Sigma^{r,s}_{n})$ is not diffeomorphic to $(B^{4},\Sigma^{r,s}_{m})$.

######  Remark 5.4.

As in Remark 4.13, if we only have that $\Delta_{J_{n}}\neq\Delta_{J_{m}}$ up
to a unit, then we still obtain that $(B^{4},\Sigma^{r,s}_{n})$ and
$(B^{4},\Sigma^{r,s}_{m})$ are not smoothly isotopic rel. boundary.

###### Proof.

Since $C$ bounds a smooth disk into the complement of the component of
$\Sigma^{r,s}_{0}$ on which it lies, it follows from Lemma 4.6 that
$\Sigma^{r,s}_{n}$ is Brunnian.

Moreover, since $C$ bounds a locally flat disk into the complement of
$\Sigma^{r,s}_{0}$, it follows also from Lemma 4.6 that $\Sigma^{r,s}_{n}$ is
topologically isotopic rel. boundary to $\Sigma^{r,s}_{0}$ for any $n$.

By Proposition 5.2, $\Sigma^{r,s}_{0}\cup b$ is isotopic to a strongly
quasipositive Seifert surface pushed into $B^{4}$. It follows from Remark 4.3
(3) and Lemma 4.7 that $\Omega(\Sigma^{r,s}_{n}\cup b)=n$. As in Theorem 4.12,
we note that if $\Sigma^{r,s}_{n}$ and $\Sigma^{r,s}_{m}$ are smoothly
equivalent, then $\Sigma^{r,s}_{n}\cup b$ is smoothly equivalent to
$\Sigma^{r,s}_{m}\cup b^{\prime}$ for some band $b^{\prime}$, so
$\Omega(\Sigma^{r,s}_{m}\cup b^{\prime})=n$. But $(\Sigma^{r,s}_{m}\cup
b^{\prime})$ is obtained from 1-twist rim surgery using knot $J_{m}$ on
$(\Sigma^{r,s}_{0}\cup b^{\prime})$, so by Lemma 4.7 we have

$\Omega(\Sigma^{r,s}_{m}\cup
b^{\prime})=-\infty\quad\text{or}\quad\Omega(\Sigma^{r,s}_{m}\cup
b^{\prime})\geq m.$

This yields a contradiction, so we conclude that $(B^{4},\Sigma^{r,s}_{n})$ is
not diffeomorphic to $(B^{4},\Sigma^{r,s}_{m})$. ∎

##  6\. Surfaces with more than two components

In this section we construct Brunnian exotic surface pairs with more than two
components. Our aim is to prove Theorem 1.5, in which we show that taking a
pair of surface links from one of Theorem 1.3 or Theorem 1.4, and applying
iterated Bing doubling to the first component, we obtain exotic pairs.

The argument of this section uses covering space methods. It was inspired by
the work of Cha-Kim [CK08] on concordance of 1-links obtained from Bing
doubling. We extend their covering link methods to the 4–dimensional setting.
The main obstacle is to show that there is no diffeomorphism of pairs relating
Bing doubled surfaces, provided we start with a pair of surface links that
admit no diffeomorphism of pairs.

Section 6.1 introduces the notion of covering surfaces, and investigates
covering surfaces of Bing doubles. Section 6.2 recalls some of the theory of
JSJ decompositions of 3-manifolds, and studies it to Bing doubles and their
covering links. Section 6.3 applies this to show that, in a fairly general
setting, Bing doubling a pair of exotic surface links yields another pair of
exotic surface links.

###  6.1. Coverings and Bing doubles

######  Lemma 6.1.

Let $\Sigma=\Sigma_{0}\sqcup\Sigma_{1}\sqcup\cdots\sqcup S_{n}$ be a surface
link in $B^{4}$, and suppose that $\Sigma_{0}$ is an unknotted disk. Fix
$k\geq 1$. Suppose that $H_{1}(\Sigma{\smallsetminus}\Sigma_{0};\mathbb{Z})\to
H_{1}(B^{4}{\smallsetminus}\Sigma_{0};\mathbb{Z})\cong\mathbb{Z}$ has image in
$k\mathbb{Z}$. Let $p\colon B^{4}\to B^{4}$ be a $k$-fold branched covering
map with branching set $\Sigma_{0}$. Then
$p^{-1}(\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n})$ is a $kn$–component surface
link.

###### Proof.

The homological condition guarantees that each component of
$\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n}$ lifts to a $k$ component surface
link. ∎

######  Definition 6.2.

Let $\Sigma$ be a surface link. A _covering surface link_ of $\Sigma$ is a
surface link obtained from $\Sigma$ by a finite sequence of the following two
operations.

  1. (1)

Taking the pre-image of $\Sigma$ under a branched covering map $p$, as in
Lemma 6.1, with branching set an unknotted disk, and forgetting the branching
set.

  2. (2)

Passing to a sublink.

The following lemma is the key example for our use of covering surfaces.

######  Lemma 6.3.

Let $\Sigma=\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n}$ be a surface link with
$\Sigma_{1}$ a trivial disk. The surface link $\Sigma$ can be realized as a
covering surface link of $\operatorname{BD}(\Sigma)$, obtained by taking the
2-fold branched cover of $B^{4}$ with branching set the first component of
$\operatorname{BD}(\Sigma)$, and then taking precisely one lift of each
component of $\operatorname{BD}(\Sigma)$ other than the first component.

To parse Lemma 6.3, recall that $\operatorname{BD}(\Sigma)$ is defined to be
$\operatorname{BD}(\Sigma_{1})\sqcup\Sigma_{2}\sqcup\cdots\sqcup\Sigma_{n}$.
The “first” component of $\operatorname{BD}(\Sigma)$ is then one of the two
components of $\operatorname{BD}(\Sigma_{1})$. By symmetry, these two
components can be arbitrarily ordered.

###### Proof.

Let $(W,\Sigma\cap
W)=(B^{4}{\smallsetminus}\nu\Sigma_{1},\Sigma{\smallsetminus}\Sigma_{1})$.
Since $\Sigma_{1}$ is smoothly unknotted, $W\cong B^{3}\times S^{1}$. Let $V$
denote the solid torus $W\cap S^{3}$, so
$V=S^{3}{\smallsetminus}\nu(\partial\Sigma_{1})$. In Figure 20, we sketch
$\overline{S^{3}{\smallsetminus}V}$ with $\Sigma_{1}$ isotoped to lie in
$S^{3}\cup\partial W$. This will be helpful for understanding the result of
Bing doubling. The surface $\Sigma_{1}$ now intersects
$\overline{S^{3}{\smallsetminus}V}$ in an annulus; the remaining disk lies in
$\overline{(\partial W){\smallsetminus}V}$.

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\partial
V}$ at 22 132
${\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}\Sigma_{1}}$
at 170 170

Figure 20. Left: a 3-ball with a solid torus, $V$, deleted; we see an annulus
contained in $\Sigma_{1}$. Right: a 3-ball. These two 3-balls glue along their
boundaries to form $\overline{S^{3}{\smallsetminus}V}$. The remainder of
$\Sigma_{1}$ is a boundary-parallel disk contained in $\partial W$.

Now Bing double $\Sigma$. There are two distinguished components
$\Sigma^{\prime}_{1}$ and $\Sigma^{\prime}_{2}$ that form the Bing double of
$\Sigma_{1}$. Isotope $\Sigma^{\prime}_{2}$ into $S^{3}\cup(\partial
W\cap\mathring{B}^{4})$, and the interior of $\Sigma^{\prime}_{1}$ into the
interior of $B^{4}$. In Figure 21, we draw a sketch of
$\partial\Sigma^{\prime}_{1}$ and $\partial\Sigma^{\prime}_{2}$ in
$\overline{S^{3}{\smallsetminus}V}$, indicating a different disk $D$ in
$S^{3}$ bounded by $\partial\Sigma^{\prime}_{1}$. Note that the disk
$\Sigma^{\prime}_{2}$ intersects $D$ in a single arc and intersects $\partial
V$ in two circles that are meridians of $V$. It is difficult to see
$\Sigma^{\prime}_{2}\cap S^{3}$ from this perspective, but in Figure 22 we
again draw $\partial\Sigma^{\prime}_{1}$ and $\partial\Sigma^{\prime}_{2}$ in
a different decomposition of $S^{3}$ where we can draw $\Sigma^{\prime}_{2}$.

