DIAGRAMMATIC AND GEOMETRIC INVARIANTS OF HYPERBOLIC WEAKLY

GENERALIZED ALTERNATING KNOTS

By

Brandon Bavier

A DISSERTATION

Michigan State University

in partial fulfillment of the requirements

Submitted to

for the degree of

Mathematics – Doctor of Philosophy

2021

ABSTRACT

DIAGRAMMATIC AND GEOMETRIC INVARIANTS OF HYPERBOLIC WEAKLY

GENERALIZED ALTERNATING KNOTS

By

Brandon Bavier

We study the relationship between knot and link diagrams on surfaces and their invariants
coming from hyperbolic geometry. A link diagram on a surface, denoted 
(
) ⊂ 
, is a way
of projecting a link 
 in some manifold 
 onto some surface 
 ⊂ 
. This generalizes the
notion of a link diagram, and has been studied with a variety of conditions [1,5,12,21,23,31].
We will work with the weakly generalized alternating knots and links of Howie and Purcell [23],
which gives certain criteria to ensure the diagram is interesting. A cusp of a hyperbolic knot 

is a neighborhood of 
 in 
 \ 
, and the cusp volume is the Euclidean volume of a maximal
cusp. We show that the cusp volume of a weakly generalized alternating knot, with some
additional conditions, is bounded both above and below based on the twist number of 
(
)
and 
(
). This is done by constructing a new essential surface for 
 that has nice properties,
including having Euler characteristic based on the twist number as opposed to the crossing
number. Our bound on cusp volume leads to interesting bounds on other geometric properties
of 
, including slope length and volumes of Dehn surgery. The volume of a hyperbolic link 

in a manifold 
 is the hyperbolic volume of the complement 
 \ 
. We can show that volume
is also bounded below by the twist number of 
(
) and 
(
). We do this by generalizing the
Jones polynomial to weakly generalized alternating links, and showing that there is a relation
between this polynomial and the twist number, and this polynomial and the volume. Through
the course of proving this bound, we also get relations between the guts of the checkerboard
surfaces of 
 and this generalized Jones polynomial. In addition, if we are working inside a
thickened surface, the twist number becomes an isotopy invariant of 
.

Copyright by
BRANDON BAVIER
2021

ACKNOWLEDGEMENTS

I would like to thank my partner, Brad, for all the support and encouragement he has given me
these past two years. He’s kept me sane, happy, and motivated to keep moving forward.

I would also like to thank my advisor, Effie Kalfagianni, for her work and helping me in

my research. Her advice and conversations on research and academia has been invaluable.

I would also like to thank my friends, Alex, Chris, Corleone, and Mark, for giving me

something to look forward to every week for the past nine years.

The research done here was partially supported by the Dr. Williams L. Harkness Endowed

Fellowship, and NSF Grants DMS-1404754, DMS-1708249, and DMS-2004155

iv

TABLE OF CONTENTS

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.

.
INTRODUCTION .

.

.

CHAPTER 1

Introduction .

LIST OF FIGURES .

2.1 Hyperbolic Geometry .

CHAPTER 2 BACKGROUND .
.

. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . .
1.1 Essential Surfaces of Weakly Generalized Alternating Knots
1.2 Cusp Volumes of Weakly Generalized Alternating Knots
. . . . . . . . . . .
1.3 Twist Number, Jones-like polynomials, and the Volume . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.1 Hyperbolic Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.2 Guts of a Manifold . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.3 Cusp Volume and Cusp Area . . . . . . . . . . . . . . . . . . . . . .
2.2 Weakly Generalized Alternating Knots . . . . . . . . . . . . . . . . . . . . .
2.3 Kauffman States and the Jones Polynomial . . . . . . . . . . . . . . . . . . .
CHAPTER 3 ESSENTIAL SURFACES . . . . . . . . . . . . . . . . . . . . . . . .
3.1
. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Augmented Diagrams and Twisted Surfaces . . . . . . . . . . . . . . . . . .
3.3 Weakly Generalized Alternating Knots and Augmentation . . . . . . . . . . .
3.4 Colored Graphs and Twisted Surfaces
. . . . . . . . . . . . . . . . . . . . .
3.4.1 Blue and red graphs . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.2 Bigons in Ɖ


 .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.3 Blue-Blue-Red Triangles . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Triangles and Squares .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Completing the Proof of the Theorems . . . . . . . . . . . . . . . . . . . . .
Properties of Twisted Surfaces . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
CHAPTER 4 CUSP AREA .
.
. . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
.
4.2 Proofs for Calculations
. . . . . . . . . . . . . . . . . . . . . . . . . . .
.
4.3 Calculation of Cusp Area Bounds . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Additional Results .
JONES POLYNOMIAL . . . . . . . . . . . . . . . . . . . . . . . . .
.
. . . . . . . . . . . . . . . . . . . . . . . . . . .
.
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Reduced Graphs .
5.3 Coefficients and Graphs .
.
. . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Twist Number, Volume, and Coefficients . . . . . . . . . . . . . . . . . . . .

5.1
Introduction .
5.2 Background .

Introduction .

CHAPTER 5

3.6.1

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BIBLIOGRAPHY .

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v

vi

1
3
4
6

9
9
9
10
11
14
18

21
21
22
24
30
30
34
40
49
57
59

72
72
74
78
81

84
84
86
88
89
94

97

LIST OF FIGURES

Figure 1.1.

(Left): A diagram of the figure eight knot on 
2 (Right): A diagram of
the unknot on the torus.

. . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 2.1.

(Left): A manifold 
 with a single cusp. The green colored area is a
neighborhood of the cusp. (Right) A preimage of 
, with three different
horoball representatives for the cusp. Note that the upper horoball extends
to infinity.

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. .

. .

.

.

Figure 2.2.

If a disk 
 has boundary intersecting our knot at exactly two crossings,
then it either contains a twist region (left) or there is another disk 
(cid:48)
intersecting at the same crossings containing a twist region (right).

. . . .

Figure 2.3. The 
 resolution (left) and the 
 resolution (right) of a crossing (center). .

Figure 2.4.

Figure 3.1.

Figure 3.2.

(Left): The figure eight knot. (Center): A Kauffman state for our knot.
We’ve kept track of the type of resolutions by coloring the edges red for

 resolutions and blue for 
 resolutions. (Right): The reduced graph for
the state.

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

.

.

.

.

.

.

(Left): A knot diagram 
. The center and right twist regions each have
at least 


 crossings with the center having an even number and the left
an odd. (Center Left:) We adjoin circles to highly twisted regions to get

. (Center Right:) We remove crossing in pairs from encircled regions
to get 
2. (Right): We remove the crossing circles to get 
2.

. . . . . . .

(Left) 
2 with the torus around the twist region. Blue circles mark where
the torus punctures the blue surface. (Center Left) Twisting around the
torus gives us back 
. In this case, we are only adding in one full twist
(Center Right) To complete the twisted surface, attach
for simplicity.
opposite sides of the torus with a band, as represented on top and bottom.
Here this amounts to adding an Möbius band. (Right) When we have
more twists, things can become more complicated. In addition, we might
be adding two annuli instead.

. . . . . . . . . . . . . . . . . . . . . . . .

Figure 3.3. Two possible configurations to have 
(
(
(cid:48)), 
) ≤ 4. On the left, the
blue disk can be used to homotope the knot away from the compression
disk, reducing the number of intersections. On the right, homotopying
along the blue disk will cause the compression disk to not intersect a
twist region.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. .

. .

vi

2

12

17

19

20

23

24

28

Figure 3.4.

(Left) A possible configuration for Ɖ


.
(Right) How Ɖ


 would
look when imposed on the diagram. The labeled blue edges in the graph
map to the corresponding labeled arcs in the diagram.

. . . . . . . . . . .

Figure 3.5.

(Top) The four different types of adjacent triangles coming from a bigon.
(Bottom) How the bigons might look in the diagram. Note that, in the
red-green bigon case, the two red edges in the diagram are, in fact, the
same edge, and must “wrap around” the diagram to create the bigon.

. . .

Figure 3.6.

(Left) A blue-blue-red triangle as viewed on the graph Ɖ

. (Right) The
same triangle as viewed on the knot diagram, 
(

,2). . . . . . . . . . . .

Figure 3.7. The three possible ways two blue-blue-red triangles can have their images

glued together. .

.

.

.

.

. . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 3.8.

(Left) A blue rectangle with a red diagonal in the graph Ɖ

. (Right) A
blue rectangle as viewed on the diagram, where both vertices represent
the same crossing circle. . . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 3.9.

(Left) Two pairs of three adjacent blue-blue-red triangles. (Center and
Right) The two possibilities if the vertices map to the same crossing
circle, one with two crossings and one with three.

. . . . . . . . . . . . .

Figure 3.10. (Left) Focusing on the crossing circle of 
, 
, and 
, our diagram must
look something like this. (Center) Once we know that 
 is related to the
crossing circle, we can draw in some extra information about our knot.
(Right) The dotted lines are disk-bounding curves that will allow us to
show 
 and 
 are also related to the crossing circle. . . . . . . . . . . . . .

Figure 3.11. Three adjacent triangle-square pairs stacked on top of each other, with

labelings for the edges.

. . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 3.12. (Left) A blue-red rectangle disjoint from all other edges. The purple curve
represents our 
. (Center) A blue-red rectangle that might intersect a red
edge, but is disjoint from blue edges. (Right) A blue-red rectangle that
might intersect any number of blue or red edges.

. . . . . . . . . . . . . .

32

38

41

43

44

45

47

49

51

Figure 3.13. (Upper Left) The graph of the case we are considering.

(Lower Left
and Right) The possible diagram configurations of this case. The purple
curve following the crossing disk is 
, while the pink curve following 
 is 
. 54

Figure 3.14. Two ways a triangle with one red on red, one edge on blue, and one edge

on the knot can sit in the diagram.

. . . . . . . . . . . . . . . . . . . . .

Figure 3.15. The three ways a red-blue rectangle can sit in our diagram . . . . . . . . .

vii

63

64

Figure 3.16. (Left) An essential product disk with blue sides 
1 and 
2 intersected
horizontally by red edges. (Right) Two sub-rectangles whose shared red
edge meets two separate crossings. (Center) Two sub-rectangles whose
share red edge meets a single crossing. We want to prove that this is what
happens for all adjacent sub-rectangles. . . . . . . . . . . . . . . . . . . .

Figure 4.1. The subsurface associated to the twist region. When it comes from
checkerboard surfaces, the subsurface can be colored red and blue, with
one component (blue in the picture) being two disks attached with Möbius
bands at each crossing, and the other being a twisted disk.

. . . . . . . . .

Figure 5.1. A Kauffman State (right) for an alternating knot on the torus (left). The
color of the edges represent the type of resolution at each crossing. Note
that we now have states that don’t bound disks on 
2.

. . . . . . . . . . .

Figure 5.2.

(Left) The state 
1 gotten from 

 by merging two noncontractible circles
into a single contractible circle; (Center) The all 
 state, 

; (Right) The
modified state, 
(cid:48)

1, with exactly one 
 resolution.

. . . . . . . . . . . . .

69

76

87

92

viii

CHAPTER 1

INTRODUCTION

Traditonally, knot theory is the study of embeddings of 
1 into 
3. A knot, then, is the image
of this embedding, and a link a union of disjoint knots. Two knots (or links) 
1 and 
2 are
said to be equivalent if there is a homeomorphism 
 : 
3 → 
3 such that 
 (
1) = 
2.

In addition to looking at links in 
3, we can also look at embeddings of 
1 into any manifold

. We call these knots in 
, and disjoint unions of them links in 
. When it is clear we are
working in a general manifold, we will also just refer to these as knots and links.

When studying links, it is beneficial, and often easier, to work with a projection of the link
onto 
2, while keeping track of crossing information. Such a projection is called a link diagram.
We can generalize these in two different ways, both of which can occur simultaneously. First,
we can look at projections of links in 
 onto 
2. Second, we can look at projections of 
 onto
a surface 
 that are not 
2, such as the torus. In this second case, we call this projection a link
diagram on 
.

On 
2, we can use the Reidemeister moves, three local moves on a link diagram, to try
to get one from one diagram to another. However, it can take a prohibitively large number of
such moves to show two diagrams are equivalent. As such, it can be difficult, in general, to
determine if they represent the same or different links. This is not made easier when we start
dealing with diagrams on different surfaces. As such, we look to invariants to help distinguish
when two diagrams represent two different links.

An invariant of a link is a way of associating some object to a link such that, if two links are
equivalent, they are assigned the same object. A simple, but useful, invariant is the crossing
number cr(
) of a knot 
 that counts the minimum number of crossings in a diagram of 
.
Other invariants include being alternating, which tells us if a knot has an alternating diagram,
and the Jones polynomial, which associates a polynomial to a link.

Some link invariants, including the ones mentioned above, can be computed by looking

1

Figure 1.1. (Left): A diagram of the figure eight knot on 
2 (Right): A diagram of the unknot on the torus.

directly at a diagram of the link. These are called diagrammatic invariants. Another class of
invairants comes from certain geometric properties of the link.

Given a link 
 ⊂ 
, we can look at the link complement, 
 \ 
. Many complements admit
a complete finite-volume hyperbolic structure. That is, 
 \ 
 admits a Riemannian metric of
constant negative curvature. When a knot or link has such a complement in 
3, it is called a
hyperbolic knot or link. Thurston showed that a knot is either a satellite knot, a torus knot, or
a hyperbolic knot [33], so there are plenty of hyperbolic knots to study. Also, due to work of
Mostow [30] and Prasad [32], these hyperbolic structures for link complements are unique up
to isometry, and so invariants of the hyperbolic structure are also invariants of the link.

Two important hyperbolic invariants that we are interested in here are the volume of a link
complement 
 \ 
, called vol(
), and the cusp volume of a knot complement 
 \ 
, called
CV(
).

As diagrammtic invariants are computed based on just a diagram for a link, while hyper-
bolic invariants involve studying the geometric properties of the complement, diagrammatic
invariants tend to be simpler to study. However, a natural question arises: is there a relation
between diagrammatic invariants and hyperbolic invariants? For alternating knots and links,
the answer is yes. Dasbach and Lin [15], and Lackenby with Agol and Dylan Thurston [25]
showed that the volume of hyperbolic alternating links can be bounded both above and below
by the twist number, a diagrammatic invariant, and Lackenby and Purcell [26] showed that

2

such alternating knots have cusp volume bounded above and below by the twist number.

In this work, we will explore this relationship between these two hyperbolic invariants and
diagrammatic invariants for a class of knots and links, called weakly generalized alternating
links, that are either projected onto a surface that is not 
2 or are embedded in a manifold that
is not 
3. In Chapter 2, we will introduce the necessary background material, including the
exact definitions of the invariants we are working with, as well as the exact class of knots and
links we will be examining. Then, in Chapter 3, we will construct an essential surface for this
class of links. In Chapter 4, we will use this essential surface to show that cusp area for this
class of links can be bounded based on a diagrammatic invariant. Finally, in Chapter 5, we
will show that the volume of the link complement of this class of knots is also bounded by
the same diagrammatic invariant. Sections 1.1, 1.2, and 1.3 will outline the major results of
Chapters 3, 4, and 5, respectively.

1.1 Essential Surfaces of Weakly Generalized Alternating Knots

We begin our study of weakly generalized alternating knots by showing that they admit
certain essential surfaces. By work of Howie and Rubinstein in 
3 [24] and Howie and Purcell
in other manifolds [23], we know that these knots will admit at least two essential surfaces,
in particular, the checkerboard surfaces. Our goal, then, is to construct two essential surfaces
that we have more control over.

To construct the surfaces, we first augment our knot 
 by removing pairs of crossings from
twist regions to get a new link 
(cid:48). We can show that this augmented knot also remains weakly
generalized alternating:
Proposition 3.3.1. If 
 is a weakly generalized alternating knot with diagram 
(
) on a
projection surface 
 in a 3-manifold 
, with 
(
(
), 
) ≥ 4 and 
(
(
), 
) > 4, then 
(cid:48)
is a weakly generalized alternating knot on the surface 
 in 
 with 
(
(
(cid:48)), 
) ≥ 4 and

(
(
(cid:48)), 
) > 4.

This means, then, that the checkerboard surfaces for our augmented knot must be essential.

3

We can pull these back to our original knot, 
, and get what are called twisted surfaces, 

,2
and 

,2. Our goal, then, is to prove these are essential.
Theorem 3.1.1. Let 
 : 

,2 → 
 \ 
 be the immersion of our blue twisted surface 

,2 into
our knot complement 
 \ 
, where 
 has no boundary components. Then this immersion is

1-injective and boundary 
1-injective provided that 


 ≥ 121.

We prove this by examining how a disk with boundary essential on one of our surfaces
must intersect the the twisted surfaces. By visualizing these intersections as colored graphs,
we show that these graphs must have simultaneously at least five adjacent bigons and less than
five adjacent bigons.

Finally, we finish the chapter by modifying our proof of Theorem 3.1.1 to show that these

twisted surfaces have the additional nice property:
Theorem 4.2.4. Let 
 be the disjoint union of our twisted surfaces 

,2 (cid:116) 

,2. Suppose that
two distinct essential arcs in the surface 

,2 have homotopic images in 
 \ 
, but not 

,2.
Then the two arcs are homotopic in 

,2 into the same subsurface associated with some twist
region of 
2.

Here, a subsurface associated with some twist region is a neighborhood of a twist region.
This will allow us to limit how many essential arcs in either twisted surface are not homotopic
in 
 \ 
 in the following chapter, based on the twist number of 
. This contrasts to the case of
checkerboard surfaces, where such essential arcs are bounded based on the crossing number,
which grows much more quickly than the twist number.

1.2 Cusp Volumes of Weakly Generalized Alternating Knots

Given a finite volume hyperbolic manifold 
, the ends for 
 will look like a half-open
neighborhood of a torus, 
2 × [1,∞). We call this end a cusp. If 
 : H3 → 
 is the covering
map, then we can view the preimage of this cusp as a collection of horoballs. A cusp is called
maximal if this preimage is tangent to itself.

4

Given a maximal cusp, we say the cusp volume is the Euclidean volume of the cusp, and the
cusp area the euclidean area of the boundary of the cusp. We can use these, then, to examine
the lengths of slopes of a knot, as well as study how Dehn surgery affects our manifold.

The cusp volume has been studied in a couple of settings for knots, although not as
extensively as other geometric invariants such as the volume. Futer, Kalfagianni, and Purcell
gave two-sided diagrammatic bounds for the cusp volume of closed 3-braids and 2-bridge
knots [19]. Lackenby and Purcell were also able to find two sided bounds for the cusp volume
of traditional alternating knots in 
3 based on the twist number.

Using the essential surfaces we construct in Chapter 3, we generalize the work of Lackenby

and Purcell to get the following result:

Theorem 4.1.1. Let 
 be a closed, irreducible 3-manifold and 
 ⊂ 
 a closed surface with

(
) ≤ 0 and such that 
 \ 
(
) is atoroidal. Suppose that 
 is a weakly generalized
alternating knot with a projection 
(
) ⊂ 
 that is weakly twist reduced, and we have

(
(
), 
) > 2, and 
(
(
), 
) > 4. Then 
 \ 
 is hyperbolic, and we have

5.8265514 × 10−7
2 (120 − 
(
))6 (tw
(
) − 
(
)) ≤ 

(
) ≤ (360tw
(
) − 3
(
))2
where 

(
) is the cusp volume of 
 and tw
(
) the twist number of 
(
).

tw
(
)

,

By some calculus, the cusp area is exactly twice the cusp volume, and so this bound also
gives us a bound on the cusp area. Then, by using a result of Burton and Kalfagianni [11], we
get the following bounds for the slopes of our knots:

Corollary 4.1.3. Let 
 be a closed, irreducible 3-manifold and 
 ⊂ 
 a closed surface with

(
) < 0 and such that 
 \ 
(
) is atoroidal. Suppose that 
 is a weakly generalized
alternating knot with a projection 
(
) ⊂ 
 that is weakly twist reduced, and we have

(
(
), 
) > 2, and 
(
(
), 
) > 4. Let 
 be the meridian and 
 any non-meridian slope

5

on maximal cusp of 
 \ 
. Then, we have

(cid:18) 5.8265514 × 10−7
(cid:19)
ࡁ(
) ≤ 720 (tw
(
) − 
(
))
ࡁ(
) ≥

tw
(
)

720 (120 − 
(
))6

tw
(
),

where tw
(
) is the twist number of 
(
).

Then, by combining with theorems from Howie and Purcell [23], Futer, Kalfagianni, and
Purcell [18], and W. Thurston [33], we can get the following two sided bounds for the volume
of manifolds obtained by Dehn surgery of highly twisted weakly generalized alternating knots:

Theorem 4.1.4. Let 
 be a weakly generalized alternating knot on a surface 
 ࣔ 
2 in a
closed manifold 
 with conditions as in Theorem 4.1.1. Suppose that

(cid:16)

2.32928 × 1010(cid:17) (120 − 
(
))6 .

tw
(
) >

Then, any manifold 
 obtained from non-meridional surgery along 
 is hyperbolic and


8
4

(tw
(
) − 
(
)) ≤ 1
2

vol(
 \ 
) ≤ vol(
) < vol(
 \ 
).

Here 
8 ≈ 3.66386 is the volume of a regular hyperbolic ideal octahedron.

1.3 Twist Number, Jones-like polynomials, and the Volume

Given a link 
 in 
3, it is possible to manually construct the link complement from a
diagram. However, it is impractical to do this for every hyperbolic link. Instead, we often look
to generalize these constructions and find bounds for classes of links based on their diagrams.

Adams [3] was able to show that, if a hyperbolic knot has at least 
 ≥ 5 crossings, then

vol(
3 \ 
) ≤ (
 − 5)
oct + 4
tet,

where 
tet = 1.0149 . . . is the volume of a regular ideal tetrahedron and 
oct = 3.6638 . . . is
the volume of a regular ideal octahedron. This was done by decomposing the complement into
ideal tetrahedrons and octahedrons, and counting.

6

While the crossing number will give us an upper bound, it also tends to get large very
quickly. As such, we also look towards other diagrammatic invariants to give us bounds on
the volume. One of the more common ones is the twist region. By work of Lackenby, with an
improvement on the upper bound by Agol and D. Thurston [25], we get the following bounds
for alternating links:


oct
2

(tw(
) − 2)vol(
3 \ 
) ≤ 10
tet(tw(
) − 1),

with the upper bound applying to non-alternating links, as well. The upper bound is found
by looking at a polyhedral decomposition of an augmentation of 
, while the lower bound is
found by using a tool called the guts of surfaces.

There is also a way of relating the coefficients of the Jones polynomial to the volume.
Described more completely in Chapter 2, the Jones polynomial is an invariant of knots that
can be found based on the diagram of the knot. Dasbach and Lin [15] showed that, in 
3, the
first two and last two coefficients of the Jones polynomial determine the twist number. Then
we get the following bounds for the volume of alternating links


oct max(|

−1|, |

+1|) ≤ vol(
3 \ 
) ≤ 10
tet (|

−1| + |

−1| − 1)

where 

−1 is the second coefficient and 

−1 is the second to last coefficent of the Jones
polynomial. The lower bound here is also found by examining the guts of surfaces.

