Balanced algebraic unknotting, linking forms, and surfaces in three- and four-space
BBALANCED ALGEBRAIC UNKNOTTING, LINKING FORMS,AND SURFACES IN THREE- AND FOUR-SPACE
PETER FELLER AND LUKAS LEWARK
Abstract.
We provide three 3–dimensional characterizations of the Z –slicegenus of a knot , the minimum genus of a locally-flat surface in 4–space cobound-ing the knot whose complement has cyclic fundamental group: in terms ofbalanced algebraic unknotting, in terms of Seifert surfaces, and in terms ofpresentation matrices of the Blanchfield pairing. This result generalizes to aknot in an integer homology 3–sphere and surfaces in certain simply connectedsignature zero 4–manifolds cobounding this homology sphere. Using the Blanch-field characterization, we obtain effective lower bounds for the Z –slice genusfrom the linking pairing of the double branched cover of the knot. In contrast,we show that for odd primes p , the linking pairing on the first homology of the p –fold branched cover is determined up to isometry by the action of the decktransformation group on said first homology. Introduction
The main result of this paper is the following.
Theorem 1.1.
For a knot K —a smooth, oriented, non-empty, and connected1–submanifold of S —and a non-negative integer g , the following are equivalent.(1) There exists an oriented compact surface F of genus g properly embedded andlocally flat in B with boundary K ⊆ S = ∂B such that π ( B \ F ) ∼ = Z .(2) There exists a smooth oriented compact genus g surface in S with twoboundary components, one of which is the knot K and the other a knot withAlexander polynomial .(3) The knot K can be turned into a knot with Alexander polynomial bychanging g positive and g negative crossings.(4) The Blanchfield pairing of K can be presented by a Hermitian matrix A ( t ) of size g over Z [ t ± ] such that the integral symmetric matrix A (1) hassignature zero. Here, the
Alexander polynomial and the
Blanchfield pairing are the classical knotinvariants introduced by their respective eponyms [Ale28, Bla57]. The Alexanderpolynomial of a knot K is most quickly defined as the order of the Alexandermodule H ( S \ K ; Z [ t ± ]) of K , which is the Z [ t ± ]–module given as the firstintegral homology group of the infinite cyclic cover of S \ K with Z [ t ± ]–modulestructure given by t acting as the group isomorphism induced by a generator ofthe deck transformation group. The Blanchfield pairing is a Hermitian pairing on Mathematics Subject Classification.
Key words and phrases.
Blanchfield pairing, linking pairings, branched coverings, unknottingnumber, slice genus. a r X i v : . [ m a t h . G T ] M a r PETER FELLER AND LUKAS LEWARK H ( S \ K ; Z [ t ± ]) taking values in Q ( t ) / Z [ t ± ]. We refer the reader to Section 2 formore detailed definitions. By changing a positive (negative) crossing of a knot K , weunderstand the 1–framed ( − crossingdisk , i.e. a smooth closed 2–disk D ⊂ S that intersects K exactly twice, only inthe interior, transversely, and such that the two intersection points have oppositeinduced orientations.Before discussing context, applications, an outline of the proof, and a generaliza-tion to more general ambient 3– and 4–manifolds of Theorem 1.1, we note that thispaper naturally splits into two, essentially independent, parts. A first part containsthe proof of Theorem 1.1. A second part is concerned with providing obstructionsfor knots to satisfy (1) of Theorem 1.1 for small g using (4). Concretely, we provideobstructions by specializing the Blanchfield form to the linking form on the firsthomology of the p –fold branched cover of the knot for p a prime. For p = 2, weprovide an easily applicable criterion (see Proposition 1.12), which turns out to beeffective for knots in the knot tables. The same strategy does not give interestingobstructions for odd primes p . This is explained by our second result. Theorem 1.2.
Let p be an odd prime, Σ p ( K ) the p –fold branched covering of aknot K , and N p ( K ) := H (Σ p ( K ); Z [ t ] / ( t p − the first homology group of Σ p ( K ) ,on which t acts as a generator of the deck transformation group. Then, any twonon-degenerate Hermitian pairings on N p ( K ) are isometric. In particular, if fortwo knots K, K (cid:48) , the Z [ t ] / ( t p − –modules N p ( K ) and N p ( K (cid:48) ) are isomorphic, thenthe linking pairings on these modules are isometric. The techniques for this second part involve elementary number theory, which arerather different from the low-dimensional topology arguments employed in the restof the text: we use quadratic reciprocity and Dirichlet’s prime number theorem, and,for our result when p is odd, we generalize parts of the proof of Wall’s classificationof symmetric pairings on finite Abelian groups with odd order [Wal63] to modulesover Dedekind domains of a certain order. This second part is contained in Section 6,and outlined in more detail in Section 1.8. The first part can be read independentlyof it.1.1. The genus zero case.
