VOLUMES, DETERMINANTS, AND MERIDIAN LENGTHS OF HYPERBOLIC LINKS
By
Stephan D. Burton

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Mathematics - Doctor of Philosophy
2017

ABSTRACT
VOLUMES, DETERMINANTS, AND MERIDIAN LENGTHS OF HYPERBOLIC LINKS
By
Stephan D. Burton
We study relationships between link diagrams and link invariants arising from hyperbolic
geometry. The volume density of a hyperbolic link K is defined to be the ratio of the
hyperbolic volume of K to the crossing number of K. We show that there are sequences
of non-alternating links with volume density approaching v8 , where v8 is the volume of the
regular ideal hyperbolic octahedron. We show that the set of volume densities is dense
in [0, v8 ]. The determinant density of a link K is 2π log det(K)/c(K). We prove that
the closure of the set of determinant densities contains the set [0, v8 ]. We examine the
conjecture, due to Champanerkar, Kofman, and Purcell [20] that vol(K) < 2π log det(K)
for alternating hyperbolic links, where vol(K) = vol(S 3 \K) is the hyperbolic volume and
det(K) is the determinant of K. We prove that the conjecture holds for 2-bridge links,
alternating 3-braids, and various other infinite families. We show the conjecture holds for
highly twisted links and quantify this by showing the conjecture holds when the crossing
number of K exceeds some function of the twist number of K. We derive bounds on the
length of the meridian and the cusp volume of hyperbolic knots in terms of the topology of
essential surfaces spanned by the knot. We provide an algorithmically checkable criterion
that guarantees that the meridian length of a hyperbolic knot is below a given bound. As
applications we find knot diagrammatic upper bounds on the meridian length and the cusp
volume of hyperbolic adequate knots and we obtain new large families of knots with meridian
lengths bounded above by four. We also discuss applications of our results to Dehn surgery.

Dedicated to my wonderful wife, Tina.

iii

ACKNOWLEDGMENTS

I would like to thank my wife, Tina, for all the support she has given me. She kept me
motivated, and listened to me no matter what I was talking about.
I would like to thank my adviser, Effie Kalfagianni, for her work and help with preparing
my research. She helped me find tractable problems and I learned a lot by studying with
her.
I would also like to thank my Masters degree adviser, Jessica Purcell, for introducing
me to hyperbolic 3-manifolds and planting the seed of interest that lead me to continue
researching and learning about the field. I appreciate my work with Darrin Doud and Jeff
Humpherys as an undergraduate, who sparked my interest in advanced mathematics.
Conversations with Colin Adams, Abhijit Champanerkar, Dave Futer, Effie Kalfagianni,
Ilya Kofman, and Jessica Purcell were helpful in producing the research contained in this
work.

iv

TABLE OF CONTENTS

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii

Chapter 1 Introduction . . . . . . . . . . . . . .
1.1 Volume and Determinant Densities . . . . . .
1.2 Relating Determinant and Volume . . . . . . .
1.3 Meridians, Cusp Area, and Essential Spanning

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2.1 Hyperbolic Geometry . . . . . . . . . . . .
2.1.1 Hyperbolic Structures on Manifolds
2.1.2 Maximal Cusps and Dehn Filling .
2.2 Essential Surfaces . . . . . . . . . . . . . .
2.3 Twist Number and Volume . . . . . . . . .

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Chapter 3 Volume and Determinant Densities . . . .
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . .
3.2 Background . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Geometrically and Diagrammatically Maximal
3.2.2 Conway Sums and Belted Sums of Links . . .
3.2.3 Adequate Tangles . . . . . . . . . . . . . . . .
3.3 Non-alternating Geometrically Maximal Knots . . . .
3.4 The Spectrum of Volume Densities . . . . . . . . . .
3.5 The Spectrum of Determinant Densities . . . . . . .

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Chapter 4 Relating Determinant and Volume
4.1 Introduction . . . . . . . . . . . . . . . . . . .
4.2 Background . . . . . . . . . . . . . . . . . . .
4.2.1 Hyperbolic Volumes . . . . . . . . . .
4.2.2 Determinants of Links . . . . . . . . .
4.3 Two Bridge Links . . . . . . . . . . . . . . . .
4.4 Alternating Braids . . . . . . . . . . . . . . .
4.4.1 3-Braids . . . . . . . . . . . . . . . . .
4.4.2 A Family of 4-braids . . . . . . . . . .
4.5 Highly Twisted Knots . . . . . . . . . . . . .
4.5.1 Application to Pretzel Knots . . . . . .

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Chapter 5 Meridians, Cusp Areas, and Essential Spanning Surfaces
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Knots with essential checkerboard surfaces . . . . . . . . . . . .
5.1.2 Knots with meridian length bounded by four . . . . . . . . . . .
5.1.3 Slope length bounds, Dehn filling and volume . . . . . . . . . .

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5.2

5.3
5.4

5.1.4 Organization . .
Background . . . . . . .
5.2.1 Pleated Surfaces
5.2.2 Turaev Surfaces .
Lengths of Curves on the
Algorithm . . . . . . . .

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APPENDIX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

vi

LIST OF FIGURES

Figure 1.1: A diagram of the 77 knot. . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Figure 1.2: Two diagrams of the same knot. . . . . . . . . . . . . . . . . . . . . . . .

2

Figure 2.1: Left: The 77 knot. Center: Checkerboard coloring of the 77 knot
diagram. Right: The checkerboard graph associated to the black coloring. 20
Figure 2.2: A nugatory crossing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

Figure 2.3: The two resolutions of a crossing, the arcs recording them, and their contribution to state surfaces. The left frame depicts the A-resolution, and
the right depicts the B-resolution. . . . . . . . . . . . . . . . . . . . . . .

22

Figure 2.4: Left: A link L. Center: The all-A resolution. The thick lines are the
state circles, and the thin lines are arcs recording the resolutions. Right:
The all-A state surface of L. Figure taken from [30]. . . . . . . . . . . .

22

Figure 3.1: The infinite weave W. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

Figure 3.2: The dashed line represents a visible Conway sphere. . . . . . . . . . . . .

29

Figure 3.3: A tangle T , its two closures N (T ) and D(T ), and its corresponding belted
tangle B(T ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

Figure 3.4: The belted sum of two belted tangles.

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31

Figure 3.5: The Conway sum of two tangles. The tangle T1 + T2 is pictured on the
right. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

Figure 3.6: Left: The braid element L6,3 . Right: The weaving tangle W6,3 . . . . .

38

Figure 3.7: The link K(n) for the proof of Theorem 3.3.3. It contains n copies of
Wn,n and n copies of W n,n . . . . . . . . . . . . . . . . . . . . . . . . . .

38

Figure 3.8: The pretzel tangle P ,m,n has , m, and n crossings in the respective twist
regions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

42

Figure 3.9: The link K is the closure of the tangle sum of a · n copies of T1 with b · n
copies of T2 shown above. . . . . . . . . . . . . . . . . . . . . . . . . . .

45

vii

Figure 3.10: The belt circle adds no more than the two twist regions indicated in the
figure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

Figure 3.11: The link K2n is the closure of the sum of n copies of T2 as indicated above. 47
Figure 3.12: The link K is the closure of the tangle sum of a copies of T1 with b copies
of T2 shown above. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

52

Figure 4.1: A regular ideal 6-bipyramid. Figure taken from [3]. . . . . . . . . . . . .

58

Figure 4.2: Left: Initial graph. Right: Result of collapsing the orange edge. . . . .

61

Figure 4.3: Left: The knot R(3, 3, 2). Right: The link R(3, 2, 2, 3). . . . . . . . . .

62

Figure 4.4: Left: The two-bridge knot R(3, 3, 2). Center Left: The checkerboard
graph for R(3, 3, 2). Center Right: The closed alternating 3-braid
B(3, 3, 2, 3). Right: The checkerboard graph for B(3, 3, 2, 3). Note that
the graph that results from deleting the highlighted edge has the same
number of spanning trees as the checkerboard graph for R(3, 3, 2). . . .

79

Figure 4.5: Left: The link Wn . Center: The checkerboard coloring of W4 . Right:
The graph E6 corresponding to the white checkerboard surface. . . . . .

85

Figure 4.6: Left: The checkerboard graph of the Pretzel knot P (1, 2, 4, 3, 4). Deletion of the red edge produces a graph with the same number of spanning
trees as the checkerboard graph of P (1, 2, 4, 3). Contraction of the cyan
edge yields the graph on the right. Right: Result of contracting the cyan
edge of the graph on the left. . . . . . . . . . . . . . . . . . . . . . . . .

88

Figure 5.1: Saddles of F (D) corresponding to two successive over-crossings of D. The
third picture illustrates how D is is alternating on F (D). The figure is
taken from [24]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
Figure 5.2: The saddle of the Turaev surface at a crossing of a link. Taken from [33]. 102
Figure 5.3: The arcs αk are each homotopic to the meridian, and their union projects
to ∂S1 ∪ ∂S2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Figure 5.4: The intersection of the surfaces SA (red) and SB (blue) with ∂C. Taken
from [45]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
Figure 5.5: The braid of example 5.3.5 in the case n = 5. . . . . . . . . . . . . . . . 109

viii

Chapter 1
Introduction
Knot theory is the study of piecewise linear embeddings f : S 1 → S 3 of the circle into the
3-sphere. A knot K ⊂ S 3 is the image of such an embedding. A link L ⊂ S 3 is a union of
disjoint knots. We say that two links L1 and L2 are equivalent if there exists an orientation
preserving piecewise linear homeomorphism h : S 3 → S 3 such that h(L1 ) = L2 . Throughout
this work, we will consider equivalent links to be the same link. Knot theory is a classical
subject of topology and there is a large library of literature devoted to its study. See [47]
and [62] for an introduction to the subject.
A natural way of studying knots and links is via diagrams. A diagram of a link L is a
projection of L onto S 2 together with “over and under” crossing information. For example,
a diagram of the knot known as the 77 knot is given in Figure 1.1. There are many different
diagrams of the same knot. It is a difficult problem in general to determine whether or not
two diagrams represent the same link. For example, the diagrams in Figure 1.2 were long
thought to represent different knots until Perko [59] proved otherwise.
Much of knot theory is centered on the study of link invariants. A link invariant is a rule
f that assigns to each link L some mathematical object f (L) such that if L1 and L2 are
equivalent links then f (L1 ) = f (L2 ). A simple example of a link invariant is the crossing
number, denoted c(L), which is the minimum number of crossings in a diagram of L. Another
example of an invariant is whether a link has an alternating diagram – a diagram where the

1

Figure 1.1: A diagram of the 77 knot.

Figure 1.2: Two diagrams of the same knot.
over and under nature of the crossings alternate as one travels along the link. A link that
has an alternating diagram is called alternating.
Some link invariants, such as the crossing number, are defined in terms of diagrams of
the link. Another interesting class of link invariants arises from the study of hyperbolic geometry. Many knot complements S 3 \K admit a finite-volume hyperbolic structure, meaning
that S 3 \K admits a Riemannian metric of constant curvature −1. Such knots are called
hyperbolic. See Section 2.1.1 for further details. In fact, Thurston [66] showed that any
knot which is not a satellite knot or torus knot is hyperbolic. Further, a celebrated theorem
of Mostow [57] and Prasad [60] states that any two finite volume hyperbolic structures are
unique up to isometry. Therefore properties of the hyperbolic structure of S 3 \K, such as
the volume or lengths of certain geodesics, are invariants of the knot K. The hyperbolic
volume of S 3 \K is of special interest in this work, and we will denote it by vol(K).
2

The following compelling question has drawn significant attention. “How can one relate
the hyperbolic invariants of a knot to their diagrams?” Lackenby [44] showed that one may
provide bounds on the volume of a link in terms of a diagrammatic property known as the
twist number (see Theorem 2.3.2 for more details). Futer, Kalfagianni, and Purcell [30]
found bounds on the volume of a hyperbolic link that depend on coefficients of the Jones
polynomial, a polynomial that may be combinatorially defined.
In this work, we will further study the relationship between hyperbolic invariants of knots
and their diagrams. In Chapter 2, we will present necessary background material. In Chapter
3 we will explore relationships between volume, crossing number, and determinant through
invariants called volume and determinant densities. In Chapter 4 we will further study the
relationship between determinant and volume showing that the determinant may be used in
some cases to find an upper bound on the volume of a link. In Chapter 5 we will explore
how topologically and diagrammatically defined essential spanning surfaces relate with the
hyperbolic invariants of meridian length and cusp volume. Sections 1.1, 1.2, and 1.3 outline
the major results of Chapters 3, 4, and 5 respectively.

1.1

Volume and Determinant Densities

We will begin our study of the relationship between diagrams and the hyperbolic geometry of
links by examining the relationship between the hyperbolic volume of a link and its crossing
number via a quantity known as volume density dvol (K) which is defined by

dvol (K) =

vol(K)
c(K)

where vol(K) is the hyperbolic volume of S 3 \K and c(K) is the crossing number.
3

(1.1)

D. Thurston [64] showed that dvol (K) ≤ v8 for all knots K, where v8 ≈ 3.66286 is the
volume of the regular ideal hyperbolic octahedron. Champanerkar, Kofman, and Purcell
[20] showed that the upper bound of v8 on volume density is asymptotically sharp. More
precisely, there exist sequences of links {Kn }∞
n=0 such that

lim d (Kn )
n→∞ vol

= v8

(see Theorem 3.2.3). Such sequences of links are called geometrically maximal. All the
geometrically maximal sequences of links constructed in [20] were alternating, and did not
contain a cycle of tangles (see Definition 3.2.2). Theorem 3.3.1 will show that there exist
geometrically maximal sequences of links having a cycle of tangles. We then use this to
prove Theorem 3.3.3 which states that there exist geometrically maximal sequences of nonalternating links.
Let Cvol = {vol(K)/c(K) : K is a hyperbolic link} ⊆ R and let Specvol be the set of limit
points of Cvol . We call Specvol the spectrum of volume densities. A question that arose in
[20] was how one may describe the sets Cvol and Specvol . (See also [19] for discussion and
related questions.) In particular, what numbers occur as volume densities? We answer this
question with the following theorem.
Theorem 3.4.3. The set Cvol of volume densities of hyperbolic links is a dense subset of
[0, v8 ], and Specvol = [0, v8 ]. In other words, given x ∈ [0, v8 ] there exists a sequence
{Kn }∞
n=1 of hyperbolic links such that the volume densities satisfy lim dvol (Kn ) = x.
n→∞

Another invariant that was studied in [19] and [20] was the determinant density, ddet (K)

4

of a link K, which is defined by the equation

ddet (K) :=

2 π log(det(K))
c(K)

The authors of [20] conjectured that

2π log(det(K))
< v8
c(K)

(1.2)

for any knot or link K. It is shown in [20] that there are sequences {Kn }∞
n=1 of links such that
limn→∞ ddet (Kn ) = v8 (see Theorem 3.2.4 of this paper). Such sequences are called diagrammatically maximal. We define Cdet = {ddet(K) : K is a (not necessarily hyperbolic) link} ⊆
R and define Specvol to be the set of limit points of Cdet . We call Specdet the spectrum of
determinant densities.
We explore the relationship between Specvol and Specdet , thus connecting a hyperbolic
invariant to a diagrammatic invariant. While we know that v8 is an upper bound on volume
density, it is unknown whether v8 forms an upper bound for Specdet . However, it is a
conjecture [20, Conjecture 1.1] that v8 is an upper bound for Specdet . We will prove the
following theorem in Chapter 3 Section 3.5.
Theorem 3.5.3. The spectrum of determinant densities Specdet contains [0, v8 ]. In other
words, given any x ∈ [0, v8 ] there exists a sequence {Kn }∞
n=1 of links satisfying

lim d (Kn )
n→∞ det

=x

(3.26)

In [19] and [20] it is shown that 0, v8 ∈ Specvol ∩ Specdet , and the authors ask what this
intersection is. As a corollary of Theorems 3.4.3 and 3.5.3 we obtain:
5

Corollary 3.1.2. The intersection Specvol ∩ Specdet is equal to [0, v8 ].
Since Specvol considers only the densities of hyperbolic links, it is useful to know what
determinant densities are realized by hyperbolic links. We show in Theorem 3.5.4 that the
set of determinant densities of hyperbolic links is also dense in [0, v8 ].

1.2

Relating Determinant and Volume

The Alexander polynomial ∆L (t) of a link L is a classical link invariant. It can be defined by
considering the infinite cyclic covering space of S 3 \K, or combinatorially via Skein relations.
One invariant stemming from the Alexander polynomial is the determinant, which is defined
to be |∆L (−1)|. For further information about the Alexander polynomial and determinant,
see [47].
Dunfield noted a relationship between the volume and determinant of a knot in an online
post [26]. He observed that there is a nearly linear relationship between the hyperbolic
volume of an alternating knot and log(J(−1)) where J denotes the Jones polynomial. After
further study of this relationship and some experimentation, Champanerkar, Kofman and
Purcell [20] made the following conjecture.
Conjecture 4.1.1. Let K be a hyperbolic alternating knot. Then vol(K) < 2π log det(K).
Conjecture 4.1.1 is false if we do not require K to be alternating. Kalfagianni [41] showed
that there are knots with trivial Alexander polynomial (hence having determinant 1) which
have arbitrarily large volume. One can use the data from Knotscape [36] and SnapPy [22]
to verify this conjecture for all alternating knots with up to 16 crossings. Champanerkar,
Kofman and Purcell in [20] computationally verified Conjecture 4.1.1 for many examples of

6

an infinite family of links known as weaving knots. Using these weaving knots, they showed
that the constant 2π is sharp in the sense that, given α < 2π, there exists an alternating
link K with α log(det(K)) < vol(K).
In Chapter 4, we will verify that Conjecture 4.1.1 holds for various infinite families of
links. In particular, we have the following theorem.
Theorem 4.1.2. If K is a 2-bridge link or an alternating 3-braid, then

vol(K) ≤ 2π log(det(K))

(4.3)

The proof of Theorem 4.1.2 is a combination of Theorems 4.3.5 and 4.4.1. We also show
that the conjecture holds for “highly twisted” links, i.e. those having many crossings and
few twist regions (see Section 2.3 for a definition of twist region). We further quantify this
by the following theorem.
Theorem 4.5.1. Let K be an alternating hyperbolic link with t twist regions and c crossings.
If
c > t + ξ t−1 − 2γ t−1

(4.42)

where γ ≈ 1.4253 is the unique positive real number satisfying γ −5 + 2γ −4 + γ −3 − 1 = 0
and ξ = e5v4 /π ≈ 5.0296, then vol(K) < 2π log(det(K)).
It follows that if some or all of the twist regions of a link have many crossings, the link
will satisfy Conjecture 4.1.1. An immediate corollary of this is:
Corollary 4.1.3. For any given integer t0 > 0 there are only finitely many links with t0
twist regions which fail to satisfy Conjecture 4.1.1.

7

This can be used, for example (as demonstrated in Section 4.5), to show that all alternating pretzel links with a fixed number t0 of twist regions satisfy Conjecture 4.1.1. The
idea is that if an alternating pretzel link has, for example 3 twist regions, then by Theorem
4.5.1, if it also has more than 25 total crossings, it must satisfy Conjecture 4.1.1. One can
then enumerate all alternating pretzel knots with no more than 25 crossings in 3 twist regions, compute the determinants, and estimate the volumes via other methods to show that
the remaining links also satisfy Conjecture 4.1.1. Using a computer to aid in calculation,
we have shown that all alternating pretzel links with no more than 13 twist regions satisfy
Conjecture 4.1.1. See Section 4.5.1 for details.

1.3

Meridians, Cusp Area, and Essential Spanning Surfaces

Let K be a hyperbolic knot in S 3 . Then there is a well-defined notion of a maximal cusp
neighborhood C of K in S 3 homeomorphic to a solid torus (see Section 2.1.2 for a formal
definition). The area of ∂C is called the cusp area and is an invariant of K. A meridian of K
is a non-separating simple closed curve µ in ∂C such that µ bounds a disk in C. A meridian
of K is isotopic to a unique geodesic, so we will refer to µ as the meridian. The length of
the meridian (µ) is therefore a well-defined quantity and an invariant of a hyperbolic knot.
Related to the meridian is the notion of a λ-curve, which is a non-separating simple closed
curve in ∂C which intersects the meridian exactly once.
We relate the geometric invariants of meridian length and cusp area to topologically defined essential spanning surfaces. A spanning surface S of a knot K is one such that its
boundary ∂S is equal to K. A surface S in a 3-manifold M is essential if it is incompress8

ible and ∂-incompressible (see Section 2.2). Some essential spanning surfaces are defined
combinatorially via diagrams (for example the checkerboard and state surfaces of Section
2.2), so the relationships between meridian length, cusp area, and essential spanning surfaces uncover another connection between diagrams of knots and hyperbolic invariants. The
following theorem demonstrates a relationship between these invariants.
Theorem 5.1.1. Let K be a hyperbolic knot with meridian length (µ). Suppose that K
admits essential spanning surfaces S1 and S2 such that

|χ(S1 )| + |χ(S2 )| ≤

b
i(∂S1 , ∂S2 )
6

(5.1)

where b is a positive real number and i(∂S1 , ∂S2 ) is the minimal intersection number of ∂S1
and ∂S2 on ∂M . Then the meridian length satisfies (µ) ≤ b. Moreover, given a hyperbolic
knot K and b > 0, there is an algorithm to determine if there are essential surfaces S1 and
S2 satisfying (5.1)
An open question is whether meridian lengths are bounded above by 4. Agol [11] and
Purcell [61] have produced examples of knots whose meridian lengths approach 4 from below.
The best known general bound can be found as follows. A slope σ on ∂M is called exceptional
if the 3-manifold M (σ) is not hyperbolic. The Gromov-Thurston “2π-theorem” [12] asserts
that if (σ) > 2π then M (σ) admits a Riemannian metric of negative curvature. This
combined with the proof of Thurston’s geometrization conjecture [56] implies that actually
M (σ) is hyperbolic. The work of Agol [10] and Lackenby [43], that has improved 2π to 6,
asserts that exceptional slopes must have length less or equal to six. Examples of exceptional
slopes with length six are given in [10] and in [4]. Since the meridian curve of every hyperbolic
knot in S 3 is an exceptional slope, we have (µ) ≤ 6. The work of Adams, Colestock, Fowler,
9

Gillam, and Katerman [6] shows that that (µ) < 6. Examples of knots whose meridian
length approaches four from below are given in [10] and by Purcell in [61].
We will show that meridian lengths are bounded above by 4 for broad families of knots.
One broad family can be defined through an invariant called Turaev genus (see Section 5.2.2).
Theorem 5.3.3. Let K be an adequate hyperbolic knot in S 3 with crossing number c = c(K)
and Turaev genus gT . Let C denote the maximal cusp of S 3 \ K and let Area(∂C) denote
the cusp area. Finally let (µ) and (λ) denote the length of the meridian and the shortest
λ-curve of K. Then we have
1. (µ) ≤ 3 +

6gT − 6
c

2. (λ) ≤ 3c + 6gT − 6
3. Area(∂C) ≤ 9c 1 +

2gT − 2 2
c

A knot is alternating if and only if gT = 0. In this case, the bounds of Theorem 5.3.3
agree with the bounds of [6]. The technique of the proof of Theorems 5.1.1 and 5.3.3, as well
as the proof of results in [6], is reminiscent of arguments with pleated surfaces that led to
the proof of the “6-Theorem” [10, 43]. The algorithm for checking the criterion of Theorem
2.1.2 involves normal surface theory and in particular the work of Jaco and Sedgwick [39].
Note that Theorem 5.3.3 implies that whenever the Turaev genus gT ≤ 3 then (µ) ≤ 4.
Another broad family for which (µ) ≤ 4 is “highly twisted” knots, i.e. those with many
crossing per twist region (see Section 2.3 for a definition of twist region), as shown in the
following theorem.
Theorem 5.3.6. Let K be a hyperbolic knot with an adequate diagram with c crossings and

10

t twist regions. Then we have
(µ) ≤ 3 +

3t 6
− .
c
c

In particular if c ≥ 3t then we have (µ) < 4.
Let K be a hyperbolic knot with maximal cusp C and slopes σ, σ on ∂C (for definitions
of maximal cusp and slopes see Section 2.1.2). Calculating area in Euclidean geometry on
∂C (see for example the proof of [10, Theorem 8.1]), we have

(σ) (σ ) ≥ Area(∂C)∆(σ, σ )

(1.3)

where ∆(σ, σ ) denotes the absolute value of the intersection number of σ and σ . Work of
Cao and Meyerhoff [16, Proposition 5.8] shows that Area(∂C) ≥ 3.35. Given an adequate
hyperbolic knot K, we will apply (1.3) for σ = µ. Using the upper bound for (µ) from
Theorem 5.3.3, we have

(σ) >
where δ =

3.35 ∆(µ, σ)
3.35∆(µ, σ)c
=
·
3c + 6gT − 6
3
1+δ

(1.4)

2gT − 2
. Note that δ is an invariant of K that can be calculated from any
c

adequate diagram (see Theorem 5.2.4). Now (1.4) implies that if

∆(µ, σ) >

18
(1 + δ) > 5.37 (1 + δ)
3.35

then (σ) > 6 and thus σ cannot be an exceptional slope.
Note that if σ is a slope represented by p/q ∈ Q in H1 (∂C) then ∆(µ, σ) = |q|. Hence
if |q| > 6(1 + δ), inequality (5.3) implies that (σ) >

3.35
· 6 > 2π. In this case, we may
3

apply a result of Futer, Kalfagianni and Purcell [31, Theorem 1.1] to estimate the change of
11

volume under Dehn filling of adequate knots. This produces the following theorem.
Theorem 5.1.3. Let K be a hyperbolic adequate knot and let δ be as above. If |q| ≥ 6(1 + δ),
then the 3-manifold N obtained by p/q surgery along K is hyperbolic and the volume satisfies
the following

vol(S 3

\ K) > vol(N ) ≥

36(1 + δ)2
1−
q2

3/2

vol(S 3 \ K).