We are able to generalize this lower bound of Dasbach and Lin for weakly generalized
alternating links. We do this by first generalizing the Jones polynomial to these settings, and
relating the twist number to it’s coefficients:

Theorem 5.1.2. Let 
 and 
 be as in Theorem 5.1.1 Suppose that a link 
 admits a weakly
generalized alternating projection 
(
) ⊂ 
 that is twist-reduced and with all regions of

 \ 
(
) disks. Then we have the following.

1. 
(guts(

)) = −|

−1| + 1
2. 
(guts(

)) = −|

+1| + 1

2 
(
)
2 
(
)

7

3. tw
(
(
)) = |

−1| + |

+1| + 
(
).

Further, if 
(
) is a cellularly embedded, twist-reduced, reduced alternating diagram, we still
get (3).

From there, we can use the relationship to the guts to get a lower bound for our volume:

Theorem 5.1.1. Let 
 be an irreducible, compact 3-manifold with empty or incompressible
boundary. Let 
 ⊂ 
 be an incompressible, closed, orientable surface such that 
 \ 
(
) is
atoroidal and 
 − 








. Suppose that a link 
 admits a weakly generalized alternating
projection 
(
) ⊂ 
 that is reduced, twist-reduced and with all regions of 
 \ 
(
) disks.
Finally, suppose that 
(
(
), 
) > 4. Then 
 is hyperbolic and
vol(
 \ 
) ≥ 
8max{|

−1|, |

+1|} − 1
2


(
 
),

where 
8 = 3.66386 . . . is the volume of an ideal octahedron.

When we are working inside a thickened surface, we get some interesting consequences. In
general, the twist number of a weakly generalized alternating link is not an isotopy invariant.
That is, two different diagrams for such a link might have different twist numbers. However,
as discussed by Boden, Karimi, and Sikora [9], the Jones-like polynomial we define is an
invariant for oriented links in a thickened surface. As such, we get the following consequence:

Corollary 5.1.3. Let 
 be a link in 
 × [−1, 1] that admits a weakly generalized alternating
projection 
(
) ⊂ 
 that is reduced, twist-reduced and with all regions of 
 \ 
(
) disks.
Then any two projections of 
have the same twist number. That is, tw
(
(
)) is an isotopy
invariant of 
.

8

CHAPTER 2

BACKGROUND

In this chapter, we will review the relevant background information that will be used through
this work. First, in Section 2.1, we recall what a hyperbolic manifold is, as well as some
important definitions and results. This will include the guts and cusps of a manifold. Next,
in Section 2.2, we go over the concept of generalizing alternating knots; in particular, we give
the definition of a weakly generalized alternating knot, and some of it’s properties. Finally,
in Section 2.3, we review diagrammatic properties of knot diagrams, as well as how some of
these properties can be generalized to generalized knot diagrams.

2.1 Hyperbolic Geometry

In this section, we will quickly recall the definition of a hyperbolic manifold. Then, we
will recall some results that make hyperbolic manifolds appealing to work with. Finally, we
will introduce certain properties of hyperbolic manifolds, in particular the volume and the cusp
area, that we will be working with.

2.1.1 Hyperbolic Manifolds

We will work with the upper half-space model of hyperbolic 3-space. That is, we take
H3 = {(
 + 

, 
) ∈ C × R : 
 > 0}, and give it the metric 

2 = 

2+

2+

2
The isometries of
H3 correspond to elements of PSL(2, C) = SL(2, C)/±
, where SL(2, 
) is the group of 2 × 2
matrices of determinant 1 and 
 is the identity matrix.


2

Now let 
 be some manifold, and 
 a group acting on 
. We say a manifold 
 has
a (
, 
)-structure on it if, for every point 
 ∈ 
, there is a chart (
, 
), with 
 ⊂ 
 a
neighborhood of 
 and 
 : 
 → 
(
) ⊂ 
 a homeomorphism. We also require charts to
work nicely together; that is, if two charts (
, 
) and (
, 
) overlap, then the transition map

 ◦ 
−1 : 
(
 ∩ 
) → 
(
 ∩ 
) should also be an element of 
. In particular, we say a

9

manifold 
 has a hyperbolic structure if it has a (PSL(2, C), H3)-structure.

When a manifold has a hyperbolic structure, it also inherits a hyperbolic metric from H3,
allowing us to ask questions such as what is the volume of a hyperbolic manifold. In this work,
we will work exclusively with complete hyperbolic manifolds of finite volume.

An important theorem about the structures of a complete hyperbolic 3-manifold with finite
volume comes from Mostow [30] and Prassad [32]; namely, there is only one hyperbolic
structure, up to isometry:

Theorem 2.1.1 (Mostow-Prassad Rigidity). Suppose 
1 and 
2 are complete hyperbolic 3-
manifolds with finite volume. Then any isomorphism of the fundamental groups 
 : 
1(
1) →

1(
2) is realized by a unique isometry.

This allows us to use the geometric properties of complete finite volume hyperbolic mani-

folds as topological invariants, as well.

2.1.2 Guts of a Manifold
As mentioned above, we can calculate vol(
), the volume of a hyperbolic manifold 
. We
can find a lower bound on vol(
) based on the guts of 
.

Definition 2.1.2. Let 
 be a hyperbolic 3-manifold 
, and 
 ⊂ 
 some embedded surface.
Then 
 cut along 
, denoted 
\\
, is the manifold obtained by taking the completion of

 \ 
.
Definition 2.1.3. Let 
 be a hyperbolic 3-manifold and 
 ⊂ 
 a properly-embedded 
1-
essential surface. Then 
\\
 admits a JSJ-decomposition along essential tori and annuli into

-bundles, Seifert fibered solid tori, and the guts of 
\\
.

The guts of 
\\
, denoted guts(
), is exactly the parts of 
\\
 that, after the decompo-

sition, admit a hyperbolic metric.

10

The guts have been studied in how they relate to the volume of a hyperbolic manifold [6], as
well as how they can, in knot theory, relate to the volume of hyperbolic alternating links [25],
as well as the coefficients of the colored Jones polynomial [17].
In particular, we will be
concerned with the following result of Agol, Storm, and Thurston [6, Corollary 2.2]:

Lemma 2.1.4. Let 
 be a closed hyperbolic 3-manifold, and 
 ⊂ 
 be an incompressible
surface. Then

vol(
) ≥ −
8 
(guts(
\\
)),

where 
8 = 3.66396 . . . is the volume of a regular ideal octahedron.

If we choose our surface carefully, we will be able to calculate 
(guts(
\\
)) by looking

at just the surface and 
 
.

2.1.3 Cusp Volume and Cusp Area

Let 
 be a finite-volume hyperbolic manifold with 
 
 a collection of tori. Then the ends of

 are of the form 
2 × [1,∞). As 
 is hyperbolic, we have some covering map 
 : H3 → 
.
Then if we look at the preimage of these ends, they look like horoballs in H3.

For each end, we can choose a horoball representative 

. The image of this representative,

(

) = 

 is called a cusp of 
. In actuality, we get a family of such cusps as we shrink or
expand 

. In our case (a knot complement), there is only a single cusp — the place where we
removed the knot. When we only have a single cusp, we can expand it as much as possible,
until it becomes tangent to itself. In H3, this is expanding each horoball pre-image until they
begin to become tangent to each other. Once we have done this, we get what is called the
maximal cusp of 
. In the case of a single cusp, this expansion is unique. We can, of course,
generalize this to multi-cusped manifolds, however it is not quite as neat. Depending on the
order we expand the cusps, we can get several maximal cusps, and each can have different
properties.

11

Figure 2.1. (Left): A manifold 
 with a single cusp. The green colored area is a neighborhood of the cusp.
(Right) A preimage of 
, with three different horoball representatives for the cusp. Note that the upper horoball
extends to infinity.

Now, once we have our choice of cusp, 
, we can start analyzing it. As our manifold 
 has
some (hyperbolic) geometric structure to it, 
 will naturally inherit a geometry from it. And
because we know the pre-image of 
 is a horoball in H3, it must inherit a hyperbolic structure
and it’s boundary must be Euclidean.

Definition 2.1.5. The volume of a cusp 
, vol(
), is the Euclidean volume of 
. The area of
a cusp, area(

), the Euclidean area of 

. In the case that 
 = 
 \ 
 we will use 

(
) to
denote the volume of the maximal cusp of 
.

A quick aside about why we look at the Euclidean area: We can always view the horoball
preimage of our cusp as being a neighborhood at ∞ ∈ H3. This amounts to looking at the
set {(
, 
, 
) ∈ H3 | 
 ≥ 
} for some 
 > 0. When we restrict to the boundary of our cusp,


 = {
 = 
}, we see that our hyperbolic metric becomes a scaled Euclidean metric. Thus it
make sense to talk about the cusp in terms of Euclidean area and volume.

As we can view the boundary of a cusp as 
2 × 1, by some simple calculus, we can see that
2area(

) = vol(
). As the volume can be found from the area, we will often focus on just
1
finding a single value, usually the area, and seeing what that gives us.

We have several methods to help us calculate the cusp area. First, if we have a triangulation

12

of 
, we can find the area by direct computation, as has been implemented in SnapPy [14].
When we work with fully augmented links, where all crossings of a knot have been removed
and crossing circles have been added in, we can completely determine the geometry by a circle
packing. In this case, we can use the circle packing to find the cusp area [20].

Definition 2.1.6. Let 
 be the meridian of our knot, and 
 the shortest longitude. Then the
lengths of these curves, ࡁ(
) and ࡁ(
), determine the similarity class of the Euclidean structure
on 

. We call this similarity class the cusp shape.

Knowing the cusp shape can tell us about the area of the cusp, as we have Area(

) ≤

ࡁ(
)ࡁ(
).

Adams, Colestock, Fowler, Gillam, and Katerman found upper bounds of slope lengths and
cusp shapes of a knot based on it’s crossing number [2]. Burton and Kalfagianni found upper
bounds coming from essential spanning surfaces of knots [11]. We need to recall their result
[11, Theorem 4.1] as we will use it in Chapter 4.

Theorem 2.1.7 (Theorem 4.1 [11]). Let 
 be a hyperbolic knot with maximal cusp 
, in an
irreducible 3-manifold 
. Suppose that 
1 and 
2 are essential spanning surfaces in 
 = 
\
,
and let 
(

1, 

2) ࣔ 0 denote the minimal intersection number of 

1 and 

2 in 

. Let
ࡁ(
) and ࡁ(
) denote the lengths of the meridian and the shortest longitude of 
, respectively.
Then:

ࡁ(
) ≤ 6| 
(
1)| + 6| 
(
2)|

ࡁ(
) ≤ 3| 
(
1)| + 3| 
(
2)|,

and


(

1, 

2)
area(

) ≤ 18(| 
(
1)| + | 
(
2)|)2

.


(

1, 

2)

Note that the authors in [2] state their result for links in 
3. However, as noted by Burton
and Kalfagianni in the paper, the results work for any manifold 
, as long as our knot is
hyperbolic.

13

2.2 Weakly Generalized Alternating Knots

When generalizing knots to non-
2 surfaces inside of manifolds that are not 
3, we will

use the definitions of Howie and Purcell [23]. We review these definitions in this section.

Let 
 be a compact, orientable, irreducible 3-manifold, with a closed, orientable surface

, which we call a projection surface. Given a knot 
 ⊂ 
 × 
 ⊂ 
, we can get a projection

(
) by flattening 
 × 
 down to just 
, and keeping track of where the knot goes. We call

(
) a generalized diagram. As we are not working with 
 = 
2 ⊂ 
3, we need to modify
some of the usual definitions we have for knots. In the setting of 
 = 
2, one uses the fact that
all curves in 
2 bound disks to define primeness. Because we do not have this property for
general surfaces, we modify as below:
Definition 2.2.1. We say that 
(
) ⊂ 
 is weakly prime if, whenever 
 ⊂ 
 is a disk with


 intersecting 
(
) transversely exactly twice, then either

• 
 = 
2, and either 
(
) ∩ 
 is a single arc or 
(
) ∩ (
 \ 
) is;
• 
 has positive genus, and 
(
) ∩ 
 is a single arc.

In order to generalize the proofs that follow, we will want to make sure that our knots are

sufficiently complex. To do this, we will introduce two conditions on the knot:
Definition 2.2.2. The edge representativity 
(
(
), 
) is the minimum number of intersections
between 
(
) and any essential curve on 
. The representativity 
(
(
), 
) is the minimum
number of intersections between 
(
) and a compression disk of 
, taken over all compression
disks of 
. If there are no essential curves, we set 
(
(
), 
) to be ∞, while if there are no
compression disks, we set 
(
(
), 
) and 
(
(
), 
) to be ∞.

While the definition of a weakly generalized alternating knot only uses the representativity,
many of our proofs in Chapter 3 will also need the edge representativity to be high enough to
deal with some edge cases. Now, though, we have enough to give a definition for a weakly
generalized alternating knot.

14

Definition 2.2.3. Let 
 ⊂ 
 be a projection surface as above. Then the diagram 
(
) on 
 of
a knot 
 is reduced alternating if

1. 
(
) is alternating on 
,

2. 
(
) is weakly prime,

3. 
(
) ∩ 
 ࣔ ∅,

4. 
(
) has at least one crossing on 
.

If, in addition, we also have

5. 
(
) is checkerboard colorable on 
,

6. 
(
(
), 
) ≥ 4

then 
(
) is a weakly generalized alternating diagram, and 
 is a weakly generalized alter-
nating knot.

This definition generalizes to links on a single surface, as well, by taking 
 to be a link.
While we will not consider it in this work, this definition can also be generalized to links with a
projection onto a disjoint union of surfaces. This includes redefining what a projection surface

 is, as well as requiring that each surface has at least one crossing projected onto it. For the
complete definition in this more general case, see [23].

These knots have been studied in several other papers, as well as knots that satisfy subsets of
this definition. These were introduced by Howie and Rubenstein [24] in 
3 as generalizations
of Hayashi [22] and Ozawa [31], and generalized further to other manifolds 
 by Howie and
Purcell [23].

If we ignore the final two conditions of the above definition, and just look at reduced
alternating, we get as a subset generalized alternating knots in 
3, as studied by Ozawa in
[31], with the additional restrictions that 
 = 
3 and 
(
) is strongly prime, which implies
weakly prime. These knots are also checkerboard colorable, although the representativity is

15

only at least 2 [31, Theorem 2.2]. If we don’t require 
(
) to be weakly prime, and let 

to be a Heegaard torus, we get toroidally alternating knots, as studied by Adams [5]. More
broadly, this category also includes alternating knots on a Heegaard surface 
, as studied by
Hayashi [22]. This also fits the alternating projection of a knot 
(
) onto it’s Turaev surface 

which also has the property that 
 \ 
(
) are disks [16]. By contrast, for reduced alternating
and weakly generalized alternating knots, 
 does not need to be a Heegaard surface, nor does

 \ 
(
) need to be disks. In addition, the requirement that 
(
(
), 
) be large enough also
often guarantees that our diagram will be sufficiently complicated.

We will sometimes narrow our definition by requiring the representativity 
(
(
), 
) to
be strictly greater than 4, and adding in that the edge representativity 
(
(
), 
) is at least 4.
These restrictions will allow us to later rule out certain troublesome cases caused by working
on a surface with genus. We will also define twist regions and twist reduced in this general
case, following Howie and Purcell’s definitions [23]:

Definition 2.2.4. Let 
(
) be a weakly generalized alternating diagram on a surface 
. A twist
region of 
(
) is either a string of bigon regions arranged vertex to vertex that is maximal in
the sense that no larger string of bigons contains it, or a single crossing adjacent to no bigons.
Then two crossings are part of the same twist region if there exists a disk 
 ⊂ 
, with 



intersecting our knot exactly four times - transversely at each of our crossings.

Definition 2.2.5. A diagram 
(
) on 
 is weakly twist reduced if every disk 
 in 
, with 


meeting 
(
) in exactly two crossings, either contains only bigon faces of 
 \ 
(
) or 
 \ 

contains a disk 
(cid:48), where 

(cid:48) meets 
(
) in the same two crossings and bounds only bigon
disks.

Note that this differs from the traditional definition of twist reduced in the following way.
When 
 = 
2, we say a diagram is twist reduced if, whenever a simple closed curve intersects
our diagram at exactly two crossings, that curve bounds a bigon region.

16

Finally, given a diagram 
(
) on 
, we can ask about how many twist regions it might

have:
Definition 2.2.6. The twist number of a weakly twist reduced diagram 
(
) on a surface 
,
labelled tw
(
(
)), is the number of twist regions in 
(
).

Note that, while the twist number is a diagram invariant for these generalized knots and
links, it is not, in general, a link invariant. However, in the case of alternating links on 
2, by
the use of flyping, we have a series of moves that will get us from any alternating diagram to
another alternating diagram, and so the twist number is an invariant of alternating links [29].
Later, in Chapter 5, we will generalize another proof that the twist number is an invariant of
alternating links [15] to show that, if we are working in a thickened surface, we can get a
similar result for our generalized links.




⇒

...

or


(cid:48)




Figure 2.2. If a disk 
 has boundary intersecting our knot at exactly two crossings, then it either contains a twist
region (left) or there is another disk 
(cid:48) intersecting at the same crossings containing a twist region (right).

We will need the following result of Howie and Purcell that allows us to determine when
a knot or link is hyperbolic based on from a generalized alternating projection and bounds the
volume with diagrammatic quantities.
Theorem 2.2.7 (Theorem 1.1, [23]). Let 
(
) be a weakly generalized alternating projection
of a link 
 onto a generalized projection surface 
 in a 3-manifold 
. Suppose 
 is compact,
orientable, irreducible, and has empty boundary. Finally, suppose 
 \ 
(
) is atoroidal.
If 
 has genus at least one, the regions in the complement of 
(
) on 
 are disks, and the
representativity 
(
(
), 
) > 4, then

1. 
 \ 
 is hyperbolic
2. 
 \ 
 admits two checkerboard surfaces that are esssential and quasifuchsian.

17

3. The hyperbolic volume of 
 \ 
 is bounded below by a function of the twist number of


(
) and the Euler characteristic of 
:

vol(
 \ 
) ≥ 
8
2

(tw
(
) − 
(
))

In order to use this theorem, we must show that our weakly generalized alternating diagrams
have disk regions. We do this below for the cases we consider in Chapters 3 and 4, when the
edge representativity is at least 4.

Lemma 2.2.8. Let 
(
) be a weakly generalized alternating projection of 
 onto a generalized
projection surface 
. If 
(
(
), 
) ≥ 4, then the regions of 
 \ 
(
) are all disks.

Proof. Suppose not. Then one of the regions of 
 \ 
(
) either contains an annulus whose
core is essential or meets itself at a crossing. If the region has an annulus, take 
 to be the
core. Then 
 doesn’t intersect 
(
), and so 
(
(
), 
) = 0, a contradiction. If a region meets
itself at a crossing, let 
 be the curve that meets this crossing and then connects back to itself
in the region. Then 
 intersects 
(
) exactly twice (at just the crossing), so 
(
(
), 
) ≤ 2,
a contradiction. Thus if 
(
(
), 
) ≥ 4, then the regions of 
 \ 
(
) are all disks.
(cid:3)

2.3 Kauffman States and the Jones Polynomial

Finally, we will recall Kauffman states and the Jones polynomial for knots and links on


2 ⊂ 
3. We will leave, however, the generalizations of these until Chapter 5.

A well-studied tool in the field of knot theory is the Jones polynomial. In particular, it has
been used to study alternating knots and links, as in [15,17]. Of particular note, Dasbach and
Lin were able to prove that the twist number of alternating links can be recovered from the
coefficients of the Jones polynomial, offering another proof of invariance.

We will review a construction of the Jones polynomial for links with diagrams on 
2 ⊂ 
3

using a function on link diagrams called the Kauffman bracket function.

18

Before we give the definition of the Kauffman bracket function, we introduce some addi-
tional definitions that will help us. First, given a crossing 
 in a link diagram, we define the 

and 
 resolutions of 
 to be the two different smoothings of the crossing in Figure 2.3.

Figure 2.3. The 
 resolution (left) and the 
 resolution (right) of a crossing (center).

Then a Kauffman state 
 for a diagram 
 is a choice of an 
 or 
 resolution at each crossing.
For a state 
, we can count how many 
 resolutions, 
(
), or 
 resolutions, 
(
), there are. In
addition, we can ask how many state circles, or the number of disjoint simple closed curves,
there are after we apply 
 to our diagram. We denote the number of state circles by |
|. We
can now define the Kauffman bracket function.

Let D be the set of all unoriented link diagrams on 
2. Then the Kauffman bracket function,

(cid:104) (cid:105) : D → Z[ 
−1, 
]

is defined by the following relations

1. (cid:104)

(cid:105) = 
(cid:104)

(cid:105) + 
−1(cid:104)

(cid:105)
(cid:105) = (−
2 − 
−2)(cid:104)
(cid:105)

2. (cid:104)
 (cid:116)

3. (cid:104)

(cid:105) = 1.

We can use Kauffman states to make this function more explicit:

We can then define the Jones polynomial of a link diagram to be

(cid:104)
(cid:105) =


(
) =






(
)−
(
)(−
2 − 
−2)|
|−1.



(cid:16)(−1)
(
) 
−3
(
)(cid:104)
(cid:105)(cid:17) |
=
4,

19

Figure 2.4. (Left): The figure eight knot. (Center): A Kauffman state for our knot. We’ve kept track of the type
of resolutions by coloring the edges red for 
 resolutions and blue for 
 resolutions. (Right): The reduced graph
for the state.

where 
(
) is the writhe of the diagram. Despite being defined using link diagrams, the Jones
polynomial is an invariant of the link [28] itself.

Another useful tool coming from Kauffman states is the state graph and, by extension, the
reduced state graph. We define the state graph for a Kauffman state 
, 
 
, to have vertices the
state circles of 
 and edges connecting state circles that are adjacent across a crossing of the
diagram. Then the reduced state graph, 
(cid:48)

, has the same vertices as 
 
, but, if two edges of

 
 are parallel, we remove one of them in 
(cid:48)

.

These state graphs have been used to study information about the links. Dasbach and Lin
showed that certain coefficients of the Jones polynomial can determine the twist number [15].
In particular, if 
 is an alternating link , 

−1 and 

+1 are the second and second last
coefficients of the Jones polynomial for 
, then the twist number is

tw(
) = |

−1| + |

+1|.

As the Jones polynomial, and thus its coefficients, are invariants of the link, this offers

another proof that the twist number is an invariant of the link.

20

CHAPTER 3

ESSENTIAL SURFACES

3.1 Introduction

This chapter is based on work that will be published in Algebraic & Geometric Topology [7].
In this chapter, we will construct two spanning surfaces for weakly generalized alternating
knots, and then show that, under certain conditions, those surfaces are essential. This gener-
alizes the work of Lackenby and Purcell [27] for alternating knots on the sphere. In addition
to giving us new essential surfaces to work with, we will use these later in Chapter 4 to find
bounds on cusp volume.

First, in Section 2, we will go through the construction of the surfaces, called 

,2 and 

,2,
involving augmentations of our diagram. We will also go over some additional definitions we
need here. Then, in Sections 3 and 4, we look at certain graphs that will allow us to study
how these surfaces might intersect disks by studying colored graphs. Finally, in Section 5, we
prove that, under certain conditions on the knot diagram, these spanning surfaces are essential.
In particular, we will show the following theorem:
Theorem 3.1.1. Let 
 : 

,2 → 
 \ 
 be the immersion of our blue twisted surface 

,2 into
our knot complement 
 \ 
, where 
 has no boundary components. Then this immersion is

1-injective and boundary 
1-injective provided that 


 ≥ 121.

