# Flat fully augmented links are determined by their complements

Christian Millichap  Department of Mathematics
Furman University
Greenville, SC 29613<EMAIL_ADDRESS>and  Rolland Trapp
Department of Mathematics
California State University
San Bernardino, CA 92407<EMAIL_ADDRESS>
###### Abstract.

In this paper, we show that two flat fully augmented links with homeomorphic
complements must be equivalent as links in $\mathbb{S}^{3}$. This requires a
careful analysis of how totally geodesic surfaces and cusps intersect in these
link complements and behave under homeomorphism. One consequence of this
analysis is a complete classification of flat fully augmented link complements
that admit multiple reflection surfaces. In addition, our work classifies
those symmetries of flat fully augmented link complements which are not
induced by symmetries of the corresponding link.

††Date:

##  1\. Introduction

Two links, $L_{1}$ and $L_{2}$, in $\mathbb{S}^{3}$ are equivalent if there
exists an orientation-preserving homeomorphism of pairs from
$(\mathbb{S}^{3},L_{1})$ to $(\mathbb{S}^{3},L_{2})$. An equivalence of links
induces a homeomorphism between the link complements $\mathbb{S}^{3}\setminus
L_{1}$ and $\mathbb{S}^{3}\setminus L_{2}$, which shows that links determine
their complements. However, the converse of this statement is generally not
true. For instance, if a link contains an unknotted component, then a Dehn
twist along this component is a homeomorphism of the complement that will
frequently produce a non-equivalent link. Whitehead used this technique in
[26] to show that that there are infinitely many distinct links with the same
complement as the Whitehead link. Some other known constructions for producing
distinct links with the same complement can be found in Berge [4] and Gordon
[10, Section 6]. In contrast to links, the Gordon–Luecke Theorem [11] shows
that knots are determined by their complements in $\mathbb{S}^{3}$. Knots in
certain closed, oriented $3$-manifolds other than $\mathbb{S}^{3}$ are also
determined by their complements; see Rong [24], Matignon [18], Gainullin [9],
and Ichihara–Saito [14] for some examples. Moving forward, we will always
assume knots and links are embedded in $\mathbb{S}^{3}$. This contrast between
links with multiple components and knots motivated the following question
raised by Mangum and Stanford [17]:

######  Question.

Is there a set of links $\mathcal{S}$ such that if $L_{1},L_{2}\in\mathcal{S}$
and $\mathbb{S}^{3}\setminus L_{1}$ is homeomorphic to
$\mathbb{S}^{3}\setminus L_{2}$, then $L_{1}$ is equivalent to $L_{2}$?

In the same paper, Mangum–Stanford show that the set of homologically trivial
and Brunnian links (called HTB links in their paper) provide an affirmative
answer to this question [17, Theorem 3]. As far as the authors know this is
the only example in the literature of an infinite set of links, other than the
set of knots, with this property. This motivates the main goal of our paper,
which is to show that the family of flat fully augmented links (flat FALs)
also have this property.

Flat FALs are a family of hyperbolic links which can be obtained from link
diagrams, meeting certain diagrammatic conditions, in the following way. Given
a link diagram add a trivial component (called a crossing circle) enclosing
each twist region, then remove all twists so the strands of the diagram run
parallel through the crossing circles (see Figure 2). Knot circles of the
resulting flat FAL are components that lie in the projection plane of the
resulting diagram. Flat FALs, and more generally, FALs, are a rich family of
hyperbolic links, which have received much attention in the last 20 years due
to the explicit combinatorial descriptions of their geometric structures and
connections to highly twisted knots via Dehn surgery; see [5], [6], [7], [8],
[13], [15], [19], [20], [21], [22], [23] for some examples from the
literature. We direct the reader to the beginning of Section 2 for more
details on essential properties of flat FALs used in this paper. We can now
formally state our main theorem, recalling that an isotopy of links in
$\mathbb{S}^{3}$ induces an equivalence of links.

######  Theorem 1.1.

Let $\mathcal{A}$ and $\mathcal{A}^{\prime}$ be flat FALs. Then
$(\mathbb{S}^{3},\mathcal{A})$ is isotopic to
$(\mathbb{S}^{3},\mathcal{A}^{\prime})$ if and only if
$\mathbb{S}^{3}\setminus\mathcal{A}$ is homeomorphic to
$\mathbb{S}^{3}\setminus\mathcal{A}^{\prime}$.

Theorem 1.1 provides a new infinite set of links that are determined by their
complements, distinct from knots and HTB links. By definition, every flat FAL
contains at least three components, and so, no flat FALs are knots. At the
same time, there are infinitely many flat FALs that are not HTB links.
Specifically, any flat FAL with at least two knot circles is not an HTB link
since it will contain a Hopf sublink. Such a sublink violates the
homologically trivial property that the linking number is $0$ for any two
components.

Another way our work differs from both Gordon–Luecke and Mangum–Stanford is in
the techniques used. The proof of Theorem 1.1 greatly leverages the hyperbolic
structure of flat FAL complements and relies on analyzing the behavior of
totally geodesic surfaces and cusps under isometries induced by homeomorphism.
This geometric approach differs from the purely topological approaches used by
Gordon–Luecke and Magnum–Stanford, where these authors determine which Dehn
surgeries on a knot or HTB link produce $\mathbb{S}^{3}$.

We note that the _flat_ hypothesis of Theorem 1.1 is necessary, making the
result as general as possible within the class of FALs. To see this, consider
the twisted FALs of Figure 1. They differ by a Dehn twist on the top circle,
so their complements are homeomorphic. The links are distinct, however, since
all components are unknots in Figure 1$(a)$ while one component is a trefoil
in Figure 1$(b)$.

$\begin{array}[]{ccc}\includegraphics{TwistedUnknot.pdf}&&\includegraphics{TwistedTrefoil.pdf}\\\
(a)\textrm{ All components unknotted}&&(b)\textrm{ One trefoil
component}\end{array}$ Figure 1. Distinct twisted FALs with homeomorphic
complements

The proof of Theorem 1.1 breaks into two cases, depending on the number of
reflection surfaces contained in a flat FAL complement. Intuitively, a
reflection surface for a flat FAL complement
$\mathbb{S}^{3}\setminus\mathcal{A}$ is a totally geodesic surface that
corresponds with the projection plane for an FAL diagram of $\mathcal{A}$; see
Section 2 for more explicit details and properties of reflection surfaces. As
an intermediary step in the proof of Theorem 1.1, we classify flat FAL
complements with multiple reflection surfaces and show that homeomorphisms
between flat FAL complements preserve reflection surfaces. For emphasis, we
now state the classification of flat FALs with multiple reflection surfaces
(see Figure 6 and the beginning of Section 3.2 for descriptions of the links
$P_{n}$ and $O_{n}$).

######  Theorem 1.2.

Suppose $M=\mathbb{S}^{3}\setminus\mathcal{A}$ is a flat FAL complement with
multiple distinct reflection surfaces. Then either

  * •

$\mathcal{A}$ is equivalent to the Borromean rings and $M$ contains exactly
three distinct reflection surfaces, or

  * •

$\mathcal{A}$ is equivalent to $P_{n}$ with $n\geq 3$, or $O_{n}$ with $n\geq
2$, and $M$ contains exactly two distinct reflection surfaces.

A slightly more general version of this classification result is given in
Theorem 3.11, along with several useful corollaries on how reflection surfaces
behave under homeomorphism.

The proof of Theorem 1.2, relies on an analysis of how cusps and totally
geodesic surfaces behave relative to different reflection surfaces in a flat
FAL complement. With Theorem 1.2 in hand, basic properties of the links
$P_{n}$ and $O_{n}$ show that two flat FALs whose complements admit multiple
reflection surfaces are equivalent as links if and only if their respective
complements are homeomorphic. This is highlighted in Corollary 3.12.

Theorem 1.2 also implies that a flat FAL with multiple reflection surfaces
cannot be homeomorphic to a flat FAL with a single reflection surface; see
Corollary 3.13. As a result, we now only need to consider the case where there
exists a homeomorphism $h:M\rightarrow M^{\prime}$ between two flat FAL
complements, each with unique reflection surfaces. This is a far more
challenging task, and relies on the topology and geometry of thrice-punctured
spheres in a flat FAL complement that are not contained in the reflection
surface, which were classified by Morgan–Mork–Ransom–Spyropoulos–Trapp–Ziegler
in [20]. In particular, we make extensive use of sets of thrice-punctured
spheres that separate $M$, whose homeomorphic images in $M^{\prime}$ are
greatly restricted by the classification in [20]. These restrictions help us
describe how the knot and crossing circles which intersect such thrice-
punctured spheres must behave under homeomorphism.

This, in turn, allows us to show that any flat FAL complement that admits a
unique reflection surface and a non-trivial homeomorphism must have a
particular link structure, which we call a signature link; see Definition 5.1.
Furthermore, only one type of non-trivial homeomorphism is possible in this
case, which we call a full-swap. We carefully describe full-swap
homeomorphisms of signature links in terms of compositions of Dehn twists
along sets of Hopf sub-links; see Definition 5.4. In addition, the induced
action of any such full-swap homeomorphism on the corresponding flat FAL
(which must be a signature link) can be made explicit on a diagrammatic level,
which makes it easy to construct a specific isotopy between any such pair of
flat FALs with homeomorphic complements.

As partially noted in the previous paragraph, our work not only shows that
flat FALs are determined by their complements, but classifies the types of
homeomorphisms that can exist between flat FAL complements and what types of
flat FAL complements can admit certain types of homeomorphisms. These severe
geometric restrictions on self-homeomorphisms of flat FAL complements lead to
a concise comparison between their symmetry groups and those of their
complements. Recall that the symmetry group of a link $L\subset\mathbb{S}^{3}$
is the group of homeomorphisms of pairs $(\mathbb{S}^{3},L)$ up to isotopy,
which we denote by $Sym(\mathbb{S}^{3},L)$. Similarly, the symmetry group of
the corresponding link complement is the group of self-homeomorphisms of
$\mathbb{S}^{3}\setminus L$ up to isotopy, denoted by
$Sym(\mathbb{S}^{3}\setminus L)$. Note that, any self-homeomorphism of
$(\mathbb{S}^{3},L)$ induces a self-homeomorphism of $\mathbb{S}^{3}\setminus
L$, and so, $Sym(\mathbb{S}^{3},L)\subseteq Sym(\mathbb{S}^{3}\setminus L)$.
However, this can be a strict containment; see [12] for some examples. The
following theorem classifies symmetries of flat FAL complements that are not
induced by symmetries of the corresponding link.

######  Theorem 1.3.

Let $\mathcal{A}$ be a flat FAL. Then either

  * •

$\mathcal{A}$ is not a signature link and both $\mathcal{A}$ and its
complement $M=\mathbb{S}^{3}\setminus\mathcal{A}$ have the same symmetry
group, or

  * •

$\mathcal{A}$ is a signature link and full-swaps on $\mathcal{A}$ generate
symmetries of $M=\mathbb{S}^{3}\setminus\mathcal{A}$ which are not
restrictions of symmetries of $\mathcal{A}$ to $M$.

Part of Theorem 1.3 is proved in Theorem 3.14 at the end of Section 3.1. The
rest of the proof of this theorem is completed at the end of Section 7.

We now describe the organization of the rest of this paper. In Section 2 we
introduce flat FALs, the necessary terminology related to cusps and totally
geodesic surfaces contained in flat FAL complements, and review some essential
facts from the literature on the geometry of FAL complements. In Section 3, we
prove Theorem 1.2. This allows us to focus the rest of the paper on
homeomorphisms between flat FAL complements, each with a unique reflection
surface and this transition is discussed in Section 4. Then Section 5
discusses “signature link” complements, a special class of flat FAL
complements, along with a “full-swap” homeomorphism that can be performed on
any signature link complement. The image of a full-swap homeomorphism is
another signature link complement, and it is straightforward to construct an
explicit isotopy between their corresponding links. In Section 6, we prove
some useful facts about sets of thrice-punctured spheres that separate a flat
FAL complement (with a unique reflection surface) and how they behave under
homeomorphisms. Finally, in Section 7 we build off the tools from Section 6 to
show that any homeomorphism between flat FAL complements (each with a unique
reflection surface) must essentially be a “full-swap” homeomorphism between
“signature link” complements. This allows us to construct an isotopy between
the corresponding links and prove our main theorem.

We would like to thank Jeffrey Meyer for comments and suggestions on
preliminary work done for this paper.

##  2\. Totally Geodesic Surfaces and Cusps in Flat FALs

This section provides a brief introduction to flat FALs, compiles some known
results about totally geodesic surfaces in their complements, and introduces
two concepts used extensively in Section 3: reflection-like surfaces and their
induced structures on a flat FAL complement. Aside from reflection-like
surfaces and their induced structures, this section is a review of results in
the literature.

We first describe how to construct a flat FAL. For this process, start with a
link $L$ and a diagram $D(L)$. We can build a diagram $D(\mathcal{F})$ for an
FAL $\mathcal{F}$ corresponding to $L$ by augmenting each twist region in
$D(L)$ with a circle and undoing all full-twists from each respective twist
region. After this procedure, twist regions of $D(L)$ that had contained an
odd number of crossings will still contain a single crossing in
$D(\mathcal{F})$. If we remove all of these remaining single crossings, then
we will have constructed a diagram $D(\mathcal{A})$ for a flat FAL
$\mathcal{A}$, as illustrated in Figure 2.

In our work, we will solely be interested in the case where a flat FAL
$\mathcal{A}$ is hyperbolic in the sense that the complement
$\mathbb{S}^{3}\setminus\mathcal{A}$ admits a complete metric of constant
negative curvature. As noted in [21, Theorem 2.5], a (flat) FAL $\mathcal{A}$
is hyperbolic if and only if it there exists a corresponding link diagram
$D(L)$ that is non-splittable, prime, twist-reduced, and contains atleast two
twist regions, where $D(\mathcal{A})$ is obtained from $D(L)$ by the
augmentation process described above. We refer the reader to [8, Section 1]
for these diagrammatic definitions. Moving forward, we will assume any flat
FAL is hyperbolic, and refer to $D(L)$ as the corresponding diagram that was
augmented to construct $D(\mathcal{A})$.

This augmentation process partitions the components of $\mathcal{A}$ into
crossing circles, the trivial components added via augmentation, and knot
circles, components coming from the original link $L$. Observe that each
crossing circle in a flat FAL bounds a crossing disk, a disk in
$\mathbb{S}^{3}$ punctured twice by parallel knot circle arcs that replace an
original twist region of $D(L)$. We define a flat FAL diagram to be a link
diagram together with this additional structure. More precisely, we have

######  Definition 2.1.

A link diagram $D(\mathcal{A})$ is a flat FAL diagram for a (hyperbolic) link
$\mathcal{A}$ if $D(\mathcal{A})$ was constructed by fully augmenting a
corresponding diagram $D(L)$ and removing all crossings from each twist
region. In addition, we assume $D(\mathcal{A})$ carries with it the partition
of components of $\mathcal{A}$ into crossing- and knot-circles, as well as the
choice of crossing disks, relative to this augmentation of $D(L)$. A link
$\mathcal{A}$ is a flat FAL if it admits a flat FAL diagram $D(\mathcal{A})$.

\begin{overpic}[width=433.62pt]{IntroPretzels.pdf} \put(2.0,22.0){$L$}
\put(30.0,22.0){$\mathcal{F}$} \put(66.0,22.0){$\mathcal{A}$}
\put(67.0,12.0){$c_{1}$} \put(81.5,12.0){$c_{2}$} \put(95.0,12.5){$c_{3}$}
\end{overpic} Figure 2. On the left is a diagram of a link $L$ with three
twist regions. The middle diagram shows the corresponding FAL $\mathcal{F}$
obtained from fully augmenting $L$. The right diagram shows the corresponding
flat FAL $\mathcal{A}$. Crossing circles of $\mathcal{A}$ are labeled by
$c_{i}$, for $i=1,2,3$.

The geometric structures of augmented links were first studied in Adams [1]. A
particularly nice geometric decomposition of (flat) FAL complements into pairs
of identical right-angled ideal polyhedra was described by Agol–Thurston in
the appendix of [16]. This geometric decomposition has proved to be a fruitful
tool for analyzing geometric and topological properties of FALs; see [6], [7],
[8], [13], [15], and [21] for a few examples. In addition, infinite subclasses
of FALs have also been examined in the literature. For instance,
Meyer–Millichap–Trapp [19] studied the arithmeticity, invariant trace fields,
symmetries, and hidden symmetries of FALs obtained by fully augmenting pretzel
links, and Purcell examined FALs whose complements admit a decomposition into
regular ideal hyperbolic octahedra in [21]. In both cases, each subclass
contains an infinite number of flat FALs. Thus the family of flat FALs is a
large set of links with many interesting properties.

We now collect some important facts about totally geodesic surfaces and cusps
contained inside a flat FAL complement. These properties will be essential for
proving both Theorems 1.1 and 1.2. Most of the results stated here are known
in the literature and we refer the reader to [8], [20], and [21] for more
details on the geometric properties of FAL complements discussed here.