The assertion that N is hyperbolic follows immediately from the discussion above. The
left hand side inequality is due to the result of Thurston that the hyperbolic volume drops
under Dehn filling (Theorem 2.1.2). The right hand side follows by [31, Theorem 1.1].
Theorem 5.14 of [30], and its corollaries, give diagrammatic bounds for vol(S 3 \ K) in
terms any adequate diagram of K. This combined with Theorem 5.1.3 implies that the
volume of N can be estimated from any adequate diagram of K. For example, Montesinos
knots with a reduced diagrams that contains at least two positive tangles and at least two
negative tangles are adequate and have δ ≤ 0. Combining Theorem 5.1.3 with [30, Theorem
9.12] and [29, Theorem 1.2] we have the following.
Corollary 5.1.4. Let K ⊂ S 3 be a Montesinos link with a reduced diagram D(K) that
contains at least two positive tangles and at least two negative tangles. If |q| ≥ 6, then the
3-manifold N obtained by p/q surgery along K is hyperbolic and we have

2v8 t > vol(N ) ≥

36 3/2 v8
(t − 9) ,
1− 2
4
q

where t = t(D) is the twist number of D(K), and v8 = 3.6638... is the volume of a regular
ideal octahedron.
12

Chapter 2
Background
In this chapter we detail relevant background information that will be used throughout
this work. In Section 2.1 we recall definitions and results that are pertinent to the study
of hyperbolic invariants of 3-manifolds. In Section 2.2 we review some properties of knot
diagrams and some notions from the classical study of 3-manifolds. In Section 2.3 we discuss
work that has been done relating hyperbolic volume to diagrammatic properties of links.

2.1

Hyperbolic Geometry

In this section we will define what it means for a 3-manifold to be hyperbolic. We will see
that the choice of hyperbolic metric on a 3-manifold is unique provided the manifold has
finite volume. We will also recall results demonstrating how the topological operation of
Dehn filling affects the geometric properties of a 3-manifold. For an introduction and survey
of work done in the study of hyperbolic 3-manifolds, see [49].

2.1.1

Hyperbolic Structures on Manifolds

Throughout this work, we will consider the upper half-space model of hyperbolic geometry.
We let H3 = {(z, t) ∈ C × R : t > 0} and endow H3 with the metric ds2 =

|dz|2 + dt2
.
t2

It is well-known that the isometries of H3 correspond to PSL(2, C) = SL(2, C)/ ± I where

13

SL(2, C) is the group of 2 × 2 matrices of determinant one and I is the identity matrix. The
group PSL(2, C) corresponds to M¨obius transformations via


az + b
a b 
A(z) =
←→ ± 

cz + d
c d

(2.1)

These M¨obius transformations extend to maps of H3 via

(z, t) →











z + d/c
a
t
+ , 2
− 2
2
2
c (|z + d/c| + t ) c |c| (|z + d/c|2 + t2
a
b
a
t
z+
,
d
a
d

when c = 0
(2.2)
when c = 0

A group Γ of M¨obius transformations is discrete if one of the following equivalent conditions holds:
1. No infinite sequence of distinct elements of Γ converges to a M¨obius transformation.
2. Γ acts properly discontinuously on H3 , i.e. given any closed ball B ⊆ H3 , the set
{g ∈ Γ : g(B) ∩ B = ∅} is finite.
3. Γ has no limit points in H3 , i.e. given x ∈ H3 there is no point y ∈ H3 with an infinite
sequence of distinct elements {gn } in Γ such that limn→∞ gn (y) = x.
A discrete group of M¨obius transformations is called a Kleinian group. Note that while
a Kleinian has no limit points in H3 , it may have limit points on ∂H3 = S 2 . The set
Ω(Γ) = {x ∈ S 2 : x is not a limit point of Γ} is known as the ordinary set.
Given torsion free Kleinian group Γ, one may form a quotient manifold M(Γ) = (H3 ∪

14

Ω(Γ))/Γ, namely the set of equivalence classes

M(Γ) = {[x] : x ∈ H3 ∪ Ω(Γ) where x ≡ y if and only if x = g(y) for some g ∈ Γ}

(2.3)

A manifold M such that M = M(Γ) for some torsion-free Kleinian group Γ is called a
hyperbolic manifold. Note that the hyperbolic metric of H3 descends to a hyperbolic metric
on the quotient manifold M(Γ).
For a given manifold M , there may be many Kleinian groups Γ such that M = M(Γ).
Indeed, if g is a M¨obius transformation then M = M(g −1 Γg). Nor is it necessarily the case
that if M(Γ1 ) is homeomorphic to M(Γ2 ) for torsion-free Kleinian groups Γ1 and Γ2 then
M(Γ1 ) is isometric to M(Γ2 ). However, when M = M(Γ) has finite volume, the following
theorem of Mostow and Prasad shows that the hyperbolic metric on M is independent of
the choice of Γ.
Theorem 2.1.1 ([57] and [60]). Suppose for some Kleinian group Γ1 that M(Γ1 ) is has finite volume. Then any isomorphism ϕ : Γ1 → Γ2 onto another Kleinian group Γ2 is realized
by an isometry M(Γ1 ) → M(Γ2 ).

In particular, if M = M(Γ1 ) is homeomorphic to M(Γ2 ), then π1 (M ) ∼
= Γ1 and π1 (M ) ∼
=
Γ2 . A homeomorphism h : M(Γ1 ) → M(Γ2 ) induces an isomorphism ϕ : Γ1 → Γ2 , and by
Theorem 2.1.1, M(Γ1 ) is isometric to M(Γ2 ). It follows that the geometric properties of a
hyperbolic 3-manifold M do not depend on the choice of Kleinian group used to form the
quotient manifold. Therefore geometric properties of a finite volume hyperbolic 3-manifold
(such as a link complement) are invariants of the manifold. In particular, the volume of a
finite volume hyperbolic 3-manifold is well-defined. A knot K such that S 3 \K is a hyperbolic
15

manifold with finite volume is said to be hyperbolic.

2.1.2

Maximal Cusps and Dehn Filling

Suppose that M is a 3-manifold with torus boundary. One can form a closed 3-manifold
from M by gluing in a solid torus. Formally, let T be a solid torus and h : ∂T → ∂M a
homeomorphism. Then one can form the manifold N = M ∪h T by identifying x ∼ h(x).
We say that N is obtained from M by Dehn filling. It is well-known (see for example [47])
that if h and g are isotopic homeomorphisms then M ∪h T is homeomorphic to M ∪g T .
Suppose that M = S 3 \N (K) where N (K) is a regular open neighborhood of a knot K.
Recall that a meridian µ of a solid torus T is a non-separating curve in ∂T that bounds a
disk in T . Any two meridians of T are equivalent by ambient isotopy of T (see [62]) so we
can refer to the meridian of T . Let T be the closure of N (K) and let µ be the meridian of
T . There is a curve λ in ∂T , unique up to ambient isotopy of T , that is homologically trivial
in M . We can choose to orient µ and λ so that they algebraic intersection number 1. Then
any closed curve c in ∂T is isotopic in ∂T to a curve with homology class a[µ] + b[λ] for some
a, b ∈ Z. The ratio a/b is the slope of c. One may show that the homeomorphism type of
M ∪h T depends only on the slope a/b of the curve h(µ) (see [62]). We define M (a/b) to be
the manifold obtained by Dehn filling sending a meridian to a curve of slope a/b.
We would like to study the effect that Dehn filling has on the geometry of a hyperbolic
3-manifold. In order to do this, we need to define some terminology. An element


a b 
A=
 ∈ PSL(2, C)
c d

16

(2.4)

is parabolic if one of the following equivalent properties hold:


1 1
1. A is conjugate to 

0 1
2. A has exactly one fixed point in S 2 = ∂H3
3. tr(A) = ±2 and A = I where I is the identity matrix.
A horosphere is a set X ⊆ H3 such that there exists a parabolic element g of PSL(2, C) for
which g(X) = X. For example, the plane C × {1} ⊆ H3 is a horosphere since it is preserved
by the M¨obius transformation g(z) = z + 1. In general, horospheres in H3 are Euclidean
spheres tangent to a point in C and horizontal planes. The point of tangency of the former
type of horosphere is called the center of the horosphere. The center of a horosphere that is
a horizontal plane is said to be ∞.
Suppose that K is a knot and S 3 \K is a hyperbolic manifold. The boundary of a
regular neighborhood C of K is homeomorphic to T 2 × [1, ∞) where T 2 is a torus. Let
ρ : H3 → S 3 \K be a covering map. One may choose C = T 2 × [1, ∞) carefully so that
ρ−1 (T 2 ×{1}) is a collection H of disjoint horospheres in H3 . One may expand or contract C
until the horospheres of ρ−1 (C) intersect tangentially. This choice of C is called a maximal
cusp neighborhood of K.
Each horosphere H in ρ−1 (C) inherits a Euclidean structure by restricting the metric
of H3 to H. The cusp area of K, denoted by Area(∂C) is the Euclidean area of ∂C. The
cusp volume, denoted by Vol(C) is the volume of C. It is known that Area(∂C) = 2Vol(C).
Given a slope s in ∂C, the a curve with slope s is isotopic to a unique geodesic in ∂C. We
let (s) denote the length of this geodesic. We define (µ) to be the length of the meridian
of C, and (λ) to be the length of the shortest curve with geometric intersection number 1
17

with the meridian. It is known that cusp volume is bounded above by (µ) (λ).
Dehn filling of S 3 \K is performed by removing the interior of a maximal cusp neighborhood C of K and then attaching a solid torus to the boundary of S 3 \C. Depending on the
choice of slope, s, for the Dehn filling, M (s) may or may not be hyperbolic. For example,
M (1/0) = S 3 which is not hyperbolic. A slope s for which M (s) is not hyperbolic is said to
be exceptional. The Gromov-Thurston 2π theorem [12] showed that if a curve of slope s is
isotopic to a geodesic of length greater than 2π, then M (s) will be hyperbolic. This bound
was further improved by Agol [10], who showed that if the slope length exceeds 6 then M (s)
is hyperbolic. In particular, a meridian must always have length less than 6.
The following theorem, due to Thurston, details the effect that Dehn filling has on the
volume of a hyperbolic 3-manifold.
Theorem 2.1.2 ([65]). Let M be a hyperbolic manifold with torus boundary components and
N the result of Dehn filling along one of the boundary components of M . Then vol(M ) >
vol(N ).
By Theorem 2.1.2 we know that if the result of Dehn filling is a hyperbolic manifold,
then the volume must drop. A question that one may ask is how much the volume drops
after Dehn filling. Futer, Kalfagianni, and Purcell proved the following theorem in answer
to this question.
Theorem 2.1.3 ([31]). Let M be a complete, finite volume hyperbolic manifold with torus
boundary components C1 , . . . , Ck . are disjoint horoball neighborhoods of some subset of the
cusps. Let s1 , . . . , sk be slopes on ∂C1 , . . . , ∂Ck , each with length greater than 2π. Denote
the minimal slope length by min . Let M (s1 , . . . , sk ) be the result of Dehn filling with slopes

18

s1 , . . . , sk on ∂C1 , . . . , ∂Ck . Then M (s1 , . . . , sk ) is a hyperbolic manifold, and

vol(M (s1 , . . . , sk )) ≥

2.2

1−

2π

2

3/2

vol(M )

(2.5)

min

Essential Surfaces

A surface S is said to be properly embedded in a 3-manifold M if ∂S is embedded in ∂M
and the interior of S is embedded in the interior of M . A compressing disk D for S in M
is a disk embedded in M such that ∂S ∩ D = ∂D. A properly embedded surface S in M is
incompressible if, given a compressing disk D for S in M , the curve ∂D bounds a disk in S.
A ∂-compressing disk D for S in M is one such that
1. D ∩ S is an arc α in ∂D
2. D ∩ M is an arc β in ∂M
3. ∂D = α ∪ β
4. α ∩ β = ∂α = ∂β
5. α does not co-bound a disk with an arc in ∂S
A surface is ∂-incompressible if there are no ∂-compressing disks for S in M . A surface that
is both incompressible and ∂-incompressible is called essential.
We now provide some examples of essential surfaces in knot complements that will be
used in this work. One example is the checkerboard surface which is described as follows.
Given a diagram D for a knot K, one may color the regions of D black and white so that
each black (and respectively white) region meets a black (respectively white) region at the
19

Figure 2.1: Left: The 77 knot. Center: Checkerboard coloring of the 77 knot diagram.
Right: The checkerboard graph associated to the black coloring.
crossings. See for example Figure 2.1. One can form a checkerboard surface by placing a
disk in each black region of the coloring and then replacing a neighborhood of each crossing
with a twisted band. Note that the boundary of the checkerboard surface is equal to the
knot, and that the surface is not necessarily orientable. One can form a checkerboard surface
in a similar manner by considering the white regions instead of the black regions.
Related to checkerboard surfaces for a knot are checkerboard graphs. A checkerboard
graph is formed by checkerboard coloring the knot diagram, placing each vertex in each
black region (alternatively in each white region), and connecting vertices with one edge per
crossing of D connecting the corresponding shaded (respectively white) regions. See the
right of Figure 2.1 for an example.
An alternating diagram of a link is reduced if it does not contain a nugatory crossing,
that is a crossing of the form shown in Figure 2.2. A diagram D of a knot is prime if a
circle S 1 in the diagram meeting the knot transversely in two points bounds an unknotted
arc on one side. It was shown by Menasco and Thistlethwaite [51] that when D is a reduced
alternating prime diagram for a knot K, then the checkerboard surfaces are essential.
Another type of essential surface that arises is related to the notion of adequacy for links.
Let D be a diagram for a knot K. At each crossing of the diagram D one may resolve the

20

X

Y

Figure 2.2: A nugatory crossing.
crossing in one of two ways: the A-resolution and the B-resolution as depicted in Figure 2.3.
One often records the resolution by replacing the crossing with an arc, for example the red
arc in Figure 2.3. A choice of resolution at every crossing of D is called a state σ. The result
of applying the state σ to D, denoted sσ (D), is a collection of disjoint circles called state
circles. See for example Figure 2.4, where the thick lines of the center figure are the state
circles arising from the all-A resolution of the link on the left. One may then form the state
graph Gσ where vertices correspond to state circles of sσ (D) and and edges correspond to
former crossings in D.
Definition 2.2.1. A diagram D is called A-adequate (respectively B-adequate) if the state
graph of the all-A resolution (respectively all-B resolution) has no 1-edge loops. A diagram
that is both A and B adequate is called adequate. A knot is called adequate if it has an
adequate diagram.
Given a diagram D of a knot K, one may form a surface SA as follows. The state circles
of the all-A resolution of D bound disks on the projection plane. Isotope these disks slightly
off the projection plane so they become disjoint. For each crossing of D, attach a half-twisted
band so that the resulting surface SA has boundary ∂SA = K. One may form the surface
SB similarly. See the right of Figure 2.4, for an example of a state surface.
The following theorem is due to Ozawa [58]. A different proof is given by Futer, Kalfagianni, and Purcell [30, Theorem 3.19].
21

Figure 2.3: The two resolutions of a crossing, the arcs recording them, and their contribution
to state surfaces. The left frame depicts the A-resolution, and the right depicts the Bresolution.

Figure 2.4: Left: A link L. Center: The all-A resolution. The thick lines are the state
circles, and the thin lines are arcs recording the resolutions. Right: The all-A state surface
of L. Figure taken from [30].
Theorem 2.2.2. Let D(K) be an adequate link diagram of a knot K. Then the all-A state
and the all-B state surfaces corresponding to D(K) are essential in S 3 \K.

2.3

Twist Number and Volume

A topic explored by researchers is finding bounds on the hyperbolic volume of a knot in
terms of information easily read from a diagram. This section recalls some results in this
direction, in particular volume bounds in terms of the number of twist regions of a diagram,
which we now define.
Definition 2.3.1. A twist region of a diagram D is either a connected collection of bigon
regions of D arranged in a row, which is maximal in the sense that it is not part of a
longer row of bigons, or a single crossing adjacent to no bigon regions. The twist number
22

of a diagram D is the number of twist regions in the diagram. A diagram is twist reduced
whenever a simple closed curve in the diagram intersects the link projection transversely in
four points disjoint from the crossings, and two of these points are adjacent to some crossing,
and the remaining two points are adjacent to some other crossing, then this curve bounds
a subdiagram that consists of a (possibly empty) collection of bigons arranged in a row
between these two crossings.
Theorem 2.3.2 ([44]). Let K be an alternating hyperbolic link with t twist regions in a
prime alternating diagram. Then

v4
(t − 2) < vol(K) < 10v4 (t − 1)
2

(2.6)

where v4 ≈ 1.01494 is the volume of a regular ideal tetrahedron.
The upper bound of Theorem 2.3.2 can be improved in the case of Montesinos links using
work of Futer, Kalfagianni, and Purcell.
Theorem 2.3.3 ([30, Theorem 9.12]). Let K be a hyperbolic Montesinos link. Then

vol(K) < 2v8 t

(2.7)

where t is the number of twists in some diagram of K and v8 ≈ 3.66386237 is the volume of
a regular ideal hyperbolic octahedron.
Note that while the statement of Theorem 2.3.3 in [30] requires the link to have at least
three positive tangles and at least three negative tangles, this condition was only necessary
to prove the lower bound stated in that theorem.

23

Chapter 3
Volume and Determinant Densities

3.1

Introduction

Recent work of Champanerkar, Kofman, and Purcell ([19], [20], and [21]) investigated a
relationship between the volume of a knot, its determinant, and its crossing number. In this
chapter we continue to explore this relationship. We recall the definition of the following
invariant relating volume and crossing number.
Definition 3.1.1. Given a hyperbolic link K, let vol(K) be its hyperbolic volume, and let
c(K) be the crossing number of K. The volume density of K is defined to be

dvol (K) :=

vol(K)
c(K)

D. Thurston [64] showed that dvol (K) ≤ v8 for all knots K, where v8 ≈ 3.66286 is the
volume of the regular ideal hyperbolic octahedron. This was done by decomposing S 3 \K
into octahedra, placing one octahedron at each crossing, and pulling the remaining vertices
to ±∞. Adams [2] showed that vol(K) ≤ (c(K) − 5)v8 + 4v4 for any link having c(K) ≥ 5,
where v4 ≈ 1.01494 is the volume of the ideal hyperbolic tetrahedron. Therefore v8 is a
strict upper bound for the volume density of any finite link.
Champanerkar, Kofman, and Purcell [20] showed that the upper bound of v8 on volume

24

Figure 3.1: The infinite weave W.
density is asymptotically sharp. More precisely, there exist sequences of links {Kn }∞
n=0 such
that
lim d (Kn )
n→∞ vol

= v8

(see Theorem 3.2.3). Such sequences of links are called geometrically maximal. The authors
of [20] produce examples of geometrically maximal sequences of links by constructing links
approaching the infinite weave W, the infinite alternating link with the square lattice projection depicted in Figure 3.1. All the links in the examples constructed were alternating,
and do not contain a cycle of tangles (see Definition 3.2.2). In Section 3.3 we will show that
there exist geometrically maximal sequences of links having a cycle of tangles (see Theorem
3.3.1). We then use this to prove Theorem 3.3.3 which states that there exist geometrically
maximal sequences of non-alternating links.
Let Cvol = {vol(K)/c(K) : K is a hyperbolic link} ⊆ R and let Specvol be the set of
limit points of Cvol . We call Specvol the spectrum of volume densities. A question that arose
25

in [20] was how one may describe the sets Cvol and Specvol . (See also [19] for discussion
and related questions.) In particular, what numbers occur as volume densities? Since v8
forms an upper bound on volume density, it is clear that Cvol and Specvol are subsets of
[0, v8 ]. The existence of geometrically maximal knots implies that v8 ∈ Specvol . In Lemma
3.4.1, we describe geometrically minimal sequences of links, i.e. those with volume density
approaching 0. Hence 0 ∈ Specvol . It was shown by Champanerkar, Kofman, and Purcell in
[21] that 2v4 ∈ Specvol . We are able to prove the following theorem.
Theorem 3.4.3. The set Cvol of volume densities of hyperbolic links is a dense subset of
[0, v8 ], and Specvol = [0, v8 ]. In other words, given x ∈ [0, v8 ] there exists a sequence
{Kn }∞
n=1 of hyperbolic links such that the volume densities satisfy lim dvol (Kn ) = x.
n→∞

Another invariant that was studied in [19] and [20] was the determinant density, ddet (K)
of a link K, which is defined by the equation

ddet (K) :=

2 π log(det(K))
c(K)

It is shown in [20] that there are sequences {Kn }∞
n=1 of links such that limn→∞ ddet (Kn ) = v8
(see Theorem 3.2.4 of this paper). Such sequences are called diagrammatically maximal. We
define Cdet = {ddet(K) : K is a (not necessarily hyperbolic) link} ⊆ R and define Specvol to
be the set of limit points of Cdet . We call Specdet the spectrum of determinant densities.
It is interesting to study the relationships between Specvol and Specdet . While we know
that Specvol ⊆ [0, v8 ], it is unknown whether v8 forms an upper bound for Specdet . However,
it is a conjecture [20, Conjecture 1.1] that v8 is an upper bound for Specdet . We are able to
prove the following theorem.

