From zero surgeries to candidates for exotic definite four-manifolds
FFROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITEFOUR-MANIFOLDS
CIPRIAN MANOLESCU AND LISA PICCIRILLO
Abstract.
One strategy for distinguishing smooth structures on closed 4-manifolds is to pro-duce a knot K in S that is slice in one smooth filling W of S but not slice in some home-omorphic smooth filling W (cid:48) . In this paper we explore how 0-surgery homeomorphisms can beused to potentially construct exotic pairs of this form. In order to systematically generate aplethora of candidates for exotic pairs, we give a fully general construction of pairs of knotswith the same zero surgeries. By computer experimentation, we find 5 topologically slice knotssuch that, if any of them were slice, we would obtain an exotic four-sphere. We also investigatethe possibility of constructing exotic smooth structures on n CP in a similar fashion. Introduction
Ever since the work of Freedman [21] and Donaldson [15], the following strategy for disprovingthe smooth 4-dimensional Poincar´e conjecture has garnered interest: Find a homotopy 4-sphere W and a knot K ⊂ S = ∂ ( W \ ˚ B ) which bounds a smoothly embedded disk in W \ ˚ B butwhich is not slice (in B ). It would then follow that W is homeomorphic but not diffeomorphicto S . In [20], Freedman, Gompf, Morrison and Walker explicitly attempted this strategy; fora homotopy 4-sphere from the literature they found a knot K which bounds a smooth disk inthe homotopy sphere, and tried to use Rasmussen’s s invariant from [46] to show that K isnot slice in B . The s invariant is known to be zero for slice knots, but (unlike other similarinvariants [42, 32]) it is unknown whether s necessarily vanishes when K bounds a smooth diskin a homotopy 4-ball. For the example in [20], however, the s invariant was 0 and in fact thehomotopy 4-sphere was proven almost immediately to be standard [4].If one wants to more systematically pursue the strategy above, one needs a robust source ofexamples of knots that bound smooth disks in homotopy 4-balls. It is possible to use pairs ofknots K and K (cid:48) with the same 0-surgeries to produce such examples as follows: If K is sliceand S ( K ) ∼ = S ( K (cid:48) ), then by gluing the complement of the slice disk for K to the trace of the0-surgery for K (cid:48) , we obtain a homotopy 4-sphere W , such that K (cid:48) bounds a disk in W \ ˚ B . If s ( K (cid:48) ) (cid:54) = 0, then K (cid:48) is not slice and W is an exotic four-sphere.In principle, the same idea can be used to produce examples of exotic smooth structures on n CP for n ≥
1. The work of Freedman [21] and Donaldson [15] implies that every simply-connected, positive definite, smooth, closed 4-manifold is homeomorphic to n CP for some n .It is unknown if W = n CP admits exotic smooth structures. Let W ◦ = W \ ˚ B and definea knot to be H-slice in W it bounds a smoothly embedded nullhomologous disk in W ◦ . If wefound knots K and K (cid:48) such that S ( K ) ∼ = S ( K (cid:48) ) , K is H-slice in W, K (cid:48) is not H-slice in W, we would produce an exotic smooth structure on W . Note that we could obstruct K (cid:48) from beingH-slice in n CP by showing that Rasmussen’s s invariant satisfies s ( K (cid:48) ) <
0; see [38].
CM was partially supported by NSF grant DMS-2003488 and a Simons Investigator award.LP was partially supported by NSF postdoctoral fellowship DMS-1902735 and the Max Planck Institute forMathematics. a r X i v : . [ m a t h . G T ] F e b CIPRIAN MANOLESCU AND LISA PICCIRILLO
Techniques for constructing pairs of knots with homeomorphic n -surgeries first appeared inthe late 70’s, see [34, 3, 35]; for n = 0, see [9]. Other fundamentally distinct constructions weregiven in [41] and [56]. Some of these constructions always produce knots such that not onlyare the 0-surgeries homeomorphic, but in fact the traces are diffeomorphic [3, 35, 9]. Othersof these constructions sometimes produce knots with diffeomorphic traces [2]. Producing knotswith diffeomorphic traces is useful for some purposes, for example for the proof that Conway’sknot is not slice [45]. But pairs of knots with the same trace are not useful for obtaining exoticstructures in the manner described above, as the trace embedding lemma ([19], see Lemma 3.5)readily implies that if K is H-slice in some manifold W , then so is K (cid:48) .In this paper, in order to produce a wide selection of examples, we give a fully generalframework for constructing pairs of knots with homeomorphic 0-surgeries. Our framework isbased on 3-component links of the following form. Definition 1.1. An RBG link L = R ∪ B ∪ G ⊂ S is a 3-component rationally framed link,with framings r, b, g respectively, such that there exist homeomorphisms ψ B : S r,g ( R ∪ G ) → S and ψ G : S r,b ( R ∪ B ) → S and such that H ( S r,b,g ( R ∪ B ∪ G ); Z ) = Z . Theorem 1.2.
Any RBG link L has a pair of associated knots K B and K G and homeomorphism φ L : S ( K B ) → S ( K G ) . Conversely, for any 0-surgery homeomorphism φ : S ( K ) → S ( K (cid:48) ) there is an associated RBG link L φ with K B = K (cid:48) and K G = K . We will explain how particular cases of RBG links recover other constructions from the liter-ature, such as annulus twisting, dualizable patterns, and Yasui’s construction. Moreover, thereis a straightforward condition on the homeomorphism φ that guarantees that it does not extendto a homeomorphism of the corresponding traces.A simple way to construct many RBG links is to consider the following special case: We take R to be an r -framed knot with r ∈ Z , and B and G to be 0-framed unknots with linking number l such that l = 0 or rl = 2. Further, letting µ R denote a meridian for R , we ask that there existlink isotopies R ∪ B ∼ = R ∪ µ R ∼ = R ∪ G. We call RBG links of this form special . Special RBG links are easy to draw, and suffice toproduce many examples of knots with the same 0-surgeries, where the corresponding traces arenot homeomorphic.Using the computer programs
SnapPy [13],
KnotTheory` [7],
SKnotJob [49], and the
KnotFloer homology calculator [52], we investigated a 6-parameter family consisting of 3375 specialRBG links. This yielded 21 interesting pairs (
K, K (cid:48) ) for which S ( K ) ∼ = S ( K (cid:48) ), s ( K (cid:48) ) = − (cid:54) = 0,and for which we could not determine whether K is slice. Shortly after the original version ofthis paper was posted to the arXiv, Nathan Dunfield and Sherry Gong informed us that, usingthe twisted Alexander polynomial obstructions from [29], they were able to prove that 16 of our21 knots are not slice [16, 17]. Thus, we are left with 5 knots K with the following property. Theorem 1.3.
If any of the knots shown in Figure 1 are slice, then an exotic four-sphereexists. We have verified that these knots pass many of the known obstructions to sliceness. Specifi-cally: • They have Alexander polynomial 1, and are therefore topologically slice by [21]. • Their τ , (cid:15) and ν invariants from knot Floer homology vanish; • Rasmussen’s s invariant equals zero; • The variants s F and s F of Rasmussen’s invariant (from Khovanov homology over thefields F and F ) also vanish; ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 3 K K K K K Figure 1.
Candidates for slice knots. • The Lipshitz-Sarkar Sq s -invariants vanish; • For at least 3 of these knots ( K , K and K ), the given homeomorphism φ : S ( K ) → S ( K (cid:48) ) does not extend to a trace diffeomorphism, so the non-sliceness of K (cid:48) does notimmediately obstruct K from being slice.The other 16 knots from our original list are denoted K through K and shown in Figure 23.They are algebraically but not topologically slice, and satisfy(1) τ = (cid:15) = ν = s = s F = s F = s Sq = 0 . This leaves open the possibility that they could lead to exotic structures on n CP . Moreover,our computer experiments produced two other knots ( K and K in Figure 23) which are noteven algebraically slice (they fail the Fox-Milnor condition on the Alexander polynomial), buthave vanishing Levine-Tristram signature function, satisfy (1), and have a companion knot K (cid:48) with s ( K (cid:48) ) = − <
0. These two knots are additional candidates for producing exotic smoothstructures on n CP . Theorem 1.4.
If any of the knots shown in Figures 1 and 23 are H-slice in n CP for some n , then an exotic n CP exists. By contrast, one can show that all these 23 knots are H-slice in n CP for some n > n CP and n CP (for some n ) are called biprojectively H-slice ,or BPH-slice . BPH-slice knots have vanishing Levine-Tristram signature function, and satisfy τ = (cid:15) = s = 0; cf. [11], [38]. We observe in Section 2 that many of the small knots for whichthese invariants vanish can be shown to be BPH-slice.So far we have focused on pairs of knots ( K, K (cid:48) ) with the same 0-surgery, for which we knowthat K (cid:48) is not slice (or not H-slice in n CP ), and we are unsure about K . We could alsolook at pairs ( K, K (cid:48) ) with the same 0-surgery for which we know that K is slice (or H-slice in n CP ), and s ( K (cid:48) ) = 0. (If s ( K (cid:48) ) (cid:54) = 0, this would be a different paper.) There are plenty of suchexamples coming from special RBG links or from annulus twisting. In some situations, we knowthat K (cid:48) is in fact slice, and in others we are not sure. In either case, interesting homotopy 4-spheres can be constructed using the RBG link L φ from a homeomorphism φ : S ( K ) → S ( K (cid:48) ).The challenge then becomes to determine whether these homotopy 4-spheres are standard. InFigure 20 we exhibit an explicit infinite family of examples of homotopy 4-spheres constructedby this method.1.1. Organization of the paper.
