JONES-TYPE LINK INVARIANTS AND APPLICATIONS TO 3-MANIFOLD
TOPOLOGY
By
Christine Ruey Shan Lee

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Mathematics - Doctor of Philosophy
2015

ABSTRACT
JONES-TYPE LINK INVARIANTS AND APPLICATIONS TO
3-MANIFOLD TOPOLOGY
By
Christine Ruey Shan Lee
It is known that the Slope Conjecture is true for an adequate link, and that the colored
Jones polynomial of a semi-adequate link has a well-defined tail (head) consisting of stable
coefficients, which carry geometric and topological information of the link complement. We
study the colored Jones polynomial of a link that is not semi-adequate and show that a tail
(head) consisting of stable coefficients of the polynomial can also be defined. Then, we prove
the Slope Conjecture for a new family of pretzel knots which are not adequate.
We also study the relationship between the Jones polynomial and the topology of the
knot complement by relating its coefficients to the non-orientable genus of an alternating
knot. The two-sided bound we obtain can often determine the non-orientable genus. Lastly,
we develop the connection between the Jones polynomial and a knot invariant coming from
Heegaard Floer homology by using it to classify 3-braids which are L-space knots.

For my parents

iii

ACKNOWLEDGMENTS

I would like to thank Effie Kalfagianni, for being the advisor that I needed, and the best that
I could hope for. She never hesitates to remind me that I can do better. Our conversations
are always inspiring, even when my thoughts are dulled.
I would like to extend my gratitude to Matt Hedden, for many enlightening discussions
and helpful advice. I am also grateful for Ben Schmidt and Bob Bell for reading my writing
and offering suggestions.
I am happy to acknowledge my friends at/from Michigan State University, in no particular
´
order: Akos,
for always knowing more math and being my laughing buddy; Samantha, for
being silly together in and out of the office; Charlotte, for being German and smart, and
willing to read the introduction of my paper; Allison, who is not in math, but we are all
the better for it; Fery, for always knowing when to call; Adam G., for being my academic
brother; Adam C. and Andrew, for showing me what life is like after grad school. It would
be difficult to imagine surviving grad school without you.
Lastly, I am grateful to Ron Fintushel. Without his phone call and well-timed encouragement, I would not be here.

iv

TABLE OF CONTENTS

LIST OF TABLES

LIST OF FIGURES

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

vii

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

Chapter 1 Introduction . . . . . . . . . . . . . . . .
1.1 Main results and organization of dissertation . . .
1.1.1 Chapter 2 . . . . . . . . . . . . . . . . . .
1.1.2 Chapter 3 . . . . . . . . . . . . . . . . . .
1.1.3 Chapter 4 . . . . . . . . . . . . . . . . . .
1.1.4 Chapter 5 . . . . . . . . . . . . . . . . . .
1.2 Preliminaries . . . . . . . . . . . . . . . . . . . .
1.2.1 The Temperley-Lieb algebra . . . . . . . .
1.2.2 The Jones-Wenzl idempotent . . . . . . .
1.2.3 Formula for the colored Jones polynomial .

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

1
3
3
5
6
8
8
9
10
11

Chapter 2 Stability properties of the colored Jones polynomial
2.1 The colored Jones polynomial of semi-adequate links. . . . . . . .
2.1.1 The Slope Conjecture . . . . . . . . . . . . . . . . . . . . .
2.1.2 The tail of the colored Jones polynomial . . . . . . . . . .
2.2 Ribbon graphs and the Kauffman bracket . . . . . . . . . . . . . .
2.3 A cancellation lemma . . . . . . . . . . . . . . . . . . . . . . . . .
n. . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Labeling edges in HA
2.5 Equivalence classes based on G(H). . . . . . . . . . . . . . . . . .
2.6 Combinatorics of G(H) and ribbon graphs . . . . . . . . . . . . .
2.7 Proof of Theorem 2.2.3 . . . . . . . . . . . . . . . . . . . . . . . .
2.8 A worked out example . . . . . . . . . . . . . . . . . . . . . . . .
2.9 Detecting semi-adequacy using the colored Jones polynomial . . .

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

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

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

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

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

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

14
15
15
15
19
26
27
39
42
51
52
58

Chapter 3 The Slope Conjecture and 3-string pretzel knots .
3.1 Essential surfaces and branched surfaces . . . . . . . . . . . . .
3.2 Essential surfaces in 2-bridge knot complements . . . . . . . . .
3.3 Curve systems and boundary slopes for Montesinos knots . . . .
3.4 The Hatcher-Oertel algorithm . . . . . . . . . . . . . . . . . . .
3.4.1 Computing the boundary slope from an edgepath system
3.4.2 The 3-strand pretzel knots P (−1/r, 1/s, 1/t). . . . . . . .
3.5 Computation of the Jones Slope . . . . . . . . . . . . . . . . . .
3.6 Proof of Theorem 1.1.2. . . . . . . . . . . . . . . . . . . . . . .

.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.

61
62
65
70
75
78
79
84
90

v

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.
.

.
.
.
.
.
.
.
.
.

Chapter 4 Crosscap number of alternating links and the
4.1 Augmented links and estimates with normal surfaces . .
4.1.1 An angled polyhedral decomposition. . . . . . . .
4.1.2 Normal surfaces and combinatorial area. . . . . .
4.1.3 Genus estimates for spanning surfaces . . . . . . .
4.2 Crosscap numbers of Alternating links . . . . . . . . . .
4.2.1 State surfaces and a minimum genus algorithm .
4.2.2 Normalization and crosscap number estimates . .
4.3 Proof of Theorem 1.1.3 . . . . . . . . . . . . . . . . . . .
4.4 Calculations of crosscap numbers . . . . . . . . . . . . .
4.4.1 Lower exact bounds . . . . . . . . . . . . . . . . .
4.4.2 Low crossing knots . . . . . . . . . . . . . . . . .
4.5 Generalizations and questions . . . . . . . . . . . . . . .
4.5.1 Non-alternating links . . . . . . . . . . . . . . . .

Jones
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .

polynomial
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .

92
95
96
98
102
105
105
113
121
123
123
127
128
128

Chapter 5 The Jones polynomial, 3-braids, and L-space knots . . . . . . 133
5.1 The Jones polynomial, 3-braids, and the Alexander polynomial . . . . . . . . 134
5.2 Proof of Theorem 1.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
BIBLIOGRAPHY

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

vi

LIST OF TABLES
Table 2.1

Coefficients for the A-adequate knot 10154 [9]. . . . . . . . . . . . .

18

Table 2.2

The tail for a non A-adequate link. . . . . . . . . . . . . . . . . . .

18

Table 2.3

The tables detail the various possibilities for the equivalence classes
corresponding to the different cases for a(H) and their corresponding
values of G(H), v(H)−k(H), and g(H), when sa is in RΩ1 or Ω0 . The
case for when sa is in LΩ1 is analogous. . . . . . . . . . . . . . . .

56

Table 4.1

Knots where the KnotInfo upper bound agrees with our lower bound.
The crosscap number is 3. . . . . . . . . . . . . . . . . . . . . . . . 128

vii

LIST OF FIGURES
Figure 1.1

The 3-string pretzel knot P (−1/10, 1/7, 1/9). . . . . . . . . . . . . .

6

Figure 1.2

Generators of the Temperley-Lieb algebra. . . . . . . . . . . . . . .

9

Figure 1.3

The Jones Wenzl idempotent and its closure in S(R2 ) [78]. . . . . .

10

Figure 1.4

Recursion formula for the Jones-Wenzl Idempotent [78]. . . . . . . .

11

Figure 1.5

Effect of adding a twist [78]. . . . . . . . . . . . . . . . . . . . . . .

11

Figure 1.6

D2 where D is a diagram of the figure 8 knot. . . . . . . . . . . . .

12

Figure 2.1

A- and B-resolutions of a crossing. . . . . . . . . . . . . . . . . . . .

19

Figure 2.2

The figure 8 knot, its all A-resolution and the resulting all-A state
graph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20

Figure 2.3

Identical local picture of the A-resolution of the 3-cable of a crossing.

28

Figure 2.4

From left to right: an illustration of the labeling on the edges of en
when n = 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

Figure 2.5

Locally, we replace the two segments making up the state circle (in
red) and the segment recording the crossing location (dotted) of the
A-resolution by that of the segments coming from the B-resolution.

31

From left to right: an all A-state graph with 3 state circles; an edge
inclusion representing a spanning sub-graph of the ribbon graph constructed from the all A-state graph; the state graph associated to the
spanning sub-graph, which has 2 state circles. . . . . . . . . . . . .

31

On the left: At least one of the two edges smaller than m in Ω1 needs
to be included in order that m ends up with two ends on distinct
state circles in H . The same is true with the edges bigger than M .
On the right, the portion of the region Ω0 cut off by the edge 2n − 1
is shown in yellow, with the edges with one end attached to it. . . .

33

The base case i = 1.

35

Figure 2.6

Figure 2.7

Figure 2.8

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

viii

Figure 2.9

Let sa = 2 ∈ RΩ1 for this example. The point of the sequences is to
rule out edges in H that may make edges after the edge labeled 2 in
RΩ1 have two ends on distinct state circles in H. The sequence for
RΩ1 are {bj } = {2, 5, ∅, . . .} and {tj } = {2, ∅, . . .}. . . . . . . . . .

38

On the left: All the edges shown in blue here after the red edge are
candidates to be included in G(H), as long as there are no edges above
and below after the middle red edge that are included in H. In the
middle and on the right: The inclusion of the top and bottom edges
in red means that only three edges in blue after red will be included
in G(H). This is so that they will still have two ends on the same
state circle as shown on the right when we only resolve the crossings
corresponding to the red edges. . . . . . . . . . . . . . . . . . . . . .

39

From left to right: An example of edges in G(H) marked in blue; the
state graph resulting from including zero edges from G(H); the state
graph resulting from choosing one edge from G(H), note that a state
circle splits off; the state graph resulting from including two edges
from G(H) with two state circles split off. . . . . . . . . . . . . . .

41

Figure 2.12

On the left we have H and on the right we have H . Their local
pictures only differ in the inclusion of two edges e1 and e2 . . . . . .

43

Figure 2.13

The edgeset V is shown in black.

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

49

Figure 2.14

The leftmost figure depicts a possible set of bk shown in blue, and
the edge excluded from bk by the edge in RΩk−1 . Depending on the
sequences {tj }, {bj } for RΩk and RΩk+1 , we have at least an extra
edge back from bk+1 or bk+2 . . . . . . . . . . . . . . . . . . . . . .

50

Figure 2.15

From left to right: a non A-adequate link diagram, the all-A state
graph, and the all A-state graph of the 3-cable. . . . . . . . . . . . .

53

Figure 3.1

A compressing disk [52]. . . . . . . . . . . . . . . . . . . . . . . . . .

63

Figure 3.2

A ∂-compressing disk [52]. . . . . . . . . . . . . . . . . . . . . . . .

63

Figure 3.3

Models of a branched surface and a branched surface neighborhood [54]. 64

Figure 3.4

Schematics for a Montesinos knot. . . . . . . . . . . . . . . . . . . .

65

Figure 3.5

Figure from [56]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

66

Figure 3.6

Intersection of S with Sr2 [56]. . . . . . . . . . . . . . . . . . . . . .

68

Figure 2.10

Figure 2.11

ix

Figure 3.7

The 1-simplex on the Poincare disc model [56]. . . . . . . . . . . . .

69

Figure 3.8

The generators a, b and c and the corresponding set of disjoint curves
on the 4-punctured sphere with a, b, c-coordinates (3, 1, 2). . . . . .

71

Figure 3.9

The three-simplex of curve systems on a four-punctured sphere [55].

72

Figure 3.10

The four-simplex of curve systems with systems of slope 1/0 added [55]. 73

Figure 3.11

On the left: the infinite strip containing the 1-dimensional complex
D. On the right: the portion of the infinite strip from height 0/1 to
1/1 [55]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

74

Figure 3.12

A choice of edgepath for each of the three fractions 1/2, 1/8, and 3/4
are shown in red, blue, and green. . . . . . . . . . . . . . . . . . . .

80

Figure 3.13

A trivalent graph representation [78]. . . . . . . . . . . . . . . . . .

84

Figure 3.14

The fusion formula [78]. . . . . . . . . . . . . . . . . . . . . . . . . .

85

Figure 3.15

The untwisting formula [78]. . . . . . . . . . . . . . . . . . . . . . .

85

Figure 3.16

The 6j-equation [78]. . . . . . . . . . . . . . . . . . . . . . . . . . .

86

Figure 3.17

Replacing a region bounded by two edges. δad is the Kronecker delta
function [78]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

86

Figure 3.18

Collapsing a triangular region to a vertex [82]. . . . . . . . . . . . .

87

Figure 3.19

The polytope cut out by equations α + β = γ, α + γ = β, β + γ = α.
It has been rescaled by dividing by n. . . . . . . . . . . . . . . . . .

89

Figure 4.1

Twist reduced: A or B must be a string of bigons [37]. . . . . . . . .

95

Figure 4.2

From left to right: A link K, an augmented link J and a fully augmented link L [32]. . . . . . . . . . . . . . . . . . . . . . . . . . . .

96

Figure 4.3

Two ways of augmenting a twist region with a single crossing. . . . .

97

Figure 4.4

Normal disks in a truncated polyhedron [37]. . . . . . . . . . . . . .

99

Figure 4.5

The two resolutions of a crossing, the arcs recording them and their
contribution to state surfaces. . . . . . . . . . . . . . . . . . . . . . 106

x

Figure 4.6

One branch of the algorithm resolves the crossings so that the triangle
becomes a state circle. The other branch resolves them the opposite
way. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Figure 4.7

A diagram of the figure 8 knot with bigon regions labeled 1 and 2
and the diagram resulting from applying the first step of Algorithm
4.2.2. The new choices of bigon regions are labeled 1, 2, and 3. . . . 109

Figure 4.8

The state surfaces resulting from resolving the rest of the bigon regions with different choices of bigon regions. . . . . . . . . . . . . . 110

Figure 4.9

The portion of S through a twist region with more than one crossing
and the crossing circle for the twist region. . . . . . . . . . . . . . . 111

Figure 4.10

Surfaces from two branches of the splitting in Step 2b in the AdamKindred algorithm and the corresponding choice of augmentation. . 112

Figure 4.11

The knots 103 (left) and 10123 (right) [19]. . . . . . . . . . . . . . . 119

Figure 4.12

From left to right: The portion of D(K) around a vertex of G, the
corresponding potion of the augmented link and portion of the polyhedral decomposition. An ideal triangle is indicated by the red line.

Figure 5.1

125

The generator σi . Note that this convention is the opposite of that
of [14]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

xi

Chapter 1
Introduction
The colored Jones polynomial belongs to a family of quantum invariants constructed from
the ideas of representation theory and physics following the discovery of the Jones polynomial
[67,68]. Despite its efficacy at distinguishing knots, its relationship to classical knot topology
and the geometry of hyperbolic knots remains to be determined. Several conjectures are
open, such as the Slope Conjecture [41], which relates the polynomial to incompressible
surfaces in the knot complement, and the celebrated Volume Conjecture [87, 88], which
connects the values of the polynomial to the volume of the hyperbolic knot.
As much of this dissertation is motivated by the Slope Conjecture, we give its statement
here. The colored Jones polynomial assigns to a knot K in S 3 a sequence of Laurent polynon (q)}∞ , where J n (q) ∈ Z[q −1/2 , q 1/2 ]. Let d(n) and d∗ (n) denote the minimum
mials {JK
n=2
K
n (q). The formula for J n (q) based on the Kauffman
degree and maximum degree in q of JK
K

bracket gives a lower bound hn (D) ≤ d(n), and an upper bound Hn (D) ≥ d∗ (n), which can
be explicitly computed from any diagram D of K, see Definition 2.2.2.
A fraction p/q ∈ Q ∪ {1/0} is a boundary slope of a knot K ⊂ S 3 if pµ + qλ represents
the homology class of the boundary of a properly embedded, essential surface in the torus
boundary of the knot complement. Here the set {µ, λ} is the canonical meridian-longitude
basis of homology of the torus boundary. Garoufalidis’ Slope Conjecture is the following
statement.

1

Conjecture 1.0.1. [41] For a knot K ⊂ S 3 , every limit point of each of the sets

jsK :=

4d(n)
:n≥0
n2

js∗K :=

4d∗ (n)
:n≥0
n2

is a boundary slope of K.
The limit points of the sets jsK and js∗K are called the Jones slopes of K, and the
conjecture predicts that every Jones slopes is a boundary slope. By the work of Garoufalidis
[41] and Garoufalidis and Le [43], the set of Jones slopes for a knot is a finite set of rational
numbers. Hatcher [54] showed that there are only finitely many boundary slopes for a link.
Therefore, the set of boundary slopes is a way to index geometric information about a knot.
Deep relationships between the colored Jones polynomial and the geometry of the knot
complement in the spirit of these conjectures have been demonstrated and studied extensively
on the class of semi-adequate links [9–11, 25, 33–35, 44, 45, 47, 105]. These are a class of links
satisfying a diagrammatic condition. In contrast, very little is known outside the class of
semi-adequate links.
In the dissertation, we investigate the structure of the colored Jones polynomial outside
the class of semi-adequate links. Our first theorem generalizes a stability result known for
semi-adequate links to all links by studying the combinatorics of the link diagram. Our
second result, joint with R. van der Veen, proves the Slope Conjecture for a class of 3-string
pretzel knots.
In a different direction, we study the connection between the Jones polynomial and
other link invariants. With E. Kalfagianni, we give two-sided linear bounds of the crosscap
number (i.e. the non-orientable genus) of an alternating link in terms of coefficients of the
Jones polynomial of the link.
2

A rational homology 3-sphere Y is an L-space if |H1 (Y ; Z)| = rank HF (Y ), where HF (Y )
denotes the ‘hat’ version of Heegaard Floer homology. A knot in S 3 is an L-space knot if
K or its mirror image admits a positive L-space surgery. L-space knots have particularly
simple Heegaard Floer homology groups and are extensively studied for their relationship to
len-space knots. Using the Jones polynomial in collaboration with F. Vafaee, we classify the
L-space knots among 3-braids.
For the rest of the introduction, we give precise statements of the main results of this
dissertation in Section 1.1 and describe the contents of each chapter. In Section 1.2, we
give preliminaries on the colored Jones polynomial that are essential for the rest of the
dissertation.

1.1

Main results and organization of dissertation

Each chapter of the dissertation is self-contained and may be read out of order by a reader
familiar with the preliminary background on the Jones polynomial and the colored Jones
polynomial, which we cover in Section 1.2.

1.1.1

Chapter 2

We say that a link is A-adequate (resp. B-adequate) if it admits an A-adequate (resp. Badequate) diagram, see Definition 2.2.1. A link is semi-adequate if it is A-or B-adequate. It
is adequate if it is both A-and B-adequate.
As we will see in Chapter 2, the colored Jones polynomial of semi-adequate knots is
closely related to the hyperbolic structure of the knot complement. The Slope Conjecture is
true for adequate knots [33], and the coefficients of the polynomial give volume bounds for

3

a hyperbolic knot [25, 34], as well as exhibit a stability propery [9, 44]. A key feature of a
semi-adequate link is the diagrammatic condition in its definition, which greatly reduces the
complexity of computing the polynomial.
Since the Slope Conjecture and the Volume Conjecture are expected to hold for all knots,
n (q)} satisfies a recurrence relation [43], analogous
and it is known that the sequence {JK

results are expected to hold for non semi-adequate knots. In particular, we would like to
understand how the relationship to the geometry of the knot complement may be different
for knots that are not semi-adequate.
A way to approach this problem is to study how the lack of the diagrammatic feature
of semi-adequacy affects the polynomial, since diagrammatic properties of links are more
directly related to the geometry of the link complement.
n (q) and h (D) is a lower bound of d(n)
Recall that d(n) is the minimal degree of JK
n

obtained from a diagram D of the link K, see Definition 2.2.2. The main result of Chapter
2 gives a new lower bound for d(n) for a link that is not A-adequate.
Theorem 1.1.1. Let K be a link with diagram D. If D is not A-adequate, then d(n) ≥
hn (D) + n − 2 for n > 2.
Theorem 1.1.1 is a first step in understanding the structure of the colored Jones polynomial outside the class of semi-adequate links. A immediate consequence of the theorem is
that we may define stable coefficients for non A-adequate links which are analogous to the
stable coefficients of the polynomial of A-adequate links. In addition, these coefficients are
trivial for non A-adequate links. Therefore, the colored Jones polynomial can characterize
A-adequate links.
To prove Theorem 1.1.1, we use the ribbon graph construction of the Kauffman bracket

4

and a formula of the colored Jones polynomial in terms of the blackboard cable of the link
diagram. The computation for the lower bound of the degree is localized to a portion of the
diagram that is not A-adequate.

1.1.2

Chapter 3

Outside the class of adequate knots, the Slope Conjecture is often verified by computing
the Jones Slopes and boundary slopes separately, and then matching them together [41, 46].
The question of how the colored Jones polynomial selects the boundary slope of an essential
surface remains a major obstacle in understanding the conjecture. In particular, there is a
lack of examples which would provide an intrinsic explanation for the connection between
n (q)} and incompressible surfaces.
the degrees of {JK

In this chapter, we prove the Slope Conjecture for a new family of pretzel knots. This
provides a large number of new examples to study. A 3-string pretzel knot P (−1/r, 1/s, 1/t)
has a diagram obtained by connecting a negative twist region with r twists and two positive
twist regions with s and t twists, respectively, see Figure 1.1
Theorem 1.1.2 (Lee-van der Veen). Let r, s, t be nonzero integers greater than 5 such
that r is even, s, t are odd, and r > s, t. The Slope Conjecture is true for pretzel knots
P (−1/r, 1/s, 1/t).
As an example, for P (−1/10, 1/7, 1/9), the boundary slope picked out by the colored
Jones polynomial in the sense of the Slope Conjecture has slope 302/7 and Euler characteristic -56.
The diagram given for P (−1/r, 1/s, 1/t) is A-adequate but not B-adequate, therefore,
n (q). Most of the
we need only to verify the conjecture for the maximum degree d∗ (n) of JK

5

Figure 1.1: The 3-string pretzel knot P (−1/10, 1/7, 1/9).
boundary slopes for each r, s, t matched by the Jones slope of Theorem 1.1.2 are rational.
Therefore, we also have many new examples of pretzel knots which are not B-adequate, since
the slopes would be integral otherwise [33].
For Theorem 1.1.2, the Hatcher-Oertel algorithm [55] determines the set of boundary
slopes for any Montesinos knot. We use this algorithm to match a boundary slope with the
Jones slope for the class of pretzel knots in the theorem.

1.1.3

Chapter 4

Let
JK (t) = αK tr + βK tr−1 + . . . + βK ts+1 + αK ts
denote the Jones polynomial of K, so that r and s denote the highest and lowest power in
t. Set
TK := |βK | + βK ,
where βK and βK are the second and penultimate coefficients of JK (t), respectively.
6

The crosscap number of a non-orientable surface with k boundary components is defined
to be 2 − χ(S) − k. The crosscap number of a link K is the minimum crosscap number over
all non-orientable surfaces spanning K.
Theorem 1.1.3 (Kalfagianni-Lee). Let K be a non-split, prime alternating link with kcomponents and with crosscap number C(K). Suppose that K is not a (2, p) torus link. We
have

TK
3

+ 2 − k ≤ C(K) ≤ TK + 2 − k,

where TK is as above, and · is the ceiling function that rounds up to the nearest integer.
Furthermore, both bounds are sharp.
Theorem 1.1.3 gives a new relation of the Jones polynomial to a fundamental topological
knot invariant and prompts several interesting questions about the topological content of
quantum link invariants. The lower bound is easy to compute from any alternating diagram
of the knot. Combined with the information on KnotInfo [19], the lower bound determines
previously unknown values of the crosscap numbers of many alternating links.
In the proof of Theorem 1.1.3 we make use of the results of [1], which give an algorithm
to determine crosscap number of any alternating link. We show that given an alternating
diagram D(K), a surface that realizes the crosscap number of K can be taken to lie in the
complement of an augmented link obtained from D(K). Then, we use results of Agol, D.
Thurston [76, Appendix], Lackenby [75] and Futer and Purcell [37], to prove Theorem 1.1.3.
In particular, we make use of the fact that augmented link complements admit angled polyhedral decompositions with several nice combinatorial and geometric features. We employ
normal surface theory and a combinatorial version of a Gauss-Bonnet theorem to estimate
7

the Euler characteristic of surfaces that realize crosscap numbers of alternating links. Using
these techniques we show that for prime alternating links, the crosscap number is bounded
in terms of the twist number of any prime, twist-reduced, alternating projection, see Section
4.1 for definitions.

1.1.4

Chapter 5

One of the most prominent problems in relating low-dimensional topology and Heegaard
Floer homology is to give a topological characterization to L-spaces and L-space knots. As
an application of the Jones polynomial, we show that we can use it to determine which
3-braids are L-space knots.
Theorem 1.1.4 (Lee-Vafaee). Twisted (3, q) torus knots are the only knots with 3-braid
representations that admit L-space surgeries.
Our proof of Theorem 1.1.4 uses the constraints on the Alexander polynomial of an Lspace knot that have previously been studied to give the classification of L-space knots among
pretzel knots [80]. We show that, except for the twisted (3, q) torus knots, the Alexander
polynomials of all of the knots with 3-braid representations violate the constraints mentioned
for the Alexander polynomial of L-space knots.

1.2

Preliminaries

In this section we give the formula of the colored Jones polynomial, and we show how the
formula in terms of blackboard cables of diagrams can be derived from it. The colored Jones
polynomial may be defined via the representation theory of sl2 (C), see [99], [100], [109], and
the references therein. We use the definition of the colored Jones polynomial in terms of
8

the Kauffman bracket and the Jones-Wenzl idempotent of the Temperley-Lieb algebra, a
description of which may be found in [78].

1.2.1

The Temperley-Lieb algebra

Let A be a fixed complex number. The linear skein S(F ) of a surface F is the vector space
of formal linear sums over C of link diagrams in F quotiented by the relations
(i) D ∪
(ii)
Here

= (−A2 − A−2 )D.

=A

+ A−1

.

is regarded as a trivial closed curve in the diagram with no crossing.

Consider the linear skein S(D, 2n) of a disc regarded as a square with 2n points marked
on its boundary and distinguish one side with n marked points as the ‘left side’ and another
side, also with n marked points, as the ‘right side.’ We define a product d1 d2 of two elements
d1 , d2 in S(D, 2n) by juxtaposing the squares of d1 and d2 side by side and identifying the
right side of d1 with the left side of d2 . Extending this multiplication by linearity, we obtain
a bilinear map S(D, 2n) × S(D, 2n) → S(D, 2n) which makes S(D, 2n) into an algebra T Ln ,
called the Temperley-Lieb algebra, over C. T Ln is generated by n elements 1, e1 , . . . , en−1 .
Here the convention is that a strand decorated by a number n indicates n parallel strands.

1=

n

ei =

n−i+1
i−1

Figure 1.2: Generators of the Temperley-Lieb algebra.
There is a map from T Ln into S(R2 ) by placing an element of T Ln in the plane and
joining the n points on the left side and n points on the right side by parallel arcs. In the
9

linear skein of R2 , the image is reduced to an empty diagram with coefficient a polynomial
in A.

1.2.2

The Jones-Wenzl idempotent

The Jones-Wenzl idempotent is an element fn in T Ln defined by the following four properties. We represent it by an empty box with n incoming strands and n outgoing strands.
n
fn =

n

n

=

Figure 1.3: The Jones Wenzl idempotent and its closure in S(R2 ) [78].
Let

n

denote the polynomial multiplying the empty diagram of the image of fn in

S(R2 ) obtained by joining the n points on both sides by n parallel strands.
Definition 1.2.1. The Jones-Wenzl idempotent fn is the unique element in T Ln which
satisfies
(i) fn ei = 0 = ei fn for 1 ≤ i ≤ n − 1.
(ii) fn − 1 belongs to the algebra generated by {e1 , . . . , en−1 }.
(iii) fn fn = fn , and
(iv)

n

=

(−1)n (A2(n+1) −A−2(n+1) )
.
(A2 −A−2 )

The existence of this element is proven in [78]. An important property of fn is a recursion
formula, which we show diagrammatically.

10

1

1
− n−1

=
n+1

n

n

n

n−1

n

Figure 1.4: Recursion formula for the Jones-Wenzl Idempotent [78].
Adding a kink via a Type I Reidemeister move to the n-strands of f n multiplies the skein
element by a scalar [78, Lemma 14.2].

2
= (−1)n An +2n

Figure 1.5: Effect of adding a twist [78].

1.2.3

Formula for the colored Jones polynomial

If D is an unoriented diagram of a link K with k ordered components, D defines a multilinear
map

, . . . , D.
, . . . , D : S(S 1 × I) × · · · × S(S 1 × I) → S(R2 ),
k times

from the Cartesian product of linear skeins of the annulus S 1 × I to S(R2 ) in the following
way: For each component of the link diagram we obtain an annulus corresponding to the
blackboard framing, where the boundaries of the annulus run parallel to the diagram as
shown for the figure 8 knot in Figure 1.6.
We identify S 1 × I with the annulus and thus obtain a map from a skein element S 1 × I
into the annulus corresponding to a link component of D. The n + 1th-unreduced colored
11

Figure 1.6: D2 where D is a diagram of the figure 8 knot.
n+1 (q) is the polynomial obtained by substituting q 1/2 = A−1/2 into the
Jones polynomial JK

polynomial multiplying the empty diagram of
2
(−1)n An +2n

−ω(D)

n, . . . ,

n

∈ S(R2 ).

(1.1)

By induction and the recursion formula of fn depicted by Figure 1.4, We have that

,...,

n, . . . ,

= ,...,α n −

n−1 , . . . ,

,

where α is the generator of S(S 1 × I) consisting of a parallel curve encircling the annulus.
n+1 (q) is also obtained by substituting q 1/2 = A−1/2
Extending by bilinearity, we have that JK

into the polynomial GD (n + 1, A).

GD (n + 1, A) :=

2
(−1)n An +2n

−ω(D)

Sn (D) ,

(1.2)

where Sn (D) is defined recursively by

Sn+1 (D) = DSn (D) − Sn−1 (D).

We set S0 (D) = 1 to be 1 times the empty diagram, while S1 (D) = D. Note that when
12

1+1 (q) = J 2 (q) is the Jones polynomial J (q) multiplied by (−A2 − A−2 ).
n = 1, we have JK
K
K

Remark 1.2.2. The polynomials {Sn (x)} defined by the recurrence relation are Chebyshev’s
polynomials of type 2 [77].

