# Twist Positivity, L-space knots, and Concordance

Siddhi Krishna and Hugh Morton  Department of Mathematics, Columbia University
New York, NY, United States<EMAIL_ADDRESS>Department of
Mathematica Sciences, University of Liverpool
Liverpool, United Kingdom<EMAIL_ADDRESS>
Abstract. Many well studied knots can be realized as positive braid knots
where the braid word contains a positive full twist; we say that such knots
are twist positive. Some important families of knots are twist positive,
including torus knots, 1-bridge braids, algebraic knots, and Lorenz knots. We
prove that if a knot is twist positive, the braid index appears as the third
exponent in its Alexander polynomial. We provide a few applications of this
result. After observing that most known examples of L-space knots are twist
positive, we prove: if $K$ is a twist positive L-space knot, the braid index
and bridge index of $K$ agree. This allows us to provide evidence for Baker’s
reinterpretation of the slice-ribbon conjecture: that every smooth concordance
class contains at most one fibered, strongly quasipositive knot. In
particular, we provide the first example of an infinite family of positive
braid knots which are distinct in concordance, and where, as $g\to\infty$, the
number of hyperbolic knots of genus g gets arbitrarily large. Finally, we
collect some evidence for a few new conjectures, including the following: the
braid and bridge indices agree for any L-space knot.

##  1\. Introduction

By Alexander’s theorem [Ale23], every link in $S^{3}$ can be realized as the
closure of some braid. Some properties of braids can be ported into important
topological properties of their closures. For example, if a braid $\beta$ is
homogeneous (i.e. for every $i$, every $\sigma_{i}$ appearing in $\beta$ has
the same sign), then the closure is a fibered link in $S^{3}$ [Sta78]. In a
different vein, the closures of quasipositive braids are exactly the links
arising as the transverse intersections of smooth algebraic curves
$f^{-1}(0)\subset\mathbb{C}^{2}$ (here, $f$ is a non-constant polynomial) with
the unit sphere $S^{3}\subset\mathbb{C}^{2}$ [Rud83, BO01, Hay21]. Another
property of braids with important topological consequences is twist
positivity:

######  Definition 1.1.

Let $\beta$ be a positive braid braid on $n$ strands. If $\beta$ can be
factored to contain a positive full twist on $n$ strands $($i.e.
$\beta=\Delta^{2}\gamma$ where $\gamma$ is a positive braid word$)$, then
$\beta$ is a twist positive braid on $n$ strands. If $K$ can be realized as
the closure of a twist positive braid on $n$ strands, then $K$ is a twist
positive knot on $n$ strands.

Figure 1.

A full twist on three strands is seen in Figure 1; here,
$\Delta^{2}=(\sigma_{2}\sigma_{1})^{3}$. Twist positivity is a well studied
property: perhaps most notably, Morton and Franks-Williams proved that if $K$
is twist positive on $n$ strands, then the braid index of $K$ is $n$ [Mor86,
FW87].

In this paper, we show that twist positivity unifies classical results in knot
theory with modern concordance invariants. Our first theorem is about the
Alexander polynomial of twist positive knots.

######  Theorem 1.2.

Suppose $K$ is twist positive on $n$ strands. Then the Alexander polynomial of
$K$ has the form $\Delta_{K}(t)=1-t+t^{n}+t^{n+1}R(t)$, where
$R(t)\in\mathbb{Z}[t]$. In particular, the braid index of $K$ appears as an
exponent in $\Delta_{K}(t)$.

The strength and utility of 1.2 is leveraged by way of the connection between
the Alexander polynomial and knot Floer homology: not only does knot Floer
homology categorify the Alexander polynomial, but it also provides a suite of
powerful knot invariants. We recall: an L-space knot in $S^{3}$ is a non-
trivial knot which admits a positive Dehn surgery to an L-space, the closed,
connected, oriented 3-manifolds with simple Heegaard Floer homology. For
L-space knots, both the Alexander polynomial and knot Floer homology take on a
restricted form [OS05, Theorem 1.2]. This connection leads to our first
application of 1.2.

######  Theorem 1.3.

If $K$ is a twist positive L-space knot, then its braid index and bridge index
agree. Moreover, $K$ realizes the minimal braid and bridge indices of its
concordance class.

######  Remark 1.4.

As we observe in Section 3, most explicit families of L-space knots are not
only twist positive, but they also happen to be Lorenz knots.

Lorenz knots exhibit rich phenomena from the dynamical, geometric, and braid
theoretic perspectives. They have been intensely studied since their
introduction in 1963; see, for example, [BW83, BK09, CFK+11, Deh15, dPP21], or
[Bir13] for a survey. In Section 3, we observe L-space Lorenz knots include
the well-studied algebraic knots, 1-bridge braids [Gab90, GLV18], and some
twisted torus knots [Vaf15], amongst others.

Therefore, 1.3 generalizes results of both Juhász-Miller-Zemke [JMZ20] (who
prove 1.3 for torus knots) and Hom-Lidman-Park [HLP22], who use input from
bordered Floer homology to prove it for algebraic knots. Our results suggests
that twist positivity is the important topological property guiding this
behavior.

Birman-Williams [BW83, Section 5] noticed that Lorenz knots are always twist
positive on some number of strands, and conjectured that this quantity was
actually the braid index. Indeed, this conjecture follows with the subsequent
announcement of the Franks-Williams result [FW87]. To the best of our
knowledge, the bridge index of Lorenz knots has not been investigated. 1.3
proves that often, the bridge index and the braid index for a Lorenz knot
agree, hence the Lorenz presentation cannot be minimized further to decrease
the bridge index.

1.2 and 1.3 suggest that twist positivity is not only an important
3-dimensional property, but that it also provides some insights into 3.5- and
4-dimensional phenomena. Indeed, we next study concordance. Baker [Bak16]
observed a novel relationship between concordance properties of fibered knots
and Fox’s slice-ribbon conjecture, which asks whether every slice knot admits
a ribbon presentation in $S^{3}$. In particular, Baker conjectured that if two
fibered, strongly quasipositive knots are concordant, then they are isotopic.
Baker observes that either his conjecture is true, or the Slice-Ribbon
conjecture is false111Baker phrases his conjecture in terms of tight fibered
knots, but by [Hed10], this is equivalent to studying fibered, strongly
quasipositive knots.. Therefore, studying the concordance properties of
fibered, strongly quasipositive knots is of central interest. L-space knots
form a prominent class of fibered, strongly quasipositive knots. As noted by
Abe-Tagami [AT16], if Baker’s conjecture is true, then every concordance class
should have at most one L-space knot.

######  Theorem 1.5.

Let $\mathcal{S}=\mathcal{T}\cup\mathcal{L}$, where $\mathcal{T}$ is the set
of positive torus knots, and $\mathcal{L}$ is the set of L-space knots of
braid index three. Every concordance class contains at most one knot from
$\mathcal{S}$.

######  Remark 1.6.

Our proof of 1.5 is independent of Litherland’s proof that torus knots are
linearly independent in concordance [Lit79].

1.5 is proved by using the braid index, $\tau(K)$, and the Seifert genus in
concert.

Baader-Dehorney-Liechti [BDL18] proved that only finitely many positive braid
knots can be concordant to each other (c.f. [Sto15]). However, to the best of
our knowledge, this is the first explicit infinite family of positive braid
knots that we know are distinct in concordance, where as $g$ tends to
infinity, the number of hyperbolic knots of genus $g$ gets arbitrarily large.

While L-space knots are well-studied, a classification remains elusive;
nevertheless, collecting their properties remains a central goal. L-space
knots are “simple” from many perspectives: they have simplest possible knot
Floer homology [OS05], “simple” knot exteriors (their exteriors are realized
as mapping tori) [Ghi08, Ni07], and, heuristically, they seem to have “simple”
monodromy [MS21, Ni21, GS22]. The bridge index and the braid index provide
different measures of complexity of a knot, as the former is a lower bound on
the latter. The difference between these quantities is itself a measure of
complexity. We conjecture that this defect is trivial for L-space knots.

######  Conjecture 1.7.

If $K$ is an L-space knot, then its braid index and bridge index agree.

This conjecture is true for the most well known families of L-space knots:

  * •

The only L-space two-bridge knots are $T(2,n)$ torus knots. [Goo72, OS05].

  * •

The only L-space Montesinos knots are the $P(-2,3,q)$, $q\geq 1$ pretzel knots
[LM16, BM18]. They are all twist positive on either two or three strands,
hence 1.3 applies.

  * •

As proved in this article, the braid and bridge index agree for 1-bridge
braids.

  * •

Baker-Kegel [BK22] constructed an infinite family of hyperbolic L-space knots
where the first knot is provably not braid positive, but the braid positivity
status of the remaining knots remains unknown. We prove 1.7 for the Baker-
Kegel family in Section 5.

  * •

Combining work of Hedden [Hed09] and Hom [Hom11], it is known that $K_{p,q}$
is an L-space knot if and only if $K$ is and $q\geq p(2g(K)-1)$. In Section 5,
we show that if 1.7 is true for an L-space knot $K$, it is true for all of its
L-space cables.

The Alexander polynomials of L-space knots have been of interest ever since
Ozsváth-Szabó proved that the coefficients of these polynomials take values in
$\\{-1,0,1\\}$ [OS05]. Hedden-Watson [HW18, Corollary 9] later proved that
when $K$ is an L-space knot, $\Delta_{K}(t)=1-t+\ldots+t^{2g-1}+t^{2g}$, where
$g$ is the Seifert genus of $K$. The results we summarized above provide
evidence for the following:

######  Conjecture 1.8.

Suppose $K$ is a hyperbolic L-space knot with Seifert genus $g$ and braid
index $n$. Then $\Delta_{K}(t)=1-t+t^{n}+\ldots+t^{2g-n-1}-t^{2g-1}+t^{2g}$.

Hyperbolicity is a key assumption in 1.8: the conjecture is false even for the
(2,3)-cable of $T(2,3)$, which is an L-space knot. In a different direction,
we also predict:

######  Conjecture 1.9.

If $K$ is a positive braid knot, the braid index and bridge index of $K$
agree.

1.9 is true for all positive braid knots in the knot tables [LM22], but
quickly becomes false for even slightly broader classes of knots. For example,
the fibered, strongly quasipositive knot $10_{145}$ is not braid positive, and
it has braid index is four while the bridge index is three [LM22]. It is worth
noting that there are positive braid knots which do not admit a positive braid
representative on a minimal number of strands [Sto03], indicating a subtlety
to 1.9.

###  1.1. Conventions

  * •

We let $B_{n}$ denote the braid group on $n$ strands.

