A COMBINATORIAL APPROACH TO KNOT THEORY: VOLUME BOUNDS FOR
HYPERBOLIC SEMI-ADEQUATE LINK COMPLEMENTS
By
Adam Joseph Giambrone
A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Mathematics - Doctor of Philosophy
2014
ABSTRACT
A COMBINATORIAL APPROACH TO KNOT THEORY: VOLUME
BOUNDS FOR HYPERBOLIC SEMI-ADEQUATE LINK COMPLEMENTS
By
Adam Joseph Giambrone
An interesting goal in knot theory is to discover how much geometric information about a
link can be carried by a representative projection diagram of that link. To this end, we
show that the volumes of certain hyperbolic semi-adequate links can be bounded above and
below in terms of two diagrammatic quantities: the twist number and the number of special
tangles in a semi-adequate diagram of the link. Given this result, we then narrow our focus
to families of plat closures, families of closed braids, and families of links that have both
plat and closed braid aspects. By more closely studying each of these families, we can often
improve the lower bounds on volume provided by the main result. Furthermore, we show
that the bounds on volume can be expressed in terms of a single stable coefficient of the
colored Jones polynomial. By doing this, we provide new collections of links that satisfy a
Coarse Volume Conjecture. The main approach of this entire work is to use a combinatorial
perspective to study the connections among knot theory, hyperbolic geometry, and graph
theory.
ACKNOWLEDGMENTS
I would like to begin by thanking my adviser, Efstratia Kalfagianni, for her unwavering
support and guidance throughout my graduate career. I would also like to thank my mother,
Diane Giambrone, for being a continual source of encouragement and perspective. Finally,
I would like to thank my friends and colleagues David Bramer, Joshua Hallam, Faramarz
Vafaee, and Luke Williams for joining me in this adventure.
iii
TABLE OF CONTENTS
LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vi
KEY TO SYMBOLS AND ABBREVIATIONS . . . . . . . . . . . . . . . . .
xi
Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Chapter 2 Preliminaries . . . . . . . . . . . . . . .
2.1 Definitions . . . . . . . . . . . . . . . . . . . . .
2.2 Volume Bounds . . . . . . . . . . . . . . . . . .
2.3 A-Adequacy and the Colored Jones Polynomial
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Chapter 3 Volume Bounds for A-Adequate Links . . . . . . . . .
3.1 Twist Regions, State Circles, and GA . . . . . . . . . . . . . . . .
3.2 Computation of −χ(GA ) . . . . . . . . . . . . . . . . . . . . . . .
3.3 Special Circles and Special Tangles . . . . . . . . . . . . . . . . .
3.4 Volume Bounds in Terms of t(D) and st(D) (The Main Theorem)
3.5 Volume Bounds in Terms of the Colored Jones Polynomial . . . .
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Polynomial
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Chapter 4 Volume Bounds for A-Adequate Plats . . . .
4.1 Background on Braids and Plat Closures . . . . . . . .
4.2 Volume Bounds for Negative Plats in Terms of t(D) . .
4.3 Applying the Main Theorem to Negative Plats . . . . .
4.4 Volume Bounds for Mixed-Sign Plats in Terms of t(D)
4.5 Applying the Main Theorem to Mixed-Sign Plats . . .
4.6 Volume Bounds for Plats in Terms of the Colored Jones
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Chapter 5 Volume Bounds for A-Adequate Closed Braids
5.1 Braid Closure and Cyclic Reduction of Braid Words . . . .
5.2 A-Adequacy for Closed 3-Braids . . . . . . . . . . . . . . .
5.3 State Circles of A-Adequate Closed 3-Braids . . . . . . . .
5.4 Primeness, Connectedness, and Twistedness for
Closed Braid Diagrams . . . . . . . . . . . . . . . . . . . .
5.5 The Foundational Theorem for Closed Braids . . . . . . .
5.6 State Circles of A-Adequate Closed n-Braids (where n ≥ 4)
5.7 Computation of −χ(GA ) . . . . . . . . . . . . . . . . . . .
5.8 Applying the Main Theorem to Closed Braids . . . . . . .
5.9 Volume Bounds in Terms of t(D) (and t− (D)) . . . . . . .
iv
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5.10 Volume Bounds in Terms of the Colored Jones Polynomial . . . . . . . . . .
Chapter 6 Volume Bounds for A-Adequate Closed 3-Braids in Terms
the Schreier Normal Form . . . . . . . . . . . . . . . . . . . . .
6.1 The Schreier Normal Form for 3-Braids . . . . . . . . . . . . . . . . . . .
6.2 Hyperbolicity for Closed 3-Braids and Volume
Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Volume Bounds in Terms of the Schreier Normal Form . . . . . . . . . .
6.4 The Proof of Theorem 6.3.1. . . . . . . . . . . . . . . . . . . . . . . . . .
of
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Chapter 7 Volume Bounds for Hybrid Braid-Plats . . . . . . . . . . . . . . . 116
7.1 Background on Hybrid Braid-Plats . . . . . . . . . . . . . . . . . . . . . . . 116
7.2 The Foundational Theorem for Hybrid Braid-Plats . . . . . . . . . . . . . . 119
7.3 Computation of −χ(GA (H)) . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
7.4 Volume Bounds in Terms of t(D) and in Terms of the Colored Jones Polynomial123
BIBLIOGRAPHY
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v
LIST OF FIGURES
Figure 2.1
A schematic depiction of non-prime link diagram D(K). The solid
boxes, marked D1 and D2 , contain the remainder of the link diagram.
Each such box is assumed to contain at least one crossing of D(K).
The simple closed curve, C, exhibits the non-primeness of the diagram. 10
Figure 2.2
A crossing neighborhood of a link diagram (middle), along with its
A-resolution (right) and B-resolution (left). . . . . . . . . . . . . . .
11
A link diagram D(K), its all-A resolution HA , its all-A surface SA ,
its all-A graph GA , and its reduced all-A graph GA . . . . . . . . . .
12
Figure 2.4
A link diagram with t(D) = 3 twist regions. . . . . . . . . . . . . . .
13
Figure 2.5
Long and short resolutions of a twist region of D(K). . . . . . . . .
13
Figure 2.6
Special tangles of D(K) and the corresponding special circles, C, of
HA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
The exceptional link diagram consisting of two long twist regions,
each of which must contain at least three crossings. . . . . . . . . .
25
A schematic depiction of the A-resolutions of the three possible types
of twist regions incident to an other circle C of HA : one-crossing
(left), short (middle), and long (right). The edge labels 1, s, and l
are used not only to indicate the type of twist region resolution, but
also to distinguish these edges from A-segments. . . . . . . . . . . .
26
Three possibilities for an other circle C with two incident long resolutions that start and end at C. . . . . . . . . . . . . . . . . . . . .
28
The fourth possibility for an other circle C with two incident long
resolutions that start and end at C. . . . . . . . . . . . . . . . . . .
28
Three possibilities for an other circle C1 with two incident twist region
resolutions, one resolution from a long twist region that starts and
ends at C1 and the other resolution connecting to a different state
circle C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
Figure 2.3
Figure 3.1
Figure 3.2
Figure 3.3
Figure 3.4
Figure 3.5
vi
Figure 3.6
An other circle C with two incident twist region resolutions, one inside
C and one outside C. . . . . . . . . . . . . . . . . . . . . . . . . . .
29
Three possibilities for an “other” circle C with two incident twist
region resolutions, one connecting C to C1 and the other connecting
C to C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
Three remaining possibilities for an other circle C with two incident
twist region resolutions, one connecting C to C1 and the other (a
short twist region resolution) connecting C to C2 . . . . . . . . . . .
29
Figure 3.9
Three possible tangle subdiagrams of D(K). . . . . . . . . . . . . .
30
Figure 4.1
The generators σi and σi−1 of the n-braid group Bn , where 1 ≤ i ≤
n − 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42
A schematic for a 2n-plat diagram with m = 2k + 1 rows of twist
regions, where the entry in the ith row and j th column is a twist
region containing ai,j crossings (counted with sign). The twist regions
depicted above are negative twist regions. Having ai,j > 0 instead
will reflect the crossings in the relevant twist region. . . . . . . . . .
43
Figure 4.3
An example of a negative plat diagram and its all-A state.
. . . . .
44
Figure 4.4
An example of a mixed-sign plat diagram and its all-A state. Note
that the diagram above is obtained from the negative plat diagram
of Figure 4.3 by changing the first negative twist region with three
crossings to a positive twist region with a single crossing and changing
the last negative twist region with three crossings to a positive twist
region with three crossings. These changes create a “secret” small
inner circle and a special circle, respectively. . . . . . . . . . . . . .
45
Figure 5.1
The closure of an n-braid. . . . . . . . . . . . . . . . . . . . . . . . .
62
Figure 5.2
The subdiagram of a closed 3-braid diagram corresponding to a subword σ2−1 σ1−1 σ2−1 (left) and the corresponding portion of the all-A
state that exhibits the non-A-adequacy of D(K) (right). . . . . . . .
64
The subdiagram of a closed 3-braid diagram corresponding to a subp p
word σ1 1 σ2 2 σ1n , where the pi are positive integers and n is a negative
integer (left), and the corresponding portion of the all-A state that
exhibits the non-A-adequacy of D(K) (right). . . . . . . . . . . . . .
65
Figure 3.7
Figure 3.8
Figure 4.2
Figure 5.3
vii
Figure 5.4
Figure 5.5
Figure 5.6
Figure 5.7
Figure 5.8
Figure 5.9
A positive closed 3-braid diagram (left) and the corresponding all-A
state that exhibits the A-adequacy of D(K) (right). . . . . . . . . .
65
Potential non-A-adequacy coming from σ1−1 . From the perspective
of the all-A state HA , the vertical segment s coming from σ1−1 is
assumed to join a state circle C to itself (left). Also depicted are the
case where σ1−1 is followed by another copy of σ1−1 (center) and the
p
case where σ1−1 is followed a positive syllable σ2 , which must then be
followed by the letter σ1−1 (right). . . . . . . . . . . . . . . . . . . .
66
p
Potential non-A-adequacy coming from a positive syllable σ1 . From
the perspective of the all-A state HA , the horizontal A-segments comp
ing from σ1 are assumed to join a state circle C to itself. . . . . . .
67
An example of a small inner circle coming from a σ1−2 subword (left),
an example of a medium inner circle and a potential nonwandering
p
circle portion coming from a σ2−1 σ1 σ2−1 subword where p > 0 (center), and an example of a wandering circle portion coming from a
σ2−2 σ1−2 = σ2−1 (σ2−1 σ1−1 )σ1−1 subword (right). . . . . . . . . . . . . .
68
A schematic depiction of a braid β ∈ Bn that contains two disjoint
induced complete subwords γ1 and γ2 . . . . . . . . . . . . . . . . . .
72
A schematic depiction of the fact that a simple closed curve C containing a point p (between braid string positions i and i + 1) cannot
contain crossings on both sides. Each box in the figure represents the
eventual occurrence of the generator σi , the generator σi−1 (if it exists), and the generator σi+1 (if it exists). No assumptions are made
about the order in which these generators appear or the frequency
with which these generators appear. . . . . . . . . . . . . . . . . . .
73
Figure 7.1
A schematic depiction of a hybrid braid-plat diagram D(K). . . . . 117
Figure 7.2
An example of the all-A state of a hybrid braid-plat. . . . . . . . . . 118
viii
KEY TO SYMBOLS AND ABBREVIATIONS
⊆
subset
K
a link
S3
the 3-sphere
D(K)
a diagram of a link K
S2
the 2-sphere
ht
a homeomorphism from S 3 to itself
HA
the all-A state associated to the link diagram D(K)
GA
the all-A graph associated to the link diagram D(K)
GA
the reduced all-A graph associated to the link diagram D(K)
v(GA )
the number of vertices in GA
e(GA )
the number of edges in GA
e(GA )
the number of edges in GA
χ(GA )
the Euler characteristic, vA − eA , of GA
χ− (·)
t(D)
the negative part, max{−χ(·), 0}, of the Euler characteristic of ·
the number of twist regions in D(K)
TELC
two-edge loop condition
T (1, s)
a tangle sum of a vertical one-crossing twist region and a vertical short twist
region
T (s1 , s2 )
T (l, s)
a tangle sum of two vertical short twist regions
a tangle sum of a horizontal long twist region and a vertical short twist
region
st(D)
the number of special tangles in D(K)
ix
S 3 \K
SA
the complement of the link K in S 3
the all-A surface of the link diagram D(K)
v8
the volume, 3.6638 . . ., of a regular ideal octahedron
v3
the volume, 1.0149 . . ., of a regular ideal tetrahedron
∂(·)
int(·)
MA
guts(MA )
the boundary of ·
the interior of ·
the 3-manifold, S 3 \\SA , formed by cutting S 3 along SA
the guts of MA , which can be thought of as the “hyperbolic pieces” of the
annular JSJ Decomposition of MA
n (t)
JK
the nth colored Jones polynomial
βn
the penultimate coefficient of the nth colored Jones polynomial of the link K
βK
the stable penultimate coefficient of the colored Jones polynomial of the link
K
t1 (D)
tshort
≥2 (D)
long
t≥2 (D)
long
e≥2 (GA )
the number of one-crossing twist regions in D(K)
the number of short twist regions in D(K)
the number of long twist regions in D(K)
the number of edges in GA coming from long twist regions
#{·} the number of ·
SIC
small inner circle of the all-A state HA
OC
other circle of the all-A state HA
SC
special (other) circle of the all-A state HA
Bn
the n-braid group
σi
the ith generator of Bn
x
β
ai,j
a braid
the number of crossings (counted with sign) in the ith row and j th column of
twist regions in a plat diagram
t+ (D)
the number of positive twist regions in D(K)
t− (D)
the number of negative twist regions in D(K)
β
the closure of the braid β
β
a generic normal form β = C k σ1
−p1 q1
−p q
σ2 · · · σ1 s σ2s
of β
k
a parameter from the generic Schreier normal form β of β
s
a parameter from the generic Schreier normal form β of β
xi
Chapter 1
Introduction
This thesis centers around the study of knots and links, which have origins tracing back over
one hundred years. A knot K is a smooth (or piecewise-linear) embedding of a circle into the
3-sphere S 3 and a link, which will also be denoted by K, is a disjoint but possibly intertwined
collection of knots. Note that a knot can be viewed as a one-component link. We view two
links K1 and K2 as equivalent if there is a one-parameter family ht of homeomorphisms from
S 3 to itself such that h0 is the identity map and h1 sends K1 to K2 . A classical but difficult
question in knot theory is to determine whether or not two links K1 and K2 are equivalent.
For further details and a more thorough introduction to knot theory, see [19].
One approach to studying links is to project a link to the plane (or the 2-sphere S 2 ) in
such a way that the only crossings are double crossings and in such a way that over- and
under-crossing information is preserved by placing small gaps around the crossings of the
projection. What results is called a link diagram. It should be noted that the study of links
up to equivalence has been shown to be the same as the study of link diagrams up to a
diagrammatic notion of equivalence ([24]).
Another approach to studying links is to associate a mathematical objects to links in such
a way that equivalent links are associated to the exact same object. To be more precise, a
function from the set of links to a set of mathematical objects is called a link invariant if it
1
is constant on equivalence classes of links. As a result, if f (K1 ) = f (K2 ) for a link invariant
f and links K1 and K2 , then K1 and K2 cannot be equivalent links.
From the above two perspectives on the study of links, we can ask the following questions:
(1) What information can a link diagram provide about a link?
(2) What information can a link invariant provide about a link?
One of the main themes of this work is to investigate the conditions under which data
extracted from a certain link diagram or data extracted from a quantum polynomial link
invariant can be used to estimate geometric data for a given link. Tied to this theme is
the goal of improving general estimates for this geometric data by studying the structure of
specific families of links. Below we provide the background and framework needed for our
investigation.
From a combinatorial and quantum perspective, a rich family of links to study is the family
of semi-adequate (A- and B-adequate) links. By reflecting the diagram if needed, we may
focus our attention on the collection of A-adequate links. To an A-adequate link diagram we
can associate a loop-free graph. This graph, which can be cellularly embedded in a surface
whose boundary is the link, is intimately related to the Jones polynomial of the link (a
quantum link invariant). For example, the Jones polynomial of the link can be recovered
from the Bollob´as-Riordan polynomial of the associated graph ([5]). Additionally, Dasbach
and Lin ([6]) have shown that the absolute value of the penultimate coefficient of the nth
colored Jones polynomial stabilizes to a value that is independent of n > 1 and only depends
2
on a reduction of the associated graph.
From a geometric perspective, a rich family of links to study is the family of hyperbolic
links. Such links are characterized by having complements that admit a unique hyperbolic
metric ([20], [23], [27]). By this uniqueness the volume, vol(S 3 \K), of the complement of a
hyperbolic link K is a geometric link invariant.
Acting as a bridge between quantum and geometric invariants of links, the Volume Conjecture of Kashaev, Murakami, and Murakami ([7], [21]) predicts that the (asymptotic behavior
of the) colored Jones polynomial can be used to determine the hyperbolic volume of the link
complement. While it has been verified for a handful of hyperbolic links, the conjecture
remains quite open in general.
Recently, Futer, Kalfagianni, and Purcell have related the colored Jones polynomial of an
A-adequate link to both the topology of essential surfaces in the link complement and to
bounds on the hyperbolic volume of the link complement. (See [11] for the complete work
or [10] for a survey of results.) As an example, it was shown that, for sufficiently twisted
negative (positive using their conventions) braid closures and for a large family of Montesinos
links, the volume of the link complement can be bounded above and below in terms of the
twist number of an A-adequate link diagram. For the class of alternating links, similar twosided bounds were found by Lackenby ([18]), with the lower bound later improved by Agol,
Storm and W. Thurston ([1]) and the upper bound later improved by Agol and D. Thurston
([18], Appendix).
The main result of this thesis, which builds upon the work mentioned above, is given below.
3
(The definitions of connected, prime, A-adequate, twist region, special tangle, and the twoedge loop condition will be provided in Section 2.1.) Let t(D) denote the number of twist
regions in D(K), let st(D) denote the number of special tangles in D(K), let v8 = 3.6638 . . .
denote the volume of a regular ideal octahedron, and let v3 = 1.0149 . . . denote the volume
of a regular ideal tetrahedron.
Theorem 1.0.1 (Main Theorem). Let D(K) be a connected, prime, A-adequate link diagram
that satisfies the two-edge loop condition and contains t(D) ≥ 2 twist regions. Then K is
hyperbolic and the complement of K satisfies the following volume bounds:
v8
· [t(D) − st(D)] ≤ vol(S 3 \K) < 10v3 · (t(D) − 1),
3
where t(D) ≥ st(D). If t(D) = st(D), then D(K) is alternating and the lower bound of
v8
· (t(D) − 2) from Theorem 2.2 of [1] may be used.
2
To provide a family of links that satisfy the assumptions of the Main Theorem, we first
restrict attention to A-adequate plat closures of certain braids. In particular, we study
two families of plat diagrams: negative plat diagrams and mixed-sign plat diagrams. (See
Section 4.1 for background on plat closures and more details about negative and mixed-sign
plat diagrams.) By studying the structure of these two families of link diagrams, we are
often able to improve the lower bounds on volume given by the Main Theorem above.
Theorem 1.0.2. Let D(K) be a negative plat diagram. Then K is hyperbolic and the
complement of K satisfies the following volume bounds:
4
4v8
v
· (t(D) − 1) + 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
5
5
Theorem 1.0.3. Let D(K) be a mixed-sign plat diagram with at least as many negative twist
regions as positive twist regions. Then K is hyperbolic and the complement of K satisfies
the following volume bounds:
2v
v8
· (t(D) − 1) − 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
3
Furthermore, we are able to translate the volume bounds of the Main Theorem so that they
may be expressed in terms of the number of special tangles, st(D), and a stable coefficient,
βK , of the colored Jones polynomial. (For more information concerning the colored Jones
polynomial of an A-adequate link, see Section 2.3.)
Theorem 1.0.4. Let D(K) be a connected, prime, A-adequate link diagram that satisfies
the two-edge loop condition and contains t(D) ≥ 2 twist regions. Then K is hyperbolic and
the complement of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) < 30v3 ·
βK − 1 + 10v3 · (st(D) − 1) .
If we return to the family of plats considered in this work, then we are often able to improve
the above upper bounds on volume.
Theorem 1.0.5. Let D(K) be a negative plat diagram. Then K is hyperbolic and the
5
complement of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) <
5v
25v3
· ( βK − 1) − 3 .
2
2
Theorem 1.0.6. Let D(K) be a mixed-sign plat diagram with at least as many negative twist
regions as positive twist regions. Then K is hyperbolic and the complement of K satisfies
the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) < 30v3 · ( βK − 1) + 20v3 .
