Classification of links with Khovanov homology of minimal rank

If L is an oriented link with $n$ components, then the rank of its Khovanov homology is at least $2^n$. We classify all the links whose Khovanov homology with Z/2-coefficients achieves this lower bound, and show that such links can be obtained by iterated connected sums and disjoint unions of Hopf links and unknots. This gives a positive answer to a question asked by Batson and Seed.

Let L be an oriented link in S 3 and R be a ring.Khovanov homology [Kho00] assigns a bi-graded R-module Kh(L; R) to the link L. In this paper, we will take the coefficient ring to be Z/2.The Euler characteristics of Kh(L; Z/2) recover the coefficients of the unreduced Jones polynomial of L. If L has n components, then the value of the unreduced Jones polynomial at t = 1 equals (−2) n , therefore rank Z/2 Kh(L; Z/2) ≥ 2 n . (1) The rank of Kh(L; Z/2) is independent of the orientation of L. If L is the unlink, then rank Z/2 Kh(L; Z/2) = 2 n .However, the unlink is not the only case that (1) achieves equality.For example, when n = 2 the Khovanov homology of the Hopf link has rank 4.More generally, for arbitrary n we have the following construction.In graph theory, a forest is a simple graph without cycles.Given a forest G, define a link L G by placing an unknot at each vertex of G and linking two unknots as a Hopf link whenever there is an edge connecting the corresponding vertices (see Figures   1, 2 for examples).The link L G is called the forest of unknots defined by G. Every forest of unknots can be obtained by iterated connected sums and disjoint unions of Hopf links and unknots.By [AP04, Corollary 6.6] and the Künneth formula, if L G is a forest of unknots with n components, then rank Z/2 Kh(L G ; Z/2) = 2 n .
The following question was asked by Batson and Seed: Question 1.1 ([BS15, Question 7.2]).Are forests of unknots the only n-component links with Khovanov homology of rank 2 n over Z/2?
Our main result gives an affirmative answer to the above question.
Theorem 1.2.If L is an n-component link such that rank Z/2 Kh(L; Z/2) = 2 n , then L is a forest of unknots.
In other words, Theorem 1.2 states that the rank of Khovanov homology with Z/2-coefficients distinguishes the forests of unknots from other links.
The detection properties of Khovanov homology have been studied intensively in the past decade.The first breakthrough of this field was the landmark paper by Kronheimer and Mrowka [KM11a], which proved that Khovanov homology detects the unknot.Since then, several other detection results have been proved.The following list is a summary of related results in chronological order: (1) [KM11a] The rank of the reduced Khovanov homology with Q-coefficients detects the unknot; (2) [HN13] The Khovanov homology with Z/2-coefficients, together with an extra module structure, detects the unlink; (3) [BS15] The Khovanov homology with Z/2-coefficients detects the unlink; (4) [BS18] The rank of the reduced Khovanov homology with Q-coefficients detects the trefoil; (5) [BSX18] Both the Khovanov homology and the reduced Khovanov homology, with coefficients in either Z or Z/2, detect the Hopf link.
Theorem 1.2 is a generalization of (2), (3) and (5) above.To see that Theorem 1.2 generalizes (2) and (3), notice that by [AP04, Corollary 6.6] and the Künneth formula, if G is a forest such that Kh(L G ; Z/2) is isomorphic to the Khovanov homology of the unlink, then G has to be an edgeless graph.To see that Theorem 1.2 generalizes (5), notice that by [Shu14, Corollary 3.2.C], the reduced Khovanov homology Khr satisfies rank Z/2 Khr(L; Z/2) = 1 2 rank Z/2 Kh(L; Z/2), and that if L is a forest of unknots with 2 components, then L is either the Hopf link or the unlink.
The proof of Theorem 1.2 relies on Kronheimer-Mrowka's spectral sequence [KM11a] and Batson-Seed's inequality comparing the ranks of the Khovanov homology of a link and its sublinks [BS15].Under the assumption of Theorem 1.2, Batson-Seed's inequality and Kronheimer-Mrowka's unknot detection theorem imply that all the components of L are unknots.On the other hand, Kronheimer-Mrowka's spectral sequence gives an upper bound on the rank of the instanton link invariant I (L).Since all the components of L are unknots, I (L) is isomorphic to an annular instanton Floer homology introduced by the first author in [Xie18b].The annular instanton Floer homology carries a Z-grading (we call it the f-grading).In [XZ19], the authors showed that this grading detects the generalized Thurston norm of surfaces with a meridian boundary, which allows us to extract topological information from I (L).The topological properties imply that either L is a forest of unknots, or L contains a sublink with a very specific configuration.We then use a direct computation of Khovanov homology and Jones polynomial to rule out the latter case.
Acknowledgements.Part of this work was done during the 2019 Summer Program Quantum Field Theory and Manifold Invariants at PCMI.The authors would like to thank the organizers for providing us with such a great environment to carry on this work.We would like to thank John Baldwin and Zhenkun Li for many helpful conversations.We also want to thank Nathan Dowlin, Peter Ozsváth, and Zoltán Szabó for their help with knot Floer homology in the proof of Lemma 6.1.

Annular intanton Floer homology
The singular instanton Floer homology theory was introduced by Kronheimer and Mrowka in [KM11a,KM11b].Let (Y, L, ω) be a triple where Y is a closed oriented 3-manifold, L ⊂ Y is a link and ω ⊂ Y is an embedded 1-manifold such that ∂ω = ω ∩ L. The triple (Y, L, ω) is called admissible if there is an embedded closed surface Σ ⊂ Y satisfying either one of the following conditions: • Σ is disjoint from L and the intersection number of ω and Σ is odd, • The intersection number of L and Σ is odd.
If (Y, L, ω) is admissible, the instanton Floer homology I(Y, L, ω) is defined to be a Morse homology of the Chern-Simons functional on a certain space of orbifold SO(3)-connections over Y , where Y is equipped with an orbifold structure with cone angle π along L, and ω represents the second Stiefel-Whitney class of the SO(3)-bundle.In this article, we will always take C-coefficients for instanton Floer homologies.
The homology group I(Y, L, ω) carries a relative Z/4-homological grading.Given an embedded closed surface F ⊂ Y , there is an operator µ orb (F ) defined on I(Y, L, ω) with degree 2. For more details the reader may refer to, for example, [XZ19, Section 2].
The rest of this section gives a brief review of the annular instanton Floer homology introduced in [Xie18b].Let L be a link in the solid torus S 1 × D 2 .The annular instanton Floer homology AHI(L) is defined by the following procedure: (1) Let K 2 be the product link S 1 × {p 1 , p 2 } in S 1 × D 2 , and let u be an arc in S 1 × D 2 connecting S 1 × {p 1 } and S 1 × {p 2 }; (2) Form the new link L ∪ K 2 in where L lies in the first copy of S 1 × D 2 , and K 2 lies in the second copy.
Definition 2.3.A properly embedded, connected, oriented surface S ⊂ S 1 × D 2 is called a meridional surface if ∂S is a meridian of S 1 × D 2 .
The annular instanton Floer homology detects the generalized Thurston norm of meridional surfaces.
Theorem 2.4 ([XZ19, Theorem 8.2]).Given a link L in S 1 × D 2 and suppose S is a meridional surface that intersects L transversely.Let g be the genus of S and let n := |S ∩ L|.Suppose S minimizes the value of 2g + n among meridional surfaces, then we have AHI(L, i) = 0 for all |i| > 2g + n, and AHI(L, ±(2g + n)) = 0.
We also need the following result.
Proposition 2.5 ([XZ19, Corollary 8.4]).Let L be a link in S 1 × D 2 .Then L is isotopic to the closure of a braid with n strands if and only if the top f-grading of AHI(L) is n, and Although the annular instanton Floer homology is defined for links in the solid torus, it can be used to study links in S 3 .Let L be a link in S 3 and let p be a base point on L. In [KM11a], Kronheimer and Mrowka defined the link invariant where m is a small meridian circle of L around p and u is an arc joining m and p.The following result is a consequence of the excision property of instanton Floer homology.
Proposition 2.6 ([Xie18b, Section 4.3]).Suppose L has an unknotted component U and let p ∈ U .Let N (U ) be a tubular neighborhood of U , then L 0 := L − U is a link in the solid torus S 3 − N (U ).We have The above isomorphism does not preserve the f-grading of AHI(L 0 ) since there is no such grading on I (L, p).Notice that a meridional surface in the solid torus S 3 − N (U ) is a Seifert surface of U .