  * •

Unless stated otherwise, we assume that all braid closures are knots.

  * •

When drawn vertically (resp. horizontally), our braids are oriented from north
to south (resp. from west to east).

  * •

We write $\beta_{1}\approx\beta_{2}$ when braids are related by conjugation.
We write $\beta_{1}=\beta_{2}$ if they are related by braid relations. We
underline the subword of a braid that is being modified.

  * •

Throughout, we denote the braid (resp. bridge) index of a knot $K$ by $i(K)$
(resp. $br(K)$).

  * •

We always work within the smooth category. In particular, we only study knots
up to smooth concordance.

###  1.2. Organization

We prove 1.2, 1.3, and 1.5 in Section 2, Section 3, and Section 4,
respectively. Section 5 provides evidence for 1.9.

###  1.3. Acknowledgements

We thank Joan Birman, Mike Miller Eismeier, Peter Feller, and Allison N.
Miller for helpful conversations, and John Baldwin, Josh Greene, Eli Grigsby,
Kyle Hayden, Matt Hedden, Francesco Lin, and Marissa Loving for comments on a
preliminary version of this work. We used Kirby Calculator (“KLO”) and
KnotInfo during the experimental phases of this work; we express our deep
gratitude to Frank Swenton, Chuck Livingston, and Allison Moore for
maintaining these invaluable resources. SK was supported by NSF DMS-2103325.

##  2\. Twist Positivity and the Alexander Polynomial

This section establishes a relationship between twist positivity and the
Alexander polynomial.

1.2. If $K$ is twist positive, then $\Delta_{K}(t)=1-t+t^{n}+t^{n+1}R(t)$,
where $R(t)\in\mathbb{Z}[t]$.

The proof of 1.2 requires the reduced Burau matrix, $B(t)$, which is
associated to a braid $\beta\in B_{n}$. We quickly recall the relevant details
here, and recommend [Bir74] for a full account.

Let $\beta$ be a braid on $n$ strands. We write
$\displaystyle\beta=\prod_{r=1}^{m}\sigma_{i_{r}}^{\epsilon_{i_{r}}}$, i.e. as
the product of Artin generators.

To $\beta$, we can associate its reduced Burau matrix, $B(t)$, which is the
product of $(n-1)\times(n-1)$ matrices $\sigma_{i}(t)$, where $\sigma_{i}(t)$
differs from the $(n-1)\times(n-1)$ identity matrix $I_{n-1}$ only in row $i$,
as in Figure 2. (Note that this matrix is truncated appropriately when
$i=1,n-1$.) In particular, $\displaystyle
B(t)=\prod_{r=1}^{m}\sigma_{i_{r}}^{\epsilon_{i_{r}}}(t)$.

$\sigma_{i}(t)=\left(\begin{tabular}[]{ccccccc}1&&&&&&\\\ &$\ddots$&&&&&\\\
&&1&&&&\\\ &&$t$&$-t$&1&&\\\ &&&&1&&\\\ &&&&&$\ddots$&\\\
&&&&&&1\end{tabular}\right)$

Figure 2. The template for the matrices $\sigma_{i}(t)$, the building blocks
of the matrix $B(t)$.

Suppose $\widehat{\beta}=K$, a knot. The Alexander polynomial for $K$, denoted
$\Delta_{K}(t)$, is derived from $B(t)$ as follows:

$\frac{\det(I_{n-1}-B(t))}{1-t^{n}}=\frac{\Delta_{K}(t)}{1-t}$

In particular,

$\Delta_{K}(t)=(1-t)\cdot(\det(I_{n-1}-B(t)))\cdot(1+t^{n}+t^{2n}+\cdots).$

We collect some important properties of $B(t)$:

  * •

In general, the entries in a reduced Burau matrix lie in $\mathbb{Z}[t^{\pm
1}]$. However, when $\beta$ is a positive braid, the entries of $B(t)$ are
genuine polynomial entries.

  * •

Recall: the full twist on $n$ strands is denoted by $\Delta^{2}$, where
$\Delta^{2}=(\sigma_{1}\ldots\sigma_{n-1})^{n}\approx(\sigma_{n-1}\ldots\sigma_{1})^{n}$.
When $\beta=\Delta^{2}$, we have $B(t)=t^{n}I_{n-1}$. In particular, if
$\beta$ is a twist positive braid, then $\beta=\Delta^{2}\gamma$, so the
reduced Burau matrix associated to $\beta$ is
$B(t)=(t^{n}I_{n-1})(C(t))=t^{n}C(t)$, where $C(t)$ is the reduced Burau
matrix associated with $\gamma$.

  * •

Suppose $\beta$ is a positive braid on $n$ strands. We define the polynomial
$q_{B(t)}(x)$ as follows:

(2.1) $\displaystyle
q_{B(t)}(x):=\det(I_{n-1}-xB(t))=1+a_{1}(t)x+a_{2}(t)x^{2}+\ldots+x^{n-1}a_{n-1}(t)$

We claim that the values of $a_{1}(t)$ and $a_{n-1}(t)$ can be quickly
determined from $B_{(}t)$: if we let $p_{B(t)}(x):=\det(xI_{n-1}-B(t))$ denote
the characteristic polynomial of $B(t)$, then
$q_{B(t)}(x)=x^{n-1}p_{B(t)}(x^{-1})$. That is, $q_{B(t)}(x)$ is obtained by
$p_{B(t)}(x)$ by exchanging the coefficients of $x^{i}$ and $x^{(n-1)-i}$ for
each $i\leq\lfloor(n-1)/2\rfloor$. In particular, in $q_{B(t)}(t)$,
$a_{n-1}(t)=(-1)^{n-1}\det(B(t))$ and $a_{1}(t)=-\text{tr}(B(t))$.

We observe that substituting $x=t^{n}$ into Equation 2.1 yields:

$\displaystyle\det(I_{n-1}-t^{n}B(t))$
$\displaystyle=1+a_{1}(t)t^{n}+a_{2}(t)t^{2n}+\ldots+t^{n^{2}-n}a_{n-1}(t)$
(2.2) $\displaystyle=1+a_{1}(t)t^{n}+t^{n+1}R(t)$

where $R(t)\in\mathbb{Z}[t]$ collects all the higher order terms.

With these preliminaries established, we can prove the result.

###### Proof of 1.2..

Suppose $K=\widehat{\beta}$, where $\beta$ is a twist positive braid on $n$
strands. Then $\beta=\Delta^{2}\gamma$, $B(t)=t^{n}C(t)$, and by applying the
observation of Equation 2.2,

$\det(I_{n-1}-B(t))=\det(I_{n-1}-t^{n}C(t))=1+a_{1}(t)t^{n}+t^{n+1}R(t)$

with $R(t)\in\mathbb{Z}[t]$. In particular, this means that

(2.3) $\displaystyle\Delta_{K}(t)$
$\displaystyle=(1-t)\cdot\left(1+a_{1}(t)t^{n}+t^{n+1}R(t)\right)\cdot(1+t^{n}+t^{2n}+\ldots)$
(2.4) $\displaystyle=1-t+(1+a_{1}(t))t^{n}+t^{n+1}S(t)$

where Equation 2.4 is obtained by distributing and rearranging the terms in
the righthand side of Equation 2.3. Here, $S(t)\in\mathbb{Z}[t]$.

######  Claim 1.

Suppose $\widehat{\beta}$ is a knot $K$. Then the polynomial $a_{1}(t)$ has no
constant term, i.e. $a_{1}(0)=0$.

###### Proof of 1..

In the paragraph following Equation 2.1, we argued that
$a_{1}(t)=-\text{tr}(B(t))$. Therefore, to see that $a_{1}(0)=0$, we must show
that $\text{tr}(B(0))=0$.

This is straightforward: first, we note that for all $i$, $\sigma_{i}(t)$ is
an upper triangular matrix. Since the product of upper triangular matrices is
also upper triangular, the matrix $B(t)$ is also an upper triangular matrix.
Moreover, the diagonal entries of $B(t)$ are explicitly determined by the
diagonal entries of its constituent matrices; more precisely,

$\displaystyle\displaystyle[B(t)]_{ii}=\prod_{r=1}^{m}\left[\sigma_{i_{r}}^{\epsilon_{i_{r}}}(t)\right]_{ii}$

where $[M]_{ij}$ denotes the $(i,j)$ entry of the matrix $M$.

Since $\widehat{\beta}$ is a knot, then for all $1\leq i\leq n-1$,
$\sigma_{i}$ appears in $\beta$ at least once. Moreover, by the definition of
the matrix $\sigma_{i}(t)$, $\left[\sigma_{i}(0)\right]_{ii}=0$. Therefore,
for all $1\leq i\leq n-1$, $[B(t)]_{ii}=0$. We deduce that
$\text{tr}(B(t))=0$, as desired. ∎

With this claim in hand, we can finish proving 1.2. In particular, since
$a_{1}(0)=0$, then $a_{1}(t)=e_{1}t+e_{2}t^{2}+\ldots+e_{k}t^{k}$. Plugging
this into Equation 2.4, we have

$\displaystyle\Delta_{K}(t)$ $\displaystyle=1-t+(1+a_{1}(t))t^{n}+t^{n+1}S(t)$
$\displaystyle=1-t+(1+(e_{1}t+e_{2}t+\ldots+e_{k}t^{k}))t^{n}+t^{n+1}S(t)$
$\displaystyle=1-t+t^{n}+t^{n+1}V(t)$

for some $V(t)\in\mathbb{Z}[t]$. This is exactly what we wanted to show. ∎

######  Corollary 2.1.

If $K$ is twist positive on $n$ strands, the braid index of $K$ appears as an
exponent of the Alexander polynomial.

###### Proof.

If $K$ is twist positive on $n$ strands, then by [FW87], the braid index of
$K$ is $n$. On the other hand, 1.2 proves that $n$ appears as the exponent of
the third term in $\Delta_{K}(t)$. ∎

##  3\. Proof of 1.3

First, we establish that some well-studied families of knots are twist
positive:

######  Proposition 3.1.

Algebraic knots, 1-bridge braids, and many twisted torus knots are twist
positive L-space knots. In fact, something stronger is true: these knots are
all L-space Lorenz knots.

