# Thin knots and the Cabling Conjecture

Robert DeYeso III

###### Abstract.

The Cabling Conjecture of González-Acuña and Short holds that only cable knots
admit Dehn surgery to a manifold containing an essential sphere. We approach
this conjecture for thin knots using Heegaard Floer homology, primarily via
immersed curves techniques inspired by Hanselman’s work on the Cosmetic
Surgery Conjecture. We show that almost all thin knots satisfy the Cabling
Conjecture, with possible exception coming from a (conjecturally non-existent)
collection of thin, hyperbolic, L-space knots. This result serves as a reproof
that the Cabling Conjecture is satisfied by alternating knots, and also a new
proof that thin, slice knots satisfy the Cabling Conjecture.

##  1\. Introduction

For a knot $K$ in $S^{3}$, let $S^{3}_{r}(K)$ denote $r$-sloped Dehn surgery
along $K$. If $S^{3}_{r}(K)$ is a reducible manifold, meaning it contains an
essential $2$-sphere, we will call $r$ a reducing slope. The primary example
of a reducible surgery to keep in mind is when $K$ is the $(p,q)$-cable of
some knot $K^{\prime}$ and $r$ is given by the cabling annulus. In this case,
we have $S^{3}_{pq}(K)\cong L(p,q)\\#S^{3}_{\frac{q}{p}}(K^{\prime})$. The
Cabling Conjecture asserts that this is the only example of a reducible
surgery.

######  Conjecture 1.1 (Cabling Conjecture, Gonzalez-Acuña – Short [GAnS86]).

If $K$ is a knot in $S^{3}$ which has a reducible surgery, then $K$ is a
cabled knot and the reducing slope is given by the cabling annulus.

The Cabling Conjecture is satisfied by many classes of knots. Torus knots, as
cables of the unknot, were shown to satisfy the conjecture in [Mos71].
Additionally, satellite [Sch90] and alternating knots [MT92] satisfy the
conjecture as well. Then to establish this conjecture in full, it remains to
show that it is satisfied by hyperbolic knots. Our aim is shorter, as we
consider thin, hyperbolic knots.

We will present knot Floer homology in more detail in Section 2, but for now
recall that $\widehat{\textit{HFK}}(K)$ with coefficients in $\mathbb{F}_{2}$
is bigraded with Alexander and Maslov gradings, respectively $A$ and $M$. A
knot $K$ is Floer homologically thin if the generators of
$\widehat{\textit{HFK}}(K)$ all have the same $\delta=A-M$ grading. This
family contains alternating knots [OS03b], and the more generalized quasi-
alternating knots [OS05b]. We say $K$ is an $L$-space knot if it admits a
surgery to a (Heegaard Floer) $L$-space, which is a manifold with the simplest
Heegaard Floer homology. Using Heegaard Floer homology via immersed curves
techniques, we show that

######  Theorem 1.2.

If a thin, hyperbolic knot $K$ in $S^{3}$ admits a reducible surgery, then $K$
is an L-space knot and the reducing slope must be $r=2g(K)-1$ after mirroring
$K$ if necessary.

It is conjectured that the only thin, L-space knots are the torus knots
$T(2,n)$. Provided this is true, there would not exist thin, hyperbolic,
L-space knots and so Theorem 1.2 would show that all thin knots satisfy the
Cabling Conjecture. While stated for thin, hyperbolic knots, this theorem
holds more generally for non-cabled knots. This is because we use the
Matignon-Sayari genus bound, stated blow, to only need to consider $r\leq
2g(K)-1$. The case where $r>2g(K)-1$ can be handled using the techniques in
this paper to conclude that $K=T(2,n)$, but perhaps more immediate is the
result of Dey that cables of non-trivial knots are not thin [Dey19]. Since the
only alternating, $L$-space knots are the $T(2,n)$’s [OS05a], Theorem 1.2
provides an immersed curves reproof that alternating knots satisfy the Cabling
Conjecture.

######  Corollary 1.3.

Alternating knots satisfy the Cabling Conjecture.

The next corollary follows because the only thin, slice, $L$-space knot is the
unknot.

######  Corollary 1.4.

Thin, slice knots satisfy the Cabling Conjecture.

######  Corollary 1.5.

Thin knots cannot admit two reducible surgeries.

Part of the proof strategy for Theorem 1.2 involves obstructing an
$\mathbb{R}P^{3}$ connected summand, and so we get the following corollary
with identical proof to that of [HLZ15, Corollary 1.5].

######  Corollary 1.6.

If $K$ is a thin, hyperbolic knot, then $S^{3}\setminus\nu K$ does not contain
properly embedded punctured projective planes.

When $K$ is a non-trivial knot in $S^{3}$ with reducible surgery
$S^{3}_{r}(K)$, the surgery decomposes as a connected sum and the reducing
slope satisfies $r\neq 0$ [Gab87]. We saw from the cabled knot example that
the reducing slope is an integer and one of the connected summands is a lens
space. The former and latter conditions occur for all reducible surgeries due
to [GL87] and [GL89], respectively. A reducible surgery can admit at most
three connected summands due to the combined efforts of [Say98][VS99][How02],
in which case two summands are lens spaces and the remaining summand is an
integer homology sphere. Since $S^{3}_{r}(K)$ must have a non-trivial lens
space summand, the integral reducing slope $r$ satisifes $r\neq-1,0,1$. In
[MS03], Matignon and Sayari provide the following genus bound if $K$ is non-
cabled:

$1<|r|\leq 2g(K)-1.$

Heegaard Floer homology satisfies a Künneth formula for connected sums, and
has proved very useful in general for studying Dehn surgery. If surgery along
$K$ produces precisely a connected sum of two lens spaces, then $K$ must be a
cabled knot due to [Gre15]. Further, [Gre15] together with [BZ98] shows that a
hyperbolic knot in $S^{3}$ cannot admit both a lens space surgery and a
reducible surgery. Hom, Lidman, and Zufelt showed that a hyperbolic, $L$-space
knot can admit at most one reducing slope, and the slope must be $2g(K)-1$
after mirroring the knot to make the slope positive [HLZ15]. They also
established a periodicity structure to the Heegaard Floer homology of a
reducible surgery, which is invaluable to the proof strategy of Theorem 1.2.
We will involve these constraints via bordered invariants in the form of
immersed curves.

Lipshitz, Ozsváth, and Thurston introduced bordered Heegaard Floer invariants
for manifolds with torus boundary in [LOT18]. With $M_{0}\cup_{h}M_{1}$
denoting a gluing of two manifolds along their torus boundaries, they prove a
pairing theorem involving the two bordered invariants that recovers the
Heegaard Floer homology of such a gluing. Hanselman, Rasmussen, and Watson
reinterpreted these bordered invariants as collections of immersed curves in
the punctured torus, and proved an analogous pairing theorem. In [Han19],
Hanselman used this package to obtain obstructions for cosmetic surgeries
along knots in $S^{3}$, and our approach in this paper is largely inspired by
this work.

## Organization

We only consider surgeries with positive slopes, and mirror knots to achieve
this whenever necessary. All manifolds are assumed to be compact, connected,
oriented 3-manifolds, unless stated otherwise, and the coefficients in Floer
homology are taken to belong to $\mathbb{F}=\mathbb{F}_{2}$. We will denote
closed manifolds by $X$ or $Y$, and manifolds with (typically torus) boundary
by $M$. Figures containing immersed curves invariants will have the curves for
$S^{3}\setminus\nu K$ in blue and the curves for the filling solid torus in
red/purple.
Section 2 summarizes the relevant background from knot Floer homology and
Heegaard Floer homology. It also contains an overview of immersed curves
invariants, their general properties and form for thin knots, as well as their
associated pairing theorem and how to compute Maslov grading differences.
Section 3 expands on the relative Maslov grading for immersed curves
invariants of complements of thin knots. Along the way we set up formulas for
components of the grading difference formula in terms of $\tau(K)$.
Section 4 uses these relations to generate obstructions to periodicity for
various cases of $r$ in relation to $\tau(K)$ and $g(K)$. It hosts a sizable
collection of lemmas for the cases with $|\tau(K)|<g(K)$. Section 5 resolves
the remaining cases where $|\tau(K)|=g(K)$, including some that use absolute
grading information. Afterward, all lemmas are collected to handle the proof
of the theorem.

## Acknowledgements

I am grateful to Tye Lidman for their unending encouragement, patience, and
insight as an advisor. I was partially supported by NSF grant DMS-1709702.

##  2\. Background Material

We will assume the reader is familiar with the $\widehat{\textit{HF}}$ and
$\textit{HF}^{+}$ constructions of Heegaard Floer homology for 3-manifolds
[OS04d], and knot Floer homology $\widehat{\textit{HFK}}$ for knots in $S^{3}$
(with associated full knot Floer complex $\textit{CFK}^{\infty}$) [OS04b,
Ras03].

###  2.1. $\widehat{\textit{HF}}$ for reducible surgeries

Let us identify $\text{Spin}^{c}(S^{3}_{r}(K))$ with $\mathbb{Z}/r\mathbb{Z}$
as in [OS08, Subsection 2.4], and denote the correspondence using
$[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$ for $[s]\in\mathbb{Z}/r\mathbb{Z}$. We
will also choose equivalence classes for elements of $\mathbb{Z}/r\mathbb{Z}$
as centered about $0$, so that for example
$\mathbb{Z}/r\mathbb{Z}=\left\\{-\dfrac{r-1}{2},\dots,0,\dots,\dfrac{r-1}{2}\right\\}$
if $r$ is odd. As an abuse of notation, we will commonly use $s$ for the
representative of $[s]$ that falls within this range.

The following lemma is a simplified version of a more general Floer homology
periodicity result for $HF^{+}$ of a general reducible 3-manifold from
[HLZ15]. Basically, we should expect to see repeated behavior among the
$\text{spin}^{c}$ summands of $\widehat{\textit{HF}}(S^{3}_{r}(K))$ if the
surgery is reducible.

######  Lemma 2.1.

Suppose $S^{3}_{r}(K)\cong X\\#Y$, where $X$ is an L-space and
$|H^{2}(Y)|=k<\infty$. Then for any $[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$ and
$\alpha\in H^{2}(S^{3}_{r}(K))\cong\mathbb{Z}/r\mathbb{Z}$, we have
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s+k\alpha])\cong\widehat{\textit{HF}}(S^{3}_{r}(K),[s])$
as relatively-graded $\mathbb{F}$ vector spaces.

###### Proof.

Let $[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$ restrict to
$[s_{i}]\in\text{Spin}^{c}(X)$ and $[s_{j}]\in\text{Spin}^{c}(Y)$. We see that
$\widehat{\textit{HF}}(X,[s_{i}])\cong\mathbb{F}$ since $X$ is an L-space, and
so the Künneth formula for $\widehat{\textit{HF}}$ [OS04c, Theorem 1.5]
implies

$\displaystyle\widehat{\textit{HF}}(S^{3}_{r}(K),[s])$ $\displaystyle\cong
H_{\ast}(\widehat{CF}(X,[s_{i}])\otimes_{\mathbb{F}}\widehat{CF}(Y,[s_{j}]))$
$\displaystyle\cong\widehat{\textit{HF}}(Y,[s_{j}]).$

For any $\alpha\in\mathbb{Z}/r\mathbb{Z}$, we have that $[s+k\alpha]$
restricts to $[s_{j}]$ in $\text{Spin}^{c}(Y)$. Then because
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s])$ is independent of $[s_{i}]$, we
obtain

$\widehat{\textit{HF}}(S^{3}_{r}(K),[s+k\alpha])\cong\widehat{\textit{HF}}(Y,[s_{j}])\cong\widehat{\textit{HF}}(S^{3}_{r}(K),[s])$

as relatively-graded $\mathbb{F}$ vector spaces. ∎

We also need to gather some integral invariants of $K$ involved with the
Mapping Cone Formula that relates $\textit{CFK}^{\infty}(K)$ to
$\textit{HF}^{+}(S^{3}_{r}(K)$ [OS08]. For $s\in\mathbb{Z}$, recall the
subcomplexes and quotient complexes of the
$\mathbb{Z}\oplus\mathbb{Z}$-filtered full knot Floer complex
$\textit{CFK}^{\infty}(K)$

$\displaystyle\mathcal{A}^{+}_{s}$
$\displaystyle=C\left\\{\text{max}\left\\{i,j-s\right\\}\geq 0\right\\},$
$\displaystyle\mathcal{B}^{+}_{s}$ $\displaystyle=C\left\\{i\geq 0\right\\}.$

Notice $\mathcal{B}^{+}_{s}\cong\textit{CF}^{+}(S^{3})$ by definition. There
are also chain maps
$\mathfrak{v}^{+}_{s}:\mathcal{A}^{+}_{s}\rightarrow\mathcal{B}^{+}_{s}$ and
$\mathfrak{h}^{+}_{s}:\mathcal{A}^{+}_{s}\rightarrow\mathcal{B}^{+}_{s+r}$
between these subcomplexes. Take homology to obtain
$A^{+}_{s}=H_{\ast}(\mathcal{A}^{+}_{s})$ and
$B^{+}_{s}=H_{\ast}(\mathcal{B}^{+}_{s})\cong\textit{HF}^{+}(S^{3})$, and
induced maps $v^{+}_{s}$ and $h^{+}_{s}$. Let $\mathcal{T}^{+}$ denote
$\textit{HF}^{+}(S^{3})$, and notice that
$U^{N}(A^{+}_{s})\cong\mathcal{T}^{+}$ for arbitrarily large $N$. By
restricting both $v^{+}_{s}$ and $h^{+}_{s}$ to this submodule, we obtain
$\overline{v}^{+}_{s}$ and $\overline{h}^{+}_{s}$. The integral invariants of
$K$ that we desire are due to [NW15], and are defined by

$\displaystyle V_{s}$
$\displaystyle=\text{rank}(\text{ker}\overline{v}^{+}_{s}),$ $\displaystyle
H_{s}$ $\displaystyle=\text{rank}(\text{ker}\overline{h}^{+}_{s}).$

These terms have simple behaviour when $K$ is alternating because of the
“staircase” part of $\textit{CFK}^{\infty}(K)$ due to [OS03b]. This holds more
generally for thin knots due to [Pet13], but we will have an alternative
geometric way of computing these terms later in Subsection 2.2. By [HLZ15,
Lemma 2.3], the maps $v^{+}_{s}$ and $h^{+}_{-s}$ agree on homology after
identifying $A_{s}^{+}\cong A_{-s}^{+}$ (essentially reversing the roles of
$i$ and $j$ above) so that $V_{s}=H_{-s}$. They are by definition non-
negative, and also satisfy the following lemma.