Then, by a slight modification to our lemmas and proofs, we are also able to show the

following theorem:
Theorem 3.1.2. Let 
 be the disjoint union of our twisted surfaces 

,2 (cid:116) 

,2. Suppose that
two distinct essential arcs in the surface 

,2 have homotopic images in 
 \ 
, but not 

,2.
Then the two arcs are homotopic in 

,2 into the same subsurface associated with some twist
region of 
2.

21

We will go over the exact definitions of a twisted surfaces and 


 in Section 2. Then, in
Section 3, we look at a specific twisted surface, by looking at a certain augmentation of our
knot. In Sections 4 and 5, we study how a disk might intersect a twisted surface by introducing
and studying colored graphs. Finally, in Section 6, we finish the proofs of both Theorems 4.1.1
and 4.2.4.

As some of the lemmas used for alternating knots on the sphere [27] apply to this more
general case with no change, we will sometimes refer to these lemmas in this chapter. When
we do, we will mention what the lemma tell us, and why we are able to use the lemma with no
change to the proof.

3.2 Augmented Diagrams and Twisted Surfaces

In this section, we will construct the surfaces we are working with. Throughout, we will
often omit the projection when talking about a diagram; that is, we will use 
 in place of 
(
).
First, let 
(
) be a weakly generalized alternating diagram on a surface 
. As 
(
) is
checkerboard colorable, we get two checkerboard surfaces, which we color blue and red and
denote 
 and 
, respectively.

Now, let 


 be some number. We will eventually require that 


 is at least 121, but for
the initial construction, it can be any number we want. For each twist region of 
(
) with at
least 


 crossings, we add a crossing circle to get a link diagram, 
.

Next, from each encircled twist region of 
, we remove crossings in pairs until we have
only one or two crossings remaining to get a diagram 
2. Finally, we remove the crossing
circles from 
2 to get an augmented diagram 
2.

We will show in the next section that 
2 ends up being a weakly generalized alternating
diagram, as well. In particular, this means that 
2 is checkerboard colorable, and so we get
blue and red surfaces. In order to get our twisted surfaces, we will send these surfaces back to

 by working backwards.

When we go back from 
2 to 
2, that is, when we put our crossing circles back in, each

22

.7

e
v
e
n

o
d
d

Figure 3.1. (Left): A knot diagram 
. The center and right twist regions each have at least 


 crossings with
the center having an even number and the left an odd. (Center Left:) We adjoin circles to highly twisted regions
to get 
. (Center Right:) We remove crossing in pairs from encircled regions to get 
2. (Right): We remove the
crossing circles to get 
2.

crossing circle will intersect either the red or the blue surface. Furthermore, when we look
at 
 \ 
2, each crossing circle will have a regular neighborhood that intersects either the red
surface or the blue surface in a meridian disk. So then define the blue and red surfaces in the
exterior of 
2 to be the blue and red surfaces of 
2 punctured by the crossing circles of 
2.
We will denote these by 
2 and 
2, respectively.

Now we need to send 
2 and 
2 back to 
 \ 
. Adding back in the crossings is the same
as twisting along the crossing circles. Most of 
2 and 
2 will twist to give us the usual
checkerboard surface of 
. Things will change, however, where the surfaces meet the crossing
circle. If we removed 2
 crossings from a twist region to get to 
2, we will need to twist a
full 
 times around in order to get them back. That is, the meridian curves where our surface
intersects our surfaces will go to ±1/
 curves on the boundary of the crossing circle, with sign
chosen appropriately.

Finally, to get back to 
 \ 
, we need to fill in the crossing circles with a meridian Dehn
filling. So in order to get our twisted surfaces, we will need to complete them in a way that
makes sense with this filling. Note that each (meridional) cross-section of the crossing circle
intersects the punctured surfaces 2
 times on the boundary. We then connect opposite points
by an interval running between them. When we do this for the whole crossing circle, we will
either be attaching a single annulus or two Möbius bands, depending on the value of 
. In
either case, however, we now have two immersed surfaces, which we can still color blue and

23

Figure 3.2. (Left) 
2 with the torus around the twist region. Blue circles mark where the torus punctures the
blue surface. (Center Left) Twisting around the torus gives us back 
. In this case, we are only adding in one
full twist for simplicity. (Center Right) To complete the twisted surface, attach opposite sides of the torus with a
band, as represented on top and bottom. Here this amounts to adding an Möbius band. (Right) When we have
more twists, things can become more complicated. In addition, we might be adding two annuli instead.

red, and which we call twisted checkerboard surfaces. We will denote them 

,2 and 

,2
appropriately.

In order to prove that these twisted surfaces are essential, we will often only work with one
surface at a time. As the construction of 

,2 does not interfere with the construction of 

,2,
we can make our study of these surfaces simpler by looking at only one surface at a time. To
do this, we modify our construction as follows.

When we start encircling twist regions with enough crossings, we only add crossing circles
if they will intersect the blue checkerboard surfaces. We call this new link 

. Then, when
we remove the pairs of crossings from encircled regions, leaving only one or two crossings,
we get a new link 

,2. Finally, we remove the crossing circles to get 

,2.

Now, the blue surface 
2 can also be embedded in the complement of 

,2, while the red
surface 
2 for 

,2 is homeomorphic to the red checkerboard surface of 
. This allows us,
while studying our surfaces, to look at just one twisted surface at a time, say 

,2, and 
, the
usual checkerboard surface for 
.

3.3 Weakly Generalized Alternating Knots and Augmentation

In this section, our main goal is to show that, if we start with a weakly generalized alternating
knot, with some additional properties, and then augment it by removing crossings, we will still
have a weakly generalized alternating knot with those same properties.

24

Let 
 be a weakly generalized alternating knot with a closed, orientable projection surface

 in a 3-manifold 
. Fix a (twist-reduced) diagram 
(
) on 
 with edge representativity,

(
(
), 
) ≥ 4, and representativity 
(
(
), 
) > 4.

Let 
 be the link obtained from 
 where we add a crossing circle to some number of twist
regions. Also let 
(cid:48) the link where we remove, in pairs, all but one or two crossings from twist
regions with a crossing circle in 
, and 
(cid:48) the knot obtained from 
(cid:48) by removing the crossing
circles. Our first goal is to prove the following which assures that the altered knot 
(cid:48) continues
to be a weakly generalized alternating and has the same properties as 
.

Proposition 3.3.1. If 
 is a weakly generalized alternating knot with diagram 
(
) on a
projection surface 
 in a 3-manifold 
, with 
(
(
), 
) ≥ 4 and 
(
(
), 
) > 4, then 
(cid:48)
is a weakly generalized alternating knot on the surface 
 in 
 with 
(
(
(cid:48)), 
) ≥ 4 and

(
(
(cid:48)), 
) > 4.

Proposition 3.3.1 allows us to look at the simpler augmented knot 
(cid:48) as opposed to the

original 
. The proof of the proposition requires a few lemmas that we will prove next.
Lemma 3.3.2. 
(
(cid:48)) is reduced alternating on 
.
Proof. There are four conditions we need to show about 
(
(cid:48)) for it to be reduced alternating.
First, as 
(
) is alternating on 
, we must also have 
(
(cid:48)) alternating on 
—we removed
crossings in pairs, and so could not have introduced any non-alternating into our diagram.
Next, we show that 
(
(cid:48)) is weakly prime, as stated in Definition 2.2.1. If it isn’t, then we can
find a disk 
 ⊂ 
 such that 

 intersects 
(
(cid:48)) exactly twice and 
(
(cid:48)) ∩ 
 is not a single
embedded arc. We can isotope 
 to be disjoint from crossing circles. When we put crossings
back in to 
(
(cid:48)), we won’t add any crossings to 
—we only add crossings at crossing circles.
But then we have a disk 
 whose boundary intersects 
(
) exactly twice with 
(
) ∩ 
 not
a single embedded arc. As 
(
) is reduced alternating, this is a contradiction.

Finally, we must show 
(
(cid:48))∩ 
 ࣔ ∅, and that there is at least one crossing on 
. However,
this follows from construction—
(
(cid:48)) still lives on 
, so the first condition is satisfied. For

25

the second condition, as we never remove all crossings from a twist region, if 
(
(cid:48)) doesn’t
have a crossing on 
, neither will 
(
), once again a contradiction. So then 
(
(cid:48)) must be
(cid:3)
reduced alternating on 
.
Lemma 3.3.3. 
(
(cid:48)) is checkerboard colorable.
Proof. As 
(
) is weakly generalized alternating, we know that 
(
) must be checkerboard
colorable. We can use this coloring to obtain a coloring of 
(
(cid:48)). Because we remove all
but one or two crossings from a region, we can obtain 
(
(cid:48)) from 
(
) by removing bigons
from the same twist region. These bigons must have the same color, so removing them will
not change the checkerboard colorability of the diagram. Using this coloring will then give us
a coloring of 
(
(cid:48)).
(cid:3)
Lemma 3.3.4. If 
(
(
), 
) ≥ 4, then 
(
(
(cid:48)), 
) ≥ 4.

Proof. Suppose not. Then we can find an essential curve ࡁ in 
 that crosses our knot diagram,

(
(cid:48)), a minimal number of times—zero, one, two, or three times. By isotopy, we can
assume ࡁ intersects 
(
(cid:48)) transversely away from crossings. First, we show that the number
of intersections can’t be odd. Because 
(
(cid:48)) is checkerboard colorable, we can look at the
colored regions ࡁ crosses through. Every time ࡁ crosses 
(
(cid:48)), it must switch colored regions:
either from red to blue or from blue to red. If ࡁ intersects 
(
(cid:48)) an odd number of times, we
would then have a curve that started in one color and ended in the other color. As it is a closed
curve and our diagram is checkerboard colorable, this is impossible, so 
(
(
(cid:48)), 
) must be
zero or two.

If 
(
(
(cid:48)), 
) = 0, then ࡁ is disjoint from 
(
(cid:48)). This means, in particular, that we can
isotope ࡁ to be disjoint from any twist region without changing the number of intersections.
Each twist region in 
(
(cid:48)) has at least one crossing; if ࡁ enters the twist region through one
side, it must exit through the same side in order to not intersect 
(
(cid:48)). But then, as 
(
)
differs from 
(
(cid:48)) only in twist regions, when we put the twists back in our diagram, we will
have an essential curve that does not intersect 
(
), contradicting the fact that 
(
(
), 
) = 4.

26

If 
(
(
(cid:48)), 
) = 2, then ࡁ must intersect 
(
(cid:48)) exactly twice. First, suppose ࡁ doesn’t
intersect a twist regions. Then, as before, when we go back to 
(
), we will have an essential
curve intersecting it twice, a contradiction. So then ࡁ must intersect 
(
(cid:48)) in a twist region.
By the definition of 
(
(
(cid:48)), 
), we assume that ࡁ intersects 
(
(cid:48)) transversely away from
crossings. As such, in order for ࡁ to intersect a twist region, it must intersect the twist region
exactly twice, so we will focus on just this region. There are two options for how ࡁ intersects
the strands in this region: either ࡁ intersects the same strand twice, or ࡁ intersects both strands
once. First, suppose it is the former. Then the arc of ࡁ intersection the bigon of the twist region
bounds a disk with the strand ࡁ intersects. Using this disk, we can isotope ࡁ away from the
twist region, giving us an essential curve that intersects 
(
(cid:48)) zero times, a contradiction.

On the other hand, if ࡁ intersects both strands of the twist region, we may isotope ࡁ away
from any crossing disks. Once we do this, when we put crossings back in to get 
(
), we will
have an essential arc ࡁ intersecting 
(
) exactly twice (as ࡁ does not cross any crossing disks).
(cid:3)
This is a contradiction, and so we are done.

We are now ready to give the proof of Proposition 3.3.1 .

Proof. All but the last part follows directly from the previous lemmas. Thus, all that we need to
show is that 
(
(
(cid:48)), 
) > 4. As any curve that bounds a compression disk must be essential,
we can say, by Lemma 3.3.4, that we at least have 
(
(
(cid:48)), 
) ≥ 4. So suppose we can find
some 
 ⊂ 
 such that 
 bounds a compression disk and intersects 
(
(cid:48)) exactly four times.
When we put the crossings back to get 
(
), we must have 
 intersecting 
(
) more than
4 times, and so 
 must be between a twist region. However, if 
 doesn’t intersect the knot
strands associated to the twist region, then we have a region with zero crossings, contradicting
our construction of 
(cid:48). So at least two of the intersections of 
 with 
(
(cid:48)) are part of the same
twist region. But then, no matter how they intersect, we can then homotope 
 to be outside of
the twist region without introducing any more intersections (but possibly reducing them), as
follows. Because we are working on a twist region, there must be a disk on 
 such that the

27

Figure 3.3. Two possible configurations to have 
(
(
(cid:48)), 
) ≤ 4. On the left, the blue disk can be used
to homotope the knot away from the compression disk, reducing the number of intersections. On the right,
homotopying along the blue disk will cause the compression disk to not intersect a twist region.

boundary passes through each of the crossings of our twist region. We can then use part of
that disk to homotope away from the compression disk, as shown in Figure 3.3. But then, by
putting crossings back in, we must have 
 intersect 
(
) four or less times, a contradiction to

(
(
), 
) > 4, and so we are done.
(cid:3)

Now we can apply Proposition 3.3.1 to our augmented knot 

,2, and we get that

(

,2) must also be a weakly generalized alternating diagram on 
 with 
(
(
), 
) > 4 and

(
(
), 
) ≥ 4.

We will finish up this section by using Proposition 3.3.1 to prove two simple, but important,
facts about 

,2, as we now know some nice properties stay under augmentation. The first
result is the following.
Lemma 3.3.5. Label the regions of the complement of the diagram 
(

,2) blue or red,
depending on whether they meet the blue or red surface.

(1) The blue regions on opposite sides of a crossing of 

,2 cannot agree.

(2) The red regions on opposite sides of a crossing of 

,2 cannot agree.

(3) The two blue regions that meet a single crossing circle of 

,2 cannot agree.

Proof. To start, we assume that blue regions on the opposite side of a crossing do agree. Then
we can create a simple closed curve on our surface 
 by drawing an arc in the blue region
from one side of the crossing to the other, then drawing an arc across the red region near this

28

crossing. Now we have a simple closed curve that intersects 
(

,2) exactly twice. If this
curve bounds a disk in 
, then we contradict the fact that 
(

,2) is weakly prime. If it doesn’t
bound a disk, then it is an essential curve intersecting our diagram exactly twice, contradicting

(
(

,2), 
) ≥ 4. A similar proof works to prove the second statement by switching the blue
and red regions.

Now, to show the third statement, assume that two blue regions meeting a single crossing
circle do agree. Then we can draw an arc entirely in the blue region from one side of the
crossing circle to the other. We then finish it into a simple closed curve by attaching the arc
of intersection of the crossing disk. As above, we then have a curve intersecting our diagram
exactly twice, contradicting either 
(

,2) being weakly prime or having edge representativity
(cid:3)
at least 4.

This generalizes Lemma 3.9 of [27]. However, there is a fourth part in [27], stating that if
distinct blue regions meeting a single crossing circle meet at the same crossing of 

,2, then
that crossing is associated with the crossing circle. This part does not carry through to the
generalization of weakly generalized alternating knots. To prove part four the authors in [27]
rely on a lemma which states that, in the case of usual alternating knots, 
2 

,2 are prime.
In our case 
2 or 

,2 are not prime, while this case can give us a simple closed curve that
goes through both the crossing circle and the crossing, because it intersects our knot exactly 4
times. This curve it could very well be an essential curve, and thus tell us nothing about how
the crossing relates to the crossing circle.

With 

,2 defined in our general context, we need to introduce a new definition to generalize
a diagram being blue twist reduced. As we already have a notion of weakly twist reduced, we
will model our definition off of this.

Definition 3.3.6. A diagram of a link is weakly blue twist reduced if every disk 
 in 
 with


 meeting 
(
) in exactly two crossings, with sides on the blue checkerboard surface, either
bounds a string of red bigons, or there is a disk 
(cid:48) in 
 meeting the diagram at the same

29

crossings that bounds a string of red bigons.

Next, we will prove our augmented diagram is weakly blue twist reduced.

Lemma 3.3.7. 
(

,2) is weakly blue twist reduced.
Proof. Suppose a disk 
 in 
 with 

 meeting only the blue regions, and intersecting 
(

,2)
in exactly two places. We can isotope 

 away from any crossing disk so that when we add
back in crossings to get 
(
), 

 remains disjoint from the red surface. Because 
(
) is
weakly twist reduced, either 
 bounds a string of bigons, or there is another disk 
(cid:48) in 
 − 
,
with 

(cid:48) meeting the diagram in the same two crossings, that bounds a string of bigons. If it
is the latter, then note that 

(cid:48) also only meets the blue region. In either case, these bigons
must all be red bigons. Without loss of generality, suppose 
 contains the red bigons.

Because 
(
) is weakly twist reduced, all of the red bigons must come from the same twist
region. When we remove some bigons to get back to 
(

,2), we either don’t touch any of the
bigons inside 
, in which case we are done, or we remove all but one or two crossings from
the region. If this is the case, then we have to have removed all but two crossings - 
 intersects
two distinct crossings, so there must be at least two distinct crossings left in the region. But
(cid:3)
then 
 contains a single red bigon, and we are done.

3.4 Colored Graphs and Twisted Surfaces

In this section, we will prove a series of technical lemmas that we will need for the proof
of Theorem 3.1.1. The lemmas aim to analyze how certain disks might intersect our twisted
surfaces. The combinatorics of these intersections will be encoded by certain colored planar
graphs. Before we can begin our analysis, we need to establish some notation and terminology.

3.4.1 Blue and red graphs

Recall that we need to prove that the immersed blue and red surfaces 

,2 and 

,2 are

1-injective and boundary 
1-injective in 
 \ 
.

30

Working with both 

,2 and 

,2 simultaneously will be difficult. Instead, we will take
advantage of the fact that the construction of one surface doesn’t involve the other. That is, we
could have constructed, say, 

,2, the blue twisted surface, without ever having mentioned the
red surface. The only part that would change is where we put the crossing circles—in order to
not disturb the red surface, a crossing circle can only be added to a highly twisted region if it
will intersect the blue surface. Thus in our proofs, we will often be working with the 

,2 and

, where 
 is the usual red checkerboard surface for 
(
).

Recall the map 


: 

,2 → 
 \ 
 from the statement of Theorem 3.1.1.

If 
 is not

1−injective, then we have a disk 
 : 
 → 
 \ 
 such that 
 |

= 
 ◦ 
, where 
 is an essential
loop on 

,2.

Let Ɖ
 = 
−1( 
 (

,2)) be the pre-image of 
 (

,2) under 
. By working with Ɖ
, we
can study the intersection 

,2 intersects the disk 
(
). By Lemma 2.1 of [27], which also
applies in our situation without changes, we have the following:

• Ɖ
 consists of a graph and a collection of simple closed curves.

• The vertices of the graph are points mapped to crossing circles of 
.

• Each interior vertex of the graph has valence a multiple of 2
 
 , where 2
 
 is the number
of crossings removed from the twist region of 
(
) that corresponds to the twist region
associated to the relevant crossing circle. Each boundary vertex has valence 
 
 + 1.

While the arguments from [27] are given in there for 
 = 
3 and 
 = 
2, as long as our 

has no boundary components, many of the proofs require no modification. This is because, in
many cases, the proofs are “local,‘’ and only reference a small neighborhood of where we are
working. Instead of reproving everything, we will instead focus on the cases where the same
proof doesn’t work, requiring us to do something different, often either a completely different
proof, or showing that the cases that occur when 
 ࣔ 
3 or 
 ࣔ 
2 cannot happen.

To continue, recall the links 

, 

,2, and 

,2 that we introduced in Section 2. We now
have three surfaces to consider—
2 and 

,2 in 
 \ 

,2 and 
 \ 
, respectively; 
2, the red

31































Figure 3.4. (Left) A possible configuration for Ɖ


. (Right) How Ɖ


 would look when imposed on the
diagram. The labeled blue edges in the graph map to the corresponding labeled arcs in the diagram.

checkerboard surface for 

,2; and the crossing disks bounded by the crossing circles of 

,2,
which we color green. This also gives us three graphs to look at.
Definition 3.4.1. The blue graph, Ɖ
 = 
−1( 
 (

,2)), is defined as above. Next, if we remove
the vertices of Ɖ
 (which map to crossing circles) from the disk 
, we get an embedding of the
punctured disk 
(cid:48) : 
(cid:48) → 
 \ 

,2. Then the blue-red-green graph Ɖ


 is the pull-back of
the union of all three surfaces—

,2, 
2, and the crossing disks— via 
(cid:48). Finally, by ignoring
the edges and vertices of Ɖ


 coming from the green surface, we get the blue-red graph,
Ɖ

. Note that Ɖ
 ⊂ Ɖ

 ⊂ Ɖ


.

We will be looking for what are called trivial bigons:

Definition 3.4.2. An edge of Ɖ


 is trivial if it is a blue arc disjoint from the red edges with
endpoints on distinct vertices of Ɖ
 corresponding to the same crossing circle in 

.

An important lemma relating to trivial arcs is the following:

Lemma 3.4.3 (Lemma 4.11 [27]). If all but one of the edges of a region in Ɖ
 are trivial, then
the remaining edge is also trivial

In particular, this means if one edge of a blue bigon is trivial, then the other one must be
as well. This will allow us to divide families of adjacent bigons into two cases, trivial and
non-trivial. Then, as shown in Lemma 2.6 of [27], Ɖ
 must have more than (


/18) − 1

32

adjacent non-trivial bigons. So, if we can show that there must be less than 5 ≤ (121/18) − 1
adjacent non-trivial bigons, then we get a contradiction. As our only assumption was that 

,2
was not essential, we will have that our twisted surface must be essential.
Lemma 3.4.4 (Lemma 2.6, [27]). The graph Ɖ
 must have more than (


/18) − 1 adjacent
non-trivial bigons, where 


 is the minimum number of crossings removed from a twist region.

Proof. The proof of Lemma 2..6 in [27] does not actually rely on either our knot or twisted
surface, but is based purely on facts about the graph itself, and involves restricting to subdisks
and subgraphs and removing trivial bigon families until we can get our result. As such, the
(cid:3)
proof and the lemma will continue to work in our setting, as well.


 to be the pull-back of 
 (

,2)) via 
(cid:48). That is, Ɖ(cid:48)

To continue recall that we must prove that 
 : 

,2 → 
 \ int(
(
)) in the statement of
Theorem 3.1.1 is boundary 
1-injective: If not, then we can get a disk 
(cid:48) : 
(cid:48) → 
 \ int(
),
where 

(cid:48) is a concatenation of two arcs, one mapped into 

(
) and the other essential in


,2 [27, Lemma 2.2].
Definition 3.4.5. Define Ɖ(cid:48)


 = 
(cid:48)−1( 
 (

,2)).
Once again the properties of Ɖ(cid:48)

 remain the same if we replace 
3 \ 
 from [27] with 
 \ 
.
That is, the interior vertices of Ɖ(cid:48)

 have valence 2
 
, while exterior vertices have valence 
 
 +1
(if they come from a crossing circle) or 1 (if the vertex maps to 

(
)). In particular, the
following lemma from [27] works also in our setting.
Lemma 3.4.6 (Lemma 2.7 [27]). The graph Ɖ(cid:48)

 must have more than (


/18) − 1 adjacent
non-trivial bigons or more than (


/18) − 1 adjacent triangles, where one edge of each
triangle lies on 
(cid:48)−1(

(
)).