${\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}\partial\Sigma^{\prime}_{2}}$
at 45 100 $D$ at 190 170

Figure 21. The solid torus $\overline{S^{3}{\smallsetminus}V}$ contains $D$
and $\partial\Sigma^{\prime}_{1}$, $\partial\Sigma^{\prime}_{2}$. The surface
$\Sigma^{\prime}_{2}$ intersects $\overline{S^{3}{\smallsetminus}V}$ in a disk
with two smaller disks deleted; we draw this in Figure 22 but it is difficult
to draw here. The remainder of $\Sigma^{\prime}_{2}$ lies in $(\partial
W)\cap\mathring{B}^{4}$.

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\partial
V}$ at 20 195
${\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}\Sigma^{\prime}_{2}}$
at 45 100 $D$ at 190 170 $D$ at 530 170
${\color[rgb]{.75,.5,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,.5,.25}\partial\Sigma^{\prime}_{1}}$
at 190 50

Figure 22. Left: a 3-ball in $S^{3}$ with $V$ removed; we see
$\Sigma^{\prime}_{2}$ intersecting $D$ in a single arc and $\partial V$ in two
circles. Right: the remaining 3-ball in $S^{3}$.

The disks $D$ and $\Sigma^{\prime}_{1}$ cobound a 3-ball; we form the double
cover of $B^{4}$ branched over $\Sigma^{\prime}_{1}$ by cutting along this
3-ball. Lift all other components of $\operatorname{BD}(\Sigma)$ to the cover;
we write $\widetilde{\Sigma}_{2}^{1}$ and $\widetilde{\Sigma}_{2}^{2}$ to
indicate the two lifts of $\Sigma^{\prime}_{2}$. In Figure 23, we draw the
resulting boundary. In Figure 24, we draw the whole cover of $B^{4}$ over
$\Sigma^{\prime}_{1}$, with $\widetilde{\Sigma}_{2}^{2}$ omitted (as it is a
disk parallel to $\widetilde{\Sigma}_{1}^{1}$, which is in the picture). In
the cover, $(W,\Sigma\cap W)$ lifts to two disjoint copies
$(\widetilde{W},\widetilde{\Sigma})$ and
$(\widetilde{W}^{\prime},\widetilde{\Sigma}^{\prime})$ of $(W,\Sigma\cap W)$.
Each of $\widetilde{W}$ and $\widetilde{W}^{\prime}$ intersects
$\widetilde{\Sigma}_{2}^{1}$ in a disk contained in the boundary. If we forget
$\widetilde{\Sigma}^{\prime}$, we can then (via isotopy) retract
$\widetilde{\Sigma}^{1}_{2}$ to a disk in $\partial\widetilde{W}$ that is
parallel to a disk in $S^{3}$. Forgetting $\widetilde{\Sigma}_{2}^{2}$, we
thus are left with $\widetilde{\Sigma}$ and $\widetilde{\Sigma}^{1}_{2}$,
which together form a copy of $(W,W\cap\Sigma)$ and a boundary-parallel disk
in $B^{4}$ contained in $\partial W$. This is exactly our original link
$\Sigma$ in $B^{4}$ with $\widetilde{\Sigma}^{1}_{2}$ identified with
$\Sigma_{1}$ (compare with Figure 20).

In summary: after Bing doubling $\Sigma_{1}$, we double branch cover $B^{4}$
with branch locus one component of $\operatorname{BD}(\Sigma_{1})$ and lift
all other components of $\Sigma$. In the cover, which is $B^{4}$, the surface
lifts of $\Sigma_{2}\sqcup\cdots\sqcup\Sigma_{n}$ form two split copies
$\widetilde{\Sigma}$ and $\widetilde{\Sigma}^{\prime}$ of
$\Sigma_{2}\sqcup\cdots\sqcup\Sigma_{n}$. The two lifts of the other component
of $\operatorname{BD}(\Sigma_{1})$ are the disks $\widetilde{\Sigma}_{2}^{1}$
and $\widetilde{\Sigma}_{2}^{2}$, which have the property that the union of
either one of $\\{\widetilde{\Sigma},\widetilde{\Sigma}^{\prime}\\}$ with
either one of $\\{\Sigma_{2}^{1},\Sigma_{2}^{2}\\}$ forms a copy of $\Sigma$.

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\partial\widetilde{V}}$
at 25 195
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\partial\widetilde{V}}$
at 210 135
${\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}\widetilde{\Sigma}^{1}_{2}}$
at 45 100
${\color[rgb]{0,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{0,.5,.5}\widetilde{\Sigma}^{2}_{2}}$
at 70 245

Figure 23. We cut both solids of Figure 22 along $D$ to form a double branched
cover along $\partial D=\partial\Sigma^{\prime}_{1}$. There is a copy of $V$
glued to each of the two tori in the left of the picture; the left and right
3-balls are then glued along their boundaries.

${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\partial\widetilde{W}^{1}}$
at 52 188
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\partial\widetilde{W}^{2}}$
at 420 188 $\widetilde{\Sigma}^{1}_{2}$ at 200 290 $B^{4}$ at -15 160 $S^{3}$
at -15 270

Figure 24. We cut $B^{4}$ along a 3-ball cobounded by $\Sigma^{\prime}_{1}$
and $D$ to obtain the double branched cover of $B^{4}$ over
$\Sigma^{\prime}_{1}$. We omit drawing $\widetilde{\Sigma}_{2}^{2}$. The top
of this picture is the resulting boundary, and agrees with Figure 23. Copies
of $W$ (called $\widetilde{W}^{1}$ and $\widetilde{W}^{2}$) are glued to each
of the two pictured $D^{2}\times S^{1}$’s. The lifts of components of
$\operatorname{BD}(\Sigma)$ other than $\Sigma^{\prime}_{1}$ and
$\Sigma^{\prime}_{2}$ all lie in the two copies of $W$.

∎

Our newfound understanding of coverings of Bing doubles allows us to study
other knots and surfaces in the complement of $\Sigma$.

######  Corollary 6.4.

Let $\Sigma$ be a surface link with first component a trivial disk. Let $J$ be
a knot in $S^{3}{\smallsetminus}\nu(\partial\Sigma)$. Suppose $J$ bounds a
smooth disk $D$ into the complement of $\operatorname{BD}(\Sigma)$. Then $J$
bounds a smooth disk into the complement of $\Sigma$.

###### Proof.

Let $\Sigma_{D}:=\operatorname{BD}(\Sigma)\cup D$. As usual, call the
components of $\operatorname{BD}(\Sigma)$ by the names
$\Sigma_{1},\ldots,\Sigma_{n+1}$, where $\Sigma_{1},\Sigma_{2}$ are the
components that Bing double the first component of $\Sigma$. Consider the
preimage of $\Sigma_{D}{\smallsetminus}\Sigma_{1}$ in the 2-fold cyclic cover
of $B^{4}$ branched along $\Sigma_{1}$, as in Lemma 6.3.

For $i=2,\ldots,n+1$ let $\Sigma_{i}^{1}$ and $\Sigma_{i}^{2}$ denote the two
lifts of $\Sigma_{i}$, and let $D^{1}$ and $D^{2}$ denote the two lifts of
$D$. From 6.3, we know that (perhaps after exchanging the names of
$\Sigma_{i}^{1}$ and $\Sigma_{i}^{2}$ for some values of $i$) the surface link
$\widetilde{\Sigma}:=\sqcup_{i=2}^{n+1}\Sigma_{i}^{1}$ is equivalent to
$\Sigma$. Moreover, since $\partial\Sigma_{1}$ is an unknot split from $J$,
the link $\partial\widetilde{\Sigma}\cup\partial D^{1}$ is equivalent to
$\partial\Sigma\cup J$ (perhaps after switching the roles of $D^{1}$ and
$D^{2}$). Then $D^{1}$ is a slice disk for $J$ into the complement of
$\Sigma$, as desired.∎

###  6.2. JSJ trees

We recall some 3-manifold theory that will be useful in the upcoming
arguments.

Every compact irreducible 3-manifold $X$ with toroidal boundary admits a JSJ
decomposition into finitely many codimension zero submanifolds, each of which
are again compact, irreducible 3-manifolds with toroidal boundary. Each of
these submanifolds is either Seifert fibered or has the property that every
incompressible torus is boundary parallel [JS79, Joh79]. This decomposition is
obtained by cutting the 3-manifold $X$ along incompressible, non-boundary-
parallel tori. A minimal collection of these tori are called _JSJ tori_. The
closures of the connected components of the complement of the JSJ tori are
called _JSJ pieces_. The JSJ tori are unique up to isotopy, and are preserved
(although possibly permuted) by any diffeomorphism from $X$ to itself.