######  Lemma 2.2 ([NW15, Lemma 2.4]).

The $V_{s}$ form a non-increasing sequence and the $H_{s}$ form a non-
decreasing sequence, so that

$V_{s}\geq V_{s+1}\,\,\text{and}\,\,H_{s}\leq H_{s+1}\,\,\text{for
all}\,\,s\in\mathbb{Z}.$

For a rational homology sphere $Y$, we can write
$\textit{HF}^{+}(Y,\mathfrak{s})\cong\mathcal{T}^{+}\oplus\textit{HF}_{red}(Y,\mathfrak{s})$,
where $\mathcal{T}^{+}\cong\mathbb{F}[U,U^{-1}]/\mathbb{F}[U]$ denotes the
“tower” submodule. The $d$-invariants $d(Y,\mathfrak{s})$, sometimes called
the Heegaard Floer correction terms, record the smallest absolutely graded
element of $\mathcal{T}^{+}\subseteq\textit{HF}(Y,\mathfrak{s})$ [OS03a].
These invariants satisfy a few symmetries, such as $\text{spin}^{c}$
conjugation symmetry $d(Y,\mathfrak{s})=d(Y,\overline{\mathfrak{s}})$ and
orientation-reversal $d(-Y,\mathfrak{s})=-d(Y,\mathfrak{s})$, as well as
additivity for connected sums. It is normalized so that
$d(S^{3},\mathfrak{s}_{0})=0$, and is recursively determined for lens spaces
in [OS03a, Proposition 4.8]. In [NW15], the $d$-invariants of rational
surgeries are related to those of $d(L(p,q),[s])$ and the $H$’s and $V$’s. We
state a special case of the more general result for our purposes.

######  Proposition 2.3 ([NW15, Proposition 1.6]).

Suppose $r$ is integral and positive, and fix $0\leq s<r-1$. Then

$d(S^{3}_{r}(K),[s])=d(L(r,1),[s])-2\,\text{max}\left\\{V_{s},V_{r-s}\right\\}.$

Among many of its applications, this result enables the following lemma.

######  Lemma 2.4 ([HLZ15, Lemma 2.5]).

For all $s\in\mathbb{Z}$, the integers $V_{s}$ and $H_{s}$ are related by

$H_{s}-V_{s}=s.$

We will involve the $d$-invariants later in Section 5.1 when necessary.

###  2.2. $\widehat{\textit{HF}}$ via immersed curves

Figure 1. Edges of a grading arrow either follow or oppose the orientations of
the attached curve components.

Bordered Heegaard Floer homology, introduced by Lipshitz, Ozsváth, and
Thurston, provides a cut-and-paste style of computing $\widehat{\textit{HF}}$
for a 3-manifold. This is done by decomposing along a surface, and then
recovering Floer homology by a suitable means of pairing the relative Floer
invariants for the decomposed pieces [LOT18]. Dubbed the pairing theorem, we
will invoke it in immersed curves form due to Hanselman, Rasmussen, and Watson
[HRW16, HRW18]. Let $M$ be an orientable 3-manifold with torus boundary.
Choose $\alpha,\beta$ in $\partial M$ so that $(\alpha,\beta)$ forms a
parameterization of $\partial M$, and fix a basepoint $z\in\partial M$. The
invariant $\widehat{\textit{HF}}(M)$ is a collection of immersed curves in
$T_{M}=\partial M\setminus z$, possibly decorated with local systems, and up
to regular homotopy of the curves. When $M=S^{3}\setminus\nu K$, we will use a
preferred meridian-longitude basis $\left\\{\mu,\lambda\right\\}$.

The manifolds in this paper all happen to be loop type (see [HW15]), which
means that their associated immersed curves invariant has trivial local
systems. If the invariant has multiple curve components, then they are
connected by pairs of edges which we denote with a grading arrow. These are
presented in Figure 1, and they carry an integral weight $m$ useful for
determining Maslov grading differences. Let $M=S^{3}\setminus\nu K$. We can
lift $\widehat{\textit{HF}}(M)$ to the infinite cylindrical cover
$\overline{T}_{M}$, where each lifted marked point resides within a
neighborhood of the lift of the meridian $\overline{\mu}$. One of the curves
wraps around the cylinder, and we will use $\overline{\gamma}$ to denote this
component. Figure 2 shows a centered lift of the invariant for the complement
of a hypothetical example of a thin knot $K$ with $g(K)=2$ and $\tau(K)=1$.

2pt $\overline{\gamma}$ at 30 120 $\mathbf{m}_{1}$ at 105 205
$\mathbf{m}_{-1}$ at 23 35

Figure 2. An example of $\widehat{\textit{HF}}(M)$ for $g(K)=2$ and
$\tau(K)=1$.

The lifts of the marked points will be taken to lie at purely half-integral
heights, so that curve components cross at integral heights. With this at
hand, the lifted curve invariant also encodes a few numerical and concordance
invariants of $K$. For example, the Seifert genus is given by the height of
the tallest curve component. Additionally, the height around which
$\overline{\gamma}$ wraps is precisely the Oszváth-Szabó invariant $\tau(K)$.
Hom’s $\epsilon$ invariant may also be determined by observing what
$\overline{\gamma}$ does next. It curve turns downwards, upwards, or continues
straight corresponding to $\epsilon(K)$ being 1, -1, and 0, respectively.
These two invariants determine the slope $\overline{\gamma}$ outside of a thin
vertical strip surrounding the lifts of the marked point, given by
$2\tau(K)-\epsilon(K)$.

Recall that $\widehat{\textit{HFK}}(K)$ detects $g(K)$ due to [OS04a]. Looking
in $\overline{T}_{M}$, genus detection manifests itself in
$\widehat{\textit{HF}}(M)$ by ensuring that some curve component crosses at
height $g(K)$. The immersed curves also satisfy a very powerful constraint
related to a conjugation symmetry. For invariants of knot complements of
$S^{3}$, this means that the curves are invariant under rotation by $\pi$.

######  Theorem 2.5 ([HRW18, Theorem 7]).

The invariant $\widehat{\textit{HF}}(M)$ is symmetric under the elliptic
involution of $\partial M$. Here, the involution is chosen so that $z$ is a
fixed point.

If $K$ admits either a horizontally or vertically simplified basis for
$\textit{CFK}^{-}(K)$, then the procedure of [HRW18, Proposition 47] allows
one to construct $\widehat{\textit{HF}}(M)$ from $\textit{CFK}^{-}(K)$. The
special case when $\textit{CFK}^{-}(K)$ is both horizontally and vertically
simplified enables us to quickly generate these curves. This condition holds
if $K$ is a thin knot, and in particular we have that every arrow encoding the
differential in $\textit{CFK}^{-}(K)$ has length one. These properties imply
two features of $\widehat{\textit{HF}}(M)$, which are effectively the immersed
curves analog of [Pet13, Lemma 7]:

  * •

The essential component $\overline{\gamma}$ winds between adjacent basepoints
determined by $\tau(K)$, before ultimately wrapping around the cylinder.

  * •

Every other component is a simple figure-eight, enclosing vertically adjacent
basepoints.

######  Definition 2.6.

Let $e_{n}$ denote the number of simple figure-eight components at height $n$
of $\widehat{\textit{HF}}(M)$, viewed in $\overline{T}_{M}$.

We have $e_{-n}=e_{n}$ due to Theorem 2.5, and Figure 2 provides an example
with $e_{0}=0$ and $e_{-1}=e_{1}=1$. With individual properties of the curve
invariants handled, we turn to the main reason for their involvement.

######  Theorem 2.7 ([HRW18, Theorem 2]).

Consider the gluing $M_{1}\cup_{h}M_{2}$, where the $M_{i}$ are compact,
oriented 3-manifolds with torus boundary and $h:\partial
M_{2}\rightarrow\partial M_{1}$ is an orientation reversing homeomorphism for
which $h(z_{2})=z_{1}$. Then

$\widehat{\textit{HF}}(M_{1}\cup_{h}M_{2})\cong\textit{HF}(\widehat{\textit{HF}}(M_{1}),h(\widehat{\textit{HF}}(M_{2}))),$

where intersection Floer homology is computed in $T_{M_{1}}$ and the
isomorphism is one of relatively graded vector spaces that respects the
$\text{Spin}^{c}$ decomposition.

More precisely,
$\textit{HF}(\widehat{\textit{HF}}(M_{1}),h(\widehat{\textit{HF}}(M_{2})))$
decomposes over $\text{spin}^{c}$ structures and carries a relative Maslov
grading on each $\text{spin}^{c}$ summand. Theorem 2.7 places these in
correspondence with the $\text{spin}^{c}$ decomposition on
$\widehat{\textit{HF}}(M_{1}\cup_{h}M_{2})$, and also ensures the relative
Maslov gradings agree. This is best seen when viewing Dehn surgery as such a
gluing, continuing to use $M$ for $S^{3}\setminus\nu K$. We have
$S^{3}_{r}(K)=M\cup_{h_{r}}(D^{2}\times S^{1})$ with $h_{r}$ the slope-$r$
gluing map. Then Theorem 2.7 provides

$\widehat{\textit{HF}}(S^{3}_{r}(K))\cong\textit{HF}(\widehat{\textit{HF}}(M),h_{r}(\widehat{\textit{HF}}(D^{2}\times
S^{1})).$

The $\text{spin}^{c}$ decomposition is recovered by using $r$ vertically-
adjacent lifts of $h_{r}(\widehat{\textit{HF}}(D^{2}\times S^{1})$, which is
the precise number required to lift every intersection from $T_{M}$ to
$\overline{T}_{M}$ without duplicates. This is motivated by the example in
Figure 3, showing the pairing of curves that recovers
$\widehat{\textit{HF}}(S^{3}_{4}(T(2,5)))$. The invariant for the solid torus
simply consists of a horizontal essential curve, and so
$h_{4}(\widehat{\textit{HF}}(D^{2}\times S^{1})$ is a slope $4$ curve in the
punctured torus. We have four lifts of
$h_{4}(\widehat{\textit{HF}}(D^{2}\times S^{1})$, each generating
intersections in correspondence with the four $\text{spin}^{c}$ summands of
$\widehat{\textit{HF}}(S^{3}_{4}(K))$. These lifts are selected at heights in
correspondence with the selected representatives of $\mathbb{Z}/r\mathbb{Z}$
from Section 2. These are $-1,0,1$, and $2$ for the example in Figure 3, and
motivate the following definition when lifting further to the tiled-plane
cover $\widetilde{T}$.

2pt $l^{2}_{4}$ at 305 235 $l^{1}_{4}$ at 340 235 $l^{0}_{4}$ at 375 210
$l^{-1}_{4}$ at 380 150 $\xrightarrow{\text{lift to}\,\,\overline{T}_{M}}$ at
190 120

Figure 3. The pairing of $\widehat{\textit{HF}}(S^{3}\setminus\nu T(2,5))$ and
$h(\widehat{\textit{HF}}(D^{2}\times S^{1}))$, whose intersection Floer
homology is $\widehat{\textit{HF}}(S^{3}_{4}(T(2,5))$.

######  Definition 2.8.

Let $l^{s}_{r}=h_{r}(\widehat{\textit{HF}}(D^{2}\times S^{1})$ denote the
slope-$r$ line in $\widetilde{T}$ that crosses between lifts of basepoints at
heights congruent to $s\,\,(\text{mod}\,\,r)$. These are selected so that each
$l^{s}_{r}$ crosses at height $s$ in the same column of $\widetilde{T}$, with
$s$ taken to be the representative of $[s]$ that falls within the
$\mathbb{Z}/r\mathbb{Z}$ range.

In this way, Theorem 2.7 implies

$\widehat{\textit{HF}}(S^{3}_{r}(K),[s])\cong\textit{HF}(\widehat{\textit{HF}}(S^{3}\setminus\nu
K),l^{s}_{r}).$

After using regular homotopy of curves to remove any possible intersections
that do not contribute to homology, we end up dim
$(\widehat{\textit{HF}}(S^{3}_{r}(K),[s])$ amounting to counting intersections
between $\widehat{\textit{HF}}(M)$ and $l^{s}_{r}$. However, we have yet to
incorportate the relative Maslov grading. We can compute grading differences
between generators belonging to the same $\text{spin}^{c}$ structure using a
formula from [Han19].

Suppose $x$ and $y$ are two intersections belonging to the same
$[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$, arising from intersections between
$\widehat{\textit{HF}}(M)$ and $l^{s}_{r}$. Further, let $P$ be the bigon from
$y$ to $x$ whose boundary consists of a (not necessarily smooth) path from $y$
to $x$ in $\widehat{\textit{HF}}(M)$, concatenated with a path from $x$ to $y$
in $l^{s}_{r}$. Defined this way, the boundary of $P$ is a closed path that is
smooth apart from right corners at $x$ and $y$, and possibly one or more
cusps. The following formula follows from the conversion of bordered
invariants into immersed curves, keeping track of grading contributions from
relevant Reeb chords [HRW18, Section 2.2].