For g = 0, the conditions (2), (3) and (4) of Theorem 1.1are all immediately seen to be equivalent to K having Alexander polynomial 1. Inother words, for g = 0, Theorem 1.1 is simply restating the following celebratedapplication of Freedman’s disk embedding theorem [Fre82b]: Theorem 1.3 ([Fre82b, Theorem 1.13]; see also [FQ90, 11.7B] and [GT04, Ap-pendix]) . A knot K has Alexander polynomial if and only if there exists a prop-erly embedded locally-flat disk D in B with boundary K ⊆ S = ∂B such that π ( B \ D ) ∼ = Z . (cid:3) We do not claim to reprove Theorem 1.3; in fact, its only-if-direction (the partbased on the disc embedding theorem) is the main input for the proof of (2) ⇒ (1).We understand Theorem 1.1 as a quantitative version of Theorem 1.3 in that itcharacterizes the existence of a genus g surface in B (rather than a disk) in termsof classical, 3–dimensional knot invariants. It does so in three a priori different ways:via the 3D–cobordism distance (2), balanced algebraic unknotting distance (3), anda condition on presentation matrices of the Blanchfield pairing (4). ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 3 changenegativecrossing changepositivecrossing
Figure 1.
The genus-one pretzel knot (3 , ,
3) (on the left) can beturned into a knot with Alexander polynomial 1 (on the right) bychanging one negative and one positive crossing.1.2.
Three-dimensional consequences.
While the principal motivation for ourmain result is the 3D–characterization of an a priori 4D–quantity, the different3D–characterizations (2), (3), and (4) of (1) in Theorem 1.1 yield 3D–consequencessuch as the following.
Corollary 1.4.
A knot of genus one, i.e. a knot arising as the boundary of aonce-punctured torus embedded in S , can be turned into a knot with Alexanderpolynomial by changing one positive and one negative crossing.More generally, this can be achieved for any knot that arises as one of theboundary components of a twice-punctured torus embedded in S , whose otherboundary component has Alexander polynomial .Proof. The first claim is indeed a special case of the second claim: puncturing agenus minimizing Seifert surface of a knot K of genus one yields a twice puncturedtorus in S with boundary consisting of K and an unknot (which of course hasAlexander polynomial 1). The second claim follows from Theorem 1.1 (2) ⇒ (3) for g = 1. (cid:3) Compare this to related results in [Ohy94, Liv19]. Corollary 1.4 stands incontrast with the existence of knots with genus one that cannot be unknotted bychanging one positive and one negative crossing, such as P ( p, q, r ) pretzel knotswith p ≥ , q ≥ , r ≥ P (3 , , P (3 , ,
3) is known to have unknotting number u ( P (3 , , > Four-dimensional consequences.
To facilitate the discussion, let us definethe Z –slice genus g Z ( K ) of a knot K as the smallest integer arising as the genusof a Z –slice surface for K , i.e. an oriented compact surface F , properly embeddedand locally flat in B with boundary K ⊆ S and π ( B \ F ) ∼ = Z . In other words, g Z ( K ) is the minimal g such that (1) holds.In previous work [FL18], the authors defined the algebraic genus g alg ( K ) of aknot K as the smallest non-negative integer g satisfying the following condition: (5) The knot K admits a g + h ) –dimensional Seifert matrix M for some h ≥ such that the upper-left square h –dimensional submatrix N of M satisfies det( t · N − N (cid:62) ) = t h . One motivation for this definition was to obtain easily calculable upper boundsfor g Z and the topological slice genus g top . Indeed, it was shown in [FL18] that PETER FELLER AND LUKAS LEWARK (5) ⇔ (2) ⇒ (1), i.e. g alg ≥ g Z . Now, as a consequence of Theorem 1.1, (5) isequivalent to (1) – (4), and we have the following Corollary 1.5.
