\subsection*{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. 
	\end{section}
	
	
	
	\begin{section}{A Very Brief Review of Link Floer homology}\label{A 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 Ozsv\'ath-Szab\'o in~\cite{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, is a multi-graded ${\mathbb{Z}}/2[U_1,U_2,\dots U_n]$ chain complex with underlying vector space; $$\CFL^-(L,M)\cong\bigoplus_{m,A_1,A_2,\dots A_n}\CFL^-_m(L,M)(A_1,A_2\dots A_n).$$ 

The complex can also be defined with ${\mathbb{Z}}$ coefficients~\cite{manolescu2007combinatorial},\cite{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 \emph{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 \emph{Maslov grading}, which is integer valued. $\CFL^-(L)$ 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 $\CFL^-(L)$ is an invariant of $L$, which we refer to as the \emph{link Floer complex of $L$}. The homology of $\CFL^-(L)$ is denoted $\HFL^-(L)$. $\CFL^-(L)$ 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 $\widehat{\CFL}(L)$, and has homology denoted by $\widehat{\HFL}(L)$, which we will call the \emph{link Floer homology of $L$}. The \emph{link Floer homology polytope} of $L$ is defined as the convex hull of the multi-Alexander gradings of $\widehat{\HFL}(L)$ with non-trivial support. The link Floer polytope determines the Thurston polytope of the link exterior by a result of Ozv\'ath-Szab\'o~\cite{ozsvath2008linkFloerThurstonnorm}. We note that $\widehat{\HFL}(\overline{L})$, where $\overline{L}$ is $L$ with the orientation reversed, can be recovered from $\CFL^-(L)$ by only counting the disks which cross the appropriate basepoints.

For each component $L_i$ of $L$ there is a spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L-L_i)\otimes V[\frac{\ell\text{k}(L_i,L_j)}{2}]$. 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 $\widehat{\HFL}(L)$ is of rank $2^{n-1}$ in the maximal $A_i$ grading of non-trivial support by a result of Martin~\cite{GageT26}.
	 
	 The \emph{knot Floer homology} of $L$, $\widehat{\HFK}(L)$, is an oriented link invariant due independently to Rasmussen \cite{Rasmussen} and Ozsv\'ath-Szab\'o \cite{Holomorphicdisksandknotinvariants} that can be obtained from $\widehat{\HFL}(L)$ by projecting the multi-Alexander grading onto the diagonal, and shifting the Maslov gradings up by $\frac{n}{2}$. $\widehat{\HFK}(L)$ determines the maximal Euler characteristic of a surface bounding $L$ \cite{ni2006note}, as well as whether or not $L$ fibered \cite{ghiggini2008knot},\cite{ni2007knot}.
	 
	 The \emph{Conway polynomial} of $L$ can be obtained as an appropriate decatigorification of $\widehat{\HFL}(L)$. A result of Hoste \cite{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.
	 
	 
	   
	\end{section}
	
	

\begin{section}{Legendrian link invariants in link Floer Homology}\label{linkLOSS}

In this section we briefly review Cavallo's~\cite{cavallo2017invariant} generalization of Lisca-Ozsv\'ath-Stipsicz-Szab\'o's~\cite{lisca2009heegaard} (LOSS) Legendrian knot invariant to links. We then strengthen Cavallo's invariant, as required to prove Proposition~\ref{chainlevel}. 

Cavallo defined his LOSS invariant in a version of link Floer homology called \textit{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;
\[\cCFL^{-}(L,M) = \frac{\CFL^{-}(L,M)}{U_1 = U_2 = \cdots= U_n}\]

It has differential induced by the differential on the $\CFL^-(L,M)$ 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, $\mathcal{L}(L)\in \cHFL^{-}(L,\overline{M})$, Cavallo associates to every triple $(L,M,\xi)$ an \emph{adapted triple} $(\Sigma,\phi,A)$, defined as follows;


\begin{itemize}
    \item $\Sigma$ is a surface of genus $g$ with $l$ boundary components.
    \item $(\Sigma,\phi)$ is an open book compatible with $(M,\xi)$.
    \item $L$ is contained in the page $\Sigma_1$.
    \item $\{[L_i]\}\subset H_1(\Sigma;{\mathbb{Q}})$ is linearly independent.
    \item $A$ is a \emph{system of generators for $\Sigma$ adapted to $L$.}
\end{itemize}

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;

\begin{enumerate}
    \item $\{a_1,a_2\dots,a_{2g+l-1}\}$ is a basis for $H_1(\Sigma,\partial\Sigma;\mathbb{Q})$.
    \item $|L\cap a_i|=|L_i\cap a_i|=1$ for $1\leq i\leq n$. $a_i$ is called the \emph{distinguished arc} for $L_i$.
    \item $\Sigma \setminus \underset{i\geq 2g+l}{\bigcup} a_i$ consists of $n$ disks, each containing a component of $L$.
    \item 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 $\Sigma \setminus \underset{i< 2g+l}\bigcup a_i$.
\end{enumerate}

$\mathcal{L}(L)$ is a generator of $\cCFL^-(L,\overline{M})$ 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~\ref{Adapted triple}.

\begin{figure}
    \centering
    \includegraphics[width = 10cm]{Basepoints.pdf}
    \caption{Two possible choices of planting basepoints, depending on the orientation of $L$}
    \label{Adapted triple}
\end{figure}

Cavallo then defines his Legendrian invariant at the chain level, setting $\mathcal{L}(L)$ to be the homology class, in $\cCFL^-(L,\overline{M})$ of the generator $\mathcal{L}(D)=\{a_i\cap b_i\}_{i=1}^{2g+l+n-2}\in\cCFL^-(L,\overline{M})$~\cite{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 $\partial_{\cCFL^-}(\mathcal{L}(L))=0$. It is then shown that $[\mathcal{L}(L)]$ is independent  of the choice of adapted triple (ref.\cite[Theorem 1.2]{cavallo2017invariant}). 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;
\begin{enumerate}
    
\item Global contact isotopies.

\item Stabilization of $(\Sigma_1,\phi_1)$, $(\Sigma_2,\phi_2)$.

\item \textit{Admissible} arc slides on the families $A_1$, $A_2$; that is arc slides for which the arcs do not cross basepoints.
\end{enumerate}
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 $c\HFL^{-}(-L)$ and $\widehat{\HFL}(-L)$. 

For the purposes of this paper, specifically for proving the results in section~\ref{Rank bounds in link Floer homology}, we will view the same element, $\mathcal{L}(D)$, as an element of the link Floer chain complex $\CFL^{-}(L,\overline{M})$, which still maps to the Legendrian link invariant that lives in $\widehat{\HFL}(L,\overline{M})$. 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~\cite{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 \cite{cavallo2017invariant}, since the admissibility condition ensures that the arcs never cross the $\textbf{w}$ 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 $\CFL^{-}(D_1)$ and codomain  $\CFL^{-}(D_2)$, instead of $\cCFL^{-}(D_1)$ and $\cCFL^{-}(D_2)$ 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 $\CFL^{-}(D_1)$ to $\CFL^{-}(D_2)$ 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 $\widehat{\HFK}(\overline{S^3},L)$.

We conclude this section by noting that, as a mild refinement of Theorem 5.6 of~\cite{cavallo2017invariant}, the $i$th-Alexander grading of $\textit{pb}\mathcal{L}(L)\in\widehat{\HFL}(L,\overline{M})$ is given by \begin{align}\label{alexandergradingpbL}\dfrac{\text{tb}(L_i)+\text{rot}(L_i)+\ell\text{k}(L_i,L-L_i)}{2}\end{align}

The proof of this fact follows from Cavallo's proof of~\cite[Theorem 5.6]{cavallo2017invariant}
 

\end{section}
		
		
\begin{section}{Rank bounds in link Floer homology}\label{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~\cite{dey2020note} on the non-triviality of LOSS invariant resulting in additional elements in the knot Floer homology:

\begin{proposition}
\label{pbLOSS}
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 
\begin{itemize}\leftskip-0.35in
\item $\widehat{\HFL}(L) (A_1, \cdots, A_i - 1, A_{i+1}, \cdots, A_n) \neq 0$, or 
\item $\widehat{\HFL}(L) (A_1, \cdots, A_i + 1, A_{i+1}, \cdots, A_n) \neq 0$ ,
\end{itemize} 
for all $1 \leq i \leq n$.\color{black}
\end{proposition}

	From here the theorem advertised in the introduction follows immediately;
	
	\begin{theorem}\label{HFKrank}
	    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 $\widehat{\HFK}(L)$. Then  $\rank(\widehat{\HFK}(L)(A-1))+\rank(\widehat{\HFK}(L)(A+1))\geq n$.
	\end{theorem}
	
	As a corollary we can generalise the main result of \cite{baldwin_note_2018} to links;
	
	\begin{corollary}\label{rankinnequalityforfiberedlinks}
	If $L$ is a non-trivial $n$ component fibered link in $S^3$ then; \[\rank(\widehat{\HFK}(L,\dfrac{n-2-\chi(L)}{2}))\geq n.\]
	\end{corollary}

Before proceeding, we set notation for the different differential maps in $\CFL^{-}$ and $\widehat{\CFL}$ that we use in this section.
By $\partial^{-}$, we denote the differential map in the chain complex $\CFL^{-}$, 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 $\partial_{\widehat{\CFL}}$, which is the boundary map in $\widehat{\CFL}$. By $\partial_{\widehat{\CF}}$, we denote the unfiltered version of $\partial_{\widehat{\CFL}}$ i.e. the boundary map in $\widehat{\CF}$ of the ambient manifold, without the additional filtrations induced by the link.

\vspace{.1in}

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

		
\begin{lemma}\label{chainlevel} 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 $\widehat{\mathcal{L}}(L)\in\widehat{\HFL}(\overline{L})$. Then there is a generator $\mathbf{d}\in\CFL^-(L)(A_1,\dots A_i-1,\dots A_n)$ representing a non-trivial class in $\widehat{\HFL}(\overline{L})$ such that $\partial^{-}\mathbf{d}=U_i \cdot \text{pb}{\mathcal{L}}(L)$.
	\end{lemma}

 This implies a generalization of a Theorem of Ni~\cite[Theorem A.1]{ni2020seifert};
 
 \begin{corollary}\label{Niremark}
 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 $\rank(\widehat{\HFL}(A_1,\dots A_i-1,\dots A_n))\geq 2$.
 \end{corollary}
	

	
The outline of the proof of the main technical lemma, Lemma~\ref{chainlevel}, 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 $\partial_{\widehat{\CF}}\mathbf{d}=\text{pb}\mathcal{L}(L)$ 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 \cite{baldwin_note_2018} and \cite{dey2020note}.




\begin{proof}[Proof of Lemma~\ref{chainlevel}]

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.
\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{LOSS.pdf}
    \caption{A pseudoholomorphic disk connecting the LOSS invariant. \label{Adapted triple}}
    \end{figure}

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, $\partial^{-}\mathbf{d}=U_i \cdot \textit{pb}\mathcal{L}(L)\in\CFL^-(L,\overline{S^3})$. 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 $\mathbf{d'} \in \widehat{\CFL}(L,S^3)(-A_1,-A_2,\dots,1-A_i,\dots, -A_n)$, $\mathbf{c'}$ which is an element in $ \widehat{\CFL}(L,S^3)(-A_1,-A_2,\dots,-A_i,\dots,- A_n)$ and we observe that the duality of $\widehat{\CFL}(L,\overline{S^3})$ and $\widehat{\CFL}(L,S^3)$ implies that $\partial_{\widehat{\CF}(S^3)}(\mathbf{d'}) = \mathbf{c'}$. Note that $\mathbf{c'}$ is non-trivial in $\widehat{\HFL}(L,S^3)$.

To see that $\mathbf{d'}$ represents a non-trivial homology class, observe that if there were a generator $\mathbf{e}$ such that $\partial_{\widehat{\CFL}} \mathbf{e}=\mathbf{d'}$, then $\partial_{\widehat{\CF}}\mathbf{e}=\mathbf{d'}+\mathbf{h}+\mathbf{g}$ for some $\mathbf{h}\in\widehat{\CFL}(L)(-A_1,-A_2,\dots,-A_i,\cdots -A_n)$, 
$\mathbf{g}\in\bigoplus_{j\neq i,n\geq 1}\widehat{\CFL}(L)(-A_1,-A_2,\dots -A_j-n,\dots -A_n)$. It follows that $\partial^2_{\widehat{\CF}}\mathbf{e}=\partial_{\widehat{\CF}} \mathbf{d'}+\partial_{\widehat{\CF}} \mathbf{h}+\partial_{\widehat{\CF}}\mathbf{g}=0$, whence $\partial_{\widehat{\CF}} \mathbf{h}+\partial_{\widehat{\CF}}\mathbf{g}=\mathbf{c'}$. Since $\partial_{\widehat{\CF}}$ is non-decreasing on every Alexander filtration, it follows that only the Alexander grading $(-A_1,-A_2,\dots, -A_n)$ part of $\partial_{\widehat{\CF}} \mathbf{h}+\partial_{\widehat{CF}}\mathbf{g}$ is non-zero, whence $\partial_{\widehat{\CF}} \mathbf{h}=\mathbf{c'}$, a contradiction since $\text{pb}\mathcal{L}(L)$ was assumed to be non-zero and hence so is $\mathbf{c'}$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{pbLOSS}] This follows from \ref{chainlevel}, 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.
\end{proof}  


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

\begin{proof}[Proof of Corollary~\ref{Niremark}]
Let $L$ be as in the hypothesis of the theorem. Suppose that 

$${\rank(\widehat{\HFL}(L,S^3)(A_1,\dots A_i-1,\dots A_n))=1}.$$ 
Note that each $A_i$ is determined by equation~\ref{alexandergradingpbL}. Since we are working over a ${\mathbb{F}}$, it follows that 

$${\rank(\widehat{\CFL}(L,S^3)(A_1,\dots A_i - 1,\dots A_n))=1},$$ 
up to chain homotopy equivalence. Let $\mathbf{d}$ be a generator of $\widehat{\CFL}(L,S^3)(A_1,\dots A_i-1,\dots A_n)$. Consider $\CFL^-(L,S^3)$. 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~\ref{chainlevel} 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~\ref{chainlevel}, that there is a generator $\mathbf{d}'$ such that $\partial^-\mathbf{d}'=U_i\textit{pb}\mathcal{L}(L,-S^3)\in\CFL^-(L,-S^3)\cong\CFL^-(-L,S^3)$. Note that $\textit{pb}\mathcal{L}(-L)$ is of multi-Alexander grading $(-A_1,-A_2,\dots -A_n)$.


Consider the properties of $\widehat{\HFL}$ under mirroring $L$. Notice that $U_i \cdot pb\mathcal{L}(-L) \in Im(\partial^{-})$ implies that $\langle\partial_{\widehat{\CF}}(\widehat{\mathcal{L}}(L)), \mathbf{e'}\rangle\neq 0$, where $\mathbf{e'} \in \widehat{\CFL}(L,S^3)(A_1,\dots A_i-1,\dots A_n) \cong \mathbb{F} \cong \langle\textbf{d} \rangle$. Hence the existence of distinguished left and right-veering arcs implies the existence of these two maps in $\CFL^{-}$: 
\[\partial^{-}(\textbf{d}) = U_i\cdot pb\mathcal{L}(L), \partial_{\widehat{CF}}(\widehat{\mathcal{L}}(L)) = \mathbf{d}\]

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

\end{proof}

Corollary~\ref{rankinnequalityforfiberedlinks} follows from Theorem~\ref{HFKrank} using the fact that the self linking number of a transverse link determines the Alexander grading of the LOSS invariant.


\begin{proof}[Proof of Corollary~\ref{rankinnequalityforfiberedlinks}]
Let $L$ be a fibered $n$ component link. Consider $\widehat{\HFK}(L)$. 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 \cite[Theorem 6.2]{cavallo2017invariant} 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~\ref{HFKrank}.\end{proof}

\begin{remark}
Note that the conclusion of Lemma~\ref{chainlevel} 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 $\HFK^{-}(L)$ by a result of Honda-Kazez-Matic~\cite{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 \textit{always} be right-veering, which would contradict our assumption of existence of a left-veering arc.
\end{remark}

\begin{remark}
If one is interested in finding lower bounds on $\rank(\widehat{\HFK}(L))$ for a general $n$ component link with non-vanishing LOSS invariant Theorem~\ref{HFKrank} can be strengthened. Specifically if the Alexander grading of the LOSS invariant, $\frac{\text{tb}(L)-\text{rot}(L)+n}{2}$ by \cite[Theorem 5.6]{cavallo2017invariant}, 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 $\rank(\widehat{\HFK}(L))\geq 4|L|$. This is only a better bound in general than the bound $\rank(\widehat{\HFK}(L))\geq 2^{n-1}$ 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 \cite[Theorem 5.6]{cavallo2017invariant}, of the LOSS invariant is not in the range $[1,n-2]$, and the Alexander grading is not in the range $[-1,1]$ then $\rank(\widehat{\HFK}(L))\geq 2^{n-1}+4$.
\end{remark}
	
	We conclude this section by noting that it is perhaps useful to view Corollary~\ref{rankinnequalityforfiberedlinks} in the context of the following more general proposition;

\begin{proposition}\label{generalrankbound}
Suppose $L$ is an $n>1$ component non-trivial link such that \\ $\rank(\widehat{\HFL}(L)(x_1,x_2,\dots x_n))=2m+1$ for some $m\in{\mathbb{Z}}$. Then $\rank(\widehat{\HFK}(L))\geq 2^{n}+2m$.
\end{proposition}

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 $\widehat{\HFK}(L)$ to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$.

\begin{proof}[Proof of Proposition~\ref{generalrankbound}]
   Suppose $L$ is an $n>1$ component link non-trivial such that $\rank(\widehat{\HFK}(L)(x_1,x_2,\dots x_n))$ $=2m+1$. 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ must be even. Thus we obtain at least an additional $2^n-1$ generators.
\end{proof}

Note that if $L$ is an $n$ component link and $\rank(\widehat{\HFK}(L,A))$ is odd for some $A$, then there is a multi-Alexander  grading $(A_1,A_2\dots A_n)$ such that $\rank(\widehat{\HFK}(L,A_1,A_2,\dots A_n))$ of odd rank, where $A=A_1+A_2+\dots A_n$ whence it follows that $\rank(\widehat{\HFK}(L))\geq 2^n$.

\end{section}


\begin{section}{Detection Results for $(2,2n)$-cables}\label{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;


\begin{theorem}\label{HFL2detectsRHTcables}
Link Floer homology detects $T(2,3)_{2,2n}$ for all $n$.
\end{theorem}

\begin{theorem}\label{HFLdetects(2,2n)cablesofthefigureeight}
Let $K$ be the figure eight knot. Link Floer homology detects $K_{2,2n}$ for all $n$.
\end{theorem}

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


\begin{lemma}\label{(2,0)trefoilcable}
    Suppose $L$ is a link such that $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,3)_{(2,0)})$, 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 \begin{align*}
        \widehat{\HFK}(K)\cong{\mathbb{F}}_0^2[1]\oplus{\mathbb{F}}_{-1}^3[0]\oplus{\mathbb{F}}_{-2}^2[-1]
    \end{align*}
\end{lemma}

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


    We first give a partial computation of $\widehat{\HFK}(T(2,3)_{(2,0)})$. To do so compute $\widehat{\HFK}(T(2,3)_{(2,1)})$ and $\widehat{\HFK}(T(2,3)_{(2,-1)})$ using Hanselman--Watson's cabling formula in the theory of immersed curves~\cite{hanselman2019cabling}, then apply the skein exact triangle for knot Floer homology to determine $\widehat{\HFL}(T(2,3)_{(2,0)})$, and thereby $\widehat{\HFK}(T(2,3)_{(2,0)})$.
    
    Hanselman--Watson's cabling formula implies that \begin{align*}
        \widehat{\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]
    \end{align*}
    
    \begin{align*}
        \widehat{\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]
    \end{align*}
    
    \vspace{.2in}
    
    \noindent Using the skein exact triangle from~\cite[Equation 7]{Holomorphicdisksandknotinvariants}  -- 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 $\widehat{\HFK}(T(2,3)_{(2,0)},i)$ is trivial for $i> 2$ and $\widehat{\HFK}(T(2,3)_{(2,0)},2)\cong {\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$. To determine $\widehat{\HFK}(T(2,3)_{(2,0)};1)$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)})$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)})$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)}))$ is at least four in each of the maximal $A_i$ gradings (namely $A_i=1$). Thus $\widehat{\HFK}(T(2,3)_{(2,0)};1)\cong {\mathbb{F}}_{-\frac{1}{2}}^2\oplus{\mathbb{F}}_{\frac{1}{2}}^2$ and indeed \begin{align*}\widehat{\HFL}(T(2,3)_{(2,0)};(1,0))\cong\widehat{\HFL}(T(2,3)_{(2,0)};(0,1))\cong{\mathbb{F}}_0\oplus{\mathbb{F}}_{-1}.\end{align*}
    
    To conclude the computation it suffices to determine $\widehat{\HFK}(T(2,3)_{(2,0)};0)$ by the symmetries of link Floer homology. Note again that the skein exact triangle implies that $\widehat{\HFK}(T(2,3)_{(2,0)};0)$ 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 \begin{align*}
\widehat{\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]   \end{align*}

or

\begin{align*} \widehat{\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]  \end{align*}

We can now proceed to the proof of Lemma~\ref{(2,0)trefoilcable}
\begin{proof}[Proof of Lemma~\ref{(2,0)trefoilcable}]
 Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,3)_{(2,0})$. 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 $\widehat{\HFK}(L)$ is $\frac{1}{2}$. Since $\rank(\widehat{\HFK}(L))$ 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~\cite{hoste1985firstcoefficientoftheconwaypolynomial}. Since the maximal Alexander grading of $\widehat{\HFK}(L)$ is $1$, $L$ either bounds an annulus or the disjoint union of an unknot and a genus one knot. If $\rank(\widehat{\HFK}(L))=16$ 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 \begin{align*}
    {\mathbb{F}}_0^2[1]\oplus{\mathbb{F}}_{-1}^3[0]\oplus{\mathbb{F}}_{-2}^2[-1]
\end{align*} 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 $\widehat{\HFK}(L)$ 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  $\widehat{\HFK}(T(2,-3)_{(2,0)})\cong(\widehat{\HFK}(T(2,-3)_{(2,0)}))^*\not\cong\widehat{\HFK(L)}$. 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.


\end{proof}

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

\begin{lemma}\label{20cablefigureeight}
    Let $K$ be the figure eight knot and orient $K_{2,0}$ as the boundary of an annulus. Suppose $\widehat{\HFK}(L)\cong\widehat\HFK(K_{2,0})$. Then $L$ is isotopic to $K_{2,0}$.
\end{lemma}

Again we note that if $\widehat{\HFL}(L)\cong\widehat{\HFL}(K_{2,0})$, 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 $\widehat\HFK(K_{2,0})$. Hanselman-Watson's cabling formula for knot Floer homology in terms of immersed curves~\cite{hanselman2019cabling} shows that \begin{align*}
\widehat{\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]\\
\widehat{\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]\end{align*}

Again we apply the skein exact triangle from~\cite[Equation 7]{Holomorphicdisksandknotinvariants} -- with $L_0=K_{2,0}, L_+=K_{2,1}, L_-=K_{2,1}$.This immediately implies that $\widehat{\HFK}(K_{2,0},i)\cong 0$ for $|i|>2$ and $\widehat{\HFK}(K_{2,1})\cong 
F_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{5}{2}}$. Indeed we see that $\widehat{\HFK}(K_{2,0},1)$ 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 $\widehat{\HFL}(K_{2,0})$ is supported on the lines $A_2=A_1$, $A_2=A_1\pm 1$. Thus since $\widehat{\HFL}(L)$ must be of even rank in each $A_i$ grading, and $\widehat{\HFL}(K_{2,0},(x,y))\cong {\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$ is excluded. While unnecessary we note that $\widehat{\HFK}(L)\cong({\mathbb{F}}_{\frac{1}{2}}\oplus{\mathbb{F}}_{-\frac{1}{2}})^n$ for some $0\leq n\leq 3$. In fact $n\geq 2$, as else $\widehat{\HFL}(K_{2,0})$ cannot admit a spectral to $\widehat{\HFL}(K)\otimes V$

\begin{proof}[Proof of Lemma~\ref{20cablefigureeight}]
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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,0))$, 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_1)$, where $L_1$ is the component of $L$ corresponding to $A_1$. Thus $\widehat{\HFL}(L_1)$ is of rank one in the top Alexander grading and $L_1$ is fibered~\cite{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.
\end{proof}





To prove theorems~\ref{HFL2detectsRHTcables} and \ref{HFLdetects(2,2n)cablesofthefigureeight} 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 $n$.

\begin{lemma}\label{algebraiccharachterizationofcables}
Let $L$ be a two component link with linking number $n>1$ such that;
\begin{enumerate}
        \item $\widehat{\HFL}(L)$ is of rank $2$ in the maximal non-zero $A_1$ or $A_2$ grading.
        \item The link Floer homology polytope has a vertex at multi-Alexander grading $(1+\frac{n}{2},1+\frac{n}{2})$.
        \item $\widehat{\HFL}(L)$ has support on the lines $A_2=A_1,A_2=A_1+1, A_2=A_1-1$.
         \end{enumerate}
         
Then $L$ is isotopic to $K_{2,2n}$ where $K$ is a trefoil or the figure eight knot.


\end{lemma}


\begin{proof}[Proof of Lemma~\ref{algebraiccharachterizationofcables}] 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  $g(L_i)+\frac{n}{2}\leq\max\{A_i:\widehat{\HFL}(L)(A_1,A_2)\neq 0\}$, it follows that $g(L_i)\leq 1$ for each $i$. Indeed since, $\widehat{\HFL}(L)$ 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.
\end{proof}
 
 To apply Lemma~\ref{algebraiccharachterizationofcables}, we need to show that $\widehat{\HFL}(K_{2,2n})$ satisfies the conditions of the hypothesis of Lemma~\ref{algebraiccharachterizationofcables} for $K$ a trefoil or figure eight knot. The easiest way to proceed is via the following remark
 
 \begin{remark}
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~\cite{stallings1978constructions}, and $T(n,m)$ along minimal genus Seifert surfaces~\cite{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~\cite{gabai1986detecting}. It follows that $(2,2n)$ cables of genus one fibered knots are fibered and of genus $n$, that is Euler characteristic $-2n$. \end{remark}

 If one wishes to only use Rudolph's stated result, for knots, one can alternately proceed via the following lemma;
 
 \begin{lemma}\label{genusof22ncables}
 Let $K$ be a genus one fibered knot. For $n\geq 1$, $K_{2,2n}$ is fibered and $\chi(K_{2,2n})=-2n$.
 \end{lemma}
 
 \begin{proof}[Proof of Lemma~\ref{genusof22ncables}]
 Note that, for $n>0$, $K_{2,2n+1}$ is the Murasugi sum of $K_{2,1}$ and $T(2,2n+1)$~\cite{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~\cite{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 $\rank(\widehat{\HFK}(K_{2,2n})(1+n))=1$ so that $K_{2,2n}$ is fibered, as desired.
 \end{proof}
 
 
It follows that conditions 1, 2 and 3 of Lemma~\ref{algebraiccharachterizationofcables} 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.

\begin{proof}[Proof of Theorem~\ref{HFL2detectsRHTcables}]

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~\ref{(2,0)trefoilcable}.


Suppose now that $n>0$. We first show that $\widehat{\HFL}(T(2,3)_{2,2n})$ satisfies the conditions in the hypothesis of Lemma~\ref{alexandergradingcharachterizationofunlinks}. Observe that for $n\geq 1$, each component of $T(2,3)_{2,2n}$ is a braid in the complement of the other, so $\widehat{\HFL}(T(2,3)_{2,2n}))$ must be rank two in the maximal $A_i$ grading, by braid detection ~\cite[Proposition 1]{GageT26}, so condition 1 of Lemma~\ref{algebraiccharachterizationofcables} holds. By Lemma~\ref{genusof22ncables} the maximal Alexander grading of $\widehat{\HFK}(L)$ is $2+n$. $T(2,-3)_{2,2n+1}$ is fibered by Lemma~\ref{genusof22ncables}. Therefore, since $\widehat{\HFL}(T(2,-3)_{2,2n})$ 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$ must be of Maslov gradings $0$ and $-1$.

Suppose now that $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,3)_{2,2n})$ for some $n\geq 2$. Since link Floer homology detects the linking number of two component links~\cite{hoste1985firstcoefficientoftheconwaypolynomial}, it follows that the components of $L$ have linking number $n$. Lemma~\ref{algebraiccharachterizationofcables} 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 $\widehat{\HFL}(L)$ has a Maslov index $0$ generator with $A_1$ grading $1+\frac{n}{2}$ which persists under the spectral sequence to $\widehat{\HFL}(L_i)$, 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$. Observe that for $n\geq 1$, each component of $L$ is a braid in the complement of the other, so $\widehat{\HFL}(L)$ must be rank two in the maximal $A_i$ grading, by braid detection~\cite{GageT26}, so Condition 1 of Lemma~\ref{algebraiccharachterizationofcables} holds. By Lemma~\ref{genusof22ncables} the maximal Alexander grading of $\widehat{\HFK}(L)$ is $2+n$. Since $T(2,-3)_{2,2n+1}$ is fibered, and $\widehat{\HFL}(T(2,-3)_{2,2n+1})$ 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 \ref{alexandergradingcharachterizationofunlinks} 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$ must be of Maslov gradings $1$ and $2$.

Suppose now that $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,-3)_{2,2n})$. Lemma~\ref{algebraiccharachterizationofcables} 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 $\widehat{\HFL}(L)$ must be of Maslov gradings $1$ and $2$, it follows that each component is $T(2,-3)$ and we have the desired result.
\end{proof}

The proof of Theorem~\ref{HFLdetects(2,2n)cablesofthefigureeight} is again similar.

\begin{proof}[Proof of Theorem~\ref{HFLdetects(2,2n)cablesofthefigureeight}]
The proof follows the proof of Theorem~\ref{HFLdetectsLHTcables} verbatim, aside from to note that the top $A_i$ generators must be of Maslov index $0$ and $1$ respectively.
\end{proof}













\begin{section}{Technical Results}
\label{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.

	
	
	
	
	\begin{subsection}{$E_2$ collapsed chain complexes}


We will rely on a structure theorem for \emph{$E_2$-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 ~\cite[Section 12]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}. In particular Ozsv\'ath-Szab\'o 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~\ref{Summands}. 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 $-\mathbbm{1}$

\begin{figure}[h!]
\label{Summands}
    \centering
    \includegraphics[width=7cm]{E2_Summands.pdf}
    \caption{The summands of an $E_2$-collapsed chain complex}
    \end{figure}
    
    
    \begin{proposition}\label{E2collapsed}
    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 $B_d[i,j]\oplus V_d^l[i,j],H_d^l[i,j],X_d^l[i,j], Y_d^l[i,j]$.
    \end{proposition}
    
    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, $\widehat{\CFL}(L)$ 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 $\widehat{\CFL}(L)$. For instance, since the full filtered $\widehat{\CFL}$ complex has homology ${\mathbb{F}}_0\oplus{\mathbb{F}}_{-1}$, we see that $\widehat{\CFL}(L)$ 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 $\widehat{\CFL}(L_1)$ determines the $V_d^l$ summands, while $\widehat{\CFL}(L_2)$ determines the $H_d^l$ summands. Specifically, if there are generators $x,y\in\widehat{\CFL}(L_1)$, with $\langle\partial_{\widehat{\CFL}(L_1)} x,y\rangle\neq 0$ then we must have a summands of $\widehat{\CFL}(L)$ of the form $V_{M(x)}^{A(x)-A(y)}$, as no other summand induces the correct differential on $\widehat{\CFL}(L_1)$.
    