There are a couple of important things to note about the last two lemmas. First, while 



is not necessarily equal to 


, we do know that 


 ≥ 2(cid:100)


/2(cid:101) − 2. With 


 ≥ 121, we
get 


 ≥ 120, so there must be more than 5 adjacent non-trivial bigons in Ɖ
. Second, our
choice of 


 ≥ 121 is not optimal for these lemmas. In fact, the proof that 

,2 is essential

33

only needs 


 ≥ 91, so we have at least 5 adjacent bigons, or 5 adjacent triangles (we will
prove this in the next section). The choice of 121 comes from a modification that will get us
our final proof, Theorem 4.2.4.

3.4.2 Bigons in Ɖ




Here we will begin to study the structure and combinatorial properties of the graphs Ɖ
, Ɖ


and Ɖ


 defined in the previous subsection. From the last subsection we know that these
graphs have a certain number of adjacent non-trivial (blue) bigons, we will be looking at how
the disk must intersect other surfaces near the blue surface. Our goal here is to study the
possible configurations of these bigons and successively rule out several of then. By the end
of this section, we will be left with only one outcome—the family of bigons must intersect the
red surface at least twice.

Recall that Ɖ
 is the pull back of 

,2 of a certain map 
 : 
 → 
 \ 
 and that Ɖ


 is the
pull-back of 

,2, as well as the red surface 
2 and the green surface, of a map 
 : 
(cid:48) → 
\

,2
and that Ɖ

 is obtained by deleting the green edges of Ɖ


. Below we summarize some
basic facts about these graphs that we will be using:

• As outside of the twisted parts, the twist surfaces are embedded, the edges of Ɖ
 (where
the disk meets the blue surface) can only meet each other at the twisted components
(where the crossing circles of 

,2 intersected our blue surface). As such, the vertices
of Ɖ
 are exactly the points that map to crossing circles of 

,2. This means that, if two
blue edges are adjacent to each other on a vertex, it must be that they are on opposite
sides of the crossing circle (as they must meet on the twisted part).

• In addition to what we have for Ɖ
, we also know that red edges can’t intersect red edges,
nor can green edges intersect green edges (as both red and green surfaces are embedded).

• Red edges can meet green edges (at the crossing disk). Likewise, green edges can meet
blue edges at vertices of Ɖ
 (where the crossing disk meets 

,2), and red edges can meet

34

blue edges at crossings of the diagram. In addition, as 
(cid:48) crosses the projection surface
when it meets a blue or red edge, regions of 
(cid:48) \ Ɖ

 are mapped above or below the
projection surface, switching from one to the other when they meet one of these edges.

• Using a complexity-minimizing argument, where complexity is measured by the number
of vertices in each graph, we get that there are no green edges, disjoint from blue, with
both endpoints on red; and no green edges, disjoint from red, with both endpoints on a
blue edge and one endpoint not a vertex. In addition, green edges cannot have both of
their endpoints on the same blue vertex. In all three of these cases, we can use the arcs
in question, in addition to the crossing disk, to find a homotopy that reduces how many
(non-blue) vertices are in our graphs. So as long as we assume our disk’s image, 
(
(cid:48)),
and the induced graphs are minimal, these cannot happen.

As before, the lemmas are generalizations of the case when 
 = 
2 and 
 = 
3. Most of
the generalizations are relatively minor, with only a few complications brought in by the fact
that we aren’t working on a sphere anymore. For completeness sake, we will also talk about
the proofs that still hold over from looking at the case of diagrams on 
2 [27], although in
much broader and looser terms.

First, we will look at a potential blue-red bigon—a bigon in Ɖ


 with one side colored

blue and one side colored red—and show that it cannot exist. We do this as follows:

Lemma 3.4.7. The graph Ɖ

 has no bigons with one blue side and one red, whether or not
the bigon meets the green surface.

Proof. If there is a bigon with only one red side and one blue side, then it must be mapped
completely above or completely below the projection surface. Without loss of generality, we
may assume it is mapped above. Then the union of the two sides of our bigon will form a
simple closed curve 
 meeting 
(

,2) exactly twice with a crossing on either side. If this
curve bounds a disk, we contradict 
(

,2) being weakly prime. Because we are not working
on the sphere, though, we are not guaranteed this. Instead, we must use the fact that the edge

35

representativity (see Definition 2.2.2) of 
(

,2) is at least four. If 
 does not bound a disk,
then we have an essential curve intersecting our knot transversely exactly twice, a contradiction
of 3.3.4. So then 
 must bound a disk with a crossing on either side, and so we get our desired
(cid:3)
contradiction.

Putting our observations above together, we get two important results. First, our disk

 only meets the blue surface in “interesting” places—that is, near or through the crossing
disks. Second, there are no monogons in Ɖ
: any that existed couldn’t meet either the red or
green surfaces, as mentioned above, and so would have to connect two opposite sides of the
crossing circle. However, by Lemma 3.3.5(3), this can’t happen. While this later is not used
immediately, we will use it later on, and so state it as a lemma:

Lemma 3.4.8 ( Lemma 4.7 [27]). The graph Ɖ
 has no monogons.

As we want to work eventually with non-trivial bigons, it’s important that we understand
how trivial arcs and bigons work. As stated along with the definition, if one edge of a bigon
is trivial, so is the other, giving us our trivial bigon families. In addition to this, we also find
that trivial arcs have some restrictions to them—they must have at least one endpoint on the
boundary of our disk, but cannot be a subset of the boundary. With that, we will set this aside
to start looking at non-trivial bigons.

Lemma 3.4.9. In the graph Ɖ


, there are no non-trivial bigons with two blue sides, disjoint
from red and green edges.

Proof. Suppose we had such a non-trivial bigon. Because the bigon is non-trivial, the two
vertices must correspond to separate crossing circles. When we put back in the crossings to
get 
(
), we get a simple closed curve 
 from the bigon. The blue edges of the bigon remain
disjoint from any crossings, and connect across where the crossing disks met the projection
surface. In particular, 
 intersects 
(
) in exactly two crossings. We want to show that 

bounds a disk in our projection surface 
—this will allow us to appeal to 
(
) being twist

36

reduced and will give us a contradiction. So suppose 
 doesn’t bound a disk in 
. Then the
curve must be essential.

If 
 bounds a compression disk, and 
(
(
), 
) > 4, then, after a small isotopy to take 

away from crossings, we get a curve bounding a compression disk that intersects our diagram
only four times, a contradiction. So now we need to see what happens if 
 doesn’t bound a
compression disk.

As 
 bounds a bigon disjoint from red and green edges, it must bound a disk in a region in
the pre-image of the blue surface in our disk. By our assumptions, 
 can’t bound a disk in 
,
nor can it bound a compressing disk for 
. Therefore, the region it bounds must pass through
our projection surface. But as our bigon is disjoint from red, this cannot happen (regions of
the disk can only pass through the projection surface at blue or red edges, neither of which can
be in our bigon). So then we get a contradiction of 
 not bounding a compression disk.

Putting it all together, we see that we have a simple closed curve 
 which must bound a
disk on 
. But then, as 
 meets 
(
) in exactly two crossings and bounds a disk, because

(
) is weakly twist reduced, we must have both of the crossings correspond to the same twist
(cid:3)
region, a contradiction of the bigon being non-trivial.

So blue bigons must have some other colored edge intersecting them. A bigon intersected
by another edge gives us two triangles, so it is natural to consider how adjacent triangles might
act. As red cannot intersect red and green cannot intersect green, we find ourselves with four
possible adjacent triangles, constructed as follows:

1. Blue-Red bigon intersected by green (this can’t happen, as there are no blue-red bigons);

2. Red-Green bigon intersected by blue (we will prove this can’t happen in Lemma 3.4.10

below);

3. Blue bigon intersected by green (this can’t happen, as we can’t have a blue-blue-green

triangle by minimality, by [27] Lemma 4.16);

37

4. Blue bigon intersected by red.

Figure 3.5. (Top) The four different types of adjacent triangles coming from a bigon. (Bottom) How the bigons
might look in the diagram. Note that, in the red-green bigon case, the two red edges in the diagram are, in fact,
the same edge, and must “wrap around” the diagram to create the bigon.

Lemma 3.4.10. In Ɖ


, there are no pairs of red-green-blue triangles adjacent across a
blue edge. More generally, no pair of green edges can be added to Ɖ

 in such a way that the
result is a pair of triangles adjacent across a blue edge.

Proof. We will prove the second statement.
If such an edge could be added, we can find
the innermost pair of triangular regions, so there are no additional red or blue edges in the
triangles. Then one of these triangles must be mapped above the projection plane, and the
other below. As there is only one place where red meets blue, there is only one crossing in the
image of this diagram, so both red edges must meet this crossing, and meet no other crossing
in our diagram.

Looking at the green edges, we see that both meet the same blue edge at the same point.
This means that these green edges are related to the same crossing circle. As our two red edges
meet these green edges, our red edges must both run back to the same crossing circle. So, we
can create a simple closed curve with crossings on either side by following along the red edges,
jumping from one to the next at either the crossing or the crossing circle. This curve intersects

38

our knot exactly twice, so by the representativity and edge representativity of our knot, must
(cid:3)
bound a disk. However, this contradicts weak primality, so this cannot happen.

The other two triangles we need to care about are blue-blue-green triangles (the third case
listed above), and blue-blue-red triangles (the fourth case). In fact, with no change in proof
from

Lemma 3.4.11. The graph Ɖ


 contains no blue-blue-green triangles.

Proof. Suppose we have a blue-blue-green triangle. We may assume we are working with the
innermost green edge of such a triangle, and so it is disjoint from any other green edges. There
are two cases to consider. First, the green edge can lie on the crossing circle of the vertex of
the triangle, or second, the green edge can lie on a separate crossing circle.

If the vertex the blue edges meet at corresponds to the same crossing circle the green edge
lies on, then we can construct two curves by following along one of the blue edges to the green
edge, and then following the green edge back to the start of the blue edge. Each of these curves
don’t intersect the diagram, and so must bound disks. Then our entire triangle must lie in the
union of these two disks along with a neighborhood of the green edge, which is also a disk.
Then, by a similar argument as in [27], we may homotope the triangle through the crossing
circle, thus removing a vertex from Ɖ


. By our assumption of minimality, this gives us a
contradiction.

If the vertex corresponds to a different crossing circle, we can get a curve that intersects
our knot at exactly two crossings by following a blue edge down to the green edge, following
the green edge to the other blue edge, then back up and across the vertex to get back to the
start. We can modify this curve slightly to get a curve intersecting our knot at exactly two
crossings, one associated to the vertex of the triangle and one associated to the green edge’s
crossing circle. Then, with a similar argument as in Lemma 3.4.9, we get that this curve must
bound a disk. But then this means our two crossings must be part of the same twist region.
(cid:3)
However, as they are encircled by two different crossing circles, this is a contradiction.

39

We will focus, then, on blue-blue-red triangles.

3.4.3 Blue-Blue-Red Triangles

Our ultimate goal is to show that there can only be so many adjacent blue bigons in Ɖ
,
contradicting Lemma 3.4.4 that says we need at least five. As a red edge must intersect any
family of non-trivial blue bigons, we can look at blue-blue-red triangles constructed from the
bigons and the red edge intersecting them.

One of the ways we will go about generalizing the results of

[27] is by using a sort of
“local” property—most of the proofs take place in a small subsection of the diagram, and so
we don’t necessarily need to know what’s happening beyond that for a given proof. To take
advantage of this for generalized surfaces, however, we will need to show that the subsections
we care about are similar enough to subsections on a sphere. Namely, we will need to show
that these small parts don’t wrap around the surface, and instead lie on a disk. Once we have
this, most of the proofs follow naturally. We will prove this, then, in two lemmas, dealing with
two different cases.

First up, one of the more common cases we will need to deal with is where we have a
triangle with two blue edges and a red edge. That such triangles induce a disk on 
 is split into
two parts below, depending on how the red edge might pass through the crossing circle of the
vertex. One important thing to note that will come up several times is that we don’t actually
need our triangle, or any of our future shapes, to bound a disk. In fact, looking at the case
when the red edge passes through the crossing circle, this would be impossible. Instead, we
just want it to lie within a disk, and hence induce a disk on 
.

Before we begin, we introduce one piece of terminology. We say that a group of edges in

a graph on 
 induces a disk if there is a disk on 
 with boundary containing those edges.

Lemma 3.4.12. A blue-blue-red triangle in Ɖ

 which, in the image on 
, has none of its
edges passing through the crossing circle of the blue vertex induces a disk on the projection
surface 
.

40

Figure 3.6. (Left) A blue-blue-red triangle as viewed on the graph Ɖ

. (Right) The same triangle as viewed on
the knot diagram, 
(

,2).

Proof. First, note that the triangle gives us a simple closed curve on 
, as seen in Figure 3.6:
let 
 be the curve that follows our triangle along the edges, and joins the blue edges across the
crossing disk associated to the vertex where the blue edges meet. By Lemma 4.18 in [27],
the triangle meets two distinct crossings, and so then 
 must also meet two distinct crossings.
We want to show that 
 bounds a disk on 
. If it doesn’t, there are two possibilities we will
consider.

The first case: 
 could bound a compression disk.

If it does, then note when we put
crossings back in, that 
 intersects our knot in exactly four places: once across each of
the crossings, and twice across the crossing disk of the vertex. But then we have a curve
bounding a compression disk meeting our knot four times, contradicting the assumption that

(
(
), 
) > 4.

So then if 
 is essential and doesn’t bound a compression disk, we can look at the triangular
region 
 bounds in the disk. Because 
 bounds neither a compression disk nor a disk on 
,
the triangular region must pass through the projection surface. But then our region must have
a red or blue edge in it, as it can only pass through the projection surface through an edge of
(cid:3)
Ɖ

. Then we don’t have a blue-blue-red triangle, and so we are done.

Lemma 3.4.13. A blue-blue-red triangle in Ɖ

 with the image of the red edge passing through
the crossing circle of the vertex induces a disk on the projection surface 
.

41

Proof. As above, we know that the triangle has to meet two distinct crossings. But now, as
our red edge must cross through the green edge, we are not guaranteed a single simple closed
curve. Instead, we shall construct two separate disk , and then glue them together. Call our
blue edges 
 and 
 and our red edge 
. Let 
1 be the curve that follows 
, then 
 until we get
to the crossing circle, and then the green edge back to 
. Likewise, let 
2 be the curve that
follows 
, then 
 until we get to the crossing circle, and then jumps back to 
 along the green
edge.

First, each of these curves must intersect our diagram exactly twice - once at the crossings,
and another time through the crossing disk. This allows us to immediately say that both

1 and 
2 bound disks, as 
(
(
), 
) ≥ 4. Next, note that 
1 intersects 
2 at exactly one
point—where red meets green. This is because a red edge cannot intersect itself, and blue
edges can only meet at vertices. But, as blue edges adjacent to the same vertex, they must map
to different sides of the twist region, and so are disjoint. Thus the only point where 
1 and 
2
can meet is the one they share—where the crossing circle meets the red edge.

Now, we can finally construct our entire disk. Let 
 be the union of the disk bounded by

1, the disk bounded by 
2, and a small neighborhood of the intersection of the crossing circle
and red edge. As 
1 doesn’t interfere with 
2 outside of this neighborhood, we do in fact have
(cid:3)
a disk as opposed to an annulus, and so are done.

We can use these disks to construct even larger disks, with some caveats—we need to be

careful that when we glue disks together, we aren’t creating an annulus.

Lemma 3.4.14. Families of blue-blue-red triangles, adjacent along blue edges, bound a disk.

Proof. If we have a single blue-blue-red triangle, we are done by above. So we will suppose
we have a group of them, and consider what happens when we want to add one more. As blue
edges can only intersect at vertices, and red edges can never intersect, we only have to worry
about two triangles—the one we are adding and the triangle that shares an edge with it. There
are three cases we have to worry about, as shown in Figure 3.7. First, both triangles have their

42

Figure 3.7. The three possible ways two blue-blue-red triangles can have their images glued together.

red edges not passing through the crossing circle. Then both triangles bound a disk, and share
a single continuous segment—their common blue edge and the green edge—so we are done.
Second, one of the triangles doesn’t pass through the crossing circle, but the other does. In
this case, for the triangle that does pass through the crossing circle, we get two separate disks,
say 
1 and 
2. The other triangle gives us a third disk, 
3. We will union these all together
in the right order to get what we want. First, note that 
1 and 
3 agree only on a portion of
the green edge, so gluing them together will still give us a disk. Then the boundary of 
2 only
agrees with the boundary of this new disk in one continuous segment—the remainder of the
green edge and the common blue edge. So we can add 
2 in and get our whole disk.

Finally, we could have both red edges pass through the crossing disk. Only one red region
can pass through the a crossing disk, by definition. So for both red edges to pass through the
crossing disk, they must be in the same region. But then at least one crossing, the one that
meets both red edges, has opposite (red) sides agreeing, contradicting Lemma 3.3.5. So this
(cid:3)
case cannot happen, and we are done.

Lemma 3.4.15. A blue rectangle in Ɖ

 with two vertices that correspond to the same crossing
disk and a red diagonal has both crossings associated to the crossing circle of the vertices of
the rectangle.

Proof. Our image has to look like the one in Figure 3.8—the red edge tells us we need at least
two crossings. In either case, look at the two curves 
-
- 
 and 
-
-
. Each of these curves
intersects our knot transversely exactly twice. This means, then, as both the representativity

43































Figure 3.8. (Left) A blue rectangle with a red diagonal in the graph Ɖ

. (Right) A blue rectangle as viewed on
the diagram, where both vertices represent the same crossing circle.

and the edge representativity are at least four, each of these curves must bound a disk. Now,
though, we have two disks who agree at only one segment of the boundary. As such, we can
glue the disks together without risking an annulus, and thus giving us another disk, this one
(cid:3)
with boundary 
- 
 -
-
, and so we are done.

To make sure we are on the right track, it is important to look at why we are studying
these blue-blue-red triangles. We know that we have to have so many adjacent non-trivial
blue bigons. As it turns out, given these blue bigons, we will eventually show that a red edge
must pass through these bigons. If only one red passes through, we get a two collections of
blue-blue-red triangles, with the families adjacent across the red edges. So it’s important to
see what happens here.

Before we begin studying these in earnest, there is an additional lemma that we will need
here. In summary, Lemma 4.17 from [27] tells us that, if we have three adjacent blue-blue-red
triangles that are adjacent at the vertex of the blue edges, two properties must hold. First, no
green edge can intersect an interior blue edge (so, in figure 3.9, edges 
 or 
). Second, if a
green edge meets the interior red edge (so edge 
), the green edge must run to the common
vertex of the triangles. As we have families of blue-blue-red triangles bounding a disk by
Lemma 3.4.14, the exact same proof as in [27] works. With this in mind, we can move on to
looking at our triangles.

44

Lemma 3.4.16. The graph Ɖ

 cannot contain two pairs of three adjacent blue-blue-red
triangles adjacent across the red edges.

Proof. Suppose it did. Then we look at such a group of triangles, as pictured in Figure 3.9
(left). First, we need to show that the two vertices of the triangle group can’t agree (that is,
can’t be related to the same crossing circle). If the vertices do agree, then the blue edges 
, 
 ,
and 
 must be in the same region, and the blue edges 
, 
, and 
 must be in the same region.
From there, we want to show that the endpoints at each of the edges are all associated with




























Ò
















































Figure 3.9. (Left) Two pairs of three adjacent blue-blue-red triangles. (Center and Right) The two possibilities if
the vertices map to the same crossing circle, one with two crossings and one with three.

the crossing circle of the vertices. First, look at the 
-
 crossing and the 
- 
 crossing. As

-
- 
 -
 forms a blue rectangle with two vertices and a red edge down the diagonal, by Lemma
3.4.15, the image lies in a disk with boundary blue-blue-green. We can assume the boundary
is 
-
-green. But then, by definition, the crossing at 
-
 must be associated with the crossing
disk. Then, by two applications of Lemma 3.4.15 (one with 
-
- 
 -
 and one with 
- 
 -
-
),
we get that the crossings of 
- 
 and of 
-
 must be associated to the crossing circle as well.

As we are working with 

,2, this does not automatically give us a contradiction—the
diagram can still work if we have only two distinct crossings, such as in Figure 3.9. However,
suppose there are only two distinct crossings for our triangle. Then we can get several curves
that bound disks in 
 that intersect our diagram exactly twice: 
 ∪ 
 ∪ 
 , 
 ∪ 
 ∪ 
, 
 ∪ 
 ∪ 
,
and 
 ∪ 
 ∪ 
, as follows. To get the last curve, look at the rectangle 
- 
 -
-
. It’s disk must
contain the 
 edge, so take the sub disk bounded by 
 ∪ 
 ∪ 
. Then, for the curve 
 ∪ 
 ∪ 
,
note that 
 must be parallel to 
, so use the sub disk bounded by 
 ∪ 
 ∪ 
, and replace 
 by 
.

45

In similar fashion, but working with the rectangle 
-
- 
 -
 and the edge 
, we can get the other
two curves. But then, as 
(

,2) is weakly prime , we get that there are only two crossings on 
.

We now have two possible options. First, 
(

,2) can’t be a single component knot, as
any attempt to connect two crossings will result in either the unknot (which can’t happen, as

(
(

,2), 
) ≥ 4) or a non-alternating knot on 
. On the other hand, if 
(

,2) is a link,
by following the link around the outside, we can find a curve that completely encompasses
the link. As 
(
(

,2), 
) ≥ 4, this curve must bound a disk on 
. But then we can find an
essential curve that does not intersect our knot diagram at all, a contradiction. So there cannot
be only two crossings for 
(

,2) on 
, and so the two vertices of our group of blue-blue-red
triangles cannot agree.

Now we want to see how the red edges might intersect the crossing disks. First, the red
edge 
 cannot intersect either crossing disk by Lemma 4.17 from [27], as if it did, we would
have to have a green edge intersecting either 
 or 
, contradicting Lemma 4.17 of [27]. Now,
looking at 
, we’ll see that it can only intersect one of our crossing disks. Otherwise, if the
crossing disk corresponding to 
, 
 , and 
 intersects 
, then we must have a green edge running
from 
 to 
. This means that

 and 
 are in the same region. On the other hand, if the
crossing disk corresponding to 
, 
, and 
 intersects 
, we have a green edge running from

 to 
, telling us that 
 and 
 are in the same region. Then we can create two curves, one
from 
, 
, and green, and the other from 
,

 , and green. Each intersects our knot trans-
versely at the 
-
 crossing, but also intersects our knot at either the green edge corresponding
to their respective crossing circles. But then we have a single crossing associated to two
distinct crossing circles, which cannot happen. So then 
 can only intersect, at most, one of
the crossing disks. Assume, then, it doesn’t intersect the crossing disk associated to 
, 
, and 
.