The JSJ decomposition of a 3-manifold determines a tree, by associating a
vertex to each JSJ piece and an edge between two vertices if and only if the
corresponding pieces share a JSJ torus.

The next lemma is entirely 3-dimensional, but we state it in terms of surface
links in $B^{4}$ to be able to use the terminology for Bing doubling we have
already established. Note that the exterior of the Bing double pattern in a
solid torus is diffeomorphic to the exterior of the Borromean rings (see
Figure 25), so this 3-manifold will appear frequently in our JSJ decomposition
arguments.

Figure 25. The exterior of the Borromean rings in $S^{3}$ can be identified
with the exterior of the Bing double of the unknot in the solid torus
$S^{3}{\smallsetminus}\nu(U)$, where $U$ denotes the dashed component.

######  Lemma 6.5.

Let $\Sigma$ be a surface link in $B^{4}$. Let $T$ be the tree associated with
the JSJ decomposition of $X:=S^{3}{\smallsetminus}\nu(\partial\Sigma)$. Let
$Y\subseteq X$ be the JSJ piece that contains $\partial\Sigma_{1}$ and let
$v_{Y}$ be the vertex of $T$ corresponding to $Y$. Assume that
$\partial\Sigma_{1}$ is not a split unknotted component of $\partial\Sigma$.
The JSJ tree of
$X_{\operatorname{BD}}:=S^{3}{\smallsetminus}\nu(\partial\operatorname{BD}(\Sigma))$
is obtained from $T$ by adding a new vertex $v_{E}$ corresponding to a JSJ
piece diffeomorphic to the exterior $E$ of the Borromean rings and an edge
connecting $v_{E}$ and $v_{Y}$.

$v_{Y}$ at 190 650 $v_{Y}$ at 480 650 $v_{E}$ at 530 650

Figure 26. JSJ trees for $S^{3}{\smallsetminus}\nu(\partial\Sigma)$ and
$S^{3}{\smallsetminus}\nu(\partial\operatorname{BD}(\Sigma))$.

###### Proof.

Let $K:=\partial\Sigma_{1}$ and let $F$ be the torus
$\partial\overline{\nu}K$. Note that $X_{\operatorname{BD}}\cong X\cup_{F}E$,
where the zero-framed longitude of one Borromean component is glued to the
meridian of $K$ on $F$ and the meridian of the Borromean component becomes the
longitude of $F$. We claim that $F$ is a JSJ torus of $X_{\operatorname{BD}}$.
Since $K$ is not a split unknotted component, the longitude of $F$ is
nontrivial in $\pi_{1}(X)$. In $X_{\operatorname{BD}}$, the exterior of the
satellite link, the longitude of $F$ is therefore again nontrivial. The
meridian of $F$ is nontrivial in $X$. The surface $F$ is also incompressible
in $E$. Therefore $F$ is a JSJ torus of $X_{\operatorname{BD}}$ as desired. So
to obtain the JSJ tree for $X_{\operatorname{BD}}$, we start with $T$, the JSJ
tree for $X$. Then, as shown in Figure 26, we add a new vertex corresponding
to $E$, and an edge connecting it to $v_{Y}$, since $Y$ is precisely the
component that shares a JSJ torus with $E$. This gives the JSJ tree asserted.
∎

By repeated application of Lemma 6.5, we see that iterated Bing doubling adds
a line to the JSJ tree of the link exterior, with as many edges as
applications of $\operatorname{BD}$. This feature of the tree will allow us to
restrict the possible automorphisms of the underlying 3-manifold, using the
following result which is a direct consequence of the uniqueness of JSJ
decompositions.

######  Theorem 6.6.

A diffeomorphism of a link exterior to itself induces an automorphism of the
associated JSJ tree. Moreover each vertex, corresponding to a JSJ piece $Y$
say, must be sent to another vertex corresponding to a JSJ piece $Y^{\prime}$
such that $Y\cong Y^{\prime}$.

In the next lemma we refer to the _3-keychain link_ , that is, the three-
component link consisting of an unknot and two copies of its meridian. (See
Figure 27.) The exterior of this link is diffeomorphic to the cartesian
product of a circle $S^{1}$ with a connected planar surface with three
boundary components (i.e. a pair of pants).

Figure 27. The 3-keychain link.

######  Lemma 6.7.

Let $L$ be an $n$–component link in $S^{3}$ with first component $L_{1}$
unknotted. Let $\widetilde{L}$ be the covering link obtained by taking the
2-fold branched cover with branching set the first component of
$\operatorname{BD}(L)$, forgetting the branching set, and forgetting one of
the lifts of $\operatorname{BD}(L)_{2}$ $($i.e. the other component of
$\operatorname{BD}(L_{1}))$. The JSJ tree of the exterior of $\widetilde{L}$
is obtained as follows. Let $T_{1}$ and $T_{2}$ be two copies of the JSJ tree
for $L$, and let $v_{1}$ and $v_{2}$ be the vertices corresponding to the JSJ
piece containing the boundary of a tubular neighborhood of $L_{1}$, in each
copy. Take $T_{1}\sqcup T_{2}$, add a new vertex $v_{C}$, and add two new
edges, one joining $v_{C}$ to $v_{1}$ and one joining $v_{C}$ to $v_{1}$. The
JSJ piece corresponding to $v_{C}$ is the 3-keychain link exterior.

$T_{1}$ at 160 760 $T_{2}$ at 440 760 $v_{1}$ at 255 680 $v_{C}$ at 300 680
$v_{2}$ at 350 680

Figure 28. JSJ tree for the exterior of the covering link $\widetilde{L}$ from
a JSJ tree for the exterior of $L$.

###### Proof.

Observe that $\widetilde{L}$ is isotopic to the result of taking two split
copies of $L$ in disjoint copies of $B^{3}$, and performing a trivial band sum
on the two copies of $L_{1}$. This follows from the argument in the proof of
Lemma 6.3, and restricting to $S^{3}$; see Figure 24. Therefore, the exterior
of $\widetilde{L}$ has JSJ decomposition consisting of two copies of the JSJ
decomposition for $L$ joined by a central $3$-keychain link exterior $C$. It
is a standard fact that, in the JSJ decomposition of exteriors of connected
sums, keychain link exteriors appear in between the JSJ pieces of the summands
(see e.g. [Sch53]). The conclusion on the resulting JSJ tree is immediate. ∎

$T_{1}$ at 25 560 $T_{2}$ at 565 560 $v_{1}$ at 255 480 $v_{C}$ at 295 480
$v_{2}$ at 335 480

Figure 29. JSJ tree for the exterior of a covering link of an iterated Bing
double.

###  6.3. Bing doubling exotic surface link pairs

We now reach the main goal of this section, namely combining our understanding
of JSJ decompositions with our work from Section 6.1 on covering surfaces, to
show that if two surfaces are smoothly inequivalent, then under some fairly
weak assumptions, the results of Bing doubling one disk component of each are
also smoothly inequivalent.

Combining the work in Sections 6.1 and 6.2 yields the following proposition,
which shows that not only are the disk links from Section 3 non-equivalent,
but in fact any of their iterated Bing doubles have non-diffeomorphic
complements.

######  Proposition 6.8.

Let $\Sigma_{1}=\operatorname{BD}(D_{1})$ and
$\Sigma_{2}=\operatorname{BD}(D_{2})$ be the disk links from Proposition 3.3.
For any $n\geq 0$, $\operatorname{BD}^{n}(\Sigma_{1})$ and
$\operatorname{BD}^{n}(\Sigma_{2})$ have non-diffeomorphic complements.

###### Proof.

Recall that $\Sigma_{1}$ and $\Sigma_{2}$ are each a link of two disks in
$B^{4}$ whose components are individually unknotted disks,
$\partial\Sigma_{1}=\partial\Sigma_{2}$, and $\Sigma_{1}$ and $\Sigma_{2}$ are
topologically isotopic rel. boundary.

In Proposition 3.3, we showed that $\Sigma_{1}$ and $\Sigma_{2}$ are not
smoothly equivalent by studying a knot $\gamma$ in
$S^{3}{\smallsetminus}\nu(\partial\Sigma_{i})$ which is preserved up to
isotopy by any automorphism of $(S^{3},\partial\Sigma_{i})$. The knot $\gamma$
bounds a smooth disk into the complement of $\Sigma_{2}$ but not into
$\Sigma_{1}$, implying that there is no diffeomorphism between
$(B^{4},\Sigma_{1})$ and $(B^{4},\Sigma_{2})$.