The three results above can be viewed as providing families of links that satisfy a Coarse
Volume Conjecture ([11], Section 10.4). A Coarse Volume Conjecture, roughly stated, says
that the volume of the link complement can be bounded above and below by linear functions
of the coefficients of the colored Jones polynomial. In the results above, we can actually
bound the volume above and below in terms of a single coefficient, namely βK .
In the same spirit as above, we also show that a family of closed braids fits into the same
volume bound framework. By studying the particular structure of these closed braid diagrams, we can often improve the bounds on volume. Since more can be said about closed
3-braids, we consider this case separately. (See Chapter 5 and Chapter 6 for further details.
In particular, see Section 5.6 for the definition of non-essential wandering circle.)
Theorem 1.0.7. Let D(K) denote the closure of a certain n-braid. Then K is a hyperbolic
6
link. In the case that n = 3, we get the following volume bounds:
v8
v
· (t(D) − 1) − 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
2
2
Suppose, in the case that n ≥ 4, that the all-A state contains m non-essential wandering
circles. Then we get the following volume bounds:
v8
v
· (t(D) − 1) − 8 · (2(n + m) − 5) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
2
2
Theorem 1.0.8. Let D(K) denote the closure of a certain n-braid. Then K is a hyperbolic
link. In the case that n = 3, we get the following volume bounds:
v8 · ( βK − 1) ≤ vol(S 3 \K) < 20v3 · ( βK − 1) + 10v3 .
Suppose, in the case that n ≥ 4, that the all-A state contains m non-essential wandering
circles. Then we get the following volume bounds:
v8 · ( βK − 1) ≤ vol(S 3 \K) < 20v3 · ( βK − 1) + 10v3 · (2(n + m) − 5).
Additionally, for the closed 3-braids under consideration, we are able to bound the volume
in terms of the parameter s from their Schreier normal form. (See Chapter 6 for further
details.)
Theorem 1.0.9. Let D(K) denote the closure of a certain 3-braid. Then K is a hyperbolic
7
link and K satisfies the following volume bounds:
v8 · (s − 1) ≤ vol(S 3 \K) < 4v8 · s.
Remark 1.0.1. It should be noted that volume bounds for closed 3-braids in terms of the
parameter s were previously found by Futer, Kalfagianni, and Purcell in [12]. By using the
results of their more recent work in [9] and [11], we are often able to improve their lower
bound on volume from [12].
To conclude our study of link diagrams and volume bounds we investigate a third family
of links, called hybrid braid-plats, which have both negative braid closure aspects and plat
closure aspects. Again, we show that these link diagrams fit into the above volume bound
framework.
Theorem 1.0.10. For D(K) a certain hybrid braid-plat diagram with negative plat diagram
aspects, we have that K is hyperbolic and the complement of K satisfies the following volume
bounds:
32v8
2v8
· (t(D) − 1) +
≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
15
Similarly, for D(K) a certain hybrid braid-plat diagram with mixed-sign plat diagram aspects,
we have that K is hyperbolic and the complement of K satisfies the following volume bounds:
v8
4v
· (t(D) − 1) + 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
3
Translating these volume bounds to be expressed in terms of the coefficient βK of the colored
8
Jones polynomial, we get the following results:
Theorem 1.0.11. For D(K) a certain hybrid braid-plat diagram with negative plat diagram
aspects, we have that K is hyperbolic and the complement of K satisfies the following volume
bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) < 15v3 · ( βK − 1) − 32v3 .
Similarly, for D(K) a certain hybrid braid-plat diagram with mixed-sign plat diagram aspects,
we have that K is hyperbolic and the complement of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) < 30v3 · ( βK − 1) − 40v3 .
9
Chapter 2
Preliminaries
2.1
Definitions
For the remainder of this thesis, we will let D(K) ⊆ S 2 denote a diagram of a link K ⊆ S 3 .
We call D(K) connected if the projection graph formed by replacing all crossings of D(K)
with 4-valent vertices is path-connected. We call D(K) prime if there does not exist a
simple closed curve in the projection plane that both intersects D(K) twice transversely
(away from the crossings) and also contains crossings on both sides of the curve. For a
schematic depiction of a non-prime diagram, see Figure 2.1.
To smooth a crossing of the link diagram D(K), we may either A-resolve or B-resolve this
crossing according to Figure 2.2. Note that interchanging overcrossings and undercrossings
(called reflecting the link diagram) interchanges A-resolutions and B-resolutions. By A-
D1
D2
C
Figure 2.1: A schematic depiction of non-prime link diagram D(K). The solid boxes, marked D1
and D2 , contain the remainder of the link diagram. Each such box is assumed to contain at least
one crossing of D(K). The simple closed curve, C, exhibits the non-primeness of the diagram.
10
A
B
Figure 2.2: A crossing neighborhood of a link diagram (middle), along with its A-resolution
(right) and B-resolution (left).
resolving each crossing of D(K) we form the all-A state of D(K), which we will denote
by HA . The all-A state HA consists of a disjoint collection of all-A circles and a disjoint
collection of dotted line segments, called A-segments, which are used record the locations of
crossing resolutions. We will adopt the convention throughout this thesis that any unlabeled
segments are assumed to be A-segments.
Definition 2.1.1. A link diagram D(K) is called A-adequate if HA does not contain any
A-segments that join an all-A circle to itself. A link diagram is called semi-adequate if either
it or its reflection is A-adequate. A link K is called A-adequate if it has a diagram that is
A-adequate and is called semi-adequate if it has a diagram that is semi-adequate.
Remark 2.1.1. While we will focus exclusively on A-adequate links, our results can easily
be extended to semi-adequate links by reflecting the link diagram D(K) and obtaining the
corresponding results for B-adequate links.
Definition 2.1.2. From HA we may form the all-A graph, denoted GA , by contracting the
all-A circles to vertices and reinterpreting the A-segments as edges. From this graph we can
form the reduced all-A graph, denoted GA , by replacing all multi-edges with a single edge.
Said another way, we reduce the graph by removing all redundant parallel edges that join
the same pair of vertices.
For an example of a diagram D(K), its all-A resolution HA , its all-A graph GA , and its
reduced all-A graph GA , see Figure 2.3.
11
HA
D(K)
SA
GA
GA
Figure 2.3: A link diagram D(K), its all-A resolution HA , its all-A surface SA , its all-A graph
GA , and its reduced all-A graph GA .
Notation: Since reducing GA to form GA leaves the vertices unchanged, then let v(GA )
denote the number of vertices of either GA or GA . Let e(GA ) denote the number of edges
of GA and let e(GA ) denote the number of edges of GA . Let −χ(GA ) = e(GA ) − v(GA )
denote the negative Euler characteristic of GA .
Remark 2.1.2. Note that v(GA ) is the same as the number of all-A circles in HA and
that e(GA ) is the same as the number of A-segments in HA . From a graphical perspective,
A-adequacy of D(K) can equivalently be defined by the condition that GA (or GA ) contains
no one-edge loops that connect a vertex to itself.
Definition 2.1.3. Define a twist region of D(K) to be a longest possible string of crossings
built from exactly two strands of D(K). Denote the number of twist regions in D(K) by
t(D) and call t(D) the twist number of D(K). See Figure 2.4 for an example.
Notice that it is potentially possible for a twist region to consist of a single crossing.
Definition 2.1.4. If a given twist region contains two or more crossings, then the A12
Figure 2.4: A link diagram with t(D) = 3 twist regions.
A
A
long resolution short resolution
Figure 2.5: Long and short resolutions of a twist region of D(K).
resolution of this twist region will consist of portions of two all-A circles that are either:
(1) joined by a path of A-segments and (small) all-A circles, or
(2) joined by a string of parallel A-segments.
Call a resolution in which (1) occurs a long resolution and call a resolution in which (2)
occurs a short resolution. See Figure 2.5 for depictions of long and short resolutions. We
will call a twist region long if its A-resolution is long and short if its A-resolution is short.
The following definition, which is very important, describes a key property of the link diagrams that will be considered in this work.
Definition 2.1.5. A link diagram D(K) satisfies the two-edge loop condition (TELC) if,
whenever two all-A circles share a pair of A-segments, these segments correspond to crossings
from the same short twist region of D(K).
Recall that the volume bounds from Theorem 1.0.1 are expressed in terms of two pieces
of diagrammatic data, the first of which is the twist number t(D). The second piece of
13
diagrammatic data, a count of specific types of alternating tangles in the diagram D(K), is
defined below.
Definition 2.1.6. Call an alternating tangle in D(K) a special tangle if, up to planar isotopy,
it consists of exactly one of the following:
(1) a tangle sum, T (1, s), of a vertical one-crossing twist region and a vertical short twist
region
(2) a tangle sum, T (s1 , s2 ), of two vertical short twist regions
(3) a tangle sum, T (l, s), of a horizontal long twist region and a vertical short twist region
The values of l, s, s1 , s2 ≥ 1 denote the number of crossings in the relevant twist regions
described above. To look for such tangles in D(K) ⊆ S 2 , we look for simple closed curves in
the plane that intersect D(K) exactly four times (away from the crossings) and that contain
a special tangle on one side of the curve. Equivalently, special tangles of D(K) can be found
in the all-A state HA by looking for all-A circles that are incident to A-segments from a
pair of twist regions from the tangle sums mentioned above. We will call these all-A circles
special circles of HA . See Figure 2.6 for depictions of special tangles and special circles. Let
st(D) denote the number of special tangles in D(K) (or, equivalently, the number of special
circles in HA ).
The advantage to looking for special circles in HA , as opposed to looking for special tangles
in D(K), is that special circles are necessarily disjoint. Special tangles, on the other hand,
can share one or both twist regions with another special tangle.
14
special circles
C
special tangles
s
1
A
T (1, s)
C
s1
s2
A
T (s1 , s2 )
C
l
s
A
T (l, s)
Figure 2.6: Special tangles of D(K) and the corresponding special circles, C, of HA .
2.2
Volume Bounds
The previous section provided the definitions of the terms involved in the statement of
Theorem 1.0.1. We would now like to provide the mathematical background surrounding
this theorem. First, to consider the hyperbolic volume of the link complement, we need to be
able to ensure that our links are hyperbolic. The following result, due to Futer, Kalfagianni,
and Purcell ([9]), does exactly this.
Proposition 2.2.1 ([9]). Let D(K) be a connected, prime, A-adequate link diagram that
satisfies the TELC and contains t(D) ≥ 2 twist regions. Then K is hyperbolic.
Let S 3 \K denote the complement of the link K. Since the link complement is formed by
removing a regular open neighborhood of the link K from S 3 , then we have that S 3 \K is a
3-manifold with boundary a disjoint union of tori.
15
A common practice in 3-manifold theory is to study properly embedded surfaces in 3manifolds. Our goal is now to find an essential surface in the complement S 3 \K. For
what follows, let S be a compact surface that is properly embedded in a 3-manifold M ,
where being properly embedded means that ∂S ⊆ ∂M and int(S) ⊆ int(M ). Let D be a
disk in M with the property that D ∩ S = ∂D. The disk D is called a compressing disk for
S in M if ∂D does not bound a disk in S. If S does not contain compressing disks, then S is
called incompressible in M . Let D be a disk in M with the properties that D ∩ S = α and
D ∩ ∂M = β, where α and β are arcs with α ∪ β = ∂D and α ∩ β = ∂α = ∂β, and where
α does not cobound a disk in S with another arc in ∂S. The disk D is called a boundarycompressing disk for S. If S does not contain boundary-compressing disks, then S is called
boundary-incompressible in M . The surface S is called essential in M if the boundary of a
regular neighborhood of S is both incompressible and boundary-incompressible in M . We
refer the reader to [13] for more details.
From the all-A state HA of the link diagram D(K) we form the all-A surface of D(K),
denoted SA , by associating a disk to each all-A circle and a half-twisted band to each Asegment, doing this in such a way that the boundary of SA is the link K and so that inner
disks lie above outer disks. See the top right side of Figure 2.3 for an example of an all-A
surface SA . Notice that SA is properly embedded in the 3-manifold S 3 \K. As it turns
out, the A-adequacy of a connected link diagram D(K) has a geometric interpretation that
involves the all-A surface SA . In [22], Ozawa showed that the A-adequacy of a connected
link diagram D(K) implies that the all-A surface SA is essential in S 3 \K.
For a connected A-adequate link diagram D(K) with essential all-A surface SA in S 3 \K,
we form the 3-manifold MA = (S 3 \K)\\SA by cutting the the link complement along the
16
all-A surface. As proved in Lemma 2.4 of [11], the 3-manifold MA is homeomorphic to a
handlebody and therefore contains no essential tori. By the annulus version of the JacoShalen-Johannson Decomposition Theorem ([14], [15]) and by Lemma 4.1 of [11], we can
cut MA along annuli (disjoint from the parabolic locus) to produce the following types of
pieces:
(1) I-bundles over subsurfaces of SA
(2) solid tori
(3) guts (which are the hyperbolic pieces with totally geodesic boundary)
Let χ− (guts(MA )) = max{−χ(guts(MA )), 0}. With the lower bound provided by Agol,
Storm, and W. Thurston ([1]) and the upper bound provided by Agol and D. Thurston ([18],
Appendix), we get the following volume bounds for the hyperbolic link complement.
Proposition 2.2.2 ([1], [18]). Let D(K) be a connected, prime, A-adequate diagram of a
hyperbolic link K. Then:
v8 · χ− (guts(MA )) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1),
where v8 = 3.6638 . . . denotes the volume of a regular ideal octahedron and v3 = 1.0149 . . .
denotes the volume of a regular ideal tetrahedron.
Assumption: We will assume throughout this work that guts(MA ) is nonempty. Otherwise,
we would have that χ− (guts(MA )) = 0 and, consequently, our lower bounds on volume would
not be valuable.
17
By Remark 5.15 and Corollary 5.19 of [11], we get the following result:
Theorem 2.2.1. Let D(K) be a connected, prime, A-adequate link diagram that satisfies
the TELC. Then:
−χ(GA ) = χ− (guts(MA )).
By combining the above results together, we are able to state the theorem that will form
the basis for our work. Specifically, Theorem 2.2.2 below forms a crucial foundation for
Theorem 1.0.1.
Theorem 2.2.2. Let D(K) be a connected, prime, A-adequate link diagram that satisfies
the TELC and contains t(D) ≥ 2 twist regions. Then K is hyperbolic and:
−v8 · χ(GA ) ≤ v8 · χ− (guts(MA )) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
Proof. By the combining Proposition 2.2.1 with Proposition 2.2.2 and Theorem 2.2.1, we
get the desired results.
Note that the above theorem uses the reduced all-A graph GA to get a lower bound on volume. This fact provides the key opportunity for a combinatorial perspective to be used.
2.3
A-Adequacy and the Colored Jones Polynomial
Now that the combinatorial and geometric approaches to the study of links have been investigated, we would like to introduce a third and final approach: quantum polynomial
18
invariants. To an oriented link we can associate a number of polynomial invariants, each of
which can be used to distinguish links from each other. Some of the most recent such link
polynomials are the Jones polynomial ([16], [17]) and the colored Jones polynomials ([28]),
which form a sequence of Laurent polynomials indexed by the natural numbers.
Notation: Denote the nth colored Jones polynomial of a link K by:
n (t) = α tmn + β tmn −1 + · · · + β trn +1 + α trn ,
JK
n
n
n
n
where n ∈ N and where the degree of each monomial summand decreases from left to
right.
The following result, due to Dasbach and Lin and due to Stoimenow, relates the penultimate
coefficient βn of the colored Jones polynomial of an A-adequate link to the reduced all-A
graph GA associated to an A-adequate diagram of that link.
Theorem 2.3.1 ([6], [26]). Let D(K) be a connected, A-adequate link diagram. Then βn
is independent of n for n > 1. Specifically, for n > 1, we have that:
βK := βn = 1 − χ(GA ).
The above result has been and will be key in translating from volume bounds in terms of the
twist number to volume bounds in terms of coefficients of the colored Jones polynomial.
19
Chapter 3
Volume Bounds for A-Adequate
Links
Let D(K) be a connected, A-adequate link diagram that satisfies the two-edge loop condition
(TELC). We begin this chapter by closely investigating the twist regions of D(K), the all-A
state HA of D(K), and the reduction of the all-A graph GA to the reduced all-A graph GA .
This is done in order to express −χ(GA ) in terms of the twist number t(D) and the number
of certain all-A circles. By studying the all-A circles and making use of graph theory, we are
able to bound −χ(GA ) below in terms of the twist number t(D) and the number, st(D), of
special circles in HA (or the corresponding number of special tangles in D(K)). This lower
bound, combined with Theorem 2.2.2, produces the Main Theorem (Theorem 1.0.1) of this
work.
3.1
Twist Regions, State Circles, and GA
We begin with a study of twist regions. Because long and short resolutions are not distinguishable when there is only one crossing in a twist region, we have defined long and short
resolutions (Definition 2.1.4) as coming from twist regions that contain at least two crossings.
Consequently, we will begin by considering the case of one-crossing twist regions.
20
See Figure 2.2 for a one-crossing twist region and its A-resolution. Let C1 and C2 denote
the (portions of the) relevant all-A circles in HA . Since D(K) is A-adequate, then C1 = C2 .
Since D(K) satisfies the TELC, then there can be no other additional A-segments between
C1 and C2 . Thus, the edge of GA corresponding to this one-crossing twist region can never
be a redundant parallel edge and, therefore, will always appear in GA .
Notation: Let t1 (D) denote the number of one-crossing twist regions in D(K).
Remark 3.1.1. By what was said in the above paragraph, t1 (D) is also the number of edges
in GA that come from the one-crossing twist regions in D(K).
Let us now consider twist regions that have at least two crossings. Such twist regions will
either have a short resolution or a long resolution. See the right side of Figure 2.5 for a twist
region and its short all-A resolution. Again using C1 and C2 to denote the (portions of the)
the relevant all-A circles, the A-adequacy of D(K) implies that C1 = C2 and the TELC
implies that there can be no other A-segments between C1 and C2 (besides those of the
short resolution). Furthermore, note that a short twist region will always create redundant
parallel edges in GA since the parallel A-segments of HA join the same pair of state circles.
Thus, all but one of these edges is removed when forming GA . Said another way, there will
be one edge of GA per short twist region of D(K).
Notation: Let ts (D) denote the number of short twist regions in D(K).
Remark 3.1.2. By what was said in the above paragraph, ts (D) is also the number of edges
in GA that come from the short twist regions in D(K).
21
See the left side of Figure 2.5 for a twist region and its long all-A resolution. We will use C1
and C2 to denote upper and lower (portions of the) relevant state circles. If there are three
or more crossings in the twist region being considered, then it must necessarily be the case
that none of the corresponding edges in GA are lost in the reduction to form GA . If there are
two crossings in the twist region, then the TELC implies that C1 = C2 because, otherwise,
we would have a two-edge loop in GA coming from a long twist region. As a result, we have
that no edges of GA coming from long resolutions are removed when forming GA .
Recall that a long resolution will consist of (portions of) two state circles joined by a path
of A-segments and (small) all-A circles.
Definition 3.1.1. We call each (small) all-A circle in the interior of the long resolution a
small inner circle (SIC). The remaining all-A circles (coming from the all-A circle portions
in the long resolutions, from the all-A circle portions in the short resolutions, and from the
all-A circle portions in the A-resolutions of one-crossing twist regions) will simply be called
other circles (OCs).
Notation: Let tl (D) denote the number of long twist regions in D(K) and let el (GA ) denote
the number of edges in GA coming from long twist regions.
By inspection, it can be seen that the number of A-segments in the long resolution is always one greater than the number of small inner circles in the long resolution. Since this
phenomenon occurs for each long resolution, then we have that:
# {SICs} = el (GA ) − tl (D).
22
(3.1)
3.2
Computation of −χ(GA)
Lemma 3.2.1. Let D(K) be a connected, A-adequate link diagram that satisfies the TELC.
Then we have that:
−χ(GA ) = t(D) − # {OCs} .
Proof. By Remark 2.1.2 and Definition 3.1.1, we get:
−χ(GA ) = e(GA ) − v(GA )
= e(GA ) − # {all − A state circles}
= e(GA ) − # {SICs} − # {OCs} .
(3.2)
Looking at how the twist regions of D(K) were partitioned in Section 3.1, we get:
t(D) = t1 (D) + ts (D) + tl (D).
(3.3)
Next, Remark 3.1.1 and Remark 3.1.2 imply that:
e(GA ) = t1 (D) + ts (D) + el (GA ).
(3.4)
By substituting Equation 3.4 and Equation 3.1 into Equation 3.2 and then using Equation 3.3, we get:
23
−χ(GA ) = e(GA ) − # {SICs} − # {OCs}
= [t1 (D) + ts (D) + el (GA )] − [el (GA ) − tl (D)] − [# {OCs}]
= t1 (D) + ts (D) + tl (D) − # {OCs}
= t(D) − # {OCs} .
