Topologically and rationally slice knots
Abstract.
A knot in is topologically slice if it bounds a locally flat disk in . A knot in is rationally slice if it bounds a smooth disk in a rational homology ball. We prove that the smooth concordance group of topologically and rationally slice knots admits a subgroup. All previously known examples of knots that are both topologically and rationally slice were of order two. As a direct consequence, it follows that there are infinitely many topologically slice knots that are strongly rationally slice but not slice.
Key words and phrases:
Topologically slice, rationally slice, involutive knot Floer homology2020 Mathematics Subject Classification:
57K10, 57K181. Introduction
The smooth concordance group of topologically slice knots contains a subgroup; the existence of a subgroup was first proved by Endo in 1995 using gauge theory [End95], while the existence of a subgroup was not proved until nearly 20 years later, using Heegaard Floer homology [HKL16]. Both of these proofs rely on Freedman’s landmark result [Fre82] that any knot with Alexander polynomial 1 is topologically slice.
Similarly, the smooth concordance group of rationally slice knots also contains a subgroup. However, our understanding of rationally slice knots proceeded in the opposite direction: the proof of the subgroup [Cha07] preceded the proof of the subgroup [HKPS22] by 15 years. For their obstructive side, the former proof relies on J. Levine’s algebraic concordance group [Lev69] while the latter relies on involutive Heegaard Floer homology [HM17]. For their constructive side, they use the fact that the figure-eight knot is rationally slice and generalize its proof. This fact was observed by Cochran using work of Fintushel-Stern [FS84].
What about knots that are both topologically and rationally slice?
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Topologically slice but not slice: The knots representing the subgroup of [HKL16] can be shown to be rationally slice by a straightforward modification of the proof of [Cha07, Theorem 4.16]. The generating set for Endo’s family [End95] (the pretzel knots for ) are not rationally slice. There are more topologically slice knots in the literature and many of them are known to be not rationally slice (see e.g. [Hom15, OSS17, KP18, FPR19, DHST21]) and some of them are not known to be rationally slice or not (see e.g. [HK12, CHH13, CK21, Cha21, KKK22]).
Note that all previously known examples of knots that are both topologically and rationally slice have order two in the smooth concordance group. The goal of this paper is to show that the smooth concordance group of topologically and rationally slice knots does in fact contain a subgroup. For any knot , let denote the positive Whitehead double of , and denote the -cable of .
Theorem 1.1.
Denote by the right-handed trefoil and the figure-eight knot by and , respectively. Then every knot in the family
is topologically and rationally slice, and has infinite order in the smooth concordance group. Furthermore, the given family admits a linearly independent infinite subset of the smooth concordance group.
As mentioned in Theorem 1.1, the knots are topologically and rationally slice. Indeed, up to topological concordance, we may ignore , resulting in , which is clearly slice. Similarly, up to rational concordance, we may ignore , which also results in a slice knot. Our family of knots exploits the fact that cabling does not induce a homomorphism on the concordance group; see [HPC21] for further discussion of satellite operators and the concordance group.
We may naturally interpret our main theorem as follows. Let be the smooth concordance group of topologically slice knots and be the smooth rational concordance group of topologically slice knots. Since two concordant knots are rationally concordant, we have the following natural surjective homomorphism:
Note that the main theorem of [HKL16] implies that contains infinitely many order two elements. As a direct corollary of our main theorem, we have the following:
Corollary 1.2.
The group contains a subgroup isomorphic to .
Furthermore, the knots in 1.1 can be used to prove the existence of a knot which is topologically slice and strongly rationally slice but not slice. Strongly rationally slice knots are knots which bound a smoothly embedded disk in a rational homology ball such that the inclusion induces an isomorphism
The question of distinguishing strongly rationally slice knots from slice knots is very subtle; the first example of such a knot was found in [KP22]. By directly applying the proof of [KP22, Theorem 1.4] to the knots in 1.1, we immediately get the following corollary.
Corollary 1.3.
Let be the knots in 1.1. Then each is topologically and strongly rationally slice, and has infinite order in the smooth concordance group.
Recall that there exists a natural surjective homomorphism defined from the smooth concordance group to the smooth rational concordance group. As its counterpart in the topological category, we can also consider the map
defined from the topological concordance group to the topological rational concordance group. Even though we know that there are subgroups in and in [Cha07, HKPS22], the existence of an infinite order element in is still not known. Hence, we ask:
Question 1.4.
Does there exist an infinite order element in ?