26

Theorem 3.5.3. The spectrum of determinant densities Specdet contains [0, v8 ]. In other
words, given any x ∈ [0, v8 ] there exists a sequence {Kn }∞
n=1 of links satisfying

lim d (Kn )
n→∞ det

=x

(3.26)

In [19] and [20] it is shown that 0, v8 ∈ Specvol ∩ Specdet , and the authors ask what this
intersection is. As a corollary of Theorems 3.4.3 and 3.5.3 we obtain:
Corollary 3.1.2. The intersection Specvol ∩ Specdet is equal to [0, v8 ].
Corollary 3.1.3. The intersection Specvol ∩ Specdet is equal to [0, v8 ].
Since Specvol considers only the densities of hyperbolic links, it is useful to know what
determinant densities are realized by hyperbolic links. We show in Theorem 3.5.4 that the
set of determinant densities of hyperbolic links is also dense in [0, v8 ].
The results of this chapter appeared in [14]. Shortly after the author posted the results of
this paper, Colin Adams, Aaron Calderon, Xinyi Jiang, Alexander Kastner, Gregory Kehne,
Nathaniel Mayer, and Mia Smith [5] independently proved results similar to those in this
paper. In particular, they proved that the set of volume densities is dense in [0, v8 ] and that
the set of determinant densities is also dense in [0, v8 ]. Their proof is similar to the one
found in this paper. They consider knots similar to the knots N (Wn,n ) of Definition 3.3.2
of this paper, and they cite work of Futer, Kalfagianni, and Purcell [31] and Thurston [65]
to obtain volume bounds. We likewise cite [65], and we cite a result of Futer, Kalfagianni,
and Purcell [32], which was derived from [31], to obtain our volume bounds.

27

3.2
3.2.1

Background
Geometrically and Diagrammatically Maximal Links

In this subsection we recall definitions and results from [20]. Given a link K ⊂ R3 we
define the projection graph G(K) to be the 4-valent graph obtained by projection of K onto
R2 × {0}.
Definition 3.2.1. Let G be a possibly infinite graph. For any finite subgraph H, let ∂H be
the set of vertices of H that share an edge with a vertex not in H. Let | · | denote the number
of vertices in a finite graph. An exhaustive nested sequence of finite, connected subgraphs
{Hn ⊆ G : Hn ⊆ Hn+1 , ∞
n=0 Hn = G} is a Følner sequence for G if
|∂Hn |
= 0.
n→∞ |Hn |
lim

The graph G is amenable if a Følner sequence for G exists. In particular, the infinite square
lattice (i.e. the projection graph of W) is amenable.
Given a link K in S 3 , a Conway sphere is a 2-sphere in S 3 intersecting K transversely in
four points. Using Conway spheres, we will define a cycle of tangles, which is an important
definition for the statement of Theorem 3.2.3.
Definition 3.2.2. A Conway sphere is called visible if it is parallel to one dividing the
diagram into two tangles, as in Figure 3.2. A tangle is called knotty if it is nontrivial, and
not a (portion of a) single twist region; i.e. not a rational tangle of type n or 1/n for n ∈ Z.
We will say that Kn contains a cycle of tangles if Kn contains a visible Conway sphere with
a knotty tangle on each side.

28

Figure 3.2: The dashed line represents a visible Conway sphere.
The following theorem was proved by Champanerkar, Kofman, and Purcell.
Theorem 3.2.3 ([20, Theorem 1.4]). Let Kn be any sequence of hyperbolic alternating link
diagrams that contain no cycle of tangles, such that
1. there are subgraphs Gn ⊆ G(Kn ) that form a Følner sequence for G(W), and
|Gn |
= 1.
n→∞ c(Kn )

2. lim

vol(Kn )
Then {Kn }∞
lim
= v8 .
n=0 is geometrically maximal: n→∞
c(Kn )
The following theorem, proved in [20, Theorem 1.5], gives a similar statement for diagrammatically maximal links.
Theorem 3.2.4. Let {Kn }∞
n=0 be a sequence of alternating link diagrams such that
1. there are subgraphs Gn ⊆ G(Kn ) that form a Følner sequence for G(W), and
|Gn |
= 1.
n→∞ c(Kn )

2. lim

det(Kn )
Then {Kn }∞
lim
= v8 .
n=0 is diagrammatically maximal: n→∞
c(Kn )
29

T
N (T )
T

D(T )
T

B(T )
T
C

Figure 3.3: A tangle T , its two closures N (T ) and D(T ), and its corresponding belted tangle
B(T ).
Note that the cycle of tangles condition is not necessary in the construction of diagrammatically maximal links. This is one reason motivating Theorem 3.3.1, which states that
there are geometrically maximal sequences of links containing a cycle of tangles.

3.2.2

Conway Sums and Belted Sums of Links

Let T be a 2-tangle. There are two ways to close the tangle T to form a link. Denote
these closures N (T ) and D(T ) as depicted in Figure 3.3. The closure N (T ) is called the
numerator closure and D(T ) is called the denominator closure. Next, one may add an extra
component, C, called the belt component, to the link N (T ) in the following manner. Let C
be an unknotted circle that lies in a plane orthogonal to the projection plane and encircles
the arcs added to T to form N (T ), as in Figure 3.3. The resulting link, B(T ), is the belted
link corresponding to T or simply the belted tangle.
Given two tangles T1 and T2 such that B(T1 ) and B(T2 ) are hyperbolic, one may form
the belted sum of B(T1 ) and B(T2 ) as follows. Let Mi be the link complement S 3 \B(Ti ),
for i = 1, 2. Work of Adams [8] shows that the belt component Ci of Ti bounds a totally
geodesic, twice-punctured disk in Mi . Cut each Mi along this twice-punctured disk to form
a manifold Mi . By Adams [8], there is a unique hyperbolic structure on a twice-punctured

30

T1

T2

T1

T2

Figure 3.4: The belted sum of two belted tangles.

T1

T2

T1

T2

Figure 3.5: The Conway sum of two tangles. The tangle T1 + T2 is pictured on the right.
disk. Therefore we may glue M1 to M2 via an isometry of twice-punctured disks that maps
C1 to C2 . The result is a link, denoted B(T1 ) + B(T2 ), which we call the belted sum of
B(T1 ) and B(T2 ). See Figure 3.4.
The following theorem about belted sums of links will be important in the proofs of
Theorems 3.4.3, 3.3.1, 3.3.3.
Lemma 3.2.5 ([8, Corollary 5.2]). Suppose that n belted tangles B(T1 ), . . . B(Tn ) are hyperbolic with finite volume. Then the belted sum B(T1 ) + . . . + B(Tn ) is also a hyperbolic link
and has volume vol(B(T1 )) + . . . + vol(B(Tn )).
In addition to the belted sum of belted tangles, we can sum tangles in the manner
described below.
Definition 3.2.6. Let T1 and T2 be 2-tangle diagrams. One may connect these 2-tangle
diagrams as indicated in Figure 3.5 to form the 2-tangle T1 + T2 . We call T1 + T2 the tangle
sum, or Conway sum of T1 and T2 .
Lemma 3.2.5 showed that volume is additive under belted sums of tangles. Similarly, we
have the following lemma describing the determinant of a Conway sum of tangles.

31

Lemma 3.2.7. Let T1 , . . . , Tn be tangles and let T = T1 + . . . + Tn be the tangle sum. Then
n

det(D(T )) =

det(D(Ti )).
i=1

Proof. When n = 2 we have the following well-known fact [47, Proposition 6.12]:

det(D(T1 + T2 )) = det(D(T1 )#D(T2 )) = det(D(T1 )) det(D(T2 ))

The result now follows by induction.

We now recall a result of Futer, Kalfagianni, and Purcell [32] which will play an important
role in the proofs of Theorems 3.3.1, 3.3.3, and 3.4.3. We begin with the following definition.
A tangle T is an east-west twist if N (T ) is the standard diagram of a (2, q)-torus link. To
simplify notation throughout, we define ξn as follows:

ξn :=

1−

8π

√
11.524 + n 4 2

2

3/2

(3.1)

Note that lim ξn = 1.
n→∞

Theorem 3.2.8. Let T1 , . . . , Tn , n ≥ 12, be tangles admitting prime, alternating diagrams,
none of which is an east-west twist. Let K be the numerator closure of the Conway sum
N (T1 + . . . + Tn ). Let L be the belted sum B(T1 ) + . . . + B(Tn ). Then K is hyperbolic and

vol(K) ≥ ξn vol(L)

(3.2)

Proof. This follows directly from the proof of [32, Theorem 1.5] as follows. In the second to
32

last line of that proof, the authors showed that

vol(K) ≥

1−

2π

2

3/2

vol(L)

(3.3)

min

where min is the minimum possible length of the meridian of the belt. They also showed in
√
11.524 + n 4 2
, hence vol(K) ≥ ξn vol(L).
the same proof that min ≥
4

3.2.3

Adequate Tangles

We state some results about adequate link diagrams that will play an important role in the
proofs of Theorems 3.3.1, 3.3.3, 3.4.3, and 3.5.3.
Definition 3.2.9. A link diagram D ⊂ S 2 is prime if any simple closed curve in S 2 that
meets D transversely at two points bounds, on one side of it, a disk that intersects D in a
diagram U of the unknotted ball-arc pair. The diagram D is strongly prime if, in addition,
U is the zero-crossing diagram.
We will use the following theorem, which follows from [30, Corollary 3.21] to show that
a link diagram is prime.
Theorem 3.2.10. Suppose that K is a non-split, prime link. Then every adequate diagram
of K without nugatory crossings is prime.
The following proposition gives a way to show that a link is non-alternating. We will use
this in the proof of Theorem 3.3.3 where we show that there exist geometrically maximal
sequences of non-alternating links.
Proposition 3.2.11. If L has a prime, non-alternating, adequate diagram, then L is nonalternating.
33

Proof. Let L be a link and D a diagram for L having n crossings. Let breadth(L) be the
breadth of the Jones polynomial for L, in other words, the difference between the maximal
and minimal degrees of the Jones polynomial for L. We recall the following facts which may
be found in [47, Proposition 5.3, Theorem 5.9, and Corollary 5.14].
1. If D is reduced and alternating, then D is adequate.
2. If D is an adequate diagram, then c(L) = n.
3. If D is reduced and alternating, then breadth(L) = n.
4. If D is non-alternating and prime, then breadth(L) < n.
Suppose D1 is a prime, non-alternating, adequate diagram for L, and that D2 is an
alternating diagram for L. By removing nugatory crossings, we may assume that D2 is a
reduced, alternating diagram. Therefore both D1 and D2 are adequate diagrams and have
the same number of crossings by facts (1) and (2) above. Let n be the number of crossings
in D1 (or D2 ). Since D1 is non-alternating, fact (4) above implies breadth(L) < n. On the
other hand, since D2 is reduced and alternating, fact (3) implies that breadth(L) = n, a
contradiction.
Definition 3.2.12. A tangle diagram T is adequate if the diagrams of both N (T ) and D(T )
are adequate. In the event that both closures produce reduced, alternating diagrams, the
tangle T is said to be strongly alternating. A strongly alternating tangle is adequate.
Proposition 3.2.13. The tangle sum of n adequate tangles is adequate for n ≥ 2.
Proof. It follows from work of Lickorish and Thistlethwaite [48, Proposition 4] that the tangle
sum of two adequate tangles is adequate. The proposition readily follows by induction on
n.
34

We close this section with the following corollary showing that crossing number of adequate tangles is additive. This plays a key role in the proof of Theorems 3.4.3 and 3.5.3.
Corollary 3.2.14. Let T1 , . . . , Tn be adequate tangle diagrams and let c(N (Ti )) be the crossing number of the closure for i = 1, . . . n. Let T1 + . . . + Tn be the tangle sum. Then the
crossing numbers satisfy

c(N (T1 + . . . + Tn )) = c(N (T1 )) + . . . + c(N (Tn ))
c(D(T1 + . . . + Tn )) = c(D(T1 )) + . . . + c(D(Tn ))

Proof. The natural diagram representing the link N (T1 + . . . + Tn ) as a Conway sum has
c(N (T1 )) + . . . + c(N (Tn )) crossings. Since T1 , . . . , Tn are adequate, we know that N (T1 +
. . . + Tn ) is adequate by Proposition 3.2.13. Note that an adequate diagram of a link has
the minimal number of crossings by Proposition 3.2.11. Therefore c(N (T1 + . . . + Tn )) =
c(N (T1 )) + . . . + c(N (Tn )). The proof that c(D(T1 + . . . + Tn )) = c(D(T1 )) + . . . + c(D(Tn ))
is similar.

3.3

Non-alternating Geometrically Maximal Knots

We now address the question of whether all the conditions in Theorem 3.2.3 are necessary.
We begin with a construction of a geometrically maximal sequence of links for which each
link contains a cycle of tangles.
Theorem 3.3.1. Let {Ti }∞
i=1 be a sequence of tangles admitting prime, alternating diagrams,
none of which is an east-west twist. Let Ki = N (Ti ) for all i. Suppose that each Ki is
hyperbolic and that the sequence {Ki }∞
i=1 is geometrically maximal. For each n ∈ N, let
35

T (n) be the Conway sum of n copies of Tn , and define K(n) = N (T (n)). Then K(n)
is hyperbolic whenever n ≥ 12, and the sequence {K(n)}∞
n=1 is a geometrically maximal
sequence of links, all but one of which contains a cycle of tangles.
Proof. By construction, whenever n ≥ 2, we have that K(n) contains a cycle of tangles.
Moreover, whenever n ≥ 12 we have from Theorem 3.2.8 that K(n) is hyperbolic. We now
show that the sequence {K(n)}∞
n=1 is geometrically maximal. Let Li = B(Ti ), the belted
link corresponding to Li , and for each n ∈ N let L(n) be the belted sum of n copies of Ln .
By Theorem 3.2.8 we have that vol(K(n)) ≥ ξn vol(L(n)) where ξn was defined in (3.1). We
obtain
n

vol(L(n)) =

vol(Ln )

by Lemma 3.2.5

i=1

= n vol(Ln )
> n vol(Kn )

by Theorem 2.1.2

(3.4)

Observe that c(K(n)) ≤ n c(Kn ) since the natural diagram demonstrating K(n) as a Conway
sum has n c(Kn ) crossings. Therefore

vol(K(n))
vol(K(n))
≥
c(K(n))
n c(Kn )
≥ ξn

vol(L(n))
n c(Kn )

by Theorem 3.2.8

> ξn

n vol(Kn )
n c(Kn )

by (3.4)

36

(3.5)

Note that limn→∞ ξn = 1. Since {Kn }∞
n=1 is geometrically maximal, we know that
vol(Kn )
= v8
n→∞ c(Kn )
lim

hence by (3.5) we obtain

vol(K(n))
vol(Kn )
lim dvol (K(n)) = lim
≥ lim ξn
= v8
n→∞
n→∞ c(K(n))
n→∞
c(Kn )
Moreover, dvol (K(n)) ≤ v8 for all n, as this is true for any link. Therefore {K(n)}∞
n=1 is
geometrically maximal.
We can use a method similar to the proof of Theorem 3.3.1 to prove that there exist
geometrically maximal sequences of non-alternating links. We will construct a specific example of such a sequence. We begin by considering a family of weaving tangles in Definition
3.3.2. By taking a Conway sum of these weaving tangles and their reflections (i.e. the result of changing all the over-crossings to under-crossings and vice versa), we will obtain a
geometrically maximal sequence of non-alternating links.
Definition 3.3.2. Let Bm be the braid group on m strings. For i = 1, . . . m−1, let σi ∈ Bm
be the the ith Artin generator, i.e. the group element corresponding to twisting the ith strand
(−1)m−1 n
)

under the (i + 1)st strand. Let Lm,n be the element (σ1 σ2−1 σ3 σ4−1 . . . σm−1

∈ Bm

depicted on the left hand side of Figure 3.6. Let Wm,n be the result of closing all but the
second and third strands of the braid as in the right hand side of Figure 3.6. We will call
Wm,n the weaving tangle of order n on m strands. The knots N (Wm,n ) are the weaving
knots of [21].
Theorem 3.3.3. There exist geometrically maximal sequences of non-alternating links.
37

Figure 3.6: Left: The braid element L6,3 . Right: The weaving tangle W6,3 .

Wn,n

W n,n

Wn,n

W n,n

Wn,n

W n,n

Figure 3.7: The link K(n) for the proof of Theorem 3.3.3. It contains n copies of Wn,n and
n copies of W n,n .

38

Proof. Let Wk,k be the weaving tangle and let W k,k be its reflection, i.e. the result of
changing each over-crossing to an under-crossing and vice versa. For n ≥ 1 define

K(n) = N (Wn,n + W n,n + . . . + Wn,n + W n,n )

to be the closure of the Conway sum of n copies of Wn,n with n copies of W n,n as indicated in
Figure 3.7. Let Kn = N (Wn,n ) and K n = N (W n,n ), and Ln = B(Wn,n ) and Ln = B(W n,n ).
Note that Kn and K n are Dehn fillings of Ln and Ln , respectively.
We prove that the sequence {K(n)}∞
n=1 is geometrically maximal. First, observe that
when n ≥ 6, Theorem 3.2.8 implies that K(n) is hyperbolic. Let

L(n) = B(Wn,n ) + B(W n,n ) + . . . + B(Wn,n ) + B(W n,n )

be the belted sum of n copies of B(Wn,n ) with n copies of B(W n,n ). Similar to the proof of
Theorem 3.3.1 we have
n

(vol(Ln ) + vol(Ln ))

vol(L(n)) =

by Lemma 3.2.5

i=1

= n[vol(Ln ) + vol(Ln )]
> n[vol(Kn ) + vol(K n )]

by Theorem 2.1.2

(3.6)

Since Kn and Kn are reduced and alternating, by counting the number of crossings in Figure
3.6 and using Proposition 3.2.11, we find that c(Kn ) = c(K n ) = n(n − 1). Now the diagram
of K(n) depicted in Figure 3.7 has 2n2 (n − 1) crossings, implying c(K(n)) ≤ 2n2 (n − 1).

39

Therefore

vol(K(n))
vol(K(n))
≥ 2
c(K(n))
2n (n − 1)
≥ ξ2n

vol(L(n))
2n2 (n − 1)

by Theorem 3.2.8

> ξ2n

n[vol(Kn ) + vol(K n )]
2n2 (n − 1)

by (3.6)

ξ
= 2n
2

vol(Kn ) vol(K n )
+
c(Kn )
c(K n )

ξ
= 2n [d(Kn ) + d(K n )]
2

(3.7)
(3.8)

Both Kn and K n are geometrically maximal by Theorem 3.2.3, so

lim d(Kn ) = lim d(K n ) = v8

n→∞

n→∞

(3.9)

Moreover, limn→∞ ξ2n = 1, so (3.8) implies that

lim d(K(n)) = v8 .

n→∞

We show that K(n) is non-alternating. Let D be the diagram shown in Figure 3.7.
Observe that D is non-alternating. Since Wk,k and W k,k are strongly alternating (hence
adequate) tangle diagrams, Proposition 3.2.13 implies that D is adequate. Next, we establish
the conditions needed to apply Theorem 3.2.10. Since an adequate diagram has the minimal
number of crossings, there are no nugatory crossings in D. An adequate (hence minimal
crossing) diagram of a split link must also be split. Since D is non-split, it follows that K(n)
is non-split. Now K(n) is hyperbolic when n ≥ 6, as noted above. Therefore K(n) is a prime
link. It now follows from Theorem 3.2.10 that the diagram D is prime. Proposition 3.2.11
40

now implies that K(n) is non-alternating.
Observe that the sequence of links {K(n)}∞
n=1 in the proof of Theorem 3.3.3 is geometrically maximal and each link of the sequence contains a cycle of tangles.

3.4

The Spectrum of Volume Densities

We now turn to the proof that the set of volume densities is dense in [0, v8 ]. In Lemma 3.4.1,
we prove the existence of tangles whose closures form a geometrically minimal sequence of
links, i.e. a sequence links with volume density approaching 0. On the other hand, Lemma
3.4.2 produces a sequence of tangles whose closures form a geometrically maximal sequence
of links.
The idea of the proof of Theorem 3.4.3 is to combine geometrically maximal tangles
with geometrically minimal tangles via Conway sum. Key to the proof is the fact that
hyperbolic volume is additive under belted sum (see Lemma 3.2.5), and that the crossing
number of adequate tangles is additive under Conway sum (see Corollary 3.2.14). The ratio
of geometrically maximal to geometrically minimal tangles may then be controlled so that
one may find a sequence of links with volume density approaching any number in the interval
[0, v8 ].
Lemma 3.4.1. There exists a sequence of strongly alternating tangles {Tm }∞
m=1 such that

lim d (N (Tm ))
m→∞ vol

=0

and N (Tm ) is hyperbolic for all m ≥ 7.
Proof. Let P ,m,n be the pretzel tangle shown in Figure 3.8. It is well-known that many
41

m

n

Figure 3.8: The pretzel tangle P ,m,n has , m, and n crossings in the respective twist regions.
pretzel links are hyperbolic. For example, whenever , m, n ≥ 7, work of Futer, Kalfagianni,
and Purcell [31, Theorem 1.2] implies that N (P ,m,n ) is hyperbolic. Moreover, P ,m,n is
strongly alternating whenever , m, and n ≥ 2. Fix

≥ 7 and n ≥ 7 and let Km =

N (P ,m,n ). By work of Lackenby, Agol, and D. Thurston [44], we know that vol(Km ) ≤
10v4 (tw(Km ) − 1), where tw(Km ) is the number of twist regions in a diagram of Km . In
particular, tw(Km ) = 3 for all values of m so vol(Km ) ≤ 20v4 . It follows that

lim d (Km )
m→∞ vol

= lim

m→∞

vol(Km )
20v4
≤ lim
=0
m→∞
c(Km )
+m+n

Lemma 3.4.2. There exists a sequence of strongly alternating (hence adequate) tangles
∞
{Tn }∞
n=1 such that {N (Tn )}n=1 is geometrically maximal.

Proof. Consider the weaving tangles Wm,n from Definition 3.3.2 and depicted in Figure 3.6.
Let Tn = Wn,n . It was shown in [21] that {N (Tn )}∞
n=1 is a geometrically maximal sequence of
links. Note that N (Tn ) and D(Tn ) have reduced, alternating diagrams, whenever n ≥ 4.