First, we describe reflection surfaces, which are an important type of totally
geodesic surface contained in every flat FAL complement. Given a flat FAL
$\mathcal{A}$ in $\mathbb{S}^{3}\cong\mathbb{R}^{3}\cup\\{\infty\\}$, position
the crossing circles and disks so that they are orthogonal to the projection
plane and embed the knot circles in the projection plane. Such an isotopy is
always possible based on the diagrammatic definition of a flat FAL. Then
reflection in the projection plane maps every component of $\mathcal{A}$ to
itself and fixes the projection plane point-wise. Mostow–Prasad rigidity
implies that this projection plane corresponds with a totally geodesic surface
$R\subset M=\mathbb{S}^{3}\setminus\mathcal{A}$ and there exists an
orientation-reversing involution $\iota_{R}:M\rightarrow M$ that fixes $R$
point-wise corresponding with reflection in the projection plane. Since
$\mathcal{A}$ could admit FAL diagrams where different surfaces play the role
of the projection plane, we provide the following definition.

######  Definition 2.2.

Let $M=\mathbb{S}^{3}\setminus\mathcal{A}$ be a flat FAL complement and let
$R\subset M$ be an embedded totally geodesic surface. If there exists an FAL
diagram $D(\mathcal{A})$ in the projection plane $P$ for which
$R=P\setminus\mathcal{A}$, then we say that $R$ is a reflection surface of $M$
(relative to the diagram $D(\mathcal{A})$).

An FAL diagram $D(\mathcal{A})$ partitions the cusps of
$M=\mathbb{S}^{3}\setminus\mathcal{A}$ into crossing circle cusps and knot
circle cusps, corresponding with crossing circles and knot circles of
$\mathcal{A}$, respectively. We also say that this partition is induced by a
reflection surface $R$, though this ultimately depends on the FAL diagram
since reflection surfaces are defined relative to the projection plane for a
diagram. As we will see in Section 3, it is possible for a flat FAL complement
to admit distinct reflection surfaces that induce different partitions on the
components of $\mathcal{A}$. At the same time, it is possible for a flat FAL
complement to admit one reflection surface that induces different partitions
on the components of $\mathcal{A}$; this phenomenon will be examined in
Section 5. Here, we say a cusp of $M$ is an $R$-knot circle cusp
(respectively, $R$-crossing circle cusp) if the corresponding link component
of $\mathcal{A}$ is a knot circle (resp. crossing circle) relative to $R$. In
this paper, we frequently use the same notation to refer to a component of
$\mathcal{A}$ and the corresponding cusp on $M$. In addition, each
$R$-crossing circle $C$ of $\mathcal{A}$ bounds an $R$-crossing disk $D$,
which is a disk twice punctured by the two (not necessarily distinct) knot
circles going through $C$. As noted in [21, Lemma 2.1], each such $R$-crossing
disk is an embedded totally geodesic thrice-punctured sphere in $M$.
Furthermore, $R$ intersects $D$ orthogonally in the set of three simple, non-
separating geodesics on $D$; see Figure 3 for a visual of simple geodesics on
a thrice-punctured sphere.

A flat FAL diagram $D(\mathcal{A})$ (or, alternatively, its reflection surface
$R$) determines more structure than the partitioning of cusps of $M$ into knot
and crossing circle cusps. It also determines a meridian and longitude on each
boundary torus of a cusp neighborhood. In this context, meridians and
longitudes will be considered slopes, or unoriented isotopy classes of simple
closed curves on this boundary torus.

Each component $L$ of a flat FAL is topologically an unknot in
$\mathbb{S}^{3}$, so the torus boundary $T_{L}$ of a tubular neighborhood
$V_{L}$ of $L$ has a natural choice of meridional and longitudinal slopes. The
unknotted torus $T_{L}\subset\mathbb{S}^{3}$ bounds a solid torus on each
side. A meridian $m$ is the slope of $T_{L}$ that bounds a disk in $V_{L}$,
while a longitude $\ell$ is a slope of $T_{L}$ that bounds a disk in
$\mathbb{S}^{3}\setminus V_{L}$. The unoriented curves $m$ and $\ell$ are
well-defined up to ambient isotopy on $T_{L}$ , so give well-defined slopes on
$T_{L}$.

From a geometric perspective, the slopes $m$ and $\ell$ can be described as
the intersection of $T_{L}$ with totally geodesic surfaces. More precisely,
let $\mathcal{A}$ be a flat FAL with reflection surface $R$. If $K$ is a knot
circle of $\mathcal{A}$ with torus boundary $T_{K}$ of a cusp neighborhood of
$K$, then $R\cap T_{K}$ is a pair of simple closed curves that represent
longitudes on $T_{K}$. Similarly, for any crossing circle $C$ of
$\mathcal{A}$, the set $R\cap T_{C}$ is a pair of simple closed curves that
represent meridians on $T_{C}$. Crossing disks relative to the reflection
surface $R$ are orthogonal to it, and so intersect $T_{K}$ in meridians and
$T_{C}$ in longitudes. See [8, Lemma 2.3] for more details.

The longitudes and meridians just described will be referred to as
$R$-meridians and $R$-longitudes when we want to emphasize the FAL diagram
(and corresponding reflection surface) used to determine them. Since each
torus boundary of a cusp $T$ of $M$ is rectangular (see [8, Lemma 2.3]),
curves orthogonal to $R\cap T$ determine the second generator for the
fundamental group of each such torus. We refer to the basis for each
$\pi_{1}(T)$ described above as the peripheral structure relative to the FAL
diagram $D(\mathcal{A})$.

The following proposition summarizes the essential features of a reflection
surface in a flat FAL complement that we just discussed.

######  Proposition 2.3.

Let $M=\mathbb{S}^{3}\setminus\mathcal{A}$ be a flat FAL complement with an
FAL diagram $D(\mathcal{A})$ which partitions the components of $\mathcal{A}$
into crossing circles and knot circles. Then $M$ contains an embedded totally
geodesic $R$ with the following features:

  * •

$R$ corresponds with $P\setminus\mathcal{A}$, where $P$ is the projection
plane for $D(\mathcal{A})$,

  * •

$R$ is fixed point-wise by an orientation reversing involution
$i_{R}:M\rightarrow M$,

  * •

$R$ intersects each $R$-crossing disk of $M$ orthogonally in its three simple
non-separating geodesics,

  * •

$R$ intersects every cusp of $M$ in two parallel curves,

  * •

$R$ intersects the boundary torus $T_{C}$ of each crossing circle cusp $C$ in
a pair of meridians and provides a peripheral structure on $\pi_{1}(T_{C})$
relative to $D(\mathcal{A})$, and

  * •

$R$ intersects the boundary torus $T_{K}$ of each knot circle cusp $K$ in a
pair of longitudes and provides a peripheral structure on $\pi_{1}(T_{K})$
relative to $D(\mathcal{A})$.

Our main goal is to show that flat FALs are determined by their complements
among the set of all flat FALs. For this reason we will need to consider
homeomorphic images of reflection surfaces and the structures associated with
them that are highlighted in the previous proposition. By Mostow–Prasad
rigidty, any homeomorphism between flat FALs induces an isometry between them,
and so, we can restrict our analysis to the isometric image of a reflection
surface.

Let $M,M^{\prime}$ be flat FAL complements with reflection surfaces
$R,R^{\prime}$, respectively, and let $\rho:M^{\prime}\rightarrow M$ be an
isometry between these hyperbolic $3$-manifolds. Since the definition of a
reflection surface is diagram dependent, we can not immediately assume that
$\rho(R^{\prime})$ is a reflection surface for $M$. This motivates the
following definition.

######  Definition 2.4.

We say that $S$ is a reflection-like surface for a flat FAL complement $M$ if
there exists an isometry $\rho:M^{\prime}\rightarrow M$ between flat FAL
complements such that $S=\rho(R^{\prime})$, where $R^{\prime}$ is a reflection
surface for $M^{\prime}$.

Note that, any reflection surface for $M$ is also a reflection-like surface
for $M$ via the identity map. In addition, any homeomorphic flat FAL
complements have the same number of reflection-like surfaces. Indeed, a
homeomorphism $h:M\rightarrow M$’ between flat FAL complements $M$ and
$M^{\prime}$ induces a unique isometry $\rho_{h}:M\rightarrow M^{\prime}$. By
Definition 2.4, the image of each reflection-like surface in M is a
reflection-like surface in M’. At the same time, $\rho_{h}^{-1}$ preserves the
property of being reflection-like, so M and M’ have the same number of
reflection-like surfaces. Similarly, equivalent flat FALs have the same number
of reflection-like surfaces since an equivalence of links induces a
homeomorphism between their respective complements.

Many important topological and geometric properties of reflection surfaces
will be preserved under isometry, which we highlight below. These directly
follow from Proposition 2.3.

######  Proposition 2.5.

Let $\rho:M^{\prime}\rightarrow M$ be an isometry between flat FAL
complements, let $R^{\prime}\subset M^{\prime}$ be a reflection surface, and
let $S=\rho(R^{\prime})$ denote the corresponding reflection-like surface in
$M$. Then

  * •

$S$ is fixed point-wise by an orientation reversing involution
$i_{S}:M\rightarrow M$,

  * •

$S$ intersects the image of each $R^{\prime}$-crossing disk of $M^{\prime}$
orthogonally in its three simple non-separating geodesics,

  * •

$S$ intersects every cusp of $M$ in two parallel curves,

  * •

Let $T$ be the boundary torus of a cusp of $M$. Then $S$ determines a
peripheral structure on $T$, namely, the image of the peripheral structure
that $R^{\prime}$ determines on $\rho^{-1}(T)$.

Our work in Section 3 will show that every reflection-like surface in an FAL
complement is a reflection surface. This is highlighted in Corollary 3.13.
After proving this result, we will drop the term reflection-like surface, and
instead, only use reflection surface.

A reflection-like surface $S\subset M$ determines a partition of the
components of $\mathcal{A}$ into $S$-crossing circles and $S$-knot circles,
coming from the partition of $\mathcal{A}^{\prime}$ into $R^{\prime}$-crossing
circles and $R^{\prime}$-knot circles. We use the term $S$-structure on $M$ to
refer to this partition of the components of $\mathcal{A}$, the peripheral
structures on each cusp determined by $S$, and a fixed choice of images of
$R^{\prime}$-crossing disks in $M$. Suppose $M$ contains distinct reflection-
like surfaces $R$ and $S$. A component $L$ of $\mathcal{A}$ changes type if
$L$ is an $R$-knot circle and an $S$-crossing circle (or vice-versa). This
terminology will be useful when discussing flat FAL complements that admit
multiple reflection surfaces.

In addition to reflection surfaces, (totally geodesic) thrice-punctured
spheres contained in flat FAL complements will serve as powerful tools for
analyzing homeomorphisms between these manifolds. We now review properties of,
and results regarding, embedded, totally geodesic, thrice-punctured spheres in
flat FAL complements.

Any thrice-punctured sphere has precisely six simple geodesics, three of which
are separating, three of which are nonseparating, and none of which are
closed, as depicted in Figure 3. Any intersection of a totally geodesic
thrice-punctured sphere $D$ with another totally geodesic surface in a flat
FAL complement must be some subset of these six simple geodesics on $D$ that
are pairwise disjoint. This puts strong restrictions on the behavior of
intersections between $D$ and other totally geodesic surfaces, which we shall
exploit.

Figure 3. A thrice-punctured sphere with separating geodesics labeled $x,y,z$
and nonseparating geodesics labeled $a,b,c$.

Every FAL complement contains many totally geodesic thrice-punctured spheres
that are not contained in a reflection surface, which we will call non-
reflection, thrice-punctured spheres (relative to a designated reflection
surface). Crossing disks in $M$, for example, constitute one category of such
spheres. Results in [20] imply there are three types of non-reflection thrice-
punctured spheres in flat FAL complements, classified in terms of their
intersections with $R$ and their punctures. Here, a puncture refers to the
intersection of a totally geodesic surface in $M$ with a torus boundary of a
cusp of $M$, which will produce a set of simple closed curves, each of which
represents a slope on this torus. We sometimes use the term longitudinal
puncture to refer to a puncture that is a representative for a longitude on
this torus and we also use the term meridional puncture for a puncture that is
a representative for a meridian on this torus (here meridian and longitude
refer to the peripheral structure induced by a reflection surface $R\subset
M$). We now define the three types of non-reflection, thrice-punctured spheres
in flat FAL complements.

######  Definition 2.6.

Let $D$ be a non-reflection thrice-punctured sphere in the flat FAL complement
$M=\mathbb{S}^{3}\setminus\mathcal{A}$ with reflection surface $R$.

  * •

$D$ is a crossing disk if $D\cap R$ consists of the three non-separating
geodesics of $D$ and the punctures of $D$ are one crossing circle longitude
and two knot circle meridians,

  * •

$D$ is a longitudinal disk if $D\cap R$ consists of the three non-separating
geodesics of $D$ and the punctures of $D$ are longitudes of three distinct
crossing circles, and

  * •

$D$ is a singly-separated disk if $D\cap R$ consists of one separating
geodesic of $D$ and the punctures of $D$ are a longitude of one crossing
circle $C$ and two meridians of another crossing circle $C^{\prime}$.

Collectively, we refer to crossing disks and longitudinal disks as $N$-disks
since their intersections with $R$ both consist of a non-separating geodesics
of $D$.

Note that this definition of crossing disk broadens the typical meaning of the
term. The crossing circle $C$ illustrated in Figure 4(a) bounds two distinct
crossing disks, the one interior to $C$ as well as the shaded disk. The reason
for extending the meaning of “crossing disk” is that any such disk $D$ for a
crossing circle $C$ could be chosen as a crossing disk in $\mathcal{A}$ by
replacing the current crossing disk with $D$. The two crossing disk structures
come from fully augmenting diagrams of the same link obtained by flyping one
twist region to a different part of the link diagram (see [20]).

Parts (b) and (c) of Figure 4 illustrate a longitudinal and singly-separated
disk, respectively. The following theorem from [20] classifies non-reflection
thrice-punctured spheres in flat FAL complements.

$\begin{array}[]{ccc}\includegraphics[width=93.95122pt]{TopB.pdf}&\includegraphics[width=72.26999pt]{A3cp.pdf}&\includegraphics[width=93.95122pt]{TopA.pdf}\\\
(a)\textrm{ A crossing disk}&(b)\textrm{ A longitudinal disk}&(c)\textrm{ A
singly-separated disk}\end{array}$ Figure 4. Types of non-reflection, thrice-
punctured spheres

######  Theorem 2.7.

[20, Theorem 3.1] Let $D$ be a non-reflection thrice-punctured sphere in the
flat FAL complement $M=\mathbb{S}^{3}\setminus\mathcal{A}$ with reflection
surface $R$. Then $D$ is orthogonal to $R$ in $M$ and $D$ is either a
crossing, longitudinal, or singly-separated disk.

In our work, we will frequently make use of Theorem 2.7 and the qualifications
of non-reflection thrice-punctured spheres given in Definition 2.6. The
properties of separating and non-separating are topological, so if a
reflection surface intersects a thrice-punctured sphere in its separating
geodesic, so will any homeomorphic image of them. Thus, if $D$ is singly-
separated by the reflection surface $R$, then $h(D)$ is also separated by
$h(R)$ along one separating geodesic. Similarly, the homeomorphic image of an
$N$-disk relative to the reflection surface $R$ must intersect the reflection-
like surface $h(R)$ in its non-separating geodesics. In the case when $h(R)$
is again a reflection surface, we say that a homeomorphism preserves the type
of a singly-separated disk, and preserves the property of being an $N$-disk.
The only type changes possible when $h(R)$ is a reflection surface, then, are
between crossing and longitudinal disks. These can be recognized by analyzing
the images of punctures under this homeomorphism.

Finally, we define a separating pair to be a pair of disjoint thrice-punctured
spheres that separate $M$. The following theorem from [20] tells us exactly
which pairs of thrice-punctured spheres are separating pairs in FAL
complements.

######  Theorem 2.8.

[20, Theorem 4.1] Let $\\{S_{1},S_{2}\\}$ be a pair of disjoint essential
thrice-punctured spheres in the complement
$M=\mathbb{S}^{3}\setminus\mathcal{A}$ of the FAL $\mathcal{A}$. The pair
$\\{S_{1},S_{2}\\}$ is a separating pair if and only if each is either a
crossing disk or a singly-separated disk and their longitudinal slopes
coincide.

It is clear that any two crossing and/or singly-separated disks that share
longitudinal crossing-circle punctures separate a flat FAL complement. Theorem
2.8 states that these are the only separating pairs in FAL complements.

##  3\. Flat FAL Complements with multiple reflection surfaces

Proposition 2.3 shows that a reflection surface determines a lot of the
geometric structure of a flat FAL complement. In this section, we will
determine exactly how many reflection surfaces can exist in a flat FAL
complement and exactly which flat FAL complements admit multiple reflection
surfaces. This will be a useful first step towards our main goal of showing
that flat FALs are determined by their complements. We begin with some
preliminary observations which specify how distinct reflection-like surfaces
can intersect boundary tori of cusp neighborhoods.