13

Chapter 2
Stability properties of the colored
Jones polynomial
The colored Jones polynomial of a link can be defined and studied in terms of combinatorial
properties of link diagrams. One approach to this is through the Kauffman bracket construction [72] as we have seen in Section 1.2. Studying the combinatorics of the Kauffman bracket,
Lickorish and Thistlethwaite [79] showed that the extreme degrees of the Jones polynomial
are bounded by concrete data from any diagram. If the diagram is adequate, which means
that it is A-adequate and B-adequate, see Definition 2.2.1, then the bounds are sharp. This
was one of the very first results utilizing the notion of semi-adequacy.
We summarize below the results known for semi-adequate links that are relevant to this
chapter. A diagram is B-adequate if its mirror image is A-adequate, and the effect of taking
the mirror image of the diagram on the colored Jones polynomial is to substitute q −1 for q.
Therefore, for the rest of the chapter, we will only refer to A-adequacy. Recall that d(n) is
n (q).
the minimum degree of JK

14

2.1
2.1.1

The colored Jones polynomial of semi-adequate links.
The Slope Conjecture

The degrees of the colored Jones polynomial, and hence the Jones slopes, are difficult to compute in general. However, if a knot is adequate, the computation boils down to determining
hn (D), see Definition 2.2.2. If D is A-adequate, then d(n) = hn (D) [33]. Futer, Kalfagianni,
and Purcell [33] use this fact and the results on essential surfaces in the complement of
semi-adequate knots to prove the Slope Conjecture for adequate knots.

2.1.2

The tail of the colored Jones polynomial

n (q) exhibit a stability behavior for semi-adequate
It is also known that the coefficients of JK

links. The stable coefficients directly relate to the geometry of hyperbolic knot complements.
i (q). If K
Theorem 2.1.1. [9, 44] For i ≥ 2, let βi be the coefficient of q d(i)+(i−1) of JK
n (q) is
admits an A-adequate diagram, then for all n ≥ i, the coefficient of q d(n)+(i−1) of JK

equal to βi .
n (q) from h (D) are stable for a link with an
That is, the last n − 1 coefficients of JK
n

A-adequate diagram D, extending the results by Dasbach and Lin [25] on the last and
n (q). Based on this result, they defined the notion of a
penultimate stable coefficients of JK

tail of the colored Jones polynomial of an A-adequate link.
Definition 2.1.2. For a link admitting an A-adequate diagram, the power series
∞

TK (q) =

i=2

15

βi q i

is a tail of the colored Jones polynomial.
n (q), and hence the last two coefficients of the
The last and penultimate coefficients of JK

tail, have been shown by the work of Futer, Kalfagianni, and Purcell to carry information
on the geometric structure of the complement of an A-adequate knot, and to provide sharp
volume bounds on the complement of hyperbolic links. See [34] for the results and a detailed
survey.
Rozansky has shown that stability behavior also occurs in the categorification of the
colored Jones polynomial [101]. The idea of this theory, developed independently by Frenkel,
Stroppel and Sussan, Cooper and Krushkal, and Rozansky, is as follows: Given a link K, for
each n, one assigns a chain complex C Kh (K, n) for which the graded Euler characteristic of
n (q).
its homology groups H Kh (K, n) is the n-th colored Jones polynomial JK

Theorem 2.1.3. [101] If K is an A-adequate link, then there is a sequence of maps

˜ Kh (K, n) → H
˜ Kh (K, n + 1),
fn : H

˜ Kh for i ≤ n − 1, where i is the grading of the chain
such that fn is an isomorphism on H
i
complex coming from the number of state circles of a Kauffman state.
The tilde indicates the appropriate degree shifts made for the chain complex so that fn
is a degree-preserving map for each n. In other words, for all n > i, the homology groups
of C Kh (K, n) of grading less than i − 1 are the same as the homology groups of C Kh (K, i)
of grading less than i − 1. As a result, he defines a tail homology of the colored Jones
˜ Kh (K, n) determined by fn .
polynomial which is the direct limit of the direct system of H
The tail homology contains the stable homology groups of C Kh (K, n) as n increases. The

16

last n − 1 coefficients of the graded Euler characteristic of the tail homology agree with the
n (q)). Therefore,
last n − 1 coefficients of TK (q) (hence the last n − 1 stable coefficients of JK

we have a categorification of the tail in the case of A-adequate links, extending the results
by Armond [9] and Garoufalidis and Le [44].
It is natural to ask whether similar results can be obtained outside the class of semiadequate links. Specifically, we are interested in the following questions.
Question 2.1. In view of Conjecture 1.0.1, what can we say about the degrees of the colored
Jones polynomial when the link is not semi-adequate?
Question 2.2. Can we define a tail for the colored Jones polynomial for all links as in
Definition 2.1.2?
Question 2.3. What is the tail homology as defined in Theorem 2.1.3 for the categorification
of the colored Jones polynomial for non semi-adequate links?
For Question 2.1, Manchon [81] has constructed an infinite family of non A-adequate
knots with diagrams D for which d(2) = h2 (D). An example of such a knot is 12n706,
see Knotinfo [19] for a diagram of the knot. For this knot, d(2) = h2 (D) = −4. However,
this knot is not A-adequate since the last coefficient of its Jones polynomial is 2, and it is
known that for an A-adequate link, the last coefficient of its Jones polynomial is ±1 [25]. In
n (q)}∞ is q-holonomic [43]. In
general, Garoufalidis and Le showed that the sequence {JK
n=2

other words, the entire sequence is determined by finitely many initial terms. This can be
viewed as a weaker form of stability in view of Question 2.2.
We study the effect that a diagram being non A-adequate has on the last n−2 coefficients
n (q). We reprint Theorem 1.1.1 for the convenience of the reader.
of JK

17

Theorem 1.1.1. Let K be a link with diagram D. If D is not A-adequate, then d(n) ≥
hn (D) + n − 2 for n > 2.
A special case of Theorem 1.1.1 is proven in [69]. If K is not A-adequate, then any link
diagram D = D(K) will not be A-adequate. Let h(n) be the maximum of hn (D) taken over
all diagrams D of K. By Theorem 1.1.1, d(n) ≥ h(n) + n − 2 for n > 2. We use this to
A (q), which is constant if K is not
obtain a link invariant in the form of a power series JK

A-adequate as in [69]. In other words, the stable coefficients βi are identically zero for i > 2.
This allows us to extend the construction of a tail of an A-adequate link to all links.
Figure 2.1 and 2.2 illustrate the difference in coefficients of the tail of an A-adequate
n (q) from
links and non A-adequate links. The symbol a(d(n) + i) is the ith-coefficient of JK

d(n) and a(h(n) + i) is similarly defined. If K is A-adequate, then d(n) = h(n).
n=2
n=3
n=4
n=5
n=6

a(d(n)) a(d(n) + 1)
1
-2
1
-2
1
-2
1
-2
1
-2

a(d(n) + 2)
2
-1
-1
-1
-1

a(d(n) + 3)
-3
5
2
2
2

Table 2.1: Coefficients for the A-adequate knot 10154 [9].

n=2
n=3
n=4
n=5
n=6

a(h(n))

a(h(n)+1)

a(h(n)+2)

a(h(n)+3)

0
0
0
0

0
0
0

0
0

0

Table 2.2: The tail for a non A-adequate link.
Thereby, we give an affirmative answer to Question 2.2 and make partial progress on
understanding Question 2.1 and 2.3.
With respect to the tail homology, Rozansky made the following conjecture:
18

Conjecture 2.1.4. If K is not A-adequate, then the tail homology for the categorification
of the colored Jones polynomial for K vanishes.
Thus, the last n − 2-coefficients of the nth-colored Jones polynomial of a non A-adequate
link should be identically 0, whence Theorem 1.1.1 provides partial evidence towards Conjecture 2.1.4.
We cover the preliminaries on ribbon graphs in Section 2.2. We reduce the theorem to
proving Theorem 2.2.3, phrased entirely in terms of the Kauffman bracket of a link diagram.
We organize the intermediary lemmas between Section 2.3 - Section 2.6, prove Theorem 2.2.3
in Section 2.7 and show how the algorithm in the proof works on an example in Section 2.8.
In Section 2.9, we discuss applications of Theorem 1.1.1, including the construction of a tail
for all links, and how the colored Jones polynomial can be used to detect semi-adequate
links.

2.2

Ribbon graphs and the Kauffman bracket

We review the notion of A-adequacy, which we will use to obtain an upper bound of
d∗ Sn−1 (D) . This will be the lower bound hn (D) of d(n) in the statement of Theorem
1.1.1 after substituting q 1/2 = A−1/2 . Let D be an oriented link diagram and x a crossing
of D. Associated to D and x are two link diagrams, each with one fewer crossing than D,
called the A-resolution and B-resolution of the crossing, see Figure 2.1.

A-resolution

B-resolution

Figure 2.1: A- and B-resolutions of a crossing.
19

A Kauffman state σ of D is a choice of A-resolution or B-resolution at each crossing
of D. Applying a Kauffman state to D, we obtain a crossing-free diagram consisting of a
disjoint collection of simple closed curves on the projection plane, see Figure 2.2. We call
these curves state circles. The all-A state chooses the A-resolution at each crossing of D. We
denote the union of the corresponding state circles by sA (D), and let |sA (D)| be the number
of state circles. For an arbitrary state σ, we similarly denote the union of the corresponding
state circles by sσ (D) and let |sσ (D)| be the number of state circles in the union.

Figure 2.2: The figure 8 knot, its all A-resolution and the resulting all-A state graph.

Definition 2.2.1. To the union of state circles sσ (D) resulting from applying a Kauffman
state σ, we attach one edge for each crossing which records the original location of the
crossing on D. (These edges are dashed in Figure 2.1.) We will call the union of the state
circles and the edges the graph of the σ-resolution, or the σ-state graph, and denote it by
Hσ (D). If σ is the all-A state then we call it the graph of the A-resolution, or the all-A state
graph and denote it by HA (D). A diagram D is A-adequate if HA (D) does not contain any
edges with both ends on the same state circle. Whenever it is clear which diagram we are
referring to, we will just write Hσ for Hσ (D) and HA for HA (D) to simplify notation.
Definition 2.2.2. Let c(D) be the number of crossings of D, ω(D) be the writhe of D, and

20

|sA (D)| be the number of state circles in the all A-state of D. We define

M (D) := c(D) + 2|sA (D)|,

and
1
hn (D) := − (M (Dn−1 ) + 4 − 2(n − 1) − ω(D)((n − 1)2 + 2(n − 1)).
4
It is known that for any link diagram D, the maximum degree d∗ D of D in A is less
than or equal to M (D). Moreover, d∗ D = M (D) if D is A-adequate [79].
n+1 (q) is obtained by substituting q 1/2 = A−1/2 into
Recall that JK
2
(−1)n An +2n

−ω(D)

Sn (D) .

(2.1)

An important corollary of the recursive formula for Sn (D) is that

Sn (D) = Dn + (1 − n) Dn−2 + lower degree cablings of D .

Therefore, d∗ Sn (D) ≤ M (Dn ) and d(n) ≥ hn (D) for all n > 2. This shows that hn (D) is a
lower bound for d(n). If D is A-adequate, then Dn is also A-adequate and thus hn (D) = d(n)
[33].
We shall prove an upper bound on the maximum degree of Dn , which will imply
Theorem 1.1.1.
Theorem 2.2.3. Let D be a link diagram that is not A-adequate. Then for n > 2, we have

d∗ Dn ≤ M (Dn ) − 4(n − 1).

21

We show that Theorem 2.2.3 implies Theorem 1.1.1:
Proof. The n-cable Dn has n2 c(D) crossings and n|sA (D)| number of state circles in the
all-A state, therefore
M (Dn ) = n2 c(D) + 2n|sA (D)|.
Note that

M (Dn ) − M (Dn−2 ) = c(D)(n2 − (n − 2)2 ) + |sA (D)|(2n − 2(n − 2)) > 4(n − 1).

Assuming Theorem 2.2.3, so M (Dn ) − d∗ Dn ≥ 4(n − 1), then M (Dn ) − d∗ Sn (D) ≥
4(n − 1). Plugging this back into the expressions for d(n) and hn (D) gives us that d(n) ≥
hn (D) + n − 2.
The Kauffman bracket of any diagram can be computed by means of ribbon graph expansions. As defined in [24], a ribbon graph is a multi-graph (i.e. loops and multiple edges
are allowed) equipped with a cyclic order on the edges at every vertex. A ribbon graph can
be embedded on an orientable surface such that every region in the complement of the graph
is a disk. We called the regions the faces of the ribbon graph.

22

Definition 2.2.4. For a ribbon graph G we define the following combinatorial quantities:

v(G) = the number of vertices of G.
e(G) = the number of edges of G.
f (G) = the number of faces of G.
k(G) = the number of connected components of G.
g(G) =

2k(G) − v(G) + e(G) − f (G)
, the genus of G.
2

For each Kauffman state σ of a link diagram, a ribbon graph Gσ is constructed as follows:
We orient the collection of non-intersecting state circles sσ (D) according to whether a circle
is inside an odd or even number of circles, respectively. The vertices of Gσ correspond
to the collection of state circles and the edges to the crossings. The orientation of the
circles defines the orientation of the edges around vertices. An equivalent definition of an Aadequate diagram is to say that a diagram is A-adequate if the ribbon graph GA constructed
from the all-A state of the diagram has no one-edge loops. For more details, see [23], where
the word dessin is used instead of ribbon graph.
Definition 2.2.5. A spanning sub-graph H ⊆ GA is a ribbon graph obtained by removing
edges from GA .
From the ribbon graph setting we obtain the spanning sub-graph expansion of the Kauffman bracket, originally introduced in [23] and proven in [24].
Theorem 2.2.6. Let GA be the ribbon graph constructed from the all-A state of a link
diagram D. Then
Ae(GA )−2e(H) (−A2 − A−2 )f (H) ,

D =
H⊆GA

23

where H ranges over all spanning sub-graphs of GA .
The term XH defined by
XH := Ae(GA )−2e(H) (−A2 − A−2 )f (H)

is the contribution of a spanning sub-graph H to D . Let the first sth-coefficient of D
from M (D) be the coefficient of AM (D)−4(s−1) in D . A spanning sub-graph H contributes
to the first sth coefficient of D if and only if v(H) − k(H) + g(H) ≤ s − 1, see [24, Theorem
n (q) from
6.1]. This is used in [24] to give a formula of the penultimate coefficient of JK

hn (D) when K has an A-adequate diagram. Theorem 2.2.3 can be rephrased as saying that
the first n − 1 coefficients of Dn from M (Dn ) are equal to zero. Our proof will depend
heavily on Theorem 2.2.6 and the restriction on v(H) − k(H) for a spanning sub-graph H
that contributes to the first n − 1 coefficients of Dn .
Definition 2.2.7. Let H ⊆ GA be a spanning sub-graph and σ be the Kauffman state that
chooses the B-resolution on a crossing corresponding to an edge in H and the A-resolution
on a crossing corresponding to an edge not in H. We associate to H the state graph H of σ.
The following important lemma counts the faces of a spanning sub-graph by the number
of state circles in its associate state graph.
Lemma 2.2.8. The number of state circles in H is equal to the number of faces of H.
Proof. Let σ be a Kauffman state of a link diagram D and σ
ˆ be its dual state which resolves
each crossing of the link the opposite way from σ. This means that if σ chooses the Aresolution at one crossing, then σ
ˆ chooses the B-resolution at that crossing, and vice versa.

24

The lemma follows from [23, Lemma 4.2], which says that f (Gσ ) = |sσˆ D|, where Gσ is a
ribbon graph constructed from σ.
A crossing of a link diagram is nugatory if there is a closed curve in the projection
plane meeting the diagram transversely at that crossing and does not intersect the diagram
anywhere else. A link diagram is reduced if it contains no nugatory crossings. Throughout
this section, D denotes a reduced, non A-adequate link diagram, Dn is its n-cable, and Gn
A
is the ribbon graph constructed as in Section 2.2 from the all-A state of Dn . A spanning
n
sub-graph H ⊆ Gn
A is obtained from GA by deleting edges, with the associated state graph
n associated
H as in Definition 2.2.7. We will mainly be interested in the all-A state graph HA

to the spanning sub-graph of no edges.
By Theorem 2.2.6, we have

Dn =

e(Gn
A )−2e(H)

H⊆Gn
A

A

(−A2 − A−2 )f (H)−1 ,

where H ranges through all spanning sub-graphs of Gn
A . Recall that
e(Gn
A )−2e(H)

XH = A

(−A2 − A−2 )f (H)−1

is the contribution of H to Dn . We will first prove a lemma that allow us to sum the terms
XH in a certain way to show the cancellation of the first n − 1 coefficients of Dn .

25

2.3

A cancellation lemma

Lemma 2.3.1. Let c, d be integers and δ = (−A2 − A−2 ), the maximum degree of the sum
k

k
Ac−2i δ d+i
i

i=0

in A is less than or equal to c + 2d − 4k.
Proof.

We factor out Ac δ d to obtain

Ac δ d

k

k
A−2i δ i .
i

i=0

Note first that A−2i δ i has the same degree 0 for each i, and each product A−2i δ i only has
non-zero coefficients for terms of degree A−4j for j a nonnegative integer. For j > 0, the
jth coefficient of the polynomial Ac δ d

k
k
i=0 i

A−2i δ i of the term Ac+2d−4(j−1) is given by

the following equation.
k

The jth coefficient =

(−1)i

i≥j−1

26

k
i

i
.
j−1

To prove the lemma, it suffices to show that the jth coefficient is zero for all j ≤ k. Let p(x)
be the polynomial x(x − 1)(x − 2) · · · (x − j + 2). This is a polynomial of degree less than k.
The sum on the right for the jth coefficient becomes
k

(−1)i

i=0

k
p(i)
.
i (j − 1)!

In order to show that this is zero, consider the binomial expansion of (1 + x)k :

(1 + x)k

k

=
i=0

k i
x.
i

Taking the derivative with respect to x (j − 1)-times on both sides yields

k · (k − 1) · · · (k − j

+ 2)(1 + x)k−j+1

k

=
i=0

k
i(i − 1) · · · (i − j + 2)xi−j+1 ,
i

where we have p(i) on the right side multiplying xi−j+1 . Setting x = −1 gives
k

0=
i=0

(−1)i−j+1

k
p(i).
i

Then we just multiply by (−1)j−1 /(j − 1)! to obtain the lemma.

2.4

Labeling edges in HAn .

By the remark immediately following Theorem 2.2.6, XH contributes to the first n − 1
coefficients of Dn if and only if v(H) − k(H) ≤ n − 2. Therefore, we may restrict to this

27

set of spanning sub-graphs and show that




d∗ 

XH  ≤ M (Dn ) − 4(n − 1),
v(H)−k(H)≤n−2

to prove Theorem 2.2.3. Our strategy is to divide up the sum over equivalence classes where
Lemma 2.3.1 can be repeatedly applied. In order to define this equivalence relation, we
introduce labels and define sequences for each H that satisfies v(H) − k(H) ≤ n − 2 in this
section.
Since D is not A-adequate, there is an edge in the all-A state graph of D with two ends on
the same state circle S, such that it lies in a region bounded by S on the projection 2-sphere.
We name the crossing in D corresponding to this edge by e, and denote the n2 crossings
corresponding to the cable of e in Dn by en . The set of crossings in en corresponding to n
n will be denoted by en .
loops in the all-A state graph HA

See Figure 2.3 for the local picture of the resulting state graph where we choose the
A-resolution at all the crossings in en .

Figure 2.3: Identical local picture of the A-resolution of the 3-cable of a crossing.
n is the all-A state graph of D n associated to the spanning sub-graph of
Recall that HA
n
Gn
A with no edges. We detail below labeling conventions for the edges corresponding to e
n , and we will just refer to these edges by en . For an illustration of the labels, where
in HA

28

we locally straighten the strands, see Figure 2.4.
n there will be n state circles corresponding to the cabling of S in D n . We denote
(a) In HA

the state circle where the two ends of edges in en are by S0 . Then, Si is the state
circle where one of its complementary regions on S 2 contains only en and S0 , . . . , Si−1 .
The state circles S0 , S1 , . . . , Sn−1 divide the sphere into n + 1 regions. We denote the
region with boundary Si−1 and Si for i = 1, 2, . . . , n − 1 by Ωi and the region with
boundary S0 containing en by Ω0 . With this labeling we can refer to edges in en by
which regions they are in.
(b) For each i, an edge in Ωi corresponding to a crossing in en will be labeled by odd or even
integers. First we label the edges in en by consecutive odd integers 1, 3, 5, . . . 2n − 1.
For i = 0, an edge in Ωi with an end on Si−1 and an end on Si is labeled with the
integer m if the end of the edge on Si−1 is between the ends of two edges labeled m − 1
and m + 1.
(c) Drawing a curve through all the edges in Ω0 and dividing the projection sphere in
half, an edge of en in Ωi for i ≥ 1 will be labeled L or R depending on whether it
is on one side of the curve or the other side, respectively, so LΩi and RΩi refer to
all the edges in en in region Ωi on the left side or on the right side, and we just let
LΩ0 = RΩ0 = Ω0 . This notation, along with the numbering of edges of en , specifies
each edge in en uniquely.
n , so that for
The crossings in en inherit the labels from the corresponding edges in HA

any state graph of Dn , a segment in the state graph resulting from choosing a resolution
at a crossing of en will also have the same label as the crossing. The numbers labeling en

29

Ω2

Ω2

Ω1

Ω1

Ω0

Ω0

Ω1

Ω1

Ω2

Ω2

3
2
1

4
3

2

RΩ1
5

4
3

3

RΩ2
Ω0
LΩ1
LΩ2

2
1

4
3

2

5
4

3

Figure 2.4: From left to right: an illustration of the labeling on the edges of en when n = 3.

induce an ordering as well. We say that an edge with label a is smaller (resp. greater ) than
an edge with label b if a < b (resp. b > a). Two edges are equal if their labeling numbers
are equal. An edge with label c is between edges with label a and b if a < c < b. As long
as it is clear which region Ωi we are referring to, we will refer to the edges by their labeling
numbers and by which side they are on. For example, La ∈ Ωi and a ∈ LΩi for a an integer
both indicate the edge in LΩi with label a.
n uniquely corresponds to an edge of Gn . We say that an edge in H n is
Each edge in HA
A
A
n
included in a spanning sub-graph H ⊆ Gn
A , if H includes the corresponding edge in GA . We
n are included in H. Schematically, the edges
represent H by indicating which edges of HA
n . This representation will allow us to compare
that are included will be shown in red on HA
n . We do this
the number of state circles in H, the state graph associated to H, to that of HA
n locally at each crossing which corresponds to
by modifying the state circles sA (Dn ) of HA

an edge included in H, replacing the A-resolution by the B-resolution. This will be crucial
n is modified
to proving the lemmas in this section. See Figure 2.5 and 2.6 for exactly how HA

to a different state graph when one changes from the A-resolution to the B-resolution at a
crossing.
Recall that our goal is to organize the spanning sub-graphs H with v(H) − k(H) ≤ n − 2
into equivalence classes where Lemma 2.3.1 can be applied. In [69], this was done for spanning

30

B

A

Figure 2.5: Locally, we replace the two segments making up the state circle (in red) and the
segment recording the crossing location (dotted) of the A-resolution by that of the segments
coming from the B-resolution.

Figure 2.6: From left to right: an all A-state graph with 3 state circles; an edge inclusion
representing a spanning sub-graph of the ribbon graph constructed from the all A-state
graph; the state graph associated to the spanning sub-graph, which has 2 state circles.
sub-graphs H with v(H) − k(H) ≤ 1. Two spanning sub-graphs are in the same equivalence
class if they only differ by what they include from en . The inclusion of each edge from en
increases the number of faces by one, since en are one-edge loops for all of their associated
state graphs. As a result, we set up sums of the form
en

e(Gn
A )−2i

A

(−A2 − A−2 )f (H0 )+i−1 ,

i=0

where H0 is the member of the equivalence class that does not include any edges from en ,
and Lemma 2.3.1 can then be applied.
When v(H) − k(H) > 1, the situation is more complicated. We can no longer set up
equivalence classes based on edge inclusions in en , because it is no longer true that inclusion
of an edge in en increases the number of faces by one. Instead, we generalize this specific
example to defining a set of edges G(H) for a spanning sub-graph H, where the increase in

31

the number of faces is the same as the number of edges included. In order to control the
change in the number of faces from edge inclusions, we make use of the explicit labelings
just introduced, and we use Lemma 2.2.8 to compute f (H) from the state graph H.
We first define a(H), which determines the set of edges in en we can add to H in any
combination while splitting off as many state circles as the edges added.
Definition 2.4.1. Given a spanning sub-graph H ⊂ Gn
A , we define H to be the spanning
sub-graph obtained from H by deleting from H all the edges outside of the region Ω1 .
Consider the associated state graph H . Let m ∈ en be the smallest edge that has two ends
on distinct state circles in H , and M ∈ en be the largest. We define a(H) to be the subset
of edges c ∈ en such that m ≤ c ≤ M , and the empty set if every edge in en has two ends
on the same state circle in H . For example, a(H) is empty for the spanning sub-graph with
no edges and sub-graphs for which only a single edge in Ω1 is included.
If a(H) is empty, then we can group them with other spanning sub-graphs where the
only difference is which edges they include from en and apply Lemma 2.3.1. For spanning
sub-graphs where a(H) is non-empty, we define sequences which controls the quantity v(H)−
k(H) + g(H). The following lemma allows us to define the first terms of the sequences.
Lemma 2.4.2. Suppose a(H) is not empty and v(H) − k(H) ≤ n − 2. Let m, M be the
smallest and largest edge in a(H), respectively.
(a) If 1 ∈
/ a(H), at least one of the pair of edges R(m − 1), L(m − 1) in Ω1 is included in
H. If 2n − 1 ∈
/ a(H), at least one of the pair of edges R(M + 1), L(M + 1) is included
in H. If 2n − 1 ∈ a(H), then an edge in Ω1 with an end attached to the portion of Ω0
cut off by 2n − 1 is included in H. See Figure 2.7 for an illustration.

32

(b) If 1 ∈ a(H): Let He be the sub-graph obtained from H by deleting all the edges in en
and in the region Ωn−1 . There is an integer 0 ≤ k ≤ n − 2 such that the edge with
the smallest labeling number in RΩk has two ends on the same state circle in the state
graph H e associated to He .
Proof.
(a) The proof for the cases 1 ∈
/ a(H), 2n − 1 ∈
/ a(H), or 2n − 1 ∈ a(H) are similar, and we
treat them together here.
This follows from the fact that the edges m and M need to be attached to separate
state circles in H , which has a simple picture, see Figure 2.7, only involving modifying
n based on edge inclusions of H in Ω . Requisite edges of
the state circles S0 , S1 of HA
1

Ω1 need to be included in H for m, M to have two ends on distinct state circles in the
associated state graph.

1
m
M
2n − 1

Figure 2.7: On the left: At least one of the two edges smaller than m in Ω1 needs to
be included in order that m ends up with two ends on distinct state circles in H . The
same is true with the edges bigger than M . On the right, the portion of the region Ω0
cut off by the edge 2n − 1 is shown in yellow, with the edges with one end attached to
it.

(b) For the case 1 ∈ a(H).
We show that there exists an index 0 ≤ i ≤ n − 2 such that the edge with label i + 1 in
RΩi has two ends on the same state circle in H e . For each edge labeled i + 1 in RΩi ,
33

we let the pieces of state circles where the two ends of the edge are on in H e inherit
n , so the top and the bottom pieces will be denoted as S
the same labeling as in HA
i

and Si−1 , respectively. If the two boundaries are joined to each other in H e , then the
edge i + 1 will have two ends on the same state circle in H e .
We show the following statement by induction on i, assuming that a(H) is non-empty
and contains the edge labeled 1.
◦ Let 0 < i ≤ n − 2. If for all 0 ≤ r ≤ i, the edge labeled r in RΩr has two ends
on distinct state circles in He , then two edges ei ∈ Ωi and ei+1 ∈ Ωi+1 that are
not in en , are included in He . Moreover, when we traverse the piece of the state
n in the clockwise direction, we meet an end of e before we meet
circle Si in HA
i

an end of ei+1 , and no edges between ei and ei+1 in Ωi+1 are included in He .
This statement implies the lemma for the case 1 ∈ a(H) since, if the k as in the lemma
does not exist, then the index n − 3 satisfies the criteria for the statement. We then
have that there is an edge en−3 and en−2 as in the statement, but in He no edge is
included in Ωn−1 . In RΩn−2 , the edge with label n − 2 will have two ends on the same
state circle since Sn−3 would have no where to exit except to join back with Sn−2 .
This is a contradiction.
For i = 1, if the edge labeled 1 in Ω0 has two ends on distinct state circles in H ,
an edge e1 ∈ Ω1 that is not in en must be included in H. Otherwise, the boundaries
on S0 where the two ends of edge 1 lie will join together and become the boundary
of a single state circle in H . If now the edge labeled 2 in RΩ1 also has two ends on
distinct state circles in He , we show that an edge e2 ∈ RΩ2 fitting the description of
the statement must also be included in He . Since applying a Kauffman state to obtain
34

a state graph gives disjoint state circles, the boundary of S1 in H e can only exit the
region enclosed by the state circle consisting of S0 by having He include an edge e2 in
Ω2 . If we traverse S1 in the clockwise direction we would meet an end of e1 before we
meet an end of e2 . Take the closest e2 to e1 in the counter-clockwise direction to be
e2 . This proves the base case. See Figure 2.8 for an illustration where we straighten
the strands locally.
Clockwise direction

RΩ2
RΩ1

4

e2

3

e1

2
1

Ω0

5

4 6
3 5

Figure 2.8: The base case i = 1.
Now for the induction step: Assume that the statement is true for i − 1. We already
have edges e1 , . . . , ei , each successive er , er+1 fitting the description of the statement
by the induction hypothesis. So we only need to show that ei+1 as described in the
statement exists. The state circle with boundary Si−1 will enclose a region in H e where
the boundary Si must exit through an edge ei+1 ∈ Ωi+1 to avoid joining with Si+1 .
Thus, an edge in RΩi+1 must be included in H e . Again we pick the one furtherest in
the counter-clockwise direction to be ei+1 . This completes the proof of the statement.

n,
Definition 2.4.3. For each H where a(H) is non-empty, we define sa , an edge in HA

according to the following rule:
35

• If a(H) contains 1: Pick the smallest index k ≥ 0 for which the edge labeled k + 1 in
RΩk has two edge on the same state circle in He . This exists by Lemma 2.4.2. Let sa
be the edge labeled k + 1 in RΩk if the edge labeled k in RΩk−1 is not included in H,
and let sa be the edge labeled k in RΩk−1 otherwise.
• If a(H) does not contain 1: Let m be the edge with the smallest label in a(H), we
define

sa :=





 R(m − 1) in Ω1 , if R(m − 1) in Ω1 is included in H.



 L(m − 1) in Ω1 , if R(m − 1) in Ω1 is not included in H.

The edge sa provides us with a starting point to find an edge with two ends on the same
state circle in the state graph associated to H, that is not included in H originally. If such
an edge is included in H, the spanning sub-graph H resulting from this inclusion will have
f (H ) = f (H)+1, and v(H )−k(H )+g(H ) = v(H)−k(H)+g(H) as desired. The sequences
{tj }, {bj } to be defined below will allow us to keep track of any edge-inclusions of H that
may result in an edge that has ends on distinct state circles in the state graph associated to
H.
Definition 2.4.4. Consider a spanning sub-graph H with sa as in Definition 2.4.3. Let
i = 0, 1, 2, . . . n − 1, we will define for each i and each side L, R, sequences of edges {tj } and
n . Recall that m, M are the smallest and largest labels of the edges
{bj } in LΩi , RΩi in HA

in a(H). See Figure 2.9 for an example.

36

• If sa is in RΩk , we define the sequences {tj }, {bj } for RΩi first.





sa





t0 = b0 := the smallest edge in RΩ greater than
i








the smallest edge in RΩi greater than

if i = k
b0 of RΩi−1

if i > k

b0 of RΩi+1

if i < k.

