Rank Bounds in Link Floer Homology and Detection Results

Viewing the BRAID invariant as a generator of link Floer homology we generalise work of Baldwin-Vela-Vick to obtain rank bounds on the next to top grading of knot Floer homology. These allow us to classify links with knot Floer homology of rank at most eight, and prove a variant of a classification of links with Khovanov homology of low rank due to Xie-Zhang. In another direction we use a variant of Ozsv\'ath-Szab\'o classification of $E_2$ collapsed $\mathbb{Z} \oplus\mathbb{Z}$ filtered chain complexes to show that knot Floer homology detects $T(2,8)$ and $T(2,10)$. Combining these techniques with the spectral sequences of Batson-Seed, Dowlin, and Lee we can show that Khovanov homology likewise detects $T(2,8)$ and $T(2,10)$.


Introduction
Link Floer homology is a powerful invariant of links in S 3 due to Ozsváth-Szabó, taking value in the category of multi-graded vector spaces [35].While powerful, it is not a complete invariant; there exist non-isotopic links with isomorphic link Floer homology [20].One of the main goals of this paper is to exhibit links which are fully determined -"detected" -by their link Floer homology.More generally we address several variants of the "geography" question for link Floer homology -namely which vector spaces arise as the link Floer homology of some link -as well as the "botany" question -which links have a prescribed link Floer homology.
One of our approaches to the geography problem goes through a result that uses a contact geometric invariant which lives in Heegaard Floer homology.Namely we study the BRAID invariant, an invariant of transverse links due to Baldwin-Vela-Vick-Vértesi [8] which is a version of Ozsváth-Szabó's contact class [37].In particular we generalize the rank bound from Baldwin-Vela-Vick [7,Theorem 1.1].The most concise version of our result is the following: Theorem 4.1.Let L be a non-trivial n component fibered link in a rational homology sphere.Then rank( HFK(L; )) ≥ n.
Here χ(L) denotes the maximal Euler characteristic of a potentially disconnected surface bounding L. We may take coefficients to be Z, Z 2 -the field with two elements -or Q, as we do for the remainder of this paper unless otherwise specified.Some care is required when working with Z or Q, see sections 2 and 3 for details.Note that n − 2 − χ(L) 2 is the next to top Alexander grading of knot Floer homology.See Lemma 4.3 and Corollary 4.2 for stronger versions of Theorem 4.1 under appropriate additional hypotheses.
Theorem 4.1 allows us to prove a number of detection results: of CFL − (L) is an invariant of L, which we refer to as the 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 CFL(L), and has homology denoted by HFL(L), which we will call the link Floer homology of L. The link Floer homology polytope of L is defined as the convex hull of the multi-Alexander gradings of HFL(L) with non-trivial support.The link Floer polytope determines the Thurston polytope of the link exterior by a result of Ozsváth-Szabó [36].Note that HFL(L) is the dual -in the sense of [2, Equation 3] -of HFL(L), where L is the mirror of L.
For each component L i of L there is a spectral sequence from HFL(L) to HFL(L ].Here V ∼ = F 0 ⊕ F −1 , supported in multi-Alexander grading zero, while [ ] indicates a shift in the A j grading, corresponding to L j which is a component in L other than L i .Indeed, HFL(L) can be viewed as the graded part of a multi-filtered chain complex with total homology HF(M ) ⊗ V n−1 , where HF(M ) is the Heegaard Floer homology of M as defined by Ozsváth-Szabó [34].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 HFL(L) is of rank 2 n−1 in the maximal A i grading of non-trivial support by a result of Martin [28].
The knot Floer homology of L, HFK(L), is an oriented link invariant due independently to J. Rasmussen [39] and Ozsváth-Szabó [33] that can be obtained from HFL(L) by projecting the multi-Alexander grading onto the diagonal, and shifting the Maslov gradings up by n−1 2 .For an n component link L, HFK(L) can also be thought as the knot Floer homology of the knotified link κ(L) ⊂ (# n−1 (S 2 × S 1 ))#M [34].Since the knot Floer chain complex is a filtered version of Heegaard Floer chain complex, we have that dim( HF K(L)) ≥ dim( HF(M (# n−1 (S 2 × S 1 ))) = 2 n−1 dim( HF(M )).It likewise follows that dim( HFK(L)) ≡ dim( HF(M )) • 2 n−1 mod 2, so that dim( HF K(L)) is odd only if L has a single component.Also note that HFK(L) determines the maximal Euler characteristic of a surface bounding L [30], as well as whether or not L fibered [16], [31].
Versions of HFL(L, M ) and HFK(L, M ) can also be defined with Z or Q coefficients [41].In fact there are 2 n−1 versions of each theory, each corresponding to a coherent system of orientations on the moduli space of pseudo-holomorphic disks.We will only have to be careful with which version of these theories we are using when applying work of Dowlin [12] which gives a spectral sequence from Khovanov homology to Knot Floer homology endowed with the orientation given in [1].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 [30], as well as whether or not L is fibered.This follows from work of Juhász [22].
The Conway polynomial of L can be obtained as an appropriate decategorification of HFL(L).A result of Hoste [21] 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.