######  Proposition 2.9.

Suppose $x,y$, and $P$ are defined as above. Let $\text{Rot}(P)$ denote
$\frac{1}{2\pi}$ times the total counterclockwise rotation along the smooth
sections of $P$, let $\text{Wind}(P)$ denote the net winding number of $P$
around enclosed basepoints, and finally let $\text{Wght}(P)$ be the sum of
weights (counted with sign) of all grading arrows traversed by $P$. Then

$M(x)-M(y)=2\text{Wind}(P)+2\text{Wght}(P)-2\text{Rot}(P).$

If $l^{s}_{r}$ intersects a simple figure-eight component at height $n$, it
generates a right intersection $y^{n}$ and a left intersection $x^{n}$. Figure
4 shows off the three types of bigons that will typically appear. The first
type has $P$ connecting a right and left intersection of the same simple
figure-eight. The bigon encloses a single basepoint with positive winding
number, total counterclockwise rotation along smooth sections as $\pi$, and no
contribution from traversed grading arrows. These traits imply
$M(x^{n})-M(y^{n})=1$. The second and third types are the more interesting
ones, and have the same winding number of enclosed basepoints, but the
rotation and grading arrow contributions to $M(y^{n})-M(a^{s})$ initially
appear to be different. We will see later that for these bigons, the
$2\text{Wght}(P)-2\text{Rot}(P)$ component of the grading difference is the
same.

2pt $(a)$ at 35 -10 $x^{n}$ at -2 60 $y^{n}$ at 78 60 $(b)$ at 155 -10 $a^{s}$
at 105 87 $y^{n}$ at 160 157 $(c)$ at 285 -10 $a^{s}$ at 235 87 $y^{n}$ at 290
157

Figure 4. Bigons used to determine the relative Maslov grading. Example (a)
does not involve a grading arrow, while (b) and (c) (with a cusp) do.

##  3\. Thin knots and Maslov grading differences

Throughout this section, let $K$ be a thin, hyperbolic knot and let $M$ denote
$S^{3}\setminus\nu K$. To enable swift grading comparisons later on, let us
designate a reference intersection associated to
$[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$. We will define a vertical intersection
to be an intersection between $l^{s}_{r}$ and a vertical segment of
$\overline{\gamma}$ within the neighborhood of $\overline{\mu}$, provided they
exist. If $s$ satisfies $0\leq|s|<|\tau(K)|$, then such an intersection occurs
and we will denote it using $a^{s}$. Alternatively, if $|s|\geq\tau(K)\geq 0$
then any intersection between $l^{s}_{r}$ and $\overline{\gamma}$ is outside
any neighborhood of the lifts of the marked point in $\overline{T}_{M}$. In
this case $l^{s}_{r}$ intersects $\overline{\gamma}$ once if $\tau(K)\geq 0$,
and so $a^{s}$ will denote this lone intersection. When $\tau(K)<0$ and $s\geq
0$, we let $a^{s}$ denote the intersection between $l^{r}_{s}$ and
$\overline{\gamma}$ to the left of $\overline{\mu}$. Analogously when
$\tau(K)<0$ and $s<0$, we will have $a^{s}$ be the intersection between
$l^{r}_{s}$ and $\overline{\gamma}$ to the right of $\overline{\mu}$. It is
likely helpful to reference Figure 5 for these different possibilities. While
cumbersome, this scheme allows us to label the interesction that often
corresponds via the Pairing theorem to a generator with the the least Maslov
grading.

2pt $(a)$ at 67 -20 $(b)$ at 245 -20 $(c)$ at 438 -20

Figure 5. The possibilities for the reference intersection $a^{s}$. (a) has
$\tau(K)=0$, (b) has $\tau(K)>0$ and $|s|<\tau(K)$, and (c) has $\tau(K)<0$
with two curves representing $s\geq 0$ in red and $s<0$ in purple. The case
when $\tau(K)>0$ and $|s|\geq\tau(K)$ is similar to (a).

It will also be particularly useful to know the winding number of enclosed
lifts of the marked point of specific regions. Consider the neighborhood of
$\overline{\mu}$ in $\overline{T}_{M}$ that contains the lifts of the marked
points, which is also wide enough to enclose the vertical segments of
$\overline{\gamma}$. Intersect $\overline{\gamma}$ with a horizontal line
$l^{s}$ slightly longer than this neighborhood at height $s$, so that these
segments together bound regions enclosing basepoints. We will define
$\widetilde{H}_{s}$ to be the number of enclosed lifts of the marked point in
the region bounded above by $l^{s}$, on the side(s) by the neighborhood of
$\overline{\mu}$, and elsewhere by $\overline{\gamma}$. If the region is
empty, then $\widetilde{H}_{s}=0$. This number coincides with the invariant
$H_{s}$ from the mapping cone formula for all examples encountered by the
author. However, we will use this makeshift definition in lieu of a
$\textit{HF}^{+}$ immersed curves theory. Analogously, there is often a region
where $l^{s}$ bounds from below and the number of enclosed lifts of the marked
point of such a region will be denoted by $\widetilde{V}_{s}$. These are
depicted in Figure 6. Due to Theorem 2.5, we recover both
$\widetilde{H}_{-s}=\widetilde{V}_{s}$ and
$\widetilde{H}_{s}-\widetilde{V}_{s}=\widetilde{H}_{s}-\widetilde{H}_{-s}=\frac{1}{2}(s-(-s))=s$.

2pt $(a)$ at 46 -20 $(b)$ at 207 -20 $(c)$ at 368 -20 $(d)$ at 529 -20

Figure 6. $\widetilde{H}_{s}$ is the number of marked points enclosed in green
regions, and $\widetilde{V}_{s}$ is the number of marked points enclosed in
pink regions. (a) shows $\tau(K)=0$, (b) shows $\tau(K)>0$, (c) shows
$\tau(K)<0$, and (d) shows $\widetilde{H}_{s}-\widetilde{V}_{s}=s$.

From the discussion in the previous section, we know that the form of
$\widehat{\textit{HF}}(M)$ is very restricted. Our goal is to leverage this to
constrain gradings on
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s])\cong\textit{HF}(\widehat{\textit{HF}}(M),l^{s}_{r})$
to obstruct reducible surgeries. We use multisets, which are sets with
repitition allowed, to collect these relative Maslov gradings.

######  Definition 3.1.

Let $[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$ be arbitrary with reference
intersection $a^{s}$. For any intersection $y$ of
$\textit{HF}(\widehat{\textit{HF}}(M),l^{s}_{r})$, let $M_{rel}(y)$ denote the
grading difference $M(y)-M(a^{s})$. We define the desired multiset by

$MR^{[s]}\coloneqq\left\\{M_{rel}(y)\mid
y\in\widehat{\textit{HF}}(M)\pitchfork l^{s}_{r}\right\\}.$

Further, let $\text{Width}(MR^{[s]})$ denote the difference between the
largest and smallest elements of this multiset.

As defined, $MR^{[s]}$ is only an invariant up to translation, which means its
width is an invariant. Next, we establish lemmas that enable us to swiftly
compute grading differences. For a bigon $P$ used to calculate a grading
difference, we will first determine the contribution due to
$2\text{Wght}(P)-2\text{Rot}(P)$ using the knot Floer homology of $K$. This is
performed using an analogous bigon $P_{K}$ for $\widehat{\textit{HFK}}$, and
then we show that the same properties hold for the bigon $P$ for
$\widehat{\textit{HF}}$.

######  Lemma 3.2.

Let $y^{n}$ be a right intersection belonging to a simple figure-eight at
height $n$ of $\widehat{\textit{HF}}(M)$, let $a$ be an intersection from a
different component of $\widehat{\textit{HF}}(M)$ and $l^{s}_{r}$, and suppose
$P$ is a bigon between them. If $K$ is thin, then
$2\text{Wght}(P)-2\text{Rot}(P)=-1-\tau(K)-|n|$.

###### Proof.

In the infinite cylinder $\overline{T}_{M}$, we can represent
$\overline{\mu}$, the lift of the meridian of $T_{M}$, as the vertical line
that pierces each lift of the marked point in $\overline{T}_{M}$. Let
$a^{-\tau(K)}$ be the last intersection that $\overline{\gamma}$ makes with
$\overline{\mu}$ before wrapping around $\overline{T}_{M}$. Because
$\widehat{\textit{HF}}(M)$ is invariant under the action by the hyperelliptic
involution, the weights of the grading arrows connecting $\overline{\gamma}$
to the simple figure-eights at heights $n$ and $-n$ are equivalent. From this
we can assume that $n$ is non-negative, and use $|n|$ in future formulas
otherwise.

Lift $\widehat{\textit{HF}}(M)$ to $\overline{T}$ for convenience, and
intersect it with $\overline{\mu}$. If we place $z$ and $w$ basepoints to the
left and right, respectively, of every lift of the marked point, then
$\widehat{\textit{HFK}}(K)\cong\textit{HF}(\widehat{\textit{HF}}(M),\overline{\mu})$
due to [HRW18, Theorem 51]. This pairing is depicted in Figure 7. The formula
in Definition 2.9 still holds with the adjustment that Wind is modified to
count the net winding number of enclosed $w$ basepoints, denoted
$\text{Wind}_{w}$.

2pt $(a)$ at 105 -20 $(b)$ at 365 -20 $(c)$ at 620 -20 $\eta$ at 90 530
$a^{-\tau(K)}$ at 60 5 $\eta$ at 350 530 $a^{-\tau(K)}$ at 320 400 $\eta$ at
585 250 $a^{-\tau(K)}$ at 575 400

Figure 7. The bigon $P_{K}$ between $a^{-\tau(K)}$ and $\eta$, formed from
path components in $\widehat{\textit{HF}}(M)$ and $\widetilde{\mu}$. $(a)$
shows this for $\tau(K)\geq 0$ and $(b)$ shows this for $A(\eta)>-\tau(K)>0$.
However for $(c)$ with $A(\eta)\geq-\tau(K)>0$, the bigon $P_{K}$ runs from
$\eta$ to $a^{-\tau(K)}$.

Since $\widehat{\textit{HF}}(M)$ has a simple figure-eight component at height
$n$, there must be a generator $\eta$ of $\widehat{\textit{HFK}}(K)$ with
$A(\eta)=n+1$. Let $P_{K}$ be the bigon from $a^{-\tau(K)}$ to $\eta$ that
traverses the grading arrow connecting the relevant components of
$\widehat{\textit{HF}}(M)$, visible in Figure 7 with $\tau(K)\geq 0$ and
$\tau(K)<0$, respectively. To determine $\text{Wght}(P_{K})$ directly would
require care for the orientations of the grading arrow. However since we are
after a different term, we can abuse notation by having every grading arrow
connect to the right side of a simple figure-eight, regardless of its
orientation. Essentially, any change that $\text{Wght}(P_{K})$ experiences
between the two ways of attaching the grading arrow is inverted and absorbed
by $\text{Rot}(P)$, so that $2\text{Wght}(P_{K})-2\text{Rot}(P_{K})$ remains
unchanged.

If $\tau(K)\geq 0$ so that $A(a^{-\tau(K)})<n$, we have

$M(\eta)-M(a^{-\tau(K)})=2\text{Wind}_{w}(P_{K})+2\text{Wght}(P_{K})-2\text{Rot}(P_{K}).$

However since $K$ is thin, it follows that

$M(\eta)-M(a^{-\tau(K)})=A(\eta)-A(a^{-\tau(K)})=A(\eta)+\tau(K).$

Then
$2\text{Wght}(P_{K})-2\text{Rot}(P_{K})=A(\eta)-2\text{Wind}(P_{K})+\tau(K)$.
Since $\text{Wind}(P_{K})=A(\eta)+\tau(K)$, we have
$2\text{Wght}(P_{K})-2\text{Rot}(P_{K})=-A(\eta)-\tau(K)=-1-\tau(K)-n$.

If $\tau(K)<0$, the above computation follows through for $A(\eta)>-\tau(K)$,
but the case for $A(\eta)\leq-\tau(K)$ differs slightly. In this situation
$P_{K}$ is a bigon from $\eta$ to $a^{-\tau(K)}$ that also traverses the
grading arrow in reverse, visible in Figure 7. Traveling the grading arrow in
reverse means that we have
$M(a^{-\tau(K)})-M(\eta)=2\text{Wind}_{w}(P_{K})-2\text{Wght}(P_{K})-2\text{Rot}(P_{K})$,
and so

$\displaystyle-2\text{Wght}(P_{K})-2\text{Rot}(P_{K})$
$\displaystyle=M(a^{-\tau(K)})-M(a^{\eta})-2\text{Wind}_{w}(P)$
$\displaystyle=-\tau(K)-(n+1)-2(-\tau(K)-(n+1))$ $\displaystyle=1+\tau(K)+n.$

Due to the shape of $P_{K}$, the bigon has a cusp near the grading arrow
regardless of how it connects these components, and so $\text{Rot}(P_{K})=0$.
Then we have
$2\text{Wght}(P_{K})-2\text{Rot}(P_{K})=2\text{Wght}(P_{K})+2\text{Rot}(P_{K})=-1-\tau(K)-n$,
as claimed.

With the formula established for $P_{K}$, we will now show that it is
satisfied for a bigon between generators of
$\textit{HF}(\widehat{\textit{HF}}(M),l^{s}_{r})$ with similar attributes. Let
$y^{n}$ be a right intersection from the simple figure-eight at height $n$ and
let $a$ be an intersection from a vertical segment of $\overline{\gamma}$ and
$l^{s}_{r}$. With $P$ denoting the bigon from $a$ to $y^{n}$, we see that $P$
must traverse the same grading arrow that $P_{K}$ traversed, and so
$\text{Wght}(P)=\text{Wght}(P_{K})$. Additionally, it is straightforward to
see that $\text{Rot}(P)=\text{Rot}(P_{K})$ after tilting the bigons as well,
with visual given in Figure 8. This completes the proof. ∎

2pt $(a)$ at 35 -15 $(b)$ at 200 -15 $\eta$ at 5 185 $a^{-\tau(K)}$ at 0 25
$y^{n}$ at 240 140 $a$ at 142 80

Figure 8. Tilting bigons to show they have equivalent net clockwise rotation
along their boundaries. (a) The bigon $P_{K}$ from $a^{-\tau(K)}$ to $\eta$.
(b) The bigon $P$ from $a$ to $y^{n}$.

The following proposition considers left and right intersections of a simple
figure-eight whose height $n$ is less than $|\tau(K)|$. There is then a nearby
vertical intersection $a^{n}$, and we will see that these three intersections
have little difference in grading.