Local coefficients
This section reviews the singular instanton Floer homology theory with local coefficients, which was introduced in [KM11b, Section 3.9] (see also [KM13,Section 3]).Let B(Y, L, ω) be the space of gauge-equivalence classes of orbifold connections over (Y, L, ω).Let R be the ring Γ µ α .
To make C µ a chain complex, we need to define a differential on it.For α, β ∈ Crit(CS), let M d (α, β) be the d-dimensional moduli space of trajectories of CS from α to β.This space carries an R-action and we denote the quotient space by Md (α, β) is well-defined.We define an R-module homomorphism by The differential D on C µ is then given by and the instanton Floer homology with local coefficients is defined by If F ⊂ Y is an embedded closed surface, the operator µ orb (F ) can be defined in the setting with local coefficients.Roughly speaking, the surface F defines a two dimensional cohomology class on B(Y, L, ω) (see [XZ19, Section 2]), and its Poincaré dual is given by a linear combination of divisors on B(Y, L, ω) as There is a map from M d (α, β) to B(Y, L, ω) by restricting the trajectories at time 0. The divisors V i 's are generic in the sense that they are transverse to the restriction map M d (α, β) → B(Y, L, ω) for all α, β ∈ Crit(CS) and d ∈ N. We define an R-module homomorphism by A standard argument shows that the map is a chain map.The map H induces the operator µ orb (F ) on I(Y, L, ω; Γ µ ).The tensor products recover the ordinary Floer chain complex (C, d) and the ordinary operator µ orb (F ) on I(Y, L, ω) with C-coefficients.Suppose there is a component K ⊂ L such that K ∩ ω = ∅.Fix an orientation and a framing of K, we can define a continuous map by taking the limit holonomy of the orbifold connections along the longitude of K.The map µ K then gives a local system.The local systems defined by different framings of K are isomorphic via multiplications by powers of t, therefore the choice of the framing is not important.More generally, suppose there is a sublink we can choose a framing for each K j and define the map µ Kj as above, hence we obtain a local system Γ associated with L defined by If L is the empty link, then µ L = 1, thus the local system Γ is the trivial system with coefficient R. In this case, we have (5) Suppose (Y 0 , L 0 , ω 0 ) and (Y 1 , L 1 , ω 1 ) are two admissible triples with local systems Γ 0 and Γ 1 associated with oriented sublinks L 0 ⊂ L 0 and L 1 ⊂ L 1 respectively.Let ).This makes the instanton Floer homology with local coefficients a functor.By the definition of cobordism of triples, S and η are required to be embedded surfaces in W .We can also consider the situation where S is an immersed surface with transverse double points, as discussed in [KM11b, Section 5] and [Kro97].In this situation, one can blow up W at the self-intersection points of S to resolve the double points and obtain an ordinary cobordism ( W , S, η), and then define I(W, S, η) := I( W , S, η).Now suppose S = S 0 S 1 and S = S 0 S 1 are two immersed surfaces with transverse double points in W such that η ∩ S = η ∩ S = ∅, ∂S = ∂S = L 0 ∪ L 1 , and ∂S 0 = ∂S 0 = L 0 ∪ L 1 .We consider the following 5 situations: (i) S is obtained from S by an ambient isotopy; (ii) S 1 = S 1 , and S 0 is obtained from S 0 by a twist move introducing a positive double point (see [FQ90,Section 1.3] for the definition of twist move); (iii) S 1 = S 1 , and S 0 is obtained from S 0 by a twist move introducing a negative double point; (iv) S 1 = S 1 , and S 0 is obtained from S 0 by a finger move introducing two double points of opposite signs (see Figure 3 for a schematic picture and see [FQ90, Section 1.5] for the precise definition of finger move); (v) S is obtained from S by a finger move introducing two double points of opposite signs in S 0 ∩ S 1 .
Proposition 3.1.Let (Y 0 , L 0 , ω 0 ), (Y 1 , L 1 , ω 1 ), W, S, S , η, Γ 0 , Γ 1 be as above, let ) be the induced cobordism maps.For the 5 cases listed above, the following equations hold respectively: (i) I(W, S , η) = I(W, S, η); Proof.Part (i) is trivial.(ii), (iii) and (iv) are from [KM13, Proposition 3.1].The proof of (v) is similar to (iv).We first review Kronheimer and Mrowka's proof of (iv) briefly.For simplicity, consider a special case that (W, S, η) is closed, thus it can be viewed as a cobordism from the empty set to the empty set.We also assume S 1 = S 1 = ∅.In this case, is the singular Donaldson invariants (from 0-dimensional moduli spaces) introduced by Kronheimer and Mrowka in [KM95].More precisely, we have where q k,l denotes the singular Donaldson invariant defined by counting the number of points in the 0-dimensional moduli spaces over all orbifold bundles with instanton number k and monopole number l.The restriction of an orbifold SO(3)-bundle to S has a reduction to K ⊕ R where K is an SO(2)-bundle.By definition, the monopole number l is given by If (W , S ) is obtained from (W, S) by a finger move, [Kro97, Proposition 3.1] used a gluing argument to prove that q k,l (W , S , η) = q k,l (W, S, η) − q k−1,l+2 (W, S, η).
When (W, S, η) is closed and S 1 = S 1 = ∅, Part (iv) follows immediately from (6) and (7).Since the gluing argument only depends on the local structure of the finger move, it is straightforward to extend the argument to the general (relative) case.
To prove Part (v), we first assume (W, S, η) is closed and S 1 = ∅.We give a refined definition of the monopole number l by taking It is clear from the definition that l = l 0 +l 1 .With this definition, we have a refined singular Donaldson invariant q k,l0,l1 .Similar to (6), we define the polynomial If S is obtained by a finger move that introduces intersection points between S 0 and S 1 , then the proof of [Kro97, Equation ( 23)] shows that there exist universal constants a i,j ∈ Z such that By the Uhlenbeck compactness theorem, only finitely many a i,j 's are non-zero.This implies that Q(W , S , η)(t 0 , t 1 ) = P (t 0 , t 1 )Q(W, S, η)(t 0 , t 1 ) (8) for a universal polynomial We also have P (t 0 , t 1 ) = P (t 1 , t 0 ) because there is no difference between the roles of S 0 and S 1 in the finger move.We claim that P (t, 1) = P (1, t) = 0. (10) In fact, suppose the contrary, then (t 0 − 1)|P (t 0 , t 1 ), (t 1 − 1)|P (t 0 , t 1 ), thus we have (t 0 − 1)(t 1 − 1)|P (t 0 , t 1 ), therefore (t − 1)(t − 1)|P (t, t), which contradicts (9), hence the claim is proved.Now let θ(t) := P (t, 1), we have therefore in the closed case, Part (v) of the proposition follows from (8) and (10).By the gluing argument, the same result holds for the non-closed case.
Suppose (Y, L 0 , ω) is an admissible triple, and let L 0 be a sublink of L 0 such that L 0 ∩ ω = ∅.Fix an orientation of L 0 .By the previous discussion, L 0 defines a local system Γ 0 with coefficient R. Suppose L 1 is obtained from L 0 by a local crossing change in Y − ω, where the crossing is either within L 0 or between L 0 and L 0 − L 0 , and let Γ 1 be the local system of (Y, L 1 , ω) associated with the image of L 0 after the crossing change.
The crossing change induces an immersed cobordism S : L 0 → L 1 , where S is an immersed surface in [0, 1] × Y with one double point.Reversing S, we obtain an immersed cobordism S : L 1 → L 0 with one double point.The composition S ∪ S ⊂ [0, 2] × Y can be obtained from the product cobordism [0, 2] × L 0 by a finger move decribed by case (iv) or case (v) of Proposition 3.1.Therefore by Proposition 3.1, the map is equal to (1 − t 2 ) id or θ(t) id.Similarly, the map is equal to (1 − t 2 ) id or θ(t) id.As a consequence, we have the following result.
Proposition 3.2.Suppose (Y, L 0 , ω) is an admissible triple and L 0 ⊂ L 0 is a sublink with L 0 ∩ ω = ∅.Fix an orientation on L 0 and let Γ 0 be the local system of (Y, L 0 , ω) defined by L 0 .Suppose L 1 is an oriented link that is homotopic to L 0 in Y − ω and is disjoint from Let Γ 1 be the local system of (Y, L 1 , ω) defined by L 1 .Then we have where T is the multiplicative system generated by (1 − t 2 )θ(t).
Proof.Since L 0 is homotopic to L 1 in Y − ω, the link L 1 can be obtained from L 0 by a finite sequence of crossing changes in Y − ω, such that the sublink L 0 − L 0 remains fixed.Without loss of generality, we may assume that L 1 is obtained from L 0 by one such crossing change.Let S ⊂ Y × [0, 1] be the immersed cobordism from L 0 to L 1 given by the crossing change, and let S be the reverse of S. By (11), (12), and the functoriality of the instanton Floer homology with local coefficients, we have ) id or θ(t) id on I(Y, L 0 , ω; Γ 0 ), and on I(Y, L 1 , ω; Γ 1 ).Therefore T −1 I(Y, L 0 , ω; Γ 0 ) and T −1 I(Y, L 1 , ω; Γ 1 ) are isomorphic.
Corollary 3.3.Let Y, ω, L 0 , L 1 , Γ 0 , Γ 1 be as in Proposition 3.2, we have Given an oriented link L in S 1 × D 2 , we define the annular instanton Floer homology with local coefficients by where Γ is the local system associated with L. The operator µ orb (S 2 ) on AHI(L; Γ) is now an R-module homomorphism instead of a C-linear map, therefore AHI(L; Γ) no longer carries the f-grading.The torus excision theorem ([KM11a, Theorem 5.6]) still holds for instanton Floer homology with local coefficients, as long as the exicion surface is disjoint from the sublink defining the local system.Therefore, Proposition 2.1 still holds for the annular instanton Floer homology with local coefficients, except that there is no f-grading anymore.
Example 3.4.By Example 2.2, the critical points of the (perturbed) Chern-Simons functional for AHI(U 1 ) (or AHI(K 1 )) consist of two points whose homological degrees differ by 2. Therefore there are no differentials in the Floer chain complex, and we have By Proposition 2.1, we have Proposition 2.6 follows from the torus excision theorem, therefore it also works in the case with local coefficients.Let L ⊂ S 1 × D 2 be a link with n components and view S 1 × D 2 as the complement of a neighborhood of the unknot U in S 3 , let p ∈ U and let Γ L be the local system associated with L, we have Suppose the annular link L has n components, then the embedded image of L in S 3 is homotopic to the embedded image of U n in S 3 .By Corollary 3.3, we have By Proposition 2.6 again, we have In conclusion, we obtain rank R AHI(L; Γ) = 2 n .(13) By the universal coefficient theorem, we have