The BRAID Invariant
In this section we review the BRAID invariant, an invariant of (generalized) braids due to Baldwin-Vela-Vick-Vértesi [8].We find two different perspectives on braids to be useful.First we define a braid in an open book (Σ, ϕ) to be a link which intersects every page transversely, up to isotopy through families of such links.Any such braid can be recovered from the data of a pointed open book (Σ, ϕ, {p i }), where ϕ : Σ → Σ fixes the points {p i } ⊆ Σ as a set.We assume henceforth that (Σ, ϕ) is an open book decomposition for some rational homology sphere.
The BRAID invariant of a braid L represented by a pointed open book (Σ, ϕ, {p i }) is defined concretely from a Heegaard diagram determined by the pointed book.We recall the construction here.Take a basis of arcs {a j } for Σ − {p i }.Here a basis of arcs is a maximal collection of properly embedded arcs in Σ − {p i } that are homologically independent in H 1 (Σ − {p i }, ∂Σ).An example of such is shown in red in the lower half of the surface shown in Figure 1.Consider the basis of arcs b j obtained by pushing the ends of the a j curves around ∂Σ in the direction dictated by the orientation of ∂Σ and isotoping so that a j ∩ b j consists of a single point, c j .In Figure 1 the points c j are represented by green dots.Form a multi-pointed Heegaard diagram consisting of Σ ∪ ∂Σ (−Σ) with α curves consisting of the union of the a arcs in Σ and −Σ, and β curves consisting of the union of the b arcs in Σ and ϕ(b) in −Σ.In Figure 1 we have not shown ϕ(b j ) for all j to avoid clutter.The w i basepoints consist of {p i } ⊂ −Σ while the z i base points consist of {p i } ⊂ Σ.In Figure 1 the two black dots in the upper half of the diagram are w i basepoints, while the two black dots in the lower half of the diagram represent z i basepoints.We call diagrams constructed in this manner Heegaard diagrams adapted to (Σ, ϕ, {p i }), {a j }.The BRAID invariant of L is defined as the homology class of {c i }.This can be viewed as a class in either CFL(−M, L; Z 2 ), or CFL − (−M, L; Z 2 ).We denote these classes by t(L) or t(L) respectively.Baldwin-Vela-Vick-Vértesi show that t(L) and t(L) are braid invariants, when viewed as elements of HFK(−M, L; Z 2 ) and HFK − (−M, L; Z 2 ) respectively.Tovstopyat-Nelip further notes that the BRAID invariant can be thought of as a well defined class in HFK(−M, L; Z 2 ), a version of knot Floer homology defined for links encoded by multiple basepoints [44].We note that t(L) is in fact well defined in HFL(−M, L; Z 2 ) complete with all Alexander gradings.We note also that we may conflate t(L) and the associated generator of the chain complex in the section 4.
The BRAID invariant has some strong non-vanishing properties.The strongest of these, a result due to Tovstopyat-Nelip, is that t(B ∪ K) ̸ = 0 where B is a link braided about a fibered link K [44, Theorem 1.3].Note that for t(B ∪ K) to be well defined, we implicitly push K transversely off itself, so that B ∪ K is braided with respect to (Σ, ϕ).Note that Tovstopyat-Nelip's result generalises earlier work of Vela-Vick [45] and Vela-Vick-Etnyre [13].In the case that B = ∅, the Alexander grading of t(B ∪ K) is readily computed [44,Lemma 4.2].
We now discuss some mapping-class group theoretic properties of braids which will be of importance in Section 4. Given a pointed open book (Σ, ϕ, {p i }) there are two notions of an arc being sent to the right.We introduce a new terminology to distinguish these two notions.We say an arc a ⊆ Σ is relatively sent to the right by ϕ if after ϕ(a) has been isotoped in Σ−{p i } so that ϕ(a) and a intersect minimally, ϕ(a) is to the right of a.This is the notion of being sent to the right used in [8].Note that if a pointed open book (Σ, ϕ, {p i }) representing a braid L has an essential arc which is relatively sent to the left then t(L) and t(L) are both trivial in the homology ( [8,Theorem 1.4]).We say that a pointed open book is relatively right veering on a component of ∂Σ if every arc which intersects that component of ∂Σ is relatively sent to the right.
We say that a is absolutely sent to the right if a is sent to the right after isotoping ϕ(a) in Σ -in particular ϕ(a) is allowed to pass over basepoints under the isotopy.We say that a pointed open book is absolutely right veering on a component of ∂Σ if every arc which intersects that component of ∂Σ is absolutely sent to the right.The distinction between these two concepts plays an important role in the next section.Similar definitions also apply for notions of left veering.
For the purposes of this paper we do not count arcs that are fixed up to isotopy as right or left veering.That is we have the following trichotomy: arcs are either (relatively) sent to the left, right, or fixed up to isotopy.
We conclude this section by noting that, while the braid invariant was originally defined in homology theories with coefficients in Z 2 , it is in fact well defined up to sign with coefficients in Z.In particular, if L is a fibered link, the BRAID invariant generates the top Alexander grading of HFK(L; Z) or HFK(L; Q).

