# A Rank Bound in Link Floer Homology and Detection Results

Fraser Binns  and  Subhankar Dey  Department of Mathematics, Boston College
<EMAIL_ADDRESS>Department of Mathematics, University of Alabama<EMAIL_ADDRESS>
###### Abstract.

Using the Heegaard Floer invariant of Legendrian links we obtain lower bounds
on the rank of certain Alexander gradings in the knot Floer homology of links,
generalising work of the second author. These allow us to classify links with
knot Floer homology of rank at most eight, and prove a variant of a
classification of links with Khovanov homology of low rank due to Xie-Zhang.
In another direction we use a variant of Ozváth-Szabó classification of
$E_{2}$ collapsed ${\mathbb{Z}}\oplus{\mathbb{Z}}$ filtered chain complexes to
show that knot Floer homology detects $T(2,8)$ and $T(2,10)$. Combining these
techniques with the spectral sequences of Batson-Seed, Dowlin, and Lee we can
show that Khovanov homology likewise detects $T(2,8)$ and $T(2,10)$.

##  1\. Introduction

Link Floer homology is a powerful invariant of links in $S^{3}$ due to
Ozsváth-Szabó, taking value in the category of multi-graded vector spaces
[HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial]. In
this paper we address several variants of the “geography” question for link
Floer homology – namely which vector spaces arise as the link Floer homology
of links as well as the “botany” question – which links have a prescribed link
Floer homology.

We take two approaches to the geography problem. The first of these is via
contact geometry. There exist a number of bridges between Heegaard Floer
theory and contact geometry, the original of which was Ozsváth-Szabó’s contact
class [ozsvath2005heegaard]. We will be interested primarily in the Heegaard
Floer Legendrian knot invariant due to Lisca-Ozsváth-Stipsicz-Szabó
[lisca2009heegaard] (the LOSS invariant)– which takes values in knot Floer
homology – as well as its generalization to links due to Cavallo
[cavallo2017invariant] – which takes value in collapsed link Floer homology.
Indeed, in section 4 we will show that Cavallo’s invariant can be thought of
as taking value in link Floer homology, complete with multi-Alexander grading.

This result enables us to generalize the second author’s main result in
[dey2020note], which is in turn a generalization of the main theorem of
Baldwin–Vela-Vick in [baldwin_note_2018]. The most readily stated version of
this generalization is;

######  Theorem 4.2.

Let $L$ be a non-trivial $n$ component Legendrian link in $(S^{3},\xi)$.
Suppose $\hat{\mathcal{L}}(L)$ is non-zero of Alexander grading $A$ in
$\widehat{\operatorname{HFK}}(L)$. Then

$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(A-1))+\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(A+1))\geq
n$

Here we may take coefficients in ${\mathbb{Z}},{\mathbb{Z}}/2$ or
${\mathbb{Q}}$, as we do for the remainder of this paper unless the
coefficients are specified. We note that $A$ is determined by the classical
invariants of the Legendrian link $L$. We note also that this result, as well
as the other results in section 4 hold in rational homology spheres. See
Theorem 4.4 for a stronger version, and Corollary 4.5 for a still stronger
version under additional geometric hypotheses.

Theorem 4.2 allows us to prove a number of detection results;

######  Theorem 8.1.

If $\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=4$ then $L$ is a Hopf
link or the three component unlink.

######  Theorem 8.2.

If $\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=6$ then $L$ is a
disjoint union of an unknot and a trefoil.

######  Theorem 8.3.

If $\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=8$ then $L$ is
$T(2,4)$, $T(2,-4)$, a four component unlink, or the disjoint union of a Hopf
link and an unknot.

######  Theorem 8.6.

Knot Floer homology detects the link consisting of $T(2,3)$ and a meridian.

These results can be viewed as extensions of some of the detection results in
[BG]. We note also that while this manuscript was in preparation, an
independent proof of the fact that the disjoint union of the Hopf link and the
unknot is the only three component link with knot Floer homology of rank eight
appeared in [kim2020links].

Applying Dowlin’s spectral sequence from Khovanov homology to knot Floer
homology, Theorems 8.1, 8.2 and 8.3 allow us to prove variants of results of
Xie-Zhang [xie2020links].

######  Corollary 9.1.

Suppose $L$ is a two component pointed link with
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L,p;{\mathbb{Q}}))\leq 4$.
Then $L$ is one of;

  * •

an unlink;

  * •

a Hopf link;

  * •

$T(2,4)$ or $T(2,-4)$.

######  Corollary 9.2.

Suppose $L$ is a three component pointed link. Then
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L,p;{\mathbb{Q}}))>2$.

Here $\widetilde{\operatorname{Kh}}(L,p;{\mathbb{Q}})$ is the reduced Khovanov
homology of the pointed link $(L,p)$. Lidman has pointed out to the authors
that Corollary 9.2 can be obtained from the spectral sequence from reduced
Khovanov homology of $L$ to the double branched cover of $L$ due to Ozsváth-
Szabó [ozsvath2005heegaard].

We can show link Floer homology detection results for a number of other
infinite families of links. Let $K_{2,2n}$ denote the cable link with pattern
$T(2,2n)$ and companion $K$. We show that link Floer homology detection
results for the $(2,2n)$ cables of the remaining knots that knot Floer
homology is currently known to detect; namely the trefoils and the figure
eight.

######  Theorem 5.1.

Link Floer homology detects $T(2,3)_{2,2n}$.

Note here that the $(2,2n)$-cable of $T(2,3)$ is the $(2,-2n)$-cable of
$T(2,-3)$.

######  Theorem 5.2.

Let $K$ be the figure eight knot. Link Floer homology detects $K_{2,2n}$ for
every $n$.

Our second approach to the geography problem via algebra, namely a version of
Ozsváth-Szabó’s classification of $E_{2}$ collapsed chain complexes
[HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial, Section
12]. We can obtain the following results;

######  Theorem 7.1.

Knot Floer homology detects $T(2,8)$.

######  Theorem 7.2.

Knot Floer homology detects $T(2,10)$.

Here we have given $T(2,8)$ and $T(2,10)$ the braid orientation, contrary to
the case in [binns2020knot], where knot Floer homology was shown to detect
$T(2,2n)$, oriented as the boundary of an annulus, for all $n$.

Combining these techniques with the Dowlin [dowlin2018spectral], Lee
[lee2005endomorphism], and Batson-Seed [batson2015link] spectral sequences we
obtain the corresponding detection results for Khovanov homology.

######  Theorem 9.3.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,8);{\mathbb{Z}})$.
Then $L$ is isotopic to $T(2,8)$.

######  Theorem 9.4.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,10);{\mathbb{Z}})$.
Then $L$ is isotopic to $T(2,10)$.

Khovanov homology was previously known to detect $T(2,2n)$ for $n=\pm 3$ by
work of Martin [GageT26], $n=\pm 2$ by Zhang-Xie [xie_instanton_2019], $n=\pm
1$ by Baldwin-Sivek-Xie [baldwin2018khovanovhopf], and $n=0$ by a combination
of Hedden-Watson [hedden2010manifolds] and Kronheimer-Mrowka
[kronheimer2011khovanov].

The outline of this paper is as follows: in section 2 we briefly review
pertinent properties of link Floer homology. After discussing generalization
of LOSS invariant in section 3, we use it to prove several rank bounds results
in link Floer homology results in section 4. In section 5 we collect our
detection results for $(2,2n)$-cables In section 6, we prove various technical
results necessary for subsequent sections. In section 7 we show knot Floer
homology detects $T(2,8)$ and $T(2,10)$, in section 8 we give botany results
for the rank of Knot Floer homology and in Section 9 we give detection results
for Khovanov homology.

### Acknowledgements

We would like to thank the organisers of the “Trends in Low Dimensional
Topology” seminar series for providing the platform which led to this project,
as well as Tye Lidman and Jen Hom for comments on an earlier draft. We would
especially like to thank Lev Tovstopyat-Nelip for pointing out an error in an
earlier version of this paper. The first author would also like to thank John
Baldwin and Gage Martin for numerous helpful conversations, as well as Tao Li.

##  2\. A Very Brief Review of Link Floer homology

In this section we review link Floer homology, partially to fix conventions
and notation, with an emphasis on the structural properties that we will be
using in the subsequent sections.

Let $L$ be an oriented $n$ component link in a $3$-manifold $M$. The pair
$(L,M)$ can be encoded as a Heegaard diagram for $M$ with $n$ pairs of
basepoints $\\{w_{i},z_{i}\\}$. The link Floer complex of $L$, defined by
Ozsváth-Szabó in
[HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial], is a
multi-graded ${\mathbb{Z}}/2[U_{1},U_{2},\dots U_{n}]$ chain complex with
underlying vector space;

$\operatorname{CFL}^{-}(L,M)\cong\bigoplus_{m,A_{1},A_{2},\dots
A_{n}}\operatorname{CFL}^{-}_{m}(L,M)(A_{1},A_{2}\dots A_{n}).$

The complex can also be defined with ${\mathbb{Z}}$ coefficients
[manolescu2007combinatorial],[sarkar2011note] – and therefore
${\mathbb{Q}}$-coefficients. The three manifold $M$ is often apparent from the
context and duly suppressed in the notation. The $A_{i}$ gradings are called
the _Alexander gradings_ and can be thought of as elements in
${\mathbb{Z}}+\frac{\ell\text{k}(L_{i},L-L_{i})}{2}$, while $m$ is called the
_Maslov grading_ , which is integer valued. $\operatorname{CFL}^{-}(L)$ is
endowed with a differential which counts pseudoholomorphic disks in a certain
auxiliary symplectic manifold endowed with an appropriate almost complex
structure which do not intersect submanifolds $V_{z_{i}}$ determined by
$z_{i}$, weighted by a count of intersections with the manifolds $V_{w_{i}}$,
determined by $w_{i}$. The filtered chain homotopy type of
$\operatorname{CFL}^{-}(L)$ is an invariant of $L$, which we refer to as the
_link Floer complex of $L$_. The homology of $\operatorname{CFL}^{-}(L)$ is
denoted $\operatorname{HFL}^{-}(L)$. $\operatorname{CFL}^{-}(L)$ has a
quotient defined by setting $U_{i}=0$ for all $i$, whose filtered chain
homotopy type is also an invariant of $L$. This complex is denoted by
$\widehat{\operatorname{CFL}}(L)$, and has homology denoted by
$\widehat{\operatorname{HFL}}(L)$, which we will call the _link Floer homology
of $L$_. The _link Floer homology polytope_ of $L$ is defined as the convex
hull of the multi-Alexander gradings of $\widehat{\operatorname{HFL}}(L)$ with
non-trivial support. The link Floer polytope determines the Thurston polytope
of the link exterior by a result of Ozváth-Szabó
[ozsvath2008linkFloerThurstonnorm]. We note that
$\widehat{\operatorname{HFL}}(\overline{L})$, where $\overline{L}$ is $L$ with
the orientation reversed, can be recovered from $\operatorname{CFL}^{-}(L)$ by
only counting the disks which cross the appropriate basepoints.

For each component $L_{i}$ of $L$ there is a spectral sequence from
$\widehat{\operatorname{HFL}}(L)$ to
$\widehat{\operatorname{HFL}}(L-L_{i})\otimes
V[\frac{\ell\text{k}(L_{i},L_{j})}{2}]$. Here
$V\cong{\mathbb{F}}_{0}\oplus{\mathbb{F}}_{-1}$, supported in multi-Alexander
grading zero, while $[\frac{\ell\text{k}(L_{i},L_{j})}{2}]$ indicates a shift
in the $A_{j}$ grading. If a component $L_{i}$ of $L$ is fibered and $L$ is
non-split, then $L-L_{i}$ is braided about $L_{i}$ exactly when
$\widehat{\operatorname{HFL}}(L)$ is of rank $2^{n-1}$ in the maximal $A_{i}$
grading of non-trivial support by a result of Martin [GageT26].

The _knot Floer homology_ of $L$, $\widehat{\operatorname{HFK}}(L)$, is an
oriented link invariant due independently to Rasmussen [Rasmussen] and
Ozsváth-Szabó [Holomorphicdisksandknotinvariants] that can be obtained from
$\widehat{\operatorname{HFL}}(L)$ by projecting the multi-Alexander grading
onto the diagonal, and shifting the Maslov gradings up by $\frac{n}{2}$.
$\widehat{\operatorname{HFK}}(L)$ determines the maximal Euler characteristic
of a surface bounding $L$ [ni2006note], as well as whether or not $L$ fibered
[ghiggini2008knot],[ni2007knot].

The _Conway polynomial_ of $L$ can be obtained as an appropriate
decatigorification of $\widehat{\operatorname{HFL}}(L)$. A result of Hoste
[hoste1985firstcoefficientoftheconwaypolynomial] implies that the Conway
polynomial detects the linking number of two component links. It follows that
knot Floer homology and link Floer homology detect the linking number of two
component links.

##  3\. Legendrian link invariants in link Floer Homology

In this section we briefly review Cavallo’s [cavallo2017invariant]
generalization of Lisca-Ozsváth-Stipsicz-Szabó’s [lisca2009heegaard] (LOSS)
Legendrian knot invariant to links. We then strengthen Cavallo’s invariant, as
required to prove Proposition 4.4.

Cavallo defined his LOSS invariant in a version of link Floer homology called
collapsed link Floer homology, which has only a single Alexander grading. The
collapsed link Floer homology of a link $L$ in a manifold $M$ is an analogue
of the single variable $\mathbb{F}[U]$ module for knot Floer homology for a
knot;

$\operatorname{cCFL}^{-}(L,M)=\frac{\operatorname{CFL}^{-}(L,M)}{U_{1}=U_{2}=\cdots=U_{n}}$

It has differential induced by the differential on the
$\operatorname{CFL}^{-}(L,M)$ complex and a single Alexander grading.

Let $L$ be an $n$ component Legendrian link in Legendrian link $L$ in
$(M,\xi)$ with components $L_{i}$. To define the LOSS invariant,
$\mathcal{L}(L)\in\operatorname{cHFL}^{-}(L,\overline{M})$, Cavallo associates
to every triple $(L,M,\xi)$ an _adapted triple_ $(\Sigma,\phi,A)$, defined as
follows;

  * •

$\Sigma$ is a surface of genus $g$ with $l$ boundary components.

  * •

$(\Sigma,\phi)$ is an open book compatible with $(M,\xi)$.

  * •

$L$ is contained in the page $\Sigma_{1}$.

  * •

$\\{[L_{i}]\\}\subset H_{1}(\Sigma;{\mathbb{Q}})$ is linearly independent.

  * •

$A$ is a _system of generators for $\Sigma$ adapted to $L$._

A system of generators for $\Sigma$ adapted to $L$ is a collection of arcs
$\\{a_{1},a_{2},\dots a_{2g+l+n-2}\\}$ in $\Sigma$ that satisfies the
following conditions;

  1. (1)

$\\{a_{1},a_{2}\dots,a_{2g+l-1}\\}$ is a basis for
$H_{1}(\Sigma,\partial\Sigma;\mathbb{Q})$.

  2. (2)

$|L\cap a_{i}|=|L_{i}\cap a_{i}|=1$ for $1\leq i\leq n$. $a_{i}$ is called the
_distinguished arc_ for $L_{i}$.

  3. (3)

$\Sigma\setminus\underset{i\geq 2g+l}{\bigcup}a_{i}$ consists of $n$ disks,
each containing a component of $L$.

  4. (4)

For $i\geq 2g+l$, the two endpoints of $a_{i}$ lie on distinct boundary
components, and they separate a unique pair of components of the disks
$\Sigma\setminus\underset{i<2g+l}{\bigcup}a_{i}$.

$\mathcal{L}(L)$ is a generator of $\operatorname{cCFL}^{-}(L,\overline{M})$
coming from a Heegaard diagram induced from triple adapted to $L$. To obtain
that Heegaard diagram from an adapted triple $(\Sigma,\phi,A)$, form a set of
arcs $\\{b_{i}\\}$ by pushing each of the $\\{a_{i}\\}$ arcs by a small
isotopy along the orientation of the boundary components such that there is a
unique intersection point between $a_{i}$ and $b_{j}$ if $i=j$ and
$\beta_{j}\cap\alpha_{i}=\emptyset$ otherwise. From here, double the surface
$\Sigma$ to obtain a Heegaard surface for $Y$. The $\alpha_{i}$ curves are
obtained by doubling the $a_{i}$ curve while the $\beta_{i}$ curves are
obtained by doubling $b_{i}$ with $-\phi(b_{i})$. The $z_{i}$ and $w_{i}$
basepoints are placed around the distinguished arcs of $L$ according to the
orientation of the relevant component, see figure 2.

Figure 1. Two possible choices of planting basepoints, depending on the
orientation of $L$

Cavallo then defines his Legendrian invariant at the chain level, setting
$\mathcal{L}(L)$ to be the homology class, in
$\operatorname{cCFL}^{-}(L,\overline{M})$ of the generator
$\mathcal{L}(D)=\\{a_{i}\cap
b_{i}\\}_{i=1}^{2g+l+n-2}\in\operatorname{cCFL}^{-}(L,\overline{M})$
[cavallo2017invariant]. To show that $\mathcal{L}(D)$ yields a well defined
homology class Cavallo shows that the location of the basepoints
$\\{z_{i}\\},\\{w_{i}\\}$ ensures that
$\partial_{\operatorname{cCFL}^{-}}(\mathcal{L}(L))=0$. It is then shown that
$[\mathcal{L}(L)]$ is independent of the choice of adapted triple
(ref.[cavallo2017invariant, Theorem 1.2]). Cavallo shows that any two pairs of
adapted triples $(\Sigma_{1},\phi_{1},A_{1})$, $(\Sigma_{2},\phi_{2},A_{2})$
are related by the following open book isotopies;

  1. (1)

Global contact isotopies.

  2. (2)

Stabilization of $(\Sigma_{1},\phi_{1})$, $(\Sigma_{2},\phi_{2})$.

  3. (3)

Admissible arc slides on the families $A_{1}$, $A_{2}$; that is arc slides for
which the arcs do not cross basepoints.

Thus proving independence of the adapted triple amounts to proving that the
above operations induce chain maps sending the LOSS invariant of the first
diagram to the LOSS invariant of the second diagram. Cavallo’s argument shows
that the homology class $[\mathcal{L}(D)]$ is an element in both
$c\operatorname{HFL}^{-}(-L)$ and $\widehat{\operatorname{HFL}}(-L)$.