By applying Lemma 3.2.1 to Theorem 2.2.2, we get the following corollary:
Corollary 3.2.1. Let D(K) be a connected, prime, A-adequate link diagram that satisfies
the TELC and contains t(D) ≥ 2 twist regions. Then K is hyperbolic and:
v8 · [t(D) − # {OCs}] ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
Rather than count the number, # {OCs}, of other circles in HA , we would instead like to
find lower bounds on volume that have a more intrinsic and visual connection to the link
diagram D(K) itself. Consequently, our goal is now to bound −# {OCs} below in terms of
quantities that can be read off of D(K) directly.
24
Figure 3.1: The exceptional link diagram consisting of two long twist regions, each of which must
contain at least three crossings.
3.3
Special Circles and Special Tangles
As in the previous section, let D(K) be a connected, A-adequate link diagram that satisfies
the TELC. We will now also require that D(K) is prime and contains t(D) ≥ 2 twist regions.
For such a link diagram D(K), we are able to identify the simplest other circles of HA with
certain alternating tangles in D(K).
Lemma 3.3.1. Let D(K) be a connected, prime, A-adequate link diagram that satisfies the
TELC and contains t(D) ≥ 2 twist regions. Furthermore, assume that D(K) is not the link
diagram depicted in Figure 3.1. Then:
(1) there are no other circles of HA that are incident to A-segments from one or fewer
twist region resolutions, and
(2) the other circles of HA that are incident to A-segments from exactly two twist region
resolutions are the special circles of HA .
Remark 3.3.1. In the case that D(K) is the exceptional link diagram depicted in Figure 3.1, we have that t(D) = 2 and st(D) = 0. Therefore, by applying the Main Theorem
25
1
C
s
C
l
C
Figure 3.2: A schematic depiction of the A-resolutions of the three possible types of twist regions
incident to an other circle C of HA : one-crossing (left), short (middle), and long (right). The edge
labels 1, s, and l are used not only to indicate the type of twist region resolution, but also to
distinguish these edges from A-segments.
(Theorem 1.0.1), we get that the two-sided volume bounds for this link are:
2v8
≤ vol(S 3 \K) < 10v3 .
3
Remark 3.3.2. In the future, it will be important to notice that st(D) = 0 for link diagrams
whose other circles all have at least three incident twist region resolutions.
Proof of Lemma. See Figure 3.2 for schematic depictions of the A-resolutions of the twist
regions of D(K). Let C denote an other circle of the all-A state HA . Such a circle must exist
because, otherwise, we would have that HA is a cycle of small inner circles and A-segments.
Since this all-A state corresponds to the standard (2, p)-torus link diagram, then we would
get a contradiction of the assumption that t(D) ≥ 2.
Case 1: Suppose C is an other circle with no incident twist region resolutions. Then C corresponds to a standard unknotted component of D(K). Thus, either D(K) is not connected
or D(K) is the standard unknot diagram. In either case we get a contradiction, given the
assumptions that D(K) is connected and contains t(D) ≥ 2 twist regions.
Case 2: Suppose C is an other circle with a single incident twist region resolution.
26
Subcase 1: Suppose the twist region resolution starts and ends at C. Similar to Case 1,
this case is impossible given the assumptions that D(K) is connected and contains t(D) ≥ 2
twist regions.
Subcase 2: Suppose the twist region resolution joins C to a different state circle C . Consider
the portion of HA corresponding to Figure 3.2 and recall that C and C are closed curves.
If C is incident to no other additional twist region resolutions, then (as in Case 1) we get
a contradiction of the assumptions that D(K) is connected and contains t(D) ≥ 2 twist
regions. If C is incident to additional twist region resolutions, then we get a contradiction
of the primeness of the corresponding diagram D(K).
Case 3: Suppose C is an other circle with exactly two incident twist region resolutions.
If one of the twist region resolutions starts and ends at C, then (recalling Figure 2.2 and
Figure 2.5 if needed) this twist region resolution can only correspond to a long resolution.
This is because, otherwise, we would get a contradiction of the assumption that D(K) is
A-adequate.
Subcase 1: Suppose both (long) twist region resolutions start and end at C. The first three
possibilities are depicted in Figure 3.3. As the rectangular dashed closed curves in the figure
indicate, we get a contradiction of the primeness of the corresponding diagram D(K). The
fourth and final possibility is depicted in Figure 3.4. Translating back to D(K), we get the
exceptional link diagram depicted in Figure 3.1. Note that, by the TELC, it must be the
case that there are at least three crossings per (long) twist region. This is because, otherwise,
we would have a two-edge loop whose edges do not correspond to crossings of a short twist
region. Recall that the link diagram of Figure 3.1 has been excluded from consideration.
27
l
l
l
l
l
l
C
C
C
Figure 3.3: Three possibilities for an other circle C with two incident long resolutions that start
and end at C.
l
l
C
Figure 3.4: The fourth possibility for an other circle C with two incident long resolutions that
start and end at C.
Subcase 2: Suppose one (long) twist region resolution starts and ends at C = C1 and
the other twist region resolution connects C1 to another circle C2 . The three possibilities
are depicted in Figure 3.5. As the rectangular dashed closed curves indicate, we get a
contradiction of the primeness of the corresponding diagram D(K).
Subcase 3: Suppose both twist region resolutions connect C to different other circles C1
and C2 , where C1 = C2 is possible in some cases. The first possibility, that C1 and C2 are
on opposite sides of C, is depicted in Figure 3.6. As the rectangular dashed closed curve
indicates, we get a contradiction of the primeness of the corresponding diagram D(K). Thus,
both twist region resolutions incident to C must be on the same side of C. The choice of
l
C1
1
l
C2
s
C1
l
C2
C1
l
C2
Figure 3.5: Three possibilities for an other circle C1 with two incident twist region resolutions, one
resolution from a long twist region that starts and ends at C1 and the other resolution connecting
to a different state circle C2 .
28
C1
C2
C
Figure 3.6: An other circle C with two incident twist region resolutions, one inside C and one
outside C.
1
C1
1
C2 C1
C
l
1
l
C2 C1
C
l
C2
C
Figure 3.7: Three possibilities for an “other” circle C with two incident twist region resolutions,
one connecting C to C1 and the other connecting C to C2 .
side is irrelevant, however, since D(K) ⊆ S 2 .
Given that both twist region resolutions must be on the same side of C, the first three possibilities are depicted in Figure 3.7. None of these cases are possible because, by translating
back to D(K) (and recalling Figure 2.2 and Figure 2.5 if needed), the two seemingly distinct
twist region resolutions are actually part of the same long resolution. This makes C a small
inner circle rather than an other circle, a contradiction.
The three remaining possibilities are depicted in Figure 3.8.
Remark 3.3.3. Note that by the TELC, it must be the case that C1 = C2 in the left and
middle diagrams of Figure 3.8. However, because a long resolution involves a path of at least
two A-segments, then it is possible that C1 = C2 in the right diagram of Figure 3.8. It is
1
C1
s
C
s
s
C2 C1
C
l
C2 C1
s
C
C2
Figure 3.8: Three remaining possibilities for an other circle C with two incident twist region
resolutions, one connecting C to C1 and the other (a short twist region resolution) connecting C
to C2 .
29
Figure 3.9: Three possible tangle subdiagrams of D(K).
also important to note that, in all three diagrams, the remaining twist region resolutions and
all-A circles (not depicted) must somehow join C1 to C2 in a second way. This is because,
otherwise, there would exist a simple closed curve that cuts C in half and separates C1 from
C2 , a contradiction of the primeness of the corresponding diagram D(K).
Assuming the conditions laid out in the above remark are satisfied, it is important to notice
that the three possibilities in Figure 3.8 do not give a contradiction of the assumptions of
Lemma 3.3.1. Equally as important, notice that these three possibilities correspond exactly
to the special circles depicted on the right side of Figure 2.6.
Note that, by translating from the portions of HA depicted in Figure 3.8 to the link diagram
D(K), we get the three special subdiagrams depicted in Figure 3.9. But these are exactly
the three special tangles T (1, s), T (s1 , s2 ), and T (l, s) of the link diagram D(K)! Therefore,
we have have shown that the special circles of HA (which correspond to the special tangles in
D(K)) are the least complicated other circles in HA with respect to the number of incident
twist region resolutions.
30
3.4
Volume Bounds in Terms of t(D) and st(D) (The
Main Theorem)
In this section, we will shift perspective from the link diagram D(K) and its all-A state HA
to the reduced all-A graph GA . By combining some graph theory with both our previous
computation of −χ(GA ) (Lemma 3.2.1) and our newly acquired knowledge about special
circles (Lemma 3.3.1), we will prove the Main Theorem, which we restate below. Notice that
the novel content of this theorem lies in the lower bound on volume.
Theorem 3.4.1 (Main Theorem). Let D(K) be a connected, prime, A-adequate link diagram
that satisfies the TELC and contains t(D) ≥ 2 twist regions. Then K is hyperbolic and the
complement of K satisfies the following volume bounds:
v8
· [t(D) − st(D)] ≤ vol(S 3 \K) < 10v3 · (t(D) − 1),
3
where t(D) ≥ st(D). If t(D) = st(D), then D(K) is alternating and the lower bound of
v8
· (t(D) − 2) from Theorem 2.2 of [1] may be used.
2
Remark 3.4.1. Note that the coefficients of t(D) in the upper and lower bounds differ by a
multiplicative factor of 8.3102 . . ., a factor that we would like to reduce by studying specific
families of links.
Definition 3.4.1. Let G be a graph. We call G simple if it contains neither one-edge loops
connecting a vertex to itself nor multiple edges connecting the same pair of vertices.
Notation: For G a simple graph, let V (G) denote its vertex set and let E(G) denote its
31
edge set. Furthermore, let deg(v) denote the degree of the vertex v, that is, the number of
edges incident to v.
Proof of the Main Theorem. We will begin by using Theorem 2.1 of [4] which states that,
for G a simple graph:
deg(v) = 2 |E(G)| .
v∈V (G)
To see why this result is true, note that the edges of G are double counted as we sum the
number of edges incident to the vertices. This is because each edge is shared by two distinct
vertices (where we get distinct by the assumption that G is simple). Our strategy will be
to apply this result to the reduced all-A graph GA . We can do this because A-adequacy of
D(K) implies that GA will not contain any loops and the fact that GA is reduced implies
that GA will not contain any multiple edges.
By Remark 2.1.2, Definition 3.1.1, and Lemma 3.3.1, we may partition V (GA ) into three
types of vertices:
(1) those corresponding to small inner circles (SICs),
(2) those corresponding to special circles (SCs), which are other circles that are incident
to exactly two twist region resolutions, and
(3) those corresponding to the other circles (remaining OCs) that are incident to three or
more twist region resolutions.
Recall that, as said in the paragraph following Remark 3.1.2, all edges corresponding to a long
32
resolution survive the reduction to GA . Thus, we have that deg(v) = 2 for v corresponding
to a small inner circle. To visually verify this claim, see the left side of Figure 2.5. Also
recall that, as said in the paragraph preceding Remark 3.1.1, the edge corresponding to a
one-crossing twist region survives the reduction to GA . Finally, as said in the paragraph
following Remark 3.1.1, only a single edge coming from a short twist region survives the
reduction to GA . By applying this knowledge to Figure 3.8, we see that deg(v) = 2 for v
corresponding to a special circle. Similarly, we can see that deg(v) ≥ 3 for v corresponding
to a remaining other circle. Therefore, translating Theorem 2.1 of [4] (stated above) to our
setting, we get:
2 · e(GA ) =
deg(v) +
SICs
deg(v) +
deg(v)
remaining
OCs
SCs
= 2 · [# {SICs}] + 2 · st(D) +
deg(v).
(3.5)
remaining
OCs
Substituting Equation 3.4 into the left-hand side of Equation 3.5 gives:
2 · t1 (D) + 2 · ts (D) + 2 · el (GA ).
(3.6)
Substituting Equation 3.1 into the right-hand side of Equation 3.5 gives:
2 · el (GA ) − tl (D) + 2 · st(D) +
deg(v).
remaining
OCs
33
(3.7)
By equating Expression 3.6 with Expression 3.7, canceling, rearranging terms, and using
Equation 3.3, we end up with the following:
2 · st(D) +
deg(v) = 2 · t1 (D) + 2 · ts (D) + 2 · tl (D)
remaining
OCs
= 2t(D).
(3.8)
Recall that deg(v) ≥ 3 for v corresponding to a remaining other circle. Thus, we get:
2 · st(D) + 3 · [# {remaining OCs}] ≤ 2t(D).
Adding st(D), the number of other circles that are not remaining other circles, to both sides
allows us to write the above inequality in terms of the total number of other circles as:
3 · [# {OCs}] ≤ 2t(D) + st(D).
This simplifies to:
# {OCs} ≤
2
1
· t(D) + · st(D).
3
3
Combining this inequality with Lemma 3.2.1 gives:
−χ(GA ) = t(D) − # {OCs} ≥
34
1
· [t(D) − st(D)] .
3
(3.9)
Finally, by applying Inequality 3.9 to either Theorem 2.2.2 or Corollary 3.2.1, we get the
desired volume bounds.
Furthermore, notice that Equation 3.8 implies that t(D) ≥ st(D). Thus, we have that the
lower bound on volume is always nonnegative and is positive precisely when there exists at
least one remaining other circle. Looking at Equation 3.8 from another perspective, note
that if t(D) = st(D), then there are can be no remaining other circles in the all-A state
HA . Hence, the only types of other circles in this case are special circles. Since each special
circle is incident to exactly two twist region resolutions (and since the conditions mentioned
in Remark 3.3.3 must be satisfied), then the all-A state HA must form a cycle alternating
between special circles and twist region resolutions. But recall that special tangles (which
correspond to special circles) are alternating tangles. Hence, by cyclically fusing these tangles
together, we form an alternating link diagram. Consequently, in the case that t(D) = st(D)
(which forces the lower bound of the Main Theorem to be zero), Theorem 2.2 of [1] can be
v
used to provide a lower bound of 8 · (t(D) − 2) on volume.
2
Corollary 3.4.1. Let D(K) satisfy the hypotheses of the Main Theorem (Theorem 3.4.1).
Furthermore, assume that each other circle of HA has at least m ≥ 3 incident twist region
resolutions. Then st(D) = 0 and:
−χ(GA ) ≥
m−2
· t(D).
m
Furthermore, K is hyperbolic and the complement of K satisfies the following volume bounds:
35
m−2
· v8 · t(D) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
m
Remark 3.4.2. Notice that, as m → ∞, the lower bound above approaches v8 ·t(D). Hence,
the coefficients of t(D) in the upper and lower bounds differ by a multiplicative factor of
2.7701 (in the limit).
Proof. We will prove this result by modifying what needs to be modified in the above proof
of the Main Theorem. First, the assumption that each other circle has at least m ≥ 3
incident twist region resolutions implies, by Lemma 3.3.1, that special circles cannot exist,
so st(D) = 0. This assumption also implies that deg(v) ≥ m for v corresponding to an
other circle (which must be a remaining other circle). Consequently, this new assumption
can be used to produce a sharper lower bound on volume. Specifically, by incorporating the
conditions that st(D) = 0 and deg(v) ≥ m into Equation 3.8, we get:
m · # {OCs} ≤ 2 · st(D) +
deg(v) = 2t(D).
remaining
OCs
Hence, we have that:
# {OCs} ≤
2
· t(D).
m
Combining this inequality with Lemma 3.2.1 gives:
−χ(GA ) = t(D) − # {OCs} ≥
36
m−2
· t(D).
m
(3.10)
Finally, by applying the above inequality to either Theorem 2.2.2 or Corollary 3.2.1, we get
the desired volume bounds.
3.5
Volume Bounds in Terms of the Colored Jones
Polynomial
We will now express the volume bounds of the Main Theorem (Theorem 3.4.1) in terms of
coefficients of the colored Jones polynomial (a quantum link invariant) and st(D) rather than
in terms of t(D) (a diagrammatic quantity) and st(D). In particular, when st(D) = 0 or
when st(D) can be bounded above by a positive constant, we will have volume bounds only
in terms of coefficients of the colored Jones polynomial. In particular, our volume bounds
will be in terms of the stable penultimate coefficient βK of the colored Jones polynomial. In
doing this, we will show that such links satisfy a Coarse Volume Conjecture ([11], Section
10.4).
Theorem 3.5.1. Let D(K) be a connected, prime, A-adequate link diagram that satisfies
the TELC and contains t(D) ≥ 2 twist regions. Then K is hyperbolic and the complement
of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) < 30v3 ·
βK − 1 + 10v3 · (st(D) − 1) .
Remark 3.5.1. As for t(D) in the Main Theorem, note that the coefficients of βK in the
37
upper and lower bounds above differ by a multiplicative factor of 8.3102 . . ., a factor that we
would like to reduce by studying specific families of links.
Proof of Theorem. Since D(K) is connected and A-adequate, then Theorem 2.3.1 gives that
βK = 1 − χ(GA ), which implies that −χ(GA ) = βK − 1. Applying this to Theorem 2.2.2,
we get the desired lower bound on volume. Theorem 2.2.2 also provides an upper bound of
10v3 · (t(D) − 1) on volume, which means that we will have proven the theorem once this
quantity is bounded above in terms of βK and st(D). First, note that Inequality 3.9 gives:
βK − 1 = −χ(GA ) ≥
1
· [t(D) − st(D)] .
3
Solving for t(D), we get:
t(D) ≤ 3 · ( βK − 1) + st(D),
which implies that:
10v3 · (t(D) − 1) ≤ 30v3 · ( βK − 1) + 10v3 · (st(D) − 1).
Corollary 3.5.1. Let D(K) satisfy the hypotheses of the Main Theorem (Theorem 3.4.1).
Furthermore, assume that each other circle of HA has at least m ≥ 3 incident twist region
resolutions. Then K is hyperbolic and the complement of K satisfies the following volume
38
bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) <
m
· 10v3 ·
m−2
βK − 1 − 10v3 .
Proof. First, notice that we get the lower bound on volume from Theorem 3.5.1. To get the
upper bound on volume, we use the proof of Corollary 3.4.1. According to Inequality 3.10:
−χ(GA ) = t(D) − # {OCs} ≥
m−2
· t(D).
m
Combining this inequality with Theorem 2.3.1 gives:
m−2
· t(D) ≤ −χ(GA ) = βK − 1.
m
Since Theorem 2.2.2 provides an upper bound of 10v3 · (t(D) − 1) on volume, then our final
step is to express this upper bound in terms of βK . Solving for t(D), we get:
t(D) ≤
m
·
m−2
βK − 1 ,
which implies that:
10v3 · (t(D) − 1) ≤
m
· 10v3 · ( βK − 1) − 10v3 .
m−2
39
Remark 3.5.2. Notice that, as m → ∞, the upper bound above approaches:
10v3 · ( βK − 1) − 10v3 .
Hence, the coefficients of βK in the upper and lower bounds differ by a multiplicative factor
of 2.7701 (in the limit).
40
Chapter 4
Volume Bounds for A-Adequate
Plats
To provide a collection of links that satisfy the hypotheses of the Main Theorem (Theorem 3.4.1) and to seek to improve the lower bounds on volume, we now investigate a certain
family of A-adequate plat diagrams. We begin by finding volume bounds solely in terms of
the twist number t(D). We then translate these volume bounds to be expressed in terms of
the stable penultimate coefficient βK of the colored Jones polynomial, thus showing that the
hyperbolic A-adequate plats under consideration satisfy a Coarse Volume Conjecture.
4.1
Background on Braids and Plat Closures
Recall that the braid group on n strings, called the n-braid group for short, has Artin
presentation:
Bn = σ1 , · · · , σn−1 | σi σj = σj σi and σi σi+1 σi = σi+1 σi σi+1 ,
where the first relation (sometimes called far commutativity) occurs whenever |i − j| ≥ 2
and the second relation occurs whenever 1 ≤ i ≤ n − 2.
41
1
i
i+1
n
i
1
i+1
n
σi−1
σi
Figure 4.1: The generators σi and σi−1 of the n-braid group Bn , where 1 ≤ i ≤ n − 1.
The braid group and its generators can also be represented geometrically. See Figure 4.1.
The multiplication operation in the braid group is concatenation. Furthermore, we will adopt
the convention that a braid word, read from left to right, is visually represented by stacking
the braid generators of Figure 4.1 vertically (reading the braid word visually from the top
down).
In this chapter, we will work with braids that have an even number of strings. For clarity,
we give the presentation for the even-stringed braid group B2n below.