A Rank Bound from the BRAID invariant
Baldwin-Vela-Vick proved that the knot Floer homology of a fibered knot is non-trivial in the next to top Alexander grading [7].In this section we generalise their techniques to obtain a number of related results.In particular we show: Theorem 4.1.Let L be a non-trivial n component fibered link in a rational homology sphere, then: This implies a generalization of a Theorem of Ni [32, Theorem A.1]: Corollary 4.2.Suppose L is a non-trivial fibered n-component link in a rational homology sphere, with monodromy neither relatively right veering nor relatively left veering on each component.Then )) ≥ 2n.
Before proceeding to the technical lemma which underlies these results, we fix some notation.∂ T shall denote the component of the total differential on CFL(L), namely the one for which H * ( CFL(L), ∂ T ) ∼ = HF(M ) ⊗ V n−1 , where M is the underlying rational homology sphere and n is the number of components of L. ∂ CFL will denote the link Floer Homology differential; i.e.H * ( CFL(L), ∂ CFL ) ∼ = HFL(L).To prove this Lemma 4.3 we explicitly find the generator d.We then show that ∂ T d = t(L) using a diagrammatic argument, and note that it is easy to determine the multi-Alexander grading of d relative to that of t(L).
Proof of Lemma 4.3.Fixing notation as in the statement of the lemma, we extend a to a basis of arcs {a j } for (Σ, ϕ, {p i }).Then we isotope ϕ(a) to a curve b in −Σ so that a and b intersect minimally.Consider the surface Σ ∪ −Σ, and the alpha and beta curves -α, β -associated to a in the Heegaard diagram adapted to ((Σ, ϕ), {a j }).Note that since a is absolutely sent to the left there is an intersection point d ∈ α ∩ β ∩ −Σ and a bigon in Σ ∪ (−Σ) with corners at c ∈ α ∩ β, ∈ t(L) and d and edges contained in α and β.
Consider now b ′ , the arc formed after isotoping ϕ(a) in (−Σ, {p i }) so that a and b ′ intersect minimally.Let β ′ be the associated beta curve in the adapted Heegaard diagram.We claim that there is a map → Σ be an isotopy from the arc b 0 = b to the arc b 1 = b ′ (where we view arcs as smooth embedding b, b ′ : [0, 1] → Σ).Note that there is a continuous map [0, 1] → {0, 1} which takes the value 0 if the image of b t , counted with multiplicity is a linear combination of regions (Σ ∪ −Σ)\(α ∪ β t ), where β t are the β curves obtained from b t .Examples of such maps of bigons can be seen shaded in orange in Figure 1, or in Figure 2. The claim follows from continuity.We note that the image of B may have multiplicity strictly greater than one at certain points.
Let d ′ = B(i) ∈ −Σ, as shown in Figure 2. Consider the generator of CFL(−M, L), d, given by {d ′ , c 2 , . . .c 2g+n−1 }.Here the intersection points c i are intersection points of the remaining α i and β i curves in Σ. ⟨∂ T d, t(L)⟩ ̸ = 0, as witnessed by the bigon B, which is the shadow of a pseudo-holomorphic disk in Sym |α| (Σ ∪ −Σ), where |α| is the number of α curves.There is no other pseudo-holomorphic disk from d to t(L) -emanating to the north east of d ′ as opposed to south west, in Figure 1, or in Figure 2 -as this would imply that a and ϕ(a) were isotopic, contradicting the hypothesis that a is absolutely sent the left.
Finally ∂ CFL d = 0, as else a and b ′ would not intersect minimally as arcs in (−Σ, {p i }).□ We note that in Lemma 4.3, the multi-Alexander grading of d is determined by the multi-Alexander grading of t(L) and the number of w i basepoints and their respective multiplicity, n i , in the image of the map B. Let x k denote the kth Alexander grading of the BRAID invariant of a braid L. It follows that d has A k grading given by x k + n k for all k.
for all j, and ∃i such that and n is the number of components of L and we take coefficients in a field.
Proof.We proceed using the same notation as in Lemma 4.3.Note that d is a cycle in ( CFL(L), ∂ CFL ).If d ̸ = ∂ CFL e for any e, the result follows.Suppose then that there is an e with ∂ CFL e = d.
Observe that since we are working with coefficients in a field ( CFL(L), ∂ T ) has a model in which CFL(L) ∼ = HFL(L) (See [20,Reduction Lemma]) in every Alexander grading except for that of d, t(L).Since ∂ T (d) = t(L) ̸ = 0 and ∂ 2 T e = 0 we have that ∂ T (e) − d ̸ = 0 ∈ x∈J CFL(L, x) where J = {(A 1 , A 2 , . . ., A n )|x j ≤ A j ≤ x j + n j , for all j, and ∃i such that Suppose towards a contradiction that rank( x∈I HFL(L, x)) = 0. Then in fact we have that ∂ T (e) − d ∈ HFL(L, (x 1 , x 2 , . . . Suppose there exists an i for which rank( HFL(L; (A 1 , . . .A i − 1, . . .A n ))) = 1.Consider CFL − (L), together with the differential ∂ − T which counts all pseudo-holomorphic disks.Note that CFL − (L) can be thought of as a ((A 1 , A 2 , . . .A n ), (j 1 , j 2 , . . .j n ))-graded complex where the j i gradings are the U i gradings, so that the action of U i decreases the j i and A i grading by 1 and preserves the other gradings.Since L is fibered, there is a unique generator x of CFL − (L) with maximal A 1 + A 2 + • • • + A n grading and j i = 0 for all i.Since rank( HFL(L; (A 1 , . . .A i − 1, . . .A n ))) = 1, we may take there to be a unique generator y of CFL − (L) with Alexander multi-grading (A 1 , A 2 , . . .A i − 1, . . ., A n ), and again j i = 0 for all i.
Since ϕ sends an arc in Σ with endpoint on L i to the left there is a component of ∂ − T from x to y.Note that this property is preserved by filtered chain homotopy.Since ϕ sends an arc in Σ with endpoint on L i to the right there is a component of ∂ − T from y to U i x.Note again that this property is preserved by filtered chain homotopy.It follows from the proof of Theorem 4.1 that ⟨(∂ − T ) 2 (x), U x⟩ ̸ = 0, a contradiction.Thus rank( HFL(L; (A 1 , . . .A i − 1, . . .A n ))) ≥ 2 for all i 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.□ For Section 7 and Section 9 it will in fact be helpful to have the following algebraic generalization of Corollary 4.2: Proposition 4.6.Suppose L is a link such that: (1) rank( HFL(L, ( Then rank( HFL(L; Here ∂ T denotes the total differential on CFL(L).We have stated the proposition for A 1 only for ease of notation, a more general version readily follows by permuting the ordering of the components of L.Here recall that CFL − (L) can be thought of as a ((A 1 , A 2 , . . .A n ), (j 1 , j 2 , . . .j n ))graded complex, where the j i gradings are the U i gradings, where the action of U i decreases the j i and A i gradings by 1 and preserves the other gradings.Recall that one can view CFL(L) as the sub-complex of CFL − (L) with j i = 0 for i = 1, 2, • • • n or alternately as the sub-complex of CFL − (L) with (A 1 , A 2 , . . .A n ) fixed, if all A i are sufficiently negative.Let ∂ − T again denote the total differential on CFL − (L), that counts all pseudo-holomorphic disks.Proof.Consider a reduced model of CFL − (L) i.e. one such that the grading preserving component of ∂ − T is trivial.This is permissible because the filtered chain homotopy type of CFL − (L) is a link invariant and one can apply a filtered change of basis to ensure that the grading preserving component of ∂ − T is trivial -see [20,Reduction Lemma] in the knot case and note that the proof carries through to the link case.Observe that there is a generator

Suppose towards a contradiction that rank( HFL(L; (
is rank one with a generator ȳ.Observe that by assumption 2) there is a component of ∂ − T from x to ȳ.Note that this property is preserved under filtered chain homotopy.
We claim that there is also a component of ∂ − T from ȳ to U 1 x as a byproduct of the assumption (3) in the statement of the Proposition.
To verify the claim, note that in any fixed Alexander multi-grading (k, k, . . .k), with k sufficiently negative, CFL − (L) is given by a filtered chain homotopy equivalent copy of CFL(L), with the role of the (positive) A i gradings in CFL(L) is interchanged with the roles of the (negative) j i gradings in CFL − (L; {a i = k for all Alexander gradings a i }) -see the proof of [35,Proposition 8.1].Forgetting ∂ − T , the isomorphism can be given at the level of the underlying vector space by mapping a generator in Alexander multigrading CFL − (L; It follows in turn that there is a component of the differential from ȳ to U 1 x, as desired.Now, since rank( HFL(L; We conclude this section by noting that it is perhaps useful to view Theorem 4.1 in the context of the following more general proposition: Proposition 4.7.Suppose L is an n > 1 component non-trivial link in a rational homology sphere such that rank( HFL(L, ( 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 rank( HFL(L)) ≥ 2 n−1 coming from the spectral sequence from HFL(L) to Proof of Proposition 4.7.Suppose L is an n > 1 component link non-trivial such that rank( HFK(L, (x 1 , x 2 , . . .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 ̸ = j, and A i ̸ = x j , since 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, . . .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 HFL(L) must be even.Thus we obtain at least an additional 2 n − 1 generators.□ Note that if L is an n component link and rank( HFK(L, A)) is odd for some Alexander grading A, then there is a multi-Alexander grading