Limits of chain complexes
This section discusses a simple observation from linear algebra and its consequences in instanton Floer homology.
Suppose {C n } n∈Z is a sequence of finite dimensional complex vector spaces.For each n ∈ Z and k ≥ 0, let be an endomorphism of C n .Suppose for each pair (n, k), we have n .Moreover, suppose for each n, the limits The maps f to be the direct sum of the generalized eigenspaces of f (k) n with eigenvalues in Λ.Similarly, define E n,Λ ⊂ H n to be the direct sum of the generalized eigenspaces of f n with eigenvalues in Λ.
(3) If we further assume that dim H Suppose there exists a closed set Λ ⊂ C such that the statement of Part (1) does not hold.After taking a subsequence, we may assume that the dimensions of n are independent of k, and that they are convergent in the corresponding Grassmannians as k → ∞.The spectrum of f n (with multiplicities) on is the limit of the spectra of f be the direct sum of the generalized eigenspaces of f On the other hand, we have contradicting the assumption.
(2) Let 0 be sufficiently small such that (3) Let 0 be sufficiently small such that Hence by Part (1), for k sufficiently large we have dim As a consequence, for k sufficiently large, dim Since the above equation holds for all < 0 , Part (3) of the lemma is proved.
Recall that given an admissible triple (Y, L, ω) and a continuous function µ : B(Y, L, ω) → R/Z, there is a local system Γ µ on B(Y, L, ω) defined by µ.The Floer chain complex C µ is a finitely generated free R-module, where we identify C h with C using the above isomorphism.The differentials d h become a continuous family of linear maps on C. Given an embedded surface F ⊂ Y , define then µ orb (F ) h is continuous with respect to h and is a chain map on (C, d h ).Therefore, for each h ∈ C − {0}, the map µ orb (F ) h induces a map on the Floer homology To simplify notations, we will use Γ µ (h) to denote Γ µ ⊗ R R/(t − h) for the rest of this article.If h = 1, then I(Y, L, ω; Γ µ (1)) is the ordinary instanton Floer homology without local coefficients, and µ orb (F ) 1 coincides with the ordinary µ map.Proposition 4.2.Let Y, ω, L 0 , L 1 , L 0 , L 1 , Γ 0 , Γ 1 be as in Proposition 3.2.Let θ(t) be the polynomial given by Part (v) of Proposition 3.1.Suppose h ∈ C−{0} satisfies then we have Proof.Let T ⊂ R be the multiplicative system generated by (1 − t 2 )θ(t) as in Proposition 3.2.By ( 16), the elements of On the other hand, since localization is an exact functor, we have Therefore by Proposition 3.2, Since R is a principal ideal domain, the localization T −1 R is also a principal ideal domain, hence (17) follows from the universal coefficient theorem and the isomorphisms ( 18) and ( 19).It remains to prove that (17) intertwines with with µ orb (F ) h .Since the isomorphism ( 19) is induced by a cobordism in which the two copies of the surface F on the two ends are homologous, it intertwines the µ orb (F ) h on the in-coming end with the µ orb (F ) h on the out-going end, hence the statement is proved.Lemma 4.1 and Proposition 4.2 have the following application.
Proposition 4.3.Suppose L ⊂ S 1 × D 2 is an oriented link such that every component of L has winding number 0 or ±1.Assume there are k components with winding number 0 and l components with winding number ±1, then we have Proof.For λ ∈ C, let N (λ, ) be the closed -neighborhood of λ in C. Given a vector space V over C, a linear map f : V → V , and a subset Λ ⊂ C, we use E(V, f, Λ) to denote the direct sum of the generalized eigenspaces of f with eigenvalues in Λ.
Recall that AHI(L; Γ) is defined to be the instanton Floer homology where Γ is the local coefficient system associated with L.
and the proposition is proved.
Corollary 4.4.Suppose L ⊂ S 1 ×D 2 is an oriented link such that every component of L has winding number 0 or ±1.Assume there are k components with winding number 0 and l components with winding number ±1.Moreover, assume Then there exists a meridional disk S in S 1 × D 2 , such that S intersects every component of L with winding number ±1 transeversely at one point, and S is disjoint from every component of L with winding number 0.
By Theorem 2.4, there exists a meridional surface S with genus g, such that S intersects L transversely at n points, and 2g + n = l.On the other hand, every component with a non-zero winding number must intersect S, therefore we have g = 0 and n = l, and the surface S is the desired meridional disk.

Linking numbers and forests of unknots
This section proves a weaker version of Theorem 1.2: (2) there exists a forest of unknots Before starting the proof, we need to make some preparations.Notice that Batson and Seed's work [BS15] implies the following useful result.

Proposition 5.2 ([BS15]
). Suppose L is a link in S 3 with n components and Hence the inequalities above achieve equality, and we have The above result together with Kronheimer-Mrowka's unknot detection theorem in [KM11a] imply the following proposition.rank Z/2 Khr(L, p; By the universal coefficient theorem, Let L be the mirror image of L. By [Kho00, Corollary 11], Using Kronheimer-Mrowka's spectral sequence ([KM11a, Theorem 8.2]) whose E 2page is Khr(L, p; Z) and which converges to I (L, p; Z), we obtain On the other hand, Proposition 5.3 and ( 14) Proof of Theorem 5.1.We prove the theorem by induction on n.When n = 1, it is the unknot-detection theorem of Kronheimer and Mrowka [KM11a].
Assume the theorem holds if the number of components is smaller than n.Since G is a forest, we can find a vertex of G with degree less than or equal to 1.We discuss two cases.
Case 1: There is a vertex of G with degree 1.Without loss of generality, assume this vertex corresponds to the component K n of L G .By the assumption of Theorem 5.1, there exists According to Proposition 5.3, K n is an unknot.Let N (K n ) be a tubular neighborhood of K n , then L can be viewed as a link in the solid torus S 3 − N (K n ).By Proposition 2.6 and Proposition 5.4 we have According to Proposition 4.4, we can find a meridional disk S in the solid torus S 3 − N (K n ) which intersects K i at a single point and is disjoint from the other components.The meridional disk S is a Seifert disk of K n .By the induction hypothesis, L is a forest of unknots.We can shrink K n via S to a small meridian of K i .Therefore L is also a forest of unknots.Since the linking numbers uniquely determine a forest of unknots, we conclude that L is isotopic to L G .
Case 2: There is a vertex of G with degree 0. Without loss of generality, assume this vertex corresponds to the component K n of L G .By the assumption of Theorem 5.1, we have lk We can view L as a link in the solid torus S 3 − N (K n ), and the same argument as above gives AHI(L ) ∼ = C 2 n−1 .By Proposition 4.4, we can find a meridional disk S in the solid torus S 3 − N (K n ) which is disjoint from L .Therefore L is the disjoint union of L and the unknot, and the result follows from the induction hypothesis on L .