######  Proposition 3.3.

Let $K$ be thin and have $M$ denote $S^{3}\setminus\nu K$. Further, let
$x^{n}$ and $y^{n}$ be left and right intersections belonging to a simple
figure-eight of $\widehat{\textit{HF}}(M)$ with height $0\leq n<|\tau(K)|$,
and let $a^{n}$ be the nearby vertical generator. Then $-1\leq
M(y^{n})-M(a^{n})\leq 0$ and $0\leq M(x^{n})-M(a^{n})\leq 1$.

###### Proof.

If $P$ is the bigon between $a^{n}$ and $y^{n}$, we have
$2\text{Wght}(P)-2\text{Rot}(P)=-1-\tau(K)-|n|$ due to Lemma 3.2. Due to the
hyperelliptic involution invariance of $\widehat{\textit{HF}}(M)$, we can take
$0\leq n<|\tau(K)|$. We have $\text{Wind}(P)$ is $\widetilde{H}_{n}$ if
$\tau(K)\geq 0$ or $\widetilde{V}_{n}$ if $\tau(K)<0$, the values of which
depend on the parity of $n$ and $\tau(K)$ when $K$ is thin. The simple
structure of $\overline{\gamma}$ for a thin knot together with a counting
argument for $\tau(K)>0$ yields

$\displaystyle\widetilde{H}_{n}=\left\\{\begin{array}[]{l
c}\dfrac{n+\tau(K)}{2}&\text{parity}(n)=\,\text{parity}(\tau(K))\\\
\dfrac{n+\tau(K)+1}{2}&\text{parity}(n)\neq\,\text{parity}(\tau(K)).\\\
\end{array}\right.$

Then for $\tau(K)>0$ we have
$M(y^{n})-M(a^{n})=2\widetilde{H}_{n}-1-\tau(K)-n$ implies $M(y^{n})-M(a^{n})$
is either -1 or 0. Since $M(x^{n})-M(y^{n})=1$, we see that
$M(x^{n})-M(a^{n})$ is either $0$ or $1$, handling the $\tau(K)>0$ case.

When $\tau(K)<0$, the bigon $P$ runs from $y^{n}$ to $a^{n}$, encloses
$\widetilde{V}_{n}$ lifts of the marked points, traverses the grading arrow in
reverse, and has $\text{Rot}(P)=0$. Figure 6 shows that $\widetilde{V}_{n}$
with $\tau(K)<0$ is the same as $\widetilde{V}_{n}=\widetilde{H}_{-n}$ with
$\tau(K)\geq 0$, except using $-\tau(K)$ or $-\tau(K)-1$ in the formula above.
Using Lemma 3.2 and the $-\tau(K)$ modified formula for $\widetilde{H}_{-n}$,
we have $M(a^{n})-M(y^{n})=2\widetilde{H}_{-n}+1+\tau(K)+n$. This is either
$1$ or $0$, and so $M(y^{n})-M(a^{n})$ is either $-1$ or $0$ and analogously
$M(x^{n})-M(a^{n})$ is either $0$ or $1$. ∎

Because $M(x^{n})-M(y^{n})=1$, these possibilities happen in pairs. A simple
figure-eight at height $n<|\tau(K)|$ contributes either
$\left\\{M_{rel}(a^{n}),M_{rel}(a^{n})-1,M_{rel}(a^{n})\right\\}\subseteq
MR^{[s]}$ or
$\left\\{M_{rel}(a^{n}),M_{rel}(a^{n}),M_{rel}(a^{n})+1\right\\}\subseteq
MR^{[s]}$. An example of this to keep in mind is when looking at large surgery
on the figure-eight knot $4_{1}$. In this situation we have
$\left\\{0,-1,0\right\\}=MR^{[0]}$, and the right intersection contributing
$-1$ to $MR^{[0]}$ actually has the smallest relative Maslov grading. This
proposition then allows us to determine which intersection associated to
$[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$ has the smallest relative Maslov grading
depending on $\text{parity}(\tau(K))$:

  * •

If $\tau(K)\geq 0$, $\text{parity}(s)=\text{parity}(\tau(K))$, and there is a
right intersection $y^{s}$, then $M_{rel}(y^{s})=-1$ is the smallest relative
grading of $MR^{[s]}$.

  * •

If $\tau(K)\geq 0$, $\text{parity}(s)=\text{parity}(\tau(K))$, and there is no
simple figure-eight at height $s$, then $M_{rel}(a^{s})=0$ is the smallest
relative grading of $MR^{[s]}$.

  * •

If $\tau(K)\geq 0$ and $\text{parity}(s)\neq\text{parity}(\tau(K))$, then
$M_{rel}(a^{s})=0$ is the smallest relative grading of $MR^{[s]}$.

  * •

If $\tau(K)<0$, then $M_{rel}(a^{s})=0$ is the smallest relative grading of
$MR^{[s]}$.

The last component of the grading difference formula to handle is
$\text{Wind}(P)$. Lift both $\widehat{\textit{HF}}(M)$ and each $l^{s}_{r}$ to
the tiled plane $\widetilde{T}$, and let the $0$th column be the neighborhood
of the lift $\widetilde{\mu}$ for which each $l^{s}_{r}$ intersects
$\widetilde{\mu}$ at height $[s]$. For $[s]\in\mathbb{Z}/r\mathbb{Z}$ define
$w_{s}=\dfrac{n-[s]}{r}$, with $n$ the largest natural number satisfying
$0\leq n\leq g(K)-1$ and $n\equiv[s]\,\,(\text{mod}\,\,r)$. This number
represents the number of columns of marked points in $\widetilde{T}$ between
$a^{s}$ and a potential furthest right intersection $y^{n}$. Further, because
the slopes we consider satisfy $r\leq 2g(K)-1$, we have $w_{s}\geq 0$. While
it is certainly possible that a simple figure-eight component may not exist at
this height, it is still sufficient for the following strategy to suppose
otherwise.

######  Proposition 3.4.

For a given $[s]\in\mathbb{Z}/r\mathbb{Z}$, let $a^{s}$ be the chosen
reference intersection and $y^{n}$ be a right intersection of a furthest
possible figure-eight component. If $\tau(K)\geq 0$, then

$\displaystyle\text{Wind}(P)=\widetilde{H}_{s}+\sum_{i=1}^{w_{s}}(s+ir).$

If $\tau(K)<0$, then

$\displaystyle\displaystyle\text{Wind}(P)=\left\\{\begin{array}[]{l
c}\displaystyle\sum_{i=0}^{w_{s}}(s+ir)&[s]\geq 0\\\
\displaystyle\sum_{i=1}^{w_{s}}(s+ir)&[s]<0,\\\ \end{array}\right.$

where all sums are taken to be zero if empty.

When $\tau(K)\geq 0$, the contribution to $\text{Wind}(P)$ from the $0$th
column of $\widetilde{T}$ is $\widetilde{H}_{s}$. The contribution from the
$i$th column is $\widetilde{H}_{s+ir}-\widetilde{V}_{s+ir}=s+ir$, and is shown
in Figure 9. When $\tau(K)<0$, we have the different choices for $a^{s}$
depending on $s$ influencing whether the contribution from the $0$th column is
non-trivial. However in every column, the contribution to $\text{Wind}(P)$ is
$\widetilde{H}_{s+ir}-\widetilde{V}_{s+ir}=s+ir$. Since these terms are always
non-negative, it follows that the smallest relative grading belongs to an
intersection in the $0$th column.

2pt $a^{s}$ at 90 320 $y^{n}$ at 200 600 $a^{s}$ at 380 330 $y^{n}$ at 575 600
$(a)$ at 110 -15 $(b)$ at 470 -15

Figure 9. Example bigons $P$ between $a^{s}$ and $y^{n}$, showing the
contributions from each column to $\text{Wind}(P)$ for $(a)\,\,\tau(K)\geq 0$
and $(b)\,\,\tau(K)<0$ with $s\geq 0$.

##  4\. Cases with $|\tau(K)|<g(K)$

Our objective is to build a collection of lemmas required to prove the main
theorem. These vary depending on $r$ in relation to $g(K)$, and on $\tau(K)$
and its parity. The primary technique involves comparing the various
$\text{Width}(MR^{[s]})$ to obstruct periodicity, typically done by showing
that $\text{Width}(MR^{[s^{\prime}]})$ is maximal if $[s^{\prime}]$ is the
$\text{spin}^{c}$ structure associated to the line that crosses height
$g(K)-1$. At other times the widths will agree up to translation, but the
multiplicity of specific elements of the grading multisets will not.

2pt $E\circlearrowright$ at 210 150

Figure 10. The correspondence between intersections of
$\widehat{\textit{HF}}(M)$ and $l^{s}_{r}$ in negative columns of
$\widetilde{T}$ and intersections of $\widehat{\textit{HF}}(M)$ and
$l^{-s}_{r}$ in positive columns.

Use Theorem 2.7 to identify
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s])\cong\textit{HF}(\widehat{\textit{HF}}(M),l^{s}_{r})$.
In order to halve the amount of comparisons to make, we leverage the fact that
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s])\cong\widehat{\textit{HF}}(S^{3}_{r}(K),[-s])$
[OS04c]. In immersed curves form, Theorem 2.5 implies that intersections
between $\widehat{\textit{HF}}(M)$ and $l^{s}_{r}$ in negative columns of
$\widetilde{T}$ are in correspondence with intersections of
$\widehat{\textit{HF}}(M)$ and $l^{-s}_{r}$ that belong to positive columns of
$\widetilde{T}$ (see Figure 10). Also, the self-conjugate $\text{spin}^{c}$
structure(s) $[0]$ (and possibly $[r/2]$) are symmetric in this way by
default.

Recall that the smallest element of $MR^{[s]}$ is the relative grading of an
intersection belonging to the $0$th column of $\widetilde{T}$, which is either
the reference intersection $a^{s}$ or a nearby right/left intersection. This
means that we can capture $\text{Width}(MR^{[s]})$ by considering non-negative
intersections associated to both $[s]$ and $[-s]$. Note that since
$\text{parity}([s])=\text{parity}([-s])$, the need to translate a multiset by
1 is consistent if it arises.

######  Definition 4.1.

The multiset $MR^{[s]}_{+}$ consists of the relative gradings of intersections
between $\widehat{\textit{HF}}(M)$ and $l^{s}_{r}$ that belong to non-negative
columns of $\widetilde{T}$. We define $MR^{[s]}_{-}$ analogously, and notice
that
$\text{Width}(MR^{[s]})=\text{max}\left\\{\text{Width}(MR^{[s]}_{+}),\text{Width}(MR^{[s]}_{-})\right\\}$.

Due to how genus detection is expressed by $\widehat{\textit{HF}}(M)$, either
$\overline{\gamma}$ achieves height $g(K)$ (equivalent to $|\tau(K)|=g(K)$),
or only a simple figure-eight at height $g(K)-1$ achieves this desired height
(equivalent to $|\tau(K)|<g(K))$. We will divide the problem among these two
cases, starting with the latter.

Case A: $\mathbf{|\tau(K)|<g(K)}$. Since $|\tau(K)|<g(K)$, there exists a
simple figure-eight component at height $g(K)-1$. Let $[s^{\prime}]$ be the
$\text{spin}^{c}$ structure for which $l^{s^{\prime}}_{r}$ intersects this
simple figure-eight, which means $w_{s^{\prime}}=\frac{g(K)-1-s^{\prime}}{r}$.
Our potential reducing slopes divide this case into two subcases. We see that
$[s^{\prime}]=g(K)-1$ when $r\geq 2(g(K)-1)$, which implies
$w_{s^{\prime}}=0$. Otherwise $w_{s^{\prime}}>0$, which is the easier starting
point.

Case A1: $\mathbf{w_{s^{\prime}}>0}$. In this situation, we will show that
$\text{Width}(MR^{[s^{\prime}]})$ is maximal.

######  Lemma 4.2.

Suppose $K$ is thin, $|\tau(K)|<g(K)$, and $1<r<2(g(K)-1)$. Then there exists
an $[s^{\prime}]\in\text{Spin}^{c}(S^{3}_{r}(K))$ for which every $[s]\neq[\pm
s^{\prime}]$ satisfies $MR^{[s]}\not\cong MR^{[s^{\prime}]}$ up to
translation.

###### Proof.

For some $[s]\neq[\pm s^{\prime}]$, the largest possible relative grading that
$MR^{[s]}_{+}$ can achieve is associated to an intersection of some
hypothetical simple figure-eight at largest height. Looking at the terms in
the grading difference formula for a bigon from $a^{s}$ to such a generator,
we see that the $2\text{Wind}(P)$ term satisifes $2\text{Wind}(P)\geq 2n$
while the other term is $-1-\tau(K)-n$. For this reason, we will suppose that
$\widehat{\textit{HF}}(M)$ has a simple figure-eight at height $n$, taken to
be the largest integer satisfying both $n<g(K)-1$ and $n\equiv
s\,\,(\text{mod}\,\,r)$. Let $P^{\prime}$ be the bigon between
$a^{s^{\prime}}$ and $y^{g-1}$, and $P$ the bigon between $a^{s}$ and $y^{n}$.
Because the choice of $a^{s^{\prime}}$ depends on $\tau(K)$, we will handle
the $\tau(K)\geq 0$ subcase first before handling the $\tau(K)<0$ subcase.