    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 $\partial_2$.
    \begin{proof}[Proof of Proposition~\ref{E2collapsed}]
    
    
    By a bi-filtered version of the Reduction Lemma~\cite[Lemma 4.1]{hedden2013khovanov} we can assume that the grading preserving part of $\partial$ is trivial. We proceed by induction on $\rank(C)$, 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. 
    \end{proof}
    
    
    
    


\begin{comment}
\color{red}
\begin{lemma}\label{cubes}
Suppose $\widehat{\HFL}(L;{\mathbb{F}})$ is $\delta$-thin. Then, as a vector space, $\widehat{\HFL}(L;{\mathbb{F}})\cong\underset{i}{\bigoplus}C_{d_i}[x_i,y_i]$.
\end{lemma}

Here $C_d(0,0)$ is the complex ${\mathbb{F}}_d(0,0)\oplus {\mathbb{F}}_{d+1}(1,0)\oplus{\mathbb{F}}_{d+1}(0,1)\oplus{\mathbb{F}}_{d+2}(1,1)$, and ${\mathbb{F}}$ is either ${\mathbb{Z}}/2$ or ${\mathbb{Q}}$.

\begin{proof}[Proof of Lemma~\ref{cubes}]
As observed in the proof of~\cite[Theorem 12.1]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, $\widehat{\HFL}(L)$ 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 $B_d$. As vector spaces, each of these complexes decomposes as a sum of $C_d(x,y)$.
\end{proof}

\end{comment}


\color{black}

    
   

	\end{subsection}
	
	
	\begin{subsection}{The Dual Thurston Polytope}
	
	

The four detection results also require the following proposition;

\begin{proposition}\label{Braidpolytope}
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
\end{proposition}
 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.

\begin{proof}[Proof of Proposition~\ref{Braidpolytope}]

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.

\end{proof}



	
	\end{subsection}
	


\end{section}










\end{section}




\begin{section}{Knot Floer Homology Detects $T(2,8),T(2,10)$}
\label{Knot Floer Homology Detects $T(2,8),T(2,10)$}

In this section we prove the following detection results;



\begin{theorem}\label{HFKT28}
	Knot Floer homology detects $T(2,8)$.
\end{theorem}

	\begin{theorem}\label{HFKT210}
	Knot Floer homology detects $T(2,10)$.
\end{theorem}


 
For the readers convenience first we recall that:
    $$\widehat{\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}
    
    \end{cases}$$


We start by showing that in general if $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,2n))$ then $L$ consists of two unknotted components and $\wiehat{\CFL}(L)$ is of an especially simple form.
\begin{lemma}\label{T22nunknottedcomponents}
    Suppose $\widehat{\HFK}(L;{\mathbb{F}})\cong\widehat{\HFK}(T(2,2n);{\mathbb{F}})$. Then $(\widehat{\CFL}(L),\partial)$ 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.
\end{lemma}




\begin{proof}[Proof of Lemma~\ref{T22nunknottedcomponents}]
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 $\widehat{\HFK}(L)$ is $\frac{1}{2}$, so since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{m-1}S^1\times S^2)$, $L$ can have at most two components. $L$ then has exactly $2$ components since the Maslov grading of $\widehat{\HFK}(L)$ 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 $\widehat{\HFL}(L)$. 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~\cite[Theorem 12.1]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, $\widehat{\HFL}(L)$ 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 $\widehat{\HFK}(L_1)$ 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 $\widehat{\HFK}(L_1)\otimes V$ then there must be a generator of odd Maslov index in $A_1$ grading $\frac{n}{2}-x-1$, since the summands of $\widehat{\HFL}(L)$ that persist under the spectral sequence to $\widehat{\HFK}(L_1)\otimes V$ 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.
\end{proof}



\begin{comment}
\color{red}
\begin{lemma}\label{T22nsummand}
	Let $n>1$. Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,2n))$. Then $L$ is a two component link with linking number $n$ and $\widehat{\HFL}(L)$ has a summand $C_{-2}[\frac{n-2}{2},\frac{n-2}{2}]\oplus C_{-2n}[\frac{-n}{2},\frac{-n}{2}]$.
\end{lemma}

The $n=1$ case is that of the Hopf link, which knot Floer homology is known to detect.

\begin{proof}[Proof of Lemma~\ref{T22nsummand}]
	Note that $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{|L|-1}(S^1\times S^2))$, where $|L|$ is the number of components of $L$. $\widehat{\HF}(\#^{|L|-1}(S^1\times S^2))$ has support in Maslov grading $\dfrac{|L|-1}{2}$, while the maximal Maslov grading of $\widehat{\HFK}(L)$ is $\frac{1}{2}$, so $L$ can have at most $2$ components. $L$ cannot have one component since the Maslov gradings of the generators of $\widehat{\HFK}(L)$ take values in ${\mathbb{Z}}+\frac{1}{2}$. $L$ has linking number $n$ because $\widehat{\HFK}(L)$ determines the Conway polynomial of two component links, and the Conway polynomial in turn detects the linking number of two component links~\cite{hoste1985firstcoefficientoftheconwaypolynomial}.
	
	Consider $\widehat{\HFL}(L)$. Let $A_i(0)$ be the Alexander index of the unique Maslov index $0$ generator in $\widehat{\HFL}(L_i)$ for each $i$. Note that $A_i(0)\geq 0$ since there are no positive Maslov index generators. Note that $A_1(0)+\frac{n}{2}+A_2(0)+\frac{n}{2}\leq n$, so $A_i(0)=0$, and the Maslov index $0$ generator is of bi-Alexander grading $(\frac{n}{2},\frac{n}{2})$. Lemma~\ref{cubes} implies that $\widehat{\HFL}(L)$ has a summand $C_{-2}[\frac{n-2}{2},\frac{n-2}{2}]$. The symmetry of $\widehat{\HFL}(L)$ then implies that $\widehat{\HFL}(L)$ has another summand $C_{-2n}[\frac{-n}{2},\frac{-n}{2}]$.
\end{proof}

\end{comment}

\color{black}


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


\begin{proof}[Proof of Theorem~\ref{HFKT28}]
Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nunknottedcomponents} 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 \cite{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~\ref{Braidpolytope}, and the result follows from the fact that link Floer homology detects $T(2,8)$~\cite{binns2020knot}.

\begin{comment}
\color{red}
Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nsummand} implies that there is a summand of $\widehat{\HFL}(L)$ given by $C_{-2}[1,1]\oplus C_{-8}[-2,-2]$. Indeed, since 
\[ \widehat{\HFL}_m(L)(A_1,A_2)\cong\widehat{\HFL}_{m-2A_1-2A_2}(L)(-A_1,-A_2)\]
Lemma~\ref{cubes} implies that the remainder of $\widehat{\HFL}(L)$ is given by $C_{-4}(x,-x)\oplus C_{-6}[-x-1,x-1]$. We seek to control $x$. We first show that $L$ has unknotted components. Suppose that there is a generator from $C_{-8}[-2,-2]\oplus C_{-2}[1,1]$ which is not of $A_i$ grading $2$ persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V(2)$. Then there must be a generator of Maslov grading at least $-1$ with $A_i$ grading at least $3$, which there is not. It follows that only a generator from $C_{-4}[x,-x]\oplus C_{-6}[-x-1,x-1]$ can possibly persist under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V[2]$, but if such a generator persists with $A_i$ grading at least $2+g(L_i)$ with $g(L_i)>0$ then there must be a generator which persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V[2]$ of Maslov grading at most $-2-2g(L_i)<-2$ of $A_i$ grading $2-g(L_i)$, a contradiction.
\end{comment}


	\begin{figure}[h]
    \centering
    \includegraphics[width=8cm]{T(2_8)pseudo.pdf}
    \caption{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.}
    \end{figure}
\begin{comment}
Thus both components of $L$ are unknotted. Therefore all of the hypothesis of Proposition~\ref{combinedHFKKHT28detection} hold and the result follows.


Since the rank in the maximal $A_i$ grading is $2$ for at least one $i$, at least one component is a braid axis for the other. Without loss of generality we may take $i=1$ and thereby deduce that the maximal $A_1$ grading is strictly less than $2$. It follows that $x\in\{-1,0\}$. The $x=-1$ case is excluded by Lemma~\ref{Braidpolytope}. Thus $x=0$, $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,8))$, and the result follows from the corresponding detection result in~\cite{binns2020knot}.
\end{comment}
\end{proof}


\color{black}
The proof of Theorem~\ref{HFKT210} is virtually identical

\begin{proof}[Proof of Theorem~\ref{HFKT210}]

Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,10))$. 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~\ref{Braidpolytope}, and the result follows from the fact that Link Floer homology detects $T(2,10)$.

\begin{comment}
	Suppose $L$ satisfies $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nsummand} implies that $L$ is a two component link with linking number $5$ and that there is a summand of $\widehat{\HFL}(L)$ of the form $C_{-10}[-\frac{5}{2},-\frac{5}{2}]\oplus C_{-2}[\frac{3}{2},\frac{3}{2}]$. Symmetry implies that there is another of the form $C_{-6}(\frac{-1}{2},\frac{-1}{2})$. There are two summands remaining $C_{-4}(x,1-x)$ and $C_{-8}(-x,x-3)$.
	
	We first note that if there were a generator from $C_{-6}(\frac{-1}{2},\frac{-1}{2})\oplus C_{-10}(-\frac{5}{2},-\frac{5}{2})\oplus C_{-2}(\frac{3}{2},\frac{3}{2})$ that persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V$ then there would be a generator of Maslov grading at least $-1$ in $A_i$ grading at least $\frac{7}{2}$, a contradiction.
	\end{comment}
	
	
    \begin{comment}
	 It follows that only the generators from $C_{-4}(x+\frac{1}{2},\frac{1}{2}-x)$ and $C_{-8}(-x-\frac{3}{2},x-\frac{3}{2})$ can possibly persist in $A_i$ gradings higher than $\frac{5}{2}$, but if such a generator persists with $A_i$ grading at least $\frac{5}{2}+g(L_i)$ with $g(L_i)>0$ then there must be a generator of Maslov grading at most $-2-2g(L_i)$ of $A_i$ grading $2-g(L_i)$ that persists under the spectral sequence to $\widehat{\HFL}(L_i)$, a contradiction.
	 
	 Thus both components of $L$ are unknotted. It follows that all of the hypotheses of Proposition~\ref{combinedHFKKHT210detection} hold and the result follows.
	 
Note that the rank of $\widehat{\HFL}(L)$ is two in at least one of the top $A_i$ gradings. We may thus, without loss of generality, take $L_1$ to be a braid axis, whence the maximal $A_1$ grading is $\frac{5}{2}$~\cite{GageT26}. We are therefore left only with the possibility that $|x|<2$. Note however that unless $x=0$ Lemma~\ref{Braidpolytope} is violated. Thus $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,8))$ and the result follows from the corresponding detection result in~\cite{binns2020knot}.
	 
\end{comment}
\end{proof}




    
    
\end{section}









\begin{section}{Rank Detection Results for Knot Floer Homology }
\label{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;
\begin{theorem}\label{rank4}
	If $\rank(\widehat{\HFK}(L))=4$ then $L$ is the Hopf link or the three component unlink.
\end{theorem}


\begin{theorem}\label{rank6}
	Suppose $L$ is a link with $\rank(\widehat{\HFL}(L))=6$. Then $L$ is a disjoint union of a trefoil and an unknot.
\end{theorem}

\begin{theorem}\label{rank8}
The only links $L$ with $\rank(\widehat{\HFL}(L))=8$ are $T(2,4)$, $T(2,-4)$, the four component unlink, and the disjoint union of a Hopf link and an unknot.
\end{theorem}

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

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

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

\begin{proposition}\label{3componentrank12}
There is no three component link with $\rank(\widehat{\HFK}(L))=10$.
\end{proposition}



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

\begin{theorem}\label{HFKdetcetstrefoilmeridian}
Let $L_0$ be $T(2,3)$ with a meridian. Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(L_0)$. Then $L$ is isotopic to $L_0$.
\end{theorem}
     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~\cite{binns2020knot}.
     





The proofs of Theorems \ref{rank4}, \ref{rank6}, \ref{rank8} and \ref{3componentrank12} 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 \cite{binns2020knot}.



We first prove Lemma~\ref{unlinkdetection}, as it is useful in proving the other theorems.

We will find the following characterization of unlinks useful;

\begin{lemma}[Ni~\cite{ni2006note}]\label{alexandergradingcharachterizationofunlinks}
Suppose $\widehat{\HFK}(L)$ is supported in a single Alexander grading. Then $L$ is an unlink.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{alexandergradingcharachterizationofunlinks}]
Observe that if $\widehat{\HFK}(L)$ is supported in a single Alexander grading then that grading must be zero, by the symmetry of $\widehat{\HFK}(L)$. 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.
\end{proof}

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 $-1$.



\begin{proof}[Proof of Lemma~\ref{unlinkdetection}]
    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 $\rank(\widehat{\HFK}(L))=2^n$ then for any component $L_i$ of $L$, $L-L_i$ is a link with $\rank(\widehat{\HFK}(L-L_i))=2^{n-1}$. By inductive hypothesis $L-L_i$ is the $n$ component unlink. It follows that the support of $\widehat{\HFL}(L)$ is contained in the line defined by $A_j=0$ for $j\neq i$. But $i$ is arbitrary, so $\widehat{\HFL}(L)$ is supported in multi-Alexander grading $\mathbf{0}$, and the result follows from Lemma~\ref{alexandergradingcharachterizationofunlinks}.
\end{proof}



We now prove Theorem~\ref{rank4}.


\begin{proof}[Proof of Theorem~\ref{rank4}]
Let $L$ be an $n$ component link with $\rank(\widehat{\HFK}(L))=4$. Since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$, it follows that $n\leq 3$. Indeed, since $\rank(\widehat{\HFK}(L))$ 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~\ref{unlinkdetection}. It therefore remains only to consider the case that $n=2$. Let $g$ be the maximal Alexander grading of $\widehat{\HFK}(L)$. We consider the following three cases; $\rank(\widehat{\HFK}(L)(g))$ is either four, two or one, in which case the link is fibered by \cite{ni2007knot}.\begin{enumerate}
       
       \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=4$}.} In this case $g=0$, whence $\widehat{\HFK}(L)$ does not admit a spectral sequence to $\widehat{\HF}(S^1\times S^2)$.
          \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=2$}.} In this case we see that 
         $$\widehat{\HFK}(L)\cong ({\mathbb{F}}_{m_1}\oplus{\mathbb{F}}_{m_2})[g]\oplus ({\mathbb{F}}_{m_1-2g}\oplus{\mathbb{F}}_{m_2-2g})[-g].$$
        
        Observe that $g>0$. Consider the spectral sequence from $\widehat{\HFK}(L)$ to $\widehat{\HF}(S^1\times S^2)$. 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.
        
       \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=1$}.} Observe that $g\geq 1$. Let $(x,g-x)$ be the bi-Alexander grading of the contact class in $\widehat{\HFL}(L)$.
       
       If $g>1$, then Theorem~\ref{HFKrank} implies that $\rank(\widehat{\HFK}(L))\geq 6$. Thus $g=1$, and $$\widehat{\HFL}(L)\cong {\mathbb{F}}_{m}(\frac{1}{2},\frac{1}{2})\oplus {\mathbb{F}}_{m-1}(-\frac{1}{2},\frac{1}{2})\oplus {\mathbb{F}}_{m-1}(\frac{1}{2},-\frac{1}{2})\oplus {\mathbb{F}}_{m-2}(-\frac{1}{2},-\frac{1}{2})$$
       Since $\widehat{\HFL}(L)$ 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~\cite{BG}.
    \end{enumerate}
    

\end{proof}

We now prove Theorem~\ref{rank6}, 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.





\begin{lemma}\label{rank6fibered}
There is no fibered link $L$ such that $\rank(\widehat{\HFK}(L))=6$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{rank6fibered}]
Suppose $L$ is an $n$ component fibered link with $\rank(\widehat{\HFK}(L))=6$. Since $\rank(\widehat{\HFL}(L))$ is even, $L$ cannot be a knot. Moreover, $L$ can have at most $3$ components, since $\widehat{\HFL}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$.

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 $\widehat{\HFK}(L)$ has support is at least $2$. By Theorem~\ref{HFKrank}, the next to top Alexander grading is of rank three, whence $\widehat{\HFK}(L)$ 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 $\widehat{\HFK}(L)$ 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 $\rank(\widehat{\HFK}(L))=\rank(\widehat{\HFK}(T(2,\pm 3)\otimes V)$. 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~\ref{chainlevel} 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 $\widehat{\HFL}(L)$, 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 \cite{GageT26}, a contradiction.
          

\end{proof}

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


\begin{proof}[Proof of Theorem~\ref{rank6}]
Let $L$ be an $n$ component link with link Floer homology of rank $6$. Lemma~\ref{rank6fibered} implies that $L$ is not fibered under any orientation. The fact that $2^{n-1}\leq\rank(\widehat{\HFK}(L))$ implies that $L$ has at most three components. Since $\rank(\widehat{\HFK}(L))$ 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 $\widehat{\HFK}(L)$ under some orientation of $L$.


Lemma~\ref{alexandergradingcharachterizationofunlinks}, 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ in the plane $A_i=p_i$ is exactly $p$, it follows that the spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L_i)\otimes V^{n-1}$ is trivial. Thus the rank of $\widehat{\HFL(L)}$ at $p$ is at least $2^{n-1}$. Since the polytope is non-degenerate, the symmetry properties of $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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 $\rank(\widehat{\HFL}(L))\geq 8$, a contradiction. \end{proof}



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

\begin{proposition}\label{rank8fibered}
Suppose $L$ is fibered and $\rank(\widehat{\HFK}(L))=8$. Then $L\cong T(2,4)$ or $T(2,-4)$.
\end{proposition}
\begin{proof}
Let $L$ be a fibered link with link Floer homology of rank eight. Since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$, 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 $\rank(\widehat{\HFK}(L))$ is even. Suppose $L$ has four components. Then by Theorem~\ref{unlinkdetection}, $L$ 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 $\widehat{\HFK}(L)$ has supported is at least $2$. By Theorem~\ref{chainlevel}, $\rank(\widehat{\HFK}(L)(g-1))\geq3$. Thus $$\widehat{\HFK}(L)\cong{\mathbb{F}}_m[g]\oplus{\mathbb{F}}_{m-1}^3[g-1]\oplus {\mathbb{F}}_{m+1-2g}^3[1-g]\oplus{\mathbb{F}}_{m-2g}[-g]$$ 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 $\widehat{\HFK}(L)$ is at least one. Suppose $g=1$. Then $\rank(\widehat{\HFK}(L;0))=6$. Such a complex clearly does not admit a spectral sequence to $\widehat{\HF}(S^1\times S^2)$, a contradiction. Thus $g\geq 2$. Consider $\widehat{\HFL}(L)$. Suppose the top Alexander grading generator is in multi-Alexander grading $(x,g-x)$ and of Maslov index $m$. Theorem~\ref{chainlevel} 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 $\widehat{\HFK}(L)$ to $\widehat{\HFK}(L_1)\otimes V$ 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 $\rank(\widehat{\HFK}(T(2,2n))=8$ are $n=\pm2$.

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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_2)$ do not have admissible Maslov gradings. Thus the complex splits as two squares, and again must be located on the diagonal, a contradiction.

\end{proof}



\begin{proof}[Proof of Theorem~\ref{rank8}]


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 $\rank(\widehat{\HFK}(L))\geq 2^{n-1}$ and $L$ is even, $L$ has either two, three or four components. If $L$ has four components, then Lemma~\ref{unlinkdetection} implies that $L$ is an unlink. We deal with the two remaining cases separately;\begin{enumerate}
   
    \item\textbf{$L$ has two components}
Consider $\widehat{\HFL}(L)$. 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 $\widehat{\HFK}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks} 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_1)$, 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~\ref{rank2face} 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~\ref{rank2face}.


	\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{Rank_8_HFL.pdf}
    \caption{A rank eight bigraded vector space not arising as the link Floer homology of a link.\label{rank2face}}
    \end{figure}


The only remaining case is that on each face the two generators sit in the same multi-Alexander grading, see figure~\ref{rank2face}. 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 $\widehat{\HFL}(L)$ 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 $\partial_1x=y$. Since $y$ does not persist under the spectral sequence to $\widehat{\HFL}(L_2)$, $\partial_2 y=z$. 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.

   
   
   



\item \textbf{$L$ 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 $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L-L_3)$ is just projection onto the $A_3=0$ plane. Suppose $\widehat{\HFL}(L)$ is supported in the $A_3=0$ plane, then $L$ consists of a disjoint union of a Hopf link and an unknot. If $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks}.
\end{enumerate}

\end{proof}


\begin{proof}[Proof of Proposition~\ref{3componentrank12}]
Let $L$ be a $3$ component link with $\rank(\widehat{\HFK}(L)=10$. Suppose $L$ is fibered. The maximal Alexander grading, $A$, of $\widehat{\HFK}(L)$ is given by $2+g$ for some $g\geq 0$. Note that $\rank(\widehat{\HFK}(L)(A))=1$. Suppose the generator is of Maslov index $m$. Theorem~\ref{HFKrank} implies that there are at least three generators of Maslov index $m-1$ in Alexander grading $A-1$. The symmetry of $\widehat{\HFK}(L)$ 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 $\widehat{\HFK}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks}. 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 $\rank(\widehat{\HFK}(L))\geq 16$, 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 $\widehat{\HFL}(L)$ on the edge $-A_1-A_2=\dfrac{g_1+g_2}{2}$, persists under the spectral sequence to $\widehat{\HFL}(L-L_3)\otimes V[\frac{\ell\text{k}(L_1,L_3)}{2},\frac{\ell\text{k}(L_2,L_3)}{2}]$, and is indeed of maximal (respectively minimal $A_1+A_2$ grading). Now, there are only two 2 component links with $\rank(\widehat{\HFK}(L))\leq 5$; 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 $i$, $\rank(\widehat{\HFK}(L-L_i)\leq 5$, 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.


\end{proof}

We conclude with a proof of  Theorem~\ref{HFKdetcetstrefoilmeridian}


\begin{proof}[Proof of Proposition~\ref{HFKdetcetstrefoilmeridian}]

Let $L_0$ be the trefoil together with a meridian. It suffices to show that if  $\widehat{\HFK}(L)\cong\widehat{\HFK}(L_0)$ then  $\widehat{\HFL}(L)\cong\widehat{\HFL}(L_0)$ by the corresponding link Floer homology detection result in~\cite{binns2020knot}.

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

Observe that $\widehat{\HFK}(L)$ detects the linking number of two component links, so the linking number is one. Consider $\widehat{\HFL}(L)$. Let $A_i(0)$ be the Alexander grading of the Maslov index $0$ generator in $\widehat{\HFL}(L_i)$. Since there are no positive Maslov index generators in $\widehat{\HFL}(L)$, 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~\ref{chainlevel}, 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 $\widehat{\HFL}(L)$ admit a spectral sequence to $\widehat{\HFL}(L_1)\otimes V$, 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 $\widehat{\HFL}(L_i)\otimes V$, there must be Maslov index $-2$ generators in bi-Alexander  gradings $(x-1,1-x)$ and $(x,-x)$. In order that $\widehat{\HFL}(L)$ be symmetric, we must then have that $x=\frac{1}{2}$. We have thus determined that $\widehat{\HFL}(L)\cong\widehat{\HFL}(L_0)$, as desired.
\end{proof}

\end{section}


\begin{section}{Detection Results for Khovanov Homology}
\label{Detection Results for Khovanov Homology}

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







 



\begin{subsection}{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 \emph{Khovanov chain complex of $L$}, $(\CKh(L,R),\partial)$, is a finitely generated ${\mathbb{Z}}\oplus{\mathbb{Z}}$-filtered chain complex over $R$~\cite{khovanov2000categorification}.

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

Here $i$ is called the \emph{homological grading}, while $j$ is called the \emph{quantum grading}. The filtered chain homotopy type of $L$ is an invariant of $L$. The $R$-module $\CKh(L,R)$ 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 \emph{Khovanov homology of $L$} is obtained by taking the homology of $\CKh(L;R)$. A choice of basepoint $p\inL$ induces an action on $\CKh(L;R)$, which commutes with the differential. Taking the quotient of $\CKh(L;R)$ by this action and then taking homology yields a bigraded $R$-module called the \emph{reduced Khovanov homology} of $L$, denoted $\widetilde{\Kh}(L,p;R)$\cite{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 \emph{pointed Khovanov homology}, $\Kh(L,\mathbf{p};R)$ due to Baldwin-Levine-Sarkar~\cite{baldwin2017khovanov}.

$\Kh(L;R)$ 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 $\Kh(L;R)$ admits a spectral sequence to $\Kh(L_1;R)\otimes\Kh(L_2;R)$, called the \emph{Batson-Seed spectral sequence}~\cite{batson2015link}. Indeed this spectral sequences respects the $i-j$ grading on $\Kh(L;R)$ in the sense that;
\begin{align}\label{GradedBatsonSeed}
    \rank^{i-j=l}(\Kh(L;R)\geq \rank^{i-j=l+2\ell\text{k}(L_1,L_2)}(\Kh(L_1;R)\otimes\Kh(L_2;R))
\end{align}

Suppose now that $L$ is an $n$ component link. There is another spectral sequence, called the \emph{Lee spectral sequence}, from $\Kh(L;{\mathbb{Q}})$ which abuts to an invariant called the \emph{Lee Homology} of $L$, $\LLL(L):=\bigoplus_i\LLL^i(L)$~\cite{lee2005endomorphism}. This spectral sequence respects the $i$ gradings in the sense that $\rank(\Kh^i(L;{\mathbb{Q}}))\geq \rank(\LLL^i(L))$. Lee showed that $\LLL(L)\cong\bigoplus_{i=1}^n{\mathbb{Q}}^2_{a_i}$ where $a_i$ are integers. Indeed if $L$ has two components $L_1,L_2$ then $\LLL(L)\cong{\mathbb{Q}}^2_{0}\oplus{\mathbb{Q}}^2_{\ell\text{k}(L_1,L_2)}$~\cite[Proposition 4.3]{lee2005endomorphism}. Moreover, there is an invariant, $s(L)$ -- due to Rasmussen~\cite{Rasmussen} in the knot case and Beliakova-Wehrli in the link case~\cite{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~\cite{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 $\Kh(L,\mathbf{p};{\mathbb{Q}})$ to $\widehat{\HFK}(L;{\mathbb{Q}})$ that respects the relative $\delta$-gradings. As a corollary, Dowlin shows that if $L$ has $n$ components then; \begin{align}
    2^{n-1}\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\geq \rank(\widehat{\HFK}(L;{\mathbb{Q}}))\label{DowlinReduced}\end{align}



\end{subsection}

\begin{subsection}{Rank Detection Results for Khovanov homology}

Applying Dowlin's spectral sequence and the rank detection results for knot Floer homology from Section~\ref{Rank Detection Results for Knot Floer homology}, we have the following result;
\begin{corollary}\label{KHr2components}
Suppose $L$ is a two component pointed link with $\rank(\widetilde{\Kh}(L,p;{\mathbb{Z}}/2))\leq 4$. Then $L$ is one of;\begin{itemize}
    \item an unlink;
\item a Hopf link;
\item $T(2,4)$ or $T(2,-4)$.
\end{itemize}
\end{corollary}

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

\begin{corollary}\label{KHr3components}
Suppose $L$ is a three component pointed link. Then $\rank(\widetilde{\Kh}(L,p;{\mathbb{Q}}))> 2$.
\end{corollary}

\begin{proof}[Proof of Corollary~\ref{KHr2components}]
Suppose $L$ is a two component link with $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 4$. Equation~\ref{DowlinReduced} implies that $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))\leq 8$, and the result follows from Theorems~\ref{rank4}, \ref{rank6} and \ref{rank8}.
\end{proof}



\begin{proof}[Proof of Corollary~\ref{KHr3components}]
Suppose $L$ is a three component link with $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 2$. Equation~\ref{DowlinReduced} implies that $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))\leq 8$. Theorems~\ref{rank4}, \ref{rank6} and \ref{rank8} 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.
\end{proof}




\end{subsection}

\begin{subsection}{Khovanov Homology Detects $T(2,8)$, $T(2,10)$}
We prove the following theorems;



	\begin{theorem}\label{KHTT28}
		Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,8);{\mathbb{Z}})$. Then $L$ is isotopic to $T(2,8)$
	\end{theorem}
	
\begin{theorem}\label{KHT210}
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then $L$ is isotopic to $ T(2,10)$.
\end{theorem}


$$\begin{array}{c|c c c c c c c c c}
\label{Table 1}
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
&0&1&2&3&4&5&6&7&8


\end{array}

\caption{Table 1. $\Kh(T(2,8);{\mathbb{Z}})$: The horizontal axis indicates the $i$ grading, while the vertical axis denotes the $j$ grading}




$$

$$\begin{array}{c|c c c c c c c c c c c}
\label{Table 2}
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
&0&1&2&3&4&5&6&7&8&9&10


\end{array}

\caption{Table 2. $\Kh(T(2,10);{\mathbb{Z}})$: The horizontal axis indicates the $i$ grading, while the vertical axis denotes the $j$ grading}

$$


    



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


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

To prove Theorems~\ref{KHTT28} and~\ref{KHT210} we first show that any link $L$ with $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,2n);{\mathbb{Z}})$ 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.

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

We prove this Lemma using Lee's spectral sequence~\cite{lee2005endomorphism}, which has $\Kh(L;{\mathbb{Q}})$ as $E_1$ page and $E_\infty$ page a complex of rank $2^{n}$ where $n$ is the number of components of $L$.

\begin{proof}[Proof of Lemma~\ref{KHcomponentsgenuslinking}]
Suppose $L$ is as in the Theorem. By the universal coefficient theorem there are exactly two homological gradings in which the rank of $\Kh^i(L;{\mathbb{Q}})$ is at least two, namely $i=0$ and $i=2n$. Since $\rank(\Kh^i(L;{\mathbb{Q}}))\geq\rank(\LLL^i(L;{\mathbb{Q}}))$, and $\LLL^i(L;{\mathbb{Q}})\cong\bigoplus_{i=1}^m{\mathbb{Q}}^2_{a_i}$, 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 $\Kh(L;{\mathbb{Z}})$ is supported in even quantum gradings~\cite[Proposition 24]{khovanov2000categorification}.

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 ~\cite[Proposition 4.3]{lee2005endomorphism}, 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 $\chi(L)\geq 2-2n$.
\end{proof}

We can now prove the following two lemmas;

\begin{lemma}\label{KHt228components}
		Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,8);{\mathbb{Z}})$. Then $L$ has unknotted components.
	\end{lemma}
	
	
\begin{lemma}\label{KHT2102components}
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then $L$ has unknotted components.
\end{lemma}

Both of these use the Batson-Seed link splitting spectral sequence~\cite{batson2015link} to bound the rank of the components of $L$ in certain combinations of $i$ and $j$ gradings. We use Corollary 3.4 of~\cite{batson2015link} which states that if $L$ is a link consisting of two components $L_1,L_2$ then
\begin{align}\label{GradedBatsonSeed}
    \rank^l(\Kh(L;F)\geq \rank^{l+2\ell\text{k}(L_1,L_2)}(\Kh(L_1;F)\otimes\Kh(L_2;F))
\end{align}
 
 Here $F$ is a field -- we will use both ${\mathbb{Z}}/2$ and ${\mathbb{Q}}$ accordingly,  and $\rank^l$ denotes the $i-j=l$ grading part of the relevant vector space.
 