The case of 2-component links
This section proves that Condition (2) of Theorem 5.1 is implied by Condition (1) when n = 2.The main result of this section is the following lemma.
Combining this lemma with Theorem 5.1, we have the following corollary.Corollary 6.2.Suppose L is a link with 2 components and rank Z/2 Kh(L; Z/2) = 4, then L is either the 2-component unlink or the Hopf link.
We start the proof of Lemma 6.1 with the following lemma.Lemma 6.3.Suppose L = K 1 ∪ K 2 satisfies the assumption of Lemma 6.1, and suppose L is not the unlink.Then K 1 is an unknot, and K 2 is a braid closure with axis K 1 .Similarly, K 2 is an unknot, and K 1 is a braid closure with axis K 2 .
Proof.Proposition 5.3 implies that K 1 and K 2 are both unknots.Let N (K 1 ) be a tubular neighborhood of K 1 , then K 2 is a knot in the solid torus S 3 − N (K 1 ).Proposition 5.4 yields dim C I (L, p) = 2 for every p ∈ L. By Proposition 2.6, If AHI(K 2 ) is supported at f-degree 0, then by Theorem 2.4, there exists a meridional disk which is disjoint from K i .This means K i is included in a 3-ball in the solid torus S 3 − N (K 1 ), hence K 1 and K 2 are split, therefore the link L is the unlink, contradicting the assumption.Therefore AHI(K 2 ) is supported at f-degrees ±l for l > 0. By (21), we have AHI(K 2 , ±l) ∼ = C and AHI(K 2 ) vanishes at all the other f-degrees.According to Proposition 2.5, K 2 is the closure of an l-braid in S 3 − N (K 1 ).
The same argument for S 3 − N (K 2 ) proves the second half of the lemma.
Remark 6.4.A link described by the conclusion of Lemma 6.3 is called an exchangeably braided link.This concept was first introduced and studied by Morton in [Mor85].
Let l > 1 be an integer.Recall that the braid group B l is given by The reduced Burau representation (see [Bir74]) is a group homomorphism for l > 2, while for l = 2 it is defined by ρ(σ 1 ) := (−t).Notice that for every β ∈ B l , there exists an integer a such that We also need the follow result by Morton.
Theorem 6.5 ([Mor85, Theorem 3]).Suppose L = U ∪ β is a 2-component link where U is the unknot and β is the closure of a braid β ∈ B l with axis U .Then the multi-variable Alexander polynomial ∆ L (x, t) of L is given by where x and t are variables corresponding to U and β respectively.
Remark 6.6.The sign " .=" in Theorem 6.5 means the two sides are equal up to a multiplication by ±x a t b .This is necessary because the multi-variable Alexander polynomial is only defined up to a multiplication by ±x a t b .Lemma 6.7.Suppose L = K 1 ∪ K 2 is an exchangeably braided link with linking number l ≥ 2. Let ∆ L (x, y) be the multi-variable Alexander polynomial of L. Then the expansion of the Laurent polynomial (x − 1)(y − 1)∆ L (x, y) has (strictly) more than 4 terms.
Proof.Without loss of generality, assume x and y are the variables corresponding to K 1 and K 2 respectively.Let β ∈ B l be the braid whose closure is isotopic to K 2 as a link in the solid torus S 3 − K 1 .By ( 22) and Theorem 6.5, we have Switching the roles of K 1 and K 2 , we have for some b ∈ Z, By (23), we have hence we have the following expansion in order of increasing powers of x: The right-hand-side has at least 4 terms after expansion, which come from the lowest and highest powers of x.Suppose it has only 4 terms in total, then all the terms in between must vanish, thus we have Plugging in x = 1, we have 0 = ±(y − 1)y a + (y − 1), therefore a = 0, and (25) gives which contradicts (24) when l ≥ 2.
The link Floer homology was originally defined only for Z/2-coefficients, and was generalized to Z-coefficients in [Sar11].It is known that but HFL carries more refined gradings.By [Dow18, Corollary 1.7], we have On the other hand, let ∆ L (x, y) be the multi-variable Alexander polynomial of L, it was proved in [OS08] that the graded Euler characteristic of HFL(L; Q) satisfies By Lemma 6.7, we have which contradictions (26).
We introduce the following condition on a link L ⊂ S 3 : Condition 6.8.
(1) L has n ≥ 3 connected components, (2) the rank of Kh(L; Z/2) is 2 n , (3) the components of L can be arranged as a sequence K 1 , • • • , K n , such that the linking number of K i and K j (i = j) is ±1 when |i − j| = 1 or n − 1, and is zero otherwise.
Theorem 5.1 and Lemma 6.1 have the following consequence.
Lemma 6.9.If L 0 is an m-component link with rank Z/2 Kh(L 0 ; Z/2) = 2 m , then either L 0 is a forest of unknots, or L 0 contains a sublink L satisfying Condition 6.8.
Proof.Let K 1 , • • • , K m be the components of L 0 .By Proposition 5.2 and Lemma 6.1, for each pair i = j, the linking number of K i and K j is equal to 0 or ±1.Let G be a simple graph with m vertices p 1 , • • • , p m , such that p i and p j are connected by an edge if and only if If G is a forest, then Theorem 5.1 implies that L 0 is a forest of unknots.If G contains a cycle, then the vertices of the shortest cycle of G corresponds to a sublink of L 0 satisfying Condition 6.8.
The rest of this article proves that there is no link satisfying Condition 6.8, therefore Theorem 1.2 follows from Lemma 6.9.