The algebraic genus and the Z –slice genus agree for all knots. (cid:3) The algebraic genus and the Seifert matrix techniques related to (5) will not beused here and the paper at hand can be read without any knowledge of [FL18].On the subject of Seifert matrices, we point out the following consequence of ourmain result. It follows from Corollary 1.5 and the fact (see [FL18, Proposition 10])that the algebraic genus is a classical knot invariant. A knot invariant is a classicalknot invariant , if for every knot K it is determined by the isometry class of theBlanchfield pairing of K (or equivalently, by the S -equivalence class of the Seifertforms of K ). Corollary 1.6.
The Z –slice genus of a knot K is a classical knot invariant . (cid:3) This distinguishes g Z from other knot genera such as the three-dimensional knotgenus, the smooth slice genus, or the topological slice genus, none of which areclassical.A topological slice surface F ⊂ B is a (topological) superslice surface if its double S ⊂ S (the surface S in S given by gluing together ( B , F ) with an oppositelyoriented copy of itself) is unknotted, i.e. the boundary of an embedded locally flathandlebody. Combining Theorem 1.1 with a recent argument by Chen [Che18], wefind that whenever a knot K has a Z –slice surface of genus g , then it has a superslicesurface of genus g . Put differently, for the topological superslice genus of a knot K ,the smallest integer that arises as the genus of a superslice surface of K , we havethe following. Corollary 1.7.
For all knots K , the Z –slice genus of K equals the topologicalsuperslice genus of K .Proof. By the Seifert-van Kampen theorem, superslice surfaces are Z –slice surfaces.Hence, the superslice genus of a knot is greater than or equal to its Z –slice genus.We discuss the other inequality.If a knot K has a Z –slice surface of genus g , then by Theorem 1.1(1) ⇒ (2), wefind a Seifert surface C of genus g with two boundary components, one of which is K and the other is a knot J with Alexander polynomial 1. One may further arrange C to be H –null (see Lemma 5.3) and hence one finds a Seifert surface F J for J that is disjoint from C away from its boundary. Since the knot J has Alexanderpolynomial 1, it has a Z –slice disk D .We now follow Chen to see that C ∪ D becomes a superslice surface when properlypushed into B . For details on the following argument, we refer to his text. Let B J be a 3–ball in S that intersects S transversely with boundary the 2–spheregiven as the double of D (known to exist by [FQ90, 11.7A]). Crucially, Chen showsthat B J can be isotoped (fixing its boundary) such that the transverse intersectionof B J with S is a surface F := B J ∩ S consisting of a Seifert surface F (cid:48) J that isa stabilization of F J and several closed components such that C ∩ F = J [Che18,Lemma 3.8]. With the isotoped B J , denoted again B J by abuse of notation, athand, one can check that C ∪ D is (after being properly pushed into B ) a superslicesurface of genus g . Indeed, let H be the handlebody in S given as thick( C ) ∪ B J ,where thick( C ) is the image of an embedding of C × [ − ε, ε ] → S mapping C × { } ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 5 to C and J × [ − ε, ε ] to ∂B J . The boundary ∂H is the double of (a properly pushedinto B copy of) C ∪ D ; compare [Che18, Proof of Theorem 1.5]. (cid:3) We note the following subtlety in the above proof. One is tempted to show thestronger statement that every Z –slice surface is a superslice surface. Promisingly, aSeifert-van Kampen calculation reveals that the double S ⊂ S of a Z –slice surfacesatisfies π ( S \ S ) ∼ = Z ; however, as far as the authors know, if S is not a sphere,it remains open whether such S are always unknotted. So, the use of our maintheorem (switching to a Z –slice surface given as the union of a Z –slice disk and asurface as in (2), which allows to reduce to the sphere case [FQ90, Theorem 11.7A])appears to be unavoidable for now.Next, we discuss applications of (1) ⇔ (4) of Theorem 1.1. As mentioned above,we use (1) ⇒ (4) and then specializations of (4) to give lower bounds for g Z ; seeSection 1.8. In a different direction, let us make a note of the following heuristic: ifsome knot theoretic construction has a well-understood effect on the Blanchfieldpairing, then one can use Theorem 1.1 to better understand its effect on g Z . Asa specific example, the following inequality for satellite knots, the main result of[FMP19], has a rather pleasing and fast proof using (1) ⇔ (4). Theorem 1.8 ([FMP19, Theorem 1.2]) . Every satellite knot P ( K ) satisfies g Z ( P ( K )) ≤ g Z ( P ( U )) + g Z ( K ) , where P and K denote the pattern knot and companion knot of P ( K ) , respectively. We omit precise definitions of the satellite operation and also keep the followingproof short (but complete), since Theorem 1.8 is not an original result of this text.