For the purposes of this paper, specifically for proving the results in
section 4, we will view the same element, $\mathcal{L}(D)$, as an element of
the link Floer chain complex $\operatorname{CFL}^{-}(L,\overline{M})$, which
still maps to the Legendrian link invariant that lives in
$\widehat{\operatorname{HFL}}(L,\overline{M})$. To distinguish our version of
the LOSS invariant for links from Cavallo’s we shall refer to it as the
“pulled back LOSS invariant”, and denote it by $\textit{pb}\mathcal{L}(L)$.

In order to ensure that $\textit{pb}\mathcal{L}(L)$ is well defined, we need
to check that it is independent of the choices of adapted triple; i.e. that
$\textit{pb}\mathcal{L}(L)$ is invariant under the moves 1,2 and 3 above. This
can be done by reapplying Cavallo’s techniques. The proofs of Lemmas 4.3 and
4.4 from [cavallo2017invariant] can be used verbatim to show invariance of
$\textit{pb}\mathcal{L}$ under global isotopies and stabilization of open
books. Invariance under admissible arc slides follows from Lemma 4.5 of
[cavallo2017invariant], since the admissibility condition ensures that the
arcs never cross the w basepoints. In each of the steps, Cavallo constructs
chain maps that relate two such compatible diagrams $D_{1}$ and $D_{2}$ and
the corresponding Legendrian link invariants. The same chain maps work in our
context as well, but now with domain $\operatorname{CFL}^{-}(D_{1})$ and
codomain $\operatorname{CFL}^{-}(D_{2})$, instead of
$\operatorname{cCFL}^{-}(D_{1})$ and $\operatorname{cCFL}^{-}(D_{2})$ as in
Cavallo’s work. This change does not affect the map since the original map is
$U$-equivariant and the relevant isotopies of open books do not cross the
basepoints. Therefore our map is $U_{i}$ equivariant for all $i$. Hence, in a
nutshell, we have that, given two diagrams $D_{1},D_{2}$, for the Legendrian
isotopic links, there is a chain map from $\operatorname{CFL}^{-}(D_{1})$ to
$\operatorname{CFL}^{-}(D_{2})$ sending $\textit{pb}\mathcal{L}(D_{1})$ to
$\textit{pb}\mathcal{L}(D_{2})$, as required.

It is worth noting that both of the Legendrian link invariants mentioned i.e.
$\mathcal{L}(L)$ and $\textit{pb}\mathcal{L}(L)$, map to the
$\widehat{\mathcal{L}}(L,M,\xi)$, living in
$\widehat{\operatorname{HFK}}(\overline{S^{3}},L)$.

We conclude this section by noting that, as a mild refinement of Theorem 5.6
of [cavallo2017invariant], the $i$th-Alexander grading of
$\textit{pb}\mathcal{L}(L)\in\widehat{\operatorname{HFL}}(L,\overline{M})$ is
given by

(1)
$\displaystyle\dfrac{\text{tb}(L_{i})+\text{rot}(L_{i})+\ell\text{k}(L_{i},L-L_{i})}{2}$

The proof of this fact follows from Cavallo’s proof of [cavallo2017invariant,
Theorem 5.6]

##  4\. Rank bounds in link Floer homology

With the various versions of the LOSS invariant described in the previous
section at hand, we can prove a generalization of the main technical result of
[dey2020note] on the non-triviality of LOSS invariant resulting in additional
elements in the knot Floer homology:

######  Proposition 4.1.

Let $L$ be a non-trivial Legendrian link in $S^{3}$ with respect to some
contact structure such that the Legendrian link invariant
$\widehat{\mathcal{L}}(L)(A_{1},A_{2},\cdots,A_{n})$ is non-zero. Then either

  * •

$\widehat{\operatorname{HFL}}(L)(A_{1},\cdots,A_{i}-1,A_{i+1},\cdots,A_{n})\neq
0$, or

  * •

$\widehat{\operatorname{HFL}}(L)(A_{1},\cdots,A_{i}+1,A_{i+1},\cdots,A_{n})\neq
0$ ,

for all $1\leq i\leq n$.

From here the theorem advertised in the introduction follows immediately;

######  Theorem 4.2.

Let $L$ be an $n$ component Legendrian link in $(S^{3},\xi)$. Suppose
$\hat{\mathcal{L}}(L)$ is non-zero and of Alexander grading $A$ in
$\widehat{\operatorname{HFK}}(L)$. Then
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(A-1))+\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(A+1))\geq
n$.

As a corollary we can generalise the main result of [baldwin_note_2018] to
links;

######  Corollary 4.3.

If $L$ is a non-trivial $n$ component fibered link in $S^{3}$ then;

$\operatorname{rank}(\widehat{\operatorname{HFK}}(L,\dfrac{n-2-\chi(L)}{2}))\geq
n.$

Before proceeding, we set notation for the different differential maps in
$\operatorname{CFL}^{-}$ and $\widehat{\operatorname{CFL}}$ that we use in
this section. By $\partial^{-}$, we denote the differential map in the chain
complex $\operatorname{CFL}^{-}$, which counts the pesudoholomorphic disks
with appropriate $U_{i}$ multiplicities, for $i=1,\cdots,n$, where $L$ has
$n$-many components. By plugging in $U_{i}=0$ in $\partial^{-}$, for
$i=1,\cdots,n$, we get $\partial_{\widehat{\operatorname{CFL}}}$, which is the
boundary map in $\widehat{\operatorname{CFL}}$. By
$\partial_{\widehat{\operatorname{CF}}}$, we denote the unfiltered version of
$\partial_{\widehat{\operatorname{CFL}}}$ i.e. the boundary map in
$\widehat{\operatorname{CF}}$ of the ambient manifold, without the additional
filtrations induced by the link.

With this notation in place, the main technical result in this section, is the
following;

######  Lemma 4.4.

Let $L$ be an $n$ component Legendrian link $L$ in $(S^{3},\xi)$ with a
component $L_{i}$. Suppose $\widehat{\mathcal{L}}(L)$ is non-zero, and that
there is an adapted triple $(B,\pi,A)$ for $(S^{3},\xi)$ such that $a_{i}$,
the distinguished arc for $L_{i}$, is sent to the left. Let $(A_{1},A_{2}\dots
A_{n})$ be the multi-Alexander grading of
$\widehat{\mathcal{L}}(L)\in\widehat{\operatorname{HFL}}(\overline{L})$. Then
there is a generator $\mathbf{d}\in\operatorname{CFL}^{-}(L)(A_{1},\dots
A_{i}-1,\dots A_{n})$ representing a non-trivial class in
$\widehat{\operatorname{HFL}}(\overline{L})$ such that
$\partial^{-}\mathbf{d}=U_{i}\cdot\text{pb}{\mathcal{L}}(L)$.

This implies a generalization of a Theorem of Ni [ni2020seifert, Theorem A.1];

######  Corollary 4.5.

Let $L_{i}$ be a component of an $n$ component Legendrian link $L$ in
$(S^{3},\xi)$. Suppose $\widehat{\mathcal{L}}(L)$ is non-zero and of multi-
Alexander grading $(A_{1},A_{2}\dots A_{n})$ and that there is a triple
$(B_{1},\pi_{1},A_{1})$ adapted to $(S^{3},\xi)$ such that $a_{i}^{1}$, the
the distinguished arc for $L_{i}$, is sent to the left, and that there is a
triple $(B_{2},\pi_{2},A_{2})$ adapted to $(S^{3},\xi)$ such that the
distinguished arc for $L_{i}$ is sent to the right. Then
$\operatorname{rank}(\widehat{\operatorname{HFL}}(A_{1},\dots A_{i}-1,\dots
A_{n}))\geq 2$.

The outline of the proof of the main technical lemma, Lemma 4.4, is as
follows: First we explicitly find a generator $\mathbf{d}$ of the correct
filtration level and Maslov grading using the associated adapted Heegaard
diagram. We then show that
$\partial_{\widehat{\operatorname{CF}}}\mathbf{d}=\text{pb}\mathcal{L}(L)$
using a diagrammatic argument. We then conclude that $\mathbf{d}$ is non-
trivial in link Floer homology using the same algebraic argument as the
corresponding step in [baldwin_note_2018] and [dey2020note].

###### Proof of Lemma 4.4.

Let $L,L_{i},a_{i},A$ be as in the hypothesis of the theorem. Let
$\alpha_{1},\beta_{1}$ be the curves associated to $a$. After isotoping
$\alpha_{1},\beta_{1}$ so that that they intersect minimally, $\alpha_{1}$ and
$\beta_{1}$ form a single bigon with corners at the intersection points
$d_{1},c_{1}\in\alpha_{1}\cap\beta_{1}$ as shown in the figure.

Figure 2. A pseudoholomorphic disk connecting the LOSS invariant.

It is readily observed that this can be the sole contribution to
$\partial^{-}\mathbf{d}$, where $\mathbf{d}=\\{d_{1},c_{2},c_{3}\dots
c_{g+n-1}\\}$. That is,
$\partial^{-}\mathbf{d}=U_{i}\cdot\textit{pb}\mathcal{L}(L)\in\operatorname{CFL}^{-}(L,\overline{S^{3}})$.
Since this disk only passes over the $w_{i}$ basepoint, we see that
$\mathbf{d}$ is of grading $(A_{1},A_{2},\dots,A_{i}-1,\dots,A_{n})$. Note
that $\partial^{-}(U_{i}\textit{pb}(\mathcal{L}))=0$

Now for the sake of simplicity of the completion of the proof, we work with
the corresponding generator
$\mathbf{d^{\prime}}\in\widehat{\operatorname{CFL}}(L,S^{3})(-A_{1},-A_{2},\dots,1-A_{i},\dots,-A_{n})$,
$\mathbf{c^{\prime}}$ which is an element in
$\widehat{\operatorname{CFL}}(L,S^{3})(-A_{1},-A_{2},\dots,-A_{i},\dots,-A_{n})$
and we observe that the duality of
$\widehat{\operatorname{CFL}}(L,\overline{S^{3}})$ and
$\widehat{\operatorname{CFL}}(L,S^{3})$ implies that
$\partial_{\widehat{\operatorname{CF}}(S^{3})}(\mathbf{d^{\prime}})=\mathbf{c^{\prime}}$.
Note that $\mathbf{c^{\prime}}$ is non-trivial in
$\widehat{\operatorname{HFL}}(L,S^{3})$.

To see that $\mathbf{d^{\prime}}$ represents a non-trivial homology class,
observe that if there were a generator $\mathbf{e}$ such that
$\partial_{\widehat{\operatorname{CFL}}}\mathbf{e}=\mathbf{d^{\prime}}$, then
$\partial_{\widehat{\operatorname{CF}}}\mathbf{e}=\mathbf{d^{\prime}}+\mathbf{h}+\mathbf{g}$
for some
$\mathbf{h}\in\widehat{\operatorname{CFL}}(L)(-A_{1},-A_{2},\dots,-A_{i},\cdots-
A_{n})$, $\mathbf{g}\in\bigoplus_{j\neq i,n\geq
1}\widehat{\operatorname{CFL}}(L)(-A_{1},-A_{2},\dots-A_{j}-n,\dots-A_{n})$.
It follows that
$\partial^{2}_{\widehat{\operatorname{CF}}}\mathbf{e}=\partial_{\widehat{\operatorname{CF}}}\mathbf{d^{\prime}}+\partial_{\widehat{\operatorname{CF}}}\mathbf{h}+\partial_{\widehat{\operatorname{CF}}}\mathbf{g}=0$,
whence
$\partial_{\widehat{\operatorname{CF}}}\mathbf{h}+\partial_{\widehat{\operatorname{CF}}}\mathbf{g}=\mathbf{c^{\prime}}$.
Since $\partial_{\widehat{\operatorname{CF}}}$ is non-decreasing on every
Alexander filtration, it follows that only the Alexander grading
$(-A_{1},-A_{2},\dots,-A_{n})$ part of
$\partial_{\widehat{\operatorname{CF}}}\mathbf{h}+\partial_{\widehat{CF}}\mathbf{g}$
is non-zero, whence
$\partial_{\widehat{\operatorname{CF}}}\mathbf{h}=\mathbf{c^{\prime}}$, a
contradiction since $\text{pb}\mathcal{L}(L)$ was assumed to be non-zero and
hence so is $\mathbf{c^{\prime}}$. ∎

###### Proof of Proposition 4.1.

This follows from 4.4, and noticing that any distinguished arc for $L_{i}$ in
any adapted triple is sent either to the left or the right, and the properties
of link Floer homology under mirroring and orientation changes. ∎

To prove Corollary 4.5, we work at the chain level in
$\operatorname{CFL}^{-}(L)$, and proceed towards a contradiction of the fact
that $\partial_{\operatorname{CF}^{-}}^{2}=0$. Note that we will work in both
$(S^{3},\xi)$ and $(-S^{3},-\xi)$.

###### Proof of Corollary 4.5.

Let $L$ be as in the hypothesis of the theorem. Suppose that

${\operatorname{rank}(\widehat{\operatorname{HFL}}(L,S^{3})(A_{1},\dots
A_{i}-1,\dots A_{n}))=1}.$

Note that each $A_{i}$ is determined by equation 1. Since we are working over
a ${\mathbb{F}}$, it follows that

${\operatorname{rank}(\widehat{\operatorname{CFL}}(L,S^{3})(A_{1},\dots
A_{i}-1,\dots A_{n}))=1},$

up to chain homotopy equivalence. Let $\mathbf{d}$ be a generator of
$\widehat{\operatorname{CFL}}(L,S^{3})(A_{1},\dots A_{i}-1,\dots A_{n})$.
Consider $\operatorname{CFL}^{-}(L,S^{3})$. Since $\widehat{\mathcal{L}}(L)$
is non-zero and there is a triple $(B_{1},\pi_{1},A_{1})$ such that
$a_{i}^{1}$ is sent to the left, Theorem 4.4 implies that
$\partial^{-}\mathbf{d}=U_{i}\cdot pb{\mathcal{L}}(L)$. Now view $L$ as a
Legendrian link in $(-S^{3},-\xi)$. Note that $(-B_{2},-\pi_{2},A_{2})$, is a
a triple adapted to the link $L$ in $(-S^{3},-\xi)$, and that the arc
associated to $L_{i}$ is sent to the left. Since $\widehat{\mathcal{L}}(L)$ is
non-zero, it follows from Proposition 4.4, that there is a generator
$\mathbf{d}^{\prime}$ such that
$\partial^{-}\mathbf{d}^{\prime}=U_{i}\textit{pb}\mathcal{L}(L,-S^{3})\in\operatorname{CFL}^{-}(L,-S^{3})\cong\operatorname{CFL}^{-}(-L,S^{3})$.
Note that $\textit{pb}\mathcal{L}(-L)$ is of multi-Alexander grading
$(-A_{1},-A_{2},\dots-A_{n})$.

Consider the properties of $\widehat{\operatorname{HFL}}$ under mirroring $L$.
Notice that $U_{i}\cdot pb\mathcal{L}(-L)\in Im(\partial^{-})$ implies that
$\langle\partial_{\widehat{\operatorname{CF}}}(\widehat{\mathcal{L}}(L)),\mathbf{e^{\prime}}\rangle\neq
0$, where
$\mathbf{e^{\prime}}\in\widehat{\operatorname{CFL}}(L,S^{3})(A_{1},\dots
A_{i}-1,\dots A_{n})\cong\mathbb{F}\cong\langle\textbf{d}\rangle$. Hence the
existence of distinguished left and right-veering arcs implies the existence
of these two maps in $\operatorname{CFL}^{-}$:

$\partial^{-}(\textbf{d})=U_{i}\cdot
pb\mathcal{L}(L),\partial_{\widehat{CF}}(\widehat{\mathcal{L}}(L))=\mathbf{d}$

Now the fact that $\partial^{-}(\widehat{\mathcal{L}}(L))=0$ and
$\partial^{2}_{\operatorname{CF}^{-}}=0$ implies a contradiction to the
existence of the two differential maps mentioned above.

∎

Corollary 4.3 follows from Theorem 4.2 using the fact that the self linking
number of a transverse link determines the Alexander grading of the LOSS
invariant.

###### Proof of Corollary 4.3.

Let $L$ be a fibered $n$ component link. Consider
$\widehat{\operatorname{HFK}}(L)$. Let $\xi$ be the contact structure on
$S^{3}$ induced by the open book associated to $L$. $L$ is a transverse link
in $\xi$ with self linking number $-\chi(L)$. By [cavallo2017invariant,
Theorem 6.2] the Alexander grading of the LOSS invariant is given by
$\frac{\text{sl}(L)+n}{2}$. Thus the LOSS invariant is of maximal Alexander
grading, and the result follows from Theorem 4.2.∎

######  Remark 4.6.

Note that the conclusion of Lemma 4.4 holds even if $pb\mathcal{L}(L)$ is not
in the image of the horizontal boundary map. For simplicity of the argument,
let $|L|=1$ and $A(\hat{\mathcal{L}}(L))=-g(L)$ or if $\hat{\mathcal{L}}(L)$
generates the $H_{*}\\{j=0\\}$ complex. Then the non-existence of such a
horizontal boundary map and the property that
$\partial^{-}(\hat{\mathcal{L}}(L))=0$ implies that $\mathcal{L}(L)$ is a non-
torsion element in $\operatorname{HFK}^{-}(L)$ by a result of Honda-Kazez-
Matic [honda2007right]. Thus the properties of LOSS invariant implies that the
ambient contact structure is tight, hence the open book decompositions
corresponding to that contact structure will always be right-veering, which
would contradict our assumption of existence of a left-veering arc.

######  Remark 4.7.

If one is interested in finding lower bounds on
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))$ for a general $n$
component link with non-vanishing LOSS invariant Theorem 4.2 can be
strengthened. Specifically if the Alexander grading of the LOSS invariant,
$\frac{\text{tb}(L)-\text{rot}(L)+n}{2}$ by [cavallo2017invariant, Theorem
5.6], is not $-1,0$ or $1$, then the fact that the Alexander polynomial of a
link evaluates to zero implies that there must be at least another $2n-2$
generators, whence $\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq
4|L|$. This is only a better bound in general than the bound
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq 2^{n-1}$ when $n\leq
5$, but these two different bounds can often be combined. For instance if the
Maslov grading, $\text{tb}(L)-\text{rot}(L)+1$ by [cavallo2017invariant,
Theorem 5.6], of the LOSS invariant is not in the range $[1,n-2]$, and the
Alexander grading is not in the range $[-1,1]$ then
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq 2^{n-1}+4$.