As described in Section 2, reflection-like surfaces induce a peripheral
structure on boundary tori of cusp neighborhoods. Recall that the peripheral
structure consists of two orthogonal, unoriented isotopy classes of simple
closed curves (slopes), one labeled meridian and the other longitude. Let
$M=\mathbb{S}^{3}\setminus\mathcal{A}$ denote a flat FAL complement with
distinct reflection-like surfaces $R$ and $S$, and let $T$ denote the boundary
torus of a cusp of $M$. Our first lemma shows that the meridian-longitude
pairs determined by $R$ and $S$ are the same set-wise.

######  Lemma 3.1.

Let $M$ be a flat FAL complement with two distinct reflection-like surfaces
$R$ and $S$ and let $T$ be the boundary torus of a cusp of $M$. Then an
$R$-meridian of $T$ is either an $S$-meridian or $S$-longitude. Likewise, an
$R$-longitude of $T$ is either an $S$-meridian or $S$-longitude.

###### Proof.

The result will follow from the fact that these meridian-longitude pairs are
orthogonal.

Let $T$ denote the boundary torus of a cusp neighborhood of $M$ for some cusp
expansion of $M$. On this torus, let $m,\ell$ denote the $R$-meridian and
$R$-longitude, with geodesic lengths $\mu,\lambda$, respectively. Likewise,
let $m^{\prime},\ell^{\prime}$ denote the $S$-meridian and $S$-longitude with
lengths $\mu^{\prime},\lambda^{\prime}$, respectively. Since meridian-
longitude pairs are orthogonal

$\mu\lambda=Area(T)=\mu^{\prime}\lambda^{\prime}.$

Given an arbitrary orientation on the slopes $\\{m,\ell\\}$ and
$\\{m^{\prime},\ell^{\prime}\\}$, there are integers $p,q,r,s$ with
$m^{\prime}=pm+q\ell$ and $\ell^{\prime}=rm+s\ell$. By orthogonality we have

$\mu^{\prime}=\sqrt{(p\mu)^{2}+(q\lambda)^{2}},\textrm{ and
}\lambda^{\prime}=\sqrt{(r\mu)^{2}+(s\lambda)^{2}}.$

At least one of $p$ or $q$ is non-zero, which implies
$\mu^{\prime}\geq\textrm{min}\\{\mu,\lambda\\}$, with equality exactly when
$m^{\prime}$ is the shortest curve of the set $\\{m,\ell\\}$. Moreover, when
neither $p$ nor $q$ are zero, $\mu^{\prime}>\textrm{max}\\{\mu,\lambda\\}$.
Similar inequalities hold for $\lambda^{\prime}$. If both $p$ and $q$ are non-
zero, then

$\mu^{\prime}\lambda^{\prime}>\textrm{max}\\{\mu,\lambda\\}\textrm{min}\\{\mu,\lambda\\}=\mu\lambda.$

Since $\mu\lambda=\mu^{\prime}\lambda^{\prime}$, however, this implies the
oriented and simple $m^{\prime}$ is one of $\pm m$ or $\pm\ell$. The slope
$m^{\prime}$, then, is one of the slopes $m$ or $\ell$. Similarly,
$\ell^{\prime}$ equals the other slope and the sets
$\\{m^{\prime},\ell^{\prime}\\}$ and $\\{m,\ell\\}$ are equal. ∎

Thus, if a flat FAL complement admits multiple reflection-like surfaces
$\\{R_{i}\\}$, each such $R_{i}$-structure still induces the same basis on a
boundary torus of a cusp, though their induced peripheral structures may
differ since meridians and longitudes might switch roles (i.e. some components
of $\mathcal{A}$ could change type).

######  Remark.

The proof of Lemma 3.1 would not work if we assumed $M$ was just an FAL
complement. If the corresponding FAL has some number of half-twists, then
certain meridian-longitude pairs might no longer be orthogonal, and so, our
arguments used above would no longer apply. We point this since multiple
essential results for this paper rely on Lemma 3.1, and we want to make sure
the reader knows why our work doesn’t immediately apply more broadly to the
class of FALs and not just flat FALs.

The next two propositions clarify how a different reflection-like surface $S$
behaves relative to a given $R$-structure by considering $R$-knot circle and
$R$-crossing circle cusps separately.

######  Proposition 3.2.

Let $M$ be a flat FAL complement with two distinct reflection-like surfaces
$R$ and $S$. If $K$ is an $R$-knot circle and $T_{K}$ the boundary torus of a
cusp corresponding to $K$, then $S\cap T_{K}$ is a pair of $R$-meridians of
$K$.

###### Proof.

Lemma 3.1 implies that $S\cap T_{K}$ is either a pair of $R$-meridians or
$R$-longitudes. For the sake of contradiction, suppose that $S\cap T_{K}$ is a
pair of $R$-longitudes.

Since $K$ is an $R$-knot circle, $R\cap T_{K}$ is a pair of $R$-longitudes.
First suppose that $R\cap T_{K}=S\cap T_{K}$. Then the composition
$\iota_{R}\circ\iota_{S}$ acts as the identity on $T_{K}$ and preserves the
normal direction. Therefore $\iota_{R}\circ\iota_{S}$ is the identity on $M$
by [3, Proposition A.2.1], and $R=S$. This is a contradiction, which implies
$S\cap T_{K}\neq R\cap T_{K}$.

Now consider the case where $S\cap T_{K}$ are $R$-longitudes distinct from
$R\cap T_{K}$, and let $D$ be an $R$-crossing disk punctured by $K$. Note that
$D$ intersects $T_{K}$ in $R$-meridians, which are orthogonal to the
$R$-longitudes $S\cap T_{K}$. Thus $S$ intersects $D$ orthogonally. Moreover,
$R$ intersects $D$ in the non-separating geodesics of $D$, which implies that
$S\cap D$ consists of a single separating geodesic with both “ends” on $K$.
The other two punctures of $D$ must be on distinct cusps because one is an
$R$-knot circle puncture, and the other an $R$-crossing circle puncture. Since
$S$ intersects $D$ orthogonally along a separating geodesic, the reflection
$\iota_{S}$ preserves $D$ and interchanges the two punctures on distinct
cusps. This contradicts the fact that $\iota_{S}$ preserves all cusps, and
$S\cap T_{K}$ cannot be $R$-longitudes distinct from $R\cap T_{K}$.

Since $S$ cannot meet $T_{K}$ in $R$-longitudes, the pair of curves $S\cap
T_{K}$ are $R$-meridians. ∎

######  Proposition 3.3.

Let $M$ be a flat FAL complement with two distinct reflection-like surfaces
$R$ and $S$. Let $C$ be an $R$-crossing circle bounding the $R$-crossing disk
$D$. Then either

  1. i.

$D$ is a component of $S$, or

  2. ii.

$S$ intersects $D$ orthogonally along the separating geodesic of $D$ with
punctures on $C$. In this case the $R$-crossing circle $C$ is also an
$S$-crossing circle.

###### Proof.

Let $C$ be the $R$-crossing circle bounding $D$. By Lemma 3.1 we know that
$S\cap T_{C}$ is either two $R$-meridians or two $R$-longitudes of $C$.

Consider the case where $S\cap T_{C}$ consists of two $R$-meridians of $C$.
Since $C$ is an $R$-crossing circle the intersection $R\cap T_{C}$ consists of
$R$-meridians as well. If the surfaces $R$ and $S$ intersect $T_{C}$ in the
same curves then, as in the proof of Proposition 3.2, one arrives at the
contradiction that $R=S$. Thus the $R$-meridians $S\cap T_{C}$ are distinct
from those of $R\cap T_{C}$. Now $S$ and $D$ are an embedded totally geodesic
surfaces in $M$, so their intersection is a union of disjoint, complete,
simple geodesics. This follows from the observation that the local picture of
$S\cap D$, as seen in the universal cover, consists of two planes intersecting
along a geodesic. The only complete, simple geodesics of $D$ that intersect
$T_{C}$ are two non-separating geodesics, say $\gamma_{1},\gamma_{2}$, and one
separating geodesic, label it $\gamma$. Since $R$ contains the geodesics
$\gamma_{1},\gamma_{2}$, their intersection with $T_{C}$ lies in $R\cap
T_{C}$. This implies that $S$ cannot contain $\gamma_{1},\gamma_{2}$, since
$S$ and $R$ intersect $T_{C}$ in distinct curves. Thus $S\cap D$ contains the
separating geodesic $\gamma$ and neither $\gamma_{1}$ nor $\gamma_{2}$. All
other complete, simple geodesics on $D$ intersect $\gamma$, which implies
$\gamma=S\cap D$ and we are in case $(ii)$. Orthogonality follows from the
fact that the $R$-meridians $S\cap T_{C}$ are orthogonal to the $R$-longitude
$D\cap T_{C}$.

It remains to prove that $C$ is an $S$-crossing circle as well. Since $S$ is
reflection-like, there is a flat FAL $M^{\prime}$ with reflection surface
$R^{\prime}$ and an isometry $h:M^{\prime}\to M$ with $S=h(R^{\prime})$. Note
that $D^{\prime}=h^{-1}(D)$ is not contained in $R^{\prime}$ since $D$ is not
contained in $S$. Moreover, since $S$ intersects $D$ in a separating geodesic,
the geodesic $R^{\prime}\cap D^{\prime}$ separates $D^{\prime}$ as well. The
classification of non-reflection thrice punctured spheres given in Theorem 2.7
shows that $D^{\prime}$ must be a singly-separated disk, and all punctures of
singly-separated disks are crossing-circle punctures (see Figure 4). Thus all
punctures of $D$ are $S$-crossing circles, and $C$ must be an $S$-crossing
circle.

Now suppose $S\cap T_{C}$ consists of two $R$-longitudes of $T_{C}$. We will
show $D\subset S$. Since $C$ is an $R$-crossing circle we know $R\cap T_{C}$
is an $R$-meridian, and since $D$ is an $R$-crossing disk for $C$ it
intersects $T_{C}$ is a single $R$-longitude of $T_{C}$. Thus $D\cap T_{C}$ is
a single curve parallel to, or equal to one of, the curves in $S\cap T_{C}$.

First consider the case where $D\cap T_{C}$ equals one of the curves of $S\cap
T_{C}$, call it $\gamma$. Then $S$ and $D$ are embedded totally geodesic
surfaces whose intersection contains $\gamma$ which, despite being geodesic in
the induced Euclidean metric on $T_{C}$, is not a geodesic in $M$. If $S$ and
$D$ met transversally, their intersection could only contain geodesics. As $D$
is connected and intersects $S$ in a non-geodesic curve, it must be a subset
of $S$.

Now consider the case where $D\cap T_{C}$ is parallel to, and distinct from,
the curves of $S\cap T_{C}$. We will show this case leads to a contradiction.
As above, let $M^{\prime}$ be a flat FAL with reflection surface $R^{\prime}$
that admits an isometry $h:M^{\prime}\to M$ for which $S=h(R^{\prime})$.
Further, let $D^{\prime}=h^{-1}(D)$ and $T^{\prime}=h^{-1}\left(T_{C}\right)$.

Since we are assuming $D\cap T_{C}$ is disjoint from $S\cap T_{C}$, so also
$D^{\prime}\cap T^{\prime}$ and $R^{\prime}\cap T^{\prime}$ are disjoint and
$D^{\prime}$ is not a subset of $R^{\prime}$. Then $D^{\prime}$ is a non-
reflection, thrice-punctured sphere in $M^{\prime}$ and must be one of the
types described in Theorem 2.7 (see Figure 4). The classification shows that
punctures of non-reflection, thrice-punctured spheres are either knot
meridians, crossing longitudes or crossing meridians. Of these, only the
crossing meridians of singly-separated disks are disjoint from the reflection
surface (see Figure 4$(c)$). Thus $D^{\prime}$ is a singly-separated disk in
$M^{\prime}$ and $D^{\prime}\cap T^{\prime}$ consists of two crossing circle
meridians. This, however, contradicts the fact that $D\cap T_{C}$ is a single
component.

Therefore, if $S\cap T_{C}$ consists of two $R$-longitudes of $T_{C}$, then
$D$ is a component of $S$ and we are in case $(i)$ of the theorem. ∎

###  3.1. Three Reflection Surfaces

In this subsection we classify flat FAL complements that contain more than two
distinct reflection-like surfaces.

######  Theorem 3.4.

Suppose a flat FAL complement $M=\mathbb{S}^{3}\setminus\mathcal{A}$ admits at
least three distinct reflection-like surfaces. Then $\mathcal{A}$ is
equivalent to the Borromean Rings and $M$ contains exactly three reflection-
like surfaces, all of which are reflection surfaces.

###### Proof.

First, we show that a flat FAL complement $M$ can admit at most 3 distinct
reflection-like surfaces. This follows from the observation that at most one
reflection-like surface can satisfy each case of Proposition 3.3. To see this,
let $D$ be a crossing disk with respect to a reflection-like surface $R$, and
suppose $S_{1}$ and $S_{2}$ are reflection-like surfaces distinct from $R$ in
$M$. If both $S_{1}$ and $S_{2}$ contain $D$ (so satisfy Proposition
3.3$(i)$), then $\iota_{S_{2}}\circ\iota_{S_{1}}$ is the identity on $D$ and
preserves the normal direction. By [3, Proposition A.2.1] we have
$S_{1}=S_{2}$. On the other hand, if $S_{1}$ and $S_{2}$ both satisfy
Proposition 3.3$(ii)$, then both intersect $D$ orthogonally along the same
separating geodesic. This implies $\iota_{S_{2}}\circ\iota_{S_{1}}$ fixes any
point on this geodesic as well as the tangent space there, so again
$S_{1}=S_{2}$.

Thus a flat FAL complement has at most 3 distinct reflection-like surfaces,
and now suppose $M$ has $3$ distinct reflection-like surfaces: $R$, $S$, and
$P$. Moreover, let $D$ be an $R$-crossing disk bounded by the $R$-crossing
circle $C$. Then the argument given above shows that (up to relabeling) $P$
contains $D$ while $S$ and $D$ satisfy Proposition 3.3$(ii)$. In particular,
$R$ and $S$ intersect $T_{C}$ in disjoint pairs of $R$-meridians. Our goal is
to show that $P$ consists of exactly two (disjoint) thrice-punctured spheres,
which allows us to quickly classify all such flat FAL complements with such a
reflection surface.

Since $S$ and $D$ satisfy Proposition 3.3$(ii)$, $C$ is an $S$-crossing
circle. We will let $D^{\prime}$ be an $S$-crossing disk for $C$. Note that,
$D^{\prime}\neq D$ since $S$ intersects $D$ in a separating geodesic so $D$
cannot be an $S$-crossing disk. Now consider the surfaces $R$ and $P$ relative
to $D^{\prime}$. Since $D^{\prime}$ intersects $S$ orthogonally, and $R\cap
T_{C}$ is parallel to $S\cap T_{C}$, we have $R$ intersects $D^{\prime}$
orthogonally. Proposition 3.3 applied to $D^{\prime}$ and $R$ implies that
they satisfy case $(ii)$. Hence $P$ and $D^{\prime}$ must satisfy case $(i)$.
Thus $D^{\prime}$ is a component of $P$, and $D\cup D^{\prime}\subset P$.

Now consider $D\cup D^{\prime}$ relative to the reflection-like surface $S$.
The disk $D$ is an $S$-singly-separated disk which shares a longitudinal
crossing-circle puncture with the $S$-crossing disk $D^{\prime}$. Thus
$D^{\prime}$ and $D$ form a separating pair by Theorem 2.8. Since no proper
subset of a reflection-like surface separates, we have $P=D\cup D^{\prime}$.

The reflection-like surface $P=D\cup D^{\prime}$ has a total of six punctures,
and must intersect the boundary of each cusp in two curves; so there are three
cusps in $M$. The only three-component flat FAL is the Borromean rings, so
$\mathcal{A}$ must be the Borromean rings.

Figure 5. Three reflection surfaces in the Borromean rings complement

Finally, note that the Borromean rings complement contains three distinct
reflection surfaces, as depicted in Figure 5. Each shaded plane is a
reflection surface with the link component it contains serving as the one knot
circle.

∎

###  3.2. Two Reflection Surfaces

Let $M=\mathbb{S}^{3}\setminus\mathcal{A}$ be a flat FAL complement with
reflection-like surface $R$ and an additional (distinct) reflection-like
surface $R^{\prime}$. Our goal in this subsection is to show any such flat FAL
must be equivalent to either the Borromean Rings, $P_{n}$ with $n\geq 3$, or
$O_{n}$ with $n\geq 2$. See Figure 6 for FAL diagrams of $P_{4}$ and $O_{4}$.

\begin{overpic}[width=433.62pt]{B_P4_O4.pdf} \put(0.0,28.0){$B$}
\put(30.0,28.0){$P_{4}$} \put(66.0,28.0){$O_{4}$} \put(34.0,20.0){$c_{1}$}
\put(50.0,25.5){$c_{2}$} \put(57.5,10.5){$c_{3}$} \put(41.0,4.0){$c_{4}$}
\put(88.0,15.0){$c_{0}$} \end{overpic} Figure 6. On the left, the Borromean
Rings, $B$, is depicted as a flat FAL. In the middle, $P_{4}$ is depicted with
its four crossing circles labeled. On the right, $O_{4}$ is depicted, which
can be constructed by adding the crossing circle $C_{0}$ to $P_{4}$.