In Section 2 we introduce BPH-slice knots and give ex-amples. In Section 3 we present the general RBG construction and prove Theorem 1.2, anddiscuss when 0-surgery homeomorphisms extend to trace homeomorphisms or diffeomorphisms.In Section 4 we restrict attention to special RBG links, and introduce a concept (small RBGlinks) that ensures the resulting diagrams for K B and K G are manageable. In Section 5 we CIPRIAN MANOLESCU AND LISA PICCIRILLO describe our computer experiments, and explain how we arrived at the knots in Figures 1 and23. In Section 6 we review how annulus twisting gives rise to 0-surgery homeomorphisms, andgive examples of homotopy 4-spheres arising from this construction. Finally, in Section 7 werelate RBG links to other known ways to produce 0-surgery homeomorphisms: annulus twisting,Yasui’s construction, and dualizable patterns.1.2.
Conventions.
All manifolds are smooth and oriented and all homeomorphisms are ori-entation preserving. Boundaries are oriented with outward normal first. Slice refers to theexistence of a smooth disk, and topologically slice to that of a locally flat disk. Homology hasintegral coefficients. The symbol ν denotes a tubular neighborhood, and U denotes the unknot.1.3. Acknowledgements.
We would like to thank the organizers of the CRM 50th AnniversaryProgram in Low-Dimensional Topology (Montr´eal, 2019), where this collaboration started. Wethank Dror Bar-Natan, Kyle Hayden, Chuck Livingston, Marco Marengon, and Allison Millerfor helpful comments on previous versions of this paper. In particular, we are grateful to KyleHayden for pointing out that some of the homotopy 4-spheres we previously considered were infact standard, and to Nathan Dunfield and Sherry Gong for checking that some of the knots inour list were not topologically slice. 2.
BPH-slice knots
Let W be a closed, smooth, oriented 4-manifold. We let W ◦ := W \ ˚ B . Definition 2.1.
We say that a knot K ⊂ S is H-slice in W ◦ if it bounds a smooth, properlyembedded disk ∆ in W ◦ , such that [∆] = 0 ∈ H ( W ◦ , ∂W ◦ ) . For convenience, we will alsosometimes use the terminology
H-slice in W to mean H-slice in W ◦ . Observe that if a knot is slice in the usual sense, then it is H-slice in any W . Remark . It is shown in [37, Corollary 1.5] and that the set of H-slice knots can detect exoticsmooth structures on some 4-manifold with indefinite intersection form. See [37, 31] for moreobstructions to H-sliceness in such manifolds.Knots that are H-slice in some simply connected 4-manifold with a positive definite (or nega-tive definite, resp.) intersection form are called 0 -positive (0 -negative , resp.) in [11]. Note that K is 0-negative iff the mirror K is 0-positive. Several obstructions to 0-positivity are collectedin [11, Proposition 1.1]. If K is 0-positive, then • The signature of the knot satisfies σ ( K ) ≤ • More generally, the Levine-Tristram signature function σ LT ( K ) (evaluated away fromthe roots of the Alexander polynomial) is non-positive; • The Ozsv´ath-Szab´o concordance invariant satisfies τ ( K ) ≥
0; cf [42, Theorem 1.1].There are additional obstructions from the Heegaard Floer correction terms of cyclic branchedcovers or ± K , and from Yang-Mills theory; see [11], [32, Corollary 5.5] and [14,Theorem 4.1].When W = n CP , another obstruction comes from Khovanov homology: From [38, Corol-lary 1.9], it follows that if K is H-slice in n CP for some n , then Rasmussen’s s invariantsatisfies(2) s ( K ) ≥ . Examples of knots that are H-slice in n CP can be easily constructed using the followingwell-known lemma. ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 5 c∂ − X γ ∂ + X ∂ + X ∼ = S Figure 2.
An annular cobordism in CP \ ( ˚ B (cid:116) ˚ B ) from K to K (cid:48) Lemma 2.3.
Let
K, K (cid:48) ⊂ S be knots such that K (cid:48) is obtained from K by changing a negativecrossing to a positive crossing in a diagram of K . If K is H-slice in n − CP , then K (cid:48) is H-slicein n CP .Proof. Let c be a negative-to-positive crossing change in a diagram of K which yields K (cid:48) . Con-sider K ⊂ S × { } ⊂ S × I and attach a 1-framed 2-handle to S × { } along a curve γ whichlinks c as in Figure 2. This handle attachment yields the cobordism X = CP \ ( ˚ B (cid:116) ˚ B ) andthere is a natural nullhomologous annular cobordism from K ⊂ ∂ − X to the knot K + ⊂ ∂ + X depicted in the center frame of Figure 2. When we identify ∂ + X with the standard diagram of S (via, say, a Rolfsen twist) as in the right frame of Figure 2, we can identify K + as K (cid:48) . Theclaim follows by stacking X on top of n − CP . (cid:3) More generally, the conclusion of Lemma 2.3 also holds when K (cid:48) is obtained from K by addinga generalized positive crossing in the sense of [12, Definition 2.7].The following concept will be of particular interest to us. Definition 2.4.
A knot K ⊂ S is called biprojectively H-slice (or BPH-slice ) if it is H-slicein both n CP and n CP , for some n ≥ . In [11], knots that are both 0-positive and 0-negative are called 0-bipolar. BPH-slice knotsare 0-bipolar. Moreover, note that every simply connected, positive definite, smooth closed4-manifold is homeomorphic to n CP for some n , and there are no known exotic smoothstructures on such manifolds. Thus, 0-bipolar and BPH-slice might be the same notion.By applying the obstructions above for both K and K , we see that for BPH-slice knots theybecome equalities instead of inequalities. Therefore, if K is BPH-slice then: σ ( K ) = 0 , σ LT ( K ) = 0 , τ ( K ) = 0 , s ( K ) = 0 . Further, Hom’s (cid:15) invariant from knot Floer homology [30] also has to vanish for BPH-slice knots;see [11, Proposition 4.10].Slice knots are BPH-slice. We see that many of the obstructions that vanish for ordinary sliceknots also vanish for BPH-slice knots. The main difference is the Fox-Milnor condition on theAlexander polynomial, which does not need to hold for BPH-slice knots.The following is an immediate consequence of Lemma 2.3.
Lemma 2.5.
Suppose that K ⊂ S is a knot with the following properties: • There exists a diagram of K and a negative crossing in that diagram, such that when wechange it to a positive crossing, we get a BPH-slice knot; • There exists a (possibly different) diagram of K and a positive crossing in that diagram,such that when we change it to a negative crossing, we get a BPH-slice knot.Then, K is BPH-slice. CIPRIAN MANOLESCU AND LISA PICCIRILLO
Figure 3.
Left: the BPH-slice knot 4 . Right: The knot 8 , whose BPH-sliceness is unknown. Example . The simplest nontrivial BPH-slice knot is the figure-eight 4 . Its standard diagram(shown in Figure 3) has two negative and two positive crossings, and changing the sign of anyof the crossings produces the unknot.Using Lemma 2.5, we found that BPH-slice knots are significantly more common than sliceknots among small knots. Indeed, most small knots with σ = 0 are BPH-slice. Here is a list ofall the prime knots with at most 9 crossings and σ = 0: • slice knots : 6 , , , , , , ; • amphichiral non-slice knots : 4 , , , , , ; • non-amphichiral, non-slice knots : 7 , , , , , , , , , , .Of these, all except 8 can be shown to be BPH-slice by starting with the diagrams foundin knot tables, checking which crossing changes produce smaller BPH-slice knots, and applyingLemma 2.5.We could not determine if 8 is BPH-slice. Changing any of the crossings in its minimaldiagram (shown in Figure 3) gives a trefoil knot, whose signature is nonzero. Observe, however,that for composite knots such as 3 − ), it is also true that changing any crossing gives atrefoil; nevertheless, the knot is slice.3. A general RBG construction
In this section we give a fully general framework for describing homeomorphisms betweenmanifolds arising as 0-surgeries on knots. Using this framework we discuss when a 0-surgeryhomeomorphism can be extended to a trace homeomorphism or diffeomorphism. In Section 4 wewill make some simplifying assumptions that lead to more user-friendly results and examples.Our construction is based on certain three-component links, called
RBG links , which generalizethose already considered in [44, Section 2].3.1.
RBG links.
Let (cid:126)f denote a finite ordered list with values in Q ∪{∞}∪{∗} . We will use thenotation S (cid:126)f ( L ) to denote the (cid:126)f surgery on a framed link L , where an ∗ denotes a complement(i.e., removing a neighborhood of that component and not filling it in). Given a 3-manifoldhomeomorphism φ : M → N we will sometimes abusively still use φ to refer to a restriction of φ to some codimension zero submanifold of M . Definition 1.1. An RBG link L = R ∪ B ∪ G ⊂ S is a 3-component rationally framed link,with framings r, b, g respectively, such that there exist homeomorphisms ψ B : S r,g ( R ∪ G ) → S and ψ G : S r,b ( R ∪ B ) → S and such that H ( S r,b,g ( R ∪ B ∪ G ); Z ) = Z . For examples of RBG links, see Section 4.
ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 7
Remark . Since there is only one orientation preserving homeomorphism of S up to isotopy[10], we do not have to record the particular homeomorphisms ψ B and ψ G in the data of anRBG link L . We have named them in the definition for convenience in later proofs. Theorem 1.2.