Next, we need to look at the crossings at the endpoints of 
 and 
, which we’ll call 
, and

46

at the endpoints of 
 and 
, which we’ll call 
. We know that 
 must be distinct from the
crossing at the endpoints of 
 and 
 , and 
 must be distinct from this third crossing as well.
So we now want to show that 
 and 
 are distinct from each other. If they are not, then look at
the curve 
 ∪ 
. This will give us a closed curve meeting both crossing disks, a contradiction
to the crossing disk at 
, 
, and 
 not meeting either 
 or 
.









































































Figure 3.10. (Left) Focusing on the crossing circle of 
, 
, and 
, our diagram must look something like this.
(Center) Once we know that 
 is related to the crossing circle, we can draw in some extra information about our
knot. (Right) The dotted lines are disk-bounding curves that will allow us to show 
 and 
 are also related to the
crossing circle.

We want to show that this can’t happen by showing the twist region associated to the shown
crossing circle has three crossings—namely, the three drawn crossings. Note that edges 
 and

 must lie in the same blue region, as they share a vertex (the crossing circle of 
, 
 , and 
)
and are mapped to the same side of that vertex, but also 
 is bounded by 
 ∪ 
 ∪ 
∪ the green
edge of the crossing disk, while 
 lies outside of it, so we should figure out how they meet. As

 and 
 are at opposite sides of a crossing, by Lemma 3.3.5, 
 and 
 are not in the same region
as 
, so they can’t cross over 
. Because 
 is red, neither 
 nor 
 can meet it in the graph. So
the only remaining options are that one of 
 or 
 meets the blue edge 
 or the crossing circle
associated to 
, 
, and 
. In either case, this will give us that 
 and 
 are in the same region as

.

Now we will show that the crossings 
 (at 
-
), 
 (at 
-
), and 
 (at Ò- 
 ) are all associated
to the same crossing circle. As we have already shown that each of these crossings are distinct,
we will get our desired contradiction. First, we show that 
 is associated with the crossing
circle of 
, 
, and 
. To do this, we create a curve 
 starting at 
, following along 
, using

47

the crossing disk’s green edge to get to 
, and then switching over from 
 to 
 to get back to

. Because 
 is bounded by 
 ∪ 
 ∪ 
∪ the green edge, we can take 
 to lie inside the disk on

 bound by the triangle 
 ∪ 
 ∪ 
 so that 
 too bounds a disk on 
. But now, when we put
crossings back in to get 
, we have a disk on 
 whose boundary intersects our knot at exactly
two crossings, and so both must be related to the same crossing circle - namely, the crossing
circle of 
, 
, and 
. So then 
 is associated to this crossing circle.

In order to work with the crossings 
 and 
, we will need to know a bit more about 
 —
namely, that it is in the same region as 
 and 
. Clearly, if 
 crosses 
 or 
, then they must be
in the same region as 
 . So then let’s suppose it doesn’t. Then note that the edge 
 is bounded
by 
 ∪ 
 ∪ 
 ∪ 
. As 
 is blue, it can’t cross the red edges 
 or 
. Likewise, neither 
 nor 

can cross 
 or 
. So, in order for these three blue edges to form the triangles in our graph, the
crossing disk of 
, 
 , and 
 must enclose either 
 or 
. By the same reason the crossing disk
of 
, 
, and 
 can’t meet 
, the crossing disk of 
, 
 , and 
 can’t meet 
 either. So then the
crossing disk must intersect 
. But then, as shown in Lemma 4.17 of [27], we must have the
green edge of the crossing circle intersect 
. Now, with the green edge how it is, we must have

 be in the same region as 
 and 
.


 ∪ 1

2 
 ∪ 1

2 
 ∪ 1

2 
 ∪ 1
2 
;

2 
; and 
 ∪ 1

Our last step is to show that 
, 
, and 
 connect to form bigons, thus are three distinct
crossings in 
(

,2) associated to the same crossing circle, a contradiction. To do this,
2 
 ∪ 1
note that we can form four closed curves intersecting our knot exactly twice: 
 ∪ 1
2 
;

 ∪ 1
2 
 , as seen in Figure 3.10 (right). As we are
assuming that 
(
(

,2), 
) ≥ 4, each of these curves must bound a disk. But then, because

(

,2) is a weakly prime diagram, and because there are crossings outside of these disks, the
intersection of each disk and 
(

,2) must be a single embedded arc. But then, drawing the
result, we see that these three crossings form bigons, and so must be associated to the same
crossing circle. As we are working with 

,2, and at most two crossings can be associated
(cid:3)
with a crossing circle, we’re done.

As mentioned earlier, our want to study adjacent families of blue-blue-red triangles came

48































Ò










Figure 3.11. Three adjacent triangle-square pairs stacked on top of each other, with labelings for the edges.

from a red edge intersecting blue bigons. As this can’t happen, we can’t have just one red edge
intersect our bigons. So, the next case to consider is two or more red edges intersecting the
bigons.

3.5 Triangles and Squares

To review where we are, we know that any non-trivial blue bigon families we have must
be intersected by at least two (parallel) red edges. We can then break these bigons up into
blue-blue-red triangles and red-blue-red-blue rectangles. The triangles come from the ends of
the bigons, and must meet the vertex of the bigon, while the squares are from the interior, and
have a red edge on either end. In this section, we will focus on adjacent triangles and squares,
such as the three in Figure 3.11. Our goal now is to show that we can only have so many of
these triangle-square pairs can be stacked on each other:
Lemma 3.5.1. There cannot be five adjacent triangle-square pairs for the diagram 
(

,2).

Once we have proved this, we can combine it with Lemma 3.4.4, which tells us we must
have at least five adjacent non-trivial blue bigons, to get a contradiction and show that 

,2
is essential. To do this, we will first prove that these triangle-square pairs have to sit inside a
disk on 
, and hence can’t have any topology to them. Next, we prove that, in a few specific
cases, if two blue sides of a crossing meet the same crossing circle, then that crossing must be

49

associated to that crossing circle. With these, we can then use the proof of [27, Section 5],
which we will review below, with little modifications to prove Lemma 3.5.1.

Before we begin, we introduce another definition. We say that a crossing is associated to
a crossing circle if the crossing is in the twist region the crossing circle encircles. There are
two primary ways we can show that a crossing is associated to a crossing circle. First, we can
show that a crossing 
 is part of the same twist region as crossing 
, where we know that 
 is
associated to a crossing circle. We do this in the typical way of showing two crossings belong
to the same twist region, by finding a disk with boundary only intersecting the diagram at 
 and

. The second way we can show a crossing is associated to a crossing circle is by finding a disk
with boundary intersecting the diagram exactly four times: twice transverse to the crossing in
question, and twice where the crossing circle meets the diagram. That is, we find a disk whose
boundary intersects the crossing and follows the crossing circle when it meets the surface 
.
As before, where, when working with triangles, we saw that graphs induced disks on the
projection surface, we will want to do something similar when working with triangle-square
pairs, that is, when we glue one edge of a triangle to an edge of a square.

Lemma 3.5.2. A blue-red rectangle with interior disjoint from the vertices of Ɖ
 induces a
disk on the projection surface 
. Furthermore, a triangle-square pair, with interior disjoint
from the vertices of Ɖ
 induces a disk on the projection surface 
.

Proof. We will prove this by starting with the simplest case, a blue-red rectangle with interior
disjoint from blue and red edges, and expand from there. So assume our rectangle has interior
disjoint from blue and red edges. Then, as we have done previously, we can form a simple
closed curve 
 by taking the boundary of the rectangle, as below.

As red edges can’t intersect red edges, neither of our two red edges can cross any additional
crossings. Likewise, as blue edges can only meet blue edges at vertices, neither of our blue
edges can meet additional crossings. So then 
 intersects our knot exactly four times. As

(
(
), 
) is strictly greater than four, 
 cannot bound a compression disk of 
. On the other
hand, if 
 is an essential curve, the disk it bounds must pass through the projection surface 
.

50

























Figure 3.12. (Left) A blue-red rectangle disjoint from all other edges. The purple curve represents our 
. (Center)
A blue-red rectangle that might intersect a red edge, but is disjoint from blue edges. (Right) A blue-red rectangle
that might intersect any number of blue or red edges.

But such a disk can only pass through 
 at a red or blue edge, which our rectangle is disjoint
from. So then 
 must bound a disk on 
.

Next, suppose our rectangle has interior disjoint from blue edges, but not necessarily from
red edges. We first want to show that our red edges must run entirely through the rectangle,
without any crossings in the interior. First, if any of the red edges meets a crossing, it must then
either meet a blue edge at that crossing or meet no other edges at that crossing (as red edges
can’t meet red edges, and can only meet green at crossing circles). If it meets a blue edge, as
our interior is disjoint from blue edges, that means the crossing must be on the boundary blue
edge, and we are good. If it meets no other edges, then our crossing is in the interior. But then,
we can go from one side of the red edge to the other by going around the crossing, with this
path never intersecting our graph. But then, as the projection of the disk switches from above
to below the projection plane at blue and red edges, we have a single region of our disk that
must be both above and below this plane, which cannot happen. So any crossings our red edge
meets must be on the boundary, and we may proceed.

Now we will proceed by induction. We already know how to proceed if we have no red
edges. Now suppose we can get a disk if our rectangle has at most 
 − 1 red edges in the
interior. Label the blue boundary edges 
, 
 and the red boundary edges 
, 
. Given a rectangle
with 
 red edges in the interior, pick one of them and label it 
. As red edges cannot intersect
red edges, this red edge must be parallel to all of the other red edges, including those on the

51

boundary. Then we can come up with two smaller blue-red rectangles, each with less than

 red edges in the interior, by splitting along 
. The boundaries of these rectangles will be

1 = 
 ∪ 1
2 
 . Each of these curves, by induction, will give
us disks on 
. To get a disk on 
 bounded by 
 ∪ 
 ∪ 
 ∪ 
 , we can glue our two smaller disks
along their common edge 
, and we are done with this case.

2 
 and 
2 = 
 ∪ 1

2 
 ∪ 
 ∪ 1

2 
 ∪ 
 ∪ 1

Finally, we deal with the broad case of a rectangle that is only disjoint from the vertices
of Ɖ
. As Ɖ
 is a connected graph, and blue edges only meet blue edges at vertices/crossing
circles, our blue edges must run all the way from one boundary (red) edge to the other. As it
turns out, this doesn’t actually matter for our proof - we just need the blue edges to run from
one red edge to the next. But if the blue edges didn’t run all the way from one end to the
other, we could find a region where our disk was both above and below the projection plane, a
contradiction.

Once again, we proceed by induction. If our rectangle has no blue edges in the interior, we
can use what we’ve shown above to find a disk. Now suppose we can get a disk if our rectangle
has at most 
 − 1 blue edges, and suppose our rectangle has 
 blue interior edges. Pick a blue
edge and call it 
. As blue edges can only meet blue edges at vertices, 
 must run parallel to
both the boundary edges 
 and 
 , as well as the other interior blue edges. So we can construct
2 
 ∪ 
 ∪ 1
two smaller blue-red rectangles with boundary 
1 = 
 ∪ 1
2
,
with each rectangle having less than 
 blue edges. By induction, we can then find disks on 

with boundaries 
1 and 
2. To get a disk with boundary 
 ∪ 
 ∪ 
 ∪ 
, we can glue the disks
for 
1 and 
2 along 
, and we are done.

2
 and 
2 = 
 ∪ 1

2 
 ∪ 
 ∪ 1

Finally, we will show that a triangle-square pair with no interior Ɖ
 vertices induces a
disk on 
. We can split the triangle-square pair into a blue-blue-red triangle and a blue-red
rectangle, whose boundaries share a red edge. The triangle bounds a disk by 3.4.14, while the
square bounds a disk by above. So we can form a disk for the triangle-square pair by gluing
(cid:3)
these two disks together along their common red side.

Now we will examine what three adjacent triangle-square pairs might look like. As we now

52

know that the triangle-square pairs must all lie on a disc, and thus there is no genus involved, we
don’t have to worry about possibilities beyond the four drawn in Figure 17 of [27]. Likewise,
many of the lemmas proven for 
 = 
2 will also hold for a general projection surface. We
only have to be careful when we are referencing [27, Lemma 3.2(4)], as this part of the lemma
relied on the whole projection surface as opposed to just a small (local) portion of it. So now
we will prove [27, Lemma 3.2(4)] for the two cases in a triangle-square pair.

Before we begin, there is something that must be noted. When the lemmas analogue to the
ones we prove are applied in [27], they are to families of adjacent bigons. As such, the interior
of these bigons will not have any vertices—if not, then any interior vertex must have a blue
edge extending to either vertex of the bigon, breaking the fact that we have adjacent bigons.
So the condition that the triangle-square pairs are disjoint from vertices holds for the ones we
care about, allowing us to apply the following two lemmas in place of [27, Lemma 3.2(4)]:

Lemma 3.5.3. Suppose that, in a triangle-square pair with interior disjoint from vertices, two
blue edges, 
 and 
, meet such that the other endpoint of 
 is the vertex of the triangle-square
pair, and 
 is in the same region as the crossing circle associated to the vertex. Then the
crossing corresponding to this endpoint must be associated to the crossing circle.

Proof. First, note that, by Lemma 3.5.2, our triangle-square pair bounds a disk on 
. Let 

be a blue edge adjacent to 
 that also has an endpoint on the vertex, and 
 the red edge from

 to 
. Then, as 
 and 
 must be in distinct blue regions, and 
 and 
 must be in distinct blue
region, we must have that 
 and 
 are in the same blue region.

There is a simple closed curve connecting the crossing at the endpoint of 
 and 
 and the
crossing circle. Let 
 be the curve that follows 
 from the crossing to the crossing circle, goes
through the crossing circle to 
, and then jumps from 
 to 
 to make it back to the crossing.
We claim that 
 bounds a disk. If 
 is contained entirely within the disk corresponding to
the triangle-square pair, then we are done. Otherwise, at some point, we would have to leave
this disk. By construction, the only time we could have done this is the jump from 
 to 
.

53

That would mean that, although 
 and 
 are in the same blue region, we have to leave the
triangle-square pair to realize this.

While this could be a problem, we are going to show that we can add to our triangle-square
pair disk, and get a disk that includes all of 
. Let 
 be the simple closed curve that follows
the jump from 
 to 
, then the red edge 
 back to 
, and then follows 
 back to where 
 started.





































Figure 3.13. (Upper Left) The graph of the case we are considering. (Lower Left and Right) The possible diagram
configurations of this case. The purple curve following the crossing disk is 
, while the pink curve following 

is 
.

Then 
 must intersect our knot in exactly two places, and so bounds a disk in 
. Further,

 is either entirely contained in the disk of the triangle square pair (as in the lower left picture
of Figure 3.13), or can be split into two arcs, one of which is entirely contained within the
disk of the triangle square pair (as in the right picture of Figure 3.13). This will allow us to
glue the 
 disk to the triangle-square pair disk, and still get a disk. This, then, will give us

 living inside a disk, and so then it must bound a disk. Then we have a disk with boundary
intersecting the crossing and with part of the boundary running along where the crossing disk
meets the surface 
. This means that the crossing must be part of the twist region the crossing
(cid:3)
circle encloses, and so the crossing is associated to the twist region, and we are done.

Lemma 3.5.4. Suppose that, in a group of triangle-square pairs with interior disjoint from
vertices, two blue edges, 
 and 
, both have an endpoint at the vertex, and in the image of the

54

knot, have their other endpoint on opposite sides of the same crossing. Then that crossing is
associated to the crossing circle of the vertex.

Proof. As with above, we can use Lemma 3.5.2 to get a disk that the triangle-square pair
lives on. In particular, both 
 and 
 live on the disk, as does the green edge associated with
the crossing circle. Also, as 
 and 
 are on opposite sides of a crossing, they must also be on
opposite sides of the crossing circle associated to the vertex. So we will create a simple closed
curve that intersects our knot twice at the crossing circle and twice at the crossing of 
 and 
 by
taking 
 to be the curve that follows 
 from the crossing to the crossing circle, then the green
edge across, then 
 back down to the crossing. As 
 and 
 live in distinct blue regions, they
cannot intersect except at the crossing, so 
 is simple. And, as 
 lives in a disk, it must also
then bound a disk. Then we have 
 intersecting our diagram twice at the crossing at twice at
the crossing circle and bounding a disk, and so the crossing must be associated to the crossing
(cid:3)
circle, and we are done.

We may now prove Lemma 3.5.1. The proof does not change much from it’s version of

[27], but we outline the method here.

Proof. The proofs of Lemmas 5.1 through 5.5 of
[27] all work as intended, where we use
the above two lemmas in place of [27, Lemma 3.2 (4)]. To summarize, these lemmas tell us
how three adjacent triangle-square pairs must be placed in our diagram. Of particular note,
the last three of these, Lemmas 5.3 through 5.5, state that, of the four red-blue crossings of the
triangles, three of them must correspond to the vertex of the crossing circle, and they cannot
all be adjacent to each other. That is, either the top two crossings and the bottom crossing are
associated to the vertex, or the bottom two and the top crossing are.

Now, by starting with three adjacent triangle-square pairs, we can start adding additional
triangle-square pairs, and keep track of if the crossings are associated to the vertex of the
crossing circle or not. Suppose we have three adjacent triangle square pairs, and the top two
red-blue crossings of the triangles are associated ot the vertex, as is the bottom crossing. To add

55

a fourth triangle-square pair, we must add it to the bottom, and the crossing we add must also be
associated to the crossing. If we try to add it to the top, then the three top triangle-square pairs
would have either three adjacent crossings all associated to the vertex, or only two crossings
associated to the vertex, both contradicting Lemmas 5.3 through 5.5 of [27]. Next, when we
try to add a fifth triangle-square pair, we must add it to the bottom agagin. However, no matter
if the added crossing is associated to the vertex or not, the botttom three pairs will either have
only two associted crossings or three adjacent associated crossings, both contradictions. So
(cid:3)
then we can’t have five adjacent triangle-square pairs, and so we are done.

We may use Lemma 3.5.1 to get the following result.

Proposition 3.5.5. The graph Ɖ
 cannot contain five or more adjacent non-trivial bigons.

Proof. Suppose Ɖ
 contains more than five adjacent non-trivial bigons. Then, by Lemma
3.4.9, the bigons cannot be disjoint from green and red. If the bigons are disjoint red, then they
must intersect a green edge. However, as we cannot have blue-green bigons or blue-blue-green
triangles ( [27, Lemmas 4.4, 4.16]), so the green edge must go from one vertex to another.
If the green edge meets only one vertex, we have a green monogon, which cannot happen
( [27, Lemma 4.16] says we can remove such monogons), while if it meets two distinct vertices
of our bigon, then the blue edges must be trivial by definition, a contradiction. So the blue
bigons must meet the red surface.

As there are no blue-red bigons (Lemma 3.4.7), any red edge must run straight through
all five bigons, giving us a collection of blue-blue-red triangles. By Lemma 3.4.16, we can’t
have three or more adjacent such triangles, so another red edge must run parallel, giving
us adjacent triangles and squares instead. However, by Lemma 3.5.1, we cannot have five
adjacent triangle-square pairs, so we get a contradiction, and thus Ɖ
 cannot have five adjacent
(cid:3)
non-trivial bigons.

56

3.6 Completing the Proof of the Theorems

We are now ready to prove Theorem 4.1.1. Once we do, we will then introduce some

additional lemmas to prove Theorem 4.2.4

We begin by showing that 
 : 

,2 → 
 \ 
 is 
1-injective:

Theorem 3.6.1. Let 
 : 

,2 → 
 \
 be the immersion of 

,2 into 
 \
. Then this immersion
is 
1-injective, provided 


 ≥ 121.
Proof. If not, then we can get a disk 
 : 
 → 
 \ 
, such that 
 |

= 
 ◦ ࡁ, for some essential
loop ࡁ, and graph Ɖ
 = 
−1( 
 (

,2)), with vertices corresponding to crossing circles. These
vertices have valence 2
 
 if in the interior, or 
 
 +1 in the boundary, where 2
 
 is the number of
crossings removed from the twist region when constructing 

,2. In particular, 2
 
 ≥ 120. By
Lemma 3.4.4, as the minimum number of crossings removed is 


 ≥ 2(cid:100)


/2(cid:101) − 2 ≥ 120,
we must then have more than five adjacent non-trivial bigons. This, however, contradicts
(cid:3)
Lemma 3.5.5 above, and so 
 must be 
1-injective.

Likewise, the proof of boundary-
1-injectivity can be proved as in [27]. As before, instead
of reproving every lemma we need, we will instead prove [27, Lemma 3.2(4)] in the specific
case we need, and replace the uses of Lemma 3.2(4) with this new one:

Lemma 3.6.2. Suppose a triangle with two blue edges and one edge on 

(
) does not meet
the red edge. Then the two blue edges must be on opposite sides of a crossing, and that crossing
must be associated to the crossing circle of the vertex.

Proof. As the two blue edges of the triangle meet at the same vertex, they must map opposite
sides of the crossing circle. However, as the third side is the knot strand, and the triangle does
not meet red, the strand must run through a single crossing, proving the first part. To see that
the crossing is associated to the crossing circle, we can construct a closed curve intersecting
the knot exactly four times: Take one of the blue edges from the crossing circle down to the
crossing, then follow the other blue edge up to the crossing circle, and pass over to the start

57

through the green edge. We can view the interior of this disk as being the interior of our triangle.
We know immediately that this curve cannot bound a compressing disk—
(
(

,2), 
) > 4.
So the other option is that it is essential. However, if this curve is essential and does not
bound a compression disk, the disk, and thus the interior of the triangle, must pass through the
projection surface. It can only do this where red and blue meet. As the triangle is disjoint from
red, this cannot happen, so the curve must bound a disk on the projection surface. But then
we have a disk whose boundary passes only through the crossing in question and the crossing
(cid:3)
circle, so the crossing is related to the crossing circle, and we are done.

From here, the remaining lemmas follow nicely. To summarize, first, by looking at how
blue-blue-

(
) triangles can be placed in the diagram, we get that, if we have three adjacent
triangles in Ɖ
, they must meet the red surface. This is the lemma that references [27, Lemma
3.2(4)], and so we can use the above lemma in place of it. Further, by careful examination,
[27, Lemma 6.2] tells us that three adjacent such triangles must meet the red surface at least
twice. This, then, gives us blue-red triangle square pairs, and we can use Lemma 3.5.1 to
say that we can’t have five adjacent triangle-square pairs, and thus, can’t have five adjacent
blue-blue-

(
) triangles.

Now, we can prove the 
-
1-injective:

Theorem 3.6.3. Let 
 : 

,2 → 
 \ int(
(
)) be the immersion of 

,2 into 
 \ int(
(
)).
Then this immersion is 
-
1-injective, provided 


 ≥ 121.
Proof. If not, then we can get a disk 
 : 
 → 
 \int(
(
)), such that 

 is a concatenation of
two arcs, one mapped into 

(
), and the other an essential arc in 

,2 [27, Lemma 2.2]. By
Lemma 3.4.6, as the minimum number of crossings removed is 


 ≥ 2(cid:100)


/2(cid:101) − 2 ≥ 120,
we must have more than five adjacent non-trivial bigons or five adjacent triangles with an
edge on 
−1(

(
)). However, as mentioned above, by [27, Lemma 6.3], we cannot have
five adjacent such triangles. Also, by Lemma 3.5.5, we cannot have five adjacent non-trivial
(cid:3)
bigons. This gives us a contradiction, so 
 must be 
-
1-injective.