###### Proof.

Birman-Williams proved that every Lorenz knot is twist positive [BW83].
Moreover, Birman-Kofman [BK09] showed that Lorenz knots are in correspondence
with T-links, which are defined to be the closures of positive braids of a
particular form:

(3.1)
$\displaystyle\tau=(\sigma_{1}\sigma_{2}\ldots\sigma_{p_{1}-1})^{q_{1}}(\sigma_{1}\sigma_{2}\ldots\sigma_{p_{2}-1})^{q_{2}}\ldots(\sigma_{1}\sigma_{2}\ldots\sigma_{p_{s}-1})^{q_{s}}$

where $2\leq p_{1}\leq p_{2}\leq\ldots\leq p_{s}$, $0<q_{i}$ for all $i$, and
$\tau$ is a braid in $B_{p_{s}}$. Thus, to prove 3.1, it suffices to show that
algebraic knots, 1-bridge braids, and twisted torus knots are $T$-links, and
also confirm their L-space status.

  * •

Birman-Williams showed that algebraic knots are Lorenz knots [BW83, Theorem
6.3]. It is well known that algebraic knots are particular iterated cables of
torus knots [Bra28]; applying Hedden’s results on cabling [Hed09], we deduce
they are L-space knots.

  * •

A 1-bridge braid is a knot $K\subset S^{3}$ realized as the closure of a braid
$\beta$, where:

$\beta=(\sigma_{b}\sigma_{b-1}\ldots\sigma_{2}\sigma_{1})(\sigma_{w-1}\sigma_{w-2}\ldots\sigma_{2}\sigma_{1})^{t}$

[GKL+22] proved that all 1-bridge braids admit $T$-link presentations (hence
they are twist positive), and [GLV18] proved they are L-space knots.

  * •

In [Vaf15], Vafaee presents twisted torus knots as the closures of positive
braids on $w$ strands with the following form:
$\rho=(\sigma_{w-1}\sigma_{w-2}\ldots\sigma_{1})^{t}(\sigma_{w-1}\sigma_{w-2}\ldots\sigma_{w-k})^{sk}$,
i.e. they are built from a torus link by inserting some full twists into $k$
adjacent strands. [GKL+22] showed that all twisted torus knots admit $T$-link
presentations, so they are twist positive; [Vaf15] proved that for certain
parameters, they are L-space knots.

∎

###  3.1. Twist positive L-space knots

We begin by proving 1.3.

1.3. If $K$ is a twist positive L-space knot, then its braid index and bridge
index agree. Moreover, $K$ realizes the minimal braid and bridge indices of
its concordance class.

###### Proof.

If $K$ is a twist positive L-space knot, then by 1.2, we have

$\Delta_{K}(t)=1-t+t^{n}+t^{n+1}R(t)$

By [FW87], we know the braid index of $K$ is $n$. Moreover, we see that $n-1$
appears as a gap in the exponents of $\Delta_{K}(t)$. Juhász-Miller-Zemke
[JMZ20] established a relationship between knot Floer homology and the bridge
index:

######  Theorem 3.2.

[JMZ20, Corollary 1.9] For $K$ is a knot in $S^{3}$, $Ord_{v}(K)\leq br(K)-1$
$($where $Ord_{v}(K)$ denotes the torsion order of $K$, an invariant from knot
Floer homology$)$. $\Box$

Let $\alpha_{0},\ldots,\alpha_{2m}$ be the non-zero degrees appearing in
$\Delta_{K}(t)$, written in decreasing order. Juhász-Miller-Zemke showed that
for L-space knots, $Ord_{v}(K)=\max\\{\alpha_{i-1}-\alpha_{i}|1\leq i\leq
2m\\}$ [JMZ20, Lemma 5.1]. Since $n-1$ appears as a gap in the degrees of the
exponents of $\Delta_{K}(t)$, we have $n-1\leq Ord_{v}(K)$. The bridge index,
$br(K)$, is a lower bound on the braid index, $i(K)$, so:

(3.2) $\displaystyle n-1$ $\displaystyle\leq Ord_{v}(K)\leq br(K)-1\leq
i(K)-1=n-1$

But this means we must have equalities throughout, hence $br(K)=n=i(K)$, as
desired.

Next, we prove that twist positive L-space knots realize the minimal braid
index of their concordance classes.

The proof of this is nearly identical to the proof of [JMZ20, Corollary 1.10].
Suppose the knot $J$ is concordant to a twist positive L-space knot $K$, and
let $N(K)$ denote the concordance invariant defined by Dai-Hom-Truong-
Stoffregen [DHST21]. We have the following string of inequalities
(justification for the respective (in)equalities are in the paragraph
following):

(3.3) $\displaystyle i(J)\geq br(J)$ $\displaystyle\geq Ord_{v}(J)+1$ (3.4)
$\displaystyle\geq N(J)+1$ (3.5) $\displaystyle=N(K)+1$ (3.6)
$\displaystyle=Ord_{v}(K)+1$ (3.7) $\displaystyle=br(K)=i(K)$

The first equality of (3.2) is standard; the second is a restatement of 3.2.
Dai-Hom-Truong-Stoffregen [DHST21] showed that $N(K)$ is a lower bound on
$Ord_{v}(K)$, yielding the inequality in (3.3). Since $N(K)$ is a concordance
invariant, we have the equality in (3.4). [JMZ20] showed that for L-space
knots, the invariants $N(\star)$ and $Ord_{v}(\star)$ agree; this yields the
equality in (3.5). The equalities of (3.5) follow from those in (3.2). ∎

##  4\. Proof of 1.5

In this section, we prove that L-space knots of braid index three lie in
distinct concordance classes, and these classes are distinct from those of
torus knots.

###  4.1. Preliminaries

We begin with some preliminaries.

######  Definition 4.1.

$T(3,k;2m)$ denotes the twisted torus knot which is the closure of
$(\sigma_{2}\sigma_{1})^{k}(\sigma_{2})^{2m}$, where $k\geq 4$, $k\not\equiv
0\mod 3$, and $m\geq 0$.

We briefly justify why we may assume that $k$ is at least 4: if $k=1,2$, then
$T(3,k;2m)$ is isotopic to a $T(2,r)$ torus knot. Additionally, for all $m\geq
0$ and $k\equiv 0\mod 3$, $T(3,k;2m)$ is a link. Thus, we may assume $k\geq
4$. We note that when $k\geq 4$, $T(3,k;2m)$ is twist positive on three
strands.

######  Lemma 4.2.

Every L-space knot of braid index three is either a torus knot or a hyperbolic
knot.

###### Proof.

Thurston proved that every knot is either a torus knot, a satellite knot, or
hyperbolic. Suppose $K$ is an L-space knot of braid index three. By Vafaee
[Vaf15, Corollary 3.3] and Lee-Vafaee [LV19], $K$ is a twisted torus knot on
three strands.

Suppose $K$ is not a torus knot. To show that it is hyperbolic, we must show
that it is not a satellite knot, i.e. $K$ does not contain an essential torus
in its exterior. Birman and Menasco [BM94] proved that if $L$ is a link in
$S^{3}$ of braid index three, and $L$ contains an essential, non-peripheral
torus in its exterior, then $L$ is the closure of a braid $\beta$ which is
conjugate to
$\gamma_{p,q}=(\sigma_{2})^{p}(\sigma_{1}\sigma_{2}\sigma_{2}\sigma_{1})^{q}$,
where $|p|\geq 2$ and $|q|\geq 1$. These braid closures are always links: the
cycle in the symmetric group $S_{3}$ induced by $\gamma_{p,q}$ contains a
1-cycle. We deduce the non-torus twisted torus knots are hyperbolic. ∎

In the next lemma, we see that it suffices to study $T(3,k;2m)$ where $k\equiv
1\mod 3$.

######  Lemma 4.3.

Suppose $q\equiv 1\mod 3$ and $m\geq 1$. Then the twisted torus knots
$T(3,q;2m)$ and $T(3,q+1;2m-2)$ are isotopic.

###### Proof.

We show that the braid words for $T(3,q+1;2m-2)$ and $T(3,q;2m)$ are related
by braid relations and conjugation. As $q\equiv 1\mod 3$, we will write
$q=3r+1$.

First, recall that conjugating a braid $\beta$ in $B_{n}$ by the Garside
element
$\Delta=(\sigma_{1})(\sigma_{1}\sigma_{2})\ldots(\sigma_{1}\sigma_{2}\ldots\sigma_{n-1})$
has the effect of exchanging $\sigma_{i}$ and $\sigma_{n-i}$ for all $1\leq
i\leq n-1$. In particular, when $n=3$, conjugating a 3-braid by the Garside
element corresponds to swapping $\sigma_{1}$ and $\sigma_{2}$. This
conjugation procedure justifies the equivalence in the first line of the
sequence of braid moves below.

$\displaystyle(\sigma_{2}\sigma_{1})^{q}(\sigma_{2})^{2m}$
$\displaystyle\approx(\sigma_{1}\sigma_{2})^{q}(\sigma_{1})^{2m}$
$\displaystyle=(\sigma_{1}\sigma_{2})^{3r}(\sigma_{1}\sigma_{2})\sigma_{1}(\sigma_{1})^{2m-1}$
$\displaystyle=\underline{(\sigma_{1}\sigma_{2})^{3r}}\
(\underline{\sigma_{1}\sigma_{2}\sigma_{1}})(\sigma_{1})^{2m-1}$ Since
$(\sigma_{1}\sigma_{2})^{3}=(\sigma_{2}\sigma_{1})^{3}$, we have:
$\displaystyle=(\sigma_{2}\sigma_{1})^{3r}(\sigma_{2}\sigma_{1}\sigma_{2})(\sigma_{1})^{2m-1}$
$\displaystyle=(\sigma_{2}\sigma_{1})^{3r}(\sigma_{2}\sigma_{1}\sigma_{2})\sigma_{1}(\sigma_{1})^{2m-2}$
$\displaystyle=(\sigma_{2}\sigma_{1})^{3r+2}(\sigma_{1})^{2m-2}$
$\displaystyle\approx(\sigma_{1})^{2m-2}(\sigma_{2}\sigma_{1})^{3r+2}$
$\displaystyle=(\sigma_{1})^{2m-3}\underline{(\sigma_{1})(\sigma_{2}\sigma_{1})}(\sigma_{2}\sigma_{1})^{3r+1}$
$\displaystyle=(\sigma_{1})^{2m-3}(\sigma_{2}\sigma_{1}\sigma_{2})(\sigma_{2}\sigma_{1})^{3r+1}$
$\displaystyle=(\sigma_{1})^{2m-4}(\underline{\sigma_{1}\sigma_{2}\sigma_{1}})\sigma_{2}(\sigma_{2}\sigma_{1})^{3r+1}$
$\displaystyle=(\sigma_{1})^{2m-4}(\sigma_{2}\sigma_{1}\sigma_{2})\sigma_{2}(\sigma_{2}\sigma_{1})^{3r+1}$
$\displaystyle=(\sigma_{1})^{2m-4}(\sigma_{2}\sigma_{1})\sigma_{2}^{2}(\sigma_{2}\sigma_{1})^{3r+1}$
We continue extracting a $\sigma_{1}$ term from the leftmost parenthetical,
and then applying a braid relation to the leftmost subword
$\sigma_{1}\sigma_{2}\sigma_{1}$. At the penultimate application of this
process, we get:
$\displaystyle=(\sigma_{1})(\sigma_{2}\sigma_{1}\sigma_{2})\sigma_{2}^{2m-4}(\sigma_{2}\sigma_{1})^{3r+1}$
$\displaystyle=(\underline{\sigma_{1}\sigma_{2}\sigma_{1}})\sigma_{2}^{2m-3}(\sigma_{2}\sigma_{1})^{3r+1}$
$\displaystyle=(\sigma_{2}\sigma_{1}\sigma_{2})\sigma_{2}^{2m-3}(\sigma_{2}\sigma_{1})^{3r+1}$
$\displaystyle\approx(\sigma_{2}\sigma_{1})^{3r+2}\sigma_{2}^{2m-2}$