Subcase A1a: $\mathbf{\tau(K)\geq 0}$. Due to Lemma 3.2, Proposition 3.3, and
Proposition 3.4, $\text{Width}(MR^{[s]}_{+})$ is nearly determined by
$M_{rel}(y^{n})$. We have $M_{rel}(y^{n})\leq\text{Width}(MR^{[s]}_{+})\leq
M_{rel}(y^{n})+1$, with either equality depending on whether $a^{s}$ is the
smallest relatively graded intersection. To compare widths, we compute

$M_{rel}(y^{g-1})=2\left(\widetilde{H}_{s^{\prime}}+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)\right)-1-(s^{\prime}+w_{s^{\prime}}r),$

and likewise

$M_{rel}(y^{n})=2\left(\widetilde{H}_{s}+\sum_{i=1}^{w_{s}}(s+ir)\right)-1-(s+w_{s}r).$

Their difference is then

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=2\left(\widetilde{H}_{s^{\prime}}+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\left(\widetilde{H}_{s}+\sum_{i=1}^{w_{s}}(s+ir)\right)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}+w_{s^{\prime}}r-(s+w_{s}r))$
$\displaystyle=2\left((\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\sum_{i=1}^{w_{s}}(s+ir)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}-s)-r(w_{s^{\prime}}-w_{s}).$

If $s<s^{\prime}$ so that $w_{s}=w_{s^{\prime}}$, then

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=2((\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})+w_{s^{\prime}}(s^{\prime}-s))-(s^{\prime}-s)$
$\displaystyle=2(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})+(2w_{s^{\prime}}-1)(s^{\prime}-s)$
$\displaystyle\geq 1,$

since $w_{s^{\prime}}>0$ and $s^{\prime}>s$ implies that
$\widetilde{H}_{s^{\prime}}\geq\widetilde{H}_{s}$.

If $s>s^{\prime}$ so that $w_{s}=w_{s^{\prime}}-1$, then shifting $P$ one
column to the right in $\widetilde{T}$ (see Figure 11) provides

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=2\left((\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\sum_{i=1}^{w_{s}}(s+ir)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}-s)-r(w_{s^{\prime}}-w_{s})$
$\displaystyle(\text{column shift})\,\,\,\,\,\,$
$\displaystyle=2\left((\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\sum_{i=2}^{w_{s^{\prime}}}(s+(i-1)r)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}+r-s)$
$\displaystyle=2\left((\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})+(s^{\prime}+r)+\sum_{i=2}^{w_{s}^{\prime}}(s^{\prime}+ir)-\sum_{i=2}^{w_{s^{\prime}}}(s+(i-1)r)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}+r-s)$
$\displaystyle=2\left((\widetilde{H}_{s^{\prime}}+s-\widetilde{H}_{s})+(s^{\prime}+r-s)+(w_{s^{\prime}}-1)(s^{\prime}+r-s)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}+r-s)$

$\displaystyle=2(\widetilde{H}_{s^{\prime}}+(s-\widetilde{H}_{s}))+(2w_{s^{\prime}}-1)(s^{\prime}+r-s)$
$\displaystyle=2(\widetilde{H}_{s^{\prime}}-\widetilde{V}_{s})+(2w_{s^{\prime}}-1)(s^{\prime}+r-s)$
$\displaystyle=2(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{-s})+(2w_{s^{\prime}}-1)(s^{\prime}+r-s).$

Notice that $s^{\prime}+s-1\leq
2(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{-s})\leq s^{\prime}+s$ depending
on the parities of $s$ and $s^{\prime}$ together with $s>s^{\prime}$. Then we
have

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=2(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{-s})+(2w_{s^{\prime}}-1)(s^{\prime}+r-s)$
$\displaystyle\geq s^{\prime}+s-1+(2w_{s^{\prime}}-1)(s^{\prime}+r-s)$
$\displaystyle\geq s^{\prime}+s-1+s^{\prime}+r-s$
$\displaystyle=2s^{\prime}-1+r$ $\displaystyle>1,$

since $w_{s^{\prime}}>0$ and $s^{\prime}<\frac{r-1}{2}$ if there exists an
$s>s^{\prime}$.

2pt $(a)$ at 115 0 $(b)$ at 490 0 $a^{1}$ at 15 35 $a^{0}$ at 50 35 $y^{5}$ at
205 400 $y^{4}$ at 205 335 $a^{1}$ at 390 35 $a^{0}$ at 467 35 $y^{5}$ at 580
400 $y^{4}$ at 580 335

Figure 11. Example bigons $P^{\prime}$ (split-shaded green and pink) and $P$
(shaded pink) when $w_{s^{\prime}}=1$. $(a)$ has $s<s^{\prime}$, while $(b)$
has $s>s^{\prime}$ together with the single column shift to the right.

In both situations, we see that $M_{rel}(y^{g-1})-M_{rel}(y^{n})\geq 1$. If
this difference is greater than one, then

$\text{Width}(MR^{[s^{\prime}]})\geq\text{Width}(MR^{[s^{\prime}]}_{+})\geq
M_{rel}(y^{g-1})>M_{rel}(y^{n})+1\geq\text{Width}(MR^{[s]}_{+}).$

This already handles the possibility where we need to translate $MR^{[s]}_{+}$
by 1, so suppose $M_{rel}(y^{g-1})-M_{rel}(y^{n})=1$. This is possible only if
$\widetilde{H}_{s}=\widetilde{H}_{s^{\prime}}$, $w_{s^{\prime}}=1$, and
$s=s^{\prime}-1$, which altogether imply that $s=\tau(K)$. However, the widths
only match if $\text{Width}(MR^{[s]}_{+})=M_{rel}(y^{n})+1$. This condition is
equivalent to having $\text{parity}(s)\neq\text{parity}(\tau(K))$, which is a
contradiction. Therefore
$\text{Width}(MR^{[s^{\prime}]})>\text{Width}(MR^{[s]}_{\pm})$, which
completes the $\tau(K)\geq 0$ subcase.

Subcase A1b: $\mathbf{\tau(K)<0}$. Recall that the reference intersection
$a^{s}$ has no nearby left/right intersections belonging to a simple figure-
eight. This means that $a^{s}$ has the smallest relative grading of
$MR^{[s]}$, and so $\text{Width}(MR^{[s]}_{+})=M_{rel}(y^{n})+1$. From
Proposition 3.4 we see

$\displaystyle\text{Wind}(P)=\left\\{\begin{array}[]{l c}\displaystyle
s+\sum_{i=1}^{w_{s}}(s+ir)&s\geq 0,\\\
\displaystyle\sum_{i=1}^{w_{s}}(s+ir)&s<0.\end{array}\right.$

If $0\leq s<s^{\prime}$, then proceeding as before we have

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=2\left(s^{\prime}+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\left(s+\sum_{i=1}^{w_{s}}(s+ir)\right)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}-s)-r(w_{s^{\prime}}-w_{s})$
$\displaystyle=2(s^{\prime}-s+w_{s^{\prime}}(s^{\prime}-s))-(s^{\prime}-s)$
$\displaystyle=(2w_{s^{\prime}}+1)(s^{\prime}-s)$ $\displaystyle\geq 3.$

If $s<s^{\prime}\leq 0$, then

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=(2w_{s^{\prime}}-1)(s^{\prime}-s)$ $\displaystyle\geq 1.$

If $s>s^{\prime}$, then as before we have $w_{s}=w_{s^{\prime}}-1$. If
$s>s^{\prime}\geq 0$, then

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=2\left((s^{\prime}-s)+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\sum_{i=1}^{w_{s}}(s+ir)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(s^{\prime}-s)-r(w_{s^{\prime}}-w_{s})$
$\displaystyle(\text{column shift})\,\,\,\,\,\,$
$\displaystyle=2\left((s^{\prime}-s)+\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\sum_{i=2}^{w_{s^{\prime}}}(s+(i-1)r)\right)-(s^{\prime}+r-s)$
$\displaystyle=2(s^{\prime}+(s^{\prime}+r-s)+(w_{s^{\prime}}-1)(s^{\prime}+r-s))-(s^{\prime}+r-s)$
$\displaystyle=2s^{\prime}+(2w_{s^{\prime}}-1)(s^{\prime}+r-s)$
$\displaystyle\geq 1.$

In the event that $0\geq s>s^{\prime}$, we get

$\displaystyle M_{rel}(y^{g-1})-M_{rel}(y^{n})$
$\displaystyle=2\left(\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\sum_{i=1}^{w_{s}}(s+ir)\right)-(s^{\prime}-s)-r(w_{s^{\prime}}-w_{s})$
$\displaystyle(\text{column shift})\,\,\,\,\,\,$
$\displaystyle=2\left(\sum_{i=1}^{w_{s^{\prime}}}(s^{\prime}+ir)-\sum_{i=2}^{w_{s^{\prime}}}(s+(i-1)r)\right)-(s^{\prime}+r-s)$
$\displaystyle=2(s^{\prime}+r+(w_{s^{\prime}}-1)(s^{\prime}+r-s))-(s^{\prime}+r-s)$
$\displaystyle=2(s+w_{s^{\prime}}(s^{\prime}+r-s))-(s^{\prime}+r-s)$
$\displaystyle=2s+(2w_{s^{\prime}}-1)(s^{\prime}+r-s)$ $\displaystyle\geq
2s+s^{\prime}+r-s$ $\displaystyle\geq(s^{\prime}+s)+r$ $\displaystyle\geq 1.$

In every inequality we have $M_{rel}(y^{g-1})>M_{rel}(y^{n})$. Then

$\text{Width}(MR^{[s^{\prime}]})\geq\text{Width}(MR^{[s^{\prime}]}_{+})=M_{rel}(y^{g-1})+1>M_{rel}(y^{n})+1=\text{Width}(MR^{[s]}_{+}),$

for each $[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$. This completes the $\tau(K)<0$
subcase, and the proof. ∎

Case A2: $\mathbf{w_{s^{\prime}}=0}$. Recall that in this case we have $r\geq
2(g(K)-1)$, so let us consider $r=2g(K)-1$ first. When $\tau(K)\geq 0$, the
surgery slope is large enough so that every intersection lies in the $0$th
column of $\widetilde{T}$. Width alone as an invariant won’t be enough, so we
will also need to appeal to the multiplicities of the elements of the relative
grading multisets. They will be used to show that only $\text{spin}^{c}$
structures with the same parity are unobstructed. When we assume that
$S^{3}_{r}(K)$ is reducible later on, the fact that $r$ is odd will provide a
contradiction with periodicity. When $\tau(K)<0$, we need far less sublety.

######  Lemma 4.3.

Suppose $K$ is thin, $0\leq\tau(K)<g(K)$, and $r=2g(K)-1$. Then there exists
an $[s^{\prime}]\in\text{Spin}^{c}(S^{3}_{r}(K))$ for which $[s]\neq[\pm
s^{\prime}]$ satisfies $MR^{[s]}\cong MR^{[s^{\prime}]}$ up to translation
only if $\text{parity}([s])=\text{parity}([s^{\prime}])$.

###### Proof.

The $\text{spin}^{c}$ structure $[s^{\prime}]$ we want to consider has
$[s^{\prime}]=g(K)-1$. Suppose for the sake of contradiction that some
$[s]\neq[\pm s^{\prime}]$ satisfies $MR^{[s]}\cong MR^{[s^{\prime}]}$ up to
translation and $\text{parity}(s)\neq\text{parity}(s^{\prime})$. We know that
$r>1$ forces $g(K)>1$, and also that each $l^{s}_{r}$ intersects
$\widehat{\textit{HF}}(M)$ exactly once due to this large surgery slope.
Because the choice of reference generator $a^{s^{\prime}}$ depends on
$\tau(K)$, let us split into two cases: $\tau(K)\geq 0$ and $\tau(K)<0$.

Assume $\tau(K)\geq 0$. Because all intersections lie within the 0th column of
$\widetilde{T}$, we will instead use the hyperelliptic involution invariance
of $\widehat{\textit{HF}}(M)$ to only consider $s\geq 0$. If
$\widehat{\textit{HF}}(M)$ has no simple figure-eight at height $s$, then
$\text{Width}(MR^{[s]})=0$ immediately does not match
$\text{Width}(MR^{[s^{\prime}]})\geq 1$, so we may as well assume that there
is a simple figure-eight at height $s$. We have
$M_{rel}(y^{s^{\prime}})=2\widetilde{H}_{s^{\prime}}-1-\tau(K)-s^{\prime}=s^{\prime}-1-\tau(K)$
by Lemma 3.2 and Proposition 3.4, since
$\widetilde{H}_{s^{\prime}}=s^{\prime}$ when $\tau(K)\leq g(K)-1=s^{\prime}$.
Further,

$\displaystyle M_{rel}(y^{s^{\prime}})-M_{rel}(y^{s})$
$\displaystyle=2\widetilde{H}_{s^{\prime}}-1-\tau(K)-s^{\prime}-(2\widetilde{H}_{s}-1-\tau(K)-s)$
$\displaystyle=2(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})-(s^{\prime}-s).$

If $s>\tau(K)$, then $\widetilde{H}_{s}=s$ implies that
$M_{rel}(y^{s^{\prime}})-M_{rel}(y^{s})=s^{\prime}-s\geq 1$. But then

$\text{Width}(MR^{[s^{\prime}]}=M_{rel}(y^{s^{\prime}})+1>M_{rel}(y^{s})+1\geq\text{Width}(MR^{[s]}),$

so we must have $s\leq\tau(K)$ together with $\text{Width}(MR^{[s]})=1$.
Notice that
$\text{Width}(MR^{[s^{\prime}]})=M_{rel}(y^{s^{\prime}})+1=s^{\prime}-\tau(K)>1$
if $\tau(K)<s^{\prime}-1$, and so we are also forced to have either
$\tau(K)=s^{\prime}-1$ or $\tau(K)=s^{\prime}$. In both cases we have
$\text{Width}(MR^{[s^{\prime}]})=1$. Since using width as an invariant has
been exhausted, let us count multiplicities of elements of the $MR^{[s]}$’s
next.

Recall that $e_{n}$ denotes the number of simple figure-eights at height $n$
of $\widehat{\textit{HF}}(M)$. Further, we need $e_{s}=e_{s^{\prime}}$ in
order to have $|MR^{[s]}|=|MR^{[s^{\prime}]}|$. We have assumed that
$\text{parity}([s])\neq\text{parity}([s^{\prime}])$, so one of these two
multisets contains $-1$ and must be translated by 1 to make 0 the smallest
element. This translated multiset will then contain $0$ with multiplicity
$e_{s^{\prime}}$, while the other multiset will contain $0$ with multiplicity
$e_{s^{\prime}}+1$. This is the desired contradiction. ∎

When $r=2(g(K)-1)$, we will end up having $\text{Width}(MR^{[s]})=1$ for every
$[s]$ if $\tau(K)$ is large enough. This means relative grading information
alone will not be enough, and so we will return to such cases in Section 5.