42

We have established examples of geometrically minimal and geometrically maximal sequences of adequate links. We restate and prove Theorem 3.4.3.
Theorem 3.4.3. The set Cvol of volume densities of hyperbolic links is a dense subset of
[0, v8 ], and Specvol = [0, v8 ]. In other words, given x ∈ [0, v8 ] there exists a sequence
{Kn }∞
n=1 of hyperbolic links such that the volume densities satisfy lim dvol (Kn ) = x.
n→∞

Proof. It suffices to show that for any non-negative integers a and b with a and b not both
zero, and any

> 0, there exists a link K such that

b
b
v8 − < dvol (K) <
v + .
a+b
a+b 8

Let > 0 be arbitrary. Let P ,m,n be the pretzel tangles from Lemma 3.4.1. Let T1 = P7,m,7
and K1 = N (T1 ), where m is chosen to be large enough that
40 a v4
<
(a + b)(m + 14)
2

(3.10)

In Lemma 3.4.2, it was pointed out that the sequence of links {N (Wk,k )}∞
k=1 , where Wk,k is
the weaving tangle, is a geometrically maximal sequence of links. Therefore we may choose
T2 = Wk,k and K2 = N (T2 ) with k sufficiently large that
vol(K2 )
> v8 −
c(K2 )

a+b
b

2

and

c(K2 ) ≥ c(K1 )

(3.11)

Since K1 and K2 have reduced alternating diagrams, by counting the number of crossings
in these diagrams, we see that

k(k − 1) = c(K2 ) ≥ c(K1 ) = m + 14.
43

(3.12)

Let m = k(k − 1) − 14. Define T1 = P7,m,7 and K1 = N (T1 ). This choice of m implies that

c(K1 ) = c(K2 )

(3.13)

Observe that m ≥ m, tw(K1 ) = 3 and c(K1 ) = m + 14, so (3.10) guarantees that
10 a v4 (tw(K1 ) + 1)
40 a v4
40 a v4
=
≤
<
(a + b) c(K1 )
(a + b)(m + 14)
(a + b)(m + 14)
2

(3.14)

Choose n ≥ 12 so that ξn as defined in (3.1) satisfies

ξn > max 1 −

2 b v8
(a + b)
,
2 b v8 − (a + b) 2 v8 + (a + b)

(3.15)

Since 0 < ξn < 1, we need to check that such a choice of n is possible. Since a, b, v8 , and
are positive we know that
0<

2bv8
<1
2bv8 + (a + b)

Moreover, as long as < 2 v8 /(a + b) we have 2 b v8 > (a + b), hence

1−

(a + b)
<1
2 b v8 − (a + b)

Let L1 = B(T1 ) and L2 = B(T2 ) be the belted links corresponding to T1 and T2 respectively.
Form the link L by taking the belted sum of a · n copies of L1 with b · n copies of L2 . Let K
be the result of filling the belt of L via the meridional Dehn filling. Note that K is simply
the result of taking the closure of the tangle sum of a · n copies of T1 with b · n copies of T2
as shown in Figure 3.9. We obtain

44

T1

. . . a · n copies . . .

T1

T2

. . . b · n copies . . .

T2

Figure 3.9: The link K is the closure of the tangle sum of a · n copies of T1 with b · n copies
of T2 shown above.

vol(K) ≥ ξn vol(L)

by Theorem 3.2.8

(3.16)

= ξn (a · n · vol(L1 ) + b · n · vol(L2 ))

by Lemma 3.2.5

(3.17)

> ξn (a · n · vol(K1 ) + b · n · vol(K2 ))

by Theorem 2.1.2

(3.18)

Since T1 and T2 are strongly alternating tangles, it follows from Corollary 3.2.14 that

c(K) = n · a · c(K1 ) + n · b · c(K2 )

45

(3.19)

This produces the following inequalities to form a lower bound for the density dvol (K).

d(K)vol =

vol(K)
c(K)

>

ξn (a · n · vol(K1 ) + b · n · vol(K2 ))
c(K)

by (3.18)

=

ξn (a · n · vol(K1 ) + b · n · vol(K2 ))
a · n · c(K1 ) + b · n · c(K2 )

by (3.19)

= ξn

vol(K1 ) a
vol(K2 ) b
+ ξn
c(K1 ) a + b
c(K2 ) a + b

≥ ξn

vol(K2 ) b
c(K2 ) a + b

>

1−

(a + b)
2 b v8 − (a + b)

v8 −

=

1−

(a + b)
2 b v8 − (a + b)

v8

a+b
b 2

since c(K1 ) = c(K2 )

b
a+b

by (3.11) and (3.15)

b
−
a+b 2

= v8

2 (a + b)
b
v8 b
−
+
−
a + b 2 b v8 − (a + b) 2(2 b v8 − (a + b)) 2

= v8

b
+
a+b 2

= v8

b
−
a+b

−2 v8 b
(a + b)
+
2 b v8 − (a + b) 2 b v8 − (a + b)

−

2

We now find an upper bound for dvol (K). Since K is obtained from L by Dehn filling,
Theorem 2.1.2 implies that vol(K) < vol(L). Moreover, Lemma 3.2.5 implies that vol(L) =
a · n · vol(L1 ) + b · n · vol(L2 ). Therefore

vol(K) < a · n · vol(L1 ) + b · n · vol(L2 )

(3.20)

Theorem 2.3.2 showed that a · n · vol(L1 ) < 10 a · n · v4 (tw(L1 ) − 1). Consider the diagrams
of L1 and K1 . Adding the belt circle to K1 to form L1 adds no more than two twist regions,
as indicated in Figure 3.10. Therefore tw(L1 ) ≤ tw(K1 ) + 2, hence

46

Figure 3.10: The belt circle adds no more than the two twist regions indicated in the figure.

T2

T2

T2

Figure 3.11: The link K2n is the closure of the sum of n copies of T2 as indicated above.
a · n · vol(L1 ) < 10 a n v4 (tw(K1 ) + 1)

(3.21)

n
Define Ln
2 to be the belted sum of n copies of L2 and let K2 be the result of filling the belt
n
of Ln
2 via the meridional filling. Then K2 = N (T2 + . . . + T2 ) where + denotes Conway

sum and the sum consists of n copies of T2 (see Figure 3.11). Then Lemma 3.2.5 implies
n
n
vol(Ln
2 ) = n vol(L2 ). Theorem 3.2.8 implies vol(L2 ) ≤ vol(K2 )/ξn . Therefore

b · n · vol(L2 ) = b · vol(Ln
2) ≤

b
vol(K2n )
ξn

(3.22)

Since v8 forms an upper bound on volume density, we know that vol(K2n ) ≤ v8 c(K2n ). It

47

follows that
b
b
vol(K2n ) ≤
v c(K2n )
ξn
ξn 8

(3.23)

The diagram for K2n shown in Figure 3.11 has n·c(K2 ) crossings. Therefore c(K2n ) ≤ n·c(K2 ).
This yields
b
b n v8
v8 c(K2n ) ≤
c(K2 )
ξn
ξn

(3.24)

Summarizing (3.22), (3.23), and (3.24) we see that

b · n · vol(L2 ) ≤

b n v8
c(K2 ).
ξn

(3.25)

This allows us to produce an upper bound on the volume density d(K) as follows.

dvol (K) =

vol(K)
a · n · vol(L1 ) + b · n · vol(L2 )
<
c(K)
c(K)

by (3.20)

a · n · vol(L1 ) + b · n · vol(L2 )
a · n · c(K1 ) + b · n · c(K2 )

by (3.19)

=
<
=

10 a n v4 (tw(K1 ) + 1) + ξb (n v8 c(K2 ))
n

a · n · c(K1 ) + b · n · c(K2 )
10 a v4 (tw(K1 ) + 1)
b v8 c(K2 )
+
(a + b)c(K1 )
ξn (a + b) c(K2 )

1 bv8
2 ξn a + b
2bv8 + (a + b) bv8
< +
2
2bv8
a+b
2bv8 + (a + b)
= +
2
2(a + b)
<

=

+

b
v +
a+b 8

48

by (3.21) and (3.25)
since c(K1 ) = c(K2 )
by (3.14)
by (3.15)

Therefore
b
b
v8 − < dvol (K) <
v +
a+b
a+b 8
as desired.

3.5

The Spectrum of Determinant Densities

We now turn to studying the spectrum of determinant densities. The proof that Specdet
contains [0, v8 ] will follow a similar method as the proof of Theorem 3.4.3. Namely, we note
that there exist sequences of diagrammatically maximal links with determinant density approaching v8 , and there exist sequences of diagrammatically minimal links, i.e. those with
determinant density near 0. We combine these diagrams via Conway sum, but instead of
taking the numerator closure as in the proof of Theorem 3.4.3, we consider the denominator
closure. We will then use the facts that log det is additive under the denominator closure of
Conway sums (see Lemma 3.2.7), and that the crossing number is additive under Conway
sums of adequate diagrams (see Corollary 3.2.14). Then by controlling the ratio of diagrammatically minimal links to diagrammatically maximal links, we may obtain a link that has
determinant density near any number x ∈ [0, v8 ].
Lemma 3.5.1. There exists a sequence of adequate tangles {Tm }∞
m=1 such that the sequence
of closures {D(Tm )}∞
m=1 satisfies limn→∞ ddet (D(Tn )) = 0.
Proof. Let P ,m,n be the pretzel tangles shown in Figure 3.8. Let Tm = P3,m,3 and Km =
D(Tm ). Then the tangles are strongly alternating, hence adequate. It follows from [47, Page
100] (see also work of Dasbach, Kalfagianni, Futer, Lin, and Stoltzfus [25, Example 4.3]1 )
1 The author thanks Jessica Purcell for pointing out this reference.

49

that det(Km ) = 6m + 9. Now

2π log(6m + 9)
2π log det(Km )
= lim
=0
m→∞
m→∞
c(Km )
m+6

ddet (Km ) = lim

Lemma 3.5.2. There exists a sequence of adequate tangles {Tm }∞
m=1 such that the sequence
of closures {D(Tm )}∞
m=1 satisfies limn→∞ ddet (D(Tn )) = v8 .
Proof. Let Wm,m be the weaving tangles from Definition 3.3.2 and depicted in Figure 3.6.
Then the sequence {D(Wm,m )}∞
m=1 satisfies the conditions of Theorem 3.2.4, hence is diagrammatically maximal.
Having now established examples of diagrammatically minimal and diagrammatically
maximal sequences of links, we can now restate and prove Theorem 3.5.3.
Theorem 3.5.3. The spectrum of determinant densities Specdet contains [0, v8 ]. In other
words, given any x ∈ [0, v8 ] there exists a sequence {Kn }∞
n=1 of links satisfying

lim d (Kn )
n→∞ det

=x

(3.26)

Proof. It suffices to show that for any non-negative integers a and b with a and b not both
zero, and any

> 0, there exists a link K such that

b
b
v8 − < ddet (K) <
v + .
a+b
a+b 8

Let > 0 be arbitrary. Let P ,m,n be the pretzel tangles. Let T1 = P3,m,3 and K1 = D(T1 ),

50

where m is chosen large enough that

2π log(6m + 9)
<
m+6

a+b
a

(3.27)

2

Since the sequence {D(Wk,k )}∞
k=1 is diagrammatically maximal, we may choose T2 = Wk,k
and K2 = D(T2 ) with k sufficiently large that
2π log det(K2 )
− v8 <
c(K2 )

a+b
b

2

and

c(K2 ) ≥ c(K1 )

(3.28)

The choice of K2 ensures that the crossing numbers c(K1 ) and c(K2 ) satisfy the inequality

k(k − 1) = c(K2 ) ≥ c(K1 ) = m + 6

(3.29)

Let m = k(k − 1) − 6. Define T1 = P3,m,3 and K1 = D(T1 ). This choice of m implies that

c(K1 ) = c(K2 )

Using the fact that m ≥ m we see that

ddet (K1 ) =

2π log(6m + 9)
2π log(6m + 9)
≤
<
m+6
m+6

a+b
a

2

(3.30)

Let T be the tangle sum of a copies of T1 with b copies of T2 , and let K = D(T ) (see Figure
3.12). Then by Lemma 3.2.7 we have that det(K) = (det K1 )a (det K2 )b . Since T1 and T2 are
strongly alternating tangles, we know from Corollary 3.2.14 that c(K) = a · c(K1 ) + b · c(K2 ).

51

. . . a copies . . .

T1

T1

. . . b copies . . .

T2

T2

Figure 3.12: The link K is the closure of the tangle sum of a copies of T1 with b copies of
T2 shown above.
Therefore, we obtain the upper bound on ddet (K):

ddet (K) =

2π log det(K)
c(K)

=

2π log[(det K1 )a (det K2 )b ]
a · c(K1 ) + b · c(K2 )

=

b 2π log det(K2 )
a 2π log det(K1 )
+
a+b
c(K1 )
a+b
c(K2 )

since c(K1 ) = c(K2 )

<

a
a+b

by (3.28) and (3.30)

=

b
v +
a+b 8

a+b
a

2

+

b
v +
a+b 8

a+b
b

2

(3.31)

Now we obtain a lower bound on ddet(K) :

ddet (K) =

a 2π log det(K1 )
b 2π log det(K2 )
+
a+b
c(K1 )
a+b
c(K2 )

≥

b 2π log det(K2 )
a+b
c(K2 )

>

b
a+b

>

b
v −
a+b 8

v8 −

a+b
b

by (3.31)

by (3.28)

2

Note that the links constructed in the proof of Theorem 3.5.3 are not hyperbolic. It is

52

interesting to study the relationship between Specdet and Specvol . Since Specvol considers
only the densities of hyperbolic links, it is useful to know what determinant densities are
realized by hyperbolic links. To this end, we have the following result.
Theorem 3.5.4. The set of determinant densities of hyperbolic links is dense in [0, v8 ].
In other words, given x ∈ [0, v8 ], and

> 0, there exists a hyperbolic link K such that

|ddet (K) − x| < .
Proof. Let > 0 be arbitrary, and let a and b be non-negative integers with a and b not both
zero. Let W m,n denote the reflection of the weaving tangle of Definition 3.3.2. Note that
the sequence {W , }∞=1 is diagrammatically maximal by Theorem 3.2.4. Then by following
an argument similar to the proof of Theorem 3.5.3, there exist integers k and

such that

the link L = D(P3,k,3 + . . . + P3,k,3 + W , + . . . + W , ), i.e. the denominator closure of the
tangle sum of a copies of P3,k,3 with b copies of W , , satisfies

ddet (L) −

b
v <
a+b 8
2

(3.32)

Let T be the tangle obtained by the tangle sum of a copies of P3,k,3 with b copies of W , .
For each positive integer m, let Km = N (T + . . . + T ) be the numerator closure of the tangle
sum of m copies of T , and let Lm = D(T + . . . + T ) be the denominator closure of this tangle
sum. Note that T is prime, alternating, and not an east-west twist. Therefore Theorem
3.2.8 implies Km is hyperbolic for m ≥ 12. Moreover, the links Km and Lm are reduced
and alternating. Now the proof of [19, Theorem 1.7] implies that

2π log det(Km )
2π log det(Lm )
2π m log det(L1 )
= lim
= lim
= ddet (L1 ) (3.33)
m→∞
m→∞
m→∞
c(Km )
c(Lm )
m · c(L1 )
lim

53

Since L1 = L, it follows that we may choose m ≥ 12 sufficiently large that |ddet (Km ) −
b
ddet (L)| < /2. Since ddet (L) −
v < by (3.32), it follows that
a+b 8
2

ddet (Km ) −

b
v <
a+b 8

54

(3.34)

Chapter 4
Relating Determinant and Volume

4.1

Introduction

In this chapter, we explore the relationship between the hyperbolic volume vol(K) =
vol(S 3 \K) of an alternating hyperbolic knot and its determinant det(K). Dunfield noted
a relationship between the volume and determinant of a knot in an online post [26]. He
observed that there is a nearly linear relationship between the hyperbolic volume of an
alternating knot and log(J(−1)) where J denotes the Jones polynomial. After further study
of this relationship and some experimentation, Champanerkar, Kofman and Purcell [20]
made the following conjecture.
Conjecture 4.1.1. Let K be a hyperbolic alternating knot. Then vol(K) < 2π log det(K).
One can use the data from Knotscape [36] and SnapPy [22] to verify this conjecture for
all alternating knots with up to 16 crossings. Champanerkar, Kofman and Purcell in [20]
computationally verified Conjecture 4.1.1 for many examples of an infinite family of links
known as weaving knots. Using these weaving knots, they showed that the constant 2π is
sharp, in the sense that given α < 2π there exists an alternating link K with 2π log(det(K)) <
vol(K).
Stoimenow [63] also explored the relationship between volume and determinant, and

55

showed that if K is a non-trivial, non-split, alternating hyperbolic link then

det(K) ≥ 2(1.0355)vol(K)

(4.1)

He further demonstrated that there exist constants C1 , C2 > 0 such that for any hyperbolic
link K
det(K) ≤

C1 c(K) C2 vol(K)
vol(K)

(4.2)

where c(K) denotes the crossing number of K.
In this chapter, we will verify that Conjecture 4.1.1 holds for various infinite families of
knots including 2-bridge knots and 3-braids. In particular, we have the following theorem.
Theorem 4.1.2. If K is a 2-bridge link or an alternating 3-braid, then

vol(K) ≤ 2π log(det(K))

(4.3)

The proof of Theorem 4.1.2 is a combination of Theorems 4.3.5 and 4.4.1. To obtain
upper bounds on the volumes of knots we largely rely on work of Adams [3] who gave an
upper bound in terms of volumes of bipyramids. Adams et al. [7] used this upper bound
to study the volume densities of 2-bridge knots. Other useful upper bounds in the case of
highly twisted knots are due to Lackenby, Agol and Thurston [44] and Futer, Kalfagianni
and Purcell [30].
To study the determinant of knots, we will count the number of spanning trees of a graph
associated to the checkerboard coloring of a diagram of the knot. We rely on a recurrence
equation due to Kauffman [42] as well as two well-known combinatorial theorems for counting
spanning trees. In the case of highly twisted knots, we utilize work of Stoimenow [63] who
56

provided a lower bound on the number of spanning trees in certain graphs.
We also show that the conjecture holds for “highly twisted” links, i.e. those having many
crossings and few twist regions (see Section 2.3 for a definition of twist region). We further
quantify this by the following theorem.
Theorem 4.5.1. Let K be an alternating hyperbolic link with t twist regions and c crossings.
If
c > t + ξ t−1 − 2γ t−1

(4.42)

where γ ≈ 1.4253 is the unique positive real number satisfying γ −5 + 2γ −4 + γ −3 − 1 = 0
and ξ = e5v4 /π ≈ 5.0296, then vol(K) < 2π log(det(K)).
It follows that if some or all of the twist regions of a link have many crossings, the link
will satisfy Conjecture 4.1.1. An immediate corollary of this is:
Corollary 4.1.3. For any given integer t0 > 0 there are only finitely many links with t0
twist regions which fail to satisfy Conjecture 4.1.1.
An outline of the chapter is as follows. In Section 4.2, we outline the technology used
in this paper to estimate volumes and determinants of knots. In Section 4.3, we will prove
Conjecture 4.1.1 for 2-bridge links. In Section 4.4, we will prove the conjecture for alternating
3-braids and an infinite family of 4-braids. We discuss a general result about highly twisted
links in Section 4.5 and include an application to alternating pretzel links.

4.2

Background

In this section, we discuss the relevant theorems that will be used throughout this chapter.
We begin with a discussion of how one may find upper bounds on volumes of alternating
57

Figure 4.1: A regular ideal 6-bipyramid. Figure taken from [3].
hyperbolic links, and conclude with results on how one may calculate the determinant.

4.2.1

Hyperbolic Volumes

Adams in [3] developed a method for finding an upper bound of the volume of an alternating
hyperbolic link given an alternating diagram of the link. We will recall his notation and
results.
Definition 4.2.1. The regular ideal n-bipyramid can be formed as follows. Begin with n
ideal tetrahedra each having dihedral angles 2π
n,

(n−2)π (n−2)π
2n , 2n .

Let e be an edge running

from a point in ∂H3 to ∞. Glue an edge of each ideal tetrahedron with dihedral angle 2π/n
to the edge e. The resulting polyhedron is called a regular ideal n-bipyramid and will be
denoted by Bn .
An example of a regular ideal n-bipyramid is shown in Figure 4.1. The volume of Bn is given
by

vol(Bn ) = n 

2π
n

π(n−2)
2n

− ln |2 sin(θ)| dθ + 2

0

0


− ln |2 sin(θ)| dθ .

(4.4)

Adams [3] proved the following theorem about the volumes of regular ideal n-bipyramids.
58

Theorem 4.2.2 ([3]). The volume of a regular ideal n-bipyramid satisfies the inequality

vol(Bn ) < 2π log

n
.
2

Moreover, this inequality is asymptotically sharp.
Adams used regular ideal n−bipryamids to give an upper bound on the volume of hyperbolic, alternating links. The following theorem directly follows from [3, Theorem 4.1].
Theorem 4.2.3. Let K be a hyperbolic link with a reduced alternating projection D. Let bn
be the number of faces of D having n edges. Suppose that there are two distinct faces of D
having respectively r and s edges. Then

vol(K) ≤ −vol(Br ) − vol(Bs ) +

bn vol(Bn )

(4.5)

Combining Theorems 4.2.2 and 4.2.3 we get the following corollary.
Corollary 4.2.4. Let K be a hyperbolic knot having an alternating projection D. Let bn
be the number of faces of D having n edges. Suppose that there are two distinct faces of D
having respectively r and s edges. Then

vol(K) < 2π log

n bn 4
2m rs

(4.6)

Proof. This is a straightforward calculation obtained by inserting the inequality of Theorem

59

4.2.2 into Theorem 4.2.3.

vol(K) ≤ −vol(Br ) − vol(Bs ) +

bn vol(Bn )

= (br − 1)vol(Br ) + (bs − 1)vol(Bs ) +

bn vol(Bn )
n=r,s





s
n 
r
+ (bs − 1) log
+
bn log
2
2
2
n=r,s


b
−1
b
b
−1
s
r
n
r
s
n 
= 2π log  b −1 b −1
2r 2s
2bn
< 2π (br − 1) log

n=r,s

= 2π log

nbn 4
2m rs

The volume bound of Corollary 4.2.4 is insufficient in certain cases involving links with
a large number of crossings in a twist region. To handle this case, we appeal to Theorems
2.3.2 and 2.3.3.

4.2.2

Determinants of Links

The determinant of a link K is defined by det(K) = |∆K (−1)| where ∆K (t) is the Alexander
polynomial. It is well-known when K is alternating, the determinant is equal to the number
of spanning trees of any of the checkerboard graphs for K. See, for example, [63, Lemma
3.14] for a proof of this fact, and see Section 2.2 for a definition of the checkerboard graph.
We recall two methods that one may use to compute the number of spanning trees of a
graph. First we present the Matrix Tree Theorem proved by Kirchoff in 1847. One can find
a modern proof in [18].

60

Figure 4.2: Left: Initial graph. Right: Result of collapsing the orange edge.
Theorem 4.2.5 (Matrix Tree Theorem). Let G be a graph and let v1 , . . . , vn be the vertices
of G. Let α(i, j) be the number of edges with endpoints on both of the vertices vi and vj .
Define L to be the matrix (known as the Laplacian) where the (i, j) entry ij of L is given
by


 deg(vi ) − 2α(i, i)

ij =



−α(i, j)

if j = i
(4.7)
if j = i

Then τ (G) is given by the determinant of any of the (n − 1) × (n − 1) minors of L.
For the next lemma, we introduce some notation. Let G be a graph and e an edge of
the graph. We define G − e to be the graph obtained by removing the edge e from G. We
define G/e to be the graph obtained by contracting the edge e and identifying the endpoints
of e to a single vertex as shown in Figure 4.2. With this notation, we recall the following
well-known result.
Lemma 4.2.6. Let G be a graph. Then τ (G) = τ (G − e) + τ (G/e).
Using spanning trees, Stoimenow [63] was able to give a lower bound on the determinant
of an alternating knot.
Theorem 4.2.7 ([63, Theorem 4.3]). Let t be the number of twist regions in a twist-reduced
61

Figure 4.3: Left: The knot R(3, 3, 2). Right: The link R(3, 2, 2, 3).
alternating diagram D of a link K. Then

det(K) ≥ 2 · γ t−1

where γ ≈ 1.4253 is the unique positive real number satisfying γ −5 + 2γ −4 + γ −3 − 1 = 0.