• We set the first term of the sequences of LΩ0 equal to the first term of the sequences
of RΩ0 . For LΩi , the first terms of the sequences are given by
t0 = b0 := the smallest edge in LΩi greater than b0 of LΩi−1 .

• If sa is in LΩk , we replace R by L and L by R in the above definitions and we get the
definition for the first terms of the sequences for this case.
• The rest of the sequences is given recursively by:

For t0 ∈ RΩi (resp. LΩi ),
tj := the smallest edge ≤ M in RΩi+j (mod 2) (resp. LΩi+(j mod 2) ) included in H
greater than tj−1 . If there is no such edge then it is empty.

For b0 ∈ RΩi (resp. LΩi ),
bj := the smallest edge ≤ M in RΩi−j (mod 2) (resp. LΩi−j (mod 2) ) included in H
greater than bj−1 . If there is no such edge then it is empty.

37

When i = 0, only the sequences {tj } are defined for RΩ0 and LΩ0 by the definitions
above. One has terms from RΩ1 and another has terms from LΩ1 , respectively. We
let {bj } for RΩ0 be the same sequence as {tj } of LΩ0 and {bj } for LΩ0 be the same
sequence as {tj } of RΩ0 .
3

RΩ2
RΩ1
Ω0

2
1

5

RΩ1

RΩ1

3

RΩ2

4
3

3

RΩ2
2

4

Ω0

2
1

4
3

RΩ1
5

Ω0

2
1

4
3

5

Figure 2.9: Let sa = 2 ∈ RΩ1 for this example. The point of the sequences is to rule out
edges in H that may make edges after the edge labeled 2 in RΩ1 have two ends on distinct
state circles in H. The sequence for RΩ1 are {bj } = {2, 5, ∅, . . .} and {tj } = {2, ∅, . . .}.
Definition 2.4.5. Let the sequences {tj } and {bj } for LΩi or RΩi be given, and let 0 ≤
n.
i ≤ n − 2. We define subsets ti and bi which we use to define G(H), a subset of edges in HA

Recall that an edge with label b is between edges labeled a and c if a < b < c. See Figure
2.10 for what kind of edges this picks out relative to the sequences.
For t0 ∈ RΩi (resp. LΩi ),
• ti := union of all edges of en in RΩi (resp. LΩi ) between tj and tj+1 for j even and tj
j
non-empty. If tj+1 is empty then we take all edges after tRi and ≤ M in RΩi (resp.

LΩi ).
• bi := union of all edges of en in RΩi (resp. LΩi ) between bj and bj+1 for j even and bj
j
non-empty. If bj+1 is empty then we take all edges after bRi and ≤ M in RΩi (resp.

LΩi ).
Finally, we define

G(H) := ∪t0 ∈RΩ (ti ∩ bi ) ∪ ∪t0 ∈LΩ (ti ∩ bi ) ∪ sa
i
i
38

Figure 2.10: On the left: All the edges shown in blue here after the red edge are candidates
to be included in G(H), as long as there are no edges above and below after the middle red
edge that are included in H. In the middle and on the right: The inclusion of the top and
bottom edges in red means that only three edges in blue after red will be included in G(H).
This is so that they will still have two ends on the same state circle as shown on the right
when we only resolve the crossings corresponding to the red edges.
if 1 ∈ a(H), and
G(H) := ∪t0 ∈RΩ (ti ∩ bi ) ∪ ∪t0 ∈LΩ (ti ∩ bi )
i

i

if 1 ∈
/ a(H). Lastly, If a(H) = {∅}, then G(H) := {∅}.

2.5

Equivalence classes based on G(H).

The point of the string of definitions and lemmas leading to the definition of a(H) and
G(H) is that it gives us a way to understand how the number of faces changes for a new
spanning sub-graph obtained by adding edges. We define an equivalence relation that will
put spanning sub-graphs into equivalence classes where we can control the differences in the
combinatorial quantities e(H) and f (H), directly related to v(H) − k(H) + g(H), from one
member to another.
Definition 2.5.1. We define an equivalence relation R on the set of spanning sub-graphs
H ⊆ Gn
A where v(H) − k(H) ≤ n − 2. We say that
H∼
=R H

39

if
1. a(H) = a(H ).
2. G(H) = G(H ), and H only differs from H by what edge it includes from G(H) ∪
n.
(en a(H)). Recall that en is the set of loops with both ends on S0 in HA

This is an equivalence relation because a(H) and G(H) are well-defined on H. Within an
equivalence class C of R, the members of C only differ from each other by edge inclusion in
G(H) ∪ (en a(H)). Let H be a member of an equivalence class C and consider the spanning
sub-graph H0 obtained from H by deleting all the edges in G(H)∪(en a(H)). This spanning
sub-graph is well-defined for C since a(H) and G(H) are the same across C.
Lemma 2.5.2. Let C be an equivalence class of the relation R, let H be a member of C and
H0 be the spanning sub-graph obtained from H by deleting all the edges in G(H)∪(en a(H)).
Then
(a) H0 is also in C.
(b) Adding any combination of k edges in G(H) ∪ (en a(H)) to H0 gives another member
H of C. In addition, f (H ) = f (H0 ) + k and

v(H ) − k(H ) + g(H ) = v(H0 ) − k(H0 ) + g(H0 ).

Proof. For (a), we need to show that a(H0 ) = a(H) and G(H0 ) = G(H). Consider the
spanning sub-graph H0 obtained from H0 by deleting all the edges in Ω0 and outside of Ω1
as in Definition 2.4.1. Let m and M be the smallest edge and the largest edge in a(H). The
set a(H0 ) would only differ from a(H) if we delete edges in Ω1 that are not between m and
40

M . However, the algorithm for G(H) requires that all edges in G(H) are between m and M .
Adding or deleting edges in G(H) ∪ (en a(H)) does not affect He , as used in Lemma 2.4.2
to define sa , and the sequences, since G(H) ∪ (en a(H)) ⊂ en , so G(H0 ) = G(H) and we
have that H and H0 are in the same equivalence class.
For (b), an edge is in G(H0 ) if and only if it is between a pair of edges e and e where
e ∈ RΩi and e ∈ RΩi+1 or e ∈ RΩi−1 , and there are no edges in RΩi−1 or RΩi+1 included
between e and e , or it is sa . We have the same condition for an edge in G(H0 ) on the left
side by switching all the labels from R to L. See Figure 2.11 for an example.

Figure 2.11: From left to right: An example of edges in G(H) marked in blue; the state graph
resulting from including zero edges from G(H); the state graph resulting from choosing one
edge from G(H), note that a state circle splits off; the state graph resulting from including
two edges from G(H) with two state circles split off.
By design and since we are beginning the sequences at sa which are especially chosen
through using Lemma 2.4.2, all the edges in G(H0 ) ∪ (en a(H)) have two ends on the same
state circle in state graph H0 associated to H0 , with the identical local picture as shown in
Figure 2.11: they all have two ends on the same state circle in H0 , and none of the pairs of
edges from G(H0 )∪(en a(H)) lies embedded on distinct sides of a state circle such that only
one end of an edge lies between the two ends of another. Thus, including any combination
of k edges to H0 splits off k additional state circles in the new spanning sub-graph H, and
this shows the first part of (b).
Since we cannot add an edge in region Ωi without there already being an edge included in
Ωi from the definition of the sequences which define G(H), we also have that v(H )−k(H ) =

41

v(H0 ) − k(H0 ). This also shows that g(H ) = g(H0 ) since the increase in the number of faces
is the same as the number of edges included.

2.6

Combinatorics of G(H) and ribbon graphs

By Lemma 2.5.2, if C is an equivalence class of the relation R, and r = |G(H) ∪ (en a(H))|
is the cardinality of C, we may write
r
H∈C

XH =

i=0

r e(Gn )−2e(H0 )−2r
A A
(−A2 − A−2 )f (H0 )+r−1 .
i

For each equivalence class C, we are almost ready to use Lemma 2.3.1 to show the cancellation
of the top coefficients of this sum. However, we need an estimate on r so that we know how
many coefficients are cancelled. We also need to know how the resulting upper bound on the
maximum degree compares with M (Dn ). Our main result of this section is the following:
Lemma 2.6.1. If a spanning sub-graph H ⊂ Gn
A satisfies v(H) − k(H) + g(H) ≤ n − 2, then

|G(H)| + v(H) − k(H) + g(H) ≥ |a(H)| − 1.

This lemma connects the cardinality of G(H) to the combinatorial ribbon graph quantities
v(H), k(H), and g(H) and is the last important piece of the proof of Theorem 2.2.3. The
reader may skip ahead to Section 2.7 to see how it is applied and come back to this section
later. To prove this lemma, we make use of the explicit labeling from Section 2.4.
Lemma 2.6.2. Let e1 be an edge in RΩi included in a spanning sub-graph H. Suppose
another spanning sub-graph H only differs from H by including two additional edges e2 in
RΩi and e3 in RΩi+1 that satisfy the following criteria:
42

• The edges e2 and e3 are not included in H.
• The edge e2 is between e1 and e3 .
• None of the edges in RΩi and RΩi−1 between e1 and e3 are included in H, and none
of the edges in RΩi+1 between e1 and e2 are included in H.
Then v(H ) − k(H ) + g(H ) = v(H) − k(H) + g(H) + 1. The same statement for LΩi can be
obtained by replacing all the labels from R to L.

e2
e1

e1

e3

Figure 2.12: On the left we have H and on the right we have H . Their local pictures only
differ in the inclusion of two edges e1 and e2 .

Proof. An example satisfying the hypotheses of the lemma is shown in Figure 2.12. By the
definition of g(H),
2v(H) − 2k(H) + 2k(H) − v(H) + e(H) − f (H)
2
v(H) + e(H) − f (H)
=
.
2

v(H) − k(H) + g(H) =

Compare the state graph H and H . Including the edge e3 will first split off a state circle.
Then, including e2 joins the state circle split off by e1 , e3 to another state circle. The end
result is that the number of states circles of H and H are the same. By Lemma 2.2.8, this

43

means f (H ) = f (H). Therefore,
v(H ) + e(H ) − f (H )
2
v(H) + e(H) + 2 − f (H)
=
2

v(H ) − k(H ) + g(H ) =

= v(H) − k(H) + g(H) + 1.

We define ga (H), which counts the occurrences of edge inclusions described in Lemma
2.6.2 from the sequences {tj }, {bj } in Definition 2.4.4.
Definition 2.6.3. If sa as in Definition 2.4.3 is in RΩk , we define ga (H) by taking the union
of tgi and bgi defined for LΩi and RΩi below:
For RΩi (resp. LΩi ) and the sequences {tj }, {bj }.
• For all i, tgi is the number of even ’s such that t is non-empty and between two
non-empty terms bj and bj+1 where j is even.
• For i = k, bgi is the number of even ’s such that b is non-empty and between two
non-empty terms tj and tj+1 where j is even.
• bgk is

#{bj }
2

where #{bj } is the number of non-empty terms in the sequence {bj },

and · is the floor function which rounds down to the nearest integer.
Lastly,

ga (H) :=

bgi +
0≤i≤k for each RΩi

tgi +
k≤i≤n−2 for each RΩi

44

tgi .
1≤i≤n−2 for each LΩi

If sa is on the left side, we replace L by R and R by L in the above definition to get the
definition of ga (H) in this case.
By Lemma 2.6.2, ga (H) counts the quantity v(H) − k(H) + g(H) locally. Now we are
ready to show Lemma 2.6.1.
Proof. We assume first that sa ∈ RΩk and M is the largest label of the edges in a(H). The
case where sa ∈ LΩk is the same by reflection. Recall that G(H) is defined in Definition
2.4.5 by taking the union of the intersection bi ∩ ti for all i ranging over RΩi and LΩi . The
cardinality of the intersection bi ∩ ti is related to tgi , bgi , bi+1 , and ti−1 when i ≤ n − 2 for
RΩi (resp. LΩi ) as follows:

|bi ∩ ti | = |bi | − |tgi | − |bi |

(2.2)

|bi ∩ ti | = |ti | − |bgi | − |ti |,
where bi consists of edges in RΩi between bj and bj+1 in the sequence {bj } for RΩi+1 where
j is even and bj is non-empty. Similarly, ti consists of edges in RΩi between tj and tj+1
in the sequence {tj } for RΩi−1 where j is even and tj is non-empty. These are the edges
excluded from bi and ti by the intersection bi ∩ ti , respectively.
We also have

|bi+1 | ≥ |bi | or |bi+1 | = |bi | − 1.

(2.3)

|ti−1 | ≥ |ti | or |ti | = |ti−1 | − 1.
By Definition 2.4.5, an edge ≤ M of RΩi (resp. LΩi ) with sequence {bj } and {tj } is in
bi , if and only if it is between bj and bj+1 where j is even and bj is non-empty. An edge is
45

excluded from bi by the intersection bi ∩ ti if and only if it is between tj and tj+1 where j
is odd and bj is non-empty. The sequence {bj } for RΩi+1 is the sequence {tj } for RΩi with
the first term removed. Therefore, bi+1 contains edges excluded from bi with one added to
all the labels. Since RΩi+1 has one less edge than RΩi , we have the two possibilities for
|bi+1 | and |bi |. Similarly, we have the statement for |ti | and |ti−1 | and for LΩi .
We define the following quantities.
• C := #{i : k ≤ i < n − 2, |bi+1 | = |bi | − 1} for each RΩi such that the edge lost from
bi is an edge coming from bRk+1 in the following sense: the edge - r in RΩr is in br
for each k + 1 ≤ r ≤ i.
• C := #{i : 0 ≤ i < n − 2, |bi+1 | = |bi | − 1}, for each LΩi such that the edge lost from
bi is an edge coming from tRk−1 in the following sense: the edge - r is in br for each
0 ≤ r ≤ i of LΩr and in tr for each 0 ≤ r ≤ k − 1 of RΩr .
• E := #{i : k < i < n − 2, i is counted by C and i + k is counted by C}.
As discussed, an edge in bi+1 is an edge in bi with one added to the label, so if |bi+1 | = |bi |−1
is counted by C, an edge with label n − i is included in bi . Similarly, an edge with label
n−i−1 is included in bi−1 , and so on. We see that an edge with label R(n−i−(i−k −1)) =
R(n − 2i + k + 1) is included in bk+1 . Similarly, an index i counted by C corresponds to an
edge with label R(n−(i−k)−(k +i−k −1)) = R(n−2i+k +1) included in tk−1 . Therefore,
E counts the number of edges with the same labeling numbers in bRk+1 and tRk−1 .
To complete our estimate of |G(H)|, we show

|bk ∩ tk | + |bk+1 | + |tk−1 | ≥ |sa | − |tgk | − |bgk | + δk + E,

46

(2.4)

where |sa | is defined to be the number of edges greater than sa and ≤ M in RΩk , and δk is
a quantity counted by C. To begin with, we have |bk | = |sa | − |bgk | − |tk−1 | since edges of
the sequence {bj } for RΩk counted by bgk are not included in bk , and there are additional
edges excluded from |sa | that are replaced by edges in tk−1 . Then, |bk ∩ tk | + |bk+1 | ≥
|sa | − |bgk | − |tk−1 | − |tgk | − δk + E, where δk = 1 if |bk+1 | = |bk | − 1 and 0 otherwise. So,
δk is counted by C. The edges greater than sa that are part of the sequence {tj } of RΩk+1 ,
counted by tgk , will not be included in bk ∩ tk . In addition, extra edges with the same labels
as those in tk−1 may be included in bk+1 . These are counted by E. Adding |tk−1 | to the
left hand side of the inequality produces the inequality (2.4).
Putting the statements (2.2) and (2.3) together for each 0 ≤ i ≤ n−2 and using the definition
of G(H), we have

|G(H)| ≥ |bk ∩ tk | +

|bi ∩ ti | +
i=k for each RΩi

|bi ∩ ti |
i=0 for each LΩi

≥ |bk ∩ tk | + |bk+1 | + |tk−1 |
−C −C −

bgi −
0≤i<k for each RΩi

tgi −
k<i≤n−2 for each RΩi

tgi .
1≤i≤n−2 for each LΩi

(2.5)
The inequality (2.4) then gives

|G(H)| ≥ |sa | − ga (H) + E − (C + C ).

(2.6)

Without loss of generality, assume that the highest index counted by C or C is in C. To

47

link the combinatorial estimate of |G(H)| to the quantity v(H) − k(H) + g(H), we show

v(H) − k(H) ≥ C + k − 1,

(2.7)

and that there are additional contributions

v (H) + |G (H)| + ga (H) ≥ C − E

(2.8)

to |G(H)| + v(H) − k(H) + g(H). If the highest index is counted by C instead, then we
switch the roles of C and C in (2.7) and (2.8).
If H is obtained from H by adding edges, then v(H )−k(H )+g(H ) ≥ v(H)−k(H)+g(H)
because f (H ) − f (H) ≤ the number of edges added [78, Lemma 5.6]. This shows that we
can compute a lower bound of v(H) − k(H) + g(H) for any spanning sub-graph H by starting
with the spanning sub-graph of no edges, Hn
A , add a subset of edges of H to it to obtain
H for which we can compute v(H ) − k(H ) + g(H ), then v(H ) − k(H ) + g(H ) will be
a lower bound for v(H) − k(H) + g(H) since adding any additional edges will not decrease
v(H) − k(H) + g(H).
n
Recall that Hn
A is the spanning sub-graph of no edges. We will be adding to HA two
n
n
subsets of edges of H in order to estimate the quantities v(Hn
A ) − k(HA ) and g(HA ):

(a) By definition and the proof of Lemma 2.4.2, the edge sa is in RΩk if for each 0 < i ≤
k − 1, an edge in Ωi that is not in en is included in H. Denote this set of edges by
V . The first set will be the union of V , the first terms of the sequences {bj } for RΩi
where 0 ≤ i ≤ k, and the largest term of the sequences {bj } and {tj }, going through
RΩi for each i > k where i ≤ the highest index counted by C.

48

(b) The second set are the rest of the edges in all the sequences {tj }, {bj } defined for H.
1
We denote the spanning sub-graph obtained by adding the set (a) to Hn
A by H , and the

spanning sub-graph obtained by adding the set (b) to H1 by H2 . The spanning sub-graph
H is obtained by adding the rest of the edges included in H to H2 .
We first show that

v(H1 ) − k(H1 ) ≥ C + k − 1.

(2.9)

In H1 , the edges from V are edges between state circles Si , Si+1 for 0 ≤ i ≤ k − 2 in
n , the all-A state graph associated to Hn , which increases as i increases. Adding each
HA
A

edge counted by C also increases v(H1 ) − k(H1 ) by 1, since it merges different state circles
from previous edge-inclusions, see Figure 2.13.

RΩk
RΩ1
Figure 2.13: The edgeset V is shown in black.
Adding the set (b) of edges from the rest of the sequences to get H2 , we apply Lemma
2.6.2 repeatedly. The increase in g(H) is given by ga (H) since we only counted the increase
to v(H1 ) − k(H1 ) + g(H1 ) from increasing sequence of edges in set (a). We get the inequality
(2.7).
For (2.8), we consider the set of edges e in RΩk+1 which do not belong to bk+1 ∩ tk+1 .
This is a set with cardinality ≥ C − E. Let s be the highest index of RΩs counted by C.
By definition, the edge cs lost from bi comes from bRk+1 . Let cr be the edge corresponding
49

to cs − r in RΩs+1−r . Now we have a set of increasing edges all less than cs − r for each r
in H2 . We also have
|bk+1 | ≥ |e | + |bk |,
or
|bk+2 | ≥ |e | + |bk+1 |,
since otherwise, cr would not belong to br for some r, see Figure 2.14.
RΩk+2
RΩk+1
RΩk
RΩk−1
Figure 2.14: The leftmost figure depicts a possible set of bk shown in blue, and the edge
excluded from bk by the edge in RΩk−1 . Depending on the sequences {tj }, {bj } for RΩk
and RΩk+1 , we have at least an extra edge back from bk+1 or bk+2 .
Combing this with (2.5), we note that any additional loss of edges from e as we increase
r increases one of v(H2 ) − k(H2 ), g(H2 ), or G(H2 ) by 1 independent of (2.9) and ga (H). So

v(H) − k(H) + g(H) + |G(H)| ≥ |sa | + E − (C + C ) + C + k − 1 + C − E,

and we are done.

50

2.7

Proof of Theorem 2.2.3

To summarize, by Theorem 2.2.6, we have that

Dn =

H⊆Gn
A

Ae(GA )−2e(H) (−A2 − A−2 )f (H) .

n is given by
The contribution of a spanning sub-graph H ⊂ Gn
A to D
e(Gn
A )−2e(H)

XH := A

(−A2 − A−2 )f (H) .

Analysis of the monomials involved in the expansion (see [24], Theorem 6.1) show that a
spanning sub-graph H contributes to the first n − 1 coefficients if and only if v(H) − k(H) +
g(H) ≤ n − 2. We organize the contributions of these spanning sub-graphs by putting them
into equivalence classes defined by R as in Definition 2.5.1. Let C denote an equivalence
class of R. Recall that d∗ of a polynomial p(A) in A is the maximum degree of p(A). We
have

d∗ D n = d∗
C

XC ,

where
XC :=

H∈C

and C ranges over equivalence classes of R.

51

XH ,

Let r = |G(H0 ) ∪ (en a(H0 ))|, where H0 is the member in C which does not include any
edges in G(H0 ). By Lemma 2.6.1, r ≥ n − 1 − (v(H0 ) − k(H0 ) + g(H0 )). We also have, by
Lemma 2.5.2,
r

XC =

i=0

r e(Gn )−2e(H0 )−2r
A A
(−A2 − A−2 )f (H0 )+r .
i

Applying Lemma 2.3.1 to this sum and writing f (H0 ) in terms of v(H0 ), g(H0 ), and k(H0 )
using the definition of genus, we get

d∗ XC ≤ c(Dn ) − 2e(H0 ) + 2(−2g(H0 ) + 2k(H0 ) − v(H0 ) + e(H0 )) − 4r
≤ c(Dn ) − 2e(H0 ) + 2(−2g(H0 ) + 2k(H0 ) − v(H0 ) + e(H0 ))
− 4(n − 1 − (v(H0 ) − k(H0 ) + g(H0 )))
≤ c(Dn ) + 2v(H0 ) − 4(n − 1)
≤ n2 c(D) + 2|sA (Dn )| − 4(n − 1),

where n2 c(D) + 2|sA (Dn )| = M (Dn ). Since degXC ≤ M (Dn ) − 4(n − 1) for all the equivalence classes C containing all spanning sub-graphs H with v(H) − k(H) ≤ n − 2, we see that
d∗ Dn ≤ M (Dn ) − 4(n − 1).

2.8

A worked out example

The proof of Theorem 2.2.3 uses the labeling on cables of a non A-adequate link diagram
to separate the contributing spanning sub-graphs into equivalence classes. There is then
cancellation of coefficients summing over the contributing polynomial of each member of the

52

equivalence class. In this section, we illustrate this process on an example.

Figure 2.15: From left to right: a non A-adequate link diagram, the all-A state graph, and
the all A-state graph of the 3-cable.
The link diagram shown on the left in Figure 2.15 is not A-adequate because there is an
edge with both ends on the same state circle S in the all-A state graph of the diagram. This
link may have an A-adequate diagram. However, our aim is to show the division of spanning
sub-graphs into equivalence classes for this non A-adequate diagram. We consider the case
n = 3 and apply the construction for the proof of Theorem 2.2.3. The theorem in this case
says that M (D3 ) − d∗ D3 ≥ 4 · (3 − 1).
The first step is to restrict to spanning sub-graphs H ⊂ Gn
A with v(H) − k(H) ≤ 1 by
the remark following Theorem 2.2.6. Following the conventions of Section 2.2, e3 denotes
the edges in the all A-state graph of D3 with two ends on a state circle S0 , S1 is the state
circle coming from cabling S in D3 that contains S0 and no other state circles coming from
cabling S in one of its complementary regions on S 2 . The edges in the all-A state graph of
D3 corresponding to e3 have their labelings. We also label all the regions Ω0 and Ω1 . See
Figure 2.16.
The restriction v(H)−k(H) ≤ 1 means that only edges between a pair of distinct vertices,
and one-edge loops e3 in Gn
A , may be included in H. There are two cases, one where a(H)
is empty and another where a(H) is non-empty. The case where a(H) is empty includes
the case where the pair of distinct vertices with an edge included between them is not S0
53

and S1 , v(H) − k(H) = 0, and when a single edge or two edges with the same label in Ω1
are included between S0 and S1 . The spanning sub-graph H of each of these cases belongs
to an equivalence class where the only difference between the members are which edges are
included from e3 . Since |e3 | = 3, applying Lemma 2.3.1 will show that the contribution of
H to terms with power greater than M (D3 ) − 8 in D3 is equal to zero.
RΩ1
Ω0

2 4

1 3 5

LΩ1 2 4
Figure 2.16
We restrict to the local picture of Gn
A involving only S0 and S1 . The case when a(H) is
non-empty is what motivates the technicalities in the proof of Theorem 2.2.3. We cannot
put H in an equivalence class where the only difference between the members is what edges
it includes from e3 , since an edge in e3 may belong to a(H) where it has two edges on
distinct state circles in the state graph associated to H. Inclusion of the edge may then give
a spanning sub-graph H with a different V (H ) − k(H ) + g(H ), which determines the power
of the term (−A2 − A−2 ) of the contribution XH . Lemma 2.3.1 will not apply to the sum
of the contributions in this case.
On the other hand, if a(H) is non-empty, then at least one edge must be included in H
between S0 and S1 and not between any other pairs of vertices which makes our restriction
to the local picture valid. If only one edge between S0 and S1 is included then a(H) is empty
and we have already addressed this case. When more than one edge between S0 and S1 is
included in H, we look to the edges in region Ω1 that have two ends on the same state circle

54

in the state graph associated to H to set up a equivalence class where we can apply Lemma
2.3.1. This is the primary motivation for our algorithm in the proof of Theorem 2.2.3.
The general case is more complicated, but we can list all the possibilities for a(H) for
the specific case we have here where n = 3. When a(H) is non-empty, the possible subsets
are a(H) = {1}, {3}, {5}, {1, 3}, {1, 5}, {3, 5}, and {1, 3, 5}. In Table 2.3 we list all the
possible sequences {bj }, {tj } and the corresponding G(H), v(H)−k(H), and g(H) for a(H) =
{∅}, {1}, {3}, {5}, {1, 3}, {1, 5}, {3, 5}.

55

en a(H) {bj } for RΩ1

{tj } for RΩ0

{tj } for LΩ0

{bj } for LΩ1

{∅}

{1, 3, 5}

{∅}

{∅}

{∅}

{∅}

{1}

{3, 5}

{3}

{1, 5}

{5}

{1, 3}

{∅}
{∅}
{R2}
{R2, 3}
{R4}
{R4, 5}
{R2}
{R2, 3}

{∅}
{1}
{∅}
{3}
{∅}
{5}
{∅}
{3}

{∅}

{1}

{R2}

{1, R2}

{R2, 3}

{1, R2, 3}

{R2}

{∅}

{R2, 3}

{3}

{R2, 5}

{5}

{R2, 3, R4}

{3, R4}

{R2, 3, R4, 5}

{3, R4, 5}

{∅}
{1}
{∅}
{3}
{∅}
{5}
{∅}
{3}
{1}
{1, L2}
{1, L2, 3}
{1}
{1, L2}
{1}
{1, L2, 3}
{∅}
{3}
{3, L4}
{3, L4, 5}
{5}
{3}
{3, L4}
{3}
{3, L4, 5}

{∅}
{∅}
{∅}
{∅}
{∅}
{∅}
{∅}
{∅}
{∅}
{L2}
{L2, 3}
{∅}
{L2}
{∅}
{L2, 3}
{∅}
{∅}
{L4}
{L4, 5}
{∅}
{∅}
{∅}
{∅}
{∅}

{1, 3}

{3, 5}

{5}

{1}

56

a(H)

G(H) v(H) − k(H)
0
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{R2} 1
{R2} 1
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{∅}
1
{R4} 1
{5}
1
{∅}
1
{∅}
1
{R4} 1
{∅}
1
{∅}
1
{∅}
1
{∅}
1

g(H)
0
0
0
1
0
1
0
1
0
1
1
1
2
1
1
2
2
0
1
1
2
1
1
1
1
2

Table 2.3: The tables detail the various possibilities for the equivalence classes corresponding to the different cases for a(H)
and their corresponding values of G(H), v(H) − k(H), and g(H), when sa is in RΩ1 or Ω0 . The case for when sa is in LΩ1 is
analogous.
56

For example, for a(H) to contain edges labeled 1 and 3, an edge attached to the portion
of the state circle S0 cut off by 1 must be included in H, and one of the edges labeled 4 in
LΩ1 and RΩ1 must be included. There will be various combinations for these edges that cut
off state circles so that 1 ends up with two ends on different state circles in the state graph
associated to H, but the only two cases that we need to deal with are
• 1 is not included in H.
• 1 is included in H.
If 1 is not included in H, the edge with label R2 will have two ends on the same state
circle in the state graph associated to H. This is the edge included in G(H). The sequence
{bj } for RΩ1 will either be {R2} or {R2, 3} since we don’t include edges with labels larger
than or equal to 4, and these are shown in the row starting with a(H) = {1, 3} in Table
2.3. On the other hand, if 1 is included, there is a corresponding increase in g(H) from
0 to 1. Taken altogether, v(H) − k(H) + g(H) ≥ 2 and so we do not need to account for
the contribution of those equivalence classes. This is the idea behind Lemma 2.6.1, which
estimates the cardinality of the set G(H) by v(H) − k(H) + g(H). The more terms there are
in the sequences, and so the smaller |G(H)| is, the bigger v(H) − k(H) + g(H) will be, so that
we will always end up with |G(H)| + v(H) − k(H) + g(H) ≥ a(H) − 1. As shown in Section
2.7, this controls the degree of the contribution of each equivalence class so that even if there
is only a single element H in an equivalence class, deg XH ≤ M (Dn ) − 4(n − 1).

57

2.9

Detecting semi-adequacy using the colored Jones
polynomial

We use Theorem 1.1.1 to define a link invariant as in [69]. Let D be a diagram of an oriented
link K. Let c− (D) be the number of negative crossings of D, and recall that |sA (D)| is the
number of state circles in the all A-resolution of D, c(D) is the number of crossings in D,
and w(D) is the writhe. We consider the complexity

(c− (D), c(D), |sA (D)| − w(D)),

ordered lexicographically. Let D(K) be the set of diagrams which minimizes this complexity.
Let GD (n, A) be as defined in Section 1.2, see equation (1.2). We let
M (GD (n, A)) := M (Dn−1 ) + 4 − 2(n − 1) − w(D)((n − 1)2 + 2(n − 1)).

Recall that a lower bound hn (D) (See Definition 2.2.2.) of the minimum degree d(n) of
JK (n, q) is − 41 M (GD (n, A)).
Definition 2.9.1. Let K be a link and D an oriented link diagram in D(K). For i ≥ 1, let
βi = βi (D) be the coefficient of AM (GD (i+2,A))−4(i−1) in GD (i + 2, A). Define
A (q)
JD

∞

:=
i=1

βi q i−1 .