More generally, $P_{n}$ is the minimally-twisted chain with $2n$ components.
Thus $P_{n}$ is a flat FAL that admits an FAL diagram with $n$ crossing
circles and $n$ knot circles, linked together in a chain alternating between
crossing circles and knot circles. This link can also be described as fully
augmenting a pretzel link with $n$ twist regions with an even number of
crossings in each twist region, and performing a homeomorphism of the
complement to undo all twists from each twist region. The link $O_{n}$ is a
flat FAL that admits an FAL diagram with $n+1$ crossing circles and $n$ knot
circles, where this FAL diagram can be obtained by adding a single crossing
circle to the FAL diagram for $P_{n}$, just as depicted in Figure 6.

Figure 7. The second reflection-like surface in $P_{6}$.

Each of $P_{n}$ and $O_{n}$ contain (at least) two distinct reflection
surfaces: one corresponding with the projection plane in the following figures
and one corresponding with a $2$-sphere containing $C_{1},\ldots,C_{n}$ and
meeting $K_{1},\ldots,K_{n}$ orthogonally in these figures; in the case of
$O_{n}$, this $2$-sphere meets the link component $C_{0}$ orthogonally. Note
that a $90^{\circ}$ rotation of $\mathbb{S}^{3}$ about the axis of the chain
interchanges the two reflection surfaces (see Figure 7 for the second
reflection-like surface in $P_{6}$).

We now prove a useful fact about about thrice-punctured spheres contained in a
reflection surface, which we will make use of throughout this subsection.

######  Lemma 3.5.

Let $M$ be a flat FAL complement with reflection-like surface $R$, and let $D$
be a thrice-punctured sphere component of $R$. Then $D$ has an odd number of
$R$-knot circle punctures.

###### Proof.

The properties involved are topological, so without loss of generality we
assume that $R$ is the reflection surface of the flat FAL complement
$M=\mathbb{S}^{3}\setminus\mathcal{A}$. Further let $\mathbb{S}^{2}$ be the
projection two-sphere in $M$, so that $R=\mathbb{S}^{2}\setminus\mathcal{A}$
is the reflection surface. Finally, let $D$ be a thrice-punctured sphere
component of $R$. Since each component of $R$ has at least one knot circle
puncture, we must show that $D$ does not have two knot circle punctures.

Suppose, on the contrary, that exactly two knot circles, labeled $J$ and $K$,
puncture $D$. We will show that one of $J$ or $K$ has at most one crossing
circle linking it, contradicting the fact that every knot circle is linked by
at least two crossing circles. Some notation will be useful.

Note that $\mathbb{S}^{2}\setminus\left(J\cup K\right)$ consists of two disks
and an annulus. Let $D_{J}$ and $D_{K}$ denote the disk components bounded by
$J$ and $K$, respectively, and note that $D$ is the annular component. Then
$D$ is a once-punctured annulus and the additional puncture, label it the
point $p$, comes from a crossing circle $C$. Now $C$ punctures
$\mathbb{S}^{2}$ twice, and only once in $D$, so (possibly after relabeling)
$C$ also punctures $D_{K}$. Now $D$ is a thrice-punctured sphere, so there is
a unique geodesic $\gamma_{pK}$ in $D$ joining the punctures $p$ and $K$.
Thus, if $D_{C}$ is a crossing disk bounded by $C$ then $\gamma_{pK}=D_{C}\cap
D$.

Now let $C^{\prime}$ be a crossing circle other than $C$ with a crossing disk
$D^{\prime}$ punctured by $J$. As $C$ is the only crossing circle puncturing
$D$, we know $C^{\prime}$ is disjoint from $D$. However, since $J$ punctures
$D^{\prime}$ we know $D^{\prime}\cap D$ is a nonempty subset of geodesics on
$D$ with at least one endpoint on $J$. The separating geodesic $\gamma_{J}$ of
$D$ with both endpoints on $J$ intersects $\gamma_{pK}$ non-trivially, since
$\gamma_{pK}$ is the non-separating geodesic of $D$ opposite $J$. Since
$\gamma_{pK}=D_{C}\cap D$ and crossing disks are disjoint, $D^{\prime}$ cannot
intersect $D$ in $\gamma_{J}$. Further, $D^{\prime}\cap D$ cannot be the non-
separating geodesic of $D$ joining $J$ and $p$ since $p$ is on the crossing
circle $C$ and $C^{\prime}\neq C$. Thus $D^{\prime}\cap D$ must be the non-
separating geodesic $\gamma_{JK}$ of $D$ joining $J$ and $K$.

We have shown that any crossing disk punctured by $J$ intersects $D$ in the
geodesic $\gamma_{JK}$. Since crossing disks are orthogonal to the reflection
surface, at most one crossing disk intersects $D$ along a given geodesic. Thus
at most one crossing circle bounds a crossing disk punctured by $J$,
contradicting the fact that each knot circle in an FAL is linked by at least
two crossing circles. ∎

Each reflection-like surface partitions the cusps corresponding to components
of $\mathcal{A}$ into a set of knot circle cusps and a set of crossing circle
cusps. If some component of $\mathcal{A}$ changes type, then the geometry of
$M$ is greatly restricted. It is also possible that no “swapping” occurs,
i.e., $R$ and $R^{\prime}$ induce the same partition on the components of
$\mathcal{A}$ into knot circle (cusps) and crossing circle (cusps). We now
show that the latter case can not happen when $M$ has two distinct reflection
surfaces.

######  Lemma 3.6.

Suppose $M=\mathbb{S}^{3}\setminus\mathcal{A}$ is a flat FAL complement with
two reflection-like surfaces $R$ and $R^{\prime}$ that induce the same
partitions on the cusps of $M$ into crossing circle cusps and knot circle
cusps. Then $R=R^{\prime}$.

###### Proof.

Proposition 4.6 of [13] shows that there exists some $R$-crossing circle $C$
of $\mathcal{A}$ such that the corresponding $R$-crossing disk $D$ is the
unique totally geodesic thrice-punctured sphere in $M$ that intersects the
torus boundary of the cusp corresponding to $C$ in a longitude and intersects
the boundary of two $R$-knot circle cusps (not necessarily distinct) in a
meridian on each of these cusps. By assumption, $C$ is also an
$R^{\prime}$-crossing circle and the two $R$-knot circle cusps intersecting
$D$ are also $R^{\prime}$-knot circle cusps. By Lemma 3.5, $D$ can not be a
subset of $R^{\prime}$, and so, $D$ is a non-reflection thrice-punctured
sphere for $R^{\prime}$. Since $D$ has two $R^{\prime}$-knot circle and one
$R^{\prime}$-crossing circle punctures, Theorem 2.7 implies $D$ must also be
the corresponding $R^{\prime}$-crossing disk for $C$. Let $T_{K}$ be the
boundary torus of a cusp neighborhood for one of the knot circles that
intersect $D$. Then $D\cap T_{K}$ is orthogonal to both $R\cap T_{K}$ and
$R^{\prime}\cap T_{K}$. However, if $R\neq R^{\prime}$, then Proposition 3.2
implies that $R\cap T_{K}$ and $R^{\prime}\cap T_{K}$ are orthogonal, which
provides a contradiction. Thus, $R=R^{\prime}$ under these conditions. ∎

In the rest of this subsection, we consider the case where a component of
$\mathcal{A}$ changes type. First, we give a useful lemma.

######  Lemma 3.7.

Let $M=\mathbb{S}^{3}\setminus\mathcal{A}$ be a flat FAL complement with two
distinct reflection-like surfaces $R$ and $R^{\prime}$. Suppose $L$ is an
$R$-knot circle and an $R^{\prime}$-crossing circle. If $D^{\prime}$ is an
$R^{\prime}$-crossing disk for $L$, then $D^{\prime}\subset R$; moreover,
$D^{\prime}$ is unique.

###### Proof.

To see this, note that since $L$ is an $R$-knot circle, the reflection
surfaces $R^{\prime}$ and $R$ intersect $T_{L}$ in orthogonal pairs of curves
by Proposition 3.2. As an $R$-knot circle there are at least two $R$-crossing
disks that link $L$. Then $\partial D^{\prime}$ is a single curve on $T_{L}$
orthogonal to $R^{\prime}\cap T_{L}$, so it must be an $R$-longitude parallel
to $R\cap T_{L}$. As such it intersects $R$-crossing disks linking $L$, and
that intersection must be in a non-separating geodesic with one puncture on
$L$. To see this let $D$ be an $R$-crossing disk punctured by $L$. If
$D^{\prime}$ intersected $D$ in a separating geodesic of $D$ with endpoints on
$L$ then $\partial D^{\prime}$ would consist of two $R$-longitudes of $L$, but
$\partial D^{\prime}\cap T_{L}$ is a single $R$-longitude since $D^{\prime}$
is an $R^{\prime}$-crossing disk.

Then $R$ and $D^{\prime}$ have a common intersection with $R$-crossing disks
that link $L$, and both are orthogonal to such crossing disks, so
$D^{\prime}\subset R$.

To see that $D^{\prime}$ is unique, assume is a second $R^{\prime}$-crossing
disk $D^{\ast}$ for $L$. By the above argument, $D^{\ast}\subset R$ as well.
Then $D^{\prime}\cup D^{\ast}$ form a separating pair contained in $R$. Since
a proper subset of $R$ cannot separate, $R$ must equal $D^{\prime}\cup
D^{\ast}$. As in the proof of Theorem 3.4, a cusp count shows manifold has
three cusps and must be Borromean rings. The Borromean rings, however, do not
have a crossing circle that bounds two crossing disks in same
$R^{\prime}$-structure; therefore, $D^{\prime}$ is unique. ∎

Suppose $K$ is a component of $\mathcal{A}$ that corresponds with an $R$-knot
circle cusp and an $R^{\prime}$-crossing circle cusp. Let $D^{\prime}$ be the
$R^{\prime}$-crossing disk bound by $K$. Then Lemma 3.7 guarantees that
$D^{\prime}$ is a component of $R$. Since $K$ is an $R$-knot circle, there
exists some $R$-crossing circle $C$ that links $K$, along with an $R$-crossing
disk $D$ which $K$ punctures. We now consider the intersection patterns for
$D\cap D^{\prime}$. The work of Yoshida [25, Proposition 3.1] shows that there
are three possible cases to consider: (i) $D\cap D^{\prime}$ is a single non-
separating geodesic on both $D$ and $D^{\prime}$, (ii) $D\cap D^{\prime}$ is a
pair of non-separating geodesics on both $D$ and $D^{\prime}$, and (iii)
$D\cap D^{\prime}$ is a separating geodesic on one and non-separating on the
other. Note that since $D^{\prime}\subset R$ and $D\cap R$ consists of non-
separating geodesics on $D$, the curve $D\cap D^{\prime}$ must be non-
separating on $D$. Hence, in case (iii), the separating geodesic must be on
$D^{\prime}$. See Figure 8 for corresponding diagrams. These cases are
considered separately in the next three propositions.

\begin{overpic}[width=433.62pt]{3cases.pdf} \put(4.0,41.0){Case (i)}
\put(28.0,50.0){Case(ii)} \put(71.0,50.0){Case (iii)}
\put(7.0,31.0){$D^{\prime}$} \put(12.0,24.0){$K$} \put(18.0,35.0){$D$}
\put(25.5,38.5){$C$} \put(43.0,24.0){$D^{\prime}$} \put(61.5,8.0){$K$}
\put(52.0,31.0){$D$} \put(58.0,35.0){$C$} \put(90.0,14.0){$D^{\prime}$}
\put(99.0,8.0){$K$} \put(90.0,31.0){$D$} \put(97.0,35.0){$C$} \end{overpic}
Figure 8. The three cases for intersection patterns for $D^{\prime}\cap D$
where the projection plane here is determined by $R$. In each figure, the
thrice-punctured spheres $D$ and $D^{\prime}$ are shaded blue and grey,
respectively. Only the top half of the crossing disk $D$ is drawn in each
diagram. Intersections between $D$ and $D^{\prime}$ are highlighted in green.

######  Proposition 3.8.

Suppose $M=\mathbb{S}^{3}\setminus\mathcal{A}$ is a flat FAL complement with
two distinct reflection-like surfaces $R$ and $R^{\prime}$. Let $K$ be a
component of $\mathcal{A}$ that is an $R^{\prime}$-crossing circle and an
$R$-knot circle, and let $D^{\prime}$ be the $R^{\prime}$-crossing disk $K$
bounds. Let $C$ be an $R$-crossing circle bounding the $R$-crossing disk $D$,
with $K$ puncturing $D$. If the intersection $D^{\prime}\cap D$ is a single
non-separating geodesic on each thrice-punctured sphere (case $(i)$ of Figure
8), then $\mathcal{A}$ is either $P_{n}$ with $n\geq 3$, or $O_{n}$ with
$n\geq 2$.

###### Proof.

Let $K$, $D^{\prime}$, $C$ and $D$ be as in the statement of the proposition.
By Lemma 3.7 the disk $D^{\prime}$ is a component of $R$ and note that, since
$D^{\prime}\cap D$ is a single non-separating geodesic on both, $D^{\prime}$
punctured once each by $K$ and $C$. Lemma 3.5 implies that the remaining
puncture of $D^{\prime}$ comes from an $R$-crossing circle, say $J$.

Since $D^{\prime}$ has punctures along three distinct cusps, the $R$-crossing
disk $D_{J}$ that $J$ bounds intersects $D^{\prime}$ in a single non-
separating geodesic for both disks (case $(i)$ of Figure 8). In the
$R$-structure, then, the cusps $J,K,C$ form a chain of three components.

Now consider $J$, $K$ and $C$ in the $R^{\prime}$-structure. We know that $K$
is an $R^{\prime}$-crossing circle and that $D^{\prime}$ is its corresponding
$R^{\prime}$-crossing disk. Thus the other two punctures, $J$ and $C$, must be
$R^{\prime}$-knot circles. By Lemma 3.7, then, the $R$-crossing disks $D$ and
$D_{J}$ are components of the reflection surface $R^{\prime}$.

Applying Lemma 3.5 to the disks $D$ and $D_{J}$ in the $R^{\prime}$-structure
implies the third puncture in each is an $R^{\prime}$-crossing circle
puncture. Translating the picture to the $R$-structure, and applying Lemma 3.7
to each end of the chain produces a chain of four or five components. There
are four components if the punctures unaccounted for in $D$ and $D_{J}$
correspond to the same $R^{\prime}$-crossing circle.

Since $\mathcal{A}$ has finitely many components, iterating this argument
terminates in a sublink of $\mathcal{A}$ isotopic to $P_{n}$, and which we
denote $\mathcal{P}$. Note that the $R$-crossing disks for $\mathcal{P}$ are
components of $R^{\prime}$, and vice-versa. For convenience, let
$K_{1},\dots,K_{n}$ denote the $R$-knot circles of $\mathcal{P}$, and let
$S_{1},\dots,S_{n}$ be the corresponding thrice-punctured sphere components of
$R$ that they bound (so the $S_{i}$ are $R^{\prime}$-crossing disks). Further
let $C_{1},\dots,C_{n}$ denote the $R$-crossing circles of $\mathcal{P}$, and
let $D_{1},\dots,D_{n}$ be the respective $R$-crossing disks that they bound.

Consider the action of $\iota_{R^{\prime}}$ on the components of the
reflection surface $R$. Each of the $S_{i}$ reflect to themselves since they
are $R^{\prime}$-crossing disks. Now let
$\widehat{R}=R\setminus\left(\cup_{i=1}^{n}S_{i}\right)$. Our goal is to show
that the $K_{i}$ are the only $R$-knot circle punctures of $\widehat{R}$. Note
that cutting $\widehat{R}$ along its intersection with the $D_{i}$ separates
$\widehat{R}$ into two subsets $\widehat{R}_{0}$ and $\widehat{R}_{1}$, which
are $R^{\prime}$-reflections of each other. Let $N$ be an $R$-knot circle
distinct from the $K_{i}$ that punctures $\widehat{R}$. Then $N$ cannot pass
through the $D_{i}$, which are already punctured twice by the $K_{i}$. Thus
$N$ is contained entirely in one of $\widehat{R}_{0}$ or $\widehat{R}_{1}$,
and $\iota_{R^{\prime}}(N)$ is in the other. This contradicts the fact that
$\iota_{R^{\prime}}$ preserves each cusp; therefore, no such $N$ exists and
the $K_{i}$ are the only $R$-knot circle punctures of $\widehat{R}$. Since
$R=\widehat{R}\cup\left(\cup_{i=1}^{n}S_{i}\right)$ and the only $R$-knot
circle puncturing $S_{i}$ is $K_{i}$ for $i=1,\ldots,n$, we have that the
$K_{i}$ are the only $R$-knot circles puncturing $R$, and so, they are also
the only $R$-knot circles of $\mathcal{A}$.