 For Lemma~\ref{KHt228components} we use the fact that the only knots $K$ with $\rank(\Kh(K;{\mathbb{Z}}/2))\leq 8$, are the trefoils and unknot. This follows from Kronheimer-Mrowka's unknot detection result \cite{kronheimer2011khovanov}, as  well as Baldwin-Sivek's trefoil detection results \cite{baldwin2018khovanov} and the fact that for knots $K$ $\rank_{{\mathbb{Z}}/2}(\Kh(K;{\mathbb{Z}}/2))$ is of the form $2+4n$ for some $n\geq 0$.
 
 \begin{proof}[Proof of Lemma~\ref{KHt228components}]
Suppose $L$ is as in the hypothesis of the Lemma. Lemma~\ref{KHcomponentsgenuslinking} implies that $L$ consists of two components, $L_1$ and $L_2$ of linking number $4$. From $\Kh(L;{\mathbb{Z}})$, as shown in Table 1, we deduce that $\rank_{{\mathbb{Z}}/2}(\Kh(L;{\mathbb{Z}}/2))=16$ by the universal coefficient theorem, so the Batson-Seed spectral sequence implies that $\rank(\Kh(L_i;{\mathbb{Z}}/2))\leq 8$, whence each component of $L$ is an unknot or a trefoil~\cite{baldwin2018khovanov}. Indeed, we readily see that at least one component is unknotted.
	
	
	We now consider $\Kh(L;{\mathbb{Q}})$, which we can determine from $\Kh(L;{\mathbb{Z}})$ via the universal coefficient theorem. Suppose $L$ has a right hand trefoil component. Then \\
	$\rank_{i-j=-4}(\Kh(T(2,3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})))=2$, but $\rank_{i-j=-12}(\Kh(L;{\mathbb{Q}}))=1$ contradicting equation~\ref{GradedBatsonSeed}. Similarly, $\rank_{i-j=2}(\Kh(T(2,-3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}}))=3$ which contradicts $\rank_{i-j=-6}(\Kh(L;{\mathbb{Q}}))=1$ by Equation~\ref{GradedBatsonSeed}. Thus both components of $L$ are unknotted.
\end{proof}

Lemma~\ref{KHT2102components} requires some extra analysis since there is no complete classification of knots with $\rank(\Kh(L;{\mathbb{Z}}/2))=10$. To deal with this latter case we  use Dowlin's spectral sequence and Proposition~\ref{E2collapsed}.


\begin{proof}[Proof of Lemma~\ref{KHT2102components}]
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then from Table 2 and the universal coefficient theorem we can see that $\rank(\Kh(T(2,10);{\mathbb{Z}}/2))= 20$. Batson-Seed's spectral sequence implies that at least one component of $L$ is unknotted, and the other is a knot $K$ with $\rank(\Kh(K;{\mathbb{Z}}/2))\leq10$. If $\rank(\Kh(K;{\mathbb{Z}}/2))\leq10$ then $\rank(\widetilde{\Kh}(L;{\mathbb{Z}}/2))\leq 5$, so that $K$ is an unknot, trefoil, figure eight or $\rank(\widetilde{\Kh}(K;{\mathbb{Z}}/2))=5$. 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;
\begin{enumerate}
	\item $L_2$ is $T(2,3)$, in which case; \begin{align*}\Kh(T(2,3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})&\cong({\mathbb{Q}}_{-6}\oplus{\mathbb{Q}}_{-3}^2\oplus{\mathbb{Q}}_{-1})\otimes({\mathbb{Q}}_{-1}\oplus{\mathbb{Q}}_1)\\&\cong{\mathbb{Q}}_{-7}\oplus{\mathbb{Q}}_{-5}\oplus{\mathbb{Q}}_{-4}^2\oplus{\mathbb{Q}}_{-2}^3\oplus{\mathbb{Q}}_0\end{align*}
		But $\rank(\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction. 
	
	\item $L_2$ is $T(2,-3)$. Then \begin{align*}\Kh(T(2,-3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})&\cong({\mathbb{Q}}_{6}\oplus{\mathbb{Q}}_{3}^2\oplus{\mathbb{Q}}_{1})\otimes({\mathbb{Q}}_{1}\oplus{\mathbb{Q}}_{-1})\\&\cong {\mathbb{Q}}_{7}\oplus{\mathbb{Q}}_{5}\oplus{\mathbb{Q}}_{4}^2\oplus{\mathbb{Q}}_{2}^3\oplus{\mathbb{Q}}_0\end{align*}
	
	But $\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction.
	
	
	
	\item $\rank(\widetilde{\Kh}(L_2;{\mathbb{Z}}/2))=5$. In this case we have that $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 5$, whence $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}}))\leq 5$. If $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}})<5$, then $L$ is a trefoil or the unknot, cases which we have already dealt with. Thus we may take $\rank(\widehat{\HFK})(L_2;{\mathbb{Q}})=5$. We show that $L_2$ is a figure eight or has the same knot Floer homology as $T(2,\pm5)$. Suppose $L_2$ is of genus $g>0$. Note that $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}},g)\leq 2$ We proceed by cases.\begin{enumerate}
	    \item Suppose $\rank(\widehat{\HFK}(L;{\mathbb{Q}},g))= 2$. Then $\widehat{\CFL}(L;{\mathbb{Q}})$ contains a differential of even length $2g$. From Proposition~\ref{E2collapsed}, we see that this yields a  summand of $(\widehat{\CFL}(L;Q),\partial)$ rank $8g$. Thus $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 8g+6$. It immediately follows that $g=1$. The symmetry of $\widehat{\HFL}(L)$ implies that $\rank(\widehat{\HFL}(L))\geq 16+12$, a contradiction.
	    
	    \item Suppose $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}},g)= 1$. Consider $\widehat{\CFL}(L_2;{\mathbb{Q}})$. If $g=1$, then $L_2$ is the figure eight knot. If $g>1$ there are two possible forms for $\widehat{\HFK}(L_1)$, as shown in Figure~\ref{rank5hfk}. To see this we need to use the fact that if $\rank(\widehat{\HFK}(L_2,g-1))=1$, 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~\cite{NiPersonalCommunication}.
	    
	    
					\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{HFKL2.pdf}
    \caption{Different possibilities for $\widehat{\HFK}(L_2;{\mathbb{Q}})$.\label{rank5hfk}}
    \end{figure}
    
    In either case, since there is both a length one differential and a length $g-1$ differential, $\rank(\widehat{\HFL}(L_2;{\mathbb{Q}}))\geq 4(g-1)+4+4=4g+4$, 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 $\rank(\widehat{\HFL}(L_2;{\mathbb{Q}}))\geq 8(g-1)+4+4+4$, so that $g=2$, and $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong \widehat{\HFK}(T(2,5);{\mathbb{Q}})$.
	    
	    Suppose we are in the second case. We have that $g\in\{2,3,4\}$. If $g=2$ then $\widehat{\HFK}(L_2)\cong\widehat{\HFK}(T(2,-5))$. If $g=4$ using the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ we find $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 24$, a contradiction.
	    
	    If $g=3$, symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ implies that the $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 24$, a contradiction.
	\end{enumerate}
	
	
	\item $L_2$ is the figure eight. In this case;
	\begin{align*}\Kh(U;{\mathbb{Q}})\otimes\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})\end{align*}
	
	But $\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction.
	
	
	\item $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong\widehat{\HFK}(T(2,5);{\mathbb{Q}})$, or $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong\widehat{\HFK}(T(2,-5);{\mathbb{Q}})$. Since $\ell\text{k}(L_1,L_2)=\pm 5$, there is a summand $\widehat{\HFK}(T(2,\pm5);{\mathbb{Q}})\otimes V[\pm\frac{5}{2},a]$ of $\widehat{\HFL}(L;{\mathbb{Q}})$, for some $a\in\pm\frac{5}{2}+{\mathbb{Z}}$. The symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ yields another $10$ generators. There can be no additional generators, as this would violate the rank inequality from Dowlin's spectral sequence - $\rank(\widehat{\HFL}(L);{\mathbb{Q}})\leq\rank(\Kh(T(2,10);{\mathbb{Q}}))=20$, 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 $\widehat{\HFL}(L;{\mathbb{Q}})$ is supported only in $A_1$ gradings $\pm\frac{1}{2}$. However since $L_1$ is unknotted and there is a spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L_1)\otimes V[\frac{5}{2}]$ there must be generators in $A_1$ grading $\frac{5}{2}$, a contradiction.
\end{enumerate}
\end{proof}


	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~\ref{E2collapsed}. From here it suffices to prove the following pair of propositions.
	
	\begin{proposition}\label{combinedHFKKHT28detection}
	Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link satisfying the following conditions;
	\begin{enumerate}
	\item $L$ has two exactly two components, $L_1$ and $L_2$, which are both unknots.
	\item $\ell\text{k}(L_1,L_2)=4$
	\item $\chi(L)\leq-6$
	    \item $\rank(\widehat{\HFK}(L;F))\leq 16$.
	    \item $\widehat{\HFK}(L;F)$ is $\delta$-thin.
	\end{enumerate}
	
	Then $L$ is $T(2,8)$.
	\end{proposition}
	
		\begin{proposition}\label{combinedHFKKHT210detection}
	Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link satisfying the following conditions;
	\begin{enumerate}
	\item $L$ has two exactly two components, $L_1$ and $L_2$, which are both unknots.
	\item $\ell\text{k}(L_1,L_2)=5$
	\item $\chi(L)\leq-8$
	    \item $\rank(\widehat{\HFK}(L;F))\leq 20$.
	    \item $\widehat{\HFK}(L;F)$ is $\delta$-thin.
	\end{enumerate}
	
	Then $L$ is $T(2,10)$.
	\end{proposition}
	
Before proving these propositions, we apply them to obtain Theorem's~\ref{KHTT28} and~\ref{KHT210} respectively.
	

	\begin{proof}[Proof of Theorem~\ref{KHTT28}]
	It suffices to show that $L$ satisfies the conditions in Proposition~\ref{combinedHFKKHT28detection}. Conditions 1,2 and 3 are implied by Lemmas~\ref{KHcomponentsgenuslinking} and~\ref{KHt228components}. To see that conditions $4$ holds, observe that $\rank(\Kh(L;{\mathbb{Z}}/2)=16$ by the universal coefficient theorem. It follows that $\rank(\widetilde{\Kh}(L;{\mathbb{Z}}/2))=8$ by~\cite{shumakovitch2014torsion}, so that $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 8$ by the universal coefficient theorem. $\rank(\Kh(L,p,{\mathbb{Q}}))\leq\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))$ by \cite[Theorem 1.2]{baldwin2017khovanov}, whence $\widehat{\HFK}(L;{\mathbb{Q}})\leq 16$.

	
	To see that condition 5 holds, note that that $\Kh(L;{\mathbb{Q}})$ is $\delta$-thin, so that $\Kh(L,p;{\mathbb{Q}})$ is $\delta$-thin by \cite[Lemma 2.11]{baldwin2017khovanov}.
	\end{proof}
	
	\begin{proof}[Proof of Theorem~\ref{KHT210}]
	    This follows by the same argument as in the proof of Theorem~\ref{KHTT28}.
	\end{proof}
	
We now prove the two requisite lemmas~\ref{combinedHFKKHT28detection}, and ~\ref{combinedHFKKHT210detection}






		\begin{proof}[Proof of proposition~\ref{combinedHFKKHT28detection}]
	Suppose $L$ satisfies the conditions given in the Proposition. Consider $\widehat{\CFL}(L;{\mathbb{Q}})$. Since $\widehat{\HFK}(L)$ is $\delta$-thin, $\widehat{\CFL}(L;{\mathbb{Q}})$ is an $E_2$ collapsed chain complex. Likewise since $\chi(L)\leq -6$ the maximal Alexander grading of $\widehat{\HFK}(L;{\mathbb{Q}})$ is at least $3$. 
	
	Consider $\widehat{\CFL}(L)$. Since $L$ has unknotted components, it follows that $(\widehat{\CFL}(L;{\mathbb{Q}}),\partial)$ 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 $(\widehat{\CFL}(L),\partial)$ has summands $X_{-1}^{l-4}[2,2]\oplus X_0^{l-5}[1,1]$. From the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$, and the fact that $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\leq 16$ we see immediately that $l\leq 6$.
			
		
			
			
			
				
					\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{Pseduo98.pdf}
    \caption{One potential link Floer homology complex for a link satisfying the conditions in Propositions~\ref{combinedHFKKHT28detection}}
    \end{figure}
				
					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 $\rank(\widehat{\HFK}(L))>16$. Similarly if $l\in\{1,2\}$ would violate the rank bound after accounting for the symmetry of $\widehat{\HFL}(L)$. If $l=0$ then $\rank(\widehat{\HFL}(L))\geq 16$. 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 $\widehat{\HFL}(L;{\mathbb{Q}})$. 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~\ref{Braidpolytope}, or that $\chi(L)>-6$.
				
				If $l=5$ we readily see that -- apart from in the case that $\widehat{\CFL}(L;{\mathbb{Q}})$ 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 $\widehat{\CFL}(L)$ 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~\ref{Braidpolytope} implies that $\widehat{\HFL}(L;{\mathbb{Q}})\cong\widehat{\HFL}(T(2,8);{\mathbb{Q}})$, whence the result follows from the corresponding detection result in~\cite{binns2020knot}.
			
				\end{proof}
	

The proof of~\ref{combinedHFKKHT210detection} is virtually the same, but with some additional case analysis.

\begin{proof}[Proof of Proposition~\ref{combinedHFKKHT210detection}]
	Suppose Suppose $L$ is as in the hypotheses of the Proposition. Since $L$ is delta-thin, $\widehat{\CFL}(T(2,2n);{\mathbb{Q}})$ is $E_2$-collapsed. Since $\chi(L)\leq -8$ the maximal Alexander grading is at least $n$. Since the components of $L$ are unknotted $\widehat{\CFL}(L;{\mathbb{Q}})$ 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 $\widehat{\CFL}(L;{\mathbb{Q}})$ has summands $X_0^{l-5}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{l-4}[2,2]$. Observe that the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ together with the fact that $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\leq 20$, imply that $l\leq 7$. 
	\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{KhT(2_10).pdf}
    \caption{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.}
    \end{figure}
	
	If $l=7$ then either $\widehat{\CFL}(L)\cong X_0^{2}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{3}[2,2]\oplus B_{-16}[\frac{-9}{2},\frac{-7}{2}]\oplus B_{-16}[\frac{-7}{2},\frac{-9}{2}]$ or there is an additional $B_{-8}[\frac{-1}{2},\frac{-1}{2}]$ summand, by the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$. 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~\cite{GageT26}, a contradiction.
	
	If $l=6$, then either $\widehat{\CFL}(L)\cong X_0^{1}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{2}[2,2]\oplus B_{-11}[\frac{-7}{2},\frac{-7}{2}]$ 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 $\widehat{\HFL}(L;{\mathbb{Q}})$ 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 $T(2,10)$, a contradiction since $T(2,10)$ has distinct link Floer homology. The $C_{-7}[\frac{5}{2},\frac{-7}{2}]\oplus C_{-7}[\frac{-7}{2},\frac{5}{2}]$ case is excluded by computing the Conway polynomial. First note that the Alexander polynomial is given by $(t^{-1/2}-t^{1/2})(t^{-5}+t^5-2t^{-4}-2t^4+t^{-3}+t^3+2t^{-1}+2t-4)$. Thus $\nabla(t-t^{-1})=-t^{11}+t^{-11}+3t^{9}-3t^{-9}-3t^{7}+3t^{-7}+t^{5}-t^{-5}-2t^{3}+2t^{3}+6t^{1}-6t^{-1}$. Therefore $\nabla(u)=-u^11-8u^9-20u^7-16u^5-4u^3$. It follows that the linking number is four, a contradiction. If there are three cubes then we note that in order to satisfy the symmetry properties there must be a summand $B_{-7}[\frac{-1}{2},\frac{-1}{2}]$. Indeed unless the reaming components are $B_{-1}[\frac{5}{2},\frac{5}{2}]\oplus B_{-13}[\frac{-7}{2},\frac{-7}{2}]$, or $B_{-8}[-\frac{5}{2},\frac{7}{2}]$ $B_{-8}[\frac{3}{2},\frac{-5}{2}]$. In the first case, we see that $L$ bounds an annulus upon reversing its orientation, a contradiction, since no link $T(2,2n)$ has the requisite homology. In the second case we see that $\Delta_L(t)=(t^{-1/2}-t^{1/2})(-t^7-t^{-7}+2t^6+2t^{-6}-t^5-t^{-5}+2t-4+2t^{-1})$ Thus $\nabla_L(t-t^{-1})=t^{15}-t^{-15}-3t^{13}+3t^{-13}+3t^{11}-3t^{-11}-t^9+t^{-9}-2t^3+2t^{-3}-2t^3+2t^{-3}-2t+2t^{-1}$
	Therefore $\nabla_L(u)=-u^{15}+12u^{13}+54u^{11}+112u^9+105u^7+36u^5-8u$ and the linking number is $-8$, a contradiction.
	
	Similarly, if $l\leq 5$ then there are summands $Y_0^{5-l}[3-l,3-l]\oplus Y_{-1}^{4-l}[2-l,2-l]$. Considering the symmetry properties of $\widehat{\HFL}(L)$ readily see that either $l=0$ or $l\geq 4$, as else $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))>20$. If $l=0$ then there are no additional generators, and we see that $L$ bounds an annulus, and hence is $T(2,10)$. This is a contradiction since $T(2,10)$ has distinct link Floer homology. If $l=4$ then $\widehat{\HFL}(L;{\mathbb{Q}})$ either contains no extra generators or an cube $C_{-6}[\dfrac{-1}{2},\dfrac{-1}{2}]$ by symmetry. We see that after reversing the orientation of a component of $L$, it bounds an annulus, whence $L$ is $T(2,10)$, a contradiction since $T(2,10)$ has distinct knot Floer homology.
	
	
We are thus left only with the case that $l=5$. In which case $\widehat{\CFL}(L;{\mathbb{Q}})$ has a summand $X_0[\frac{5}{2},\frac{5}{2}]\oplus Y_{-1}[\frac{3}{2},\frac{3}{2}]$. By symmetry $\widehat{\CFL}(L;{\mathbb{Q}})$ has a summand $B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ $\widehat{\CFL}(L;{\mathbb{Q}})$ can have at most three other $B_d$ summands. If there is one or three more cubes then the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ imply that one such summand must be of the form $B_{-6}[\frac{-1}{2},\frac{-1}{2}]$. Thus if $\widehat{\HFL}(L;{\mathbb{Q}})$ has $0$ or $1$ extra $B_d$ summands we see that after reversing the orientation of a component of $L$, $L$ bounds an annulus, and is therefore $T(2,10)$, a contradiction since $T(2,10)$ has distinct link Floer homology.

If $L$ has two extra $B_d$ summands then unless they are of the form $B_{-6}[\frac{3}{2},\frac{-5}{2}]\oplus[\frac{-5}{2},\frac{5}{2}]$, or $B_{-2}[\frac{3}{2},\frac{3}{2}]\oplus B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ then at least one component of $L$ is a braid axis and the result follows from Lemma~\ref{Braidpolytope} and the corresponding detection result for link Floer homology in~\cite{binns2020knot}. The first case is dealt with by computing the Conway polynomial from $\widehat{\HFL}(L;{\mathbb{Q}})$. Explicitly, $\Delta_L(t)=(t^{-1/2}-t^{1/2})(t^5+t^{-5}-2t^{4}-2t^{-4}+t^{3}+t^{-3}+2t-4+2t^{-1})$, whence $\nabla(t-t^{-1})=-t^11+t^{-11}+3t^9-3t^{-9}-3t^7+3t^{-7}+t^5-t^{-5}-2t^3+2t^{-3}+6t-6t$, and $\nabla_L(u)=-u^{11}-8u^9-20u^7-16u^5-5u^3$, whence the linking number is $-5$, a contradiction. In the second case we see that after reversing the orientation of a component, $L$ bounds an annulus, and is therefore $T(2,10)$, a contradiction since this link has distinct link Floer homology.

If $L$ has three extra $B_d$ summands, the by the symmetry of $\widehat{\HFL}(L)$, there must be a $B_{-6}[\frac{-1}{2},\frac{-1}{2}]$ summand. Unless the remaining summands are of the form $B_{-6}[\frac{3}{2},\frac{-5}{2}]\oplus[\frac{-5}{2},\frac{5}{2}]$, or $B_{-2}[\frac{3}{2},\frac{3}{2}]\oplus B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ then at least one component of $L$ is a braid axis and the result follows from Lemma~\ref{Braidpolytope} and the corresponding detection result for Link Floer homology~\cite{binns2020knot}.
For the $B_{-6}(\frac{3}{2},\frac{-5}{2})\oplusB_{-6}(\frac{-5}{2},\frac{5}{2})$ case we see that $\Delta_L(t)=(t^{-1/2}-t^{1/2})(t^5+t^{-5}-2t^{4}-2t^{-4}+t^{3}-t^{-3}+3t-6+3t^{-1})$, whence $\nabla_L(t-t^{-1})=-t^{11}+t^{-11}+3t^9-3t^{-9}-3t^7+3t^{-7}+t^5-t^{-5}-3t^3+3t^{-3}+9t-9t^{-1}$ and $\Nabla_L(u)=-u^{11}-8u^{9}-20u^7-16u^5-5u^3$, whence the linking number is $-5$, a contradiction. For the $B_{-2}(\frac{3}{2},\frac{3}{2})\oplus B_{-10}(\frac{-5}{2},\frac{-5}{2})$ case we see that $L$ bounds an annulus upon reversing the orientation of a component of $L$ and the desired result follows.


Thus at least one component is a braid axis, and Lemma~\ref{Braidpolytope} implies that $\widehat{\HFL}(L;{\mathbb{Q}})\cong\widehat{\HFL}(T(2,10);{\mathbb{Q}})$ and the result follows from the corresponding detection result for link Floer homology.
\end{proof}


We conclude by noting that Martin's detection result for $T(2,6)$~\cite{GageT26}, $T(2,4)$ detection~\cite{xie2020links} and $T(2,2)$ detection~\cite{baldwin2017khovanov} could have been obtained by similar methods to those presented in this paper.


\end{subsection}
\end{section}




\bibliographystyle{plain}

\subsection*{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 are especially grateful to 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. 
	\end{section}
	
	
	
	\begin{section}{A Very Brief Review of Link Floer homology}\label{A 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 use in the subsequent sections.
	
Let $L$ be an oriented $n$ component link in a rational homology sphere $M$. 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 Ozsv\'ath-Szab\'o in~\cite{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, is a multi-graded ${\mathbb{Z}}/2[U_1,U_2,\dots U_n]$ chain complex with underlying vector space; $$\CFL^-(L,M)\cong\bigoplus_{m,A_1,A_2,\dots A_n}\CFL^-_m(L,M)(A_1,A_2\dots A_n).$$ 

The three manifold $M$ is often apparent from the context and duly suppressed in the notation. The $A_i$ gradings are called the \emph{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 \emph{Maslov grading}, which is integer valued. $\CFL^-(L)$ 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 $\CFL^-(L)$ is an invariant of $L$, which we refer to as the \emph{link Floer complex of $L$}. The homology of $\CFL^-(L)$ is denoted $\HFL^-(L)$. $\CFL^-(L)$ 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 $\widehat{\CFL}(L)$, and has homology denoted by $\widehat{\HFL}(L)$, which we will call the \emph{link Floer homology of $L$}. The \emph{link Floer homology polytope} of $L$ is defined as the convex hull of the multi-Alexander gradings of $\widehat{\HFL}(L)$ with non-trivial support. The link Floer polytope determines the Thurston polytope of the link exterior by a result of Ozsv\'ath-Szab\'o~\cite{ozsvath2008linkFloerThurstonnorm}. We note that $\widehat{\HFL}(\overline{L})$, where $\overline{L}$ is $L$ with the orientation reversed, can be recovered from $\CFL^-(L)$ by only counting the disks which cross the appropriate basepoints.


For each component $L_i$ of $L$ there is a spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L-L_i)\otimes V[\frac{\ell\text{k}(L_i,L_j)}{2}]$. 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 $\widehat{\HFL}(L)$ is of rank $2^{n-1}$ in the maximal $A_i$ grading of non-trivial support by a result of Martin~\cite{GageT26}.
	 
	 The \emph{knot Floer homology} of $L$, $\widehat{\HFK}(L)$, is an oriented link invariant due independently to Rasmussen~\cite{Rasmussen} and Ozsv\'ath-Szab\'o \cite{Holomorphicdisksandknotinvariants} that can be obtained from $\widehat{\HFL}(L)$ by projecting the multi-Alexander grading onto the diagonal, and shifting the Maslov gradings up by $\frac{n}{2}$. $\widehat{\HFK}(L)$ determines the maximal Euler characteristic of a surface bounding $L$ \cite{ni2006note}, as well as whether or not $L$ fibered~\cite{ghiggini2008knot},~\cite{ni2007knot}.
	 
	 Versions of $\widehat{\HFL}(L,M)$ and $\widehat{\HFK}(L,M)$ can also be defined with ${\mathbb{Z}}$ or ${\mathbb{Q}}$ coefficients~\cite{sarkar2011note}. In fact there are $2^{n-1}$ versions of each theory, each corresponding to a \emph{coherent system of orientations} on the moduli space of pseudoholomorphic disks. We will only have to be careful with which version of these theories we are using when applying work of Dowlin~\cite{dowlin2018spectral} which gives a spectral sequence from Khovanov homology to Knot Floer homology endowed with the orientation given in~\cite{alishahi2015refinement}. As is customary, we suppress the dependence of knot Floer homology on the coherent system of orientations in our notation. Each of the knot Floer homology theories detects the maximal Euler characteristic of a surface bounding $L$~\cite{ni2006note}, as well as whether or not $L$ fibered. This follows from work of Juh\'asz~\cite{juhasz2008floer}.
	 
	 The \emph{Conway polynomial} of $L$ can be obtained as an appropriate decatigorification of $\widehat{\HFL}(L)$. A result of Hoste~\cite{hoste1985firstcoefficientoftheconwaypolynomial} implies that the Conway polynomial detects the linking number of two component links. It follows that knot Floer homology and link Floer homology also detect the linking number of two component links.
	 
	 
	   
	\end{section}
	
	\begin{section}{The BRAID Invariant}\label{BRAIDsection}
	In this section we review the BRAID invariant, due to Baldwin-Vela-Vick-V\'ertesi~\cite{baldwin2013equivalence}. We find two different perspectives on generalised braids to be useful. First we may define a braid in an open book $(\Sigma,\phi)$ to be a link which intersects every page transversely. Any such braid can be recovered from the data of a \emph{pointed open book} $(\Sigma,\phi,\{p_i\})$, where $\phi:\Sigma\to\Sigma$ fixes the points $\{p_i\}\subseteq\Sigma$ as a set. We assume henceforth that $(\Sigma,\phi)$ is an open book decomposition for some rational homology sphere.
	
	
	The BRAID invariant of a braid $B$ represented by a pointed open book $(\Sigma,\phi,\{p_i\})$ is defined as follows. Form a Heegaard diagram by taking a basis of arcs $\{a_i\}$ for $\Sigma-\{p_i\}$. Consider the basis of arcs $b_i$ obtained by pushing the ends of the $a_i$ curves around $\partial \Sigma$ and isotoping so that $a_i\cap b_i$ consists of a single point. Form a multi-pointed Heegaard diagram consisting of $\Sigma\cup_{\partial\Sigma}(-\Sigma)$ with $\alpha$ curves consisting of the union of the $a$ arcs in $\Sigma$ and $-\Sigma$, and $\beta$ curves consisting of the union of the $b$ arcs in $\Sigma$ and $\phi(b)$ in $-\Sigma$. The $w_i$ basepoints consist of $\{p_i\} \subseteq-\Sigma$ while the $z_i$ base points consist of $\{p_i\}\cap\Sigma$. The BRAID invariant of $B$ is defined as the homology class of $\{a_i\cap b_i\}$. This can be viewed as a class in either $\widehat{\CFL}(L;{\mathbb{Z}}/2)$, or $\CFL^-(L;{\mathbb{Z}}/2)$. We denote the classes by $\widehat{t}(B)$ or $t(B)$ respectively.  Baldwin-Vela-Vick-V\'ertesi show this is an invariant of transverse Braids, when viewed as element of $\widehat{\HFK}(L)$ or $\HFK^-(L;{\mathbb{Z}}/2)$. Tovstopyat-Nelip further notes that the BRAID invariant can be thought of as a well defined class in $\widetilde{\HFK}(L;{\mathbb{Z}}/2)$, a version of knot Floer homology defined for links encoded by multiple basepoints~\cite{tovstopyat2018transverse}. We note that it has not been proven that $\widehat{t}(B)$ is well defined in $\widehat{\HFL}(L)$ complete with all Alexander gradings, and we may conflate $\widehat{t}(B)$ and the associated generator of the chain complex in the section~\ref{linkcontact}.
	
	
\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{BRAID.pdf}
    \caption{A pseudoholomorphic disk connecting the BRAID invariant.}\label{BRAIDDISK}
    \end{figure}
	
	The BRAID invariant has some some strong non-vanishing properties. The strongest of these, a result due to Tovstopyat-Nelip, is that $\widehat{t}(B\cup K)\neq 0$ where $B$ is a braid in the component of a fibered link $K$~\cite[Theorem 1.3]{tovstopyat2018transverse}. This generalises earlier work of Vela-Vick~\cite{vela2011transverse} and Vela-Vick-Etnyre~\cite{etnyre2010torsion}. In the case that $B=\emptyset$, the Alexander grading of $\widehat{t}(B\cup K)$ is readily computed~\cite[Lemma 4.2]{tovstopyat2018transverse}. 
	
 Given a pointed open book $(\Sigma,\phi,\{p_i\})$ there are two notions of an arc being sent to the right(left). We introduce new terminology to distinguish these two notions. We say an arc $a\subseteq\Sigma$ is \emph{relatively sent to the right} by $\phi$ if after $\phi(a)$ has been isotoped in $\Sigma-\{p_i\}$ so that $\phi(a)$ and $a$ intersect minimally, $\phi(a)$ is to the right of $a$. This is notion of being sent to the right used in~\cite{baldwin2013equivalence}. We say that $a$ is \emph{absolutely sent to the right} if $a$ is sent to the right after isotoping $\phi(a)$ in $\Sigma$ - in particular $\phi(a)$ is allowed to pass over basepoints under the isotopy. The distinction between these two concepts plays an important role in the next section.
 
 We conclude this section by noting that, while the braid invariant was originally defined in homology theories with coefficients in ${\mathbb{Z}}/2$, it is in fact well defined up to sign with coefficients in ${\mathbb{Z}}$. In particular, if $L$ is a fibered link, the BRAID invariant generates the top Alexander grading of $\widehat{\HFK}(L;{\mathbb{Q}})$.
	\end{section}

\begin{section}{A Rank Bound from the BRAID invariant}\label{linkcontact}

	Baldwin-Vela-Vick proved that the knot Floer homology of a fibered knot is non-trivial in the next to top Alexander grading \cite{baldwin_note_2018}. In this section we generalise their techniques to obtain a number of related results. In particular we show; 
	
	\begin{theorem}\label{rankinnequalityforfiberedlinks}
	If $L$ is a non-trivial $n$ component fibered link in $S^3$ then; \[\rank(\widehat{\HFK}(L,\dfrac{n-2-\chi(L)}{2}))\geq n.\]
	\end{theorem}
	
	This implies a generalization of a Theorem of Ni~\cite[Theorem A.1]{ni2020seifert};
	
	\begin{corollary}\label{Nigeneralized}
	Suppose $L$ is fibered, with modromy neither right veering nor left veering on each component. Then \[\rank(\widehat{\HFK}(L,\dfrac{n-2-\chi(L)}{2}))\geq 2n.\]
	\end{corollary}



These results follow from the from the following more technical lemma


\begin{lemma}\label{chainlevel}
    Suppose $L$ is a braid represented by a pointed open book $(\Sigma,\phi,\{p_i\})$ and that $\widehat{t}(L)\neq0$. Suppose there is an arc $a$ which is absolutely sent to the left by $\phi$. Then there is a generator   $\mathbf{d}\in\widehat{\CFL}(-Y,-L,[\Sigma])$ with $\partial \mathbf{d}=\widehat{t}(\phi)$ with multi-Alexander grading determined by the number of $w_i$ in a certain domain of the Heegaard diagram, as well as the multi-Alexander grading of $\widehat{t}(D)$.
\end{lemma}

To prove this Lemma~\ref{chainlevel} we explicitly find the generator $\mathbf{d}$. We then show that $\partial_{\widehat{\CF}}\mathbf{d}=\widehat{t}(L)$ using a diagrammatic argument, and note that it is easy to determine the multi-Alexander grading of $\mathbf{d}$.

\begin{proof}[Proof of Lemma~\ref{chainlevel}]
Suppose $L$, $a$ are as in the theorem. Since $a$ is absolutely sent to the left, after isotoping $\phi(a)$ to a curve $b_1$ in $-\Sigma$ so that $a$ and $b_1$ intersect minimally, there is an intersection point $a\cap b_1\in-\Sigma$ and a domain $D$ as shown in figure~\ref{BRAIDDISK}. Let $b_2$ be the image of the curve $\phi(a)$ upon isotoping relative to the basepoints $\{p_i\}$ so that $a$, $b_2$ intersect minimally. Note that $b_2$ is to the right of $a$, as else $\widehat{t}(\phi)=0$. There is an isotopy $f:-\Sigma\times[0,1]\to-\Sigma$ which takes $b_1$ to $b_2$. Consider the image of $D$ under the isotopy $f$. Note that points in $f(D)$ may have multiplicity strictly greater than $1$. Consider the class $\mathbf{d}=\{d_1,c_2,\dots c_{2g+|L|-1}\}$. There is a component of a differential from $\mathbf{d}$ to $\widehat{t}(\phi)$, as witnessed by $f(D)$. There can be no other components of $\partial d$, as else $a$ and $b_2$ would not intersect minimally as arcs in $(-\Sigma,\mathbf{p})$. The difference in multi-Alexander gradings between $\mathbf{d}$ and $\widehat{t}(L)$ is determined by the number of $w$ basepoints of each type in $f(D)$.