But this is exactly the braid whose closure is $T(3,q+1;2(m-1))$. Thus, when
$q\equiv 1\mod 3$ and $m\geq 1$, the twisted torus knots $T(3,q;2m)$ and
$T(3,q+1;2(m-1))$ are isotopic. ∎

######  Corollary 4.4.

Every positive twisted torus knot on three strands can be presented as a
positive twisted torus knot $T(3,q;2m)$ where $q\geq 4$, $q\equiv 1\mod 3$ and
$m\geq 0$. $\Box$

######  Proposition 4.5.

Fix a torus knot $K=T(3,q)=T(3,q;0)$ with $q\geq 4$. There are
$\lfloor\frac{q}{3}\rfloor-1$ other twisted torus knots $T$ of braid index 3
with $\tau(T)=\tau(K)$ where $\tau$ denotes the Ozsváth-Szabó concordance
invariant.

In fact, our proof of 1.5 will show that the $\lfloor\frac{q}{3}\rfloor-1$
twisted torus knots are pairwise distinct from $T(3,q)$ and from each other.

###### Proof.

Since $T$ and $K$ are both positive braid knots, they are fibered and strongly
quasipositive [Sta78]. By Hedden, $g_{4}(T)=g(T)=\tau(T)$ and
$g_{4}(K)=g(K)=\tau(K)$ [Hed10]. To deduce the statement, it suffices to
determine which twisted torus knots have the same Seifert genus as $K$.

For positive braid knots, the standard Bennequin surface is a fiber surface
for the knot [Sta78]. It is therefore straightforward to compute the Seifert
genus of any positive braid knot $P=\widehat{\rho}$, where $\rho$ is a
positive braid on $n$ strands: $2g(P)-1=wr(\rho)-n$. If two positive
$n$-braids have the same writhe, then they have the same Seifert genus, hence
the same $\tau$. Therefore, our problem reduces to counting the number of
twisted torus knots on 3 strands with the same writhe as $K$.

First, suppose that $q\equiv 1\mod 3$, so $q=3r+1$, where $r\geq 1$. The
writhe of $T(3,q)=T(3,q;0)$ is $2q=2(3r+1)$. Note that a twisted torus knot of
the form $T(3,q+\epsilon,2m)$ with $\epsilon\geq 1,m\geq 0$ can never be
concordant to $T(3,q)$, as these knots do not have the same writhe. However,
it is elementary to check that for all $s$ with $0\leq s\leq r-1$, the knots
$T(3,3(r-s)+1;6s)$ all have writhe $6r+2$; thus, they have the same Seifert
genus. By Lemma 4.3, the twisted torus knots $T(3,3(r-s)+1;6s)$ and
$T(3,3(r-s)+2;6s-2)$ are isotopic knots. Therefore, to determine the number of
positive twisted torus knots $T$ with $\tau(T)=\tau(K)$, it suffices to count
how many twisted torus knots of the form $T(3,3(r-s)+1;6s)$ there are, where
$r-1\geq s\geq 1$. A simple counting argument allows us to deduce that there
are $\frac{q-4}{3}=\lfloor\frac{q}{3}\rfloor-1$ such knots.

Next, suppose $q\equiv 2\mod 3$, so we are studying torus knots of the form
$T(3,q)=T(3,3r+2)$. By Lemma 4.3, the knots $T(3,3r+2)$ and $T(3,3r+1;2)$ are
isotopic. Applying the same argument as in the previous paragraph, every
twisted torus knot of the form $T(3,3(r-s)+1;2+6s)$, where $r-1\geq s\geq 1$,
will have the same writhe as $T(3,3r+1;2)$. There are
$\lfloor\frac{q}{3}\rfloor-1$ such knots.

Therefore, there are $\lfloor\frac{q}{3}\rfloor-1$ many twisted torus knots
$T$ with $\tau(T)=\tau(K)$, regardless of the congruence class of $q\mod 3$. ∎

Our long-term goal is to show that, despite there being many twisted torus
knots with the same $\tau$ invariants, these knots represent distinct
concordance classes. This is proved by computing the signatures of 3-stranded
twisted torus knots.

###  4.2. Computing the signature via the Goeritz matrix

To compute the signature, we use the Gordon-Litherland [GL78] method, which
now we recall: let $\mathcal{K}$ be a knot in $S^{3}$, and fix a regular
projection $K$ of $\mathcal{K}$. First, we checkerboard color $K$, where the
checkerboard has black and white regions (without loss of generality, we
assume that the unbounded region is white). Each white region of the diagram
is labelled $X_{i}$, and we set $X_{0}$ to be the unbounded white region. To
each double point $D$ in $K$, we assign a value $\eta(D)=\pm 1$, as in Figure
3. Each double point $D$ is also assigned a type, i.e. it is either Type I or
Type II; see Figure 3.

This information is organized into a symmetric, integral matrix
$G^{\prime}(K)$, with entries $g_{ij}$, where:

$\displaystyle g_{ij}=\begin{cases}-\displaystyle\sum\eta(D)&\text{summed over
double points $D$ incident to $X_{i}$ and $X_{j}$, where $i\neq j$}\\\ \\\
\displaystyle-\sum_{k,k\neq i}g_{ik}&\text{if $i=j$}\end{cases}$

The Goeritz matrix $G(K)$ is constructed from $G^{\prime}(K)$ by deleting the
$0^{th}$ row and column of $G^{\prime}(K)$. Thus, $G$ is a symmetric,
integral, $n\times n$ matrix, where $n$ is the number of bounded white regions
in $K$. Gordon-Litherland [GL78] proved that

(4.1) $\displaystyle\sigma(\mathcal{K})=\text{sign}(G(K))-\mu(K)$

where $\mu(K)=\sum\eta(D)$, where $D$ is a Type II double point in $K$.

$\eta(D)=+1$ at 120 -30 $\eta(D)=-1$ at 440 -30 Type I at 1050 -30 Type II at
1380 -30

Figure 3. Conventions for $\eta(D)$ are on the left, and assignments for Type
I and II double points are on the right.

We are now ready to compute the signature of $T(3,q;2m)$ where $q\geq 4$ and
$m\geq 0$.

######  Proposition 4.6.

Let $K=T(3,q;2m)$ denote a twisted torus knot with $q=3k+1\equiv 1\mod 3$,
where $k\geq 1$ and $m\geq 0$. There is a $q\times q$ Goeritz matrix $G(K)$
for $K$, where all the non-zero entries are either along the tridiagonal, or
in the entries $(1,q)$ or $(q,1)$.

###### Proof.

Throughout, we conflate a knot and a regular projection of it.

Let $K$ be the specified knot. Rather than presenting $K$ as the closure of
$(\sigma_{2}\sigma_{1})^{q}(\sigma_{2})^{2m}$, we conjugate the braid by the
Garside element (as in Lemma 4.3), and view $K$ as the closure of
$(\sigma_{1}\sigma_{2})^{q}(\sigma_{1})^{2m}$. Next, we modify the braid:

$\displaystyle(\sigma_{1}\sigma_{2})^{q}(\sigma_{1})^{2m}$
$\displaystyle=(\underline{\sigma_{1}}\sigma_{2})(\sigma_{1}\sigma_{2})^{q-2}(\sigma_{1}\sigma_{2})(\sigma_{1})(\sigma_{1})^{2m-1}$
$\displaystyle\approx(\sigma_{2})(\sigma_{1}\sigma_{2})^{q-2}(\sigma_{1}\sigma_{2})(\sigma_{1})(\sigma_{1})^{2m-1}(\sigma_{1})$
$\displaystyle=(\sigma_{2}\sigma_{1})^{q}(\sigma_{1})^{2m}$
$\displaystyle=:\beta$

This presentation is well adapted to find a spanning surface. Indeed, we can
produce a checkerboard coloring with exactly $q+1$ white regions: present $K$
as the closure of $\beta$, and read $\beta$ from left to right. For $i\neq
0,q$, the white region $X_{i}$ lies between the $i^{th}$ and $(i+1)^{th}$
occurrences of $\sigma_{2}$ in $\beta$. $X_{q}$ is the white region whose
boundary contains the following double points: the first $\sigma_{2}$ in
$\beta$, the last $\sigma_{2}$ in $\beta$, and double points from
$(\sigma_{1})^{2m}$. See Figure 4 for this labelling demonstrated on
$K=T(3,7;4)$.

$X_{1}$$X_{2}$$X_{3}$$X_{4}$$X_{5}$$X_{6}$$X_{7}$ Figure 4. The twisted torus
knot $T(3,7;4)$ and a well-adapted spanning surface used to identify a Goeritz
matrix. The label for the unbounded region, $X_{0}$, is suppressed. In this
figure, $\sigma_{1}$ is closer to the bottom of the page, and $\sigma_{2}$ is
closer to the top of the page.

We now study the double points/crossings, and understand $\eta(D)$ and the
type of each crossing. By inspection, we see that each double point coming
from a $\sigma_{1}$ has $\eta(D)=+1$, while each double point coming from a
$\sigma_{2}$ has $\eta(D)=-1$. Additionally, each $\sigma_{1}$ corresponds to
a Type II double point, while each $\sigma_{2}$ corresponds to a Type I double
point. In particular, for $T(3,q;2m)$, we have:

(4.2) $\displaystyle\mu(K)$ $\displaystyle=\sum_{\begin{subarray}{c}\text{Type
II}\\\ \text{double
points}\end{subarray}}\eta(D)=\sum_{\sigma_{1}\in\beta}(+1)=q+2m$

We now construct the $(q+1)\times(q+1)$ matrix $G^{\prime}(K)$. We begin by
collecting some observations about the incidence of the white regions, and
what that tells us about the non-diagonal entries $g_{ij}$. For clarity of
exposition, we let $D_{i}$ denote a double point arising from $\sigma_{i}$. As
we observed above, $\eta(D_{1})=1$ and $\eta(D_{2})=-1$.