Since $\gamma$ bounds a smooth disk in the complement of $\Sigma_{2}$,
$\gamma$ bounds a smooth disk (e.g. the same disk) into the complement of
$\operatorname{BD}^{n}(\Sigma_{2})$ for any $n\geq 0$. Since $\gamma$ does not
bound a smooth disk into the complement of $\Sigma_{1}$, it follows from
Corollary 6.4 that $\gamma$ does not bound a smooth disk into the complement
of $\operatorname{BD}^{n}(\Sigma_{1})$ for any $n\geq 0$. Therefore, if there
is a diffeomorphism from the complement of $\operatorname{BD}^{n}(\Sigma_{1})$
to the complement of $\operatorname{BD}^{n}(\Sigma_{2})$, it must not preserve
the isotopy class of $\gamma$.

In Figure 30, we show a JSJ decomposition for
$S^{3}{\smallsetminus}\nu(\operatorname{BD}^{n}(\Sigma_{i}))$. Recall that
$\partial\Sigma_{i}$ is a Bing double of a hyperbolic knot $K$ (cf. Lemma
3.4), so $S^{3}{\smallsetminus}\nu(\partial\Sigma_{i})$ has a JSJ
decomposition of one essential torus cutting the manifold into two pieces: the
complement of $K$ and a Borromean ring complement. Then
$S^{3}{\smallsetminus}\nu(\operatorname{BD}^{n}(\Sigma_{i}))$ admits a JSJ
decomposition with $n+2$ pieces in a line, as illustrated in Figure 30, with
essential tori $T_{1},\ldots,T_{n+1}$ in between. The first piece (bounded by
$T_{1}$) is the complement of $K$, and every piece thereafter is a Borromean
ring complement.

The boundary $M^{3}$ of
$B^{4}{\smallsetminus}\nu(\operatorname{BD}^{n}(\Sigma_{i}))$ is obtained from
$S^{3}{\smallsetminus}\nu(\partial\operatorname{BD}^{n}(\Sigma_{i}))$ by Dehn
filling every torus boundary component with the 0-filling. Note that this Dehn
filling on the Borromean rings exterior yields the 3-torus $T^{3}$. In this
closed 3-manifold $M^{3}$, $T_{n+1}$ bounds a copy of
$(T^{2}{\smallsetminus}\mathring{D}^{2})\times S^{1}$, $T_{1}$ bounds a copy
of $S^{3}{\smallsetminus}\nu(K)$ and $T_{i}\sqcup T_{i+1}$ bounds

$T^{3}{\smallsetminus}\nu(\text{two independent primitive curves})$

for $i=1,\ldots,n$. This is again a JSJ decomposition for $M^{3}$. Since the
complement of $K$ is not homeomorphic to any other piece in this JSJ
decomposition, we conclude that any automorphism of $M^{3}$ preserves $T_{1}$
up to isotopy and induces an automorphism of $S^{3}{\smallsetminus}\nu(K)$.
Since $S^{3}{\smallsetminus}\nu(K)$ has trivial automorphism group (Lemma
3.4), we conclude that $\gamma$ is preserved up to isotopy by any automorphism
of $M^{3}$ and thus there is no diffeomorphism from the complement of
$\operatorname{BD}^{n}(\Sigma_{1})$ to the complement of
$\operatorname{BD}^{n}(\Sigma_{2})$, as desired. ∎

$n+1$ vertices, each representing at 102 -5 a Borromean ring complement at 102
-12 $S^{3}{\smallsetminus}\nu(K)$ at 208 8 $T_{1}$ at 191 23 $T_{2}$ at 156 23
$T_{n+1}$ at 53 23

Figure 30. A JSJ decomposition of $S^{3}{\smallsetminus}\nu(\partial L_{i})$
as in Proposition 6.8. The bold red edges represent essential tori. The black
vertices are the components of the JSJ decomposition. The white vertices
represent torus boundary components; grey edges between white and black
vertices show to which JSJ component the corresponding boundary belongs.

Proposition 6.8 is immediately enough to conclude that
$\operatorname{BD}^{n}(\Sigma_{1})$ and $\operatorname{BD}^{n}(\Sigma_{2})$
are an exotic pair of $(n+1)$–component Brunnian links for any $n\geq 0$. But
we may now prove the following main technical result on covering surfaces.
This result has the advantage that it applies to a much more general family of
surface links, e.g. all of the families in Section 4 and those in Theorem 5.3.

######  Proposition 6.9.

Let $\Sigma=\Sigma_{1}\sqcup\Sigma_{2}$ and
$\Sigma^{\prime}=\Sigma_{1}^{\prime}\sqcup\Sigma_{2}^{\prime}$ be
$2$–component surface links in $B^{4}$ with $\Sigma_{1}$ and
$\Sigma_{1}^{\prime}$ trivial disks. Suppose that
$\partial\Sigma=\partial\Sigma^{\prime}\subseteq S^{3}$, and let $X$ be the
link exterior in $S^{3}$. Suppose that either:

  1. (i)

$X$ has no Borromean rings exterior and no 3-keychain link exterior in its JSJ
decomposition; or

  2. (ii)

$X$ has no 3-keychain link exterior and exactly one Borromean rings exterior
$E$ in its JSJ decomposition, but two of the three boundary components of $E$
are $\partial X$.

Then, if the iterated Bing doubles $\operatorname{BD}^{n}(\Sigma)$ and
$\operatorname{BD}^{n}(\Sigma^{\prime})$ are smoothly equivalent $($Bing
doubling the first component at each iteration$)$, then so are $\Sigma$ and
$\Sigma^{\prime}$.

###### Proof.

Let $S:=\operatorname{BD}^{n}(\Sigma)$ and
$S^{\prime}:=\operatorname{BD}^{n}(\Sigma^{\prime})$. The first components
$S_{1}$ and $S^{\prime}_{1}$ are both unknotted disks, since they arose from
Bing doubling. Consider the 2-fold branched coverings $p\colon B^{4}\to B^{4}$
and $p^{\prime}\colon B^{4}\to B^{4}$ with branching sets the first components
$S_{1}$ and $S^{\prime}_{1}$ of each surface link respectively. Since first
components are unknotted the branched covering spaces are indeed again
$B^{4}$.

By Lemma 6.3, $\operatorname{BD}^{n-1}(\Sigma)$ is a covering surface link of
$S=\operatorname{BD}^{n}(\Sigma)$. Moreover, $\operatorname{BD}^{n-1}(\Sigma)$
is a sublink of the inverse image $p^{-1}(S{\smallsetminus}S_{1})$.
Analogously $\operatorname{BD}^{n-1}(\Sigma^{\prime})$ is a sublink of the
inverse image $p^{-1}(S^{\prime}{\smallsetminus}S^{\prime}_{1})$.

Let $F\colon(B^{4},S)\to(B^{4},S^{\prime})$ be a diffeomorphism of pairs. We
claim that it must send $S_{1}$ to $S^{\prime}_{1}$ or $S^{\prime}_{2}$.

To prove the claim we consider the JSJ decomposition and tree for
$S^{3}{\smallsetminus}\nu(\partial S)$. There is a unique univalent vertex in
the JSJ tree for $S^{3}{\smallsetminus}\nu(\partial S)$ that corresponds to a
copy of the exterior $E$ of the Borromean rings. This vertex must be sent to
the analogous vertex in the JSJ tree for $S^{3}{\smallsetminus}\nu(\partial
S^{\prime})$ by Theorem 6.6. The two boundary components of
$S^{3}{\smallsetminus}\nu(\partial S^{\prime})$ that lie in this copy of $E$
are $S^{\prime}_{1}$ and $S^{\prime}_{2}$. The claim that $F(S_{1})$ is
$S^{\prime}_{1}$ or $S^{\prime}_{2}$ then follows. By relabelling we can and
shall assume, without loss of generality, that the diffeomorphism $F$ in fact
sends $S_{1}$ to $S^{\prime}_{1}$.

Now we know that the diffeomorphism $F\colon(B^{4},S)\to(B^{4},S^{\prime})$
preserves the branching set, i.e. $F(S_{1})=S^{\prime}_{1}$, we may lift $F$
to a diffeomorphism of pairs

$\widetilde{F}\colon(B^{4},p^{-1}(\operatorname{BD}^{n}(\Sigma)))\to(B^{4},p^{-1}(\operatorname{BD}^{n}(\Sigma^{\prime}))).$

In the inverse image $p^{-1}(\operatorname{BD}^{n}(\Sigma))$ we see two copies
of each component of $\operatorname{BD}^{n-1}(\Sigma)$. Restrict the
diffeomorphism $\widetilde{F}$ to

$\operatorname{BD}^{n-1}(\Sigma)\subseteq
p^{-1}(S{\smallsetminus}S_{1})\subseteq
p^{-1}(S)=p^{-1}(\operatorname{BD}^{n}(\Sigma)).$

Its image is a sublink of $p^{-1}(S^{\prime}{\smallsetminus}S^{\prime}_{1})$.
We claim that the image is a copy of
$\operatorname{BD}^{n-1}(\Sigma^{\prime})$. Given this claim, we are done,
because by induction we see that $\Sigma$ and $\Sigma^{\prime}$ are smoothly
equivalent, which was the desired statement.