\end{proof}
Note that there must be at least one $w_i$ component in the domain, as else the BRAID invariant vanishes.
\begin{lemma}\label{generalnonzero}
   Suppose the domain from Lemma~\ref{chainlevel} contains basepoints $\{w_i\}$ each with multiplicity $n_i$, then $\rank(\bigoplus_{i\in I}\widehat{\HFL}(B))\geq 1$, where $$I=\{(A_1,A_2,\dots A_i,\dots, A_{|B|}):A_i(\widehat{t}(B))\leq A_i\leq A_i(\widehat{t}(B))+n_i:\exists i:A_i\neq A_i(\widehat{t}(B))\}.$$
\end{lemma}

\begin{proof}
    We use the same notation as in Lemma~\ref{generalnonzero}. Note that $\mathbf{d}$ is a cycle in $\widehat{\CFL}(B)$. If $\mathbf{d}$ is not a boundary in $\widehat{\CFL}(B)$ the result follows. Suppose then that $\mathbf{d}$ is a boundary in $\widehat{\CFL}(B)$. Reduce $\widehat{\CFL}(B)$ in every Alexander grading except for that of $\mathbf{d}$. Since $\partial^2=0$ there is a homogeneous generator $\mathbf{d}'$ of $\widehat{\HFL}(L)$ with Alexander multi-grading $(A_1,\dots A_{|B|})$ such that $A_i(\widehat{t}(B))\leq A_i(\mathbf{d}')\leq A_i(\widehat{t}(B))+n_i$ satisfying the property that $\langle\partial\mathbf{d}',\widehat{t}(B)\rangle\neq0$. If $A_i(\widehat{t}(B))= A_i(\mathbf{d}')$ for all $i$ we have a contradiction since we are working with a reduced complex. Thus there is some $i$ for which $A_i(\mathbf{d}')\neq A_i(\widehat{t}(B))$ and the result follows.
\end{proof}




\begin{proof}[Proof of Theorem~\ref{rankinnequalityforfiberedlinks}]

A fibered link $L$ arising as the boundary of an open book $(\Sigma,\phi)$ may be perturbed to yield an $|L|$-braid $B$ in the complement of $L$, with associated pointed monodromy $(\Sigma,\phi')$.  Consider the $m$th boundary component of $\Sigma$. After mirroring we may assume that $\phi$ sends some non-separating arc absolutely to the left. Extend this arc to a basis for $(\Sigma,\phi')$ and obtain $\widehat{t}(B)$. Now, $\widehat{t}(B)\neq 0$ by~\cite[Theorem 1.1]{tovstopyat2018transverse}. Suppose $\widehat{t}(B)$ has multi-Alexander grading $(A_1,A_2,\dots A_{|L|})$. The bigon obtained via~\ref{chainlevel}, and pictured in Figure~\ref{fiberedBRAID} in this special case, contains a single $w_m$ basepoint. Thus applying Lemma~\ref{generalnonzero} yields a non-trivial generator in multi-Alexander grading $(A_1,A_2\dots A_{m-1},\dots A_{|L|})$. The result follows from repeating this process for each boundary component and symmetry properties of Link Floer homology under mirroring.
\end{proof}

\begin{figure}[h]
    \centering
    \includegraphics[width=5cm]{braidfib.pdf}
    \caption{A pseudoholomorphic disk and the BRAID invariant of a braid representing the binding of an open book}\label{fiberedBRAID}
    \end{figure}


\begin{proof}[Proof of Corollary~\ref{Nigeneralized}]
 View $L$ as an $|L|$-braid as in the proof of Theorem~\ref{rankinnequalityforfiberedlinks}. Suppose $\widehat{t}(L)$ is of multi-Alexander grading $(A_1,A_2,\dots A_n)$. Suppose $\rank(\widehat{\HFL}(Y,L,[\Sigma];(A_1,\dots A_i-1,\dots A_n)))=1$. Consider $\CFL^-(Y,L,[\Sigma])$. Since there is an arc that is sent to the left and an arc that is sent to the right we find that $\partial_{-}^2(\widehat{t}(L))\neq 0$, a contradiction. Thus $\rank(\widehat{\HFL}(Y,L,[\Sigma];(A_1,\dots A_i-1,\dots A_n)))\geq 2$ and the result follows from the fact that Knot Floer homology is obtained from Link Floer homology by collapsing the multi-Alexander grading to a single Alexander grading.
\end{proof}


		
	

	
	We conclude this section by noting that it is perhaps useful to view Theorem~\ref{rankinnequalityforfiberedlinks} in the context of the following more general proposition;

\begin{proposition}\label{generalrankbound}
Suppose $L$ is an $n>1$ component non-trivial link such that \\ $\rank(\widehat{\HFL}(L)(x_1,x_2,\dots x_n))=2m+1$ for some $m\in{\mathbb{Z}}$. Then $\rank(\widehat{\HFK}(L))\geq 2^{n}+2m$.
\end{proposition}

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 $\widehat{\HFK}(L)$ to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$.

\begin{proof}[Proof of Proposition~\ref{generalrankbound}]
   Suppose $L$ is an $n>1$ component link non-trivial such that $\rank(\widehat{\HFK}(L)(x_1,x_2,\dots x_n))$ $=2m+1$. 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ must be even. Thus we obtain at least an additional $2^n-1$ generators.
\end{proof}

Note that if $L$ is an $n$ component link and $\rank(\widehat{\HFK}(L,A))$ is odd for some $A$, then there is a multi-Alexander  grading $(A_1,A_2\dots A_n)$ such that $\rank(\widehat{\HFK}(L,A_1,A_2,\dots A_n))$ of odd rank, where $A=A_1+A_2+\dots A_n$ whence it follows that $\rank(\widehat{\HFK}(L))\geq 2^n$.

\end{section}


\begin{section}{Detection Results for $(2,2n)$-cables}\label{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;


\begin{theorem}\label{HFL2detectsRHTcables}
Link Floer homology detects $T(2,3)_{2,2n}$ for all $n$.
\end{theorem}

\begin{theorem}\label{HFLdetects(2,2n)cablesofthefigureeight}
Let $K$ be the figure eight knot. Link Floer homology detects $K_{2,2n}$ for all $n$.
\end{theorem}

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


\begin{lemma}\label{(2,0)trefoilcable}
    Suppose $L$ is a link such that $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,3)_{(2,0)})$, 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 \begin{align*}
        \widehat{\HFK}(K)\cong{\mathbb{F}}_0^2[1]\oplus{\mathbb{F}}_{-1}^3[0]\oplus{\mathbb{F}}_{-2}^2[-1]
    \end{align*}
\end{lemma}

Here, and for the remainder of this section we take ${\mathbb{F}}$ to be ${\mathbb{Z}},{\mathbb{Z}}/2$ or $Q$.

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


    We first give a partial computation of $\widehat{\HFK}(T(2,3)_{(2,0)})$. To do so compute $\widehat{\HFK}(T(2,3)_{(2,1)})$ and $\widehat{\HFK}(T(2,3)_{(2,-1)})$ using Hanselman--Watson's cabling formula in the theory of immersed curves~\cite{hanselman2019cabling}, then apply the skein exact triangle for knot Floer homology to determine $\widehat{\HFL}(T(2,3)_{(2,0)})$, and thereby $\widehat{\HFK}(T(2,3)_{(2,0)})$.
    
    Hanselman--Watson's cabling formula implies that \begin{align*}
        \widehat{\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]
    \end{align*}
    
    \begin{align*}
        \widehat{\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]
    \end{align*}
    
    \vspace{.2in}
    
    \noindent Using the skein exact triangle from~\cite[Equation 7]{Holomorphicdisksandknotinvariants}  -- 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 $\widehat{\HFK}(T(2,3)_{(2,0)},i)$ is trivial for $i> 2$ and $\widehat{\HFK}(T(2,3)_{(2,0)},2)\cong {\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$. To determine $\widehat{\HFK}(T(2,3)_{(2,0)};1)$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)})$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)})$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)}))$ is at least four in each of the maximal $A_i$ gradings (namely $A_i=1$). Thus $\widehat{\HFK}(T(2,3)_{(2,0)};1)\cong {\mathbb{F}}_{-\frac{1}{2}}^2\oplus{\mathbb{F}}_{\frac{1}{2}}^2$ and indeed \begin{align*}\widehat{\HFL}(T(2,3)_{(2,0)};(1,0))\cong\widehat{\HFL}(T(2,3)_{(2,0)};(0,1))\cong{\mathbb{F}}_0\oplus{\mathbb{F}}_{-1}.\end{align*}
    
    To conclude the computation it suffices to determine $\widehat{\HFK}(T(2,3)_{(2,0)};0)$ by the symmetries of link Floer homology. Note again that the skein exact triangle implies that $\widehat{\HFK}(T(2,3)_{(2,0)};0)$ is either ${\mathbb{F}}_{-\frac{1}{2}}\oplus{\mathbb{F}}_{-\frac{3}{2}}$ or $ {\mathbb{F}}_{-\frac{1}{2}}^2\oplus{\mathbb{F}}_{-\frac{3}{2}}^2$.
    
    Thus we have that \begin{align*}
\widehat{\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]   \end{align*}

or

\begin{align*} \widehat{\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]  \end{align*}

We can now proceed to the proof of Lemma~\ref{(2,0)trefoilcable}
\begin{proof}[Proof of Lemma~\ref{(2,0)trefoilcable}]
 Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,3)_{(2,0})$. 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 $\widehat{\HFK}(L)$ is $\frac{1}{2}$. Since $\rank(\widehat{\HFK}(L))$ 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~\cite{hoste1985firstcoefficientoftheconwaypolynomial}. Since the maximal Alexander grading of $\widehat{\HFK}(L)$ is $1$, $L$ either bounds an annulus or the disjoint union of an unknot and a genus one knot. If $\rank(\widehat{\HFK}(L))=16$ 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 \begin{align*}
    {\mathbb{F}}_0^2[1]\oplus{\mathbb{F}}_{-1}^3[0]\oplus{\mathbb{F}}_{-2}^2[-1]
\end{align*} 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 $\widehat{\HFK}(L)$ 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  $\widehat{\HFK}(T(2,-3)_{(2,0)})\cong(\widehat{\HFK}(T(2,-3)_{(2,0)}))^*\not\cong\widehat{\HFK(L)}$. 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.


\end{proof}

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

\begin{lemma}\label{20cablefigureeight}
    Let $K$ be the figure eight knot and orient $K_{2,0}$ as the boundary of an annulus. Suppose $\widehat{\HFK}(L)\cong\widehat\HFK(K_{2,0})$. Then $L$ is isotopic to $K_{2,0}$.
\end{lemma}

Again we note that if $\widehat{\HFL}(L)\cong\widehat{\HFL}(K_{2,0})$, 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 $\widehat\HFK(K_{2,0})$. Hanselman-Watson's cabling formula for knot Floer homology in terms of immersed curves~\cite{hanselman2019cabling} shows that \begin{align*}
\widehat{\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]\\
\widehat{\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]\end{align*}

Again we apply the skein exact triangle from~\cite[Equation 7]{Holomorphicdisksandknotinvariants} -- with $L_0=K_{2,0}, L_+=K_{2,1}, L_-=K_{2,1}$.This immediately implies that $\widehat{\HFK}(K_{2,0},i)\cong 0$ for $|i|>2$ and $\widehat{\HFK}(K_{2,1})\cong 
F_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{5}{2}}$. Indeed we see that $\widehat{\HFK}(K_{2,0},1)$ 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 $\widehat{\HFL}(K_{2,0})$ is supported on the lines $A_2=A_1$, $A_2=A_1\pm 1$. Thus since $\widehat{\HFL}(L)$ must be of even rank in each $A_i$ grading, and $\widehat{\HFL}(K_{2,0},(x,y))\cong {\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$ is excluded. While unnecessary we note that $\widehat{\HFK}(L)\cong({\mathbb{F}}_{\frac{1}{2}}\oplus{\mathbb{F}}_{-\frac{1}{2}})^n$ for some $0\leq n\leq 3$. In fact $n\geq 2$, as else $\widehat{\HFL}(K_{2,0})$ cannot admit a spectral to $\widehat{\HFL}(K)\otimes V$

\begin{proof}[Proof of Lemma~\ref{20cablefigureeight}]
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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,0))$, 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_1)$, where $L_1$ is the component of $L$ corresponding to $A_1$. Thus $\widehat{\HFL}(L_1)$ is of rank one in the top Alexander grading and $L_1$ is fibered~\cite{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.
\end{proof}





To prove theorems~\ref{HFL2detectsRHTcables} and \ref{HFLdetects(2,2n)cablesofthefigureeight} 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 $n$.

\begin{lemma}\label{algebraiccharachterizationofcables}
Let $L$ be a two component link with linking number $n>1$ such that;
\begin{enumerate}
        \item $\widehat{\HFL}(L)$ is of rank $2$ in the maximal non-zero $A_1$ or $A_2$ grading.
        \item The link Floer homology polytope has a vertex at multi-Alexander grading $(1+\frac{n}{2},1+\frac{n}{2})$.
        \item $\widehat{\HFL}(L)$ has support on the lines $A_2=A_1,A_2=A_1+1, A_2=A_1-1$.
         \end{enumerate}
         
Then $L$ is isotopic to $K_{2,2n}$ where $K$ is a trefoil or the figure eight knot.


\end{lemma}


\begin{proof}[Proof of Lemma~\ref{algebraiccharachterizationofcables}] 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  $g(L_i)+\frac{n}{2}\leq\max\{A_i:\widehat{\HFL}(L)(A_1,A_2)\neq 0\}$, it follows that $g(L_i)\leq 1$ for each $i$. Indeed since, $\widehat{\HFL}(L)$ 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.
\end{proof}
 
 To apply Lemma~\ref{algebraiccharachterizationofcables}, we need to show that $\widehat{\HFL}(K_{2,2n})$ satisfies the conditions of the hypothesis of Lemma~\ref{algebraiccharachterizationofcables} for $K$ a trefoil or figure eight knot. The easiest way to proceed is via the following remark
 
 \begin{remark}
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~\cite{stallings1978constructions}, and $T(n,m)$ along minimal genus Seifert surfaces~\cite{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~\cite{gabai1986detecting}. It follows that $(2,2n)$ cables of genus one fibered knots are fibered and of genus $n$, that is Euler characteristic $-2n$. \end{remark}

 If one wishes to only use Rudolph's stated result, for knots, one can alternately proceed via the following lemma;
 
 \begin{lemma}\label{genusof22ncables}
 Let $K$ be a genus one fibered knot. For $n\geq 1$, $K_{2,2n}$ is fibered and $\chi(K_{2,2n})=-2n$.
 \end{lemma}
 
 \begin{proof}[Proof of Lemma~\ref{genusof22ncables}]
 Note that, for $n>0$, $K_{2,2n+1}$ is the Murasugi sum of $K_{2,1}$ and $T(2,2n+1)$~\cite{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~\cite{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 ${\rank(\widehat{\HFK}(K_{2,2n})(1+n))=1}$ so that $K_{2,2n}$ is fibered, as desired.
 \end{proof}
 
 
It follows that conditions 1, 2 and 3 of Lemma~\ref{algebraiccharachterizationofcables} 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.

\begin{proof}[Proof of Theorem~\ref{HFL2detectsRHTcables}]

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~\ref{(2,0)trefoilcable}.


Suppose now that $n>0$. We first show that $\widehat{\HFL}(T(2,3)_{2,2n})$ satisfies the conditions in the hypothesis of Lemma~\ref{alexandergradingcharachterizationofunlinks}. Observe that for $n\geq 1$, each component of $T(2,3)_{2,2n}$ is a braid in the complement of the other, so $\widehat{\HFL}(T(2,3)_{2,2n}))$ must be rank two in the maximal $A_i$ grading, by braid detection ~\cite[Proposition 1]{GageT26}, so condition 1 of Lemma~\ref{algebraiccharachterizationofcables} holds. By Lemma~\ref{genusof22ncables} the maximal Alexander grading of $\widehat{\HFK}(L)$ is $2+n$. $T(2,-3)_{2,2n+1}$ is fibered by Lemma~\ref{genusof22ncables}. Therefore, since $\widehat{\HFL}(T(2,-3)_{2,2n})$ 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$ must be of Maslov gradings $0$ and $-1$.

Suppose now that $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,3)_{2,2n})$ for some $n\geq 2$. Since link Floer homology detects the linking number of two component links~\cite{hoste1985firstcoefficientoftheconwaypolynomial}, it follows that the components of $L$ have linking number $n$. Lemma~\ref{algebraiccharachterizationofcables} 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 $\widehat{\HFL}(L)$ has a Maslov index $0$ generator with $A_1$ grading $1+\frac{n}{2}$ which persists under the spectral sequence to $\widehat{\HFL}(L_i)$, 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$. Observe that for $n\geq 1$, each component of $L$ is a braid in the complement of the other, so $\widehat{\HFL}(L)$ must be rank two in the maximal $A_i$ grading, by braid detection~\cite{GageT26}, so Condition 1 of Lemma~\ref{algebraiccharachterizationofcables} holds. By Lemma~\ref{genusof22ncables} the maximal Alexander grading of $\widehat{\HFK}(L)$ is $2+n$. Since $T(2,-3)_{2,2n+1}$ is fibered, and $\widehat{\HFL}(T(2,-3)_{2,2n+1})$ 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 \ref{alexandergradingcharachterizationofunlinks} 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$ must be of Maslov gradings $1$ and $2$.

Suppose now that $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,-3)_{2,2n})$. Lemma~\ref{algebraiccharachterizationofcables} 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 $\widehat{\HFL}(L)$ must be of Maslov gradings $1$ and $2$, it follows that each component is $T(2,-3)$ and we have the desired result.
\end{proof}

The proof of Theorem~\ref{HFLdetects(2,2n)cablesofthefigureeight} is again similar.

\begin{proof}[Proof of Theorem~\ref{HFLdetects(2,2n)cablesofthefigureeight}]
The proof follows the proof of Theorem~\ref{HFL2detectsRHTcables} verbatim, aside from to note that the top $A_i$ generators must be of Maslov index $0$ and $1$ respectively for $n\geq 0$.
\end{proof}













\begin{section}{Technical Results}
\label{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.

	
	
	
	
	\begin{subsection}{$E_2$ collapsed chain complexes}


We will rely on a structure theorem for \emph{$E_2$-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 ~\cite[Section 12]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}. In particular Ozsv\'ath-Szab\'o 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~\ref{Summands}. 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 $-\mathbbm{1}$

\begin{figure}[h!]
\label{Summands}
    \centering
    \includegraphics[width=7cm]{E2_Summands.pdf}
    \caption{The summands of an $E_2$-collapsed chain complex}
    \end{figure}
    
    
    \begin{proposition}\label{E2collapsed}
    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 $B_d[i,j]\oplus V_d^l[i,j],H_d^l[i,j],X_d^l[i,j], Y_d^l[i,j]$.
    \end{proposition}
    
    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, $\widehat{\CFL}(L)$ 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 $\widehat{\CFL}(L)$. For instance, since the full filtered $\widehat{\CFL}$ complex has homology ${\mathbb{F}}_0\oplus{\mathbb{F}}_{-1}$, we see that $\widehat{\CFL}(L)$ 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 $\widehat{\CFL}(L_1)$ determines the $V_d^l$ summands, while $\widehat{\CFL}(L_2)$ determines the $H_d^l$ summands. Specifically, if there are generators $x,y\in\widehat{\CFL}(L_1)$, with $\langle\partial_{\widehat{\CFL}(L_1)} x,y\rangle\neq 0$ then we must have a summands of $\widehat{\CFL}(L)$ of the form $V_{M(x)}^{A(x)-A(y)}$, as no other summand induces the correct differential on $\widehat{\CFL}(L_1)$.
    

    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 $\partial_2$.
    \begin{proof}[Proof of Proposition~\ref{E2collapsed}]
    
    
    By a bi-filtered version of the Reduction Lemma~\cite[Lemma 4.1]{hedden2013khovanov} we can assume that the grading preserving part of $\partial$ is trivial. We proceed by induction on $\rank(C)$, 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 $\mathbf{w}\in \langle x_i\rangle$ 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. 
    \end{proof}
    
    
    
    



    
   

	\end{subsection}
	
	
	\begin{subsection}{The Dual Thurston Polytope}
	
	

The four detection results also require the following proposition;

\begin{proposition}\label{Braidpolytope}
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
\end{proposition}
 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.

\begin{proof}[Proof of Proposition~\ref{Braidpolytope}]

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.

\end{proof}



	
	\end{subsection}
	


\end{section}










\end{section}




\begin{section}{Knot Floer Homology Detects $T(2,8),T(2,10)$}
\label{Knot Floer Homology Detects $T(2,8),T(2,10)$}

In this section we prove the following detection results;



\begin{theorem}
\label{HFKT28}
	Knot Floer homology detects $T(2,8)$.
\end{theorem}

	\begin{theorem}\label{HFKT210}
	Knot Floer homology detects $T(2,10)$.
\end{theorem}

From here the following corollary follow immediately from work in~\cite{binns2020knot}.

\begin{corollary}
Annular Khovanov homology detects the closure of the braids $\sigma_1\sigma_2\dots\sigma_7$ and $\sigma_1\sigma_2\dots\sigma_{9}$, where $\sigma_i$ are the standard Artin generators of the braid group.
\end{corollary}

Here Annular Khovanov homology is a combinatorial invariant of links in a thickened annulus. For details we refer the reader to~\cite{binns2020knot}.
 
For the readers convenience first we recall that:
    $$\widehat{\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}
    
    \end{cases}$$
    
    Here ${\mathbb{F}}$ can be ${\mathbb{Q}},{\mathbb{Z}}$ or ${\mathbb{Z}}/2$.

We start by showing that in general if $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,2n))$ then $L$ consists of two unknotted components and $\widehat{\CFL}(L)$ is of an especially simple form.
\begin{lemma}\label{T22nunknottedcomponents}
    Suppose $\widehat{\HFK}(L;{\mathbb{F}})\cong\widehat{\HFK}(T(2,2n);{\mathbb{F}})$. Then $(\widehat{\CFL}(L),\partial)$ 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.
\end{lemma}

Here ${\mathbb{F}}$ is either ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$.




\begin{proof}[Proof of Lemma~\ref{T22nunknottedcomponents}]
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 $\widehat{\HFK}(L)$ is $\frac{1}{2}$, so since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{m-1}S^1\times S^2)$, $L$ can have at most two components. $L$ then has exactly $2$ components since the Maslov grading of $\widehat{\HFK}(L)$ 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 $\widehat{\HFL}(L)$. 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~\cite[Theorem 12.1]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, $\widehat{\HFL}(L)$ 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 $\widehat{\HFK}(L_1)$ 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 $\widehat{\HFK}(L_1)\otimes V$ then there must be a generator of odd Maslov index in $A_1$ grading $\frac{n}{2}-x-1$, since the summands of $\widehat{\HFL}(L)$ that persist under the spectral sequence to $\widehat{\HFK}(L_1)\otimes V$ 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.
\end{proof}



\begin{comment}
\color{red}
\begin{lemma}\label{T22nsummand}
	Let $n>1$. Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,2n))$. Then $L$ is a two component link with linking number $n$ and $\widehat{\HFL}(L)$ has a summand $C_{-2}[\frac{n-2}{2},\frac{n-2}{2}]\oplus C_{-2n}[\frac{-n}{2},\frac{-n}{2}]$.
\end{lemma}

The $n=1$ case is that of the Hopf link, which knot Floer homology is known to detect.

\begin{proof}[Proof of Lemma~\ref{T22nsummand}]
	Note that $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{|L|-1}(S^1\times S^2))$, where $|L|$ is the number of components of $L$. $\widehat{\HF}(\#^{|L|-1}(S^1\times S^2))$ has support in Maslov grading $\dfrac{|L|-1}{2}$, while the maximal Maslov grading of $\widehat{\HFK}(L)$ is $\frac{1}{2}$, so $L$ can have at most $2$ components. $L$ cannot have one component since the Maslov gradings of the generators of $\widehat{\HFK}(L)$ take values in ${\mathbb{Z}}+\frac{1}{2}$. $L$ has linking number $n$ because $\widehat{\HFK}(L)$ determines the Conway polynomial of two component links, and the Conway polynomial in turn detects the linking number of two component links~\cite{hoste1985firstcoefficientoftheconwaypolynomial}.
	
	Consider $\widehat{\HFL}(L)$. Let $A_i(0)$ be the Alexander index of the unique Maslov index $0$ generator in $\widehat{\HFL}(L_i)$ for each $i$. Note that $A_i(0)\geq 0$ since there are no positive Maslov index generators. Note that $A_1(0)+\frac{n}{2}+A_2(0)+\frac{n}{2}\leq n$, so $A_i(0)=0$, and the Maslov index $0$ generator is of bi-Alexander grading $(\frac{n}{2},\frac{n}{2})$. Lemma~\ref{cubes} implies that $\widehat{\HFL}(L)$ has a summand $C_{-2}[\frac{n-2}{2},\frac{n-2}{2}]$. The symmetry of $\widehat{\HFL}(L)$ then implies that $\widehat{\HFL}(L)$ has another summand $C_{-2n}[\frac{-n}{2},\frac{-n}{2}]$.
\end{proof}

\end{comment}

\color{black}


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


\begin{proof}[Proof of Theorem~\ref{HFKT28}]
Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nunknottedcomponents} 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 \cite{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~\ref{Braidpolytope}, and the result follows from the fact that link Floer homology detects $T(2,8)$~\cite{binns2020knot}.

\begin{comment}
\color{red}
Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nsummand} implies that there is a summand of $\widehat{\HFL}(L)$ given by $C_{-2}[1,1]\oplus C_{-8}[-2,-2]$. Indeed, since 
\[ \widehat{\HFL}_m(L)(A_1,A_2)\cong\widehat{\HFL}_{m-2A_1-2A_2}(L)(-A_1,-A_2)\]
Lemma~\ref{cubes} implies that the remainder of $\widehat{\HFL}(L)$ is given by $C_{-4}(x,-x)\oplus C_{-6}[-x-1,x-1]$. We seek to control $x$. We first show that $L$ has unknotted components. Suppose that there is a generator from $C_{-8}[-2,-2]\oplus C_{-2}[1,1]$ which is not of $A_i$ grading $2$ persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V(2)$. Then there must be a generator of Maslov grading at least $-1$ with $A_i$ grading at least $3$, which there is not. It follows that only a generator from $C_{-4}[x,-x]\oplus C_{-6}[-x-1,x-1]$ can possibly persist under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V[2]$, but if such a generator persists with $A_i$ grading at least $2+g(L_i)$ with $g(L_i)>0$ then there must be a generator which persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V[2]$ of Maslov grading at most $-2-2g(L_i)<-2$ of $A_i$ grading $2-g(L_i)$, a contradiction.
\end{comment}


	\begin{figure}[h]
    \centering
    \includegraphics[width=8cm]{T_2_8_pseudo.pdf}
    \caption{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.}
    \end{figure}
\begin{comment}
Thus both components of $L$ are unknotted. Therefore all of the hypothesis of Proposition~\ref{combinedHFKKHT28detection} hold and the result follows.


Since the rank in the maximal $A_i$ grading is $2$ for at least one $i$, at least one component is a braid axis for the other. Without loss of generality we may take $i=1$ and thereby deduce that the maximal $A_1$ grading is strictly less than $2$. It follows that $x\in\{-1,0\}$. The $x=-1$ case is excluded by Lemma~\ref{Braidpolytope}. Thus $x=0$, $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,8))$, and the result follows from the corresponding detection result in~\cite{binns2020knot}.
\end{comment}
\end{proof}


\color{black}
The proof of Theorem~\ref{HFKT210} is virtually identical

\begin{proof}[Proof of Theorem~\ref{HFKT210}]

Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,10))$. Lemma~\ref{T22nunknottedcomponents} 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~\ref{Braidpolytope}, and the result follows from the fact that Link Floer homology detects $T(2,10)$.

\begin{comment}
	Suppose $L$ satisfies $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nsummand} implies that $L$ is a two component link with linking number $5$ and that there is a summand of $\widehat{\HFL}(L)$ of the form $C_{-10}[-\frac{5}{2},-\frac{5}{2}]\oplus C_{-2}[\frac{3}{2},\frac{3}{2}]$. Symmetry implies that there is another of the form $C_{-6}(\frac{-1}{2},\frac{-1}{2})$. There are two summands remaining $C_{-4}(x,1-x)$ and $C_{-8}(-x,x-3)$.
	
	We first note that if there were a generator from $C_{-6}(\frac{-1}{2},\frac{-1}{2})\oplus C_{-10}(-\frac{5}{2},-\frac{5}{2})\oplus C_{-2}(\frac{3}{2},\frac{3}{2})$ that persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V$ then there would be a generator of Maslov grading at least $-1$ in $A_i$ grading at least $\frac{7}{2}$, a contradiction.
	\end{comment}
	
	
    \begin{comment}
	 It follows that only the generators from $C_{-4}(x+\frac{1}{2},\frac{1}{2}-x)$ and $C_{-8}(-x-\frac{3}{2},x-\frac{3}{2})$ can possibly persist in $A_i$ gradings higher than $\frac{5}{2}$, but if such a generator persists with $A_i$ grading at least $\frac{5}{2}+g(L_i)$ with $g(L_i)>0$ then there must be a generator of Maslov grading at most $-2-2g(L_i)$ of $A_i$ grading $2-g(L_i)$ that persists under the spectral sequence to $\widehat{\HFL}(L_i)$, a contradiction.
	 
	 Thus both components of $L$ are unknotted. It follows that all of the hypotheses of Proposition~\ref{combinedHFKKHT210detection} hold and the result follows.
	 
Note that the rank of $\widehat{\HFL}(L)$ is two in at least one of the top $A_i$ gradings. We may thus, without loss of generality, take $L_1$ to be a braid axis, whence the maximal $A_1$ grading is $\frac{5}{2}$~\cite{GageT26}. We are therefore left only with the possibility that $|x|<2$. Note however that unless $x=0$ Lemma~\ref{Braidpolytope} is violated. Thus $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,8))$ and the result follows from the corresponding detection result in~\cite{binns2020knot}.
	 