Any RBG link L has a pair of associated knots K B and K G and homeomorphism φ L : S ( K B ) → S ( K G ) . Conversely, for any 0-surgery homeomorphism φ : S ( K ) → S ( K (cid:48) ) there is an associated RBG link L φ with K B = K and K G = K (cid:48) .Proof. For the first claim, define K B to be the knot in S satisfying ψ B : S r, ∗ ,g ( L ) → S ∗ ( K B )and let f b denote the rational satisfying ψ B : S r,b,g ( L ) → S f b ( K B ). Similarly take K G to be theknot satisfying ψ G : S r,b, ∗ ( L ) → S ∗ ( K G ) and f g denote the rational satisfying ψ B : S r,b,g ( L ) → S f g ( K G ). Then ψ B ◦ ψ − G gives the desired homeomorphism φ L , and the homology assumptionon S r,b,g ( L ) implies f b = f g = 0.For the second claim, take B := K (cid:48) and b = 0. Let µ K be the meridian for K , and let( R, r ) be the framed curve given as the image of ( µ K ,
0) under the homeomorphism φ . Sincethere is the slam-dunk homeomorphism h : S , ( K, µ K ) → S , there is a homeomorphism h ◦ φ − : S ,r ( B, R ) → S . (Note that r is an integer, for homology reasons.) Now define L = B ∪ R ∪ G where G is the 0-framed meridian of R . Observe that L is an RBG link where ψ B is the slam dunk homeomorphism from S r, ( R, G ) to S and ψ G is h ◦ φ − . (cid:3) Remark . It is possible to produce the same 0-surgery homeomorphism from different RBGlinks. In the second part of Theorem 1.2 we constructed a particular RBG link, having b = g = 0and r ∈ Z .There are three primary techniques presently in the literature for constructing a 0-surgeryhomeomorphism: dualizable patterns [3, 6, 9, 24, 35], annulus twisting [41], and Yasui’s con-struction [56]. In Section 7 we will give explicit RBG links for each of these constructions.3.2. H-slice knots from 0-surgery homeomorphisms.
In order to build knots K (cid:48) that areH-slice in an exotic copy of a simply connected four-manifold W , we will begin instead with aknot K which is H-slice in W and then construct a K (cid:48) with S ( K ) ∼ = S ( K (cid:48) ). That K (cid:48) is thenslice in a homotopy W follows from the following folklore: Lemma 3.3.
Let W be a smooth, closed, oriented, simply connected four-manifold. If there is ahomeomorphism φ : S ( K ) → S ( K (cid:48) ) and K is H-slice in W , then K (cid:48) is H-slice in a 4-manifold X with the homotopy type of W . To prove the lemma, we require a definition.
Definition 3.4.
The trace of a knot K , denoted X ( K ) , is the 4-manifold obtained by attachinga single 0-framed 2-handle to B along K .Proof of Lemma 3.3. Since K is H-slice in W , we can choose a slice disk for K in W ◦ andconsider the 4-manifold V obtained by excising an open tubular neighborhood of that diskfrom W ◦ . It is routine to confirm that ∂V ∼ = S ( K ) and that π ( V ) is normally generated by ι ∗ ( π ( ∂V )), where ι : ∂V → V is the inclusion.Now consider the 4-manifold X := X ( − K (cid:48) ) ∪ φ V, where φ is the assumed homeomorphism from − ∂ ( X ( − K (cid:48) )) = S ( K (cid:48) ) to ∂V ∼ = S ( K ). Let µ K (cid:48) denote the meridian of K (cid:48) in S ( K (cid:48) ). By thinking of gluing an upside down X ( − K (cid:48) ) ontoa rightside up V , we see that X has a handle diagram obtained from that of V by addingan additional 0-framed 2-handle along φ ( µ K (cid:48) ), followed by a 4-handle. As such, π ( X ) = CIPRIAN MANOLESCU AND LISA PICCIRILLO π ( V ) / (cid:104) [ φ ( µ K (cid:48) )] (cid:105) . Since π ( S ( K (cid:48) )) / (cid:104) [ µ K (cid:48) ] (cid:105) = 1 we have π ( ∂V ) / (cid:104) [ φ ( µ K (cid:48) )] (cid:105) = 1. Since π ( V ) isnormally generated by π ( ∂V ), we have that π ( X ) = 1. It is routine to confirm that X has thehomology type of W . It is then a consequence of Whitehead’s Theorem (see [40, p.103, Theorem1.5] or [25, Theorem 1.2.25]) that X is homotopy equivalent to W . (In fact, X is homeomorphicto W by Freedman’s theorem [21].)To see that K (cid:48) is H-slice in X , let X ◦ ( − K (cid:48) ) denote X ( − K (cid:48) ) with an open ball removed, andobserve that K (cid:48) ⊂ S ⊂ ∂ ( X ◦ ( − K (cid:48) )) bounds a disk D in X ◦ ( − K (cid:48) ) made up of the productcobordism in S × I and the core of the 2-handle. This slice disk survives into X , and it is againroutine to check that [ D ] = 0 ∈ H ( X, ∂X ). (cid:3) Trace homeomorphisms.
To build an exotic copy of some closed simply connected four-manifold W , we will want to start with a knot K that is H-slice in W and construct a knot K (cid:48) with S ( K ) ∼ = S ( K (cid:48) ) such that K (cid:48) is hopefully not H -slice in W . We now observe that thefollowing lemma implies that it is unproductive to construct a K (cid:48) with the stronger propertythat X ( K ) is diffeomorphic to X ( K (cid:48) ): Lemma 3.5 (Trace Embedding Lemma, originally [19], cf. [28] Lemma 3.3) . Let W be a smooth,closed 4-manifold. Then K ⊂ ∂S is H-slice in W if and only if there is a smooth embedding of X ( − K ) in W which induces the 0-map on H . Hence, in this paper we are particularly interested in knots with homeomorphic zero-surgerieswhich do not have diffeomorphic traces. In full generality, it is a subtle problem to demonstratethat a pair of knot traces with homomorphic boundaries are not diffeomorphic, see [56, 27].In fact, even the easier problem of determining whether given zero surgery homeomorphismcan be extended to a trace diffeomorphism is open in general. There is some luck though:many knots have that the mapping class group of S ( K ) is just a single element, hence if therewas a trace diffeomorphism it would restrict to any given boundary homeomorphism. That M CG ( S ( K )) = 1 for a particular knot K can be verified in SnapPy and
Sage [13, 50]. Therefore, we will be especially interested in zero-surgery homeomorphisms which do notextend to trace diffeomorphisms. First, we remind the reader that the zero-surgery homeomor-phisms which do not extend to trace homeomorphisms are well understood:
Definition 3.6.
A 0-surgery homeomorphism φ : S ( K ) → S ( K (cid:48) ) is even if the 4-manifold Z := X ( K (cid:48) ) ∪ φ − X ( K ) has even intersection form, and is odd otherwise. An RBG link is even (resp. odd ) if the associated 0-surgery homeomorphism is even (resp odd). Theorem 3.7 ([8] Theorem 0.7 and Proposition 0.8) . A 0-surgery homeomorphism φ : S ( K ) → S ( K (cid:48) ) extends to a trace homeomorphism Φ : X ( K ) → X ( K (cid:48) ) if and only if φ is even.Proof. For the reader’s convenience, we include a proof of the easy ‘only if’ direction.Suppose for a contradiction that φ is not even, and that there is some homeomorphismΦ : X ( K ) → X ( K (cid:48) ) extending φ . Let W denote the 4-manifold obtained from X ( K ) byattaching a 0-framed 2-handle along µ K followed by a 4-handle, and observe that W has evenintersection form. Define Z := X ( K (cid:48) ) ∪ φ − X ( K ), which has odd intersection form by hypothesis.Observe that Φ gives a natural homeomorphism from W to Z , a contradiction. (cid:3) Remark . We emphasize that Theorem 3.7 only shows that a particular boundary homeomor-phism does not extend, not that the traces are not homeomorphic. Indeed, there are for exampleboundary homeomorphisms X ( U ) → X ( U ) that don’t extend to trace homeomorphisms [23]. SnapPy computes the symmetry group of a hyperbolic manifold by finding a canonical cellulation. This isdone using numerical methods. If one wants a mathematical proof, then the
SnapPy answer needs to be certifiedrigorously, e.g. using interval arithmetic as in [18].
ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 9
We observe that the knots whose 0-surgery admits an odd homeomorphism are somewhatrestricted.
Lemma 3.9.
If a homeomorphism φ : S ( K ) → S ( K (cid:48) ) is odd then Arf( K ) = Arf( K (cid:48) ) = 0 .Proof. Robertello [47] showed that if X is a simply connected smooth 4-manifold with S bound-ary and K ⊂ S bounds a smooth disk D in X such that [ D ] ∈ H ( X, ∂X ) ∼ = H ( X ) ischaracteristic then Arf( K ) = [ D ] · [ D ] − σ ( X )2 mod 2.Consider the 4-manifold Z obtained by gluing X ( K ) (upside down) to X ( K (cid:48) ) (right side up)along S ( K (cid:48) ) via φ . Remove the 0-handle of X ( K (cid:48) ) to get X with S boundary, and observethat K (cid:48) ⊂ S bounds a 0-framed disk D in X (the core of the 2-handle). Further, this handledecomposition of X gives a natural presentation of H with basis { α, β } and intersection form Q X = (cid:18) n (cid:19) . Further, we see that [ D ] = α and, since n is odd, α is characteristic. Then Robertello’s resultapplies and we get Arf( K (cid:48) ) = 0. We can obtain the same conclusion about K by turning Z upside down. (cid:3) Remark . The converse is false, as can be seen by considering the identity homeomorphismon S ( K ) for any knot with Arf( K ) = 0.We remark that Lemma 3.9 pairs with Theorem 3.7 to demonstrate that Corollary 3.11. If Arf( K ) = 1 then every 0-surgery homeomorphism φ : S ( K ) → S ( K (cid:48) ) extends to a trace homeomorphism. Trace diffeomorphisms.
It remains well out of reach to classify when 0-surgery homeo-morphisms extend to trace diffeomorphisms. We give a sufficient condition for a homeomorphismto extend.
Definition 3.12.
Let φ : S ( K ) → S ( K (cid:48) ) be a -surgery homeomorphism. Let γ ⊂ S ( K (cid:48) ) bethe framed knot given by the image of the 0-framed meridian of K under φ . We say that φ has property U if there is some diagrammatic choice of γ in the standard diagram of S ( K (cid:48) ) where γ is framed and appears unknotted in the diagram. Theorem 3.13. If φ has property U then there exists a diffeomorphism Φ : X ( K ) → X ( K (cid:48) ) with Φ | ∂ = φ .Remark . This theorem follows from a now-standard technique due to Akbulut, called carv-ing [3], which was developed in fact exactly for this purpose. Versions of the theorem alreadyappear in the literature; see for example [2]. Here we eliminate some hypotheses by making useof a theorem of Gabai in the last step of the proof.