4.3

Two Bridge Links

It is known that any 2-bridge link has an alternating projection of one of the forms shown
in Figure 4.3. We will denote 2-bridge links by R(a1 , a2 , . . . , an ) where the sequence
a1 , a2 , . . . , an denotes the number of half-twists in each crossing region. Examples of R(3, 3, 2)
and R(3, 2, 2, 3) are provided in Figure 4.3. We begin the proof that Conjecture 4.1.1 holds
for 2-bridge links by studying the determinant of a 2-bridge link. Kauffman and Lopes [42]
gave the following recursive method of calculating the determinant of rational links.
Theorem 4.3.1 ([42]). Let K = R(a1 , a2 , . . . , an ) be a two-bridge link. Then det(K) = T (n)

62

where T (n) is defined by the recursion














T (0)

= 1

T (1)

= a1

(4.8)

T (k + 1) = ak+1 T (k) + T (k − 1)

It is interesting to note that when ak = 1 for all k, then the recursion yields the Fibonacci
sequence. We now introduce some notation that will aid the exposition. Define
n

V (a1 , . . . , an ) =
i=1

(ai + 2)
2

(4.9)

Let K = R(a1 , . . . , an ). Note that by Corollary 4.2.4 we have

vol(K) < 2π log

(a1 + 1)(an + 1)
4

n−1
i=2

(ai + 2)
2

< 2π log(V (a1 , . . . , an ))

(4.10)

We will obtain a lower bound on T (n) from the the recurrence of Theorem 4.3.1 and then
show that it exceeds V (a1 , . . . , an ). The most problematic cases for obtaining a lower bound
on T (n) are when ai = 1 for many values of i. The following lemma allows us to reduce to
the case where a1 , . . . , an contains no long sequences of consecutive ones.
Lemma 4.3.2. Let K = R(a1 , a2 , . . . , an ) and let T (i) be the recurrence (4.8). Fix k ≥ 2.
Let
K = R(a1 , . . . , ak−1 , ak+m , . . . , an )

63

and define another recurrence T (i) by

T (i) =



 T (i)

if


 a T (i − 1) + T (i − 2)
i+1

if

i<k
(4.11)
i≥k

Then the following are true:
(a) If T (k) > 32 T (k − 1) and T (k − 1) > 32 T (k − 2), then T (n) > 23 T (n − 1) and

det(R(a1 , . . . , an )) >

3
det(R(a1 , . . . , ak−1 , ak+1 , . . . , an ))
2

(b) Suppose k ≥ 4 and ak−2 = ak−1 = ak = . . . = ak+m−1 = 1 for some m ≥ 1. Then

det(K) >

3 m
det(K )
2

(c) Suppose k ≥ 4 and ak−2 = ak−1 = ak = . . . = ak+m−1 = 1 for some m ≥ 1. If

2π log(V (a1 , . . . , ak−1 , ak+m , . . . , an )) ≤ 2π log(det(K ))

then vol(K) < 2π log(det(K)).
Proof. To begin the proof of part (a) by proving the following claim:

T (k + i + 1) > (3/2)T (k + i) for all i ≥ −1

64

(4.12)

The case where i = −1 holds since

3
3
T (k) > T (k − 1) = T (k − 1)
2
2

To prove the case where i = 0, note that

T (k + 1) = ak+1 T (k) + T (k − 1)
3
> [ak+1 T (k − 1) + T (k − 2)]
2
3
= T (k)
2
We now proceed by induction. Assume that T (k + i + 1) > 23 T (k + i) and T (k + i) >
3 T (k
2

+ i − 1). Then

T (k + i + 2) = ak+i+2 T (k + i + 1) + T (k + i)
3
> [ak+i+2 T (k + i) + T (k + i − 1)]
2
3
= T (k + i + 1)
2

This proves (4.12). Observe that

T (n) = det(R(a1 , . . . , an ))
T (n − 1) = det(R(a1 , . . . , ak−1 , ak+1 , . . . , an ))

thus completing the proof of part (a).

65

We prove part (b) by induction on m. Note that

T (k − 1) = T (k − 2) + T (k − 3)
= 2T (k − 3) + T (k − 4)

since T (k − 2) = T (k − 3) + T (k − 4)

3
3
≥ T (k − 3) + T (k − 4)
2
2
3
= T (k − 2)
2

since T (k − 3) ≥ T (k − 4)

Similarly, T (k) ≥ 32 T (k − 1). Then by part (a)
3
3
det(R(a1 , . . . , an )) = T (n) > T (n − 1) = det(R(a1 , . . . , ak−1 , ak+1 , . . . , an ))
2
2

(4.13)

This proves the case where m = 1. Assume that

det(K) >

3 m−1
det(R(a1 , . . . , ak−1 , ak+m−1 , . . . , an )
2

(4.14)

Since ak−2 = ak−1 = ak+m−1 = 1 we may use part (a) to obtain

det(K) >
>

3 m−1
det(R(a1 , . . . , ak−1 , ak+m−1 , ak+m , . . . , an ))
2

by (4.14)

3 m
det(R(a1 , . . . , ak−1 , ak+m , . . . , an ))
2

by part (a)

which finishes the proof of part (b). To prove part (c), observe that if

2π log(V (a1 , . . . , ak−1 , ak+m , . . . an )) < det(R(a1 , . . . , ak−1 , ak+m , . . . , an ))

66

(4.15)

then

vol(K) < 2π log(V (a1 , . . . , an ))
= 2π log

3 m
V (a1 , . . . , ak−1 , ak+m , . . . , an )
2

< 2π log

3 m
det(R(a1 , . . . , ak−1 , ak+m , . . . , an ))
2

< 2π log(det(K))

by (4.15)
by part (b)

Using Lemma 4.3.2 we can reduce to the case where we do not have ak−2 = ak−1 = ak = 1
for any k ≥ 4, i.e. there is no subsequence of three or more consecutive ones. Next we will
prove Lemma 4.3.3 which empowers us to bound det(R(a1 , . . . , an )) by breaking up the
sequence a1 , . . . , an into shorter subsequences.
Lemma 4.3.3. Let 1 ≤ k ≤ n − 1. Then

det R(a1 , . . . , an ) > det R(a1 , . . . , ak ) det(R(ak+1 , . . . , an ))

Proof. Let T (i) be the recursion defined in Theorem 4.3.1. Define the following recursive

67

sequences T (i) and T (i):



T




T





 T



T




T





 T

(0) = 1
(4.16)

(1) = ak+1
(i) = ak+i T (i − 1) + T (i − 2)
(0) = 1

(4.17)

(1) = ak+2
(i) = ak+i+1 T (i − 1) + T (i − 2)

Note that
T (n − k) = det(R(ak+1 , ak+2 , . . . , an ))

(4.18)

T (k + m) = T (m)T (k) + T (m − 1)T (k − 1)

(4.19)

We will show that

for m ≥ 1. We proceed by induction on m. When m = 1 we have

T (k + 1) = ak+1 T (k) + T (k − 1)

by (4.8)

(4.20)

= T (1)T (k) + T (0)T (k − 1) since T (1) = ak+1 and T (0) = 1 by definition
(4.21)

68

When m = 2 we have

T (k + 2) = ak+2 T (k + 1) + T (k)

by (4.8)

= ak+2 T (1)T (k) + ak+2 T (0)T (k − 1) + T (0)T (k) by (4.21) and T (0) = 1
= [ak+2 T (1) + T (0)]T (k) + ak+2 T (k − 1)

since T (0) = 1

= T (2)T (k) + T (1)T (k − 1)

by (4.16) and T (1) = ak+2

Now assume that
T (k + m) = T (m)T (k) + T (m − 1)T (k − 1)

(4.22)

for m ≥ 2. Then

T (k + m + 1) = ak+m+1 T (k + m) + T (k + m − 1)

Which by applying (4.22) to T (k + m) and T (k + m − 1) becomes

T (k + m + 1) = ak+m+1 T (m)T (k) + ak+m+1 T (m − 1)T (k − 1)

(4.23)

+ T (m − 1)T (k) + T (m − 2)T (k − 1)

By collecting like terms (4.23) simplifies to

T (k + m + 1) = [ak+m+1 T (m) + T (m − 1)]T (k)
+ [ak+m+1 T (m − 1) + T (m − 2)]T (k − 1)

69

(4.24)

By (4.16) we have that

ak+m+1 T (m) + T (m − 1) = T (m + 1)

(4.25)

and by (4.17) we also know that

ak+m+1 T (m − 1) + T (m − 2) = T (m)

(4.26)

Combining (4.24), (4.25), and (4.26) we see that

T (k + m + 1) = T (m + 1)T (k) + T (m)T (k − 1)

Finally, we observe that

det(R(a1 , . . . , an )) = T (n)
= T (k + n − k)
= T (n − k)T (k) + T (n − k − 1)T (k − 1)

by (4.19)

> T (n − k)T (k)
= det(R(ak+1 , . . . , an )) det(R(a1 , . . . , ak ))

by (4.18)

Using Lemmata 4.3.2 and 4.3.3 we will break up the sequence a1 , . . . , an into smaller
subsequences which will have one of the eleven special types listed in the following lemma.
Lemma 4.3.4. Let a1 , . . . , an be a sequence of one of the following eleven types:
70

1. a1 where a1 ≥ 2
2. 1, a2 where a2 ≥ 2
3. a1 , 1 where a1 ≥ 2
4. 1, 1, a3 where a3 ≥ 2
5. a1 , 1, 1 where a1 ≥ 2
6. 1, 1, 1, a4 where a4 ≥ 2
7. 1, a2 , 1, 1 where a2 ≥ 2
8. 1, 1, a3 , 1 where a3 ≥ 2
9. 1, 1, a3 , 1, 1 where a3 ≥ 2
10. 1, a2 , 1, a4 , 1 where a2 ≥ 2 and a4 ≥ 2
11. 1, 1, a3 , 1, a5 , 1 where a3 ≥ 2 and a5 ≥ 2
Let T (i) be the recurrence of Theorem 4.3.1. Then V (a1 , . . . , an ) ≤ T (n).
Proof. For type (1), one readily obtains that a1 + 2 ≤ 2a1 implying
a +2
2a
V (a1 ) = 1
≤ 1 = T (1)
2
2

For type (2), we see that T (2) = a2 + 1 and

V (1, a2 ) =

3(a2 + 2)
4

71

When a2 ≥ 2, one readily obtains V (1, a2 ) < T (2).

For type (3), we have T (2) = a1 + 1 and

V (a1 , 1) =

3(a1 + 2)
4

and the proof proceeds in a similar manner to type (2).

For type (4), T (3) = 2a3 + 1 and

V (1, 1, a3 ) =

9(a3 + 2)
8

When a3 ≥ 2 one readily obtains
9(a3 + 2)
≤ 2a3 + 1
8

For type (5), we have T (3) = 2a1 + 1 and

V (a1 , 1, 1) =

and the proof proceeds similarly to type (4).

72

9(a1 + 2)
8

For type (6), we have T (4) = 3a4 + 2 and

V (1, 1, 1, a4 ) =

27(a4 + 2)
16

When a4 ≥ 2, one may show that
27(a4 + 2)
≤ 3a4 + 2
16

For type (7), we have T (4) = 2a2 + 3 while

V (1, a2 , 1, 1) =

27
(a + 2)
16 2

Then
T (4) − V (1, a2 , 1, 1) =

5
3
a2 − ≥ 0
16
8

since a2 ≥ 2.

For type (8), we have T (4) = 2a3 + 3 and

V (1, 1, a3 , 1) =

The proof is now similar to type (7).

73

27
(a + 2)
16 3

For type (9), we have T (5) = 4a3 + 4 and

V (1, 1, a3 , 1, 1) =

81
(a + 2)
32 3

Then
T (5) − V (1, 1, a3 , 1, 1) =

17
47
a3 −
≥0
32
16

since a3 ≥ 2.

For type (10), we have T (5) = a2 a4 + 2a2 + 2a4 + 3 and

V (1, a2 , 1, a4 , 1) =

27
(a a + 2a2 + 2a4 + 4)
32 2 4

Then
T (5) − V (1, a2 , 1, a4 , 1) =

5
3
(a2 a4 + 2a2 + 2a4 ) − ≥ 0
32
8

since a2 ≥ 2 and a4 ≥ 2.

For type (11), we have T (6) = 2a3 a5 + 4a3 + 3a5 + 4 an

V (1, 1, a3 , 1, a5 , 1) =

81
(a a + 2a3 + 2a5 + 4)
64 3 5

Then
T (6) − V (1, 1, a3 , 1, a5 , 1) =

47
47
15
17
a3 a5 + a3 + a5 −
≥0
64
32
32
16

since a3 ≥ 2 and a5 ≥ 2.

74

We are now prepared to present the proof of Theorem 4.3.5.
Theorem 4.3.5. Let K be the 2-bridge link R(a1 , a2 , . . . , an ). Then vol(K) < 2π log det(K).
Proof. We consider three cases:
1. n = 3, a1 = 1, a2 ≥ 2, a3 = 1
2. a1 = a2 = . . . = an = 1
3. ai = 1 for some i and not case 1
Case 1:
We can calculate that det(K) = a2 + 2. Using Corollary 4.2.4 we see that

vol(K) < 2π log

a2 + 2
2

< 2π log(a2 + 2) = 2π log(det(K))

(4.27)

For the remaining cases, it suffices to show that V (a1 , . . . , an ) ≤ det(K).

Case 2:
If n = 1, 2, or 3 then K is R(1), R(1, 1), or R(1, 1, 1) respectively, none of which is hyperbolic.
If n = 4 then det(K) = T (4) = 5. Corollary 4.2.4 implies that

vol(K) < 2π log

2·3·3·2
24

= 2π log

9
4

< 2π log(5) = 2π log(det(K))

If n = 5 then det(K) = T (5) = 8 and while V (1, 1, 1, 1, 1) = 243/32 < 8. If n ≥ 6 then we
use Lemma 4.3.2 part (c). We can let k = 6 and then the link K defined in Lemma 4.3.2
part (c) will be R(1, 1, 1, 1, 1). The result now follows from the case n = 5 above.

75

Case 3:
We may use Lemma 4.3.2 part (c) to assume that we do not have ak−2 = ak−1 = ak = 1
for any k ≥ 4, i.e. a1 , . . . , an has no subsequences of three or more consecutive ones except
possibly a1 = a2 = a3 = 1. Let m be the cardinality |{ak ∈ {ai }n
i=1 : ak ≥ 2}|. We
(1)

(1)

(2)

(2)

(m)

will partition the sequence {a1 , . . . , an } = {b1 , . . . , bn1 , b1 , . . . bn2 , . . . , b1

(m)

, . . . bnm } into

subsequences of the types found in Lemma 4.3.4 according to one of the cases below.

Case 3a: a1 ≥ 2
(k)

Let b1

be the kth element of the sequence a1 , . . . , an that is greater than or equal to 2.
(k)

(k)

Then each subsequence b1 , . . . , bnk is either type 1, 3, or 5 from Lemma 4.3.4.

Case 3b: a1 = 1 and an ≥ 2
(k)

Let bnk be the kth element of a1 , . . . , an that is greater than or equal to 2. Then each
(k)

(k)

b1 , . . . , bnk is either of type 1, 2, 4, or 6 in 4.3.4.

Case 3c: a1 = an−1 = an = 1
(k)

For 1 ≤ k ≤ m − 1 let bnk be the kth element of a1 , . . . , an that is greater than or equal to
(k)

(k)

(m)

2. Then each b1 , . . . , bnk is either of type 1, 2, 4, or 6 in Lemma 4.3.4. Let b1
(m)

be the remaining elements of the sequence a1 , . . . , an . Then b1

(m)

, . . . , b nm

(m)

, . . . , bnm is either of type

7 or 9 from Lemma 4.3.4.

Case 3d : a1 = an = 1 and an−1 = 1
(k)

For 1 ≤ k ≤ m − 2 let bnk be the kth element of a1 , . . . , an that is greater than or equal to
(k)

(k)

(m)

2. Then each b1 , . . . , bnk is either of type 1, 2, 4, or 6 in Lemma 4.3.4. Let b1
76

(m)

, . . . , b nm

be the remaining elements of the sequence a1 , . . . , an . Since case 3 excludes cases 1 and
(m)

2, {a1 , a2 , a3 } = {1, a2 , 1}. Therefore b1

(m−1)

, . . . , b nm
(m−1)

Lemma 4.3.4. For notational purposes, take b1

is either of type 8, 10, or 11 from
(m−1)

, . . . , bnm−1 to be the empty sequence.

Now that we have partitioned the sequence

(1)

(1)

(2)

(2)

(m)

{a1 , . . . , an } = {b1 , . . . , bn1 , b1 , . . . bn2 , . . . , b1

(m)

, . . . bnm }

into subsequences of the types found in Lemma 4.3.4, we observe that

(1)

(1)

(2)

(2)

(m)

det(K) > det(R(b1 , . . . , bn1 )) det(R(b1 , . . . , bn2 )) . . . det(R(b1
(1)

(1)

(2)

(2)

(m)

≥ V (b1 , . . . , bn1 )V (b1 , . . . , bn2 ) . . . V (b1

(m)

, . . . , bnm ))

(m)

, . . . , b nm )

= V (a1 , . . . , an )

where the first line follows by Lemma 4.3.3 and the second follows by Lemma 4.3.4.
The proof of Theorem 4.3.5 also verified the following fact which will be used in Section 4.4.
Corollary 4.3.6. Let K = R(a1 , . . . , an ) be a 2-bridge link. If K = R(1, 1, 1, 1), R(1, a2 , 1),
R(1, 1) or R(1) for any a2 ≥ 1 then

det(R(a1 , . . . , an )) ≥ V (a1 , . . . , an )

77

4.4

Alternating Braids

We will show in this section that Conjecture 4.1.1 holds for alternating 3-braids and for an
infinite family of 4-braids. The former fact will rely on the result of Theorem 4.3.5, while
the latter fact will be proved by bounding the hyperbolic volume and explicitly computing
the determinant.

4.4.1

3-Braids

Let B(a1 , b1 , . . . , an , bn ) denote the alternating 3-braid with a1 positive crossings in the first
twist region, b1 negative crossings in the second twist region, and so on. See for example
Figure 4.4. Note that up to reflection, this considers all alternating 3-braids, and that up
to isomorphism all alternating 3-braids have an even number of twist regions. We state the
main theorem for this section.
Theorem 4.4.1. If K is an alternating 3-braid then vol(K) < 2π log det(K).
The proof of Theorem 4.4.1 will follow immediately from from Lemmata 4.4.2 and
4.4.3. Note that by using Corollary 4.2.4 it is sufficient to show that V (a1 , b1 , . . . , an , bn ) ≤
det(B(a1 , b1 , . . . , an , bn )).
Lemma 4.4.2. Let K = B(a1 , b1 , . . . , an , bn ). If ai = 1 or bi = 1 for some i then vol(K) <
2π log det(K).
Proof. Suppose bi = 1. Let σ1 and σ2 be generators of the 3-braid, where σi denotes a posia −b
tive half-twist of the ith and (i+1)st strands. Then K is the closure of σ1 1 σ2 1 . . . σ1an σ2−bn .
b
−a
b
a
Let α = (σ2bn σ1−an σ2n−1 σ1 n−1 . . . σ2i+1 σ1 i+1 ). Then

a
b
a −b
a −b
a −b
α−1 (σ1 1 σ2 1 . . . σ1an σ2−bn )α = σ1 i−1 σ2i−1 . . . σ1an σ2−bn σ1 1 σ2 1 . . . σ1 i σ1 i

78

(4.28)

Figure 4.4: Left: The two-bridge knot R(3, 3, 2). Center Left: The checkerboard graph
for R(3, 3, 2). Center Right: The closed alternating 3-braid B(3, 3, 2, 3). Right: The
checkerboard graph for B(3, 3, 2, 3). Note that the graph that results from deleting the
highlighted edge has the same number of spanning trees as the checkerboard graph for
R(3, 3, 2).
The braid closure of the right hand side of (4.28) corresponds to the three-braid

K = B(ai−1 , bi−1 , . . . , an , bn , a1 , b1 , . . . , ai , bi )

so K is equivalent to K . Therefore we may assume in this case that bn = 1. We will
now reduce the problem to the case of 2-bridge links, and the result will follow from Theorem 4.3.5.

Observe that one of the checkerboard graphs for B(a1 , b1 , . . . , an , bn ) and

R(a1 , b1 , . . . , bn−1 , an ) will be of the form shown in Figure 4.4. Apply Lemma 4.2.6 by
contracting and deleting the highlighted right edge in the far right of Figure 4.4. The result
of deleting the edge yields a graph with the same number of spanning trees as one of the
checkerboard graphs of R(a1 , b1 , . . . , bn−1 , an ). If bn ≥ 2 then contraction of the highlighted
edge is the checkerboard graph of B(a1 , b1 , . . . , an , bn − 1). Further, contracting the highlighted edge of a checkerboard graph of B(a1 , b1 , . . . , an , 1) yields a graph isomorphic to a

79

checkerboard graph of B(a1 + an , b1 , a2 , b2 , . . . , an−1 , bn−1 ). Therefore we see inductively
that

det(B(a1 , b1 , . . . , an , bn )) = det(R(a1 , b1 , . . . , bn−1 , an )) + det(B(a1 , b1 , . . . , an , bn − 1))
= 2 det(R(a1 , b1 , . . . , bn−1 , an )) + det(B(a1 , b1 , . . . , an , bn − 2))
..
.
= (bn − 1) det(R(a1 , b1 , . . . , bn−1 , an )) + det(B(a1 , b1 , . . . , an , 1))
= bn det(R(a1 , b1 , . . . , bn−1 , an ))
+ det(B(a1 + an , b1 , . . . , an−1 , bn−1 ))
≥ bn det(R(a1 , b1 , . . . , bn−1 , an ))

(4.29)
(4.30)

Suppose that {a1 , b1 , . . . , an , bn } = {1, b1 , 1, b2 } or {1, b1 }. Then since bn ≥ 2, we have
bn ≥ (bn + 2)/2 and we can apply Corollary 4.3.6:

det(B(a1 , b1 , . . . , an , bn )) ≥ bn det(R(a1 , b1 , . . . , bn−1 , an ))
≥ bn V (a1 , b1 , . . . , bn−1 , an )
= bn
≥

by 4.30
by Corollary 4.3.6

(4.31)

n
i=1 (ai

+ 2) n−1
i=1 (bi + 2)
22n−1
n
n
i=1 (ai + 2) i=1 (bi + 2)
22n

= V (a1 , b1 , . . . , an , bn )

If {a1 , b1 , . . . , an , bn } = {1, b1 , 1, b2 } or {1, b1 } then we cannot apply Corollary 4.3.6 to obtain
(4.31). If {a1 , b1 , . . . , an , bn } = {1, b1 } then B(1, b1 ) is seen to be the (2, b1 )-torus link which

80

is not hyperbolic. If {a1 , b1 , . . . , an , bn } = {1, b1 , 1, b2 } then using (4.29) we see that

det(B(1, b1 , 1, b2 )) = b2 det(R(1, b1 , 1)) + det(B(2, b1 ))
= b2 (b1 + 2) + 2b1

(4.32)
(4.33)

On the other hand
V (1, b1 , 1, b2 ) =

9
(b b + 2b1 + 2b2 + 4)
16 1 2

(4.34)

Since b2 ≥ 2 we see that
7
36
(b1 b2 + 2b1 + 2b2 ) −
16
16
56 36
≥
−
16 16

det(B(1, b1 , 1, b2 )) − V (1, b1 , 1, b2 ) =

>0

If bi = 1 for all i, then ai = 1 for some i. Then K is equivalent to the link

B(ai , bi , . . . , an , bn , a1 , b1 , . . . , ai−1 , bi−1 )

and therefore we assume that a1 = 1. Then we can repeat the argument above by considering
the other checkerboard surface (i.e. considering the checkerboard graph obtained from the
white regions instead of the shaded regions).
Lemma 4.4.3. Let K = B(1, 1, . . . , 1, 1) where there are 2n copies of 1. Then vol(K) <
2π log det(K).
Proof. We will use the Matrix Tree Theorem, Theorem 4.2.5, to compute the number of
spanning trees of the checkerboard graph, and hence the determinant of K. If n ≥ 3, the
81

associated checkerboard graph has Laplacian



 n −1 −1 −1 −1 −1 . . . −1 −1 −1




−1 3 −1 0
0
0 ... 0
0 −1






−1 −1 3 −1 0
0 ... 0
0
0






−1
0
−1
3
−1
0
.
.
.
0
0
0




L=

−1 0

0
−1
3
−1
0
0
0





 .
..
..
..
..
..
..
..
 ..