Lastly, we make a remark that the positive Whitehead double of the figure-eight knot is also a good candidate that is likely to have infinite order in . To the best of the authors’ knowledge it is not even known if this knot is slice or not. Here, we ask:
Question 1.5.
Is the subgroup generated by in isomorphich to ? More specifically, is slice?
1.1. Proof outline
For the reader’s convenience, we now provide a brief overview of the proof strategy. The proof relies on bordered and involutive knot Floer homology, using several tricks and techniques along the way to ease the calculations as much as possible.
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(1)
Up to -local equivalence, we can replace by (Lemma 4.3).
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(2)
There is a (non-involutive) inclusion map
Using Zhan’s bordered Floer homology calculator [Zha], we determine that certain type-D morphisms are nullhomotopic (Lemmas 4.1 and 4.2), allowing us to conclude that induces an almost -local map from the almost -complex of the -cable of to that of (or equivalently that of , as in Equation 4.1).
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(3)
We combine the result from (2) with results from [KP22] to compare our knots to certain standard complexes , with respect to the partial order on the horizontal almost -local equivalence group (described in Section 2). Using properties of how the interact with the partial order, we arrive at the desired linear independence result.
Acknowledgements
The authors would like to thank Gary Guth for his help on using Zhan’s bordered Floer homology calculator. JH was partially supported by NSF grant DMS-2104144 and a Simons Fellowship. SK was supported by the Institute for Basic Science (IBS-R003-D1). JP was partially supported by Samsung Science and Technology Foundation (SSTF-BA2102-02) and the POSCO TJ Park Science Fellowship.
2. The horizontal almost -local equivalence group
We assume that the reader is familiar with involutive knot Floer homology [HM17] and the -local equivalence group [HMZ18]. See [HKPS22, Section 2] for an expository overview of these concepts. In this paper, we focus on the horizontal almost -local equivalence group, which has the advantage that, modulo the image of the figure eight knot, it is totally ordered. What follows is a short summary of [KP22, Section 2].
A horizontal almost -complex is a pair of a bigraded complex of finitely generated free modules over and a chain homotopy equivalence , where is the hat-flavored truncation of , such that the following conditions are satisfied.
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The formal variable has bigrading .
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is skew-graded, i.e. maps a -bigraded element to a -bigraded one.
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There exists a chain map whose hat-flavored truncation is homotopic to .
Here, denotes the formal derivative of the differential of with respect to the formal variable , which is a chain map which is well-defined up to homotopy.
A degree-preserving chain map between horizontal almost -complexes and is almost -local map if it satisfies the following conditions.
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is local, i.e. the localized map is a homotopy equivalence.
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, where is the hat-flavored truncation of .
Furthermore, if there exist almost -local maps and , then we say that and are almost -locally equivalent.
We denote the almost -local equivalence classes of horizontal almost -complexes by . We endow this set with a tensor product operation as , where is defined as . Although this operation is nonsymmetric, the group is indeed abelian [KP22, Proposition 2.6].
The involutive knot Floer homology package associates to a knot a well-defined element in , given by the chain complex , together with the -action on . For simplicity, we denote this element as . It is clear that this defines a group homomorphism
where is the smooth concordance group. The image of the unknot is denoted by .
Notice that the existence of almost -local map gives a partial order on . In particular, we say that if there exists an almost -local map . It turns out that this is actually a total order , modulo the figure-eight complex , i.e. the 2-torsion element induced by the involutive knot Floer homology of the figure-eight knot . In other words, two elements are incomparable if and only if [KP22, Theorem 2.11]. We will exploit this ordering to prove the main theorem.
3. Lemmas from involutive bordered Floer homology
We assume that the reader is familiar with standard materials in bordered Heegaard Floer homology, in particular the materials in [LOT18].
For simplicity, from now on, we will denote the -framed solid torus, as a bordered manifold, as . Given a pattern , we can define the minus-flavored type-A module , which is a type-A structure over the torus algebra with the coefficient ring . We denote its truncation by as . Furthermore, we can also remove a tubular neighborhood of from and endow the newly created torus boundary with the -framing; this defines a bordered manifold with two torus boundaries, whose type-DA bordered Floer homology is denoted as . Note that we have
via standard gluing formulae.