Topological properties from instanton Floer homology
From now on, let L = K 1 ∪• • •∪K n be a hypothetical link that satisfies Condition 6.8.The goal is to deduce a contradiction from Condition 6.8.This section derives several topological properties of L using instanton Floer homology.7.1.Seifert surfaces of K i .Proposition 7.1.For each K i , there exists an embedded disk D i such that (1) Proof.Pick a base point p ∈ K i .By Proposition 5.4, we have dim I (L, p) = 2 n−1 .By Proposition 5.3, K i is an unknot.Let N (K i ) be a tubular neighborhood of K i , and view L − K i as a link in the solid torus By Corollary 4.4, there exists a meridional disk Di which is disjoint from K j if |i − j| = 1 or n − 1 and intersects K j transversely at one point if |i − j| = 1 or n − 1.
The meridional disk Di extends to the desired Seifert disk of K i .
Definition 7.2.Let D 1 , • • • , D n be a sequence of immersed disks in R 3 such that ∂D i = K i for all i.We say that the sequence D 1 , • • • , D n is generic, if every selfintersection point of D i is locally diffeomorphic to one of the following models in R 3 at (0, 0, 0): (1) the intersection of {(x, y, z)|z = 0, y ≥ 0} and the yz-plane, (2) the intersection of the xy-plane and the yz-plane, (3) the intersection of the xy-, yz-, and xz-planes.Notice that if D is generic, then the complexity of D is greater than or equal to 1 2 #Σ 1 (D) which is at least n.Definition 7.4.We say that the sequence Remark 7.5.In the definitions above, the disks {D i } are only required to be immersed.Condition (2) in the definition above is equivalent to the following statement: for each j = i, if |i − j| = 1 or n − 1, then D i intersects K j transversely at one point; otherwise, the immersed disk D i is disjoint from K j .Moreover, the interior of D i is disjoint from K i .
Proposition 7.6.There exists a sequence of disks  We first show that all the D i 's are embedded.Suppose there exists i such that D i is not embedded, then by admissibility, D i does not have triple self-intersections, and the self-intersection locus of D i is a disjoint union of circles.Let γ ⊂ D i be a circle in the self-intersection of D i .
Let B 2 be the unit disk in R 2 , and let f i : B 2 → R 3 be an immersion that parametrizes D i .Then f −1 i (γ) is a double cover of γ.There are three possibilities: Case 1. f −1 i (γ) is a disjoint union of two circles, and they bound disjoint disks B 1 and B 2 .In this case, take a diffeomorphism ι from B 1 to B 2 , such that By smoothing f i , we obtain an immersed disk with the same boundary as D i but has fewer self-intersection components.Figure 4 shows a local picture of f i after the smoothing.Replacing D i by the image of the smoothed f i decreases the complexity of D and preserves the admissibility condition.
Case 2. f −1 i (γ) is a disjoint union of two circles, and they bound disks B 1 and B 2 with B 1 ⊃ B 2 .In this case, take a diffeomorphism ι from B 1 to B 2 , such that Replacing f i by the smoothed version of f i gives an admissible configuration with smaller complexity.
Case 3. f −1 i (γ) is one circle, and it bounds a disk B 1 .In this case, f i | ∂B1 is a non-trivial covering map.Take a diffeomorphism ι from B 1 to B 1 , such that its restriction to ∂B 1 is the deck transformation.Define Replacing f i by the smoothed version of f i gives an admissible configuration with smaller complexity.Since D is assumed to have minimal complexity among admissible configurations, we conclude that D i has to be embedded.Now we show that the complexity of D is n.In fact, since all the disks D i are embedded, the intersection of D i and D j (i = j) is a disjoint union of compact 1-manifolds possibly with boundary.If the complexity of D is greater than n, then there exists i = j such that the intersection of D i and D j contains a circle γ.The circle γ bounds a disk B 1 in D i , and bounds a disk B 2 in D j .Let Replace D i , D j by D i and D j and smooth the corners, we obtain a generic configuration with smaller complexity.Since neither D i nor D j has triple self-intersection points, the new configuration is still admissible, contradicting the definition of D. In conclusion, the complexity of D is n.
6 has the following corollary.
Corollary 7.7.The link L is a connected sum of n − 2 Hopf links as given by Figure 5.The link L has a diagram described by Figure 6.
Proof.By Proposition 7.6, there exists a sequence of disks   respectively at an arc, and is disjoint from hence L is isotopic to a diagram described by Figure 6.7.2.Seifert surfaces of L .We recall the following property of fibered links.
Lemma 7.8.Suppose L 1 and L 2 are two oriented fibered links with oriented Seifert surfaces S 1 and S 2 respectively.Let f 1 : S 1 → S 1 and f 2 : S 2 → S 2 be the monodromies.Take p 1 ∈ L 1 , p 2 ∈ L 2 , and form the connected sum L 1 #L 2 and the boundary connected sum S 1 # b S 2 with respect to (p 1 , p 2 ).Then L 1 #L 2 is a fibered link with Seifert surface S 1 # b S 2 and monodromy f 1 # b f 2 .
Proof.Given a compact surface S with boundary, and given a diffeomorphism f : S → S that restricts to the identity on a neighborhood of ∂S, define where ∼ is defined by (x, 0) ∼ (f (x), 1) for x ∈ S, and (x, t 1 ) ∼ (x, t 2 ) for x ∈ ∂S, t 1 , t 2 ∈ [0, 1].By the assumptions of the lemma, and the images of ∂S 1 and ∂S 2 are isotopic to L 1 and L 2 respectively.Therefore, Notice that the Hopf link is fibered.Depending on the orientations of the components, the corresponding Seifert surface is given by Figure 7 or Figure 8.Both Seifert surfaces are diffeomorphic to the annulus, and the monodromies are Dehn twists along the core circles.
Let S 1 and S 2 be the Seifert surfaces of L given by Figure 9 and Figure 10 respectively.By Lemma 7.8, the link L is fibered with respect to both S 1 and S 2 .For each j = 1, 2, endow the components K 1 , • • • , K n−1 with the boundary orientation of S j and choose an arbitrary orientation for K n , then the algebraic intersection number of K n and S j is equal to the sum of linking numbers  Therefore, part (3) of Condition 6.8 implies that there exists exactly one element j ∈ {1, 2} such that the algebraic intersection number of S j and K n is zero.The main result of this subsection is the following proposition.Proposition 7.9.Suppose j ∈ {1, 2} and the algebraic intersection number of K n with S j is zero.Then there exists a knot K n , such that K n is disjoint from S j , and Before proving Proposition 7.9, we need to prove some results on instanton Floer homology.Let U be an unknot included in a 3-ball which is disjoint from L , let m i be a small meridian circle around K i (1 ≤ i ≤ n − 1) and u i be a small arc joining K i and m i .
Lemma 7.10.We have and where Γ U is the local system associated with U .
Proof.Pick a crossing between m 1 and K 1 and apply Kronheimer-Mrowka's unoriented skein exact triangle in [KM11a, Section 6], we obtain a 3-cyclic exact sequence Section 3] for more details.The above exact triangle implies Repeating this argument to the other meridians, we obtain By Proposition 5.2 and Proposition 5.4, A similar earrings-removal argument yields We recall some properties of the instanton knot Floer homology KHI for oriented links, which was introduced in [KM10, Definition 2.4].Given an oriented link M ⊂ S 3 , the homology group KHI(M ) carries an Alexander Z-grading and a homological Z/2-grading.The rank of KHI(M ) does not depend on the orientation of M .We use KHI(M, i) to denote the summand of KHI(M ) with Alexander degree i, and use χ(KHI(M, i)) to denote its Euler characteristic with respect to the homological grading.Recall that we always take coefficients in C for instanton Floer homology in this article.According to [KM10, Theorem 3.6 and (14)], we have where ∆ M (t) denotes the single-variable Alexander polynomial of M .Notice that the Alexander polynomial for L satisfies |∆ L (−1)| = 2 n−2 for every orientation of L .Therefore, taking M = L , we have By [Xie18a, Proposition 5.1], we have Equations ( 30) and (32) imply (27).Consider the two admissible triples where v is an arc joining the two components of S 1 × {p 1 , p 2 }.Let N (K 1 ) be a small tubular neighborhood of K 1 in the first triple, and deform U into N (K 1 ) by an isotopy.Let N (S 1 × {p 1 }) be a small tubular neighborhood of S 1 × {p 1 } in the second triple.Cut out N (K 1 ) and N (S 1 × {p 1 }), exchange them, and glue back, we obtain two new triples where U is an unknot included in a 3-ball disjoint from S 1 × {p 1 , p 2 }.By the torus excision theorem and and the definition of AHI, we have where Γ ∅ is the trivial local system with coefficient R. By (5) and Example 3.4, AHI(∅; Γ) ∼ = R and AHI(U ; Γ) ∼ = R 2 .By ( 5) and ( 27), Therefore by (33), we have This completes the proof of (29).Let Γ Kn be the local system on R(S 3 , L ∪ n−1 i=1 m i , n−1 i=1 u i ) associated with K n .By Corollary 3.3 and the universal coefficient theorem, we have The above inequality together with (31) implies (28).
Choose j ∈ {1, 2} such that the algebraic intersection number of S j and K n is zero.Choose an orientation of S j , and endow L with the boundary orientation.For each i = 1, • • • , n − 1, let N (K i ) be a sufficiently small tubular neighborhood of K i that is disjoint from m i , and apply a Dehn surgery on N (K i ) that glues the meridian of N (K i ) to S j ∩ ∂N (K i ).Since S 3 − L is fibered over S 1 with fiber S j , the manifold obtained from the Dehn surgeries is fibered over S 1 with fiber S 2 .Since the orientation-preserving mapping class group of S 2 is trivial, the resulting manifold is diffeomorphic to S 1 × S 2 with the product fibration.Let K1 , be the union of the earrings, and let m := m1 ∪ • • • ∪ mn−1 be the image of m.We further require that the surgery on N (K i ) fixes u i ∩ ∂N (K i ) for all i = 1, • • • , n − 1, hence the image of u i is an arc connecting Ki and mi , and we denote the image of u i by ûi .
Given a C-vector space V , a linear map f : V → V , and λ ∈ C, we will use E(V, f, λ) to denote the generalized eigenspace of f with eigenvalue λ.
Lemma 7.11.For all λ ∈ C, we have where the operator µ orb (S 2 ) is the µ-map defined by {p} × S 2 ⊂ S 1 × S 2 for an arbitrary p ∈ S 1 .
Proof.By the torus excision theorem and Lemma 7.10, we have and where Γ Û is the local system associated with Û , and Γ U is the local system associated with U .Since the algebraic intersection number of K n and S j is zero, we conclude that Kn is homotopic to Û in S 1 × S 2 − n−1 i=1 ûi .Let Γ Kn be the local system associated with Kn .By Proposition 4.2, we have for every h ∈ C − {0} satisfying (1 − h 2 )θ(h) = 0, and this isomorphism commutes with µ orb (S 2 ).As a consequence, for every λ ∈ C and h (37) When h(1 − h 2 )θ(h) = 0, the universal coefficient theorem and (35), (36) imply On the other hand, notice that when h = 1, the local systems Γ Kn (h) and Γ Û (h) become the trivial system with coefficient C, hence by ( 34) and (35), Therefore the desired result follows from (37) by taking the limit h → 1 and invoking Part (3) of Proposition 4.1 .
Notices that L ∪ m is a braid in S 1 × S 2 .In fact, the projection of L ∪ m to S 1 is a diffeomorphism on each component.Therefore, after an isotopy, we may write (1) K1 , m1 are included in A 0 and are given by S 1 × {p 1 } and S 2 × {p 2 } with p 1 , p 2 ∈ D 2 , (2) û1 is an arc connecting K1 and m1 , and û1 is included in then L 0 and L 1 are two annular links in A 1 .By the definition of annular instanton Floer homology, we have Lemma 7.12.Assume there exists a connected oriented Seifert surface S ⊂ S 3 of L , such that S is compatible with the orientation of L , and S has genus g and is disjoint from K n .Then we have for all integers i > 2g + 2n − 4.
Proof.After an isotopy, we may assume that S intersects each m i transversely at one point.The image of S − Recall that K1 and m1 are contained in A 0 ∼ = S 1 ×D 2 and are given by S 1 ×{p 1 } and S 2 × {p 2 } for p 1 , p 2 ∈ D 2 .Take a point p 0 ∈ D 2 − {p 1 , p 2 }, and let K0 ⊂ A 0 be the knot S 1 × {p 0 }.After a further isotopy, we may assume that Ŝ intersects K0 transversely at one point.Let c be a simple closed curve on D 2 such that p 0 , p 1 are inside of c and p 2 is outside.let T 1 ⊂ A 0 be the torus given by T 1 := S 1 × c.
Notice that Ŝ is homologous to the slice of S 2 in S 1 × S 2 , therefore we have µ orb ( Ŝ) = µ orb (S 2 ).The surface Ŝ intersects L ∪ m ∪ Kn ∪ K0 transversely at 2n − 1 points.Apply [XZ19, Theorem 6.1] to the surface Ŝ, we deduce that the set of eigenvalues of µ orb (S 2 ) on Consider the triple (S 1 × S 2 , S 1 × {q 1 , q 2 }, v), where q 1 , q 2 ∈ S 2 and v is an arc connecting S 1 × {q 1 } and S 1 × {q 2 }.Let T 2 be a torus given by the boundary of a tubular neighborhood of S 1 × {q 1 }.Recall that T 1 ⊂ A 0 is the torus S 1 × c as defined above.Applying the torus excision on the triple and for all integers i > 2g + 2n − 4.
Similarly, applying torus excision on the triple Let {z 1 , • • • , z 2n−1 } ⊂ Ŝ be the intersection of Ŝ with L ∪ m ∪ Kn ∪ K0 .Apply the singular excision theorem [XZ19, Theorem 6.4] on the following triple along Ŝ in the first component, and a slice of Ŝ in the second component, and invoke [XZ19, Proposition 6.7], we obtain Since µ orb ( Ŝ0 ) = µ orb (S 2 ), the first part of the lemma is proved by ( 42), ( . The second part of the lemma is proved by (43).
Lemma 7.13.Let L 0 ⊂ A 1 be the annular link defined by (38).Suppose there exists a meridional surface S (cf.Definition 2.3) with genus g such that S intersects L 0 transversely at m points.Then there exsits a connected Seifert surface Ŝ of L in S 3 , such that Ŝ is compatible with the orientation of L and is disjoint from K n , and the genus of Ŝ is equal to g + m/2 − n + 2.
Proof.Suppose there is a component K of L 0 whose intersection with S has different signs, then we can attach a tube to S along a segment of K to decrease the value of m by 2 and increase the value of g by 1. Repeating this process until the number of intersection points of S with each component of L 1 equals the absolute value of their algebraic intersection number, we obtain a new meridional surface S ⊂ A 1 , such that (1) the genus of S equals g + (m Since S is a meridional surface, by attaching a disk in A 0 , we can complete the surface S to a closed surface with the same genus that intersects each of K1 , • • • , Kn−1 transversely at one point and is disjoint from Kn , therefore it gives rise to a Seifert surface of L in S 3 with the same genus that is disjoint from K n , hence the lemma is proved.
Corollary 7.14.Let g 0 be the smallest integer with the following property.There exists a connected oriented Seifert surface S ⊂ S 3 of L that is compatible with the orientation of L , such that S has genus g 0 and is disjoint from K n .Then we have for all integers i > 2g 0 + 2n − 4.
Proof.By Theorem 2.4 and (40), there are integers g, m, such that there exists a meridional surface in A 1 with genus g and intersects L 1 transversely at m points, such that Let g := g + m/2 − n + 2. By Lemma 7.13, we may choose g, m such that there exists a connected oriented Seifert surface of L in S 3 that is compatible with the orientation of L , has genus g , and is disjoint from K n .Since 2g +m = 2g + 2n−4, by Lemma 7.12 and (46), we have By the definition of g 0 , we have g 0 ≤ g .On the other hand, the second part of Lemma 7.12 implies that for all integers i > 2g 0 + 2n − 4. Therefore by (47) and (48), we have g 0 ≥ g .In conclusion, we must have g = g 0 , and the lemma follows from (47) and (48).
Replacing Kn with Û in the previous arguments, we also have the following lemma.
Lemma 7.15.Let g 1 be the smallest integer with the following property.There exists a connected oriented Seifert surface S ⊂ S 3 of L that is compatible with the orientation of L , such that S has genus g 1 and is disjoint from U .Then we have Proof of Proposition 7.9.It is obvious that the minimal genus g 1 in Lemma 7.15 is zero, therefore by Lemma 7.11, Corollary 7.14, and Lemma 7.15, the genus g 0 in Corollary 7.14 is also zero.As a result, there exists a connected oriented Seifert surface S ⊂ S 3 for L with genus zero that is disjoint from K n and is compatible with the orientation of L .Since the minimal-genus Seifert surface for an oriented fibered link is unique up to isotopy, we conclude that there exists an ambient isotopy of S 3 that fixes L and takes S to S j .This ambient isotopy gives the desired isotopy from 8. The fundamental group of R 3 − L This section takes a detour to study the properties of π 1 (R 3 − L ).The results of this section will be used in the proof of the non-existence of L.
By the Wirtinger presentation, π 1 (R 3 − L ) is generated by as shown in Figure 11, where the base point is taken to be a point above and far away from the diagram.Notice that g i and g i are homotopic relative to the base point because one can shrink , and the Wirtinger presentation To simplify the notation, for the rest of this section we will use m to denote n − 1, and use G to denote the group π 1 (R 3 − L ).For i = 1, • • • , m, define the set The first part of this section solves the word problem for G.
Definition 8.3.Define an equivalent relation ∼ on the set of words using the following relations as generators: It is straightforward to verify that if two words are equivalent and one of them is reduced, then the other is also reduced.Therefore ∼ defines an equivalence relation on the set of reduced words.
Every word (x 1 , x 2 , • • • , x N ) represents an element of G by taking the product If the word [x 1 , • • • , x N , y] is not reduced, then there exists u such that x u y = 1 and every letter in (x u+1 , • • • , x N ) is commutative with both x u and y.In this case, define For different choices of x u , the right-hand side of (50) gives the same equivalence class.Moreover, if we take a different representative of [x 1 , • • • , x N ], the right-hand sides of ( 49) and (50) remain the same.It is also straightforward to verify that if y 1 and y 2 are commutative generators of G, then Therefore, we obtain a well-defined product operator on G defined inductively by The associativity of the product operator is clear from the definition.For an element , hence every element in G has an inverse, therefore G is a group.By the universal property, there is a unique homomorphism ϕ from G to G defined by ϕ(g i ) := [g i ].We also have a map ψ from G to G defined by the map ψ is a group homomorphism.It is obvious from the definitions that ϕ and ψ are inverse to each other, therefore ϕ and ψ are isomorphisms, hence the proposition is proved.
Definition 8.6.For w ∈ G, define length(w) to be the length of a reduced presentation of w.
By Proposition 8.4, length(•) does not depend on the choice of the reduced presentation, hence it is well-defined.
is not reduced, there exist letters y u and y v such that y u y v = 1 and every letter between y u and y v are commutative with both y u and y v .Removing y u and y v from the word yields a shorter word representing the same element of G. Repeating this process, we obtain a reduced word representing w which is a sub-word of (y 1 , • • • , y M ).By Proposition 8.4, this word is equivalent to The other direction of the lemma is obvious.
Lemma 8.8.For each i, the centralizer of g i in G is generated by C i .
Proof.Suppose there exists an element w in the centralizer of g i that is not generated by C i , choose such a w such that N := length(w) is as small as possible.
Let w = x 1 • • • x N be a reduced presentation of w, then there exists u such that x N is an element in the centralizer of g i , and by Lemma 8.7, the element x 2 • • • x N is not generated by C i , which contradicts the minimality of N .Therefore x 1 / ∈ C i .Similarly, x N / ∈ C i .Moreover, the same property holds for every reduced word that is equivalent to ( ) is a reduced word.By Proposition 8.4, w is not in the centralizer of g i , contradicting the assumption.Lemma 8.9.Suppose m ≥ 4, then the only element that is commutative to both g 1 and g m is 1.
Proof.The lemma is an immediate consequence of Proposition 8.4, Lemma 8.7 and Lemma 8.8.Lemma 8.10.Suppose (x 1 , • • • , x N ) and (y 1 , • • • , y N ) are reduced words such that x 1 , y 1 are not commutative, then Proof.By Proposition 8.4, we only need to show that the two words (x 1 , • • • , x N ) and (y 1 , • • • , y N ) are not equivalent.Assume the contrary, let w x and w y be the sub-words of (x 1 , • • • , x N ) and (y 1 , • • • , y N ) respectively, consisting of all the letters in {x 1 , y 1 }.Since x 1 and y 1 are not commmutative, w x has to be equal to w y if (x 1 , • • • , x N ) and (y 1 , • • • , y N ) are equivalent.On the other hand, w x starts with x 1 , and w y starts with y 1 , hence w x = w y , which is a contradiction.Lemma 8.11.Suppose m ≥ 4, then the centralizer of g 1 g m is generated by g 1 g m .
Proof.Let w = x 1 • • • x N be an element in the centralizer of g 1 g m , and assume (x 1 , • • • , x N ) is a reduced word.We use induction on N to show that w is a power of g 1 g m .If N = 0, then w = 1 and the property is trivial.From now, assume N > 0, and assume that the claim is proved when length(w) < N .
By the assumptions on w, we have g 1 g m wg −1 m g −1 1 = w, hence the word (g 1 , g m , x 1 , • • • , x N , g −1 m , g −1 1 ) is not reduced.Therefore there are three possibilities: have a presentation of the form u = g 1 û, where length(û) = length(u)−1.Similarly by the assumption that v = v 1 , there is a presentation of v given by v = g −1 m v, where length(v) = length(v) − 1. Equation (51) gives By the induction hypothesis, σ(v) = (g 1 g m ) k v , σ(û) = (g 1 g m ) k û , where k ∈ Z, v is in the centralizer of g 1 , and û is in the centralizer of g m .Therefore is in the centralizer of g 1 , and g −1 m σ(v ) is in the centralizer of g m , the desired result is proved.
Corollary 8.14.Suppose m ≥ 4. The solutions to the equation where k ∈ Z, u is in the centralizer of g 1 , and v is in the centralizer of g m .
Proof.Notice that there is an isomorphism σ : G → G defined by σ(g i ) := g i for i < m, and σ(g m ) := g −1 m .Apply σ to Lemma 8.12 yields this result.