We conclude this section by noting that it is perhaps useful to view Corollary
4.3 in the context of the following more general proposition;

######  Proposition 4.8.

Suppose $L$ is an $n>1$ component non-trivial link such that
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L)(x_{1},x_{2},\dots
x_{n}))=2m+1$ for some $m\in{\mathbb{Z}}$. Then
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq 2^{n}+2m$.

For instance this proposition yields non-trivial lower bounds on the ranks of
fibered links of two or more components – they must have rank at least
$2^{n}$. Note that this improves the lower bound coming from the spectral
sequence from $\widehat{\operatorname{HFK}}(L)$ to
$\widehat{\operatorname{HF}}(\\#^{n-1}S^{1}\times S^{2})$.

###### Proof of Proposition 4.8.

Suppose $L$ is an $n>1$ component link non-trivial such that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(x_{1},x_{2},\dots
x_{n}))$ $=2m+1$. For each choice of $i$ there exist an odd number of
generators with $A_{j}$ grading $x_{j}$ for every $i\neq j$, and $A_{i}\neq
x_{j}$, since $\widehat{\operatorname{HFL}}(L)$ must have even rank in each
such hyperplane. Recursively, we can find additional generators for each
choice of hyperplane defined by $A_{i}=x_{i}$ for $i$ in any subset of
$\\{1,2,\dots n\\}$. Since each of the prior stages have odd number of
generators, we obtain an additional odd numbers of generators at each stage -
again since the rank of $\widehat{\operatorname{HFL}}(L)$ must be even. Thus
we obtain at least an additional $2^{n}-1$ generators. ∎

Note that if $L$ is an $n$ component link and
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L,A))$ is odd for some $A$,
then there is a multi-Alexander grading $(A_{1},A_{2}\dots A_{n})$ such that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L,A_{1},A_{2},\dots A_{n}))$
of odd rank, where $A=A_{1}+A_{2}+\dots A_{n}$ whence it follows that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq 2^{n}$.

##  5\. Detection Results for $(2,2n)$-cables

In this section we give detection results for $2$ component cables of the the
trefoils and figure eight knot;

######  Theorem 5.1.

Link Floer homology detects $T(2,3)_{2,2n}$ for all $n$.

######  Theorem 5.2.

Let $K$ be the figure eight knot. Link Floer homology detects $K_{2,2n}$ for
all $n$.

Each of these results is proven in three cases; where $n>0$, where $n<0$ and
where $n=0$.

######  Lemma 5.3.

Suppose $L$ is a link such that
$\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)})$,
where $T(2,3)_{(2,0)}$ is oriented as the boundary of an annulus. Then $L$ is
either isotopic to $T(2,3)_{(2,0)}$ or the disjoint union of an unknot and a
knot $K$ with knot Floer homology given by

$\displaystyle\widehat{\operatorname{HFK}}(K)\cong{\mathbb{F}}_{0}^{2}[1]\oplus{\mathbb{F}}_{-1}^{3}[0]\oplus{\mathbb{F}}_{-2}^{2}[-1]$

Note that this implies that link Floer homology detects $T(2,3)_{(2,0)}$,
since if a link $L$ satisfies
$\widehat{\operatorname{HFL}}(L)\cong\widehat{\operatorname{HFL}}(T(2,3)_{(2,0)})$,
then $L$ does not contain an unknotted component.

We first give a partial computation of
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)})$. To do so compute
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,1)})$ and
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,-1)})$ using Hanselman–Watson’s
cabling formula in the theory of immersed curves [hanselman2019cabling], then
apply the skein exact triangle for knot Floer homology to determine
$\widehat{\operatorname{HFL}}(T(2,3)_{(2,0)})$, and thereby
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)})$.

Hanselman–Watson’s cabling formula implies that

$\displaystyle\widehat{\operatorname{HFK}}(T(2,3)_{(2,1)})\cong{\mathbb{F}}_{0}[2]\oplus({\mathbb{F}}_{-1}\oplus{\mathbb{F}}_{0})[1]\oplus{\mathbb{F}}_{-1}[0]\oplus({\mathbb{F}}_{-2}\oplus{\mathbb{F}}_{-3})[-1]\oplus{\mathbb{F}}_{-4}[2]$

$\displaystyle\widehat{\operatorname{HFK}}(T(2,3)_{(2,-1)})\cong{\mathbb{F}}_{2}[2]\oplus({\mathbb{F}}_{1}\oplus{\mathbb{F}}_{0})[1]\oplus({\mathbb{F}}_{-1}^{2}\oplus{\mathbb{F}}_{0})[0]\oplus({\mathbb{F}}_{-2}\oplus{\mathbb{F}}_{-1})[-1]\oplus{\mathbb{F}}_{-2}[2]$

Using the skein exact triangle from [Holomorphicdisksandknotinvariants,
Equation 7] – taking $L_{-}$ to be $T(2,3)_{(2,-1)}$, $L_{+}$ to be
$T(2,3)_{(2,1)}$ and $L_{0}$ to be $T(2,3)_{(2,0)}$ – we can deduce that
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)},i)$ is trivial for $i>2$ and
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)},2)\cong{\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$.
To determine $\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)};1)$ we note that it
follows immediately from the skein exact triangle that it is either
${\mathbb{F}}_{\frac{1}{2}}\oplus{\mathbb{F}}_{-\frac{1}{2}}$ or
${\mathbb{F}}_{\frac{1}{2}}^{2}\oplus{\mathbb{F}}_{-\frac{1}{2}}^{2}$. To
exclude the former case first note that
$\widehat{\operatorname{HFL}}(T(2,3)_{(2,0)})$ is supported on the lines
$A_{2}=A_{1}$, $A_{2}=A_{1}+1$, $A_{2}=A_{1}-1$, since $T(2,3)_{(2,0)}$ bounds
an annulus. It follows that the component of
$\widehat{\operatorname{HFL}}(T(2,3)_{(2,0)})$ with Alexander gradings
satisfying $A_{1}+A_{2}=2$ is ${\mathbb{F}}_{0}\oplus{\mathbb{F}}_{-1}[1,1]$.
Since $T(2,3)$ is fibered, but each component of $T(2,3)_{(2,0)}$ is not
braided with respect to the other (since the linking number is zero), it
follows that the rank of $\widehat{\operatorname{HFL}}(T(2,3)_{(2,0)}))$ is at
least four in each of the maximal $A_{i}$ gradings (namely $A_{i}=1$). Thus
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)};1)\cong{\mathbb{F}}_{-\frac{1}{2}}^{2}\oplus{\mathbb{F}}_{\frac{1}{2}}^{2}$
and indeed

$\displaystyle\widehat{\operatorname{HFL}}(T(2,3)_{(2,0)};(1,0))\cong\widehat{\operatorname{HFL}}(T(2,3)_{(2,0)};(0,1))\cong{\mathbb{F}}_{0}\oplus{\mathbb{F}}_{-1}.$

To conclude the computation it suffices to determine
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)};0)$ by the symmetries of link
Floer homology. Note again that the skein exact triangle implies that
$\widehat{\operatorname{HFK}}(T(2,3)_{(2,0)};0)$ is either
$F_{-\frac{1}{2}}\oplus{\mathbb{F}}_{-\frac{3}{2}}$ or
$F_{-\frac{1}{2}}^{2}\oplus{\mathbb{F}}_{-\frac{3}{2}}^{2}$.

Thus we have that

$\displaystyle\widehat{\operatorname{HFK}}(T(2,3)_{2,0})\cong({\mathbb{F}}_{\frac{1}{2}}^{2}\oplus{\mathbb{F}}_{-\frac{1}{2}}^{2})[1]\oplus({\mathbb{F}}_{-\frac{1}{2}}^{3}\oplus{\mathbb{F}}_{-\frac{3}{2}}^{3})[0]\oplus({\mathbb{F}}_{-\frac{3}{2}}^{2}\oplus{\mathbb{F}}_{-\frac{5}{2}}^{2})[-1]$

or

$\displaystyle\widehat{\operatorname{HFK}}(T(2,3)_{2,0})\cong({\mathbb{F}}_{\frac{1}{2}}^{2}\oplus{\mathbb{F}}_{-\frac{1}{2}}^{2})[1]\oplus({\mathbb{F}}_{-\frac{1}{2}}^{4}\oplus{\mathbb{F}}_{-\frac{3}{2}}^{4})[0]\oplus({\mathbb{F}}_{-\frac{3}{2}}^{2}\oplus{\mathbb{F}}_{-\frac{5}{2}}^{2})[-1]$

We can now proceed to the proof of Lemma 5.3

###### Proof of Lemma 5.3.

Suppose
$\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(T(2,3)_{(2,0})$.
It suffices to show that $L$ bounds an annulus, has linking number $0$ and has
a trefoil component.

We first note that $L$ has at most two components since the maximal Maslov
grading of $\widehat{\operatorname{HFK}}(L)$ is $\frac{1}{2}$. Since
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))$ is even it follows that
$L$ cannot be a knot, so $L$ is a two component link. The fact that the
linking number is zero follows from the fact that the Conway polynomial
detects the linking number [hoste1985firstcoefficientoftheconwaypolynomial].
Since the maximal Alexander grading of $\widehat{\operatorname{HFK}}(L)$ is
$1$, $L$ either bounds an annulus or the disjoint union of an unknot and a
genus one knot. If $\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=16$
this is impossible as in this case the genus one knot would have to have knot
Floer homology of even rank, a contradiction. Thus the genus one knot must
have knot Floer homology of the form

$\displaystyle{\mathbb{F}}_{0}^{2}[1]\oplus{\mathbb{F}}_{-1}^{3}[0]\oplus{\mathbb{F}}_{-2}^{2}[-1]$

as required. Suppose now that $L$ bounds an annulus. Note that a two component
link containing a knot of genus $g$ must have knot Floer homology containing
generators of Maslov grading which differ by at least $1+2g$. Since the Maslov
gradings of $\widehat{\operatorname{HFK}}(L)$ differ by at most $3$ it follows
that each component of $L$ is of genus at most one. Note that the only link
with unknotted components bounding an annulus and of linking number zero is
the two component unlink, which is of the incorrect Euler characteristic. Thus
$L$ contains genus one components. To see that each of these components is
fibered, we note that if $L$ has a component $K$ with $g(K)=1$ which is not
fibered then $L$ contains at least two pairs of generators which differ in
Maslov grading by $3$. There are only two such pairs, namely the generators of
Maslov grading $\frac{1}{2}$ and $-\frac{5}{2}$. Since the linking number of
$L$ is zero we see that in link Floer homology the two Maslov index $0$
generators are supported in, without loss of generality $A_{1}=1$, so we have
that they are supported in bi-Alexander grading $(1,0)$. But since $L$ bounds
an annulus, both component of $L$ are isotopic and we must have two generators
of $A_{2}$ grading $1$, a contradiction. Thus $L$ has genus one fibered
components. It thus suffices to show these components are neither $T(2,-3)$
nor the figure eight. Note that $T(2,-3)_{(2,0)}$ is the mirror of
$T(2,3)_{(2,0)}$, therefore
$\widehat{\operatorname{HFK}}(T(2,-3)_{(2,0)})\cong(\widehat{\operatorname{HFK}}(T(2,-3)_{(2,0)}))^{*}\not\cong\widehat{\operatorname{HFK}(L)}$.
Similarly, the $(2,0)$ cable of the figure eight is isotopic to the
$(2,0)$-cable of the mirror of the figure eight, so if $L$ where the
$(2,0)$-cable of the figure eight then its knot Floer homology would be
isomorphic to it’s dual, which is not the case. Thus $L$ is isotopic to
$T(2,3)_{(2,0)}$ as desired.

∎

We now prove a similar result for The $(2,0)$ cable of the figure eight.

######  Lemma 5.4.

Let $K$ be the figure eight knot and orient $K_{2,0}$ as the boundary of an
annulus. Suppose
$\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(K_{2,0})$.
Then $L$ is isotopic to $K_{2,0}$.

Again we note that if
$\widehat{\operatorname{HFL}}(L)\cong\widehat{\operatorname{HFL}}(K_{2,0})$,
then $L$ cannot have an unkotted component, whence the preceding lemma implies
that link Floer homology detects the $(2,0)$-cable of the figure eight knot.

As before we begin with a partial computation of
$\widehat{\operatorname{HFK}}(K_{2,0})$. Hanselman-Watson’s cabling formula
for knot Floer homology in terms of immersed curves [hanselman2019cabling]
shows that

$\displaystyle\widehat{\operatorname{HFK}}(K_{2,1})\cong{\mathbb{F}}_{1}[2]\oplus({\mathbb{F}}_{1}\oplus{\mathbb{F}}_{0})[1]\oplus({\mathbb{F}}_{0}^{3})[0]\oplus({\mathbb{F}}_{-2}\oplus{\mathbb{F}}_{-1})[-1]\oplus{\mathbb{F}}_{-3}[-2]$
$\displaystyle\widehat{\operatorname{HFK}}(K_{2,-1})\cong{\mathbb{F}}_{3}[2]\oplus({\mathbb{F}}_{2}\oplus{\mathbb{F}}_{1})[1]\oplus({\mathbb{F}}_{0}^{3})[0]\oplus({\mathbb{F}}_{-1}\oplus{\mathbb{F}}_{0})[-1]\oplus{\mathbb{F}}_{-1}[-2]$

Again we apply the skein exact triangle from
[Holomorphicdisksandknotinvariants, Equation 7] – with
$L_{0}=K_{2,0},L_{+}=K_{2,1},L_{-}=K_{2,1}$.This immediately implies that
$\widehat{\operatorname{HFK}}(K_{2,0},i)\cong 0$ for $|i|>2$ and
$\widehat{\operatorname{HFK}}(K_{2,1})\cong
F_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{5}{2}}$. Indeed we see that
$\widehat{\operatorname{HFK}}(K_{2,0},1)$ is either
${\mathbb{F}}_{\frac{3}{2}}^{2}\oplus{\mathbb{F}}_{\frac{1}{2}}^{2}$ or
${\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$. To rule out the
latter case note that $K_{2,0}$ bounds an annulus upon reversing the
orientation of a component, so that $\widehat{\operatorname{HFL}}(K_{2,0})$ is
supported on the lines $A_{2}=A_{1}$, $A_{2}=A_{1}\pm 1$. Thus since
$\widehat{\operatorname{HFL}}(L)$ must be of even rank in each $A_{i}$
grading, and
$\widehat{\operatorname{HFL}}(K_{2,0},(x,y))\cong{\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$
is excluded. While unnecessary we note that
$\widehat{\operatorname{HFK}}(L)\cong({\mathbb{F}}_{\frac{1}{2}}\oplus{\mathbb{F}}_{-\frac{1}{2}})^{n}$
for some $0\leq n\leq 3$. In fact $n\geq 2$, as else
$\widehat{\operatorname{HFL}}(K_{2,0})$ cannot admit a spectral to
$\widehat{\operatorname{HFL}}(K)\otimes V$

###### Proof of Lemma 5.4.

Suppose $L$ is as in the statement of the Lemma. Then $L$ is a two component
link with linking number $0$. Note that upon reversing the orientation of a
component, $L$ bounds a surface of Euler characteristic $0$. Thus either $L$
consists of the disjoint union of an unknot and a genus one knot, or $L$
bounds an annulus. The former case is excluded since
$\widehat{\operatorname{HFL}}(L)$ is not supported along a single $A_{i}=0$
grading. Thus $L$ bounds an annulus, and is therefore the $(2,0)$-cable of
some knot $K$.

To see that $K$ is the figure eight knot, note that $K$ can be of genus at
most one since the maximal $A_{i}$ grading is $1$ for each $i$. Indeed, it is
plainly not the case that
$\widehat{\operatorname{HFL}}(L)\cong\widehat{\operatorname{HFL}}(T(2,0))$, so
$K$ must be a genus one knot. To see that $K$ is fibered note that the Maslov
gradings of $A_{1}=1$ part of $\widehat{\operatorname{HFL}}(L)$ is given by
${\mathbb{F}}_{0}\oplus{\mathbb{F}}_{2}\oplus{\mathbb{F}}_{1}^{2}$ while the
$A_{i}=-1$ part is given by
${\mathbb{F}}_{-1}\oplus{\mathbb{F}}_{-2}^{2}\oplus{\mathbb{F}}_{-3}$. It
follows that the Maslov index $2$ generator, as well as the Maslov index $-3$
generator cannot persist under the spectral sequence to
$\widehat{\operatorname{HFL}}(L_{1})$, where $L_{1}$ is the component of $L$
corresponding to $A_{1}$. Thus $\widehat{\operatorname{HFL}}(L_{1})$ is of
rank one in the top Alexander grading and $L_{1}$ is fibered
[ghiggini2008knot]. Thus $L$ is a $(2,0)$-cable of a trefoil or the figure
eight. But the $(2,0)$-cables of the trefoils have distinct knot Floer
homology, as shown previously. Thus $L$ is $K_{2,0}$ as desired. ∎

To prove theorems 5.1 and 5.2 we first show that if the link Floer homology of
a link satisfies certain properties then the link is a $(2,2n)$ cable of a
trefoil or the figure eight knot. Then we show that the link Floer homology of
each of these three links satisfies these properties for the given $n$.

######  Lemma 5.5.

Let $L$ be a two component link with linking number $n>1$ such that;

  1. (1)

$\widehat{\operatorname{HFL}}(L)$ is of rank $2$ in the maximal non-zero
$A_{1}$ or $A_{2}$ grading.

  2. (2)

The link Floer homology polytope has a vertex at multi-Alexander grading
$(1+\frac{n}{2},1+\frac{n}{2})$.

  3. (3)

$\widehat{\operatorname{HFL}}(L)$ has support on the lines
$A_{2}=A_{1},A_{2}=A_{1}+1,A_{2}=A_{1}-1$.

Then $L$ is isotopic to $K_{2,2n}$ where $K$ is a trefoil or the figure eight
knot.

###### Proof of Lemma 5.5.

Suppose $L$ is as in the hypotheses of the theorem. Condition $3$ implies that
$L$ bounds an annulus upon reversing the orientation of either component – it
bounds an Euler characteristic zero surface with two boundary components,
neither of which is a disk. Since the maximal Alexander grading is $2+n$, it
follows that the maximal $A_{i}$ gradings are $1+\frac{n}{2}$.