B2n = σ1 , · · · , σ2n−1 | σi σj = σj σi and σi σi+1 σi = σi+1 σi σi+1 ,
where the first relation occurs whenever |i − j| ≥ 2 and the second relation occurs whenever
1 ≤ i ≤ 2n − 2. Within B2n , define the following braid words:
•
j
βe
j
=
j
r2n−2
r2j r4j
σ2 σ4 · · · σ2n−2 ,
rj rj
rj
j
j
where either ri ≤ −3 or ri ≥ 1.
j
2n−1
• βo = σ11 σ33 · · · σ2n−1
, where ri ≤ −2.
The types of 2n-braids that we will consider in this chapter are of the form:
β = βe1 βo1 βe2 βo2 · · · βek βok βek+1 ∈ B2n ,
42
a1,1
a1,2
a1,n−2
a1,n−1
a2,1
a2,2
a2,n−1
a2,n
a2k,1
a2k,2
a2k,n−1
a2k,n
a2k+1,1
a2k+1,2
a2k+1,n−2 a2k+1,n−1
Figure 4.2: A schematic for a 2n-plat diagram with m = 2k + 1 rows of twist regions, where the
entry in the ith row and j th column is a twist region containing ai,j crossings (counted with sign).
The twist regions depicted above are negative twist regions. Having ai,j > 0 instead will reflect the
crossings in the relevant twist region.
j
where k ≥ 1 and n ≥ 3. Call such a braid β of negative-type if ri < 0 for all i and j in the
braid word β and call β of mixed-type otherwise.
For an even-stringed braid β ∈ B2n of the type above, we form the plat closure of β (the
result of which is called a 2n-plat) by connecting the ends of consecutive braid strings by
trivial semi-circular arcs. See Figure 4.2 for a schematic depiction of the plat closure of
β.
Definition 4.1.1. A negative plat diagram is the plat closure of a negative-type braid and
a mixed-sign plat diagram is the plat closure of a mixed-type braid. See the left side of
Figure 4.3 for an example of a negative plat diagram and see the left side of Figure 4.4 for
an example of a mixed-sign plat diagram.
43
A
Figure 4.3: An example of a negative plat diagram and its all-A state.
Let D(K) denote a negative or mixed-sign plat diagram. Furthermore, let D(K) have
m = 2k + 1 rows of twist regions. Specifically, if we number the rows of twist regions from
the top down, then there are k + 1 odd-numbered rows, each of which contains n − 1 twist
regions, and k even-numbered rows, each of which contains n twist regions. Index the twist
regions according to row and column (where by column we really mean the left-to-right
ordering of twist regions in a given row). Denote the number of twist regions in row i and
column j (counted with sign) by ai,j , where 1 ≤ i ≤ m and:
1 ≤ j ≤ n − 1 if i is odd
1≤j≤n
if i is even.
Refer back to Figure 4.2 to see this notation in use.
Remark 4.1.1. For the remainder of this chapter, the term “for all i and j” will be assumed
to apply to i and j that satisfy the above conditions.
Remark 4.1.2. When n = 1 we have that D(K) must be a (twisted) diagram of the
44
“secret” small inner circle
A
special circle
Figure 4.4: An example of a mixed-sign plat diagram and its all-A state. Note that the diagram
above is obtained from the negative plat diagram of Figure 4.3 by changing the first negative twist
region with three crossings to a positive twist region with a single crossing and changing the last
negative twist region with three crossings to a positive twist region with three crossings. These
changes create a “secret” small inner circle and a special circle, respectively.
unknot, which is not hyperbolic and will not be considered. When n = 2 we have that
D(K) represents a two-bridge link K. Using the fact that two-bridge links are alternating,
let D = Dalt (K) denote a reduced alternating diagram of K. It will be shown later that the
plats considered in this work are all hyperbolic. Therefore, by Theorem B.3 of [8], we get
the following volume bounds:
2v3 · t(D) − 2.7066 < vol(S 3 − K) < 2v8 · (t(D) − 1) .
Note that the coefficients of t(D) in the upper and lower bounds above differ by a multiplicative factor of 3.6100 . . .. Since we have the above volume bounds when n = 2, then will
assume for the remainder of this chapter that n ≥ 3.
Remark 4.1.3. When k = 0 (and n ≥ 3) we have that D(K) represents a connected sum of
45
two or more standard torus link diagrams, which is not hyperbolic and will not be considered.
Hence, the assumption that k ≥ 1 is a natural one to make.
4.2
Volume Bounds for Negative Plats in Terms of
t(D)
In this section, we show that negative plat diagrams satisfy the assumptions of the Main
Theorem (Theorem 3.4.1). By studying the specific structure of these diagrams, we are able
to improve the lower bound on volume provided by the theorem.
Theorem 4.2.1. Let D(K) be a negative plat diagram (where k ≥ 1, where n ≥ 3, where
ai,j ≤ −3 in odd-numbered rows, and where ai,j ≤ −2 in even-numbered rows). Then D(K)
is a connected, prime, A-adequate link diagram that satisfies the TELC, contains t(D) ≥ 7
twist regions, and contains st(D) = 0 special tangles. Furthermore, K is hyperbolic and the
complement of K satisfies the following volume bounds:
4v8
v
· (t(D) − 1) + 8 ≤ vol(S 3 − K) < 10v3 · (t(D) − 1) .
5
5
For an example of a plat diagram that satisfies the assumptions above, see the left side of Figure 4.3. Having such a figure in mind will help when considering the proof of Theorem 4.2.1
(which is given below).
Proof of Theorem 4.2.1. Since ai,j = 0 for all i and j, then D(K) must be a connected link
diagram. See Figure 4.2 for visual support.
46
Since we have that k ≥ 1, that n ≥ 2, and that ai,j = 0 for all i and j, then by careful
and methodical inspection we get that D(K) is prime. To see this, let C denote a simple
closed curve in the projection plane that intersects D(K) twice transversely (away from the
crossings) and let p be an arbitrary base point for C. Considering the possible locations of
p in S 2 \D(K) (perhaps using Figure 4.3 to assist in visualization), it can be seen that it is
impossible for C to both close up and contain crossings on both sides. Thus, D(K) is indeed
a prime link diagram.
By inspecting HA , we get that D(K) is A-adequate since ai,j ≤ −2 in odd-numbered rows.
To see this, first notice that the vertical A-segments between (the necessarily distinct) other
circles can never contribute to non-A-adequacy. Second, notice that the vertical A-segments
within a given other circle either connect distinct small inner circles or connect an other
circle to a small inner circle. Therefore, since no A-segment connects a circle to itself, then
D(K) is A-adequate.
The assumptions that ai,j ≤ −3 in odd-numbered rows and ai,j ≤ −2 in even-numbered
rows guarantee that D(K) satisfies the TELC. To be specific, having ai,j ≤ −2 in evennumbered rows forces there to always be at least one small inner circle to act as a buffer
between adjacent other circles, making it is impossible for two given other circles to share
any (let alone two) A-segments. Furthermore, notice that a small inner circle from an evennumbered row must always connect to a pair of distinct circles. Next, having ai,j ≤ −3 in
odd-numbered rows guarantees that there are at least two inner circles for each odd-rowed
twist region, which prevents an other circle from connecting to an interior small inner circle
and then back to itself along another A-segment. Finally, by construction, it is impossible
for a pair of small inner circles to share more than one A-segment. Since we have just shown
47
that no two all-A circles share more than one A-segment, then the TELC is trivially satisfied.
By inspection, we have that D(K) contains t(D) ≥ 7 ≥ 2 regions. Combining this fact with
what was shown above and using Proposition 2.2.1, we can conclude that K is hyperbolic.
Inspection also shows that st(D) = 0, because each other circle is incident to at least five
twist region resolutions.
It remains to show that K satisfies the desired volume bounds. Since there is one other circle
(OC) of HA corresponding to each odd row of twist regions in D(K), then we have that:
# {OCs} = k + 1.
Applying Lemma 3.2.1 gives:
−χ(GA ) = t(D) − # {OCs} = t(D) − k − 1.
We would now like to eliminate the dependence of −χ(GA ) on k. Expand t(D) as:
t(D) = #(odd-numbered rows) · #(twist regions per odd row)
+#(even-numbered rows) · #(twist regions per even row)
= (k + 1)(n − 1) + kn.
Since n ≥ 3, then:
48
t(D) = (k + 1)(n − 1) + kn = 2kn − k + n − 1 ≥ 5k + 2,
which implies that:
k≤
t(D) − 2
.
5
Thus, we get the following:
−χ(GA ) = t(D) − k − 1 ≥ t(D) −
t(D) − 2
5
−1=
1
4
· (t(D) − 1) + .
5
5
(4.1)
By applying Theorem 2.2.2, we have the desired volume bounds.
4.3
Applying the Main Theorem to Negative Plats
In this section, we apply the Main Theorem (actually we apply Corollary 3.4.1) to obtain
volume bounds for the family of negative plats considered in this work. Furthermore, we
compare these volume bounds to those found in Theorem 4.2.1.
Proposition 4.3.1. Applying Corollary 3.4.1 to the negative plats of Theorem 4.2.1, we get
that:
3v8
· t(D) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1)
5
49
Proof. By the conclusions of Theorem 4.2.1, we see that we can apply Theorem 3.4.1. In fact,
inspection of the all-A state HA shows that each other circle is incident to at least m ≥ 5
twist region resolutions. Using either this observation or using Theorem 4.2.1, we have that
st(D) = 0 for D(K) a negative plat diagram. Therefore, by applying Corollary 3.4.1, we get
the desired volume bounds.
Let us now compare the lower bounds of Proposition 4.3.1 to those of Theorem 4.2.1. We
would like to determine exactly when the lower bounds given by our study of negative plats
are an improvement over those that result from applying Corollary 3.4.1 (using Proposition 4.3.1).
For the negative plats considered in this thesis, we have that:
4v8
v
3v
· (t(D) − 1) + 8 ≥ 8 · t(D)
5
5
5
is equivalent to the condition that t(D) ≥ 3. But this condition is always satisfied because we
have shown in the proof of Theorem 4.2.1 that t(D) ≥ 7. Therefore, the lower bound found in
Theorem 4.2.1 is always sharper than the lower bound provided by applying Corollary 3.4.1
(using Proposition 4.3.1).
50
4.4
Volume Bounds for Mixed-Sign Plats in Terms of
t(D)
In this section, we first show that we can alter the negative plat diagrams considered in this
work to get new plat diagrams that enjoy the same nice properties as the original negative
plat diagrams. Specifically, we show that our new plat diagrams (which will be mixed-sign
plat diagrams) satisfy the assumptions of the Main Theorem (Theorem 3.4.1). By studying
the specific structure of these diagrams, we are often able to improve the lower bound on
volume provided by the theorem.
Beginning with a negative plat diagram that satisfies the hypotheses of Theorem 4.2.1, we
may iteratively change any of the negative twist regions in the odd-numbered rows to positive
twist regions. Furthermore, the positive twist regions need only contain at least one crossing.
For an example of this process, see Figure 4.3 and Figure 4.4.
Notation: Let t+ (D) and t− (D) denote the number of positive and negative twist regions
in D(K), respectively.
Remark 4.4.1. Changing an arbitrary negative twist region of an odd-numbered row to a
positive twist region will break the relevant other circle into two all-A circles. This is because
a long twist region is changed to a one-crossing or short twist region. In the relevant part of
the new all-A state, all but one of the new horizontal A-segments correspond to redundant
parallel edges of GA . Thus, this entire new positive twist region corresponds to a single edge
of GA . These remarks hold true during every iteration of the procedure mentioned above.
51
Theorem 4.4.1. Let D(K) be a mixed-sign plat diagram (where k ≥ 1, where n ≥ 3, where
ai,j ≤ −3 or ai,j ≥ 1 in odd-numbered rows, and where ai,j ≤ −2 in even-numbered rows).
Then D(K) is a connected, prime, A-adequate link diagram that satisfies the TELC, contains
t(D) ≥ 3 twist regions, and contains st(D) ≤ 4 special tangles. Furthermore, K is hyperbolic
and the complement of K satisfies the following volume bounds:
v8
2v
· 2t− (D) − 1 − 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
3
If we also have that D(K) contains at least as many negative twist regions as it does positive
twist regions, then:
v8
2v
· (t(D) − 1) − 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
3
Proof of Theorem. The proofs of the connectedness and primeness of D(K) are the same as
those found in the proof of Theorem 4.2.1 and the proof that D(K) is A-adequate is very
similar. The only new observation that is needed is that any horizontal A-segments coming
from positive twist regions necessarily connect distinct all-A circles. The proof that D(K)
satisfies the TELC is also similar to that found in the proof of Theorem 4.2.1, but two-edge
loops may now exist. The new possibility that ai,j ≥ 1 in odd-numbered rows will give rise
to two-edge loops whenever ai,j ≥ 2. These two-edge loops come from the same short twist
region and are, therefore, allowed by the TELC.
Inspection of HA shows that the mixed-sign plat diagrams under consideration contain at
least 7 − 4 = 3 twist regions. This is because having ai,j = 1 in any of the four corners
52
of D(K) means that the corresponding state circles of HA in those corners will be “secret”
small inner circles rather than other circles and, consequently, we may have that potentially
different twist regions are actually part of a single twist region. See Figure 4.4 for an example.
Using what was shown above, we can apply Proposition 2.2.1 to conclude that K is hyperbolic. Furthermore, inspection of HA shows that, unlike for negative plat diagrams, special
circles can actually occur in mixed-sign plat diagrams. However, special circles can only
possibly occur at the four corners of the link diagram. This is because, by the assumption
that ai,j = 0 for all i and j, all but at most the four corner other circles must be incident to
three or more twist region resolutions. See Figure 4.4 for an example. Therefore, we have
that st(D) ≤ 4. It remains to show that K satisfies the desired volume bounds.
Recall the observation that we may start with a negative plat diagram and iteratively change
any of the negative twist regions in the odd-numbered rows to positive twist regions. Specifically, such an alteration creates either a new other circle (OC) or a new “secret” small inner
circle. Thus, after changing some or all of the allowed negative twist regions to positive twist
regions, we have that:
# {OCs} ≤ k + 1 + t+ (D).
Applying Lemma 3.2.1 gives:
−χ(GA ) = t(D) − # {OCs} ≥ t(D) − k − 1 − t+ (D) = t− (D) − k − 1.
Recall that, by construction, we can only have positive twist regions in odd-numbered rows.
53
Thus, all of the even-numbered rows must still contain only negative twist regions. Said
another way:
t− (D) ≥ #(even-numbered rows) · #(twist regions per even row)
= k · n.
(4.2)
We would now like to eliminate the dependence of −χ(GA ) on k. Since t− (D) ≥ kn, then
the assumption that n ≥ 3 gives:
k≤
t− (D)
t− (D)
≤
.
n
3
Therefore, we get:
−χ(GA ) ≥ t− (D) − k − 1 ≥ t− (D) −
t− (D)
1
2
− 1 = · (2t− (D) − 1) − .
3
3
3
Now suppose that D(K) contains at least as many negative twist regions as it does positive
twist regions, so that we have t− (D) ≥ t+ (D). This implies that:
2t− (D) = t− (D) + t− (D) ≥ t− (D) + t+ (D) = t(D),
which then implies that:
54
−χ(GA ) ≥
1
2
1
2
· (2t− (D) − 1) − ≥ · (t(D) − 1) − .
3
3
3
3
(4.3)
By applying Theorem 2.2.2, we have the desired volume bounds.
Remark 4.4.2. Note that the lower bounds on volume in terms of t− (D) will be sharper
than those in terms of t(D) in the case that D(K) is a mixed-sign plat with more negative
twist regions than positive twist regions. Furthermore, as the disparity between the number
of negative and positive twist regions increases, the lower bound on volume in terms of t− (D)
will become increasingly sharper than the lower bound on volume in terms of t(D).
Remark 4.4.3. To conclude our study of mixed-sign plats, we would like to find a sufficient
condition to guarantee that the mixed-sign plats considered in this work also contain at least
as many negative twist regions as positive twist regions.
For negative plats, recall that Equation 4.1 gives that t(D) = (k + 1)(n − 1) + kn. Since (as
mentioned in the above proof of Theorem 4.4.1) the process to change a negative plat into a
mixed-sign plat may create a situation where potentially different twist regions are actually
part of a single twist region, then Equation 4.1 becomes the inequality:
t(D) ≤ (k + 1)(n − 1) + kn
for mixed-sign plats. Also, since the even-numbered rows of a mixed-sign plat must contain
only negative twist regions, then Inequality 4.2 gives that t− (D) ≥ kn. Combining this
55
information, we get:
t− (D) + t+ (D) = t(D) ≤ (k + 1)(n − 1) + kn ≤ (k + 1)(n − 1) + t− (D),
which implies that:
t+ (D) ≤ (k + 1)(n − 1) = kn + n − k − 1 ≤ t− (D) + n − k − 1.
Thus, to guarantee that t− (D) ≥ t+ (D), we need that n − k − 1 ≤ 0. But this condition is
equivalent to k ≥ n − 1 is equivalent to m = 2k + 1 ≥ 2n − 1. Therefore, to guarantee that
the mixed-sign plats considered in this work contain at least as many negative twist regions
as positive twist regions, we need that the mixed-sign plats must contain at least 2n − 1 rows
of twist regions.
4.5
Applying the Main Theorem to Mixed-Sign Plats
In this section, we apply the Main Theorem (Theorem 3.4.1) to obtain volume bounds for the
family of mixed-sign plats considered in this work. Furthermore, we compare these volume
bounds to those found in Theorem 4.4.1 (at least for the mixed-sign plats that contain at
least as many negative twist regions as positive twist regions).
Proposition 4.5.1. Applying Theorem 3.4.1 to the mixed-sign plats of Theorem 4.4.1, we
get that:
56
v8
· (t(D) − 4) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1)
3
Proof. By the conclusions of Theorem 4.4.1, we see that we can apply Theorem 3.4.1. Theorem 4.4.1 also shows that st(D) ≤ 4 for D(K) a mixed-sign plat diagram. Therefore, by
applying Theorem 3.4.1, we get the desired volume bounds.
Let us now compare the lower bounds of Proposition 4.5.1 to those of Theorem 4.4.1, at
least for the mixed-sign plats that contain at least as many negative twist regions as positive
twist regions. We would like to determine exactly when the lower bounds given by our
study of such mixed-sign plats are an improvement over those that result from applying
Theorem 3.4.1 (using Proposition 4.5.1).
For the mixed-sign plats considered in this work (that contain at least as many negative
twist regions as positive twist regions), we have that:
v8
2v
v
· (t(D) − 1) − 8 ≥ 8 · (t(D) − 4)
3
3
3
4
is equivalent to the condition that −1 ≥ − . But this condition is always satisfied. There3
fore, the lower bound found in Theorem 4.4.1 is always (slightly) sharper than the lower
bound provided by applying Theorem 3.4.1 (using Proposition 4.5.1).
57
4.6
Volume Bounds for Plats in Terms of the Colored
Jones Polynomial
To conclude our study of volume bounds for hyperbolic A-adequate plats, we will translate
our diagrammatic volume bounds (in terms of t(D)) to volume bounds in terms of the stable
penultimate coefficient βK of the colored Jones polynomial. We shall prove the following
theorems:
Theorem 4.6.1. Let D(K) be a negative plat diagram (where n ≥ 3). Then D(K) is
a connected, prime, A-adequate link diagram that satisfies the TELC, contains t(D) ≥ 7
twist regions, and contains st(D) = 0 special tangles. Furthermore, K is hyperbolic and the
complement of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 − K) <
25v3
5v
· ( βK − 1) − 3 .
2
2
Theorem 4.6.2. Let D(K) be a mixed-sign plat diagram (where n ≥ 3) that contains at least
as many negative twist regions as it does positive twist regions. Then D(K) is a connected,
prime, A-adequate link diagram that satisfies the TELC, contains t(D) ≥ 3 twist regions,
and contains st(D) ≤ 4 special tangles. Furthermore, K is hyperbolic and the complement
of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 − K) < 30v3 · ( βK − 1) + 20v3 .
Combined Proof of Both Theorems. To begin, note that Theorem 4.2.1 and Theorem 4.4.1
58
combine to give all but the volume bounds. Theorem 3.5.1 gives the desired lower bounds
on volume. Since Theorem 2.2.2 provides 10v3 · (t(D) − 1) as an upper bound on volume,
then we will have written the volume bounds in terms of the coefficients of the colored Jones
polynomial (specifically, in terms of βK ) if we are able to bound t(D) above by a function
of βK .
By Theorem 2.3.1 and Inequality 4.1 for negative plat diagrams, we get that:
βK − 1 = −χ(GA ) ≥
1
4
· (t(D) − 1) + ,
5
5
which implies that:
t(D) − 1 ≤
5
1
· ( βK − 1) − ,
4
4
which implies that:
10v3 · (t(D) − 1) ≤
25v3
5v
· ( βK − 1) − 3 .
2
2
This gives the desired result for negative plats.