For the claim we again argue by considering the effect of the diffeomorphism
restricted to the link exteriors in $S^{3}$

$\widetilde{F}|\colon
S^{3}{\smallsetminus}\nu(\partial(p^{-1}(S{\smallsetminus}S_{1})))\to
S^{3}{\smallsetminus}\nu(\partial(p^{-1}(S^{\prime}{\smallsetminus}S^{\prime}_{1}))).$

Since the links on the boundary coincide, so do their covering links. We will
restrict how the diffeomorphism can permute components using the relative
rigidity of $3$-manifold diffeomorphisms, in particular by studying the JSJ
decomposition and applying Theorem 6.6 again.

Lemma 6.7 provides a description of the JSJ tree of the covering link obtained
by forgetting the branching set and one of the lifts of $\partial S_{2}$ and
$\partial S_{2}^{\prime}$. As above we chose a lift of $S_{2}$, and then the
lift of $S_{2}^{\prime}$ was determined by the diffeomorphism $F$. Now the
automorphisms of the JSJ tree of this covering link are given by composing
automorphisms of the two copies of the JSJ tree for the exterior of $\partial
S$ with either:

  1. (i)

the identity, or

  2. (ii)

a flip map that switches the two sub-trees that are copies of the JSJ tree for
the exterior of $\partial S$.

To see this, we consider the structure of the JSJ trees. The JSJ tree $T(n-1)$
of the exterior of $\partial\operatorname{BD}^{n-1}(\Sigma)$ is obtained by
iterating the procedure pictured in Figure 26 and consists of the JSJ tree for
$X$ with a tail of $(n-1)$ Borromean exteriors attached to the vertex
corresponding to the JSJ piece containing $\partial\Sigma_{1}$. The JSJ tree
for the covering link is pictured in Figure 29 and is obtained from two copies
of $T(n-1)$ by adding one more vertex corresponding to the 3-keychain link
exterior, with two edges connecting this new vertex to the ends of both tails,
as described in Lemma 6.7 and shown in Figure 28. By our assumptions on the
JSJ decomposition of $X$, the Borromean and 3-keychain link exteriors cannot
be replicated anywhere in the tree but at its centre, and the tails must get
either fixed or swapped with each other by $\widetilde{F}$. The link
components which have boundaries of tubular neighborhoods that lie in JSJ
pieces corresponding to one half of the tree form a copy of
$\operatorname{BD}^{n-1}(\Sigma)$ in the domain and form a copy of
$\operatorname{BD}^{n-1}(\Sigma^{\prime})$ in the codomain. Since to decide
how $\widetilde{F}$ acts on connected components of our surface links it is
enough to control how $\widetilde{F}$ acts on their boundary components, this
completes the proof of the claim that the image
$\widetilde{F}(\operatorname{BD}^{n-1}(\Sigma))$ is a copy of
$\operatorname{BD}^{n-1}(\Sigma^{\prime})$. As stated above, this proves the
inductive step and therefore completes the proof of the proposition. ∎

Finally, we prove the following theorem.

######  Theorem 1.5.

Let $\Sigma$ and $\Sigma^{\prime}$ be an exotic pair of Brunnian 2–component
surface links constructed as in Theorem 1.3 or Theorem 1.4, with orderings
such that the first components are disks. For all $n\geq 3$, the $n$-component
surface links $\operatorname{BD}^{n-2}(\Sigma)$ and
$\operatorname{BD}^{n-2}(\Sigma^{\prime})$ are an exotic pair.

The disk links in Theorem 1.3 arose by taking two slice disks $D_{1}$ and
$D_{2}$ in $B^{4}$ for a fixed knot $K$ in $S^{3}$, and taking the Bing
doubles $\Sigma:=\operatorname{BD}(D_{1})$ and
$\Sigma^{\prime}:=\operatorname{BD}(D_{2})$. This is an exotic pair of
Brunnian disk links. Then the theorem states that applying iterated Bing
doubling both of these links produces an exotic pair with any specified number
of components.

The surface disk links in Theorem 1.4 were constructed by taking a particular
2–component surface link $\Sigma$ consisting of a disk and a genus one
surface, and applying 1-twisted rim surgery (using a knot $J$) along a curve
$\alpha$ that bounds a locally flat disk in the exterior of $\Sigma$ but does
not bound any smoothly embedded disk in the exterior. By varying $J$, we
obtain exotic surface links. Theorem 1.5 shows that applying Bing doubling to
the disk component of $\Sigma$ produces $n$–component surface links such that
the same rim surgeries give rise to an infinite family of pairwise exotic
surface links.

###### Proof.

By Lemma 2.2, the surface links $\operatorname{BD}^{n-2}(\Sigma)$ and
$\operatorname{BD}^{n-2}(\Sigma^{\prime})$ are still Brunnian. Moreover, Lemma
2.1 (with $\operatorname{CAT}=\operatorname{TOP}$) ensures they are still
topologically isotopic (we may assume the doubling is done such that the
boundaries are the same link in $S^{3}$).

If $\operatorname{BD}^{n-2}(\Sigma)$ and
$\operatorname{BD}^{n-2}(\Sigma^{\prime})$ were smoothly equivalent, then if
it applies, Proposition 6.9 ensures that $\Sigma$ and $\Sigma^{\prime}$ would
be smoothly equivalent, a contradiction. We need to check the hypotheses of
the proposition. For the disk links of Theorem 1.3, the JSJ decomposition of
the exterior of $\partial\Sigma=\partial\Sigma^{\prime}$ consists of a
Borromean exterior and the exterior of a hyperbolic knot (Lemma 3.4), and so
((ii))(ii) is satisfied, and therefore the proposition applies. For the
surface links of Theorem 1.4, we note that their boundary is a satellite of
the knot $K$, as shown in Figure 31 (left). In this case, the JSJ
decomposition of $\partial S=\partial S^{\prime}$ consists of the exterior of
the hyperbolic (verified by SnapPy [CDGW]) 3–component link $L$ illustrated in
Figure 31 (right) along with a JSJ decomposition of the exterior of $K$ in
$S^{3}$. Since $L$ has pairwise zero linking numbers, the exterior of $L$ is
not the exterior of a 3-keychain link. Moreover, as verified by SnapPy [CDGW],
the exterior of $L$ has symmetry group $\mathbb{Z}/2$, so is also not a
Borromean rings exterior. Then taking $K$ to be e.g. the positive untwisted
Whitehead double of the trefoil, whose exterior admits a JSJ decomoposition
consisting of a trefoil exterior and a Whitehead link exterior, we find that
((i)))(i) is satisfied and therefore the proposition applies. See Appendix B
for further documentation of this use of SnapPy. ∎

Figure 31. Left: an essential torus in the exterior of $\partial\Sigma$, where
$\Sigma$ is the surface link from Theorem 1.4. This torus splits the exterior
of $\partial\Sigma$ into the exterior of $K$ glued to the exterior of a
hyperbolic 3–component link $L$. Right: the link $L$.

##  7\. Links of other Brunnian types

In this section, we show how to use the techniques developed previously in
this paper to construct exotic surfaces of any Brunnian type, in the language
of Debrunner [Deb61].

######  Definition 7.1.

Let $\Sigma$ be an $n$–component surface link in $B^{4}$. Fix an integer $k$
with $1\leq k\leq n$.

We say that $\Sigma$ is _$(n,k)$ -Brunnian_ if every sublink of $\Sigma$ with
fewer than $k$ components is an unlink of surfaces, while every sublink of
$\Sigma$ with at least $k$ components is nontrivial.

Now we discuss how to form $(n,k)$-Brunnian links, given that we know how to
produce Brunnian links in the usual sense.

######  Procedure 7.2.

To construct an $(n,k)$-Brunnian surface link, let $D$ be a disk in $B^{4}$
whose boundary is a nontrivial knot. To simplify some notation, let $n_{k}$
denote $\binom{n}{k}$. Let $\Sigma_{n,k}(D)$ be the $n$–component surface link
constructed as follows (see also Figures 32 and 33).