We will need the following lemma from [69].
Lemma 2.9.2. [69, Lemma 3.4] Suppose that for a link K, there is D ∈ D(K) that is
58

A-adequate. Then, all the diagrams in D(K) are A-adequate.
Applying Theorem 1.1.1, we have the following corollaries.
A (q) = 0 if and only if D is A-adequate.
Corollary 2.9.3. JD

Proof. By definition, the coefficient βi of AM (GD (i+2,A))−4(i−1) in GD (i + 2, A) is the coefficient of q hi+2 (D)+i−1 in JK (i + 2, q). If D is not A-adequate, then Theorem 1.1.1 says
that d(i + 2) ≥ hi+2 (D) + i, so βi = 0 for all i. This shows the forward direction. For the
converse, β1 is the coefficient of q h3 (D) in JK (3, q). If D is A-adequate, then h3 (D) is equal
A (q) = 0.
to the minimum degree d(3) of JK (3, q), so β1 = 0, and this shows that JD
A (q) defined above is independent of the diagram D ∈
Corollary 2.9.4. The power series JD
A (q).
D(K), thus it is an invariant of K, which we denote by JK

Proof. If K is not A-adequate, then any diagram in D(K) is not A-adequate. Let D be
A (q) = 0. If K is A-adequate, then an Aa diagram in D(K), then by Theorem 1.1.1, JD

adequate diagram of K minimizes the complexity (c− (D), c(D), |sA (D) − w(D)|), thus it
belongs to D(K), and all the diagrams in D(K) are A-adequate by Lemma 2.9.2. Let D
A (q)
be a diagram in D(K). As shown in [9], the colored Jones polynomial has a tail and JD
n (q)}∞ , therefore it is also independent
records the stable coefficients of the sequence {JK
n=2

of the diagram D.
Definition 2.9.5. Consider the graph GA with vertices the state circles, and edges the
segments from the graph of the A-resolution HA of D, we denote by χA (D) the Euler
characteristic of GA .
Corollary 2.9.6. Suppose D is an A-adequate diagram of a link K and D is another
diagram of K. Then D is A-adequate if and only if c− (D) = c− (D ) and χA (D) = χA (D ).
59

Proof. If D is A-adequate, then c− (D) and |sA (D)|−w(D) are invariants of K [78, Theorem
5.13]. Thus, χA (D) = |sa (D)| − c(D) = |sA (D)| − w(D) − 2c− (D) is also an invariant of
K. For the converse, since D is A-adequate, the minimum degree d(n + 1) of JK (n + 1, q)
is equal to hn+1 (D), which we rewrite here slightly differently:
1
hn+1 (D) = − (n2 c(D) + 2n|sA (D)| − 2 + 4 − 2n − ω(D)(n2 + 2n))
4
1
= − (2c− (D)n2 + 2n(|sA (D)| − w(D)) + 2 − 2n).
4
If D is not A-adequate and c− (D) = c− (D ), χA (D) = χA (D ), then sA (D) − ω(D) =
sA (D ) − ω(D ), so hn+1 (D) = hn+1 (D ). Theorem 1.1.1 applied to D will imply that
d(n + 1) < hn+1 (D), which is a contradiction.

60

Chapter 3
The Slope Conjecture and 3-string
pretzel knots
Outside the class of semi-adequate links, the Slope Conjecture has been verified for the
class of 3-string pretzel knots P (−1/2, 1/3, 1/p) in the paper [41] by Garoufalidis where he
introduced the Slope Conjecture and also verified it for alternating knots and knots with
crossing number ≤ 9. In [28], Dunfield and Garoufalidis gave criteria for a spun-normal
surface in an ideal triangulation to be incompressible. Based on this work, Garoufalidis and
van der Veen proved the Slope Conjecture for 2-fusion knots [46]. Suppose that Kp,q is the
(p, q)-cable of a knot for which the Slope Conjecture is true, Kalfagianni and Tran [70, 71]
recently showed that the Slope Conjecture is also true for Kp,q provided certain conditions
on the degrees are met. Using this, they proved the Slope Conjecture for iterated cables of
torus knots and iterated torus knots.
The work of this chapter is motivated by the following question.
Question 3.1. How does the Jones slope “pick” the corresponding boundary slope for non
semi-adequate links?
To provide more examples to study Question 3.1, we prove:
Theorem 1.1.2. Let r, s, t be nonzero integers greater than 5 such that r is even, s, t are
odd, and r > s, t. The Slope Conjecture is true for pretzel knots P (−1/r, 1/s, 1/t).
61

We begin with the background information on essential surfaces in Section 3.1, whose
boundary slopes may be enumerated by classifying branched surfaces. As we will see in
Section 3.2, this can be applied to a 2-bridge knot to find all of its boundary slopes. This is
the building block for the Hatcher-Oertel algorithm, which combines incompressible surfaces
locally defined for each tangle into an incompressible surface for the Montesinos knot. We
discuss how the surfaces are fitted together in Section 3.3 and describe the algorithm in
Section 3.4. We will apply the algorithm to 3-string pretzel knots and compute the boundary
slope of a specific incompressible surface. In Section 3.5, we describe the computation of the
Jones slope for the pretzel knot, and show that it agrees with the boundary slope in Section
3.6.

3.1

Essential surfaces and branched surfaces

For the rest of this chapter, we assume that M is a compact, orientable and irreducible
3-manifold with boundary ∂M and that all surfaces S ⊂ M are properly embedded, so
∂S ⊂ ∂M . A surface S is two-sided if its normal bundle in M is trivial.
Definition 3.1.1. A properly embedded, two-sided, and connected surface S ⊂ M is incompressible if for each disk D ⊂ M with D ∩ S = ∂D there is a disk D ⊂ S with ∂D = ∂D.
See Figure 3.1. The disk D is called a compressing disk for S if there is no D ⊂ S with
∂D = ∂D.
S is ∂-incompressible if for each disk D ⊂ M such that ∂D is the union of two arcs α
and β where α ⊂ S and β ⊂ ∂M , intersecting only at their endpoints, there is a disk D ⊂ S
with boundary α ∪ β such that β ⊂ ∂S. The disk D is called a ∂-compressing disk for S if
there is no D ⊂ S as above. See Figure 3.2.
62

D
D

D

S

S
Figure 3.1: A compressing disk [52].
α

α

S
D

S
D

∂M

D

∂M
Figure 3.2: A ∂-compressing disk [52].

Definition 3.1.2. A properly embedded surface S ⊂ M is essential if the boundary of its
normal bundle in M is incompressible and ∂-incompressible.
Remark 3.1.3. This definition of essential surface is consistent with the usual definition for
two-sided surfaces, since the boundary of the normal bundle of S consists of two copies of S
if S is two-sided.
Definition 3.1.4. A branched surface B ⊂ M is a compact surface locally modeled on
Figure 3.3a. i.e., B is locally homeomorphic to two planes R2 with the half-planes identified.
A branched surface neighborhood of B, see Figure 3.3b, in M is a codimension 0, compact
submanifold N (B) ⊂ M 3 foliated by interval fibers locally isomorphic to the model, where
the branched surface is obtained by identifying the fibers to a point. A surface S ⊂ M is
carried by B if S can be isotoped into the interior of N (B) so that it intersects the fibers
transversely.
Floyd and Oertel showed that there are only finitely many branched surfaces in M and
63

The proof of the theorem will follow fairly easily from a fundamental result of
manifolds,
which
are aclosed
subsets
locally
[FO] about
branched
The proof
of thesurfaces
theorem in
will3follow
fairly easily
from
fundamental
result
of diffeowhich are closed subsets locally diffeo[FO] about
branched
3 manifolds,
morphic
to the
modelsurfaces
in the in
first
figure below.
morphic to the model in the first figure below.

(a) A branched surface.

(b) A branched surface neighborhood.

A branched surface B is said to carry a surface S if S lies in a fibered regular neighA branched
surface
B is
to carry
a and
surface
S if surface
S liesneighborhood
in a fibered
regular neighFigure
3.3: of
Models
of asaid
branched
surface
a branched
[54].
borhood
N(B)
B , indicated
in the
second
figure,
and is
transverse to all the
fibers

borhood
of B ,all
indicated
second
figure,
is transverse
toresult
all the fibers
of N(B)N(B)
. If S meets
the fibersin
of the
N(B)
it is said
to haveand
positive
weights. The

that .itIfsuffices
to enumerate
branched
for
thesaid
computation
ofpositive
boundaryweights.
slopes of aThe result
of of
N(B)
S meets
all the fibers
ofsurfaces
N(B)
it is
to have
M with
incompressible
boundary
[FO] is that
in a compact
irreducible
3 manifold
with
boundary
of [FO]
isexist
thatfinitely
in a compact
irreducible
3 manifold
∂M)incompressible
such that the surthere
many branched
surfaces
(Bi , ∂Bi ) ⊂M(M,
knot
[29].

are, ∂B
exactly
all the incompressible,
faces
carried
with positive
weights bysurfaces
these Bi ’s(B
there
exist
finitely
many branched
i
i ) ⊂ (M, ∂M) such that the sur-

inweights
M ,1]upLet
toby
isotopy.
A refinement
[O] is that
the Bi ’s
∂ incompressible
Theorem
[29, Theorem
M
be a Haken
3-manifoldin(compact,
orientable,
faces
carried 3.1.5.
with surfaces
positive
these
B
i ’s are exactly all the incompressible,
can be chosen so that all the surfaces they carry, whether of positive weights or not,

surfaces
in Ma , two-sided
up to isotopy.
A refinement
[O] is that the Bi ’s
∂ incompressible
irreducible 3-manifold
containing
incompressible
surface) with in
incompressible
are incompressible and ∂ incompressible.

can be chosen so that all the surfaces they carry, whether of positive weights or not,
boundary.
arethese
a finite
number surfaces
of properlyBembedded
branched surfaces B1 , . . . ,track,
Bk ,
Let B beThere
one of
branched
i . Then ∂B = B ∩ ∂M is a train

are incompressible and ∂ incompressible.

∂M with
twoin
key
or such
branched
1 manifold,
that every
two-sided in
essential
surface
M properties:
is isotopic to a surface carried by one of

Let B be one of these branched surfaces Bi . Then ∂B = B ∩ ∂M is a train track,

(1)theThere
Bi ’s. is no smooth disk D ⊂ ∂M with D ∩ ∂B = ∂D .

or branched 1 manifold, in ∂M with two key properties:

(2) There is no disk D ⊂ ∂M , smooth except for one outward cusp point in ∂D , such
Their
[29]
also
proof
Hatcher’s
result
that
the. list of boundary slopes
that
Dpaper
∩
= ∂D
. contains
(1) There
is
no∂Bsmooth
disk Da ⊂
∂Mof with
D∩
∂B =
∂D

is finite
for
a link
using
surfaces
[54].
(2)The
There
no
disk
D
⊂
∂M , smooth
except
one outward
cusp
point
latteris
condition
is branched
explicitly
given in
[FO]. Iffor
condition
(1) failed,
then
any in
sur-∂D , such

face
carried
that
D ∩ by
∂B B=with
∂D .positive weights would have a boundary circle which was conTheorem
3.1.6.
Corollary to Theorem
For awould
knot Kbound
⊂ S 3 , athere
only finitelyof
tractible
in ∂M
. By[54,
incompressibility,
this 1]
circle
diskare
component

Thethelatter
condition
explicitly
givenofinB [FO].
condition
thenasany sursurface,
contrary is
to the
construction
in [FO].IfCondition
(2)(1)
canfailed,
be phrased
3
many slopes realized by boundary curves of essential surfaces in S K.
saying
thatby
theBtrain
track
∂B hasweights
no monogons.
train tracks
are required
face
carried
with
positive
wouldSometimes
have a boundary
circle
which was con-

to have
no
digons
well,
but we
have
to allow
these
here.
tractible
∂M
. Byasincompressibility,
this
circle
would bound
a disk
of
Wein
study
branched
surfaces
in the
specific
setting
of Montesinos
knots, which
are component
knots
Let S be
a surface
by B with positive
weights.
No component
of ∂S
the surface,
contrary
tocarried
the construction
of B in
[FO]. Condition
(2) can
be can
phrased as

obtained by connecting rational tangles in a cyclic pattern, see Figure 3.4. Let B be a 3be
in ∂M
, since
there would be
a smooth disk
⊂ ∂M are
withrequired
sayingcontractible
that the train
track
∂Botherwise
has no monogons.
Sometimes
trainDtracks
with
foursomewhere
points marked
on itsthis
boundary.
A (2-)tangle
is (2)
a pair
of properly
embedded
∂Dball
⊂ ∂B
, and
inside
disk condition
(1) or
would
be violated.
Thus

to have no digons as well, but we have to allow these here.

in each component torus Ti of ∂M which B meets, ∂S consists of a number of parallel

Let S be a surface carried by B with positive weights. No component of ∂S can
circles.
64
be contractible
∂M , since
otherwise
there
be a
smooth
disk
one of the
tori
of ∂M we
let D
∂i B⊂=∂M with
To simplifyinnotation
in what
follows, if
Ti is would

∂D∂B
⊂∩
∂BTi, ,and
somewhere
inside
disk condition (1) or (2) would be violated. Thus
and similarly
∂i S =
∂S ∩ Tthis
i for any surface S carried by B .

arcs into B each endpoint of the arcs is sent to a marked point on the boundary. A tangle
consisting of two unlinked arcs is a trivial tangle.
Definition 3.1.7. A tangle is a rational tangle if one can obtain a trivial tangle after
applying a finite number of consecutive twists to neighboring endpoints of the tangle, see
Figure 3.4a. To each rational tangle we associate a slope p/q. See [21] for Conway’s original
treatment on rational tangles, where he proves that two rational tangles are isotopic if and
only if their slopes are the same.

p1
q1

(a) A rational tangle.

p2
q2

pn−1
qn−1

pn
qn

(b) A Montesinos knot K(p1 /q1 , p1 /q2 , ..., pn /qn ).

Figure 3.4: Schematics for a Montesinos knot.
We denote a Montesinos knot by K(p1 /q1 , p2 /q2 , . . . , pn /qn ) which indicates that the link
is obtained by connecting n rational tangles of slope pi /qi . A pretzel knot is a Montesinos
knot of the form K(1/q1 , 1/q2 , . . . , 1/qn ). We will just write P (1/q1 , 1/q2 , . . . , 1/qn ) for
pretzel knots from now on.
The Hatcher-Oertel algorithm is obtained by patching together essential surfaces for each
rational tangle, which we discuss in the next section.

3.2

Essential surfaces in 2-bridge knot complements

Definition 3.2.1. A rational, or 2-bridge knot K(p/q), is a knot obtained by closing a single
rational tangle of slope p/q with two trivial arcs.
65

.... ,n~_l)
of n parallel
sheets running
the vertical
porS.(n~ ....S.(n~
,n~_l)
consistsconsists
of n parallel
sheets running
close toclose
the to
vertical
porn d 1of
1.....
bk], bifurcate
which bifurcate
n~ parallel
copies
tions oftions
each ofb aeach
n d ofb aE[-b
.....E[-b
bk],
which
into n~ into
parallel
copies of
the of the
b i n g square
n - n i parallel
outer plumbing
i'h inneri'h pinner
l u m b i npgl u m
square
and n - nandi parallel
copies copies
of the of
ith the
outerith plumbing
F o r example,
n = surfaces
1, the surfaces
( h i =1)0 are
or 1) are
square. square.
F o r example,
when nwhen
= 1, the
S l(n 1. . .S. l(n
, n k 1- .1.). . (, nhki =- 10) or
2 k- ~ plumbings
of the original
just thejust
2 k- the
~ plumbings
of the original
k bands.k bands.
the surfaces
,nk_different
0 for different
1.... ,bk]'s
In orderIn toorder
havetoallhave
the all
surfaces
S,(n~ ....S,(n~
,nk_....
0 for
Z'[b 1....Z'[b
,bk]'s
in a copy
singleofcopy
S 3 -weKp/q,
we choose
fixed position
forsay
Kp/q,
lying inlying
a single
S 3 - of
Kp/q,
choose
a fixed aposition
for Kp/q,
the say the

Incompressible surfaces in 2-bridge knot complements

A. Hatcher 1 and W. Thurston 2
Department of Mathematics, Cornell University, Ithaka, NY 14853, USA
Department of Mathematics, Princeton University, Princeton NJ 08544, USA

To each rational number
Kp/q shown in Fig. 1.

QI

p/q, with

q odd, there is associated the 2-bridge knot

bl

(a) The “two bridges.”

(b) 4-plat.Fig. 2

Fig. 1. The 2-bridge knot Kp/q

Fig.(c)
2

A
branched
surface
obtained by
plumbing.

In (a), the central grid consists of lines of slope +p/q, which one can
magine as being drawn on a square "pillowcase". In (b) this "pillowcase" is
punctured and flattened out onto a plane, making the two "bridges" more
vident. The knot drawn is K3/5, which happens to be
the figure
knot.
Figure
3.5: eight
Figure
from [56].
We assume q odd in order to get a knot rather than a two-component link.)
The double cover of S 3 branched along Kp/q is the lens space Lq,p. With this
observation, attributed in [16] to Seifert, the isotopy classification of 2-bridge
knots follows easily from
the classification
[14]well-studied.
of oriented lens
spaces:
K~/q
Two-bridge
knots are
They
have
a 4-plat representation, see Figure 3.5b,
=gp,/q, if and only if q'=q and p,_p+_l (modq). Basic references for 2-bridge Fig. 3 Fig. 3
knots are [2, 16, 17].
and they have been classified by Schubert [103] via the classification of lens spaces. In [56],
We shall derive in this paper the isotopy classification of the incompressible
urfaces, orientable or not, i n S3-Kp/q. As an application, we obtain some
and Thurston
presents
an algorithm
enumerate all boundary slopes of a 2-bridge
nformation aboutHatcher
the manifold
resulting from
Dehn surgery
o n Kp/q:to
Excludng the cases when Kp/q is a torus knot (Dehn surgery on torus knots was
completely analyzed
in [10]),
every
surgery
K m yields
an irreducible
knot
K(p/q)
byDehn
finding
all on
possible
continued
fraction expansions of p/q, each of which
manifold, and all but finitely many Dehn surgeries yield non-Haken (i.e., not
ufficiently large) corresponds
non-Seifert-fibered
manifolds. surface
The caseinofthe
the complement
figure eight of K(p/q).
to a branched

I I

Let p/q = r + [b1 , . . . , bk ] correspond to a continued fraction expansion of p/q.
1

p/q = r +
b1 −

.
1

b2 − · · ·

From each continued fraction expansion, we get a branched surface Σ[b1 , . . . , bk ] obtained
by plumbing k bands, each having bi twists (right-handed if bi > 0 and left-handed if

66

bi > 0), using both the inner and outer plumbing squares. See Figure 3.5c. We denote by
Sm (n1 , . . . , nk−1 ) the surface carried by Σ[b1 , . . . , bk ] consisting of m parallel sheets running
along vertical portions of each band, which split into ni parallel copies of the ith inner
plumbing square and n − ni parallel copies of the ith outer plumbing square.
Theorem 3.2.2. [56, Theorem 1(b)] An essential surface with boundary in S 3 K(p/q) is
isotopic to one of the surfaces Sm (n1 , . . . , nk−1 ) carried by Σ[b1 , . . . , bk ].
We describe part of the proof of the theorem, particularly the notion of edgepaths which
will be relevant for the description of the Hatcher-Oertel algorithm.
To prove Theorem 3.2.2, Hatcher and Oertel use the diagram in Figure 3.5a with the
height function regarded as a natural projection S 3 → R, so that the knot lies in S 2 × [0, 1].
˚2
The complement of the knot in each level sphere Sr2 = S 2 × {r} is a 4-punctured sphere S
r
for 0 < r < 1. Consider the quotient R2 /Γ, where Γ is the set of 180◦ rotations about the
integer lattice. This covering of S 2 by R2 branches at the four punctures. We may regard
(a) isotopy classes of smooth circles in Sr2 separating the four punctures into pairs, and
(b) isotopy classes of smooth arcs in Sr2 joining one given puncture to any of the other
three punctures
as lines in R2 , by considering its preimage in the covering. Thus we can assign the slope of
˚2 . Now K(p/q) ∩ S 2 with the diagram
the respective line in R2 to each circle and arc on S
r
r
that we have chosen consists of two arcs of slope 1/0 for r = 1 and two arcs of slope p/q for
r = 0.
The height function on an essential surface S may be regarded as a Morse function with
critical points at distinct r’s. Using the incompressibility and the ∂-incompressibility of S,

67

the intersection of S with each level sphere Sr2 consist of parallel copies of circles and arcs
of the same slopes joining a pairs of punctures, or two sets of arcs, each of a different slope,
joining different pairs of punctures. See Figure 3.6.
n
n

n

n

m
Figure 3.6: Intersection of S with Sr2 [56].
We represent a curve system of arcs of two different slopes a/b and c/d by
k a
m−k
+
m b
m

c
,
d

so that there are 2k arcs of S ∩ Sr2 of slope a/b and 2(k − m) arcs of slope c/d.
Definition 3.2.3. Let S be a connected and properly embedded surface in S 3 K. The
number of intersections of S with the meridian circle of N (K) is called the number of sheets
of S.
Since S is assumed to be properly embedded, at level r = 0, S ∩ Sr2 consists of arcs of
a single slope joining two punctures, and at level r = 1, S ∩ Sr2 are arcs of ∞ slope. Going
from r = 0 to r = 1, the surface S must pass through a sequence of saddles by which the
slopes of the arcs and circles changes to a different one. Suppose that a saddle changes arcs
of slope a/b to slope c/d. Through a linear change of coordinate, these arcs may be brought
to arcs of slope 0/1 and 1/0. Thus, ad − bc = 1.
Consider the Poincare disc model of the hyperbolic plane with rational points marked
on the boundary. We define a 1-simplex C on the hyperbolic plane by letting the vertices
68

be the rational points on the boundary. Additionally, there is an edge between fractions a/b
and c/d if ad − bc = 1. See Figure 3.7.

2/1

5/3

3/2

4/3 1/1 3/4

2/3

3/5
1/2

5/2

2/5

3/1

1/3

4/1

1/4

1/0

0/1

-4/1

-1/4
-1/3

-3/1

-2/5

-5/2
-2/1
-5/3

-2/3
-3/2 -4/3
-1/1 -3/4

-1/2
-3/5

Figure 3.7: The 1-simplex on the Poincare disc model [56].

Definition 3.2.4. An edgepath of C is a path on the 1-simplex C.
We represent the sequence of saddles of S by an edgepath, where an edge corresponds
to a saddle passing between two slopes. For S to be essential, the edgepath needs to satisfy
the following:
(E1) no three successive vertices of the edgepath lie on two different edges of the same
triangle of the diagram.
(E2) pi /qi = pi+2 /qi+2 for each i. In other words, no backtracking is allowed.
Given any continued fraction expansion p/q = r + [b1 , . . . , bk ], we have that the sequence
69

of partial sums pi /qi = r + [b1 , . . . , bi ], 1 ≤ r ≤ k satisfies

pi qi+1 − qi pi+1 = ±1.

Conversely, suppose that we have a sequence of fractions {pi /qi } satisfying the equation
for pi /qi and pi+1 /qi+1 , the conditions (E1) and (E2), and such that pk /qk = p/q and
p1 /q1 = {−1, 0, 1}. Then, we can construct a partial fraction decomposition of p/q. Thus
an essential surface for K(p/q) corresponds to a continued fraction expansion of p/q.
To find the branched surface carrying S, take the continued fraction expansion r +
[b1 , . . . , bk ] of p/q given by the edgepath of S. Isotope the branched surface Σ[b1 , . . . , bk ] so
that its boundary is the diagram 3.5a of K(p/q). It is clear that S is carried by Σ[b1 , . . . , bk ].

3.3

Curve systems and boundary slopes for Montesinos
knots

Viewing S 3 as the join of two circles A and B, let the circle B be subdivided as an n-sided
polygon. The join of A with the ith edge of B is then a ball Bi . We choose Bi so that each
of them contains a tangle of slope pi /qi . These n balls Bi cover S 3 , meeting each other only
in their boundary spheres. The union of the tangles in each ball with the trivial tangle gives
K(p1 /q1 , p2 /q2 , . . . , pn /qn ).
Within each ball Bi , we can still view the tangle as lying in S 2 × [0, 1], as before. The
intersection of an essential surface S of K(p1 /q1 , p2 /q2 , . . . , pn /qn ) with a level sphere Sr2
will determine an edgepath for each fraction pi /qi satisfying the same criteria E(1), E(2).
Since we can determine the essential surface for each tangle, we can reverse the process and

70

look for conditions that will allow them to fit together to make an essential surface for the
Montesinos knot. The key difference between the case for 2-bridge knots and for Montesinos
knots now is that at r = 1, the surface S determines a curve system on the 4-punctured
S12 which does not necessarily consist solely of slope 1/0 arcs. With certain conditions, the
curve system of each surface may be identified with each other to give a connected, essential
surface.
Definition 3.3.1. On a compact surface (with or without boundary), a curve system is a
finite collection of disjoint, embedded curves which are either
• circles not bounding disks in the surface and not isotopic to boundary components of
the surface.
• arcs with endpoints on the boundary components of the surface and not isotopic rel
endpoints to arcs in the boundary.
We will consider the curves a, b, and c on the 4-punctured sphere, where a is an arc
joining two punctures, b is a circle containing the two punctures joined by a, and c is a circle
containing one of the puctures contained by b and one of the other punctures.
c
a

a

c

b

c

a
b

c

Figure 3.8: The generators a, b and c and the corresponding set of disjoint curves on the
4-punctured sphere with a, b, c-coordinates (3, 1, 2).

71

˚2 denotes the 4-punctured S 2 . We consider the subgroup of H1 (S
˚2 ) generRecall that S
ated by a, b, and c, quotiented out by multiples of the coefficients of a, b, and c. This is a
˚2 .
subset of the projective lamination space on S
As discussed in the previous section, a curve system determined by an essential surface on
˚2 can be uniquely expressed in the form na a + nb b + nc c, where na are the coefficients of the
S
˚2 ). We will write (na , nb , nc ) to indicate the curve system. We have the
curve system in H1 (S
planar three-simplex shown below where each point represents, in barycentric coordinate, a
curve system (na , nb , nc ), where nc > 0.
c
3/1

3/1

2/1

2/1
1/1

1/1

1/2
0/1
a

0/1
b

Figure 3.9: The three-simplex of curve systems on a four-punctured sphere [55].
˚2 in R2 Γ, where Γ is the integer
Recall that by considering the preimage of a curve on S
lattice in the plane, we may assign a slope to the curve. Choose coordinates on R2 so that
a and b have slop 0/1 and c has slope 1/0. For a curve system represented by (na , nb , nc ),
the slope is given by nc /(na + nb ), which we mark next to each vertex. There is also a
mirror image of the simplex with c < 0 carrying curve systems of negative slopes. Joining
the left-hand edges of both of those 2-simplices to the point representing the slope 1/0 arcs
joining the other pairs of punctures gives a planar four-simplex.
Now we see the familiar figure of Figure 3.7 in the form of a planar four-simplex. In
72

1/1
3/2
2/1
3/1

2/3
1/2
1/3

1/0

0/1

−3/2

−1/3
−1/2
−2/3

−2/1
−3/1

Figure 3.10: The four-simplex of curve systems with systems of slope 1/0 added [55].
addition to edges between slopes p/q and r/s where ps − rq = ±1, an edge between two
points with the same slope p/q is allowed. An edgepath of Figure 3.7 is also an edgepath on
this new diagram.
For each ball Bi containing the tangle of slope pi /qi , the axis A is a slope 1/0 circle, and
the circle which is the joint of two points of B ∩ ∂Bi with two antipodal points of A is a
slope 0 circle. In the complement of K(p1 /q1 , p2 /q2 , . . . , pn /qn ), the right hemisphere of Bi
as divided by the axis is identified with the left hemisphere of Bi+1 , forming a new curve
system on the boundary sphere of the union. Suppose we have the curve system (na , nb , nc )
on ∂Bi and the curve system (na , nb , nc ) on ∂Bi+1 . In view of homology, the necessary
condition to add them is that na = na , nb = nb . The new curve system will have the same
coefficients for a and b while the coefficients for c are added.
Now we slit the 1-simplex C open along the slope 1/0 edge, see Fig 3.11. The 1-simplex
now forms an infinite strip D on the plane. Each point (na , nb , nc ) now has horizontal
coordinate nb /(na + nb ) and vertical coordinate nc /(na + nb ). An edgepath from the point

73

1/1

2/1

3/4
2/3
3/5
1/2

1/1
⇒
1/0

1/3
1/6
1/5

0/1

−1/1

1/1

0/1

0/1

Figure 3.11: On the left: the infinite strip containing the 1-dimensional complex D. On the
right: the portion of the infinite strip from height 0/1 to 1/1 [55].
of slope p/q to the point of slope r/s corresponding to a saddle within a ball Bi from curve
systems of slope p/q to the system of slope r/s is now represented as

p/q − r/s .

We will decorate the point corresponding to the curve system (0, q, p) with slope p/q with
the symbol ◦ to distinguish it from the curve system (1, q − 1, p) with the same slope.
For an edge p/q − r/s , let k/m p/q + (m − k)/m r/s denote the curve system
consisting of k parallel copies of the pair of arcs of slope p/q joining the four punctures
together with m − k parallel copies of the pair of arcs of slope r/s. This is a point on
p/q − r/s . In terms of a, b, and c, the curve system has coordinates

(m, k(q − 1) + (m − k)(s − 1), kp + (m − k)r).

As before, we isotope an essential surface so that its height function is a Morse function.
74

The sequence of saddles in each ball Bi correspond to an edgepath on the complex D.
Therefore, we may construct a candidate surface by considering the choices of edgepath for
the tangle in each ball, and ask whether the surfaces associated to the edgepath glue together
to form a connected, properly embedded, and essential surface. The endpoint of an edgepath
will correspond to a curve system on the 4-punctured sphere which we glue with the curve
system on the other 4-punctured sphere.
In order for the pieces of these surfaces to form a connected surface after identifying Bi ’s,
we need to have na = na , nb = nb as before. This means that the horizontal coordinates of
the endpoints of each edgepath in D need to agree, and the vertical coordinates need to sum
to zero.
We can now describe the Hatcher-Oertel algorithm for enumerating all candidate surfaces
for boundary slopes.

3.4

The Hatcher-Oertel algorithm

For each fraction pi /qi , an edgepath γi is a piecewise linear path in the 1-skeleton of D,
starting and ending at points which may not be vertices of D. It satisfies the following
conditions:
(E1) The starting point of γi lies on the edge pi /qi − pi /qi◦ . If this starting point is not
a vertex, then γi is constant.
(E2) γi never stops and retraces itself or go along two sides of the same triangle in D in
succession. This is the same local condition coming from incompressibility and ∂incompressibility as in Section 3.2 for 2-bridge knots.