Since
$g(L_{i})+\frac{n}{2}\leq\max\\{A_{i}:\widehat{\operatorname{HFL}}(L)(A_{1},A_{2})\neq
0\\}$, it follows that $g(L_{i})\leq 1$ for each $i$. Indeed since,
$\widehat{\operatorname{HFL}}(L)$ is of rank $2$ in $A_{1}$ grading
$\frac{n}{2}+1$ $L_{i}$ is genus one and fibered or genus zero, and hence an
unknot, a trefoil or the figure eight knot. It follows that $L\cong K_{2,2n}$
where $K$ is the unknot, a trefoil, or the figure eight knot. $(2,2n)$-cables
of the unknot are the torus links $T(2,2n)$ which do not satisfy property $2$.
Thus $L$ is a $(2,2n)$ cable of a trefoil knot or the figure eight knot, as
desired. ∎

To apply Lemma 5.5, we need to show that
$\widehat{\operatorname{HFL}}(K_{2,2n})$ satisfies the conditions of the
hypothesis of Lemma 5.5 for $K$ a trefoil or figure eight knot. The easiest
way to proceed is via the following remark

######  Remark 5.6.

Rudolph shows that $K_{n,m}$ can be written as a Murasugi sum of
$K_{n,\text{sign}(m)}$, which is fibered by a result of Stallings
[stallings1978constructions], and $T(n,m)$ along minimal genus Seifert
surfaces [rudolph2002non]. Note that Rudolph’s result is only stated for $n,m$
coprime, but the proof works for arbitrary $n,m$. Results of Gabai show that
Murasugi sums of fiber surfaces are fibered, and that Murasugi sums of minimal
genus Seifert surfaces are minimal genus Seifert surfaces
[gabai1986detecting]. It follows that $(2,2n)$ cables of genus one fibered
knots are fibered and of genus $n$, that is Euler characteristic $-2n$.

If one wishes to only use Rudolph’s stated result, for knots, one can
alternately proceed via the following lemma;

######  Lemma 5.7.

Let $K$ be a genus one fibered knot. For $n\geq 1$, $K_{2,2n}$ is fibered and
$\chi(K_{2,2n})=-2n$.

###### Proof of Lemma 5.7.

Note that, for $n>0$, $K_{2,2n+1}$ is the Murasugi sum of $K_{2,1}$ and
$T(2,2n+1)$ [neumann1987unfoldings, rudolph2002non] along their fiber
surfaces, which are minimal genus. Let $g$ be a genus one fibered knot. It
follows from a result of Gabai [gabai1986detecting] that $K_{2,2n+1}$ is
fibered and that $g(K_{2,2n+1})=2+n$. From here, using the Skein exact
triangle for knot Floer homology, we see that
$\dfrac{2-\chi(K_{2,2n})}{2}=1+n$ for all $n\geq 1$ – that is
$\chi(K_{2,2n})=-2n$ – and that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(K_{2,2n})(1+n))=1$ so that
$K_{2,2n}$ is fibered, as desired. ∎

It follows that conditions 1, 2 and 3 of Lemma 5.5 hold for $(2,2n)$ cables of
the trefoils, and figure eight. To see that condition $3$ holds notes that
$(2,2n)$ cables of the three links bounds an annuli under reversing the
orientation of one of the components.

We can now proceed to the detection results.

###### Proof of Theorem 5.1.

Suppose $L$ is as in the statement of the Theorem. We proceed by cases; when
$n=0$, $n>0$ and when $n<0$. The $n=0$ case was dealt with in Lemma 5.3.

Suppose now that $n>0$. We first show that
$\widehat{\operatorname{HFL}}(T(2,3)_{2,2n})$ satisfies the conditions in the
hypothesis of Lemma 8.7. Observe that for $n\geq 1$, each component of
$T(2,3)_{2,2n}$ is a braid in the complement of the other, so
$\widehat{\operatorname{HFL}}(T(2,3)_{2,2n}))$ must be rank two in the maximal
$A_{i}$ grading, by braid detection [GageT26, Proposition 1], so condition 1
of Lemma 5.5 holds. By Lemma 5.7 the maximal Alexander grading of
$\widehat{\operatorname{HFK}}(L)$ is $2+n$. $T(2,-3)_{2,2n+1}$ is fibered by
Lemma 5.7. Therefore, since $\widehat{\operatorname{HFL}}(T(2,-3)_{2,2n})$ is
rank one in the maximal Alexander grading, and must be symmetric under
interchanging the components, it follows that the link Floer homology polytope
of $L$ has a vertex at $(1+\frac{n}{2},1+\frac{n}{2})$, so that condition 2
holds. We note that the two generators of the top $A_{i}$ grading of
$\widehat{\operatorname{HFL}}(T(2,-3)_{2,2n})$ must be of Maslov gradings $0$
and $-1$.

Suppose now that
$\widehat{\operatorname{HFL}}(L)\cong\widehat{\operatorname{HFL}}(T(2,3)_{2,2n})$
for some $n\geq 2$. Since link Floer homology detects the linking number of
two component links [hoste1985firstcoefficientoftheconwaypolynomial], it
follows that the components of $L$ have linking number $n$. Lemma 5.5 implies
that $L$ is the $(2,2n)$ cable of a trefoil or the figure eight. To see that
$L$ is the cable of $T(2,3)$, observe that the
$\widehat{\operatorname{HFL}}(L)$ has a Maslov index $0$ generator with
$A_{1}$ grading $1+\frac{n}{2}$ which persists under the spectral sequence to
$\widehat{\operatorname{HFL}}(L_{i})$, whence $L_{i}$ is $T(2,3)$.

We now proceed to the case $n<0$, i.e. we show that Link Floer homology
detects $T(2,-3)_{2,2n}$ for $n>0$

Consider $\widehat{\operatorname{HFL}}(T(2,-3)_{2,2n})$. Observe that for
$n\geq 1$, each component of $L$ is a braid in the complement of the other, so
$\widehat{\operatorname{HFL}}(L)$ must be rank two in the maximal $A_{i}$
grading, by braid detection [GageT26], so Condition 1 of Lemma 5.5 holds. By
Lemma 5.7 the maximal Alexander grading of $\widehat{\operatorname{HFK}}(L)$
is $2+n$. Since $T(2,-3)_{2,2n+1}$ is fibered, and
$\widehat{\operatorname{HFL}}(T(2,-3)_{2,2n+1})$ must be symmetric under
interchanging the components, it follows that the link Floer homology polytope
of $L$ has a vertex at $(1+\frac{n}{2},1+\frac{n}{2})$, so that condition 2 of
Lemma 8.7 holds. Condition 3 holds since $T(2,-3)_{2,2n}$ bounds an annulus
upon reversing the orientation of a component. Note that the two generators of
the top $A_{i}$ grading of $\widehat{\operatorname{HFL}}(T(2,-3)_{2,2n})$ must
be of Maslov gradings $1$ and $2$.

Suppose now that
$\widehat{\operatorname{HFL}}(L)\cong\widehat{\operatorname{HFL}}(T(2,-3)_{2,2n})$.
Lemma 5.5 implies that $L$ is a $(2,2n)$ cable of a trefoil or the figure
eight knot. Since the two generators of the top $A_{i}$ grading of
$\widehat{\operatorname{HFL}}(L)$ must be of Maslov gradings $1$ and $2$, it
follows that each component is $T(2,-3)$ and we have the desired result. ∎

The proof of Theorem 5.2 is again similar.

###### Proof of Theorem 5.2.

The proof follows the proof of Theorem LABEL:HFLdetectsLHTcables verbatim,
aside from to note that the top $A_{i}$ generators must be of Maslov index $0$
and $1$ respectively. ∎

##  6\. Technical Results

In this section we prepare the ground for our detection results for $T(2,8)$
and $T(2,10)$ in subsequent sections. We will prove a purely algebraic result
for $E_{2}$ collapsed chain complexes and a result regarding the Thurston
polytope.

###  6.1. $E_{2}$ collapsed chain complexes

We will rely on a structure theorem for _$E_{2}$ -collapsed chain complexes
over ${\mathbb{Q}}$_ – that is ${\mathbb{Z}}\oplus{\mathbb{Z}}$ graded chain
complexes with coefficients in ${\mathbb{Q}}$ and differentials of length at
most one. The ${\mathbb{Z}}/2$ coefficient version is provided in
[HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial, Section
12]. In particular Ozsváth-Szabó note that each $E_{2}$-collapsed spectral
sequences over ${\mathbb{Z}}/2$. split as the direct sum of complexes of one
of five forms, $B_{d},V_{d}^{l},H_{d}^{l},X_{d}^{l}$ and $Y_{d}^{l}$. These
are shown in the Figure 3. For the ${\mathbb{Q}}$ version the only necessary
modification is that the arrows correspond to maps
$\mathbbm{1}:{\mathbb{Q}}\to{\mathbb{Q}}$, save for the top arrow of the
$B_{d}$ complex, which is given
by $-\mathbbm{1}$

Figure 3. The summands of an $E_{2}$-collapsed chain complex

######  Proposition 6.1.

Suppose $(C,\partial)$ is a bi-filtered $E_{2}$ collapsed chain complex with
coefficients in ${\mathbb{Q}}$. Then $(C,\partial)$ is bi-filtered chain
homotopy equivalent to a complex which decomposes as a direct sum of copies of
chain complexes of the type $B_{d}[i,j]\oplus
V_{d}^{l}[i,j],H_{d}^{l}[i,j],X_{d}^{l}[i,j],Y_{d}^{l}[i,j]$.

Here and henceforth, $[i,j]$ indicates a shift up in $A_{1}$ grading by $i$
and a shift up in $A_{2}$ grading by $j$. Since if $L$ is a two component
$\delta$-thin link, $\widehat{\operatorname{CFL}}(L)$ is $E_{2}$ collapsed,
this result is of great utility. Indeed, the filtered chain homotopy types of
the components of $L$ yield significant restraints on the filtered chain
homotopy type of $\widehat{\operatorname{CFL}}(L)$. For instance, since the
full filtered $\widehat{\operatorname{CFL}}$ complex has homology
${\mathbb{F}}_{0}\oplus{\mathbb{F}}_{-1}$, we see that
$\widehat{\operatorname{CFL}}(L)$ has a summand $X_{0}^{l}$ or $Y_{0}^{l}$ and
a summand $X_{-1}^{l}$ or $Y_{-1}^{l}$, and no other summands of the form
$X_{d}^{l},Y_{d}^{l}$. Similarly, it is readily seen that the chain homotopy
type of $\widehat{\operatorname{CFL}}(L_{1})$ determines the $V_{d}^{l}$
summands, while $\widehat{\operatorname{CFL}}(L_{2})$ determines the
$H_{d}^{l}$ summands. Specifically, if there are generators
$x,y\in\widehat{\operatorname{CFL}}(L_{1})$, with
$\langle\partial_{\widehat{\operatorname{CFL}}(L_{1})}x,y\rangle\neq 0$ then
we must have a summands of $\widehat{\operatorname{CFL}}(L)$ of the form
$V_{M(x)}^{A(x)-A(y)}$, as no other summand induces the correct differential
on $\widehat{\operatorname{CFL}}(L_{1})$.

The proof is entirely elementary. We proceed in two stages, the first
corresponding to the case that $(C,\partial)$ has $B_{d}$ summands of a
certain form and the second that it does not. We denote the horizontal
component of $\partial$ by $\partial_{1}$ and the vertical component by
$\partial_{2}$.

###### Proof of Proposition 6.1.

By a bi-filtered version of the Reduction Lemma [hedden2013khovanov, Lemma
4.1] we can assume that the grading preserving part of $\partial$ is trivial.
We proceed by induction on $\operatorname{rank}(C)$, splitting of summands of
$(C,\partial)$ of the requisite form to reduce to the rank.

We first show that if
$\partial_{1}\circ\partial_{2}(C(x,y))\cap\partial_{2}\circ\partial_{1}(C(x,y))\neq\mathbf{0}$
then there is a $B_{d}$ summand. Suppose
$\mathbf{v}\in\partial_{1}\circ\partial_{2}(C(x,y))\cap\partial_{1}\circ\partial_{2}(C(x,y)$
is non-zero. We may pick a basis for $\partial_{1}(C(x-1,y))$
$\\{y_{i}\\}_{i=1}$ such that $\langle\partial_{2}y_{j},\mathbf{v}\rangle=0$
for $j\neq 1$, $\partial y_{1}=\mathbf{v}$. Likewise we may pick a basis for
$\partial_{2}(C(x,y-1))$ for $C(x,y)$ such that $\\{z_{i}\\}_{i=1}$ such that
$\langle\partial_{2}z_{j},\mathbf{v}\rangle=0$ for $i\neq 1$, $\partial
z_{1}=\mathbf{v}$. Finally we may find a basis $\\{w_{i}\\}_{i=1}$ such that
$\langle\partial_{1}w_{i},z_{1}\rangle=0$ for $i\neq 1$, $\partial
w_{1}=z_{1}$. Observe that $\partial_{2}\circ\partial_{1}(w_{1})=\mathbf{v}$,
so that $\partial_{2}\circ\partial_{1}(w_{i})=-\mathbf{v}$. to guarantee that
$\partial^{2}=0$ Thus $\partial_{2}(w_{i})=-y_{i}$

We seek to show that the complex $\langle
w_{i},\partial_{2}(w_{i}),\partial_{1}(w_{i}),\mathbf{v}\rangle$ is a summand
of $(C,\partial)$. By construction if $\mathbf{w}\in C(x-1,y)$ and
$\partial_{2}\mathbf{w}\in\langle\mathbf{v}\rangle$ then
$\matyhbf{w}\in\langle x_{i}\rangle$ as required. Likewise if $\mathbf{w}\in
C(x,y-1)$ and $\partial_{1}\mathbf{w}\in\langle\mathbf{v}\rangle$. Similarly
we have that if $\mathbf{w}\in C(x,y)$ and
$\partial_{1}\mathbf{w}\in\langle\partial_{1}v_{i}\rangle$ then
$\mathbf{w}\in\langle w_{i}\rangle$ as else
$\partial_{2}\mathbf{w}=\partial_{2}w_{i}$ to insure that $\partial^{2}=0$, a
contradiction.

The remaining cases are obvious.

We are therefore left with the case that there is no generator $\mathbf{v}\in
C(x,y)$ with
$(\partial_{1}\circ\partial_{2}\mathbf{v})\cap(\partial_{1}\circ\partial_{2}\mathbf{v})\neq\langle\mathbf{0}\rangle$.
Suppose $\mathbf{v}\in C(x,y)$. If
$\partial_{1}(\mathbf{v})=\partial_{2}(\mathbf{v})=\mathbf{0}$ then either we
have a summand of the form $Y_{0}$ and we are done, or we can find a
$\mathbf{w}$ with either $\partial_{1}\mathbf{w}\neq\mathbf{0}$ or
$\partial_{2}\mathbf{w}\neq\mathbf{0}$.

Suppose then that $\partial_{2}(\mathbf{v})\neq 0$. Then since
$\partial^{2}=0$,
$\partial_{1}\partial_{2}(C(x,y)\cap\partial_{1}\partial_{2}(C(x,y))=\langle\mathbf{0}\rangle$,
we have that $\partial_{1}\circ\partial_{2}(\mathbf{v})=\mathbf{0}$. Indeed,
without loss of generality, we may take $\mathbf{0}=C(x+1,y-1)$, so that
$\partial_{1}(\mathbf{v})=\mathbf{0}$. We
$\partial(\partial_{1}(\mathbf{v}))=0$. Consider
$\langle\partial_{1}(\mathbf{v})\rangle$. If there is no $\mathbf{z}$ with
$\partial_{1}\mathbf{z}=\partial_{2}\mathbf{v}$, then we have a summand.
Suppose that
$\langle\partial_{2}\mathbf{w}\rangle=\langle\partial_{1}\mathbf{v}\rangle$
then after a change of basis for $C(x-1,y+1)$ we may assume that $\mathbf{w}$
is the unique element with that property, and we can iterate this process to
find a staircase summand. ∎

###  6.2. The Dual Thurston Polytope

The four detection results also require the following proposition;

######  Proposition 6.2.

Let $L$ be a two component link, with components $L_{1},L_{2}$. Suppose
$L_{1}$ is an unknotted braid axis for $L_{2}$ and that
$|\ell\text{k}(L_{1},L_{2})|>1$. Then the dual Thurston polytope has support
contained in the interval $(-1,1)$ when intersected with the $A_{2}$ axis

Note that if $|\ell\text{k}(L_{1},L_{2})|=1$ in the above proposition then $L$
is a Hopf link. Since the link Floer homology of $L$ determines the dual
Thurston polytope, this proposition will yield restraints on the link Floer
homology of links with given knot Floer homology.

###### Proof of Proposition 6.2.

Let $L$ be as in the hypotheses of the Proposition. After potentially
reversing the orientation of $L_{2}$ we can take
$\ell\text{k}(L_{1},L_{2})>1$. Set $n=\ell\text{k}(L_{1},L_{2})$. There is a
disk bounding $L_{1}$ that intersects $L_{2}$ exactly $n$ times positively. It
follows that there is an Euler characteristic $1-n$ surface embedded in the
exterior of $L$, $X_{L}$, representing the homology classes $(0,\pm n)\in
H_{1}(X_{k},\partial X_{L})$. Let $x$ denote the Thurston norm. Then $x(0,\pm
n)\leq n-1$. It follows that $x(0,\frac{\pm n}{n-1})\leq 1$. Thus $x^{*}(0,\pm
n)=\sup_{x(\beta)\leq 1}((0,\pm n)\cdot\beta)\geq(0,\pm n)\cdot(0,\frac{\pm
n}{n-1})=\frac{n^{2}}{n-1}$. Thus $x^{*}(0,\dfrac{\pm
n(n-1)}{n^{2}})=x^{*}(0,\dfrac{\pm(n-1)}{n})\geq 1$. Thus $x^{*}(0,\pm 1)>1$,
and the result follows.

∎

##  7\. Knot Floer Homology Detects $T(2,8),T(2,10)$

In this section we prove the following detection results;

######  Theorem 7.1.

Knot Floer homology detects $T(2,8)$.

######  Theorem 7.2.

Knot Floer homology detects $T(2,10)$.

For the readers convenience first we recall that:

$\widehat{\operatorname{HFK}}(T(2,2n),i)\cong\begin{cases}{\mathbb{F}}_{\frac{1+2i-2n}{2}}&\text{
for }i=\pm n\\\ {\mathbb{F}}^{2}_{\frac{1+2i-2n}{2}}&\text{ for }-n<i<n\\\
0&\text{ otherwise}\par\end{cases}$

We start by showing that in general if
$\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(T(2,2n))$
then $L$ consists of two unknotted components and
$\wiehat{\operatorname{CFL}}(L)$ is of an especially simple form.