Detection Results for (2, 2n)-cables
In this section we provide detection results for the simplest 2 component cables of the trefoils and figure eight knot.Some care with coefficients is required.
Theorem 5.2.Let K be the figure eight knot.Link Floer homology with Z 2 coefficients detects K 2,2n for all n.
Each of these results is proven in three cases: where n > 0, n < 0 and n = 0.Only the proof of the n = 0 cases requires Z 2 coefficients.Throughout this section we will use K 2,2n to indicate the (2, 2n)-cable of K oriented so that the two components of K are oriented in parallel, and K (2,2n) to indicate the (2, 2n)-cable of K oriented so that it bounds an annulus.We take F to be Z, Z 2 or Q unless otherwise stated.
We first note that L has at most two components since the maximal Maslov grading of HFK(L) is 1  2 .Since rank( 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 of two component links [21].Since the maximal Alexander grading of HFK(L) is 1, L either bounds an annulus or it is the disjoint union of an unknot and a genus one knot.If rank( HFK(L)) = 16 then L cannot be a two component link with a split, unknotted component 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: 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 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 pairs of generators of Maslov grading 1 2 and − 5 2 .Since the linking number of L is zero we see that in link Floer homology the two Maslov index 0 generators have, without loss of generality, A 1 grading equal to 1, so we have that they are supported in bi-Alexander grading (1, 0).But since L bounds an annulus, both components 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 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 its dual, which is not the case.Thus L is isotopic to T (2, 3) (2,0) as desired. □ We now prove a similar result for the (2, 0) cable of the figure eight knot.
As before we begin with a partial computation of HFK(K 2,0 ).
Lemma 5.6.Let F be Z 2 and K be the figure eight knot.
for i = 1, and (F ) n for i = 0 and some n ≤ 3.
Proof.Hanselman-Watson's cabling formula for knot Floer homology in terms of immersed curves [17] shows that: Again we apply the skein exact triangle from [33, Equation 7] -with To rule out the latter case, we apply similar tactics as in after stating Lemma 5.3.In brief, note that K 2,0 bounds an annulus, so that HFL(K 2,0 ) is supported on the lines Thus since HFL(K 2,0 ) must be of even rank in each A i grading, and in the gradings A i = 1 the rank of the link Floer homology of L must be at least 4 since L is not fibered, the case that HFK(K 2,0 , 1) ) n for some 0 ≤ n ≤ 3.In fact n ≥ 2, as else HFK(K 2,0 ) cannot admit a spectral sequence to HFL(K) ⊗ V .□ Proof of Lemma 5.5.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 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 HFL(K 2,0 ) ∼ = HFL(T (2, 0)), so K must be a genus one knot.To see that K is fibered, note that the Maslov gradings of the A 1 = 1 part of HFL(L) is given by It follows that the Maslov index 2 generator, as well as the Maslov index −3 generator cannot persist under the spectral sequence to HFL(L 1 ), where L 1 is the component of L corresponding to A 1 .Thus HFL(L 1 ) is of rank one in the top Alexander grading and L 1 is fibered [16].Thus L is a (2, 0)-cable of a trefoil or the figure eight.But the (2, 0)-cables of the trefoils have distinct knot Floer homology, as shown previously.Thus L is K 2,0 as desired.□ To prove Theorem 5.1 and Theorem 5.2 we first show that if the link Floer homology of a link satisfies certain properties then the link is a (2, 2n) cable of a trefoil or the figure eight knot.Then we show that the link Floer homology of each of these three links satisfies these properties for the given n.Lemma 5.7.Let L be a two component link with linking number n > 1 such that: (1) HFL(L) is of rank 2 in the maximal non-zero A 1 or A 2 grading.
(2) The link Floer homology polytope has a vertex at multi-Alexander grading (1+ n 2 , 1+ n 2 ).( 3) HFL(L) has support on the lines Then L is isotopic to K 2,2n where K is a trefoil or the figure eight knot.
Proof of Lemma 5.7.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 + n 2 .Since g(L i ) + n 2 ≤ max{A i : HFL(L, (A 1 , A 2 )) ̸ = 0}, it follows that g(L i ) ≤ 1 for each i.Indeed since, HFL(L) is of rank 2 in A 1 grading 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 ∼ = K 2,2n where K is the unknot, a trefoil, or the figure eight knot.(2, 2n)-cables of the unknot are the torus links T (2, 2n) which do not satisfy property 2. Thus L is a (2, 2n) cable of a trefoil knot or the figure eight knot, as desired.□ To apply Lemma 5.7, we need to show that HFL(K 2,2n ) satisfies the conditions of the hypothesis of Lemma 5.7 for K a trefoil or figure eight knot.The easiest way to proceed is via the following remark: Remark 5.8.Rudolph shows that K n,m can be written as a Murasugi sum of K n,sign(m) , which is fibered by a result of Stallings [43], and T (n, m) along minimal genus Seifert surfaces [40].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 [15].It follows that (2, 2n) cables of genus one fibered knots are fibered and of genus n, that is Euler characteristic −2n.
If one wishes to only use Rudolph's stated result, for knots, one can alternately proceed via the following lemma: Lemma 5.9.Let K be a genus one fibered knot.For n ≥ 1, K 2,2n is fibered and χ(K 2,2n ) = −2n.
We can now proceed to the detection results.
Proof of Theorem 5.1.Suppose L is as in the statement of the Theorem.We proceed by cases: when n = 0, n > 0 and when n < 0. The n = 0 case was dealt with in Lemma 5.3.Suppose now that n > 0. We first show that HFL(T (2, 3) 2,2n ) satisfies the conditions in the hypothesis of Lemma 5.7.Observe that for n ≥ 1, each component of T (2, 3) 2,2n is a braid in the complement of the other, so HFL(T (2, 3) 2,2n )) must be rank two in the maximal A i grading, by braid detection [28, Proposition 1], so condition 1 of Lemma 5.7 holds.By Lemma 5.9 the maximal Alexander grading of HFK(L) is 2 + n.T (2, −3) 2,2n+1 is fibered by Lemma 5.9.Therefore, since 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 + n 2 , 1 + n 2 ), so that condition 2 holds.We note that the two generators of the top A i grading of HFL(T (2, −3) 2,2n ) must be of Maslov gradings 0 and −1.
Suppose now that HFL(L) ∼ = HFL(T (2, 3) 2,2n ) for some n ≥ 1.Since link Floer homology detects the linking number of two component links [21], it follows that the components of L have linking number n. Lemma 5.7 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 HFL(L) has a Maslov index 0 generator with A 1 grading 1 + n 2 which persists under the spectral sequence to HFL(L i ), whence L i is T (2, 3).
Consider HFL(T (2, −3) 2,2n ).Observe that for n ≥ 1, each component of L is a braid in the complement of the other, so HFL(L) must be rank two in the maximal A i grading, by braid detection [28], so Condition 1 of Lemma 5.7 holds.By Lemma 5.9 the maximal Alexander grading of HFK(L) is 2 + n.Since T (2, −3) 2,2n+1 is fibered, and 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+ n 2 , 1+ n 2 ), so that condition 2 of Lemma 8.7 holds.Condition 3 holds since T (2, −3) 2,2n bounds an annulus upon reversing the orientation of a component.Note that the two generators of the top A i grading of HFL(T (2, −3) 2,2n ) must be of Maslov gradings 1 and 2.
Suppose now that HFL(L) ∼ = HFL(T (2, −3) 2,2n ).Lemma 5.7 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 HFL(L) must be of Maslov gradings 1 and 2, it follows that each component is T (2, −3) and we have the desired result.□ The proof of Theorem 5.2 is again similar.
Proof of Theorem 5.2.The proof follows the proof of Theorem 5.1 verbatim, aside from to note that the top A i generators must be of Maslov index 0 and 1 respectively for n ≥ 0. □

