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Ribbon concordances and doubly slice

Ruppik, Benjamin Matthias Advisors: Arunima Ray & Peter Teichner

The concordance group Glossary Definition 1. A knot K ⊂ S3 is called (topologically/smoothly) slice if it arises as the equatorial slice of a (locally flat/smooth) embedding of a 2- in S4. – Abelian invariants: Algebraic invari- Proposition. Under the equivalence relation “K ∼ J iff K#(−J) is slice”, the commutative monoid ants extracted from the Alexander module Z 3 −1 A (K) = H1( \ K; [t, t ]) (i.e. from the  n isotopy classes of o  S Z K := 1 3 , # := oriented knots S ,→S of knots homology of the universal abelian cover of the becomes a group C := K/ ∼, the knot concordance group. ). The neutral element is given by the (equivalence class) of the , and the inverse of [K] is [rK], i.e. the – Blanchfield linking form: Sesquilin- mirror image of K with opposite orientation. ear form on the Alexander module ( ) sm top Bl: AZ(K) × AZ(K) → Q Z , the definition Remark. We should be careful and distinguish the smooth C and topologically locally flat C version, Z[Z] because there is a huge difference between those: Ker(Csm → Ctop) is infinitely generated! uses that AZ(K) is a Z[Z]-torsion module. – Levine-Tristram-signature: For ω ∈ S1, Some classical structure results: take the signature of the Hermitian matrix sm top ∼ ∞ ∞ ∞ T – There are surjective homomorphisms C → C → AC, where AC = Z ⊕ Z2 ⊕ Z4 is Levine’s algebraic (1 − ω)V + (1 − ω)V , where V is a Seifert concordance group. Open question: Does this split? Nobody has observed 4-torsion in the concordance matrix for K. For any algebraically sm top groups C or C ; the currently best conjecture is that there is none! K and ω not a root of ∆K (t), this is zero. – Derivative: On a genus g a – Every knot K with ∆K (t) = 1 is topologically slice by Freedman’s work [Fre82]. But derivative is any nonseparating g-component not all topologically slice knots are explained by this theorem, i.e. there are plenty of topologically slice with linking and self-linking zero. knots not smoothly concordant to any Alexander polynomial = 1 knot [HLR12]. Figure 1: A slice knot appears when one uses a hyperplane to cut a 2-sphere in S4 in half. The lower (or upper) hemisphere – Metabelian invariants: Their definition uses 2 Still, many natural questions about the structure of C remain a mystery. of S forms a slice disk for the knot. In the picture we see a finer quotients of π1 of the complement and sig- 4 connected sum of a trefoil with its mirror image; there are also Application. We can use any topologically, but not smoothly slice knot to construct an exotic R (i.e. a 4- nature defects of the associated covers, exam- 4 n many slice prime knots. manifold which is homeomorphic, but not diffeomorphic to Rstd). In any other dimension n 6= 4, R has a ples are the Casson-Gordon invariants. unique smooth structure!