We will denote the longitudinal knot inside by . In particular, we have a gluing formula [LOT18, Theorem 11.29]:
where denotes the truncation of by . Note that we can also apply the gluing formula to get an identification
Given a bordered manifold with one torus boundary, its type-D bordered involution takes the form
where denotes the Auroux-Zarev piece, defined in [LOT11, Section 4]. Although is not well-defined up to homotopy due to the lack of naturality in bordered Floer homology, it is still a homotopy equivalence, and we will not need its uniquenss anyway. We denote the set of homotopy classes of all possible type-D bordered involutions of by .
The following lemma follows from the proof of [KP22, Theorem 4.5]. Its proof is straightforward from the discussions preceding the proof of [Kan22, Theorem 1.2], so we omit it for the sake of simplicity.
Lemma 3.1.
Given two knots , let be a type-D morphism. Given a pattern , consider the induced type-D morphism , defined as
Then we have
for suitable choices of bordered involutions and for each .
4. Proof of the main theorem
We start with several explicit computations of bordered Floer homology modules. Given any knot in , the type-D module can be easily computed from the knot Floer chain complex . For example, for the right-handed trefoil , the type-D module for the 0-framed complement of can be described as follows:
Similarly, we can also describe for the 0-framed complement of the figure-eight knot , as shown below. We will call its summand generated by as the square-module, and denote it by .
We will also need the (hat-flavored) type-A module of the -cabling pattern in the -framed solid torus , which is shown below. Note that was computed originally in [OSS17, Lemma 8.3]; we simply truncated their computation by taking .
Based on these computations, we will now prove two computational lemmas, which play a crucial role in the proof of 1.1.
Lemma 4.1.
Any type-D morphism from to the square-summand becomes nullhomotopic after box-tensored with the identity morphism of for any positive integer .
Proof.
Zhan’s bordered Floer homology calculator [Zha] tells us that the space of homotopy classes of type-D morphisms from to is six-dimensional, generated over by , which are defined as follows.
Thus, to prove the lemma, we only have to show that these six morphisms become nullhomotopic after box-tensored with . For simplicity, we will use the following convention: a chain complex over has an acyclic summand if and generate a direct summand of and .
We start with the map . The only simple tensors (of basis elements) on which takes nontrivial values are and . In particular, we have
However, is contained in the acyclic summand and is contained in the acyclic summand . Thus is nullhomotopic.
For , the map takes nontrivial values only on , (which are both mapped to ), and . But and are contained in the acyclic summands and , respectively. Thus is nullhomotopic.
Similarly, the map takes nontrivial values on and , which are contained in the acylic summands and . Hence is also nullhomotopic.
For the maps and , the only simple tensor on which takes a nontrivial value for is , on which takes the value for and for . But the former is contained in the acyclic summand and the latter is contained in the acyclic summand . Hence is nullhomotopic for .
It remains to show that is nullhomotopic. The complete list of its nontrivial values is given as follows:
But for each , we have that is contained in the acyclic summand , and thus is also nullhomotopic. ∎
Recall that bordered Heegaard Floer homology comes with gradings by nonabelian groups; given a bordered 3-manifold with boundary , the associated type-D module is graded by a transitive -set. Hence the notion of degree-preserving endomorphisms of is well-defined. See [LOT18, Chapter 10] for a detailed description of gradings in bordered Floer homology.
Lemma 4.2.
Let be a degree-preserving endomorphism of . Then is either nullhomotopic or homotopic to the identity morphism.
Proof.
One can use Zhan’s bordered Floer homology calculator to show that the space is six-dimensional, generated by the identity morphism and the morphisms described below.
It is straightforward to see that are not degree-preserving. Therefore the space of degree-preserving type-D endomorphisms of is one-dimensional and generated by the identity morphism. ∎
We need one more lemma regarding the -local equivalence class of the involutive knot Floer homology of , which we will denote as for simplicity.
Lemma 4.3.
The -complex of is -locally equivalent to the knot Floer complex of the twist knot , or equivalently, .
Proof.
It follows from the proof of [HKL16, Lemma A.1] that we have
where is a unit box summand, as shown below.
We follow the general strategy of the proof of [HM17, Proposition 8.1], noting that in this case, our knot is not thin. In particular, the proof uniquely determines up to chain homotopy, which we summarize in the following table. Note that we have three copies of ; we denote their generators as for .