E 2 collapsed chain complexes
In this section we prepare the ground for our detection results for T (2, 8) and T (2, 10) in subsequent sections by proving a purely algebraic result for E 2 collapsed chain complexes.
We will rely on a structure theorem for E 2 -collapsed chain complexes over Q i.e.Z ⊕ Z graded finitely generated chain complexes with coefficients in Q and differentials of length at most one.The Z 2 coefficient version is provided in [35, Section 12], see [38,Lemma 7] for a similar result.In particular Ozsváth-Szabó note that each E 2 -collapsed spectral sequences over Z 2 split as the direct sum of complexes of one of five forms, B d , V l d , H l d , X l d and Y l d .These are shown in the Figure 3.For the Q coefficient version the only necessary modification is that the arrows correspond to maps 1 : Q → Q, save for the top arrow of the B d complex, which is given by −1.Proposition 6.1.Suppose (C, ∂) is a bi-filtered E 2 collapsed chain complex with coefficients in Q.Then (C, ∂) is bi-filtered chain homotopy equivalent to a complex which decomposes as a direct sum of copies of chain complexes of the type 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 for a two component δ-thin link L 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 CFL(L).For instance, since the full filtered CFL complex has homology F 0 ⊕ F −1 , we see that CFL(L) has a summand X l 0 or Y l 0 and a summand X l −1 or Y l −1 , and no other summands of the form X l d , Y l d .Similarly, it is readily seen that the chain homotopy type of CFL(L 1 ) determines the V l d summands, while CFL(L 2 ) determines the H l d summands.Specifically, if there are generators x, y ∈ CFL(L 1 ), with ⟨∂ CFL(L 1 ) x, y⟩ ̸ = 0 then we must have a summands of CFL(L) of the form V , as no other summand induces the correct differential on CFL(L 1 ).The proof of Proposition 6.1 is entirely elementary.
Proof of Proposition 6.1.We prove this by showing that (C, ∂) admits a graded basis {v i } for C such that ∂v j is either trivial, a single generator in {v i } ∩ C(x − 1, y) ⊕ C(x, y − 1), or the sum of a generator in {v i } ∩ C(x, y − 1) and a generator in {v i } ∩ C(x − 1, y).For the purposes of this proof we call a chain complex with a choice of such a basis simplified.The result then follows.By a bi-filtered version of the reduction lemma [19, Lemma 4.1] we can assume that the grading preserving part of ∂ is trivial.
Indeed, by repeatedly changing basis we may assume that each summand C(x, y) splits as a direct sum A(x, y) ⊕ B(x, y) ⊕ D(x, y), where ∂ 2 | A(x,y) : A(x, y) → B(x, y − 1) is represented by the identity matrix under a choice of basis and ∂ 2 is trivial when restricted to B(x, y) ⊕ D(x, y).
We now proceed to produce a basis for which x,y C(x, y) is simplified.Starting with the basis as above, we proceed by a double induction: we assume the subcomplex (x,y):x<n C(x, y) is simplified and then in turn assume that each summand x<n,y=m C(x, y) is simplified then proceeding by induction on n then m.
Consider z a basis element in C(n, m).Suppose ∂ 1 (z) = q i a i + r j b j + s k d k , where a i are basis elements for A(n − 1, m), b j are basis elements for B(n − 1, m) and d k are basis elements for D(n − 1, m) and q i , r j , s k ∈ Q.
Suppose r j = 0 for all j.Then we can change basis for A(n − 1, m) ⊕ D(n − 1, m) ⊕ B(n − 1, m − 1) so that ∂ 1 z = q for q a basis element in A(n − 1, m), as desired.
Likewise if q i = 0 for all i.We can change basis for B(n−1, m)⊕D(n−1, m)⊕A(n−1, m+1) so that ∂ 1 z = q for q a basis element in B(n − 1, m), as desired.
In general, we may assume after making a change of basis for with a a basis vector in A(n, m), b a basis vector in B(n, m), ϵ 1 , ϵ 2 ∈ {0, 1}.If either ϵ 1 or ϵ 2 is non-zero then we are done.Suppose they are both 1.Note that in order that ∂ 2  1 ̸ = 0 we must have that z is not in the image of ∂ 1 .Moreover, In order that ∂ 2 = 0, we must have that z ∈ A(n, m).Indeed by induction we have that From here the following corollary follows immediately from work in [11].
Here Annular Khovanov homology is a combinatorial invariant of links in a thickened annulus.For details we refer the reader to [11].
For the readers convenience first we recall that: Here F can be Q, Z or Z 2 .We start by showing that in general if HFK(L) ∼ = HFK(T (2, 2n)) then L consists of two unknotted components and CFL(L) is of an especially simple form.
summand, and L has unknotted components.
Proof of Lemma 7.4.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 HFK(L) is 1  2 , so since HFK(L) admits a spectral sequence to HF(# m−1 S 1 × S 2 ), L can have at most two components.L then has exactly 2 components since the Maslov grading of HFK(L) is Z + 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 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 [ n 2 , n 2 ] and since there are two Maslov index −1 generators there is a summand As observed in the proof of [35,Theorem 12.1], HFL(L) consists of pairs of summands , 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 summands with maximal x.Note that x > n 2 .Amongst these consider the summands with minimal d.It follows that there is a generator in HFK(L 1 ) of minimal Alexander grading with Maslov grading d − 2(x − 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 n 2 − x which persists under the spectral sequence to HFK(L 1 ) ⊗ V then there must be a generator of odd Maslov index in A 1 grading n 2 − x − 1, since the summands of HFL(L) that persist under the spectral sequence to HFK(L 1 ) ⊗ V are of the form We can now prove the two Knot Floer homology detection results.
Proof of Theorem 7.1.Suppose HFK(L) ∼ = HFK(T (2, 8)).Lemma 7.4 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 [y, −y], 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 [28,Proposition 1] 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 ∈ {1, 0, −1}.The y = 1 case can be obstructed using Proposition 4.6.Specifically observe that y can be taken to be the generator of Alexander multi-grading (2, −1).The y = −1 can be obstructed similarly.The result then follows from the fact that Link Floer homology detects T (2, 8) [11,Theorem 3.2].□ The proof of Theorem 7.2 is virtually identical: Proof of Theorem 7.2.Suppose HFK(L) ∼ = HFK(T (2, 10)).Lemma 7.4 implies that L is a two component link with unknotted components, and indeed determines the complex up to a choice of y in a summand B −4 [y + 1 2 , 1 2 − y], 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 y ∈ {1, 0, −1}.The y = 1 case can be obstructed using Proposition 4.6.Specifically observe that y can be taken to be the generator of Alexander multi-grading ( 52 , − 1 2 ).The y = −1 can be obstructed similarly.The result then follows from the fact that Link Floer homology detects T (2, 10) [11,Theorem 3.2].□