  * •

When $i\neq q$, the white regions $X_{0}$ and $X_{i}$ meet in exactly one
double point corresponding to the $i^{th}$ occurrence of $\sigma_{1}$ in
$\beta$. Therefore, for $i\not\in\\{0,q\\}$,
$g_{0,i}=g_{i,0}=-\eta(D_{1})=-1$.

  * •

The white regions $X_{0}$ and $X_{q}$ meet in $2m+1$ double points; each
double point arises from a single $\sigma_{1}$ which has $\eta(D_{1})=1$).
Thus, $g_{0,q}=g_{q,0}=-(2m+1)$.

  * •

For distinct, non-zero $i$ and $j$, the regions $X_{i}$ and $X_{j}$ meet in
exactly one double point (coming from a single occurrence of $\sigma_{2}$) if
and only if $|i-j|\equiv\pm 1\mod q$. Therefore, in this case,
$g_{ij}=g_{ji}=-\eta(D_{2})=+1$.

This is enough information to tell us about the diagonal entries of
$G^{\prime}(K)$:

  * •

$g_{0,0}=-(g_{1,0}+g_{2,0}+\ldots+g_{q-1,0}+g_{q,0})=-\left((q-1)(-1)+(-1)(2m+1)\right)=2m+q$.

  * •

For $i\not\in\\{0,q\\}$,
$g_{i,i}=-(g_{0,i}+g_{i-1,i}+g_{i+1,i})=-(-1+1+1)=-1$.

  * •

$g_{q,q}=-(g_{0,q}+g_{1,q}+g_{q-1,q})=-((-1)(2m+1)+1+1)=2m-1$.

Since $G(K)$ is obtained by deleting the zero-th row and column of
$G^{\prime}(K)$, the matrix $G_{k}$ is the $(3k+1)\times(3k+1)$ matrix with
the following entries:

$\displaystyle g_{ij}=\begin{cases}-1&i=j\neq q\\\ 2m-1&i=j=q\\\
+1&|i-j|\equiv\pm 1\mod q\\\ 0&\text{otherwise}\end{cases}$

As claimed, $G(K)$ has non-zero entries in the tridiagonal, and in the entries
$(1,q)$ and $(q,1)$. ∎

The matrices $G^{\prime}(K)$ and $G(K)$ for the twisted torus knot $T(3,7;6)$
from Figure 4 are below.

$G^{\prime}(K)=\blockarray{ccccccccc}&\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}0}}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}1}}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}2}}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}3}}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}4}}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}5}}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}6}}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\block{c(cccccccc)}\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\scriptscriptstyle{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color<EMAIL_ADDRESS>\qquad G(K)=\begin{pmatrix}\begin{array}[]{ccccccc}-1&+1&0&0&0&0&+1\\\
+1&-1&+1&0&0&0&0\\\ 0&+1&-1&+1&0&0&0\\\ 0&0&+1&-1&+1&0&0\\\
0&0&0&+1&-1&+1&0\\\ 0&0&0&0&+1&-1&+1\\\ +1&0&0&0&0&+1&+3\\\
\end{array}\end{pmatrix}$

Our goal is to compute the signature of $G(K)$. We will need a powerful
theorem, typically refereed to as “Sylvester’s Law of Inertia”: two symmetric,
$n\times n$ matrices have the same number of positive, negative, and zero
eigenvalues if and only if they are congruent [Car17, Theorem 9.13]. (Recall
that two matrices $A$ and $B$ are congruent if there exists a non-singular
matrix $S$ such that $B=SAS^{T}$.) We will prove that $G(K)$ is congruent to a
diagonal matrix $D$. Since, for diagonal matrices, the eigenvalues are exactly
the diagonal entries, we can compute the signature of $G(K)$ from $D$.

######  Proposition 4.7.

Let $K$ be the positive twisted torus knot $T(3,3k+1;2m)$, where $k\geq
1,m\geq 0$. Then the Goeritz matrix $G(K)$ is congruent to the diagonal matrix
$D$, where

$\displaystyle d_{ii}=\begin{cases}(2m-1)\pm 2&i=q\\\ 1&i\equiv 0\mod 3\\\
-1&\text{otherwise}\end{cases}$

To prove this result, we will first establish some preliminaries.

######  Definition 4.8.

The symmetric $(3\ell+1)\times(3\ell+1)$ matrices $N_{\ell,\epsilon}=(n_{ij})$
and $P_{\ell,\epsilon}=(p_{ij})$ have entries as defined in Figure 5 (left).

We note: the first parameter in the subscript identifies the dimension of the
matrix, while the second index records the value $\epsilon$. Figure 5 (right)
displays $N_{1,\epsilon}$ and $P_{1,\epsilon}$. We note that the entries of
$P_{\ell,\epsilon}$ and $N_{\ell,\epsilon}$ agree in all but two entries:
$n_{1,3\ell+1}=n_{3\ell+1,1}=-1$ while $p_{1,3\ell+1}=p_{3\ell+1,1}=+1$
(indeed, $N$ and $P$ stand for “negative” and “positive”, respectively).

$\displaystyle n_{ij}$ $\displaystyle=\begin{cases}-1&i=j\neq 3\ell+1,\text{
or }\\{i,j\\}=\\{1,q\\}\\\ \epsilon&i=j=3\ell+1\\\ +1&|i-j|=\pm 1,\text{ and
either }i\geq 2\text{ or }j\geq 2\\\ 0&\text{otherwise}\end{cases}\hskip
42.67912ptN_{1,\epsilon}=\begin{pmatrix}\begin{array}[]{ccc|c}-1&+1&0&-1\\\
+1&-1&+1&0\\\ 0&+1&-1&+1\\\ \hline\cr-1&0&+1&\epsilon\\\
\end{array}\end{pmatrix}$ $\displaystyle p_{ij}$
$\displaystyle=\begin{cases}-1&i=j\neq q\\\ \epsilon&i=j=q\\\
+1&|i-j|\equiv\pm 1\mod q\\\ 0&\text{otherwise}\end{cases}\hskip
128.0374ptP_{1,\epsilon}=\begin{pmatrix}\begin{array}[]{ccc|c}-1&+1&0&+1\\\
+1&-1&+1&0\\\ 0&+1&-1&+1\\\ \hline\cr+1&0&+1&\epsilon\\\
\end{array}\end{pmatrix}$ Figure 5. Defining the matrices $N_{\ell,\epsilon}$
and $P_{\ell,\epsilon}$.

######  Definition 4.9.

The matrix $B$ is the $3\times 3$ diagonal matrix with $d_{1,1}=d_{2,2}=-1$
and $d_{3,3}=+1$.

######  Lemma 4.10.

For all $q\geq 1$, $P_{q,\epsilon}$ is congruent to the block diagonal matrix
with $q$-many $B$ factors, and a single $1\times 1$ factor with entry
$\epsilon\pm 2$.

###### Proof.

We will need (strong) induction to prove the claim. To begin, we prove:

######  Claim 2.

$N_{1,\epsilon}$ and $P_{1,\epsilon}$ are congruent to $4\times 4$ block
diagonal matrices containing $B$ as a factor, as in Figure 6.

$\displaystyle N_{1,\epsilon}$
$\displaystyle\sim\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\ 0&-1&0&0\\\
0&0&+1&0\\\ \hline\cr 0&0&0&\epsilon+2\\\ \end{array}\end{pmatrix}\hskip
42.67912ptP_{1,\epsilon}\sim\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&0&0\\\ 0&0&+1&0\\\ \hline\cr 0&0&0&\epsilon-2\\\
\end{array}\end{pmatrix}$ Figure 6. The matrices $N_{1,\epsilon}$ and
$P_{1,\epsilon}$ are congruent to block diagonal matrices.

Proof of 2: We begin by showing $N_{1,\epsilon}$ can be diagonalized into the
claimed form. The sequence of simultaneous row and column operations is
written out explicitly in Figure 7. In Step 1 and 2, we simultaneously clear
the non-zero, non-diagonal entries in the first row/column. In Step 3, we
simultaneously swap the second and third rows and columns. In Steps 4 and 5,
we clear the non-zero, non-diagonal entries of the second row/column.

The analogous sequence of operations used to diagonalize $P_{(}1,\epsilon)$
are in Figure 8. $\Box$

$\displaystyle
N_{1,\epsilon}=\begin{pmatrix}\begin{array}[]{ccc|c}-1&+1&0&-1\\\
+1&-1&+1&0\\\ 0&+1&-1&+1\\\ \hline\cr-1&0&+1&\epsilon\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{1}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&-1\\\
0&0&+1&-1\\\ 0&+1&-1&+1\\\ \hline\cr-1&-1&+1&\epsilon\\\
\end{array}\end{pmatrix}\overset{2}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&0&+1&-1\\\ 0&+1&-1&+1\\\ \hline\cr 0&-1&+1&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{3}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&+1&+1\\\ 0&+1&0&-1\\\ \hline\cr 0&+1&-1&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{4}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&0&+1\\\ 0&0&+1&0\\\ \hline\cr 0&+1&0&\epsilon+1\\\
\end{array}\end{pmatrix}\overset{5}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&0&0\\\ 0&0&+1&0\\\ \hline\cr 0&0&0&\epsilon+2\\\
\end{array}\end{pmatrix}$ Figure 7. $N_{1,\epsilon}\sim B\oplus L$, where $L$
is the $1\times 1$ matrix $(\epsilon+2)$. The steps are enumerated, with the
step number recorded above the $``\sim"$. $\displaystyle
P_{1,\epsilon}=\begin{pmatrix}\begin{array}[]{ccc|c}-1&+1&0&+1\\\
+1&-1&+1&0\\\ 0&+1&-1&+1\\\ \hline\cr+1&0&+1&\epsilon\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{1}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&0&+1&+1\\\ 0&+1&-1&+1\\\ \hline\cr 0&+1&+1&\epsilon+1\\\
\end{array}\end{pmatrix}\overset{2}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&+1&+1\\\ 0&+1&0&+1\\\ \hline\cr 0&+1&+1&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{3}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&0&+1\\\ 0&0&+1&+2\\\ \hline\cr 0&+1&+2&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{4}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&0&0\\\ 0&0&+1&+2\\\ \hline\cr 0&0&+2&\epsilon+2\\\
\end{array}\end{pmatrix}\overset{5}{\sim}\begin{pmatrix}\begin{array}[]{ccc|c}-1&0&0&0\\\
0&-1&0&0\\\ 0&0&+1&0\\\ \hline\cr 0&0&0&\epsilon-2\\\
\end{array}\end{pmatrix}$ Figure 8. $P_{1,\epsilon}\sim B\oplus L$, where $L$
is the $1\times 1$ matrix $(\epsilon-2)$.