\end{comment}
\end{proof}




    
    
\end{section}









\begin{section}{Rank Detection Results for Knot Floer Homology }
\label{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;
\begin{theorem}\label{rank4}
	If $\rank(\widehat{\HFK}(L))=4$ then $L$ is the Hopf link or the three component unlink.
\end{theorem}


\begin{theorem}\label{rank6}
	Suppose $L$ is a link with $\rank(\widehat{\HFL}(L))=6$. Then $L$ is a disjoint union of a trefoil and an unknot.
\end{theorem}

\begin{theorem}\label{rank8}
The only links $L$ with $\rank(\widehat{\HFL}(L))=8$ are $T(2,4)$, $T(2,-4)$, the four component unlink, and the disjoint union of a Hopf link and an unknot.
\end{theorem}

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

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

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

\begin{proposition}\label{3componentrank12}
There is no three component link with $\rank(\widehat{\HFK}(L))=10$.
\end{proposition}



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

\begin{theorem}\label{HFKdetcetstrefoilmeridian}
Let $L_0$ be $T(2,3)$ with a meridian. Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(L_0)$. Then $L$ is isotopic to $L_0$.
\end{theorem}
     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~\cite{binns2020knot}.
     





The proofs of Theorems \ref{rank4}, \ref{rank6}, \ref{rank8} and \ref{3componentrank12} 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 \cite{binns2020knot}. Throughout this section, we take ${\mathbb{F}}$ to be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$.



We first prove Lemma~\ref{unlinkdetection}, as it is useful in proving the other theorems.

We will find the following characterization of unlinks useful;

\begin{lemma}[Ni~\cite{ni2006note}]\label{alexandergradingcharachterizationofunlinks}
Suppose $\widehat{\HFK}(L)$ is supported in a single Alexander grading. Then $L$ is an unlink.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{alexandergradingcharachterizationofunlinks}]
Observe that if $\widehat{\HFK}(L)$ is supported in a single Alexander grading then that grading must be zero, by the symmetry of $\widehat{\HFK}(L)$. 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.
\end{proof}

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 $-1$.



\begin{proof}[Proof of Lemma~\ref{unlinkdetection}]
    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 $\rank(\widehat{\HFK}(L))=2^n$ then for any component $L_i$ of $L$, $L-L_i$ is a link with $\rank(\widehat{\HFK}(L-L_i))=2^{n-1}$. By inductive hypothesis $L-L_i$ is the $n$ component unlink. It follows that the support of $\widehat{\HFL}(L)$ is contained in the line defined by $A_j=0$ for $j\neq i$. But $i$ is arbitrary, so $\widehat{\HFL}(L)$ is supported in multi-Alexander grading $\mathbf{0}$, and the result follows from Lemma~\ref{alexandergradingcharachterizationofunlinks}.
\end{proof}



We now prove Theorem~\ref{rank4}.


\begin{proof}[Proof of Theorem~\ref{rank4}]
Let $L$ be an $n$ component link with $\rank(\widehat{\HFK}(L))=4$. Since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$, it follows that $n\leq 3$. Indeed, since $\rank(\widehat{\HFK}(L))$ 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~\ref{unlinkdetection}. It therefore remains only to consider the case that $n=2$. Let $g$ be the maximal Alexander grading of $\widehat{\HFK}(L)$. We consider the following three cases; $\rank(\widehat{\HFK}(L)(g))$ is either four, two or one, in which case the link is fibered by \cite{ni2007knot}.\begin{enumerate}
       
       \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=4$}.} In this case $g=0$, whence $\widehat{\HFK}(L)$ does not admit a spectral sequence to $\widehat{\HF}(S^1\times S^2)$.
          \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=2$}.} In this case we see that 
         $$\widehat{\HFK}(L)\cong ({\mathbb{F}}_{m_1}\oplus{\mathbb{F}}_{m_2})[g]\oplus ({\mathbb{F}}_{m_1-2g}\oplus{\mathbb{F}}_{m_2-2g})[-g].$$
        
        Observe that $g>0$. Consider the spectral sequence from $\widehat{\HFK}(L)$ to $\widehat{\HF}(S^1\times S^2)$. 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.
        
       \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=1$}.} Observe that $g\geq 1$. Let $(x,g-x)$ be the bi-Alexander grading of the contact class in $\widehat{\HFL}(L)$.
       
       If $g>1$, then Theorem~\ref{rankinnequalityforfiberedlinks} implies that $\rank(\widehat{\HFK}(L))\geq 6$. Thus $g=1$, and $$\widehat{\HFL}(L)\cong {\mathbb{F}}_{m}(\frac{1}{2},\frac{1}{2})\oplus {\mathbb{F}}_{m-1}(-\frac{1}{2},\frac{1}{2})\oplus {\mathbb{F}}_{m-1}(\frac{1}{2},-\frac{1}{2})\oplus {\mathbb{F}}_{m-2}(-\frac{1}{2},-\frac{1}{2})$$
       Since $\widehat{\HFL}(L)$ 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~\cite{BG}.
    \end{enumerate}
    

\end{proof}

We now prove Theorem~\ref{rank6}, 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.





\begin{lemma}\label{rank6fibered}
There is no fibered link $L$ such that $\rank(\widehat{\HFK}(L))=6$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{rank6fibered}]
Suppose $L$ is an $n$ component fibered link with $\rank(\widehat{\HFK}(L))=6$. Since $\rank(\widehat{\HFL}(L))$ is even, $L$ cannot be a knot. Moreover, $L$ can have at most $3$ components, since $\widehat{\HFL}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$.

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 $\widehat{\HFK}(L)$ has support is at least $2$. By Theorem~\ref{rankinnequalityforfiberedlinks}, the next to top Alexander grading is of rank three, whence $\widehat{\HFK}(L)$ 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 $\widehat{\HFK}(L)$ 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 $\rank(\widehat{\HFK}(L))=\rank(\widehat{\HFK}(T(2,\pm 3)\otimes V)$. 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~\ref{chainlevel} 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 $\widehat{\HFL}(L)$, 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 \cite{GageT26}, a contradiction.
          

\end{proof}

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


\begin{proof}[Proof of Theorem~\ref{rank6}]
Let $L$ be an $n$ component link with link Floer homology of rank $6$. Lemma~\ref{rank6fibered} implies that $L$ is not fibered under any orientation. The fact that $2^{n-1}\leq\rank(\widehat{\HFK}(L))$ implies that $L$ has at most three components. Since $\rank(\widehat{\HFK}(L))$ 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 $\widehat{\HFK}(L)$ under some orientation of $L$.


Lemma~\ref{alexandergradingcharachterizationofunlinks}, 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ in the plane $A_i=p_i$ is exactly $p$, it follows that the spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L_i)\otimes V^{n-1}$ is trivial. Thus the rank of $\widehat{\HFL(L)}$ at $p$ is at least $2^{n-1}$. Since the polytope is non-degenerate, the symmetry properties of $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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 $\rank(\widehat{\HFL}(L))\geq 8$, a contradiction. \end{proof}



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

\begin{proposition}\label{rank8fibered}
Suppose $L$ is fibered and $\rank(\widehat{\HFK}(L))=8$. Then $L\cong T(2,4)$ or $T(2,-4)$.
\end{proposition}
\begin{proof}
Let $L$ be a fibered link with link Floer homology of rank eight. Since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$, 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 $\rank(\widehat{\HFK}(L))$ is even. Suppose $L$ has four components. Then by Theorem~\ref{unlinkdetection}, $L$ 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 $\widehat{\HFK}(L)$ has supported is at least $2$. By Theorem~\ref{chainlevel}, $\rank(\widehat{\HFK}(L)(g-1))\geq3$. Thus $$\widehat{\HFK}(L)\cong{\mathbb{F}}_m[g]\oplus{\mathbb{F}}_{m-1}^3[g-1]\oplus {\mathbb{F}}_{m+1-2g}^3[1-g]\oplus{\mathbb{F}}_{m-2g}[-g]$$ 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 $\widehat{\HFK}(L)$ is at least one. Suppose $g=1$. Then $\rank(\widehat{\HFK}(L;0))=6$. Such a complex clearly does not admit a spectral sequence to $\widehat{\HF}(S^1\times S^2)$, a contradiction. Thus $g\geq 2$. Consider $\widehat{\HFL}(L)$. Suppose the top Alexander grading generator is in multi-Alexander grading $(x,g-x)$ and of Maslov index $m$. Theorem~\ref{chainlevel} 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 $\widehat{\HFK}(L)$ to $\widehat{\HFK}(L_1)\otimes V$ 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 $\rank(\widehat{\HFK}(T(2,2n))=8$ are $n=\pm2$.

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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_2)$ do not have admissible Maslov gradings. Thus the complex splits as two squares, and again must be located on the diagonal, a contradiction.

\end{proof}



\begin{proof}[Proof of Theorem~\ref{rank8}]


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 $\rank(\widehat{\HFK}(L))\geq 2^{n-1}$ and $L$ is even, $L$ has either two, three or four components. If $L$ has four components, then Lemma~\ref{unlinkdetection} implies that $L$ is an unlink. We deal with the two remaining cases separately;\begin{enumerate}
   
    \item\textbf{$L$ has two components}
Consider $\widehat{\HFL}(L)$. 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 $\widehat{\HFK}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks} 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_1)$, 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~\ref{rank2face} 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~\ref{rank2face}.


	\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{Rank_8_HFL.pdf}
    \caption{A rank eight bigraded vector space not arising as the link Floer homology of a link.\label{rank2face}}
    \end{figure}


The only remaining case is that on each face the two generators sit in the same multi-Alexander grading, see figure~\ref{rank2face}. 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 $\widehat{\HFL}(L)$ 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 $\partial_1x=y$. Since $y$ does not persist under the spectral sequence to $\widehat{\HFL}(L_2)$, $\partial_2 y=z$. 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.

   
   
   



\item \textbf{$L$ 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 $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L-L_3)$ is just projection onto the $A_3=0$ plane. Suppose $\widehat{\HFL}(L)$ is supported in the $A_3=0$ plane, then $L$ consists of a disjoint union of a Hopf link and an unknot. If $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks}.
\end{enumerate}

\end{proof}


\begin{proof}[Proof of Proposition~\ref{3componentrank12}]
Let $L$ be a $3$ component link with $\rank(\widehat{\HFK}(L)=10$. Suppose $L$ is fibered. The maximal Alexander grading, $A$, of $\widehat{\HFK}(L)$ is given by $2+g$ for some $g\geq 0$. Note that $\rank(\widehat{\HFK}(L)(A))=1$. Suppose the generator is of Maslov index $m$. Theorem~\ref{rankinnequalityforfiberedlinks} implies that there are at least three generators of Maslov index $m-1$ in Alexander grading $A-1$. The symmetry of $\widehat{\HFK}(L)$ 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 $\widehat{\HFK}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks}. 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 $\rank(\widehat{\HFK}(L))\geq 16$, 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 $\widehat{\HFL}(L)$ on the edge $-A_1-A_2=\dfrac{g_1+g_2}{2}$, persists under the spectral sequence to $\widehat{\HFL}(L-L_3)\otimes V[\frac{\ell\text{k}(L_1,L_3)}{2},\frac{\ell\text{k}(L_2,L_3)}{2}]$, and is indeed of maximal (respectively minimal $A_1+A_2$ grading). Now, there are only two 2 component links with $\rank(\widehat{\HFK}(L))\leq 5$; 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 $i$, $\rank(\widehat{\HFK}(L-L_i)\leq 5$, 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.


\end{proof}

We conclude with a proof of  Theorem~\ref{HFKdetcetstrefoilmeridian}


\begin{proof}[Proof of Proposition~\ref{HFKdetcetstrefoilmeridian}]

Let $L_0$ be the trefoil together with a meridian. It suffices to show that if  $\widehat{\HFK}(L)\cong\widehat{\HFK}(L_0)$ then  $\widehat{\HFL}(L)\cong\widehat{\HFL}(L_0)$ by the corresponding link Floer homology detection result in~\cite{binns2020knot}.

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

Observe that $\widehat{\HFK}(L)$ detects the linking number of two component links, so the linking number is one. Consider $\widehat{\HFL}(L)$. Let $A_i(0)$ be the Alexander grading of the Maslov index $0$ generator in $\widehat{\HFL}(L_i)$. Since there are no positive Maslov index generators in $\widehat{\HFL}(L)$, 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~\ref{chainlevel}, 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 $\widehat{\HFL}(L)$ admit a spectral sequence to $\widehat{\HFL}(L_1)\otimes V$, 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 $\widehat{\HFL}(L_i)\otimes V$, there must be Maslov index $-2$ generators in bi-Alexander  gradings $(x-1,1-x)$ and $(x,-x)$. In order that $\widehat{\HFL}(L)$ be symmetric, we must then have that $x=\frac{1}{2}$. We have thus determined that $\widehat{\HFL}(L)\cong\widehat{\HFL}(L_0)$, as desired.
\end{proof}

\end{section}


\begin{section}{Detection Results for Khovanov Homology}
\label{Detection Results for Khovanov Homology}

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







 



\begin{subsection}{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 \emph{Khovanov chain complex of $L$}, $(\CKh(L,R),\partial)$, is a finitely generated ${\mathbb{Z}}\oplus{\mathbb{Z}}$-filtered chain complex over $R$~\cite{khovanov2000categorification}.

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

Here $i$ is called the \textit{homological grading}, while $j$ is called the \textit{quantum grading}. The filtered chain homotopy type of $L$ is an invariant of $L$. The $R$-module $\CKh(L,R)$ 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 \textit{Khovanov homology of $L$} is obtained by taking the homology of $\CKh(L;R)$. A choice of basepoint $p\in L$ induces an action on $\CKh(L;R)$, which commutes with the differential. Taking the quotient of $\CKh(L;R)$ by this action and then taking homology yields a bigraded $R$-module called the~\textit{reduced Khovanov homology} of $L$, denoted $\widetilde{\Kh}(L,p;R)$~\cite{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 \textit{pointed Khovanov homology}, $\Kh(L,\mathbf{p};R)$ due to Baldwin-Levine-Sarkar~\cite{baldwin2017khovanov}.

$\Kh(L;R)$ 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 $\Kh(L;R)$ admits a spectral sequence to $\Kh(L_1;R)\otimes\Kh(L_2;R)$, called the \emph{Batson-Seed spectral sequence}~\cite{batson2015link}. Indeed this spectral sequences respects the $i-j$ grading on $\Kh(L;R)$ in the sense that;
\begin{align}\label{GradedBatsonSeed}
    \rank^{i-j=l}(\Kh(L;R)\geq \rank^{i-j=l+2\ell\text{k}(L_1,L_2)}(\Kh(L_1;R)\otimes\Kh(L_2;R))
\end{align}

Suppose now that $L$ is an $n$ component link. There is another spectral sequence, called the \emph{Lee spectral sequence}, from $\Kh(L;{\mathbb{Q}})$ which abuts to an invariant called the \emph{Lee Homology} of $L$, $\LLL(L):=\bigoplus_i\LLL^i(L)$~\cite{lee2005endomorphism}. This spectral sequence respects the $i$ gradings in the sense that $\rank(\Kh^i(L;{\mathbb{Q}}))\geq \rank(\LLL^i(L))$. Lee showed that $\LLL(L)\cong\bigoplus_{i=1}^n{\mathbb{Q}}^2_{a_i}$ where $a_i$ are integers. Indeed if $L$ has two components $L_1,L_2$ then $\LLL(L)\cong{\mathbb{Q}}^2_{0}\oplus{\mathbb{Q}}^2_{\ell\text{k}(L_1,L_2)}$~\cite[Proposition 4.3]{lee2005endomorphism}. Moreover, there is an invariant, $s(L)$ -- due to Rasmussen~\cite{Rasmussen} in the knot case and Beliakova-Wehrli in the link case~\cite{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~\cite{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 $\Kh(L,\mathbf{p};{\mathbb{Q}})$ to $\widehat{\HFK}(L;{\mathbb{Q}})$ that respects the relative $\delta$-gradings. Here $\widehat{\HFK}(L;{\mathbb{Q}})$ uses the coherent system of orientations given in~\cite{alishahi2015refinement}. We use this version of knot Floer homology, and the corresponding knot Floer homology for the remainder of the paper. As a corollary, Dowlin shows that if $L$ has $n$ components then; \begin{align}
    2^{n-1}\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\geq \rank(\widehat{\HFK}(L;{\mathbb{Q}}))\label{DowlinReduced}\end{align}



\end{subsection}

\begin{subsection}{Rank Detection Results for Khovanov homology}

Applying Dowlin's spectral sequence and the rank detection results for knot Floer homology from Section~\ref{Rank Detection Results for Knot Floer homology}, we have the following result;
\begin{corollary}\label{KHr2components}
Suppose $L$ is a two component pointed link with $\rank(\widetilde{\Kh}(L,p;{\mathbb{Z}}/2))\leq 4$. Then $L$ is one of;\begin{itemize}
    \item an unlink;
\item a Hopf link;
\item $T(2,4)$ or $T(2,-4)$.
\end{itemize}
\end{corollary}

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

\begin{corollary}\label{KHr3components}
Suppose $L$ is a three component pointed link. Then $\rank(\widetilde{\Kh}(L,p;{\mathbb{Q}}))> 2$.
\end{corollary}

\begin{proof}[Proof of Corollary~\ref{KHr2components}]
Suppose $L$ is a two component link with $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 4$. Equation~\ref{DowlinReduced} implies that $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))\leq 8$, and the result follows from Theorems~\ref{rank4}, \ref{rank6} and \ref{rank8}.
\end{proof}



\begin{proof}[Proof of Corollary~\ref{KHr3components}]
Suppose $L$ is a three component link with $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 2$. Equation~\ref{DowlinReduced} implies that $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))\leq 8$. Theorems~\ref{rank4}, \ref{rank6} and \ref{rank8} 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.
\end{proof}




\end{subsection}

\begin{subsection}{Khovanov Homology Detects $T(2,8)$, $T(2,10)$}
We prove the following theorems;



	\begin{theorem}\label{KHTT28}
		Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,8);{\mathbb{Z}})$. Then $L$ is isotopic to $T(2,8)$
	\end{theorem}
	
\begin{theorem}\label{KHT210}
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then $L$ is isotopic to $ T(2,10)$.
\end{theorem}


\[ \begin{array}{c|c c c c c c c c c}
\label{Table 1}
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
&0&1&2&3&4&5&6&7&8


\end{array}
\]

Table 1. $\Kh(T(2,8);{\mathbb{Z}})$: The horizontal axis indicates the $i$ grading, while the vertical axis denotes the $j$ grading



\[\begin{array}{c|c c c c c c c c c c c}
\label{Table 2}
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
&0&1&2&3&4&5&6&7&8&9&10


\end{array}
\]

Table 2. $\Kh(T(2,10);{\mathbb{Z}})$: The horizontal axis indicates the $i$ grading, while the vertical axis denotes the $j$ grading


\vspace{.5in}

This requires the Lee, Batson-Seed, and Dowlin spectral sequences, as well as an analysis similar to that of section~\ref{Knot Floer Homology Detects $T(2,8),T(2,10)$}.


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

To prove Theorems~\ref{KHTT28} and~\ref{KHT210} we first show that any link $L$ with $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,2n);{\mathbb{Z}})$ 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.

\begin{lemma}\label{KHcomponentsgenuslinking}
Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,2n);{\mathbb{Z}})$. Then $L$ is a two component link with linking number $n$ and $\chi(L)\leq 2-2n$.
\end{lemma}

We prove this Lemma using Lee's spectral sequence~\cite{lee2005endomorphism}, which has $\Kh(L;{\mathbb{Q}})$ as $E_1$ page and $E_\infty$ page a complex of rank $2^{n}$ where $n$ is the number of components of $L$.

\begin{proof}[Proof of Lemma~\ref{KHcomponentsgenuslinking}]
Suppose $L$ is as in the Theorem. By the universal coefficient theorem there are exactly two homological gradings in which the rank of $\Kh^i(L;{\mathbb{Q}})$ is at least two, namely $i=0$ and $i=2n$. Since $\rank(\Kh^i(L;{\mathbb{Q}}))\geq\rank(\LLL^i(L;{\mathbb{Q}}))$, and $\LLL^i(L;{\mathbb{Q}})\cong\bigoplus_{i=1}^m{\mathbb{Q}}^2_{a_i}$, 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 $\Kh(L;{\mathbb{Z}})$ is supported in even quantum gradings~\cite[Proposition 24]{khovanov2000categorification}.

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 ~\cite[Proposition 4.3]{lee2005endomorphism}, 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 $\chi(L)\leq 2-2n$~\cite{beliakova2008categorification}.
\end{proof}

We can now prove the following two lemmas;

\begin{lemma}\label{KHt228components}
		Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,8);{\mathbb{Z}})$. Then $L$ has unknotted components.
	\end{lemma}
	
	
\begin{lemma}\label{KHT2102components}
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then $L$ has unknotted components.
\end{lemma}

Both of these use the Batson-Seed link splitting spectral sequence~\cite{batson2015link} to bound the rank of the components of $L$ in certain combinations of $i$ and $j$ gradings. We use Corollary 3.4 of~\cite{batson2015link} which states that if $L$ is a link consisting of two components $L_1,L_2$ then
\begin{align}\label{GradedBatsonSeed}
    \rank^l(\Kh(L;F)\geq \rank^{l+2\ell\text{k}(L_1,L_2)}(\Kh(L_1;F)\otimes\Kh(L_2;F))
\end{align}
 
 Here $F$ is a field -- we will use both ${\mathbb{Z}}/2$ and ${\mathbb{Q}}$ accordingly,  and $\rank^l$ denotes the $i-j=l$ grading part of the relevant vector space.
 
 For Lemma~\ref{KHt228components} we use the fact that the only knots $K$ with $\rank(\Kh(K;{\mathbb{Z}}/2))\leq 8$, are the trefoils and unknot. This follows from Kronheimer-Mrowka's unknot detection result \cite{kronheimer2011khovanov}, as  well as Baldwin-Sivek's trefoil detection results \cite{baldwin2018khovanov} and the fact that for knots $K$ $\rank_{{\mathbb{Z}}/2}(\Kh(K;{\mathbb{Z}}/2))$ is of the form $2+4n$ for some $n\geq 0$.
 
 \begin{proof}[Proof of Lemma~\ref{KHt228components}]
Suppose $L$ is as in the hypothesis of the Lemma. Lemma~\ref{KHcomponentsgenuslinking} implies that $L$ consists of two components, $L_1$ and $L_2$ of linking number $4$. From $\Kh(L;{\mathbb{Z}})$, as shown in Table 1, we deduce that $\rank_{{\mathbb{Z}}/2}(\Kh(L;{\mathbb{Z}}/2))=16$ by the universal coefficient theorem, so the Batson-Seed spectral sequence implies that $\rank(\Kh(L_i;{\mathbb{Z}}/2))\leq 8$, whence each component of $L$ is an unknot or a trefoil~\cite{baldwin2018khovanov}. Indeed, we readily see that at least one component is unknotted.
	
	
	We now consider $\Kh(L;{\mathbb{Q}})$, which we can determine from $\Kh(L;{\mathbb{Z}})$ via the universal coefficient theorem. Suppose $L$ has a right hand trefoil component. Then \\
	$\rank_{i-j=-4}(\Kh(T(2,3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})))=2$, but $\rank_{i-j=-12}(\Kh(L;{\mathbb{Q}}))=1$ contradicting equation~\ref{GradedBatsonSeed}. Similarly, $\rank_{i-j=2}(\Kh(T(2,-3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}}))=3$ which contradicts $\rank_{i-j=-6}(\Kh(L;{\mathbb{Q}}))=1$ by Equation~\ref{GradedBatsonSeed}. Thus both components of $L$ are unknotted.
\end{proof}

Lemma~\ref{KHT2102components} requires some extra analysis since there is no complete classification of knots with $\rank(\Kh(L;{\mathbb{Z}}/2))=10$. To deal with this latter case we  use Dowlin's spectral sequence and Proposition~\ref{E2collapsed}.


\begin{proof}[Proof of Lemma~\ref{KHT2102components}]
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then from Table 2 and the universal coefficient theorem we can see that $\rank(\Kh(L;{\mathbb{Z}}/2))= 20$. Batson-Seed's spectral sequence implies that at least one component of $L$ is unknotted, and the other is a knot $L_2$ with $\rank(\Kh(L_2;{\mathbb{Z}}/2))\leq10$. If $\rank(\Kh(L_2;{\mathbb{Z}}/2))\leq10$ then $\rank(\widetilde{\Kh}(L_2;{\mathbb{Z}}/2))\leq 5$, so that $K$ is an unknot, trefoil, figure eight or a knot with $\rank(\widetilde{\Kh}(L_2;{\mathbb{Z}}/2))=5$. Let $U$ denote the unknot. To see that $L_2$ must also be unknotted we proceed by cases;
\begin{enumerate}
	\item $L_2$ is $T(2,3)$. In this case; \begin{align*}\Kh(T(2,3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})&\cong({\mathbb{Q}}_{-6}\oplus{\mathbb{Q}}_{-3}^2\oplus{\mathbb{Q}}_{-1})\otimes({\mathbb{Q}}_{-1}\oplus{\mathbb{Q}}_1)\\&\cong{\mathbb{Q}}_{-7}\oplus{\mathbb{Q}}_{-5}\oplus{\mathbb{Q}}_{-4}^2\oplus{\mathbb{Q}}_{-2}^3\oplus{\mathbb{Q}}_0\end{align*}
		But $\rank(\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction. 
	
	\item $L_2$ is $T(2,-3)$. Then \begin{align*}\Kh(T(2,-3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})&\cong({\mathbb{Q}}_{6}\oplus{\mathbb{Q}}_{3}^2\oplus{\mathbb{Q}}_{1})\otimes({\mathbb{Q}}_{1}\oplus{\mathbb{Q}}_{-1})\\&\cong {\mathbb{Q}}_{7}\oplus{\mathbb{Q}}_{5}\oplus{\mathbb{Q}}_{4}^2\oplus{\mathbb{Q}}_{2}^3\oplus{\mathbb{Q}}_0\end{align*}
	
	But $\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction.
	
	
	
	\item $\rank(\widetilde{\Kh}(L_2;{\mathbb{Z}}/2))=5$. In this case we have that $\rank(\widetilde{\Kh}(L_2;{\mathbb{Q}}))\leq 5$, whence $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}}))\leq 5$. If $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}})<5$, then $L_2$ is a trefoil or the unknot, cases which we have already dealt with. Thus we may take $\rank(\widehat{\HFK})(L_2;{\mathbb{Q}})=5$. We show that $L_2$ is a figure eight or has the same knot Floer homology as $T(2,\pm5)$. Suppose $L_2$ is of genus $g>0$. Note that $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}},g))\leq 2$. Note also that since $\Kh(L;{\mathbb{Z}})$ is Khovanov thin, $\Kh(L,p;{\mathbb{Q}})$ is $\delta$-thin~\cite{baldwin2017khovanov}, whence $\widehat{\HFK}(L)$ is $\delta$-thin. We proceed by cases.\begin{enumerate}
	    \item Suppose $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}},g))= 2$. Then $\widehat{\CFK}(L_2;{\mathbb{Q}})$ contains a differential of length $2g$ and one of length $g$. From Proposition~\ref{E2collapsed}, we see that this yields a summand of $(\widehat{\CFL}(L;{\mathbb{Q}}),\partial)$ of rank $8g$ and another of rank $4g$. Thus $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 12g$. It immediately follows that $g=1$. The symmetry of $\widehat{\HFL}(L)$ implies that $\rank(\widehat{\HFL}(L))\geq 24$, a contradiction.
	    
	    \item Suppose $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}},g))= 1$. Consider $\widehat{\CFL}(L_2;{\mathbb{Q}})$. If $g=1$, then $L_2$ is the figure eight knot. If $g>1$ there are two possible forms for $\widehat{\HFK}(L_2)$, as shown in Figure~\ref{rank5hfk}. To see this we need to use the fact that if $\rank(\widehat{\HFK}(L_2,g-1))=1$, 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~\cite{baldwin_note_2018}.
	    
	    
					\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{HFKL2.pdf}
    \caption{Different possibilities for $\widehat{\CFK}(L_2;{\mathbb{Q}})$.\label{rank5hfk}}
    \end{figure}
    
    In either case, since there is both a length one differential and a length $g-1$ differential, $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 4(g-1)+4+4=4g+4$, whence $g\leq 4$. Suppose we are in the first case.  Indeed, since the minimal $A_2$ grading of these generators is $\frac{-3}{2}$, symmetry yields at least another $4(g-1)$ generators, so that $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 8(g-1)+4+4+4$, so that $g=2$, and $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong \widehat{\HFK}(T(2,5);{\mathbb{Q}})$.
	    
	    Suppose we are in the second case. We have that $g\in\{2,3,4\}$. If $g=2$ then $\widehat{\HFK}(L_2)\cong\widehat{\HFK}(T(2,-5))$. If $g=4$ using the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ we find $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 24$, a contradiction.
	    
	    If $g=3$, symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ implies that the $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 24$, a contradiction.
	\end{enumerate}
	
	
	\item $L_2$ is the figure eight. In this case;
	\begin{align*}\Kh(U;{\mathbb{Q}})\otimes\Kh(L_2;{\mathbb{Q}})\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})\end{align*}
	
	But $\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction.
	
	
	\item $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong\widehat{\HFK}(T(2,5);{\mathbb{Q}})$, or $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong\widehat{\HFK}(T(2,-5);{\mathbb{Q}})$. Since $\ell\text{k}(L_1,L_2)=\pm 5$, there is a summand $\widehat{\HFK}(T(2,\pm5);{\mathbb{Q}})\otimes V[\pm\frac{5}{2},a]$ of $\widehat{\HFL}(L;{\mathbb{Q}})$, for some $a\in\pm\frac{5}{2}+{\mathbb{Z}}$. The symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ yields another $10$ generators. There can be no additional generators, as this would violate the rank inequality from Dowlin's spectral sequence - $\rank(\widehat{\HFL}(L);{\mathbb{Q}})\leq\rank(\Kh(T(2,10);{\mathbb{Q}}))=20$, 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 $\widehat{\HFL}(L;{\mathbb{Q}})$ is supported only in $A_1$ gradings $\pm\frac{1}{2}$. However since $L_1$ is unknotted and there is a spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L_1)\otimes V[\frac{5}{2}]$ there must be generators in $A_1$ grading $\frac{5}{2}$, a contradiction.
\end{enumerate}
\end{proof}


	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~\ref{E2collapsed}. From here it suffices to prove the following pair of propositions.
	
	\begin{proposition}\label{combinedHFKKHT28detection}
	Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link satisfying the following conditions;
	\begin{enumerate}
	\item $L$ has two exactly two components, $L_1$ and $L_2$, which are both unknots.
	\item $\ell\text{k}(L_1,L_2)=4$
	\item $\chi(L)\leq-6$
	    \item $\rank(\widehat{\HFK}(L;F))\leq 16$.
	    \item $\widehat{\HFK}(L;F)$ is $\delta$-thin.
	\end{enumerate}
	
	Then $L$ is $T(2,8)$.
	\end{proposition}
	
		\begin{proposition}\label{combinedHFKKHT210detection}
	Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link satisfying the following conditions;
	\begin{enumerate}
	\item $L$ has two exactly two components, $L_1$ and $L_2$, which are both unknots.
	\item $\ell\text{k}(L_1,L_2)=5$
	\item $\chi(L)\leq-8$
	    \item $\rank(\widehat{\HFK}(L;F))\leq 20$.
	    \item $\widehat{\HFK}(L;F)$ is $\delta$-thin.
	\end{enumerate}
	
	Then $L$ is $T(2,10)$.
	\end{proposition}
	
Before proving these propositions, we apply them to obtain Theorem's~\ref{KHTT28} and~\ref{KHT210} respectively.
	

	\begin{proof}[Proof of Theorem~\ref{KHTT28}]
	It suffices to show that $L$ satisfies the conditions in Proposition~\ref{combinedHFKKHT28detection}. Conditions 1,2 and 3 are implied by Lemmas~\ref{KHcomponentsgenuslinking} and~\ref{KHt228components}. To see that conditions $4$ holds, observe that $\rank(\Kh(L;{\mathbb{Z}}/2)=16$ by the universal coefficient theorem. It follows that $\rank(\widetilde{\Kh}(L;{\mathbb{Z}}/2))=8$ by~\cite{shumakovitch2014torsion}, so that $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 8$ by the universal coefficient theorem. $\rank(\Kh(L,p,{\mathbb{Q}}))\leq\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))$ by \cite[Theorem 1.2]{baldwin2017khovanov}, whence $\widehat{\HFK}(L;{\mathbb{Q}})\leq 16$.