Now assume that D(K) is a mixed-sign plat diagram with at least as many negative twist
regions as positive twist regions. By Theorem 2.3.1 and Inequality 4.3 for mixed-sign plat
diagrams, we get:
59
βK − 1 = −χ(GA ) ≥
1
2
· (t(D) − 1) − ,
3
3
which implies that:
t(D) − 1 ≤ 3 · ( βK − 1) + 2,
which implies that:
10v3 · (t(D) − 1) ≤ 30v3 · ( βK − 1) + 20v3 .
This gives the desired result for mixed-sign plats with at least as many negative twist regions
as positive twist regions.
60
Chapter 5
Volume Bounds for A-Adequate
Closed Braids
To provide a second collection of links that satisfy the hypotheses of the Main Theorem
(Theorem 3.4.1) and to seek to improve the lower bounds on volume, we now investigate
a certain family of A-adequate closed braid diagrams. We begin by finding volume bounds
solely in terms of the twist number t(D). We then translate these volume bounds to be
expressed in terms of the penultimate coefficient βK of the colored Jones polynomial, thus
showing that the hyperbolic A-adequate closed braids under consideration satisfy a Coarse
Volume Conjecture.
5.1
Braid Closure and Cyclic Reduction of Braid Words
We now return to n-braids and consider a second type of closure operation (viewing the
the plat closure of even-stringed braids as the first type of closure). Let β ∈ Bn denote an
n-braid. To form the braid closure of β means that, for all braid string positions 1 ≤ i ≤ n,
we identify the top of the string in the ith position with the bottom of string in the ith
position. We denote the braid closure of β by β. See Figure 5.1 for a schematic depiction of
a braid closure.
61
2
1
n
n-braid
1
2
n
Figure 5.1: The closure of an n-braid.
Definition 5.1.1. Call a subword γ of β ∈ Bn (cyclically) induced if the letters of γ are
all adjacent up to cyclic permutation of the braid word β and if these letters appear in the
same order as in the braid word β.
As an example, given the braid word:
β = σ13 σ2−3 σ12 σ3−2 σ2 σ3 = σ13 σ2−1 (σ2−2 σ12 σ3−1 )σ3−1 σ2 σ3 ,
the subword γ = σ2−2 σ12 σ3−1 is an induced subword of β but the subword γ = σ2−2 σ12 σ2 is
not an induced subword of β.
Because conjugate braids have isotopic braid closures and because our focus will ultimately be
on links that are braid closures, we will work within conjugacy classes of braids. Furthermore,
note that cyclic permutation is a special case of conjugation.
62
r
r1 r2
l
Definition 5.1.2. As in [26], call a braid β = σm
1 σm2 · · · σml ∈ Bn cyclically reduced into
syllables if the following hold:
(1) for all j, we have that rj = 0
(2) for all j, (looking up to cyclic permutation) we have that there are no occurrences of
cyclically induced subwords of the form σj σj−1 or σj−1 σj
(3) for all j (modulo l), we have that mj = mj+1
r
r
Definition 5.1.3. Let β = σir1 σjr2 · · · σi 2l−1 σj 2l denote a 3-braid that has been cyclically
reduced into syllables, where {i, j} = {1, 2}. As was done in [26], form the exponent vector
(r1 , r2 , . . . , r2l ) of β.
Definition 5.1.4. Call an n-braid β ∈ Bn positive if all of its exponents are positive and
call β negative if all of its exponents are negative.
5.2
A-Adequacy for Closed 3-Braids
We now present Stoimenow’s ([26]) classification of A-adequate closed 3-braids. We also
present the proof of this result (which was left to the reader) since the methodology and
perspective will be useful in what follows.
Proposition 5.2.1 ([26], Lemma 6.1). Let D(K) = β denote the closure of a 3-braid
r
r
β = σir1 σjr2 · · · σi 2l−1 σj 2l , where {i, j} = {1, 2} and where β has been cyclically reduced
into syllables. Furthermore, assume that l ≥ 2 (which says that the exponent vector has
length at least four). Then D(K) is A-adequate if and only if either:
(1) β is positive, or
63
non-A-adequacy
Figure 5.2: The subdiagram of a closed 3-braid diagram corresponding to a subword σ2−1 σ1−1 σ2−1
(left) and the corresponding portion of the all-A state that exhibits the non-A-adequacy of D(K)
(right).
(2) β does not contain σ1−1 σ2−1 σ1−1 = σ2−1 σ1−1 σ2−1 as a cyclically induced subword and
β also has the property that all positive entries of the exponent vector are cyclically
isolated (meaning they must be adjacent to negative entries on both sides).
Remark 5.2.1. The condition that positive entries of the exponent vector be cyclically
isolated can equivalently be phrased as the condition that positive syllables in the braid
p
word (of the form σi for p > 0) must be cyclically adjacent to negative syllables (of the form
σjn for n < 0) on both sides, where {i, j} = {1, 2}.
Proof of Proposition. (⇒) We shall proceed by contraposition.
Case 1: Suppose β is not positive and that β contains σ1−1 σ2−1 σ1−1 = σ2−1 σ1−1 σ2−1 as a
cyclically induced subword. From the portion of the all-A state HA corresponding to this
subword, it can be seen that D(K) is not A-adequate. (See Figure 5.2.)
Case 2: Suppose β is not positive and that there exist cyclically adjacent positive syllables
in the braid word. Since β is not positive, then there must exist at least one negative
syllable in the braid word. Choose a pair of cyclically adjacent positive syllables that are
are immediately followed by a negative syllable. Hence, β must have a subword of the form
p
p
σi 1 σj 2 σin , where p1 , p2 > 0 are positive integers, where n < 0 is a negative integer, and
64
p1
non-A-adequacy
p1
p2
p2
n
n
Figure 5.3: The subdiagram of a closed 3-braid diagram corresponding to a subword σ1p1 σ2p2 σ1n ,
where the pi are positive integers and n is a negative integer (left), and the corresponding portion
of the all-A state that exhibits the non-A-adequacy of D(K) (right).
p1
p1
p2
p2
p2l
p2l
Figure 5.4: A positive closed 3-braid diagram (left) and the corresponding all-A state that exhibits
the A-adequacy of D(K) (right).
where {i, j} = {1, 2}. From the portion of the all-A state HA corresponding to this subword,
it can be seen that D(K) is not A-adequate. (See Figure 5.3.)
(⇐)
Case 1: Assume D(K) is a positive closed 3-braid. From the corresponding all-A state HA
it can be seen that D(K) is A-adequate. (See Figure 5.4.)
Case 2: Assume β does not contain σ1−1 σ2−1 σ1−1 = σ2−1 σ1−1 σ2−1 as a cyclically induced
subword and assume β has the property that all positive entries of the exponent vector are
cyclically isolated (which implies that β cannot be positive). For a contradiction, suppose
65
C
s
C
C
s
C
C
s
C
p
Figure 5.5: Potential non-A-adequacy coming from σ1−1 . From the perspective of the all-A state
HA , the vertical segment s coming from σ1−1 is assumed to join a state circle C to itself (left). Also
depicted are the case where σ1−1 is followed by another copy of σ1−1 (center) and the case where
σ1−1 is followed a positive syllable σ2p , which must then be followed by the letter σ1−1 (right).
D(K) is not A-adequate.
Subcase 1: Suppose non-A-adequacy comes from the occurrence of σ1−1 (or, symmetrically,
σ2−1 ) in the braid word. Since the argument is very similar, we will not consider the symmetric case. Having non-A-adequacy in this setting means that when we A-resolve the crossing
of D(K) corresponding to σ1−1 , the corresponding vertical A-segment of HA , call it s, will
connect an all-A circle, call it C, to itself. (See the left side of Figure 5.5.) Let us now
consider what other letters can surround σ1−1 in the braid word.
First, notice that σ1−1 cannot be surrounded by other copies of σ1−1 on either side, as this
will contradict how C must close up. (See the center of Figure 5.5.) Second, σ1−1 cannot be
surrounded on both sides by σ2−1 , as this would imply the existence of a forbidden subword
σ2−1 σ1−1 σ2−1 , a contradiction. Thus, it must be the case that σ1−1 is surrounded on at least
p
one side by a positive syllable σ2 . Furthermore, since the exponent vector is assumed to
have length at least four and since positive entries are assumed to be cyclically isolated, then
p
the next adjacent letter after the positive syllable σ2 must be σ1−1 . But then it can be seen
p
that the A-resolution of the subword σ1−1 σ2 σ1−1 contradicts how C must close up. (See the
right side of Figure 5.5.)
66
C
p
C
C
Figure 5.6: Potential non-A-adequacy coming from a positive syllable σ1p . From the perspective
of the all-A state HA , the horizontal A-segments coming from σ1p are assumed to join a state circle
C to itself.
p
Subcase 2: Suppose non-A-adequacy comes from the occurrence of a positive syllable σ1 (or,
p
symmetrically, σ2 ) in the braid word. Since the argument is very similar, we will not consider
the symmetric case. Having non-A-adequacy in this setting means that when we A-resolve
p
the twist region of D(K) corresponding to σ1 , the corresponding horizontal A-segments will
connect a state circle, call it C, to itself. Since the exponent vector is assumed to have
p
length at least four and since positive entries are assumed to be cyclically isolated, then σ1
must be surrounded by the same letter σ2−1 on both sides. But then it can be seen that the
p
A-resolution of the subword σ2−1 σ1 σ2−1 contradicts how C must close up. (See Figure 5.6.)
5.3
State Circles of A-Adequate Closed 3-Braids
The goal of this section is to classify the possible types of all-A circles that can arise in the
all-A state of an A-adequate closed 3-braid.
r
r
Proposition 5.3.1. Let β = σir1 σjr2 · · · σi 2l−1 σj 2l denote a 3-braid that has been cyclically
reduced into syllables, where l ≥ 2 and {i, j} = {1, 2}. Assume that D(K) = β is an Aadequate closed 3-braid. Then we may categorize the all-A circles of HA into the following
67
Figure 5.7: An example of a small inner circle coming from a σ1−2 subword (left), an example
of a medium inner circle and a potential nonwandering circle portion coming from a σ2−1 σ1p σ2−1
subword where p > 0 (center), and an example of a wandering circle portion coming from a
σ2−2 σ1−2 = σ2−1 (σ2−1 σ1−1 )σ1−1 subword (right).
types:
(1) small inner circles that come from negative exponents ri ≤ −2 in the braid word β
(2) medium inner circles that come from cyclically isolated positive syllables in the braid
word β
(3) wandering circles whose wandering arises from adjacent negative syllables in the braid
word β
(4) nonwandering circles that come from the case when a given generator σi occurs with
only positive exponents in the braid word β
For depictions of the types of circles mentioned above (or portions of the types of circles
mentioned above), see Figure 5.7.
Remark 5.3.1. From the 3-braid perspective, it will be important to note that the left
half of Figure 2.5 depicts the A-resolution of a negative syllable (excluding a portion of a
third trivial braid string). Similarly, the right half of Figure 2.5 depicts the A-resolution of
a positive syllable (excluding a portion of a third trivial braid string).
68
Proof of Proposition. Recall that A-adequate closed 3-braids D(K) = β were classified by
Proposition 5.2.1 into two main types.
Case 1: Suppose β is positive. Then each all-A circle in the all-A state HA of D(K) is a
nonwandering circle. (See Figure 5.4.)
Case 2: Suppose β does not contain σ1−1 σ2−1 σ1−1 = σ2−1 σ1−1 σ2−1 as a cyclically induced
subword and suppose β also has the property that all positive syllables are cyclically isolated.
Since positive syllables of β are cyclically isolated, then (using cyclic permutation if needed)
we may decompose β as β = N1 (in the case that β is a negative braid) or β = P1 N1 · · · Pt Nt ,
where Pi denotes a positive syllable of β and Ni denotes a maximal length negative induced
subword of β.
Small Inner Circles: Let σin denote a negative syllable. Then, except for the additional
vertical trivial braid string portion to the right or left (depending on whether i = 1 or
i = 2), the A-resolution of this syllable will look like the left side of Figure 5.7 if n = −2
and look like the left side of Figure 2.5 in general. In particular, having n ≤ −2 is equivalent
to the existence of small inner circles.
Wandering Circles: Let σin1 σjn2 = σin1 +1 σi−1 σj−1 σjn2 +1 denote a pair of adjacent negative
syllables, where {i, j} = {1, 2}. Then the corresponding portion of HA will resemble the
right side of Figure 5.7 (except that the long resolutions may possibly consist of longer paths
of A-segments and small inner circles). The key feature of this figure is the fact that we see
a portion of a wandering circle, where the wandering behavior corresponds to the existence
of the σi−1 σj−1 induced subword.
69
p
Medium Inner Circles and Nonwandering Circles: Let σi−1 σj σi−1 denote a positive syllable
that is cyclically surrounded by negative letters, where {i, j} = {1, 2}. Then the corresponding portion of HA will resemble the center of Figure 5.7. In particular, the existence of a
(cyclically isolated) positive syllable corresponds to the existence of a medium inner circle.
Furthermore, a generator σ1 or σ2 occurring with only positive exponents (again, see the
center of Figure 5.7) will correspond to a nonwandering circle.
Since the A-resolutions of all portions of the closure of β = N1 and β = P1 N1 · · · Pt Nt have
been considered locally and since gluing such portions together joins wandering and potential
nonwandering circle portions together, then we have the desired result.
5.4
Primeness, Connectedness, and Twistedness for
Closed Braid Diagrams
Assume that an n-braid β ∈ Bn has been cyclically reduced into syllables and let D(K) = β.
The main goal of this section is to find sufficient conditions for the diagram D(K) to be
prime.
Definition 5.4.1. Call two induced subwords γ1 and γ2 of a given braid word β disjoint if
the subwords share no common letters when viewed as part of β.
As an example, given the braid word:
70
β = σ13 σ2−3 σ12 σ3−2 σ2 σ3 = σ12 (σ1 σ2−3 σ1 )σ1 (σ3−2 σ2 )σ3 ,
the subwords γ1 = σ1 σ2−3 σ1 and γ2 = σ3−2 σ2 are disjoint induced subwords of β.
Definition 5.4.2. Call a subword γ ∈ Bn of a braid word β ∈ Bn complete if it contains,
at some point, each of the generators σ1 , . . . , σn−1 of Bn .
It is important to note that the generators σ1 , . . . , σn−1 need not occur in any particular
order and that repetition of some or all of these generators is allowed.
Proposition 5.4.1. Let β ∈ Bn be cyclically reduced into syllables. If β contains two disjoint
induced complete subwords γ1 and γ2 , then D(K) = β is a prime link diagram.
Remark 5.4.1. Note that, in the case when n = 3, the condition that β contains two disjoint
induced complete subwords is equivalent to the condition that σ1 and σ2 each occur at least
twice nontrivially in the braid word β is equivalent to the condition that the exponent vector
of β has length at least four.
Proof of Proposition. Let γ1 and γ2 be two disjoint induced complete subwords of β ∈ Bn .
See Figure 5.8 for a schematic depiction of this situation. Let C be a simple closed curve
that intersects D(K) = β exactly twice (away from the crossings). We need to show that
C cannot contain crossings of D(K) on both sides. Note that we may view a closed braid
diagram as lying in an annular region of the plane. (See Figure 5.1.)
Suppose C contains a point p that lives outside of this annular region. Since C only intersects
D(K) twice and must start and end at p, then it must be the case that C intersects D(K)
71
1
n−1n
2
γ1
γ2
1
n−1n
2
Figure 5.8: A schematic depiction of a braid β ∈ Bn that contains two disjoint induced complete
subwords γ1 and γ2 .
twice in braid string position 1 or twice in braid string position n. In either case, it is
impossible for such a closed curve C to contain crossings on both sides. This is because C
would need to intersect D(K) more than twice to be able to close up and surround crossings
on both sides, and this would contradict the assumption that C intersects D(K) exactly
twice.
Suppose C contains a point p between braid string positions i and i+1 for some 1 ≤ i ≤ n−1.
See Figure 5.9. Since C intersects D(K) exactly twice and since C is a simple closed curve,
then it must be the case that C either intersects D(K) twice in braid string position i or
twice in braid string position i + 1. Since β contains two disjoint induced complete subwords,
then the generator σi , the generator σi−1 (if it exists), and the generator σi+1 (if it exists)
must each occur at least twice, once cyclically before p and once cyclically after p. Since
σi−1 and σi+1 occur both before and after p, then it is impossible for C to both close up and
72
σi−1 , σi , σi+1
i
i+1
C
p
σi−1 , σi , σi+1
Figure 5.9: A schematic depiction of the fact that a simple closed curve C containing a point p
(between braid string positions i and i + 1) cannot contain crossings on both sides. Each box in
the figure represents the eventual occurrence of the generator σi , the generator σi−1 (if it exists),
and the generator σi+1 (if it exists). No assumptions are made about the order in which these
generators appear or the frequency with which these generators appear.
contain crossings on both sides. This is because C is forced to cross a braid string position
±1
before encountering σi±1 (cyclically above and below), and because the occurrences of σi−1
±1 (cyclically above and below) block C from being able to close up in a way that will
and σi+1
contain crossings on both sides.
The assumptions about subwords of β in Proposition 5.4.1 also force D(K) to have other
desired properties, as seen in the proposition below.
Proposition 5.4.2. Let β ∈ Bn be cyclically reduced into syllables. If β contains two
disjoint induced complete subwords γ1 and γ2 , then D(K) = β is a connected link diagram
with t(D) ≥ 2(n − 1) twist regions.
73
Proof. Recall that D(K) is connected if and only if the projection graph of D(K) is connected. Since β ∈ Bn contains a complete subword, then each generator of Bn (each of which
corresponds to a crossing of D(K) between adjacent braid string positions) must occur at
least once. This fact implies that the closed n-braid diagram D(K) must be connected. (See
Figure 5.1.)
Since β contains two disjoint complete subwords, then it must be that each of the generators
σ1 , . . . , σn−1 of Bn must occur at least twice in (two distinct syllables of) the braid word.
Since syllables of the cyclically reduced braid word β correspond to twist regions of D(K),
then we have that t(D) ≥ 2(n − 1).
Because the majority of the braids considered in this paper will satisfy the same set of
assumptions, we make the following definition.
Definition 5.4.3. Call a braid β ∈ Bn nice if the following hold:
(1) β is cyclically reduced into syllables.
(2) β contains two disjoint induced complete subwords γ1 and γ2 .
5.5
The Foundational Theorem for Closed Braids
r
r1
l
Theorem 5.5.1. Let D(K) = β denote the closure of a nice n-braid β = σm
1 · · · σml , where
1 ≤ m1 , . . . , ml ≤ n − 1. If β satisfies the conditions:
(1) all negative exponents ri < 0 in β satisfy the stronger requirement that ri ≤ −3, and
74
(2) when ri > 0, we have that ri−1 < 0 and ri+1 < 0 and either mi−1 = mi+1 = mi + 1
or mi−1 = mi+1 = mi − 1,
then D(K) is a connected, prime, A-adequate link diagram that satisfies the TELC and
contains t(D) ≥ 2(n − 1) twist regions. Furthermore, we have that K is a hyperbolic link.
Remark 5.5.1. Condition (2) above says that positive syllables in the letter σmi are cyclically adjacent to negative syllables in the same adjacent letter (either σmi +1 or σmi −1 ). This
condition extends to n-braids Stoimenow’s 3-braid condition (see Proposition 5.2.1) that positive entries in the exponent vector are cyclically isolated. Condition (2) also implies that β
cannot be a positive braid. Furthermore, Condition (1) above trivially satisfies Stoimenow’s
condition that the 3-braid does not contain σ1−1 σ2−1 σ1−1 = σ2−1 σ1−1 σ2−1 as an induced subword. Hence, to conclude A-adequacy (among other things) for braids with n ≥ 3 strings, we
generalize one of Stoimenow’s two conditions and impose further restrictions on the other.
Definition 5.5.1. Since β ∈ Bn is cyclically reduced into syllables, then the positive syllap
bles σi of β (where p > 0) correspond to what we will call positive twist regions of D(K)
and the negative syllables σin of β (where n < 0) correspond to what we will call negative
twist regions of D(K).
Proof of Theorem. From Proposition 5.4.1 and Proposition 5.4.2, we have already seen that
D(K) is connected and prime with t(D) ≥ 2(n − 1) twist regions. Furthermore, once it
is shown that D(K) is A-adequate and satisfies the TELC, then the conclusion that K is
hyperbolic will follow from Proposition 2.2.1.
To see that D(K) is A-adequate, it suffices to show that no A-segment of the all-A state
HA joins an all-A circle to itself. Note that positive syllables in β (positive twist regions of
75
D(K)) A-resolve to give horizontal segments and that negative syllables in β (negative twist
regions of D(K)) A-resolve to give vertical segments. Suppose, for a contradiction, that an
A-segment joins an all-A circle to itself.