Arcs on compact surfaces
This section collects several results about arcs that will later be used in the proof of the non-existence of L.
Let S be a smooth compact surface with a non-empty boundary, let γ 1 , γ 2 be two smoothly embedded arcs in S such that γ i ∩ ∂S = ∂γ i , and γ i intersects ∂S transversely for i = 1, 2. The following result is well-known, but we need to review the proof for later reference.
The proof of Proposition 9.1 follows from a bigon argument.We start by giving the following definition.Definition 9.2.Let M be a (not necessarily compact) smooth surface with nonempty boundary.For i = 1, 2, let u i be a locally finite disjoint union of smoothly embedded arcs in M such that u i ∩ ∂M = ∂u i , and u i intersects ∂M transversely.A bigon with respect to (M, u 1 , u 2 ) is a closed domain D ⊂ M homeomorphic to a disk, such that ∂D is the union of two arcs, one is contained in u 1 , the other is contained in u 2 .See Figure 12 for an example of a bigon.Definition 9.3.Let M , u 1 , u 2 be as in Definition 9.2.A bigon D with respect to (M, u 1 , u 2 ) is called innermost, if the interior of D is disjoint from u 1 and u 2 .Lemma 9.4.Let M , u 1 , u 2 be as in Definition 9.2.Suppose u 1 is transverse to u 2 .If there exists a bigon with respect to (M, u 1 , u 2 ), then there exists an innermost bigon with respect to (M, u 1 , u 2 ).
Proof.Fix an area form on M .Suppose D is a bigon, let D 0 be a bigon that is included in D and has the smallest area.We claim that D 0 is an innermost bigon.In fact, assume the contrary, then the interior of D 0 intersects u 1 or u 2 .Without loss of generality, assume the interior of D 0 intersects u 1 .Since u 1 does not have closed components, every component of interior(D 0 ) ∩ u 1 bounds a bigon with ∂D 0 ∩ u 2 , see Figure 13 for an illustration.This contradicts the minimality of the area of D 0 .Lemma 9.5.Let M , u 1 , u 2 be as in Definition 9.2 such that u 1 and u 2 intersect transversely in M .Suppose M is a planar surface.If there is a component c 1 of u 1 and a component c 2 of u 2 such that ∂c 1 = ∂c 2 , then there exists an innermost bigon with respect to (M, u 1 , u 2 ).
Proof.Since M is planar, every simple closed curve in M bounds a disk.Let {p 1 , p 2 } be the boundary of c 1 and c 2 , let q be the element of (c 1 ∩ c 2 ) − {p 1 } that is closest to p 1 along c 1 .Then p 1 and q bound two segments in c 1 and c 2 .These two segments form a simple closed curve in M , hence they bound a bigon with respect to (M, c 1 , c 2 ).See Figure 14 for an illustration.By Lemma 9.4, there exists an innermost bigon with respect to (M, u 1 , u 2 ).
Proof of Proposition 9.1.Perturb γ 1 such that γ 1 and γ 2 intersect transversely.Let S be the universal cover of S, and let γ1 and γ2 be the pre-images of γ 1 and γ 2 in S respectively.Since ∂S = ∅, the surface S is planar.By the assumption, γ 1 and γ 2 are homotopic in S relative to the boundary.Therefore, for every component c 1 of γ1 , there exists a component c 2 of γ2 such that ∂c 1 = ∂c 2 .By Lemma 9.5, there exists an innermost bigon D with respect to ( S, γ1 , γ2 ).Let τ be a non-trivial deck transformation on S, we claim that the interior of D is disjoint from τ (D).In fact, assume interior(D) ∩ τ (D) = ∅, since D is an innermost bigon, we have D ⊂ τ (D), hence by Brower's fixed point theorem τ has a fixed point, which yields a contradiction.As a consequence, the interior of D maps injectively into S, hence it is the interior of an innermost bigon D 0 with respect to (S, γ 1 , γ 2 ).
If γ 1 and γ 2 do not intersect in the interior, then ∂D 0 = γ 1 ∪ γ 2 , and D 0 gives an isotopy from γ 1 to γ 2 relative to the boundary.If γ 1 and γ 2 have a non-empty intersection in the interior, then D 0 gives an isotopy relative to the boundary that reduces the number of interior intersection points.By induction on the number of interior intersection points of γ 1 and γ 2 , we conclude that γ 1 and γ 2 are isotopic relative to the boundary.
Using similar techniques, we can prove the following lemma.Proof.Since M is a planar surface, Lemma 9.5 implies that there exists an innermost bigon D with respect to (M, γ 1 , γ 2 ).Since the partition of {• • • , D 0 , D 1 , • • • } by γ 1 and γ 2 are the same, the bigon D has to be disjoint from ∪ k D k .
If γ 1 and γ 2 do not intersect in the interior, then ∂D = γ 1 ∪ γ 2 , and D gives an isotopy from γ 1 to γ 2 relative to the boundary in M − ∪ k D k .If γ 1 and γ 2 have a non-empty intersection in the interior, then D gives an isotopy relative to the boundary in M − ∪ k D k that reduces the number of interior intersection points.By induction on the number of interior intersection points of γ 1 and γ 2 , we conclude that γ 1 and γ 2 are isotopic relative to the boundary in Let f 1 : S → S be a Dehn twist along a curve parallel to S 1 × {0}, and let f 2 : S → S be a Dehn twist along a curve parallel to S 1 × {1}.Suppose γ is an arc on S from (p, 0) to (p, 1), then there exist integers u, v such that γ is isotopic to f u 1 f v 2 (γ 0 ) in S relative to the boundary.Proof.Notice that R × [0, 1] is the universal cover of S 1 × [0, 1], and the deck transformation group is generated by (x, y) → (x + 1, y).  (1) τ (x, t) is independent of t for x ∈ ∂S, (2) τ (x, 0) = x for all x ∈ S, (3) τ (x, 1) = f (x) for all x ∈ S. The map τ induces an isotopy from K(S, γ) to K(S, f (γ)) in R 3 − L 0 by the family of knots K(τ (S, t), τ (γ, t)).