Proof.
Let X ⊂ X ( K ) be a (closed) tubular neighborhood of the cocore disk of the 2-handle,which is naturally identified with D × D , and let X (cid:48) ⊂ X be the open neighborhood of thecocore disk, which is a D × ˚ D . Since the image φ ( µ K ) appears unknotted in the standardhandle diagram of X ( K (cid:48) ), we can identify the standard slice disk D for φ ( µ K ) in B (thought ofas the 0-handle). Define Y := ν ( D ) ∼ = D × D and Y (cid:48) ∼ = D × ˚ D similarly. Since φ preservesthe 0-framing on µ K , the natural bundle diffeomorphism F (cid:48) : X → Y has that F (cid:48) | ∂ ( X ( K )) agreeswith f | ν ( µ K ) .Let X denote X ( K ) \ X (cid:48) and Y denote X ( K (cid:48) ) \ Y (cid:48) . It is evident that X is diffeomorphicto B . We will argue momentarily that Y is also diffeomorphic to B . Assuming this for now, we will finish the argument. Observe that φ | ∂X ( K ) \ ν ( µ K ) and F (cid:48) | D × ∂ ( D ) give a piecewisehomeomorphism f from ∂X ∼ = S to ∂Y ∼ = S . Since there is only one homeomorphism of S up to isotopy [10], and it extends smoothly over B , we can extend f to a diffeomorphism F : X → Y . By construction F and F (cid:48) together produce a diffeomorphism from X ( K ) to X ( K (cid:48) ) extending φ .Now we argue that Y is diffeomorphic to B . Observe that Y has S boundary and a handledecomposition given by dotting φ ( µ K ) in the standard handle decomposition of X ( K (cid:48) ). So ∂Y is naturally described as (0 ,
0) surgery on the 2-component link K (cid:48) ∪ φ ( µ K ) ⊂ S , and one linkcomponent is an unknot. Then, by performing the 0-surgery on φ ( µ K ) first, we can think of ∂Y as S obtained by surgery on some knot (cid:96) in S × S . By Gabai’s proof of property R [22], (cid:96) isisotopic to S × { pt } . Performing this isotopy on the attaching sphere of our 2-handle (slidingover the 1-handle as needed) yields a handle decomposition of Y which is just a canceling 1-2pair, hence Y is diffeomorphic to B . (cid:3) Question 3.15.
Is the converse of Theorem 3.13 true?
We comment now about these properties (parity and property U ) for some constructions of0-surgery homeomorphisms from the literature. (These constructions are discussed further inSection 7.) Dualizable pattern homeomorphisms [34, 9, 24] are all even and have property U .Annulus twisting [41] can be odd, and sometimes has property U ; see Figure 19 and Remark 6.6.Yasui’s homeomorphisms [56] are constructed such that the boundary diffeomorphism extendsto a homeomorphism, hence are all even. Yasui’s homeomorphism sometimes does not haveproperty U , in fact it may never have property U ; see Section 7.3.4. Special and small RBG links
We are interested in constructing many explicit pairs K and K (cid:48) with homeomorphic 0-surgeries, such that K and K (cid:48) are both simple enough that we have some hope of computingtheir concordance invariants or constructing slice disks for them in practice. Towards these ends,in this section we collect several user-friendly results about certain subclasses of RBG links.4.1. Special RBG links.Definition 4.1.
Let µ R denote a meridian of R . An RBG link L is special if b = g = 0 , r ∈ Z ,and there exist link isotopies R ∪ B ∼ = R ∪ µ R ∼ = R ∪ G. Notice that for a special RBG link, H ( S r,b,g ( L ); Z ) ∼ = Z if and only if the determinant of theframing matrix is zero. Let l denote the linking number of B with G . Then we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r l l (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 l − rl = 0 . Therefore, special RBG links have either l = 0 or rl = 2.For a special RBG link, the homeomorphism ψ G : S r, ( R, B ) → S is given by the slam-dunkhomeomorphism of B over R . Therefore, to exhibit the knot K G one should slide G over R until G no longer intersects the disk ∆ B bounded by B . The knot K B can be exhibited in the samemanner, everywhere reversing the roles of B and G . See Figure 4 for an example where theseslides are marked and performed.It is easy to recognize the parity of a special RBG link: Lemma 4.2.
A special RBG link L is even if and only if r is. ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 11
Figure 4.
A small RBG link and the associated knots K G and K B . R B ... µ G r K B ... γ r ... ... Figure 5.
The curve γ obtained from µ G after some handle slides and a slam-dunk. Proof.
Let γ ⊂ S ( K B ) be the framed knot which is the image of the 0-framed meridian of K G under φ L . Observe that X ( K B ) ∪ φ L X ( − K G ) admits a handle decomposition given by attachingtwo 2-handles to B along the framed link K B ∪ γ followed by a single 4-handle. Since K B is0-framed and [ γ ] generates H ( S ( K B )), we have lk ( K B , γ ) = 1. So X ( K B ) ∪ φ L X ( − K G ) haseven intersection form if and only if the framing on γ is even.Now we will argue that γ is r -framed. First we observe that in the S r,b,g ( R, B, G ) surgerydiagram of S ( K G ), the 0-framed meridian of K G is represented by the 0-framed meridian µ G of G . Now we will consider the framed image of this in the surgery diagram of S ( K B ). Observethat the image of µ G under the slam dunk homeomorphism S r, ( R, G ) ∼ = S is isotopic (as aframed curve) to the curve you get from sliding µ G over R to clear the (single) intersection of µ G with the disk G bounds. See Figure 5. As such, γ ⊂ S ( K B ) looks like the (framed) ghostof R , in particular the framing on γ is r . (cid:3) r Figure 6.
The leftmost strand may have any orientation. r rr r r Figure 7.
A homeomorphism from S ( K ) to surgery on an RBG link L . Lemma 4.3.
If a special RBG link has R = U and r = 0 then the associated homeomorphismhas property U.Proof. Let L be a special RBG link and let γ ⊂ S ( K B ) be the framed knot given by image ofthe 0-framed meridian of K G under φ L . In the proof of Lemma 4.2 we argued that γ ⊂ S ( K B )looks like the (framed) ghost of R , in particular γ has the knot type and framing of R . (cid:3) Ideally, we would like to produce knots with homeomorphic zero surgeries such that exactlyone of them is slice. Thus, one might like to start with their favorite slice knot K and producea distinct knot K (cid:48) with the same 0-surgery. It is not presently well-understood for which (slice)knots this is possible, but we give some lemmas that sometimes help: Lemma 4.4 ([44] Proposition 3.2) . If K has unknotting number one then there exists a specialRBG link L with K B isotopic to K .Remark . In [44] it is not emphasized that the resulting RBG link is special, but one quicklychecks that the proof given there does indeed produce special RBG links.
Lemma 4.6. If K can be unknotted by performing a tangle replacement as depicted in Figure6, then there exists a special RBG link L with K G isotopic to K .Proof. In Figure 7 we demonstrate a homeomorphism from S ( K ) to S r,b,g ( R, B, G ) for an RBGlink L , and it is evident from the figure that K is K G . (cid:3) Remark . In fact, all knots can be unknotted by performing a tangle replacement as depictedin Figure 6 with r = 0. (This can be readily deduced from [51, Lemma 1], for example.) Howeverwhen r = 0 one can check that the RBG link from the proof of Lemma 4.6 yields knots with K B = K G . ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 13 R
000 = −→ Figure 8.
Crossing changes of B with R to change L (left frame) into L (cid:48) (rightframe). R Figure 9.
Banding B to itself to turn L into J (cid:48) (right frame). R R
000 0
Figure 10.
The dual bands from J (cid:48) give a diagram of L from which it is easyto spot that K B is slice.We now give a criterion under which special RBG links produce pairs of slice knots; for ourpurposes this will be a setting we will want to avoid. The following lemma is analogous toTheorem 2.3 of [44]. Lemma 4.8. If L is a special RBG link and B ∪ G is split then K B and K G are ribbon.Proof. Since B and G are split, there is some sequence of crossing changes of B with R whichturns L into the link L (cid:48) = R ∪ µ R ∪ µ R in the right frame of Figure 8. Each of these crossingchanges can instead be exhibited by banding B to itself, as in the left frame of Figure 9, atthe expense of generating an additional (blue) meridian of R . Therefore, by adding bands fromthe blue component to itself L can be turned into a link J (cid:48) as in the right frame of Figure 9(with some number of blue meridians). Thus we can isotope the link L into the form shown inthe left frame of Figure 10; i.e. so that L looks like J (cid:48) with (dual) bands connecting the bluecomponents.From this picture of L , we can easily cancel G ∪ R so that we are left with a surgery diagramof K B ; to perform the cancellation we just first have to slide every band that geometrically links G across R . As such, we see that K B has a diagram of the form shown in the right frame ofFigure 10. That K B is ribbon follows from the picture. The claim that K G is ribbon follows bysymmetry of hypothesis. (cid:3) Small RBG links.Definition 4.9.
An RBG link L is small if L is special and in addition ±
20 00 0
Figure 11.