Case 1: Suppose the A-segment is a vertical segment. Condition (1), that negative exponents
are at least three in absolute value, implies that it is impossible for a vertical A-segment to
join an all-A circle to itself. This is because all vertical A-segments either join distinct small
inner circles or join a small inner circle to a medium or a wandering circle. (See the left side
of Figure 2.5.)
Case 2: Suppose the A-segment is a horizontal segment. Condition (2), that positive syllables are cyclically surrounded by negative syllables in the same adjacent generator, implies
that it is impossible for a horizontal A-segment to join an all-A circle to itself. This is because all horizontal A-segments join a medium inner circle to either a wandering circle or a
nonwandering circle. (See Figure 5.6.)
We will now show that D(K) satisfies the TELC. Let C1 and C2 be two distinct all-A circles
that share a pair of distinct A-segments, call them s1 and s2 .
Case 1: Suppose one of s1 and s2 is a horizontal segment. By Conditions (1) and (2), it is
impossible for the other segment to be a vertical segment. This is because, as in Figure 5.6,
one of C1 and C2 must be a medium inner circle. Let us say that C1 is a medium inner
circle. This circle is adjacent, via horizontal A-segments, to the second circle C2 , which will
either be a wandering circle or a nonwandering circle. The circle C1 is also adjacent to small
inner circles above and below. Therefore, since a small inner circle is neither wandering nor
76
nonwandering, then the second segment cannot be vertical. Thus, if one of the segments s1
and s2 is horizontal, then the other segment must also be horizontal. Since the horizontal
segments incident to the medium inner circle C1 necessarily belong to the same short twist
region of D(K), then the TELC is satisfied.
Case 2: Suppose both s1 and s2 are vertical segments. Since all vertical A-segments either
join distinct small inner circles or join a small inner circle to a medium inner circle or a
wandering circle, then one of C1 and C2 must be a small inner circle. But, by construction,
it is impossible for a small inner circle to share more than one A-segment with another circle.
Therefore, this case cannot occur.
5.6
State Circles of A-Adequate Closed n-Braids (where
n ≥ 4)
The goal of this section is to classify the possible types of all-A circles that can arise in the
all-A state of an A-adequate closed n-braid, where n ≥ 4 and where the braid is of the type
described in Theorem 5.5.1.
r
r1
l
Proposition 5.6.1. Let D(K) = β denote the closure of a nice n-braid β = σm
1 · · · σml ,
where 1 ≤ m1 , . . . , ml ≤ n − 1. Furthermore, assume that β satisfies the assumptions of
Theorem 5.5.1 and assume that n ≥ 4. View D(K) as lying in an annular region of the
plane. Then we may categorize the all-A circles of HA into the following types:
(1) small inner circles that come from negative exponents ri ≤ −2 in the braid word β
77
(2) medium inner circles that come from cyclically isolated positive syllables in the braid
word β
(3a) essential wandering circles that are essential in the annulus and have wandering that
arises from adjacent negative syllables in the braid word β
(3b) non-essential wandering circles that are non-essential (contractible) in the annulus and
have wandering that arises from adjacent negative syllables in the braid word β
(4) nonwandering circles that come from the cases when σ1 or σn−1 or an adjacent pair
of generators σi and σi+1 for 2 ≤ i ≤ n − 3 occur with only positive exponents in the
braid word β.
Remark 5.6.1. Note that, as compared with the n = 3 case of Section 5.3, we now have
that a new type of wandering circle, called a non-essential wandering circle, is possible. Also
note that neither the pair of generators σ1 and σ2 nor the pair of generators σn−2 and σn−1
can occur with only positive exponents. This is due to Condition (2) of Theorem 5.5.1 (the
condition that positive syllables are cyclically surrounded by negative syllables in the same
adjacent generator).
Proof of Proposition. With some exceptions, this proof is similar to the proof of Proposition 5.3.1. Recall that, as noted in Remark 5.5.1, β cannot be positive. Because positive
syllables of β are cyclically isolated, we may decompose β as β = N1 (in the case that β is
a negative braid) or β = P1 N1 · · · Pt Nt , where Pi denotes a positive syllable of β and Ni
denotes a maximal length negative induced subword of β.
Small Inner Circles: Let σin denote a negative syllable. Then, except for the additional
78
n − 2 vertical trivial braid string portions to the right and/or left of the twist region, the
A-resolution of this syllable will look like the left side of Figure 5.7 if n = 2 and look like the
left side of Figure 2.5 in general. In particular, having n ≤ −2 is equivalent to the existence
of small inner circles.
Wandering Circles: Let σin1 σjn2 = σin1 +1 (σi−1 σj−1 )σjn2 +1 denote a pair of adjacent negative
syllables.
Suppose {i, j} = mq , mq+1 . Then the pair of adjacent negative syllables involve adjacent
letters (generators of Bn ). Consequently, the corresponding portion of HA will resemble the
right side of Figure 5.7 (except that the long resolutions may possibly consist of longer paths
of A-segments and small inner circles). The key feature of this figure is the fact that we see
a portion of a wandering circle, where the wandering behavior corresponds to the existence
of the σi−1 σj−1 induced subword.
Suppose {i, j} =
mq , mq+r , where r ≥ 2. In this case, the pair of adjacent negative
syllables involve far commuting generators of Bn . Then, except for the additional n − 4
vertical trivial braid string portions around the two negative twist regions, we get that the
A-resolution will look like two copies of the left side of Figure 2.5.
p
Medium Inner Circles and Nonwandering Circles: Let σi−1 σj σi−1 denote a positive syllable
of β that is cyclically surrounded by negative letters, where {i, j} = mq , mq+1 . Note that
the letters involved in this induced subword are adjacent generators of Bn . Consequently, by
adding n − 3 trivial braid string portions, the corresponding portion of HA will resemble the
center of Figure 5.7. In particular, the existence of an isolated positive syllable corresponds
79
to the existence of a medium inner circle. Furthermore, a nonwandering circle will occur in
HA precisely when σ1 or σn−1 or an adjacent pair of generators σi and σi+1 for 2 ≤ i ≤ n−3
occur with only positive exponents.
Note that we may classify the wandering circles in the annulus into essential and nonessential
circles. Since the A-resolutions of all portions of the closure of β = N1 and β = P1 N1 · · · Pt Nt
have been considered locally and since gluing such portions together joins wandering and
potential nonwandering circle portions together, then we have the desired result.
5.7
Computation of −χ(GA)
Recall that an other circle (OC) is an all-A circle that is not a small inner circle (so it must
be a medium inner circle, a wandering circle, or a nonwandering circle). Also, recall that
t+ (D) denotes the number of positive (short) twist regions in D(K) and t− (D) denotes the
number of negative (long) twist regions in D(K). To compute −χ(GA ), we will consider the
cases n = 3 and n ≥ 4 separately. This is because it is only in the case that n ≥ 4 that
non-essential wandering circles can exist.
Case 1: Suppose n = 3.
Lemma 5.7.1. Let D(K) = β denote the closure of a nice 3-braid β ∈ B3 . Furthermore,
assume that the positive syllables of β are cyclically isolated. Then the all-A state, HA , of
D(K) satisfies precisely one of the following two properties:
(1) HA contains exactly one nonwandering circle and no wandering circles.
80
(2) HA contains exactly one wandering circle and no nonwandering circles.
Proof. Either β is alternating or β is nonalternating.
Suppose β is alternating. Then one of the braid generators must always occur with positive
exponents and the other generator must always occur with negative exponents. Hence, by
Proposition 5.6.1, since a generator occurs with only positive exponents, then there will be
a nonwandering circle. Since only one generator occurs with only positive exponents, then
there is only one such nonwandering circle. Also, since adjacent negative syllables cannot
occur in β, then wandering circles cannot occur in HA .
Suppose β is nonalternating. Since positive syllables are cyclically isolated, then the nonalternating behavior of β must come from a pair of adjacent negative syllables. This implies
the existence of a wandering circle in HA and prevents the existence of a nonwandering circle
(since both generators occur once with negative exponent). Finally, since a wandering circle
(which is always essential in the case that n = 3) “uses up” a braid string from the braid
closure and since a wandering circle must wander from braid string position 1 to braid string
position 3 and back before closing up, then the existence of a second such wandering circle
is impossible.
Theorem 5.7.1. Let D(K) = β, where β ∈ B3 satisfies the assumptions of Theorem 5.5.1.
Then:
−χ(GA ) = t− (D) − 1 ≥
81
1
1
· (t(D) − 1) − .
2
2
Proof. Since Theorem 5.5.1 guarantees that D(K) is connected, A-adequate, and satisfies the
TELC, then by Lemma 3.2.1 we have that −χ(GA ) = t(D)−# {OCs}. By Proposition 5.3.1,
we have classified the types of all-A circles. By Condition (2) of Theorem 5.5.1 (that positive
syllables are cyclically isolated) and by inspecting the A-resolution of a positive twist region
(see Figure 5.6), we see that the number of medium inner circles in HA is exactly the number
of positive twist regions in D(K), which we have denoted by t+ (D). By Lemma 5.7.1, we
know that the total number of wandering circles and nonwandering circles is one. Again
using the assumption that positive syllables are cyclically isolated, we have that at least half
of the twist regions in D(K) must be negative twist regions. Therefore, using all of what
was said above (in order) gives:
−χ(GA ) = t(D) − # {OCs}
= t(D) − # {medium inner circles} − # {wandering circles}
−# {nonwandering circles}
= t(D) − t+ (D) − 1
= t− (D) − 1
t(D)
−1
2
1
1
=
· (t(D) − 1) − .
2
2
≥
82
Case 2: Suppose n ≥ 4.
Theorem 5.7.2. For n ≥ 4, let D(K) = β, where β ∈ Bn satisfies the assumptions of
Theorem 5.5.1. Let m denote the number of non-essential wandering circles in the all-A
state HA of D(K). Then:
−χ(GA ) ≥ t− (D) − (n + m − 2) ≥
1
1
· (t(D) − 1) − · (2(n + m) − 5).
2
2
Proof. By applying Lemma 3.2.1, we have that:
−χ(GA ) = t(D) − # {OCs} .
By Proposition 5.6.1, we have classified the types of all-A circles. As discussed in the proof of
Theorem 5.7.1, the number of medium inner circles in HA is exactly the number of positive
twist regions in D(K) and at least half of the twist regions in D(K) must be negative twist
regions. Since Condition (2) of Theorem 5.5.1 implies that β is not a positive braid, then
there must be at least one negative syllable in the braid word. Also, note that both essential
wandering circles and nonwandering circles “use up” a braid string from the braid closure.
The previous two facts imply that:
# {essential wandering circles} + # {nonwandering circles} ≤ n − 2.
Therefore, using all of what was said above gives:
83
−χ(GA ) = t(D) − # {OCs}
= t(D) − # {medium inner circles} − # {non-essential wandering circles}
−# {essential wandering circles} − # {nonwandering circles}
≥ t(D) − t+ (D) − m − (n − 2)
= t− (D) − (n + m − 2)
t(D)
− (n + m − 2)
2
1
1
=
· (t(D) − 1) − · (2(n + m) − 5).
2
2
≥
Remark 5.7.1. Recall that, by Proposition 5.4.2, we have that t(D) ≥ 2(n − 1). Since at
least half of the twist regions of D(K) are negative, then:
t− (D) ≥
t(D)
≥ n − 1,
2
which gives that:
−χ(GA ) ≥ t− (D) − (n + m − 2) ≥ 1 − m.
Note in particular that, for m = 0, we are able to conclude that the lower bound, −v8 ·χ(GA ),
on volume given by Theorem 2.2.2 will be positive.
84
5.8
Applying the Main Theorem to Closed Braids
In this section, we apply the Main Theorem (Theorem 3.4.1) to obtain volume bounds for
the family of closed braids considered in Theorem 5.5.1.
Proposition 5.8.1. Applying Theorem 3.4.1 to the closed braids of Theorem 5.5.1, we get
that:
v8
· (t(D) − 1) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1)
3
in the case that n = 3 and get that:
v8
v
· (t(D) − 1) − 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1)
3
3
in the case that n ≥ 4.
Proof. Recall that special circles in the all-A state HA are other circles that are incident to
A-segments from one of three possible pairs of twist region resolutions. (See Definition 2.1.6
and Figure 2.6.) Given the assumptions of Theorem 5.5.1, the all-A state circles of HA
have been classified by Proposition 5.3.1 and Proposition 5.6.1. Furthermore, we claim that
medium circles, essential wandering circles, and non-essential wandering circles (which can
only exist if the number of braid strings is n ≥ 4) must all be incident to A-segments from
three or more distinct twist region resolutions. To see why this is true for medium inner
circles, see the center of Figure 5.7. To see why this is true for wandering circles, note that
wandering circles must wander at least twice (wandering and wandering back) to be able to
85
return to the same braid string position and close up. Therefore, the image on the right side
of Figure 5.7 occurs at least twice per wandering circle and the claim can now seen to be true
for wandering circles. Thus, we have that medium inner circles and wandering circles cannot
be special circles. It is possible, however, for a nonwandering circle to be a special circle.
Given the classification of nonwandering circles in Proposition 5.3.1 and Proposition 5.6.1,
we see that there are two main cases to consider.
Case 1: Suppose that σ1 (or, symmetrically, σn−1 ) occurs in the braid word with only
positive exponents. Then braid string 1 (or, symmetrically, braid string n) is a nonwandering
circle. If this is the case, to be a special circle, the generator σ1 (or, symmetrically, σn−1 )
must appear exactly twice (as a positive syllable) in the braid word.
Case 2: Suppose that an adjacent pair of generators σi and σi+1 for 2 ≤ i ≤ n − 3 occur
with only positive exponents. Then braid string i + 1 is a nonwandering circle. To be a
special circle, this braid string must be incident to A-segments from exactly two (positive)
twist regions of D(K) (which are positive syllables in β). Since β is nice, then each generator
must occur at least twice nontrivially in the braid word. Therefore, since we need to have
two positive syllables in σi and two positive syllables in σi+1 , then we get a total of at
least four positive twist region resolutions incident to braid string i + 1. This prevents the
nonwandering circle (braid string i + 1) from being a special circle.
To summarize, we have shown that at most two special circles (namely braid string 1 and
braid string n) are possible, so st(D) ≤ 2. In the case that n = 3, Lemma 5.7.1 gives that
there can only be at most one nonwandering circle and, therefore, at most one special circle.
Thus, st(D) ≤ 1 when n = 3. Notice that the conclusions of Theorem 5.5.1 allow us to
86
apply Theorem 3.4.1. Applying Theorem 3.4.1 in the case that n = 3, we get that:
v8
· (t(D) − 1) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
3
Applying Theorem 3.4.1 in the case that n ≥ 4, we get that:
v8
v
· (t(D) − 1) − 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
3
3
5.9
Volume Bounds in Terms of t(D) (and t−(D))
We will now use our study of A-adequate closed braids to obtain bounds on volume. Additionally, we will compare the lower bounds on volume we find to those that were found
in Proposition 5.8.1 by applying Theorem 1.0.1). In doing this, we show that studying the
specific structure of the closed braids considered in this paper often provides sharper lower
bounds on volume.
r
r1
l
Theorem 5.9.1. Let D(K) = β denote the closure of a nice n-braid β = σm
1 · · · σml , where
1 ≤ m1 , . . . , ml ≤ n − 1. If β satisfies the conditions:
(1) all negative exponents ri < 0 in β satisfy the stronger requirement that ri ≤ −3, and
(2) when ri > 0, we have that ri−1 < 0 and ri+1 < 0 and either mi−1 = mi+1 = mi + 1
or mi−1 = mi+1 = mi − 1,
87
then D(K) is a connected, prime, A-adequate link diagram that satisfies the TELC and
contains t(D) ≥ 2(n − 1) twist regions. Furthermore, we have that K is a hyperbolic link.
In the case that n = 3, we get the following volume bounds:
v8
v
· (t(D) − 1) − 8 ≤ v8 · (t− (D) − 1) ≤ vol(S 3 \K) < 10v3 · (t(D) − 1).
2
2
Suppose, in the case that n ≥ 4, that the all-A state HA contains m non-essential wandering
circles. Then we get the following volume bounds:
v
v8
·(t(D)−1)− 8 ·(2(n + m) − 5) ≤ v8 ·(t− (D)−(n+m−2)) ≤ vol(S 3 \K) < 10v3 ·(t(D)−1).
2
2
Proof. By combining the results of Theorem 5.5.1, Theorem 2.2.2, Theorem 5.7.1, and Theorem 5.7.2, we get the desired results.
Let us now compare the lower bounds of Proposition 5.8.1 to those of Theorem 5.9.1. We
would like to determine exactly when the lower bounds given by our study of A-adequate
closed braids are an improvement over those that result from applying Theorem 3.4.1 (using
Proposition 5.8.1).
In the case that n = 3, we have that:
88
v8
v
v
· (t(D) − 1) − 8 ≥ 8 · (t(D) − 1)
2
2
3
is equivalent to the condition that t(D) ≥ 4 = 2(3 − 1) = 2(n − 1). But this condition is
always satisfied because the assumption that β is nice allows us to use Proposition 5.4.2.
Therefore, the lower bound found in Theorem 5.9.1 is always the same or sharper than the
lower bound provided by applying Proposition 5.8.1.
In the case that n ≥ 4, we have that:
v
v
v
v8
· (t(D) − 1) − 8 · (2(n + m) − 5) ≥ 8 · (t(D) − 1) − 8
2
2
3
3
is equivalent to the condition that t(D) ≥ 6(n + m) − 16. Therefore, the lower bound found
in Theorem 5.9.1 is the same or sharper than the lower bound provided by Proposition 5.8.1
effectively when the number of twist regions is large compared to the sum, n + m, of the
number of braid strings and number of non-essential wandering circles.
Remark 5.9.1. It is also worth mentioning that, at the cost of keeping track of both twist
regions and negative twist regions, there is further potential to improve the lower bounds on
volume by using the lower bounds of Theorem 5.9.1 that are expressed in terms of t− (D).
This improvement will be especially noticeable when t− (D) − t+ (D) ≥ 0 is large.
89
5.10
Volume Bounds in Terms of the Colored Jones
Polynomial
To conclude our study of volume bounds for hyperbolic A-adequate closed braids, we will
translate our diagrammatic volume bounds (in terms of t(D)) to volume bounds in terms of
the stable penultimate coefficient βK of the colored Jones polynomial.
r
r1
l
Theorem 5.10.1. Let D(K) = β denote the closure of a nice n-braid β = σm
1 · · · σml ,
where 1 ≤ m1 , . . . , ml ≤ n − 1. If β satisfies the conditions:
(1) all negative exponents ri < 0 in β satisfy the stronger requirement that ri ≤ −3, and
(2) when ri > 0, we have that ri−1 < 0 and ri+1 < 0 and either mi−1 = mi+1 = mi + 1
or mi−1 = mi+1 = mi − 1,
then D(K) is a connected, prime, A-adequate link diagram that satisfies the TELC and
contains t(D) ≥ 2(n − 1) twist regions. Furthermore, we have that K is a hyperbolic link.
In the case that n = 3, we get the following volume bounds:
v8 · ( βK − 1) ≤ vol(S 3 \K) < 20v3 · ( βK − 1) + 10v3 .
Suppose, in the case that n ≥ 4, that the all-A state HA contains m non-essential wandering
circles. Then we get the following volume bounds:
v8 · ( βK − 1) ≤ vol(S 3 \K) < 20v3 · ( βK − 1) + 10v3 · (2(n + m) − 5).
90
Proof. The first conclusions of the theorem follow from Theorem 5.5.1. Combining Theorem 2.3.1 with Theorem 2.2.2, we get that (for all n ≥ 3):
v8 · ( βK − 1) = −v8 · χ(GA ) ≤ vol(S 3 \K).
Consider the case when n = 3. Combining Theorem 2.3.1 with Theorem 5.7.1 gives:
βK = 1 − χ(GA ) ≥ 1 +
1
t(D)
1
· (t(D) − 1) − =
,
2
2
2
which implies that:
t(D) − 1 ≤ 2 · ( βK − 1) + 1.
Applying this inequality to Theorem 2.2.2, we get:
vol(S 3 \K) < 10v3 · (t(D) − 1) ≤ 20v3 · ( βK − 1) + 10v3 .
This gives the first desired set of volume bounds. Now consider the case when n ≥ 4.
Combining Theorem 2.3.1 with Theorem 5.7.2 gives:
βK = 1 − χ(GA ) ≥ 1 +
1
1
t(D) − 2(n + m) + 6
· (t(D) − 1) − · (2(n + m) − 5) =
,
2
2
2
which implies that:
91
t(D) − 1 ≤ 2 · ( βK − 1) + 2(n + m) − 5.