The link L u,v
This section defines a family of links L u,v and computes their Jones polynomials at t = −1.The computation will be used in the proof of the non-existence of the hypothetical link L satisfying Condition 6.8.Definition 10.1.For a pair of integers (u, v) with u ≥ 3, we define a link L u,v as follows.If v ≥ 0, define L u,v to be the link given by Figure 15 with u components such that there are v crossings in the dotted rectangle.If v < 0, define L u,v to be the link given by Figure 16 with u components such that there are |v| crossings in the dotted rectangle.
The only difference between Figure 15 and Figure 16 is that the crossings in the dotted rectangles are reversed.Notice that L u,v is alternating if v ≥ 0. Let V (L u,v ) be the (reduced) Jones polynomial of L u,v with the orientation given by Figure 17.Let V u,v be the value of V (L u,v ) when plugging in t 1/2 = −i.
Notice that the Hopf link with linking number 1 has Jones polynomial −t 1/2 − t 5/2 ; and the Hopf link with linking number −1 has Jones polynomial −t −1/2 −t −5/2 .Moreover, the Jones polynomial of the connected sum of links is the product of the Jones polynomial of each summand.Therefore, if v is even, by the skein relation at the dotted circle in Figure 17, we have On the other hand, if v is even, the skein relation at a crossing in the dotted box in Figure 17 yields If v is odd, then the skein relation gives It can be directly computed that Combining the computations above, we have