######  Lemma 7.3.

Suppose
$\widehat{\operatorname{HFK}}(L;{\mathbb{F}})\cong\widehat{\operatorname{HFK}}(T(2,2n);{\mathbb{F}})$.
Then $(\widehat{\operatorname{CFL}}(L),\partial)$ consists of $B_{d}$ summands
and a single $Y_{0}^{0}[\frac{n}{2},\frac{n}{2}]\oplus
Y_{-1}^{1}[\frac{n-1}{2},\frac{n-1}{2}]$ summand, and $L$ has unknotted
components.

###### Proof of Lemma 7.3.

Let $L$ be as in the Lemma, and $m$ be the number of components of $L$. We
first show that $m=2$. To see this note that the maximal Maslov grading of a
generator of $\widehat{\operatorname{HFK}}(L)$ is $\frac{1}{2}$, so since
$\widehat{\operatorname{HFK}}(L)$ admits a spectral sequence to
$\widehat{\operatorname{HF}}(\\#^{m-1}S^{1}\times S^{2})$, $L$ can have at
most two components. $L$ then has exactly $2$ components since the Maslov
grading of $\widehat{\operatorname{HFK}}(L)$ is ${\mathbb{Z}}+\frac{m-1}{2}$
valued.

Let $L_{1}$ and $L_{2}$ be the two components of $L$. It follows that there is
a unique Maslov grading $0$ generator. Consider
$\widehat{\operatorname{HFL}}(L)$. Since the linking number is $n$ and the
Alexander grading of the Maslov index $0$ generator is $n$, it follows that
there is a summand $Y_{0}^{0}[\frac{n}{2},\frac{n}{2}]$ and since there are
two Maslov index $-1$ generators there is a summand
$Y_{-1}^{1}[\frac{n-1}{2},\frac{n-1}{2}]$. As observed in the proof of
[HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial, Theorem
12.1], $\widehat{\operatorname{HFL}}(L)$ consists of pairs of summands
$V^{l}_{d}(x,y)\oplus V^{l}_{d-1}(x,y-1)$, $H^{l}_{d}(x,y)\oplus
H^{l}_{d-1}(x,y-1)$, $Y_{0}^{l}(x,y)\oplus Y_{-1}^{l-1}(x+1,y+1)$,
$X_{0}^{l}(x,y)\oplus X^{l-1}_{-1}(x,y)$, as well as a collection of copies of
$B_{d}$. Note that for the case at hand the rank in each Alexander grading is
at most two, so we have that the rest of the complex is given by summands of
the form $B_{d}$ or $V_{d}^{1}[x,y]\oplus V_{d-1}^{1}[x,y-1]$ or
$H_{d}^{1}[x,y]\oplus H_{d-1}^{1}[x-1,y]$. Indeed we readily see that $d$ is
even.

Suppose $L_{1}$ is not unknotted. Consider the $V_{d}^{1}[x,y]\oplus
V_{d-1}^{1}[x,y-1]$ summands with maximal $x$. Note that $x>\frac{n}{2}$.
Amongst these consider the summands with minimal $d$. It follows that there is
a generator in $\widehat{\operatorname{HFK}}(L_{1})$ of minimal Alexander
grading with Maslov grading $d-2(x-\frac{n}{2})$. Note that this is an even
number. But this is a contradiction, since if there is a generator of even
Maslov index in $A_{1}$ grading $\frac{n}{2}-x$ which persists under the
spectral sequence to $\widehat{\operatorname{HFK}}(L_{1})\otimes V$ then there
must be a generator of odd Maslov index in $A_{1}$ grading $\frac{n}{2}-x-1$,
since the summands of $\widehat{\operatorname{HFL}}(L)$ that persist under the
spectral sequence to $\widehat{\operatorname{HFK}}(L_{1})\otimes V$ are of the
form $V_{d}^{1}[x,y]\oplus V_{d-1}^{1}[x,y-1]$ with $d$ even.

Thus there are no $V_{d}^{1}[x,y]\oplus V_{d-1}^{1}[x,y-1]$ summands. A
similar proof shows that there are no $H_{d}^{1}[x,y]\oplus
H_{d-1}^{1}[x-1,y]$ summands. Thus the remaining summands are all of the form
$B_{d}$ and each component of $L$ is unknotted, as desired. ∎

We can now prove the two Knot Floer homology detection results.

###### Proof of Theorem 7.1.

Suppose
$\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(T(2,8))$.
Lemma 7.3 implies that $L$ is a two component link with unknotted components,
and indeed determines the complex up to a choice of $x$ in a summand
$B_{-4}[x,-x]$, by the symmetry of link Floer homology. Also since the link
Floer homology is of rank $2$ in at least one of the maximal $A_{i}$ gradings,
by [GageT26] it follows that at least one component is a braid axis, and
therefore that the maximal $A_{i}$ grading is exactly $2$. It follows that
$x\in\\{1,0,-1\\}$. The $x=1$ and $x=-1$ cases are then excluded by Lemma 6.2,
and the result follows from the fact that link Floer homology detects $T(2,8)$
[binns2020knot].

Figure 4. One potential link Floer homology for a link with the same knot
Floer homology as $T(2,8)$. Here the red one denotes the link Floer polytope
and the blue one denotes the dual Thurston polytope for the link.

∎

The proof of Theorem 7.2 is virtually identical

###### Proof of Theorem 7.2.

Suppose
$\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(T(2,10))$.
Lemma 6.3 implies that $L$ is a two component link with unknotted components,
and indeed determines the complex up to a choice of $x$ in a summand
$B_{-4}[x+\frac{1}{2},\frac{1}{2}-x]$, by the symmetry of link Floer homology.
Since the link Floer homology is of rank $2$ in at least one of the maximal
$A_{i}$ gradings, it follows that at least one component is a braid axis, and
therefore that the maximal $A_{i}$ grading is exactly $2$. It follows that
$x\in\\{1,0,-1\\}$. The $x=1$ and $x=-1$ cases are then excluded by Lemma 6.2,
and the result follows from the fact that Link Floer homology detects
$T(2,10)$.

∎

##  8\. Rank Detection Results for Knot Floer Homology

In this section we show that there are only a small number of links with knot
Floer homology of low rank. In particular we show the following sequence of
theorems;

######  Theorem 8.1.

If $\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=4$ then $L$ is the
Hopf link or the three component unlink.

######  Theorem 8.2.

Suppose $L$ is a link with
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L))=6$. Then $L$ is a
disjoint union of a trefoil and an unknot.

######  Theorem 8.3.

The only links $L$ with
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L))=8$ are $T(2,4)$,
$T(2,-4)$, the four component unlink, and the disjoint union of a Hopf link
and an unknot.

Indeed, this yields a complete list of links with at least two components of
rank at most eight, in combination with the following lemma;

######  Lemma 8.4.

Suppose $L$ is an $n+1$ component link with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=2^{n}$. Then $L$ is an
unlink.

We can also provide a partial classification of links with knot Floer homology
of rank ten;

######  Proposition 8.5.

There is no three component link with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=10$.

By similar methods we can show a detection result for knot Floer homology;

######  Theorem 8.6.

Let $L_{0}$ be $T(2,3)$ with a meridian. Suppose
$\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(L_{0})$.
Then $L$ is isotopic to $L_{0}$.

Here we take the linking number of the right handed trefoil and its meridian
to be one. Again this result should be compared to the corresponding link
Floer homology detection result in [binns2020knot].

The proofs of Theorems 8.1, 8.2, 8.3 and 8.5 each essentially follow from case
analysis in which we use algebraic and geometric properties of link Floer
homology to exclude the unwanted cases, as well as the detection results for
Knot Floer homology found in [binns2020knot].

We first prove Lemma 8.4, as it is useful in proving the other theorems.

We will find the following characterization of unlinks useful;

######  Lemma 8.7 (Ni [ni2006note]).

Suppose $\widehat{\operatorname{HFK}}(L)$ is supported in a single Alexander
grading. Then $L$ is an unlink.

###### Proof of Lemma 8.7.

Observe that if $\widehat{\operatorname{HFK}}(L)$ is supported in a single
Alexander grading then that grading must be zero, by the symmetry of
$\widehat{\operatorname{HFK}}(L)$. It follows that
$0=n-\sum_{i=1}^{m}(2-2g-n_{i})$, where $n$ is the number of components of
$L$, and the surface that $L$ bounds has $m$ components, the $i$th of which
has $n_{i}$ boundary components. Thus $0=n-m+\sum_{i}g_{i}$. But $n\geq m$, so
$0=n-m+\sum_{i=1}^{m}g_{i}\geq\sum_{i=1}^{m}g_{i}\geq 0$. Thus
$\sum_{i=1}^{m}g_{i}=0$ and $n=m$. It follows that $L$ bounds $n$ disks,
whence $L$ is an unlink with $n$ components. ∎

Throughout this section we use $V$ to denote the rank two vector space of
multi-Alexander grading $0$, with one generator of Maslov index $0$ and the
other $-1$.

###### Proof of Lemma 8.4.

We proceed by induction. For the base, $n=0$, the result follows from the fact
that knots of genus at least one have rank at least two, since they have rank
at least one in Alexander gradings $g$ and $-g$, where $g$ is the genus of the
knot. For the inductive step, note that if $L$ is an $n+1$ component link with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=2^{n}$ then for any
component $L_{i}$ of $L$, $L-L_{i}$ is a link with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L-L_{i}))=2^{n-1}$. By
inductive hypothesis $L-L_{i}$ is the $n$ component unlink. It follows that
the support of $\widehat{\operatorname{HFL}}(L)$ is contained in the line
defined by $A_{j}=0$ for $j\neq i$. But $i$ is arbitrary, so
$\widehat{\operatorname{HFL}}(L)$ is supported in multi-Alexander grading
$\mathbf{0}$, and the result follows from Lemma 8.7. ∎

We now prove Theorem 8.1.

###### Proof of Theorem 8.1.

Let $L$ be an $n$ component link with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=4$. Since
$\widehat{\operatorname{HFK}}(L)$ admits a spectral sequence to
$\widehat{\operatorname{HF}}(\\#^{n-1}S^{1}\times S^{2})$, it follows that
$n\leq 3$. Indeed, since
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))$ is even, it follows
that $L$ has either two or three components. We evaluate each case separately.

If $L$ has three components then $L$ is the three component unlink by lemma
8.4. It therefore remains only to consider the case that $n=2$. Let $g$ be the
maximal Alexander grading of $\widehat{\operatorname{HFK}}(L)$. We consider
the following three cases;
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(g))$ is either four, two
or one, in which case the link is fibered by [ni2007knot].

  1. (1)

$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(g))=4$. In this case
$g=0$, whence $\widehat{\operatorname{HFK}}(L)$ does not admit a spectral
sequence to $\widehat{\operatorname{HF}}(S^{1}\times S^{2})$.

  2. (2)

$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(g))=2$. In this case we
see that

$\widehat{\operatorname{HFK}}(L)\cong({\mathbb{F}}_{m_{1}}\oplus{\mathbb{F}}_{m_{2}})[g]\oplus({\mathbb{F}}_{m_{1}-2g}\oplus{\mathbb{F}}_{m_{2}-2g})[-g].$

Observe that $g>0$. Consider the spectral sequence from
$\widehat{\operatorname{HFK}}(L)$ to $\widehat{\operatorname{HF}}(S^{1}\times
S^{2})$. There must be exactly one generator with non-trivial differential.
Without loss of generality we may then take $m_{1}-1=m_{2}-2g$, but in that
case we cannot have that the remaining generators from the set
$\\{\frac{1}{2},\frac{-1}{2}\\}$, a contradiction.

  3. (3)

$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(g))=1$. Observe that
$g\geq 1$. Let $(x,g-x)$ be the bi-Alexander grading of the contact class in
$\widehat{\operatorname{HFL}}(L)$.

If $g>1$, then Theorem 4.2 implies that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq 6$. Thus $g=1$, and

$\widehat{\operatorname{HFL}}(L)\cong{\mathbb{F}}_{m}(\frac{1}{2},\frac{1}{2})\oplus{\mathbb{F}}_{m-1}(-\frac{1}{2},\frac{1}{2})\oplus{\mathbb{F}}_{m-1}(\frac{1}{2},-\frac{1}{2})\oplus{\mathbb{F}}_{m-2}(-\frac{1}{2},-\frac{1}{2})$

Since $\widehat{\operatorname{HFL}}(L)$ admits a spectral sequence to $V$ it
follows that $m\in\\{0,1\\}$. These two cases correspond to the two oriented
link types of $T(2,2)$, which link Floer homology is known to detect per [BG].

∎

We now prove Theorem 8.2, proceeding with a method different from the rank
four case; we first analyse the case that $L$ is fibered under some
orientation, and the remaining cases separately.

######  Lemma 8.8.

There is no fibered link $L$ such that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=6$.

###### Proof of Lemma 8.8.

Suppose $L$ is an $n$ component fibered link with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=6$. Since
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L))$ is even, $L$ cannot be
a knot. Moreover, $L$ can have at most $3$ components, since
$\widehat{\operatorname{HFL}}(L)$ admits a spectral sequence to
$\widehat{\operatorname{HF}}(\\#^{n-1}S^{1}\times S^{2})$.

If $L$ has three components, then since $L$ is fibered, the maximal Euler
characteristic Seifert surface for $L$ is connected, whence $\chi(L)\leq-1$,
and the maximal Alexander grading in which $\widehat{\operatorname{HFK}}(L)$
has support is at least $2$. By Theorem 4.2, the next to top Alexander grading
is of rank three, whence $\widehat{\operatorname{HFK}}(L)$ is of rank at least
eight, a contradiction.

Thus $L$ has two components. Again, as $L$ is fibered, the maximal Euler
characteristic Seifert surface for $L$ is connected, whence $\chi(L)\leq 0$,
and the maximal Alexander grading, $g$, in which
$\widehat{\operatorname{HFK}}(L)$ has support is at least $1$.

If $g=1$ then $L$ bounds an annulus, and hence $L$ is a $(2,2n)$-cable. Note
that each component of $L$ has knot Floer homology of rank at most four and is
therefore an unknot or a trefoil. None of the links $T(2,2n)$ has knot Floer
homology of rank $6$, while if $L$ is the $2$-cable of a trefoil, we can see
that the linking number of $L$ must be zero since the spectral sequence to
either component must be trivial, since
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=\operatorname{rank}(\widehat{\operatorname{HFK}}(T(2,\pm
3)\otimes V)$. However, we can see that in the top $A_{1}$ or $A_{2}$ grading
the link Floer homology is of rank $2$ whence at least one component is
braided about the other – contradicting the fact that the linking number is
zero – or $L$ is a split link, which is a contradiction since split links are
not fibered.

Suppose now that $g>1$. Let $[x,g-x]$ be bi-Alexander grading of the top
Alexander grading generator. Theorem 4.4 implies that the complex is given by

${\mathbb{F}}_{m}[x,g-x]\oplus{\mathbb{F}}_{m-1}[x-1-x,x-g]$
$\oplus{\mathbb{F}}_{m+1-2g}[1-x,x-g]\oplus{\mathbb{F}}_{m+1-2g}[-x,x+1-g]$

To ensure that each Alexander grading is of even rank we must have that
$g=2,x=1$, i.e. the complex is

${\mathbb{F}}_{m}[1,1]\oplus{\mathbb{F}}_{m-1}[1,0]\oplus{\mathbb{F}}_{m-1}[0,1]\oplus{\mathbb{F}}_{m-4}[-1,-1]\oplus{\mathbb{F}}_{m-3}[-1,0]\oplus{\mathbb{F}}_{m-3}[0,-1]$

By the symmetry of $\widehat{\operatorname{HFL}}(L)$, we can see that the link
obtained by reversing the orientation of one of the complements bounds an
annulus, or the disjoint union of a disk and a punctured torus. Since split
links cannot be fibered, the latter option is impossible.

Suppose $L$ bounds an annulus. Again since there is no torus link $T(2,2n)$ of
rank $6$, $L$ must have trefoil components. Since the spectral sequence
corresponding to either component is trivial, it follows that the linking
number is zero. But from the complex we can see that either $L$ is split –
which is again impossible since we are assuming $L$ is fibered – or the
components are exchangeably braided by braid detection [GageT26], a
contradiction.

∎

Having eliminated the case that $L$ is fibered, we can proceed to the
remaining cases.

###### Proof of Theorem 8.2.

Let $L$ be an $n$ component link with link Floer homology of rank $6$. Lemma
8.8 implies that $L$ is not fibered under any orientation. The fact that
$2^{n-1}\leq\operatorname{rank}(\widehat{\operatorname{HFK}}(L))$ implies that
$L$ has at most three components. Since
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))$ is even, it follows
that $L$ has two or three components.

Consider the Link Floer homology polytope of $L$. Observe that it is contained
in the polytope defined by $\sum_{j=1}^{n}A_{j}=\pm g_{i}$ for each $g_{i}$,
where $g_{i}$ is maximal Alexander grading of
$\widehat{\operatorname{HFK}}(L)$ under some orientation of $L$.

Lemma 8.7, together with the fact that there is no unlink with Knot Floer
homology of rank $6$, implies that we may take $g_{j}>0$ for all $i$.

Suppose that $\widehat{\operatorname{HFL}}(L)$ has support at a vertex
$p=(p_{1},p_{2},\dots p_{n})$ of the link Floer polytope. Since there exists
an Alexander grading $A_{i}$ for which the support of
$\widehat{\operatorname{HFL}}(L)$ in the plane $A_{i}=p_{i}$ is exactly $p$,
it follows that the spectral sequence from $\widehat{\operatorname{HFL}}(L)$
to $\widehat{\operatorname{HFL}}(L_{i})\otimes V^{n-1}$ is trivial. Thus the
rank of $\widehat{\operatorname{HFL}(L)}$ at $p$ is at least $2^{n-1}$. Since
the polytope is non-degenerate, the symmetry properties of
$\widehat{\operatorname{HFL}}(L)$ implies that $2^{n}\leq 6$, whence $n=2$.
The linking number is zero, as the range of the Alexander gradings cannot
change under the spectral sequence to the relevant component, because of
symmetry properties of knot Floer homology of knots. The symmetry of
$\widehat{\operatorname{HFL}}(L)$ implies that the rank at $p$ of the Link
Floer homology is four so that $L$ is a split link – and the only such link
with link is the disjoint union of an unknot and a trefoil – or $L_{i}$ is a
braid axis – which is impossible since the linking number of $L_{i}$ and the
other component is zero.