This proves 2. Next, we establish the base cases required to perform strong
induction.

######  Claim 3.

$P_{2,\epsilon}$ is congruent to a block matrix containing $B$ and
$N_{1,\epsilon}$ as factors, as in Figure 9 (left). $N_{2,\epsilon}$ is
congruent to a block matrix containing $B$ and $P_{1,\epsilon}$ as factors, as
in Figure 9 (right).

$\displaystyle P_{2,\epsilon}\sim\begin{pmatrix}\begin{array}[]{c|c}B&\\\
\hline\cr&N_{1,\epsilon}\end{array}\end{pmatrix}\hskip
56.9055ptN_{2,\epsilon}\sim\begin{pmatrix}\begin{array}[]{c|c}B&\\\
\hline\cr&P_{1,\epsilon}\end{array}\end{pmatrix}$ Figure 9. We claim that
$P_{2,\epsilon}\sim B\oplus N_{1,\epsilon}$ and $N_{2,\epsilon}\sim B\oplus
P_{1,\epsilon}$.

Proof of 3: We first prove the claim for $P_{2,\epsilon}$. We perform a
sequence of elementary matrix operations in Figure 10. In Step 1, we
simultaneously clear the non-zero, non-diagonal entries of the first row and
column. In Step 2, we swap the second and third rows and columns. In Step 3,
we clear the non-zero, non-diagonal entries of the second row and column. In
Step 4, we clear the non-zero, non-diagonal entries of the third row and
column. The lower right block of the final matrix is exactly $N_{1}$.

The proof for $N_{2,\epsilon}$ is nearly identical, so we suppress the steps
for brevity. $\Box$

$\displaystyle P_{2,\epsilon}$
$\displaystyle=\begin{pmatrix}\begin{array}[]{ccc|ccc|c}-1&+1&0&0&0&0&+1\\\
+1&-1&+1&0&0&0&0\\\ 0&+1&-1&+1&0&0&0\\\ \hline\cr 0&0&+1&-1&+1&0&0\\\
0&0&0&+1&-1&+1&0\\\ 0&0&0&0&+1&-1&+1\\\ \hline\cr+1&0&0&0&0&+1&\epsilon\\\
\end{array}\end{pmatrix}\overset{1}{\sim}\begin{pmatrix}\begin{array}[]{ccc|ccc|c}-1&0&0&0&0&0&0\\\
0&0&+1&0&0&0&+1\\\ 0&+1&-1&+1&0&0&0\\\ \hline\cr 0&0&+1&-1&+1&0&0\\\
0&0&0&+1&-1&+1&0\\\ 0&0&0&0&+1&-1&+1\\\ \hline\cr 0&+1&0&0&0&+1&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{2}{\sim}\begin{pmatrix}\begin{array}[]{ccc|ccc|c}-1&0&0&0&0&0&0\\\
0&-1&+1&+1&0&0&0\\\ 0&+1&0&0&0&0&+1\\\ \hline\cr 0&+1&0&-1&+1&0&0\\\
0&0&0&+1&-1&+1&0\\\ 0&0&0&0&+1&-1&+1\\\ \hline\cr 0&0&+1&0&0&+1&\epsilon+1\\\
\end{array}\end{pmatrix}\overset{3}{\sim}\begin{pmatrix}\begin{array}[]{ccc|ccc|c}-1&0&0&0&0&0&0\\\
0&-1&0&0&0&0&0\\\ 0&0&+1&+1&0&0&+1\\\ \hline\cr 0&0&+1&0&+1&0&0\\\
0&0&0&+1&-1&+1&0\\\ 0&0&0&0&+1&-1&+1\\\ \hline\cr 0&0&+1&0&0&+1&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{4}{\sim}\begin{pmatrix}\begin{array}[]{ccc|ccc|c}-1&0&0&0&0&0&0\\\
0&-1&0&0&0&0&0\\\ 0&0&+1&0&0&0&0\\\ \hline\cr 0&0&0&-1&+1&0&-1\\\
0&0&0&+1&-1&+1&0\\\ 0&0&0&0&+1&-1&+1\\\ \hline\cr 0&0&0&-1&0&+1&\epsilon\\\
\end{array}\end{pmatrix}=\begin{pmatrix}\begin{array}[]{c|c}B&\\\
\hline\cr&N_{1,\epsilon}\end{array}\end{pmatrix}$ Figure 10. $P_{2,\epsilon}$
is a block diagonal matrix with $P_{2,\epsilon}\sim B\oplus N_{1,\epsilon}$.

2 and 3 establish the base cases for our induction. To prove the lemma, we
will use strong induction: we assume that for all $2\leq i\leq k$, we have
that $P_{i,\epsilon}$ and $N_{i,\epsilon}$ are congruent to block diagonal
matrices of the form in Figure 11. Therefore, to prove the claim, it suffices
to establish the inductive step.

$\displaystyle P_{i,\epsilon}\sim\begin{pmatrix}\begin{array}[]{c|c}B&\\\
\hline\cr&N_{i-1,\epsilon}\end{array}\end{pmatrix}\hskip
56.9055ptN_{i,\epsilon}\sim\begin{pmatrix}\begin{array}[]{c|c}B&\\\
\hline\cr&P_{i-1,\epsilon}\end{array}\end{pmatrix}$ Figure 11. We claim
$P_{i,\epsilon}\sim B\oplus N_{i-1,\epsilon}$ and $N_{i,\epsilon}\sim B\oplus
P_{i-1,\epsilon}$.

######  Claim 4.

For $k\geq 2$, the matrix $P_{k+1,\epsilon}$ (resp. $N_{k+1,\epsilon}$) is
congruent to the block diagonal matrix with factors $B$ and $N_{k,\epsilon}$
(resp. $P_{k,\epsilon}$).

Proof of 4: We prove the claim for $P_{k+1,\epsilon}$; the steps are in Figure
12. We entries that stay unchanged throughout the process are labelled with a
"$\star$". The suppressed block with the "$\star$" entries is a
$((3k-1)\times(3k-1))$ matrix. In Step 1, we simultaneously clear the non-
zero, non-diagonal entries of the first row/column. In Step 2, we swap the
second and third rows and columns. In Step 3, we clear the non-zero, non-
diagonal entries of the second row/column. In Step 4, we clear the non-zero,
non-diagonal entries of the third row/column. Finally, we redraw our
guidelines to see that the resulting matrix contains $B$ and $N_{k,\epsilon}$
as factors.

The analogous manipulation on $N_{k+1,\epsilon}$ yields the result. $\Box$

$\displaystyle P_{k+1,\epsilon}$
$\displaystyle=\begin{pmatrix}\begin{array}[]{cccc|ccc|c}-1&+1&&&&&&+1\\\
+1&-1&+1&&&&\\\ &+1&-1&+1&&&\\\ &&+1&-1&+1&&\\\
\hline\cr&&&+1&\star&\star&\star\\\ &&&&\star&\star&\star\\\
&&&&\star&\star&\star&+1\\\ \hline\cr+1&&&&&&+1&\epsilon\\\
\end{array}\end{pmatrix}\overset{1}{\sim}\begin{pmatrix}\begin{array}[]{cccc|ccc|c}-1&&&&&&&\\\
&&+1&&&&&+1\\\ &+1&-1&+1&&&\\\ &&+1&-1&+1&&\\\
\hline\cr&&&+1&\star&\star&\star\\\ &&&&\star&\star&\star\\\
&&&&\star&\star&\star&+1\\\ \hline\cr&+1&&&&&+1&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{2}{\sim}\begin{pmatrix}\begin{array}[]{cccc|ccc|c}-1&&&&&&&\\\
&-1&+1&+1&&&&\\\ &+1&&&&&&+1\\\ &+1&&-1&+1&&\\\
\hline\cr&&&+1&\star&\star&\star\\\ &&&&\star&\star&\star\\\
&&&&\star&\star&\star&+1\\\ \hline\cr&&+1&&&&+1&\epsilon+1\\\
\end{array}\end{pmatrix}\overset{3}{\sim}\begin{pmatrix}\begin{array}[]{cccc|ccc|c}-1&&&&&&&\\\
&-1&&&&&&\\\ &&+1&+1&&&&+1\\\ &&+1&&+1&&\\\
\hline\cr&&&+1&\star&\star&\star\\\ &&&&\star&\star&\star\\\
&&&&\star&\star&\star&+1\\\ \hline\cr&&+1&&&&+1&\epsilon+1\\\
\end{array}\end{pmatrix}$
$\displaystyle\overset{4}{\sim}\begin{pmatrix}\begin{array}[]{cccc|ccc|c}-1&&&&&&&\\\
&-1&&&&&&\\\ &&+1&&&&&\\\ &&&-1&+1&&&-1\\\ \hline\cr&&&+1&\star&\star&\star\\\
&&&&\star&\star&\star\\\ &&&&\star&\star&\star&+1\\\
\hline\cr&&&-1&&&+1&\epsilon\\\
\end{array}\end{pmatrix}=\begin{pmatrix}\begin{array}[]{ccc|cccc|c}-1&&&&&&&\\\
&-1&&&&&&\\\ &&+1&&&&&\\\ \hline\cr&&&-1&+1&&&-1\\\ &&&+1&\star&\star&\star\\\
&&&&\star&\star&\star\\\ &&&&\star&\star&\star&+1\\\
\hline\cr&&&-1&&&+1&\epsilon\\\ \end{array}\end{pmatrix}$ Figure 12. Proving
that $P_{i,\epsilon}\sim B\oplus N_{i-1,\epsilon}$.

We can now finish the proof of Lemma 4.10: for all $q\geq 1$, $P_{q,\epsilon}$
is congruent to a diagonal matrix with $q$-many $B$ factors, and a single
$1\times 1$ factor with entry $\epsilon\pm 2$. We begin by diagonalizing the
first three rows and columns of $P_{q,\epsilon}$ to produce a block diagonal
matrix with factors $B$ and $N_{q-1,\epsilon}$. Then, we factor
$N_{q-1,\epsilon}$ into a block diagonal matrix with factors $B$ and
$P_{q-2,\epsilon}$. Exhausting this procedure eventually yields a
factorization with $q$-many $B$ blocks. The remaining block is a $1\times 1$
matrix. When $q$ is even, this entry will be $\epsilon+2$, and when $q$ is
odd, the entry will be $\epsilon-2$. ∎

This concludes the proof of Lemma 4.10. We can now prove 4.7, which claimed
that $G(K)$ can be diagonalized.