The non-existence of L
This section combines the results from Sections 7, 8, 9 and 10 to prove that the hypothetical link L satisfying Condition 6.8 does not exist.We will proceed by showing more properties of L and eventually deduce a contradiction.By Lemma 6.9, this will finish the proof of Theorem 1.2.
Recall that the components of L are K 1 , • • • , K n , and L = K 1 ∪ • • • ∪ K n−1 .We have defined S 1 and S 2 to be the Seifert surfaces of L given by Figure 9 and Figure 10 respectively.By the conditions on the linking numbers of L, there are two possibilities: Case 1.The algebraic intersection number of S 1 and K n is zero; Case 2. The algebraic intersection number of S 2 and K n is zero.By Proposition 7.9, for j ∈ {1, 2}, if the algebraic intersection number of S j and K n is zero, then K n can be isotopically deformed in R 3 − L into R 3 − S j .The first half of this section will focus on Case 1.The argument for Case 2 is similar and will be sketched afterwards.
Let γ 0 be the arc on S 1 as shown in Figure 18, where γ 0 starts from a point p 1 ∈ K 1 and travels from left to right, goes through the crossings of L in an alternating way, and ends at a point p 2 ∈ K n−1 .
Lemma 11.1.Suppose Case 1 holds, then there exists an arc γ ⊂ S 1 from p 1 to p 2 such that K n is isotopic to K(S 1 , γ) in R 3 − L .
Proof.By Proposition 7.9, there exists a knot K n ⊂ R 3 −S 1 such that K n is isotopic to K n in R 3 − L .By Proposition 7.1, K n bounds a disk D n such that D n intersects K 1 and K n−1 respectively at one point, and is disjoint from K 2 ∪ • • • ∪ K n−2 .After a further isotopy, we may assume that D n ∩ L = {p 1 , p 2 }, and that D n intersects S 1 transversely.Therefore D n ∩ S 1 consists of an arc γ ⊂ S 1 from p 1 to p 2 and a union of circles.By Lemma 9.10, K n is isotopic to K(S 1 , γ) in R 3 − L .
Lemma 11.2.Suppose Case 1 holds.Fix an orientation on S 1 , let f 1 , f 2 be the Dehn twists on S 1 along an oriented curve parallel to K 1 and an oriented curve parallel to K n−1 respectively, and let f 3 : S 1 → S 1 be the monodromy of the fibered structure of L .Let γ be the arc given by Lemma 11.1, then there exist integers a, b, c such that γ is isotopic to f a 1 f b 2 f c 3 (γ 0 ) relative to {p 1 , p 2 } on S 1 .
Proof.If n = 3, then S 1 is an annulus, and every arc from p 1 to p 2 is isotopic to f a 1 γ 0 for some integer a.If n = 4, then S 1 is an annulus with a disk removed, and the result follows from Lemma 9.7 with c = 0. From now we assume n ≥ 5.
Fix a point q in the interior of γ 0 as shown in Figure 18.Let γ 1 be the sub-arc of γ 0 from p 1 to q, and let γ 2 be the sub-arc of γ 0 from q to p 2 .Then there exists a closed curve w in the interior of S 1 bases on q, such that γ is homotopic to γ 1 • w • γ 2 relative to {p 1 , p 2 } on S 1 .The loop w is not necessarily simple.
Let g 1 , • • • , g n−1 be the generators of π 1 (R 3 − L , q ) defined in Section 8, where q is their base point.Fix an arc t from q to q as given by Figure 18, let t −1 be the same arc with the reversed orientation, let [w] ∈ π 1 (R 3 − L , q ) be the homotopy class of t • w • t −1 .
We construct a set of generators of π 1 (interior(S 1 ), q) as follows.Let u 1 , • • • , u n−2 be the oriented simple closed curves on S 1 as given by Figure 19.Each u i intersects γ 0 at one point near one of the crossings of L .Let q i be the intersection point of u i and γ 0 , let v i be the sub-arc of γ 0 from q to q i , let v −1 i be the same arc with the reversed orientation.Let u i be the loop based at q defined by v i • u i • v −1 i .
... By Corollary 10.2, if n ≥ 4 then rank Z/2 Kh(L n,1−n ; Z/2) > 2 n .It can be directly verified that L 3,−2 is isotopic (up to mirror image) to the link L6n1 in the Thistlethwaite link table, and the rank of Kh(L 3,−2 ; Z/2) equals 12. Therefore all the links L n,1−n fail to satisfy Part (2) of Condition 6.8.This proves the nonexistence of L for Case 1.
To prove the result for Case 2, let γ 0 be the arc on S 2 given by Figure 23.Then the same argument as in Lemma 11.2 and Corollary 11.3 shows that K n is isotopic to K(S 2 , γ 0 ) in R 3 − L .To see that L is isotopic to L n,C (n) for some function C (n), we use the diagram of (S 2 , γ 0 ) given by Figure 24.Similar to Corollary 11.6, the function C (n) satisfies C (n + 1) = C (n) − 1 .Moreover, when n = 3,   the link L ∪ K(S 2 , γ 0 ) is isotopic to L 3,−1 , see Figure 25.Therefore equation (58) implies C (3) = −1 or −2.On the other hand, the condition of Case 2 implies that C (n) has the same parity as n, therefore C (3) = −1, hence C (n) = 2 − n.By Corollary 10.2, when n ≥ 6 we have rank Z/2 Kh(L n,2−n ; Z/2) > 2 n .The link L 3,−1 is isotopic up to mirror image to the link L6n1 in the Thistlethwaite link table, and the rank of Kh(L 3,−1 ; Z/2) equals 12.The link L 4,−2 is isotopic up to mirror image to L8n8, and the rank of Kh(L 4,−2 ; Z/2) equals 24.The link L 5,−3 is isotopic up to mirror image to L10n113, and the rank of Kh(L 5,−3 ; Z/2) equals 60.This proves the desired result for Case 2. In conclusion, we have proved that the link L satisfying Condition 6.8 does not exist, therefore Theorem 1.2 follows from Lemma 6.9.
numbers and forests of unknots 15 6.The case of 2-component links 16 7.Topological properties from instanton Floer homology 19 8.The fundamental group of R 3 − L 32 9. Arcs on compact surfaces 37 10.The link L u,v 41 11.The non-existence of L

Figure 1 .
Figure 1.Example of a forest of unknots.
Figure 1.Example of a forest of unknots.

Figure 2 .
Figure 2. Example of a forest of unknots.
and suppose µ : B(Y, L, ω) → R/Z is a continuous function.For each a ∈ B(Y, L, ω), define a rank-1 free R-module by the formal multiplication Γ µ a := t μ(a) • R, where μ(a) is a lift of µ(a) in R. Let Crit(CS) be the set of critical points of the (perturbed) Chern-Simons functional CS, define a free R-module C µ by C µ := α∈Crit(CS)

Figure 3 .
Figure 3.A schematic picture of the finger move.
Proposition 5.3 ([BS15, Proposition 7.1]).Suppose L is a link in S 3 with n components and rank Z/2 Kh(L; Z/2) = 2 n , then each component of L is an unknot.Proposition 5.4.Suppose L is a link in S 3 with n components and rank Z/2 Kh(L; Z/2) = 2 n , then for every point p ∈ L, we have dim C I (L, p) = 2 n−1 .Proof.Given a point p ∈ L, we use Khr(L, p) to denote the reduced Khovanov homology with base point p.By [Shu14, Corollary 3.2.C], be the set of selfintersection points described by (1), (2), (3) above respectively.Definition 7.3.If D = (D 1 , • • • , D n ) is generic, define the complexity of D to be the number of components of Σ 2 (D).
and D is admissible with complexity n.Proof.By Proposition 7.1, there exists a sequence of disks D = ( D1 , • • • Dn ) such that for all i, Di is embedded, ∂ Di = K i , and #Σ 1 ( D) = 2n.Perturb D1 , • • • Dn such that they are generic.Since all the disks are embedded, every point in Σ 3 ( D) is contained in three different disks.Therefore D is admissible, hence admissible configurations exist.Let D = (D 1 , • • • , D n ) be an admissible configuration with minimal complexity.

Figure 4 .
Figure 4.The local construction of f i after smoothing.
the disks D i and D j intersect at an arc, (3) if i = j and |i − j| = 1, the disks D i and D j are disjoint.It follows that L is isotopic to a connected sum of n − 2 Hopf links as given by Figure5.Moreover, we may choose the sequence of disks D 1 , • • • , D n−1 in such a way that K n bounds an embedded disk D n which intersects D 1 and D n−1

Figure 7 .
Figure 7. Seifert surface of the Hopf link with linking number −1.

Figure 8 .
Figure 8. Seifert surface of the Hopf link with linking number 1.
is a connected surface with n − 1 boundary components, where the boundary components are give by the meridians of K1 , • • • , Kn−1 .Therefore we can glue disks to the boundary of the image of S − n−1 i=1 N (K i ) and obtain a connected closed surface in S 1 × S 2 with genus g, that is disjoint from Kn and intersects each of K1 , • • • , Kn−1 , m1 , • • • , mn−1 transversely at one point.Denote this surface by Ŝ.After a further isotopy, we may assume that the arcs m1 , • • • , mn−1 lie on Ŝ.
By the definition of G, equivalent words represent the same element.Proposition 8.4.Every element of G is represented by a reduced word.Two reduced words represent the same element if and only if they are equivalent.Proof.Define another group G as follows.The elements of G are the equivalence classes of reduced words.

Figure 17 .
Figure 17.An orientation of L u,v .