At this stage, any components of $\mathcal{A}$ that are not in $\mathcal{P}$
must be $R$-crossing circles that puncture $\widehat{R}$ twice. Our goal is to
show there is at most one such $R$-crossing circle. Let $\widetilde{C}$ be an
$R$-crossing circle of $\mathcal{A}$ that is not a component of $\mathcal{P}$,
and let $\widetilde{D}$ be an $R$-crossing disk bounded by $\widetilde{C}$.
Since the $K_{i}$ are the only $R$-knot circles of $\mathcal{A}$, the disk
$\widetilde{D}$ intersects some $K_{j}$ and the thrice-punctured sphere
$S_{j}\subset R$ that it bounds. The crossing disks of $\mathcal{P}$ intersect
$S_{j}$ in the two non-separating geodesics on $S_{j}$ with endpoints on
$K_{j}$, so $\gamma_{j}=S_{j}\cap\widetilde{D}$ must be a separating geodesic
on $S_{j}$ since $R$-crossing disks must be disjoint.

Suppose there are at least two $R$-crossing circles in
$\mathcal{A}\setminus\mathcal{P}$, and label a second one
$\widetilde{C}^{\prime}$ with $R$-crossing disk $\widetilde{D}^{\prime}$. We
now describe how to construct an open, embedded, essential annulus $A\subset
M$, with boundary (in $\mathbb{S}^{3}$) curves $\widetilde{C}$ and
$\widetilde{C}^{\prime}$. Consider the standard genus one Heegaard
decomposition of $\mathbb{S}^{3}=T_{1}\cup T_{2}$, where $T_{1}$ and $T_{2}$
are solid tori. A sufficiently small closed neighborhood of
$\displaystyle{\cup_{i=1}^{n}\left(K_{i}\cup S_{i}\cup C_{i}\cup
D_{i}\right)}$ produces a solid torus in $\mathbb{S}^{3}$ that is isotopic to
one of these solid tori, say $T_{1}$, and disjoint from $\widetilde{C}$,
$\widetilde{C}^{\prime}$, as well as any additional $R$-crossing circles not
in $\mathcal{P}$. In addition, by taking appropriate neighborhoods, we can
assume each of $\widetilde{D}$ and $\widetilde{D}^{\prime}$ intersects $T_{1}$
in a meridional disk of $T_{1}$. Note that, $\partial
T_{1}\setminus(\widetilde{D}\cup\widetilde{D}^{\prime})$ produces two embedded
annuli in $M$, and label one of these $A^{\prime}$. Then
$A=A^{\prime}\cup(\widetilde{D}\setminus
int(T_{1}))\cup(\widetilde{D}^{\prime}\setminus int(T_{1}))$ provides the
desired annulus, contradicting hyperbolicity. Thus,
$\mathcal{A}\setminus\mathcal{P}$ is either empty or contains exactly one
$R$-crossing circle that intersects $\widehat{R}$ twice, once in
$\widehat{R}_{0}$ and once in $\widehat{R}_{1}$, as needed. ∎

######  Proposition 3.9.

Suppose $M=\mathbb{S}^{3}\setminus\mathcal{A}$ is a flat FAL complement with
two distinct reflection-like surfaces $R$ and $R^{\prime}$. Let $K$ be a
component of $\mathcal{A}$ that is an $R^{\prime}$-crossing circle and
$R$-knot circle, and let $D^{\prime}$ be the $R^{\prime}$-crossing circle disk
$K$ bounds. Let $C$ be an $R$-crossing circle bounding the $R$-crossing disk
$D$, with $K$ puncturing $D$. If the intersection $D^{\prime}\cap D$ is a pair
of non-separating geodesics on each thrice-punctured sphere (case $ii$ in
Figure 8), then $\mathcal{A}$ is the Borromean Rings.

###### Proof.

By the work of Yoshida [25, Lemma 3.7], any hyperbolic $3$-manifold with two
thrice-punctured spheres intersecting in this manner must be a (possibly
empty) Dehn filling of one of three manifolds: a certain double cover of the
Whitehead link complement, the Borromean rings complement, or the minimally
twisted hyperbolic 4-chain link complement. See Figure 17 in [25] for diagrams
of these links. All three of these manifolds have the common hyperbolic volume
of $2v_{8}$, where $v_{8}$ denotes the volume of a regular ideal hyperbolic
octahedra. At the same time, the work of Purcell [21, Proposition 3.6] shows
that the Borromean rings complement is the unique minimal volume flat FAL
complement. Since non-empty Dehn filling strictly decreases volume, the only
flat FAL complement with thrice-punctured spheres intersecting in this manner
is the Borromean rings complement, as needed. ∎

######  Proposition 3.10.

Suppose $M=\mathbb{S}^{3}\setminus\mathcal{A}$ is a flat FAL complement with
two distinct reflection-like surfaces $R$ and $R^{\prime}$. Let $K$ be a
component of $\mathcal{A}$ that is an $R^{\prime}$-crossing circle and
$R$-knot circle, and let $D^{\prime}$ be the $R^{\prime}$-crossing circle disk
$K$ bounds. Let $C$ be an $R$-crossing circle bounding the $R$-crossing disk
$D$, with $K$ puncturing $D$. If the intersection $D^{\prime}\cap D$ is a
separating geodesic on $D^{\prime}$ and a non-separating geodesic on $D$ (case
$(iii)$ in Figure 8), then $\mathcal{A}$ is either the Borromean Rings or
$O_{n}$ with $n\geq 2$.

###### Proof.

Let $K$, $D^{\prime}$, $C$ and $D$ be as in the statement of the proposition.
By Lemma 3.7 the disk $D^{\prime}$ is a component of $R$, and note that it is
punctured once by $K$. Since $D^{\prime}$ is a thrice-punctured sphere, we
know that it must have two more punctures. We break down this proof into cases
depending on whether those punctures come from $R$-crossing circles or
$R$-knot circles. Note that, since $D\cap D^{\prime}$ is a separating geodesic
$\gamma_{K}$ on $D^{\prime}$, the thrice-punctured sphere $D$ partitions
$D^{\prime}$ into two regions separated by $\gamma_{K}$.

Case I: Suppose the other two punctures of $D^{\prime}$ come from $R$-crossing
circles $C_{1}$ and $C_{2}$. Further, suppose $C_{1}=C_{2}$. Then $C_{1}$ and
$C_{2}$ must puncture different regions of $D^{\prime}\setminus D$ since
$D^{\prime}$ must have a puncture on each side of the separating geodesic
$\gamma_{k}$. In this case, $\mathcal{A}$ contains a Borromean Rings sub-link,
$L_{B}=K\cup C\cup C_{1}$. Let $D_{1}$ designate the $R$-crossing disk
corresponding to $C_{1}$. As an $R$-crossing disk, $D_{1}$ intersects $R$ in
its three nonseparating geodesics. Two of these geodesics must be contained in
$D^{\prime}\subset R$ since $C_{1}$ punctures $D^{\prime}$ in two different
regions and $D^{\prime}$ can not intersect $\gamma_{K}$. Since $D^{\prime}\cap
D_{1}$ is a pair of non-separating geodesics on $D_{1}$, it follows that this
intersection is also a pair of non-separating geodesics on $D^{\prime}$ by the
work of Yoshida [25, Proposition 3.1]. Following the proof of Proposition 3.9
with $D_{1}$ replacing $D$ implies that $A=L_{B}$, i.e., $A$ is the Borromean
rings.

Now suppose that $C_{1}$ and $C_{2}$ are distinct $R$-crossing circles, each
of which puncture $D^{\prime}$. Then $C_{1}$ has an $R$-crossing disk $D_{1}$,
which is punctured by $K$ and some distinct $R$-knot circle $K_{1}$. This
implies that the intersection $D^{\prime}\cap D_{1}$ is a single non-
separating geodesic on each of these thrice-punctured spheres. Then
Proposition 3.8 with $D_{1}$ replacing $D$ shows that $L$ is either $P_{n}$ or
$O_{n}$. However, $P_{n}$ does not contain a crossing circle $C$ whose
crossing disk $D$ separates a region of $R$, which implies that $\mathcal{A}$
must be $O_{n}$ here.

Case II: Suppose that at least one of the other two punctures of $D^{\prime}$
comes from an $R$-knot circle, $K_{1}$. We will show this case leads to a
contradiction. By Lemma 3.5, $D^{\prime}$ must have an odd number of $R$-knot
circle punctures, and so, the third puncture of $D^{\prime}$ comes from an
$R$-knot circle distinct from $K$ and $K_{1}$, which we label $K_{2}$.
Furthermore, since $\gamma_{K}$ is a separating geodesic on $D^{\prime}$,
$K_{1}$ and $K_{2}$ must puncture different regions of $D^{\prime}\setminus
D$. Any $R$-crossing disk punctured by $K_{1}$, call it $D_{1}$, is disjoint
from $D$ and so must intersect $D^{\prime}$ in geodesic(s) disjoint from
$\gamma_{K}=D\cap D^{\prime}$. The only geodesics of $D^{\prime}$ disjoint
from $D\cap D^{\prime}$ are the non-separating geodesics
$\\{\gamma_{1},\gamma_{2}\\}$ joining $K$ to the other punctures $K_{1}$ and
$K_{2}$ respectively. Since $D_{1}$ is punctured by $K_{1}$, it’s one other
$R$-knot circle puncture must come from $K$, and so, $\gamma_{1}=D_{1}\cap
D^{\prime}$. Now there is at most one connected, embedded, totally geodesic
surface orthogonal to $R$ and containing $\gamma_{1}$; therefore, at most one
crossing disk intersects $D^{\prime}$ along $\gamma_{1}$. This implies at most
one crossing circle links $K_{1}$, contradicting the fact that every knot
circle must be linked by at least two crossing circles in an FAL. ∎

We can now give the following classification of flat FAL complements that
admit multiple reflection surfaces. A slightly less general version of this
theorem was originally stated in Theorem 1.2 in Section 1.

######  Theorem 3.11.

Suppose $M=\mathbb{S}^{3}\setminus\mathcal{A}$ is a flat FAL complement with
multiple distinct reflection-like surfaces. Then either

  * •

$\mathcal{A}$ is equivalent to the Borromean rings, and $M$ contains exactly
three distinct reflection-like surfaces, all of which are reflection surfaces,
or

  * •

$\mathcal{A}$ is equivalent to $P_{n}$ with $n\geq 3$, or $O_{n}$ with $n\geq
2$, and $M$ contains exactly two distinct reflection-like surfaces, both of
which are reflection surfaces.

###### Proof.

Suppose $M$ is a flat FAL complement with at least two distinct reflection-
like surfaces. Then Lemma 3.6 shows that some $R^{\prime}$-crossing circle $K$
of $\mathcal{A}$ must switch to become an $R$-knot circle (or vice versa). Let
$D^{\prime}$ be the $R^{\prime}$-crossing disk corresponding to $K$. By Lemma
3.7, $D^{\prime}\subset R$. At the same time, since $K$ is an $R$-knot circle,
it punctures at least two $R$-crossing disks, one of which we label $D$. Then
$D$ and $D^{\prime}$ are both totally geodesic thrice-punctured spheres in $M$
that intersect non-trivially since $D^{\prime}$ must intersect $R$ on each
side of $K$ (thinking of $K$ as a simple closed curve in the projection
plane), and one of these components is $D^{\prime}$. The work of Yoshida [25,
Proposition 3.1], shows that there are exactly three possibilities for such
intersection patterns, which are covered in Proposition 3.8, Proposition 3.9,
and Proposition 3.10. Combined, these propositions tell us that $\mathcal{A}$
is equivalent to either the Borromean rings, $P_{n}$ with $n\geq 3$, or
$O_{n}$ with $n\geq 2$. Theorem 3.4 distinguishes the Borromean rings as the
only flat FAL whose complement admits at least three distinct reflection-like
surfaces. In this case, this FAL complement has exactly three reflection-like
surfaces, all of which are reflection surfaces, as noted in Theorem 3.4; see
Figure 5 for a visualization of the three reflection surfaces. If an FAL
complement has exactly two reflection-like surfaces, then the corresponding
link is either $P_{n}$ with $n\geq 3$ or $O_{n}$ with $n\geq 2$. The
discussion at the beginning of Section 3.2 shows that the corresponding FAL
complements for these links each admit two distinct reflection surfaces, and
so, every reflection-like surface is also a reflection surface in all of these
cases. ∎

This classification theorem implies that within the family of flat FALs, those
whose complements admit multiple reflection surfaces are determined by their
complements. We highlighted this result in the following corollary.

######  Corollary 3.12.

Let $\mathcal{A}$ and $\mathcal{A^{\prime}}$ be flat FALs, and suppose the
complement of $\mathcal{A}$ admits multiple distinct reflection surfaces. Then
$\mathbb{S}^{3}\setminus\mathcal{A}$ is homeomorphic to
$\mathbb{S}^{3}\setminus\mathcal{A^{\prime}}$ if and only if $\mathcal{A}$ and
$\mathcal{A^{\prime}}$ are equivalent links.

###### Proof.

If $\mathcal{A}$ and $\mathcal{A}^{\prime}$ are equivalent links, the
orientation-preserving homeomorphism between the pairs
$(\mathbb{S}^{3},\mathcal{A})$ and $(\mathbb{S}^{3},\mathcal{A}^{\prime})$
induces one between their complements. So, suppose
$\mathbb{S}^{3}\setminus\mathcal{A}$ and
$\mathbb{S}^{3}\setminus\mathcal{A}^{\prime}$ are homeomorphic flat FAL
complements where $\mathbb{S}^{3}\setminus\mathcal{A}$ admits multiple
reflection surfaces. Then these flat FAL complements are isometric and
$\mathbb{S}^{3}\setminus A^{\prime}$ also contains multiple reflection-like
surfaces. By Theorem 3.11, $A^{\prime}$ is equivalent to either the Borromean
rings $B$, $P_{n}$ with $n\geq 3$, or $O_{n}$ with $n\geq 2$. Thus, we just
need to consider the cases where $\mathbb{S}^{3}\setminus\mathcal{A}$ is
homeomorphic to the complement of one of these links. Note that $B$ has three
components, each $P_{n}$ has $2n$ components with $n\geq 3$, and $O_{n}$ has
$2n+1$ components with $n\geq 2$. Thus the number of components distinguishes
the links $B$, $\\{P_{n}\\}_{n=3}^{\infty}$ and $\\{O_{n}\\}_{n=2}^{\infty}$.
Since there is a one-to-one correspondence between components of a link and
cusps of the corresponding link complement, this shows that the number of
cusps of one of these link complements determines the corresponding link,
completing the proof. ∎

The work from this section places an important restriction on the behavior of
homeomorphisms between flat FAL complements.

######  Corollary 3.13.

Let $M$ and $M^{\prime}$ be flat FAL complements and suppose there exists a
homeomorphism $h:M\rightarrow M^{\prime}$, which induces an isometry
$\rho_{h}$. Then $R$ is a reflection-like surface for $M$ if and only if $R$
is a reflection surface for $M$. In particular, $\rho_{h}$ provides a one-to-
one correspondence between reflection surfaces.

###### Proof.

By the comments immediately following Definition 2.4, the isometry $\rho_{h}$
produces a one-to-one correspondence between reflection-like surfaces. If we
show that every reflection-like surface is actually a reflection surface, we
will be done.

Theorem 3.11 covers the multiple reflection-like surface case. On the other
hand, the reflection surface in a flat FAL is one reflection-like surface.
Therefore, if a flat FAL has a unique reflection-like surface it must be the
reflection surface, completing the proof. ∎

Before moving on, we make two useful observation that follow from Corollary
3.13. First off, we will no longer use the term reflection-like surface since
a reflection-like surface is a reflection surface. In addition, if
$h:M\rightarrow M^{\prime}$ is a homemorphism, $M$ has a unique reflection
surface $R$, and $M^{\prime}$ contains a reflection surface $R^{\prime}$, then
$M^{\prime}$ also has a unique reflection surface and
$\rho_{h}(R)=R^{\prime}$.

Our final result from this section shows that in most cases, every symmetry of
a flat FAL complement with multiple reflection surfaces is induced by a
symmetry of that link. In particular, the following theorem proves the first
statement from Theorem 1.3 in the introduction.

######  Theorem 3.14.

Let $\mathcal{A}$ be a flat FAL, other than $P_{3}$, whose complement admits
multiple reflection surfaces. Then both $\mathcal{A}$ and its complement
$M=\mathbb{S}^{3}\setminus\mathcal{A}$ have the same symmetry group.

###### Proof.

Since every symmetry of $(\mathbb{S}^{3},\mathcal{A})$ induces one of its
complement, it’s enough to show that every homeomorphism $h:M\to M$ extends to
an isotopy of $\mathbb{S}^{3}$.

Since $\mathcal{A}$ is not $P_{3}$, Theorem 3.11 implies it is either $P_{n}$
with $n\geq 4$, $O_{n}$ with $n\geq 2$, or the Borromean rings, and we
consider the cases separately.

Let $\mathcal{A}=O_{n}$, with $n\geq 2$, and let $R$ be a reflection surface
in $M=\mathbb{S}^{3}\setminus O_{n}$. Note that $R$ consists of $n$ thrice-
punctured spheres and one $(n+2)$-punctured sphere $S_{n+2}^{2}$ whose
punctures are $n$ longitudes along the $R$-knot circles of $O_{n}$ and $2$
punctures by the same $R$-crossing circle $C_{0}$. Since $n\geq 2$,
$S^{2}_{n+2}$ is the only component of $R$ with more than three punctures.