A small RBG link with ∆ B ∩ ∆ G a single arc in each disk. • B bounds a properly embedded disk ∆ B that intersects R in exactly one point, and in-tersects G in at most points. • G bounds a properly embedded disk ∆ G that intersects R in exactly one point, and inter-sects B in at most points.(All intersections are required to be transverse.)Example . Consider the small RBG link L and associated knots K B and K G in Figure 4.Our natural diagrams of K B and K G have 20 and 16 crossings, respectively. Using SnapPy toidentify the complements, we find that in fact K B is the 12 crossing knot 12 n
309 and K G is the14 crossing knot 14 n c ( K ) + c ( K (cid:48) ) amongall pairs of distinct knots with S ( K ) ∼ = S ( K (cid:48) ) in the literature.Notice that the diagrammatic conditions making a special RBG link small ensure that oneonly has to perform at most two slides of G over R in order to exhibit the knot K G (resp. twoslides of B over R to exhibit K B ). This helps keep the crossing number of K G (resp. K B )somewhat small.In fact, for a small RBG link to produce K B (cid:54) = K G we show that two slides are necessary. Proposition 4.11. If L is a small RBG link with ∆ B intersecting G in less than points then K B = K G . To prove the proposition, we require two lemmas.
Lemma 4.12.
Let L be a small RBG link and ∆ B and ∆ G be disks as in Definition 4.9. If ∆ B ∩ ∆ G is a single non-proper arc in each of ∆ B and ∆ G , then K B = K G .Proof. Since we also know that both ∆ B and ∆ G intersect R in exactly one point, we can isotope L in S so that a neighborhood of ∆ B ∪ ∆ G looks as in the middle frame of Figure 11. Performingthe slam dunks to exhibit K B and K G from this picture of L yields diagrams of K B and K G asin the left and right frames of Figure 11; these diagrams are identical outside the neighborhoodshown and isotopic inside the neighborhood shown. (cid:3) Lemma 4.13.
Let L be a small RBG link and ∆ B and ∆ G be disks as in Definition 4.9. If | ∆ B ∩ G | = 1 then, after an isotopy of ∆ G rel boundary, we can arrange so that ∆ B ∩ ∆ G is asdescribed in Lemma 4.12.Proof. Since ∆ B intersects G once geometrically, any disk bounded by G must intersect B oncealgebraically. Thus, the hypothesis that L is small implies that ∆ G intersects B exactly oncegeometrically.Consider the intersection ∆ B ∩ ∆ G (cid:44) → ∆ B ; by general position this is a 1-dimensional subman-ifold of ∆ B (with a finite number of components, which are not necessarily properly embedded).Since G ∩ ∆ B is a point, there must be exactly one arc of intersection that has exactly oneendpoint in int (∆ B ). The same analysis of ∆ B ∩ ∆ G (cid:44) → ∆ G shows that there is exactly one arcin ∆ G that has exactly one endpoint in int (∆ G ). Endpoints of arcs in int (∆ G ) correspond to ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 15 ∆ B ∆ B ∩ ∆ G G R
Figure 12.
The disk ∆ B for a small RBG link with ∆ B ∩ G a single point.endpoints of arcs in ∂ ∆ B . Thus there is exactly one arc in ∆ B that has exactly one endpoint in ∂ ∆ B . We deduce that there is exactly one arc of intersection in ∆ B , and it has one endpoint in int (∆ B ) and one endpoint in ∂ ∆ B .The other intersections are circles on ∆ B . See Figure 12. Recall that ∆ B ∩ R = { pt } , so eachcircle either contains { pt } or does not. Consider an innermost circle γ ⊂ ∆ B which does notcontain { pt } . Then γ bounds a disk ∆ (cid:48) B ⊂ ∆ B with ∆ (cid:48) B ∩ R = ∅ . Since γ is also a circle in∆ G , it bounds a disk ∆ (cid:48) G ⊂ ∆ G . Note that ∆ (cid:48) G does not contain { pt (cid:48) } := ∆ G ∩ R (since if itdid the disks ∆ (cid:48) B and ∆ (cid:48) G would give an S (cid:35) S with R ∩ S = { pt } , which is impossible forhomology reasons). Thus we can replace ∆ (cid:48) G ⊂ ∆ G with a parallel copy of ∆ (cid:48) B ; this yields anew ∆ G that has fewer circles of intersection with ∆ B . This replacement does not change thearc intersections of ∆ B with ∆ G nor the intersections of either disk with R . Continue until nosuch γ remain.Now consider an innermost circle γ on ∆ B which does contain { pt } . By a similar analysis, γ ⊂ ∆ G bounds a disk ∆ (cid:48) G ⊂ ∆ G which does contain { pt (cid:48) } . Now replacing ∆ (cid:48) G ⊂ ∆ G with aparallel copy of ∆ (cid:48) B does change ∆ G ∩ R , but the new ∆ G ∩ R is still a single point. Continueuntil no such γ remain. (cid:3) Proof of Proposition 4.11.
The case that ∆ B ∩ G in a single point is a consequece of Lemmas4.13 and 4.12. In the case that ∆ B ∩ G is empty, use ∆ B to isotope B to a meridian of R whichdoes not link G . Since R ∪ G ∼ = R ∪ µ G , we can isotope B ∪ R ∪ G to µ R ∪ R ∪ µ R . From herethe slam dunk homeomorphisms give K B = B = U = G = K G . (cid:3) Remark . When L is small and ∆ B intersects G in two points, it is possible that ∆ B ∩ ∆ G contains no circles of intersection and yet the knots K B and K G are distinct; the reader cancheck that the RBG link in Example 4.10 has this property.In view of Proposition 4.11 (and its analogue with B and G switched), we see that if wewant to produce K B (cid:54) = K G from a small RBG link, we should only consider the cases when | ∆ B ∩ G | = | ∆ G ∩ B | = 2. 5. Computer experiments
A six-parameter family of RBG links.
To generate many examples of pairs of knotswith the same 0-surgery, we studied the family of small RBG links shown in Figure 13. The linkdepends on 6 parameters ( a , b , c , d , e , f ), corresponding to the numbers of full twists in eachbox. The first parameter a represents full twists between R and the 2-handle framing curve of R . We also have twists between R and its 2-handle framing in the box with b full twists, sooverall the red curve has framing r = a + b. Thus, in view of Lemma 4.2, the parity of the RBG link is given by a + b mod 2.The RBG link from Figure 13 produces the knots K G ( a, b, c, d, e, f ) and K B ( a, b, c, d, e, f )with the same 0-surgery, as shown in Figure 14. We investigated these knots for values of the a b cd e fa + b Figure 13.
The 6-parameter RBG link which generated our 3375 examples of0-surgery homeomorphisms. a b cd e f a b cd e f
Figure 14.
Knots K G ( a, b, c, d, e, f ) and K B ( a, b, c, d, e, f ) share a 0-surgery.parameters ranging between − − a, c, e ∈ [ − , , b, d, e ∈ [ − , . These parameters ensure that the crossing numbers of the knots K B and K G are at most 55.We obtained a family of 3375 pairs of knots. We used SnapPy [13] to generate a list of theseknots, compute their hyperbolic volumes, and identify some of them with knots from knot tables.Our family is sufficiently general that it includes many small knots; for example, out of the 31hyperbolic knots with at most 8 crossings, SnapPy recognized 19 among our data. In fact, onecan show that all the knots in our family have Seifert genus at most 2 and four-ball genus atmost 1; our knots include 19 of the 21 hyperbolic knots with at most 8 crossings that have theseproperties.The results of our investigations are described below, and supporting files can be found onlineat http://web.stanford.edu/ ∼ cm5/RBG.html .5.2. Methodology.
We searched our list for promising pairs of knots, in particular pairs suchthat one knot has s < n CP ), whereas the other has s = σ = 0, so has a chance of being H-slice in some n CP (or perhaps even slice in B ). (For ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 17 ed + fc −
10 0 a − Figure 15.
The RBG link from Figure 13, when b = − s < s >
0, see Remark 5.3.) From the start, we couldexclude some such pairs from the promising sublist using the following lemma.
Lemma 5.1. (a) If b = − , then K G ( a, b, c, d, e, f ) = K B ( a, b, c, d, e, f ) .(b) If a + b = 0 , then the knots K G ( a, b, c, d, e, f ) and K B ( a, b, c, d, e, f ) have diffeomorphictraces. Item (b) is relevant because if the knots have diffeomorphic traces and one has s < n CP ; cf. Lemma 3.5. Proof of Lemma 5.1. (a) One can check that for b = −
1, the RBG link from Figure 13 is isotopicto the one shown in Figure 15, which has a symmetry exchanging the B and G components.Therefore, the resulting K B and K G knots are isotopic.(b) By Lemma 4.3, the fact that r = 0 implies that the 0-surgery homeomorphism has property U , and therefore (by Theorem 3.13) it extends to a trace diffeomorphism. (cid:3) We then computed the signatures and the s invariants of the remaining 1800 pairs of knots.To do this, we wrote a general formula for the DT (Dowker-Thistlethwaite) codes of the knots K G ( a, b, c, d, e, f ) and K B ( a, b, c, d, e, f ), depending on the six parameters. We plugged it into Mathematica [55] and used the functions
KnotSignature and sInvariant from the
KnotTheory` package [7]. (Note that the signature is a 0-surgery invariant, so the signature of K B alwaysequals that of K G .) For the s invariant, for some values of the parameters, the program tooktoo long to compute (more than a few minutes). In such cases we had the option of simplifyingthe knot diagram in SnapPy with
Sage [13, 50], using the command K . simplify ( (cid:48) global (cid:48) ). Wethen plugged it back into either KnotTheory` or SKnotJob [49], where tried to decrease girth asin [20, Section 5.1] and then re-compute.We also used the following trick to reduce the number of knots for which we had to explic-itly compute s : If we found two knots of the same type ( K B or K G ) corresponding to values( a, b, c, d, e, f ) and ( a (cid:48)(cid:48) , b (cid:48)(cid:48) , c (cid:48)(cid:48) , d (cid:48)(cid:48) , e (cid:48)(cid:48) , f (cid:48)(cid:48) ) with a ≤ a (cid:48)(cid:48) , b ≤ b (cid:48)(cid:48) , c ≤ c (cid:48)(cid:48) , d ≤ d (cid:48)(cid:48) , e ≤ e (cid:48)(cid:48) , f ≤ f (cid:48)(cid:48) and with the same value of s , then we knew this value of s also holds for all the knots of thesame type with intermediate values ( a (cid:48) , b (cid:48) , c (cid:48) , d (cid:48) , e (cid:48) , f (cid:48) ), i.e. such that a ≤ a (cid:48) ≤ a (cid:48)(cid:48) , b ≤ b (cid:48) ≤ b (cid:48)(cid:48) , c ≤ c (cid:48) ≤ c (cid:48)(cid:48) , d ≤ d (cid:48) ≤ d (cid:48)(cid:48) , e ≤ e (cid:48) ≤ e (cid:48)(cid:48) , f ≤ f (cid:48) ≤ f (cid:48)(cid:48) . This is a consequence of the monotonicity of the s -invariant under generalized crossing changes,which was proved in [38, Theorem 1.1]. An exclusion.