.
.
.
.
.
.
.






−1 0
0
0
0
0 . . . −1 3 −1




−1 −1 0
0
0
0 . . . 0 −1 3

(4.35)

where L is an (n + 1) × (n + 1) matrix. Let L be the minor of L obtained by eliminating
the first row and first column. Then it is known (see for example [55]) that


3 −1

det(L ) = −2 + tr 


i=1 1 0


n

(4.36)

By diagonalizing, we can compute an explicit formula:
√

n 
3+ 5
3 −1


 = 2
1 0
1
√

3+ 5
 2


=


1

√ 
√
3− 5
3+ 5

2 
 2
1
√ 
3− 5 
2 





1

√
√ −1
3+ 5 3− 5
0  
2 
√   2

3− 5
0
1
1
2
  3 + √5 3 − √5 −1
√ n
3+ 5
0

2
2 


2


√ n




3 − 5 

0
2
1
1

82

n 

Therefore
n 
3 −1 
det(L ) = −2 + tr 
 


1 0
√ n
√ n
3− 5
3+ 5
+
= −2 +
2
2


The volume of K is bounded above by

V (1, 1, . . . , 1, 1) =

3 2n
=
2

9 n
4

(4.37)

It is straightforward to check that

−2 +

√ n
3+ 5
+
2

√ n
3− 5
>
2

9 n
4

if n ≥ 3

(4.38)

If n = 1, then K is not hyperbolic. If n = 2 then K is the figure-eight knot, which is a
2-bridge knot and therefore satisfies Conjecture 4.1.1 by Theorem 4.3.5.

4.4.2

A Family of 4-braids

Let σ1 , σ2 , σ3 be the generators of the 4-braid, where σi denotes a positive half-twist of the
ith and (i + 1)st strands. Let Wn be the closure of (σ1 σ3 σ2−1 )n . We note that these links
correspond to the weaving links W (4, n) of [20] and [21]. A checkerboard graph associated
with Wn is the maximal planar lantern graph En+2 on (n + 2) vertices as shown in Figure
4.5. Work of Modabish, Lotfi, and El Marraki [54] shows that

τ (En+2 ) =

√
√
√
n
n+2
√ (2 + 3)n − (2 − 3)n ≤ √ (2 + 3)n
2+2 3
2 3
83

(4.39)

On the other hand, Corollary 4.2.4 shows that

vol(Wn ) ≤ 2π log

Since 2 +

32n 4n
23n

= 2π log

9 n
2

(4.40)

√
3 < 9/2 it follows from equations (4.39) and (4.40) that the volume bound of

Corollary 4.2.4 is insufficient to prove Conjecture 4.1.1 for these links. However, one may
instead use Theorem 4.2.3 to find that

exp

vol(Wn )
2π

≤ exp [2n vol(B3 ) + n vol(B4 )]
≤ (3.418677233748620053022)n

This bound may also be obtained from [21, Theorem 1.1]. On the other hand, equation
√
√
(4.39) and the fact that (2 − 3)n < 12 (2 + 3)n for n ≥ 1 together imply that

τ (En+2 ) =

√
√
√
n+2
n+2
√ (2 + 3)n − (2 − 3)n ≥ √ (2 + 3)n
2 3
4 3

(4.41)

It is straightforward to show that

(3.418677)n <

√
n+2
√ (2 + 3)n for n ≥ 4
4 3

so Conjecture 4.1.1 holds for all Wn with n ≥ 4. Note that the case n ≤ 3 has been verified
in [20].
One can use this method to find many more infinite families of links for which Conjecture
4.1.1 holds. Given a planar graph G, one may create an alternating link K for which G is the
checkerboard graph of K. This is done by replacing each edge with a crossing and connecting
84

Figure 4.5: Left: The link Wn . Center: The checkerboard coloring of W4 . Right: The
graph E6 corresponding to the white checkerboard surface.
ends of crossings so that each vertex is on the shaded part of the checkerboard surface. One
can then calculate the volume estimates and then if the number of spanning trees of the graph
is known test whether the conjecture holds. This method works for the wheel, fan, crystal,
star-flower graphs of [53] and [54] as well as the grid graphs and triangulated grid-graphs of
[52].

4.5

Highly Twisted Knots

We consider the situation where a link has a twist region with many crossings. It is known by
[44] that the volume of an alternating link is bounded by the number of twist regions in the
diagram. Therefore, increasing the number of crossings in a twist region of an alternating
hyperbolic link has a bounded effect on the hyperbolic volume. On the other hand, the
number of spanning trees in the checkerboard graph will increase by adding crossings to a
twist region. It follows that highly twisted links must satisfy Conjecture 4.1.1. We quantify
this in the following theorem.

85

Theorem 4.5.1. Let K be an alternating hyperbolic link with t twist regions and c crossings.
If
c > t + ξ t−1 − 2γ t−1

(4.42)

where γ ≈ 1.4253 is the unique positive real number satisfying γ −5 + 2γ −4 + γ −3 − 1 = 0
and ξ = e5v4 /π ≈ 5.0296, then vol(K) < 2π log(det(K)).
Proof. Let a1 , . . . , at be the crossing numbers of the t twist regions of K. Let G(x1 , . . . , xt )
be the checkerboard graph of the link obtained by placing xi crossings in the ith twist region
of K. Since a checkerboard graph has the same number of spanning trees as its dual, we may
assume that the first twist region of K corresponds to a path on a1 vertices in G(a1 , . . . , at ).
By Lemma 4.2.6 we then obtain

τ (G(a1 , a2 , . . . , at )) = τ (G(a1 − 1, a2 . . . , at )) + τ (G(0, a2 , . . . , at ))
= τ (G(1, a2 , . . . , at )) + (a1 − 1)τ (G(0, . . . , at ))
≥ τ (G(1, a2 , . . . , at )) + (a1 − 1)
t

(ai − 1)

≥ τ (G(1, 1, . . . , 1)) +
i=1

= τ (G(1, 1, . . . , 1)) + c − t

Theorem 4.2.7 then implies that

det(K) = τ (G(a1 , . . . , at )) ≥ 2γ t−1 + c − t

(4.43)

By Theorem 2.3.2 we know that vol(K) ≤ 10v4 (t − 1). It is then straightforward to check

86

that if (4.42) holds then

vol(K) < 10v4 (t − 1) ≤ 2π log(2γ t−1 + c − t) ≤ 2π log(det(K))

(4.44)

Corollary 4.5.2. Let K be an alternating hyperbolic Montesinos link with t twist regions
and c crossings. If
c > t + ζ t − 2γ t−1

(4.45)

where ζ = ev8 /π ≈ 3.2099 then vol(K) > 2π log(det(K)).
Proof. The proof is the same as for Theorem 4.5.1, except we replace the upper bound on
volume with the bound 2v8 t of Theorem 2.3.3.

4.5.1

Application to Pretzel Knots

We give an application of Theorem 4.5.1 and Corollary 4.5.2 to alternating pretzel links. We
begin by calculating the determinant of an alternating pretzel knot.
Proposition 4.5.3. Let P (a1 , a2 , . . . , an ) be the alternating pretzel knot having a1 , a2 , . . . , an
crossings in the first, second, and so on to the nth twist region. Then
n

det(P (a1 , a2 , . . . , an )) =

aj

(4.46)

i=1 j=i

Proof. We will use Lemma 4.2.6. An example checkerboard graph for P (a1 , a2 , . . . , an ) is
given on the left of Figure 4.6. Deleting an edge in the nth twist region produces a graph
with the same number of spanning trees as the checkerboard graph of P (a1 , . . . , an−1 ). For
87

Figure 4.6: Left: The checkerboard graph of the Pretzel knot P (1, 2, 4, 3, 4). Deletion of
the red edge produces a graph with the same number of spanning trees as the checkerboard
graph of P (1, 2, 4, 3). Contraction of the cyan edge yields the graph on the right. Right:
Result of contracting the cyan edge of the graph on the left.
example one may delete the red edge in Figure 4.6. On the other hand, if an ≥ 2, then
contracting that edge results in the checkerboard graph of P (a1 , a2 . . . , an − 1). If an = 1
then the resulting graph is the join of (n − 1) cycles. For example, contracting the blue
edge in Figure 4.6 produces the graph on the right of Figure 4.6) which has a1 a2 . . . an−1
spanning trees. Therefore by Lemma 4.2.6 we see that

det(P (a1 , . . . , an )) = a1 . . . an−1 + an det(P (a1 , a2 , . . . , an−1 ))

(4.47)

Applying the above method to P (a1 , . . . , an−1 ) et cetera we obtain the desired result.
Given a fixed number t of twist regions and a pretzel link K with t twist regions, Corollary
4.5.2 can be used to say that if K has more than t +ζ t − 2γ t−1 crossings in any twist region,
then it satisfies Conjecture 4.1.1. Therefore for a given t, there are only finitely many links
which may fail to satisfy Conjecture 4.1.1. We may enumerate these links, use Theorem 4.2.3

88

to compute an upper bound on volume, and check that this upper bound is less than the
determinant. Using this method, we have shown with computer assistance that Conjecture
4.1.1 holds for all alternating pretzel links with no more than 13 twist regions.
The number of links for which we need to evaluate the determinant and use the upper
bound of Theorem 4.2.3 grows exponentially in terms of the number of twist regions. It
is therefore prudent to use further techniques to make the computation feasible. We now
explain the techniques used to verify the conjecture for alternating pretzel links with up to
13 twist regions. Let P (a1 , . . . , at ) denote the alternating pretzel link with ai crossings in
the ith twist region. A volume bound that proves useful in this context is due to Dasbach
and Tsvietkova [23].
Theorem 4.5.4 ([23]). Given a diagram D of a hyperbolic alternating link K, with ti twist
regions of precisely i half-twists, and gi twist regions of at least i half-twists, then

Vol(S 3 \K) ≤ (10g4 + 8t3 + 6t2 + 4t1 − a)v4

(4.48)

where a = 10 if g4 is non-zero, a = 7 if t3 is non-zero, and a = 6 otherwise.
Let DT(a1 , . . . , at ) be the right hand side of (4.48). This bound proves particularly useful,
since if ai ≥ 4 then
DT(a1 , . . . , at ) = DT(a1 , . . . , ai + k, . . . , at )

(4.49)

for any integer k. Moreover DT(a1 , . . . , at ) remains unchanged if the ai ’s are permuted.
One can use the arithmetic-geometric mean inequality in conjunction with Theorem 4.2.3
to derive another bound on the volume of P (a1 , . . . , at ) that is unchanged under permutation
of the ai ’s.
89

Lemma 4.5.5. The following inequality is true:
t

t
i=1 ai

vol(P (a1 , . . . , at )) ≤

(4.50)

t

Proof. This follows from Theorem 4.2.3 and the geometric-arithmetic mean inequality:

a + at
vol(P (a1 , . . . , at )) ≤ 1
2
≤

=

a1 +at
2

t−1
i=1

ai + ai+1
2
t−1 ai +ai+1
2
i=1

+

t

t
t
i=1 ai

t

t

We consequently define

AG(a1 , . . . , at ) =

t
i=1 ai

t

t

(4.51)

We can use Corollary 4.5.2 to find the maximum number of crossings in any twist region.
Suppose ai = 1 for i ≥ 2. Then if c is the number of crossings in P (a1 , . . . , at ) then
c = a1 + t − 1, hence by Corollary 4.5.2 a1 > 1 + ζ t−1 − 2γ t−1 . Let

M = 1 + ζ t−1 − 2γ t−1

(4.52)

By Corollary 4.5.2, we will only need to verify that P (a1 , . . . , at ) satisfies Conjecture 4.1.1
if ai ≤ M for all i.
Combining all the above, we produce Algorithm 1 to check Conjecture 4.1.1 for alter90

nating pretzel links with 3 twist regions. If the algorithm does not return any links, then
it has verified the conjecture. Otherwise it returns a problem case. The algorithm works
by using the AG and DT bounds to verify P (a1 , . . . , at ) satisfies Conjecture 4.1.1 for all
permutations of the ai ’s. It utilizes the fact that DT remains unchanged when a twist region
goes from k to k + 1 crossings for k ≥ 4 to eliminate many cases. When the AG and DT
bounds are insufficient, it computes the volume bound of Corollary 4.2.4. We do this only
as a last resort since it requires t! computations because we must consider all permutations
of the ai ’s. The algorithm utilizes the fact that det(P (a1 , . . . , an )) < det(P (a1 , . . . , an + k))
for any integer k ≥ 1. The algorithm can be generalized to check for t twist regions by
increasing the number of for loops, and generalizing line 17 to compute the volume bound
of Corollary 4.2.4.
Algorithm 1 Verify Conjecture 4.1.1 for Pretzel Links
1: Let t ≥ 3 be an integer representing the number of twist regions in the pretzel link.
2: Set M = 1 + ζ t − 2γ t−1 .
3: for x ∈ {1, 2, . . . , M } do
4:
if DT(x, 4, 4) < det(P (x, x, x)) then
5:
break
6:
for y ∈ {x, x + 1, . . . , M } do
7:
if DT(x, y, 4) < det(P (x, y, y)) then
8:
break
9:
for z ∈ {y, y + 1, . . . , M } do
10:
if DT(x, y, z) < det(P (x, y, z)) then
11:
P (x, y, z) satisfies Conjecture 4.1.1
12:
else if AG(x, y, z) < det(P (x, y, z)) then
13:
P (x, y, z) satisfies Conjecture 4.1.1
14:
else
15:
Let S be the set of all permutations of (x, y, z)
16:
for (s1 , s2 , s3 ) ∈ S do
17:
if det(P (s1 , s2 , s3 )) ≥ (s1 + s2 )(s2 + s3 )(s3 + s1 )/8 then
18:
Algorithm fails
19:
return P (s1 , s2 , s3 )

91

Chapter 5
Meridians, Cusp Areas, and Essential
Spanning Surfaces

5.1

Introduction

In this chapter we derive bounds of slope lengths on the maximal cusp and of the cusp area
of hyperbolic knots in terms of the topology of essential surfaces spanned by the knots. Our
results are partly motivated by the open question of whether there exist hyperbolic knots in
S 3 whose meridian length exceeds four. We show that there is an algorithmically checkable
criterion to decide whether a hyperbolic knot has meridian length less than a given bound,
and we use it to we obtain large families of knots with meridian lengths bounded above
by four. Our results are particularly interesting in the case of knots that project on closed
embedded surfaces in an alternating fashion and admit essential checkerboard surfaces. In
this case, our bounds are purely combinatorial and can be read directly from a knot diagram.
We also discuss applications of our results to Dehn surgery.
Given a hyperbolic knot K in S 3 , there is a well-defined notion of a maximal cusp C
of the complement M = S 3 \ K (see Section 2.1.2). The interior of C is neighborhood of
the missing K and the boundary ∂C is a torus that inherits a Euclidean structure from the
hyperbolic metric. Each slope σ on ∂C has a unique geodesic representative. The length of

92

σ, denoted by (σ), is the length of its geodesic representative. By Mostow-Prasad Rigidity
(Theorem 2.1.1), these lengths are topological invariants of K.
By abusing notation and terminology, we will also refer ∂C as the boundary of M , and
we will sometimes use the alternative notation ∂M . For a slope σ on ∂M let M (σ) denote
the 3-manifold obtained by Dehn filling M along σ. By the knot complement theorem of
Gordon and Luecke [35], there is a unique slope µ, called the meridian of K, such that M (µ)
is S 3 . A λ-curve of K is a slope on ∂M that intersects µ exactly once and a spanning surface
of K is a properly embedded surface in M whose boundary is a λ-curve.
Theorem 5.1.1. Let K be a hyperbolic knot with meridian length (µ). Suppose that K
admits essential spanning surfaces S1 and S2 such that

|χ(S1 )| + |χ(S2 )| ≤

b
i(∂S1 , ∂S2 )
6

(5.1)

where b is a positive real number and i(∂S1 , ∂S2 ) is the minimal intersection number of ∂S1
and ∂S2 on ∂M . Then the meridian length satisfies (µ) ≤ b. Moreover, given a hyperbolic
knot K and b > 0, there is an algorithm to determine if there are essential surfaces S1 and
S2 satisfying (5.1)
A slope σ on ∂M is called exceptional if the 3-manifold M (σ) is not hyperbolic. The
Gromov-Thurston “2π-theorem” [12] asserts that if (σ) > 2π then M (σ) admits a Riemannian metric of negative curvature. This combined with the proof of Thurston’s geometrization conjecture [56] implies that actually M (σ) is hyperbolic. The work of Agol [10] and
Lackenby [43], that has improved 2π to 6, asserts that exceptional slopes must have length
less or equal to six. Examples of exceptional slopes with length six are given in [10] and in
[4]. Since the meridian curve of every hyperbolic knot in S 3 is an exceptional slope we have
93

(µ) ≤ 6. The work of Adams, Colestock, Fowler, Gillam, and Katerman [6] shows that that
(µ) < 6. Examples of knots whose meridian length approaches four from below are given in
[10] and by Purcell in [61]. An open conjecture in the area is that (µ) ≤ 4 for all hyperbolic
knots in S 3 .
Theorem 5.1.1 provides a criterion for checking algorithmically whether a given knot
satisfies this conjecture. Indeed, given a hyperbolic knot K there is an algorithm using
normal surface theory to decide whether K admits essential spanning surfaces S1 , S2 such
that
|χ(S1 )| + |χ(S2 )| ≤

4
· i(∂S1 , ∂S2 ),
6

and thus whether (µ) ≤ 4.
We will then discuss applications of Theorem 5.1.1. As an example, we first mention
the hyperbolic 3-pretzel knots P (a, −b, −c) with a, b, c > 1 and all odd. For these knots,
Theorem 5.1.1 applies to give (µ) ≤ 3. See Example 5.3.2 for details and for generalizations.

5.1.1

Knots with essential checkerboard surfaces

Theorem 5.1.1 can be applied to knots that admit alternating projections on closed surfaces
so that they define essential checkerboard surfaces. A large such class of knots is the class
of adequate knots, that admit alternating projections with essential checkerboard surfaces
on certain Turaev surfaces. In this case we have the following theorem, where the terms
involved are defined in detail in Chapter 2, Section 5.2.1, and Section 5.2.2.
Theorem 5.3.3. Let K be an adequate hyperbolic knot in S 3 with crossing number c = c(K)
and Turaev genus gT . Let C denote the maximal cusp of S 3 \ K and let Area(∂C) denote
the cusp area. Finally let (µ) and (λ) denote the length of the meridian and the shortest
94

λ-curve of K. Then we have
1. (µ) ≤ 3 +

6gT − 6
c

2. (λ) ≤ 3c + 6gT − 6
2gT − 2 2
3. Area(∂C) ≤ 9c 1 +
c
A knot is alternating if and only if gT = 0. In this case, the bounds of Theorem 5.3.3
agree with the bounds of [6]. The technique of the proof of Theorems 5.1.1 and 5.3.3, as well
as the proof of results in [6], is reminiscent of arguments with pleated surfaces that led to
the proof of the “6-Theorem” [10, 43]. The algorithm for checking the criterion of Theorem
2.1.2 involves normal surface theory and in particular the work of Jaco and Sedgwick [39].

5.1.2

Knots with meridian length bounded by four

As mentioned earlier, it has been conjectured that the meridian length of every hyperbolic
knot in S 3 is at most four. The conjecture is known for several classes of knots. Adams [9]
showed that the meridian of a 2-bridge hyperbolic knot has length less than 2. By [6], when
K is an alternating hyperbolic knot then (µ) < 3. Agol [10] found families of knots whose
meridian lengths approach four from below and Purcell [61] generalized his construction
to construct families of knots whose meridian length approach four from below. She also
showed that “highly twisted” knots meridian lengths are less than four. The results of this
chapter allow us to verify the meridian length conjecture for additional broad classes of
hyperbolic knots. We provide two results for adequate knots. Note that, by Theorem 5.3.3,
if c ≥ 6gT − 6 then (µ) ≤ 4. Thus, for every Turaev genus there can be at most finitely
many adequate knots with (µ) > 4. In particular if gT ≤ 3, then (µ) ≤ 4 unless c ≤ 12.

95

Since the knots with up to 12 crossings are known to have meridian lengths less than two
(see [17]), in fact, we have the following result.
Corollary 5.1.2. Given gT > 0, there can be at most finitely many hyperbolic adequate
knots of Turaev genus gT and with (µ) > 4. In particular, if K is a hyperbolic adequate
knot with gT ≤ 3, then we have (µ) < 4.
Note that for gT = 1, we actually get (µ) ≤ 3. This case includes Conway sums of
strongly alternating tangles (see Section 3.2.2). Therefore, if a knot K is a Conway sum of
strongly alternating links, then the length of the meridian of K is less or equal to three.
Another instance where our length bounds work well is to show that knots admitting
diagrams with large ratio of crossings to twist regions have small meridian length. We have
the following result which in particular applies to closed positive braids. See Corollary 5.3.7.
Theorem 5.3.6. Let K be a hyperbolic knot with an adequate diagram with c crossings and
t twist regions. Then we have
(µ) ≤ 3 +

3t 6
− .
c
c

In particular if c ≥ 3t then we have (µ) < 4.

5.1.3

Slope length bounds, Dehn filling and volume

Let K be a hyperbolic knot with maximal cusp C and slopes σ, σ on ∂C. Calculating area
in Euclidean geometry on ∂C (see for example the proof of [10, Theorem 8.1]), we have

(σ) (σ ) ≥ Area(∂C)∆(σ, σ )

(5.2)

where ∆(σ, σ ) denotes the absolute value of the intersection number of σ and σ . Work of
96

Cao and Meyerhoff [16, Proposition 5.8] shows that Area(∂C) ≥ 3.35. Given an adequate
hyperbolic knot K, we will apply (5.2) for σ = µ. Using the upper bound for (µ) from
Theorem 5.3.3, we have

(σ) >
where δ =

3.35∆(µ, σ)c
3.35 ∆(µ, σ)
=
·
3c + 6gT − 6
3
1+δ

(5.3)

2gT − 2
. Note that δ is an invariant of K that can be calculated from any
c

adequate diagram (see Theorem 5.2.4). Now (5.3) implies that if

∆(µ, σ) >

18
(1 + δ) > 5.37 (1 + δ)
3.35

then (σ) > 6 and thus σ cannot be an exceptional slope.
Note that if σ is a slope represented by p/q ∈ Q in H1 (∂C) then ∆(µ, σ) = |q|. Hence
if |q| > 6(1 + δ), inequality (5.3) implies that (σ) >

3.35
· 6 > 2π. In this case, we may
3

apply a result of Futer, Kalfagianni and Purcell [31, Theorem 1.1] to estimate the change of
volume under Dehn filling of adequate knots. This produces the following theorem.
Theorem 5.1.3. Let K be a hyperbolic adequate knot and let δ be as above. If |q| ≥ 6(1 + δ),
then the 3-manifold N obtained by p/q surgery along K is hyperbolic and the volume satisfies
the following

vol(S 3

\ K) > vol(N ) ≥

36(1 + δ)2
1−
q2

3/2

vol(S 3 \ K).