Proof.
By applying (1) ⇒ (4) to the knots K and P ( U ), we know that there existpresentation matrices A K ( t ) and A P ( t ) of K and P ( U ), respectively, satisfying (4),i.e. A K ( t ) and A P ( t ) are of size 2 g Z ( K ) and 2 g Z ( P ( U )), respectively, and σ ( A K (1)) = σ ( A P (1)) = 0. By a result of Livingston and Melvin [LM85, Theorem 2], theBlanchfield pairing of P ( K ) is presented by the matrix A P ( K ) ( t ) := A K ( t ) ⊕ A P ( t w ),where w denotes the algebraic winding number of the pattern knot P . However, A P ( K ) ( t ) is a presentation matrix of size 2 g Z ( K ) + 2 g Z ( P ( U )) satisfying (4), since σ ( A P ( K ) (1)) = σ ( A K (1)) + σ ( A P (1)) = 0. Thus, g Z ( P ( K )) ≤ g Z ( P ( U )) + g Z ( K )by (4) ⇒ (1). (cid:3) Finally, we discuss two four-dimensional knot invariants, the slice genus andthe stabilizing number, showing that their natural Z –analogs coincide. In [CN19],Conway and Nagel introduce the stabilizing number sn( K ) of a knot K with Arfinvariant zero, the minimal c such that a null-homologous locally flat slice disk for K exists in B S × S ) c , and they show the following. For all knots K with Arf( K ) , sn( K ) ≤ g top ( K ) [CN19] . While the classical Levine-Tristram signature lower bounds on g top also hold for sn,Casson-Gordon invariants (which are not classical in the sense that they do notonly depend on the S -equivalence class of knots) may be employed to show that g top and sn differ substantially [CN19]. As a consequence of a generalization of ourmain result (stated below as Theorem 1.10), we find that the corresponding notionof Z –stabilizing number is equal to the Z –slice genus for all knots with Arf invariantzero. Here, we define the Z –stabilizing number , denoted sn Z ( K ), like sn with theadditional assumption that the slice disk D is a Z –slice disk in B S × S ) c . PETER FELLER AND LUKAS LEWARK
The condition that the disc is null-homologous in B S × S ) c turns out tobe superfluous: it is implied by its complement having fundamental group Z ; seeLemma 5.1. Corollary 1.9.
For all knots K with Arf( K ) = 0 , sn Z ( K ) = g Z ( K ) . (cid:3) It was surprising to the authors that two different 4-dimensional knot invariants( g top and sn) with very subtle behavior, become the same if a regularity assumption(complements have fundamental group Z ) is added. We also note that g top ( K ) = 0if and only if sn( K ) = 0, by definition. So, the difference between sn and g top manifests itself only for the “non-genus zero case”.1.4. Perspective: The Z –slice genus is a balanced algebraic unknottingnumber. By Theorem 1.1, the Z –slice genus of a knot K can be seen as a variationof the algebraic unknotting number u a ( K ) introduced by Murakami [Mur90] andstudied amongst others by Fogel [Fog93], Saeki [Sae99] and by Borodzik and Friedl[BF15, BF14]. Indeed, compare to (3) that u a ( K ) equals the minimum number ofcrossing changes necessary to convert K into a knot with Alexander polynomial 1.One could say u a ( K ) and g Z ( K ) are respectively the unsigned and balanced Gordiandistance between K and Alexander polynomial 1 knots. From this perspective, theinequalities g Z ( K ) ≤ u a ( K ) ≤ g Z ( K ) ≤ deg(∆ K ) first proven in [FL18] are nowquite evident.Of course, one may also more generally consider signed algebraic unknotting,and ask whether for given p, n ≥
0, the knot K can be turned into an Alexanderpolynomial 1 knot by changing p positive and n negative crossings. This questionadmits an answer in terms of presentation matrices of the Blanchfield pairing [BF14]generalizing (4), which we cite and use in this text as Theorem 4.1. However, thebalanced setting is of special interest, because there seems to be no analog in theunsigned or signed setting of the characterizations (1) and (2) of the Z –slice genusin terms of surfaces in 3– and 4–space.1.5. Proof of Theorem 1.1.
We provide the structure of the proof of Theorem 1.1.The details of the argument will be given in Sections 3 and 4.