As a result of our calculations, we found 24 pairs of knots with σ = 0, suchthat one knot in the pair has s = 0 and the other s = −
2. If the knot with s = 0 were H-slicein n CP , this would produce an exotic n CP . For one of the 24 knots with s = 0, namely K = K G (0 , , , − , , K is not H-slice in n CP , using the following method.If the knot K G (0 , , , − , ,
1) were H-slice in some n CP , then K G ( − , , , − , , n +1 CP by Lemma 2.3. However, K G ( − , , , − , ,
1) has the same trace as K B ( − , , , − , ,
1) byLemma 5.1(b). Therefore, K B ( − , , , − , ,
1) would also be H-slice in n +1 CP , according toLemma 3.5. Direct computation using KnotTheory` gives that s ( K B ( − , , , − , , − , which contradicts (2).5.4. Interesting examples.
We were left with 23 promising pairs of knots. For the knot in eachpair with s = 0 (i.e. each candidate for being H-slice in n CP ), we simplified the diagrams in SnapPy with
Sage , using K . simplify ( (cid:48) global (cid:48) ). The resulting diagrams are shown in Figures 1and 23. In Table 1 we list the apparent number of crossings (in the simplified diagram), volumeand Alexander polynomials of these knots.The knots K through K have trivial Alexander polynomial, and are therefore topologicallyslice by Freedman’s theorem [21]; they are candidates for being slice. The knots K through K satisfy the Fox-Milnor condition on the Alexander polynomial, but were shown to not betopologically slice by Dunfield and Gong [16], using their program [17] for computing twistedAlexander polynomials. The last two knots K and K do not satisfy the Fox-Milnor condition.Still, all of these knots are candidates for being H-slice in n CP . Proofs of Theorems 1.3 and 1.4. If K i ( i = 1 , . . . ,
23) is a knot from our list, then there is acompanion knot K (cid:48) i such that S ( K ) ∼ = S ( K (cid:48) i ) and s ( K (cid:48) i ) = −
2. Therefore, K (cid:48) i is not H-slice in n CP by the inequality (2). Suppose K i were H-slice in W = n CP for some n . (The case n = 0 corresponds to S .) Then Lemma 3.3 would show that K (cid:48) is H-slice in a 4-manifold X that is homotopy equivalent to W , and therefore homeomorphic to it by Freedman’s theorem[21]. On the other hand, X could not be diffeomorphic to W , because K (cid:48) i is not H-slice in W . (cid:3) Note that if any of the 23 knots were H-slice in n CP , they would actually be BPH-slice, inview of the following lemma. Lemma 5.2.
The knots in Figures 1 and 23 are all H-slice in CP .Proof. Observe that for all our knots, we have b + c + e ∈ { , , } . When b + c + e = 0, theRBG link in Figure 13 has the property that its B and G components are split. By Lemma 4.8,the corresponding knots K B and K G (with b + c + e = 0) are slice. Increasing b , c or e by onecorresponds to an annular cobordism in CP \ ( ˚ B (cid:116) ˚ B ) between the respective knots. Therefore,if n := b + c + e >
0, the knots K B ( a, b, c, d, e, f ) and K G ( a, b, c, d, e, f ) are H-slice in n CP . (cid:3) Remark . We also searched in our data for pairs of knots such that one has s > s = σ = 0 (which would be relevant for H-sliceness in n CP instead of n CP ), but wefound no pairs of this type. Of course, we can obtain such pairs by taking the mirrors of the 23knots in our table.We analyzed the 23 knots from Table 1 further, to make sure they pass some well-knownobstructions to being slice (or BPH-slice, as the case may be).First, we looked at algebraic obstructions coming from the Seifert matrix. ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 19
Name Identifier K K B (0 , , , , , −
1) 29 15.451403388 1 K K B (1 , , , , , −
1) 29 14.698440095 1 K K G (1 , , , − , ,
1) 32 20.930658865 1 K K B (2 , , − , , , −
1) 29 15.552102256 1 K K G (2 , , − , − , ,
1) 32 21.888892554 1 K K B (1 , , − , , , −
1) 29 16.583603453 t − t + 3 − t − + t − K K B (1 , , , , , −
1) 31 18.694759676 t − t + 11 − t − + t − K K B (1 , , , , − , −
1) 36 21.768651216 4 t − t + 33 − t − + 4 t − K K G (1 , , − , − , ,
1) 32 21.917877366 t − t + 3 − t − + t − K K G (1 , , , − , ,
1) 32 23.276452522 t − t + 11 − t − + t − K K G (1 , , , − , − ,
1) 41 25.720923264 4 t − t + 33 − t − + 4 t − K K G (1 , , , , − ,
1) 20 20.032239211 2 t − t + 21 − t − + 2 t − K K B (2 , , − , , , −
1) 35 18.623983982 2 t − t + 9 − t − + 2 t − K K B (2 , , , , , −
1) 31 16.662235002 − t + 5 − t − K K B (2 , , , , − , −
1) 33 20.505101934 2 t − t + 21 − t − + 2 t − K K B (2 , , , , − , −
1) 37 22.919098178 6 t − t + 49 − t − + 6 t − K K G (2 , , − , − , ,
1) 37 23.396805316 2 t − t + 9 − t − + 2 t − K K G (2 , , − , , ,
1) 16 17.009749601 t − t + 3 − t − + t − K K G (2 , , , − , ,
1) 34 22.384645541 − t + 5 − t − K K G (2 , , , − , − ,
1) 36 24.655381040 2 t − t + 21 − t − + 2 t − K K G (2 , , , − , − ,
1) 42 26.731842490 2 t − t + 21 − t − + 2 t − K K G (2 , , , , − ,
1) 18 19.113083865 t − t + 15 − t − + t − K K G (2 , , , , − ,
1) 22 21.642574192 5 t − t + 43 − t − + 5 t − Table 1.
Examples coming out of our computer experiments.
Proposition 5.4.
The knots K through K from Figure 1 are algebraically slice.Proof. Recall that the algebraic concordance class of a knot is given by its Seifert matrix up toS-equivalence [33, 54]. Since the Seifert matrix can be read from the 0-surgery, any two knotswith the same 0-surgery are algebraically concordant. Thus, for the knots in our table of theform K G , it suffices to check algebraic sliceness for their companions K B .A genus 2 Seifert surface for the knot K B ( a, b, c, d, e, f ) is shown in Figure 16. The associatedSeifert matrix with respect to the basis ( α, β, γ, δ ) is(3) A = − a + c + d + f c + 1 f + c + d + 10 c b + c + e b + c + 10 f + c + d b + c + 1 b + c + d + f . To show that a knot K B ( a, b, c, d, e, f ) is algebraically slice, we will explicitly find a two-dimensional subspace V on which the Seifert form A restricts to the zero matrix. In other ab cd e fα βγδ Figure 16.
A Seifert surface for the knot K B ( a, b, c, d, e, f ).words, if we form a 4 × S whose columns are the basis vectors of V , we should have S T A S = 0 . Looking at the knots K through K in Table 1, we see that with the exception of K and K , all the other values ( a, b, c, d, e, f ) satisfy b = 1 , d + f = 0 , a + c + e = 2. In these cases, wecan take V = Span { (0 , , , − , ( e − , , , } . For K we have ( a, b, c, d, e, f ) = (1 , , , , − ,
1) and we choose V = Span { (2 , , , − , (2 , − , , } . For K we have ( a, b, c, d, e, f ) = (2 , , − , , ,
1) and we choose V = Span { (1 , , , , (2 , , , } . (cid:3) Recall from Section 2 that a necessary condition for a knot to be BPH slice is that it itsLevine-Tristram signature function σ LT vanishes. Algebraically slice knots satisfy this, so Propo-sition 5.4 ensures that K through K have σ LT = 0. Of course, we also have σ LT = 0 for thetopologically slice examples K through K . For the two remaining knots K and K , theAlexander polynomial has no roots on the unit circle, and therefore σ LT is a constant function.Using the Seifert matrix (3), we checked that σ LT ( −
1) = σ = 0 and hence σ LT = 0.Second, for each of the 23 knots, we computed the knot Floer homology using the Knot Floerhomology calculator [52]. The concordance invariants τ from [42], ν from [43] and (cid:15) from [30]vanish. As an aside, the program indicated that all the knots from our list are non-fibered,non-L-space, and have Seifert genus equal to 2.Third, we used the program SKnotJob [49] to minimize the girth of the diagrams, and computeseveral Ramussen-type concordance invariants. Apart from the usual s (which is defined fromKhovanov homology over Q ), the program computed s F and s F (from Khovanov homologyover F and F ), as well as the Lipshitz-Sarkar s Sq invariant (from the first Steenrod squareon Khovanov homology [36]). All the knots had girth at most 10, and each calculation lastedonly a few seconds. All the invariants turned out to be 0. (On the other hand, computing theLipshitz-Sarkar s Sq ± invariants using a program such as [48] did not seem feasible, because ourknots have at least 16 crossings.)Finally, from Lemma 4.2 we see that 14 knots in Table 1 (namely, K , K , K , and K through K ) correspond to odd RBG links, because r = a + b is odd. In such a case, Theorem 3.7 showsthat the 0-surgery homeomorphism relating the knot and its companion does not extend totraces. Therefore, for those 14 knots, sliceness and BPH-sliceness cannot be excluded by the ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 21 fact that we have obstructed their companion knot (which we knew not to be BPH-slice because s = − , SnapPy indicates that the 0-surgeries on allour 23 knots are hyperbolic and have trivial isometry group (and hence trivial homeomorphismgroup, by Mostow rigidity). Hence, we expect that for the 14 examples with r odd, the trace ofthe knot is not even homeomorphic to that of its companion knot.We also attempted (unsuccessfully) to show the 5 topologically slice knots are slice. Wesearched for ribbon bands using Gong’s program [26], but we could not found any simple bandsthat produce a strongly slice link. For one example, namely K = K , we also tried to find aslice derivative as follows: we wrote down a (genus 2) Seifert surface F for K and the associatedSeifert matrix A , and then classified all dimension 2 subspaces of H ( F ) on which A restrictsto the 0-matrix. For each such half-basis, we drew a 2-component link L ⊂ S embedded on F representing that basis; such a link is usually called a derivative of K . It is well known (see[33]) that if K admits a strongly slice derivative then K is slice. Unfortunately, none of the links L that came out of this example were strongly slice; this was checked using Levine-Tristramsignatures or covering link calculus.Furthermore, for all 23 knots, we tried to use Lemma 2.5 to prove BPH-sliceness. However,changing any positive to a negative crossing in our diagrams resulted in knots with σ = 2, whichcannot be BPH-slice.6. Homotopy -spheres from annulus twisting Annulus twisting is a construction of 0-surgery homeomorphisms which stands out for its abil-ity to naturally produce infinitely many knots with the same 0-surgery. In view of Theorem 1.2,there are RBG links associated to annulus twisting; see Section 7.2 for their description. Inthis section we will discuss some homotopy 4-spheres that arise from annulus twisting, withoutexplicit reference to the corresponding RBG links.6.1.