######  Lemma 4.4.

Suppose $K$ is thin, $0\leq\tau(K)<g(K)-2$, and $r=2(g(K)-1)$. Then there
exists an $[s^{\prime}]\in\text{Spin}^{c}(S^{3}_{r}(K))$ for which every
$[s]\neq[\pm s^{\prime}]$ satisfies $MR^{[s]}\not\cong MR^{[s^{\prime}]}$ up
to translation.

###### Proof.

We again use $s^{\prime}=g(K)-1$, and notice that when $\tau(K)<g(K)-2$, we
have

$M_{rel}(y^{s^{\prime}})=2(g(K)-1)-1-\tau(K)-(g(K)-1)=g(K)-2-\tau(K)>0.$

This shows that $\text{Width}(MR^{[s^{\prime}]})=M_{rel}(y^{s^{\prime}})+1>1$.
Any $[s]\neq[\pm s^{\prime}]$ with $|s|\leq\tau(K)$ has
$\text{Width}(MR^{[s]})=1$ due to Proposition 3.3, so suppose
$\tau(K)<|s|<s^{\prime}$. In this case, $\text{Width}(MR^{[s]})\leq
M_{rel}(y^{s})+1$, but we also have
$M_{rel}(y^{s^{\prime}})-M_{rel}(y^{s})=s^{\prime}-|s|>0$. Then
$\text{Width}(MR^{[s]})<\text{Width}(MR^{[s^{\prime}]})$, which completes the
proof. ∎

When $\tau(K)<0$, the fact that the reference intersection $a^{s}$ lies
outside of the neighborhood of $\widetilde{\mu}_{0}$ is very convenient. This
is an example of a non-vertical intersection, which is an intersection between
$l^{s}_{r}$ and $\overline{\gamma}$ that lies outside of a neighborhood of a
lift $\widetilde{\mu}$.

######  Lemma 4.5.

Suppose $K$ is thin, $-g(K)<\tau(K)<0$, and $r\geq 2(g(K)-1)$. Then there
exists an $[s^{\prime}]\in\text{Spin}^{c}(S^{3}_{r}(K))$ for which every
$[s]\neq[\pm s^{\prime}]$ satisfies $MR^{[s]}\not\cong MR^{[s^{\prime}]}$ up
to translation.

###### Proof.

Since $w_{s^{\prime}}=0$, we again have $[s^{\prime}]=g(K)-1$. Notice that
each $l^{s}_{r}$ gives rise to only two non-vertical intersections around the
$0$th column and intersections at height $s$ when $[s]\neq[\pm s^{\prime}]$.
We have $s^{\prime}$ maximal when $w_{s^{\prime}}=0$, so use hyperelliptic
involution invariance to assume $0\leq s<s^{\prime}$. Recall that
$\text{Width}(MR^{[s]})=M_{rel}(y^{s})+1$ under the assumptions that
$\tau(K)<0$. The formula for $\text{Wind}(P)$ does not depend on $\tau(K)$,
which means

$\displaystyle M_{rel}(y^{s^{\prime}})-M_{rel}(y^{s})$
$\displaystyle=2s^{\prime}-1-\tau(K)-s^{\prime}-(2s-1-\tau(K)-s)$
$\displaystyle=s^{\prime}-s.$

Then
$\text{Width}(MR^{[s^{\prime}]})=M_{rel}(y^{s^{\prime}})+1>M_{rel}(y^{s})+1=\text{Width}(MR^{[s]})$,
which implies $MR^{[s]}\not\cong MR^{[s^{\prime}]}$. ∎

In the following section we address the remaining cases involving
$|\tau(K)|=g(K)$, as well as the few unresolved cases of this section. In
particular, the cases with $g(K)-2\leq\tau(K)<g(K)$ and $r=2(g(K)-1)$ are
handled in Lemma 5.5.

##  5\. Remaining Cases and Absolute Gradings

With the case analysis for $|\tau(K)|<g(K)$ out of the way, we turn to the
more difficult part.

Case B: $\mathbf{|\tau(K)|=g(K)}$. When $|\tau(K)|$ is at its largest, the
essential curve $\overline{\gamma}$ suffices to indicate $g(K)$ and we are not
guaranteed a simple figure-eight at height $g(K)-1$. For these cases we still
choose $[s^{\prime}]$ so that $g-1\equiv s^{\prime}\,\,(\text{mod}\,\,r)$ and
continue to use $w_{s}$, except now modifying it to just be the largest
multiple of $r$ so that $s+w_{s}r<g(K)$. The $\tau(K)=-g(K)$ case is easier,
so we start there.

Case B1: $\mathbf{\tau(K)=-g(K)}$.

######  Lemma 5.1.

Suppose $K$ is thin with $\tau(K)=-g(K)$, and let $1<r\leq 2g(K)-1$. Then
there exists an $[s^{\prime}]\in\text{Spin}^{c}(S^{3}_{r}(K))$ for which every
$[s]\neq[\pm s^{\prime}]$ satisfies $MR^{[s]}\not\cong MR^{[s^{\prime}]}$ up
to translation.

###### Proof.

Recall the labeling scheme from Figure 5. The reference intersection $a^{s}$
is immediately to the left of the $0th$ column if $s\geq 0$, and is similarly
immediately to the right of the $0th$ column if $s>0$. In general we will
label these generators $x_{s}^{l}$ and $x_{s}^{r}$, respectively. Let us
dispense with the $w_{s^{\prime}}=0$ case first.

Notice that each $MR^{[s]}$ contains two elements whose difference is
precisely $2|s|$. These arise from
$M_{rel}(x_{s}^{r})-M_{rel}(x_{s}^{l})=2\widetilde{H}_{s}-1-(2\widetilde{V}_{s}-1)=2(\widetilde{H}_{s}-\widetilde{V}_{s})=2s$.
We also see that $2\widetilde{H}_{s}-1\leq\text{Width}(MR^{[s]})\leq
2\widetilde{H}_{s}$ if $s\leq 0$ and
$2\widetilde{V}_{s}-1\leq\text{Width}(MR^{[s]})\leq 2\widetilde{V}_{s}$ if
$s>0$, with the even equalities achieved if an appropriate generator from a
simple figure-eight exists at height $s$. So if some $[s]\neq[\pm s^{\prime}]$
is to achieve $MR^{[s]}\cong MR^{[s^{\prime}]}$ up to translation, we should
see that the widths of these multisets agree and that there exist pairs with
grading differences $2|s|$ and $2|s^{\prime}|$. These are only possibly
simultaneously true if $\widetilde{V}_{s}=\widetilde{H}_{s^{\prime}}$ and
$\widetilde{H}_{s}=\widetilde{V}_{s^{\prime}}$, which forces $s=-s^{\prime}$
with $K$ thin. Thus, $w_{s^{\prime}}>0$.

If $w_{s^{\prime}}>0$, we can appeal to $MR^{[s^{\prime}]}$ achieving maximal
width once again. Due to the formula for $\text{Wind}(P)$ when $\tau(K)<0$,
the grading difference between consecutive vertical intersections between
$\overline{\gamma}$ and $l^{s}_{r}$ around the $i$th column is $2(s+ir)$. As
before this happens as
$2\widetilde{H}_{s+ir}-1-(2\widetilde{V}_{s+ir}-1)=2(\widetilde{H}_{s+ir}-\widetilde{V}_{s+ir})=2(s+ir)$,
and is positive. For this reason it is often the case that the vertical
intersection on the left side of the $w_{s}$th column, which we now denote by
$b^{s}$, has the largest relative grading in $MR^{[s]}$. As before, by
appealing to the hyperelliptic involution invariance of
$\widehat{\textit{HF}}(M)$ we see that $\text{Width}(MR^{[s]})$ is either
$M_{rel}(b^{s})$ or $M_{rel}(b^{-s})$ when $w_{s}>0$. We will obtain our
desired contradiction by comparing $M_{rel}(b^{s^{\prime}})$ to every
$M_{rel}(b^{s})$ with $s\neq\pm s^{\prime}$, just as in the lemmas of the
previous section.

Chaining the grading differences of vertical intersection pairs from $b^{s}$
back to $a^{s}$, we see that

$\displaystyle\displaystyle M_{rel}(b^{s})=\left\\{\begin{array}[]{l
c}\displaystyle
2\left(\sum_{i=0}^{w_{s}-1}(s+ir)+\widetilde{H}_{s+w_{s}r}\right)-1&\,\text{if}\,s\geq
0\\\ \displaystyle
2\left(\sum_{i=1}^{w_{s}-1}(s+ir)+\widetilde{H}_{s+w_{s}r}\right)-1&\,\text{if}\,s\leq
0,\\\ \end{array}\right.$

with empty sums taken to be zero as before. Since it can be hectic determining
when such a sum is empty, we break into more cases.

When $s<s^{\prime}$ we have $w_{s}=w_{s^{\prime}}$, and it is straightforward
to check that

$M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})\geq
2(\widetilde{H}_{s^{\prime}+w_{s^{\prime}}r}-\widetilde{H}_{s+w_{s^{\prime}}r})>0.$

This follows because the various multiples of $(s^{\prime}-s)$ are positive if
they appear, and because
$\widetilde{H}_{s^{\prime}+w_{s^{\prime}}r}>\widetilde{H}_{s+w_{s^{\prime}}r}$
when $s<s^{\prime}$.

Let us begin the $s^{\prime}<s$ cases with $w_{s^{\prime}}=1$. For $0\leq
s^{\prime}<s$ we can once again use a column shift to see

$M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})=2(s^{\prime}+H_{s^{\prime}+r}-H_{s})>0,$

since $s^{\prime}\geq 0$ and $H_{s^{\prime}+r}>H_{s}$. The same inequality
holds if $s^{\prime}\leq 0<s$, together with dropping the $s^{\prime}$ term.
For $s^{\prime}<s\leq 0$ with $w_{s}=0$, we are forced to have
$\text{Width}(MR^{[s]})=2\widetilde{H}_{s}-1$ if $s\geq 0$ and
$\text{Width}(MR^{[s]})=2\widetilde{V}_{s}-1$ if $s\leq 0$, since
$\text{Width}(MR^{[s^{\prime}]})$ is guaranteed to be odd. For the former we
get

$M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})=2(\widetilde{H}_{s^{\prime}+r}-\widetilde{H}_{s})>0,$

since $s<s^{\prime}+r$. The latter yields

$M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})=2(\widetilde{H}_{s^{\prime}+r}-\widetilde{V}_{s})>0,$

since $s>s^{\prime}$.

Finally we are left with $w_{s^{\prime}}>1$ with $s^{\prime}<s$. If we have
$0\leq s^{\prime}<s$, then the fact that
$\widetilde{H}_{s^{\prime}+w_{s^{\prime}}r}$ is maximal ensures

$\displaystyle M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})$
$\displaystyle=2\left(\sum_{i=0}^{w_{s^{\prime}}-1}(s^{\prime}+ir)-\sum_{i=0}^{w_{s}-1}(s+ir)\right)+2(\widetilde{H}_{s+w_{s^{\prime}}r}-\widetilde{H}_{s+(w_{s^{\prime}}-1)r})$
$\displaystyle(\text{column shift})\,\,\,\,\,\,$
$\displaystyle=2\left(s^{\prime}+\sum_{i=1}^{w_{s^{\prime}}-1}(s^{\prime}+ir)-\sum_{i=1}^{w_{s^{\prime}}-1}(s+(i-1)r)\right)$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2(\widetilde{H}_{s+w_{s^{\prime}}r}-\widetilde{H}_{s+(w_{s^{\prime}}-1)r})$
$\displaystyle=2s^{\prime}+2(w_{s^{\prime}}-1)(s^{\prime}+r-s)+2(\widetilde{H}_{s+w_{s^{\prime}}r}-\widetilde{H}_{s+(w_{s^{\prime}}-1)r})$
$\displaystyle>0,$

Analogously, the same inequality holds true if $s^{\prime}\leq 0<s$ by
dropping the $2s^{\prime}$ term. For $s^{\prime}<s\leq 0$ a single
$(s^{\prime}+r-s)$ term disappears, but the inequality holds since
$s^{\prime}+r-s>0$ and
$\widetilde{H}_{s+w_{s^{\prime}}r}-\widetilde{H}_{s+(w_{s^{\prime}}-1)r}>0$.

Then since $M_{rel}(b^{s^{\prime}})>M_{rel}(b^{s})$ for every configuration of
$s$ relative to $s^{\prime}$ for $w_{s^{\prime}}>0$, we have
$\text{Width}(MR^{[s^{\prime}]})>\text{Width}(MR^{[s]})$. Together with the
argument for $w_{s^{\prime}}=0$, this completes the proof. ∎

Case B2: $\mathbf{\tau(K)=g(K)}$. Let us consider $1<r<2(g(K)-1)$ first,
delaying the penultimate slope to Lemma 5.5 and the maximal slope to Lemma
5.3. If $r<2(g(K)-1)$, then $l^{s^{\prime}}_{r}$ intersects
$\overline{\gamma}$ more than once for $s^{\prime}\equiv
g(K)-1\,\,(\text{mod}\,\,r)$. Our approach involves different arguments
depending on whether $l^{s^{\prime}}_{r}$ makes vertical intersections on both
sides of the $0$th column.

######  Lemma 5.2.

Suppose $K$ is thin with $\tau(K)=g(K)$, the surgery slope satisfies
$1<r<2(g(K)-1)$, and that there exists a $k$ properly dividing $r$ so that
every $[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$ satisfies $MR^{[s]}\cong
MR^{[s+k]}$ up to translation.

  * •

If $r<g(K)-1$, then $MR^{[s]}\cong MR^{[s^{\prime}]}$ up to translation only
if $[s]=[-s^{\prime}]$.

  * •

If $r\geq g(K)-1$, then $\tau(K)=g(K)=r=3$.