Now let $h:M\to M$ be a homeomorphism with induced isometry $\rho_{h}:M\to M$.
Corollary 3.13, applied to the case where $M^{\prime}=M$, implies that
$R^{\prime}=\rho_{h}(R)$ is a reflection surface for $O_{n}$. Since
$R^{\prime}$ is a reflection surface for $O_{n}$, it has a unique
$(n+2)$-punctured sphere component ${S^{2}_{n+2}}^{\prime}$. The unique cusp
punctured twice by ${S^{2}_{n+2}}^{\prime}$ corresponds to an
$R^{\prime}$-crossing circle while the remaining $n$ cusps punctured by
${S^{2}_{n+2}}^{\prime}$ correspond to the $R^{\prime}$-knot circles of
$O_{n}$. Since $\rho_{h}$ preserves the topology of components of $R$ we have
$\rho_{h}\left({S^{2}_{n+2}}\right)={S^{2}_{n+2}}^{\prime}$, and $R$-knot
circles must map to $R^{\prime}$-knot circles. Moreover,
$C_{0}^{\prime}=\rho_{h}(C_{0})$ must be the $R^{\prime}$-crossing circle of
$O_{n}$ puncturing ${S^{2}_{n+2}}^{\prime}$ twice. The remaining components of
$O_{n}$ are $R$-crossing circles and, since all $R^{\prime}$-knot circles are
accounted for, $\rho_{h}$ must map them to $R^{\prime}$-crossing circles.

Thus $\rho_{h}$ preserves the type of each component of $O_{n}$–crossing
circles map to crossing circles, and similarly with knot circles. Let $L$ be a
component of $O_{n}$ with image $L^{\prime}=\rho_{h}(L)$, and let
$\\{m,\ell\\}$ and $\\{m^{\prime},\ell^{\prime}\\}$ denote the respective
$R$\- and $R^{\prime}$\- meridian-longitude pairs. Lemma 3.1 implies
$\rho_{h}$ maps the set $\\{m,\ell\\}$ to the
$\\{m^{\prime},\ell^{\prime}\\}$, and we must show it takes meridians to
meridians.

If $K$ is an $R$-knot circle of $O_{n}$, then $T_{K}\cap R$ is a pair of
simple closed curves, each representing an $R$-longitude, which we denoted by
$\ell$ earlier. As above, let $\\{m^{\prime},\ell^{\prime}\\}$ denote the
$R^{\prime}$ meridian and longitude for $T_{K^{\prime}}=\rho_{h}(T_{K})$. Then
$\rho_{h}(T_{K}\cap R)=T_{K^{\prime}}\cap R^{\prime}$, and so maps
longitudinal slopes of $K$ to those of $K^{\prime}$. As above, Lemma 3.1
implies that $\rho_{h}$ preserves meridians as well. Thus $\rho_{h}$ preserves
peripheral structures on all $R$-knot circles.

To see that $h$ preserves peripheral structures on $R$-crossing circles, let
$C$ be an $R$-crossing circle with $R^{\prime}$-crossing circle image
$C^{\prime}=\rho_{h}(C)$. Using notation similar to the above, we have the
calculation $\rho_{h}(T_{C}\cap R)=T_{C^{\prime}}\cap R^{\prime}$ so
$\rho_{h}$ preserves meridional slopes, and Lemma 3.1 implies it preserves
peripheral structures on crossing circles of $O_{n}$ as well.

Thus $\rho_{h}$, and therefore our original $h:M\to M$, preserves peripheral
structures on all components and extends to an isotopy of $\mathbb{S}^{3}$.
This implies, of course, that a symmetry of $M$ is the restriction of a
symmetry of $(\mathbb{S}^{3},O_{n})$ to its complement.

The proof for $P_{n}$, with $n\geq 4$, follows similarly, but is a little more
direct. In this case the reflection surface $R$ consists of $n$ thrice
punctured spheres and one $n$-punctured sphere $S_{n}^{2}$. Since $n\geq 4$,
the sphere $S^{2}_{n}$ is the unique component of $R$ with more than three
punctures. The proof follows as above, with the simplification that punctures
of $S^{2}_{n}$ correspond to the distinct $R$-knot circles of $P_{n}$.

We have proven the theorem for flat FALs with two reflection surfaces and the
only case remaining is the flat FAL with three reflection surfaces–the
Borromean rings. In this case, a quick check using SnapPy confirms the
symmetry group of the Borromean rings and its complement coincide. ∎

Note that the proof of Theorem 3.14 does not extend to $P_{3}$. For $P_{3}$
all components of the reflection surface $R$ are thrice-punctured spheres,
with one punctured by three knot circles. A simple puncture count, then, does
not guarantee that the isometry $\rho_{h}$ preserves the component of $R$
punctured by knot circles. Section 5 will show that this is more than just a
shortcoming of the above proof. In fact $P_{3}$ is a signature link, and it
will be seen that signature link complements have more symmetries than the
links themselves.

##  4\. Transition to the unique reflection surface case

Corollary 3.13 tells us that homeomorphic flat FAL complements must have the
same number of reflection surfaces. This allows us to break down the proof of
our main result, Theorem 1.1, into two cases:

(1) There exists a homeomorphism between flat FAL complements, each with
multiple reflection surfaces. This case is already covered by Corollary 3.12.

(2) There exists a homeomorphism between flat FAL complements, each with a
unique reflection surface.

We can actually put a more narrow focus on the homeomorphisms we need to
analyze in case (2). Let $M$ and $M^{\prime}$ be the complements of the flat
FALs $\mathcal{A}$ and $\mathcal{A}^{\prime}$, respectively, each containing
unique reflection surfaces denoted by $R$ and $R^{\prime}$. If there exists a
homeomorphism $h:M\to M^{\prime}$, which induces isometry $\rho_{h}$, then we
must have $R^{\prime}=\rho_{h}(R)$. Recall that the reflection surfaces
determine peripheral structures on each component of $\mathcal{A}$ and
$\mathcal{A}^{\prime}$, respectively. Thus if $h$ preserves both knot circles
and crossing circles, then it preserves peripheral structures and extends to
an isotopy of $\mathbb{S}^{3}$, making the links $\mathcal{A}$ and
$\mathcal{A}^{\prime}$ equivalent. For this reason we will be mainly
interested in homeomorphisms that change a knot circle $K$ into a crossing
circle, or vice-versa. In this case we will say that $h$ changes the type of a
component $K$ of $\mathcal{A}$, and will call $h$ a type-changing
homeomorphism.

The rest of this paper is dedicated to analyzing type-changing homeomorphisms
of flat FAL complements containing unique reflection surfaces. Section 5 will
introduce a particular class of type-changing homeomorphisms where we can
easily find an isotopy in $\mathbb{S}^{3}$ between the corresponding FALs.
Section 6 examines how separating sets (partially introduced in Section 2)
behave under type-changing homeomorphisms to help restrict the behavior of
such homeomorphisms. Combining the work of these two sections, we then prove
our main result in Section 7.

##  5\. Signature links and full-swap homeomorphisms

In this section we define signature links, which are a special class of FALs,
and show that they admit a particular type-changing homeomorphisms, which we
call a full-swap, on their complements. Furthermore, we will show that
signature links whose complements are homeomorphic via a full-swap exhibit an
explicit isotopy in $\mathbb{S}^{3}$.

######  Definition 5.1.

A signature link is a flat FAL $\mathcal{L}$ whose components can be
partitioned into four non-empty sets

$\mathcal{L}=\\{K_{f}\\}\cup\mathcal{K}\cup\mathcal{C}\cup\mathcal{C_{K}},$

with the following properties. The first set consists solely of a knot circle
$K_{f}$ which cuts the projection plane into two disks, one of which contains
the remaining knot circles $\mathcal{K}=\left\\{K_{1},\dots,K_{n}\right\\}$,
which we call the inside of $K_{f}$. Each $K_{i}\in\mathcal{K}$ is linked with
$K_{f}$ by a unique crossing circle $C_{i}$, and
$\mathcal{C}=\left\\{C_{1},\dots,C_{n}\right\\}$. The set of remaining
crossing circles is denoted $\mathcal{C_{K}}$, and components in
$\mathcal{C_{K}}$ link two distinct knot circles in $\mathcal{K}$. A crossing
circle of $\mathcal{C_{K}}$ that links $K_{i},K_{j}\in\mathcal{K}$ will be
denoted by $C_{ij}$.

Throughout this section, we let $D_{i}$ designate a crossing disk for
$C_{i}\in\mathcal{C}$ and we let $D_{ij}$ designate a crossing disk for
$C_{ij}\in\mathcal{C_{K}}$.

$\begin{array}[]{ccc}\includegraphics{SignatureSublink.pdf}&&\includegraphics{VertSigLink.pdf}\\\
(a)\textrm{ Aesthetic diagram}&&(b)\textrm{ Pragmatic diagram}\end{array}$
Figure 9. Two diagrams of a signature link $\mathcal{L}$.

Figure 9$(a)$ depicts a signature link. The knot circles in $\mathcal{K}$ are
all inside $K_{f}$, and are numbered counterclockwise around $K_{f}$ (only
$K_{1}$ is labeled in Figure 9$(a)$). The diagram of Figure 9$(b)$ will be
helpful in visualizing a full-swap, and is obtained by isotoping $K_{f}$ until
it is vertical. We remark that there can be more than one way to decompose
$\mathcal{L}$ as a signature link. The knot circle $K_{4}$ (unlabeled in
Figure 9) could have been designated $K_{f}$ instead since it is linked to
every other knot circle.

According to Definition 5.1 the link $P_{3}$ is a signature link, but a quick
check verifies that this is the only link whose complement contains multiple
reflection surfaces which is a signature link. Thus all other signature link
complements have a unique reflection surface.

The fact that $\mathcal{L}$ is hyperbolic places restrictions on the set
$\mathcal{C_{K}}$. The set $\mathcal{C_{K}}$, for example, can not be empty.
More can be said, of course, about properties of $\mathcal{C_{K}}$ resulting
from the hyperbolicity of $\mathcal{L}$, but we content ourselves with a
result about longitudinal disks which requires the following technical lemma.

######  Lemma 5.2.

Two $N$-disks in an FAL complement are either identical or disjoint.

###### Proof.

To see this, we show that if two $N$-disks intersect, then they are identical.
Let $D_{1},D_{2}$ be $N$-disks in an FAL complement $M$, and so, they each
intersect the reflection surface $R$ in their non-separating geodesics. Lemma
3.4 of [25] states that thrice-punctured spheres in orientable three-manifolds
cannot intersect along a geodesic that is separating in both. Thus, if
$\gamma\in D_{1}\cap D_{2}$, then $\gamma$ is non-separating in at least one
disk, say $D_{1}$. The disk $D_{1}$ is an $N$-disk so $\gamma$ is contained in
the reflection surface $R$ and $D_{2}$ intersects $D_{1}$ along a geodesic in
$R$. By Theorem 2.7 both $D_{1}$ and $D_{2}$ are orthogonal to $R$ along
$\gamma$. Therefore, since $D_{1}$ and $D_{2}$ are both embedded totally
geodesic surfaces that intersect in a common geodesic and are orthogonal to
$R$, we can conclude that they must be equal, completing the proof. ∎

######  Lemma 5.3.

Let $\mathcal{L}$ be a signature link. Then every crossing circle
$C_{ij}\in\mathcal{C_{K}}$ bounds a totally geodesic, longitudinal disk with
crossing circles $C_{i}$ and $C_{j}$.

###### Proof.

Let $\mathcal{L}$ be a signature link with complement
$M=\mathbb{S}^{3}\setminus\mathcal{L}$. Given a crossing circle
$C_{ij}\in\mathcal{C_{K}}$, we will construct a longitudinal disk with
punctures $C_{i},C_{j},C_{ij}$ by gluing two disks, which are essentially
topological descriptions of the geodesic disks described in [20]. Afterwards,
we will show this longitudinal disk is totally geodesic.

A knot circle $K_{i}\in\mathcal{K}$ of a signature link $\mathcal{L}$ bounds
two disks in the projection plane, and we define the _inside_ of $K_{i}$ to be
the disk $\mathcal{P}_{i}$ That does not contain $K_{f}$. Thus
$\mathcal{P}_{i}$ is punctured once by $C_{i}$ and once by each crossing
circle of $\mathcal{C_{K}}$ that links $K_{i}$.

Now consider how crossing disks intersect $\mathcal{P}_{i}$. Since all
crossing circles in $\mathcal{L}$ link distinct knot circles, and since
$K_{i}$ is the only knot circle puncture of $\mathcal{P}_{i}$, no crossing
disk intersects $\mathcal{P}_{i}$ in a geodesic arc with both endpoints on
$K_{i}$. Crossing disks that intersect $\mathcal{P}_{i}$, then, do so in a
geodesic joining a crossing circle puncture to the boundary curve $K_{i}$.
Further, since crossing disks are disjoint they intersect $\mathcal{P}_{i}$ in
disjoint arcs (see Figure 10$(a)$). The complement of crossing disks in
$\mathcal{P}_{i}$ is then connected and there is an embedded arc, disjoint
from crossing disks, between any two crossing circle punctures of
$\mathcal{P}_{i}$.

$\begin{array}[]{ccc}\includegraphics{InIsectD.pdf}&&\includegraphics{SigLinkProjPlane.pdf}\\\
(a)\textrm{ Crossing disks in }\mathcal{P}_{i}&&(b)\ \alpha\textrm{ curves for
}D_{\ell}\end{array}$ Figure 10. Crossing disks and a longitudinal disk
intersecting the reflection surface.

Similarly, the knot circle $K_{f}$ divides the projection plane into two
topological disks, one punctured by the knot circles of $\mathcal{K}$ and the
other by the crossing circles of $\mathcal{C}$. The _outside_ of $K_{f}$,
denoted $\mathcal{P}_{f}$, refers to the disk punctured by the crossing
circles of $\mathcal{C}$. The argument of the previous paragraph shows that
there is an embedded arc in $\mathcal{P}_{f}$ joining any two crossing circle
punctures that is disjoint from crossing disks of $\mathcal{L}$.

Given a crossing circle $C_{ij}\in\mathcal{C_{K}}$ we construct a longitudinal
disk $D_{\ell}$, with punctures $C_{i},C_{j},C_{ij}$, by gluing a disk $D_{+}$
above the reflection surface to its reflection $D_{-}$. We begin by describing
the boundary of $D_{+}$. Let $\alpha_{i}\subset\mathcal{P}_{i}$ be an embedded
arc disjoint from crossing disks that joins $C_{i}$ and $C_{ij}$ punctures,
and define $\alpha_{j}$ similarly. Also let $\alpha_{ij}$ denote an embedded
arc in $\mathcal{P}_{f}$ which is disjoint from crossing disks and joins the
$C_{i}$ and $C_{j}$ punctures. Figure 10$(b)$, for example, illustrates the
$\alpha$ arcs corresponding to the crossing circle $C_{12}$ of the signature
link in Figure 9.

The arcs $\alpha_{i},\alpha_{j},\alpha_{ij}$, together with the top halves of
the crossing circles $C_{i},C_{j},C_{ij}$, form a simple closed curve $\gamma$
in $\mathbb{S}^{3}$. Let $M^{+}$ be the region of $M$ above the reflection
surface and we wish to show that $\gamma$ bounds a disk in $M^{+}$. First,
$M^{+}$ is a handlebody, since it is a three-ball with arcs removed for each
crossing circle. Further, the top half of each crossing disk is a meridional
disk for each handle. Thus removing an open neighborhood of each crossing disk
from $M^{+}$ results in a three ball with $\gamma$ in its boundary. The curve
$\gamma$, then, bounds a disk $D_{+}$ in $M^{+}$. The disk $D_{+}$ is disjoint
from crossing disks and intersects the reflection surface along the $\alpha$
arcs in its boundary. Let $D_{-}$ be the reflection of $D_{+}$, and let
$D_{\ell}=D_{+}\cup D_{-}$. Then the crossing circles $C_{i},C_{j},C_{ij}$
form the boundary of $D_{\ell}$ in $\mathbb{S}^{3}$, and the interior of
$D_{\ell}$ is an embedded thrice-punctured sphere in $M$ with longitudinal
punctures along the crossing circles $C_{i},C_{j},C_{ij}$, as desired.

To see that $D_{\ell}$ is totally geodesic it is enough to show that it is
incompressible and boundary incompressible, by Theorem 3.1 of [2]. The proof
given here is essentially that of [21, Lemma 2.1], but slightly simpler
because the punctures of $D_{\ell}$ are distinct. Suppose $\alpha$ is a curve
in $D_{\ell}$ that bounds a compressing disk $D\subset M\setminus D_{\ell}$.
Then $\alpha$ separates $D_{\ell}$ into two pieces, one of which contains a
single puncture $C$ of $D_{\ell}$. Thus $\alpha\cup C$ bound an annulus in
$D_{\ell}$ whose union with $D$ is a boundary compressing disk for the
crossing circle $C$, contradicting the fact that $M$ is hyperbolic.