58

Putting Theorems 3.6.1 and 3.6.3 together, we get:

Theorem 3.1.1. Let 
 : 

,2 → 
 \ 
 be the immersion of our blue twisted surface 

,2 into
our knot complement 
 \ 
, where 
 has no boundary components. Then this immersion is

1-injective and boundary 
1-injective provided that 


 ≥ 121.
Proof. We first prove that 
 : 

,2 → 
 \ 
 is 
1-injective. If not, then, by [27, Lemma 2.1]
(and the remarks at the beginning of section 5 about why this carries over to our general case),
we get a map of a disk 
 : 
 → 
 \ 
 with 
 |

= 
 ◦ ࡁ for some essential loop ࡁ in 

,2,
with Ɖ
 = 
−1( 
 (

,2)) a collection of embedded closed curves and an embedded graph in

. Each vertex in the interior of 
 has valence a non-zero multiple of 2
 
 > 121, where 2
 

is the number of crossings removed, and each vertex in the exterior has valence 
 
 + 1 > 60.
Then, as we are working with a graph with exactly the same properties as the graph in Section
2 of [27], we can get that Ɖ
 has more than (


/18) − 1 > 5 adjacent non-trivial bigons. (cid:3)

3.6.1 Properties of Twisted Surfaces
Now that we have proven that our twisted surfaces are essential surfaces in 
 \ 
, we can get a
bit more out of them as well. Ultimately, we want to show the following:
Theorem 4.2.4. Let 
 be the disjoint union of our twisted surfaces 

,2 (cid:116) 

,2. Suppose that
two distinct essential arcs in the surface 

,2 have homotopic images in 
 \ 
, but not 

,2.
Then the two arcs are homotopic in 

,2 into the same subsurface associated with some twist
region of 
2.

This theorem is proven in a series of lemmas, all but one of which require little to no
alteration in statement and proof. We will go through each of the lemmas, and give a sketch of
why they still hold for our case. Recall that the subsurface associated to a twist region of 
2
is the intersection of a surface with a regular neighborhood of the twist region.

The first step in our proof of Theorem 4.2.4 is to get a disk to work on.

59

Lemma 3.6.4. Suppose homotopically distinct essential arcs 
1 and 
2 in 

,2 map by 
 :


,2 → 
 \ 
 to homotopic arcs 
1 and 
2 in 
 \ 
. Then there is a map of a disk

 : 
 → 
 \ int(
(
)) with 

 expressed as four arcs, with opposite arcs mapping by 
 to

1 and 
2, and the other two arcs mapping to 

(
). Moreover, Ɖ
 = 
−1( 
 (

,2)) forms a
graph on 
 whose edges have endpoints either at vertices where 
(
) meets a crossing circle,
or on 
−1(

(
)) on 

. Each vertex has valence either a multiple of 2
 (if in the interior
of 
), 
 + 1 (if in the interior of an arc on 

 that maps to 

,2), or 1 (if on an arc that maps
to 

(
)), where 2
 is the number of crossings removed from the twist region.
Proof. As 
1 and 
2 are homotopic, we immediately get a disk 
 : 
 → 
 \ int(
(
)) with
the correct edges. We then modify this disk to get the remaining properties.

First, we lift 
1 and 
2 to the orientable double cover of 

,2 By pushing 
1 and 
2 in
the transverse direction, and using the fact that 

(
) is transversely orientable, we can get
the map of 
 on a neighborhood of 

. Then we can extend 
 over the rest of 
, making it
transverse to all crossing circles and to 
 (

,2).

Now look at Ɖ
 = 
−1( 
 (

,2)). As 

,2 is embedded in 
 \ 
 except as crossing circles,
Ɖ
 must consist of embedded closed curves (where 

,2 intersects 
(
) away from crossing
circles or 

(
)), arcs with endpoints on vertices (where 

,2 intersects a crossing circle),
and arcs with endpoints on 

(
).

As a vertex of Ɖ
 in the interior of 
 comes from where 
(
) intersects a crossing circle,
the valence is equal to the number of crossings removed, by construction of the twisted surface


,2. If we have a vertex on an arc in 

, it maps either to 

,2 or 

(
). If it maps to


,2, then a neighborhood of the arc must map to half a meridian disk of a crossing circle, and
so has valence 
 + 1, where we removed 2
 crossings from the corresponding twist region in
our construction of 

,2. If it maps to 

(
), however, because the arc must be transverse to
(cid:3)


,2, the vertex must have valence 1.

We can also take our disk to be minimal, in the sense that it is a disk that gives us the fewest

60

number of vertices of Ɖ
 (the graph coming from how the disk intersects our blue surface) and
Ɖ

 (the graph coming from how the disk intersects both our blue and red surfaces). Next,
by examining graphs in a disk that satisfy the properties above, we get an analogue to Lemma
3.4.6:

Lemma 3.6.5. Suppose Ɖ
 contains at least one blue vertex. Then either Ɖ
 has more than
(


/24) − 1 adjacent non-trivial bigons, or more than (


/24) − 1 adjacent triangles, each
with one arc on 
−1(

(
)).

The proof of this lemma is a graph theoretic proof, and so does not rely on the ambient

space 
 or the twisted surfaces 

,2 or 

,2.
Proof (Sketch). First, suppose we have a graph Ɖ on 
 × 
 with no monogons and only valence
one vertices on 
× 

, with interior vertices with valence at least 
 and vertices on 

×(
− 

)
have valence at least 
/2 + 1. Then Ɖ must have more than 
/8 − 1 adjacent bigons or more
than 
/8 − 1 adjacent triangles with an edge on 
 × 

.

Next, we can modify our graph Ɖ
 coming from Lemma 3.6.4 by looking at subdisks,
doubling along edges, and collapsing certain bigons and triangles so that we satisfy the above
statement, with 
 = 


 is the minimal number of crossings removed from a twist region.
When we collapsed bigons or triangles, we made it so at most three bigons or triangles in Ɖ

are associated to a bigon or triangle in this new graph. As such, we get (


/24) − 1 adjacent
non-trivial bigons in Ɖ
, or more than (


/24) − 1 adjacent triangles, and are done.
(cid:3)

Using this, we can prove:

Lemma 3.6.6 (Lemma 7.6 [27]). If 


 ≥ 121, then the graph Ɖ
 contains no blue vertices.
That is, 
(
) meets no crossing circles.

Proof. As shown in Lemma 3.4.8, Ɖ
 has no monogons. If Ɖ
 has any trivial arcs in 

,
then that arc must bound a disk 
 that is a subset of 

,2. By sliding 

 along 
, we can
remove such an arc, so we may assume that Ɖ
 has no trivial arcs in 

. Then, if there is a

61

blue vertex in Ɖ
, we can use Lemma 3.6.5, and get more than ((


 − 1)/24) − 1 adjacent
non-trivial bigons or adjacent triangles with an arc on 
−1(

(
)). When 


 ≥ 121, we
get at least 5 such bigons or triangles. By Lemma 3.5.5, we cannot have five such bigons, and,
as discussed before, by [27, Lemma 6.3], we cannot have five adjacent triangles, and so Ɖ

(cid:3)
cannot have any blue vertices.

So now we can start looking at how exactly the surfaces 

,2 and 
2 (the red surface arising
from 
2) intersects our disk. First, with no blue vertices, Ɖ
 consists only of blue arcs with
end points on the north and south edges of the disk, corresponding to 

(
). If both of the
edges are on the same edge, by Theorem 3.6.3, because the disk is 

1-injective, the blue
edge is trivial and we can find a disk that doesn’t meet that edge. So blue edges must run from
north to south. Likewise, the red edges of Ɖ

 are only arcs that go from one distinct edge
to another, from Lemma 3.4.7 (no blue red bigons), and the fact that the red checkerboard
surface 
2 is essential.

Now we need to show that the red arcs run along opposite edges, either east to west (
1 to


2), or north to south (

(
) to 

(
)):

Lemma 3.6.7. The graph Ɖ

 consists of red and blue arcs with endpoints on opposite sides
of 
 (north-south or east-west). Further, blue arcs run north to south.

Proof. In the case of blue edges, we are done, as we know the arcs run from north to south as
discussed above. For red arcs, if they don’t have endpoints on opposite sides, then they must
form a triangle with one endpoint on either the north or south side, and the other on the east
or west side. This gives us a blue-red-

(
) triangle. The blue and red arcs must meet at a
crossing of our arc, so, assuming the triangle region maps above the projection plane, there are
two ways this could happen. Either the red edge runs from one crossing (where it meets the
blue arc) to near another crossing, or runs from one crossing to 

(
), with the blue arc on
the opposite side of this meeting.

62

Figure 3.14. Two ways a triangle with one red on red, one edge on blue, and one edge on the knot can sit in the
diagram.

Before we can use primality arguments to show neither of the cases can happen, we need
to show that we can get a disk from this triangle. By joining the blue and red arcs together at
the endpoints that don’t meet, either by sliding along 
 or connecting across a crossing, we get
a closed curve that intersects our knot diagram twice. As 
(
(
), 
) ≥ 4, this curve cannot
be essential, and so must bound a disk on 
.

From here, the remainder of the proof follows as in [27]. If we are in the left case of
Figure 3.14, connecting the endpoints will give us a blue-red bigon. But then, as in the proof
of Lemma 3.4.7, we can get a contradiction to primality. On the other hand, if we are in
the other case, because 

,2 is weakly prime, there must be no crossings in the interior of
the closed curve we constructed. As such, both the red and blue arcs must be homotopic to a
portion of 

(
). We can then use a homotopy to slide our disk 
 away from where the blue
and red arc intersect. This will give us one less vertex on 
, and so contradicts the minimality
of 
. In either situation, we get a contradiction, so the red arcs must run either north-south or
(cid:3)
east-west.

As red arcs cannot intersect red arcs, we must also have that all red arcs run in the same

direction—either all north-south or all east west. We consider the first case now:

Lemma 3.6.8. If the graph Ɖ

 cuts 
 into a subrectangle with two opposite sides mapped
to 

(
), one side on blue, one side on red, and the interior disjoint from blue and red,
then the blue and red sides of that rectangle are homotopic to the same crossing arc, and the
homotopies can be taken to lie entirely in 

,2 and 
2.

Proof. First, note that such a rectangle, with interior disjoint from both blue and red, must be

63

mapped either completely above or below the projection surface. Suppose it is above. Then
there are three cases to consider. First, the blue and red arcs could straddle over crossings at
both ends. Second, one set of endpoints could straddle an over crossing, while the other set
lies on opposite sides of a strand. Third, both sets of endpoints lie on opposite ends of strands.

Figure 3.15. The three ways a red-blue rectangle can sit in our diagram

Note that in all three of these cases, by connecting the red and blue arcs, we get a curve
that intersects our knot exactly twice and so, as before, must bound a disk on 
. The first case,
with two over crossings, as in Lemma 3.4.7 (there are no blue-red bigons), violates the fact
that our diagram is Weakly Prime. The third case, with two strands, can be homotoped away
from as in Lemma 3.6.7. That leaves us with just our second case. In this case, both the blue
and red arc are homotopic to the same crossing arc (the over crossing they straddle), and the
homotopies lie entirely in their corresponding surfaces, either 

,2 or 
2, so we are done. (cid:3)
From here, with no change to proof except to use 
(
(
), 
) ≥ 4 to show the curves in
question bound a disk, by carefully examining how rectangles with two colored sides (either
both blue, both red, or blue and red) and two 

(
) sides can give us homotopies, we get the
following result:

Lemma 3.6.9. If the graph Ɖ

 consists of disjoint red and blue arcs on 
, all running north
to south, then 
1 and 
2 are each homotopic in the blue surface to arcs in the same subsurface
associated with a twist region of 
2.

The 
1 and 
2 here are the two homotopic arcs in 
 \ 
 from Lemma 3.6.4.

Proof (Sketch). Our arcs split our disk into several sub-rectangles, with east and west sides
colored either red or blue, and the north and south side on 

(
).
If both colored sides

64

are blue, the subrectangle must be mapped to a single side of the projection surface, and the
north and south sides must run over crossings (or else we could homotope 
 away to remove
intersections). This means our subrectangle gives us a simple closed curve that meets our
diagram at two crossings with the subrectangle on only one side of the projection surface, and
thus gives us a disk. Because 

,2 is weakly blue-twist reduced, this disk bounds a collection
of red bigons, and so we can isotope the blue arcs to lie in the same subsurface. In a similar
fashion, having to be a bit more careful as we don’t know 

,2 is weakly red-twist reduced,
we get the same result for when both sides are red.

If the two sides are colored differently (one red, one blue), then Lemma 3.6.8 allows us to
homotope the two sides into the same subsurface. Putting it all together, two successive arcs
are homotopic into the same subsurface, in either 

,2 or 
2, as appropriate. Expanding this
out, we get that 
1 and 
2 must be homotopic in the blue surface into the same subsurface. (cid:3)

For the last lemma, we need an analogue to the polyhedron decomposition of alternating

knots. We will use the chunk decomposition of Howie and Purcell [23]:

Definition 3.6.10. A chunk 
 is a compact, oriented, irreducible 3-manifold with boundary


 containing an embedded non-empty graph with all vertices having valence at least 3. The
graph separates 

 into regions called faces. If the region is disjoint from the graph, it is
called an exterior face, otherwise, it is an interior face.

A truncated chunk is a chunk where a regular neighborhood of each vertex of the edge

graph has been removed. This leaves a boundary face surrounded by boundary edges.

A chunk decomposition for a 3-manifold 
 is a decomposition of 
 into chunks, such
that 
 is obtained by gluing chunks by homeomorphisms of non-exterior, non-boundary faces
with edges mapping to edges homomorphically.

Howie and Purcell showed in Proposition 3.1 in the same paper that weakly generalized

alternating knots admit a chunk decomposition:

65

Theorem 3.6.11. [Proposition 3.1 [23]] Let 
 be a compact, orientable, irreducible 3-manifold
with no boundary components, containing a generalized projection surface 
. Let 
 be a
weakly generalized alternating knot with a diagram 
(
) on 
. Then 
 \ 
 can be decomposed
into pieces such that:

1. Pieces are homeomorphic to components of 
 \ 
(
), except each piece has a finite set

of points removed from 
(
 \ 
(
)), namely the ideal vertices below.

2. On each copy of 
, there is an embedded graph with vertices, edges, and regions

identified with the diagram graph 
(
). All vertices are ideal and 4-valent.

3. To obtain 
 \ 
, glue pieces as follows. Each region of 
 \ 
(
) is glued to the
corresponding region on the opposite copy of 
 by a homeomorphism that is the identity
composed with a rotation along the boundary. The rotation takes an edge of the boundary
to the nearest edge in the direction of that boundary component’s orientation.

4. Edges correspond to crossing arcs, and are glued in fours. At each ideal vertex, two

opposite edges are glued together.

With this in mind, we can generalize the following lemma, initially proven by Lackenby
[25]. Before we begin, note that an essential square is a simple closed curve, intersecting

(
) exactly four times, thus dividing the curve into four sides, each of which intersects a
different edge of the diagram. As we will be working in Ɖ
, and essential squares are blue-red
rectangles, we can use Lemma 3.5.2 to assume that these squares must bound a disk in 
.

Lemma 3.6.12. Let 
 be a chunk in a chunk decomposition for 
 \ 
, where 
 is a weakly
generalized alternating knot on 
 with 
(
(
), 
) ≥ 4. Then, if 
 and 
 are essential squares,
each bounding a disk on 
 and isotoped to minimize |
 ∩ 
|, they intersect either zero or two
times. If they intersect two times, then the points of intersection lie in distinct regions of the
diagram with the same color.

66

Proof. We can color the faces of our chunk blue or red, depending on how it meets the
checkerboard coloring of 
 \ 
. An essential square intersects the knot projection at four
distinct points, and so the square must have two blue edges and two red edges, each opposite
each other. As 
 and 
 have been isotoped to have minimal intersection, there are at most four
points of intersection, one at each edge. If there are four points of intersection, then 
 and

 are isotopic to each other, and can be pushed off to have no intersection. In addition, they
cannot intersect an odd number of times, as both bound disks on 
. If they do, look at how

 as a curve intersects 
. It must cross the boundary of 
 exactly three times. As 
 bounds a
disk on 
, however, this is impossible. So 
 and 
 intersect either zero or two times.

Suppose they intersect exactly twice, and one of those intersection points happens on red,
and the other on blue. Because 
 and 
 must intersect the knot where blue meets red, if we
look at the arcs of 
 \ 
 and 
 \ 
, they must intersect our knot either once or three times.
Look at the two arcs that intersect the knot once, each, and glue them together to get a curve
intersecting our knot exactly twice. As 
(
(
), 
) ≥ 4, this curve must bound a disk on 
.
Then, as 
 is weakly prime, because the curve intersects the knot exactly twice and bounds a
disk, the disk must intersect our knot in a single arc with no crossings. We can then use this
arc to homotope 
 and 
 away from each other, contradicting the fact that our intersection was
minimal. So, if 
 and 
 intersect, they must intersect in distinct regions of the same color. (cid:3)

As a consequence to this, we get the following result:

Lemma 3.6.13. Let 
 be a chunk in a chunk decomposition for 
 \ 
, where 
 is a weakly
generalized alternating knot on 
 with 
(
(
), 
) ≥ 4. Let 
 and 
 be essential squares in

, each bounding a disk on 
 and moved by normal isotopy to minimize |
 ∩ 
|. Suppose 

and 
 pass through the same red face 
, and that edges 
 ∩ 
 and 
 ∩ 
 differ by a single
rotation of 
. Then exactly one of the following two conclusions holds:

1. Each of 
 and 
 cuts off a single ideal vertex in 
, and 
 and 
 are disjoint.

67

2. Neither 
 nor 
 cuts off a single vertex in 
. The two essential squares intersect in 


and in another face 
(cid:48).

Proof. First, if 
 cuts off a single ideal vertex, then so must 
, and so do not intersect in 
. By
Lemma 3.6.12, then, if 
 and 
 intersect, it must be in a blue face. As 
(
(
), 
) ≥ 4, 
 can
only meet a given blue face once (otherwise we could find an essential curve on 
 intersecting
our knot exactly twice). Then, if 
 and 
 do intersect in a blue face (and thus in two blue
faces), then 
 ∩ 
 and 
 ∩ 
 must be parallel to each other. We could then isotope 
 and 

away from each other in the blue faces so that 
 and 
 are disjoint.

If 
 does not cut off a single vertex, then neither does 
. As 
 ∩ 
 and 
 ∩ 
 differ by a
single rotation, they must intersect in 
. By Lemma 3.6.12, then, they must intersect in two
red faces, 
 and 
(cid:48), and so we are done.
(cid:3)

We can use this to get our final needed lemma. There is one last definition to introduce.

Definition 3.6.14. An essential product disk for the blue checkerboard surface 
 of a knot 

is an essential disk properly embedded in 
3 \ int(
(
)), with boundary a rectangle with two
oppostie sides on 
(
) and two opposite sides on 

(
).

Then note that the disk of Lemma 3.6.4 is an essential product disk, so if we can prove the
following lemma for essential product disks in general, it will also apply to the specific case
we are concerned with.

Lemma 3.6.15. Let 
1 and 
2 denote the boundary arcs on the blue surface in an essential
product disk for the blue checkerboard surface of 

,2. Then there is a subsurface associated
with a twist region of the diagram of 
2, and arcs 
1 and 
2 in that subsurface, such that 
1
is homotopic in 

,2 to 
1, and 
2 is homotopic in 

,2 to 
2.

Proof. Let 
 be the essential product disk. If it is disjoint from the red surface, then the blue
edges of 
 are homotopic to arcs in a subsurface associated to one twist region of 

,2. This
follows from [27, Lemma 7.10]. To summarize, if we have a rectangle made up of two blue

68

arcs and two edges on 

(
), and interior disjoint from red or blue, the rectangle must be
mapped to a single side of the projection surface. If the 

(
) edges don’t run over a crossing,
then we could homotope the rectangle away from the projection surface, and thus Ɖ

 was
not minimal, a contradiction. So then the edges on 

(
) must run over crossings, and thus
we can get a curve that intersects the diagram at exactly two crossings. Then this curve must
bound a series of red bigons, and so we may isotope the blue arcs into a neighborhood of these
bigons, and are done.

If 
 does meet the red surface, then, by Lemma 3.6.7, either the red edges all run north-
south (

(
) to 

(
)) or east-west (from blue edge to blue edge).
In the former case,
Lemma 3.6.8 gives us our result. So we may assume we are in the latter case, and red edges
run from blue edge to blue edge, cutting the rectangle into blue-red-blue-red sub-rectangles.

Figure 3.16. (Left) An essential product disk with blue sides 
1 and 
2 intersected horizontally by red edges.
(Right) Two sub-rectangles whose shared red edge meets two separate crossings. (Center) Two sub-rectangles
whose share red edge meets a single crossing. We want to prove that this is what happens for all adjacent
sub-rectangles.

By Theorem 3.6.11, we get a chunk decomposition for our knot, with pieces homeomorphic
to components of 
 \ 
(
), except for ideal vertices corresponding to crossings. Each face of
the chunk can be colored either red or blue, depending on what color it meets on 
. As 
 is
a connected surface, 
 \ 
(
) has either one or two components. Each of our sub-rectangles
that make up 
 must be contained entirely in a single chunk. Further, if two rectangles glue
to each other along a red edge, the rectangles are either not in the same chunk, if we have two
chunks, or meet two separate red faces that glue to each other, if there is only one chunk. We
proceed based on whether we have only one chunk or two.

If we only have a single chunk, then look at the first two rectangles in 
, 
1 the north

69

most one that meets 

(
), and 
2, the one below that which glues to the red edge, 
 of

1. Looking at the side of 
1 that meets 

(
), it must run through an ideal vertex of the
chunk decomposition, and so we can push it off into the red face so that it cuts off a single
vertex. Consider the other side of 
1 now, which glues by a clockwise or counterclockwise
turn to 
2. While the red side of 
1 and the red side of 
2 are not on the same face, we can
create a new rectangle ¯
2 by copying 
 in 
2 into the same face of 
 in 
1, after rotating by
the corresponding turn, and then copying the rest of 
2 over. By Lemma 3.6.12, if the two
copies of 
 (for 
1 and ¯
2) intersect, then 
1 and ¯
2 must also intersect in the other red face.
However, as the other edge of 
1 cuts of an ideal vertex, we can homotope it so it does not
intersect the other edge of ¯
2. This means, then, that 
1 and ¯
2 cannot intersect in a red face.
By Lemma 3.6.13, then, both 
 in 
1 and in ¯
2 must cut off a single vertex, and so 
 in 
2
must also cut off a single vertex. Continuing by induction, each rectangle making up 
 must
have sides in the red faces cutting off a single vertex.

Now suppose we have two chunks. Once again, we start by looking at the first two rectangles
of 
, 
1 and 
2, which glue by a single turn along a red edge 
. The first step is to get both
of these rectangles into the same chunk. Like before, we can homotope the north end of 
1 so
that it cuts off a single vertex. As 
1 and 
2 are in separate chunks, we can impose 
2 into
the same chunk as 
1 by copying 
 in 
2 into the chunk containing 
1 under a clockwise turn,
and continuing for the rest of 
2. We call this new rectangle ¯
2. Now, with both rectangles in
the same chunk, we can use the same argument as in the single chunk case, and get that 
 of
both 
1 and 
2 must cut off a single vertex, and, by induction, each red edge of 
 must cut off
a single ideal vertex.