	
	To see that condition 5 holds, note that that $\Kh(L;{\mathbb{Q}})$ is $\delta$-thin, so that $\Kh(L,p;{\mathbb{Q}})$ is $\delta$-thin by \cite[Lemma 2.11]{baldwin2017khovanov}.
	\end{proof}
	
	\begin{proof}[Proof of Theorem~\ref{KHT210}]
	    This follows by the same argument as in the proof of Theorem~\ref{KHTT28}.
	\end{proof}
	
We now prove the two requisite lemmas~\ref{combinedHFKKHT28detection}, and ~\ref{combinedHFKKHT210detection}






		\begin{proof}[Proof of proposition~\ref{combinedHFKKHT28detection}]
	Suppose $L$ satisfies the conditions given in the Proposition. Consider $\widehat{\CFL}(L;{\mathbb{Q}})$. Since $\widehat{\HFK}(L)$ is $\delta$-thin, $\widehat{\CFL}(L;{\mathbb{Q}})$ is an $E_2$ collapsed chain complex. Likewise since $\chi(L)\leq -6$ the maximal Alexander grading of $\widehat{\HFK}(L;{\mathbb{Q}})$ is at least $3$. 
	
	Consider $\widehat{\CFL}(L)$. Since $L$ has unknotted components, it follows that $(\widehat{\CFL}(L;{\mathbb{Q}}),\partial)$ 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 $(\widehat{\CFL}(L),\partial)$ has summands $X_{-1}^{l-4}[2,2]\oplus X_0^{l-5}[1,1]$. From the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$, and the fact that $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\leq 16$ we see immediately that $l\leq 6$.
			
		
			
			
			
				
					\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{Pseduo98.pdf}
    \caption{One potential link Floer homology complex for a link satisfying the conditions in Propositions~\ref{combinedHFKKHT28detection}}
    \end{figure}
				
					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 $\rank(\widehat{\HFK}(L))>16$. Similarly if $l\in\{1,2\}$ would violate the rank bound after accounting for the symmetry of $\widehat{\HFL}(L)$. If $l=0$ then $\rank(\widehat{\HFL}(L))\geq 16$. 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 $\widehat{\HFL}(L;{\mathbb{Q}})$. 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~\ref{Braidpolytope}, or that $\chi(L)>-6$.
				
				If $l=5$ we readily see that -- apart from in the case that $\widehat{\CFL}(L;{\mathbb{Q}})$ 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 $\widehat{\CFL}(L)$ 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~\ref{Braidpolytope} implies that $\widehat{\HFL}(L;{\mathbb{Q}})\cong\widehat{\HFL}(T(2,8);{\mathbb{Q}})$, whence the result follows from the corresponding detection result in~\cite{binns2020knot}.
			
				\end{proof}
	

The proof of~\ref{combinedHFKKHT210detection} is virtually the same, but with some additional case analysis.

\begin{proof}[Proof of Proposition~\ref{combinedHFKKHT210detection}]
	Suppose Suppose $L$ is as in the hypotheses of the Proposition. Since $L$ is delta-thin, $\widehat{\CFL}(T(2,2n);{\mathbb{Q}})$ is $E_2$-collapsed. Since $\chi(L)\leq -8$ the maximal Alexander grading is at least $n$. Since the components of $L$ are unknotted $\widehat{\CFL}(L;{\mathbb{Q}})$ 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 $\widehat{\CFL}(L;{\mathbb{Q}})$ has summands $X_0^{l-5}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{l-4}[2,2]$. Observe that the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ together with the fact that $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\leq 20$, imply that $l\leq 7$. 
	\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{KhT_2_10.pdf}
    \caption{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.}
    \end{figure}
	
	If $l=7$ then either $\widehat{\CFL}(L)\cong X_0^{2}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{3}[2,2]\oplus B_{-16}[\frac{-9}{2},\frac{-7}{2}]\oplus B_{-16}[\frac{-7}{2},\frac{-9}{2}]$ or there is an additional $B_{-8}[\frac{-1}{2},\frac{-1}{2}]$ summand, by the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$. 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~\cite{GageT26}, a contradiction.
	
	If $l=6$, then either $\widehat{\CFL}(L)\cong X_0^{1}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{2}[2,2]\oplus B_{-11}[\frac{-7}{2},\frac{-7}{2}]$ 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 $\widehat{\HFL}(L;{\mathbb{Q}})$ 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 $T(2,10)$, a contradiction since $T(2,10)$ has distinct link Floer homology. The $C_{-7}[\frac{5}{2},\frac{-7}{2}]\oplus C_{-7}[\frac{-7}{2},\frac{5}{2}]$ case is excluded by computing the Conway polynomial. First note that the Alexander polynomial is given by $(t^{-1/2}-t^{1/2})(t^{-5}+t^5-2t^{-4}-2t^4+t^{-3}+t^3+2t^{-1}+2t-4)$. Thus $\nabla(t-t^{-1})=-t^{11}+t^{-11}+3t^{9}-3t^{-9}-3t^{7}+3t^{-7}+t^{5}-t^{-5}-2t^{3}+2t^{3}+6t^{1}-6t^{-1}$. Therefore $\nabla(u)=-u^11-8u^9-20u^7-16u^5-4u^3$. It follows that the linking number is four, a contradiction. If there are three cubes then we note that in order to satisfy the symmetry properties there must be a summand $B_{-7}[\frac{-1}{2},\frac{-1}{2}]$. Indeed unless the reaming components are $B_{-1}[\frac{5}{2},\frac{5}{2}]\oplus B_{-13}[\frac{-7}{2},\frac{-7}{2}]$, or $B_{-8}[-\frac{5}{2},\frac{7}{2}]$ $B_{-8}[\frac{3}{2},\frac{-5}{2}]$. In the first case, we see that $L$ bounds an annulus upon reversing its orientation, a contradiction, since no link $T(2,2n)$ has the requisite homology. In the second case we see that $\Delta_L(t)=(t^{-1/2}-t^{1/2})(-t^7-t^{-7}+2t^6+2t^{-6}-t^5-t^{-5}+2t-4+2t^{-1})$ Thus $\nabla_L(t-t^{-1})=t^{15}-t^{-15}-3t^{13}+3t^{-13}+3t^{11}-3t^{-11}-t^9+t^{-9}-2t^3+2t^{-3}-2t^3+2t^{-3}-2t+2t^{-1}$
	Therefore $\nabla_L(u)=-u^{15}+12u^{13}+54u^{11}+112u^9+105u^7+36u^5-8u$ and the linking number is $-8$, a contradiction.
	
	Similarly, if $l\leq 5$ then there are summands $Y_0^{5-l}[3-l,3-l]\oplus Y_{-1}^{4-l}[2-l,2-l]$. Considering the symmetry properties of $\widehat{\HFL}(L)$ readily see that either $l=0$ or $l\geq 4$, as else $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))>20$. If $l=0$ then there are no additional generators, and we see that $L$ bounds an annulus, and hence is $T(2,10)$. This is a contradiction since $T(2,10)$ has distinct link Floer homology. If $l=4$ then $\widehat{\HFL}(L;{\mathbb{Q}})$ either contains no extra generators or an cube $C_{-6}[\dfrac{-1}{2},\dfrac{-1}{2}]$ by symmetry. We see that after reversing the orientation of a component of $L$, it bounds an annulus, whence $L$ is $T(2,10)$, a contradiction since $T(2,10)$ has distinct knot Floer homology.
	
	
We are thus left only with the case that $l=5$. In which case $\widehat{\CFL}(L;{\mathbb{Q}})$ has a summand $X_0[\frac{5}{2},\frac{5}{2}]\oplus Y_{-1}[\frac{3}{2},\frac{3}{2}]$. By symmetry $\widehat{\CFL}(L;{\mathbb{Q}})$ has a summand $B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ $\widehat{\CFL}(L;{\mathbb{Q}})$ can have at most three other $B_d$ summands. If there is one or three more cubes then the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ imply that one such summand must be of the form $B_{-6}[\frac{-1}{2},\frac{-1}{2}]$. Thus if $\widehat{\HFL}(L;{\mathbb{Q}})$ has $0$ or $1$ extra $B_d$ summands we see that after reversing the orientation of a component of $L$, $L$ bounds an annulus, and is therefore $T(2,10)$, a contradiction since $T(2,10)$ has distinct link Floer homology.

If $L$ has two extra $B_d$ summands then unless they are of the form $B_{-6}[\frac{3}{2},\frac{-5}{2}]\oplus[\frac{-5}{2},\frac{5}{2}]$, or $B_{-2}[\frac{3}{2},\frac{3}{2}]\oplus B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ then at least one component of $L$ is a braid axis and the result follows from Lemma~\ref{Braidpolytope} and the corresponding detection result for link Floer homology in~\cite{binns2020knot}. The first case is dealt with by computing the Conway polynomial from $\widehat{\HFL}(L;{\mathbb{Q}})$. Explicitly, $\Delta_L(t)=(t^{-1/2}-t^{1/2})(t^5+t^{-5}-2t^{4}-2t^{-4}+t^{3}+t^{-3}+2t-4+2t^{-1})$, whence $\nabla(t-t^{-1})=-t^11+t^{-11}+3t^9-3t^{-9}-3t^7+3t^{-7}+t^5-t^{-5}-2t^3+2t^{-3}+6t-6t$, and $\nabla_L(u)=-u^{11}-8u^9-20u^7-16u^5-5u^3$, whence the linking number is $-5$, a contradiction. In the second case we see that after reversing the orientation of a component, $L$ bounds an annulus, and is therefore $T(2,10)$, a contradiction since this link has distinct link Floer homology.

If $L$ has three extra $B_d$ summands, the by the symmetry of $\widehat{\HFL}(L)$, there must be a $B_{-6}[\frac{-1}{2},\frac{-1}{2}]$ summand. Unless the remaining summands are of the form $B_{-6}[\frac{3}{2},\frac{-5}{2}]\oplus[\frac{-5}{2},\frac{5}{2}]$, or $B_{-2}[\frac{3}{2},\frac{3}{2}]\oplus B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ then at least one component of $L$ is a braid axis and the result follows from Lemma~\ref{Braidpolytope} and the corresponding detection result for Link Floer homology~\cite{binns2020knot}.
For the $B_{-6}(\frac{3}{2},\frac{-5}{2})\oplus B_{-6}(\frac{-5}{2},\frac{5}{2})$ case we see that $\Delta_L(t)=(t^{-1/2}-t^{1/2})(t^5+t^{-5}-2t^{4}-2t^{-4}+t^{3}-t^{-3}+3t-6+3t^{-1})$, whence $\nabla_L(t-t^{-1})=-t^{11}+t^{-11}+3t^9-3t^{-9}-3t^7+3t^{-7}+t^5-t^{-5}-3t^3+3t^{-3}+9t-9t^{-1}$ and $\nabla_L(u)=-u^{11}-8u^{9}-20u^7-16u^5-5u^3$, whence the linking number is $-5$, a contradiction. For the $B_{-2}(\frac{3}{2},\frac{3}{2})\oplus B_{-10}(\frac{-5}{2},\frac{-5}{2})$ case we see that $L$ bounds an annulus upon reversing the orientation of a component of $L$ and the desired result follows.


Thus at least one component is a braid axis, and Lemma~\ref{Braidpolytope} implies that $\widehat{\HFL}(L;{\mathbb{Q}})\cong\widehat{\HFL}(T(2,10);{\mathbb{Q}})$ and the result follows from the corresponding detection result for link Floer homology.
\end{proof}


We conclude by noting that Martin's detection result for $T(2,6)$~\cite{GageT26}, $T(2,4)$ detection~\cite{xie2020links} and $T(2,2)$ detection~\cite{baldwin2017khovanov} could have been obtained by similar methods to those presented in this paper.


\end{subsection}
\end{section}




\bibliographystyle{plain}

\subsection*{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 are especially grateful to 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. 
	\end{section}
	
	
	
	\begin{section}{A Very Brief Review of Link Floer homology}\label{A 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 Ozsv\'ath-Szab\'o in~\cite{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, is a multi-graded ${\mathbb{Z}}/2[U_1,U_2,\dots U_n]$ chain complex with underlying vector space; $$\CFL^-(L,M)\cong\bigoplus_{m,A_1,A_2,\dots A_n}\CFL^-_m(L,M)(A_1,A_2\dots A_n).$$ 

The complex can also be defined with ${\mathbb{Z}}$ coefficients~\cite{manolescu2007combinatorial},\cite{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 \emph{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 \emph{Maslov grading}, which is integer valued. $\CFL^-(L)$ 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 $\CFL^-(L)$ is an invariant of $L$, which we refer to as the \emph{link Floer complex of $L$}. The homology of $\CFL^-(L)$ is denoted $\HFL^-(L)$. $\CFL^-(L)$ 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 $\widehat{\CFL}(L)$, and has homology denoted by $\widehat{\HFL}(L)$, which we will call the \emph{link Floer homology of $L$}. The \emph{link Floer homology polytope} of $L$ is defined as the convex hull of the multi-Alexander gradings of $\widehat{\HFL}(L)$ with non-trivial support. The link Floer polytope determines the Thurston polytope of the link exterior by a result of Ozv\'ath-Szab\'o~\cite{ozsvath2008linkFloerThurstonnorm}. We note that $\widehat{\HFL}(\overline{L})$, where $\overline{L}$ is $L$ with the orientation reversed, can be recovered from $\CFL^-(L)$ by only counting the disks which cross the appropriate basepoints.

For each component $L_i$ of $L$ there is a spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L-L_i)\otimes V[\frac{\ell\text{k}(L_i,L_j)}{2}]$. 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 $\widehat{\HFL}(L)$ is of rank $2^{n-1}$ in the maximal $A_i$ grading of non-trivial support by a result of Martin~\cite{GageT26}.
	 
	 The \emph{knot Floer homology} of $L$, $\widehat{\HFK}(L)$, is an oriented link invariant due independently to Rasmussen \cite{Rasmussen} and Ozsv\'ath-Szab\'o \cite{Holomorphicdisksandknotinvariants} that can be obtained from $\widehat{\HFL}(L)$ by projecting the multi-Alexander grading onto the diagonal, and shifting the Maslov gradings up by $\frac{n}{2}$. $\widehat{\HFK}(L)$ determines the maximal Euler characteristic of a surface bounding $L$ \cite{ni2006note}, as well as whether or not $L$ fibered \cite{ghiggini2008knot},\cite{ni2007knot}.
	 
	 The \emph{Conway polynomial} of $L$ can be obtained as an appropriate decatigorification of $\widehat{\HFL}(L)$. A result of Hoste \cite{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.
	 
	 
	   
	\end{section}
	
	

\begin{section}{The contact class and a rank bound}\label{linkcontact}

	In this section we generalise the main result of \cite{baldwin_note_2018} to links;
	
	\begin{theorem}\label{rankinnequalityforfiberedlinks}
	If $L$ is a non-trivial $n$ component fibered link in $S^3$ then; \[\rank(\widehat{\HFK}(L,\dfrac{n-2-\chi(L)}{2}))\geq n.\]
	\end{theorem}
	
	Similarly this implies a generalization of a Theorem of Ni~\cite[Theorem A.1]{ni2020seifert};
	
	\begin{corollary}\label{Nigeneralized}
	Suppose $L$ is fibered, with modromy neither right veering nor left veering on each component. Then \[\rank(\widehat{\HFK}(L,\dfrac{n-2-\chi(L)}{2}))\geq 2n.\]
	\end{corollary}
 

Before proceeding, we set notation for the different differential maps in $\CFL^{-}$ and $\widehat{\CFL}$ that we use in this section.
By $\partial^{-}$, we denote the differential map in the chain complex $\CFL^{-}$, 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 $\partial_{\widehat{\CFL}}$, which is the boundary map in $\widehat{\CFL}$. By $\partial_{\widehat{\CF}}$, we denote the unfiltered version of $\partial_{\widehat{\CFL}}$ i.e. the boundary map in $\widehat{\CF}$ of the ambient manifold, without the additional filtrations induced by the link.


The outline of the proof of the main technical lemma, Lemma~\ref{chainlevel}, 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 $\partial_{\widehat{\CF}}\mathbf{d}=\text{pb}\mathcal{L}(L)$ 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 \cite{baldwin_note_2018}.


We now review Honda-Kazez-Matic's reinterpretation of Ozsv\'ath-Szab\'o's contact class~\cite{honda2006contact}, as well as their generalization of the contact class to the sutured manifold setting~\cite{honda2009contact}, and explain how it can be viewed as a generator of link Floer homology. We closely follow Baldwin-Vela-Vick's analysis of the one component case.

Honda-Kazez-Matic showed that the contact class of a contact $3$-manifold $(Y,\xi)$ can be defined as follows. Take an open book $(\Sigma,\phi)$ representing $(Y,\xi)$. Note that $\Sigma$ need not have connected boundary. Take a \emph{basis of arcs} for $\Sigma$, that is a collection of oriented arcs $\{a_i\}$ in $\Sigma$ whose complement is a polygon. Form another collection of arcs $\{b_i\}$ by pushing $\{a_i\}$ via a Hamiltonian isotopy so that each $b_i$ intersects $a_i$ transversely in the interior of $\Sigma$ with positive intersection number. Obtain a Heegaard diagram for $Y$ by doubling $\Sigma$ along its boundary with $\alpha$ curves given by doubling $\{a_i\}$ and $\beta$ curves given by gluing $b_i$ to $\phi(a_i)$. The contact class is given by $\{a_i\cap b_i\}$.

Suppose now that $L$ is a fibered link with at least two components. The $L$ is the binding of an open book $(\Sigma, \phi)$. Note that we can push $L$ of itself we obtain a new link $L'$, which intersects every page exactly once. Indeed there is an associated partial open book $(\Sigma',\phi')$ where $\Sigma'=\Sigma-\nu(L')$ and $\phi'$ is the restriction of $\phi$ to $\Sigma'$. Suppose $\phi$ sends an arc to the right $a_1$ to the right. Pick an arc $a_2$ which loops around the hole corresponding to $L_i$. We note that $\phi(a_2)$ can be isotoped as shown in the figure. Note that we may take $a_2$ to be disjoint from $a_1$. Extend $\{a_1,a_2\}$ to a basis of arcs for the surface. Our sutured Heegaard diagram is then given by taking the $\alpha$ curves to be the doubles of the $a$ arcs and $\beta$ curves to be push offs of the $a$ arcs in $\Sigma_0$ with $\phi(a)$ in $\Sigma_1$.

We thus obtain the $EH(Y(L))$ class $a$ arcs in $\Sigma_0-\nu(L')$. The intersection point shown is then the $EH(Y(L))$ class.

\begin{figure}
    \centering
    \includegraphics[width = 10cm]{EHclass.jpeg}
    \caption{The contact class in sutured Floer homology. The circles are boundary components of $\Sigma'$ coming from the intersection of $\Sigma$ with $L$.}
    \label{Adapted triple}
\end{figure}


\begin{lemma}
    $EH(Y(L))$ is non-zero in $\SFH(-Y(L))\cong\widehat{\HFL}(-L,-Y)$.
\end{lemma}
\begin{proof}
    There is a sutured manifold decomposition of the link exterior $Y(L)$ along the fiber surface yields the product sutured manifold $Y(\Sigma)$. The contact class of the associated contact structure is non-zero~\cite{honda2009contact}. It follows that the contact class of the original thing is non-zero~\cite[Theorem 6.2]{honda2009contact}
\end{proof}


Let $v_0$ be a non-vanishing vector field along $\nu(L)$ that points into $Y(L)$ along $R_-(\gamma)$, out of $Y(L)$ along $R_+(\gamma)$ and restricts to the gradient of a height function $s(\gamma)\times I\to I$ along $\gamma$.
\begin{lemma}
    $EH(Y(L))$ is of Alexander grading $-\dfrac{|L|-\chi(L)}{2}$ in $\widehat{\HFK}(-L,-Y,[\Sigma])$.
\end{lemma}
For this we follow the proof of ~\cite[Theorem 1.5]{juhasz2008floer}.
\begin{proof}
Decompose $Y(L)$ along a Seifert surface for $L$. Let $\xi$ be a tangent vector field along $\partial\nu(L)$ pointing in the meridional direction. Then $\xi$ lies in $v_0^\perp$. Since $Y(\Sigma)$ is strongly balanced we can apply~\cite[Theorem 3.11]{juhasz2008floer}. Note that $Y(\Sigma)$ is a product sutured manifold, so we have that the summand of $\widehat{\SFH}(Y(L))\cong\widehat{\HFL}(L)$ that is outer with respect to $\Sigma$ is just ${\mathbb{Z}}/2$. The outer $\text{spin}^c$ structures are exactly those for which $\langle c_1(\mathfrak{s},t_0),[S]\rangle=\chi(\Sigma)+I(\Sigma)-r(\Sigma,t_0)$. Now $\chi(\Sigma)=\chi(L)$, $I(T)=-|L|$ by ~\cite[Theorem 3.9]{juhasz2008floer}, and $r(\Sigma,t_0)=0$ since the components of $\partial \Sigma$ are longitudes. Thus the $\text{spin}^c$ structures that are outer with respect to $\Sigma$ are exactly those with with $\langle c_1(\mathfrak{s},t_0),[S]\rangle=\chi(L)-|L|$, i.e those with maximal Alexander grading.
\end{proof}


We need to work with the filtered complex, so pinch boundary components and replace with $z_i$, $w_i$.

\begin{lemma}\label{chainlevel}
    Suppose $L$ is a fibered link represented by an open book $(\Sigma,\phi)$ whose monodromy sends an arc $a\subseteq \Sigma$ with $a\cap L_i\neq\emptyset$ to the right. Suppose that the multi-Alexander grading of $EH(Y(L)$ is $(A_1,A_2,\dots A_N)$. Then there is a generator $d\in\widehat{\CFL}(-Y,-L,[\Sigma])$ with $\partial d=EH(Y(L))$ of multi-Alexander grading $(A_1,A_2,\dots A_i-1,\dots A_N)$.
\end{lemma}
We follow Baldwin-Vela-Vick's proof of the knot case
\begin{proof}
Suppose $L$, $a$ are as in the theorem. Isotope $\phi(a)$ so that $a$ and $\phi(a)$ intersect minimally. Consider the class $d=\{d_1,c_2,\dots c_{2g+|L|-1}\}$. There is clearly a component of a differential from $d$ to $EH(Y(L))$, as shown in the figure. There can be no other components, as else $a$ and $\phi(a)$ would not intersect minimally.

\begin{figure}
    \centering
    \includegraphics[width = 10cm]{EHdisk.jpeg}
    \caption{A disk contributing to the total differential of $\CFL(L)$. $\{c_1,c_2\}$ is part of the contact class, $\{d_1,c_2\}$ part of the class with differential the contact class.}

\end{figure}
\end{proof}


\begin{proof}[Proof of Theorem~\ref{rankinnequalityforfiberedlinks}]

$L$ is a fibered link represented by an open book $(\Sigma,\phi)$ whose monodromy sends an arc $a\subseteq \Sigma$ with $a\cap L_i\neq\emptyset$ to the right. Then Lemma~\ref{chainlevel} implies that there is a generator $d$ with $\partial d=EH(Y(L))$. $d$ is clearly a cycle in $\widehat{\HFL}(L)$. Suppose it were a boundary, i.e that there is some $y\in\widehat{\CFL}(L)$ of the same multi-Alexander grading such that $\partial y=d$. Then $\partial^2y=c$, contradicting the condition that $\partial^2=0$.


If the monodromy does not send an arc abutting to $L_i$ to the right, then the mirror of $L$ will have such an arc, and we can reapply the same argument. The rank inequality follows from symmetry properties of $\widehat{\HFL}(L)$. The stated rank inequality then follows from the fact that Knot Floer homology is obtained from Link Floer homology by collapsing the multi-Alexander grading to a single Alexander grading.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{Nigeneralizsed}]
 Suppose $EH(Y(L))$ is of multi-Alexander grading $(A_1,A_2,\dots A_n)$. Suppose $\rank(\widehat)(Y,L,[\Sigma];(A_1,\dots A_i-1,\dots A_n))=1$. Consider $\CFL^-(Y,L,[\Sigma])$. Since there is an arc that is sent to the left and an arc that is sent to the right we find that $\partial_{-}^2(EH(L))\neq 0$, a contradiction. Thus $\rank(\widehat)(Y,L,[\Sigma];(A_1,\dots A_i-1,\dots A_n))\geq 2$ and the result follows from the fact that Knot Floer homology is obtained from Link Floer homology by collapsing the multi-Alexander grading to a single Alexander grading.
\end{proof}


		
	

	
	We conclude this section by noting that it is perhaps useful to view Theorem~\ref{rankinnequalityforfiberedlinks} in the context of the following more general proposition;

\begin{proposition}\label{generalrankbound}
Suppose $L$ is an $n>1$ component non-trivial link such that \\ $\rank(\widehat{\HFL}(L)(x_1,x_2,\dots x_n))=2m+1$ for some $m\in{\mathbb{Z}}$. Then $\rank(\widehat{\HFK}(L))\geq 2^{n}+2m$.
\end{proposition}

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 $\widehat{\HFK}(L)$ to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$.

\begin{proof}[Proof of Proposition~\ref{generalrankbound}]
   Suppose $L$ is an $n>1$ component link non-trivial such that $\rank(\widehat{\HFK}(L)(x_1,x_2,\dots x_n))$ $=2m+1$. 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ must be even. Thus we obtain at least an additional $2^n-1$ generators.
\end{proof}

Note that if $L$ is an $n$ component link and $\rank(\widehat{\HFK}(L,A))$ is odd for some $A$, then there is a multi-Alexander  grading $(A_1,A_2\dots A_n)$ such that $\rank(\widehat{\HFK}(L,A_1,A_2,\dots A_n))$ of odd rank, where $A=A_1+A_2+\dots A_n$ whence it follows that $\rank(\widehat{\HFK}(L))\geq 2^n$.

\end{section}


\begin{section}{Detection Results for $(2,2n)$-cables}\label{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;


\begin{theorem}\label{HFL2detectsRHTcables}
Link Floer homology detects $T(2,3)_{2,2n}$ for all $n$.
\end{theorem}

\begin{theorem}\label{HFLdetects(2,2n)cablesofthefigureeight}
Let $K$ be the figure eight knot. Link Floer homology detects $K_{2,2n}$ for all $n$.
\end{theorem}

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


\begin{lemma}\label{(2,0)trefoilcable}
    Suppose $L$ is a link such that $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,3)_{(2,0)})$, 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 \begin{align*}
        \widehat{\HFK}(K)\cong{\mathbb{F}}_0^2[1]\oplus{\mathbb{F}}_{-1}^3[0]\oplus{\mathbb{F}}_{-2}^2[-1]
    \end{align*}
\end{lemma}

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


    We first give a partial computation of $\widehat{\HFK}(T(2,3)_{(2,0)})$. To do so compute $\widehat{\HFK}(T(2,3)_{(2,1)})$ and $\widehat{\HFK}(T(2,3)_{(2,-1)})$ using Hanselman--Watson's cabling formula in the theory of immersed curves~\cite{hanselman2019cabling}, then apply the skein exact triangle for knot Floer homology to determine $\widehat{\HFL}(T(2,3)_{(2,0)})$, and thereby $\widehat{\HFK}(T(2,3)_{(2,0)})$.
    
    Hanselman--Watson's cabling formula implies that \begin{align*}
        \widehat{\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]
    \end{align*}
    
    \begin{align*}
        \widehat{\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]
    \end{align*}
    
    \vspace{.2in}
    
    \noindent Using the skein exact triangle from~\cite[Equation 7]{Holomorphicdisksandknotinvariants}  -- 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 $\widehat{\HFK}(T(2,3)_{(2,0)},i)$ is trivial for $i> 2$ and $\widehat{\HFK}(T(2,3)_{(2,0)},2)\cong {\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$. To determine $\widehat{\HFK}(T(2,3)_{(2,0)};1)$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)})$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)})$ 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 $\widehat{\HFL}(T(2,3)_{(2,0)}))$ is at least four in each of the maximal $A_i$ gradings (namely $A_i=1$). Thus $\widehat{\HFK}(T(2,3)_{(2,0)};1)\cong {\mathbb{F}}_{-\frac{1}{2}}^2\oplus{\mathbb{F}}_{\frac{1}{2}}^2$ and indeed \begin{align*}\widehat{\HFL}(T(2,3)_{(2,0)};(1,0))\cong\widehat{\HFL}(T(2,3)_{(2,0)};(0,1))\cong{\mathbb{F}}_0\oplus{\mathbb{F}}_{-1}.\end{align*}
    
    To conclude the computation it suffices to determine $\widehat{\HFK}(T(2,3)_{(2,0)};0)$ by the symmetries of link Floer homology. Note again that the skein exact triangle implies that $\widehat{\HFK}(T(2,3)_{(2,0)};0)$ 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 \begin{align*}
\widehat{\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]   \end{align*}

or

\begin{align*} \widehat{\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]  \end{align*}

We can now proceed to the proof of Lemma~\ref{(2,0)trefoilcable}
\begin{proof}[Proof of Lemma~\ref{(2,0)trefoilcable}]
 Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,3)_{(2,0})$. 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 $\widehat{\HFK}(L)$ is $\frac{1}{2}$. Since $\rank(\widehat{\HFK}(L))$ 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~\cite{hoste1985firstcoefficientoftheconwaypolynomial}. Since the maximal Alexander grading of $\widehat{\HFK}(L)$ is $1$, $L$ either bounds an annulus or the disjoint union of an unknot and a genus one knot. If $\rank(\widehat{\HFK}(L))=16$ 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 \begin{align*}
    {\mathbb{F}}_0^2[1]\oplus{\mathbb{F}}_{-1}^3[0]\oplus{\mathbb{F}}_{-2}^2[-1]
\end{align*} 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 $\widehat{\HFK}(L)$ 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  $\widehat{\HFK}(T(2,-3)_{(2,0)})\cong(\widehat{\HFK}(T(2,-3)_{(2,0)}))^*\not\cong\widehat{\HFK(L)}$. 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.