Ribbon concordances ribbon Multi-infection Definition 2 (Ribbon concordance). J is smoothly ribbon concordant From the handle decomposition of the concordance exterior we immedi- We can take different parts of a knot R and tie them into the shape of to K, written J ≥ K, if there is a smooth concordance1 C from J sm ately see that a string link P , the resulting knot is called the infection of R with the to K such that the restriction of the projection map 3 × [0, 1] → [0, 1] S pattern P . The following picture shows a very special example where to C yields a Morse function without maxima. Note that this is not a π ( 3 \ νJ) π (( 3 × [0, 1]) \ νC) (1) 1 S  1 S the infection sites don’t interact with each other3: symmetric relation! is surjective. Gordon showed that on the other end, the induced map Index slice disk ribbon disk 3 3 Maximum 2   π1(S \ νK) ,→ π1((S × [0, 1]) \ νC) (2) Saddle 1   is injective. This motivates the following homotopy analogue of the notion of a smooth ribbon concordance: Minimum 0   3 Definition 4. An oriented knot J in S is homotopy ribbon concor- Table 1: Concordance vs. ribbon concordance dant to a knot K if there is a locally flatly embedded concordance C ∼= 1 ×[0, 1] in 3 ×[0, 1], restricting to J ⊂ 3 ×{1} and K ⊂ 3 ×{0} Conjecture (Gordon, [Gor81]). ≥sm is a partial ordering on the set of S S S S 3 such that the induced maps on fundamental groups of exteriors satisfy (isotopy classes of) knots in S . Remark. The difficult part is the anti-symmetry, i.e. showing that from (1) and (2). We write J ≥top K. Figure 5: The boxes labelled with J and K indicate that the strands going the existence of two ribbon concordances J ≥ K and K ≥ J we get through them are tied into the knots J and K (all parallel to each other). sm sm The extra full twists on the left arise from untwisting the of the that the knots J and K are already isotopic. projection of the trefoil J. 3 Definition 3. K ⊂ S is called a if K ≥sm unknot. Conjecture (Slice-ribbon, [Fox, 1960s]). Every smoothly slice knot is If there is only a single box, this is a satellite construction. The op- a ribbon knot. eration is well-defined on concordance classes (“trace out the concor- Remark. This is probably the most famous open question in knot con- dance annulus with the pattern”), and so any satellite operator P de- cordance. It is subtle, in part because there are examples of slice disks scends to a function (usually not a homomorphism!) P : C → C, where which are not even isotopic to ribbon disks (e.g. start with a nontrivial [K] 7→ [P (K)]. 2-knot and take a connected sum with the trivial disk on the unknot). – With this, one can construct (n)-solvable knots, i.e. knots lying in the n-th level Fn of the Cochran–Orr–Teichner solvable filtration [COT03] of the knot concordance group C. The interesting but difficult part is to find examples which are not slice (or not even in Fn.5)! – Large classes of satellite operators are (weakly) injective on concor- dance classes. There are several ways of making C into an interesting metric space, and one can investigate how these metrics play together Figure 3: Going upwards in a ribbon concordance, we at first see the birth of with the operations on the concordance group. some , which subsequently are connected via band-sums at the saddles. I would like to look at string link infection in the context of doubly slice Recently, Friedl and Powell observed that a ribbon concordance between knots: Many results should have an analog formulation for the double J and K implies the divisibility of Alexander polynomials: 3 concordance group. For example, there is already work on the algebraic Figure 2: A ribbon disk for the stevedore knot 61, immersed into S with ribbon singularities. J ≥top K ⇒ ∆K | ∆J double concordance group ACds, and Kim [Kim06] extended the COT filtration to topologically doubly slice knots. Recently, there has been an influx of papers dealing with the maps in- Maybe this statement can be extended to an injection of Blanchfield duced by ribbon concordances on various knot homology theories: For forms in the topologically locally flat setting. example, ribbon concordances yield injective maps on knot Floer ho- References mology [Zem19] and on [LZ19]. Doubly slice knots Proposition. A handle decomposition of the concordance annulus C [CFT09] Tim D. Cochran, Stefan Friedl, and Peter Teichner. New constructions of slice links. Com- mentarii Mathematici Helvetici, 84(3):617–638, 2009. https://arxiv.org/abs/0709. translates into a recipe for a handle decomposition of the concordance Question ([Fox, 1960s]). Which slice knots are cross sections of un- 2148. 3 3 exterior XC := (S × [0, 1]) \ νC relative to XJ := S \ νJ: Whenever we knotted2 2- S2 ,→ S4? [COT03] Tim D. Cochran, Kent E. Orr, and Peter Teichner. Knot concordance, Whitney towers pass through a level 3 ×{t} corresponding to a critical point of index k, and L2-signatures. Annals of Mathematics, pages 433–519, 2003. https://arxiv.org/ S Definition 5. We call such slices of a 2-unknot doubly slice. abs/math/9908117. a k-handle is added to the concordance cylinder C and a (k + 1)-handle Warning. As in the case of slice knots, there is an important distinction [Fre82] Michael H. Freedman. A surgery sequence in dimension four; the relations with knot is added to the exterior XC . between the smooth and the topologically locally flat category. concordance. Inventiones mathematicae, 68(2):195–226, 1982. Examples of doubly slice knots: [Gor81] C. McA. Gordon. Ribbon concordance of knots in the 3-sphere. Mathematische Annalen, XC = XJ × [0, 1] ∪ (2-handles) ∪ (3-handles) – Any knot of the form K#(−K) (this is a slice 257(2):157–170, 1981. of the (±1)-spin of K) [HLR12] Matthew Hedden, Charles Livingston, and Daniel Ruberman. Topologically slice knots and dually with nontrivial Alexander polynomial. Advances in Mathematics, 231(2):913–939, 2012. –9 46 (perform a saddle move on one of the two https://arxiv.org/abs/1001.1538. XC = XK × [0, 1] ∪ (1-handles) ∪ (2-handles) “arms”, in any case this yields two unlinked un- [Kim06] Taehee Kim. New obstructions to doubly slicing knots. , 45(3):543–566, 2006. knotted circles which can be capped off by disks. https://arxiv.org/abs/math/0411126. Figure 4: Two Conjecture. Ribbon concordance exteriors are aspherical. Claim: These slice disks fuse to an unknotted [LZ19] Adam Simon Levine and Ian Zemke. Khovanov homology and ribbon concordance. arXiv band-moves for 946. preprint, 2019. . Remark. Since the exterior of a ribbon disk is homotopy equivalent to a S2 in S4) https://arxiv.org/abs/1903.01546 2-complex, the asphericity is equivalent to π2(XC ) = 0. The Whitehead Warning. We have to be careful when we want to define a double con- [Zem19] Ian Zemke. Knot Floer homology obstructs ribbon concordance. arXiv preprint, 2019. https://arxiv.org/abs/1902.04050. conjecture (Conjecture: Any subcomplex of an aspherical 2-complex cordance group Cds. A natural choice would be to declare that K and J is aspherical.) would imply this, because adding a meridian 2-disk of C are doubly concordant iff K#(−J) is doubly slice, but it is not known to XC produces a contractible space. For example, we know that the whether this gives a transitive relation! exterior is a K(π, 1) for ribbon disks with a single saddle, in this case Open question (Stably doubly slice =? doubly slice). Suppose K and 2019 Max Planck Institute for Mathematics, Bonn the statement follows from the one-relator-theorem. J#K are doubly slice. Then, must J be doubly slice? [email protected]

1A concordance is a smoothly embedded annulus S1 × [0, 1] in S3 × [0, 1] with boundary −{0} × J ∪ {1} × K. 2A sphere S2 ,→ S4 is by definition unknotted if it extends to an embedding of a disk D3 ,→ S4. 3In the general situation of multi-infection one should start with a proper embedded multi-disk, thicken this and replace with the string link