75

(E3) γi proceeds monotonically from right to left where motion along vertical edges are
permitted.
We are looking for an edgepath system {γ1 , . . . , γn } satisfying the additional properties
from the conditions on identifying curve systems.
(E4) The endpoints of all the γi ’s are rational points of D which have the same horizontal
coordinates, and whose vertical coordinates add up to zero.
We associate to an edgepath system {γ1 , . . . , γn } a candidate surface as follows. Viewing
each tangle pi /qi as lying in S 2 × [0, 1] in Bi as before. There are two cases for γi .
• If γi is constant: We associate the surface Si meeting each level Sr2 in the curve system
with m arcs of slope p/q coming into each puncture. The circles in the intersection
Si ∩ S02 are capped off with disks.
• If γi is not constant: The surface is associated just as before in the case for 2-bridge
knot.
If an edgepath ends at the point with slope 1/0, then all the other edgepaths in the system
also have to end at the same point.
To enumerate all the edgepath systems given a Montesinos knot K(p1 /qi , . . . , pn /qn ),
one finds all possible edgepaths.
1. Each γi either correspond to a continued fraction expansion of pi /qi or has a constant
edgepath p/q − p/q ◦ added at the beginning.
2. Given a choice of edgepath system {γ1 , . . . , γn }, solve for endpoints satisfying condition
(E4).
76

Each of the edgepath systems with the appropriate endpoints then give a candidate
egdepath system. Depending on the fractions pi /qi , some edgepath systems will not have
endpoints satisfying (E4), and they will be ruled out.
Hatcher and Oertel shows that the algorithm gives a complete list of candidate surfaces
for boundary slopes.
Theorem 3.4.1. [55, Proposition 1.1] Every essential surface in S 3 − K(p1 /q1 , . . . , pn /qn )
having nonempty boundary of finite slope is isotopic to one of the candidate surfaces.
Just as in the case for 2-bridge knots, we obtain a branched surface in the complement
of the Montesinos knot by gluing branch surfaces in the complement of each tangle. Every
essential surface is carried by a branch surface constructed in this way.
Thus the surfaces of the edgepaths systems are similar to that of the case for two-bridge
knots, except that a constant edge is allowed at the beginning of an edgepath. The endpoint
of an edgepath in a candidate edgepath system is also not required to end at a curve system
of slope 1/0.
To check that a given candidate surface is essential, Hatcher and Oertel used a technical
idea of the r-values of the edgepath system, which we will not describe in more detail in this
dissertation. The idea is to examine the intersection of a compressing or ∂-compressing disk
with the boundary sphere of each ball Bi , which will determine an r-value for each edgepath
γi . If a candidate surface does not have a list of r-values from its edgepath system that
would result from the existence of a compressing disk, then it is incompressible.
We state the criterion in terms of quantities thare are easily computed given an edgepath
system.
Definition 3.4.2. The r-value for an edge p/q - r/s where p/q = r/s is s−q. If p/q = r/s
77

or the path is vertical, then the r-value is 0.
Then the r-value for an edgepath γi is just the r-value of the final edge of γ.
Theorem 3.4.3. [55, Corollary 2.4] A candidate surface is incompressible unless the cycle
of r-values of {γi } is one of the folloing types:
• (0, r2 , . . . , rn ), or
• (1, . . . , 1, rn ).
Dunfield has implemented the algorithm completely in a program [27] that will determine
the list of boundary slopes given any Montesinos’ knot.

3.4.1

Computing the boundary slope from an edgepath system

Given an edgepath system {γi } corresponding to an essential candidate surface, we describe
how to compute its boundary slope. Within a ball Bi , all surfaces look alike near S02 , since it
must have slope p/q at the bottom. The number of times the boundary of the surface winds
around the longitude is given by m, the number of sheets of the surface. Each time the
surface passes through a slope-changing saddle, the boundary twists around the meridian
once. Thus the total number of twists for a candidate surface S in arcs away from the level
r = 0 is
τ (s) = 2(s− − s+ ),
where s− is the number of slope-decreasing saddles and s+ is the number of slope-increasing
saddles. In terms of egdepaths, we have

τ (S) = 2(e− − e+ ),
78

where e− is the number of edges of γi that decreases slope, and e+ is the number of edges
that increases slope. If γi ends at a point on the segment
k
m

p
q

+

m−k
m

r
,
s

then the final edge is counted as a fraction k/m. We add back the twists happening around
the knot at level r = 0 by subtracting the twist number of a Seifert surface S0 obtained from
the algorithm, since the Seifert surface will always have zero boundary slope. The boundary
slope of a candidate surface S is
τ (S) − τ (S0 ).

3.4.2

The 3-strand pretzel knots P (−1/r, 1/s, 1/t).

For Theorem 1.1.2, we shall restrict the Hatcher-Oertel algorithm to 3-string pretzel knots
P (−1/r, ±1/s, ±1/t), where r is even and s, t are odd. We restrict the algorithm to this case
due to the relative ease in computing the corresponding Jones slope, which we will describe
in the next section.
For each fraction of the form 1/p, there are two choices of edgepaths that satisfy conditions
(E1)-(E3). They correspond to the two continued fraction expansions of 1/p:
1
= 0 + [p] gives edgepath
p

1
p

− 0,

79

and
1
= ±1 + [±2, ±2, . . . , ±2] gives edgepath
p
p − 1 times

±1
p

−

±1
p−1

− · · · − ±1 .

where it is a plus or minus sign for the slope of each vertex for the second type of continued fraction expansion depending on whether p is positive or negative, respectively. These
edgepaths are shown on the diagram D.
1/1
3/4◦
1/2◦
1/4◦
1/8◦
0/1
Figure 3.12: A choice of edgepath for each of the three fractions 1/2, 1/8, and 3/4 are shown
in red, blue, and green.
Thus for the 3-strings pretzel knots there are 23 choices of edgepath systems, which
we may augment with constant edgepaths at the beginning. Finding a candidate edgepath
system amounts to solving the equations for the horizontal and vertical coordinates imposed
by condition (E4).
We consider the following edgepath system for P (−1/r, 1/s, 1/t) when r > s, t > 0.
• For 1/r:
−1
r

−

−1
r−1

80

− · · · − −1 .

• For 1/s:
1
s

− 0 .

1
t

− 0 .

• For 1/t: 0 − 1/t

Condition (E3) requires that we set the horizontal coordinates of points on the path of form
−1/(q + 1) − −1/q , and 1/s − 0 , and 1/t − 0 equal to each other: That is, we
have the following equation.
m(q − 1) + k
k (s − 1)
k (t − 1)
=
=
.
mq + k
m + k (s − 1)
m + k (t − 1)
Recall that for the curve system represented by each point, the number k represents the
number of arcs coming into each puncture with one slope and m − k represents the number
of arcs coming into each punctre of a different slope. The number of sheets m will be the
same. Dividing the top and the bottom by m, we get
k /m(s − 1)
k /m(t − 1)
(q − 1) + k/m
=
=
.
q + k/m
1 + k /m(s − 1)
1 + k /m(t − 1)

(3.1)

We simplify the notation by setting A = k/m, B = k /m, and C = k /m.
q−1+A
B(s − 1)
C(t − 1)
=
=
.
q+A
1 + B(s − 1)
1 + C(t − 1)
1
1
1
=
=
.
q+A
1 + B(s − 1)
1 + C(t − 1)
q + A = 1 + B(s − 1) = 1 + C(t − 1).

81

(3.2)

The sum of the vertical coordinates are equal to zero, which means we have
k
k
−m
+
+
= 0.
mq + k m + k (s − 1) m + k (t − 1)

(3.3)

Since all the denominators are the same, we also get

−m + k + k = 0,

and

−1 + B + C = 0.

(3.4)

By equation (3.2),

B=

C(t − 1)
.
s−1

(3.5)

We can then solve for c in terms of given quantities s, t using equation (3.4), obtaining

C=

s−1
.
t+s−2

(3.6)

82

Now the boundary slope is given by τ (S) − τ (S0 ), where S0 is a candidate surface that is
a Seifert surface. We can describe the edgepath system for S0 .
• For −1/r: −1/r − 0 .
• For 1/s: 1/s − 1/(s − 1) − · · · − 1/2 − 1 .
• For 1/t: 1 − 1/2 − · · · 1/(t − 1) − 1/t .
corresponding to the edgepath system with the number of sheets m = 1, and each γi turning
across an even number of triangles at each vertex in order for S0 to be orientable. For
a general algorithm for determining a Seifert surface which is a candidate surface, see the
discussion in [55], where the formula for the boundary slope is given. So τ (S0 ) = −2(−1 +
s + t). To compute τ (S), note that the edges are all decreasing. We add up A, B, C and the
number of paths for τ (S). We have,

τ (S) = 2(r − q − A + 1 − B + 1 − C) = 2(r − q − A + 1).

Given that q + A = 1 + C(t − 1), we get

τ (S) = 2(r − (1 + C(t − 1)) + 1),

so then

τ (S) − τ (S0 ) = 2 r + t +

(−1 + s)2
−2 + s + t

.

(3.7)

The cycle of r-values is (1, s−1, t−1). Therefore by Theorem 3.4.3, the surface is essential
83

as long as s > 2.

3.5

Computation of the Jones Slope

We describe the computation for the Jones slope done by Roland van der Veen, which is
relatively simple to determine for the class of 3-string pretzel knots P (−1/r, 1/s, 1/t) where
r > s, t and r, s, t > 5. We assume that r is even and s, t are odd as in the premise of Theorem
1.1.2. Recall from Section 1.2 that we may obtain the colored Jones polynomial by evaluating
the Kauffman bracket on a link diagram decorated with the Jones-Wenzl idempotent. We
list here a few important formulas for the graphical calculus with the idempotent. See [78]
and [82] for more details and proofs.
First we represent a special skein element, shown in Figure 3.13, by a trivalent graph on
the right.
a

b

a

b

z
x

y

c

c

Figure 3.13: A trivalent graph representation [78].
The next few figures give formulas for replacing certain skein elements with a sum and
the 6j-symbol. For k an integer, recall that we have

k =

(−1)k+1 (A2(k+2) − A−2(k+2))
.
A2 − A−2

This is the skein element in S(R2 ) obtained by closing up fk with k parallel strands. We
84

also define
x+y+z !

θ(α, β, γ) =
where

k!

=

k

k−1 · · ·

1

y+z−1 !

x−1 !

y−1 !

z+x−1 !

z−1 !

x+y−1 !

,

and is interpreted as 1 if k is 0.

We define the 6j-symbol, see [78] and [82] for references. Let
a+b+e a+c+f b+d+f c+d+e
,
,
,
2
2
2
2

zmin := max

a+d+b+c a+d+e+f b+c+e+f
,
,
2
2
2

zmax := min




 a b c 


 d e f 


zmin ≤z≤zmax

!
z− a+b+e
2

z+1 !

a+c+f !
z− 2

b+d+f !
z− 2

1
!
b+c+e+f
−z
2

!
a+d+e+f
−z
2

b+c+a+d −z !
2

=

c
c:(a,b,c) θ(a,b,c)
admissible

c

b

Figure 3.14: The fusion formula [78].
a
b

= (−1)

a+b−c a+b−c+ a2 +b2 −c2
2 A
2

Figure 3.15: The untwisting formula [78].
85

!
z− c+d+e
2

.

a

a
b

c

.

:=

(−1)z

·

, and

a
b

·

b

c

b

a

b
c

d

a

=

i

c
b

a b ⇢
i
P
c d j a b i
=
i
c d j

d

c

i

a

i
d
a

d

Figure 3.16: The 6j-equation [78].
Figure 3.16: The 6j-equation [78].
c

a

d

a

ad ✓(a,b,c)
4a

=

b
Figure 3.17: Replacing a region bounded by two edges.
is the Kronecker delta function
Figure 3.17: Replacing a region bounded by two edges. δad is the ad
Kronecker delta function
[78].
[78].
Since fn2 = fn , we may insert as many projectors for a link component. We may arrange
Since fn2 = fn , we
may insert as many projectors for a link component. We may arrange
them so that for twist region, there is a Jones-Wenzl idempotent on each outgoing strands.
them so that for twist region, there is a Jones-Wenzl idempotent on each outgoing strands.
The e↵ect of replacing a positive crossing with a negative crossing is to replace A by A 1 .
The effect of replacing a positive crossing with a negative crossing is to replace A by A−1 .
Let d(↵, n) = 2n ↵ + n2 ↵2 and d( ) and d( ) be defined similarly. Using the fusion and
Let d(α, n) = 2n − α + n2 − α2 and d(β)
and d(γ) be defined similarly. Using the fusion and

the untwisting formulas, the n-th colored Jones polynomial before substituting q 1/2 = A 1/2
the untwisting formulas, the n-th colored Jones polynomial before substituting q 1/2 = A−1/2
is then a sum:
is then a sum:
✓

n2 +2n
(2 +2n
1)n A−ω(D)
n
n
(−1) A

◆ !(D)

X

4
4
4↵
·
β
γ
· n, )
✓(n, n, ↵) ✓(n, n, ) ✓(n,
θ(n, n, α) θ(n, n, β) θ(n, n, γ)
↵, , admissible
α

α,β,γ admissible

*
⇣
⌘ ⇣
⌘ ⇣
⌘
↵ +s n
r
n
+t
n
2 A rd(↵,n)+sd( ,n)+td( ,n)
· ( α2 1)+s n−2β2 +t n− 2γ2 −rd(α,n)+sd(β,n)+td(γ,n)
r n−

· (−1)

α

A

86

86

β

+
↵

γ

.

.
(3.8)

(3.8)

A
B
F

E
D

=

A B E
D C F

A

1
· θ(A,F,C)

C

F

C

Figure 3.18: Collapsing a triangular region to a vertex [82].
The sum is taken over positive integers α, β, and γ which are admissible. This means that
each of them satisfy the admissibility conditions with respect to n. We remark that this is
a restriction of the more general admissibility condition defined by Lickorish in [78] shown
below

α, β, and γ are even,
0 ≤ α, β, and γ ≤ 2n,

and

|α − β| ≤ γ ≤ α + β.

(3.9)

87

For evaluating the colored Jones polynomial of general links, we may use the 6j-equation
to reduce the number of edges around a complementary region of the trivalent graph. If a
region is bounded by just two edges, we use the identity of Figure 3.17. There is a simple
tetrahedral shape of the resulting trivalent graph for the case that we consider for the 3-string
pretzel knot. We use the identity of Figure 3.18 to change the previous sum to

2
(−1)n An +2n

−ω(D)
α,β,γ admissible

β
γ
r n− α
2 +s n− 2 +t n− 2

· (−1)

·

β

α

γ

θ(n, n, α) θ(n, n, β) θ(n, n, γ)

·

A−rd(α,n)+sd(β,n)+td(γ,n) ·


2

 α n n 


α
1
2

 n β γ 
 θ(α, β, γ)

γ

β

(3.10)

.

n before
Finally, using the symmetry relation of the 6j-symbol [18, 82], we have that JK

subsituting q is

2
(−1)n An +2n

ω(D)

α

α,β,γ admissible

β

γ

θ(n, n, α) θ(n, n, β) θ(n, n, γ)

2
2 2 2
· (−1)6n−(α+β+γ)/2 A6n+3n −(α +β +γ )/2

·


2

 n n n 


1
.

 α β γ 
 θ(α, β, γ)

n . We maximizes MD
Let M D(α, β, γ) be the degree of a single term in the sum for JK

over the 3-dimensional polytope defined by admissible parameters α, β, and γ. The solutions
will be integer lattice points of the polytope, see Figure 3.19.

88

1.0
0.5
0.0
1.0

0.0
0.0
0.5
1.0
Figure 3.19: The polytope cut out by equations α + β = γ, α + γ = β, β + γ = α. It has
been rescaled by dividing by n.
In the event that there are more than one set of parameters which maximizes M D,
cancellations are possible. However, the restrictions placed on r, s, and t from Theorem
1.1.2 guarantees that there is a single maximum degree term.
The formula for the 6j-symbol is complicated, however, analyzing M D with certain
restrictions on α, β, and γ enables us to write down explicit formulas for the restrictions.
We rescale α, β, and γ by replacing them with nα, nβ, and nγ.





(−4 + 2r)α2 − 2sβ 2 − 2tγ 2 + 4αβ + 4αγ − r + s + t for α ≥ β, γ.





M D(α, β, γ) = (−4 − 2s)β 2 + 2rα2 − 2tγ 2 + 4αβ + 4βγ − 4αγ − r + s + t for β ≥ α, γ.








(−4 − 2t)γ 2 − 2sβ 2 − 2tα2 + 4γα + 4βγ − 4αβ − r + s + t for γ ≥ α, β.
We call the restrictions of M to these three domains M α, M β and M γ. We see that

89

∂M α
∂α

>0

∂M α
∂β

<0

∂M β
∂α

>0

∂M β
∂β

<0

∂M γ
∂α

>0

∂M γ
∂β

<0

∂M α
∂γ
∂M β
∂c

< 0,

< 0, and

∂M γ
∂γ

< 0.

Therefore, the maximum must lie in the triangle α = β + γ, β + γ ≤ 1. Thus we need only
to find the integer lattice maximum of M α(b + c, b, c) over the triangle. The determinant of
the Hessian of M α is H(b, c) = st − r(−2 + s + t) − 1.
If H < 0, then the maximum is attained at one of the edges of the triangle, so the
problem reduces to maximizing function over the edges of the triangle. The critical point on
−1+t and the slope is
the diagonal is b0 = −2+s+t

2 − 2s + s2 − 2t + t2 + r(−2 + s + t)
+ W (r, s, t),
−2 + s + t
where W (r, s, t) is the writhe of the pretzel knot P (−1/r, 1/s, 1/t).

3.6

Proof of Theorem 1.1.2.

Proof. From the previous section, the Jones slope is
2 − 2s + s2 − 2t + t2 + r(−2 + s + t)
+ r + s + t,
−2 + s + t
after ignoring all terms of order less than n2 and multiplying by 4. The slope of an essential
surface in the complement of P (−1/r, 1/s, 1/t) is equal to
(−1 + s)2
2 r+t+
−2 + s + t

90

,

as we have seen in Section 3.4. These are the same, and we are done.

91

Chapter 4
Crosscap number of alternating links
and the Jones polynomial
Clark [20] showed that the crosscap number of any knot K, is bounded above in terms of the
orientable genus of K, and Murakami and Yasuhara [89] showed that it is bounded above
in terms of the crossing number of K. For an alternating link K, C(K) > 1, unless K is
a (2, p) torus link. A lower bound for the 4-dimensional crosscap number of all knots, and
thus for the 3-dimensional crosscap number, was given by Batson [13]. For alternating knots,
however, this bound is non-positive.
We restate Theorem 1.1.3, which is the main result of this chapter.
Let
JK (t) = αK tr + βK tr−1 + . . . + βK ts+1 + αK ts
denote the Jones polynomial of K, so that r and s denote the highest and lowest power in
t. Set
TK := |βK | + βK ,
where βK and βK denote the second and penultimate coefficients of JK (t), respectively.
Also let sK = r − s, denote the degree span of JK (t).
Theorem 1.1.3. Let K be a non-split, prime alternating link with k-components and with

92

crosscap number C(K). Suppose that K is not a (2, p) torus link. We have
TK
3

+ 2 − k ≤ C(K) ≤ TK + 2 − k,

where TK is as above, and · is the ceiling function that rounds up to the nearest integer.
Furthermore, both bounds are sharp.
Combining Theorem 1.1.3 with the results of [89] and [72], we have the following.
Theorem 4.0.1. Let K be an alternating, non-torus knot with crosscap number C(K) and
let TK be as above. We have
TK
3

+ 1 ≤ C(K) ≤ min TK + 1,

sK
2

where sK denotes the degree span of JK (t), and · is the floor function that rounds up to
the nearest integer. Furthermore, both bounds are sharp.
By a result of Menasco [83] a link with a connected, prime alternating diagram that is
not the standard diagram of a (2, p) torus link is non-split, prime and non-torus link. Hence
the hypotheses of Theorem 1.1.3 are easily checked from alternating diagrams.
Crowell [22] and Murasugi [90] have independently shown that the orientable genus of an
alternating knot is equal to half the degree span of the Alexander polynomial of the knot.
Theorem 4.0.1 can be thought of as the non-orientable analogue of this classical result.
The oriented link genus has been well studied, and a general algorithm for calculation
is known [5, 51]. For knots with low crossing number, effective computations can also be
made from genus bounds coming from invariants such as the Alexander polynomial and
the Heegaard Floer homology [19]. Crosscap numbers, however, are harder to compute.
93

Although the crosscap numbers of several special families of knots are known, see [61,64,107],
no effective general method of calculation is known. Some progress in this direction was made
by Burton [17] using normal surface theory and integer programming. Despite that, for the
majority of prime knots up to twelve crossings the crosscap numbers are listed as unknown
in KnotInfo [19].
For alternating links, a method to compute crosscap numbers was given by Adams and
Kindred [1]. They showed that to compute the crosscap number of an alternating link it
is enough to find the minimal crosscap number realized by state surfaces corresponding to
alternating link diagrams.
Note that for a link with n crossings, there correspond 2n state surfaces that, a priori one
has to search and select one with minimal crosscap number. The Adams-Kindred algorithm
of [1] cuts down significantly this number, but the number of surfaces that one has to deal
with still grows as the number of the “non-bigon” regions of the alternating link does.
The advantage of Theorem 1.1.3 is that it provides estimates that are easy to calculate
from any alternating diagram and, as mentioned above, these estimates often compute the
exact crosscap number. Indeed, the quantity TK is particularly easy to calculate from any
alternating knot diagram as each of βK , βK can be calculated from the checkerboard graphs
of the diagram [24]. Our lower bound improves the lower bound given in KnotInfo [19] for
1472 prime alternating knots for which the crosscap number is listed as unknown and for 262
of these knots our bounds determine the exact value of the crosscap number. See Section
4.4 for some examples of these knots.
In Section 4.1 we define augmented links and state definitions and results from [37, 75]
that are necessary to this chapter. In Section 4.2, first we define state surfaces and we recall
the results of [1] that we use. Then we prove Theorems 1.1.3 and 4.0.1 in Section 4.3. In
94

Section 4.4, we calculate the crosscap numbers of infinite families of alternating knots for
which the lower bound is sharp as well as for several knots up to 12 crossings. Finally, in
Section 4.5, we discuss generalizations of our results outside the class of alternating links.
The overall technique used to prove Theorem 1.1.3 goes beyond the class of alternating links
and allow us to generalize Theorem 1.1.3 for large classes of non-alternating links. We also
state some questions that arise from this work.

4.1

Augmented links and estimates with normal surfaces

Consider a link diagram D(K) as a 4–valent graph in the plane, with over–under crossing
information associated to each vertex. A bigon region is a region of the graph bounded by
only two edges. A twist region of a diagram consists of maximal collections of bigon regions
arranged end to end. We will assume that the crossings in each twist region occur in an
alternating fashion. A single crossing adjacent to no bigons is also a twist region.
A or B

A

B

Figure 4.1: Twist reduced: A or B must be a string of bigons [37].

Definition 4.1.1. A link diagram D(K) is prime if any simple closed curve which meets
95

two edges of the diagram transversely bounds a region of the diagram with no crossings.
The diagram D(K) is called twist reduced, if any simple closed curve that meets the
diagram transversely in four edges, with two points of intersection adjacent to one crossing
and the other two adjacent to another crossing, bounds a (possibly empty) collection of
bigons arranged end to end between the crossings. See Figure 4.1, borrowed from [32].

4.1.1

An angled polyhedral decomposition.

For a link K ⊂ S 3 , let E(K) denote the exterior of K; that is E(K) = S 3 η(K).
Let D(K) be a prime, twist–reduced diagram of a link K. For every twist region of D(K),
we add an extra link component, called a crossing circle, that wraps around the two strands
of the twist region. A choice of crossing links for each twist region of D(K) gives a new link
J, called an augmented link. The link L obtained by removing all full twists from the twist
regions of J is called a fully augmented link. The exteriors E(J), E(L) are homeomorphic
3-manifolds and E(K) can be expressed as a Dehn filling of E(L) ∼
= E(J). See Figure 4.2,
borrowed from [32].

Figure 4.2: From left to right: A link K, an augmented link J and a fully augmented link
L [32].
For twist regions consisting of a single crossing the addition of a crossing circle can be
done in two ways. See Figure 4.3.
The geometry of augmented links is well understood. Below, we will summarize some
96

Figure 2. A link K, an augmented link J and a fully augmented link L.

Figure 3. Two ways of augmenting a twist region with a single crossing.

Figure 4.3: Two ways of augmenting a twist region with a single crossing.
the polyhedral decomposition and combinatorial length of simple closed curves that lie
resultsto and
properties we need; for more details see Purcell’s expository article [98] and
the boundary @E(L). The general setting was discussed by Lackenby in [28] and applied
by Futer and Purcell [17] to study the geometry of augmented highly twisted links.
references
therein. We will consider the non-crossing circle components of L flat on the proStart with a prime, twist reduced diagram of a link K and let L be a fully augmented
link obtained from it. The manifold E(L) can be subdivided into two identical, convex
jection
plane
and theP1crossing
bound crossing
disks
vertical
to the projection
plane.
ideal
polyhedra
and P2 circles
in the hyperbolic
3-space.
The
ideal vertices
of each polyhedron P 2 {P1 , P2 } are 4-valent and they correspond to crossing circles or to arcs of
As explained
by Agol
and
Thurston
thecrossing
appendix
of [76],
E(L)
has an angled
KrD, where
D is
theD.union
of allinthe
disks.
After
truncating
all thepolyhedral
ideal
vertices of P we have:

decomposition with nice geometric and combinatorial properties. We are particularly inter-

• Dihedral angles at each edge of P are ⇡/2. In particular E(L) is hyperbolic.
• There are two types of edges of @P : these that are created by truncation called
ested in the combinatorial
decomposition:
define
combinatorial
boundary edges structure
(edges of Pof\this
@E(L))
and the ones one
that can
come
from the
edges
existing
before truncation, called interior edges. The interior edges come from intersecarea of surfaces
that are
in normal
with plane.
respect to the polyhedral decompotionsinofE(L)
the crossing
discs
with theform
projection
• There are two types of faces on @P : these that are created by truncation called
sition and combinatorial
closedand
curves
thatthat
lie come
the boundary
boundary faceslength
(faces of
of simple
P \ @E(L)),
the ones
from faces∂E(L).
existing The
before truncation, called interior faces.
• Each
boundary
facebyisLackenby
a rectangleinthat
four interior
edgesand
at the
vertices
general setting
was
discussed
[75]meets
and applied
by Futer
Purcell
[37] to
of the rectangle. The boundary faces of P1 , P2 subdivide @E(J) into rectangles.
• The interior
faces of P highly
can be twisted
colored with
study the geometry
of augmented
links.two colors (shaded and white) in a
checkerboard fashion so that at each rectangular boundary face of P opposite
side
interior twist
faces have
the same
color.
shaded
faces
to crossing
Start with
a prime,
reduced
diagram
of The
a link
K and
letcorrespond
L be a fully
augmented
disks while the white faces correspond to regions of the projection plane. See
Figure
andmanifold
Figure 12.
link obtained[29,
from
it. 15]
The
E(L) can be subdivided into two identical, convex
The complement of all faces (interior and boundary) on the projection plane is a graph

⇢ @P with
at least three
and edges
edges
of P .of each polyhedron
ideal polyhedra
P1vertices
and P2ofinvalence
the hyperbolic
3-space.
The the
ideal
vertices
A polyhedral P with the above properties is called a rectangular-cusped polyhedron.

P ∈ {P1 , P2 } are 4-valent and they correspond to crossing circles or to arcs of K D, where
D is the union of all the crossing disks. After truncating all the ideal vertices of P we have:
• Dihedral angles at each edge of P are π/2. In particular E(L) is hyperbolic.
• There are two types of edges of ∂P : these that are created by truncation called boundary edges (edges of P ∩ ∂E(L)) and the ones that come from edges existing before
truncation, called interior edges. The interior edges come from intersections of the
97

crossing discs with the projection plane.
• There are two types of faces on ∂P : these that are created by truncation called boundary faces (faces of P ∩ ∂E(L)), and the ones that come from faces existing before
truncation, called interior faces.
• Each boundary face is a rectangle that meets four interior edges at the vertices of the
rectangle. The boundary faces of P1 , P2 subdivide ∂E(J) into rectangles.
• The interior faces of P can be colored with two colors (shaded and white) in a checkerboard fashion so that at each rectangular boundary face of P opposite side interior
faces have the same color. The shaded faces correspond to crossing disks while the
white faces correspond to regions of the projection plane. See [76, Figure 15] and
Figure 4.12.
The complement of all faces (interior and boundary) on the projection plane is a graph
Γ ⊂ ∂P with vertices of valence at least three and edges the edges of P .
A polyhedral P with the above properties is called a rectangular-cusped polyhedron.

4.1.2

Normal surfaces and combinatorial area.

We are interested in surfaces with boundary that are properly embedded in E(L) and are in
normal form with respect to the above polyhedral decomposition. We recall the following
definition.
Definition 4.1.2. A properly embedded surface (F, ∂F ) ⊂ (E(L), ∂E(L)) is said to be in
normal form with respect to the polyhedra decomposition, if for any P ∈ {P1 , P2 } we have
the following:
98

1. F ∩ P consists of properly embedded disks (D, ∂D) ⊂ (P, ∂P ).
2. F ∩ ∂P is a collection of simple closed curves none of which lies entirely in a single face
6

E. KALFAGIANNI AND C. LEE

of P .

2.2. Normal surfaces and combinatorial area. In this paper we are interested in
surfaces with boundary that are properly embedded in E(L) and are in normal form
respect to
the of
above
decomposition.
We recall the
following
3. with
F intersects
faces
P inpolyhedral
a collection
of properly embedded
arcs
none ofdefinition.
which passes
Definition 2.2. A properly embedded surface (F, @F ) ⇢ (E(L), @E(L)) is said to be in
throughform
vertices
Γ. Furthermore,
none decomposition,
of these arcs runs
from
normal
withofrespect
to the polyhedra
if for
anyan
P edge
2 {P1of
, PF2 }to
weitself
have the following:
or from
edge
to an adjacent
boundary
(1) Fan
\ Pinterior
consists
of properly
embedded
disks (D,edge.
@D) ⇢ (P, @P ).
(2) F \ @P is a collection of simple closed curves none of which lies entirely in a
single face of P .
4. A component of F ∩ P can intersect each boundary face in at most one arc.
(3) F intersects faces of P in a collection of properly embedded arcs none of which
passes through vertices of . Furthermore, none of these arcs runs from an edge
of F to itself or from an interior edge to an adjacent boundary edge.
The components
of F ∩ Pofare
An example
normal
disks in a
(4) A component
F \called
P cannormal
intersectdisks.
each boundary
face of
in three
at most
one arc.
The components
\ P areincalled
normal
disks. An example
of three
in athere
truncated
polyhedronofisFshown
Figure
4.4, borrowed
from [37].
Note normal
that P disks
shown
truncated polyhedron is shown in Figure 4, borrowed from [17]. Note that P shown there
a rectangular-cusped
polyhedron
as not
boundaryfaces
facesare
are rectangles.
rectangles.
is notisa not
rectangular-cusped
polyhedron
as not
allall
thethe
boundary

Figure
4.4: Normal
disks
in aintruncated
polyhedron
Figure
4. Normal
disks
a truncated
polyhedron.[37].
We recall the definition of the combinatorial area of a surface in normal form.
Let (D,
(P, @P ) be a normal
in a polyhedron
2 {P1 , P2 }.
WeDefinition
recall the 2.3.
definition
of@D)
the ⇢combinatorial
area ofdisk
a surface
in normalP form.