In either case, each rectangle of 
 maps to a region of the diagram meeting 
(

,2) exactly
four times, adjacent to two crossings (one for each red edge). As 
(

,2) is weakly blue twist
reduced by Lemma 3.3.7, this region must bound a string of red bigons, and so the boundaries
all lie in the same twist region. By applying this to all the rectangles in 
, we get that the two
(cid:3)
blue edges must lie in a neighborhood of the same twist region, and so are done.

70

Now Theorem 4.2.4 follows as a consequence to several of the above lemmas.

Proof of Theorem 4.2.4. Call our two arcs 
1 and 
2. Then, by Lemma 3.6.4, we get a disk

 : 
 → 
 \ 
 and a graph Ɖ
. Then, by Lemma 3.6.6, there are no blue vertices in our
graph. Next, Lemma 3.6.7 tells us that blue edges must run north to south, while red edges
either run all north to south or all east to west. If they all run north to south, then Lemma 3.6.9
tells us 
1 and 
2 are homotopic into the same subsurface. If the red edges all run east to west,
(cid:3)
then Lemma 3.6.15 gives us the same result, and we are done.

71

CHAPTER 4

CUSP AREA

4.1 Introduction

This chapter is based on work that will be published in Algebraic & Geometric Topology [7].
The aim of this chapter is to show that weakly generalized alternating knots, under certain
conditions, have two-sided bounds for the cusp area based on the twist number of a diagram
and 
(
), thus generalizing Lackenby and Purcell’s results for alternating knots on 
2 ⊂ 
3
[26].
Theorem 4.1.1. Let 
 be a closed, irreducible 3-manifold and 
 ⊂ 
 a closed surface with

(
) ≤ 0 and such that 
 \ 
(
) is atoroidal. Suppose that 
 is a weakly generalized
alternating knot with a projection 
(
) ⊂ 
 that is weakly twist reduced, and we have

(
(
), 
) > 2, and 
(
(
), 
) > 4. Then 
 \ 
 is hyperbolic, and we have

5.8265514 × 10−7
2 (120 − 
(
))6 (tw
(
) − 
(
)) ≤ 

(
) ≤ (360tw
(
) − 3
(
))2
where 

(
) is the cusp volume of 
 and tw
(
) the twist number of 
(
).

tw
(
)

,

This result, then, allows us to investigate certain properties of the knot by looking at the

cusp, as we talk about next.

Given a hyperbolic knot complement 
 = 
 \ 
 there is a well defined notion of its
maximal cusp 
. The boundary of C, denoted by 

, inherits a Euclidean structure from the
hyperbolic metric. Given an essential simple closed curve 
 ⊂ 

 (a.k.a. a slope), the length
of 
 is the Euclidean length of the unique geodesic in the homotopy class of 
. We well
denote this length by ࡁ(
).

Definition 4.1.2. If 
 is a knot in a 3-manifold 
, take a regular torus neighborhood of 
. The
meridian 
 is the curve that bounds a disk in 
 that intersects 
 exactly once.

72

Using Theorem 4.1.1 and a result of Burton and Kalfagianni

[11], we can obtain the

following corollary:

Corollary 4.1.3. Let 
 be a closed, irreducible 3-manifold and 
 ⊂ 
 a closed surface with

(
) < 0 and such that 
 \ 
(
) is atoroidal. Suppose that 
 is a weakly generalized
alternating knot with a projection 
(
) ⊂ 
 that is weakly twist reduced, and we have

(
(
), 
) > 2, and 
(
(
), 
) > 4. Let 
 be the meridian and 
 any non-meridian slope
on maximal cusp of 
 \ 
. Then, we have

(cid:19)
(cid:18) 5.8265514 × 10−7
ࡁ(
) ≤ 720 (tw
(
) − 
(
))
ࡁ(
) ≥

tw
(
)

720 (120 − 
(
))6

tw
(
),

where tw
(
) is the twist number of 
(
).

In addition, we are able to use Theorem 4.1.1 with a theorem of Howie and Purcell and a

theorem of Futer, Kalfagianni, and Prucell to prove the following result:

Theorem 4.1.4. Let 
 be a weakly generalized alternating knot on a surface 
 ࣔ 
2 in a
closed manifold 
 with conditions as in Theorem 4.1.1. Suppose that

(cid:16)

2.32928 × 1010(cid:17) (120 − 
(
))6 .

tw
(
) >

Then, any manifold 
 obtained from non-meridional surgery along 
 is hyperbolic and


8
4

(tw
(
) − 
(
)) ≤ 1
2

vol(
 \ 
) ≤ vol(
) < vol(
 \ 
).

Here 
8 ≈ 3.66386 is the volume of a regular hyperbolic ideal octahedron.

We will proceed as follows. First, in Section 2, we will prove the theorems and lemmas
we will need to show our main result. Then, in Section 3, we will put these all together to
prove our main result, Theorem 4.1.1. Finally, in Section 4, we will go through the proofs of
Corollary 4.1.3 and Theorem 4.1.4.

73

4.2 Proofs for Calculations

Many of the statements below will hold as long as we have an essential spanning surface
for our knot, including our checkerboard surfaces. Occasionally, we will specialize to the
essential twisted surfaces introduced in the previous chapter. This will allow us use numbers
specific to the twisted surfaces that will give us better results.

For the lower bound we will adapt the approach of Lackenby and Purcell

[26] to our
setting. We note that some of the results we reference from [26], and used in there for knot
complements in 
3, apply immediately to knot complements in any closed 
. We will restate
them and we will review them and their proofs briefly below.

Before we start the proofs, there is an important definition that we will use throughout.
Most geodesics in a hyperbolic manifold are bi-infinite, but we will want to be able to talk
about their lengths. As such, we use the following definition to get around this complication:

Definition 4.2.1. Let 
 be a (bi-)infinite geodesic in a hyperbolic surface 
, with both ends in
a horoball neighborhood of the cusps of 
, 
. Then 
 ∩ 
 is a collection of closed intervals
(if 
 enters and exits 
) along with two helf-open intervals (the ends). The length of 
 with
respect to 
 is the hyperbolic length of 
 minus the two half-open intervals.

If 
 is an arc with boundary on 

, we first find a geodesic 
 such that 
 is homotopic to


 in 
, and define the length of 
 with respect to 
 to be the length of 
 with respect to 
.

The next result concerns an estimate of geodesic arcs on hyperbolic surfaces. We will

apply the result specifically to our twisted surfaces, but it works in more general settings.

Theorem 4.2.2 (Theorem 2.7 [26]). Let 
 be a (possibly disconnected) finite-area hyper-
bolic surface. Let 

 be an embedded horoball neighborhood of the cusps of 
. Let

 = Area(

)/Area(
) and let 
 > 0. Then there is a collection of at least

(


 − 1)


(

 − 1)(sinh(
) + 2
) | 
(
)|

74

embedded disjoint bi-infinite geodesic arcs, each with both ends in 

, and each having length
at most 2
 with respect to 

.

We do not need to modify this proof for our more general setting, as it does not rely on the
ambient manifold, and instead only involves the surface. This is done by looking at how much
“area” disjoint geodesics take up of the cusp, and then using this to find a lower bound for the
area of 

. We can then rearrange the equation to get our result. For a complete proof, refer
to [26].

The next lemma that holds without change deals with estimating the cusp area of a manifold,

based on essential arcs of bounded length:

Lemma 4.2.3. Suppose that a one-cusped hyperbolic 3-manifold 
 contains at least 
 ho-
motopically distinct essential arcs, each with length at most 
 measured with respect to the
maximal cusp 
 of 
. Then the cusp area Area(

) is at least 


√
3
−2
.

Proof (Sketch). The proof is similar to that of Lemma 2.8 in [26]: Each arc in our collection
has a geodesic representative in 
, which then lifts to two geodesics in the universal cover
H3. Look at the “shadows” of these geodesics on the lift of the maximal cusp 

 (that is, a
neighborhood of where the geodesic intersects 

). For the maximal cusp to be embedded,
each of these neighborhoods must be disjoint, and, because the length is at most 
, the diameter
of this neighborhood must be at least 
−
. This means that the total area of the disks must be



−2
/2. Finally, combining with a disk packing argument, we get the area of the cusp must
(cid:3)
be at least 


√
3
−2
.

While these two lemmas suggest that we should use them one after the other, we must be
careful. The cusp in Theorem 4.2.2 is a two-dimensional cusp, while the cusp in Lemma 4.2.3
is three-dimensional. In addition, the geodesic arcs of Theorem 4.2.2 live on a surface, while
the essential arcs of Lemma 4.2.3 are in the whole manifold. To get from one to the other, we
will need the following theorem, proved in the previous chapter:

75

Figure 4.1. The subsurface associated to the twist region. When it comes from checkerboard surfaces, the
subsurface can be colored red and blue, with one component (blue in the picture) being two disks attached with
Möbius bands at each crossing, and the other being a twisted disk.

Theorem 4.2.4. Let 
 be the disjoint union of our twisted surfaces 

,2 (cid:116) 

,2. Suppose that
two distinct essential arcs in the surface 

,2 have homotopic images in 
 \ 
, but not 

,2.
Then the two arcs are homotopic in 

,2 into the same subsurface associated with some twist
region of 
2.

Here, a subsurface associated to a twist region is the intersection of one of the checkerboard

surfaces with a regular neighborhood of the twist region. See Figure 4.1.

As a corollary to Theorem 4.2.4 we get the following:

Corollary 4.2.5. Suppose 
 ≥ 121. Let C be a collection of disjoint embedded essential
non-parallel arcs in 

,2 (cid:116) 

,2 which are all homotopic in 
 \ 
. Then the number of arcs in
C is at most 2(3
 − 2).
Proof. By Theorem 4.2.4, all such arcs in C that lie in 

,2 can be homotoped in 

,2 to lie
in a subsurface associated with a twist region of 
2. This homotopy keeps theses arcs disjoint,
essential, and non-parallel. Because of this, we can use the fact that the number of disjoint,
essential, and non-parallel arcs in a subsurface is at most 3
 − 2, where 
 is the number of
crossings in the twist region associated to that subsurface. Then, as we are working in 
2,

 ≤ 
, so there are at most 3
 − 2 such arcs in the blue twisted surface. By switching out blue
for red, we get that there are also at most 3
 − 2 such arcs in 

,2. Adding these together, we
get that there are at most 2(3
 − 2) such arcs total.
(cid:3)

76

From Corollary 4.2.5 we can then limit how many disjoint arcs we get from Theorem

4.2.2.

In addition to these,

[26, Proposition 3.5] gives us an embedded cusp in our twisted

surfaces with a lower bound on area, which we will use with Theorem 4.2.2:

Lemma 4.2.6 (Proposition 3.5 [26]). If 
 is the disjoint union of our twisted surfaces with an
immersion 
 : 
 → 
 \ 
, and 
 is the maximal cusp for 
 \ 
, then there is an embedded
cusp 

 for 
 contained in 
 −1(
) with area at least 25/4tw
(
), as long as 
 is neither the
figure-eight knot or the 52 knot.

We can find such a cusp by pleating 
 and giving it a hyperbolic structure, and then letting
the embedded cusp be the union of the cusps of the two twisted surfaces. The lower bound on
area then comes from carefully examining the boundaries of our twisted surfaces. By resolving
any intersections between the boundaries, we will get a number of curves with length the same
as the union of the two boundaries. The exact number is based on the intersection number of
our surfaces; in our case, 

,2 and 

,2 intersect 2cr(
2) ≥ tw
(
) times. Then, as long as

 isn’t the figure-eight knot or the 52 knot, by Adams [4], we get that each curve must have
length at least 25/4. Putting this together, we then get the desired lower bound on area for our
lemma.

Finally, we will need the following lemma computing the Euler characteristic of the disjoint

union of our twisted surfaces:

Lemma 4.2.7. Suppose 
 is a weakly generalized alternating Knot on a projection surface

. Let twN(
) denote the number of twist regions of 
(
) with at least 
 crossings, and let
cr(
2) be the number of crossings of the diagram of 
(
2), the knot where we remove all but
1 or 2 crossings from twist regions with more than 
 crossings in even pairs. Then the Euler
characteristics of the blue and red twisted surfaces satisfy

| 
(

,2) + 
(

,2)| = cr(
2) + 2twN(
) − 
(
).

77

Proof. Note that the union of the two (non-twisted) checkerboard surfaces in 
2 will have each
crossing arc of 
2 appear twice, once in the red surface, and once in the blue surface. Then, by
splitting the crossing arcs of 
2 into two homotopic arcs and attaching the blue regions to one arc
and the red regions to the other, we get that the Euler characteristic for the disjoint union of these
non-twisted surfaces is cr(
2)− 
(
). To obtain 
2 and 
2, the punctured surfaces, we remove
two disks from each crossing circle, and get | 
(
2) + 
(
2)| = cr(
2) + 2tw
(
) − 
(
).
Then, finally, to obtain the twisted surfaces, we connect these punctures by annuli or Möbius
(cid:3)
bands. This doesn’t change the Euler characteristic, and so we get our desired result.

4.3 Calculation of Cusp Area Bounds

We are now ready to give the proof of Theorem 4.1.1.

Proof. We will begin with the upper bound, using Theorem 2.1.7 (from [11]) to almost
immediately get our result. We know that | 
(
)| = cr(
2) +2tw
(
) − 
(
) ≤ 120tw
(
) −

(
). Then, note that, by construction, our twisted surfaces must intersect at least once for
every twist region in our projection, so 
(


,2, 


,2) ≥ tw
(
). Finally, as 
(

,2) and

(

,2) are both negative, we get

| 
(

,2)| + | 
(

,2)| = | 
(

,2) + 
(

,2)| ≤ 120tw
(
) − 
(
).

So, combining with Theorem 2.1.7, we get:

Area(

) ≤ 18(| 
(

,2)| + | 
(

,2)|)2


(


,2, 


,2)
which finishes the proof of the upper bound.
To proceed with the proof of the lower bound define

≤ 18(120tw
(
) − 
(
))2

,

tw
(
)

23/4
(120 − 
(
))6 + 4(120 − 
(
))5
To prove the lower bound of 

(
) in Theorem 4.1.1 we will show that

.

1


(
) =

181/4
5776
4


(
) (tw
(
) − 
(
)) ≤ area(

), 


 
(
) ≥ 5.8265514 × 10−7
(120 − 
(
))6

.

78

Since 2

(
) = area(

), we obtain the desired bound.

Let 
 be the disjoint union of the twisted blue and red surfaces, 

,2 (cid:116) 

,2, with an
immersion 
 into 
 \ 
. Then let 
 be the maximal cusp of 
 \ 
, and 

 the embedded
cusp for 
 contained in 
 −1(
) with area at least 25/4tw
(
) from Lemma 4.2.6. Note that
cr(
2) is less than or equal to 120 times the number of twist regions with less than 121 twists.
Thus cr(
2) + 2tw
(
) ≤ 120tw
(
). Then, by Lemma 4.2.7, we have

Area(
) = 2
(cr(
2) + 2tw
(
) − 
(
)) ≤ 2
(120tw
(
) − 
(
)).

Recall that


 = Area(

)/Area(
) ≥ 4√

2



tw
(
)

120tw
(
) − 
(
) .

The inequality comes from combining our upper bound on Area(
) and the lower bound
on Area(

) of 25/4tw
(
) from Lemma 4.2.6. Then, let 
 = log(2/
), so 


 = 2,

 ≤ 2−1/4
(120 − 
(
)), so we can say the
sinh(
) = 1
following:

4, and 

 = 2


 . Note that 1


 − 


2



− 1 < 23/4

sinh(
) + 2
 ≤ 2−1/4

(cid:19)
(cid:19)

(cid:18)
(cid:18)
120 − 
(
)
tw
(
)
120 − 
(
)
tw
(
)
< 2−1/4
(
, 
) + 2


= 23/4

(
, 
)
− 

4

+ 2


(sinh(
) + 2
) < 2−1/4
(23/4

(
, 
)2 + 4
(
, 
)),

(cid:19)

− 1

(cid:18) 2




where 
(
, 
) = 120 − 
(
)

tw
(
) . As a side note, while we will leave in 
(
, 
) for our
calculations, once we choose a projection surface, we immediately get an upper and lower
bound: 
(
, 
) is between 120 and 120 − 
(
).

Putting this together, by Theorem 4.2.2, we have at least
21/4

(


 − 1)


(

 − 1)(sinh(
) + 2
) | 
(
)| >

23/4

(
, 
)2 + 4
(
, 
) | 
(
)|

79

embedded disjoint bi-infinite geodesics, with both ends in 

, of length at most 2
. Now we
can map these arcs into 
 \ 
. As 
 is essential, each arc remains essential under this mapping.
Furthermore, the length of these arcs can only decrease, so the upper bound of length 2

remains. Some of these arcs may be homotopic now in 
 \ 
, however, by Corollary 4.2.5,
there are at most 2(3
 − 2) = 722 such arcs. So, dividing out, we have at least

21/4

722(23/4

(
, 
)2 + 4
(
, 
)) | 
(
)|

disjoint embedded geodesics with both ends in 
 and length at most 2
 with respect to 
.

Now we can use Lemma 4.2.3 to say that, if the above number is 
,

Before we put all the math together, to make the numbers look nicer, we will find an upper
bound for 
−4
:

3
−2(2
).

Area(

) ≥ √
(cid:18) 2
(cid:19)4
23/4

(
, 
)(cid:17)4
≤(cid:16)





4
 =


−4
 ≥

1

8
4
(
, 
)4

= 8
4
(
, 
)4

Now, combining this, with all of our other bounds, we get that our cusp volume satisfies

21/4√

3| 
(
)|

Area(

) ≥

≥

722(23/4

(
, 
)2 + 4
(
, 
)) 
−4

722(23/4

(
, 
)2 + 4
(
, 
))

181/4

1

8
4
(
, 
)4 (tw
(
) − 
(
)) .

Because [26] gives us a lower bound for the case when 
 = 
2, we will focus on the other
cases. When 
(
) ࣔ 2, 
(
, 
) ≤ 120 − 
(
), so we can replace 
(
, 
) in our equation
with 120 − 
(
), and we get a lower bound of 
(
)(tw
(
) − 
(
)), and so are done with
(cid:3)
the lower bound, and our proof.

Remark 4.3.1. In the course of proving the lower bound in Theorem 4.1.1, we proved that we
can replace 
(
) with a function 
(
, 
) depending on the Euler characteristic of 
 and and

80

tw
(
). In particular, we can replace 
(
) with the following function for a more precise
lower bound:


(
, 
) =

181/4
5776
4

1

23/4

(
, 
)6 + 4
(
, 
)5


Ò


 
(
, 
) = 120 − 
(
)
tw
(
) .

We can get a better lower bound if we replace 
(
) with 
(
, 
). Below are some value for

(
, 
) and 
(
) for different values of 
(
) and tw
(
), showing that, while as 
(
) de-
creases, both our bounds get worse, as tw
(
) increases, the bound 
(
, 
) starts to improve.


(
)
0
−2
−4
−100


(
)


(
, 
) when tw
(
) = 10 
(
, 
) when tw
(
) = 100

2.3059 × 10−19
2.0884 × 10−19
1.8945 × 10−19
6.0903 × 10−21

2.3059 × 10−19
2.2830 × 10−19
2.2604 × 10−19
1.4272 × 10−19

2.3059 × 10−19
2.3036 × 10−19
2.3013 × 10−19
2.1941 × 10−19

4.4 Additional Results

Now we can apply Theorem 4.1.1 to obtain estimates of slope lengths on the maximal cusp
of 
. We discuss an upper bound on the length on meridian curves for K and lower length
bounds for non-meridian slopes.
Corollary 4.1.3. Let 
 be a closed, irreducible 3-manifold and 
 ⊂ 
 a closed surface with

(
) < 0 and such that 
 \ 
(
) is atoroidal. Suppose that 
 is a weakly generalized
alternating knot with a projection 
(
) ⊂ 
 that is weakly twist reduced, and we have

(
(
), 
) > 2, and 
(
(
), 
) > 4. Let 
 be the meridian and 
 any non-meridian slope
on maximal cusp of 
 \ 
. Then, we have

(cid:18) 5.8265514 × 10−7
(cid:19)
ࡁ(
) ≤ 720 (tw
(
) − 
(
))
ࡁ(
) ≥

tw
(
)

720 (120 − 
(
))6

tw
(
),

where tw
(
) is the twist number of 
(
).

81

Proof. As shown in the proof of Theorem 4.1.1 above, if 

,2 and 

,2 are our two twisted
essential surfaces, then:

| 
(

,2)| + | 
(

,2)| ≤ 120tw
(
) − 
(
) ≤ 120(tw
(
) − 
(
)),

where the right inequality comes from the fact that 
(
) ≤ 0, so −
(
) ≤ −120
(
). Also,

(


,2, 


,2) ≥ tw
(
). Using these two results with Burton and Kalfagianni
[11] on
upper bounds for slope lengths, we get our desired upper bound on ࡁ(
):


(


,2, 


,2)

ࡁ(
) ≤ 6| 
(

,2)| + | 
(

,2)|
≤ 6(120tw
(
) − 
(
))
≤ 720(tw
(
) − 
(
))

tw
(
)
tw
(
)

.

Next, Theorem 4.1.1 tells us that 
(
)(tw
(
) − 
(
)) ≤ Area(

). Then, as in [13],

ࡁ(
)ࡁ(
) ≥ Area(

)Ɗ(
, 
).

Then, using our bound for Area(

), and that 
 and 
 must intersect at least once, we get

ࡁ(
)ࡁ(
) ≥ 
(
)(tw
(
) − 
(
)).

Next, we use our upper bound for ࡁ(
) to get:

720(tw
(
) − 
(
))

tw
(
)

ࡁ(
) ≥ ࡁ(
)ࡁ(
) ≥ 
(
)(tw
(
) − 
(
)).

Finally, we rearrange:

ࡁ(
) ≥ 
(
)tw
(
) − 
(
)

= 
(
) tw
(
)
720

,

tw
(
)

720(tw
(
) − 
(
))

giving the desired inequality.

(cid:3)

For our final result, recall that Dehn surgery along a knot 
 ⊂ 
 involves removing a torus
neighborhood of the knot from 
 and replacing it with a torus glued along a slope 
 of the
knot to get a manifold 
. An important result related to such surgery is as follows.

82

Theorem 4.4.1 (Theorem 1.1 [18]). Let 
 be a complete, finite-volume hyperbolic manifold
with a single cusp 
. Let 
 be a slope on 

 with length ࡁ(
) greater than 2
, and let 
(
)
denote the 3-manifold obtained by Dehn filling 
 along 
. Then, 
(
) is hyperbolic and we
have

vol(
(
)) ≥

1 −

vol(
).

(cid:32)

(cid:19)2(cid:33)3/2

(cid:18) 2


ࡁ(
)

This, combined with Theorem 2.2.7 from [23], gives us our other result.

Theorem 4.1.4. Let 
 be a weakly generalized alternating knot on a surface 
 ࣔ 
2 in a
closed manifold 
 with conditions as in Theorem 4.1.1. Suppose that

(cid:16)

2.32928 × 1010(cid:17) (120 − 
(
))6 .

tw
(
) >

Then, any manifold 
 obtained from non-meridional surgery along 
 is hyperbolic and


8
4

(tw
(
) − 
(
)) ≤ 1
2

vol(
 \ 
) ≤ vol(
) < vol(
 \ 
).