  1. (1)

For $i=1,2,\ldots,{n_{k}}$, let $\Sigma_{i}$ be a copy of
$\operatorname{BD}^{k-1}(D)$. Arrange $\Sigma_{1},\ldots,\Sigma_{n_{k}}$ from
left to right, as ${n_{k}}$ split copies of $\operatorname{BD}^{k-1}(D)$.

  2. (2)

Let $\mathcal{C}=\\{C_{1},\ldots,C_{n}\\}$ indicate a set of $n$ distinct
colors. For each $i$, color every component of $\Sigma_{i}$ a color in
$\mathcal{C}$, so that no two components of $\Sigma_{i}$ are the same color
and so that for $i\neq j$, the surface links $\Sigma_{i}$ and $\Sigma_{j}$ are
not colored with the same size-$k$ subset of $\mathcal{C}$. See Figure 32.

  3. (3)

Let $B_{1},\ldots,B_{n_{k}}$ be disjoint 4-balls containing
$\Sigma_{1},\ldots,\Sigma_{n_{k}}$, respectively. In words, we now band all
components of $\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n_{k}}$ of each single
color together to form an $n$–component link of disks, with the bands chosen
to be trivial near their ends in the $B_{i}$’s and disjoint from other
$B_{j}$’s.

In different language, for each $i$ and $j$: when $\Sigma_{i}$ includes a
color-$C_{j}$ component and there is a color-$C_{j}$ component in $\Sigma_{s}$
for some $s>i$, let $i_{j}=\min\\{s>i\mid\Sigma_{s}$ has a color-$C_{j}$
component$\\}$ and band the color-$C_{j}$ components in $\Sigma_{i}$ and
$\Sigma_{i_{j}}$ together by a band that is trivial in $B_{i}$ and $B_{i_{j}}$
and does not intersect any other $B_{l}$. See Figure 33.

$\Sigma_{1}$ at 25 -8 $\Sigma_{2}$ at 80 -8 $\Sigma_{n_{k}}$ at 247 -8

Figure 32. To build $\Sigma_{n,k}(D)$, we start with a split union of $n_{k}$
copies $\Sigma_{1},\ldots,\Sigma_{n_{k}}$ of $\operatorname{BD}^{k-1}(D)$.
Here we draw the boundaries of $\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n_{k}}$,
where $K=\partial D$. In addition, we color every component in each
$\Sigma_{i}$ with a color in $\mathcal{C}$, with the rules that no two
components of $\Sigma_{i}$ may be the same color and for $i\neq j$, the links
$\Sigma_{i}$ and $\Sigma_{j}$ may not use the same subset of colors in
$\mathcal{C}$. In this picture, $(n,k)=(5,4)$.

$B_{1}$ at 15 82 $B_{2}$ at 70 82 $B_{n_{k}}$ at 237 82

Figure 33. Starting from Figure 32, to complete our construction of
$\Sigma_{n,k}(D)$ we band together all $C_{i}$-colored components of
$\Sigma_{1}\sqcup\cdots\sqcup\Sigma_{n_{k}}$ for each $C_{i}\in\mathcal{C}$.
The bands are taken to be trivial in the 4-balls $B_{1},\ldots,B_{n_{k}}$ but
otherwise may be chosen freely so long as they are disjoint. The result is an
$n$–component link of disks that we call $\Sigma_{n,k}(D)$.

When restricted to $S^{3}$, this construction differs slightly from the
original one of Debrunner [Deb61], simply because we have already worked to
understand Bing doubling in this paper so we elect to use Bing doubles to
construct these generalized Brunnian links.

In the case that $n=k$, $n_{n}=1$ and so
$\Sigma_{n,n}(D)=\operatorname{BD}^{n-1}(D)$. In the case $k=1$,
$\Sigma_{n,1}(D)$ is a split union of $n$ copies of $D$.

######  Proposition 7.3.

The surface link $\Sigma:=\Sigma_{n,k}(D)$ from Procedure 7.2 is
$(n,k)$-Brunnian.

###### Proof.

The proof of Proposition 7.3 naturally divides into two claims.

######  Claim.

Let $\Sigma^{\prime}$ be a $(k-1)$–component sublink of $\Sigma$. Then
$\Sigma^{\prime}$ is an unlink of disks.

Proof of Claim. By construction, each component of $\Sigma$ is colored by a
different element of a size-$n$ set $\mathcal{C}$. Say $\Sigma^{\prime}$ is
obtained from $\Sigma$ by deleting the components of colors
$\mathcal{C}_{r}\subseteq\mathcal{C}$, where $|\mathcal{C}_{r}|=n-k+1$. Since
$\mathcal{C}$ has $n$ colors, we are deleting $n-k+1$ of them, and each
$\Sigma_{i}$ has $k$ distinct colors, we conclude that $\Sigma^{\prime}$ is
obtained from $\Sigma=\cup_{\text{bands}}\Sigma_{i}$ by deleting a nonzero
number of components from each $\Sigma_{i}$.

Note that

$\Sigma^{\prime}=\bigcup_{\text{along bands}}^{i\geq
1}\Sigma_{i}{\smallsetminus}(\text{components of colors }\mathcal{C}_{r}).$

In Figure 34 (left), we depict the ball neighborhood $B_{1}$ of $\Sigma_{1}$
(i.e. the leftmost copy of $\operatorname{BD}^{k-1}(D)$ in the construction of
$\Sigma$). The ball $B_{1}$ meets bands in $\Sigma^{\prime}$ attached to
$\Sigma_{1}$ trivially, with all bands extending from $\Sigma_{1}$ toward the
right. The surface $B_{1}\cap\Sigma^{\prime}$ is obtained from
$B_{1}\cap\Sigma_{1}$ by deleting at least one component of $\Sigma_{1}$ (and
any attached bands). Then in $B_{1}$ we may isotope $B_{1}\cap\Sigma^{\prime}$
rel. $(\partial\overline{B_{1}})\cap\mathring{B}^{4}$ to obtain trivial disks
as indicated in Figure 34. (Note that while we only draw the boundary of
$\Sigma^{\prime}$ in Figure 34, once any component of $\Sigma^{\prime}\cap
B_{1}$ is deleted the remaining components are determined up to smooth isotopy
rel. boundary as trivial disks in $B^{4}$.) We conclude

$\Sigma^{\prime}=\bigcup_{\text{along bands}}^{i\geq
2}\Sigma_{i}{\smallsetminus}(\text{components of colors }\mathcal{C}_{r})$

up to smooth isotopy.

Now proceed by induction, assuming

$\Sigma^{\prime}=\bigg{(}\bigcup_{\text{along bands}}^{i\geq
s}\Sigma_{i}{\smallsetminus}(\text{components of colors
}\mathcal{C}_{r})\bigg{)}\cup(\text{an unlink of disks})$

for some $s>1$ (up to smooth isotopy). In Figure 34 (right), we depict the
ball neighborhood $B_{s}$ of $\Sigma_{s}$. The ball $B_{s}$ meets bands in
$\Sigma^{\prime}$ attached to $\Sigma_{s}$ trivially, with all bands extending
from $\Sigma_{s}$ toward the right. Note that some components of $\Sigma_{s}$
may not meet any bands – in the construction of $\Sigma$, this means that such
a component is the last of its color. The surface $\Sigma^{\prime}\cap B_{s}$
is obtained from $\Sigma_{s}\cup$(bands extending to the right$)$ by deleting
at least one component (and any attached bands). Then in $B_{s}$ we may
isotope $B_{s}\cap\Sigma^{\prime}$ rel.
$(\partial\overline{B}_{s})\cap\mathring{B}^{4}$ to obtain trivial disks as
indicated in Figure 34. Note that some disks might be split in the interior of
$B_{s}$. We conclude

$\Sigma^{\prime}=\bigg{(}\bigcup_{\text{along bands}}^{i\geq
s+1}\Sigma_{i}{\smallsetminus}(\text{components of colors
}\mathcal{C}_{r})\bigg{)}\cup(\text{an unlink of disks}).$

Inductively, we obtain

$\displaystyle\Sigma^{\prime}$
$\displaystyle=\bigg{(}\Sigma_{n_{k}}{\smallsetminus}(\text{components of
colors }\mathcal{C}_{r})\bigg{)}\cup(\text{an unlink of disks})$
$\displaystyle=\text{an unlink of disks},$

since $\Sigma_{n_{k}}$ is Brunnian and includes at least one color in
$\mathcal{C}_{r}$. This completes the proof of the claim.