When discussing $\partial$-incompressibility it will be convenient to think of
$M$ as the interior of a orientable, closed three-manifold $\overline{M}$ with
torus boundary components. In this case, $D_{\ell}$ is properly embedded in
$\overline{M}$ with longitudinal boundary curves along $T_{i},T_{j},T_{ij}$,
the boundary tori of $\overline{M}$ corresponding to cusps
$C_{i},C_{j},C_{ij}$ of $M$. For convenience, and by an abuse of notation, we
let $C_{i},C_{j},C_{ij}$ denote the boundary curves of $D_{\ell}$ as well.

Now suppose that $D$ is a $\partial$-compressing disk for $D_{\ell}$. Then
$\partial D=\alpha\cup\beta$ where $\alpha=D\cap D_{\ell}$ is an arc in
$D_{\ell}$ with both endpoints on the same boundary curve, say $C_{i}$, and
$\beta$ is an arc in $T_{i}$. The arc $\alpha$ is isotopic to a separating
geodesic of $D_{\ell}$, which decomposes $D_{\ell}$ into two annuli, and we
let $A$ denote the one containing $C_{j}$. The other boundary component of the
annulus $A$ consists of the arc $\alpha$ together with a subarc of $C_{i}$,
call it $\delta$. Gluing the $\partial$-compressing disk $D$ to $A$ along
their intersection $\alpha$ yields another annulus $A\cup D$ with $C_{j}$ as
one boundary component. The other boundary component of $A\cup D$ is the
simple closed curve $\beta\cup\delta$ on the boundary torus $T_{i}$.

If $\beta\cup\delta$ bounds a disk on $T_{i}$, a copy of it in $M$ caps off
one boundary component of $A\cup D$, creating a $\partial$-compressing disk
for $C_{j}$, which is impossible. On the other hand, suppose $\beta\cup\delta$
is non-trivial on $T_{i}$. Then $A\cup D$ is an incompressible annulus which
is not boundary parallel since its boundary curves are on separate boundary
components of $\overline{M}$. Again, this contradicts the fact that $M$ is
hyperbolic.

Thus $D_{\ell}$ is incompressible and $\partial$-incompressible and it has a
totally geodesic representative by [2, Theorem 3.1]. ∎

The remainder of this section is devoted to constructing a “full-swap”, which
is a type-changing homeomorphism of the complement of a signature link
$\mathcal{L}$. Full-swaps, despite changing the types of some components of
$\mathcal{L}$, will be shown to produce a link equivalent to $\mathcal{L}$.

To begin, we define an $ml$-swap homeomorphism on a a flat FAL complement
$M=\mathbb{S}^{3}\setminus\mathcal{A}$ in terms of Dehn twists on a Hopf
sublink $\mathcal{H}\subset\mathcal{A}$. In particular, the Hopf sublink will
consist of a knot and crossing circle pair that are linked. Zevenbergen, in
[27], first constructed a product that exchanged meridional and longitudinal
slopes on each component of $\mathcal{H}$ and was able to analyze the effect
on the other components of $\mathcal{A}$. We review his construction here,
then provide an alternative cut-and-paste construction which highlights how
$ml$-swaps effect reflection surfaces and crossing disks.

Figure 11 illustrates an $ml$-swap on the Hopf sublink determined by the knot-
and crossing-circle pair labeled $\\{K,C\\}$. Before describing the Dehn
twists we observe some features of $\mathcal{A}$ relative to $\mathcal{H}$.
Since $\mathcal{H}=K\cup C$ is a Hopf link, the crossing circle $C$ must link
distinct knot circles and we label the other one $J$. Let $D$ be a crossing
disk for $C$ and let $N$ be an open regular neighborhood of the cell complex
$K\cup C\cup D$ in $\mathbb{S}^{3}$. Then $N$ is an unknotted open solid
torus, so $W=\mathbb{S}^{3}\setminus N$ is a closed unknotted solid torus in
$\mathbb{S}^{3}$. The neighborhood $N$ can be chosen so that
$N\cap\mathcal{A}$ contains the components $K$ and $C$, and an arc of $J$ that
intersects $D$ (see Figure 11$(a)$). Then $W$ contains all components of
$\mathcal{A}\setminus(K\cup C\cup J)$ together with one arc of $J$.

Figure 11. An $ml$-swap as a product of Dehn twists

We now describe Zevenbergen’s Dehn twists that make up an $ml$-swap and,
abusing notation, we refer to components by their original labels throughout
the Dehn twist process. First perform a Dehn twist along $K$, which adds a
full twist to $W$ and links $C$ around $W$ as in Figure 11$(b)$. For
simplicity, isotope $C$ so that it is flat, twisting $J$ and $K$ vertical in
the process, to get Figure 11$(c)$. Now perform a Dehn twist along $C$ that
untwists $W$ so that it is returned to its original form. This twist unlinks
$K$ and $W$ while linking the arc of $J$ with both $W$ and $K$, as in Figure
11$(d)$. Finally, perform a Dehn twist on $K$ that unlinks $C$ and $J$,
obtaining the link depicted in Figure 11$(e)$. This composition of Dehn twists
is an $ml$-swap. The result will not be another flat FAL in general, as
observed in [27], but we will see that performing multiple $ml$-swaps on
signature links can produce a flat FAL.

Note that this product of Dehn twists is ultimately a local operation in the
sense that changes to the link occur within a 3-ball containing the Hopf
sublink. Moreover, if there are multiple Hopf sublinks that are contained in
disjoint three balls, then the result of performing Dehn twists on each Hopf
sublink is independent of the order in which they are done. With this
background in place we make the following definition.

######  Definition 5.4.

An $ml$-swap on a Hopf sublink $\mathcal{H}$ of a flat FAL $\mathcal{A}$ is
the homeomorphism resulting from performing the Dehn twists just described
above on the components of $\mathcal{H}$. A full-swap on a signature link
$\mathcal{L}$ is the composition of all $ml$-swaps on Hopf sublinks $K_{i}\cup
C_{i}$, for $i=1,\ldots,n$.

Dehn twists provide a convenient description of an $ml$-swap, and we now
consider an alternative description that highlights the effect of an $ml$-swap
on the reflection surface and crossing disks involved. Figure 12$(a)$
highlights a crossing circle $C$ and knot circle $K$ whose meridians and
longitudes will be swapped. The other knot circle linked by $C$ is included to
emphasize how the reflection surface moves, but the rest of the link is not
pictured. The homeomorphism maps $C$ to the knot circle $C^{\prime}$ and $K$
to the crossing circle $K^{\prime}$ depicted in Figure 12$(g)$. Note that an
$ml$-swap preserves the reflection surface, while moving the location of the
component $R_{0}$ to that of $R_{0}^{\prime}$.

Figure 12. An $ml$-swap

Let us walk through the homeomorphism one step at a time. In Figure 12$(b)$,
torus neighborhoods of $C$ and $K$ are pictured to emphasize what happens to
meridians and longitudes. Now slice the manifold along the reflection surface
and consider the top half pictured in Figure 12$(c)$, which is a handlebody
$H_{+}$ (the homeomorphism on the bottom half is the reflection of that
pictured). In Figure 12$(c)$, the half-tori around $K$ and $C$, as well as the
copy of $R_{0}$, form three annuli. The middle row of Figure 12 depicts an
isotopy sliding the three annuli along a handle of $H_{+}$. Note that in the
process meridians and longitudes of $K$ and $C$ are swapped. The final row
depicts regluing the isotoped $H_{\pm}$ along $R$, and finally removing the
torus neighborhoods of $K^{\prime}$ and $C^{\prime}$. In terms of peripheral
structures on cusps of $M$, an $ml$-swap swaps meridians and longitudes on
cusps corresponding to $C$ and $K$, and leaves the remaining structures the
same.

It is instructive to consider the image under an $ml$-swap of crossing disks
punctured by the knot circle involved. The image of a crossing disk $D$
bounded by $C$ is the natural first choice to consider. Figure 13 illustrates
that $D$ is sliced in half, each half rotated by “a third”, then re-glued
along its non-separating geodesics. A similar rotation is done on the bottom
half, so swapping the types of $C$ and $K$ has the effect of “rotating” $D$.

Figure 13. Rotating a crossing disk in an $ml$-swap

Let $C^{\ast}$ be a crossing circle other than $C$ linking $K$. Let $D^{\ast}$
a crossing disk bounded by $C^{\ast}$, and consider the image of $D^{\ast}$
under the $ml$-swap. Since the $K$-puncture of $D^{\ast}$ becomes a crossing
circle puncture, its image ${D^{\ast}}^{\prime}$ has two crossing circle
punctures. This is illustrated in Figure 14. Since an $N$-disk in an FAL
complement has either one or three crossing circle punctures, we know
${D^{\ast}}^{\prime}$ is not a crossing disk, assuming the image of
$M=\mathbb{S}^{3}\setminus\mathcal{A}$ under this $ml$-swap is a flat FAL
complement.

Figure 14. Sliding a crossing disk in an $ml$-swap

In fact, an $ml$-swap on a flat FAL can result in a link that is not an FAL
(see [27]). Performing $ml$-swaps on all possible $(C_{i},K_{i})$ pairs in a
signature link (i.e. a full-swap), however, does produce another flat FAL.
Consider, for example, the simplest signature link: the chain $P_{3}$ in
Figure 15$(a)$. A full-swap homeomorphism $h_{f}$ is realized by performing
successive $ml$-swaps on the Hopf sublinks $C_{1}\cup K_{1}$ and $C_{2}\cup
K_{2}$, which yields the sequence of Figure 15.

$\begin{array}[]{ccc}\includegraphics{2MLswapsA.pdf}&\includegraphics{2MLswapsB.pdf}&\includegraphics{2MLswapsC.pdf}\\\
(a)\textrm{ Original FAL}&(b)\textrm{ Non-FAL after one $ml$-swap}&(c)\textrm{
FAL isotopic to original}\\\ &&\textrm{after two $ml$-swaps}\end{array}$
Figure 15. Two $ml$-swaps

Note that the image of $P_{3}$ is isotopic to $P_{3}$. Proposition 5.5 will
show that this holds more generally–that the image of a signature link under a
full-swap is always isotopic to the original link. While full-swaps produce
equivalent links, they do interchange some crossing- and longitudinal-disks.
For example, in Figure 15$(c)$, the image of the crossing disk $D_{12}$ is the
longitudinal disk $D_{12}^{\prime}$. In addition, the longitudinal disk with
punctures $\\{C_{1},C_{2},C_{12}\\}$ of Figure 15$(a)$ becomes the crossing
disk that $C_{12}^{\prime}$ bounds (neither are pictured, but reading Figure
15 backwards illustrates the change from longitudinal to crossing disk). The
proof of Proposition 5.5 will show that a full-swap on an arbitrary signature
link interchanges crossing- and longitudinal-disks for every crossing circle
in $\mathcal{C}_{\mathcal{K}}$.

The pragmatic diagram of Figure 9$(b)$ will be more convenient for the proof
of Proposition 5.5, so we assume $K_{f}$ is vertical and crossing circles of
$\mathcal{C}$ are ordered from bottom to top. Figure 16 illustrates the effect
of a full swap on such a diagram of a signature link. Do note that $h_{f}$
_does not_ map the crossing disk for $C_{24}$ to that of $C^{\prime}_{24}$ but
to the longitudinal disk with punctures $K_{2}^{\prime}$, $K_{4}^{\prime}$,
and $C^{\prime}_{24}$. This happens on a more general scale and will be
justified in the following proof.

Figure 16. The type-changing homeomorphism $h_{f}$ applied to a signature link

######  Proposition 5.5.

Let $\mathcal{L}$ be a signature link and let the homeomorphism $h_{f}$
designate the full swap on all Hopf sublinks $C_{i}\cup K_{i}$. Then
$\mathcal{L}$ and $h_{f}\left(\mathcal{L}\right)$ are isotopic links. In
particular, $h_{f}\left(\mathcal{L}\right)$ is a signature link.

###### Proof.

Let $\mathcal{L}=\\{K_{f}\\}\cup\mathcal{K}\cup\mathcal{C}\cup\mathcal{C_{K}}$
be a signature link with $\mathcal{K}$ and $\mathcal{C}$ each containing $n$
components. Since each ml-swap maps a link in $\mathbb{S}^{3}$ to a link in
$\mathbb{S}^{3}$, we have that $h_{f}\left(\mathcal{L}\right)$ is a link in
$\mathbb{S}^{3}$ by construction. We will describe how to build
$h_{f}(\mathcal{L})$ by performing a full-swap on the pragmatic diagram of
$\mathcal{L}$, where $K_{f}$ corresponds with the z-axis and so that each
$C_{j}$ is contained in a vertical translate of the $xy$-plane at height
$z=j$, as depicted on the left side of Figure 16. Here, the $yz$-plane
corresponds with the projection plane for an FAL diagram of $\mathcal{L}$ with
each $C_{ij}\in\mathcal{C_{K}}$ linking only $K_{i}$ and $K_{j}$ and meeting
the $yz$-plane orthogonally.

Now, consider the sublink
$\mathcal{L}_{s}=\\{K_{f}\\}\cup\mathcal{K}\cup\mathcal{C}$ and its image
$h_{f}(\mathcal{L}_{s})$. Recall that $h_{f}$ is the composition of $n$ ml-
swaps, one performed on each Hopf sublink
$(C_{i},K_{i})\in\mathcal{C}\times\mathcal{K}$ for $i=1,\ldots,n$; see Figure
12 for the local picture of a single ml-swap. Then $h_{f}(\mathcal{L}_{s})$ is
constructed from $\mathcal{L}_{s}$ by keeping $K_{f}$ fixed as the $z$-axis,
while each Hopf sublink $h_{f}(K_{j}\cup C_{j})$ links
$K_{f}^{\prime}=h_{f}(K_{f})$ via $K_{j}^{\prime}=h_{f}(K_{j})$. In addition,
each $K_{j}^{\prime}$ is contained in the vertical translate of the $xy$-plane
at height $z=j$, as depicted on the right in Figure 16.

We will now show that the images of the crossing circles in
$\mathcal{C}_{\mathcal{K}}$ under $h_{f}$ are unlinked unknots, which will
help us determine how $h_{f}(\mathcal{C}_{\mathcal{K}})$ behaves. Every
component of $\mathcal{L}$ is unknotted, and links those components used in
the Dehn twists of an $ml$-swap at most once. In this situation, the Dehn
twists never knot an unknotted component so the image of each component in
$\mathcal{L}$ is an unknot as well. Further, an $ml$-swap does not link two
crossing circles of $\mathcal{C_{K}}$ because the linking introduced by the
first Dehn twist along $K$ in Figure 11 is undone by the following twist along
$C$. In addition, since the ml-swaps that comprise a full-swap occur in
disjoint $3$-balls, the same applies to a full-swap.

Now unknots in $\mathbb{S}^{3}$ have a canonical peripheral structure in which
a meridian links the component once and a longitude bounds an embedded disk in
its complement. An $ml$-swap preserves this $\mathbb{S}^{3}$-peripheral
structure on all components of $\mathcal{L}$ except $K$ and $C$, for which it
swaps meridians and longitudes. This observation allows us to discuss the
_topology_ of the images of thrice-punctured spheres under a full swap by
analyzing images of their punctures.

A crossing disk $D_{ij}$ has a longitudinal puncture along $C_{ij}$ and
meridional punctures along $K_{i}$ and $K_{j}$. Since meridians of
$K_{i},K_{j}$ map to $\mathbb{S}^{3}$-longitudes of
$K_{i}^{\prime},K_{j}^{\prime}$, the image $D_{ij}^{\prime}$ of $D_{ij}$ under
a full swap is a thrice-punctured sphere with longitudinal punctures along
each of $C_{ij}^{\prime}$, $K_{i}^{\prime}$ and $K_{j}^{\prime}$. Similarly,
the longitudinal disk $D_{ij}^{\ell}$ (guaranteed by Lemma 5.3) has
longitudinal punctures along $C_{ij}$, $C_{i}$ and $C_{j}$. A full swap maps
longitudes of $C_{i}$ and $C_{j}$ to meridians in the
$\mathbb{S}^{3}$-peripheral structure of $C_{i}^{\prime}$ and
$C_{j}^{\prime}$, so ${D_{ij}^{\ell}}^{\prime}$ has a
$\mathbb{S}^{3}$-longitudinal puncture along $C_{ij}^{\prime}$, and
$\mathbb{S}^{3}$-meridional punctures along $C_{i}^{\prime}$ and
$C_{j}^{\prime}$. The component $C_{ij}^{\prime}$, then, bounds an embedded
disk in $\mathbb{S}^{3}$ punctured once by each of $C_{i}^{\prime}$ and
$C_{j}^{\prime}$. This implies $C_{ij}^{\prime}$ links only $C_{i}^{\prime}$
and $C_{j}^{\prime}$.