Here 
8 ≈ 3.66386 is the volume of a regular hyperbolic ideal octahedron.

Proof. As 
 is a weakly generalized alternating knot and, by Lemma 2.2.8, has disk regions,
we get the leftmost inequality from Theorem 2.2.7. Then, by Corollary 4.1.3, as we’ve taken
the number of twist regions to be high enough, any non-meridional slope must have length
strictly greater than 2
. This allows us to use Theorem 4.4.1 from [18] to get our center
inequality. Finally, we get vol(
) < vol(
 \ 
) from the fact that volume must decrease under
(cid:3)
surgery.w

83

CHAPTER 5

JONES POLYNOMIAL

5.1 Introduction

Parts of this chapter is based on joint work with Efstratia Kalfagianni that will be published

in Fundamenta Mathematicae [8].

The goal of this chapter is to show that the hyperbolic volume of certain weakly generalized
alternating links on a closed surface 
 in a compact, irreducible 3-manifold 
 is bounded
below in terms of a Kauffman bracket function defined on link diagrams on 
. Further, for

 = 
 × [−1, 1], this function leads to a Jones polynomial link invariant and coefficients of it
provide 2-sided linear bounds on the volume of hyperbolic alternating links.

As a corollary, we obtain that the twist-number of a twist-reduced, weakly generalized
alternating link projection with representativity 
(
(
), 
) > 4 and checkerboard disk regions,
is an invariant of the link in 
 × [−1, 1]. This generalizes work of Dasbach and Lin [15] and
Futer, Kalfagianni and Purcell [17,18], who have obtained similar results for families of links
in 
3.

Let 
 be an irreducible, compact 3-manifold with or without boundary, and let 
 ⊂ 
 be
a closed, orientable, embedded surface. Then, if we have a link 
 ⊂ 
 × [−1, 1], we can get a
projection under the map 
 : 
 × [−1, 1] → 
 = 
 × {0}. We can then define, for connected

, a Kauffman bracket function, and construct a family of Jones-like polynomials


(
(
)) = 

,
 

 + 

−1,
 

−1 + · · · + 

+1,
 

+1 + 

,
 

,

where 
 is a collection of simple disjoint closed curves on 
. We will give the precise
definitions for these polynomials in the next section. We can consider the sums of these
coefficients,

84









 =



,




 =









,
 ,

where we sum over all possible collections of simple disjoint closed curves on 
. Then we

are able to say the following:

Theorem 5.1.1. Let 
 be an irreducible, compact 3-manifold with empty or incompressible
boundary. Let 
 ⊂ 
 be an incompressible, closed, orientable surface such that 
 \ 
(
) is
atoroidal and 
 − 








. Suppose that a link 
 admits a weakly generalized alternating
projection 
(
) ⊂ 
 that is reduced, twist-reduced and with all regions of 
 \ 
(
) disks.
Finally, suppose that 
(
(
), 
) > 4. Then 
 is hyperbolic and
vol(
 \ 
) ≥ 
8max{|

−1|, |

+1|} − 1
2


(
 
),

where 
8 = 3.66386 . . . is the volume of an ideal octahedron.

Now, given an alternating link projection 
(
) as in the statement of Theorem 5.1.1, let

 
 and 

 denote the two checkerboard surfaces corresponding to 
(
), and let 

 and 


denote the manifolds obtained by cutting 
 \ 
 along 
 
 and 

, respectively. We can show
the following.

Theorem 5.1.2. Let 
 and 
 be as in Theorem 5.1.1 Suppose that a link 
 admits a weakly
generalized alternating projection 
(
) ⊂ 
 that is twist-reduced and with all regions of

 \ 
(
) disks. Then we have the following.

1. 
(guts(

)) = −|

−1| + 1
2. 
(guts(

)) = −|

+1| + 1
3. tw
(
(
)) = |

−1| + |

+1| + 
(
).

2 
(
)
2 
(
)

Further, if 
(
) is a cellularly embedded, twist-reduced, reduced alternating diagram, we still
get (3).

85

Theorem 5.1.1 follows directly from 5.1.2. We can also obtain the following corollary for

links in thickened surfaces:

Corollary 5.1.3. Let 
 be a link in 
 × [−1, 1] that admits a weakly generalized alternating
projection 
(
) ⊂ 
 that is reduced, twist-reduced and with all regions of 
 \ 
(
) disks.
Then any two projections of 
have the same twist number. That is, tw
(
(
)) is an isotopy
invariant of 
.

The rest of this chapter is outlined as follows. First, in Section 2, we introduce the exact
functions and polynomials we will be working with. Then, in Section 3, we will show that
there is a relation between certain coefficients of our polynomials and certain reduced state
graphs. Finally, in Section 4, we will prove Theorem 5.1.2, and show how Theorem 5.1.1
follows.

5.2 Background

Our definition of a Kauffman bracket function for links on general surfaces will follow, in
many regards, the definition of the Kauffman bracket function for links on 
2, as defined in
Chapter 2. Let 
 be a 3-manifold, and let 
 ⊂ 
 be an orientable surface. Given a link

 ⊂ 
 × [−1, 1] ⊂ 
, we can consider it’s image under the projection 
 : 
 × [−1, 1] → 
 =

 × {0}.

Let D(
) denote the set of all (unoriented) link diagrams on 
, taken up isotopy on 
. Also
let 

 denote the set of all collections of disjoint simple closed curves (a.k.a multi-curves) on

. We define a Kauffman bracket function

(cid:104) (cid:105) : D(
) −→ Z[ 
, 
−1, ( 
2 − 
−2)−1] 

,

by the following skein relations.
(cid:105) + 
−1(cid:104)

(cid:105) = 
(cid:104)

(cid:105)
(cid:105) = (−
2 − 
−2)(cid:104)
(cid:105)

1. (cid:104)

2. (cid:104)
 (cid:116)

86

Figure 5.1. A Kauffman State (right) for an alternating knot on the torus (left). The color of the edges represent
the type of resolution at each crossing. Note that we now have states that don’t bound disks on 
2.

3. (cid:104)

(cid:105) = −
2 − 
−2

As we are working on a surface that may have essential curves, we also require that the
unknots in the above relations bound disks on 
. We can then describe (cid:104)
(cid:105) as a function in
Z[ 
, 
−1] 

, as follows.

Recall that, given a Kauffman state 
, 
(
) is the number of 
-resolutions, 
(
) the number
of 
-resolutions, and |
| the number of state circles. We will need to make a distinction between
contractible and non-contractible state circles, so we say 

 is the collection of contractible state
circles, 


 the collection of non-contractible state circles, and |
|
 the number of contractible
state circles.

Given a link diagram 
 = 
(
) in D(
), we define

where 
 ∈ 

 is a collection of disjoint simple closed curves on 
. Then

(cid:104)
(cid:105)
 
.


∈



87



(
)−
(
)(−
2 − 
−2)|
|




(
)−
(
)(−
2 − 
−2)|
|


(cid:104)
(cid:105)0 =

(cid:104)
(cid:105)
 =

{
|


=∅}





(cid:104)
(cid:105) = (cid:104)
(cid:105)0 + 


{
|


=
}

(5.2.1)

(5.2.2)

We can then define the Jones-like polynomials as follows:

(cid:17)(cid:12)(cid:12)
=
4
(cid:16)(−1)
(
) 
−3
(
)(cid:104)
(
)(cid:105)0
(cid:16)(−1)
(
) 
−3
(
)(cid:104)
(
)(cid:105)

(cid:17)(cid:12)(cid:12)
=
4.


0(
(
)) =


(
(
)) =

We can write these out with coefficients:


0(
(
)) = 

,0

 + 

−1,0

−1+· · · + 

+1,0

+1 + 

,0




(
(
)) = 

,
 

 + 

−1,
 

−1+· · · + 

+1,
 

+1 + 

,0

.

As 
0(
(
)) and 

(
(
)) might have different highest or lowest degrees, we take 
 to
be the highest degree over all possible 

(
(
)) and 
0(
(
)), and 
 to be the lowest. Then,
if any polynomial has a lower maximum degree, we take all higher coefficients to be 0.

Note that, in general, these are not link invariants, but instead diagram invariants. As
discussed in [8, 10],these polynomials can be promoted to isotopy invariants in the special
case that 
 = 
 × [−1, 1].

5.2.1 Reduced Graphs

Recall that a Kauffman state 
 gave us two different graphs, the state graph 
 
 and the reduced
graph 
(cid:48)

. When working with generalized link diagrams, we will also get similar graphs,
however a slight modification of definition is necessary. We let 
 
 be the the graph with a
vertex for each state circle, and edges connecting state circles that are adjacent across crossings
of 
(
). Then the reduced graph 
(cid:48)

 is obtained from 
 
 by, whenever two edges of 
 
 bound
a disk on 
, we remove one of them. Then we let 
(cid:48)

. This
will mean that two edges adjacent to the same vertices that form an essential curve on 
 will
both still be in 
(cid:48)

.


 denote the number of edges of 
(cid:48)

In particular, we will be concerned with the reduced graphs for the all 
 state 

 and the

all 
 state 

. We call these 
(cid:48)


 and 
(cid:48)


, respectively.

88

5.3 Coefficients and Graphs

As a note before we begin, many of our proofs will work with a more general category
of links: ones which are cellularly embedded and with no removable nugatory crossings (a
crossing is removable nugatory if there is a disk on 
 that intersects the diagram at only the
crossing). These are called reduced diagrams Boden, Karimi, and Sikora [9].

As we will use some results for reduced diagrams, it’s important to show that weakly
generalized alternating link diagrams are also reduced diagrams. To see this, note that a
weakly generalized alternating link must be cellularly embedded, as the complement has disk
regions, and there can be no nugatory crossings, as if there were, we would have a disk violating
weakly prime.

We begin now by examining the coefficients for weakly generalized alternating diagrams.

Lemma 5.3.1. Suppose 
(
) is twist-reduced, reduced alternating diagram on a closed ori-
entable surface 
, with 
 \ 
(
) disks. Suppose also that 
(
) is checkerboard colorable.
Then |

| = |

| = 1.

Proof. As the regions of 
\ 
(
) are disks, the all 
 state, 

, has all state circles compressible
disks (in particular, we can view 

 as the collection of all regions of 
 \ 
(
) colored a single
color). Then 

’s contributions to 
(
(
)) will have highest degree (
 + 2|

|
), and the
highest term contributed will be ±1.

Now suppose we have a state 

 with exactly 
 
-resolutions, and a state 

−1 which
differs from 

 by exactly one resolution change from an 
-resolution to a 
-resolution. This
resolution change can, at most, increase the number of contractible state circles by 1. Look,
then, at any state that has exactly one 
 resolution, 
1. As 

 has only contractible state circles,
there are exactly three ways a resolution change from 

 to a state 
1 can act.

1. We could merge two contractible circles into a single contractible circle. Then |
1|
 =

|

|
 − 1.

89

2. We could split a single contractible circle into two noncontractible circles. Then |
1|
 =

|

|
 − 1.

3. We could split a single contractible circle into two contractible circles. Then |
1|
 =

|

|
 + 1.

Note that the highest degree 
1 can contribute to a polynomial is (
 + 2|
1|
 − 2). The
only way 
1 could contribute to the highest degree of 
(
(
)) is if the last case, (3), happens.
However, note that for this to happen, the resolution change must occur at a crossing that meets
a single state circle of 

 twice. Then we can construct a loop 
 that intersects the diagram
at exactly this crossing, and then follows along the boundary of the state circle. As there are
no nugatory crossings, this must mean that 
 is a non-separating curve. In particular, 
 must
be essential. After the resolution change, we will get two state circles that must have essential
boundary, and thus be noncontractible. But then we must be in the second case, (2), and have
a contradiction. So no 
1 can contribute to the highest degree.

Finally, look at any state 

 with exactly 
 
 resolutions. This must have highest degree
(
 + 2|

|
 − 2
). In order to contribute to the highest degree, we must have |

|
 = |

|
 + 
.
However, as each resolution change can, at most, introduce one contractible state circle, and
we know that the change from 

 to 
1 does not, we must have 

 not contribute to the highest
degree. So then 

 = ±1, and we are done.
(cid:3)

Before we begin, we introduce one last definition that we will use in the later half of the

proof for the second coeffient.


. We say two edges, ࡁ and ࡁ(cid:48) in 
(cid:48)

Definition 5.3.2. Let 
(
) ⊂ 
 be a reduced alternating diagram, and 
 a Kauffman state,
with reduced graph 
(cid:48)

 are parallel if they are adjacent to
the same vertices (or, equivalently, connect the same state circles) and, when placed on 
 with
the state circles, there is a disk in 
 with boundary containing ࡁ, ࡁ(cid:48), and a portion of the state
circles they connect.

90

Lemma 5.3.3. Suppose 
(
) is a twist-reduced, reduced alternating diagram on a closed
orientable surface 
, with 
 \ 
(
) disks. Suppose also that 
(
) is checkerboard colorable.
Then

1.

2.

|

−1| = 
(cid:48)
|

+1| = 
(cid:48)


 − |

|
.

 − |

|
.

Proof. First, note that 

 contributes to the second highest degree. In particular, it contributes
(−1)|

|
|

|
. Next, we also know that any state with exactly one 
 resolution, 
1, will
contribute to the second highest degree. We know this because we have one less contractible
circle than 

, by work above, one less 
 resolution, and one more 
 resolution; then we can
use either Equation 5.2.1 or 5.2.2. Now suppose we have a sequence of states 

, 
1, · · · , 

,
each differing from the previous state by a resolution change from an 
-resolution to a 
-
resolution. Then, for 

 to contribute, we must have each resolution change after the first
introducing a new contractible state circle. We know that 
1 either has all contractible state
circles, or has exactly two non-contractible state circles that are parallel to each other. Focus
first on the states 
1 and 
2. Our goal is to show that 
2 will only contribute to |

−1| if the
edge changed from 
1 is parallel to the edge changed from 

.

There are three possible ways a resolution change can act and still have 
2 contribute to the

second highest degree:

1. We split a single contractible circle into two contractible circles.

2. We merge two noncontractible circles into a single contractible circle.

3. We split a single noncontractible circle into a noncontractible circle and a contractible

circle.

We focus on each of these cases individually. In the case of (1), where we split a contractible
circle into two contractible circles and don’t change the pattern of noncontractible state circles,
the edge changed, ࡁ2 would have to, in 
1, be adjacent to a single state circle. If this edge is

91

parallel to the edge changed from 

 to 
1, ࡁ1, we are fine. Otherwise, we must have both ࡁ2
and ࡁ1 appear in 
(cid:48)

, and so ࡁ1 and ࡁ2, with a portion of some state circles, forms an essential
curve on 
. Then 
2 will introduce a noncontractible circle, and so we cannot be in the first
case.

In the case of (2), where we merge two noncontractible circles into a single contractible
circle, denote the edge changed in the resolution change from 

 to 
1 as ࡁ1, and the edge
changed from 
1 to 
2 as ࡁ2. We know that the two noncontractible circles must be parallel,
and so, to merge them to get a contractible circle, we must have ࡁ1 parallel to ࡁ2. Look at ࡁ2 in


, and construct a new state 
(cid:48)
1, which has all 
-resolutions except at ࡁ2. In Figure 5.2, the
leftmost state is 
1, the center 

, and the right 
(cid:48)
1.

ࡁ2

ࡁ1

Figure 5.2. (Left) The state 
1 gotten from 

 by merging two noncontractible circles into a single contractible
circle; (Center) The all 
 state, 

; (Right) The modified state, 
(cid:48)

1, with exactly one 
 resolution.

Note that 
(cid:48)

1 must, then, introduce a new contractible state circle from 

. In addition,
it will have the same types of non-contractible state circles as 

. Then, by Boden, Karimi,
and Sikora [9, Theorem 11], any state of a reduced alternating diagram (and thus a weakly
generalized alternating diagram) with one 
 resolution must have, at most, the same number
of contractible state circles as 

, so we get a contradiction.

Finally, in the case of (3), we split a noncontractible state into a contractible circle and a
noncontractible circle. Label the edges as before: ࡁ1 is the edge first changed in the resolution
change from 

 to 
1, and ࡁ2 the edge changed from 
1 to 
2. There are two possibilities for
ࡁ2: either it is parallel to ࡁ1 in 

, or it is not. If the two edges are not parallel, then ࡁ2 must
have it’s endpoints both to the same side of ࡁ1, and cut off a contractible circle disjoint from

92

ࡁ1. Then we can construct a new state, 
(cid:48)
have a 
 resolution. As ࡁ2 cuts off a contractible circle disjoint from ࡁ1, 
(cid:48)
contractible circles than 

, a contradiction. So then ࡁ1 and ࡁ2 must be parallel in 

.

1 that has all 
 resolutions except for at ࡁ2, where we
1 must have more

Note that, in order for 
2 to contribute to the second highest degree, the second edge must
be parallel to the first. If it isn’t, it either does not introduce a new contractible circle (like in
the first case), or we can find a state with 1 B resolution that has more contractible circles than


. We can do this same process for any state 

 with 
 B resolutions, and so all states that
contribute to |

−1| must have 
 resolutions at only parallel edges.

Also note, if any two edges are parallel, they must be part of the same twist region. To
see this, as parallel edges bound a disk on 
, we get a disk whose boundary intersects the link
diagram at exactly two crossings. As the diagram is twist reduced, this must mean both of
these crossings, and thus the edges, belong to the same twist region.

So then we get a contribution to |

−1| only when we make resolution changes to edges
that, in 

, are all parallel to each other. As we are asking about |

−1|, we will leave
off the essential curve variable 
 during our calculations. Each such state will contribute

(−1)|

|−2(−1) 
, where 
 is the number of changed edges, and there are(cid:0)


(cid:1) possible states




with exactly 
 resolution changes, where 
 is the total number of parallel edges. Using the
binomial theorem, we get that each family of parallel edges must contribute (−1)|

|−1 to the
second highest coefficient. As each family of parallel edges corresponds to a single edge in

(cid:48)

, and factoring in the contribution to the second coefficient from 

, we get that

|

−1| = |(−1)|

|
−1
(cid:48)
= |(−1)|

|
−1(
(cid:48)
= 
(cid:48)


 − |

|



 + (−1)|

|
 (|

|
)|

 − |

|
)|

and so we are done with part (1) of the theorem.

To prove part (2), we can apply the same argument to the diagram 
∗.

(cid:3)

93

5.4 Twist Number, Volume, and Coefficients

Finally, in this section, we will prove Theorems 5.1.1 and 5.1.2 using Lemmas 5.3.1 and
5.3.3. We will start by proving the following relation between the coefficients, the twist number,
and the Euler characteristic of 
:

Theorem 5.4.1. Suppose that 
(
) is a twist-reduced, reduced alternating link diagram with
twist number 

(
(
)) on a closed orientable surface 
 ⊂ 
 of genus at least 1. Suppose
also that the regions of 
 \ 
(
) are all disks. Then

|

−1| + |

+1| = 

(
(
)) − 
(
)

Proof. First, note that we have



(
(
)) = 
 − (
 − 
(cid:48)


) − (
 − 
(cid:48)


) = 
(cid:48)


 + 
(cid:48)


 − 
,

where 
 is the number of crossings of 
(
). We can see this as follows: we call a twist
region an 
 (or 
) twist region if the edges in 
 
 (or 
 
) are all parallel. Then 
(cid:48)

 (or 
(cid:48)

)
will have exactly one edge for each 
 (or 
) twist region, as well as edges for each crossing in
a 
 (or 
) twist region. Note that 
 − 
(cid:48)

 counts exactly the number of edges in 
 
 that aren’t
in 
(cid:48)

. We can think of this as counting each edge that is an 
 twist region, except for one for
each such region. Likewise, 
 − 
(cid:48)

 counts the number of edges in a 
 twist region, except one

) + (
 − 
(cid:48)
for each such region. Then (
 − 
(cid:48)


) = 
 − 

(
(
)).

Next, as the regions of 
 \ 
(
) are disks, we also have

|

| + |

| = 
 + 2 − 2
(
) = 
 + 
(
).

It is important to note that, as 
 \ 
(
) are all disks, then |

| = |

|
 and |

| = |

|
, as

94

there are only contractible circles. Then we get
|

−1| + |

+1| = 
(cid:48)
= 
(cid:48)
= 

(
(
)) + 
 − 
 − 
(
)
= 

(
(
)) − 
(
),


 − |

|


 − |

| − |

|


 − |

|
 + 
(cid:48)

 + 
(cid:48)

and we are done.

(cid:3)

We can now use this to prove Theorem 5.1.2:

Theorem 5.1.2. Let 
 and 
 be as in Theorem 5.1.1 Suppose that a link 
 admits a weakly
generalized alternating projection 
(
) ⊂ 
 that is twist-reduced and with all regions of

 \ 
(
) disks. Then we have the following.

1. 
(guts(

)) = −|

−1| + 1
2. 
(guts(

)) = −|

+1| + 1
3. tw
(
(
)) = |

−1| + |

+1| + 
(
).

2 
(
)
2 
(
)

Further, if 
(
) is a cellularly embedded, twist-reduced, reduced alternating diagram, we still
get (3).

Proof. We have already proved the twist number result above, in Theorem ??, as well as the
last last statement about reduced alternating diagrams. So we now focus on the other two
statements.

Let 
 
 and 

 be the checkerboard surfaces of 
(
), 
 = 
 \ 
, and 

 = 
\\
 
 and



 = 
\\

. As 
(
) is a weakly generalized alternating diagram, [23] gives us


(guts(

)) = 
(
) + 1
2


(
 
) − |
(cid:48)

|,

(5.4.1)

where |
(cid:48)


| is the number of non-bigon 
 regions of 
 
. Then note that


(
) = |

| − 
(cid:48)


 + |
(cid:48)

|.

95

In particular, by rearranging, we get


(
) − |
(cid:48)


| = |

| − 
(cid:48)




Finally, by Lemma 5.3.3, we get


(
) − |
(cid:48)


| = −|

−1|.

(5.4.2)

Then by combining Equations 5.4.1 and 5.4.2, we immediately get our result. The proof for
guts(

) follows similarly, by swapping 
 and 
.

(cid:3)

Theorem 5.1.1. Let 
 be an irreducible, compact 3-manifold with empty or incompressible
boundary. Let 
 ⊂ 
 be an incompressible, closed, orientable surface such that 
 \ 
(
) is
atoroidal and 
 − 








. Suppose that a link 
 admits a weakly generalized alternating
projection 
(
) ⊂ 
 that is reduced, twist-reduced and with all regions of 
 \ 
(
) disks.
Finally, suppose that 
(
(
), 
) > 4. Then 
 is hyperbolic and
vol(
 \ 
) ≥ 
8max{|

−1|, |

+1|} − 1
2


(
 
),

where 
8 = 3.66386 . . . is the volume of an ideal octahedron.
Proof. Let 
 
 and 

 be the checkerboard surfaces of 
(
), 
 = 
 \ 
 the link complement,
and 

 = 
\\
 
 and 

 = 
\\

. Then, by [23]Theorem 9.1, we have

vol(
) ≥ −
8 
(guts(

))

and

vol(
) ≥ −
8 
(guts(

)).

Then, by Theorem 5.1.2, we immediately get our result, and are done.

(cid:3)

96

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