###### Proof of 4.7..

Let $K$ be the specified knot, and construct a spanning surface and Goeritz
matrix for $K$ as in 4.6. Notice that $G(K)=P_{k,2m-1}$. By Lemma 4.10,
$P_{k,2m-1}$ is congruent to a diagonal matrix $D$, where
$D=(\bigoplus_{i=1}^{k}B)\oplus L$, where $L$ is a $1\times 1$ matrix with
entry $(2m-1)\pm 2$. This is what we wanted to show. ∎

######  Theorem 4.11.

Let $K=T(3,3k+1;2m)$ where $k\geq 1$ and $m\geq 0$. Then

$\displaystyle\sigma(K)=\begin{cases}-4k-2m-2&\qquad k\equiv 1\mod 2,\ \
m=0,1\\\ -4k-2m&\qquad\text{otherwise}\end{cases}$

###### Proof.

For the specified $K$, 4.7 tells us that the Goeritz matrix $G(K)$ is
congruent to $D=(\bigoplus_{i=1}^{k}B)\oplus L$, where $L$ is a $1\times 1$
matrix with entry $(2m-1)\pm 2$. The entry of $L$ is completely determined by
$m$ and the parity of $k$. In particular, $(2m-1)\pm 2<0$ if and only if
$m=0,1$ and $k$ is odd. Therefore, by applying Sylvester’s theorem of inertia,
we have:

(4.3)
$\displaystyle\text{sign}(G(K))=\text{sign}(P_{k,2m-1})=\text{sign}(D)=\begin{cases}-k-1&\qquad
k\equiv 1\mod 2,\ \ m=0,1\\\ -k+1&\qquad\text{otherwise}\end{cases}$

By Equation 4.2, $\mu(K)=(3k+1)+2m$. Combining this with
$\sigma(K)=\text{sign}(G(K))-\mu(K)$, we deduce the result. ∎

######  Remark 4.12.

Suppose $k\geq 1$ and $m\geq 0$. If $\sigma(T(3,3k+1;2m))=-4k-2m-2$, then by
4.11, $K$ is a torus knot: either $K\approx T(3,3k+1)$ or $K\approx
T(3,3k+1;2)$, which, by Lemma 4.3, is isotopic to $T(3,3k+2)$.

###  4.3. Signatures and the Seifert genus

The following lemmas study how the signatures of twisted torus knots of a
fixed genus are related.

######  Lemma 4.13.

Let $K_{1}=T(3,3k+1;2m)$ and $K_{2}=T(3,3(k-1)+1;2(m+3))$ where $m\geq 2$ and
$(3(k-1)+1)\geq 4$. Then $g(K_{1})=g(K_{2})$, but
$\sigma(K_{1})-\sigma(K_{2})=2$.

###### Proof.

As in the proof of 4.5, to prove that $g(K_{1})=g(K_{2})$, it suffices to
check that $K_{1}$ and $K_{2}$ have the same writhe:

$wr(K_{1})=2(3k+1)+2m=2(3k+1+3-3)+2m=2(3(k-1)+1)+6+2m=wr(K_{2})$

$K_{1}$ and $K_{2}$ both satisfy the hypotheses of 4.11. Since we assumed that
$m\geq 2$, the signatures of $K_{1}$ and $K_{2}$ have forms in the second case
of 4.11. Thus,

$\displaystyle\sigma(K_{1})-\sigma(K_{2})=(-4k-2m)-(-4(k-1)-2(m+3))=2$

This is what we wanted to show. ∎

######  Lemma 4.14.

Suppose $k\geq 1$. For $0\leq i\leq k-1$, let $K_{i}=T(3,3(k-i)+1;6i)$. The
knots $K_{i}$ have the same Seifert genus. Additionally, if $k\equiv 0\mod 2$,
then for all $i$, $\sigma(K_{i})-\sigma(K_{i+1})=2$, but if $k\equiv 1\mod 2$,
then $\sigma(K_{0})=\sigma(K_{1})$, and for all $i\geq 1$,
$\sigma(K_{i})-\sigma(K_{i+1})=2$.

###### Proof.

It is elementary to check that all the knots have the same writhe. Since they
are all positive 3-braid closures, it follows they have the same Seifert
genus.

Suppose $k$ is even. By applying the 4.11, we see that $\sigma(K_{0})=-4k$,
$\sigma(K_{1})=-4k-2$, and $\sigma(K_{2})=-4k-4$. In particular,
$\sigma(K_{0})-\sigma(K_{1})=2$ and $\sigma(K_{1})-\sigma(K_{2})=2$. Applying
Lemma 4.13 for $2\leq i\leq s-1$ yields the desired result.

Now suppose $k$ is odd. 4.11 tells us that $\sigma(K_{0})=-4k-2$,
$\sigma(K_{1})=-4k-2$, and $\sigma(K_{2})=-4k-4$. In particular,
$\sigma(K_{0})=\sigma(K_{1})$ and $\sigma(K_{1})-\sigma(K_{2})=2$. Applying
Lemma 4.13 for $2\leq i\leq s-1$ concludes the proof. ∎

######  Lemma 4.15.

Suppose $k\geq 1$. For $0\leq i\leq k-1$, let $K_{i}=T(3,3(k-i)+1;2+6i)$. The
knots $K_{i}$ have the same Seifert genus. Additionally, if $k\equiv 0\mod 2$,
then for all $i$, $\sigma(K_{i})-\sigma(K_{i+1})=2$, but if $k\equiv 1\mod 2$,
then $\sigma(K_{0})=\sigma(K_{1})$, and for all $i\geq 1$,
$\sigma(K_{i})-\sigma(K_{i+1})=2$.

###### Proof.

The proof is nearly identical to the one in Lemma 4.14. Note that if $k\equiv
1\mod 2$, then $\sigma(K_{0})=\sigma(K_{1})=-4k-4$. ∎

To prove 1.5, we need one last lemma which analyzes the case where a twisted
torus knot and a torus knot have the same Seifert genus and signature.

######  Lemma 4.16.

Suppose $K_{1}=T(3,q)$ is a torus knot, and $K_{2}$ is a twisted torus knot on
three strands such that $g(K_{1})=g(K_{2})$ and $\sigma(K_{1})=\sigma(K_{2})$.
The knots $K_{1}$ and $K_{2}$ are not concordant.

###### Proof.

Suppose $K_{1}$ and $K_{2}$ are a pair of knots satisfying the hypothesis of
the lemma. By Lemma 4.14 and Lemma 4.15, we must be in one of the following
two scenarios:

  1. (1)

$k$ is odd, with $K_{1}=T(3,3k+1;0)$ and $K_{2}=T(3,3(k-1)+1;6)$.

  2. (2)

$k$ is odd, with $K_{1}=T(3,3k+1;2)$ and $K_{2}=T(3,3(k-1)+1;8)$.

We first prove the lemma for Case (1). If $K_{1}$ is concordant to $K_{2}$,
then $K=K_{1}\\#m(K_{2}^{r})$ is slice. The Fox-Milnor criterion says that the
Alexander polynomial of $K$ would factor in a particular way: in particular,
$\Delta_{K}(t)=f(t)f(t^{-1})$. To obstruct a concordance between $K_{1}$ and
$K_{2}$, we will show that the putative factorization of Alexander polynomial
does not exist.

We recall that torus knots have simple Alexander polynomials [Rol76]: in
general,

$\displaystyle\Delta_{T(p,q)}(t)=\displaystyle\prod_{\begin{subarray}{c}h|p,\ell|q\\\
h,\ell\neq 1\end{subarray}}\phi_{\ell h}(t)$

where $\phi_{n}(t)$ denotes the $n$-th cyclotomic polynomial. Thus,
$\Delta_{K_{1}}(t)=\Delta_{T(3,q)}(t)=\displaystyle\prod_{\begin{subarray}{c}\ell|q,\
\ell\neq 1\end{subarray}}\phi_{3\ell}(t)$.

In particular, $\Delta_{K_{1}}(t)$ factors into a product of cyclotomic
polynomials, where there are no repeated factors. Therefore, to prove that
$\Delta_{K}(t)$ cannot factor as described by the Fox-Milnor criterion, it
suffices to prove that $\Delta_{K_{2}}(t)\neq\Delta_{K_{1}}(t)$. We will use
the reduced Burau matrix, which we described in Section 2. In particular, when
a knot is realized as the closure of some braid $B$, we have

$\frac{\det(I_{n-1}-B(t))}{1-t^{n}}=\frac{\Delta_{\widehat{\beta}}(t)}{1-t}$

$K_{1}$ and $K_{2}$ are realized on $n=3$ strands. So, to show that
$\Delta_{K_{1}}(t)\neq\Delta_{K_{2}}(t)$, we must prove that:

(4.4) $\displaystyle\det(I_{2}-B_{1}(t))\neq\det(I_{2}-B_{2}(t))$

where $B_{1}(t)$ and $B_{2}(t)$ denote the reduced Burau matrices of
$\beta_{1}$ and $\beta_{2}$, respectively, where $\widehat{\beta_{i}}\approx
K_{i}$. Additionally, we notice that
$\det(I_{2}-B_{i}(t))=q_{B_{i}(t)}(1)=1-\text{tr}(B_{i}(t))+\det(B_{i}(t))$,
where $q_{B(t)}(x)$ is the characteristic-type polynomial defined in Equation
2.1. Since $\beta_{1}$ and $\beta_{2}$ are both positive 3-braids with the
same writhe, $\det(B_{1}(t))=\det(B_{2}(t))$. Therefore, to deduce that (4.4)
holds, it suffices to prove that $\text{tr}(B_{1}(t))\neq\text{tr}(B_{2}(t))$.

Recall that

$\displaystyle\sigma_{1}(t)$
$\displaystyle=\begin{pmatrix}\begin{array}[]{cc}-t&1\\\
0&1\end{array}\end{pmatrix}\hskip 56.9055pt\text{ and }\hskip
56.9055pt\sigma_{2}(t)=\begin{pmatrix}\begin{array}[]{cc}1&0\\\
t&-t\end{array}\end{pmatrix}$

Therefore, we know:

$\displaystyle\sigma_{2}(t)\sigma_{1}(t)=\begin{pmatrix}\begin{array}[]{cc}-t&1\\\
-t^{2}&0\end{array}\end{pmatrix}\text{, }\hskip
10.00002pt(\sigma_{2}(t)\sigma_{1}(t))^{3}=\begin{pmatrix}\begin{array}[]{cc}t^{3}&0\\\
0&t^{3}\end{array}\end{pmatrix}\text{, \ \ }\hskip
5.0pt(\sigma_{1})^{6}=\begin{pmatrix}\begin{array}[]{cc}t^{6}&-t^{5}+t^{4}-t^{3}+t^{2}-t+1\\\
0&1\end{array}\end{pmatrix}$

This allows us to compute $B_{1}(t)$ and $B_{2}(t)$.