Suppose that D crosses m interior edges of P . The combinatorial area of D, denoted by
a(D), is defined by
m⇡
Definition 4.1.3. Let (D, ∂D)
⊂ (P,
a\
normal
in a polyhedron P ∈ {P1 , P2 }.
a(D)
= ∂P )+be
⇡|D
@E(L)|disk2⇡,
2
where
|D D
\ @E(L)|
the number
of @D
running on boundary
faces
of P . by
Suppose
that
crosses denotes
m interior
edges ofof Parcs
. The
combinatorial
area of D,
denoted
For an embedded surface (F, @F ) ⇢ (E(L), @E(L)), in normal form, the combinatorial
area a(F ) is defined by summing over all the normal disks of F in the polyhedra P1 , P2 .

a(D), is defined by

Next we recall the following form the combinatorial version of the Gauss-Bonnet themπ
orem due to Lackenby [28]
a(D) =
+ π|D ∩ ∂E(L)| − 2π,

2

Proposition 2.4. [28, Proposition 4.3] Let (F, @F ) ⇢ (E(L), @E(L))) be a surface in
normal form of Euler characteristic (F ). We have
a(F ) =99 2⇡ (F ).

where |D ∩ ∂E(L)| denotes the number of arcs of ∂D running on boundary faces of P . For
an embedded surface (F, ∂F ) ⊂ (E(L), ∂E(L)), in normal form, the combinatorial area a(F )
is defined by summing over all the normal disks of F in the polyhedra P1 , P2 .
Next we recall the following form the combinatorial version of the Gauss-Bonnet theorem
due to Lackenby [75]
Proposition 4.1.4. [75, Proposition 4.3] Let (F, ∂F ) ⊂ (E(L), ∂E(L))) be a surface in
normal form of Euler characteristic χ(F ). We have

a(F ) = −2πχ(F ).

To continue, with P as above, consider a normal disk (D, ∂D) ⊂ (P, ∂P ) such that ∂D
intersects at least one boundary face of P . Given an arc of γ ⊂ ∂D on a boundary face of
P , define the combinatorial length of γ with respect to D by

l(γ, D) =

a(D)
.
|D ∩ ∂E(L)|

For a simple closed curve γ ⊂ ∂(E(L), that is a boundary component of a surface
(F, ∂F ) ⊂ (E(L), ∂E(L)), let H ⊂ F be the union of normal disks in F whose intersections
with the boundary faces of P1 , P2 give γ. Thus γ is the union of arcs each properly embedded
in a boundary face of the polyhedra. Now define the combinatorial length

lc (γ) := l(γ, H) =
i

l(γi , D),

where the sum is taken over all normal disks in H and all normal arcs on boundary faces.

100

The quantity lc (γ), defined above, depends on F . To obtain a well defined notion of
combinatorial length one needs to consider the infimum over all normal surfaces F and
collections H with ∂F = γ. In fact, one can define the combinatorial length of any simple
closed curve on ∂E(L) [37, 75]. This is the definition of lc used by the authors in [37]. We
will not repeat these definitions here as we don’t need them. However, the combinatorial
length estimates of simple closed curves obtained in [37] also hold for l(γ, H). We need the
following lemma that follows immediately from the above definitions.
Lemma 4.1.5. [37, Lemme 4.13] Let (F, ∂F ) ⊂ (E(L), ∂E(L))) be an embedded surface
in normal form with respect to the polyhedral decomposition and let γ1 , . . . , γk denote the
components of ∂F . Then,

k

a(F ) ≥
j=1

c (γj ) .

Finally we need the following.
Lemma 4.1.6. Suppose that E(L) is an augmented link obtained from a prime, twist-reduced
link diagram D(K). Let (F, ∂F ) ⊂ (E(L), ∂E(L))) be an embedded normal surface and let
γ be a component of ∂F that is homologically non-trivial on a component T ⊂ ∂E(L). If T
comes from a crossing circle of L, let m denote the number of crossings in the corresponding
twist region of D(K). If T comes from a component Kj of K, let m be the number of twist
regions visited by Kj , counted with multiplicity.

c (s)

≥

mπ
.
3

Proof. It is proven in the proof of [37, Corollary 5.12] using [37, Proposition 5.3].

101

4.1.3

Genus estimates for spanning surfaces

Let (S, ∂S) ⊂ (E(K), ∂E(K)) be a spanning surface of K. That is a surface where the
components of ∂S are the components of K and it contains no closed components. In
fact, often, we will use (some times implicitly) the following convenient characterization of
spanning surfaces.
Lemma 4.1.7. A properly embedded surface (S, ∂S) ⊂ (E(K), ∂E(K)), without closed components, is a spanning surface of K if and only if for each component of ∂E(K), the total
geometric intersection number of the boundary curves ∂S with the corresponding meridian
is 1.
Proof. If S is a spanning surface of K, then clearly the desired conclusion holds. Conversely,
suppose that each component of ∂S has geometric intersection number 1 with the meridian
on the component of ∂E(L) it lies. Then ∂S must have exactly one component on each
component of ∂E(K) . For, if we had more that one curves on some component of ∂E(K)
then these curves will be parallel creating more intersections of ∂S with the corresponding
meridian that would contribute to the geometric intersection number. Thus each component
of ∂S is a longitude of ∂E(L).
Let J be an augmented link obtained from a diagram of a K and let L be the corresponding fully augmented link.

A spanning surface (S, ∂S) ⊂ (E(K), ∂E(K)) gives

punctured surface (F, ∂F ) ⊂ (E(J), ∂E(J)). Now F gives a properly embedded surface
(F , ∂F ) ⊂ (E(L), ∂E(L)). Since we are only interested in the Euler characteristic of the
surface we will often choose to work with F instead of F . In fact, abusing the setting,
we will identify F with F and say we can view F as a surface in the exterior of the fully

102

augmented link E(L) ∼
= E(J). For the next theorem we will also assume that the surface F
can be isotoped into normal form with respect to the polyhedral decomposition of E(L).
Theorem 4.1.8. Let K ⊂ S 3 be a link of k components with a prime, twist-reduced diagram
D(K). Suppose that D(K) has t ≥ 2 twist regions and let τ denote the smallest number of
crossings corresponding to a twist region of D(K). Let S be a spanning surface of K and
let L be an augmented link, obtained from K, such that S intersects n crossing circles of L.
Suppose that the corresponding punctured surface F ⊂ E(L) can be isotoped to be normal
with respect to the polyhedral decomposition P1 , P2 . Then we have

−χ(S) ≥

nτ
t
+
−n .
3
6

Proof. The boundary ∂F consists of curves γ1 , . . . , γk one for each torus component of ∂E(L)
that comes from a component of K and curves γk+1 , . . . , γk+n on the components coming
from crossing circles. By assumption F can be isotoped into normal form in the polyhedra
P1 and P2 ; so we can compute its combinatorial area. By applying Proposition 4.1.4 and
Lemma 4.1.5 we have

−2π · χ(S) = −2π · χ(F ) − 2πn = a(F ) − 2πn
k

≥
i=1

n

(γi ) +

i=1

(γk+i ) − 2πn.

By Lemma 4.1.6, the total length of the curves γ1 , . . . , γk is at least 2tπ/3, because K
passes through each twist region twice. By the same lemma the total length of the curves

103

γk+1 , . . . , γk+n is at least nτ π/3. Thus from the last equation we obtain

nτ π
2tπ
+
− 2πn
3
3
t
nτ
= 2π
+
−n .
3
6

−2π · χ(S) ≥

Since the Euler characteristic is an integer, the conclusion follows.
Recall that for a non-orientable surface S, with k boundary components, the crosscap
number is C(S) = 2 − χ(S) − k. We have the following result that should be compared
with [37, Theorem 1.5].
Corollary 4.1.9. Let the notation and setting be as in Theorem 4.1.8. Suppose moreover
that D(K) has at least six crossings in each twist region and that S is non-orientable. Then
we have
C(S) ≥

t
3

+ 2 − k.

Proof. Let τ denote the smallest number of crossings corresponding to a twist region of
D(K). By assumption, τ ≥ 6. Thus Theorem 4.1.8 gives
t
nτ
+
− n +2 − k
3
6
t
t
+ n − n +2 − k =
+ 2 − k.
3
3

C(S) = −χ(S) + 2 − k ≥
≥

104

4.2

Crosscap numbers of Alternating links

Adams and Kindred [1] gave an algorithm, starting with an alternating link projection,
to construct a spanning surface of maximal Euler characteristic among all the spanning
surfaces of the link. Our goal in this section is to show that the surface obtained from the
algorithm of [1] lies in the complement of an appropriate augmented link obtained from the
link projection. In the next section we will use the techniques of Section 4.1 to estimate the
crosscap number of a prime alternating link in terms of the twist number and the crossing
number of any prime, twist-reduced alternating diagram of the link.

4.2.1

State surfaces and a minimum genus algorithm

Given a crossing on a link diagram D(K) there are two ways to resolve it. A Kauffman state
σ on D(K) is a choice of one of these two resolutions at each crossing of D(K). Given a
state σ of D(K) we obtain a spanning surface Sσ of K, as follows: The result of applying σ
to D(K) is a collection vσ (D) of non-intersecting circles in the plane, called state circles. We
record the crossing resolutions along σ by embedded segments connecting the state circles.
Each circle of vσ (D) bounds a disk in S 3 . These disks may be nested on the projection plane
but can be made disjointly embedded in S 3 by pushing their interiors at different heights
below the projection plane. For each arc recording the resolution of a crossing of D(K) in
σ, we connect the pair of neighboring disks by a half-twisted band. The result is a surface
Sσ ⊂ S 3 whose boundary is K. See Figure 4.5.
Note that there might be several ways to make the disks bounded by the circles vσ (D)
disjoint and the resulting surface may not necessarily be isotopic in S 3 . Nevertheless, this
point is not important for the results of this paper, or for those of [1] as the resulting surfaces
105

circle of v (D) bounds a disk in S 3 . These disks may be nested on the projection plane
but can be made disjointly embedded in S 3 by pushing their interiors at di↵erent heights
below the projection plane. For each arc recording the resolution of a crossing of D(K)
in , we connect the pair of neighboring disks by a half-twisted band. The result is a
surface S ⇢ S 3 whose boundary is K. See Figure 5.

Figure 5. The two resolutions of a crossing, the arcs recording them

Figure 4.5: and
Thetheir
two contribution
resolutions of
crossing,
the arcs recording them and their contribution
to astate
surfaces.
to state surfaces.
will have the same topology (i.e. Euler characteristic and orientability). The geometry of the
state surfaces, Sσ with the particular construction described above, was studied by Futer,
Kalfagianni and Purcell [34, 36]. In [1] the authors show that state surfaces of alternating
diagrams can be used to determine the crosscap numbers of alternating links. In particular,
starting with an alternating diagram D(K), they gave an algorithm to obtain a state surface
with maximal Euler characteristic among all the state surfaces of D(K). Then they showed
that this Euler characteristic is the maximum over all spanning surfaces of K, state or nonstate. To review this algorithm, recall that a diagram D(K) may be viewed as a 4-valent
graph on S 2 with over and under information at each vertex. A complementary region of
this graph on S 2 is an m-gon if its boundary consists of m vertices or edges. We will refer
to a 2-gon as a bigon and a 3-gon as a triangle. We need the following elementary lemma.
Lemma 4.2.1. Suppose that an alternating diagram D(K) contains no 1-gon or bigon. Then
it must contain at least one triangle.
Proof. Consider the 4-valent graph on S 2 defined by D(K) and let V , E, F denote the
number of its vertices, edges and complimentary regions, respectively. Then, V − E + F = 2.
We have E = 4V /2 = 2V . Hence V − 2V + F = 2, which implies that F > V . Let m be the
smallest positive integer for which the 4-valent graph contains an m-gon. If it contains no
bigon or triangle, then m > 3. This implies that F < 4V /4 = V since each face must have
at least four distinct vertices in its boundary and each vertex can only be on the boundary
106

of at most 4 distinct faces. This is a contradiction. Therefore, m > 3 and if m > 2, then
m = 3.
Observe that the Euler characteristic of a state surface, corresponding to a state σ, is
χ(Sσ ) = vσ − c, where c is the number of crossings on D(K). Thus to maximize χ(Sσ ) we
must maximize the number of state circles vσ . Now we review the algorithm of [1]:
Algorithm 4.2.2. Let D(K) be a connected, alternating diagram.
1. Find the smallest m for which the projection D(K) contains an m-gon.
2. (a) If m = 1, then resolve the corresponding crossing so the 1-gon becomes a state
circle.
Suppose that m = 2. Then D(K) contains twist regions with more than one
crossings. Pick R to be such a twist region with cR > 1 crossings and cR − 1
bigons. Resolve all the crossings of R in such a way so that all these bigons
become state circles. Create one branch of the algorithm for each bigon on D(K).

(b) Suppose m > 2. Then by Lemma 4.2.1, we have m = 3. Pick a triangle region
on D(K). Now the process has two branches: For one branch we resolve each
crossing on this triangle’s boundary so that the triangle becomes a state circle.
For the other branch, we resolve each of the crossings the opposite way. See Figure
4.6.
3. Repeat Steps 1 and 2 until each branch reaches a projection without crossings. Each
branch corresponds to a Kauffman state of D(K) for which there is a corresponding
state surface. Of all the branches involved in the process choose one that has the largest
107

CROSSCAP NUMBERS AND THE JONES POLYNOMIAL

11

Figure 6. One branch of the algorithm resolves the crossings so that

Figure 4.6: One
branchbecomes
of the algorithm
resolves
the crossings
so that
thethe
triangle
the triangle
a state circle.
The other
branch resolves
them
becomes a state
circle.
The other branch resolves them the opposite way.
opposite
way.
the largest
number corresponding
of state circles. to
The
surface
numberone
of that
statehas
circles.
The surface
this
state corresponding
has maximal toEuler
this state has maximal Euler characteristic over all the states corresponding to
D(K). Note that, a priori , more than one branches of the algorithm may lead
characteristic over all the states corresponding to D(K). Note that, a priori , more than
to surfaces of maximal Euler characteristic. That is, there might be several state
surfaces that have the same (maximal) Euler characteristic.

one branches of the algorithm may lead to surfaces of maximal Euler characteristic.

The following result allows to calculate the crosscap number of an alternating link
from
obtained
applying
Algorithm
3.2that
to any
alternating
projection
of the
Thatthe
is, surface
there might
bebyseveral
state
surfaces
have
the same
(maximal)
Euler
link.

characteristic.
Theorem
3.3. [1, Corollary 6.1] Let S be any maximal Euler characteristic surface
obtained via Algorithm 3.2 from an alternating diagram of k-component link K. Let
C(K) denote the crosscap number of K and let g(K) denote the (orientable) genus of
The
K.following
Then, weresult
have allows to calculate the crosscap number of an alternating link from
(1) If there is a surface S as above that is non-orientable then C(K) = 2
(S) k.
the surface
applying
Algorithm
to any we
alternating
projection
of thek.link.
(2)obtained
If all theby
surfaces
S as
above are4.2.2
orientable,
have C(K)
= 3
(S)
Furthermore S is a minimal genus Seifert surface of K and C(K) = 2g(K) + 1.
It should
clarified
that 6.1]
choices
as well asEuler
the order
in resolvingsurface
bigon obTheorem
4.2.3.be[1,
Corollary
LetofS branches
be any maximal
characteristic
regions following algorithm 3.2 may result in di↵erent state surfaces. If a state surface
resulting from
is orientable,
it of
is not
necessarilylink
the K.
caseLet
that
tainedfor
viaKAlgorithm
4.2.2 the
fromalgorithm
an alternating
diagram
k-component
C(K)
C(K) = 3
(S) k. If any state surface resulting from a choice of the branch and
of bigon regions to resolve is non-orientable, then C(K) = 2
(S) k. We
denoteorder
the crosscap
number of K and let g(K) denote the (orientable) genus
of K. Then,
illustrate the subtlety in the algorithm by applying it to the figure 8 knot. The figure 8
knot has non-orientable genus 2 realized by its checkerboard surfaces.
we have In the first figure above, a diagram of the figure 8 knot is shown as well as two bigon
regions of the diagram to which we can apply Step (a) of Algorithm 3.2. Suppose that
choose region 1, then for the next step of the algorithm, we now have 3 choices of
1. we
If there
is a surface S as above that is non-orientable then C(K) = 2 − χ(S) − k.
bigon regions to resolve. This is shown in the second figure above. If we choose region
1, then we obtain a checkerboard surface shown as a first figure below. If we choose
2 or
3, then Sweasobtain
surface
shown
as the
second figure.Furthermore
Both of
2. region
If all the
surfaces
above an
areorientable
orientable,
we have
C(K)
= 3−χ(S)−k.
these surfaces realize the maximal Euler characteristic of -1. It would be incorrect to
conclude based on orientable surface that the non-orientable genus is 3 since a particular
S is a minimal genus Seifert surface of K and C(K) = 2g(K) + 1.
application of the algorithm gives a non-orientable surface. For our purposes, we will not
need to know whether the resulting state surface from the algorithm is non-orientable
therefore
realize the
non-orientable
genus. Weasneed
to know
that
maximizes
It and
should
be clarified
that
choices of branches
wellonly
as the
order
in itresolving
bigon
the Euler characteristic.
The next lemma is important for the results in this paper as it will allow us to apply
regions following algorithm 4.2.2 may result in different state surfaces. If a state surface
the techniques of Section 2 to obtain bounds on crosscap numbers of alternating links.

for K resulting from the algorithm is orientable, it is not necessarily the case that C(K) =

108

3 − χ(S) − k. If any state surface resulting from a choice of the branch and order of bigon
regions to resolve is non-orientable, then C(K) = 2 − χ(S) − k. We illustrate the subtlety
in the algorithm by applying it to the figure 8 knot. The figure 8 knot has non-orientable
genus 2 realized by its checkerboard surfaces.

1

3
1
2

2

Figure 4.7: A diagram of the figure 8 knot with bigon regions labeled 1 and 2 and the
diagram resulting from applying the first step of Algorithm 4.2.2. The new choices of bigon
regions are labeled 1, 2, and 3.
In the first figure above, a diagram of the figure 8 knot is shown as well as two bigon
regions of the diagram to which we can apply Step (a) of Algorithm 4.2.2. Suppose that
we choose region 1, then for the next step of the algorithm, we now have 3 choices of bigon
regions to resolve. This is shown in the second figure above. If we choose region 1, then we
obtain a checkerboard surface shown as a first figure below. If we choose region 2 or 3, then
we obtain an orientable surface shown as the second figure. Both of these surfaces realize
the maximal Euler characteristic of -1. It would be incorrect to conclude based on orientable
surface that the non-orientable genus is 3 since a particular application of the algorithm gives
a non-orientable surface. For our purposes, we will not need to know whether the resulting
state surface from the algorithm is non-orientable and therefore realize the non-orientable
genus. We need only to know that it maximizes the Euler characteristic.

109

3

3

1

1
2

2

Figure 4.8: The state surfaces resulting from resolving the rest of the bigon regions with
different choices of bigon regions.
The next lemma is important as it will allow us to apply the techniques of Section 4.1 to
obtain bounds on crosscap numbers of alternating links.
Lemma 4.2.4. Given a prime, alternating, twist-reduced diagram D(K), let S be a maximal
Euler characteristic surface obtained by applying Algorithm 4.2.2 to D(K). There is an
augmented link J = JS such that S is in the complement E(J).
Proof. An augmented link is obtained from D(K) by adding a simple closed curve encircling
each twist region. If a twist region involves more than one crossing, then there is only one
way to add a crossing circle. Otherwise, there are two ways of adding a crossing circle as
shown in Figure 4.3. Let S be a state surface with maximal Euler characteristic obtained by
applying Algorithm 4.2.2 to D(K). We show that we can always augment D(K) by making
a choice of a crossing circle for each twist region involving a single crossing, such that S lies
in the complement of the augmented link.
For a twist region R involving more than one crossing, and thus consists of a maximum
string of complementary bigon regions arranged end to end, we augment by adding a crossing
circle CR encircling the twist region. Since D(K) is prime, none of its complementary regions
can be an 1-gon. In this case, the algorithm picks the resolution of the crossings of R so
110

that each bigon becomes a state circle following Step 2a. In any state surface which have
this resolution at the crossings of R, these bigon disks are joined with twisted strips, and
the crossing circle CR encloses the twisted strips. We may arrange so that each twisted strip
intersects the crossing disk corresponding to CR only in its interior. Hence, the portion of
CROSSCAP NUMBERS AND THE JONES POLYNOMIAL

13

any state surface obtained by the algorithm involving a twist region containing at least one
interior. Hence, the portion of any state surface obtained by the algorithm involving a
twist
at least one
lie in 4.9.
the complement of CR . See Figure 9.
bigon,
willregion
lie incontaining
the complement
of Cbigon,
. Seewill
Figure
R

Figure 9. The portion of S through a twist region with more than one

Figure 4.9: crossing
The portion
of crossing
S through
a twist
with more than one crossing and the
and the
circle
for theregion
twist region.
crossing circle for the twist region.
Continue creating a branch of the algoritm for each bigon. Examine the link diagram
D0 created after resolving all the twist regions involving more than one crossings by
Algorithm
3.2.
Continue
creating
a branch of the algorithm for each bigon. Examine the link diagram D
Each of the remaining crossings corresponds to a twist region of D(K) containing a
single
crossing.
We all
decide
on which
way involving
to add a crossing
circle
to crossings
each of these
twist
created
after
resolving
the twist
regions
more than
one
by Algorithm
regions. Since D(K) is twist-reduced, bigons in D0 can only come from triangles in
region with more than one crossings attached to them. If there
4.2.2.D(K) that had a twist
is such a bigon in D0 , Algorithm 3.2 will apply Step 2a to these crossings such that the
bigon becomes a state circle. If this bigon is adjacent to a nother one then we choose an
Each of the remaining crossings corresponds to a twist region of D(K) containing a single
augmentation for each of the two crossings by putting two crossing circles, one for each of
the two crossings resolved, so that each crossing circle encloses a twisting strip from the
crossing.
We decide
on whichThe
wayportion
to addof athe
crossing
circle coming
to eachfrom
of these
twist
regions.
resolution
at one crossing.
state surface
resolving
these
two crossings will then also be disjoint from the crossing circles. Repeat this procedure
Sincefollowing
D(K) isthe
twist-reduced,
bigons
D nocan
only
come from triangles in D(K) that had a
algorithm until
thereinare
more
bigons.
If there are no more crossings left in the projection, then we are done. Otherwise, we
twisthave
region
with more
one
crossingsn-gon
attached
to them.
If there3.1,
is such
a bigon
a projection
forthan
whom
a minimal
is a triangle
by Lemma
and Step
2b ofin D ,
the algorithm is applied to resolve each of the three crossings of the triangle, see Figure 6.
Algorithm
4.2.2 each
will apply
2a tocrossings
these crossings
that
a state
In addition,
of the Step
remaining
is a twistsuch
region
in the
the bigon
originalbecomes
projection
D(K). Let S be a maximal Euler characteristic surface obtained from the algorithm.
timebigon
that Step
2b is applied,
a branch
of thewe
algorithm
is chosen
to generate
circle.Each
If this
is adjacent
to another
one then
choose an
augmentation
for S.
each of
We augment each of the three crossings of the triangle based on which branch is chosen.
In either
branch,
we add two
a crossing
circle
for each
three
crossings,
such that
it
the two
crossings
by putting
crossing
circles,
one of
forthe
each
of the
two crossings
resolved,
encircles the crossing strip from the resolution of the chosen branch. See Figure 10.

so that each crossing circle encloses a twisting strip from the resolution at one crossing.
The portion of the state surface coming from resolving these two crossings will then also be

111
Figure 10. Surfaces from two branches of the splitting in Step 2b in the

Figure 9. The portion of S through a twist region with more than one
crossing and the crossing circle for the twist region.
the algoritm
for each bigon.
Examine
the link diagram
disjointContinue
from thecreating
crossinga branch
circles. ofRepeat
this procedure
following
the algorithm
until there

D0 created after resolving all the twist regions involving more than one crossings by
Algorithm
3.2.
are no
more bigons.
Each of the remaining crossings corresponds to a twist region of D(K) containing a
We decide
on which
way
to projection,
add a crossing
circle
of these
twist we
Ifsingle
therecrossing.
are no more
crossings
left in
the
then
we to
areeach
done.
Otherwise,
0
regions. Since D(K) is twist-reduced, bigons in D can only come from triangles in
that hadfor
a twist
with more
than
crossings
to them.
there2b of
have D(K)
a projection
whomregion
a minimal
n-gon
is aone
triangle
by attached
Lemma 4.2.1,
andIf Step
is such a bigon in D0 , Algorithm 3.2 will apply Step 2a to these crossings such that the
bigon becomes a state circle. If this bigon is adjacent to a nother one then we choose an
the algorithm is applied to resolve each of the three crossings of the triangle, see Figure 4.6.
augmentation for each of the two crossings by putting two crossing circles, one for each of
the two crossings resolved, so that each crossing circle encloses a twisting strip from the
In addition,
each
of the
remaining
a twist
the original
projection
D(K).
resolution
at one
crossing.
The crossings
portion ofisthe
state region
surface in
coming
from resolving
these
two crossings will then also be disjoint from the crossing circles. Repeat this procedure
Let Sfollowing
be a maximal
Euler until
characteristic
surface
obtained from the algorithm. Each time
the algorithm
there are no
more bigons.
If there are no more crossings left in the projection, then we are done. Otherwise, we
that have
Step a2bprojection
is applied,
branch
of the algorithm
chosenby
to Lemma
generate
S.and
WeStep
augment
forawhom
a minimal
n-gon is aistriangle
3.1,
2b of each
the algorithm is applied to resolve each of the three crossings of the triangle, see Figure 6.
of theInthree
crossings
of the
theremaining
triangle based
on which
branch
is in
chosen.
In either
branch, we
addition,
each of
crossings
is a twist
region
the original
projection
D(K). Let S be a maximal Euler characteristic surface obtained from the algorithm.
time circle
that Step
2b is of
applied,
a branch
of thesuch
algorithm
chosen tothe
generate
S. strip
add aEach
crossing
for each
the three
crossings,
that itisencircles
crossing
We augment each of the three crossings of the triangle based on which branch is chosen.
branch,ofwe
a crossing
circle
each of
the three crossings, such that it
from In
theeither
resolution
theadd
chosen
branch.
Seefor
Figure
4.10.
encircles the crossing strip from the resolution of the chosen branch. See Figure 10.

Figure 10. Surfaces from two branches of the splitting in Step 2b in the
and theofcorresponding
of augmentation.
Figure 4.10:Adam-Kindred
Surfaces fromalgorithm
two branches
the splitting choice
in Step
2b in the Adam-Kindred

algorithm and the corresponding choice of augmentation.

We repeat as in Step 3 of Algorithm 4.2.2 which will make a decision for splitting at
the rest of the crossings. If a bigon is encountered again as a minimal n-gon we repeat the
procedure for a bigon region where each of the crossings belongs to a twist region in D(K).
Otherwise we apply the procedure for when a minimal n-gon is a triangle. We stop when
there are no more twist regions in D(K) to augment.

112

4.2.2

Normalization and crosscap number estimates

Here we use the results of Section 4.1 to obtain two-sided bounds of the crosscap number of
an alternating link in terms of the twist number of any prime, twist-reduced alternating link
diagram.
We begin with the following lemma that shows that the crosscap number of an alternating
link is always bounded by the twist number of any alternating projection from above; that
is, primeness and twist-reducibility is not needed for this part.
Lemma 4.2.5. Let K ⊂ S 3 be a link of k components with an alternating diagram D(K)
that has t ≥ 2 twist regions. Let C(K) denote the crosscap number of K. We have

C(K) ≤ t + 2 − k.
Proof. Let S be a surface obtained by applying Algorithm 4.2.2 to D(K) and let σ denote
the Kaufmann state of D(K) to which S corresponds. Let vb denote the number of state
circles that are bigons and let vnb denote the non-bigon state circles. Also let c1 denote the
number of crossings in D(K) each of which forms its own twist region and let c2 denote
the remaining crossings. Since we assume that there are at least two twist regions we have
vnb ≥ 1. Thus we have

−χ(S) = = c2 − vb + c1 − vnb
≤ t − 1.