###### Proof.

If $r<g(K)-1$, then the slope of $l^{s^{\prime}}_{r}$ is small enough so that
intersecting it with $\overline{\gamma}$ produces vertical intersections in at
least 3 columns of $\widetilde{T}$. We know $w_{s^{\prime}}>0$ since
$r<2(g(K)-1)$, so suppose $w_{s^{\prime}}=1$. We have vertical intersections
with $\overline{\gamma}$ to the left and right of this column, which we can
label $c^{s^{\prime}}$ and $b^{s^{\prime}}$, respectively. Then
$M_{rel}(c^{s^{\prime}})=2\widetilde{V}_{s^{\prime}}-1$ and
$M_{rel}(b^{s^{\prime}})=2\widetilde{H}_{s^{\prime}}-1$, and so
$2\widetilde{H}_{s^{\prime}}-1\leq\text{Width}(MR^{[s^{\prime}]})\leq
2\widetilde{H}_{s^{\prime}}$ since $s^{\prime}>0$ yields
$\widetilde{H}_{s^{\prime}}>\widetilde{V}_{s^{\prime}}$. If some $[s]\neq[\pm
s^{\prime}]$ satisfies $MR^{[s]}\cong MR^{[s^{\prime}]}$ up to translation
then the parities of their widths must agree. Under the same labeling
convention for intersections associated to $[s]$, we see that
$2\widetilde{V}_{s}-1\leq\text{Width}(MR^{[s]})\leq 2\widetilde{V}_{s}$ if
$s\leq 0$ and $2\widetilde{H}_{s}-1\leq\text{Width}(MR^{[s]})\leq
2\widetilde{H}_{s}$ if $s\geq 0$.

For $0<s^{\prime}<s$, we compute for either parity of width that

$\text{Width}(MR^{[s^{\prime}]})-\text{Width}(MR^{[s]})=2(\widetilde{H}_{s^{\prime}}-\widetilde{V}_{s}).$

This implies that $\widetilde{V}_{s}=\widetilde{H}_{s^{\prime}}$, which is
impossible when $s^{\prime}>0$. Similarly, if $s<-s^{\prime}$ then the
analogous statement holds true using $\widetilde{H}_{s}$.

When $-s^{\prime}<s<s^{\prime}$, something interesting occurs. In addition to
$l^{s^{\prime}}_{r}$, we see that $l^{s}_{r}$ successfully makes two non-
vertical intersections on both sides of the $0$th column. Also since $K$ is
thin, it follows that $\widetilde{V}_{s}\leq\widetilde{H}_{s^{\prime}}$ and
$\widetilde{H}_{s}\leq\widetilde{H}_{s^{\prime}}$, with equality only possible
when $s=-s^{\prime}+1$ or $s=s^{\prime}-1$, respectively. However, this
results in the configuration shown for the latter situation in Figure 12. We
see that while the widths of $MR^{[s]}$ and $MR^{[s^{\prime}]}$ can agree if
$\widetilde{H}_{s}=\widetilde{H}_{s^{\prime}}$, this necessarily results in
$\widetilde{V}_{s}\neq\widetilde{V}_{s^{\prime}}$ since $K$ is thin. This is
true vice versa as well, and so the multisets cannot both contain the same
relative gradings for their respective vertical intersection pairs. Thus we
must consider $w_{s^{\prime}}>1$, and we will do so following similar
computations to those in Lemma 5.1.

2pt $\xrightarrow{\widetilde{V}_{s}-\widetilde{V}_{s^{\prime}}}$ at -60 230
$\xleftarrow{\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s}}$ at 180 230

Figure 12. If $\widetilde{H}_{s}=\widetilde{H}_{s^{\prime}}$ and
$s=s^{\prime}-1$, then $\widetilde{V}_{s}-\widetilde{V}_{s^{\prime}}=1$ when
$K$ is thin.

When $w_{s^{\prime}}>1$ the intersection with largest relative grading in
$MR^{[s]}$ comes from the furthest non-vertical intersection, which is $b^{s}$
when $s\geq 0$ or $c^{s}$ if $s\leq 0$. Since we can use the hyperelliptic
involution invariance of $\widehat{\textit{HF}}(M)$ to treat such a $c^{s}$ as
$b^{-s}$, let us only compare $M_{rel}(b^{s^{\prime}})$ to the various
$M_{rel}(b^{s})$. Chaining grading differences between adjacent non-vertical
intersections from $b^{s}$ back to $a^{s}$, we have

$\displaystyle
M_{rel}(b^{s})=2\left(\widetilde{H}_{s}+\sum_{i=1}^{w_{s}-1}(s+ir)\right)-1.$

If $s<s^{\prime}$ then $w_{s}=w_{s^{\prime}}$, and we compute

$\displaystyle M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})$
$\displaystyle=2\left(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s}+\sum_{i=1}^{w_{s}-1}(s+ir)-\sum_{i=1}^{w_{s}-1}(s+ir)\right)$
$\displaystyle=2(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s}+(w_{s^{\prime}}-1)(s^{\prime}-s))$
$\displaystyle\geq 1.$

We have equality only if $\widetilde{H}_{s^{\prime}}=\widetilde{H}_{s}$,
$s=s^{\prime}-1$, and $w_{s^{\prime}}=2$. In this case, we have
$\text{parity}(s^{\prime}+2r)=\text{parity}(s^{\prime})$ regardless of $r$.
However $\text{parity}(s^{\prime}+2r)\neq\text{parity}(\tau(K))$, and so
$\text{parity}(s)=\text{parity}(\tau(K))$. This implies
$\text{Width}(MR^{[s]})\leq M_{rel}(b^{s})$, which means we cannot have
$s<s^{\prime}$.

If $s>s^{\prime}$, then with $w_{s}=w_{s^{\prime}}-1$ we obtain

$\displaystyle M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})$
$\displaystyle=2\left(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s}+\sum_{i=1}^{w_{s}-1}(s+ir)-\sum_{i=1}^{w_{s}-1}(s+ir)\right)$
$\displaystyle(\text{column shift})\,\,\,\,\,\,$
$\displaystyle=2\left(s^{\prime}+r+\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s}+\sum_{i=2}^{w_{s}-1}(s+ir)-\sum_{i=2}^{w_{s^{\prime}}-1}(s+ir)\right)$
$\displaystyle=2(s^{\prime}+r)-2(\widetilde{H}_{s}-\widetilde{H}_{s^{\prime}})+2(w_{s^{\prime}}-2)(s^{\prime}+r-s).$

Now $s-s^{\prime}\leq 2(\widetilde{H}_{s}-\widetilde{H}_{s^{\prime}})\leq
s-s^{\prime}+1$ when $K$ is thin by careful inspection of these regions. This
implies

$\displaystyle M_{rel}(b^{s^{\prime}})-M_{rel}(b^{s})$
$\displaystyle=2(s^{\prime}+r)+2(\widetilde{H}_{s^{\prime}}-\widetilde{H}_{s})+2(w_{s^{\prime}}-2)(s^{\prime}+r-s)$
$\displaystyle\geq
2(s^{\prime}+r)-(s-s^{\prime}+1)+2(w_{s^{\prime}}-2)(s^{\prime}+r-s)$
$\displaystyle=2s^{\prime}+r-1+(2w_{s^{\prime}}-3)(s^{\prime}+r-s)$
$\displaystyle>1.$

Altogether, these grading comparisons are enough to see that
$MR^{[s]}\not\cong MR^{[s^{\prime}]}$ up to translation when $r<g(K)-1$. Next
we look at the cases with larger surgery slopes.

If $r=g(K)-1$, then $w_{s^{\prime}}=1$ and $s^{\prime}=0$. Due to
hyperelliptic involution invariance, we can assume $s\neq s^{\prime}$
satisfies $s<0$. Notice that $MR^{[s^{\prime}]}$ contains
$2\widetilde{H}_{s^{\prime}}-1$ with multiplicity at least two since
$l^{s^{\prime}}_{r}$ generates non-vertical intersections on both sides of the
0th column and $\widetilde{V}_{s^{\prime}}=\widetilde{H}_{s^{\prime}}$. The
only way that $MR^{[s]}$ could contain this grading with multiplicity greater
than one is if a simple figure-eight component in the 1st column has an
intersection with $l^{s}_{r}$. The nearby vertical intersection $a^{s+r}$ has
$M_{rel}(a^{s+r})=2\widetilde{H}_{s^{\prime}}-2$, and so we would require
$\text{parity}(s+r)\neq\text{parity}(\tau(K))$ in order for an intersection
with a simple figure-eight to have the desired grading. However
$s+r=\tau(K)-2$, and so $MR^{[s]}$ cannot contain
$2\widetilde{H}_{s^{\prime}}-1$ more than once. Thus, no $[s]\neq[s^{\prime}]$
satisfies $MR^{[s]}\cong MR^{[s^{\prime}]}$ up to translation when $r=g(K)-1$.

We still have $w_{s^{\prime}}=1$ if $r>g(K)-1$, but now $s^{\prime}<0$. The
crux of the argument in the previous case relied on
$\widetilde{H}_{s^{\prime}}>1$. This holds more generally when $r<2g(K)-3$,
except now $MR^{[s^{\prime}]}$ need only contain
$2\widetilde{H}_{s^{\prime}}-1$ once. Since
$\text{Width}(MR^{[s^{\prime}]})\geq 3$, the above argument still applies to
show that $MR^{[s]}\cong MR^{[s^{\prime}]}$ up to translation only if
$[s]=[s^{\prime}\pm 1]$. This forces $k=1$, which in turn forces $r\leq 3$ so
that there cannot exist $[s^{\prime}+\alpha k]$ with
$\text{Width}(MR^{[s^{\prime}+\alpha k]})=1$. The possibility $r=2$ is handled
exactly as in the $r=g(K)-1$ argument, and so we must have $r=3$. This means
$s^{\prime}=-1$, and so $\tau(K)=g(K)=r=3$.

If $r=2g(K)-3$, then only $l^{\pm s^{\prime}}_{r}$ generates non-vertical
intersections between columns of $\widetilde{T}$. All other $l^{s}_{r}$
intersect $\widehat{\textit{HF}}(M)$ only in the 0th column, which means that
every $\text{Width}(MR^{[s]})=1$. Since
$\text{parity}(s^{\prime})=\text{parity}(\tau(K))$ when $r=2g(K)-3$, we must
have $e_{s^{\prime}}=0$. This is because a simple figure-eight component at
this height would contribute an intersection with relative grading $-1$ to
$MR^{[s^{\prime}]}$, which would yield $\text{Width}(MR^{[s^{\prime}]})=2$ and
prevent periodicity. We have dim
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s^{\prime}])=3+2e_{s^{\prime}+r}$, and so
some $[s^{\prime}+k]$ satisfying $MR^{[s^{\prime}+k]}\cong MR^{[s^{\prime}]}$
up to translation forces $1+2e_{s^{\prime}+k}=3+2e_{s^{\prime}+r}$, or
$e_{s^{\prime}+k}=e_{s^{\prime}+r}+1$. If necessary, translate
$MR^{[s^{\prime}+k]}$ so that $0$ is the smallest element. The only way that
the multiplicities of $0$ and $1$ agree is if $MR^{[s]}$ contains more 0’s
than 1’s, which happens only when
$\text{parity}(s^{\prime}+k)=\text{parity}(\tau(K))$. But
$\text{parity}(s^{\prime})=\text{parity}(\tau(K))$ as well, which is a
contradiction since $k$ is odd when $r$ odd. ∎

We return to the two unhandled cases of $\tau(K)=g(K)=r=3$ and $r=2(g(K)-1)$
shortly in Subsection 5.1, and for now are left with the case where
$r=2g(K)-1$. Because $\tau(K)=g(K)$, there is no guaranteed simple figure-
eight at height $g(K)-1$. This small difference is enough of an issue if $K$
is an $L$-space knot, since each $MR^{[s]}=\left\\{0\right\\}$ means
$\widehat{\textit{HF}}$ cannot provide an obstruction. With existing
techniques, we can only show the following:

######  Lemma 5.3.

Suppose $K$ is thin with $\tau(K)=g(K)$, and let $r=2g(K)-1$. If there exists
a $k$ properly dividing $r$ such that every
$[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$ satisfies
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s])\cong\widehat{\textit{HF}}(S^{3}_{r}(K),[s+k])$,
then $K$ is an $L$-space knot.

###### Proof.

Suppose for the sake of contradiction that $K$ is not an $L$-space knot,
meaning that dim $\widehat{\textit{HF}}(S^{3}_{r}(K),[s])>1$ for some
$[s]\in\text{Spin}^{c}(S^{3}_{r}(K))$. Each $l^{s}_{r}$ intersects
$\overline{\gamma}$ precisely once since $r\geq 2g(K)-1$. In order to have dim
$\widehat{\textit{HF}}(S^{3}_{p}(K),[s])>1$ for some $[s]$, we need for
$\widehat{\textit{HF}}(M)$ to have a simple figure-eight component at height
$s$. Let $t$ be the height of the lowest simple figure-eight component. We
have $e_{t}$ many simple figure-eights at height $t$, and so we must also have
$e_{t+k}=e_{t}$ many simple figure-eight components at height $t+k$ to satisfy

$\text{dim}\,\,\widehat{\textit{HF}}(S^{3}_{r}(K),[t])=\,\text{dim}\,\,\widehat{\textit{HF}}(S^{3}_{r}(K),[t+k]).$

If $\text{parity}([t])\neq\text{parity}([t+k])$, then one of $MR^{[t]}$ or
$MR^{[t+k]}$ contains $-1$ and would need to be translated by 1 to make 0 the
smallest element by Proposition 3.3. However, this results in both multisets
having unequal multiplicities of $0$’s and $1$’s. This would lead to
$MR^{[t]}\not\cong MR^{[t+k]}$ up to translation, and so we must have
$\text{parity}([t+k])=\text{parity}([t])$. However this condition implies that
$k$ is even, which contradicts $r=2g(K)-1$ being odd. Therefore, $K$ must be
an $L$-space knot. ∎

###  5.1. Obstructions from Absolute Gradings

Until now, we have primarily appealed to information carried by
$\widehat{\textit{HF}}(S^{3}_{r}(K))$ as this is currently the only flavor of
Heegaard Floer homology computable by immersed curves techniques. When
considering $S^{3}_{r}(K)=Y\\#Z$ with $|H^{2}(Y)|=k<\infty$, we will use
properties of the $d$-invariants mentioned in Section 2 in order to obtain a
relationship between $r$, $k$, and the $V$’s associated to $K$. We initially
settle the curious $\tau(K)=g(K)=r=3$ case, and afterwards assemble the proof
of Theorem 1.2.