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: Indeed, this yields a complete list of links with at least two components of rank at most eight, in combination with the following lemma: Lemma 8.4.Suppose L is an n + 1 component link with rank( HFK(L)) = 2 n .Then L is an unlink.
We can also provide a partial classification of links with knot Floer homology of rank ten: Proposition 8.5.There is no three component link with rank( HFK(L)) = 10.
By similar methods we can show a detection result for knot Floer homology: Theorem 8.6.Let L 0 be T (2, 3) with a meridian.Suppose HFK(L) ∼ = HFK(L 0 ).Then L is isotopic to L 0 .
Here we take the linking number of the right handed trefoil and its meridian to be one.Again this result should be compared to the corresponding link Floer homology detection result in [11].
The proofs of Theorems 8.1, 8.2, 8.3 and 8.5 each essentially follow from case analysis in which we use algebraic and geometric properties of link Floer homology to exclude the unwanted cases, as well as the detection results for Knot Floer homology found in [11].Throughout this section, we take F to be Q or Z 2 .
We first prove Lemma 8.4, as it is useful in proving the other theorems.
We will find the following characterization of unlinks useful: Lemma 8.7 (Ni [30]).Suppose HFK(L) is supported in a single Alexander grading.Then L is an unlink.
Proof of Lemma 8.7.Observe that if HFK(L) is supported in a single Alexander grading then that grading must be zero, by the symmetry of HFK(L).It follows that 0 where n is the number of components of L, and the surface that L bounds has m components, the ith of which has n i boundary components.Thus 0 = n − m + i g i .But n ≥ m, so 0 = n − m + m i=1 g i ≥ m i=1 g i ≥ 0. Thus m i=1 g i = 0 and n = m.It follows that L bounds n disks, whence L is an unlink with n components.□ Proof of Lemma 8.8.Suppose L is an n component fibered link with rank( HFK(L)) = 6.Since rank( HFL(L)) is even, L cannot be a knot.Moreover, L can have at most 3 components, since HFL(L) admits a spectral sequence to HF(# n−1 S 1 × S 2 ).If L has three components, then since L is fibered, the maximal Euler characteristic Seifert surface for L is connected, whence χ(L) ≤ −1, and the maximal Alexander grading in which HFK(L) has support is at least 2. By Theorem 4.1, the next to top Alexander grading is of rank three, whence 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 χ(L) ≤ 0, and the maximal Alexander grading, g, in which 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( HFK(L)) = rank( HFK(T (2, ±3)⊗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 the bi-Alexander grading of the top Alexander grading generator.Lemma 4.4 implies that the complex is given by To ensure that each Alexander grading is of even rank we must have that g = 2, x = 1, i.e. the complex is By the symmetry of 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 [28], a contradiction.□ Having eliminated the case that L is fibered, we can proceed to the remaining cases.
Proof of Theorem 8.2.Let L be an n component link with link Floer homology of rank 6. Lemma 8.8 implies that L is not fibered under any orientation.The fact that 2 n−1 ≤ rank( HFK(L)) implies that L has at most three components.Since rank( 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 n j=1 A j = ±g i for each g i , where g i is maximal Alexander grading of HFK(L) under some orientation of L.
Lemma 8.7, together with the fact that there is no unlink with Knot Floer homology of rank 6, implies that we may take g j > 0 for all i.
Suppose that HFL(L) has support at a vertex p = (p 1 , p 2 , . . .p n ) of the link Floer polytope.Since there exists an Alexander grading A i for which the support of HFL(L) in the plane A i = p i is exactly p, it follows that the spectral sequence from HFL(L) to HFL(L i ) ⊗ V n−1 is trivial.Thus the rank of HFL(L) at p is at least 2 n−1 .Since the polytope is non-degenerate, the symmetry properties of HFL(L) implies that 2 n ≤ 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 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 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( HFL(L)) ≥ 8, a contradiction.□ For the rank eight case, we proceed as in the rank six case, dealing with fibered links first, and the remaining cases subsequently.Proposition 8.9.Suppose L is fibered and rank( HFK(L)) = 8.Then L ∼ = T (2, 4) or T (2, −4).
Proof.Let L be a fibered link with link Floer homology of rank eight.Since HFK(L) admits a spectral sequence to HF(# n−1 S 1 ×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( HFK(L)) is even.Suppose L has four components.Then by Theorem 8.4, L must be a four component unlink, which is not fibered.
Suppose L has three components.Since L is fibered, the minimal genus Seifert surface for L is the fiber surface, which is connected.It follows that the highest Alexander grading, g, in which HFK(L) has supported is at least 2. By Lemma 4.4, rank( HFK(L)(g − 1) where m ∈ 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 HFK(L) is at least one.Suppose g = 1.Then rank( HFK(L; 0)) = 6.Such a complex clearly does not admit a spectral sequence to HF(S 1 × S 2 ), a contradiction.Thus g ≥ 2. Consider HFL(L).Suppose the top Alexander grading generator is in multi-Alexander grading (x, g − x) and of Maslov index m.Lemma 4.4 implies that there are generators in Alexander gradings (x − 1, g − x), (x, g − x − 1) both of Maslov index m − 1, while symmetry implies that there are generators in multi-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 ≥ 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 ≥ g.Suppose (x, g) ̸ ∈ {( 32 , 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 the spectral sequence from HFK(L) to HFK(L 1 ) ⊗ 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 ∂ 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 ( ). 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( HFK(T (2, 2n)) = 8 are n = ±2.
Suppose (x, g) = (1, 2).Reversing the orientation of a component, yields a link that bounds an annulus (or the disjoint union of a punctured torus and a disk, but such a link is split and therefore not fibered) whence L is a 2-cable link.L cannot be the 2-cable of a trefoil, as we know that the linking number is 0, since the span of each A i grading is 3, but the rank of 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) = ( 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 1 2 to one in lower A 1 grading, as else the generators that persist under the spectral sequence to HFL(L 2 ) do not have admissible Maslov gradings.Thus the complex splits as two squares, and again must be located on the diagonal, a contradiction.□ Proof of Theorem 8.3.By the previous lemma it suffices to classify links that are not fibered under any orientation and with link Floer homology of rank eight.Suppose L is such a link, with n components.Since rank( HFK(L)) ≥ 2 n−1 and L is even, L has either two, three or four components.If L has four components, then Lemma 8.4 implies that L is an unlink.We deal with the two remaining cases separately: (1) L has two components Consider HFL(L).Since L is not fibered under any orientation, the link Floer polytope of L is a subset of the polytope with boundaries Here g 1 , g 2 are the maximal Alexander gradings of HFK(L) under its two possible orientations.This has vertices ±( g 1 +g 2 2 , g 1 −g 2 2 ), ±( g 1 −g 2 2 , g 1 +g 2 2 ).Lemma 8.7 together with the fact that the two component unlink has knot Floer homology of rank two imply that we may take g 1 , g 2 > 0.
Suppose that 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 ( . Observe that both of the generators must persist under the spectral sequence to 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 ∈ {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 4 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 4.The only remaining case is that on each face the two generators sit in the same multi-Alexander grading, see Figure 4. Suppose the generators occur in multi-Alexander gradings ±(x, g 1 − x), ±(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 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 {±( g 1 2 , g 1 2 ), (± g 1 2 , ∓ g 1 2 )}.Let m 1 ≥ m 2 be the Maslov gradings of the two generators of grading ( g 1 2 , g 1 2 ).Since A 1 is an unknot we must have that ∂ 1 x ̸ = 0. Without loss of generality, let ∂ 1 x = y.Since y does not persist under the spectral sequence to HFL(L 2 ), ∂ 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 contradiction, since these do not have the right homology, or the remaining generator of bi-Alexander grading is of Maslov grading {0, 1} ∋ m + 2g 1 − 2 ≥ m + 2. The remaining generator of bi-Alexander grading ( g 1 2 , −g 1 2 ) must be of Maslov grading m + 2g 1 − 2 ± 1.This is because the other remaining generator of bi-Alexander grading (− g 1 2 , − g 1 2 ) is of Maslov grading either m + 2g 1 − 2 or m + 2g 1 − 2. But it should also be of grading m − 2g 1 , by symmetry, a contradiction.
(2) 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 HFL(L) to HFL(L − L 3 ) is just projection onto the A 3 = 0 plane.Suppose 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 HFL(L) is not supported in the A 3 = 0 plane then since the rank in the top A 1 grading 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 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 also exist generators in multi-Alexander gradings (0, y 1 , x), (y 2 , 0, x) to ensure that the generator in multi-Alexander grading (0, 0, x) do not persist under the spectral sequences to L−L 1 , L−L 2 .It follows that there are at least 6 generators that have non-zero A 3 component, and at least another 4 with A 3 = 0, a contradiction.
The result then follows from Lemma 8.7.