113

Now the upper bound follows at once from Theorem 4.2.3.
Before we are able to estimate C(K) from below we need some preparation. Let D(K) be
a link diagram and let J be an augmented link obtained from D(K) with L the corresponding
fully augmented link. Suppose that the exterior E(J) ∼
= E(L) is hyperbolic with an angled
polyhedral decomposition {P1 , P2 } as described in Section 4.1. Let S be a spanning surface
of K that realizes C(K). Having Lemma 4.2.4 in mind, we will also assume that S is disjoint
from the crossing circles of J; hence we can view it as a surface in E(J) ∼
= E(L).
We wish to apply Theorem 4.1.8 to estimate χ(S). In order to do so we need to have
S in normal position with respect to the polyhedral decomposition {P1 , P2 }. The standard
argument of making an surface that is incompressible and ∂-incompressible (that is, essential ) normal in a triangulated 3-manifold [50], can be adjusted to work in the setting of
more general polyhedral decompositions. The argument is written down by Futer and Guritaud [30, Theorem 2.8]. An additional difficulty in our setting is that surfaces that realize
C(K) need not be essential in E(K). For instance, if all the state surfaces obtained from
Algorithm 4.2.2 applied to an alternating projection D(K) are orientable, then a spanning
surface that realizes C(K) is a obtained from a minimal genus Seifert surface by adding a
half-twisted band. Such a surface is ∂-compressible. This, for example, happens for the knot
74 [1].
For our purposes we are only interested in the question of whether S can be converted
to a spanning surface that is normal with respect to the polyhedral decomposition of E(L),
without changing χ(S) and the surface orientability. For this we examine how a spanning
surface S that realizes C(K) behaves under the general process that converts any properly
embedded surface in E(J) into a normal one with possibly different topology [66]. We

114

have the following lemma that applies beyond the class of alternating links and might be of
independent interest.
Lemma 4.2.6. Let K ⊂ S 3 be a link with a prime, twist-reduced diagram D(K). Suppose
that D(K) has t ≥ 2 twist regions. Suppose that there is a spanning surface S in the exterior
E(J) ∼
= E(L) of an augmented link of D(K) and such that C(S) = C(K). Then exactly one
of the following is true:
1. There is a non-orientable, spanning surface S ⊂ E(L) for K, that is in normal form
and such that χ(S) = χ(S ).
2. We have C(K) = 2g(K) + 1. Furthermore, there a Seifert surface of K that lies in
E(L) such that it realizes g(K) and it is in normal form.
Proof. By assumption S realizes C(K). Thus S has maximal Euler characteristic among
all non-oriented spanning surfaces of K. As discussed above, we will assume that S is not
necessarily essential and examine how compressions and ∂-compressions may interfere with
a process of converting S to a normal surface. Examining the moves required during this
process [30, Theorem 2.8], and since S contains no closed components, we see that there are
two situations to consider:
1. S admits a compression disk D, that lies in the interior of a single face of a polyhedron
P ∈ {P1 , P2 }.
2. The intersection of S with a face of a polyhedron f ⊂ ∂P is an arc γ that runs from an
edge of e ⊂ Γ ⊂ ∂P to itself, or from an interior edge to an adjacent boundary edge.
In (1) there are two cases to consider according to whether ∂D separates S or not. First
suppose that ∂D is non-separating on S. Compressing along D we obtain a spanning surface
115

S for K with χ(S ) = χ(S) + 2. Since S realizes C(K), S cannot be non-orientable.
Thus we have an orientable spanning surface of K; that is a Seifert surface. Adding a
half-twisted band to S (i.e. adding a crosscap) produces a non-orientable spanning surface
S1 of K with χ(S1 ) = χ(S ) − 1. Thus, S1 is a non-orientable spanning surface with
χ(S1 ) = χ(S) + 1 > χ(S) which contradicts the fact that S realizes C(K). Thus this case
will not happen.
Suppose now that ∂D separates S. We will look at such a disk so that ∂D is innermost
in the sense that one of the components of S ∂D lies entirely in a single polyhedron P .
Compressing along D gives two surfaces S1 and S2 with χ(S) = χ(S1 ) + χ(S2 ) − 2.
Suppose that both surfaces have non-empty boundary. Then the disjoint union of S1 , S2
is a non-orientable spanning surface of K with Euler characteristic χ(S1 ) + χ(S2 ) > χ(S),
contradicting the assumption that S realizes C(K). Thus, one of S1 , S2 , say S1 , must be
a closed surface and all of ∂S is left on S2 . Since S1 is a closed surface embedded in S 3 ,
it is orientable. Hence S2 is a non-orientable. Since S realizes C(K) and χ(S1 ) ≤ 2, it
follows that χ(S) = χ(S2 ). Thus we may ignore S2 , replace S with S2 and continue with
the normalization process.
Next we treat case (2): The arc γ cuts off a disk D ⊂ f , with ∂D consisting of γ and
an arc that lies on the boundary of f . By an innermost argument, we may assume that the
interior of D is disjoint from S. Again, following the argument in the proof of [30, Theorem
2.8], if γ runs from an interior edge to an adjacent boundary edge, the disk D will guide an
isotopy of S that can be used to eliminate the arc γ [30, Figure 2.2]. Similarly, if e is an
interior edge or e is a boundary edge and f is a boundary face, we can use D to obtain an
isotopy that will eliminate γ and decrease the number intersections of S with Γ (compare,
left panel of [30, Figure 2.1]). It follows, that the only case left to examine is when e is
116

a boundary edge and f is an interior face of S. In this case D is a ∂-compression disk of
S. Consider the arc δ := ∂D γ ⊂ e and let T denote the boundary component of ∂E(K)
containing it. There are two cases to consider: (i) δ cuts off a disk in the annulus T ∂S; and
(ii) δ runs from one component of T ∂S. We will perform surgery (∂-compression) along
D. This may cut S into more components or not according to whether we are in case (i) or
(ii). Surgery along a ∂D doesn’t change the total algebraic intersection number of ∂S with
the meridians of ∂E(J). Thus, by Lemma 4.1.7, it will produce a spanning surface of K and
possibly some (redundant) components.
First suppose that surgery along D splits S into two surfaces S1 and S2 . Suppose that
both surfaces have non-empty boundary. Then the disjoint union of S1 , S2 is a non-orientable
spanning surface of K with Euler characteristic χ(S1 ) + χ(S2 ) > χ(S), contradicting the
assumption that S realizes C(K). Thus all the components of ∂S disjoint from D must
remain on one of S1 , S2 , say on S2 and the intersections of ∂S with the meridians of ∂E(K),
that count towards algebraic intersection number, also remain on ∂S2 .
Thus S1 has one boundary component that is either homotopicaly trivial on ∂E(K) or
isotopic to a meridian of ∂E(K). In either case ∂S1 bounds a disk in S 3 . We may cap ∂S1
with this disk to produce a closed surface embedded in S 3 , which must be orientable. Hence
S2 is a non-orientable spanning surface for K. But since S realizes C(K) and χ(S1 ) ≤ 1 we
must have χ(S) = χ(S2 ). Thus we may ignore S1 , replace S with S2 and continue with the
normalization process.
Next suppose that surgery along D doesn’t disconnect S. Then we get a spanning surface
S , with χ(S ) = χ(S)+1. Since S was assumed to realize C(K), S cannot be non-orientable.
Thus we have an orientable spanning surface of K. We claim that S must be a minimal
genus Seifert surface, that is g(S ) = g(K). For, suppose that K has a Seifert surface S1 with
117

χ(S1 ) > χ(S ). Then adding a half-twisted band to S1 would give a non-oriented spanning
surface S with χ(S ) = χ(S1 ) − 1 > χ(S ) − 1 = χ(S), contradicting the fact that S realizes
C(K). Thus, S is a minimal genus Seifert surface of K that lies in E(L); that is we have
g(S ) = g(K). Now it is clear that the surface obtained by a half-twisted band to S is a nonorientable spanning surface S of K that has maximal Euler characteristic among all such
surfaces. Thus we have C(K) = C(S ) = 2g(S ) + 1 = 2g(K) + 1. Since S is minimal genus
and orientable, it is incompressible and ∂-incompressible in E(K) and thus in E(L). Hence
we may isotope S to be normal with respect to the polyhedral decomposition [30, Theorem
2.8].
Now we are ready to prove the following.
Theorem 4.2.7. Let K ⊂ S 3 be a link of k components with a prime, twist-reduced alternating diagram D(K). Suppose that D(K) has t ≥ 2 twist regions. Let C(K) denote the
crosscap number of K. We have

t
3

+ 2 − k ≤ C(K) ≤ t + 2 − k,

where · is the ceiling function that rounds up to the nearest integer. Furthermore, both
bounds are sharp.
Proof. The upper bound comes from Lemma 4.2.5. To derive the lower bound, let S be a
state surface obtained from Algorithm 4.2.2 to D(K) and let J be the augmented link of
Lemma 4.2.4, with L the corresponding fully augmented link. By Theorem 4.2.3, one of
following is true:
1. There is one such surface that is non-orientable and realizes C(K); that is we have
C(K) = C(S).
118

2. All such surfaces S are orientable and we have C(K) = 2g(S) + 1.
NUMBERS AND THE JONES POLYNOMIAL
17
Suppose we are in caseCROSSCAP
(1). Then
by Lemma 4.2.6 we may replace S by a spanning surface

⇠ ⇡

that also realizes C(K) and is normal
with respect to the polyhedral decomposition of E(L).
t
+ 2 k  C(K)  t + 2 k,
3
Theorem
where4.1.6
d·e isgives
the ceiling function that rounds up to the nearest integer. Furthermore, both
bounds are sharp.

Proof. The upper bound comes from Lemma 3.5. To derive the lower bound, let S be
t
a state surface obtained
Algorithm
3.2 to −
D(K)
k, augmented link
C(K) =from
C(S)
= 2 − χ(S)
k ≥and let+J 2be−the
3
of Lemma 3.4, with L the corresponding fully augmented link. By Theorem 3.3, one of
following is true:
(1) There is one such surface that is non-orientable and realizes C(K); that is we
and the lower
bound
have
C(K)follows.
= C(S).On the other, hand if we are in (2) then S is a minimal genus
(2) All such surfaces S are orientable and we have C(K) = 2g(S) + 1.
Seifert
surfaceweofare
K in
and
thus
is incompressible
∂-incompressible.
Thus wesurface
may again
Suppose
case
(1).itThen
by Lemma 3.6ands
we may
replace S by a spanning
that also realizes C(K) and is normal with respect to the polyhedral decomposition of
replace
S with
one into
normal form. Since S is disjoint from the crossing circle components
E(L).
Theorem
2.6 gives
⇠ ⇡
t
C(K) = C(S) = 2
(S)
k
+ 2 k,
of J, Theorem 4.1.6 gives
3
and the lower bound follows. On the other, hand if we are in (2) then S is a minimal
genus Seifert surface of K and thus it is incompressible ands @-incompressible. Thus we
may again replace
one+into
form.
S is tdisjoint
from
C(K)S=with
2g(S)
1 = normal
2 − 2χ(S)
− kSince
+1≥
+3−
k. the crossing
3
circle components of J, Theorem 2.6 gives
⇠ ⇡
t
C(K) = 2g(S) + 1 = 2 2 (S) k + 1
+3
k.
3
Hence the lower bound follows.
Hence the lower bound follows.

Figure 11. The knots 103 (left) and 10123 (right).

Figure 4.11: The knots 103 (left) and 10123 (right) [19].
It remains to prove that both bounds are sharp: Consider the alternating knot 103
of Figure 11. The twist number of the diagram shown there is t = 2 and each twist
consists
of more
than
one bounds
crossing.are
Thus
Algorithm
3.2 has
one branch
that103 of
Itregion
remains
to prove
that
both
sharp:
Consider
theonly
alternating
knot
is easily seen to give an oriented surface of genus 1. Thus, by Theorem 3.3, C(K) =
2g(K)
+ 1The
= 3 twist
= t + 1.
The same
argument
applies
to the
knots
adding
anyregion
Figure
4.11.
number
of the
diagram
shown
there
is t obtained
= 2 andbyeach
twist
even number of crossings in each twist region of the knot 103 . Hence we have an infinite
family of alternating knots with C(K) = 2g(K) + 1 = 3 = t + 1.
consists
of more than one crossing. Thus Algorithm 4.2.2 has only one branch that is easily
To discuss some examples where our lower bound is sharp, note that if a knot K
processes an alternating diagram with c = t, then we have
seen to give an oriented surface of genus 1. Thus, by Theorem 4.2.3, C(K) = 2g(K) + 1 =
lcm
jc k
1+
 C(K) 
.
3
2

119

3 = t + 1. The same argument applies to the knots obtained by adding any even number of
crossings in each twist region of the knot 103 . Hence we have an infinite family of alternating
knots with C(K) = 2g(K) + 1 = 3 = t + 1.
To discuss some examples where our lower bound is sharp, note that if a knot K processes
an alternating diagram with c = t, then we have

1+

c
3

≤ C(K) ≤

c
.
2

Now observe that, for instance, if c = t = 13, then C(K) = 6. Similarly, if c = t = 10,
then C(K) = 5. A concrete example, is the knot 10123 shown in Figure 4.11. More examples
where the lower bound is sharp are discussed in Section 4.4.
In particular, for knots we have the following:
Theorem 4.2.8. Let K ⊂ S 3 be a knot with a prime, twist-reduced alternating diagram
D(K). Suppose that D(K) has t ≥ 2 twist regions and let C(K) denote the crosscap number
of K. We have

1+

t
3

≤ C(K) ≤ min t + 1,

c
2

where c denotes the number of crossings of D. Furthermore, both bounds are sharp.
Proof. For k = 1, the inequality of Theorem 4.2.7 becomes
t
3

+ 1 ≤ C(K) ≤ t + 1.

Thus the lower bound follows. Murakami and Yasuhara [89] showed that for a knot K

120

with a connected, prime diagram of c crossings, we have

C(K) ≤

c
.
2

Thus the upper bound follows from these two inequalities.
Having related C(K) to twist numbers of alternating link projections, Theorems 1.1.3
and 4.0.1 follow by [26].

4.3

Proof of Theorem 1.1.3

Now we discuss how the results stated in the introduction follow from the above results.
Recall that for a knot K, we have TK := |βK | + βK , where βK and βK denote the second
and the penultimate coefficients of the Jones polynomial of K, respectively.
Theorem 1.1.3. Let K be a non-split, prime alternating link with k-components and with
crosscap number C(K). Suppose that K is not a (2, p) torus link. We have

TK
3

+ 2 − k ≤ C(K) ≤ TK + 2 − k.

Furthermore, both bounds are sharp.
Proof. Let D(K) be a connected, twist-reduced alternating diagram that has t ≥ 2 twist
regions. Then, by [26, Theorem 5.1], we have t = TK . A prime alternating link admits
prime, twist reduced alternating diagrams; every alternating diagram can be converted to
a twist-reduced one by flype moves [76, Lemma 4]. Thus, the result follows from Theorem
4.2.7.

121

Now we are ready prove 4.0.1 which we restate for the convenience of the reader.
Theorem 4.0.1. Let K be an alternating, non-torus knot with crosscap number C(K) and
let TK be as above. We have

1+

TK
3

≤ C(K) ≤ min TK + 1,

sK
2

where sK denotes the span of JK (t). Furthermore, both the upper and lower bounds are
sharp.
Proof. Kauffman [72] showed that the degree span of the Jones polynomial of alternating
link is equal to the crossing number of the link. Furthermore, both the degree span and
the crossing number of alternating knots are known to be additive under the operation of
connect sum [78]. Using these, the upper inequality follows at once from Theorem 4.2.8.
Furthermore, the lower inequality follows for all prime alternating knots.
To finish the proof we need to show that the lower inequality holds for connect sums of
alternating knots. To that end let K#K be such a knot. By [89], we have

C(K#K ) ≥ C(K) + C(K ) − 1.

Since the Jones polynomial is multiplicative under connect sum we have TK +T

K

Hence we have
T
T
C(K#K ) ≥ C(K) + C(K ) − 1 = K + K + 4 − 2 − 1
3
3
TK + T
K + 1.
≥
3

122

≥T

K#K

.

Hence the conclusion follows.

4.4
4.4.1

Calculations of crosscap numbers
Lower exact bounds

The lower bound of Theorem 4.2.7 gives the exact value of the crosscap number in the case
we have an alternating projection D(K), where each twist region of D(K) has at least two
crossings and the twist regions of D(K) meet in groups of three. A concrete construction
of such examples is as follows: Let G be a trivalent planar graph and let N (G) denote a
neighborhood of G on the plane. The boundary ∂N (G) is a link. For each edge of G we
have two parallel arcs belonging on different components of ∂N (G). Construct a diagram of
a new link by adding a number of half twists between these parallel arcs of the components
∂N (G) corresponding to each edge of G. Suppose that for each edge of G we add at least two
crossings on D(K). We can do this so that the resulting diagram is alternating. Depending
on the numbers of twists we put we may obtain a knot or a multi-component link. Let
D(K) denote any alternating projection obtained this way and let SG be a state surface
obtained from Algorithm 4.2.2 applied to D(K): Since each twist region contains bigons,
SG corresponds to the Kauffman state of D(K) that resolves all the crossings so that the
bigons are state circles. To analyze these surfaces further we need a definition.
Definition 4.4.1. A normal disk D in a polyhedron P ∈ {P1 , P2 } is called an ideal triangle
if ∂D intersects exactly three boundary faces of ∂E(L) and it intersects no interior edges of
P.

123

Corollary 4.4.2. Let K be a k-component link with a prime, twist-reduced alternating diagram D(K) constructed from a trivalent planar graph G as above. Then we have

C(K) =
where

t
3

+

= 2 if SG is non-orientable and

−k =

TK
3

+

− k.

= 3 otherwise.

Proof. By assumption D(K) is obtained by adding twists along components of the boundary
∂N (G) of a regular neighborhood of a planar trivalent graph. The surface SG is obtained
by N (G) by similar twisting and the augmented link of Lemma 4.2.4 is obtained by adding
a component encircling each twist region of D(K). In order to calculate χ(SG ) we need
some more detailed information about the polyhedral decomposition {P1 , P2 } of E(L) [98].
The surface SG gives rise to surface SG in E(L); we will calculate χ(SG ). Let D denote
the union of the crossing disks bounded by the crossing circles of L. Each disk intersects
the projection plane in a single arc. We may isotope the interior of SG so that it is disjoint
from the intersections of D with the projection plane. Let P ∈ {P1 , P2 }. Recall that (before
truncation) all the vertices of P are of valence four and they correspond to crossing circles
of L and to arcs of K D and the faces can be colored in a checkerboard fashion (in shaded
and white) as follows:
1. The shaded faces of P correspond to the crossing disks: Each disk D gives to two
triangular shaded faces of P meeting at an ideal vertex corresponding to the crossing
circle ∂D (“bowties”).
2. The edges of P come from the intersections of D with the projection plane.
3. The white faces of P correspond to regions of D(K) on the projection plane.
124

To a vertex v ∈ G there corresponds a triangular region of D(K) that is a neighborhood of
v and around which the three twist regions D(K), corresponding to the edges of G emanating
from v, meet. Let Dv denote the union of the three crossing disks corresponding to the three
twist regions of D(K) around v. This triangular region will become a an ideal triangular
white face, say Dv , after truncating the vertices of P . See Figure 4.12. The ideal vertices of
Dv come from the arcs of K Dv that surround v. Each of the three interior edges of P on
∂Dv is attached to a bowtie: two shaded faces meeting at an ideal vertex coming from one
of crossing disks of Dv .

G

Figure 4.12: From left to right: The portion of D(K) around a vertex of G, the corresponding
potion of the augmented link and portion of the polyhedral decomposition. An ideal triangle
is indicated by the red line.
Since the surface SG doesn’t intersect the crossing circles, SG is disjoint from the ideal
vertices corresponding to disks in Dv . Furthermore, since we arranged so that the interior
of SG is disjoint from the intersection of D with the projection plane, SG is disjoint from
the boundary faces of P corresponding to Dv . It follows that after putting SG into normal
form in P we will have a normal triangle (an ideal triangle surrounding Dv ) corresponding
to the part of SG around v. Now SG will consist of a collection of ideal triangles and the
only contributions to a(SG ) will come from these ideal triangles. Let T be such a triangle

125

and let γi , i = 1, 2, 3, denote the three arcs of ∂T on ∂E(L). By Definition 4.1.3 and the
definition of combinatorial lengths we have

a(T ) = π = 3 ·

π
= Σ3i=1 l(γi , T ).
3

The number of ideal triangles in Sg is at most 2tπ
3 . Hence the calculation in the proof of
Theorem 4.2.7 gives
a(SG ) =

2tπ
,
3

and by Theorem 4.1.4 we obtain −χ(SG ) = −χ(SG ) = 3t . Now if SG is is non-orientable
then, by Theorem 4.2.3, C(K) = C(SG ) = −χ(S) + 2 − k. If SG is orientable then
C(K) = 2g(K) + 1 = −χ(SG ) + 3 − k. In both cases the desired result follows, since D(K)
is prime and twist reduced and we have t = TK .
Example 4.4.3. Let D(K) = P (p1 , p2 , . . . , pN ) denote the standard diagram of an alternating N -string pretzel knot. Suppose that, for i = 1, . . . , N , |pi | > 1. We can see that the
surface S corresponding to Theorem 4.2.3 is the pretzel surface consisting of two disks and
N -twisted bands; one for each twist region of D(K). Augment the diagram D(K) by adding
a crossing circle at each twist region so that the surface S intersects each crossing disk in a
single arc only. This surface is disjoint from the crossing circles of the fully augmented link
L. Furthermore, the surface S is essential in E(K) and thus in E(L); see argument in the
proof of Theorem 4.5.2 below. Hence we may put it in normal form to calculate the area
a(S). Again by inspecting the combinatorics of the polyhedra decomposition as we did in
the proof of Corollary 4.4.2, that the contributions to a(S) come from two identical, normal
N -gons, D1 , D2 such that ∂Di intersects N -boundary faces of the polyhedral decomposition

126

and it intersects no interior edges. We have

a(S) = a(Di ) + a(D2 ) = 2πN − 4π = 2π(TK − 2),

and thus −χ(S) = N −2. If all, but one, pi are odd the S is non-orientable and thus C(K) =
N − 1 = TK − 1. If all the pi ’s are odd, then S is orientable and then, C(K) = N = TK .
Note that the crosscap numbers of pretzel knots have been calculated in [64].

4.4.2

Low crossing knots

Next we turn our attention to knots with small crossing numbers: The crosscap numbers
of all alternating links up to 9 crossings are known. KnotInfo [19] provides an upper and
a lower bound for the crosscap numbers of all knots up to 12 crossings for which the exact
values of crosscap numbers are not known. Note that in all, but a handful of cases, where
the crosscap number is not determined, the lower bound given in KnotInfo is 2. There
are 1778 prime, alternating knots with crossing numbers 10 ≤ c ≤ 12. For these knots
we calculated the quantity TK := |βK | + βK using the Jones polynomial value given in
KnotInfo and we compared our crosscap number lower bound with the one given in therein.
For 1472 of these knots our lower bound is better than the one given in KnotInfo and for
262 of them our lower bound agrees with the upper bound in there; thus we are able to
calculate the exact value of the crosscap number in these cases. The remaining 306 knots
are those where the exact crosscap number values are given in [19] or our lower bound
is the same as the one in there. The computer file of our calculations can be found at
http://www.math.msu.edu/∼kalfagia/crosscapdata.txt.
For example, in Table 4.1 we have all the 37 alternating knots K for which KnotInfo

127

K
1085
11a97
11a263
11a323
12a0636
12a0845
12a1031
12a1142
12a1220
12a1285

TK
6
5
4
6
5
5
5
5
6
4

K
1093
11a223
11a279
11a330
12a0641
12a0970
12a1095
12a1171
12a1240
-

TK
6
5
6
6
4
6
6
6
6
-

K
10100
11a250
11a293
11a338
12a0753
12a0984
12a1107
12a1179
12a1243
-

TK
6
5
6
4
5
6
6
6
4
-

K
11a74
11a259
11a313
11a346
12a0827
12a1017
12a1114
12a1205
12a1247
-

TK
5
5
6
6
5
6
6
6
6
-

Table 4.1: Knots where the KnotInfo upper bound agrees with our lower bound. The crosscap
number is 3.
states 2 ≤ C(K) ≤ 3, together with the value of the corresponding quantity TK . In all
the 37 cases the lower bound above is also 3; thus for all these knots we can determine the
crosscap number to be 3.

4.5
4.5.1

Generalizations and questions
Non-alternating links

A question arising from this work is the question of the extent to which the Jones polynomial
(coarsely) determines the crosscap number outside the class of alternating links. This is an
interesting question that merits further investigation. Our contribution towards an answer to
this question is to provide a generalization of Theorem 1.1.3 for some class of non-alternating
links. To state our result we need a definition.
Definition 4.5.1. For a link diagram D(K) let SA denote the state surface corresponding
to the Kauffman state where all the crossings are resolved one way and let SB denote the

128

state surface corresponding to the state where all crossings are resolved the opposite way.
A link K is called adequate if it admits a link diagram D(K) such that none of SA , SB
contains a half-twisted band with both ends attached on the same state circle.
Adequate links form a large class that contains the alternating ones but it is much wider.
See [32, 34, 79].
Theorem 4.5.2. Let K be a k-component link with crosscap number C(K) and let TK be
as above. Suppose that K admits a connected, twist-reduced, diagram D(K) that has t ≥ 2
twist regions, and such that each twist region of D(K) contains at least six crossings. We
have

t
3

+ 2 − k ≤ C(K) ≤ t + 2 − k.

If moreover D(K) is adequate then we have

TK
6

+ 2 − k ≤ C(K) ≤ 3TK − k − 1.

Proof. Let S be a spanning surface of K that is of maximal Euler characteristic over all
spanning surfaces (that is both oriented and non-oriented). Then S gives gives an incompressible and ∂-incompressible surface in E(K). For, surgery of S along a compression or
a ∂-compression disk will produce a surface of higher Euler characteristic (compare proof
of Lemma 4.2.6). Let L be a fully augmented link obtained from D(K). Recall that each
crossing circle added in this process bounds a disk D whose interior is pierced exactly twice
by K. We will isotope S in the complement of K so that S ∩ D is minimized. In this
process, since S is incompressible, if there is a simple closed curve in δ ⊂ S ∩ D that bounds
a disk in D with its interior disjoint from S, then we can eliminate δ by isotopy of S in the
129

complement of L. Similarly we can eliminate arc components of S ∩ D that cut off disks on
D with their interior disjoint from S. Finally, simple closed curves that are parallel to ∂D
can be eliminated by sliding S off the boundary of D.
The surface S gives rise to a surface F in E(L). We claim that F is incompressible and
∂-incompressible in E(L). To see that, suppose that there is an essential simple closed curve
γ ⊂ S that bounds a compressing disk in ∆ ⊂ E(L). Since S is incompressible, γ must
bound a disk ∆ ⊂ S whose interior is intersected by the crossing circles of L. Now ∆ ∪ ∆
bounds a 3-ball that can be used to produce an isotopy that reduces the intersection of S
and the crossing disks of L; contradiction.
Now we argue that F is ∂-incompressible. To that end, suppose that F admits a ∂compression disk ∆, with ∂∆ = γ ∪ δ, where γ is a spanning arc in F and δ is an arc on
a component T ⊂ ∂E(L). Assume, for a moment, that T corresponds to a component of
K. Since S is ∂-incompressible in E(K), the arc γ must cut a disk ∆ ⊂ S whose interior
is pieced by the crossing circles of L. Since S is incompressible, the boundary of the disk
∆ ∪ ∆ also bounds a disk ∆ ⊂ S. Now ∆ ∪ ∆ ∪ ∆ bounds a 3-ball that can be used to
produce an isotopy that reduces the intersections of S with the crossing disks of L. This is
a contradiction.
Suppose now that T is a component of ∂E(L) that corresponds to a crossing circle of
L. Since S intersects crossing circles an even number of times, T ∂S has at least two
components. Thus δ lies on an annulus of A ⊂ T ∂S and either it cuts off a disk on A
or it runs between different components of A. Now the usual argument that shows that
an orientable, spanning surface of a link that is incompressible, has to be ∂-incompressible
applies to obtain a contradiction (see [53, Lemma 1.10]). Thus the punctured surface S is
essential in E(L).
130

Now we may replace S by surface, of the same orientability and Euler characteristic, that
is in normal form with respect to the polyhedral decomposition of E(L) [30, Theorem 2.8].
Now Corollary 4.1.9 applies.
If S is non-orientable, we have C(S) = C(K) and by Corollary 4.1.9 we have

C(K) ≥

t
3

+ 2 − k.

If S is orientable then C(K) = 2g(S) + 1 = 2g(K) + 1 and by Theorem 4.1.8 again we have
C(K) ≥ 3t + 2 − k. Combining these inequalities with Lemma 4.2.5 we have
t
3

+ 2 − k ≤ C(K) ≤ t + 2 − k,

which proves the first part of the theorem. Suppose now that D(K) is also adequate.
Then [32, Theorem 1.5] implies that

t
3

≤ TK ≤ 2t.

Now combining the last two inequalities gives the desired result.
Theorem 4.0.1 should be compared with Murasugi’s classical result [90] that the Alexander polynomial determines the Seifert genus of alternating knots. The Alexander polynomial
doesn’t determine the genus of non-alternating knots. However the Knot Floer Homology
(the categorification of the Alexander polynomial) determines the genus of all knots [94].
A related question is the question of whether the Khovanov homology [73] of knots (the
categorification of the Jones polynomial) is related to the crosscap number and the extent to
which the former determines the later. We close the subsection with the following questions:
131

Question 4.1. For which links does the Jones polynomial (coarsely) determine the crosscap
number?
Question 4.2. Are there two sided bounds of C(K) of every link K in terms of the Khovanov
homology of K?
2 (q) and β be the
For an adequate knot K, let β2 be the coefficient of q d(2) + 1 of JK
2
∗
2 (q). Recall that these are stable coefficients of the colored Jones
coefficient of q d (2)−1 of JK

polynomial, see Section 2.1.2 for details. The proof of Theorem 1.1.3 as well as Corollary
β +|βK |
TK
= K 2π
is related, and is often
4.4.2 and Example 4.4.3 show that the quantity 2π
equal, to the combinatorial areas of a normal surfaces in augmented links of K. In view of
this, one can also ask
Question 4.3. Do the higher order stable coefficients of the colored Jones polynomial have
similar interpretations in terms of topological quantities in augmented links from K?

132

Chapter 5
The Jones polynomial, 3-braids, and
L-space knots
An L-space knot generalizes the notion of knots which admit lens space surgeries. Recall
that a rational homology 3-sphere Y is an L-space if |H1 (Y ; Z)| = rank HF (Y ), where HF
denotes the ‘hat’ version of Heegaard Floer homology, and the name stems from the fact
that lens spaces are L-spaces. Besides lens spaces, examples of L-spaces include all connected
sums of manifolds with elliptic geometry [95]. A knot, K ⊂ S 3 , is an L-space knot if K or
its mirror image admits a positive L-space surgery.
Ozsv´ath and Szab´o’s result states that L-spaces admit no co-orientable taut foliations [94,
Theorem 1.4]. It is also known that an L-space knot K ⊂ S 3 must be prime [74] and
fibered [92], and that K supports the tight contact structure on S 3 [59, Proposition 2.1]. In
addition, the Alexander polynomial

K (t)

of an L-space knot K satisfies the following:

• The absolute value of a nonzero coefficient of

K (t) is 1.

The set of nonzero coefficients

alternates in sign [95, Corollary 1.3].
• If g is the maximum degree of

K (t)

in t, then the coefficients of the term tg−1 is

nonzero and therefore ±1 [60].
In this chaper, we study which 3-braids, that close to form a knot, admit L-space surgeries.
We prove that:
133

Theorem 1.1.4. Twisted (3, q) torus knots are the only knots with 3-braid representations
that admit L-space surgeries.
We begin by computing certain coefficients of the Jones polynomials of closed 3-braids
in Section 5.1. The Alexander polynomial of a closed 3-braid may be written in terms of the
Jones polynomial [14]. This allows us to use our computation of the Jones polynomials to
rule out closed 3-braids whose Alexander polynomials violate the aforementioned conditions.
We state definitions and results from [31] on 3-braids that we need, and we use them to prove
Theorem 1.1.4 in Section 5.2.

5.1

The Jones polynomial, 3-braids, and the Alexander
polynomial

We will first derive an expression of the Alexander polynomial of a closed 3-braid in terms
of the Jones polynomial. Let Bn be the n-string braid group. The Burau representation of
Bn is a map ψ from Bn to n − 1 × n − 1 matrices with entries in Z[t, t−1 ].

ψ : Bn → GL(n − 1, Z[t, t−1 ]).

For n = 3, ψ is defined explicitly on the generators σ1 , σ2 (see Figure 5.1) as




 −t 1 
ψ(σ1−1 ) = 
,
0 1

134





 1 0 
ψ(σ2−1 ) = 
.
t −t

1

n

i

Figure 5.1: The generator σi . Note that this convention is the opposite of that of [14].
Let a be an element of B3 , a
ˆ be the closed braid, and ea be the exponent sum of a. Note
that when a
ˆ is a knot, 2 ± ea is even and therefore ea is even. The Jones polynomial Jaˆ (t)
of a
ˆ can be written in terms of ψ [68]:
√
Jaˆ (t) = (− t)−ea (t + t−1 + trace ψ(a))

(5.1)

The sign change on ea is due to the difference in convention as indicated in Figure 5.1. When
n = 3, the Alexander polynomial of a
ˆ may be written in terms of the trace of ψ. [14, Eq.
(7)]

(t−1 + 1 + t) aˆ (t) = (−1)−ea (t−ea /2 − tea /2 trace ψ(a) + tea /2 )

(5.2)

Rearranging equations (5.1) and (5.2) above, we have the symmetric Alexander polynomial
of a closed 3-braid re-written in terms of the Jones polynomial.