Annulus twisting.
Annulus twisting was defined by Osoinach [41], and extended to otherframings and to Klein bottle twisting by Abe-Jong-Luecke-Omae [1]. Since we will need todiscuss it in some detail, we reproduce the proof that annulus twisting gives rise to 0-surgeryhomeomorphisms.Let A : S × I → S be an embedding (by the standard abuse, we will conflate the embeddingand its image) and let (cid:96) ∪ (cid:96) = ∂A a framed oriented link in S , where both the framing and theorientation are inherited from A ; when A is thought of an an oriented cobordism from S → S ,we are setting (cid:96) = ∂ + A . Now let γ denote a pair of pants and let Γ : γ → S be an embeddingsuch that the framed oriented boundary of Γ is (cid:96) ∪ (cid:96) ∪ J for some 0-framed knot J . We canthink of J as obtained by joining a parallel copy of (cid:96) to a parallel copy of (cid:96) using a band. Seethe left hand side of Figure 17 for examples of links (cid:96) ∪ (cid:96) ∪ J ; there, the K box representsparallel strands that are tied in some 0-framed knot K , and the k box represents k positive fulltwists. See also Figure 18 for examples of knots J that arise this way, when K is the unknot. Theorem 6.1 (Main theorem of [41]) . Associated to such a link (cid:96) ∪ (cid:96) ∪ J there is an infinitefamily of knots ( J n ) n ∈ Z such that S ( J n ) ∼ = S ( J ) .Remark . We make no claim on the distinctness of the knots J n here, but the interested readershould consult [5], which gives some weak conditions on (cid:96) ∪ (cid:96) ∪ J that guarantee infinitely manyof the J n are pairwise distinct. Proof.
The proof follows from two claims: first we will show that S ( J ) ∼ = S , /n, − /n ( J, (cid:96) , (cid:96) ).Second we will define J n and show that S , /n, − /n ( J, (cid:96) , (cid:96) ) ∼ = S ( J n ). We remark that thesesurgery coefficients are relative to the given framing (which often differs from the Seifert framing). ∞ k ∞ m k K k m ∞ K0 0 0 ∞ { µ, } − { Φ ( µ ) , } { F (Φ ( µ )) , k + 1 } m(cid:96) (cid:96) (cid:96) (cid:96) (cid:96) (cid:96) Figure 17.
Annulus twisting a link (cid:96) ∪ (cid:96) ∪ J , with (cid:96) and (cid:96) drawn in purpleand J in green. Here n = 1 and we keep track of the image of a 0-framed meridianof J under the annulus twist homeomorphism. m k Figure 18.
A family of knots J m [ k ]. When m = 3 − k the knots are ribbon,which can be seen by performing the purple band move indicated.Towards the first claim, observe that in S ( J ) the pair of pants Γ can be capped off with thesurgery disk to give an embedded annulus Γ (cid:48) with framed oriented boundary (cid:96) ∪ (cid:96) ⊂ S ( J ).We’ll use Γ (cid:48) to define a homeomorphismΦ n : S ( J ) → S , /n, − /n ( J, (cid:96) , (cid:96) )as follows. Consider Y = S ( J ) (cid:114) ν ( (cid:96) ∪ (cid:96) ) and observe that Γ (cid:48) restricts to a properly embeddedannulus in Y . Consider ν (Γ (cid:48) ) ∼ = S × I × I , where the final I factor comes from the normaldirection. Define the homeomorphism φ n : S × I × I → S × I × I, φ n ( θ, x, t ) = ( θ + n (2 πi ) t, x, t ) . Since φ n | S × I × ∂I is the identity map, we can extend φ n to a self homeomorphism Φ n of Y bytaking the identity map outside of ν (Γ (cid:48) ).We will now extend Φ n to a homeomorphism with domain S ( J ). We can do this by simplyreattaching the excised neighborhoods ν ( (cid:96) ∪ (cid:96) ), but in the target space we must take care toreattach along the gluing map modified by Φ n . We see that Φ n | ∂Y takes the meridional curvein ∂ ( ν ( (cid:96) )) (resp. ∂ ( ν ( (cid:96) ))) to the n curve on ∂ ( ν ( (cid:96) )) (resp. − n curve on ∂ ( ν ( (cid:96) ))). Thus thefirst claim follows.To prove the second claim, observe that by hypothesis (cid:96) ∪ (cid:96) cobound an annulus A in S .We can twist S along A as in the proof of the first claim to produce a homeomorphism f n : S /n, − /n ( (cid:96) ∪ (cid:96) ) → S . Since the knot ι n ( J ) ⊂ S /n, − /n ( (cid:96) ∪ (cid:96) ) (where the embedding ι n is given by any diagram of J ∪ (cid:96) ∪ (cid:96) ) may intersect A , the image f n ( ι n ( J )) is some a priori new knot J n ⊂ S . Surgering ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 23 m k m k
Figure 19.
The knots J m [ k ] and J m [ k ] − obtained from the knots in Figure 18by annulus twisting. When m = 3 − k the knots are ribbon, which can be seenby performing the purple and yellow band moves indicated.on ι n ( J ) and its image under f n , we get a homeomorphism F n : S , /n, − /n ( J, (cid:96) , (cid:96) ) → S r n ( J n ) . In general one would now need to inspect f n to compute r n , but since H ( S , /n, − /n ( J, (cid:96) , (cid:96) )) ∼ = Z we can conclude r n = 0. (cid:3) Examples.
Consider the family of knots J m [ k ] pictured in Figure 18. (When k = 1, thisrecovers the family J m from [2, Figure 4].) We ask for m to be odd but we allow it to benegative—in which case we have | m | left-hand half twists in the picture. Similarly, we allow thenumber k of full twists on the right to be an arbitrary integer. In all figures we are taking theannulus A ∼ = S × I to be oriented such that the S factor is counterclockwise in the figure andthe I factor points radially inward. For this orientation of A , (cid:96) is the oriented curve markedin Figure 17 and in the standard orientation of S , the normal direction t to A points into thepage.Annulus twisting applied to each J = J m [ k ] (with (cid:96) and (cid:96) as in Figure 21) produces aninfinite family of knots, J m [ k ] n , with the same 0-surgeries as we vary n . Of these, we show J m [ k ] and J m [ k ] − in Figure 19.As in Section 5, in the hope of producing an exotic S or n CP , we looked for exampleswhere one of the J m [ k ] and J m [ k ] ± was slice (or H-slice in n CP ), and another was not. Forthis, we would like that the knots satisfy σ = 0, and the values of the s invariant are different.Observe that if k = 0, then J m [ k ] and J m [ k ] ± are unknotted. Furthermore, when k = ± J m [ k ] and J m [ k ] ± share the same trace; this was originally proven in [2, Theorem2.8]; the readers can see this for themselves as a consequence of Theorem 3.13 and Remark 6.6.Thus, in view of Lemma 3.5, the only hope for interesting examples comes from the cases | k | ≥ J m [ k ] once in either direction is odd if and only if k is even, so the setting k = 2 shouldgenerically yield knots which do not have homeomorphic traces.Unfortunately, we found no examples with σ = 0 and different s for small values of k and m .Table 2 displays the values of the pair ( σ, s ) for J m [ k ]. For all the knots in the table except theone marked in blue, namely J [ −
2] = 11 n
79, the value of s stays the same when we do annulustwisting in either direction. For J [ − s = 0 but both annulus twists, J [ − and J [ − − , have s = 2 instead of s = 0. (However, the signature is nonzero in this example.) m k -2 -1 0 1 2 3-5 (2, 2) (2, 2) (0, 0) (0, 0) (0, 0) (0, 0)-3 (2, 2) (2, 2) (0, 0) (0, 0) (0, 0) (0, 0)-1 (2, 2) (2, 2) (0, 0) (0, 0) (0, 0) (0, 0)1 (2, 0) (2, 0) (0, 0) (0, 0) (0, 0) (0, 0)3 (2, 0) (2, 0) (0, 0) (0, -2) (0, -2) (0, -2)5 (0, 0) (0, 0) (0, 0) (0, -2) (0, -2) (0, -2)7 (0, 0) (0, 0) (0, 0) (-2, -2) (-2, -2) (-2, -2) Table 2.