Using these representatives for $\mathcal{L}$ and $h_{f}(\mathcal{L})$, we see
that a rotation along the $z$-axis by $180^{\circ}$ provides the necessary
isotopy between $\mathcal{L}$ and $h_{f}(\mathcal{L})$ in
$\mathbb{R}^{3}\cup\\{\infty\\}\cong\mathbb{S}^{3}$. Furthermore,
$h_{f}(\mathcal{L})$ is a signature link
$\\{K_{f}^{\prime}\\}\cup\mathcal{K}^{\prime}\cup\mathcal{C}^{\prime}\cup\mathcal{C}_{\mathcal{K}}^{\prime}$,
where $K_{f}^{\prime}=h_{f}(K_{f})$,
$\mathcal{K}^{\prime}=h_{f}(\mathcal{C})$,
$\mathcal{C}^{\prime}=h_{f}(\mathcal{K})$, and
$\mathcal{C}_{\mathcal{K}}^{\prime}=h_{f}(\mathcal{C}_{\mathcal{K}})$. ∎

##  6\. Separating Sets

In this section, we will utilize two important subsets of thrice-punctured
spheres whose removal separates an FAL complement: separating pairs and
separating quadruples. The behavior of type-changing homeomorphisms between
flat FAL complements is significantly restricted by the existence of
separating sets.

Separating pairs were introduced in Section 2 as a pair of disjoint thrice-
punctured spheres whose union separates an FAL complement. Theorem 2.8 from
[20] was also introduced in that section, which states that a pair of thrice-
punctured spheres $\\{S_{1},S_{2}\\}$ is a separating pair if and only if each
is either a crossing disk or a singly-separated disk and their longitudinal
slopes coincide.

We now introduce a particular type of separating set consisting of four
thrice-punctured spheres. Let $D$ be a longitudinal disk with longitudinal
punctures along the crossing circles $C_{1},C_{2},C_{3}$, and let $D_{i}$ be a
crossing disk with crossing circle puncture $C_{i}$. Then the set
$Q=\\{D,D_{1},D_{2},D_{3}\\}$ is a separating set of four thrice-punctured
spheres. The sets we are concerned with have one additional property.

######  Definition 6.1.

Let $D$ be a longitudinal disk in a flat FAL complement $M$ with crossing
circle punctures $C_{1},C_{2}$, and $C_{3}$ that bound crossing disks
$D_{1},D_{2}$ and $D_{3}$. Then $Q=\\{D,D_{1},D_{2},D_{3}\\}$ is a separating
quadruple if each crossing disk $D_{i}$ is punctured by distinct knot circles.

The signature links of Section 5 contain separating quadruples. To see this,
note that in a signature link every crossing circle links distinct knot
circles. Moreover, Lemma 5.3 guarantees the existence of a longitudinal disk
$D^{\ell}_{ij}$ for every $C_{ij}\in\mathcal{C}_{\mathcal{K}}$. Thus, each
triple $(C_{i},C_{j},C_{ij})$ in a signature link generates a separating
quadruple $Q_{ij}$. The remark below will show that $Q_{ij}$ is unique up to a
choice of crossing disks.

We introduce some terminology. A general separating quadruple $Q$ is
illustrated in Figure 17, which depicts only those components which puncture
$Q$. In a flat FAL knot circles cannot cross each other, so knot circles
puncture adjacent crossing disks. Let $K_{i}$ denote the knot circle
puncturing $Q$ that is opposite the crossing circle $C_{i}$ in the sense that
they are not linked. By an abuse of terminology, a component of (or in) $Q$
will refer to components that puncture disks in $Q$.

Remark: We also point out that two separating quadruples with the same
crossing circle punctures have the same knot circle punctures and longitudinal
disk. Suppose $Q$ and $Q^{\prime}$ are separating quadruples with the same
crossing circle punctures. Lemma 4.2 of [20] shows that there is at most one
longitudinal disk containing any two given crossing circle punctures, let
alone three, so $Q$ and $Q^{\prime}$ have the same longitudinal disk. Now
recall that crossing circles in a separating quadruple link distinct knot
circles, and only those components of $\mathcal{A}$. This implies that every
disk they bound is punctured by the same two knot circles. Hence two
separating quadruples sharing the same crossing circle punctures can differ
only in their crossing disks.

Figure 17. A separating quadruple

We begin with the following lemma which immediately leads to a special case,
Corollary 6.3, of our main result. This lemma will also be essential for
establishing some technical results on how separating quadruples can behave
under a type-changing homeomorphism.

######  Lemma 6.2.

Let $M=\mathbb{S}^{3}\setminus\mathcal{A}$ flat FAL complement with a unique
reflection surface, and let $C$ a crossing circle in $\mathcal{A}$ that links
the same knot circle $K$ twice. If $h:M\to M^{\prime}$ is a homeomorphism of
flat FAL complements, then $h$ preserves the types of both $C$ and $K$.

###### Proof.

Let $C$ be a crossing circle in a flat FAL complement $M$ containing a unique
reflection surface $R$. Further, let $C$ bound a crossing disk $D$ which is
punctured twice by the same knot circle $K$, and suppose $h:M\to M^{\prime}$
is a homeomorphism of flat FAL complements. The assumption that the reflection
surface $R\subset M$ is unique implies that $M^{\prime}$ has a unique
reflection surface $R^{\prime}$, and that $R^{\prime}=h(R)$.

We first prove that $K^{\prime}=h(K)$ must be a knot circle in $M^{\prime}$.
Suppose, on the contrary, that $K^{\prime}$ is a crossing circle in
$M^{\prime}$. Then meridians of $K$ map to longitudes of $K^{\prime}$ because
both are perpendicular to reflection surfaces and $R^{\prime}=h(R)$. The fact
that $R^{\prime}=h(R)$ further implies that $D^{\prime}=h(D)$ is a non-
reflection thrice-punctured sphere in $M^{\prime}$, since $D$ is in $M$. The
two meridional $K$-punctures of $D$ map to two longitudinal
$K^{\prime}$-punctures of $D^{\prime}$. However, by Theorem 2.7, non-
reflection thrice-punctured spheres in a flat FAL complement do not have two
longitudinal punctures along the same crossing circle. So, $K^{\prime}$ must
be a knot circle in $M^{\prime}$.

Since $K^{\prime}$ is a knot circle puncturing the disk $D^{\prime}$ twice in
$M^{\prime}$, the remaining puncture of $D^{\prime}$ must be an $M^{\prime}$
crossing circle by the characterization of Theorem 2.7. The remaining puncture
of $D^{\prime}$ is the image $C^{\prime}=h(C)$ of $C$, and $h$ preserves the
types of both $C$ and $K$. ∎

If a flat FAL (whose complement has a unique reflection surface) has a single
knot circle, as is the case for FALs of two-bridge links with an even number
of twist regions (other than the Borromean rings, which has three reflection
surfaces), then every crossing circle links the same knot circle twice. Thus
any homeomorphism preserves the type of all components and their peripheral
structures and can be realized by an isotopy of $\mathbb{S}^{3}$. This
observation leads to the following immediate corollary of Lemma 6.2:

######  Corollary 6.3.

A flat FAL with a single knot circle is determined by its complement among all
flat FALs. In particular, flat FALs corresponding to $2$-bridge links with an
even number of twists are determined by their complements.

###### Proof.

The only flat FAL with one knot circle and multiple reflection surfaces is the
Borromean rings, which we’ve already seen to be determined by its complement.
The unique reflection surface case follows from Lemma 6.2. ∎

######  Remark.

We would like to emphasize the necessity of the unique reflection surface
hypothesis in Lemma 6.2. For instance, as noted in Theorem 3.4, the Borromean
rings complement admits three distinct reflection surfaces and the flat FAL
diagram for this link has two crossing circles and a single knot circle that
links each crossing circle twice. However, there exists homeomorphisms of the
Borromean rings complement where both a crossing circle and the knot circle
switch types.

We now prove a technical lemma considering homeomorphisms that change a
crossing disk to a longitudinal disk, or vice-versa. It turns out that such a
homeomorphism $h$ implies the existence of a separating quadruple $Q$, and the
action of $h$ on $Q$ can be made quite precise.

######  Lemma 6.4.

Let $M$ and $M^{\prime}$ be homeomorphic flat FAL complements with unique
reflection surfaces, and $h:M\to M^{\prime}$ a homeomorphism that changes the
type of an $N$-disk $D\subset M$. Then

  1. i.

The disk $D$ is part of a separating quadruple $Q$ in $M$ whose image
$Q^{\prime}=h(Q)$ in $M^{\prime}$ is also a separating quadruple,

  2. ii.

The homeomorphism $h$ fixes the types of exactly one opposite knot- and
crossing-circle pair $K_{f},C_{f}$ in $Q$.

  3. iii.

The longitudinal disk and exactly one crossing disk in $Q$ change type under
$h$. These disks share the crossing circle puncture $C_{f}$.

###### Proof.

We are given that $M$ and $M^{\prime}$ are homeomorphic flat FAL complements
with unique reflection surfaces, and that $h:M\to M^{\prime}$ is a
homeomorphism changing the type of an $N$-disk $D$. Since $M$ and $M^{\prime}$
each have unique reflection surfaces $R$ and $R^{\prime}$, we have
$R^{\prime}=h(R)$. Then the image of a non-reflection thrice-punctured sphere
in $M$ is non-reflection in $M^{\prime}$. Further, $N$-disks and $S$-disks are
distinguished by the topological property of separating, which implies that
$h$ maps $N$-disks to $N$-disks.

Consider first the case where $D$ is a longitudinal disk whose image
$D^{\prime}=h(D)$ is a crossing disk. Let $C_{1},C_{2},C_{3}$ denote the
crossing circle punctures of $D$, and let $D_{i}$ be a choice of crossing disk
bounded by $C_{i}$. To show $D$ is part of a separating quadruple we must show
that the $D_{i}$ are punctured by distinct knot circle components.

Two of the crossing circle punctures of $D$, say $C_{2},C_{3}$, change type
because $D^{\prime}$ is a crossing disk. Thus Lemma 6.2 implies $D_{2},D_{3}$
are punctured by distinct knot circle punctures. Now consider $D_{1}$, whose
image $D_{1}^{\prime}$ must be a crossing disk or longitudinal disk since
$N$-disks map to $N$-disks. If $D_{1}^{\prime}$ stays a crossing disk, then
Theorem 2.8 implies that $\\{D_{1}^{\prime},D^{\prime}\\}$ is a separating
pair since both are crossing disks sharing the crossing circle puncture
$C_{1}^{\prime}$. This cannot happen because the pair $\\{D_{1},D\\}$ does not
separate in $M$ (again by Theorem 2.8). Therefore, $D_{1}$ changes type and
$D_{1}^{\prime}$ must be a longitudinal disk. Then, by Lemma 6.2, $D_{1}$ has
distinct knot circle punctures and $Q=\\{D,D_{1},D_{2},D_{3}\\}$ is a
separating quadruple in $M$.

For convenience, label the knot circle punctures $K_{1},K_{2},K_{3}$, where
$K_{i}$ is the knot circle opposite $C_{i}$ in that it does not puncture
$D_{i}$.

Figure 18. Homeomorphic image of $Q$ when longitudinal disk changes type

To see that
$Q^{\prime}=\\{D^{\prime},D_{1}^{\prime},D_{2}^{\prime},D_{3}^{\prime}\\}$ is
also a separating quadruple, we must show that it consists of three crossing
disks, with distinct knot circle punctures, and one longitudinal disk. The
disk $D^{\prime}$ is assumed to be a crossing disk and, in this case, the disk
$D_{1}^{\prime}$ was shown to be longitudinal. Moreover, since
$C_{2}^{\prime},C_{3}^{\prime}$ are knot circles, the disks
$D_{2}^{\prime},D_{3}^{\prime}$ must be crossing disks in $M^{\prime}$ because
these are the only non-reflection disks with knot circle punctures (Theorem
2.7). Further, each of the disks in $Q$ are punctured by three distinct
components, so their images are as well, and each crossing disk in
$Q^{\prime}$ is punctured by distinct knot circle components. Thus
$Q^{\prime}$ is a separating quadruple in $M^{\prime}$.

Note that this analysis proves the third conclusion as well, since $D,D_{1}$
change type while the crossing disks $D_{2},D_{3}$ do not.

To see statement $(ii)$ note that
$C_{1}^{\prime},K_{2}^{\prime},K_{3}^{\prime}$ are the crossing circle
punctures of $Q^{\prime}$ since they are the punctures of the longitudinal
disk $D_{1}^{\prime}$. The remaining punctures of $Q^{\prime}$ must be knot
circles in $M^{\prime}$, so $K_{1}^{\prime}$ is a knot circle. Thus $h$
preserves the type of the opposite knot- and crossing-circle pair
$C_{1},K_{1}$, while changing the type of all other components. Observe that
$C_{1}$ is the crossing circle puncture shared by the disks that change type,
namely $D$ and $D_{1}$.

Now suppose $D$ is a crossing disk in $M$ that changes type to a longitudinal
disk $D^{\prime}$ in $M^{\prime}$. Apply the previous argument to $h^{-1}$ and
$D^{\prime}$, then note that if $h^{-1}$ and $D^{\prime}$ satisfy the
conclusions of the lemma, then so does $h$ and $D$. ∎

Lemma 6.4 can be applied to the full-swap homeomorphisms discussed in Section
5, revealing some of the geometric structure inherent in such homeomorphisms.
Before proceeding we remark that any homeomorphism between flat FAL
complements with unique reflection surfaces that does not preserve peripheral
structures must change the type of a knot circle. Indeed, if a crossing circle
changes type, then one of the knot circles it links changes type as well since
there are no (non-reflection) thrice-punctured spheres with three knot circle
punctures. Thus if $h$ changes the type of a component, there must be a knot
circle that changes type.

######  Lemma 6.5.

Let $h:M\to M^{\prime}$ be a homeomorphism between flat FAL complements with
unique reflection surfaces, and suppose $h$ changes the type of the knot
circle $K$ in $M$. Then $h$ changes the type of exactly one crossing circle
$C_{1}$ linked by $K$ and, of all crossing disks punctured by $K$, $h$ fixes
the type of exactly those bounded by $C_{1}$.

###### Proof.

First we show $h$ changes the type of at most one crossing circle linking $K$.
Suppose, on the contrary, that $C_{1},C_{2}$ are distinct crossing circles
linking $K$ whose images $C_{1}^{\prime},C_{2}^{\prime}$ are both knot circles
in $M^{\prime}$. Let $D_{1},D_{2}$ be a choice of crossing disks they bound
and note that, since $C_{1},C_{2}$ are distinct, the disks $D_{1}$ and $D_{2}$
do not form a separating pair by Theorem 2.8. To determine if the image of an
$N$-disk is a crossing disk, it is enough to show that it has a knot circle
puncture because homeomorphisms between flat FAL complements with unique
reflection surfaces map $N$-disks to $N$-disks. Since
$D_{i}^{\prime}=h(D_{i})$ is punctured by the knot circle $C_{i}^{\prime}$,
the disk $D_{i}^{\prime}$ must be a crossing disk in $M^{\prime}$. The disks
$D_{1}^{\prime},D_{2}^{\prime}$ also share the crossing circle puncture
$K^{\prime}$ and so form a separating pair in $M^{\prime}$, again by Theorem
2.8. The homeomorphic image of a non-separating set, however, cannot be
separating, and $h$ changes the type of at most one crossing circle linking
$K$.

Now we argue that the image of at least one crossing circle linking $K$ is a
knot circle in $M^{\prime}$. Since $K$ is linked by at least two crossing
circles, at most one of which can change type, there is a crossing circle $C$
linking $K$ whose image $C^{\prime}$ is a crossing circle in $M^{\prime}$. Let
$D$ be a crossing disk bounded by $C$ and $J$ be the knot circle other than
$K$ which punctures $D$. By Lemma 6.2, $J\neq K$. Since $D^{\prime}=h(D)$ has
two crossing circle punctures in $K^{\prime}$ and $C^{\prime}$, we find
$J^{\prime}$ must also be a crossing circle. Hence $D^{\prime}$ is a
longitudinal disk in $M^{\prime}$ (see Figure 19).

Figure 19. Crossing disk $D$ maps to longitudinal disk $D^{\prime}$

Thus $D$ is a crossing disk in $M$ that changes type, and Lemma 6.4 shows that
$D$ is part of a separating quadruple $Q=\\{D,D_{1},D_{2},D_{3}\\}$. Assume
we’ve labeled disks so that $D_{3}$ is the longitudinal disk, and $J$
punctures $D_{2}$ while $K$ punctures $D_{1}$. Finally, let $C_{1}$ be
crossing circle puncture of $D_{1}$ and $K_{f}$ the final knot circle in $Q$.
Figure 20 depicts a “schematic” diagram of this labeling in the sense that
components of the FAL which are not in $Q$ are not pictured.

Now $h$ changes the crossing disk $D$ to a longitudinal disk, so Lemma
6.4$(iii)$ implies the images of $D_{1},D_{2}$ are again crossing disks. Since
$D_{1}^{\prime}$ is a crossing disk with crossing circle puncture
$K^{\prime}$, the crossing circle $C_{1}$ changes type under $h$. Thus at
least one crossing circle linking $K$ maps to a knot circle. Combining that
with the first part of the proof shows that $h$ changes the type of exactly
one crossing circle linking $K$.

Figure 20. Schematic of $Q$

To see the last statement, let $C_{1}$ be the unique crossing circle linking
$K$ that changes type and suppose $D_{1}$ is any crossing disk punctured by
$K$ which is bounded by $C_{1}$. Since $C_{1}$ changes type, the image