Applying this inequality to Theorem 2.2.2, we get:
vol(S 3 \K) < 10v3 · (t(D) − 1) ≤ 20v3 · ( βK − 1) + 10v3 · (2(n + m) − 5).
This gives the second desired set of volume bounds.
92
Chapter 6
Volume Bounds for A-Adequate
Closed 3-Braids in Terms of the
Schreier Normal Form
6.1
The Schreier Normal Form for 3-Braids
A useful development in the history of 3-braids was the solution to the Conjugacy Problem
(and, as a corollary, the Word Problem) for the 3-braid group ([25]). As it turns out, there
is an explicit algorithm that produces from an arbitrary 3-braid word β a conjugate 3-braid
word β which is called the Schreier normal form of β. This new braid word β is the unique
representative of the conjugacy class of β. See below (or see Section 7.1 of [3]) for a modern
exposition of the algorithm.
Given such an algorithm, Birman and Menasco ([3]) gave a complete classification of links
that can be represented as 3-braid closures. To be more specific they showed that, up to an
explicit list of exceptions, each 3-braid closure comes from a single conjugacy class. Hence,
up to some exceptions, the normal form of the braid β determines the unique link type of
the closed braid β.
93
Because it will be needed in what follows, we now present the algorithm (adapted from
Section 7.1 of [3]) that, given a 3-braid β ∈ B3 , produces the Schreier normal form β of
β.
Remark 6.1.1. It is important to note that cyclic permutation (a special case of conjugacy)
may be needed during the steps of this algorithm.
Schreier Normal Form Algorithm:
(1) Let β ∈ B3 = σ1 , σ2 | σ1 σ2 σ1 = σ2 σ1 σ2 be cyclically reduced into syllables.
Introduce new variables x = (σ1 σ2 σ1 )−1 and y = σ1 σ2 . Thus we have that:
• σ1 = y 2 x
• σ2 = xy 2
• σ1−1 = xy
• σ2−1 = yx
Possibly using cyclic permutation, rewrite β as a cyclically reduced word that is positive
in x and y.
(2) Introduce C = x−2 = y 3 ∈ Z(B3 ), where Z(B3 ) denotes the center of the 3-braid
group B3 .
By using these relations and commutativity of C as much as possible, group all powers
of C at the beginning of the braid word and reduce the exponents of x and y as much
as possible.
94
Rewrite β as:
β = C k η,
where k ∈ Z and where η is a subword whose syllables in x have exponent at most one
and whose syllables in y have exponent at most two. To be more precise, by possibly
using cyclic permutation:
q
q
(xy)p1 xy 2 1 · · · (xy)ps xy 2 s for some s, pi , qi ≥ 1
(xy)p
for some p ≥ 1
q
for some q ≥ 1
xy 2
η=
y
2
y
x
1
(3) Possibly using cyclic permutation and the commutativity of C, rewrite β back in terms
of σ1 and σ2 as β = C k η , where:
η =
−p1 q1
−p q
σ2 · · · σ1 s σ2s
for some s, pi , qi ≥ 1
p
for some p ∈ Z
σ1
σ1
σ1 σ2
σ1 σ2 σ1
σ1 σ2 σ1 σ2
95
Definition 6.1.1. We call β ∈ B3 the Schreier normal form of β ∈ B3 . The braid word β
is the unique representative of the conjugacy class of β.
Definition 6.1.2. Following [12], we will call a braid β generic if it has Schreier normal
form
−p1 q1
−p q
σ2 · · · σ1 s σ2s .
β = C k σ1
6.2
Hyperbolicity for Closed 3-Braids and Volume
Bounds
Using the Schreier normal form of a 3-braid, Futer, Kalfagianni, and Purcell ([12]) classified
the hyperbolic 3-braid closures. Furthermore, given such a hyperbolic closed 3-braid, they
gave two-sided bounds on the volume of the link complement, expressing the volume in terms
of the parameter s from the Schreier normal form of the 3-braid. Because we will compare
the lower bounds on volume of this paper to those of [12], we give precise statements of the
relevant results from [12] below.
Proposition 6.2.1 ([12], Theorem 5.5). Let D(K) = β denote the closure of a 3-braid
β ∈ B3 . Then K is hyperbolic if and only if:
(1) β is generic, and
p q
(2) β is not conjugate to σ1 σ2 for any integers p and q.
Proposition 6.2.2 ([12], Theorem 5.6). Let D(K) = β denote the closure of a 3-braid
β ∈ B3 and let β denote the Schreier normal form of β. Then, assuming that K is hyperbolic,
we have that:
96
4v3 · s − 276.6 < vol(S 3 − K) < 4v8 · s.
By using the more recent machinery built by the same authors in [11], we will often obtain
a sharper lower bound on volume.
6.3
Volume Bounds in Terms of the Schreier Normal
Form
In this section, we study the Schreier normal form of the 3-braids β given in Theorem 5.5.1.
Theorem 6.3.1. Let D(K) = β denote the closure of a nice 3-braid β, which we denote by
r
r
β = σir1 σjr2 · · · σi 2l−1 σj 2l , where {i, j} = {1, 2}. Assume that β satisfies the conditions:
(1) all negative exponents ri < 0 in β satisfy the stronger requirement that ri ≤ −3.
(2) positive syllables are cyclically isolated.
−p1 q1
−p q
σ2 · · · σ1 s σ2s ,
Then β is generic with Schreier normal form β = C k σ1
where k ∈ Z and
s, pi , qi ≥ 1. Furthermore, we are able to express the parameters k and s of the Schreier
normal form β in terms of the original braid β as follows:
(a) k = −# induced products σ2n2 σ1n1 of negative syllables of β, where n1 , n2 ≤ −3 .
(b) s = t− (D) = # {negative syllables in β}.
Note that, when looking for the induced products σ2n2 σ1n1 of negative syllables of β to find
k, one must look cyclically in the braid word. As an example computation, consider the
97
3-braid:
β = σ13 σ2−3 σ1−5 σ2−3 .
Applying the Schreier Normal Form Algorithm, we get:
β = σ13 σ2−3 σ1−5 σ2−3
= (y 2 x)3 (yx)3 (xy)5 (yx)3
= y 2 (xy 2 )2 (xy)3 (x2 )y(xy)3 (xy 2 )(xy)2 x
= y 2 (xy 2 )2 (xy)3 (C −1 )y(xy)3 (xy 2 )(xy)2 x
∼
= C −1 y 2 (xy 2 )2 (xy)2 (xy 2 )(xy)3 (xy 2 )(xy)2 x
∼
= C −1 (xy 2 )3 (xy)2 (xy 2 )(xy)3 (xy 2 )(xy)2
∼
= C −1 (xy)2 (xy 2 )3 (xy)2 (xy 2 )(xy)3 (xy 2 )
= C −1 σ1−2 σ23 σ1−2 σ2 σ1−3 σ2
= β,
where ∼
= denotes that either cyclic permutation or the fact that C ∈ Z(B3 ) has been used.
Thus, we see that:
k = −1 = −# induced products σ2n2 σ1n1 of negative syllables of β, where n1 , n2 ≤ −3
98
and:
s = 3 = # {negative syllables in β} .
p
p
n
Remark 6.3.1. As a special case of the above theorem, notice that if β = σi 1 σjn1 · · · σi l σj l
is an alternating 3-braid where {i, j} = {1, 2}, then k = 0 and s = l.
Deferring the proof of the above theorem to the next section, we now relate s to the colored Jones polynomial of a hyperbolic A-adequate closed 3-braid from Theorem 5.5.1. In
particular, we relate s to the stable penultimate coefficient βK of the colored Jones polynomial.
Corollary 6.3.1. Given the assumptions of Theorem 6.3.1, we have that:
s = t− (D) = |βK |.
Thus, t− (D) and s are actually link invariants.
Proof. To begin, note that the assumptions of Theorem 6.3.1 are the same as those of Theorem 5.5.1 because Condition (2) of Theorem 5.5.1 can be expressed more simply for 3-braids.
Theorem 5.7.1 also relies on the assumptions of Theorem 5.5.1. Using the conclusions of Theorem 5.5.1, we can apply Theorem 2.3.1, Theorem 5.7.1, and Theorem 6.3.1 respectively to
get that:
99
|βK | = 1 − χ(GA )
= 1 + (t− (D) − 1)
= t− (D)
= s.
Since the colored Jones polynomial and therefore its coefficients are link invariants, then we
can conclude that t− (D) and s are link invariants.
Theorem 6.3.2. Let D(K) = β denote the closure of a nice 3-braid β, which we denote by
r
r
β = σir1 σjr2 · · · σi 2l−1 σj 2l , where {i, j} = {1, 2}. Assume that β satisfies the conditions:
(1) all negative exponents ri < 0 in β satisfy the stronger requirement that ri ≤ −3.
(2) positive syllables are cyclically isolated.
Then D(K) is a connected, prime, A-adequate link diagram that satisfies the TELC and
contains t(D) ≥ 2 twist regions. Furthermore, we have that K is a hyperbolic link and β is
generic. Moreover, we get the following volume bounds:
v8 · (s − 1) ≤ vol(S 3 \K) < 4v8 · s.
Proof. By combining Theorem 5.10.1, Proposition 6.2.1, Corollary 6.3.1, and Proposition 6.2.2,
100
we get the desired result.
Remark 6.3.2. Comparing the lower bound on volume of Theorem 6.3.2 to that of Proposition 6.2.2, we get that v8 · (s − 1) ≥ 4v3 · s − 276.6 is equivalent to the condition that
276.6 − v8
s≤
< 689. Therefore, the lower bound found in Theorem 6.3.2 is sharper than
4v3 − v8
the lower bound provided by Proposition 6.2.2 unless the parameter s from the Schreier
normal form is very large.
6.4
The Proof of Theorem 6.3.1.
Proof. By Theorem 5.5.1, we have that K is hyperbolic. By Proposition 6.2.1, this implies
that β is generic. Consider the following two cases.
Case 1: Suppose β is a negative braid. Then, possibly using cyclic permutation, we may
n
write β as β = σ1n1 σ2n2 · · · σ1 2m−1 σ2n2m , where ni ≤ −3 and the fact that β is nice forces the
condition that m ≥ 2. Applying the Schreier Normal Form Algorithm, we get:
n
β = σ1n1 σ2n2 · · · σ1 2m−1 σ2n2m
= (xy)−n1 (yx)−n2 · · · (xy)−n2m−1 (yx)−n2m
= (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (x2 )y · · · (x2 )y(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 x
= (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (C −1 )y · · · (C −1 )y(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 x
∼
= (C −1 )m−1 (xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
101
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 x
∼
= (C −1 )m−1 x(xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1
= (C −1 )m−1 (x2 )y(xy)−n1 −2 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1
= (C −1 )m y(xy)−n1 −2 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1
∼
= (C −1 )m (xy)−n1 −2 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 y
= (C −1 )m (xy)−n1 −2 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −2 (xy 2 )
= (C −1 )m σ1n1 +2 σ2 σ1n2 +2 σ2 · · ·
n
· · · σ2 σ1 2m−1
+2
σ2 σ1n2m +2 σ2
= β,
where ∼
= denotes that cyclic permutation or the fact that C ∈ Z(B3 ) has been used. Thus,
we see that:
k = −m = −# induced products σ2n2 σ1n1 of negative syllables of β, where n1 , n2 ≤ −3
and:
102
s = 2m = # {negative syllables in β} .
Recall that, when looking for the induced products σ2n2 σ1n1 of negative syllables of β to find
k, one must look cyclically in the braid word.
Case 2: Suppose β is not a negative braid. Then, possibly using cyclic permutation, we may
write β as β = P1 N1 · · · Pt Nt , where Pi denotes a positive syllable of β and where Ni denotes
a maximal length negative induced subword of β. Note that this decomposition arises as a
result of the fact that positive syllables are assumed to be cyclically isolated.
Our strategy for the proof of this case is to apply the Schreier Normal Form Algorithm to
the subwords Pi Ni , show that the theorem is locally satisfied for the Pi Ni , show that cyclic
permutation is never needed in applying the algorithm to the Pi Ni , and to finally show that
juxtaposing the subwords Pi Ni to form β allows the local conclusions of the theorem for the
Pi Ni to combine to give the global conclusion of the theorem for β.
To begin, we list the possible subwords Pi Ni . For each subword, we consider the two possible
subtypes. Let p > 0 be a positive exponent and let the ni ≤ −3 be negative exponents.
List of Types of Induced Subwords Pi Ni of β:
p
(1a) σ2 σ1n1
p
(1b) σ1 σ2n1
p
(2a) σ2 σ1n1 σ2n2
p
(2b) σ1 σ2n1 σ1n2
103
p
(3a) σ2 σ1n1 σ2n2 σ1n3
p
(3b) σ1 σ2n1 σ1n2 σ2n3
p
n
p
n
p
n
n
p
n
n
(4a) σ2 σ1n1 σ2n2 · · · σ1 2m−1 σ2n2m , where m ≥ 2
(4b) σ1 σ2n1 σ1n2 · · · σ2 2m−1 σ1n2m , where m ≥ 2
(5a) σ2 σ1n1 σ2n2 · · · σ1 2m−1 σ2n2m σ1 2m+1 , where m ≥ 2
(5b) σ1 σ2n1 σ1n2 · · · σ2 2m−1 σ1n2m σ2 2m+1 , where m ≥ 2
We now apply the Schreier Normal Form Algorithm to each (sub)type of induced subword
above, implicitly showing along the way that cyclic permutation is never used during the
application of the algorithm. Let ki denote the exponent of C in the normal form for the
subword Pi Ni . We also explicitly show that:
n
n
ki = −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni (where nj , nj+1 ≤ −3) .
The Algorithm for Type (1a):
p
Pi Ni = σ2 σ1n1
= (xy 2 )p (xy)−n1
p
= σ2 σ1n1
Thus we get that:
ki = 0
104
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (1b):
p
Pi Ni = σ1 σ2n1
= (y 2 x)p (yx)−n1
= y 2 (xy 2 )p−1 (xy)−n1 x
p−1 n1
σ1 x
= y 2 σ2
Thus we get that:
ki = 0
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (2a):
p
Pi Ni = σ2 σ1n1 σ2n2
= (xy 2 )p (xy)−n1 (yx)−n2
= (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −1 x
p
= σ2 σ1n1 +1 σ2 σ1n2 +1 x
105
Thus we get that:
ki = 0
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (2b):
p
Pi Ni = σ1 σ2n1 σ1n2
= (y 2 x)p (yx)−n1 (xy)−n2
= y 2 (xy 2 )p−1 (xy)−n1 (x2 )y(xy)−n2 −1
= y 2 (xy 2 )p−1 (xy)−n1 (C −1 )y(xy)−n2 −1
∼
= C −1 y 2 (xy 2 )p−1 (xy)−n1 −1 (xy 2 )(xy)−n2 −1
p−1 n1 +1
σ1
σ2 σ1n2 +1
= C −1 y 2 σ2
Thus we get that:
ki = −1
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (3a):
p
Pi Ni = σ2 σ1n1 σ2n2 σ1n3
= (xy 2 )p (xy)−n1 (yx)−n2 (xy)−n3
106
= (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (x2 )y(xy)−n3 −1
= (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (C −1 )y(xy)−n3 −1
∼
= C −1 (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 )(xy)−n3 −1
p
= C −1 σ2 σ1n1 +1 σ2 σ1n2 +2 σ2 σ1n3 +1
Thus we get that:
ki = −1
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (3b):
p
Pi Ni = σ1 σ2n1 σ1n2 σ2n3
= (y 2 x)p (yx)−n1 (xy)−n2 (yx)−n3
= y 2 (xy 2 )p−1 (xy)−n1 (x2 )y(xy)−n2 −2 (xy 2 )(xy)−n3 −1 x
= y 2 (xy 2 )p−1 (xy)−n1 (C −1 )y(xy)−n2 −2 (xy 2 )(xy)−n3 −1 x
∼
= C −1 y 2 (xy 2 )p−1 (xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 )(xy)−n3 −1 x
p−1 n1 +1
σ1
σ2 σ1n2 +2 σ2 σ1n3 +1 x
= C −1 y 2 σ2
Thus we get that:
ki = −1
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
107
The Algorithm for Type (4a):
n
p
Pi Ni = σ2 σ1n1 σ2n2 · · · σ1 2m−1 σ2n2m
= (xy 2 )p (xy)−n1 (yx)−n2 · · · (xy)−n2m−1 (yx)−n2m
= (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (x2 )y · · ·
· · · (x2 )y(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 x
= (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (C −1 )y · · ·
· · · (C −1 )y(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 x
∼
= (C −1 )m−1 (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 x
n
p
= (C −1 )m−1 σ2 σ1n1 +1 σ2 σ1n2 +2 σ2 · · · σ2 σ1 2m−1
+2
σ2 σ1n2m +1 x
Thus we get that:
ki = −(m − 1)
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (4b):
p
n
Pi Ni = σ1 σ2n1 σ1n2 · · · σ2 2m−1 σ1n2m
= (y 2 x)p (yx)−n1 (xy)−n2 · · · (yx)−n2m−1 (xy)−n2m
= y 2 (xy 2 )p−1 (xy)−n1 (x2 )y(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −1 (x2 )y(xy)−n2m −1
108
= y 2 (xy 2 )p−1 (xy)−n1 (C −1 )y(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −1 (C −1 )y(xy)−n2m −1
∼
= (C −1 )m y 2 (xy 2 )p−1 (xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1
n
+2
p−1 n1 +1
σ1
σ2 σ1n2 +2 σ2 · · · σ2 σ1 2m−1 σ2 σ1n2m +1
= (C −1 )m y 2 σ2
Thus we get that:
ki = −m
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (5a):
n
p
n
Pi Ni = σ2 σ1n1 σ2n2 · · · σ1 2m−1 σ2n2m σ1 2m+1
= (xy 2 )p (xy)−n1 (yx)−n2 · · · (xy)−n2m−1 (yx)−n2m (xy)−n2m+1
= (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (x2 )y · · ·
· · · (x2 )y(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 (x2 )y(xy)−n2m+1 −1
= (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −1 (C −1 )y · · ·
· · · (C −1 )y(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −1 (C −1 )y(xy)−n2m+1 −1
∼
= (C −1 )m (xy 2 )p (xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −2 (xy 2 )(xy)−n2m+1 −1
n
p
= (C −1 )m σ2 σ1n1 +1 σ2 σ1n2 +2 σ2 · · · σ2 σ1 2m−1
109
+2
n
σ2 σ1n2m +2 σ2 σ1 2m+1
+1
Thus we get that:
ki = −m
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
The Algorithm for Type (5b):
n
p
n
Pi Ni = σ1 σ2n1 σ1n2 · · · σ2 2m−1 σ1n2m σ2 2m+1
= (y 2 x)p (yx)−n1 (xy)−n2 · · · (yx)−n2m−1 (xy)−n2m (yx)−n2m+1
= y 2 (xy 2 )p−1 (xy)−n1 (x2 )y(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −1 (x2 )y(xy)−n2m −2 (xy 2 )(xy)−n2m+1 −1 x
= y 2 (xy 2 )p−1 (xy)−n1 (C −1 )y(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −1 (C −1 )y(xy)−n2m −2 (xy 2 )(xy)−n2m+1 −1 x
∼
= (C −1 )m y 2 (xy 2 )p−1 (xy)−n1 −1 (xy 2 )(xy)−n2 −2 (xy 2 ) · · ·
· · · (xy 2 )(xy)−n2m−1 −2 (xy 2 )(xy)−n2m −2 (xy 2 )(xy)−n2m+1 −1 x
n
+2
n
+1
p−1 n1 +1
σ1
σ2 σ1n2 +2 σ2 · · · σ2 σ1 2m−1 σ2 σ1n2m +2 σ2 σ1 2m+1 x
= (C −1 )m y 2 σ2
Thus we get that:
ki = −m
n
n
= −# induced products σ2 j σ1 j+1 of negative syllables of Pi Ni .
It is very important to notice that, in all the the cases above, the application of the Schreier
110
Normal Form Algorithm to each of the induced subword types Pi Ni does not ever require
cyclic permutation. This is very important because the goal is to juxtapose the normal forms
of the induced subwords Pi Ni and claim that this gives, after some minor modifications, the
normal form of the full braid word β. To summarize, below is the list of normal forms of the
induced subwords Pi Ni of β (as computed above).