######  Lemma 5.4.

Let $K$ be a thin knot with $\tau(K)=g(K)=3$. Then $S^{3}_{3}(K)$ is
irreducible.

###### Proof.

If $S^{3}_{3}(K)$ is reducible, it must admit an integer homology sphere
connected summand $Y$ since $r=3$ is prime. Using the additivity of the
$d$-invariants, we have

$d(S^{3}_{3}(K),[s])=d(L(3,\pm 1),[s])+d(Y).$

Proposition 2.3 then implies $d(Y)=-2V_{0}(K)=-2V_{1}(K)$, which in turn
forces $V_{0}(K)=V_{1}(K)$. However this is true only for thin knots with even
$\tau(K)$, which can be seen using the formula in Proposition 3.3 together
with $V_{s}=H_{-s}$. This forms the desired contradiction. ∎

######  Lemma 5.5.

Let $K$ be a thin knot with $\tau(K)\geq g(K)-2$. Then $S^{3}_{r}(K)$ is
irreducible when $r=2(g(K)-1)$.

###### Proof.

Let $\tau(K)\geq g(K)-2$, and suppose for the sake of contradiction that
$S^{3}_{r}(K)$ is reducible for $r=2(g(K)-1)$. Then $S^{3}_{r}(K)$ admits
connected summands $Y$ a lens space and $Z$ with $|H^{2}(Z)|=k<\infty$. Since
$H_{1}(S^{3}_{r}(K))$ is cyclic and $r$ is even, one of
$|H_{1}(Y)|=\frac{r}{k}$ or $k$ is even. We will show the latter must be true.

Using the immersed curves techniques of the previous section, we see that
$\text{Width}(MR^{[s]})$
$=1$ for all $[s]$ when $K$ is thin and $\tau(K)\geq g(K)-2$. Using
$s^{\prime}\equiv g(K)-1(\,\,\text{mod}\,\,r)$ again, we are guaranteed to
have dim $\widehat{\textit{HF}}(S^{3}_{r}(K),[s^{\prime}])>1$ since
$l^{s^{\prime}}_{r}$ either intersects a simple figure-eight at height
$g(K)-1$ when $\tau(K)<g(K)$ or intersects $\overline{\gamma}$ multiple times
when $\tau(K)=g(K)$. In order for some $[s^{\prime}-k]$ to satisfy
$MR^{[s^{\prime}-k]}\cong MR^{[s^{\prime}]}$ up to translation, we also
require $\text{parity}(s^{\prime}-k)=\text{parity}(s^{\prime})$ so that the
multiplicities of 0 and 1 agree. This implies $k$ is even.

Let $\pi_{Y}([s])$ and $\pi_{Z}([s])$ denote the restrictions of $[s]$ to
$\text{Spin}^{c}(Y)$ and $\text{Spin}^{c}(Z)$, respectively. Since
$\text{Spin}^{c}(S^{3}_{r}(K))\cong\mathbb{Z}/r\mathbb{Z}$ is
$\mathbb{Z}/r\mathbb{Z}$-equivariant [OS08], we have both

$\pi_{Y}([s+\tfrac{r}{k})])=\pi_{Y}([s])\,\,\text{and}\,\,\pi_{Z}([s+k])=\pi_{Z}([s]).$

The two self-conjugate $\text{spin}^{c}$ structures of $S^{3}_{r}(K)$ must
project onto the lone self-conjugate structure of $L(\tfrac{r}{k},q)$, and so

$\pi_{Y}([0])=\pi_{Y}([\tfrac{r}{2}])\in\text{Spin}^{c}(L(\tfrac{r}{k},q)).$

Their respective restrictions on $Z$ are distinct, and so let
$\pi_{Z}([0])=\mathfrak{u}_{e}$ and
$\pi_{Z}([\tfrac{r}{k}])=\mathfrak{u}_{o}$. Due to the additivity of
$d$-invariants, we have

$d(S^{3}_{r}(K),[s])=d(L(\tfrac{r}{k},q),\pi_{Y}([s]))+d(Z,\pi_{Z}([s])).$

Since $k$ is even, we may apply this to the self-conjugate structures to see

$\displaystyle d(S^{3}_{r}(K),[0])-d(S^{3}_{r}(K),[\tfrac{r}{2}])$
$\displaystyle=(d(L(\tfrac{r}{k},q),[0])+d(Z,\mathfrak{u}_{e}))-(d(L(\tfrac{r}{k},q),[0])+d(Z,\mathfrak{u}_{o}))$
$\displaystyle=d(Z,\mathfrak{u}_{e})-d(Z,\mathfrak{u}_{o}).$

Observe that $\pi_{Z}([\tfrac{k}{2}])=\mathfrak{u}_{0}$, and so
$\pi_{Z}([\tfrac{r+k}{2}])=\mathfrak{u}_{e}$. We likewise have
$\pi_{Y}([\tfrac{k}{2}])=\pi_{Y}([\tfrac{r+k}{2}])$, and so

$d(S^{3}_{r}(K),[\tfrac{k}{2}])-d(S^{3}_{r}(K),[\tfrac{r+k}{2}])=d(Z,\mathfrak{u}_{o})-d(Z,\mathfrak{u}_{e}).$

Using the inductive formula for $d(L(p,q))$ [OS03a, Proposition 4.8], it
follows that $d(L(r,1),[s])=\tfrac{s^{2}}{r}-s+\tfrac{r-1}{4}$. Summing the
prior two equations and using Proposition 2.3 (with
$V_{\frac{r-k}{2}}=\text{max}\\{V_{\frac{r+k}{2}},V_{r-\frac{r+k}{2}}\\}$)
yields

$\displaystyle
2\left(V_{0}-V_{\frac{r}{2}}+V_{\frac{k}{2}}-V_{\frac{r-k}{2}}\right)$
$\displaystyle=d(L(r,1),[0])-d(L(r,1),[\tfrac{r}{2}])+d(L(r,1),[\tfrac{k}{2}])$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-d(L(r,1),[\tfrac{r+k}{2}])$
$\displaystyle=-\left(\frac{r^{2}}{4r}-\frac{r}{2}\right)+\left(\frac{k^{2}}{4r}-\frac{k}{2}\right)-\left(\frac{(r+k)^{2}}{4r}-\frac{r+k}{2}\right)$
$\displaystyle=\frac{r-k}{2},$

Therefore, we have the following relationship between $r$, $k$, and the $V$’s
associated to $K$:

(1)
$\frac{r-k}{4}=(V_{0}-V_{\frac{r}{2}})+(V_{\frac{k}{2}}-V_{\frac{r-k}{2}}).$

Notice that when $K$ is thin and $\tau(K)\geq 0$, we have that

$\displaystyle V_{0}=\left\\{\begin{array}[]{l c}\dfrac{\tau(K)+1}{2}&\text{if
parity}(\tau(K))=1\\\ \dfrac{\tau(K)}{2}&\text{if parity}(\tau(K))=0\\\
\end{array}\right.$

We will use this to generate contradictions, and break into cases since the
values of $V_{\frac{r}{2}}$ and $V_{\frac{r-k}{2}}$ depend on $\tau(K)$. It
will also be useful to use the fact that $k\leq\frac{r}{3}$.

Case C1: $\mathbf{\tau(K)=g(K)}$. Here we have $V_{\frac{r}{2}}=1$ and
$V_{\frac{r-k}{2}}>0$. We see that $V_{\frac{k}{2}}-V_{\frac{r-k}{2}}$ is
given by half the distance between $\frac{r-k}{2}-\frac{k}{2}$ since $K$ is
thin. Thus,

$V_{\frac{k}{2}}-V_{\frac{r-k}{2}}=\frac{r}{4}-\frac{k}{2}.$

This together with Equation 1 above then yields

$V_{0}=\frac{k}{4}+1.$

If $\text{parity}(\tau(K))=1$, then we have

$\displaystyle\,\,\,\,\,\,\,\,\,\,\frac{\tau(K)+1}{2}=\frac{k}{4}+1$
$\displaystyle\Leftrightarrow\frac{\tau(K)-1}{2}=\frac{k}{4}$
$\displaystyle\Leftrightarrow\frac{\tau(K)-1}{2}\leq\frac{r}{12}$
$\displaystyle\Leftrightarrow 6(\tau(K)-1)\leq 2(g(K)-1)$
$\displaystyle\Leftrightarrow 6(\tau(K)-1)\leq 2(\tau(K)-1)$
$\displaystyle\Rightarrow 4\tau(K)\leq 4.$

However this would imply $g(K)=\tau(K)=1\Rightarrow r=0$, a clear
contradiction. If $\text{parity}(\tau(K))=0$, then similar reasoning yields
$\tau(K)\leq 2$, which forces $r=\tau(K)=g(K)=2$. We return to immersed curves
techniques to rule out this case by comparing the multiplicity of elements of
$MR^{[0]}$ and $MR^{[1]}$. Since $\text{parity}(\tau(K))=0$, we must translate
$MR^{[0]}$ by one so that $0$ is its smallest element. The multiplicity of $0$
in the translated $MR^{[0]}$ is $e_{0}$, and the multiplicity of $0$ in
$MR^{[1]}$ is $2e_{1}+2$. However if $S^{3}_{2}(K)$ is reducible then Lemma
2.1 forces $e_{0}=2e_{1}+1$ in order for dim
$\widehat{\textit{HF}}(S^{3}_{2}(K),[1])=$ dim
$\widehat{\textit{HF}}(S^{3}_{2}(K),[0])$, generating the desired
contradiction.

Case C2: $\mathbf{\tau(K)=g(K)-1}$. Once again $V_{\frac{r-k}{2}}>0$, and in
this case we obtain $V_{0}=\frac{k}{4}$ since $V_{\frac{r}{2}}=0$. Together
with $r=2\tau(K)$, the argument of the previous case yields the contradiction
$4\tau(K)\leq-4$ when $\text{parity}(\tau(K))=1$ or $4\tau(K)\leq 2$ when
$\text{parity}(\tau(K))=0$.

Case C3: $\mathbf{\tau(K)=g(K)-2}$. We still have $V_{\frac{r}{2}}=0$, and
things are more interesting here since it is possible for
$V_{\frac{r-k}{2}}=0$. This happens only if $k=2$, in which case Equation 1
becomes

$\frac{r-2}{4}=V_{0}+V_{1}.$

Curiously enough $\tau(K)=V_{0}+V_{1}$ for a thin knot, and so this would
force $\tau(K)=\frac{r-2}{4}=\frac{2(\tau(K)+1)-2}{4}\Leftrightarrow
4\tau(K)=2\tau(K)\Rightarrow\tau(K)=0$. However this forces $r=k$, a
contradiction. Then we cannot have $k=2$, and so $V_{\frac{r-k}{2}}>0$ and we
once again have $V_{0}=\frac{k}{4}$. As with the previous cases, having
$r=2(\tau(K)+1)$ would yield the contradictions $6(\tau(K)+1)\leq
2(\tau(K)+1)$ if $\text{parity}(\tau(K))=1$ or $4\tau(K)\leq 2$ if
$\text{parity}(\tau(K))=0$. This completes the proof. ∎

###  5.2. Proof of Theorem 1.2

###### Proof.

Suppose $S^{3}_{r}(K)$ is reducible for $K$ thin and hyperbolic. The Matignon-
Sayari bound implies that $1<r\leq 2g(K)-1$, after mirroring the knot if
necessary to make the surgery slope positive. Reducibility also gives
$S^{3}_{r}(K)\cong Y\\#Z$ for some lens space $Y$ and some $Z$ with
$|H^{2}(Z)|=k<\infty$. By Lemma 2.1, we have
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s+\alpha
k])\cong\widehat{\textit{HF}}(S^{3}_{r}(K),[s])$ for arbitrary
$[s],\alpha\in\mathbb{Z}/r\mathbb{Z}$. When $r<2(g(K)-1)$, Lemmas 4.2, 5.1,
and 5.2 apply to show that there exists an
$[s^{\prime}]\in\text{Spin}^{c}(S^{3}_{r}(K))$ such that either
$\widehat{\textit{HF}}(S^{3}_{r}(K),[s^{\prime}])$ is relatively-graded
isomorphic only to $\widehat{\textit{HF}}(S^{3}_{r}(K),[-s^{\prime}])$, or
$\tau(K)=g(K)=r=3$. The latter is prevented by Lemma 5.4, so we proceed with
the former. Since $Y\not\cong S^{3}$, we see that $[s^{\prime}]$ cannot be
self-conjugate and also that $|H_{1}(Y)|=2$. This implies $Y=\mathbb{R}P^{3}$,
as well as $k=|[s^{\prime}]-[-s^{\prime}]|=2|[s^{\prime}]|$. However, together
this means 4 divides $r$, which is impossible when $S^{3}_{r}(K)$ admits an
$\mathbb{R}P^{3}$ summand with $H_{1}(S^{3}_{r}(K))$ cyclic.

Therefore, we must have $r\geq 2(g(K)-1)$. Lemmas 4.4 and 4.5 cover
$-g(K)<\tau(K)<g(K)-2$ and Lemma 5.5 covers $\tau(K)\geq g(K)-2$ for the
possibility that $r=2(g(K)-1)$, and so we must have $r=2g(K)-1$. In this
situation $k$ is odd since $r$ is odd, which means periodicity will cycle
through $\text{spin}^{c}$ structures with different parities. Then Lemmas 4.3,
4.5, and 5.1 apply to fully obstruct reducibility via the above argument if
$\tau(K)\neq g(K)$. If $\tau(K)=g(K)$, our techniques have been exhausted and
leave just the conclusion of Lemma 5.3, showing that $K$ must be an $L$-space
knot. ∎

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