□
Proof of Proposition 8.5.Let L be a 3 component link with rank( HFK(L)) = 10.Suppose L is fibered.The maximal Alexander grading, A, of HFK(L) is given by 2 + g for some g ≥ 0.
Note that rank( HFK(L)(A)) = 1.Suppose the generator is of Maslov index m.Theorem 4.1 implies that there are at least three generators of Maslov index m − 1 in Alexander grading 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 HFK(L) under each of the four possible orientations of L, up to an overall reversal in the orientation of L. Note that g i > 0 by Lemma 8.7.Consider the polytope P cut out by Observe that each of these faces must contain at least two generators.If the edges of P do not contain any generators, then rank( HFK(L)) ≥ 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 Observe that any generator on this edge, as well as the respective generators yielded by the symmetry of HFL(L) on the edge , and is indeed of maximal (respectively minimal A 1 + A 2 grading).Now, there are only two 2 component links with rank( HFK(L)) ≤ 5; the Hopf link and the unlink.It follows that ℓk(L i , L j ) ∈ {−1, 0, 1} for each i, j.Thus the maximal A 1 + A 2 grading can only be 1.That is, we have that , so that g 1 = g 2 = 1.It follows that, under an appropriate orientation of L, χ(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( HFK(L − L i ) ≤ 5, so that L − L i is an unlink, or a Hopf link.Thus L is the disjoint union of a Hopf link and an unknot, but this link has knot Floer homology of rank eight, a contradiction.□ We conclude with a proof of Theorem 8.6 Proof of Theorem 8.6.Let L 0 be the trefoil together with a meridian.It suffices to show that if HFK(L) ∼ = HFK(L 0 ) then HFL(L) ∼ = HFL(L 0 ) by the corresponding link Floer homology detection result in [11].Since the maximal Maslov grading of HFK(L) is 1 2 , it follows that L has at most two components.Since rank( HFK(L)) is even, it follows that L cannot be a knot, whence L is a two component link.