Proof of Theorem 1.1. (1) ⇒ (4): This is the content of Section 3. This part of theproof is in a way the generalization of the if-part of Theorem 1.3 from g = 0 toarbitrary g ≥
0. However, while the case g = 0 is a quite straight-forward homologycalculation (see [Fre82a, last paragraph of Sec. 1.2]), the case g ≥ F as in (1), theBlanchfield pairing appears as sesquilinear intersection form of the universal coverof a slight modification of B \ F . The second homology of that cover is a free Z [ t ± ]–module of rank 2 g , which gives a Hermitian presentation matrix A ( t ) of thedesired size for the Blanchfield pairing. The signature of A (1) may be calculated bythe Novikov-Wall non-additivity theorem.(4) ⇒ (3): A result of Borodzik and Friedl implies (4) ⇒ (3) under the addedhypothesis that A (1) be congruent to a diagonal matrix [BF14, Thm. 5.1]. We showthat this hypothesis is not necessary. The details are provided in Section 4.1.(3) ⇒ (2): The Seifert surface may be constructed from the crossing changes inan explicit and geometric way; see Section 4.2.(2) ⇒ (1): This is known to be a consequence of the only-if-part of Theorem 1.3.We recall the brief argument. The surface F as in (1) is found by taking the union ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 7 of the Seifert surface in S with two boundary components as in (2) with a locallyflat disk D with boundary the Alexander polynomial 1 component as described inTheorem 1.3. After pushing the interior of F into the interior of B , it remains tocheck that π ( B \ F ) ∼ = Z ; see [FL18, Proof of Claim 20] for details of how this canbe done.Alternatively, using the setup up from Section 1.3, we note that (2) ⇔ (5) by[FL18, Proposition 17] and (5) ⇒ (1) by [FL18, Theorem 1]; however, we prefer theabove direct argument since it makes it clear that the Seifert matrix argumentsfrom [FL18] are not needed. (cid:3) Attempts at variations of the main theorem.
The reader might betempted to vary the main theorem by replacing the condition of ‘Alexander polyno-mial 1’. Such attempts will probably succeed to produce variations (1’), (2’), (3’)of conditions (1), (2), (3), and prove the implications (3’) ⇒ (2’) ⇒ (1’), but willprobably fail to produce a fitting analog to condition (4), or prove the equivalenceof (1’), (2’), (3’).For example, consider the balanced Gordian distance between K and the set ofsmoothly slice knots; this knot invariant was introduced by Livingston [Liv02], whodenoted it by U s ( K ). As in the proof of Theorem 1.1, one may prove that U s ( K )is an upper bound for the 3D–cobordism distance of K to a smoothly slice knot,which is in turn an upper bound for the smooth slice genus g smooth of K . Indeed,the resulting inequality U s ( K ) ≥ g smooth ( K ) was one of Livingston’s motivationsto study U s ( K ). However, as shown by Owens [Owe10], there exist knots K with U s ( K ) (cid:54) = g smooth ( K ).We invite the reader to try further variations, e.g. replacing the condition of‘Alexander polynomial 1’ by ‘topologically slice’, ‘smoothly Z –slice’ or ‘trivial’. Wehave found no case where the equivalence of the resulting conditions (1’), (2’), (3’)seems plausible for g > A generalization of the main theorem to integral homology spheresand Z –slice surfaces in other four-manifolds. So far, we have dealt withknots in S . However, the only property of S we ever use is its homology, and soTheorem 1.1 and its proof generalize with minimal changes to all integer homologyspheres M . By the work of Freedman [Fre82b, FQ90], there is a unique contractiblefour-manifold B with boundary M , which is the natural ambient space in whichto consider Z –slice surfaces. Perhaps more exciting than the transition from S toan integral homology sphere M is to consider Z –slice surfaces in four-manifolds V with boundary M that are different from B . This was motivated by Corollary 1.9,which one gets as an application. We note that it turns out that Z –slice surfaces inthe four-manifolds we consider are automatically null-homologous; see Lemma 5.1.This is why we do not add a null-homologous assumption below (in contrast to thedefinition of sn above).We describe a version of Theorem 1.1 that incorporates all of this. Since this willbe a bit lengthy to express, we invite the reader to first parse the statements thatfollow for the cases M = S , h = 0, and either c = 0 or c = 0. Theorem 1.10.