Values of ( σ, s ) for knots of the form J m [ k ].A few remarks are in order about the knots in Table 2. We marked in red those knots wherewe know that the knot and its annulus twists are slice. For the knots J − k [ k ] and their annulustwists, Figures 18 and 19 show the existence of ribbon bands relating them to the unlink. Notethat many of these (untwisted) examples can be recognized from knot tables. Indeed, we have: J [ −
1] = 13 n , J [0] = unknot , J [1] = 8 , J − [2] = 8 , J − [3] = 13 n . Observe also that all knots between the vertical unknot line and the diagonal slice line (thatis, J m [ k ] with k ≥ , m ≤ − k or k ≤ , m ≥ − k ) are BPH-slice, and so are their annulustwists. This is because we can get from them to a slice knot by changing crossings in eitherdirection; cf. Lemma 2.5. Remark . The knots J − [2] = 8 and J − [2] appear in the list considered in Section 5, as K G (2 , , , , , −
1) and K B (2 , , , , , − J − [2] = 8 and J − [2] − appear in the list as K G ( − , , − , , ,
1) and K B ( − , , − , , , K G (1 , , , , , − K B (1 , , , , , − Homotopy 4-spheres.
Suppose that K is a slice knot, and that K and K (cid:48) admit a 0-surgery homeomorphism φ : S ( K ) → S ( K (cid:48) ). As in the proof of Lemma 3.3, we can constructthe homotopy 4-sphere(4) X = X ( − K (cid:48) ) ∪ φ V, where V is any slice disk exterior for K . Even when we know that both K and K (cid:48) are slice, itis not clear that X is a standard 4-sphere. We present below some examples of such homotopy4-spheres, for which we could not verify that X is S . Many more homotopy 4-spheres can beconstructed via the techniques of this paper; for example by taking K to be a slice knot whichsatisfies the hypothesis of Lemma 4.6. We demonstrate via our examples how one can drawexplicit handle decompositions of homotopy spheres X built as in (4). Example . We first give examples coming from the knot K = J − [2] = 8 and K (cid:48) any of itsannulus twists. The knot 8 is slice; we have exhibited a ribbon band, and hence ribbon disk D , in Figure 18. To draw a handle decomposition of V = B \ ν ( D ) we use the rising waterprinciple; see [25, Chapter 6.2]. This decomposition is given by the black and purple curves inFigure 20. ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 25 n | n | n + 2 n n + 2 n Figure 20.
Homotopy spheres associated to annulus twisting 8 . Left n > n < X has a handle diagram obtained from that of V by adding an additional 0-framed 2-handle along φ − ( µ K (cid:48) ), followed by a 4-handle. In Figure17 we give an example identification of the framed curve φ − ( µ K (cid:48) ) in S ( K ), where φ is thehomeomorphism given by annulus twisting J m ( k ) once and µ ( K (cid:48) ) is 0-framed. In Figure 20we exhibit in red the framed curve φ − n ( µ K (cid:48) ) ⊂ ∂V where φ n is the homeomorphism given byannulus twisting 8 n times, for n (cid:54) = 0. These images are computed by inspection of the annulustwist homeomorphism, which we gave explicitly in the proof of Theorem 6.1. Remark . Kyle Hayden informed us that for n = 1, the homotopy 4-sphere shown in Figure 20is standard. Remark . By keeping track of the image of a meridian as we have done in Figure 17, it isstraightforward to check that annulus twisting once in either direction is even if and only if k isodd, and that annulus twisting has property U when K = U and k = ± Connections with other constructions
In this section we draw RBG links for some of the 0-surgery homeomorphisms in the liter-ature. Theoretically, there is no work to this; the procedure to draw an RBG link for a fixedhomeomorphism is given in the proof of Theorem 1.2. However, we find the pictures to be ahelpful reference, thus we include them here. We have also made some effort to give particularlyelegant RBG links where possible.Some relationships between annulus twisting, the methods in [44], and dualizable patternshave already been illustrated in the literature; see [39, 44, 53].7.1.
Preliminaries.
Recall that Dehn surgery diagrams can be modified by “handle sliding”without changing the homeomorphism type of the manifold described. We observe now thatcertain handle slides of RBG links are RBG preserving; this proposition will allow us to drawparticularly nice RBG links for some constructions.
Proposition 7.1.
Let L be an RBG link and L (cid:48) be a framed link obtained from L by somenumber of slides of B over R . Then L (cid:48) is an RBG link and the pair of knots K (cid:48) B and K (cid:48) G associated to L (cid:48) are pairwise isotopic to the pair of knots K B and K G associated to L .Remark . By symmetry of the RBG construction, the proposition also holds with the rolesof B and G reversed. k K k K k K k + 1 k + 1 k + 10 0 k − k − Figure 21.
Simplifying an RBG link for annulus twisting.
Proof.
First we check that L (cid:48) is an RBG link. Since R and G are unchanged, the framed link G ∪ R still surgers to S . Since the framed link B ∪ R surgered to S before the slides, the framedlink B (cid:48) ∪ R surgers to S . Finally observe that handle slides preserve the homeomorphism type,hence homology type, of the resulting manifold.To check that K B ∼ = K (cid:48) B we will exhibit a homeomorphism from S ( K B ) to S ( K (cid:48) B ) whichcarries a 0-framed meridian to a 0-framed meridian. By excising these meridians, we observethat there is a homeomorphism from S (cid:114) ν ( K B ) to S (cid:114) ν ( K B ) taking meridians to meridians.It follows that there is a homeomorphism of pairs ( S , K B ) → ( S , K (cid:48) B ), hence that the knotsare isotopic. We will show that K G ∼ = K (cid:48) G in the same manner.To define our homeomorphism S ( K B ) → S ( K (cid:48) B ) recall that the RBG construction comesequipped with homeomorphisms ψ B : S r,b,g ( L ) → S ( K B ) and ψ (cid:48) B : S r,b (cid:48) ,g ( L (cid:48) ) → S ( K (cid:48) B ). Let f : S r,b,g ( L ) → S r,b (cid:48) ,g ( L (cid:48) ) the slide homeomorphism, and consider ψ (cid:48) B ◦ f ◦ ( ψ B ) − : S ( K B ) → S ( K (cid:48) B ). Now we’ll watch a 0-framed meridian of K B under each leg of the map: under ( ψ B ) − it goes to a 0-framed meridian of B , under f to a 0-framed meridian of B (cid:48) , and under ψ (cid:48) B to a0-framed meridian of K (cid:48) B . The construction and meridian-watching for G is similar and left tothe reader. (cid:3) Annulus twisting.
To draw an RBG link for annulus twisting we follow the processoutlined in the proof of Theorem 1.2. In Figure 21 we have carried a (red) 0-framed meridian of K through the homeomorphism to S ( J ). Here the framing label on red is in terms of the usualconvention, i.e. with respect to the diagramatic Seifert framing. Thus we obtain an RBG linkfor annulus twisting from the left hand frame of Figure 21 by adding a blue 0-framed meridianto the red curve.By Proposition 7.1, we can modify this RBG by slides of B or G over R as we like. We exhibita particularly simple RBG link by considering the slide and isotopy shown in Figure 21. Weremark that if K is non-trivial or k (cid:54) = ±
1, then this RBG link is not special.7.3.
Yasui’s construction.
In [56] Yasui gives a construction of knots with homeomorphic0-surgeries which he uses to disprove the Akbulut-Kirby conjecture. While we enthusiasticallyrecommend his paper, in fact we will model our diagrams off of the (reproduced) proof of hisconstruction given in Section 2.1 of [27].We restrict to the n = 0 setting and consider a (red) 0-framed meridian of the green curve inFigure 3(a) of [27]. We then inspect the image of that red meridian under the handle calculusin Figure 3 of [27]. Part (h) of that figure, together with image of the red curve, and a green0-framed meridian of that red image, is the first frame of our Figure 22. (We remark that thecalculus described in Figure 3 of [27] is local; the knot K ⊂ ∂Z used in their Proposition 2.2 can ROM ZERO SURGERIES TO CANDIDATES FOR EXOTIC DEFINITE FOUR-MANIFOLDS 27 − − − − KK K − Figure 22.
Braid closures of these tangles give RBG links for Yasui’s homeo-morphism.go wherever it wants outside of the region shown in their Figure 3. Our Figure 22 is global andour Z ∼ = B ; thus we depict that K may be knotted with the region marked K in Figure 22.)Appealing to Proposition 7.1, we can tidy up this RBG link via the slides and isotopy markedin Figure 22 to obtain the rather nice simple RBG link in the right frame of Figure 22.7.4. Dualizable patterns.
The dualizable patterns technique was developed and utilized in[3, 35, 9, 24, 6, 39]. In [44] an early version of the
RBG construction was developed; in [44]an RBG link is required to have r = 0, B ∪ R = B ∪ µ B , G ∪ R = G ∪ µ G and lk ( B, G ) = 0.(We remark that the RBG links studied there have property U , but are not necessarily special.)In the appendix to [44], it is proven that any homeomorphism constructed by the dualizablepatterns technique may be presented by an RBG link of the type studied in [44] and the converse:any RBG link of that type gives rise to a dualizable pattern. Instead of including a reproof here,which would require recalling the dualizable patterns construction, we refer the reader to theappendix of that paper. References [1] T. Abe, I. D. Jong, J. Luecke, and J. Osoinach,
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Department of Mathematics, Stanford University, Stanford, CA 94305
Email address : [email protected] Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139