Detection Results for Khovanov Homology
In this section we briefly review Khovanov homology before proving various detection results.9.1.A Brief Review of Khovanov homology.Let L be a link, and R be the ring Z, Z 2 or Q.The Khovanov chain complex of L, (CKh(L, R), ∂), is a finitely generated Z⊕Z-filtered chain complex over R [24].
Here i is called the homological grading, while j is called the quantum grading.The filtered chain homotopy type of L is an invariant of L. The R-module CKh(L, R) has generators corresponding to decorated complete resolutions of D, while ∂ is determined by a simple TQFT.The parity of the j gradings in which Kh(L) has non-trivial support agree with the parity of the number of components of L. The Khovanov homology of L is obtained by taking the homology of CKh(L; R).A choice of basepoint p ∈ 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 reduced Khovanov homology of L, denoted Kh(L, p; R) [23].Given a collection of points p = {p 1 , p 2 , . . ., p k } ⊂ L, there is a generalization of reduced Khovanov homology called pointed Khovanov homology, Kh(L, p; R) due to Baldwin-Levine-Sarkar [3].
Kh(L; R) admits a number of useful spectral sequences.Suppose L has two components L 1 , L 2 .If R is Q or Z 2 then Kh(L; R) admits a spectral sequence to Kh(L 1 ; R) ⊗ Kh(L 2 ; R), called the Batson-Seed spectral sequence [9].Indeed this spectral sequences respects the i − j grading on Kh(L; R) in the sense that: There is another spectral sequence, called the Lee spectral sequence, from Kh(L; Q) which abuts to an invariant called the Lee Homology of L, L(L) := i L i (L) [27].This spectral sequence respects the i gradings in the sense that rank(Kh . Moreover, there is an invariant, s(L) -due to Rasmussen [39] in the knot case and Beliakova-Wehrli in the link case [10] -defined from the Lee spectral sequence -with the property that s(L) ≤ 1 − χ(L).
Finally, there is a spectral sequence, due to Dowlin [12], relating Khovanov homology and Knot Floer homology.If L is a link and p ⊆ L, with exactly one element of p in each component of L, then there is a spectral sequence from Kh(L, p; Q) to HFK(L; Q) that respects the relative δ-gradings.Here HFK(L; Q) uses the coherent system of orientations given in [1].We use this version of knot Floer homology, and the corresponding link Floer homology for the remainder of the paper.As a corollary, Dowlin shows that if L has n components then:  This requires the Lee, Batson-Seed, and Dowlin spectral sequences, as well as an analysis similar to that of section 7.
To prove Theorems 9.3 and 9.4 we first show that any link L with Kh(L; Z) ∼ = Kh(T (2, 2n); Z) has two components, with linking number n and χ(L) ≥ 2 − 2n.We then show that the two components are unknotted in the n = 4, 5 cases.After that we conclude by showing that L must bound an annulus after reversing the orientation of one of the components, thereby completing the proofs.Lemma 9.5.Suppose Kh(L; Z) ∼ = Kh(T (2, 2n); Z).Then L is a two component link with linking number n and χ(L) ≤ 2 − 2n.
We prove this Lemma using Lee's spectral sequence [27], which has Kh(L; Q) as E 1 page and E ∞ page a complex of rank 2 n where n is the number of components of L.
Proof of Lemma 9.5.Suppose L is as in the Theorem.By the universal coefficient theorem there are exactly two homological gradings in which the rank of Kh i (L; Q) is at least two, namely i = 0 and i = 2n.Since rank(Kh i (L; Q)) ≥ rank(L i (L; Q)), and L i (L; Q) ∼ = m i=1 Q 2 a i , where a i ∈ 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; Z) is supported in even quantum gradings [24,Proposition 24].
The linking number of the two components of L is half the difference in the homological gradings of the generators which persist under Lee's spectral sequence [27,Proposition 4.3], hence the linking number of L is n.
Finally we note that since L is Khovanov-thin, with generators supported in the gradings j − 2i = 2n, 2n + 2 it has s-invariant given by n + 1, whence χ(L) ≤ 2 − 2n [10].□ orientation of a component, L bounds an annulus.Since L has unknotted components and linking number 4 it follows that L is T (2, 8), as desired.
If there is only one remaining B d summand, then it must be of the form B d [−1/2, −1/2] for some d by the symmetry of Link Floer homology.But this is a contradiction, since ℓk(L) is even, so that the Alexander gradings are Z valued.
If there is no other B d summand, then again we see 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 proof of Proposition 9.9 is virtually the same, but with some additional case analysis.
Proof of Proposition 9.9.Suppose L is as in the hypotheses of the Proposition.Since L is delta-thin, HFL(T (2, 10); Q) is If l = 7 then the position of the remaining B d summands are forced by the symmetry of HFL(L).We find that the maximal A i gradings are 9  2 , and each of rank 2. It follows that each component is braided with respect to the other.But then, since each component should be unknotted, it follows that the maximal Alexander grading should be half the linking number of L, i.e. 5  2 , a contradiction.If l = 6, then either CFL(L) contains a summand X 0 ] case is excluded by noting that we can apply Proposition 4.6 with x a generator in (i, j) grading ( −7 2 , 7 2 ), x a generator of grading ( 7 2 , −7 2 ) and y a generator of grading ( 72 , −5 2 ).If there are three additional B d summands then we note that in order to satisfy the symmetry properties there must be a summand B 2 ], we have that at least one component, L i is a braid axis, so again the maximal A i grading must be 5  2 , a contradiction.In the

Figure 1 .
Figure 1.A Heegaard diagram adapted to a pointed open book (Σ, ϕ, {p i }) and basis of arcs {a j }.Note that we have only shown segments of some of the β curves.The green dots indicate the intersection points that represent the BRAID invariant.The orange region is the shadow of a pseudo-holomorphic disk which plays a role in Section 4. The blue dots in the lower half of the surface are w i basepoints, which come from the points p i in the pointed open book.The basepoints in the upper half of the surface are z i .Note that not all of the basepoints are shown in the figure.

Lemma 4 . 3 .
Suppose (Σ, ϕ, {p i }) is a pointed open book representing a braid L in a rational homology sphere M .Suppose ϕ sends an essential arc a ⊂ (Σ, {p i }) absolutely to the left.Then there is a generator d ∈ CFL(−M, L, [Σ]) with ∂ T d = t(L) and ∂ CFL (d) = 0.

Figure 2 . 1 . 4 . 4 .
Figure 2. A picture of part of the Heegaard diagram used to define the BRAID invariant for the binding of an open book, via a transverse pushoff.The BRAID invariant includes the green dots c i and c j .As described in Lemma 4.4, d ′ includes c i , but not c j , instead including d.The orange region is the shadow of a pseudo-holomorphic disk from d to the BRAID invariant.The black dots are basepoints.

Figure 3 .
Figure 3. Examples of possible summands of an E 2 -collapsed chain complex.

Theorem 8 . 1 .
If rank( HFK(L)) = 4 then L is the Hopf link or the three component unlink.Theorem 8.2.Suppose L is a link with rank( HFL(L)) = 6.Then L is a disjoint union of a trefoil and an unknot.Theorem 8.3.The only links L with rank( 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.

Figure 4 .
Figure 4.A rank eight bigraded vector space not arising as the link Floer homology of a link.

E 2 -
collapsed.Since χ(L) ≤ −8 the maximal Alexander grading is at least n.Since the components of L are unknotted CFL(L; 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 lineA i = 52 .Let l be the A = A 1 + A 2 grading of the Maslov index 0 generator.If l > 5 then CFL(L; Q) has summands X l−6 Note that for l > 5, rank(X l−5 4l − 16.Observe that the symmetry of HFL(L; Q) together with the fact that rank( HFL(L; Q)) ≤ 20, imply that l ≤ 7.

2 ] 2 ]
at most three extra B d summands.If there is one extra B d summands then must be of the form B −7 [ −1 2 , −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 L is therefore should be T (2, 10), a contradiction since T (2, 10) has distinct link Floer homology.If HFL(L; Q) contains two extra B d summands, then unless they are of the form B −1 [ ⊕ B −7 [ −7 2 , 5 2 ], we see immediately that at least one component is a braid axis, and hence the maximal A i grading is 5 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 B −7 [ 5 2 , −7 2 ] ⊕ B −7 [ −7 2 , 5 2 −7 [ −1 2 , −1 2 ].Unless the remaining B d summands are B −1 [ 5 2 , 5 2 ] ⊕ B −13 [ −7 2 , −7 2 ], or B −8 [− 5 2 , 7 2 ] B −8 [ 3 2 , −5 Proof of Theorem 4.1.An n component fibered link L arising as the binding of an open book (Σ, ϕ) may be perturbed to yield an n-braid B in the complement of L, with associated pointed monodromy (Σ, ϕ ′ ).Consider the mth boundary component of Σ, P m .After mirroring we may assume that ϕ sends some non-separating arc a with an endpoint on P m absolutely to the left or fixes some non-separating arc.Extend this arc to a basis for (Σ, ϕ ′ ), and consider t(B).Suppose a is fixed up to isotopy.Then M contains an S 1 × S 2 summand and hence cannot be a rational homology sphere, a contradiction.

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

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