######  Claim.

Let $\Sigma^{\prime\prime}$ be a $k$–component sublink of $\Sigma$. Then
$\Sigma^{\prime\prime}$ is nontrivial and in fact, $\Sigma^{\prime\prime}$ is
smoothly isotopic to $\operatorname{BD}^{k-1}(D)$.

_Proof of Claim._ Note that since $\partial D$ is a nontrivial knot,
$\partial\operatorname{BD}^{k-1}(D)$ is not an unlink so
$\operatorname{BD}^{k-1}(D)$ is not an unlinked disk link.

The proof of this claim is very similar to that of the previous one. Suppose
that $\Sigma^{\prime\prime}$ is obtained from $\Sigma$ by deleting the
components of colors $\mathcal{C}_{r}\subseteq\mathcal{C}$, where
$|\mathcal{C}_{r}|=n-k$. Since $\mathcal{C}$ has $n$ colors, we are deleting
$n-k$ of them, and each $\Sigma_{i}$ has $k$ distinct colors, we conclude that
$\Sigma^{\prime\prime}$ is obtained from
$\Sigma=\cup_{\text{bands}}\Sigma_{i}$ by deleting a nonzero number of
components from each $\Sigma_{i}$ for all $i\neq m$ for some distinguished
$m$. Specifcally, $m$ is the unique integer so that $\Sigma_{m}$ is colored
with all of the colors in $\mathcal{C}-\mathcal{C}_{r}$. Again, we write

$\Sigma^{\prime\prime}=\bigcup_{\text{along bands}}^{i\geq
1}\Sigma_{i}{\smallsetminus}(\text{components of colors }\mathcal{C}_{r}).$

By the proof of the previous claim applied to $\Sigma_{1}$ and working
rightward,

$\Sigma^{\prime\prime}=\bigg{(}\bigcup_{\text{along bands}}^{i\geq
m}\Sigma_{i}{\smallsetminus}(\text{components of colors
}\mathcal{C}_{r})\bigg{)}\cup(\text{an unlink of disks}).$

By horizontally mirroring the argument and applying it to $\Sigma_{n_{k}}$ and
working leftward, we find

$\displaystyle\Sigma^{\prime\prime}$
$\displaystyle=\Big{(}\Sigma_{m}{\smallsetminus}(\text{components of colors
}\mathcal{C}_{r})\Big{)}\cup(\text{an unlink of disks})$
$\displaystyle=\Sigma_{m}\cup\text{an unlink of disks}$
$\displaystyle=\Sigma_{m}=\operatorname{BD}^{k-1}(D),$

since $\Sigma^{\prime\prime}$ and $\Sigma_{m}$ each have $k$ components. Since
$\operatorname{BD}^{k-1}(D)$ is a nontrivial disk link, this completes the
proof of the claim.

Proposition 7.3 follows. ∎

$\sim$ at 60 35 $\sim$ at 213 35

Figure 34. Left: The ball $B_{1}$ intersecting $\Sigma^{\prime}$. If we delete
at least one component, e.g. the one highlighted, then we obtain a trivial
link of disks with trivial bands extending out of $B_{1}$. Right: The ball
$B_{s}$ intersecting $\Sigma^{\prime}$ after applying an inductive hypothesis
in the proof of Proposition 7.3. If we delete at least one component, e.g. the
one highlighted, then we obtain a trivial link of disks with trivial bands
extending from some components out of $B_{s}$.

######  Theorem 1.8.

For any integers $n$ and $k$ with $n\geq 2$ and $1\leq k\leq n$, there exists
a pair of $(n,k)$–Brunnian disk links in $B^{4}$ forming an exotic pair.

###### Proof.

Let $D_{1}$ and $D_{2}$ be the exotic pair of disks of Proposition 3.1. Recall
that $D_{1}$ and $D_{2}$ are topologically isotopic rel. boundary. Let
$\Sigma:=\Sigma_{n,k}(D_{1})$ and $\Sigma^{\prime}:=\Sigma_{n,k}(D_{2})$ be as
in Procedure 7.2, with all choices of colors and bands the same in $S^{3}$ so
that $\partial\Sigma=\partial\Sigma^{\prime}$. By Proposition 7.3, $\Sigma$
and $\Sigma^{\prime}$ are $(n,k)$-Brunnian.

Since $D_{1}$ and $D_{2}$ are topologically isotopic rel. boundary, by Lemma
2.1 so are $\operatorname{BD}^{k-1}(D_{1})$ and
$\operatorname{BD}^{k-1}(D_{2})$. Then since $\Sigma$ and $\Sigma^{\prime}$
are obtained by banding together copies of $\operatorname{BD}^{k-1}(D_{1})$
and $\operatorname{BD}^{k-1}(D_{2})$ with the same choices of bands, we
conclude that $\Sigma$ and $\Sigma^{\prime}$ are topologically isotopic rel.
boundary.

On the other hand, suppose there is a diffeomorphism
$f:(B^{4},\Sigma)\to(B^{4},\Sigma^{\prime})$. Let $\Sigma_{k}$ be a
$k$–component sublink of $\Sigma$. Then $f(\Sigma_{k})$ is a $k$–component
sublink of $\Sigma^{\prime}$. By the proof of Proposition 7.3, $\Sigma_{k}$ is
smoothly isotopic to $\operatorname{BD}^{k-1}(D_{1})$ and $f(\Sigma_{k})$ is
smoothly isotopic to $\operatorname{BD}^{k-1}(D_{2})$. This contradicts
Theorem 1.5. We conclude there is no diffeomorphism from $(B^{4},\Sigma)$ to
$(B^{4},\Sigma^{\prime})$. ∎

##  8\. Closed surfaces

We show that in some cases our exotic Brunnian surface links may be promoted
to exotic links of closed surfaces, in $4$-manifolds that have positive second
Betti number.

######  Theorem 8.1.

Let $\Sigma_{1}$ and $\Sigma_{2}$ be the 2–component exotic disk links from
Proposition 3.3. Fix a non-negative integer $n$. Let $X^{4}$ be the 4-manifold
obtained from $B^{4}$ by attaching a 0-framed 2-handle along every boundary
component of $\operatorname{BD}^{n}(\Sigma_{i})$. Let $S_{i}\subseteq X$ be
the $(n+1)$–component link of 2-spheres obtained from
$\operatorname{BD}^{n}(\Sigma_{i})$ by gluing a core of each 2-handle to the
corresponding component of $\operatorname{BD}^{n}(\Sigma_{i})$. Then $S_{1}$
and $S_{2}$ are topologically isotopic in $X$, but $(X,S_{1})$ is not
diffeomorphic to $(X,S_{2})$.

###### Proof.

Since $S_{1}$ and $S_{2}$ agree outside $\operatorname{BD}^{n}(\Sigma_{1})$
and $\operatorname{BD}^{n}(\Sigma_{2})$, which are topologically isotopic rel.
boundary, it is immediate that $S_{1}$ and $S_{2}$ are topologically isotopic.

Let $Y^{4}_{i}$ be the 4-manifold obtained from $X$ by surgering each
component of $S_{i}$ (i.e. for each component $F$ of $\Sigma_{i}$, delete
$\nu(F)\cong S^{2}\times\mathring{D}^{2}$ and reglue $B^{3}\times S^{1}$ in
order to obtain $Y_{i}$). It is a well-known fact that $Y_{i}\cong
B^{4}{\smallsetminus}\nu(\operatorname{BD}^{n}(\Sigma_{i}))$. This can be seen
via Kirby calculus: attaching a 0-framed 2-handle along an unknot and then
surgering along a 2-sphere consisting of a trivial disk glued to a core of
that 2-handle has the effect in a Kirby diagram of first adding a 0-framed
circle (the 2-handle) and then changing the 0 to a dot, representing carving
out a trivial disk with the specified boundary.

If $(X,S_{1})$ were diffeomorphic to $(X,S_{2})$, then $Y_{1}$ would be
diffeomorphic to $Y_{2}$. But Proposition 6.8 says precisely that $Y_{1}$ is
not diffeomorphic to $Y_{2}$, so we conclude that $(X,S_{1})$ and $(X,S_{2})$
are not diffeomorphic. ∎

In the 2–component case we can obtain the analogous result to Theorem 8.1 in a
closed 4-manifold $Z$.

######  Theorem 8.2.

The 2–component disk links $\Sigma_{1}=\operatorname{BD}(D_{1})$ and
$\Sigma_{2}=\operatorname{BD}(D_{2})$ from Proposition 3.3 smoothly embed into
2–component sphere links $S_{1}$ and $S_{2}$, respectively, in a closed
4-manifold $Z$ such that

  1. (i)

$S_{1}$ and $S_{2}$ are topologically isotopic in $Z$, and

  2. (ii)

$(Z,S_{1})$ is not diffeomorphic to $(Z,S_{2})$.