$\displaystyle B_{1}(t)=(\sigma_{2}(t)\sigma_{1}(t))^{3k+1}$
$\displaystyle=\begin{pmatrix}\begin{array}[]{cc}t^{3}&0\\\
0&t^{3}\end{array}\end{pmatrix}^{k}\cdot\begin{pmatrix}\begin{array}[]{cc}-t&1\\\
-t^{2}&0\end{array}\end{pmatrix}=\begin{pmatrix}\begin{array}[]{cc}-t^{3k+1}&t^{3k}\\\
-t^{3k+2}&0\end{array}\end{pmatrix}$ $\displaystyle
B_{2}(t)=(\sigma_{2}(t)\sigma_{1}(t))^{3(k-1)+1}(\sigma_{1})^{6}$
$\displaystyle=\begin{pmatrix}\begin{array}[]{cc}t^{3}&0\\\
0&t^{3}\end{array}\end{pmatrix}^{k-1}\cdot\begin{pmatrix}\begin{array}[]{cc}-t&1\\\
-t^{2}&0\end{array}\end{pmatrix}\cdot\begin{pmatrix}\begin{array}[]{cc}t^{6}&-t^{5}+t^{4}-t^{3}+t^{2}-t+1\\\
0&1\end{array}\end{pmatrix}$
$\displaystyle=t^{3(k-1)}\cdot\begin{pmatrix}\begin{array}[]{cc}-t^{7}&t^{6}-t^{5}+t^{4}-t^{3}+t^{2}-t+1\\\
-t^{8}&t^{7}-t^{6}+t^{5}-t^{4}+t^{3}-t^{2}\end{array}\end{pmatrix}$

It is immediate that $\text{tr}(B_{1}(t))\neq\text{tr}(B_{2}(t))$. Therefore,
we see that the quantities $\det(I_{2}-B_{1}(t))$ and $\det(I_{2}-B_{2}(t))$
do not agree, thus $\Delta_{K_{1}}(t)\neq\Delta_{K_{2}}(t)$, and
$\Delta_{K}(t)$ cannot factor as $f(t)f(t^{-1})$. Therefore, in Case (1), the
knots $K_{1}$ and $K_{2}$ cannot be smoothly concordant.

An analogous argument works for Case (2), which we verify below. Recall that,
by Lemma 4.3, $T(3,3k+1;2)\approx T(3,3k+2)$. So, in this case, we have:

$\displaystyle B_{1}(t)$
$\displaystyle=(\sigma_{2}(t)\sigma_{1}(t))^{3k+2}=t^{3k}\begin{pmatrix}\begin{array}[]{cc}-t&1\\\
-t^{2}&0\end{array}\end{pmatrix}^{2}=\begin{pmatrix}\begin{array}[]{cc}0&-t^{3k+1}\\\
t^{3k+3}&-t^{3k+2}\end{array}\end{pmatrix}$ $\displaystyle B_{2}(t)$
$\displaystyle=(\sigma_{2}(t)\sigma_{1}(t))^{3(k-1)+1}(\sigma_{1})^{8}=t^{3(k-1)}\cdot\begin{pmatrix}\begin{array}[]{cc}-t&1\\\
-t^{2}&0\end{array}\end{pmatrix}\cdot\begin{pmatrix}\begin{array}[]{cc}t^{8}&-t^{7}+t^{6}-t^{5}+t^{4}-t^{3}+t^{2}-t+1\\\
0&1\end{array}\end{pmatrix}$

Once again, the quantities $\text{tr}(B_{1}(t))$ and $\text{tr}(B_{2}(t))$
cannot agree, so neither can the corresponding Alexander polynomials, and the
knots cannot be concordant. ∎

We can now prove our last main result.

1.5 Let $\mathcal{S}=\mathcal{T}\cup\mathcal{L}$, where $\mathcal{T}$ is the
set of all positive torus knots, and $\mathcal{L}$ is the set of L-space knots
of braid index three. Every concordance contains at most one knot from
$\mathcal{S}$.

As we will see, the knot invariants $i(K)$, $\tau(K)$, and $\sigma(K)$, when
combined, are sufficient to show that every knot in $\mathcal{S}$ lies in a
distinct concordance class.

###### Proof.

Let $\mathcal{S}$ be the the set of knots stated above. Every knot
$K\in\mathcal{S}$ is a twist positive L-space knot, so by 1.3, they each
realize the minimal braid index of their concordance classes. Additionally,
notice that $\mathcal{T}\cap\mathcal{L}=\\{T(3,q)\ |\ q\geq 4\\}$.

First, we notice that if the positive torus knots $T_{1}$ and $T_{2}$ are
concordant to each other, then by 1.3, they have the same braid index.
Therefore, $T_{1}=T(p,q_{1})$ and $T_{2}=T(p,q_{2})$ for some $p\geq
2,q_{i}\geq 3$. However, if they are concordant, then
$\tau(T_{1})=\tau(T_{2})$. Since $T_{1}$ and $T_{2}$ are fibered, strongly
quasipositive knots, then by [Hed10], $\tau(T_{1})=g(T_{1})$ and
$\tau(T_{2})=g(T_{2})$. Since $g(T(r,s))=\frac{(r-1)(s-1)}{2}$, we deduce that
$q_{1}=q_{2}$, hence $T_{1}\approx T_{2}$. Therefore, all the elements of
$\mathcal{T}$ represent distinct concordance classes.

It remains to study how the elements of $\mathcal{L}$ interact with each
other, and the elements of $\mathcal{T}$. If $K\in\mathcal{L}$, then by 1.3,
the only knots in $\mathcal{S}$ to which $K$ could be concordant must also
have braid index 3, so $K$ could only be concordant to another knot in
$\mathcal{L}$. Thus, it suffices to show that every $K\in\mathcal{L}$
represents a distinct concordance class.

We begin by sorting all knots in $\mathcal{L}$ by their $\tau$ invariants.
Since all knots under consideration are fibered and strongly quasipositive, we
can instead recast this as a sorting by Seifert genus. For all $g\geq 3$,
there exists a unique torus knot $T(3,q)$ of genus $g$. As we proved in 4.5,
there will be $s=\lfloor\frac{q}{3}\rfloor-1$ twisted torus knots which could
be potentially concordant to it. We enumerate these knots as follows:
$K_{0}=T(3,q),\ldots,K_{s}\in\\{T(3,4;2(3s)),T(3,4;2(3s+1))\\}$, where (as in
as in Lemmas 4.14 and 4.15) $K_{s}$ is determined by the congruence class of
$q\mod 3$.

By combining Lemmas 4.13, 4.14, 4.15, we see that all the knots potentially
concordant to $K_{0}$ all have pairwise distinct signatures (hence they must
represent pairwise distinct concordance classes), or there is exactly one pair
of knots with the same signature; however, if the latter occurs, we can apply
Lemma 4.16, and deduce that none of the knots with the same $\tau$ invariant
as $K_{0}$ can be concordant to it. Thus, all the knots
$K_{0},K_{1},\ldots,K_{s}$ must lie in distinct concordance classes.

Therefore, every knot $K\in L$ lies in a unique concordance class, and these
must be distinct from the classes represented by $\mathcal{T}$. Thus, every
element of $S$ represents a distinct concordance class. ∎

######  Corollary 4.17.

$\mathcal{S}$ is an infinite family of positive braid knots where each knot
represents a distinct concordance class, and as $g\to\infty$, the number of
hyperbolic knots of genus $g$ gets arbitrarily large.

###### Proof.

This follows by combining 1.5, Lemma 4.2, and 4.5. ∎

##  5\. Discussion

In this section, we justify some of the claims we made in the introduction.
Namely, we provide evidence for 1.7: for any L-space knot, the braid index and
bridge index agree.

######  Proposition 5.1.

Let $K_{n},n\geq 1$ be the infinite family of hyperbolic L-space knots defined
by Baker-Kegel [BK22, Section 2]. Then, for all $n$, $br(K_{n})=i(K_{n})=4$.

###### Proof.

Baker-Kegel define the knot $K_{n}$ to be to be the closure of the 4-braid
$\beta_{n}$, where

$\beta_{n}=(\sigma_{2}\sigma_{1}\sigma_{3}\sigma_{2})^{2n+1}{\sigma_{1}}^{-1}\sigma_{2}\sigma_{1}\sigma_{1}\sigma_{2}$

In particular, for all $n\geq 1$, $i(K_{n})\leq 4$. By inspecting [BK22,
Proposition 3.1 (3)], we see that

$\Delta_{K_{n}}(t)=1-t+t^{4}+t^{5}+\ldots+t^{4n+2}+\ldots-t^{8n+3}+t^{8n+2}$

In particular, we see that 3 appears as a gap in the exponents of the
Alexander polynomial of these knots. Thus, as in the proof of 1.3,

$3\leq Ord_{v}(K_{n})\leq br(K_{n})-1\leq i(K_{n})-1\leq 3$

In particular, we must have equalities throughout, and $br(K_{n})=i(K_{n})=4$.
∎

To conclude, we justify why 1.7 holds for certain cables of L-space knots.

######  Proposition 5.2.

If 1.7 holds for an L-space knot $K$, then it holds for all of its L-space
cables.

###### Proof.

Schultens [Sch07] reproves a classical result of Schubert [Sch54]: suppose $S$
is a satellite knot with companion $J$ and pattern $P$ of wrapping number $k$
(the wrapping number is the minimal geometric intersection number of the
pattern with a meridional disk of the pattern torus), then $br(S)\geq k\cdot
br(J)$. In particular, if $K_{p,q}$ is a cable of $K$, then $br(K_{p,q})\geq
p\cdot br(K)$. Williams [Wil92] proved that analogous result holds the braid
index: $i(K_{p,q})=p\cdot i(K)$. Therefore, we have:

(5.1) $\displaystyle p\cdot br(K)\leq br(K_{p,q})\leq i(K_{p,q})=p\cdot i(K)$

If $K$ is an L-space knot with $i(K)=br(K)$, then we have equalities
throughout Equation 5.1, proving the claim. ∎

We hope these observations provide some motivation for investigating 1.7
further.

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