The assertion that N is hyperbolic follows immediately from the discussion above. The
left hand side inequality is due to the result of Thurston that the hyperbolic volume drops
under Dehn filling (Theorem 2.1.2). The right hand side follows by [31, Theorem 1.1].
97

Theorem 5.14 of [30], and its corollaries, give diagrammatic bounds for vol(S 3 \ K) in
terms any adequate diagram of K. This combined with Theorem 5.1.3 implies that the
volume of N can be estimated from any adequate diagram of K. For example, Montesinos
knots with a reduced diagrams that contains at least two positive tangles and at least two
negative tangles are adequate and have δ ≤ 0. Combining Theorem 5.1.3 with [30, Theorem
9.12] and [29, Theorem 1.2] we have the following.
Corollary 5.1.4. Let K ⊂ S 3 be a Montesinos link with a reduced diagram D(K) that
contains at least two positive tangles and at least two negative tangles. If |q| ≥ 6, then the
3-manifold N obtained by p/q surgery along K is hyperbolic and we have

2v8 t > vol(N ) ≥

36 3/2 v8
(t − 9) ,
1− 2
4
q

where t = t(D) is the twist number of D(K), and v8 = 3.6638... is the volume of a regular
ideal octahedron.

5.1.4

Organization

In Section 5.2.1 we recall the hyperbolic geometry terminology we need for this chapter, and
the results and facts about pleated surfaces we will use. In Section 5.2.2 we recall results and
terminology about Turaev surfaces we need in subsequent sections. In Section 5.3 we derive
the bound of the meridian length in Theorem 5.1.1 and corresponding bounds for the length
of the shortest λ-curve and cusp volume. Then we prove Theorem 5.3.3 and its corollaries.
In Section 5.4 we show that given K and b > 0 there is an algorithm which determines if
there are essential spanning surfaces S1 and S2 satisfying inequality of Theorem 2.1.2 . This
completes the proof of Theorem 5.1.1.
98

The material of this chapter was developed jointly with the author and Efstratia Kalfagianni.

5.2

Background

In this section we will review results that will be used throughout the chapter. See Section
2.1.2 for a review of the notions of maximal cusp and slope length.

5.2.1

Pleated Surfaces

Consider a (possibly non-connected) surface S (possibly with boundary) and a singular
continuous map f : S −→ M that embeds each component of ∂S in ∂C. We will say that f
is homotopically-essential
1. The image of no essential simple closed loop on S is homotopically trivial in M .
2. The image of no essential embedded arc on S can be homotoped (relative to its endpoints) on ∂C.
If S ⊂ M is an essential embedded surface, the inclusion map is homotopically-essential. We
now recall Thurston’s notion of pleated surface. See Thurston’s notes [65] or the exposition by
Canary, Epstein and Green [15] for more details. A singular continuous map f : (S, ∂S) −→
(M, ∂C) is called pleated if the following are true:
1. The components of ∂S map to geodesics on ∂C.
2. The interior of S, denoted by int(S), is triangulated so that each triangle maps under
f to a subset of M that lifts to an ideal hyperbolic geodesic triangle in H3 .

99

3. The 1-skeleton of the triangulation forms a lamination on S.
Given a pleated map f we may pull-back the path metric from M by f to obtain a
hyperbolic metric on int(S), where the 1-skeleton lamination is geodesic.
The following lemma will be used to make the essential spanning surfaces considered in
this chapter pleated. For a proof the reader is referred to [15, 65] or to [10, Lemma 4.1].
Lemma 5.2.1. Let M = S 3 \ K be a hyperbolic knot complement and let S be a surface with
boundary and χ(S) < 0. Let f : (S, ∂S) −→ (M, ∂C) be a homotopically essential map.
Suppose that each component of ∂S is mapped to a geodesic in ∂C. Then there is a pleated
map g : (S, ∂S) −→ (M, ∂C), such that g|int(S) is homotopic to f |int(S) and there exists a
hyperbolic metric on S so that g|∂S is an isometry.
Let M = S 3 \ K be a hyperbolic knot complement with maximal cusp C and let f :
(S, ∂S) −→ (M, ∂C) be a homotopically essential map that is pleated. In this paper we
are interested in the case that S is the disjoint union of spanning surfaces of K. Suppose
that ∂S has s components. The geometry of f (S) ∩ C can be understood using arguments
of [10, Theorem 5.1] and [43, Lemma 3.3]. By the argument in the proof of [10, Theorem
5.1], we can find disjoint horocusp neighborhoods H =

s
i=1 Hi

of S, such that f (Hi ) ⊂ C,

(∂Hi ) = Area(Hi ) and such that (∂Hi ) is at least as big as the length of f (∂Hi ) measured
on C. Thus we have
s
C (S) ≤

(∂Hi ) = Area(H),
i=1

where C (S) denotes the total length of the intersection curves in f (S) ∩ ∂C. Since for all
i = j, we have Hi ∩ Hj = ∅, a result of B¨or¨oczky [13] on horocycle packings in the hyperbolic

100

plane applies. Using this result one obtains
s

Area(Hi ) ≤
i=1

6
6
Area(S) =
(2π|χ(S)|),
2π
2π

where the last equation follows by the Gauss-Bonnet theorem. The above inequality is also
proved in [43, Lemma 3.3]. Combining all these leads to the following theorem which is a
special case of [10, Theorem 5.1] and [43, Lemma 3.3].
Theorem 5.2.2. Let M = S 3 \ K be a hyperbolic knot complement with maximal cusp C.
Suppose that f : (S, ∂S) −→ (M, ∂C) is a homotopically essential map that is pleated and
let C (S) denote the total length of the intersection curves in f (S) ∩ ∂C. Then we have

C (S)

5.2.2

≤ 6|χ(S)|.

Turaev Surfaces

The Turaev genus of a knot diagram D = D(K) with c crossings is defined by gT (D) =
(2 − vA − vB + c)/2, where vA , vB denotes the number of the state circles in the all-A and
all-B resolutions of D respectively. The Turaev genus of a knot K is defined by

gT (K) = min {gT (D) | D = D(K)} .

The genus gT (D) is the genus of the T uraev surface F (D) corresponding to D. This surface
is constructed as follows. Let Γ ⊂ S 2 be the planar, 4–valent graph defined by D. Thicken
the (compactified) projection plane to S 2 × [−1, 1], so that Γ lies in S 2 × {0}. Outside a
neighborhood of the vertices (crossings), Γ × [−1, 1] will be part of F (D).
101

Figure 5.1: Saddles of F (D) corresponding to two successive over-crossings of D. The third
picture illustrates how D is is alternating on F (D). The figure is taken from [24].

Figure 5.2: The saddle of the Turaev surface at a crossing of a link. Taken from [33].
In the neighborhood of each vertex, we insert a saddle, positioned so that the boundary
circles on S 2 × {1} are the components of the A–resolution and the boundary circles on
S 2 × {−1} are the components of the B–resolution. See Figure 5.2.
The following is proved in [24].
Lemma 5.2.3. The Turaev surface F (D) has the following properties:
(i) It is a Heegaard surface of S 3 .
(ii) D is alternating on F (D); in particular D is an alternating diagram if and only if

102

gT (F (D)) = 0. See Figure 5.1.
(iii) The 4-valent graph underlying D defines a cellulation of F (D) for which the 2-cells
can be colored in a checkerboard fashion.
(iv) The checkerboard surfaces defined by D on F (D) are the state surfaces SA and SB .
We note that an adequate diagram realizes the crossing number of the knot, hence it is
a knot invariant. The following result of Abe [1, Theorem 3.2] shows that the same is true
for the Turaev genus.
Theorem 5.2.4. Suppose that D is an adequate diagram of a knot K. Then,

2gT (K) = 2gT (D) = 2 − vA (D) − vB (D) + c(D).

5.3

Lengths of Curves on the Maximal Cusp Boundary

In this section, we prove the main results of this paper. We begin by giving a general bound
for lengths of curves in the boundary of a maximal cusp neighborhood of a hyperbolic knot.
We then apply this bound to the special cases of adequate knots and three-string pretzel
knots.
Theorem 5.3.1. Let K be a hyperbolic knot with maximal cusp C. Suppose that S1 and S2
are essential spanning surfaces in M = S 3 \ K and let i(∂S1 , ∂S2 ) = 0 denote the minimal
intersection number of ∂S1 , ∂S2 in ∂C. Let (µ) and (λ) denote the length of the meridian
and the shortest λ-curve of K, respectively. Then we have:
1. (µ) ≤

6(|χ(S1 )| + |χ(S2 )|)
i(∂S1 , ∂S2 )
103

2. (λ) ≤ 3(|χ(S1 )| + |χ(S2 )|)
3. Area(∂C) ≤ 18

(|χ(S1 )| + |χ(S2 )|)2
i(∂S1 , ∂S2 )

Proof. Consider S to be the disjoint union of S1 , S2 , and let f : S −→ M , where f (S) is the
union of S1 , S2 in the complement of K. Since f |Si is an embedding for i = 1, 2, and each
Si is essential, f is a homotopically essential map. Hence, by Lemma 5.2.1, we may pleat f
and then apply Theorem 5.2.2. With the notation as in that theorem we have

C (S)

≤ 6|χ(S)|,

where C (S) is the total length of the curves f (S) ∩ ∂C.
To find bounds of this total length, we orient ∂S1 , ∂S2 and µ so that ∂S1 , ∂S2 have
opposite algebraic intersection numbers with µ. Let [∂S1 ], [∂S2 ], and [µ] denote their classes
in π1 (∂C) = H1 (∂C). Since S1 is a spanning surface, we know that [∂S1 ] and [µ] generate
π1 (∂C).
Recall the covering π := ρ|RH : RH −→ ∂C, where RH is the boundary of a horoball
at infinity, say H ⊂ ∪ρ−1 (C). To fix ideas, assume that ∂S1 lifts to the horizontal lines
π −1 (∂S1 ) = {(x, n) : x ∈ R} for each n ∈ Z and where µ lifts to the vertical lines π −1 (µ) =
{(n, y) : y ∈ R} for each n ∈ Z. We may apply a homotopy to µ so that ∂S1 ∩∂S2 ∩µ = {x0 },
where π −1 (x0 ) = Z2 .
Since [∂S1 ] and [µ] generate π1 (∂C), we can write [∂S2 ] = α[µ]+β[∂S1 ] for some α, β ∈ Z.
The fact that S2 is a spanning surface implies |β| = 1 and |α| = i(∂S1 , ∂S2 ). Therefore [∂S2 ]
can be represented as a curve which lifts to the segment {(x, αx) : x ∈ [0, 1]} ⊂ R2 = RH .

104

∂S1
µ

α4

µ

α3

µ

∂S1
µ
∂S1
α2

µ

µ
∂S1

µ

α1

µ
∂S1

Figure 5.3: The arcs αk are each homotopic to the meridian, and their union projects to
∂S1 ∪ ∂S2 .
The collection of arcs

αk = {(x, αx) : x ∈ [k/α, (k + 1)/α]} ∪ {(x, k + 1) : x ∈ [k/α, (k + 1)/α]}

for k = 0, 1, . . . , α − 1 is mapped to ∂S1 ∪ ∂S2 by π. Moreover, each π(αk ) is a loop in ∂C
homotopic to a meridian. See Figure 5.3, where each αk is indicated in a different color.
Therefore ∂S1 ∪∂S2 can be decomposed into a collection of simple closed curves that contain
|α| meridians. Hence we obtain

i(∂S1 , ∂S2 ) (µ) ≤ C (S) ≤ 6|χ(S1 )| + 6|χ(S2 )|.

The decomposition of ∂S1 ∪ ∂S2 described above can be also seen by resolving all the
intersections of ∂S1 , ∂S2 in a way consistent with the orientations chosen above.
To prove part (2), once consider ∂S1 and ∂S2 oriented as above in ∂C. By resolving
the crossings of ∂S1 with ∂S2 in a manner not consistent consistent with the orientations of
105

∂S1 and ∂S2 , one obtains two -curves in ∂C. Thus 2 (λ) ≤ C (S) and Theorem 5.2.2 now
implies that

2 (λ) < 6|χ(S1 )| + 6|χ(S2 )|.
To prove part (3), observe that Area(∂C) ≤ (µ) (λ).
As an example, we apply Theorem 5.3.1 to 3-string pretzel knots. Note that nonalternating 3-string pretzel knots are not adequate as it follows from the work of Lee and
van der Veen [46].
Example 5.3.2. Let K be the pretzel knot P (a, −b, −c) with a, b, c all positive and odd.
The standard 3-pretzel diagram of K is A-adequate. Hence the corresponding all-A state
surface SA is essential in the complement of K. Moreover, the 3-pretzel surface SP is
a minimum genus Seifert surface for K and thus also essential. The boundary slope of
B
A
the spanning surface SA of K is given by s(S) = 2cB
+ − 2c− where c+ is the number of

positive crossings resolved in the B direction. We then see that s(SA ) = −2b − 2c. On the
other hand, s(SP ) = 0. The difference in slopes of two surfaces is equal to the geometric
intersection number, so we obtain that i(∂SA , ∂SP ) = 2b + 2c. An easy calculation shows
that χ(SA ) = 1 − b − c and χ(SP ) = −1. Using Theorem 5.3.1 we have (µ) ≤ 3.
The same process will apply to any knot that admits an essential state surface that has
non-zero slope. Large familes of such knots are the semi-adequate knots or more generally
the σ-adequate and σ-homogeneous knots [30, Definition 2.22].
We now consider an application of Theorem 5.3.1 to the case of adequate knots, and
we derive Theorem 5.3.3 stated in the introduction. For the convenience of the reader, we
restate the theorem.
106

Figure 5.4: The intersection of the surfaces SA (red) and SB (blue) with ∂C. Taken from
[45].
Theorem 5.3.3. Let K be an adequate hyperbolic knot in S 3 with crossing number c = c(K)
and Turaev genus gT . Let C denote the maximal cusp of S 3 \ K and let Area(∂C) denote
the cusp area. Finally let (µ) and (λ) denote the length of the meridian and the shortest
λ-curve of K. Then we have
1. (µ) ≤ 3 +

6gT − 6
c

2. (λ) ≤ 3c + 6gT − 6
3. Area(∂C) ≤ 9c 1 +

2gT − 2 2
c

Proof. Let D be an adequate diagram for K and let SA and SB be the corresponding all-A
and all-B state surfaces respectively. By Theorem 2.2.2, SA , SB are essential in M = S 3 \K.
Now ∂SA and ∂SB intersect transversely exactly twice per crossing in D. We show that this
number of intersections is in fact minimal. To do so, we use the well-known “bigon criterion”
(see for example [28, Proposition 1.7]) which states that two transverse simple closed curves
in a surface are in minimal position if and only if they do not form a bigon.
Consider the curves ∂SA and ∂SB near two consecutive crossings of D. If one crossing
is an over-crossing and the other crossing is an under-crossing in the diagram D, then the
intersection curves will be as in Figure 5.4. Note that this forms a diamond pattern on ∂C
near alternating crossings, hence there are no bigons near alternating crossings.

107

Consider the Turaev surface F (D) corresponding to D. Recall that D is alternating on
F (D) and that SA , SB are the checkerboard surfaces of this projection (Lemma 5.2.3).
We turn to the case where two consecutive crossings in D are over-crossings. The Turaev
surface T of K in a neighborhood of these two crossings may be visualized as in Figure 5.1.
The neighborhood may be straightened as shown in Figure 5.1, and we then see that the
intersection of ∂C with SA ∪ SB in a neighborhood of these two crossings is as in Figure
5.4. Therefore we get an intersection pattern similar to that of 5.4 near pairs of consecutive
over-crossings, and it follows that there are no bigons near pairs of over-crossings. Similarly
there are no bigons near pairs of under-crossings. Thus we have i(∂S1 , ∂S2 ) = 2c.
On the other hand, by construction of the state surface and using the notation of §3.2,
we have χ(SA ) = vA − c and χ(SB ) = vB − c. Note that if χ(SA ) = 0 or χ(SB ) = 0 then SA
or SB is a M¨obius band. But then D is a diagram of the (2, p) torus knot contradicting the
assumption that K is hyperbolic. Thus χ(SA ), χ(SB ) < 0. Now by the definition of gD (T )
and Theorem 5.2.4 we have

|χ(SA )| + |χ(SB )| = 2c − vA − vB = c + 2gT − 2.
Using these observations, claims (1)-(3) of the statement follow immediately from Theorem 5.3.1.
An immediate consequence of Theorem 5.3.3 is that the meridian length of a knot with
Turaev genus 1 never exceeds 3. Also as noted in Corollary 5.1.2 for every Turaev genus
there can be at most finitely many adequate knots where (µ) ≥ 4.
Theorem 5.3.3 gives a bound on meridian lengths in terms of the Turaev genus and
crossing number. We wish to understand the relationship between these invariants, so we

108

Figure 5.5: The braid of example 5.3.5 in the case n = 5.
make the following definition.
Definition 5.3.4. The Turaev genus density of a link K is defined to be

2gT − 2
where gT
c

is the Turaev genus of K and c is the crossing number of K.
We know that there are many families of knots with meridian lengths approaching 4 from
below. However the bound of Theorem 5.3.3 does not require that the meridian length of
all knots be less than 4, as shown in the following example.
Example 5.3.5. Let n ≥ 2 be an integer and consider the following braid on 2n strings:

3
3
bn =σn σn−1 σn+1 σn−2 σn σn+2 . . . σ2 σ4 . . . σ2n−4 σ2n−2 σ13 σ33 . . . σ2n−3
σ2n−1
σ22 σ42
2
2
. . . σ2n−4
σ2n−2
σ3 σ5 . . . σ2n−5 σ2n−3 . . . σn−1 σn+1 σn

See Figure 5.5.
One can readily verify that the closure Kn of bn is an adequate knot. The A-resolution
has 2n state circles. The B-resolution has (3n − 1) state circles coming from bigon regions,
and n circles coming from non-bigon regions. Thus vA (Kn ) + vB (Kn ) = 6n − 1. The number
of crossings in the braid diagram for Kn is equal to n2 + 3n − 1, and since this diagram is

109

adequate, we know that the crossing number c(Kn ) is equal to n2 + 3n − 1. Therefore
6n − 1
vA (Kn ) + vB (Kn )
= lim 2
=0
n→∞ n + 3n − 1
n→∞
c(Kn )
lim

(5.4)

Let g(Kn ) be the Turaev genus of Kn . Since the braid diagram is adequate, we know that
the Turaev genus of the braid diagram is equal to the Turaev genus of Kn by [1]. Therefore
2g(Kn ) = c(Kn ) + 2 − vA (Kn ) − vB (Kn ) implying
v (Kn ) + vB (Kn )
2g(Kn ) − 2
=1− A
c(Kn )
c(Kn )

(5.5)

Therefore the meridian bound of Theorem 5.3.3 is

3+3

2g(Kn ) − 2
c(Kn )

v (Kn ) + vB (Kn )
=3+3 1− A
c(Kn )

(5.6)

which by (5.4) approaches 6 as n approaches ∞.
On the other hand, the next result, stated in the introduction, shows that in a certain
sense “most” adequate hyperbolic knots have meridian length less than 4.
Theorem 5.3.6. Let K be a hyperbolic knot with an adequate diagram with c crossings and
t twist regions. Then we have
(µ) ≤ 3 +

3t 6
− .
c
c

In particular if c ≥ 3t then we have (µ) < 4.
Proof. Let gT be the Turaev genus of K and let vA and vB be the number of A and B state
circles arising from D. Recall that 2gT − 2 = c − vA − vB . Now vA + vB = vbi + vnb where

110

vbi is the number of bigon regions in D and vnb is the number of non-bigon regions. Then

c − vbi = t

(5.7)

Since D is adequate and hyperbolic, both the A and B resolutions must have a state circle
corresponding to a non-bigon region. For if all the regions in one of the resolutions are bigons
then D represents a (2, p) torus knots, which is not hyperbolic. Therefore vnb ≥ 2 and it
follows that

2gT − 2 = c − vbi − vnb = t − vnb ≤ t − 2

Now by Theorem 5.3.3 we see that

(µ) < 3 + 3

2gT − 2
c

≤3+3

t−2
c

≤3+

3t 6
− .
c
c

Now if c ≥ 3t, say for example if D has at least three crossings per twist region, then
3t/c ≤ 1, so we see that
(µ) < 3 + 1 −

6
< 4.
c

Theorem 5.3.6 applies to positive/negative closed braids. Let Bn be the braid group
on n strands, with n ≥ 3, and let σ1 , . . . , σn−1 be the elementary braid generators. Let
r

r

r

b = σi 1 σi 2 · · · σi k be a braid in Bn . It is straightforward to check that if either rj ≥ 2 for
1 2
k
all j, or else rj ≤ −2 for all j, then the braid closure Db of b is an adequate diagram. In
particular we have the following.

111

Corollary 5.3.7. Suppose that a knot K is represented by a braid closure Db such that either
rj ≥ 3 for all j, or else rj ≤ −3 for all j. Additionally, suppose Db is a prime diagram.
Then K is hyperbolic and the meridian length satisfies (µ) < 4.
Proof. The fact that K is hyperbolic follows by [34, Corollary 1.2] and the claim about the
meridian follows from Theorem 5.3.6.

The twist number of any diagram of a hyperbolic knot K bounds Area(∂C) from above.
More precisely, if a hyperbolic knot with maximal cusp C admits a diagram with t twist
√
regions then Area(∂C) ≤ 10 3 · (t − 1) ≈ 17.32 · (t − 1). The derivation of this bound
is explained for example in [6]. Note that if c >> t, this general bound does better than
the one of Theorem 5.3.3. On the other hand if c = t and gT is small the upper bound of
Theorem 5.3.3 is sharper than the general bound. For instance if gT ≤ 1 and c = t, then
Theorem 5.3.3 gives Area(∂C) ≤ 9t which for t ≥ 3 is sharper than the general bound.
Theorem 5.3.1 more generally applies to knots that admit alternating projections on
surfaces so that they define essential checkerboard surfaces. Specifically, let F be closed
surface that is embedded in S 3 in a standard or non-standard way. Let K be a knot and
suppose that there is a projection p : S 3 −→ F such that: (i) p(K) is alternating and it
separates F ; (ii) the components of F \ p(K) are disks that can be colored in two different
colors so that the colors at each crossing of p(K) meet in a checkerboard fashion; and (iii)
the surface F \ p(K) is essential in S 3 \ K. For instance results similar to Theorem 5.3.3 and
Corollary 5.1.2 should also hold for weakly alternating knots considered by Ozawa [58] and
further discussed in [38]. In this case one should replace gT with the genus of the surface
F and the crossing number of the knot with the number of crossings of the alternating

112

projection on F .

5.4

Algorithm

In this section we will finish the proof of Theorem 5.1.1. The proof of the first part of the
Theorem follows from part (a) of Theorem 5.3.1. That is, if a hyperbolic knot K in S 3
admits essential spanning surfaces S1 , S2 such that

|χ(S1 )| + |χ(S2 )| <

b · i(∂S1 , ∂S2 )
,
6

(5.8)

for some real number b > 0, then

(µ) <

6(|χ(S1 )| + |χ(S2 )|)
< b.
i(∂S1 , ∂S2 )

The proof of Theorem 5.1.1 will be complete once we show the following.
Theorem 5.4.1. Given any hyperbolic knot K and positive real number b, there is an algorithm which determines if there are spanning surfaces S1 and S2 satisfying inequality (5.8).