(t−1 + 1 + t) aˆ (t) = (−1)−ea (−(−1)ea tea Jaˆ (t) + tea /2+1 + tea /2−1 + t−ea /2 + tea /2 ). (5.3)

This expression allows us to compute certain coefficients of the Alexander polynomial
135

from the Jones polynomial. By Birman and Menasco’s solution [15] to the classification of
3-braids, there are finitely many conjugacy classes of B3 , and each 3-braid is isotopic to
a representative of a conjugacy class. Schreier’s work [102] puts each representative of a
conjugacy class in a normal form.
Theorem 5.1.1. (Schreier) Let b ∈ B3 be a braid on three strands, and C be the 3-braid
(σ1 σ2 )3 . Then b is conjugate to a braid in exactly one of the following forms:
p −q
p −q
1. C k σ1 1 σ2 1 · · · σ1 s σ2 s , where k ∈ Z and pi , qi and s are all positive integers,
p

2. C k σ1 , for k, p ∈ Z,
3. C k σ1 σ2 , for k ∈ Z,
4. C k σ1 σ2 σ1 , for k ∈ Z, or
5. C k σ1 σ2 σ1 σ2 , for k ∈ Z.

It suffices to study the 3-braids among the conjugacy representatives above to determine
p

which closed 3-braids is an L-space knot. It is straightforward to check that C k σ1 and
C k σ1 σ2 σ1 represent links for any k, p ∈ Z. Also, by noting that C ∼ (σ1 σ2 )3 , we get that,
for any k ∈ Z, C k σ1 σ2 and C k σ1 σ2 σ1 σ2 represent the (3, 3k + 1) and (3, 3k + 2) torus
knots, respectively. Thus we will only need to study class (1) of conjugacy representatives
of Theorem 5.1.1.
Recall that if a knot K is a L-space knot, then the absolute value of a nonzero coefficient
of the Alexander polynomial

K (t)

is 1, and the nonzero coefficients alternate in sign.

Moreover, let g be the maximum degree of

K (t)

in t, then the coefficients of the term tg−1

is nonzero and therefore ±1. Since the Alexander polynomial is symmetric, it has the two
136

possible forms given below for an L-space knot.

tg − tg−1 + · · · + terms in-between − t−(g−1) + t−g

or

−tg + tg−1 + · · · + terms in-between + t−(g−1) − t−g .

Either way, when we take the product

−1
K (t) · (t

+ 1 + t).

The result is a symmetric polynomial with coefficients in {−1, 0, 1}, which do not necessarily
alternate in sign, and the second coefficient and the second-to-last coefficient are zero.

The conjugacy representatives of class (1) in Theorem 5.1.1 are called generic 3-braids.
That is, b ∈ B3 is generic if it has the following form.
p −q
p −q
b = C k σ1 1 σ2 1 · · · σ1 s σ2 s ,

where pi , qi , k ∈ Z, with pi , qi > 0, and C = (σ1 σ2 σ1 )2 = (σ1 σ2 )3 by the braid relations
p −q
p −q
σ1 σ2 σ1 = σ2 σ1 σ2 . The braid a = σ1 1 σ2 1 · · · σ1 s σ2 s is called an alternating 3-braid. The

first three coefficients and the last three coefficients of the Jones polynomial for this class
of 3-braids, as well as the degree, are explicitly calculated in [31]. We assemble below the
results we will need.
137

p −q
p −q
Definition 5.1.2. For an alternating braid a = σ1 1 σ2 1 · · · σ1 s σ2 s , let
s

p :=
i=1

s

pi , and q :=

i=1

qi ,

so the exponent sum ea = p − q.
Lemma 5.1.3. [31, Lemma 6.2] Suppose that a link a
ˆ is the closure of an alternating 3-braid
a:
p −q
p −q
a = σ1 1 σ2 1 · · · σ1 s σ2 s ,

with pi , qi > 0 and p > 1 and q > 1, then the following holds
(a) The highest and lowest powers, M (ˆ
a) and m(ˆ
a) of Jaˆ (t) in t are

M (ˆ
a) =

q − 3p
3q − p
and m(ˆ
a) =
.
2
2

(b) The first two coefficients α, β from M (ˆ
a) , and the last two coefficients β , α in Jaˆ (t)
from m(ˆ
a) are

α = (−1)p , β = (−1)p+1 (s − q ), β = (−1)q+1 (s − p ), α = (−1)q .

where p = 1 if p = 2 and 0 if p > 2, and similarly for q .
(c) [31, in the proof of Lemma 6.2] Let γ, γ denote the third and the third-to-last coefficient
of Jaˆ (t), respectively. We have
(−1)p γ =

s2 + 3s
− #{i : pi = 1} − #{i : qi = 1} − δq=3 ,
2
138

and

(−1)q γ =

s2 + 3s
− #{i : pi = 1} − #{i : qi = 1} − δp=3 ,
2

where δq=3 is zero if q = 3 and 1 otherwise, and δp=3 is similarly defined.
The next result writes the Jones polynomial of a generic 3-braid in terms of the Jones
polynomial of an alternating braid.
Lemma 5.1.4. [31] If b is a generic braid of the form

b = C k a,

where a is an alternating 3-braid, and let Jˆb (t) denote the Jones polynomial of ˆb, then
√
Jˆb (t) = t−6k Jaˆ (t) + (− t)−ea (t + t−1 )(t−3k − t−6k ).
If K is a knot which is the closure of a generic 3-braid b = C k a, then by equation (5.3),
we have

−1
K (t) · (t

+ 1 + t)

= (−1)−eb (−(−1)eb teb Jˆb (t) + teb /2+1 + teb /2−1 + t−eb /2 + teb /2 ).

139

Putting this together with Lemma 5.1.4 and noting that eb must be even in order for ˆb to
be a knot, we get that the right side is equal to
√
= −teb (t−6k Jaˆ (t) + (− t)−ea (t + t−1 )(t−3k − t−6k ))
+ teb /2+1 + teb /2−1 + t−eb /2 + teb /2 .

Since b = C k a and eb = 6k + ea , we have

−1
K (t)(t

+ 1 + t)

√
= −t(6k+ea ) (t−6k Jaˆ (t) + (− t)−ea (t + t−1 )(t−3k − t−6k ))
+ t(6k+ea )/2+1 + t(6k+ea )/2−1 + t−(6k+ea )/2 + t(6k+ea )/2
= −tea Jaˆ (t) + tea /2−1 + tea /2+1 + t−3k−ea /2 + t3k+ea /2 .

5.2

(5.4)

Proof of Theorem 1.1.4

We are now ready to determine which closed 3-braids are L-space knots.
Proof. For K = ˆb to be an L-space knot, the right side of equation (5.4) has to be have
coefficients in ±1. This immediately restricts s to be less than 4, since s ≥ 4 implies that
there will be two coefficients β, β , the second and the penultimate, whose absolute values are
greater than or equal to 4 in Jaˆ (t) by (b) of Lemma 5.1.3. Based on (5.4), This will result in at
least one coefficient whose absolute value is greater than or equal to 2 in

−1
K (t)(t + 1 + t),

even after possible cancellation from the terms tea /2−1 , tea /2+1 , t−3k−ea /2 , and t3k+ea /2 .
Similarly, if s = 3, then p, q ≥ 3, and |β| and |β | are both greater than or equal to 3, each

140

of which would need to be cancelled out by at least two terms of tea /2−1 , tea /2+1 , t−3k−ea /2 ,
and t3k+ea /2 . In addition, part (c) of Lemma 5.1.3 gives that the absolute values of the
third and third-to-last coefficients γ, γ are greater than or equal to 2, which would also need
to be cancelled out in the sum of the right side of (5.4). This is impossible, so s = 3.
Now assume that s = 2. If p > 2, then the penultimate coefficient |β | = 2 and the
third coefficient |γ| ≥ 2. Since p, q are either both even or both odd, β and γ have
opposite signs. One of them is positive, which would not cancel out with any of the terms
tea /2−1 , tea /2+1 , t−3k−ea /2 , and t3k+ea /2 . The case is similar for q > 2, so we must have
that p = 2 and q = 2. This means that a = σ1 σ2−1 σ1 σ2−1 . The Jones polynomial of this
alternating closed braid is
Jaˆ(t) = t−2 − t−1 + 1 − t + t2 ,
for a braid b = C k a,
1
(t−1 + 1 + t) ˆb (t) = −Jaˆ (t) + + t + t−3k + t3k
t
= −(t−2 − t−1 + 1 − t + t2 ) +

1
+ t + t−3k + t3k
t

= −t−2 + 2t−1 − 1 + 2t − t2 + t−3k + t3k

For all k = 0, this shows that the product on the left side of this equation has nonzero
coefficients that are not ±1. This rules out the possibility that a closed 3-braid of this form
can be an L-space knot.
Therefore, we need only to consider the case when s = 1. Assuming that both p, q are
greater than 1, the absolute values of the third coefficient γ and the third-to-last coefficient

141

γ of −teα Jαˆ (t) have the form
s2 + 3s
− #{i : pi = 1} − #{i : qi = 1} − δq=3
2

t(q+p)/2−2

γ

and
s2 + 3s
− #{i : pi = 1} − #{i : qi = 1} − δp=3
2

t−(q+p)/2+2 ,

γ

respectively. If |γ| or |γ | is 2, they need to be canceled out by at least one term out of

tea /2−1 , tea /2+1 , t−3k−ea /2 , t3k+ea /2 .

If q > 3 and p > 1, then γ = 2, and (q + p)/2 − 2 needs to be equal to (p − q)/2 − 1,
(p − q)/2 + 1, −3k − (p − q)/2, or 3k + (p − q)/2. Similarly, we have the constraints on
−(q + p)/2 + 2. We examine the resulting equations and rule out values of p and q which
lead to contradictions. Setting (q + p)/2 − 2 equal to (p − q)/2 − 1 or (p − q)/2 + 1 gives
q = 1 or q = 3. Setting (q + p)/2 − 2 equal to −3k − (p − q)/2 or 3k + (p − q)/2 gives
p = −3k + 2 or q = 3k + 2. Similarly, if p > 3 and q > 1, then we must have p = 3, p = 1,
q = 3k + 2, or p = −3k + 2. We dismiss the cases where p or q ≤ 3 for now, and suppose
that k = 0. We cannot have that p = −3k + 2 and q = 3k + 2 since they are both supposed
to be positive. Therefore we suppose that p = −3k + 2 or q = 3k + 2. In the first case, k is

142

negative. In the second case, k is positive. Either way, we end up having, for k < 0,

−1
K (t) · (t

= (±t−
+t

+ 1 + t)

−3k+2+q
2

−3k+2−q
−1
2

∓ t−

+t

−3k+2+q
+1
2

± 0 ∓ ··· ± 0 ∓ t

−3k+2+q
−1
2

±t

−3k+2+q
2
)

−3k+2−q
+1
2
,

or, for k > 0,

−1
K (t) · (t

= (±t−
+t

+ 1 + t)

3k+2+p
2

∓ t−

−(3k+2)+p
−1
2

+t

3k+2+p
+1
2

± 0 ∓ ··· ± 0 ∓ t

3k+2+p
−1
2

±t

3k+2+p
2
)

−(3k+2)+p
+1
2

One of the conditions on the product

−1
K (t) · (t + 1 + t)

is that the second and the penulti-

−3k+2+q
−3k+2+q
+1
−1
2
2
mate coefficients are equal to zero. When k < 0, the terms t−
,t
are

the second and the penultimate coefficient which is not zero since we assume p, q > 3, this
is impossible so this case cannot happen. The same argument applies to rule out the second
case when k > 0.
When both p, q = 3, we have that the alternating 3-braid a takes the form σ13 σ2−3 . The
Alexander polynomial of this alternating 3-braid is

a (t)

1
2
= 3 + 2 − − 2t + t2 ,
t
t

obtained by multiplying the Alexander polynomial of the trefoil by itself, since this 3-braid
is a connected sum of two (right-hand and left-hand) trefoils. It is clear from the Alexander

143

polynomial that this knot cannot be an L-space knot due to the fact that several of its nonzero
coefficients are not ±1. Now we consider a generic 3-braid b = C k a with a = σ13 σ2−3 . Since
ea = 0, the highest degree and the lowest degree of the Jones polynomial of a
ˆ are 3 and −3.
By equation (4),
1
(t−1 + 1 + t) ˆb (t) = −Jaˆ (t) + + t + t−3k + t3k ,
t
where
1
1
1
Jaˆ (t) = 3 − 3 + 2 − − t + t2 − t3 .
t
t
t
When k = 0, it is clear that the constant term 3 of Jaˆ (t) will not be canceled out by the
terms 1t , t, t−3k , or t−3k . Thus none of the closure of braids of the form C k σ13 σ2−3 will be an
L-space knot. We may also rule out the case p or q = 2 since this would give a link rather
than a knot. Thus, the only generic 3-braids whose closure can be an L-space knot are given
below.
−q

C k σ11 σ2
p
C k σ1 σ2−1

for q odd.
for p odd.

p

We now claim that C k σ1 σ2−1 , for p odd and k > 0, represents an L-space knot. Note
that:

p

p

(σ1 σ2 σ1 )2k σ1 σ2−1 ∼ (σ2 σ1 )3k σ1 σ2−1
p+1

∼ (σ2 σ1 )3k−1 σ1

.

The latter braid is the twisted torus knot, K(3, 3k − 1; 2, 1), which is known to be an

144

L-space knot [110, Corollary 3.2]. Now if k < 0 then

p

p

(σ1 σ2 σ1 )2k σ1 σ2−1 ∼ (σ1−1 σ2−1 σ1−1 σ1−1 σ2−1 σ1−1 )−k σ1 σ2−1
p

∼ σ2−1 (σ1−1 σ2−1 σ1−1 σ1−1 σ2−1 σ1−1 )−k σ1

p

∼ σ2−1 (σ1−1 σ2−1 σ1−1 σ2−1 σ1−1 σ2−1 )−k σ1

p+1

∼ σ2−1 (σ1−1 σ2−1 σ1−1 σ2−1 σ1−1 σ2−1 )−k σ1−1 σ1 σ1
p+1

∼ (σ2−1 σ1−1 )−3k+1 σ1
p+1

∼ (σ1 σ2 )3k−1 σ1

.

Using [110, Corollary 3.2], we get that the closure of the latter braid represents an L-space
knot. We should point out that [110, Corollary 3.2], as stated, only holds for positive
knots. However, it turns out that every twisted (p, q) torus knot, where the twisting happens
between p − 1 strands, admits an L-space surgery from the proof of [110, Theorem 3.1]. A
−q

similar argument shows that the closure of C k σ11 σ2 , for q odd, also represents an L-space
−p−1

knot. Notice that in this case the knot is isotopic to the closure of (σ2 σ1 )1−3k σ1
mirror image admits a positive L-space surgery.

145

. So its

BIBLIOGRAPHY

146

BIBLIOGRAPHY
[1] Colin Adams and Thomas Kindred. A classification of spanning surfaces for alternating
links. Algebr. Geom. Topol., 13(5):2967–3007, 2013.
[2] Colin C. Adams. Thrice-punctured spheres in hyperbolic 3-manifolds. Trans. Amer.
Math. Soc., 287(2):645–656, 1985.
[3] Colin C. Adams. Noncompact Fuchsian and quasi-Fuchsian surfaces in hyperbolic
3–manifolds. Alebr. Geom. Topol., 7:565–582, 2007.
[4] Ian Agol. Criteria for virtual fibering. J. Topol., 1(2):269–284, 2008.
[5] Ian Agol, Joel Hass, and William Thurston. The computational complexity of knot
genus and spanning area. Trans. Amer. Math. Soc., 358(9):3821–3850, 2006.
[6] Ian Agol, Peter A. Storm, and William P. Thurston. Lower bounds on volumes of
hyperbolic Haken 3-manifolds. J. Amer. Math. Soc., 20(4):1053–1077, 2007. with an
appendix by Nathan Dunfield.
[7] E. M. Andreev. Convex polyhedra in Lobaˇcevski˘ı spaces. Mat. Sb. (N.S.), 81 (123):445–
478, 1970.
[8] E. M. Andreev. Convex polyhedra of finite volume in Lobaˇcevski˘ı space. Mat. Sb.
(N.S.), 83 (125):256–260, 1970.
[9] Cody Armond. The head and tail conjecture for alternating knots. Algebr. Geom.
Topol., 13(5):2809–2826, 2013.
[10] Cody Armond and Oliver T. Dasbach. The head and tail of the colored Jones polynomial for adequate knots. arXiv:1310.4537.
[11] Cody Armond and Oliver T. Dasbach. Rogers–Ramanujan type identities and the head
and tail of the colored Jones polynomial. arXiv:1106.3948.
[12] Kenneth L. Baker and Allison H. Moore. Montesinos knots, Hopf plumbings, and
L-space surgeries. arXiv:1404.7585.
[13] Joshua Batson. Nonorientable four-ball genus can be arbitrarily large. arXiv:1204.1985.
147

[14] Joan Birman. On the jones polynomial of closed 3-braids. Invent. Math., 81(2):287–
294, 1985.
[15] Joan S. Birman and Willam W. Menasco. Studying links via closed braids. i. Pacific
J. Math., 154(1):17–36, 1992.
[16] Gerhard Burde and Heiner Zieschang. Knots, volume 5 of de Gruyter Studies in
Mathematics. Walter de Gruyter & Co., Berlin, second edition, 2003.
[17] Benjamin A. Burton and Melih Ozlen. Computing the crosscap number of a knot using
integer programming and normal surfaces. ACM Trans. Math. Software, 39(1):Art. 4,
18, 2012.
[18] Flath Carter, J. Scott, Daniel E., and Saito Masahico. The Classical and Quantum
6j-symbols, volume 43 of Mathematical Notes. Princeton University Press, 1995.
[19] Jae Choon Cha and Charles Livingston. Knotinfo: Table of knot invariants. http://
www.indiana.edu/~knotinfo, June 14 2014.
[20] Bradd Evans Clark. Crosscaps and knots. Internat. J. Math. Math. Sci., 1(1):0161–
1712, 1978.
[21] J. H. Conway. An enumeration of knots and links, and some of their algebraic properties. In Computational Problems in Abstract Algebra (Proc. Conf., Oxford, 1967),
pages 329–358. Pergamon, Oxford, 1970.
[22] Richard Crowell. Genus of alternating link types. Ann. of Math. (2), 69:258–275, 1959.
[23] Oliver T. Dasbach, David Futer, Efstratia Kalfagianni, Xiao-Song Lin, and Neal W.
Stoltzfus. The Jones polynomial and graphs on surfaces. Journal of Combinatorial
Theory, Series B, 98:384–399, 2008.
[24] Oliver T. Dasbach, David Futer, Efstratia Kalfagianni, Xiao-Song Lin, and Neal W.
Stoltzfus. Alternating sum formulae for the determinant and other link invariants. J.
Knot Theory Ramifications, 19(6):765–782, 2010.
[25] Oliver T. Dasbach and Xiao-Song Lin. On the head and the tail of the colored Jones
polynomial. Compositio Math., 142(5):1332–1342, 2006.
[26] Oliver T. Dasbach and Xiao-Song Lin. A volume-ish theorem for the Jones polynomial
of alternating knots. Pacific J. Math., 231(2):279–291, 2007.
148

[27] Nathan Dunfield. Boundary slopes of montesinos knots. http://www.math.uiuc.edu/
~nmd/montesinos/index.html.
[28] Nathan M. Dunfield and Stavros Garoufalidis. Incompressibility criteria for spunnormal surfaces. Trans. Amer. Math. Soc., 364(11):6109–6137, 2012.
[29] W. Floyd and U. Oertel. Incompressible surfaces via branched surfaces. Topology,
23(1):117–125, 1984.
[30] David Futer and Fran¸cois Gu´eritaud. Angled decompositions of arborescent link complements. Proc. Lond. Math. Soc. (3), 98(2):325–364, 2009.
[31] David Futer, Efstratia Kalfagianni, and Jessica Purcell. Cusp areas of farey manifolds
and applications to knot theory. Int. Math. Res Not., (23):4434–4497, 2010.
[32] David Futer, Efstratia Kalfagianni, and Jessica S. Purcell. Dehn filling, volume, and
the Jones polynomial. J. Differential Geom., 78(3):429–464, 2008.
[33] David Futer, Efstratia Kalfagianni, and Jessica S. Purcell. Slopes and colored Jones
polynomials of adequate knots. Proc. Amer. Math. Soc., 139(5):1889–1896, 2011.
[34] David Futer, Efstratia Kalfagianni, and Jessica S. Purcell. Guts of surfaces and the
colored Jones polynomial, volume 2069 of Lecture Notes in Mathematics. Springer,
Heidelberg, 2013.
[35] David Futer, Efstratia Kalfagianni, and Jessica S. Purcell. Jones polynomials, volume,
and essential knot surfaces: a survey. In Proceedings of Knots in Poland III, volume
100, pages 51–77. Banach Center Publications, 2014.
[36] David Futer, Efstratia Kalfagianni, and Jessica S. Purcell. Quasifuchsian state surfaces.
Trans. Amer. Math. Soc., 366:4323–4343, 2014.
[37] David Futer and Jessica S. Purcell. Links with no exceptional surgeries. Comment.
Math. Helv., 82(3):629–664, 2007.
[38] David Gabai. The Murasugi sum is a natural geometric operation. In Low-dimensional
topology (San Francisco, Calif., 1981), volume 20 of Contemp. Math., pages 131–143.
Amer. Math. Soc., Providence, RI, 1983.
[39] David Gabai. Detecting fibred links in S 3 . Comment. Math. Helv., 61(4):519–555,
1986.
149

[40] Stavros Garoufalidis. The degree of a q-holonomic sequence is a quadratic quasipolynomial. Electron. J. Combin., 18(2):Paper 4, 23, 2011.
[41] Stavros Garoufalidis. The Jones slopes of a knot. Quantum Topology, 2:43–69, 2011.
[42] Stavros Garoufalidis and Thang T. Q. Lˆe. Nahm sums, stability and the colored jones
polynomial. Research in Mathematical Sciences, to appear.
[43] Stavros Garoufalidis and Thang T. Q. Lˆe. The colored Jones function is q-holonomic.
Geom. Topology., (9):1253–1293, 2005.
[44] Stavros Garoufalidis and Thang T. Q. Lˆe. Nahm sums, stability and the colored Jones
polynomial. Research in Mathematical Sciences, to appear.
[45] Stavros Garoufalidis, Sergei Norin, and Thao Vong. Flag algebras and the stable
coefficients of the Jones polynomial. arXiv:1309.5867.
[46] Stavros Garoufalidis and Roland van der Veen. Quadratic integer programming and
the slope conjecture. arXiv:1405.5088.
[47] Stavros Garoufalidis and Thao Vong. A stability conjecture for the colored Jones
polynomial. arXiv:1310.7143.
[48] Paolo Ghiggini. Knot Floer homology detects genus-one fibred knots. American Journal of Mathematics, 130(5):1151–1169, 2008.
[49] Fran¸cois Gu´eritaud and David Futer (appendix). On canonical triangulations of oncepunctured torus bundles and two-bridge link complements. Geom. Topol., 10:1239–
1284, 2006.
[50] Wolfgang Haken. Theorie der Normalfl¨achen. Acta Math., 105:245–375, 1961.
[51] Joel Hass, Jeffrey C. Lagarias, and Nicholas Pippenger. The computational complexity
of knot and link problems. J. ACM, 46(2):185–211, 1999.
[52] Allen Hatcher. Notes on basic 3-manifold topology.
[53] Allen Hatcher. Notes on basic 3-manifold topology. http://www.math.cornell.edu/
~hatcher/3M/3Mdownloads.html.

150

[54] Allen Hatcher. On the boundary curves of incompressible surfaces. Pacific J. Math.,
99(2):373–377, 1982.
[55] Allen Hatcher and U. Oertel.
28(4):453–480, 1989.

Boundary slopes for Montesinos knots.

Topology,

[56] Allen Hatcher and William Thurston. Incompressible surfaces in 2-bridge knot complements. Invent. Math., 79(2):225–246, 1985.
[57] Matthew Hedden. On knot Floer homology and cabling II. International Mathematics
Research Notices, pages 2248–2274, 2009.
[58] Matthew Hedden. Notions of positivity and the Oszvath-Szabo concordance invariant.
J. Knot Theory Ramifications, 19(5):617–629, 2010.
[59] Matthew Hedden. Notions of positivity and the Ozsv´ath-Szab´o concordance invariant.
J. Knot Theory Ramifications, 19(5):617–629, 2010.
[60] Matthew Hedden and Liam Watson. On the geography and botany of knot Floer
homology. preprint, page arXiv:1404.6913, 2014.
[61] Mikami Hirasawa and Masakazu Teragaito. Crosscap numbers of 2-bridge knots. Topology, 45(3):513–530, 2006.
[62] Jennifer Hom. A note on cabling and L-space surgeries. Algebraic & Geometric topology, (1):219–223, 2011.
[63] Jennifer Hom, Faramarz Vafaee, and Tye Lidman. Berge-Gabai knots and L-space
satellite operations. Algebraic & Geometric topology, to appear.
[64] Kazuhiro Ichihara and Shigeru Mizushima. Crosscap numbers of pretzel knots. Topology Appl., 157(1):193–201, 2010.
[65] Kazuhiro Ichihara, Masahiro Ohtouge, and Masakazu Teragaito. Boundary slopes of
non-orientable Seifert surfaces for knots. Topology Appl., 122(3):467–478, 2002.
[66] William Jaco and J. Hyam Rubinstein. PL equivariant surgery and invariant decompositions of 3-manifolds. Adv. in Math., 73(2):149–191, 1989.
[67] Vaughan F. R. Jones. A polynomial invariant for knots via von Neumann algebras.
Bull. Amer. Math. Soc. (N.S.), 12(1):103–111, 1985.
151

[68] Vaughan F. R. Jones. Hecke algebra representations of braid groups and link polynomials. Ann. of Math. (2), 126(2):335–388, 1987.
[69] Efstratia Kalfagianni and Christine Ruey Shan Lee. On the degree of the colored
Jones polynomial. Acta Mathematica Vietnamica (Proceedings of Quantum Topology
and Hyperbolic Geometry in Nha Trang, May 2013), to appear.
[70] Efstratia Kalfagianni and Anh T. Tran. Knot cabling and the degree of the colored
jones polynomial. http://arxiv.org/pdf/1501.01574v2.pdf.
[71] Efstratia Kalfagianni and Anh T. Tran. Knot cabling and the degree of the colored
jones polynomial ii.
[72] Louis H. Kauffman. State models and the Jones polynomial. Topology, 26(3):395–407,
1987.
[73] Mikhail Khovanov. A categorification of the Jones polynomial. Duke Mathematical
Journal, 101(3):359–426, 2000.
[74] David Krcatovich. The reduced knot Floer complex. arXiv:1310.7624.
[75] Marc Lackenby. Word hyperbolic dehn surgery. Invent. Math., 140(2):243–282, 2000.
[76] Marc 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.
[77] Thang T. Q. Lˆe. The colored Jones polynomial and AJ conjecture. http://people.
math.gatech.edu/%7Eletu/Papers/Lectures_Luminy_2014_new.pdf.
[78] W. B. Raymond Lickorish. An Introduction to Knot Theory. Springer-Verlag New
York, Inc., 1997.
[79] W. B. Raymond Lickorish and Morwen B. Thistlethwaite. Some links with nontrivial
polynomials and their crossing-numbers. Comment. Math. Helv., 63(4):527–539, 1988.
[80] Tye Lidman and Allison H. Moore.
arXiv:1306.6707.

Pretzel knots with l-space surgeries.

[81] P. M. G. Manch´on. Extreme coefficients of Jones polynomials and graph theory. J.
Knot Theory Ramifications, 13(2):277–295, 2004.
152

[82] G. Masbaum and P. Vogel. 3-valent graphs and the Kauffman bracket. Pacific J.
Math., 164(2):361–381, 1994.
[83] William W. Menasco. Closed incompressible surfaces in alternating knot and link
complements. Topology, 23(1):37–44, 1984.
[84] William W. Menasco and Morwen B. Thistlethwaite. Surfaces with boundary in alternating knot exteriors. J. Reine Angew. Math., 426:47–65, 1992.
[85] William W. Menasco and Morwen B. Thistlethwaite. The classification of alternating
links. Ann. of Math. (2), 138(1):113–171, 1993.
[86] Kimihiko Motegi. L-space surgery and twisting operation. arXiv:1405.6487.
[87] Hitoshi Murakami. An introduction to the volume conjecture. In Interactions between
hyperbolic geometry, quantum topology and number theory, volume 541 of Contemp.
Math., pages 1–40. Amer. Math. Soc., Providence, RI, 2011.
[88] Hitoshi Murakami and Jun Murakami. The colored Jones polynomials and the simplicial volume of a knot. Acta Math., 186(1):85–104, 2001.
[89] Hitoshi Murakami and Akira Yasuhara. Crosscap number of a knot. Pacific J. Math.,
171(1):261–273, 1995.
[90] Kunio Murasugi. On the Alexander polynomial of the alternating knot. Osaka Math.
J., 10:181–189; errata, 11 (1959), 95, 1958.
[91] Kunio Murasugi. Jones polynomials and classical conjectures in knot theory. Topology,
26(2):187–194, 1987.
[92] Yi Ni. Knot Floer homology detects fibred knots. Invent. Math., 170(3):577–608, 2007.
[93] Makoto Ozawa. Essential state surfaces for knots and links. J. Aust. Math. Soc.,
91(3):391–404, 2011.
[94] Peter Ozsv´ath and Zolt´an Szab´o. Holomorphic disks and genus bounds. Geom. Topol.,
8:311–334, 2004.
[95] Peter Ozsv´ath and Zolt´an Szab´o. On knot Floer homology and lens space surgeries.
Topology, 44(6):1281–1300, 2005.
153

[96] Peter Ozsvath and Zoltan Szabo. On the heegaard floer homology of branched doublecovers. Advances in Mathematics, 194(1):1–33, 2005.
[97] Peter Ozsv´ath and Zolt´an Szab´o. Link Floer homology and the Thurston norm. J.
Amer. Math. Soc., 21(3):671–709, 2008.
[98] Jessica S. Purcell. An introduction to fully augmented links. In Interactions between
hyperbolic geometry, quantum topology and number theory, volume 541 of Contemp.
Math., pages 205–220. Amer. Math. Soc., Providence, RI, 2011.
[99] N. Yu. Reshetikhin and V. G. Turaev. Ribbon graphs and their invariants derived
from quantum groups. Comm. Math. Phys., 127(1):1–26, 1990.
[100] Nikolai Reshetikhin and Vladimir G. Turaev. Invariants of 3-manifolds via link polynomials and quantum groups. Invent. Math., 103(3):547–597, 1991.
[101] Lev Rozansky. Khovanov homology of a unicolored B-adequate link has a tail.
arXiv:1203.5741.
[102] Otto Schreier. U¨ ber die gruppen aabb = 1. Abh. Math. Sem. Univ. Hamburg, (3):167–
169, 1924.
[103] Hermann Schubert. Knoten mit zwei brucken. Math. Zeitschr, 65:133–170, 1956.
[104] Alexander Stoimenow. Non-triviality of the jones polynomial and the crossing numbers
of amphicheiral knots.
[105] Alexander Stoimenow. Coefficients and non-triviality of the Jones polynomial. Journal
f¨
ur die Reine und Angewandte Mathematik, 657:1–55, 2011.
[106] Alexander Stoimenow. On the crossing number of semi-adequate links. Forum Math.,
pages DOI:10.1515/forum–2011–0121, in press.
[107] Masakazu Teragaito. Crosscap numbers of torus knots. Topology Appl., 138(1-3):219–
238, 2004.
[108] Morwen B. Thistlethwaite. On the Kauffman polynomial of an adequate link. Invent.
Math., 93(2):285–296, 1988.

154

[109] Vladimir G. Turaev. Quantum invariants of knots and 3-manifolds, volume 18 of de
Gruyter Studies in Mathematics. Walter de Gruyter & Co., Berlin, revised edition,
2010.
[110] Faramarz Vafaee. On the knot Floer homology of twisted torus knots. Int. Math. Res.
Not. IMRN, page doi: 10.1093/imrn/rnu130, 2014.

155