It thus remains only to consider the case that
$\widehat{\operatorname{HFL}}(L)$ does not have support at any of the
vertices. Since $L$ is not fibered, each face of the polytope must have at
least two generators. If $L$ has two components, then it must have at least
four faces, each of which has two generators in its interior, a contradiction.
If $L$ has three components, notice that the polytope has a collection of four
faces that meet only at vertices. Since the polytope does not have support at
the vertices, it follows that
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L))\geq 8$, a contradiction.
∎

For the rank eight case, we proceed as in the rank six case, dealing with
fibered links first, and the remaining cases subsequently.

######  Proposition 8.9.

Suppose $L$ is fibered and
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))=8$. Then $L\cong T(2,4)$
or $T(2,-4)$.

###### Proof.

Let $L$ be a fibered link with link Floer homology of rank eight. Since
$\widehat{\operatorname{HFK}}(L)$ admits a spectral sequence to
$\widehat{\operatorname{HF}}(\\#^{n-1}S^{1}\times S^{2})$, where $n$ is the
number of components of $L$, it follows that $n$ has at most four components.
Indeed, $L$ cannot be a knot since
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))$ is even. Suppose $L$
has four components. Then by Theorem 8.4, $L$ must be a four component unlink,
which is not fibered.

Suppose $L$ has three components. Since $L$ is fibered, the minimal genus
Seifert surface for $L$ is the fiber surface, which is connected. It follows
that the highest Alexander grading, $g$, in which
$\widehat{\operatorname{HFK}}(L)$ has supported is at least $2$. By Theorem
4.4, $\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(g-1))\geq 3$. Thus

$\widehat{\operatorname{HFK}}(L)\cong{\mathbb{F}}_{m}[g]\oplus{\mathbb{F}}_{m-1}^{3}[g-1]\oplus{\mathbb{F}}_{m+1-2g}^{3}[1-g]\oplus{\mathbb{F}}_{m-2g}[-g]$

where $m\in{\mathbb{Z}}$. It follows that the Alexander polynomial of this
link does not evaluate to zero on $1$, a contradiction.

Hence $L$ is a two component link. Then, as previously, the maximal Alexander
grading, $g$, of $\widehat{\operatorname{HFK}}(L)$ is at least one. Suppose
$g=1$. Then $\operatorname{rank}(\widehat{\operatorname{HFK}}(L;0))=6$. Such a
complex clearly does not admit a spectral sequence to
$\widehat{\operatorname{HF}}(S^{1}\times S^{2})$, a contradiction. Thus $g\geq
2$. Consider $\widehat{\operatorname{HFL}}(L)$. Suppose the top Alexander
grading generator is in multi-Alexander grading $(x,g-x)$ and of Maslov index
$m$. Theorem 4.4 implies that there are generators in Alexander gradings
$(x-1,g-x)$, $(x,g-x-1)$ both of Maslov index $m-1$, while symmetry implies
that there are generators in mutli-Alexander grading $(-x,x-g)$ of Maslov
index $m-2g$ as well as in bi-Alexander gradings $(1-x,x-g)$ and $(-x,g+1-x)$
of Maslov index $m+1-2g$. Since $g\geq 2$ it follows that all of these
generators are distinct. To ensure that the complex is even rank in each
$A_{i}=k$ grading, we require that there are generators in multi-Alexander
gradings $(x-1,g-x-1)$ and $(1-x,x+1-g)$. Observe that the symmetry properties
of link Floer homology, together with the existence of the spectral sequences
to complexes arising as tensor products with $V$, imply that the Maslov
gradings of these generators are $m-2$ and $m+2-2g$ respectively.

Observe that each component of $L$ has rank at most four whence each component
is an unknot or a trefoil.

Without loss of generality, after relabeling we may take $2x\geq g$. Suppose
$(x,g)\not\in\\{(\frac{3}{2},2),(1,2)\\}$. Then the chain complex splits, as
the sum of two squares, and each component is an unknot, since there do not
exist three consecutive Alexander gradings containing generators of the
correct Maslov indices to have a trefoil component. To ensure that the
spectral sequence from $\widehat{\operatorname{HFK}}(L)$ to
$\widehat{\operatorname{HFK}}(L_{1})\otimes V$ we must add three horizontal
arrows, meaning that one of the squares gets at least two. Correspondingly we
must add three vertical arrows, and to insure that $\partial^{2}=0$ we must
then have that the two arrows are added to the square with two horizontal
arrows. This means that the upper right corner, or lower left corner of one of
the squares persists under both spectral sequences and hence must be on the
diagonal, as its coordinates are then given by
$(\frac{\ell\text{k}(L_{1},L_{2})}{2},\frac{\ell\text{k}(L_{1},L_{2})}{2})$.
Reversing the orientation of a component, we see that $L$ bounds an Euler
characteristic zero surface, that is either an annulus or the disjoint union
of a trefoil an unknot. Since fibered links cannot be split, $L$ bounds an
annulus, and since each component is unknotted, it follows that $L$ is
$T(2,2n)$ for some $n$. The only $n$ for which
$\operatorname{rank}(\widehat{\operatorname{HFK}}(T(2,2n))=8$ are $n=\pm 2$.

Suppose $(x,g)=(1,2)$. Reversing the orientation of a component, yields a link
that bounds an annulus (or the disjoint union of a punctured torus and a disk,
but such a link is split and therefore not fibered) whence $L$ is a $2$-cable
link. $L$ cannot be the $2$-cable of a trefoil, as we know that the linking
number is $0$, since the span of each $A_{i}$ grading is $3$, but the rank of
$\widehat{\operatorname{HFL}}(L)$ in the top Alexander $A_{1}$ grading is $2$
telling us that $L$ is either split or each component is a braid axis, which
would contradict the fact that $L$ is fibered. $L$ cannot be split since split
links aren’t fibered. If $L$ is a $(2,2n)$-cable of the unknot, then $L$ is a
torus link $T(2,2n)$ for some $n$, but the only such torus links with link
Floer homology of rank $8$ are $T(2,4)$ and $T(2,-4)$, as required.

Suppose $(x,g)=(\frac{3}{2},2)$. It can be quickly checked that this complex
does not admit the requisite spectral sequences, irrespective of whether the
remaining component is a trefoil or an unknot. Specifically, it can readily be
seen that there cannot be a component of the differential from a generator in
$A_{1}$ grading $\frac{1}{2}$ to one in lower $A_{1}$ grading, as else the
generators that persist under the spectral sequence to
$\widehat{\operatorname{HFL}}(L_{2})$ do not have admissible Maslov gradings.
Thus the complex splits as two squares, and again must be located on the
diagonal, a contradiction.

∎

###### Proof of Theorem 8.3.

By the previous lemma it suffices to classify links that are not fibered under
any orientation and with link Floer homology of rank eight. Suppose $L$ is
such a link, with $n$ components. Since
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq 2^{n-1}$ and $L$ is
even, $L$ has either two, three or four components. If $L$ has four
components, then Lemma 8.4 implies that $L$ is an unlink. We deal with the two
remaining cases separately;

  1. (1)

$L$ has two components Consider $\widehat{\operatorname{HFL}}(L)$. Since $L$
is not fibered under any orientation, the link Floer polytope of $L$ is a
subset of the polytope with boundaries $a_{1}+a_{2}=g_{1}$,
$-a_{1}-a_{2}=g_{1}$, $a_{1}-a_{2}=g_{2}$, $a_{2}-a_{1}=g_{2}$. Here
$g_{1},g_{2}$ are the maximal Alexander gradings of
$\widehat{\operatorname{HFK}}(L)$ under its two possible orientations. This
has vertices
$\pm(\frac{g_{1}+g_{2}}{2},\frac{g_{1}-g_{2}}{2}),\pm(\frac{g_{1}-g_{2}}{2},\frac{g_{1}+g_{2}}{2})$.

Lemma 8.7 together with the fact that the two component unlink has knot Floer
homology of rank two imply that we may take $g_{1},g_{2}>0$.

Suppose that $\widehat{\operatorname{HFL}}(L)$ has support at one of the
vertices of the polytope. After relabeling the components, and considering the
symmetry of the complex, we may take them to be a generators in
$(\frac{g_{1}+g_{2}}{2},\frac{g_{1}-g_{2}}{2})$,
$(-\frac{g_{1}+g_{2}}{2},\frac{g_{2}-g_{1}}{2})$. Observe that both of the
generators must persist under the spectral sequence to
$\widehat{\operatorname{HFL}}(L_{1})$, whence $g(L_{1})=g_{1}+g_{2}$. Since
each component of $L$ is either an unknot or a trefoil it follows that
$g_{1}+g_{2}\in\\{0,1\\}$. In either case it follows that at least one of
$g_{1}$ or $g_{2}$ is zero, a contradiction.

Thus the polytope defined does not have support at its vertices. It follows
that the interior of each face contains exactly two generators, see figure 5
for an example. Consider the $a_{1}+a_{2}=g_{1}$ face. Assume the two
generators on the interior of this face are of distinct multi-Alexander
gradings. In order that each $A_{1},A_{2}$ grading is of even rank and the
complex admits the requisite symmetries and spectral sequences we must have
that $g_{1}=g_{2}$. Indeed, considering the Maslov grading $m$ of one of the
generators on the $A_{1}+A_{2}=g_{1}$ face we see that $m-2(g_{1}+g_{2})=m-2$,
whence again $g_{1}+g_{2}=1$, and we obtain a contradiction as previously, see
figure 5.

Figure 5. A rank eight bigraded vector space not arising as the link Floer
homology of a link.

The only remaining case is that on each face the two generators sit in the
same multi-Alexander grading, see figure 5. Suppose the generators occur in
multi-Alexander gradings $\pm(x,g_{1}-x)$, $\pm(y,g_{2}+y)$. In order that the
spectral sequences to each component are non-zero, in which case the rank of
one the components would be even, we must have that $x=-y$, $g_{1}-x=g_{2}+y$.
It follows that $g_{1}=g_{2}$. Since each $\widehat{\operatorname{HFL}}(L)$
has support in only two $A_{i}$, $A_{j}$ gradings, it follows that each
component is an unknot. Without loss of generality we may take the linking
number to be positive, whence it follows that $x=g_{1}-x$. It follows that the
complex has support in grading
$\\{\pm(\frac{g_{1}}{2},\frac{g_{1}}{2}),(\pm\frac{g_{1}}{2},\mp\frac{g_{1}}{2})\\}$.
Let $m_{1}\geq m_{2}$ be the Maslov gradings of the two generators of grading
$(\frac{g_{1}}{2},\frac{g_{1}}{2})$. Since $A_{1}$ is an unknot we must have
that $\partial_{1}x\neq 0$. Without loss of generality $\partial_{1}x=y$.
Since $y$ does not persist under the spectral sequence to
$\widehat{\operatorname{HFL}}(L_{2})$, $\partial_{2}y=z$. Thus $z$ is of
Maslov grading $m-2$, so that either $g_{1}=1$, in which case $L$ bounds an
annulus and is thereby $T(2,2n)$ for some $n$ a contadiction, since these do
not have the right homology, or the remaining generator of bi-Alexander
grading is of Maslov grading $\\{0,1\\}\ni m+2g_{1}-2\geq m+2$. The remaining
generator of bi-Alexander grading $(\frac{g_{1}}{2},\frac{-g_{1}}{2})$ must be
of Maslov grading $m+2g_{1}-2\pm 1$ so that the remaining generator of bi-
Alexander grading $(-\frac{g_{1}}{2},-\frac{g_{1}}{2})$ is of Maslov grading
either $m+2g_{1}-2$ or $m+2g_{1}-2$. But it should also be of grading
$m-2g_{1}$, by symmetry, a contradiction.

  2. (2)

$L$ has three components Suppose $L$ has three components. Then each two
component sublink is either a Hopf link or a two component unlink. Suppose a
two component sublink $L-L_{3}$ is a Hopf link. Then the linking number of
$L_{3}$ with $L_{1}$ and $L_{2}$ must be zero as else the rank of the complex
would be greater than eight. It follows that the spectral sequence from
$\widehat{\operatorname{HFL}}(L)$ to $\widehat{\operatorname{HFL}}(L-L_{3})$
is just projection onto the $A_{3}=0$ plane. Suppose
$\widehat{\operatorname{HFL}}(L)$ is supported in the $A_{3}=0$ plane, then
$L$ consists of a disjoint union of a Hopf link and an unknot. If
$\widehat{\operatorname{HFL}}(L)$ is not supported in the $A_{3}=0$ plane then
since the rank in the top $A_{1}$ is four, it follows that $L_{1}$ is a braid
axis – which would contradict the fact that the linking number is non-zero –
or $L$ is split, in which case $L$ is the disjoint union of the a Hopf link
and an unknot.

Suppose now that each two component sublink consists of a two component
unlink. It follows that the $\widehat{\operatorname{HFL}}(L)$ has rank four on
each plane $\\{A_{i}=0\\}_{i=1,2,3}$. $L$ must be of rank eight in multi-
Alexander grading $(0,0,0)$, as else the rank is too high. To see this note
that if there is a generator in multi-Alexander grading $(0,0,x)$ for $x>0$,
then there must exist generators too in multi-Alexander gradings
$(0,y_{1},x)$, $(y_{2},0,x)$ to ensure that the generator in multi-Alexander
grading $(0,0,x)$ do not persist under the spectral sequences to $L-L_{1}$,
$L-L_{2}$. It follows that there are at least $6$ generators that have non-
zero $A_{3}$ component, and at least another $4$ with $A_{3}=0$, a
contradiction.

The result then follows from Lemma 8.7.

∎

###### Proof of Proposition 8.5.

Let $L$ be a $3$ component link with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L)=10$. Suppose $L$ is
fibered. The maximal Alexander grading, $A$, of
$\widehat{\operatorname{HFK}}(L)$ is given by $2+g$ for some $g\geq 0$. Note
that $\operatorname{rank}(\widehat{\operatorname{HFK}}(L)(A))=1$. Suppose the
generator is of Maslov index $m$. Theorem 4.2 implies that there are at least
three generators of Maslov index $m-1$ in Alexander grading $A-1$. The
symmetry of $\widehat{\operatorname{HFK}}(L)$ yields a summand
${\mathbb{F}}_{m-4-2g}[-2-g]\oplus{\mathbb{F}}_{m-3-2g}^{3}[-1-g]$. In order
for the Alexander polynomial evaluate to $0$ on $-1$, we must add at least
another four generators, yielding a complex of rank at least $12$, a
contradiction.

We may thus assume that $L$ is not fibered under any orientation. Let $g_{i}$
be the maximal Alexander grading of $\widehat{\operatorname{HFK}}(L)$ under
each of the four possible orientations of $L$, up to an overall reversal in
the orientation of $L$. Note that $g_{i}>0$ by Lemma 8.7. Consider the
polytope $P$ cut out by $A_{1}+A_{2}+A_{3}=\pm g_{1}$, $A_{1}+A_{2}-A_{3}=\pm
g_{2}$, $A_{1}-A_{2}+A_{3}=\pm g_{3}$, $A_{1}-A_{2}-A_{3}=\pm g_{4}$. Observe
that each of these faces must contain at least two generators. If the edges of
$P$ do not contain any generators, then
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\geq 16$, a
contradiction.

We may therefore assume a generator lies on an edge. Without loss of
generality we may take this to be the edge formed by
$A_{1}+A_{2}+A_{3}=g_{1}$, $A_{1}+A_{2}-A_{3}=g_{2}$, namely
$A_{1}+A_{2}=\dfrac{g_{1}+g_{2}}{2}$. Observe that any generator on this edge,
as well as the respective generators yielded by the symmetry of
$\widehat{\operatorname{HFL}}(L)$ on the edge
$-A_{1}-A_{2}=\dfrac{g_{1}+g_{2}}{2}$, persists under the spectral sequence to
$\widehat{\operatorname{HFL}}(L-L_{3})\otimes
V[\frac{\ell\text{k}(L_{1},L_{3})}{2},\frac{\ell\text{k}(L_{2},L_{3})}{2}]$,
and is indeed of maximal (respectively minimal $A_{1}+A_{2}$ grading). Now,
there are only two 2 component links with
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))\leq 5$; the Hopf link
and the unlink. It follows that $\ell\text{k}(L_{i},L_{j})\in\\{-1,0,1\\}$ for
each $i,j$. Thus the Maximal $A_{1}+A_{2}$ grading can only be $1$. That is,
we have that $1=\dfrac{g_{1}+g_{2}}{2}$, so that $g_{1}=g_{2}=1$. It follows
that, under an appropriate orientation of $L$, $\chi(L)=1$. It follows that
$L$ bounds either an annulus and a disk, or two disks and a punctured torus.
Note that for each $i$,
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L-L_{i})\leq 5$, so that
$L-L_{i}$ is an unlink, or a Hopf link. Thus $L$ is the disjoint union of a
Hopf link and an unknot, but this link has knot Floer homology of rank eight,
a contradiction.

∎

We conclude with a proof of Theorem 8.6

###### Proof of Proposition 8.6.

Let $L_{0}$ be the trefoil together with a meridian. It suffices to show that
if $\widehat{\operatorname{HFK}}(L)\cong\widehat{\operatorname{HFK}}(L_{0})$
then $\widehat{\operatorname{HFL}}(L)\cong\widehat{\operatorname{HFL}}(L_{0})$
by the corresponding link Floer homology detection result in [binns2020knot].

Since the maximal Maslov grading of $\widehat{\operatorname{HFK}}(L)$ is
$\frac{1}{2}$, it follows that $L$ has at most two components. Since
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))$ is even, it follows
that $L$ cannot be a knot, whence $L$ is a two component link.