Proof. We now show that the condition of equation (5.8) is algorithmically checkable. Start
with a triangulation of the complement M = S 3 \ K. There is an algorithm [40] to turn
the triangulation to one that has a single vertex that lies on the boundary of M . Moreover,
by Jaco and Sedgwick [39] there is an algorithm that “layers” this triangulation so that a
meridian of K is a single edge on ∂M that is connected to the vertex of the triangulation.
Call the latter triangulation T . For normal surface background and terminology the reader
is referred to Matveev [50] or the introduction of [39].

113

Lemma 5.4.2. Suppose that there are essential spanning surfaces S1 , S2 that satisfy (5.8).
Then we can find essential spanning surfaces that satisfy condition (5.8) and, in addition,
are normal fundamental surfaces with respect to T .
Proof. Suppose that one of S1 , S2 , say S1 is not connected. Then since S1 is a spanning
surface, and hence has a single boundary component, one of the connected components must
be a closed surface F . Since K is hyperbolic and F is essential χ(F ) ≤ 0, so taking S = S1 \F
we see that |χ(S)| ≤ |χ(S1 )|, and i(∂S, ∂S2 ) = i(∂S1 , ∂S2 ). Replacing S1 with S, we may
assume S1 (and likewise S2 ) is connected.
Any essential surface in S 3 \K may be isotoped to a normal surface with respect to T .
Moreover, this normal surface may be be taken to be minimal in the sense of [50, Definition
4.1.6]. This means that the number of intersections of the surface with the edges of T is
minimal in the (normal) isotopy class of the surface. We will show that S1 and S2 may be
taken to be fundamental normal surfaces.
Suppose that S1 is not fundamental. Then S1 can be represented as a Haken sum
S1 = Σ1 ⊕ . . . ⊕ Σn ⊕ F1 ⊕ . . . ⊕ Fk where each Σi is a fundamental normal surface with
boundary, and each Fi is a closed fundamental normal surface. A theorem of Jaco and
Sedgwick [39] states that each Σi has the same slope. Since S1 is a spanning surface, and
hence it has a single boundary component, this implies that n = 1. Since K is hyperbolic,
we know that either χ(Fi ) < 0 or Fi is a boundary parallel torus for all i. In the latter case,
it is known, as noted in [37] that Σ1 ⊕ Fi is isotopic in S 3 \N (K) to Σ1 . In the event that
χ(Fi ) < 0, we note that |χ(Σ1 )| < |χ(S1 )| and equation (5.8) will hold with S1 replaced
by Σ1 . Moreover Matveev [50, Corollary 4.1.37] shows that Σ1 must be incompressible.
Therefore we can ignore the other terms of the Haken sum and assume that S1 is fundamental.

114

Similarly, we can assume that S2 is fundamental.
By Lemma 5.4.2, in order to decide whether there are spanning surfaces that satisfy
(5.8), it is enough to decide whether there are fundamental normal spanning surfaces with
the same property. Given K, there are only finitely many fundamental surfaces in M , and
there is an algorithm, due to Haken, to find them. Let F denote the list of all fundamental
surfaces. Since one of the boundary edges of the triangulation is a meridian, we may create
a subset FSpan ⊂ F of fundamental normal spanning surfaces which are spanning by finding
the surfaces that intersect the meridian exactly once. There is an algorithm to compute
χ(F ) for all surfaces F ∈ F, and to compute the minimal intersection number of two
fundamental normal surfaces [40]. The algorithm now works by computing |χ(S1 )| + |χ(S2 )|
and i(∂S1 , ∂S2 ) for all pairs of surfaces S1 , S2 ∈ FSpan and checking whether inequality (5.8)
holds. If the condition holds, then use the algorithm of Haken to check that S1 and S2 are
incompressible. If the condition fails for all pairs S1 , S2 ∈ FSpan , then inequality (5.8) does
not hold for any pair of essential spanning surfaces of K.
Knots with pairs of essential spanning surfaces S1 , S2 with i(∂S1 , ∂S2 ) = 0 are abundant.
Note however that not all knots have distinct essential spanning surfaces S1 , S2 for which
i(∂S1 , ∂S2 ) = 0. An example of such a knot is given by Dunfield in [27]. In this case, the
algorithm outlined above will return that inequality (5.8) cannot be satisfied. This may be
seen as follows. In this case, either
1. the set FSpan contains only one member, in which case there are no pairs to test, or
2. the intersection number i(∂S1 , ∂S2 ) = 0 for all pairs S1 , S2 ∈ FSpan , and inequality
(5.8) will always fail since K is hyperbolic implies |χ(S1 )| > 0.

115

APPENDIX

116

Appendix
Python Code for Algorithm 1 of
Section 4.5
This is the Python code used to verify Conjecture 4.1.1 for alternating pretzel links with 4
twist regions.
# Number of twist regions
regions = 4
import numpy as np
import itertools

# Computes the determinant
# Input: List
# Output: Integer
def det(x):
partial_sum = 0
for a in x:
partial_sum = partial_sum + np.prod(x)/a
return partial_sum

117

# Computes a lower bound on determinant
# Input: List
# Output: integer
def partial_det(x):
if len(x) == 1:
return x[0]*(regions-1) + 1
if len(x) == 2:
return x[0]*x[1]*(regions - 2) + x[0] + x[1]
else:
return np.prod(x)*(regions - len(x)) + det(x)

# Computes an upper bound of the e^(volume of an x-bipyramid/(2pi))
def bipyramid_vol(x):
if x == 2:
return 1.0
elif x == 3:
return 1.382
elif x == 4:
return 1.792
elif x == 5:
return 2.212
elif x == 6:
return 2.636
elif x == 7:
118

return 3.063
elif x == 8:
return 3.491
elif x == 9:
return 3.921
elif x == 10:
return 4.351
elif x == 11:
return 4.781
elif x == 12:
return 5.211
elif x == 13:
return 5.643
elif x == 14:
return 6.074
elif x < 20 and x > 14:
return 8.664/20*x
elif x == 20:
return 8.664
elif x > 20 and x < 100:
return .43265*x
elif x == 100:
return 43.265
elif x > 100 and x <= 1000000:
119

return .44*x
else:
return x/2.0

#Some code I took from ShreevatsaR on Stack Overflow to permute more
#intelligently. It sorts a sorted list l in place and then returns
# false only when sorted in reverse
def next_permutation(l):
#Changes a list to its next permutation, in place.
#Returns true unless wrapped around so result is lexicographically
#smaller.
n = len(l)
#Step 1: Find tail
last = n-1 #tail is from ‘last‘ to end
while last>0:
if l[last-1] < l[last]: break
last -= 1
#Step 2: Increase the number just before tail
if last>0:
small = l[last-1]
big = n-1
while l[big] <= small: big -= 1
l[last-1], l[big] = l[big], small
#Step 3: Reverse tail
120

i = last
j = n-1
while i < j:
l[i], l[j] = l[j], l[i]
i += 1
j -= 1
return last>0

# Computes the Adams volume estimate for all configurations of strings
# with lengths specified in a list x
# Input: x = list of string lengths
#

regions = number of strings (probably not really needed, oh well)

# Output: float volume = Adams volume upper bound for configuration
# with highest volume
def vol(x, regions):
# We loop over all the permutations of x to find highest volume
#configuration
volume = 0
#configurations = list(itertools.permutations(x))
perm = x
#for perm in configurations:
stay_in_loop = True
while stay_in_loop:
v = []
121

# First we add in all the polygons with twists on the sides
v.append(perm[0]+perm[len(perm)-1])
for n in range(0, len(perm)-1):
v.append(perm[n]+perm[n + 1])
# Next we add the top and bottom polygons
v.append(regions)
v.append(regions)
# Take the product of all but the two largest polygons
v = np.sort(v)
w = []
for poly in v:
w.append(bipyramid_vol(poly))
temp_volume = np.prod(w[0:len(w)-2])
if temp_volume > volume:
volume = temp_volume
stay_in_loop = next_permutation(perm)
return volume

def dt_vol(twists):
x = np.array(twists)
twist_counts = []
for n in range(1, 5):
twist_counts.append((x == n).sum())
twist_counts.append((x>4).sum())
122

if twist_counts[4] > 0:
a = 12.111
elif twist_counts[3] > 0:
a = 10.2873
elif twist_counts[2] > 0:
a = 10.088
elif twist_counts[1] > 0:
a = 11*1.01494
else:
a = 15.4972
volume = twist_counts[0]*3.6639+\
twist_counts[1]*6*1.01495+\
twist_counts[2]*7.8549+\
twist_counts[3]*9.2375+\
twist_counts[4]*10*1.01495-a
volume = np.exp(volume/(2*np.pi))
return volume

def agm_vol(twists):
return np.mean(twists)**len(twists)

# Lackenby Agol-Thurston-Storm

volume bound

twist_bound = 5.04**(regions - 1)

123

# Maximum number of twists in a region
max_twist = int(regions + 5.0297**(regions - 1) - \
2*(1.4254)**(regions-1))+1

# Number of problem cases
problem = 0
problem_list = []

# Total Adams volume checks:
total_checks = 0

# Maximum volume found and the case where it arises
max_volume = 0
max_volume_twists = []

for n in range(0, regions):
max_volume_twists.append(n)

twists = []
for x in range(1,max_twist+1):
twists.append(x)
determinant = partial_det(twists)
if determinant > twist_bound:
twists.pop()
124

break
# If the current spot is a 5 or higher then the dt_vol of all the
#successive links
# can be computed and will be constant, while determinant will
#continue to increase
last_tw = twists[len(twists)-1]
rem_tw = regions - len(twists)
if last_tw > 4:
partial_volume = dt_vol(twists+rem_tw*[5])
partial_determinant = partial_det(twists + rem_tw*[last_tw])
if partial_volume < partial_determinant:
twists.pop()
break
for x in range(twists[len(twists)-1],max_twist+1):
twists.append(x)
determinant = partial_det(twists)
if determinant > twist_bound:
twists.pop()
break
# If the current spot is a 5 or higher then the dt_vol of all the
#successive links
# can be computed and will be constant, while determinant will
#continue to increase
last_tw = twists[len(twists)-1]
125

rem_tw = regions - len(twists)
if last_tw > 4:
partial_volume = dt_vol(twists+rem_tw*[5])
partial_determinant = partial_det(twists + rem_tw*[last_tw])
if partial_volume < partial_determinant:
twists.pop()
break
for x in range(twists[len(twists)-1],max_twist+1):
twists.append(x)
determinant = partial_det(twists)
if determinant > twist_bound:
twists.pop()
break
# If the current spot is a 5 or higher then the dt_vol of all the
#successive links
# can be computed and will be constant, while determinant will
#continue to increase
last_tw = twists[len(twists)-1]
rem_tw = regions - len(twists)
if last_tw > 4:
partial_volume = dt_vol(twists+rem_tw*[5])
partial_determinant = partial_det(twists + rem_tw*[last_tw])
if partial_volume < partial_determinant:
twists.pop()
126

break
prev = 0
for x in range(twists[len(twists)-1],max_twist+1):
twists.append(x)
determinant = partial_det(twists)
if determinant > twist_bound:
twists.pop()
break
# If the current spot is a 5 or higher then the dt_vol of all
#the successive links
# can be computed and will be constant, while determinant will
#continue to increase
last_tw = twists[len(twists)-1]
rem_tw = regions - len(twists)
if last_tw > 4:
partial_volume = dt_vol(twists+rem_tw*[5])
partial_determinant = partial_det(twists + rem_tw*[last_tw])
if partial_volume < partial_determinant:
twists.pop()
break
else:
#In this case det(twists) < twist_bound
total_checks = total_checks+1
agm_volume = agm_vol(twists)
127

if agm_volume < determinant:
twists.pop()
break
volume = vol(twists,regions)
if volume > max_volume:
max_volume = volume
for n in range(0, regions):
max_volume_twists[n] = twists[n]
if volume > determinant:
# In this case the algorithm has failed
print "Algorithm failed."
problem = problem + 1
problem_twist = []
for n in range(0, len(twists)):
problem_twist.append(twists[n])
problem_list.append([problem_twist, det(twists), volume])
twists.pop()
twists.pop()
twists.pop()
twists.pop()
print "Problem cases: ", problem
print "Cases checked: ", total_checks
print "Maximum volume:", max_volume
print "Max twist volume:", max_volume_twists
128

np.save("./problem_list"+str(regions)+".npy", problem_list)
for stuff in problem_list:
print "Problem twist: ", stuff[0]
print "det: ", stuff[1], "vol: ", stuff[2]

129

BIBLIOGRAPHY

130

BIBLIOGRAPHY

[1] T. Abe. The Turaev genus of an adequate knot. Topology Appl., 156(17):2704–2712,
2009.
[2] C. Adams. Triple crossing number of knots and links. J. Knot Theory Ramifications,
22(2):1350006, 17, 2013.
[3] C. Adams. Bipyramids and bounds on volumes of hyperbolic links. arXiv:1511.02372v1,
2015.
[4] C. Adams, H. Bennett, C. Davis, M. Jennings, J. Kloke, N. Perry, and E. Schoenfeld. Totally geodesic Seifert surfaces in hyperbolic knot and link complements. II. J.
Differential Geom., 79(1):1–23, 2008.
[5] C. Adams, A. Calderon, X. Jiang, A. Kastner, G. Kehne, N. Mayer, and M. Smith.
Density spectra for knots. arXiv:1506.05841, 2015.
[6] C. Adams, A. Colestock, J. Fowler, W. Gillam, and E. Katerman. Cusp size bounds from
singular surfaces in hyperbolic 3-manifolds. Trans. Amer. Math. Soc., 358(2):727–741,
2006.
[7] C. Adams, A. Kastner, A. Calderon, X. Jiang, G. Kehne, N. Mayer, and M. Smith.
Volume and determinant densities of hyperbolic rational links. J. Knot Theory Ramifications, 26(1):1750002, 13, 2017.
[8] C. C. Adams. Thrice-punctured spheres in hyperbolic 3-manifolds. Trans. Amer. Math.
Soc., 287(2):645–656, 1985.
[9] C. C. Adams. Hyperbolic 3-manifolds with two generators. Comm. Anal. Geom., 4(12):181–206, 1996.
[10] I. Agol. Bounds on exceptional Dehn filling. Geom. Topol., 4:431–449, 2000.
[11] I. Agol. Bounds on exceptional Dehn filling II. Geom. Topol., 14(4):1921–1940, 2010.
[12] S. A. Bleiler and C. D. Hodgson. Spherical space forms and Dehn filling. Topology,
35(3):809–833, 1996.
[13] K. B¨or¨oczky. Packing of spheres in spaces of constant curvature. Acta Math. Acad. Sci.
Hungar., 32(3–4):243–261, 1978.

131

[14] S. D. Burton. The spectra of volume and determinant densities of links. Topology Appl.,
211:38–55, 2016.
[15] R. D. Canary, D. B. A. Epstein, and P. Green. Notes on notes of Thurston. In Analytical
and geometric aspects of hyperbolic space (Coventry/Durham, 1984), volume 111 of
London Math. Soc. Lecture Note Ser., pages 3–92. Cambridge Univ. Press, Cambridge,
1987.
[16] C. Cao and G. R. Meyerhoff. The orientable cusped hyperbolic 3-manifolds of minimum
volume. Invent. Math., 146(3):451–478, 2001.
[17] J. C. Cha and C. Livingston. Knotinfo:
www.indiana.edu/~knotinfo, June 14 2014.

Table of knot invariants.

http://

[18] S. Chaiken. A combinatorial proof of the all minors matrix tree theorem. SIAM J.
Algebraic Discrete Methods, 3(3):319–329, 1982.
[19] A. Champanerkar, I. Kofman, and J. S. Purcell. Density spectra for knots. J. Knot
Theory Ramifications, 25(3):1640001, 11, 2016.
[20] A. Champanerkar, I. Kofman, and J. S. Purcell. Geometrically and diagrammatically
maximal knots. Journal of the London Mathematical Society, 94(3):883–908, 2016.
[21] A. Champanerkar, I. Kofman, and J. S. Purcell. Volume bounds for weaving knots.
Algebr. Geom. Topol., 16(6):3301–3323, 2016.
[22] M. Culler, N. M. Dunfield, M. Goerner, and J. R. Weeks. Snappy, a computer
program for studying the geometry and topology of 3-manifolds. Available at
http://snappy.computop.org.
[23] O. Dasbach and A. Tsvietkova. A refined upper bound for the hyperbolic volume of
alternating links and the colored Jones polynomial. Math. Res. Lett., 22(4):1047–1060,
2015.
[24] O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin, and N. W. Stoltzfus. The Jones
polynomial and graphs on surfaces. J. Combin. Theory Ser. B, 98(2):384–399, 2008.
[25] O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin, and N. W. Stoltzfus. Alternating
sum formulae for the determinant and other link invariants. J. Knot Theory Ramifications, 19(6):765–782, 2010.
[26] N. Dunfield. http://www.math.uiuc.edu/ nmd/preprints/misc/dylan/index.html.
[27] N. M. Dunfield.
arXiv:1509.06653.

A knot without a nonorientable essential spanning surface.

132

[28] B. Farb and D. Margalit. A primer on mapping class groups, volume 49 of Princeton
Mathematical Series. Princeton University Press, Princeton, NJ, 2012.
[29] K. Finlinson and J. S. Purcell. Volumes of Montesinos links. Pacific J. Math., 286(1):63–
105, 2016.
[30] D. Futer, E. Kalfagianni, and J. Purcell. Guts of surfaces and the colored Jones polynomial, volume 2069 of Lecture Notes in Mathematics. Springer, Heidelberg, 2013.
[31] D. Futer, E. Kalfagianni, and J. S. Purcell. Dehn filling, volume, and the Jones polynomial. J. Differential Geom., 78(3):429–464, 2008.
[32] D. Futer, E. Kalfagianni, and J. S. Purcell. Symmetric links and Conway sums: volume
and Jones polynomial. Math. Res. Lett., 16(2):233–253, 2009.
[33] D. Futer, E. Kalfagianni, and J. S. Purcell. Jones polynomials, volume and essential
knot surfaces: a survey. In Knots in Poland. III. Part 1, volume 100 of Banach Center
Publ., pages 51–77. Polish Acad. Sci. Inst. Math., Warsaw, 2014.
[34] D. Futer, E. Kalfagianni, and J. S. Purcell. Hyperbolic semi-adequate links. Comm.
Anal. Geom., 23(5):993–1030, 2015.
[35] C. M. Gordon and J. Luecke. Knots are determined by their complements. J. Amer.
Math. Soc., 2(2):0894–0347, 1989.
[36] J. Hoste and M. Thistlethwaite.
Knotscape
http://www.math.utk.edu/ morwen/knotscape.html.

1.01.

Available

at

[37] J. Howie. A characterisation of alternating knot exteriors. arXiv:1511.04945v1.
[38] J. Howie. Surface-alternating knots and links. University Of Melbourne PhD. Thesis.
[39] W. Jaco and E. Sedgwick. Decision problems in the space of Dehn fillings. Topology,
42(4):845–906, 2003.
[40] W. Jaco and J. L. Tollefson. Algorithms for the complete decomposition of a closed
3-manifold. Illinois J. Math., 39(3):358–406, 1995.
[41] E. Kalfagianni. Alexander polynomial, finite type invariants and volume of hyperbolic
knots. Algebr. Geom. Topol., 4:1111–1123, 2004.
[42] L. H. Kauffman and P. Lopes. Determinants of rational knots. Discrete Math. Theor.
Comput. Sci., 11(2):111–122, 2009.
[43] M. Lackenby. Word hyperbolic Dehn surgery. Invent. Math., 140(2):243–282, 2000.
133

[44] M. Lackenby. The volume of hyperbolic alternating link complements. Proc. London Math. Soc. (3), 88(1):204–224, 2004. With an appendix by Ian Agol and Dylan
Thurston.
[45] M. Lackenby and J. Purcell. Essential twisted surfaces in alternating link complements.
arXiv:1410.6297v3, 2014.
[46] C. R. S. Lee and R. van der Veen. Slopes for pretzel knots. arXiv:1602.04546.
[47] W. B. R. Lickorish. An introduction to knot theory, volume 175 of Graduate Texts in
Mathematics. Springer-Verlag, New York, 1997.
[48] W. B. R. Lickorish and M. B. Thistlethwaite. Some links with nontrivial polynomials
and their crossing-numbers. Comment. Math. Helv., 63(4):527–539, 1988.
[49] A. Marden. Outer circles. Cambridge University Press, Cambridge, 2007. An introduction to hyperbolic 3-manifolds.
[50] S. Matveev. Algorithmic topology and classification of 3-manifolds, volume 9 of Algorithms and Computation in Mathematics. Springer, Berlin, second edition, 2007.
[51] W. Menasco and M. Thistlethwaite. The classification of alternating links. Ann. of
Math. (2), 138(1):113–171, 1993.
[52] A. Modabish and M. El Marraki. The number of spanning trees of certain families of
planar maps. Appl. Math. Sci. (Ruse), 5(17-20):883–898, 2011.
[53] A. Modabish and M. El Marraki. Counting the number of spanning trees in the star
flower planar map. Appl. Math. Sci. (Ruse), 6(49-52):2411–2418, 2012.
[54] A. Modabish, D. Lotfi, and M. El Marraki. Formulas for the number of spanning trees
in a maximal planar map. Appl. Math. Sci. (Ruse), 5(61-64):3147–3159, 2011.
[55] L. G. Molinari. Determinants of block tridiagonal matrices. Linear Algebra Appl.,
429(8-9):2221–2226, 2008.
[56] J. Morgan and G. Tian. The geometrization conjecture, volume 5 of Clay Mathematics Monographs. American Mathematical Society, Providence, RI; Clay Mathematics
Institute, Cambridge, MA, 2014.
[57] G. D. Mostow. Quasi-conformal mappings in n-space and the rigidity of hyperbolic
´
space forms. Inst. Hautes Etudes
Sci. Publ. Math., (34):53–104, 1968.
[58] M. Ozawa. Non-triviality of generalized alternating knots. J. Knot Theory Ramifications, 15(3):351–360, 2006.
134

[59] K. A. Perko, Jr. On the classification of knots. Proc. Amer. Math. Soc., 45:262–266,
1974.
[60] G. Prasad. Strong rigidity of Q-rank 1 lattices. Invent. Math., 21:255–286, 1973.
[61] J. S. Purcell. Slope lengths and generalized augmented links. Comm. Anal. Geom.,
16(4):883–905, 2008.
[62] D. Rolfsen. Knots and links, volume 7 of Mathematics Lecture Series. Publish or Perish,
Inc., Houston, TX, 1990. Corrected reprint of the 1976 original.
[63] A. Stoimenow. Graphs, determinants of knots and hyperbolic volume. Pacific J. Math.,
232(2):423–451, 2007.
[64] D.
Thurston.
Hyperbolic
volume
and
the
http://pages.iu.edu/ dpthurst/speaking/Grenoble.pdf, 2000.
11.

jones
polynomial.
Accessed: 2015-06-

[65] W. P. Thurston. The Geometry and Topology of Three-Manifolds. Princeton Univ.
Math. Dept. Notes, 1979.
[66] W. P. Thurston. Three-dimensional manifolds, Kleinian groups and hyperbolic geometry. Bull. Amer. Math. Soc. (N.S.), 6(3):357–381, 1982.

135