###### Proof.

Much as in the proof of Theorem 8.1, we will begin by attaching 0-framed
2-handles to $B^{4}$ along the two unknotted components of the link
$\operatorname{BD}(K)$, where $K$ is the underlying slice knot from the proof
of Proposition 3.3. Denote this 4-manifold by $X$. As in the proof of Theorem
8.1, the disk links $\operatorname{BD}(D_{1})$ and $\operatorname{BD}(D_{2})$
give rise to a pair of exotic sphere links $S_{1}$ and $S_{2}$ in $X$ such
that surgering $S_{i}$ turns $X$ into the disk link exterior
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{i}))$.

Our strategy will be to embed $X$ into a closed 4-manifold $Z$ by constructing
a well-chosen 4–dimensional “cap” $C$ with $\partial C=\partial X$ (but
oriented so that $\partial C=-\partial X$). The induced sphere links $S_{1}$
and $S_{2}$ will remain topologically isotopic in the larger 4-manifold $Z$.
To show that $S_{1}$ and $S_{2}$ also remain smoothly distinct, we will
distinguish the 4-manifolds
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{1}))\cup C$ and
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{2}))\cup C$ obtained by
surgering $Z$ along $S_{1}$ and $S_{2}$, respectively. The cap $C$ will be
constructed with this goal in mind.

To this end, recall from the proof of Proposition 3.3 that we can attach three
2-handles to $B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{1}))$ to produce a
4-manifold $W$ that admits a Stein structure; see Figure 7. For later use, we
note that one of these is a $-1$-framed 2-handle attached along the curve
$\gamma$ from Figure 6. By [LM97], we can then further embed the Stein domain
$W$ into a closed, minimal Kähler surface, which we denote by $Q$; here
_minimality_ implies that $Q$ contains no smoothly embedded 2-spheres of self-
intersection $-1$. We define the desired cap $C$ to be the exterior of
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{1}))$ in $Q$, and let $Z=X\cup
C$ as discussed above.

It remains to distinguish the 4-manifolds obtained from $Z$ by surgering
$S_{1}$ and $S_{2}$. The resulting 4-manifolds are
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{1}))\cup C$ (which is the
minimal Kähler surface $Q$) and
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{2}))\cup C$. To distinguish
them, we first recall that the curve $\gamma$ bounds a smoothly embedded disk
in the complement of $\operatorname{BD}(D_{2})$. After gluing $C$ to
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{2}))$, this disk can be glued
to the core of the 2-handle attached along $\gamma$ to yield a smoothly
embedded 2-sphere of self-intersection $-1$ in
$B^{4}{\smallsetminus}\nu(\operatorname{BD}(D_{2}))\cup C$. It follows that
surgering $Z$ along $S_{1}$ and $S_{2}$ yields distinct 4-manifolds, so the
pairs $(Z,S_{1})$ and $(Z,S_{2})$ are not diffeomorphic. ∎

##  Appendix A The isometry group of $K$

We began by drawing $K$ in SnapPy’s link editor, as depicted in Figure 35, and
extracted a Dowker-Thistlethwaite code for this knot projection:

$\displaystyle\text{DT: }[($
$\displaystyle-70,20,-84,-82,-80,42,-32,56,-66,2,-44,18,36,-52,40,76,-16,-64,28,-60,-78,$
$\displaystyle\quad
68,-22,-12,-30,62,-26,-74,14,50,-38,54,72,46,-4,-24,34,-58,48,-10,-8,-6)]$

Figure 35.

Running SnapPy inside Sage, we then verified that $K$ is a hyperbolic knot
whose complement has trivial isometry group.

sage: import snappy

sage: K = snappy.Manifold(‘DT:[(-70,20,-84,-82,-80,42,-32,56,-66,2,-44,18,36,

-52,40,76,-16,-64,28,-60,-78,68,-22,-12,-30,62,-26,-74,14,50,-38,54,72,

46,-4,-24,34,-58,48,-10,-8,-6)]’)

sage: K.solution_type()
‘all tetrahedra positively oriented’

sage: K.verify_hyperbolicity()
(True, [0.69085717467? + 0.50830991237?*I, -0.09695795674? + 0.91647294852?*I,
0.97074783390? + 0.28150095915?*I, 1.26374636309? + 0.55883319058?*I,
0.13653313161? + 0.57669418470?*I, 0.18676207068? + 0.42212622171?*I,
1.71203281718? + 0.96878529641?*I, 0.16077780284? + 2.5757092416?*I,
-0.35397677376? + 0.58988805254?*I, -0.03195090969? + 1.66462568439?*I,
-0.7299955531? + 1.9313492842?*I, 0.16909543612? + 0.10116074904?*I,
-1.1477637844? + 1.2694777149?*I, 0.23479520751? + 0.83739365476?*I,
-0.6690405348? + 0.68508313707?*I, 1.6959300535? + 2.5175042223?*I,
-0.9638799118? + 2.3142283467?*I, 1.11640893998? + 0.55523026497?*I,
0.7891261692? + 1.31673271484?*I, 0.98323205817? + 1.45469490101?*I,
-0.56189716658? + 0.44115682612?*I, -0.4883656525? + 2.1993281961?*I,
0.19274256226? + 0.71697136107?*I])

sage: R = K.canonical_retriangulation(verified = True)

sage: len(R.isomorphisms_to(R)) #This gives the size of the isometry group.
1

The size of the isometry group is 1, so the identity is the unique isometry of
$S^{3}{\smallsetminus}K$.

##  Appendix B The isometry group of $L$

As in Appendix A, we began by drawing $L$ in SnapPy’s link editor, as depicted
in Figure 36, and extracted a Dowker-Thistlethwaite code for this knot
projection:

$\displaystyle\text{DT: }[($ $\displaystyle
16,34,-64,54,40,68,-44,-60,32,50,42,-66,-38,-56,70,-48),$
$\displaystyle\quad(30,-58,8,-24,-12,20,52,-2,-62,-14,-36,-26,6,46,18),(-4,28,10,-22)]$

Figure 36.

Running SnapPy inside Sage, we then verified that $L$ is a hyperbolic link
whose complement has isometry group $\mathbb{Z}/2\mathbb{Z}$.

sage: import snappy

sage: L =
snappy.Manifold(‘DT:[(16,34,-64,54,40,68,-44,-60,32,50,42,-66,-38,-56,

70,-48), (30,-58,8,-24,-12,20,52,-2,-62,-14,-36,-26,6,46,18),(-4,28,10,

-22)]’)

sage: L.solution_type()
‘all tetrahedra positively oriented’

sage: L.verify_hyperbolicity()
(True, [0.63324006006? + 1.10322026006?*I, 0.10993343042? + 0.47600086029?*I,
0.16413577877? + 0.79246676123?*I, 0.66490644413? + 1.23999412266?*I,
0.38972085196? + 0.84475722622?*I, -0.02052870871? + 0.71226998786?*I,
-1.09458686743? + 1.17956342451?*I, 1.08258208868? + 0.80923399264?*I,
0.07133786888? + 0.68448305451?*I, 0.63642037823? + 0.39539813066?*I,
0.31403980641? + 1.10759837099?*I, 0.40489568556? + 0.46123907137?*I,
0.16243243753? + 0.76871625189?*I, 0.38972085196? + 0.84475722622?*I,
0.66413540183? + 0.62635898843?*I, 0.50355806795? + 0.72289062001?*I,
0.50203604063? + 0.63579926026?*I, 1.13580695569? + 1.15030349850?*I,
0.07522244611? + 0.83274531566?*I, -0.09975044831? + 0.76471739458?*I,
-0.38903974680? + 0.93113075790?*I, 0.20490074235? + 1.28331759380?*I,
1.63503890307? + 1.75198642689?*I, 0.25525121024? + 0.99618331677?*I,
0.38848526399? + 0.87511804068?*I, 0.81713558269? + 0.50449818986?*I,
0.33037527178? + 0.55709566053?*I, 0.6350389031? + 1.75198642689?*I,
0.33037527178? + 0.55709566053?*I])

sage: R = L.canonical_retriangulation(verified = True)

sage: len(R.isomorphisms_to(R)) #This gives the size of the isometry group.
2

We conclude that $L$ is hyperbolic and not the Borromean rings.

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††Epilogue Peculiar surfaces live in $B^{4}$,
delicately tangled sorts:
drab if untethered
but exotic together.
Bing doubling yields them in scores!