Observe that $\widehat{\operatorname{HFK}}(L)$ detects the linking number of
two component links, so the linking number is one. Consider
$\widehat{\operatorname{HFL}}(L)$. Let $A_{i}(0)$ be the Alexander grading of
the Maslov index $0$ generator in $\widehat{\operatorname{HFL}}(L_{i})$. Since
there are no positive Maslov index generators in
$\widehat{\operatorname{HFL}}(L)$, it follows that $A_{i}(0)\geq 0$. Since
$0\leq A_{1}(0)+A_{2}(0)+\ell\text{k}(L_{1},L_{2})=2$, we have that
$\\{A_{1}(0),A_{2}(0)\\}=\\{0,1\\}$. Without loss of generality we may take
$A_{1}(0)=1$. It follows that there is a Maslov index $0$ generator in bi-
Alexander grading $(\frac{3}{2},\frac{1}{2})$. By Theorem 4.4, we have that
there are Maslov index $-1$ generators of bi-Alexander grading
$(\frac{1}{2},\frac{1}{2}),(\frac{3}{2},\frac{-1}{-2})$. Observe that in order
that $\widehat{\operatorname{HFL}}(L)$ admit a spectral sequence to
$\widehat{\operatorname{HFL}}(L_{1})\otimes V$, there must be a Maslov index
$-2$ generator with $A_{1}$ grading $\frac{1}{2}$. Such a generator must be in
bi-Alexander grading $(\frac{1}{2},\frac{-1}{2})$. By symmetry this determines
the Alexander gradings of all but four generators. Consider the remaining
Maslov index $-1$ suppose it is of bi-Alexander grading $(x,1-x)$. Observe
that in order to admit spectral sequences to
$\widehat{\operatorname{HFL}}(L_{i})\otimes V$, there must be Maslov index
$-2$ generators in bi-Alexander gradings $(x-1,1-x)$ and $(x,-x)$. In order
that $\widehat{\operatorname{HFL}}(L)$ be symmetric, we must then have that
$x=\frac{1}{2}$. We have thus determined that
$\widehat{\operatorname{HFL}}(L)\cong\widehat{\operatorname{HFL}}(L_{0})$, as
desired. ∎

##  9\. Detection Results for Khovanov Homology

In this section we briefly review Khovanov homology before proving various
detection results.

###  9.1. A Brief Review of Khovanov homology

Let $L$ be a link, and $R$ be the ring ${\mathbb{Z}},{\mathbb{Z}}/2$ or
${\mathbb{Q}}$. The _Khovanov chain complex of $L$_,
$(\operatorname{CKh}(L,R),\partial)$, is a finitely generated
${\mathbb{Z}}\oplus{\mathbb{Z}}$-filtered chain complex over $R$
[khovanov2000categorification].

$\operatorname{CKh}(L;R):=\underset{i,j\in{\mathbb{Z}}}{\bigoplus}\operatorname{CKh}^{i,j}(L)$

Here $i$ is called the _homological grading_ , while $j$ is called the
_quantum grading_. The filtered chain homotopy type of $L$ is an invariant of
$L$. The $R$-module $\operatorname{CKh}(L,R)$ has generators corresponding to
decorated complete resolutions of $D$, while $\partial$ is determined by a
simple TQFT. The parity of the $j$ gradings in which $L$ has non-trivial
support agree with the parity of the number of components of $L$. The
_Khovanov homology of $L$_ is obtained by taking the homology of
$\operatorname{CKh}(L;R)$. A choice of basepoint $p\inL$ induces an action on
$\operatorname{CKh}(L;R)$, which commutes with the differential. Taking the
quotient of $\operatorname{CKh}(L;R)$ by this action and then taking homology
yields a bigraded $R$-module called the _reduced Khovanov homology_ of $L$,
denoted $\widetilde{\operatorname{Kh}}(L,p;R)$[khovanov2003patterns]. Given a
collection of points $\mathbf{p}=\\{p_{1},p_{2},\dots,p_{k}\\}\subset L$,
there is a generalization of reduced Khovanov homology called _pointed
Khovanov homology_ , $\operatorname{Kh}(L,\mathbf{p};R)$ due to Baldwin-
Levine-Sarkar [baldwin2017khovanov].

$\operatorname{Kh}(L;R)$ admits a number of useful spectral sequences. Suppose
$L$ has two components $L_{1},L_{2}$. If $R$ is ${\mathbb{Q}}$ or
${\mathbb{Z}}/2$ then $\operatorname{Kh}(L;R)$ admits a spectral sequence to
$\operatorname{Kh}(L_{1};R)\otimes\operatorname{Kh}(L_{2};R)$, called the
_Batson-Seed spectral sequence_ [batson2015link]. Indeed this spectral
sequences respects the $i-j$ grading on $\operatorname{Kh}(L;R)$ in the sense
that;

(2)
$\displaystyle\operatorname{rank}^{i-j=l}(\operatorname{Kh}(L;R)\geq\operatorname{rank}^{i-j=l+2\ell\text{k}(L_{1},L_{2})}(\operatorname{Kh}(L_{1};R)\otimes\operatorname{Kh}(L_{2};R))$

Suppose now that $L$ is an $n$ component link. There is another spectral
sequence, called the _Lee spectral sequence_ , from
$\operatorname{Kh}(L;{\mathbb{Q}})$ which abuts to an invariant called the
_Lee Homology_ of $L$,
$\operatorname{L}(L):=\bigoplus_{i}\operatorname{L}^{i}(L)$
[lee2005endomorphism]. This spectral sequence respects the $i$ gradings in the
sense that
$\operatorname{rank}(\operatorname{Kh}^{i}(L;{\mathbb{Q}}))\geq\operatorname{rank}(\operatorname{L}^{i}(L))$.
Lee showed that
$\operatorname{L}(L)\cong\bigoplus_{i=1}^{n}{\mathbb{Q}}^{2}_{a_{i}}$ where
$a_{i}$ are integers. Indeed if $L$ has two components $L_{1},L_{2}$ then
$\operatorname{L}(L)\cong{\mathbb{Q}}^{2}_{0}\oplus{\mathbb{Q}}^{2}_{\ell\text{k}(L_{1},L_{2})}$
[lee2005endomorphism, Proposition 4.3]. Moreover, there is an invariant,
$s(L)$ – due to Rasmussen [Rasmussen] in the knot case and Beliakova-Wehrli in
the link case [beliakova2008categorification] – defined from the Lee spectral
sequence – with the property that $s(L)\leq 1-\chi(L)$.

Finally, there is a spectral sequence, due to Dowlin [dowlin2018spectral],
relating Khovanov homology and Knot Floer homology. If $L$ is a link and
$\mathbf{p}\subseteq L$, with exactly one element of $\mathbf{p}$ in each
component of $L$, then there is a spectral sequence from
$\operatorname{Kh}(L,\mathbf{p};{\mathbb{Q}})$ to
$\widehat{\operatorname{HFK}}(L;{\mathbb{Q}})$ that respects the relative
$\delta$-gradings. As a corollary, Dowlin shows that if $L$ has $n$ components
then;

(3) $\displaystyle
2^{n-1}\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Q}}))\geq\operatorname{rank}(\widehat{\operatorname{HFK}}(L;{\mathbb{Q}}))$

###  9.2. Rank Detection Results for Khovanov homology

Applying Dowlin’s spectral sequence and the rank detection results for knot
Floer homology from Section 8, we have the following result;

######  Corollary 9.1.

Suppose $L$ is a two component pointed link with
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L,p;{\mathbb{Z}}/2))\leq
4$. Then $L$ is one of;

  * •

an unlink;

  * •

a Hopf link;

  * •

$T(2,4)$ or $T(2,-4)$.

Since
$\operatorname{rank}(\operatorname{Kh}(L;{\mathbb{Z}}/2))=2\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Z}}/2))$
by [shumakovitch2014torsion] and
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Z}}/2))\geq\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Q}}))$
by the universal coefficient theorem, this recovers Zhang-Xie’s classification
of two component links with
$\operatorname{rank}(\operatorname{Kh}(L;{\mathbb{Z}}/2))\leq 8$
[xie2020links, Corollary 1.2].

######  Corollary 9.2.

Suppose $L$ is a three component pointed link. Then
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L,p;{\mathbb{Q}}))>2$.

###### Proof of Corollary 9.1.

Suppose $L$ is a two component link with
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Q}}))\leq 4$.
Equation 3 implies that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L;{\mathbb{Q}}))\leq 8$, and
the result follows from Theorems 8.1, 8.2 and 8.3. ∎

###### Proof of Corollary 9.2.

Suppose $L$ is a three component link with
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Q}}))\leq 2$.
Equation 3 implies that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L;{\mathbb{Q}}))\leq 8$.
Theorems 8.1, 8.2 and 8.3 imply that $L$ is either the disjoint union of a
Hopf link and an unknot, or an unlink. But these links have reduced Khovanov
homology of too high a rank, a contradiction. ∎

###  9.3. Khovanov Homology Detects $T(2,8)$, $T(2,10)$

We prove the following theorems;

######  Theorem 9.3.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,8);{\mathbb{Z}})$.
Then $L$ is isotopic to $T(2,8)$

######  Theorem 9.4.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,10);{\mathbb{Z}})$.
Then $L$ is isotopic to $T(2,10)$.

$\begin{array}[]{c|c c c c c c c c c}24&&&&&&&&&{\mathbb{Z}}\\\
22&&&&&&&&{\mathbb{Z}}&{\mathbb{Z}}\\\ 20&&&&&&&&{\mathbb{Z}}/2\\\
18&&&&&&{\mathbb{Z}}&{\mathbb{Z}}\\\ 16&&&&&&{\mathbb{Z}}/2\\\
14&&&&{\mathbb{Z}}&{\mathbb{Z}}\\\ 12&&&&{\mathbb{Z}}/2\\\
10&&&{\mathbb{Z}}\\\ 8&{\mathbb{Z}}\\\ 6&{\mathbb{Z}}\\\
\hline\cr&0&1&2&3&4&5&6&7&8\par\par\end{array}\par\@@generic@caption{Table1.\operatorname{Kh}(T(2,8);{\mathbb{Z}}):Thehorizontalaxisindicatestheigrading,whiletheverticalaxisdenotesthejgrading}\par\par\par\par$

$\begin{array}[]{c|c c c c c c c c c c c}28&&&&&&&&&&&{\mathbb{Z}}\\\
26&&&&&&&&&&{\mathbb{Z}}&{\mathbb{Z}}\\\ 24&&&&&&&&&&{\mathbb{Z}}/2\\\
22&&&&&&&&{\mathbb{Z}}&{\mathbb{Z}}\\\ 22&&&&&&&&{\mathbb{Z}}/2\\\
20&&&&&&{\mathbb{Z}}&{\mathbb{Z}}\\\ 18&&&&&&{\mathbb{Z}}/2\\\
16&&&&{\mathbb{Z}}&{\mathbb{Z}}\\\ 14&&&&{\mathbb{Z}}/2\\\
12&&&{\mathbb{Z}}\\\ 10&{\mathbb{Z}}\\\ 8&{\mathbb{Z}}\\\
\hline\cr&0&1&2&3&4&5&6&7&8&9&10\par\par\end{array}\par\@@generic@caption{Table2.\operatorname{Kh}(T(2,10);{\mathbb{Z}}):Thehorizontalaxisindicatestheigrading,whiletheverticalaxisdenotesthejgrading}\par$

This requires the Lee, Batson-Seed, and Dowlin spectral sequences, as well as
an analysis similar to those from the previous sections.

$\operatorname{Kh}(T(2,m);{\mathbb{Z}})$ was computed in
[khovanov2000categorification], and is shown in Tables 1 and 2 in the $m=8$
and $m=10$ cases. Note in particular that
$\operatorname{Kh}(T(2,2n);{\mathbb{Z}})$ is _Khovanov-thin_ , meaning that
$\operatorname{Kh}(T(2,2n);{\mathbb{Z}})$ is supported in two adjacent $j-2i$
gradings.

To prove Theorems 9.3 and 9.4 we first show that any link $L$ with
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,2n);{\mathbb{Z}})$
has two components, with linking number $n$ and $\chi(L)\geq 2-2n$. We then
show that the two components are unknotted in the $n=4,5$ cases. After that we
conclude by showing that $L$ must bound an annulus after reversing the
orientation of one of the components, thereby completing the proofs.

######  Lemma 9.5.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,2n);{\mathbb{Z}})$.
Then $L$ is a two component link with linking number $n$ and $\chi(L)\geq
2-2n$.

We prove this Lemma using Lee’s spectral sequence [lee2005endomorphism], which
has $\operatorname{Kh}(L;{\mathbb{Q}})$ as $E_{1}$ page and $E_{\infty}$ page
a complex of rank $2^{n}$ where $n$ is the number of components of $L$.

###### Proof of Lemma 9.5.

Suppose $L$ is as in the Theorem. By the universal coefficient theorem there
are exactly two homological gradings in which the rank of
$\operatorname{Kh}^{i}(L;{\mathbb{Q}})$ is at least two, namely $i=0$ and
$i=2n$. Since
$\operatorname{rank}(\operatorname{Kh}^{i}(L;{\mathbb{Q}}))\geq\operatorname{rank}(\operatorname{L}^{i}(L;{\mathbb{Q}}))$,
and
$\operatorname{L}^{i}(L;{\mathbb{Q}})\cong\bigoplus_{i=1}^{m}{\mathbb{Q}}^{2}_{a_{i}}$,
where $a_{i}\in{\mathbb{Z}}$, $m$ is the number of components of $L$, it
follows that $L$ can have at most two components. To see that $L$ has exactly
two components note that $L$ has an even number of components since
$\operatorname{Kh}(L;{\mathbb{Z}})$ is supported in even quantum gradings
[khovanov2000categorification, Proposition 24].

The linking number of the two components of $L$ is half the difference in the
homological gradings of the generators which persist under Lee’s spectral
sequence [lee2005endomorphism, Proposition 4.3], hence the linking number of
$L$ is $n$.

Finally we note that since $L$ is Khovanov-thin, with generators supported in
the gradings $j-2i=2n,2n+2$ it has $s$-invariant given by $n+1$, whence
$\chi(L)\geq 2-2n$. ∎

We can now prove the following two lemmas;

######  Lemma 9.6.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,8);{\mathbb{Z}})$.
Then $L$ has unknotted components.

######  Lemma 9.7.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,10);{\mathbb{Z}})$.
Then $L$ has unknotted components.

Both of these use the Batson-Seed link splitting spectral sequence
[batson2015link] to bound the rank of the components of $L$ in certain
combinations of $i$ and $j$ gradings. We use Corollary 3.4 of [batson2015link]
which states that if $L$ is a link consisting of two components $L_{1},L_{2}$
then

(4)
$\displaystyle\operatorname{rank}^{l}(\operatorname{Kh}(L;F)\geq\operatorname{rank}^{l+2\ell\text{k}(L_{1},L_{2})}(\operatorname{Kh}(L_{1};F)\otimes\operatorname{Kh}(L_{2};F))$

Here $F$ is a field – we will use both ${\mathbb{Z}}/2$ and ${\mathbb{Q}}$
accordingly, and $\operatorname{rank}^{l}$ denotes the $i-j=l$ grading part of
the relevant vector space.

For Lemma 9.6 we use the fact that the only knots $K$ with
$\operatorname{rank}(\operatorname{Kh}(K;{\mathbb{Z}}/2))\leq 8$, are the
trefoils and unknot. This follows from Kronheimer-Mrowka’s unknot detection
result [kronheimer2011khovanov], as well as Baldwin-Sivek’s trefoil detection
results [baldwin2018khovanov] and the fact that for knots $K$
$\operatorname{rank}_{{\mathbb{Z}}/2}(\operatorname{Kh}(K;{\mathbb{Z}}/2))$ is
of the form $2+4n$ for some $n\geq 0$.

###### Proof of Lemma 9.6.

Suppose $L$ is as in the hypothesis of the Lemma. Lemma 9.5 implies that $L$
consists of two components, $L_{1}$ and $L_{2}$ of linking number $4$. From
$\operatorname{Kh}(L;{\mathbb{Z}})$, as shown in Table 1, we deduce that
$\operatorname{rank}_{{\mathbb{Z}}/2}(\operatorname{Kh}(L;{\mathbb{Z}}/2))=16$
by the universal coefficient theorem, so the Batson-Seed spectral sequence
implies that $\operatorname{rank}(\operatorname{Kh}(L_{i};{\mathbb{Z}}/2))\leq
8$, whence each component of $L$ is an unknot or a trefoil
[baldwin2018khovanov]. Indeed, we readily see that at least one component is
unknotted.

We now consider $\operatorname{Kh}(L;{\mathbb{Q}})$, which we can determine
from $\operatorname{Kh}(L;{\mathbb{Z}})$ via the universal coefficient
theorem. Suppose $L$ has a right hand trefoil component. Then
$\operatorname{rank}_{i-j=-4}(\operatorname{Kh}(T(2,3);{\mathbb{Q}})\otimes\operatorname{Kh}(U;{\mathbb{Q}})))=2$,
but $\operatorname{rank}_{i-j=-12}(\operatorname{Kh}(L;{\mathbb{Q}}))=1$
contradicting equation 4. Similarly,
$\operatorname{rank}_{i-j=2}(\operatorname{Kh}(T(2,-3);{\mathbb{Q}})\otimes\operatorname{Kh}(U;{\mathbb{Q}}))=3$
which contradicts
$\operatorname{rank}_{i-j=-6}(\operatorname{Kh}(L;{\mathbb{Q}}))=1$ by
Equation 4. Thus both components of $L$ are unknotted. ∎

Lemma 9.7 requires some extra analysis since there is no complete
classification of knots with
$\operatorname{rank}(\operatorname{Kh}(L;{\mathbb{Z}}/2))=10$. To deal with
this latter case we use Dowlin’s spectral sequence and Proposition 6.1.

###### Proof of Lemma 9.7.

Suppose
$\operatorname{Kh}(L;{\mathbb{Z}})\cong\operatorname{Kh}(T(2,10);{\mathbb{Z}})$.
Then from Table 2 and the universal coefficient theorem we can see that
$\operatorname{rank}(\operatorname{Kh}(T(2,10);{\mathbb{Z}}/2))=20$. Batson-
Seed’s spectral sequence implies that at least one component of $L$ is
unknotted, and the other is a knot $K$ with
$\operatorname{rank}(\operatorname{Kh}(K;{\mathbb{Z}}/2))\leq 10$. If
$\operatorname{rank}(\operatorname{Kh}(K;{\mathbb{Z}}/2))\leq 10$ then
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Z}}/2))\leq 5$,
so that $K$ is an unknot, trefoil, figure eight or
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(K;{\mathbb{Z}}/2))=5$. The
Batson-Seed spectral sequence implies immediately that $L$ has an unknotted
component. To see that the remaining component, $L_{2}$, must also be
unknotted we proceed by cases;

  1. (1)

$L_{2}$ is $T(2,3)$, in which case;

$\displaystyle\operatorname{Kh}(T(2,3);{\mathbb{Q}})\otimes\operatorname{Kh}(U;{\mathbb{Q}})$
$\displaystyle\cong({\mathbb{Q}}_{-6}\oplus{\mathbb{Q}}_{-3}^{2}\oplus{\mathbb{Q}}_{-1})\otimes({\mathbb{Q}}_{-1}\oplus{\mathbb{Q}}_{1})$
$\displaystyle\cong{\mathbb{Q}}_{-7}\oplus{\mathbb{Q}}_{-5}\oplus{\mathbb{Q}}_{-4}^{2}\oplus{\mathbb{Q}}_{-2}^{3}\oplus{\mathbb{Q}}_{0}$

But $\operatorname{rank}(\operatorname{Kh}(T(2,10);{\mathbb{Q}})$, is rank one
in exactly one $i-j$ grading, a contradiction.