We are left with the possibility that may contain terms that are multiples of or . Due to the bidegree reasons, such terms can arise only for the values of at for . Since , we may homotope , which changes the values of only at , to eliminate all terms arising in which have nontrivial or exponents in their coefficients. Then, since , we can further homotope , which changes the values of only at , to eliminate all terms arising in involving nontrivial -exponents. Thus we may assume that
for some element . But then should lie in bidegree , and there is no such an element in the given complex, so we should have
Then it follows from
that , and then we deduce from
that . Thus, to summarize, we have homotoped in the square summand generated by so that it acts by
Similarly, we can also homotope in the square summand generated by so that it acts by
But then the action of splits as the direct sum of its action on the summand generated by and the summand generated by , . Since the latter summand is acyclic after localizing by , we see that the given -complex is -locally equivalent to its summand generated by . However [HM17, Proposition 8.1] implies that this summand is -locally equivalent to the involutive knot Floer complex of . The lemma follows. ∎
We are now ready to prove 1.1; its proof will be divided in two parts. In the first part, we will prove that the knots
have infinite order in the smooth concordance group. Then, in the second part, we will prove that infinitely many form a linearly independent family of the smooth concordance group.
Proof of 1.1, first part.
It follows from 4.3 and [Kan22, Theorem 1.2] that
Since has as a direct summand, the type-D module also has as a direct summand; here, all knot complements are endowed with the 0-framing. Take the inclusion map
By box-tensoring with the identity morphism on the type-DA module , we get a type-D morphism
Box-tensoring this morphism further with the identity morphism on gives a -bidegree-preserving chain map
whose truncation by is given by . Then, by [KP22, Lemma 3.1] and 3.1, it follows that
for some bordered involutions and , since we have
Hence, if we consider the type-D morphism
then we have
Claim: is nullhomotopic. Assuming the claim, we immediately get
To prove the claim, we recall that ; note that the three summands lie in different bigradings, but we do not have to care about this issue here. By [LOT18, Theorem 11.26], this corresponds to the splitting
where the inclusion map for the summand is the map . Choose any one of the three -summands, and take its projection map
Then it follows from 4.1 that is nullhomotopic.
It remains to show that is nullhomotopic, were denotes the projection map to the trefoil summand, i.e.
We will actually prove a stronger assertion that is nullhomotopic. To see this, we start with the fact that is degree-preserving, which implies that is a degree-preserving type-D endomorphism of . Then it follows from 4.2 that is either nullhomotopic or homotopic to the identity morphism. However, if it were homotopic to the identity, then we should have
where denotes the projection map from to its direct summand . However, the right hand side of the above equation is the truncation (by ) of a chain endomorphism of which becomes nullhomotopic after localizing by . This is impossible since the identity map of clearly does not satisfy this property. Thus should be nullhomotopic and the claim is proved.
Now that we have
it follows that is an almost -local map by [KP22, Lemma 3.3], i.e. we have an inequality
(4.1) |
Since and , we get
Furthermore, it is shown in the proof of [KP22, Theorem 4.5] that and has infinite order in . Hence, we deduce that
and has infinite order in . Therefore has infinite order in the smooth concordance group. ∎
To prove the linear independence part of 1.1, we recall more facts about from [KP22]. Recall from [KP22, Section 4] that, for each , the horizontal almost -complex is generated by elements , where and have bidegree . The differential is given by
and the involution on its hat-flavored truncation is given by
These complexes satisfy the following inequalities in (for detailed discussion see the proofs of [KP22, Lemma 4.6 and Theorem 4.7]). First of all, for any , we have
(4.2) |
Furthermore, for any , we have
(4.3) |
Lastly, if a knot satisfies , for some acyclic summand (which is the case for each knot ), then there is a positive integer such that
(4.4) |
Now, we are ready to prove the second part of the main theorem.
Proof of 1.1, second part.
It follows from the first part of the proof of 1.1 and (4.3) that for each , we have
Furthermore, by (4.4) for each positive integer , there is a positive integer such that
Hence, if we choose an increasing sequence of integers such that and for each , then by combining the above two inequalities with (4.2) we have that
Suppose that the knots are linearly dependent in the smooth concordance group. Then there exists a finite sequence of nonzero integers such that
Without loss of generality, we may assume that . Then we have
a contradiction. Therefore the knots are linearly independent, as desired. ∎
Remark 4.4.
By applying the arguments used in the proof of [HKPS22, Lemma 3.2], it follows that we can take to be . With a more explicit calculation of for , we expect that a much smaller value for should be possible.
Remark 4.5.
The argument used in the second part of the proof of 1.1 can be summarized as follows; note that the same argument was also used in [KP22]. Let be a sequence of rationally slice knots such that for any , there exists a positive integer such that in . Then admits a linearly independent infinite subsequence in .
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