\end{proof}

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

\begin{lemma}\label{20cablefigureeight}
    Let $K$ be the figure eight knot and orient $K_{2,0}$ as the boundary of an annulus. Suppose $\widehat{\HFK}(L)\cong\widehat\HFK(K_{2,0})$. Then $L$ is isotopic to $K_{2,0}$.
\end{lemma}

Again we note that if $\widehat{\HFL}(L)\cong\widehat{\HFL}(K_{2,0})$, 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 $\widehat\HFK(K_{2,0})$. Hanselman-Watson's cabling formula for knot Floer homology in terms of immersed curves~\cite{hanselman2019cabling} shows that \begin{align*}
\widehat{\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]\\
\widehat{\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]\end{align*}

Again we apply the skein exact triangle from~\cite[Equation 7]{Holomorphicdisksandknotinvariants} -- with $L_0=K_{2,0}, L_+=K_{2,1}, L_-=K_{2,1}$.This immediately implies that $\widehat{\HFK}(K_{2,0},i)\cong 0$ for $|i|>2$ and $\widehat{\HFK}(K_{2,1})\cong 
F_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{5}{2}}$. Indeed we see that $\widehat{\HFK}(K_{2,0},1)$ 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 $\widehat{\HFL}(K_{2,0})$ is supported on the lines $A_2=A_1$, $A_2=A_1\pm 1$. Thus since $\widehat{\HFL}(L)$ must be of even rank in each $A_i$ grading, and $\widehat{\HFL}(K_{2,0},(x,y))\cong {\mathbb{F}}_{\frac{3}{2}}\oplus{\mathbb{F}}_{\frac{1}{2}}$ is excluded. While unnecessary we note that $\widehat{\HFK}(L)\cong({\mathbb{F}}_{\frac{1}{2}}\oplus{\mathbb{F}}_{-\frac{1}{2}})^n$ for some $0\leq n\leq 3$. In fact $n\geq 2$, as else $\widehat{\HFL}(K_{2,0})$ cannot admit a spectral to $\widehat{\HFL}(K)\otimes V$

\begin{proof}[Proof of Lemma~\ref{20cablefigureeight}]
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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,0))$, 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_1)$, where $L_1$ is the component of $L$ corresponding to $A_1$. Thus $\widehat{\HFL}(L_1)$ is of rank one in the top Alexander grading and $L_1$ is fibered~\cite{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.
\end{proof}





To prove theorems~\ref{HFL2detectsRHTcables} and \ref{HFLdetects(2,2n)cablesofthefigureeight} 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 $n$.

\begin{lemma}\label{algebraiccharachterizationofcables}
Let $L$ be a two component link with linking number $n>1$ such that;
\begin{enumerate}
        \item $\widehat{\HFL}(L)$ is of rank $2$ in the maximal non-zero $A_1$ or $A_2$ grading.
        \item The link Floer homology polytope has a vertex at multi-Alexander grading $(1+\frac{n}{2},1+\frac{n}{2})$.
        \item $\widehat{\HFL}(L)$ has support on the lines $A_2=A_1,A_2=A_1+1, A_2=A_1-1$.
         \end{enumerate}
         
Then $L$ is isotopic to $K_{2,2n}$ where $K$ is a trefoil or the figure eight knot.


\end{lemma}


\begin{proof}[Proof of Lemma~\ref{algebraiccharachterizationofcables}] 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  $g(L_i)+\frac{n}{2}\leq\max\{A_i:\widehat{\HFL}(L)(A_1,A_2)\neq 0\}$, it follows that $g(L_i)\leq 1$ for each $i$. Indeed since, $\widehat{\HFL}(L)$ 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.
\end{proof}
 
 To apply Lemma~\ref{algebraiccharachterizationofcables}, we need to show that $\widehat{\HFL}(K_{2,2n})$ satisfies the conditions of the hypothesis of Lemma~\ref{algebraiccharachterizationofcables} for $K$ a trefoil or figure eight knot. The easiest way to proceed is via the following remark
 
 \begin{remark}
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~\cite{stallings1978constructions}, and $T(n,m)$ along minimal genus Seifert surfaces~\cite{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~\cite{gabai1986detecting}. It follows that $(2,2n)$ cables of genus one fibered knots are fibered and of genus $n$, that is Euler characteristic $-2n$. \end{remark}

 If one wishes to only use Rudolph's stated result, for knots, one can alternately proceed via the following lemma;
 
 \begin{lemma}\label{genusof22ncables}
 Let $K$ be a genus one fibered knot. For $n\geq 1$, $K_{2,2n}$ is fibered and $\chi(K_{2,2n})=-2n$.
 \end{lemma}
 
 \begin{proof}[Proof of Lemma~\ref{genusof22ncables}]
 Note that, for $n>0$, $K_{2,2n+1}$ is the Murasugi sum of $K_{2,1}$ and $T(2,2n+1)$~\cite{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~\cite{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 $\rank(\widehat{\HFK}(K_{2,2n})(1+n))=1$ so that $K_{2,2n}$ is fibered, as desired.
 \end{proof}
 
 
It follows that conditions 1, 2 and 3 of Lemma~\ref{algebraiccharachterizationofcables} 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.

\begin{proof}[Proof of Theorem~\ref{HFL2detectsRHTcables}]

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~\ref{(2,0)trefoilcable}.


Suppose now that $n>0$. We first show that $\widehat{\HFL}(T(2,3)_{2,2n})$ satisfies the conditions in the hypothesis of Lemma~\ref{alexandergradingcharachterizationofunlinks}. Observe that for $n\geq 1$, each component of $T(2,3)_{2,2n}$ is a braid in the complement of the other, so $\widehat{\HFL}(T(2,3)_{2,2n}))$ must be rank two in the maximal $A_i$ grading, by braid detection ~\cite[Proposition 1]{GageT26}, so condition 1 of Lemma~\ref{algebraiccharachterizationofcables} holds. By Lemma~\ref{genusof22ncables} the maximal Alexander grading of $\widehat{\HFK}(L)$ is $2+n$. $T(2,-3)_{2,2n+1}$ is fibered by Lemma~\ref{genusof22ncables}. Therefore, since $\widehat{\HFL}(T(2,-3)_{2,2n})$ 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$ must be of Maslov gradings $0$ and $-1$.

Suppose now that $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,3)_{2,2n})$ for some $n\geq 2$. Since link Floer homology detects the linking number of two component links~\cite{hoste1985firstcoefficientoftheconwaypolynomial}, it follows that the components of $L$ have linking number $n$. Lemma~\ref{algebraiccharachterizationofcables} 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 $\widehat{\HFL}(L)$ has a Maslov index $0$ generator with $A_1$ grading $1+\frac{n}{2}$ which persists under the spectral sequence to $\widehat{\HFL}(L_i)$, 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$. Observe that for $n\geq 1$, each component of $L$ is a braid in the complement of the other, so $\widehat{\HFL}(L)$ must be rank two in the maximal $A_i$ grading, by braid detection~\cite{GageT26}, so Condition 1 of Lemma~\ref{algebraiccharachterizationofcables} holds. By Lemma~\ref{genusof22ncables} the maximal Alexander grading of $\widehat{\HFK}(L)$ is $2+n$. Since $T(2,-3)_{2,2n+1}$ is fibered, and $\widehat{\HFL}(T(2,-3)_{2,2n+1})$ 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 \ref{alexandergradingcharachterizationofunlinks} 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 $\widehat{\HFL}(T(2,-3)_{2,2n})$ must be of Maslov gradings $1$ and $2$.

Suppose now that $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,-3)_{2,2n})$. Lemma~\ref{algebraiccharachterizationofcables} 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 $\widehat{\HFL}(L)$ must be of Maslov gradings $1$ and $2$, it follows that each component is $T(2,-3)$ and we have the desired result.
\end{proof}

The proof of Theorem~\ref{HFLdetects(2,2n)cablesofthefigureeight} is again similar.

\begin{proof}[Proof of Theorem~\ref{HFLdetects(2,2n)cablesofthefigureeight}]
The proof follows the proof of Theorem~\ref{HFLdetectsLHTcables} verbatim, aside from to note that the top $A_i$ generators must be of Maslov index $0$ and $1$ respectively.
\end{proof}













\begin{section}{Technical Results}
\label{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.

	
	
	
	
	\begin{subsection}{$E_2$ collapsed chain complexes}


We will rely on a structure theorem for \emph{$E_2$-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 ~\cite[Section 12]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}. In particular Ozsv\'ath-Szab\'o 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~\ref{Summands}. 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 $-\mathbbm{1}$

\begin{figure}[h!]
\label{Summands}
    \centering
    \includegraphics[width=7cm]{E2_Summands.pdf}
    \caption{The summands of an $E_2$-collapsed chain complex}
    \end{figure}
    
    
    \begin{proposition}\label{E2collapsed}
    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 $B_d[i,j]\oplus V_d^l[i,j],H_d^l[i,j],X_d^l[i,j], Y_d^l[i,j]$.
    \end{proposition}
    
    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, $\widehat{\CFL}(L)$ 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 $\widehat{\CFL}(L)$. For instance, since the full filtered $\widehat{\CFL}$ complex has homology ${\mathbb{F}}_0\oplus{\mathbb{F}}_{-1}$, we see that $\widehat{\CFL}(L)$ 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 $\widehat{\CFL}(L_1)$ determines the $V_d^l$ summands, while $\widehat{\CFL}(L_2)$ determines the $H_d^l$ summands. Specifically, if there are generators $x,y\in\widehat{\CFL}(L_1)$, with $\langle\partial_{\widehat{\CFL}(L_1)} x,y\rangle\neq 0$ then we must have a summands of $\widehat{\CFL}(L)$ of the form $V_{M(x)}^{A(x)-A(y)}$, as no other summand induces the correct differential on $\widehat{\CFL}(L_1)$.
    

    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 $\partial_2$.
    \begin{proof}[Proof of Proposition~\ref{E2collapsed}]
    
    
    By a bi-filtered version of the Reduction Lemma~\cite[Lemma 4.1]{hedden2013khovanov} we can assume that the grading preserving part of $\partial$ is trivial. We proceed by induction on $\rank(C)$, 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. 
    \end{proof}
    
    
    
    


\begin{comment}
\color{red}
\begin{lemma}\label{cubes}
Suppose $\widehat{\HFL}(L;{\mathbb{F}})$ is $\delta$-thin. Then, as a vector space, $\widehat{\HFL}(L;{\mathbb{F}})\cong\underset{i}{\bigoplus}C_{d_i}[x_i,y_i]$.
\end{lemma}

Here $C_d(0,0)$ is the complex ${\mathbb{F}}_d(0,0)\oplus {\mathbb{F}}_{d+1}(1,0)\oplus{\mathbb{F}}_{d+1}(0,1)\oplus{\mathbb{F}}_{d+2}(1,1)$, and ${\mathbb{F}}$ is either ${\mathbb{Z}}/2$ or ${\mathbb{Q}}$.

\begin{proof}[Proof of Lemma~\ref{cubes}]
As observed in the proof of~\cite[Theorem 12.1]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, $\widehat{\HFL}(L)$ 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 $B_d$. As vector spaces, each of these complexes decomposes as a sum of $C_d(x,y)$.
\end{proof}

\end{comment}


\color{black}

    
   

	\end{subsection}
	
	
	\begin{subsection}{The Dual Thurston Polytope}
	
	

The four detection results also require the following proposition;

\begin{proposition}\label{Braidpolytope}
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
\end{proposition}
 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.

\begin{proof}[Proof of Proposition~\ref{Braidpolytope}]

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.

\end{proof}



	
	\end{subsection}
	


\end{section}










\end{section}




\begin{section}{Knot Floer Homology Detects $T(2,8),T(2,10)$}
\label{Knot Floer Homology Detects $T(2,8),T(2,10)$}

In this section we prove the following detection results;



\begin{theorem}
\label{HFKT28}
	Knot Floer homology detects $T(2,8)$.
\end{theorem}

	\begin{theorem}\label{HFKT210}
	Knot Floer homology detects $T(2,10)$.
\end{theorem}


 
For the readers convenience first we recall that:
    $$\widehat{\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}
    
    \end{cases}$$


We start by showing that in general if $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,2n))$ then $L$ consists of two unknotted components and $\wiehat{\CFL}(L)$ is of an especially simple form.
\begin{lemma}\label{T22nunknottedcomponents}
    Suppose $\widehat{\HFK}(L;{\mathbb{F}})\cong\widehat{\HFK}(T(2,2n);{\mathbb{F}})$. Then $(\widehat{\CFL}(L),\partial)$ 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.
\end{lemma}




\begin{proof}[Proof of Lemma~\ref{T22nunknottedcomponents}]
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 $\widehat{\HFK}(L)$ is $\frac{1}{2}$, so since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{m-1}S^1\times S^2)$, $L$ can have at most two components. $L$ then has exactly $2$ components since the Maslov grading of $\widehat{\HFK}(L)$ 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 $\widehat{\HFL}(L)$. 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~\cite[Theorem 12.1]{HolomorphicdiskslinkinvariantsandthemultivariableAlexanderpolynomial}, $\widehat{\HFL}(L)$ 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 $\widehat{\HFK}(L_1)$ 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 $\widehat{\HFK}(L_1)\otimes V$ then there must be a generator of odd Maslov index in $A_1$ grading $\frac{n}{2}-x-1$, since the summands of $\widehat{\HFL}(L)$ that persist under the spectral sequence to $\widehat{\HFK}(L_1)\otimes V$ 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.
\end{proof}



\begin{comment}
\color{red}
\begin{lemma}\label{T22nsummand}
	Let $n>1$. Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,2n))$. Then $L$ is a two component link with linking number $n$ and $\widehat{\HFL}(L)$ has a summand $C_{-2}[\frac{n-2}{2},\frac{n-2}{2}]\oplus C_{-2n}[\frac{-n}{2},\frac{-n}{2}]$.
\end{lemma}

The $n=1$ case is that of the Hopf link, which knot Floer homology is known to detect.

\begin{proof}[Proof of Lemma~\ref{T22nsummand}]
	Note that $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{|L|-1}(S^1\times S^2))$, where $|L|$ is the number of components of $L$. $\widehat{\HF}(\#^{|L|-1}(S^1\times S^2))$ has support in Maslov grading $\dfrac{|L|-1}{2}$, while the maximal Maslov grading of $\widehat{\HFK}(L)$ is $\frac{1}{2}$, so $L$ can have at most $2$ components. $L$ cannot have one component since the Maslov gradings of the generators of $\widehat{\HFK}(L)$ take values in ${\mathbb{Z}}+\frac{1}{2}$. $L$ has linking number $n$ because $\widehat{\HFK}(L)$ determines the Conway polynomial of two component links, and the Conway polynomial in turn detects the linking number of two component links~\cite{hoste1985firstcoefficientoftheconwaypolynomial}.
	
	Consider $\widehat{\HFL}(L)$. Let $A_i(0)$ be the Alexander index of the unique Maslov index $0$ generator in $\widehat{\HFL}(L_i)$ for each $i$. Note that $A_i(0)\geq 0$ since there are no positive Maslov index generators. Note that $A_1(0)+\frac{n}{2}+A_2(0)+\frac{n}{2}\leq n$, so $A_i(0)=0$, and the Maslov index $0$ generator is of bi-Alexander grading $(\frac{n}{2},\frac{n}{2})$. Lemma~\ref{cubes} implies that $\widehat{\HFL}(L)$ has a summand $C_{-2}[\frac{n-2}{2},\frac{n-2}{2}]$. The symmetry of $\widehat{\HFL}(L)$ then implies that $\widehat{\HFL}(L)$ has another summand $C_{-2n}[\frac{-n}{2},\frac{-n}{2}]$.
\end{proof}

\end{comment}

\color{black}


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


\begin{proof}[Proof of Theorem~\ref{HFKT28}]
Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nunknottedcomponents} 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 \cite{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~\ref{Braidpolytope}, and the result follows from the fact that link Floer homology detects $T(2,8)$~\cite{binns2020knot}.

\begin{comment}
\color{red}
Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nsummand} implies that there is a summand of $\widehat{\HFL}(L)$ given by $C_{-2}[1,1]\oplus C_{-8}[-2,-2]$. Indeed, since 
\[ \widehat{\HFL}_m(L)(A_1,A_2)\cong\widehat{\HFL}_{m-2A_1-2A_2}(L)(-A_1,-A_2)\]
Lemma~\ref{cubes} implies that the remainder of $\widehat{\HFL}(L)$ is given by $C_{-4}(x,-x)\oplus C_{-6}[-x-1,x-1]$. We seek to control $x$. We first show that $L$ has unknotted components. Suppose that there is a generator from $C_{-8}[-2,-2]\oplus C_{-2}[1,1]$ which is not of $A_i$ grading $2$ persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V(2)$. Then there must be a generator of Maslov grading at least $-1$ with $A_i$ grading at least $3$, which there is not. It follows that only a generator from $C_{-4}[x,-x]\oplus C_{-6}[-x-1,x-1]$ can possibly persist under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V[2]$, but if such a generator persists with $A_i$ grading at least $2+g(L_i)$ with $g(L_i)>0$ then there must be a generator which persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V[2]$ of Maslov grading at most $-2-2g(L_i)<-2$ of $A_i$ grading $2-g(L_i)$, a contradiction.
\end{comment}


	\begin{figure}[h]
    \centering
    \includegraphics[width=8cm]{T(2_8)pseudo.pdf}
    \caption{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.}
    \end{figure}
\begin{comment}
Thus both components of $L$ are unknotted. Therefore all of the hypothesis of Proposition~\ref{combinedHFKKHT28detection} hold and the result follows.


Since the rank in the maximal $A_i$ grading is $2$ for at least one $i$, at least one component is a braid axis for the other. Without loss of generality we may take $i=1$ and thereby deduce that the maximal $A_1$ grading is strictly less than $2$. It follows that $x\in\{-1,0\}$. The $x=-1$ case is excluded by Lemma~\ref{Braidpolytope}. Thus $x=0$, $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,8))$, and the result follows from the corresponding detection result in~\cite{binns2020knot}.
\end{comment}
\end{proof}


\color{black}
The proof of Theorem~\ref{HFKT210} is virtually identical

\begin{proof}[Proof of Theorem~\ref{HFKT210}]

Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,10))$. 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~\ref{Braidpolytope}, and the result follows from the fact that Link Floer homology detects $T(2,10)$.

\begin{comment}
	Suppose $L$ satisfies $\widehat{\HFK}(L)\cong\widehat{\HFK}(T(2,8))$. Lemma~\ref{T22nsummand} implies that $L$ is a two component link with linking number $5$ and that there is a summand of $\widehat{\HFL}(L)$ of the form $C_{-10}[-\frac{5}{2},-\frac{5}{2}]\oplus C_{-2}[\frac{3}{2},\frac{3}{2}]$. Symmetry implies that there is another of the form $C_{-6}(\frac{-1}{2},\frac{-1}{2})$. There are two summands remaining $C_{-4}(x,1-x)$ and $C_{-8}(-x,x-3)$.
	
	We first note that if there were a generator from $C_{-6}(\frac{-1}{2},\frac{-1}{2})\oplus C_{-10}(-\frac{5}{2},-\frac{5}{2})\oplus C_{-2}(\frac{3}{2},\frac{3}{2})$ that persists under the spectral sequence to $\widehat{\HFL}(L_i)\otimes V$ then there would be a generator of Maslov grading at least $-1$ in $A_i$ grading at least $\frac{7}{2}$, a contradiction.
	\end{comment}
	
	
    \begin{comment}
	 It follows that only the generators from $C_{-4}(x+\frac{1}{2},\frac{1}{2}-x)$ and $C_{-8}(-x-\frac{3}{2},x-\frac{3}{2})$ can possibly persist in $A_i$ gradings higher than $\frac{5}{2}$, but if such a generator persists with $A_i$ grading at least $\frac{5}{2}+g(L_i)$ with $g(L_i)>0$ then there must be a generator of Maslov grading at most $-2-2g(L_i)$ of $A_i$ grading $2-g(L_i)$ that persists under the spectral sequence to $\widehat{\HFL}(L_i)$, a contradiction.
	 
	 Thus both components of $L$ are unknotted. It follows that all of the hypotheses of Proposition~\ref{combinedHFKKHT210detection} hold and the result follows.
	 
Note that the rank of $\widehat{\HFL}(L)$ is two in at least one of the top $A_i$ gradings. We may thus, without loss of generality, take $L_1$ to be a braid axis, whence the maximal $A_1$ grading is $\frac{5}{2}$~\cite{GageT26}. We are therefore left only with the possibility that $|x|<2$. Note however that unless $x=0$ Lemma~\ref{Braidpolytope} is violated. Thus $\widehat{\HFL}(L)\cong\widehat{\HFL}(T(2,8))$ and the result follows from the corresponding detection result in~\cite{binns2020knot}.
	 
\end{comment}
\end{proof}




    
    
\end{section}









\begin{section}{Rank Detection Results for Knot Floer Homology }
\label{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;
\begin{theorem}\label{rank4}
	If $\rank(\widehat{\HFK}(L))=4$ then $L$ is the Hopf link or the three component unlink.
\end{theorem}


\begin{theorem}\label{rank6}
	Suppose $L$ is a link with $\rank(\widehat{\HFL}(L))=6$. Then $L$ is a disjoint union of a trefoil and an unknot.
\end{theorem}

\begin{theorem}\label{rank8}
The only links $L$ with $\rank(\widehat{\HFL}(L))=8$ are $T(2,4)$, $T(2,-4)$, the four component unlink, and the disjoint union of a Hopf link and an unknot.
\end{theorem}

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

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

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

\begin{proposition}\label{3componentrank12}
There is no three component link with $\rank(\widehat{\HFK}(L))=10$.
\end{proposition}



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

\begin{theorem}\label{HFKdetcetstrefoilmeridian}
Let $L_0$ be $T(2,3)$ with a meridian. Suppose $\widehat{\HFK}(L)\cong\widehat{\HFK}(L_0)$. Then $L$ is isotopic to $L_0$.
\end{theorem}
     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~\cite{binns2020knot}.
     





The proofs of Theorems \ref{rank4}, \ref{rank6}, \ref{rank8} and \ref{3componentrank12} 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 \cite{binns2020knot}.



We first prove Lemma~\ref{unlinkdetection}, as it is useful in proving the other theorems.

We will find the following characterization of unlinks useful;

\begin{lemma}[Ni~\cite{ni2006note}]\label{alexandergradingcharachterizationofunlinks}
Suppose $\widehat{\HFK}(L)$ is supported in a single Alexander grading. Then $L$ is an unlink.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{alexandergradingcharachterizationofunlinks}]
Observe that if $\widehat{\HFK}(L)$ is supported in a single Alexander grading then that grading must be zero, by the symmetry of $\widehat{\HFK}(L)$. 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.
\end{proof}

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 $-1$.



\begin{proof}[Proof of Lemma~\ref{unlinkdetection}]
    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 $\rank(\widehat{\HFK}(L))=2^n$ then for any component $L_i$ of $L$, $L-L_i$ is a link with $\rank(\widehat{\HFK}(L-L_i))=2^{n-1}$. By inductive hypothesis $L-L_i$ is the $n$ component unlink. It follows that the support of $\widehat{\HFL}(L)$ is contained in the line defined by $A_j=0$ for $j\neq i$. But $i$ is arbitrary, so $\widehat{\HFL}(L)$ is supported in multi-Alexander grading $\mathbf{0}$, and the result follows from Lemma~\ref{alexandergradingcharachterizationofunlinks}.
\end{proof}



We now prove Theorem~\ref{rank4}.


\begin{proof}[Proof of Theorem~\ref{rank4}]
Let $L$ be an $n$ component link with $\rank(\widehat{\HFK}(L))=4$. Since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$, it follows that $n\leq 3$. Indeed, since $\rank(\widehat{\HFK}(L))$ 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~\ref{unlinkdetection}. It therefore remains only to consider the case that $n=2$. Let $g$ be the maximal Alexander grading of $\widehat{\HFK}(L)$. We consider the following three cases; $\rank(\widehat{\HFK}(L)(g))$ is either four, two or one, in which case the link is fibered by \cite{ni2007knot}.\begin{enumerate}
       
       \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=4$}.} In this case $g=0$, whence $\widehat{\HFK}(L)$ does not admit a spectral sequence to $\widehat{\HF}(S^1\times S^2)$.
          \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=2$}.} In this case we see that 
         $$\widehat{\HFK}(L)\cong ({\mathbb{F}}_{m_1}\oplus{\mathbb{F}}_{m_2})[g]\oplus ({\mathbb{F}}_{m_1-2g}\oplus{\mathbb{F}}_{m_2-2g})[-g].$$
        
        Observe that $g>0$. Consider the spectral sequence from $\widehat{\HFK}(L)$ to $\widehat{\HF}(S^1\times S^2)$. 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.
        
       \item{\textbf{$\rank(\widehat{\HFK}(L)(g))=1$}.} Observe that $g\geq 1$. Let $(x,g-x)$ be the bi-Alexander grading of the contact class in $\widehat{\HFL}(L)$.
       
       If $g>1$, then Theorem~\ref{HFKrank} implies that $\rank(\widehat{\HFK}(L))\geq 6$. Thus $g=1$, and $$\widehat{\HFL}(L)\cong {\mathbb{F}}_{m}(\frac{1}{2},\frac{1}{2})\oplus {\mathbb{F}}_{m-1}(-\frac{1}{2},\frac{1}{2})\oplus {\mathbb{F}}_{m-1}(\frac{1}{2},-\frac{1}{2})\oplus {\mathbb{F}}_{m-2}(-\frac{1}{2},-\frac{1}{2})$$
       Since $\widehat{\HFL}(L)$ 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~\cite{BG}.
    \end{enumerate}
    

\end{proof}

We now prove Theorem~\ref{rank6}, 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.





\begin{lemma}\label{rank6fibered}
There is no fibered link $L$ such that $\rank(\widehat{\HFK}(L))=6$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{rank6fibered}]
Suppose $L$ is an $n$ component fibered link with $\rank(\widehat{\HFK}(L))=6$. Since $\rank(\widehat{\HFL}(L))$ is even, $L$ cannot be a knot. Moreover, $L$ can have at most $3$ components, since $\widehat{\HFL}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$.

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 $\widehat{\HFK}(L)$ has support is at least $2$. By Theorem~\ref{HFKrank}, the next to top Alexander grading is of rank three, whence $\widehat{\HFK}(L)$ 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 $\widehat{\HFK}(L)$ 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 $\rank(\widehat{\HFK}(L))=\rank(\widehat{\HFK}(T(2,\pm 3)\otimes V)$. 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~\ref{chainlevel} 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 $\widehat{\HFL}(L)$, 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 \cite{GageT26}, a contradiction.
          

\end{proof}

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


\begin{proof}[Proof of Theorem~\ref{rank6}]
Let $L$ be an $n$ component link with link Floer homology of rank $6$. Lemma~\ref{rank6fibered} implies that $L$ is not fibered under any orientation. The fact that $2^{n-1}\leq\rank(\widehat{\HFK}(L))$ implies that $L$ has at most three components. Since $\rank(\widehat{\HFK}(L))$ 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 $\widehat{\HFK}(L)$ under some orientation of $L$.


Lemma~\ref{alexandergradingcharachterizationofunlinks}, 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ in the plane $A_i=p_i$ is exactly $p$, it follows that the spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L_i)\otimes V^{n-1}$ is trivial. Thus the rank of $\widehat{\HFL(L)}$ at $p$ is at least $2^{n-1}$. Since the polytope is non-degenerate, the symmetry properties of $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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 $\rank(\widehat{\HFL}(L))\geq 8$, a contradiction. \end{proof}



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

\begin{proposition}\label{rank8fibered}
Suppose $L$ is fibered and $\rank(\widehat{\HFK}(L))=8$. Then $L\cong T(2,4)$ or $T(2,-4)$.
\end{proposition}
\begin{proof}
Let $L$ be a fibered link with link Floer homology of rank eight. Since $\widehat{\HFK}(L)$ admits a spectral sequence to $\widehat{\HF}(\#^{n-1}S^1\times S^2)$, 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 $\rank(\widehat{\HFK}(L))$ is even. Suppose $L$ has four components. Then by Theorem~\ref{unlinkdetection}, $L$ 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 $\widehat{\HFK}(L)$ has supported is at least $2$. By Theorem~\ref{chainlevel}, $\rank(\widehat{\HFK}(L)(g-1))\geq3$. Thus $$\widehat{\HFK}(L)\cong{\mathbb{F}}_m[g]\oplus{\mathbb{F}}_{m-1}^3[g-1]\oplus {\mathbb{F}}_{m+1-2g}^3[1-g]\oplus{\mathbb{F}}_{m-2g}[-g]$$ 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 $\widehat{\HFK}(L)$ is at least one. Suppose $g=1$. Then $\rank(\widehat{\HFK}(L;0))=6$. Such a complex clearly does not admit a spectral sequence to $\widehat{\HF}(S^1\times S^2)$, a contradiction. Thus $g\geq 2$. Consider $\widehat{\HFL}(L)$. Suppose the top Alexander grading generator is in multi-Alexander grading $(x,g-x)$ and of Maslov index $m$. Theorem~\ref{chainlevel} 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 $\widehat{\HFK}(L)$ to $\widehat{\HFK}(L_1)\otimes V$ 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 $\rank(\widehat{\HFK}(T(2,2n))=8$ are $n=\pm2$.

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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_2)$ do not have admissible Maslov gradings. Thus the complex splits as two squares, and again must be located on the diagonal, a contradiction.

\end{proof}



\begin{proof}[Proof of Theorem~\ref{rank8}]


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 $\rank(\widehat{\HFK}(L))\geq 2^{n-1}$ and $L$ is even, $L$ has either two, three or four components. If $L$ has four components, then Lemma~\ref{unlinkdetection} implies that $L$ is an unlink. We deal with the two remaining cases separately;\begin{enumerate}
   
    \item\textbf{$L$ has two components}
Consider $\widehat{\HFL}(L)$. 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 $\widehat{\HFK}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks} 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 $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L_1)$, 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~\ref{rank2face} 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~\ref{rank2face}.


	\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{Rank_8_HFL.pdf}
    \caption{A rank eight bigraded vector space not arising as the link Floer homology of a link.\label{rank2face}}
    \end{figure}


The only remaining case is that on each face the two generators sit in the same multi-Alexander grading, see figure~\ref{rank2face}. 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 $\widehat{\HFL}(L)$ 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 $\partial_1x=y$. Since $y$ does not persist under the spectral sequence to $\widehat{\HFL}(L_2)$, $\partial_2 y=z$. 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.

   
   
   



\item \textbf{$L$ 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 $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L-L_3)$ is just projection onto the $A_3=0$ plane. Suppose $\widehat{\HFL}(L)$ is supported in the $A_3=0$ plane, then $L$ consists of a disjoint union of a Hopf link and an unknot. If $\widehat{\HFL}(L)$ 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 $\widehat{\HFL}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks}.
\end{enumerate}

\end{proof}


\begin{proof}[Proof of Proposition~\ref{3componentrank12}]
Let $L$ be a $3$ component link with $\rank(\widehat{\HFK}(L)=10$. Suppose $L$ is fibered. The maximal Alexander grading, $A$, of $\widehat{\HFK}(L)$ is given by $2+g$ for some $g\geq 0$. Note that $\rank(\widehat{\HFK}(L)(A))=1$. Suppose the generator is of Maslov index $m$. Theorem~\ref{HFKrank} implies that there are at least three generators of Maslov index $m-1$ in Alexander grading $A-1$. The symmetry of $\widehat{\HFK}(L)$ 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 $\widehat{\HFK}(L)$ 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~\ref{alexandergradingcharachterizationofunlinks}. 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 $\rank(\widehat{\HFK}(L))\geq 16$, 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 $\widehat{\HFL}(L)$ on the edge $-A_1-A_2=\dfrac{g_1+g_2}{2}$, persists under the spectral sequence to $\widehat{\HFL}(L-L_3)\otimes V[\frac{\ell\text{k}(L_1,L_3)}{2},\frac{\ell\text{k}(L_2,L_3)}{2}]$, and is indeed of maximal (respectively minimal $A_1+A_2$ grading). Now, there are only two 2 component links with $\rank(\widehat{\HFK}(L))\leq 5$; 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 $i$, $\rank(\widehat{\HFK}(L-L_i)\leq 5$, 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.


\end{proof}

We conclude with a proof of  Theorem~\ref{HFKdetcetstrefoilmeridian}


\begin{proof}[Proof of Proposition~\ref{HFKdetcetstrefoilmeridian}]

Let $L_0$ be the trefoil together with a meridian. It suffices to show that if  $\widehat{\HFK}(L)\cong\widehat{\HFK}(L_0)$ then  $\widehat{\HFL}(L)\cong\widehat{\HFL}(L_0)$ by the corresponding link Floer homology detection result in~\cite{binns2020knot}.