  2. (2)

$L_{2}$ is $T(2,-3)$. Then

$\displaystyle\operatorname{Kh}(T(2,-3);{\mathbb{Q}})\otimes\operatorname{Kh}(U;{\mathbb{Q}})$
$\displaystyle\cong({\mathbb{Q}}_{6}\oplus{\mathbb{Q}}_{3}^{2}\oplus{\mathbb{Q}}_{1})\otimes({\mathbb{Q}}_{1}\oplus{\mathbb{Q}}_{-1})$
$\displaystyle\cong{\mathbb{Q}}_{7}\oplus{\mathbb{Q}}_{5}\oplus{\mathbb{Q}}_{4}^{2}\oplus{\mathbb{Q}}_{2}^{3}\oplus{\mathbb{Q}}_{0}$

But $\operatorname{Kh}(T(2,10);{\mathbb{Q}})$, is rank one in exactly one
$i-j$ grading, a contradiction.

  3. (3)

$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L_{2};{\mathbb{Z}}/2))=5$.
In this case we have that
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Q}}))\leq 5$,
whence
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}}))\leq 5$.
If $\operatorname{rank}(\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}})<5$,
then $L$ is a trefoil or the unknot, cases which we have already dealt with.
Thus we may take
$\operatorname{rank}(\widehat{\operatorname{HFK}})(L_{2};{\mathbb{Q}})=5$. We
show that $L_{2}$ is a figure eight or has the same knot Floer homology as
$T(2,\pm 5)$. Suppose $L_{2}$ is of genus $g>0$. Note that
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}},g)\leq 2$
We proceed by cases.

    1. (a)

Suppose
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L;{\mathbb{Q}},g))=2$. Then
$\widehat{\operatorname{CFL}}(L;{\mathbb{Q}})$ contains a differential of even
length $2g$. From Proposition 6.1, we see that this yields a summand of
$(\widehat{\operatorname{CFL}}(L;Q),\partial)$ rank $8g$. Thus
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L;{\mathbb{Q}}))\geq 8g+6$.
It immediately follows that $g=1$. The symmetry of
$\widehat{\operatorname{HFL}}(L)$ implies that
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L))\geq 16+12$, a
contradiction.

    2. (b)

Suppose
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}},g)=1$.
Consider $\widehat{\operatorname{CFL}}(L_{2};{\mathbb{Q}})$. If $g=1$, then
$L_{2}$ is the figure eight knot. If $g>1$ there are two possible forms for
$\widehat{\operatorname{HFK}}(L_{1})$, as shown in Figure 6. To see this we
need to use the fact that if
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L_{2},g-1))=1$, then there
cannot be both differentials from the top grading to the next to top grading
and from the next to bottom grading to the bottom grading. This is the
algebraic content of [NiPersonalCommunication].

Figure 6. Different possibilities for
$\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}})$.

In either case, since there is both a length one differential and a length
$g-1$ differential,
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L_{2};{\mathbb{Q}}))\geq
4(g-1)+4+4=4g+4$, whence $g\leq 4$. Suppose we are in the first case. Indeed,
since the minimal $A_{2}$ grading of these generators is $\dfrac{-3}{2}$,
symmetry yields at least another $4(g-1)$ generators, so that
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L_{2};{\mathbb{Q}}))\geq
8(g-1)+4+4+4$, so that $g=2$, and
$\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}})\cong\widehat{\operatorname{HFK}}(T(2,5);{\mathbb{Q}})$.

Suppose we are in the second case. We have that $g\in\\{2,3,4\\}$. If $g=2$
then
$\widehat{\operatorname{HFK}}(L_{2})\cong\widehat{\operatorname{HFK}}(T(2,-5))$.
If $g=4$ using the symmetry of $\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$
we find $\operatorname{rank}(\widehat{\operatorname{HFL}}(L;{\mathbb{Q}}))\geq
24$, a contradiction.

If $g=3$, symmetry of $\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$ implies
that the
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L;{\mathbb{Q}}))\geq 24$, a
contradiction.

  4. (4)

$L_{2}$ is the figure eight. In this case;

$\displaystyle\operatorname{Kh}(U;{\mathbb{Q}})\otimes\operatorname{Kh}(L_{2})\cong({\mathbb{Q}}_{3}\oplus{\mathbb{Q}}_{1}\oplus{\mathbb{Q}}_{0}^{2}\oplus{\mathbb{Q}}_{-1}\oplus{\mathbb{Q}}_{-3})\otimes({\mathbb{Q}}_{1}\oplus{\mathbb{Q}}_{-1})$

But $\operatorname{Kh}(T(2,10);{\mathbb{Q}})$, is rank one in exactly one
$i-j$ grading, a contradiction.

  5. (5)

$\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}})\cong\widehat{\operatorname{HFK}}(T(2,5);{\mathbb{Q}})$,
or
$\widehat{\operatorname{HFK}}(L_{2};{\mathbb{Q}})\cong\widehat{\operatorname{HFK}}(T(2,-5);{\mathbb{Q}})$.
Since $\ell\text{k}(L_{1},L_{2})=\pm 5$, there is a summand
$\widehat{\operatorname{HFK}}(T(2,\pm 5);{\mathbb{Q}})\otimes
V[\pm\frac{5}{2},a]$ of $\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$, for
some $a\in\pm\frac{5}{2}+{\mathbb{Z}}$. The symmetry of
$\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$ yields another $10$ generators.
There can be no additional generators, as this would violate the rank
inequality from Dowlin’s spectral sequence -
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L);{\mathbb{Q}})\leq\operatorname{rank}(\operatorname{Kh}(T(2,10);{\mathbb{Q}}))=20$,
where the final equality is obtained from the universal coefficient theorem.
Note that in order that each $A_{1}$ grading have even rank, we must have that
$\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$ is supported only in $A_{1}$
gradings $\pm\frac{1}{2}$. However since $L_{1}$ is unknotted and there is a
spectral sequence from $\widehat{\operatorname{HFL}}(L)$ to
$\widehat{\operatorname{HFL}}(L_{1})\otimes V[\frac{5}{2}]$ there must be
generators in $A_{1}$ grading $\frac{5}{2}$, a contradiction.

∎

We can now proceed to proofs to the Khovanov homology detection results. We
apply Dowlin’s spectral sequence and the classification of $E_{2}$ collapsed
chain complexes given in Lemma 6.1. From here it suffices to prove the
following pair of propositions.

######  Proposition 9.8.

Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link
satisfying the following conditions;

  1. (1)

$L$ has two exactly two components, $L_{1}$ and $L_{2}$, which are both
unknots.

  2. (2)

$\ell\text{k}(L_{1},L_{2})=4$

  3. (3)

$\chi(L)\leq-6$

  4. (4)

$\operatorname{rank}(\widehat{\operatorname{HFK}}(L;F))\leq 16$.

  5. (5)

$\widehat{\operatorname{HFK}}(L;F)$ is $\delta$-thin.

Then $L$ is $T(2,8)$.

######  Proposition 9.9.

Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link
satisfying the following conditions;

  1. (1)

$L$ has two exactly two components, $L_{1}$ and $L_{2}$, which are both
unknots.

  2. (2)

$\ell\text{k}(L_{1},L_{2})=5$

  3. (3)

$\chi(L)\leq-8$

  4. (4)

$\operatorname{rank}(\widehat{\operatorname{HFK}}(L;F))\leq 20$.

  5. (5)

$\widehat{\operatorname{HFK}}(L;F)$ is $\delta$-thin.

Then $L$ is $T(2,10)$.

Before proving these propositions, we apply them to obtain Theorem’s 9.3 and
9.4 respectively.

###### Proof of Theorem 9.3.

It suffices to show that $L$ satisfies the conditions in Proposition 9.8.
Conditions 1,2 and 3 are implied by Lemmas 9.5 and 9.6. To see that conditions
$4$ holds, observe that
$\operatorname{rank}(\operatorname{Kh}(L;{\mathbb{Z}}/2)=16$ by the universal
coefficient theorem. It follows that
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Z}}/2))=8$ by
[shumakovitch2014torsion], so that
$\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Q}}))\leq 8$ by
the universal coefficient theorem.
$\operatorname{rank}(\operatorname{Kh}(L,p,{\mathbb{Q}}))\leq\operatorname{rank}(\widetilde{\operatorname{Kh}}(L;{\mathbb{Q}}))$
by [baldwin2017khovanov, Theorem 1.2], whence
$\widehat{\operatorname{HFK}}(L;{\mathbb{Q}})\leq 16$.

To see that condition 5 holds, note that that
$\operatorname{Kh}(L;{\mathbb{Q}})$ is $\delta$-thin, so that
$\operatorname{Kh}(L,p;{\mathbb{Q}})$ is $\delta$-thin by
[baldwin2017khovanov, Lemma 2.11]. ∎

###### Proof of Theorem 9.4.

This follows by the same argument as in the proof of Theorem 9.3. ∎

We now prove the two requisite lemmas 9.8, and 9.9

###### Proof of proposition 9.8.

Suppose $L$ satisfies the conditions given in the Proposition. Consider
$\widehat{\operatorname{CFL}}(L;{\mathbb{Q}})$. Since
$\widehat{\operatorname{HFK}}(L)$ is $\delta$-thin,
$\widehat{\operatorname{CFL}}(L;{\mathbb{Q}})$ is an $E_{2}$ collapsed chain
complex. Likewise since $\chi(L)\leq-6$ the maximal Alexander grading of
$\widehat{\operatorname{HFK}}(L;{\mathbb{Q}})$ is at least $3$.

Consider $\widehat{\operatorname{CFL}}(L)$. Since $L$ has unknotted
components, it follows that
$(\widehat{\operatorname{CFL}}(L;{\mathbb{Q}}),\partial)$ cannot contain a
$V^{l}_{d}$ or $H_{d}^{l}$ summand, and that there are Maslov grading $0$ and
$-1$ generators with $A_{i}=2$. Suppose the Maslov grading $0$ generator is of
$A_{1}+A_{2}$-grading $l$. Suppose $l>4$. Since $L_{1},L_{2}$ are unknots with
linking number $2$, there must be Maslov grading $0$ and Maslov grading $-1$
generators on both the line $A_{1}=2$ and the line $A_{2}=2$. In particular
there are Maslov grading $0$ generators of bi-Alexander gradings $(2,l-2)$ and
$(l-2,2)$ and Maslov grading $-1$ generators of bi-Alexander gradings
$(2,l-3)$, $(l-3,2)$. Since these persist under the spectral sequence to $V$,
and are suitably identified there, it follows that
$(\widehat{\operatorname{CFL}}(L),\partial)$ has summands
$X_{-1}^{l-4}[2,2]\oplus X_{0}^{l-5}[1,1]$. From the symmetry of
$\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$, and the fact that
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L;{\mathbb{Q}}))\leq 16$ we
see immediately that $l\leq 6$.

Figure 7. One potential link Floer homology complex for a link satisfying the
conditions in Propositions 9.8

If $l\leq 4$ then there are summands $Y_{0}^{4-l}[l-2,l-2]\oplus
Y_{-1}^{5-l}[l-3,l-3]$. Note $l\geq 0$ as else
$\operatorname{rank}(\widehat{\operatorname{HFK}}(L))>16$. Similarly if
$l\in\\{1,2\\}$ would violate the rank bound after accounting for the symmetry
of $\widehat{\operatorname{HFL}}(L)$. If $l=0$ then
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L))\geq 16$. It follows that
there are no additional generators and we immediately see that $L$ bounds an
annulus, contradicting the fact that the maximal Alexander is at least $3$. If
$l\in\\{3,6\\}$, then the complex is determined by the symmetry of
$\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$. In particular we see that
there are copies of $B_{d}$ in specified bi-gradings. We quickly see that each
component is a braid axis, which implies that the Maximal $A_{i}$ gradings are
$2$, ruling out the $l=6$ case, while if $l=3$ we see immediately that the
dual Thurston polytope violates Lemma 6.2, or that $\chi(L)>-6$.

If $l=5$ we readily see that – apart from in the case that
$\widehat{\operatorname{CFL}}(L;{\mathbb{Q}})$ contains summands either
$B_{-6}[-3,2]\oplus B_{-6}[2,-3]$ or $B_{-1}[2,2]\oplus B_{-11}[-3,-3]$ – at
least one component is a braid axis, meaning that the maximal $A_{i}$ grading
is $2$, a contradiction. The $B_{-1}[2,2]\oplus B_{-11}[-3,-3]$ is excluded by
noting that upon reversing the orientation of a component of $L$, $L$ bounds
an annulus and hence is $T(2,8)$, a contradiction since it has the wrong link
Floer homology type. The $B_{-6}[-3,2]\oplus B_{-6}[2,-3]$ case is excluded by
noting that the Alexander polynomial of $L$ can be computed as
$(t^{-1/2}-t^{1/2})(-t^{-6}+2t^{-5}-t^{-4}+2t^{-1}-4+2t-t^{4}+2t^{5}-t^{6})$.
We then have that
$\nabla_{L}(t-t^{-1})=(t^{13}-t^{-13})-3(t^{11}-t^{-11})+3(t^{9}-t^{-9})-(t^{7}-t^{-7})-2(t^{3}-t^{-3})+6(t-t^{-1})$.
It follows that $\nabla(u)=u^{13}+10u^{11}+35u^{9}+50u^{7}+25u^{5}$, so that
the linking number is 25, a contradiction.

If $l=4$ we likewise see that apart from in the case that there is a summand
of $\widehat{\operatorname{CFL}}(L)$ given be either $B_{-5}[-2,1]\oplus
B_{-5}[1,-2]$ or $B_{-2}[1,1]\oplus B_{-8}[-2,-2]$, at least one component of
$L$ is a braid axis. The $B_{-2}[1,1]\oplus B_{-8}[-2,-2]$ case is excluded by
noting that in that case $L$ bounds an annulus after reversing the orientation
of a component and hence is $T(2,8)$ a contradiction, since $T(2,8)$ has a
distinct link Floer homology type. In the the $B_{-5}[-2,1]\oplus
B_{-5}[1,-2]$ case, we see that the Alexander polynomial is given by
$(t^{-1/2}-t^{1/2})((t^{4}+t^{-4})-2(t^{3}+t^{-3})+(t^{2}+t^{-2})-2(t+t^{-1})+4)$
whence
$\nabla_{L}(t-t^{-1})=-t^{9}+t^{-9}+3t^{7}-3t^{-7}-3t^{5}+3t^{-5}+3t^{3}-3t^{-3}-6t+6t^{-1}$,
so that $\nabla_{L}(u)=-u^{9}-6u^{7}-9u^{5}$, so that the linking number is
$-9$ a contradiction. Thus both components are braid axis and the result
follows from that the maximal $A_{i}$ gradings are $2$, and Lemma 6.2 implies
that
$\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})\cong\widehat{\operatorname{HFL}}(T(2,8);{\mathbb{Q}})$,
whence the result follows from the corresponding detection result in
[binns2020knot].

∎

The proof of 9.9 is virtually the same, but with some additional case
analysis.

###### Proof of Proposition 9.9.

Suppose Suppose $L$ is as in the hypotheses of the Proposition. Since $L$ is
delta-thin, $\widehat{\operatorname{CFL}}(T(2,2n);{\mathbb{Q}})$ is
$E_{2}$-collapsed. Since $\chi(L)\leq-8$ the maximal Alexander grading is at
least $n$. Since the components of $L$ are unknotted
$\widehat{\operatorname{CFL}}(L;{\mathbb{Q}})$ consists only of
$X^{l}_{d},Y^{l}_{d}$ and $B_{d}$ summands. Note that $L$ has Maslov index
$0$, $-1$ generators in each line $A_{i}=\frac{5}{2}$. Let $l$ be the $A$
grading of the Maslov index $0$ generator. If $l\geq 5$ then
$\widehat{\operatorname{CFL}}(L;{\mathbb{Q}})$ has summands
$X_{0}^{l-5}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{l-4}[2,2]$. Observe that
the symmetry of $\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$ together with
the fact that
$\operatorname{rank}(\widehat{\operatorname{HFL}}(L;{\mathbb{Q}}))\leq 20$,
imply that $l\leq 7$.

Figure 8. One potential component of the link Floer homology complex for a
link with the same Khovanov homology as $T(2,10)$. Note that some of the dots
correspond to rank two summands.

If $l=7$ then either $\widehat{\operatorname{CFL}}(L)\cong
X_{0}^{2}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{3}[2,2]\oplus
B_{-16}[\frac{-9}{2},\frac{-7}{2}]\oplus B_{-16}[\frac{-7}{2},\frac{-9}{2}]$
or there is an additional $B_{-8}[\frac{-1}{2},\frac{-1}{2}]$ summand, by the
symmetry of $\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$. In either case, we
see that the each component is a braid axis, whence the maximal $A_{i}$
gradings are $\frac{5}{2}$, per Remark $3$ of [GageT26], a contradiction.

If $l=6$, then either $\widehat{\operatorname{CFL}}(L)\cong
X_{0}^{1}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{2}[2,2]\oplus
B_{-11}[\frac{-7}{2},\frac{-7}{2}]$ or there are at most three extra $B_{d}$
summands. If there is one extra $B_{d}$ summands then must be of the form
$C_{-7}[\frac{-1}{2},\frac{-1}{2}]$ by symmetry, so if there are either zero
or one extra $B_{d}$ summands we see that $L$ bounds an annulus upon reversing
the orientation of a component, and is therefore $T(2,10)$ a contradiction
since $T(2,10)$ has distinct link Floer homology. If
$\widehat{\operatorname{HFL}}(L;{\mathbb{Q}})$ contains two extra cubes, then
unless they are of the form $C_{-1}[\frac{5}{2},\frac{5}{2}]\oplus
C_{-13}[\frac{-7}{2},\frac{-7}{2}]$ or $C_{-7}[\frac{5}{2},\frac{-7}{2}]\oplus
C_{-7}[\frac{-7}{2},\frac{5}{2}]$, we see immediately that at least one
component is a braid axis, and hence the Maximal $A_{i}$ grading is
$\dfrac{5}{2}$, a contradiction. The $C_{-1}[\frac{5}{2},\frac{5}{2}]\oplus
C_{-13}[\frac{-7}{2},\frac{-7}{2}]$ is excluded by noting that $L$ bounds an
annulus upon reversing the orientation of a component whence $L$ is $T(2,10)$,