List of Normal Forms of Induced Subwords Pi Ni of β:
p
(1a) σ2 σ1n1
p−1 n1
σ1 x
(1b) y2 σ2
p
(2a) σ2 σ1n1 +1 σ2 σ1n2 +1 x
p−1 n1 +1
σ1
σ2 σ1n2 +1
(2b) C −1 y2 σ2
p
(3a) C −1 σ2 σ1n1 +1 σ2 σ1n2 +2 σ2 σ1n3 +1
p−1 n1 +1
σ1
σ2 σ1n2 +2 σ2 σ1−n3 −1 x
(3b) C −1 y2 σ2
n
p
(4a) (C −1 )m−1 σ2 σ1n1 +1 σ2 σ1n2 +2 σ2 · · · σ2 σ1 2m−1
+2
σ2 σ1n2m +1 x, where m ≥ 2
n
+2
p−1 n1 +1
σ1
σ2 σ1n2 +2 σ2 · · · σ2 σ1 2m−1 σ2 σ1n2m +1 ,
(4b) (C −1 )m y2 σ2
where m ≥ 2
p
(5a) (C −1 )m σ2 σ1n1 +1 σ2 σ1n2 +2 σ2 · · ·
n
· · · σ2 σ1 2m−1
+2
n
σ2 σ1n2m +2 σ2 σ1 2m+1
+1
, where m ≥ 2
p−1 n1 +1
σ1
σ2 σ1n2 +2 σ2 · · ·
(5b) (C −1 )m y2 σ2
n
· · · σ2 σ1 2m−1
+2
n
σ2 σ1n2m +2 σ2 σ1 2m+1
+1
x, where m ≥ 2
Let us now consider which types of induced subwords Pi+1 Ni+1 can (cyclically) follow a
given induced subword Pi Ni . Since β is assumed to be cyclically reduced into syllables and
111
since the the induced subwords Pi Ni contain unbroken syllables of β, then we get that:
• Induced subwords of Types (1), (3), and (5) can be followed by any of the five subwords,
so long as they are of the same subtype.
• Induced subwords of Types (2) and (4) can be followed by any of the five subwords, so
long as they are of different subtype.
See the “List of Types of Induced Subwords Pi Ni of β” to verify these claims. As an example,
we have that an induced subword of Type (1a) can be followed by an induced subword of
Type (2a) but can not be followed by a subword of Type (2b). On the other hand, an
induced subword of Type (2a) can be followed by an induced subword of Type (5b) but can
not be followed by a subword of Type (5a).
Recall that C ∈ Z(B3 ) commutes with the generators of B3 . Thus, when juxtaposing the
normal form of Pi Ni with the normal form of Pi+1 Ni+1 , we may move the factor C ki+1 out
of the way, moving it from between the normal forms to the beginning of the normal form
of Pi Ni . This fact will be utilized below.
Looking now at the “List of Normal Forms of Induced Subwords Pi Ni of β”, notice that
half of the normal forms may potentially contain the variable x at the end of the word and
p−1
half of the normal forms may contain the expression y2 σ2
at the beginning of the word
(immediately after the C ki term that will be moved out of the way). We can now see that
juxtaposing the normal form of Pi Ni with that of Pi+1 Ni+1 either:
(1) involves neither x at the end of the normal form of Pi Ni nor y2 at the beginning of
the normal form of Pi+1 Ni+1 , or
112
(2) involves (after moving C ki+1 out of the way) both an x at the end of the normal form
p−1
of Pi Ni and a y2 σ2 at the beginning of the normal form of Pi+1 Ni+1 .
Consequently, upon juxtaposing the normal forms of all of the induced subwords together,
p−1
we see that all factors x and y2 σ2 in the list of normal forms combine to form
p−1
xy2 σ2
p−1
= σ2 σ2
p
= σ2 .
Note, in particular, that juxtaposing the normal form of Pi Ni with that of Pi+1 Ni+1 does
not create any new nontrivial powers of C. Therefore, since C commutes with the generators
of B3 , then we may group all C ki terms together at the beginning of the normal form braid
word. We also have that juxtaposing the normal form of Pi Ni with that of Pi+1 Ni+1 does
n
n
not create any new induced products σ2 j σ1 j+1 of negative syllables. Thus, we can add the
n
n
local number −ki of induced products σ2 j,i σ1 j+1,i of negative syllables in the Pi Ni together
to get the global number of such subwords. With this information, we are now able to
conclude that:
t
k =
ki
i=1
t
n
n
−# induced products σ2 j,i σ1 j+1,i of negative syllables of Pi Ni , where nj , nj+1 ≤ −3
=
i=1
= −# induced products σ2n2 σ1n1 of negative syllables of β, where n1 , n2 ≤ −3 .
This gives the first desired result, that the parameter k from the Schreier normal form β can
113
be seen in the original braid β. To conclude the proof, we need to relate the global parameter
s from the Schreier normal form to local versions of the parameter s. Let si denote the local
version of the parameter s, which comes from the normal form for the subword Pi Ni and
will be more precisely defined below.
Look again at the “List of Normal Forms of Induced Subwords Pi Ni of β”. Recall that
juxtaposing the normal forms to create a braid word groups together the x and y2 factors
in the normal forms in such a way that they are absorbed into a trivial or positive syllable
p−1
σ2
p
to form σ2 . Also recall that we will collect together all individual powers of C from
each Pi Ni normal form and use commutativity to form a single power of C at the beginning
of the normal form braid word. Thus, after juxtaposition of the normal forms of the Pi Ni ,
what results is a braid word that looks like C k W1 · · · Wt , where k is an integer and Wi is an
alternating word that is positive in σ2 , negative in σ1 , begins with a σ2 syllable, and ends
with a σ1 syllable. By cyclic permutation and the commutativity of C, we get that this
braid word is equivalent to the generic normal form of β.
p
n
p
n
Given an alternating subword Wi = σ2 1,i σ1 1,i · · · σ2 q,i σ1 q,i as described above, we define
si = q. To provide an example, for the normal form:
p
C −1 σ2 σ1n1 +1 σ2 σ1n2 +2 σ2 σ1n3 +2 σ2 σ1n4 +1 x
of Type (4a), we have that si = 4. Consider the product:
p
n
p
n
p
n
p
n
Wi Wi+1 = (σ2 1,i [σ1 1,i · · · σ2 q,i σ1 q,i ) · (σ2 1,i+1 ]σ1 1,i+1 · · · σ2 q,i+1 σ1 q,i+1 ).
114
n
p
n
p
Note that the subword [σ1 1,i · · · σ2 q,i σ1 q,i · σ2 1,i+1 ] looks like the alternating part of a generic
braid word. From this perspective of (cyclically) borrowing the first syllable of the next
subword, the local parameter si makes sense as being the local version of the global parameter
s.
Returning to the above application of the Schreier Normal Form Algorithm to each subtype
of induced subword Pi Ni , it can be seen that the number of negative syllables in Pi Ni is
equal to si for the normal form of Pi Ni . Therefore, combining this fact with what we have
learned above about how the juxtaposition works, we are now able to conclude that:
t− (D) = # {negative syllables in β}
t
# {negative syllables in Pi Ni }
=
i=1
t
si
=
i=1
= s.
115
Chapter 7
Volume Bounds for Hybrid
Braid-Plats
After studying both closures of braids and plat closures of braids, we would now like to join
these families together by considering links that have both closed braid and plat closure
aspects. Such links were studied by Birman and Kanenobu in [2]. Specifically, we will use
our volume bounds for plat closures and volume bounds for negative closed braids (studied
in Section 9.1 of [11]) to obtain volume bounds for this new family of links, which we will
call hybrid braid-plats.
7.1
Background on Hybrid Braid-Plats
rl
r1
Definition 7.1.1. Let β = σm
1 · · · σml ∈ Bn , where 1 ≤ m1 , . . . , ml ≤ n − 1, denote an
n-braid diagram. From this braid diagram, we form a hybrid braid-plat diagram D(K) by
closing up the first 2p strands of the braid in the sense of plat closure and closing up the
remaining n − 2p strands in the sense of braid closure.
For a schematic depiction of a hybrid braid-plat, see Figure 7.1. For an example of the all-A
state HA of a hybrid braid-plat D(K), see Figure 7.2. Also, it may be useful to return to
116
n-braid
1
2
p
Figure 7.1: A schematic depiction of a hybrid braid-plat diagram D(K).
Section 4.1 to recall both notation and the types of braids used to form the plat closures of
this work.
Assumption: To strategically separate the plat information from the braid information in a
hybrid braid-plat, we will need to place some restrictions on the occurrence of the generator
σ2p in the braid word β ∈ Bn . Specifically we will assume that:
(1) σ2p can only occur in the braid word with positive exponents, and that
(2) a syllable in the letter σ2p can only occur at most once after each braid subword
j
rj rj
r
j
2p−1
βo = σ11 σ33 · · · σ2p−1
of the plat portion of the diagram.
Definition 7.1.2. We will call the column of positive crossings corresponding to the occurrences of σ2p in β the transitional crossing column, and shall denote the number of (positive)
twist regions in the transitional crossing column by t(T ).
Definition 7.1.3. Call a hybrid braid-plat separable if it satisfies Condition (1) and Condition (2) above.
117
all-A state
of a negative
(n-2p)-braid
2p-plat
Figure 7.2: An example of the all-A state of a hybrid braid-plat.
With this new notation and terminology, we may now view the hybrid braid-plat diagram
D(K) as decomposed into a plat part, a transitional crossing column, and a braid part. The
reason for the restriction placed on σ2p is to ensure that the reduced all-A graph GA will
contain a plat part reduced all-A graph and a braid part reduced all-A graph that do not
share vertices and are only connected to each other via edges coming from the transitional
crossing column. Consequently, we will be able to use prior work on volume bounds for
closed braids and plat closures to obtain volume bounds for hybrid braid-plats.
Notation: Denote the subdiagram coming from the braid part of D(K) by DB and denote
the subdiagram coming from the plat part of D(K) by DP .
Notation: Denote the reduced all-A graph of the hybrid braid-plat diagram D(K) by
GA (H), the reduced all-A graph of the braid part of D(K) by GA (B), and the reduced all-A
graph of the plat part of D(K) by GA (P ).
118
7.2
The Foundational Theorem for Hybrid Braid-Plats
In what follows, we will use the assumption (from Section 9.1 of [11]) that each (negative)
twist region of the braid part contains at least three crossings. Letting rj denote the exponent
of an arbitrary generator σj in the braid part of D(K), we may rewrite this assumption as
rj ≤ −3.
Theorem 7.2.1. Let D(K) be a separable hybrid braid-plat diagram coming from a braid
β ∈ Bn with plat closure on the first 2p strands and braid closure on the last n − 2p strands.
Furthermore, assume that D(K) satisfies the following conditions:
(1a) For a negative plat part, we have that ai,j ≤ −3 in odd-numbered rows and ai,j ≤ −2
in even-numbered rows.
(1b) For a mixed-sign plat part, we have that ai,j ≤ −3 or ai,j ≥ 1 in odd-numbered rows
and ai,j ≤ −2 in even-numbered rows.
(2) For all plat parts, we have that k ≥ 1, that p ≥ 3, and that t(DP ) ≥ 2.
(3) For all (negative) braid parts, we have that the restriction of β to σ2p+1 , . . . , σn is a
nice braid, that rj ≤ −3 for all 2p + 1 ≤ j ≤ n, and that t(DB ) ≥ 2.
(4) For the transitional crossing column, we have that t(T ) ≥ 2.
Then D(K) is a connected, prime, A-adequate link diagram that satisfies the TELC and
contains t(D) ≥ 6 twist regions. Furthermore, we have that K is hyperbolic.
Proof. Since the plat part satisfies ai,j = 0 for all i and j and since the braid part satisfies
|rj | ≥ 1 for all 2p + 1 ≤ j ≤ n, then we have that D(K) is a connected link diagram.
119
Since:
(1) ai,j = 0 for all i and j,
(2) k ≥ 1,
(3) p ≥ 3,
(4) t(T ) ≥ 2, and
(5) the restriction of β to σ2p+1 , . . . , σn is a nice braid,
then we have that D(K) is a prime link diagram. The first three assumptions guarantee (as
shown in the proof of Theorem 4.2.1) that the plat part of the diagram will be prime. The
assumption that t(T ) ≥ 2 prevents nugatory crossings from occurring between the plat part
of D(K) and the braid part of D(K). The last assumption guarantees, by Proposition 5.4.1,
that the braid part of the diagram is prime. Finally, by looking globally at the link diagram,
it can be seen by inspection that the entire diagram D(K) is prime.
Since:
(1) ai,j ≤ −2 or ai,j ≥ 1 in odd-numbered rows,
(2) the twist regions in the transitional crossing column are positive, and
(3) rj ≤ −2 for all 2p + 1 ≤ j ≤ n,
then we have that D(K) is an A-adequate link diagram. The first assumption guarantees (as
shown in the proof of Theorem 4.2.1 for negative plats and the proof of Theorem 4.4.1 for
mixed-sign plats) that the plat part of the diagram is A-adequate. The second assumption
guarantees that the corresponding horizontal A-segments of HA connect pairs of distinct
120
all-A circles. Finally, the last assumption and Lemma 9.8 of [11] guarantee that the braid
part of the diagram is A-adequate. Thus, since we have inspected the A-segments of the
braid part, plat part, and transitional crossing column, then we may globally conclude that
the entire hybrid braid-plat diagram D(K) is A-adequate.
Since:
(1) ai,j ≤ −3 or ai,j ≥ 1 in odd-numbered rows,
(2) ai,j ≤ −2 in even-numbered rows,
(3) rj ≤ −3 for all 2p + 1 ≤ j ≤ n, and
(4) D(K) is a separable hybrid braid-plat,
then we have that D(K) satisfies the TELC. As seen in the proofs of Theorem 4.2.1 and
Theorem 4.4.1, the first two assumptions guarantee that the TELC is satisfied for the (negative or mixed-sign) plat part of the diagram. As seen in the proof of Lemma 9.9 of [11], the
third assumption guarantees that the braid part of the diagram trivially satisfies the TELC.
By the last assumption, each twist region of the transitional crossing column is a (positive)
short twist region and, furthermore, each other circle of the all-A state of the plat part of
D(K) is incident to the horizontal A-segments from at most one short twist region resolution
of the transitional crossing column. (See Figure 7.2 for an example.) Consequently, we now
have that D(K) globally satisfies the TELC.
Next, notice that the structure of our hybrid braid-plats gives:
t(D) = t(DB ) + t(DP ) + t(T )
121
(7.1)
Since t(DB ) ≥ 2, t(DP ) ≥ 2, and t(T ) ≥ 2, then t(D) ≥ 6. Hence, by what was shown
above and by Proposition 2.2.1, we have that K is hyperbolic.
7.3
Computation of −χ(GA(H))
Lemma 7.3.1. Given a separable hybrid braid-plat diagram D(K), we have that:
−χ(GA (H)) = −χ(GA (B)) − χ(GA (P )) + t(T ).
Proof. By definition of −χ(GA (H)) and by Remark 2.1.2, we have that:
−χ(GA (H)) = e(GA (H)) − v(GA (H))
= e(GA (H)) − # {all − A state circles} .
Since D(K) is separable, then each (positive) short twist region of the transitional crossing
column corresponds to a single edge of GA (H) and, furthermore, the edges of GA (H) coming
from the transitional crossing column are the only edges that join GA (B) to GA (P ). Let
e(T ) = t(T ) denote the number of edges of GA (H) coming from the transitional crossing
column. Putting everything together we get:
122
−χ(GA (H)) = e(GA (H)) − # {all − A state circles}
=
e(GA (B)) + e(GA (P )) + e(T ) − v(GA (B)) + v(GA (P ))
= −χ(GA (B)) − χ(GA (P )) + t(T ).
We now present a lemma from [11] that will be needed in this work:
Lemma 7.3.2 ([11], Lemma 9.1). In the notation of this work, we have that:
−χ(GA (B)) ≥
7.4
2t(DB )
.
3
Volume Bounds in Terms of t(D) and in Terms of
the Colored Jones Polynomial
We now find volume bounds for the separable hybrid braid-plats considered in this work.
Specifically, we first express these bounds in terms of the twist number t(D) and, second,
express these bounds in terms of the stable penultimate coefficient βK of the colored Jones
polynomial.
Theorem 7.4.1. For D(K) a separable hybrid braid-plat diagram that satisfies the assumptions of Theorem 7.2.1 and that contains a negative plat part, we have that the complement
of K satisfies the following volume bounds:
123
2v8
· (t(D) − 1) + v8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
Similarly, for D(K) a separable hybrid braid-plat diagram that satisfies the assumptions of
Theorem 7.2.1, that contains a mixed-sign plat part, and that has mixed-sign plat part with
at least as many negative twist regions as positive twist regions, we have that the complement
of K satisfies the following volume bounds:
4v
v8
· (t(D) − 1) + 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
3
Theorem 7.4.2. For D(K) a separable hybrid braid-plat diagram that satisfies the assumptions of Theorem 7.2.1 and that contains a negative plat part, we have that the complement
of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) < 15v3 · ( βK − 1) − 15v3 .
Similarly, for D(K) a separable hybrid braid-plat diagram that satisfies the assumptions of
Theorem 7.2.1, that contains a mixed-sign plat part, and that has mixed-sign plat part with
at least as many negative twist regions as positive twist regions, we have that the complement
of K satisfies the following volume bounds:
v8 ·
βK − 1 ≤ vol(S 3 \K) < 30v3 · ( βK − 1) − 40v3 .
124
Combined Proof of Both Theorems. We will separate into negative plat part and mixed-sign
plat part cases.
Case 1: Suppose that the plat part of D(K) comes from a negative plat diagram. Recall
that we have previously shown in Inequality 4.1 of the proof of Theorem 4.2.1 (in slightly
different notation) that:
−χ(GA (P )) ≥
4
1
· (t(DP ) − 1) + .
5
5
Then, by respectively applying Lemma 7.3.1, Lemma 7.3.2, the above inequality, Equation 7.1, and the assumptions that t(DP ) ≥ 2 and t(T ) ≥ 2, we get:
−χ(GA (H)) = − χ(GA (B)) − χ(GA (P )) + t(T )
≥
=
=
=
≥
=
2t(DB ) 4
1
+ · (t(DP ) − 1) + + t(T )
3
5
5
10t(DB ) + 12t(DP ) − 9 + 15t(T )
15
10 [t(DB ) + t(DP ) + t(T )] + 2t(DP ) + 5t(T ) − 9
15
10t(D) + 2t(DP ) + 5t(T ) − 9
15
2t(D) + 1
3
2
· (t(D) − 1) + 1.
3
(7.2)
Since D(K) satisfies the assumptions of Theorem 7.2.1, then we may apply Theorem 2.2.2
to get:
125
2v8
· (t(D) − 1) + v8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
Transitioning to the colored Jones polynomial setting, note that Theorem 3.5.1 gives the
desired lower bound on volume. Combining Theorem 2.3.1 with Inequality 7.2 gives:
βK − 1 = −χ(GA (H)) ≥
2
· (t(D) − 1) + 1.
3
Hence:
t(D) − 1 ≤
3
3
· ( βK − 1) − ,
2
2
which implies that:
10v3 · (t(D) − 1) ≤ 15v3 · ( βK − 1) − 15v3 ,
which gives the desired upper bound on volume.
Case 2: Suppose that the plat part of D(K) comes from a mixed-sign plat diagram with at
least as many negative twist regions as positive twist regions. Recall that we have previously
shown in Inequality 4.3 of the proof of Theorem 4.4.1 (in slightly different notation) that:
−χ(GA (P )) ≥
1
2
· (t(D) − 1) − .
3
3
126
Then, by respectively applying Lemma 7.3.1, Lemma 7.3.2, the above inequality, Equation 7.1, and the assumptions that t(DB ) ≥ 2 and t(T ) ≥ 2, we get:
−χ(GA (H)) =
−χ(GA (B)) − χ(GA (P )) + t(T )
2
2t(DB ) 1
+ · (t(DP ) − 1) − + t(T )
3
3
3
2t(DB ) + t(DP ) − 3 + 3t(T )
3
[t(DB ) + t(DP ) + t(T )] + t(DB ) + 2t(T ) − 3
3
t(D) + t(DB ) + 2t(T ) − 3
3
t(D) + 3
3
1
4
· (t(D) − 1) + .
(7.3)
3
3
≥
=
=
=
≥
=
Since D(K) satisfies the assumptions of Theorem 7.2.1, then we may apply Theorem 2.2.2
to get:
4v
v8
· (t(D) − 1) + 8 ≤ vol(S 3 \K) < 10v3 · (t(D) − 1) .
3
3
Transitioning to the colored Jones polynomial setting, note that Theorem 3.5.1 gives the
desired lower bound on volume. Combining Theorem 2.3.1 with Inequality 7.3 gives:
βK − 1 = −χ(GA (H)) ≥
Hence:
127
1
4
· (t(D) − 1) + .
3
3
t(D) − 1 ≤ 3 · ( βK − 1) − 4,
which implies that:
10v3 · (t(D) − 1) ≤ 30v3 · ( βK − 1) − 40v3 ,
which gives the desired upper bound on volume.
Remark 7.4.1. It should be noted that similar methods can be used to obtain volume
bounds for hybrid braid-plats where the (closed) braid part comes from the closed braids
considered in this work.
128
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