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

Observe that $\widehat{\HFK}(L)$ detects the linking number of two component links, so the linking number is one. Consider $\widehat{\HFL}(L)$. Let $A_i(0)$ be the Alexander grading of the Maslov index $0$ generator in $\widehat{\HFL}(L_i)$. Since there are no positive Maslov index generators in $\widehat{\HFL}(L)$, 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~\ref{chainlevel}, 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 $\widehat{\HFL}(L)$ admit a spectral sequence to $\widehat{\HFL}(L_1)\otimes V$, 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 $\widehat{\HFL}(L_i)\otimes V$, there must be Maslov index $-2$ generators in bi-Alexander  gradings $(x-1,1-x)$ and $(x,-x)$. In order that $\widehat{\HFL}(L)$ be symmetric, we must then have that $x=\frac{1}{2}$. We have thus determined that $\widehat{\HFL}(L)\cong\widehat{\HFL}(L_0)$, as desired.
\end{proof}

\end{section}


\begin{section}{Detection Results for Khovanov Homology}
\label{Detection Results for Khovanov Homology}

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







 



\begin{subsection}{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 \emph{Khovanov chain complex of $L$}, $(\CKh(L,R),\partial)$, is a finitely generated ${\mathbb{Z}}\oplus{\mathbb{Z}}$-filtered chain complex over $R$~\cite{khovanov2000categorification}.

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

Here $i$ is called the \emph{homological grading}, while $j$ is called the \emph{quantum grading}. The filtered chain homotopy type of $L$ is an invariant of $L$. The $R$-module $\CKh(L,R)$ 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 \emph{Khovanov homology of $L$} is obtained by taking the homology of $\CKh(L;R)$. A choice of basepoint $p\inL$ induces an action on $\CKh(L;R)$, which commutes with the differential. Taking the quotient of $\CKh(L;R)$ by this action and then taking homology yields a bigraded $R$-module called the \emph{reduced Khovanov homology} of $L$, denoted $\widetilde{\Kh}(L,p;R)$\cite{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 \emph{pointed Khovanov homology}, $\Kh(L,\mathbf{p};R)$ due to Baldwin-Levine-Sarkar~\cite{baldwin2017khovanov}.

$\Kh(L;R)$ 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 $\Kh(L;R)$ admits a spectral sequence to $\Kh(L_1;R)\otimes\Kh(L_2;R)$, called the \emph{Batson-Seed spectral sequence}~\cite{batson2015link}. Indeed this spectral sequences respects the $i-j$ grading on $\Kh(L;R)$ in the sense that;
\begin{align}\label{GradedBatsonSeed}
    \rank^{i-j=l}(\Kh(L;R)\geq \rank^{i-j=l+2\ell\text{k}(L_1,L_2)}(\Kh(L_1;R)\otimes\Kh(L_2;R))
\end{align}

Suppose now that $L$ is an $n$ component link. There is another spectral sequence, called the \emph{Lee spectral sequence}, from $\Kh(L;{\mathbb{Q}})$ which abuts to an invariant called the \emph{Lee Homology} of $L$, $\LLL(L):=\bigoplus_i\LLL^i(L)$~\cite{lee2005endomorphism}. This spectral sequence respects the $i$ gradings in the sense that $\rank(\Kh^i(L;{\mathbb{Q}}))\geq \rank(\LLL^i(L))$. Lee showed that $\LLL(L)\cong\bigoplus_{i=1}^n{\mathbb{Q}}^2_{a_i}$ where $a_i$ are integers. Indeed if $L$ has two components $L_1,L_2$ then $\LLL(L)\cong{\mathbb{Q}}^2_{0}\oplus{\mathbb{Q}}^2_{\ell\text{k}(L_1,L_2)}$~\cite[Proposition 4.3]{lee2005endomorphism}. Moreover, there is an invariant, $s(L)$ -- due to Rasmussen~\cite{Rasmussen} in the knot case and Beliakova-Wehrli in the link case~\cite{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~\cite{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 $\Kh(L,\mathbf{p};{\mathbb{Q}})$ to $\widehat{\HFK}(L;{\mathbb{Q}})$ that respects the relative $\delta$-gradings. As a corollary, Dowlin shows that if $L$ has $n$ components then; \begin{align}
    2^{n-1}\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\geq \rank(\widehat{\HFK}(L;{\mathbb{Q}}))\label{DowlinReduced}\end{align}



\end{subsection}

\begin{subsection}{Rank Detection Results for Khovanov homology}

Applying Dowlin's spectral sequence and the rank detection results for knot Floer homology from Section~\ref{Rank Detection Results for Knot Floer homology}, we have the following result;
\begin{corollary}\label{KHr2components}
Suppose $L$ is a two component pointed link with $\rank(\widetilde{\Kh}(L,p;{\mathbb{Z}}/2))\leq 4$. Then $L$ is one of;\begin{itemize}
    \item an unlink;
\item a Hopf link;
\item $T(2,4)$ or $T(2,-4)$.
\end{itemize}
\end{corollary}

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

\begin{corollary}\label{KHr3components}
Suppose $L$ is a three component pointed link. Then $\rank(\widetilde{\Kh}(L,p;{\mathbb{Q}}))> 2$.
\end{corollary}

\begin{proof}[Proof of Corollary~\ref{KHr2components}]
Suppose $L$ is a two component link with $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 4$. Equation~\ref{DowlinReduced} implies that $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))\leq 8$, and the result follows from Theorems~\ref{rank4}, \ref{rank6} and \ref{rank8}.
\end{proof}



\begin{proof}[Proof of Corollary~\ref{KHr3components}]
Suppose $L$ is a three component link with $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 2$. Equation~\ref{DowlinReduced} implies that $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))\leq 8$. Theorems~\ref{rank4}, \ref{rank6} and \ref{rank8} 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.
\end{proof}




\end{subsection}

\begin{subsection}{Khovanov Homology Detects $T(2,8)$, $T(2,10)$}
We prove the following theorems;



	\begin{theorem}\label{KHTT28}
		Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,8);{\mathbb{Z}})$. Then $L$ is isotopic to $T(2,8)$
	\end{theorem}
	
\begin{theorem}\label{KHT210}
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then $L$ is isotopic to $ T(2,10)$.
\end{theorem}


$$\begin{array}{c|c c c c c c c c c}
\label{Table 1}
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
&0&1&2&3&4&5&6&7&8


\end{array}

\caption{Table 1. $\Kh(T(2,8);{\mathbb{Z}})$: The horizontal axis indicates the $i$ grading, while the vertical axis denotes the $j$ grading}




$$

$$\begin{array}{c|c c c c c c c c c c c}
\label{Table 2}
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
&0&1&2&3&4&5&6&7&8&9&10


\end{array}

\caption{Table 2. $\Kh(T(2,10);{\mathbb{Z}})$: The horizontal axis indicates the $i$ grading, while the vertical axis denotes the $j$ grading}

$$


    



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


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

To prove Theorems~\ref{KHTT28} and~\ref{KHT210} we first show that any link $L$ with $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,2n);{\mathbb{Z}})$ 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.

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

We prove this Lemma using Lee's spectral sequence~\cite{lee2005endomorphism}, which has $\Kh(L;{\mathbb{Q}})$ as $E_1$ page and $E_\infty$ page a complex of rank $2^{n}$ where $n$ is the number of components of $L$.

\begin{proof}[Proof of Lemma~\ref{KHcomponentsgenuslinking}]
Suppose $L$ is as in the Theorem. By the universal coefficient theorem there are exactly two homological gradings in which the rank of $\Kh^i(L;{\mathbb{Q}})$ is at least two, namely $i=0$ and $i=2n$. Since $\rank(\Kh^i(L;{\mathbb{Q}}))\geq\rank(\LLL^i(L;{\mathbb{Q}}))$, and $\LLL^i(L;{\mathbb{Q}})\cong\bigoplus_{i=1}^m{\mathbb{Q}}^2_{a_i}$, 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 $\Kh(L;{\mathbb{Z}})$ is supported in even quantum gradings~\cite[Proposition 24]{khovanov2000categorification}.

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 ~\cite[Proposition 4.3]{lee2005endomorphism}, 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 $\chi(L)\geq 2-2n$.
\end{proof}

We can now prove the following two lemmas;

\begin{lemma}\label{KHt228components}
		Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,8);{\mathbb{Z}})$. Then $L$ has unknotted components.
	\end{lemma}
	
	
\begin{lemma}\label{KHT2102components}
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then $L$ has unknotted components.
\end{lemma}

Both of these use the Batson-Seed link splitting spectral sequence~\cite{batson2015link} to bound the rank of the components of $L$ in certain combinations of $i$ and $j$ gradings. We use Corollary 3.4 of~\cite{batson2015link} which states that if $L$ is a link consisting of two components $L_1,L_2$ then
\begin{align}\label{GradedBatsonSeed}
    \rank^l(\Kh(L;F)\geq \rank^{l+2\ell\text{k}(L_1,L_2)}(\Kh(L_1;F)\otimes\Kh(L_2;F))
\end{align}
 
 Here $F$ is a field -- we will use both ${\mathbb{Z}}/2$ and ${\mathbb{Q}}$ accordingly,  and $\rank^l$ denotes the $i-j=l$ grading part of the relevant vector space.
 
 For Lemma~\ref{KHt228components} we use the fact that the only knots $K$ with $\rank(\Kh(K;{\mathbb{Z}}/2))\leq 8$, are the trefoils and unknot. This follows from Kronheimer-Mrowka's unknot detection result \cite{kronheimer2011khovanov}, as  well as Baldwin-Sivek's trefoil detection results \cite{baldwin2018khovanov} and the fact that for knots $K$ $\rank_{{\mathbb{Z}}/2}(\Kh(K;{\mathbb{Z}}/2))$ is of the form $2+4n$ for some $n\geq 0$.
 
 \begin{proof}[Proof of Lemma~\ref{KHt228components}]
Suppose $L$ is as in the hypothesis of the Lemma. Lemma~\ref{KHcomponentsgenuslinking} implies that $L$ consists of two components, $L_1$ and $L_2$ of linking number $4$. From $\Kh(L;{\mathbb{Z}})$, as shown in Table 1, we deduce that $\rank_{{\mathbb{Z}}/2}(\Kh(L;{\mathbb{Z}}/2))=16$ by the universal coefficient theorem, so the Batson-Seed spectral sequence implies that $\rank(\Kh(L_i;{\mathbb{Z}}/2))\leq 8$, whence each component of $L$ is an unknot or a trefoil~\cite{baldwin2018khovanov}. Indeed, we readily see that at least one component is unknotted.
	
	
	We now consider $\Kh(L;{\mathbb{Q}})$, which we can determine from $\Kh(L;{\mathbb{Z}})$ via the universal coefficient theorem. Suppose $L$ has a right hand trefoil component. Then \\
	$\rank_{i-j=-4}(\Kh(T(2,3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})))=2$, but $\rank_{i-j=-12}(\Kh(L;{\mathbb{Q}}))=1$ contradicting equation~\ref{GradedBatsonSeed}. Similarly, $\rank_{i-j=2}(\Kh(T(2,-3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}}))=3$ which contradicts $\rank_{i-j=-6}(\Kh(L;{\mathbb{Q}}))=1$ by Equation~\ref{GradedBatsonSeed}. Thus both components of $L$ are unknotted.
\end{proof}

Lemma~\ref{KHT2102components} requires some extra analysis since there is no complete classification of knots with $\rank(\Kh(L;{\mathbb{Z}}/2))=10$. To deal with this latter case we  use Dowlin's spectral sequence and Proposition~\ref{E2collapsed}.


\begin{proof}[Proof of Lemma~\ref{KHT2102components}]
	Suppose $\Kh(L;{\mathbb{Z}})\cong\Kh(T(2,10);{\mathbb{Z}})$. Then from Table 2 and the universal coefficient theorem we can see that $\rank(\Kh(T(2,10);{\mathbb{Z}}/2))= 20$. Batson-Seed's spectral sequence implies that at least one component of $L$ is unknotted, and the other is a knot $K$ with $\rank(\Kh(K;{\mathbb{Z}}/2))\leq10$. If $\rank(\Kh(K;{\mathbb{Z}}/2))\leq10$ then $\rank(\widetilde{\Kh}(L;{\mathbb{Z}}/2))\leq 5$, so that $K$ is an unknot, trefoil, figure eight or a knot with $\rank(\widetilde{\Kh}(L;{\mathbb{Z}}/2))=5$. 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;
\begin{enumerate}
	\item $L_2$ is $T(2,3)$, in which case; \begin{align*}\Kh(T(2,3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})&\cong({\mathbb{Q}}_{-6}\oplus{\mathbb{Q}}_{-3}^2\oplus{\mathbb{Q}}_{-1})\otimes({\mathbb{Q}}_{-1}\oplus{\mathbb{Q}}_1)\\&\cong{\mathbb{Q}}_{-7}\oplus{\mathbb{Q}}_{-5}\oplus{\mathbb{Q}}_{-4}^2\oplus{\mathbb{Q}}_{-2}^3\oplus{\mathbb{Q}}_0\end{align*}
		But $\rank(\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction. 
	
	\item $L_2$ is $T(2,-3)$. Then \begin{align*}\Kh(T(2,-3);{\mathbb{Q}})\otimes\Kh(U;{\mathbb{Q}})&\cong({\mathbb{Q}}_{6}\oplus{\mathbb{Q}}_{3}^2\oplus{\mathbb{Q}}_{1})\otimes({\mathbb{Q}}_{1}\oplus{\mathbb{Q}}_{-1})\\&\cong {\mathbb{Q}}_{7}\oplus{\mathbb{Q}}_{5}\oplus{\mathbb{Q}}_{4}^2\oplus{\mathbb{Q}}_{2}^3\oplus{\mathbb{Q}}_0\end{align*}
	
	But $\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction.
	
	
	
	\item $\rank(\widetilde{\Kh}(L_2;{\mathbb{Z}}/2))=5$. In this case we have that $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 5$, whence $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}}))\leq 5$. If $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}})<5$, then $L$ is a trefoil or the unknot, cases which we have already dealt with. Thus we may take $\rank(\widehat{\HFK})(L_2;{\mathbb{Q}})=5$. We show that $L_2$ is a figure eight or has the same knot Floer homology as $T(2,\pm5)$. Suppose $L_2$ is of genus $g>0$. Note that $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}},g)\leq 2$ We proceed by cases.\begin{enumerate}
	    \item Suppose $\rank(\widehat{\HFK}(L;{\mathbb{Q}},g))= 2$. Then $\widehat{\CFL}(L;{\mathbb{Q}})$ contains a differential of even length $2g$. From Proposition~\ref{E2collapsed}, we see that this yields a  summand of $(\widehat{\CFL}(L;Q),\partial)$ rank $8g$. Thus $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 8g+6$. It immediately follows that $g=1$. The symmetry of $\widehat{\HFL}(L)$ implies that $\rank(\widehat{\HFL}(L))\geq 16+12$, a contradiction.
	    
	    \item Suppose $\rank(\widehat{\HFK}(L_2;{\mathbb{Q}},g)= 1$. Consider $\widehat{\CFL}(L_2;{\mathbb{Q}})$. If $g=1$, then $L_2$ is the figure eight knot. If $g>1$ there are two possible forms for $\widehat{\HFK}(L_1)$, as shown in Figure~\ref{rank5hfk}. To see this we need to use the fact that if $\rank(\widehat{\HFK}(L_2,g-1))=1$, 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~\cite{NiPersonalCommunication}.
	    
	    
					\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{HFKL2.pdf}
    \caption{Different possibilities for $\widehat{\CFK}(L_2;{\mathbb{Q}})$.\label{rank5hfk}}
    \end{figure}
    
    In either case, since there is both a length one differential and a length $g-1$ differential, $\rank(\widehat{\HFL}(L_2;{\mathbb{Q}}))\geq 4(g-1)+4+4=4g+4$, 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 $\rank(\widehat{\HFL}(L_2;{\mathbb{Q}}))\geq 8(g-1)+4+4+4$, so that $g=2$, and $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong \widehat{\HFK}(T(2,5);{\mathbb{Q}})$.
	    
	    Suppose we are in the second case. We have that $g\in\{2,3,4\}$. If $g=2$ then $\widehat{\HFK}(L_2)\cong\widehat{\HFK}(T(2,-5))$. If $g=4$ using the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ we find $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 24$, a contradiction.
	    
	    If $g=3$, symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ implies that the $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\geq 24$, a contradiction.
	\end{enumerate}
	
	
	\item $L_2$ is the figure eight. In this case;
	\begin{align*}\Kh(U;{\mathbb{Q}})\otimes\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})\end{align*}
	
	But $\Kh(T(2,10);{\mathbb{Q}})$, is rank one in exactly one $i-j$ grading, a contradiction.
	
	
	\item $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong\widehat{\HFK}(T(2,5);{\mathbb{Q}})$, or $\widehat{\HFK}(L_2;{\mathbb{Q}})\cong\widehat{\HFK}(T(2,-5);{\mathbb{Q}})$. Since $\ell\text{k}(L_1,L_2)=\pm 5$, there is a summand $\widehat{\HFK}(T(2,\pm5);{\mathbb{Q}})\otimes V[\pm\frac{5}{2},a]$ of $\widehat{\HFL}(L;{\mathbb{Q}})$, for some $a\in\pm\frac{5}{2}+{\mathbb{Z}}$. The symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ yields another $10$ generators. There can be no additional generators, as this would violate the rank inequality from Dowlin's spectral sequence - $\rank(\widehat{\HFL}(L);{\mathbb{Q}})\leq\rank(\Kh(T(2,10);{\mathbb{Q}}))=20$, 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 $\widehat{\HFL}(L;{\mathbb{Q}})$ is supported only in $A_1$ gradings $\pm\frac{1}{2}$. However since $L_1$ is unknotted and there is a spectral sequence from $\widehat{\HFL}(L)$ to $\widehat{\HFL}(L_1)\otimes V[\frac{5}{2}]$ there must be generators in $A_1$ grading $\frac{5}{2}$, a contradiction.
\end{enumerate}
\end{proof}


	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~\ref{E2collapsed}. From here it suffices to prove the following pair of propositions.
	
	\begin{proposition}\label{combinedHFKKHT28detection}
	Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link satisfying the following conditions;
	\begin{enumerate}
	\item $L$ has two exactly two components, $L_1$ and $L_2$, which are both unknots.
	\item $\ell\text{k}(L_1,L_2)=4$
	\item $\chi(L)\leq-6$
	    \item $\rank(\widehat{\HFK}(L;F))\leq 16$.
	    \item $\widehat{\HFK}(L;F)$ is $\delta$-thin.
	\end{enumerate}
	
	Then $L$ is $T(2,8)$.
	\end{proposition}
	
		\begin{proposition}\label{combinedHFKKHT210detection}
	Let $F$ be ${\mathbb{Q}}$ or ${\mathbb{Z}}/2$. Suppose $L$ is a link satisfying the following conditions;
	\begin{enumerate}
	\item $L$ has two exactly two components, $L_1$ and $L_2$, which are both unknots.
	\item $\ell\text{k}(L_1,L_2)=5$
	\item $\chi(L)\leq-8$
	    \item $\rank(\widehat{\HFK}(L;F))\leq 20$.
	    \item $\widehat{\HFK}(L;F)$ is $\delta$-thin.
	\end{enumerate}
	
	Then $L$ is $T(2,10)$.
	\end{proposition}
	
Before proving these propositions, we apply them to obtain Theorem's~\ref{KHTT28} and~\ref{KHT210} respectively.
	

	\begin{proof}[Proof of Theorem~\ref{KHTT28}]
	It suffices to show that $L$ satisfies the conditions in Proposition~\ref{combinedHFKKHT28detection}. Conditions 1,2 and 3 are implied by Lemmas~\ref{KHcomponentsgenuslinking} and~\ref{KHt228components}. To see that conditions $4$ holds, observe that $\rank(\Kh(L;{\mathbb{Z}}/2)=16$ by the universal coefficient theorem. It follows that $\rank(\widetilde{\Kh}(L;{\mathbb{Z}}/2))=8$ by~\cite{shumakovitch2014torsion}, so that $\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))\leq 8$ by the universal coefficient theorem. $\rank(\Kh(L,p,{\mathbb{Q}}))\leq\rank(\widetilde{\Kh}(L;{\mathbb{Q}}))$ by \cite[Theorem 1.2]{baldwin2017khovanov}, whence $\widehat{\HFK}(L;{\mathbb{Q}})\leq 16$.

	
	To see that condition 5 holds, note that that $\Kh(L;{\mathbb{Q}})$ is $\delta$-thin, so that $\Kh(L,p;{\mathbb{Q}})$ is $\delta$-thin by \cite[Lemma 2.11]{baldwin2017khovanov}.
	\end{proof}
	
	\begin{proof}[Proof of Theorem~\ref{KHT210}]
	    This follows by the same argument as in the proof of Theorem~\ref{KHTT28}.
	\end{proof}
	
We now prove the two requisite lemmas~\ref{combinedHFKKHT28detection}, and ~\ref{combinedHFKKHT210detection}






		\begin{proof}[Proof of proposition~\ref{combinedHFKKHT28detection}]
	Suppose $L$ satisfies the conditions given in the Proposition. Consider $\widehat{\CFL}(L;{\mathbb{Q}})$. Since $\widehat{\HFK}(L)$ is $\delta$-thin, $\widehat{\CFL}(L;{\mathbb{Q}})$ is an $E_2$ collapsed chain complex. Likewise since $\chi(L)\leq -6$ the maximal Alexander grading of $\widehat{\HFK}(L;{\mathbb{Q}})$ is at least $3$. 
	
	Consider $\widehat{\CFL}(L)$. Since $L$ has unknotted components, it follows that $(\widehat{\CFL}(L;{\mathbb{Q}}),\partial)$ 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 $(\widehat{\CFL}(L),\partial)$ has summands $X_{-1}^{l-4}[2,2]\oplus X_0^{l-5}[1,1]$. From the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$, and the fact that $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\leq 16$ we see immediately that $l\leq 6$.
			
		
			
			
			
				
					\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{Pseduo98.pdf}
    \caption{One potential link Floer homology complex for a link satisfying the conditions in Propositions~\ref{combinedHFKKHT28detection}}
    \end{figure}
				
					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 $\rank(\widehat{\HFK}(L))>16$. Similarly if $l\in\{1,2\}$ would violate the rank bound after accounting for the symmetry of $\widehat{\HFL}(L)$. If $l=0$ then $\rank(\widehat{\HFL}(L))\geq 16$. 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 $\widehat{\HFL}(L;{\mathbb{Q}})$. 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~\ref{Braidpolytope}, or that $\chi(L)>-6$.
				
				If $l=5$ we readily see that -- apart from in the case that $\widehat{\CFL}(L;{\mathbb{Q}})$ 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 $\widehat{\CFL}(L)$ 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~\ref{Braidpolytope} implies that $\widehat{\HFL}(L;{\mathbb{Q}})\cong\widehat{\HFL}(T(2,8);{\mathbb{Q}})$, whence the result follows from the corresponding detection result in~\cite{binns2020knot}.
			
				\end{proof}
	

The proof of~\ref{combinedHFKKHT210detection} is virtually the same, but with some additional case analysis.

\begin{proof}[Proof of Proposition~\ref{combinedHFKKHT210detection}]
	Suppose Suppose $L$ is as in the hypotheses of the Proposition. Since $L$ is delta-thin, $\widehat{\CFL}(T(2,2n);{\mathbb{Q}})$ is $E_2$-collapsed. Since $\chi(L)\leq -8$ the maximal Alexander grading is at least $n$. Since the components of $L$ are unknotted $\widehat{\CFL}(L;{\mathbb{Q}})$ 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 $\widehat{\CFL}(L;{\mathbb{Q}})$ has summands $X_0^{l-5}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{l-4}[2,2]$. Observe that the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ together with the fact that $\rank(\widehat{\HFL}(L;{\mathbb{Q}}))\leq 20$, imply that $l\leq 7$. 
	\begin{figure}[h]
    \centering
    \includegraphics[width=7cm]{KhT(2_10).pdf}
    \caption{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.}
    \end{figure}
	
	If $l=7$ then either $\widehat{\CFL}(L)\cong X_0^{2}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{3}[2,2]\oplus B_{-16}[\frac{-9}{2},\frac{-7}{2}]\oplus B_{-16}[\frac{-7}{2},\frac{-9}{2}]$ or there is an additional $B_{-8}[\frac{-1}{2},\frac{-1}{2}]$ summand, by the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$. 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~\cite{GageT26}, a contradiction.
	
	If $l=6$, then either $\widehat{\CFL}(L)\cong X_0^{1}[\dfrac{5}{2},\dfrac{5}{2}]\oplus X_{-1}^{2}[2,2]\oplus B_{-11}[\frac{-7}{2},\frac{-7}{2}]$ 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 $\widehat{\HFL}(L;{\mathbb{Q}})$ 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 $T(2,10)$, a contradiction since $T(2,10)$ has distinct link Floer homology. The $C_{-7}[\frac{5}{2},\frac{-7}{2}]\oplus C_{-7}[\frac{-7}{2},\frac{5}{2}]$ case is excluded by computing the Conway polynomial. First note that the Alexander polynomial is given by $(t^{-1/2}-t^{1/2})(t^{-5}+t^5-2t^{-4}-2t^4+t^{-3}+t^3+2t^{-1}+2t-4)$. Thus $\nabla(t-t^{-1})=-t^{11}+t^{-11}+3t^{9}-3t^{-9}-3t^{7}+3t^{-7}+t^{5}-t^{-5}-2t^{3}+2t^{3}+6t^{1}-6t^{-1}$. Therefore $\nabla(u)=-u^11-8u^9-20u^7-16u^5-4u^3$. It follows that the linking number is four, a contradiction. If there are three cubes then we note that in order to satisfy the symmetry properties there must be a summand $B_{-7}[\frac{-1}{2},\frac{-1}{2}]$. Indeed unless the reaming components are $B_{-1}[\frac{5}{2},\frac{5}{2}]\oplus B_{-13}[\frac{-7}{2},\frac{-7}{2}]$, or $B_{-8}[-\frac{5}{2},\frac{7}{2}]$ $B_{-8}[\frac{3}{2},\frac{-5}{2}]$. In the first case, we see that $L$ bounds an annulus upon reversing its orientation, a contradiction, since no link $T(2,2n)$ has the requisite homology. In the second case we see that $\Delta_L(t)=(t^{-1/2}-t^{1/2})(-t^7-t^{-7}+2t^6+2t^{-6}-t^5-t^{-5}+2t-4+2t^{-1})$ Thus $\nabla_L(t-t^{-1})=t^{15}-t^{-15}-3t^{13}+3t^{-13}+3t^{11}-3t^{-11}-t^9+t^{-9}-2t^3+2t^{-3}-2t^3+2t^{-3}-2t+2t^{-1}$
	Therefore $\nabla_L(u)=-u^{15}+12u^{13}+54u^{11}+112u^9+105u^7+36u^5-8u$ and the linking number is $-8$, a contradiction.
	
	Similarly, if $l\leq 5$ then there are summands $Y_0^{5-l}[3-l,3-l]\oplus Y_{-1}^{4-l}[2-l,2-l]$. Considering the symmetry properties of $\widehat{\HFL}(L)$ readily see that either $l=0$ or $l\geq 4$, as else $\rank(\widehat{\HFK}(L;{\mathbb{Q}}))>20$. If $l=0$ then there are no additional generators, and we see that $L$ bounds an annulus, and hence is $T(2,10)$. This is a contradiction since $T(2,10)$ has distinct link Floer homology. If $l=4$ then $\widehat{\HFL}(L;{\mathbb{Q}})$ either contains no extra generators or an cube $C_{-6}[\dfrac{-1}{2},\dfrac{-1}{2}]$ by symmetry. We see that after reversing the orientation of a component of $L$, it bounds an annulus, whence $L$ is $T(2,10)$, a contradiction since $T(2,10)$ has distinct knot Floer homology.
	
	
We are thus left only with the case that $l=5$. In which case $\widehat{\CFL}(L;{\mathbb{Q}})$ has a summand $X_0[\frac{5}{2},\frac{5}{2}]\oplus Y_{-1}[\frac{3}{2},\frac{3}{2}]$. By symmetry $\widehat{\CFL}(L;{\mathbb{Q}})$ has a summand $B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ $\widehat{\CFL}(L;{\mathbb{Q}})$ can have at most three other $B_d$ summands. If there is one or three more cubes then the symmetry of $\widehat{\HFL}(L;{\mathbb{Q}})$ imply that one such summand must be of the form $B_{-6}[\frac{-1}{2},\frac{-1}{2}]$. Thus if $\widehat{\HFL}(L;{\mathbb{Q}})$ has $0$ or $1$ extra $B_d$ summands we see that after reversing the orientation of a component of $L$, $L$ bounds an annulus, and is therefore $T(2,10)$, a contradiction since $T(2,10)$ has distinct link Floer homology.

If $L$ has two extra $B_d$ summands then unless they are of the form $B_{-6}[\frac{3}{2},\frac{-5}{2}]\oplus[\frac{-5}{2},\frac{5}{2}]$, or $B_{-2}[\frac{3}{2},\frac{3}{2}]\oplus B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ then at least one component of $L$ is a braid axis and the result follows from Lemma~\ref{Braidpolytope} and the corresponding detection result for link Floer homology in~\cite{binns2020knot}. The first case is dealt with by computing the Conway polynomial from $\widehat{\HFL}(L;{\mathbb{Q}})$. Explicitly, $\Delta_L(t)=(t^{-1/2}-t^{1/2})(t^5+t^{-5}-2t^{4}-2t^{-4}+t^{3}+t^{-3}+2t-4+2t^{-1})$, whence $\nabla(t-t^{-1})=-t^11+t^{-11}+3t^9-3t^{-9}-3t^7+3t^{-7}+t^5-t^{-5}-2t^3+2t^{-3}+6t-6t$, and $\nabla_L(u)=-u^{11}-8u^9-20u^7-16u^5-5u^3$, whence the linking number is $-5$, a contradiction. In the second case we see that after reversing the orientation of a component, $L$ bounds an annulus, and is therefore $T(2,10)$, a contradiction since this link has distinct link Floer homology.

If $L$ has three extra $B_d$ summands, the by the symmetry of $\widehat{\HFL}(L)$, there must be a $B_{-6}[\frac{-1}{2},\frac{-1}{2}]$ summand. Unless the remaining summands are of the form $B_{-6}[\frac{3}{2},\frac{-5}{2}]\oplus[\frac{-5}{2},\frac{5}{2}]$, or $B_{-2}[\frac{3}{2},\frac{3}{2}]\oplus B_{-10}[\frac{-5}{2},\frac{-5}{2}]$ then at least one component of $L$ is a braid axis and the result follows from Lemma~\ref{Braidpolytope} and the corresponding detection result for Link Floer homology~\cite{binns2020knot}.
For the $B_{-6}(\frac{3}{2},\frac{-5}{2})\oplusB_{-6}(\frac{-5}{2},\frac{5}{2})$ case we see that $\Delta_L(t)=(t^{-1/2}-t^{1/2})(t^5+t^{-5}-2t^{4}-2t^{-4}+t^{3}-t^{-3}+3t-6+3t^{-1})$, whence $\nabla_L(t-t^{-1})=-t^{11}+t^{-11}+3t^9-3t^{-9}-3t^7+3t^{-7}+t^5-t^{-5}-3t^3+3t^{-3}+9t-9t^{-1}$ and $\nabla_L(u)=-u^{11}-8u^{9}-20u^7-16u^5-5u^3$, whence the linking number is $-5$, a contradiction. For the $B_{-2}(\frac{3}{2},\frac{3}{2})\oplus B_{-10}(\frac{-5}{2},\frac{-5}{2})$ case we see that $L$ bounds an annulus upon reversing the orientation of a component of $L$ and the desired result follows.


Thus at least one component is a braid axis, and Lemma~\ref{Braidpolytope} implies that $\widehat{\HFL}(L;{\mathbb{Q}})\cong\widehat{\HFL}(T(2,10);{\mathbb{Q}})$ and the result follows from the corresponding detection result for link Floer homology.
\end{proof}


We conclude by noting that Martin's detection result for $T(2,6)$~\cite{GageT26}, $T(2,4)$ detection~\cite{xie2020links} and $T(2,2)$ detection~\cite{baldwin2017khovanov} could have been obtained by similar methods to those presented in this paper.


\end{subsection}
\end{section}




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