Let K be a knot in an integral homology sphere M . Let h and c be non-negative integers and c ∈ { , } , such that h + c ≥ if Arf( K ) = 1 . Set g = h + c + c . Then the following are equivalent: PETER FELLER AND LUKAS LEWARK (1’) There exists an oriented compact surface F of genus h properly embeddedand locally flat in V = B C P C P ) c S × S ) c with boundary K ⊆ M = ∂V such that π ( V \ F ) ∼ = Z . Here, B denotes the uniquecontractible topological four-manifold with ∂B = M .(2’) There exists a smooth oriented compact genus g surface in M with twoboundary components, one of which is the knot K and the other a knot withAlexander polynomial .(3’) The knot K can be turned into a knot with Alexander polynomial bychanging g positive and g negative crossings.(4’) The Blanchfield pairing of K can be presented by a Hermitian matrix A ( t ) of size g over Z [ t ± ] such that the integral symmetric matrix A (1) hassignature zero. Note that (2’), (3’), (4’) of Theorem 1.10 are identical to (2), (3), (4) of Theo-rem 1.1 (with S replaced by M ). Also recall that C P S × S = C P C P C P ,which is why c is restricted to { , } . The proof of Theorem 1.10 will be given indetail in Section 5. It follows the same outline as the proof of Theorem 1.1 given inSection 1.5.By now, the eager reader has spotted that the special case M = S , h = 0, c = 0of Theorem 1.10 yields Corollary 1.9, while the special case M = S , h = 0, c = 0takes the following form: Corollary 1.11.
Let K ⊂ S be a knot. Then the Z –slice genus g Z ( K ) of K equalsthe minimum c such that a Z –slice disk for K exists in B C P C P ) c . (cid:3) Linking forms of cyclic branched covers.
In a second part of this paper(Section 6), we use our new characterization of g Z given in Theorem 1.1(4) to providea criterion to obstruct knots from having g Z ≤
1. We summarize what we obtain.The Blanchfield specializes (essentially by setting t = −
1) to the linking pairing (cid:96) : H (Σ ( K ); Z ) × H (Σ ( K ); Z ) → Q / Z on the first integral homology group ofthe double branched cover Σ ( K ) of S along K . Using Theorem 1.1(4), we showthat g Z ( K ) ≤ (cid:96) admits a 2 × − H (Σ ( K ); Z ) is of odd order,the following proposition about pairings on Abelian groups of odd order provides atestable criterion whether (cid:96) has such a 2 × Proposition 1.12.
Let A be an Abelian group of odd order with two generators,equipped with a non-degenerate symmetric pairing (cid:96) : A × A → Q / Z . Then A decomposes as orthogonal sum of two subgroups generated cyclically by g , g , whichare of respective order q and q with q | q . Note q , q are odd and may equal 1.Let a i q i = (cid:96) ( g i , g i ) . Note that a i and q i are coprime. Then, for a given u ∈ {− , } ,the two following statements (A) and (B) are equivalent:(A) (cid:96) can be presented by an odd symmetric × integer matrix M with det M ≡ u (mod 4) .(B) a , a , q , q , u satisfy both of the following two conditions.(B1) ( − ( q q − u ) / a a is a square residue modulo q ,(B2) u = 1 , or q q ≡ , or the Jacobi symbol (cid:16) a q /q (cid:17) equals . ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 9
Note that criterion (B) is easy to check for a given pairing. While we find this tobe of theoretical interest, Proposition 1.12 also allows to complete the calculation ofthe Z –slice genus for all prime knots of crossing number up to 11; see Section 6.3.Proposition 1.12 can also be applied to simplify [BF15, Lemma 5.2], which canbe used to show that a knot has algebraic unknotting number at least 3, andwhich was our inspiration to implement obstructions for Z –slice genus (the balancedversion of the algebraic unknotting number). The proof of Proposition 1.12 usesWall’s classification [Wal63], quadratic reciprocity and (for the proof of (B) ⇒ (A))Dirichlet’s prime number theorem, but is elementary apart from that.In contrast to the effectiveness in obstructing Z –slice genus using the linkingpairing of the double-branched cover, we have Theorem 1.2: for odd primes p ,the linking pairing of the p –fold branched cover does not provide any additionalinformation to the Z [ Z /p ]–module structure of its first homology. Theorem 1.2partially explains a disappointing finding of Borodzik and Friedl in [BF15]: theimplementation of their obstruction [BF15, Lemma 5.1(2)] for p > Structure of the paper.
Section 2 gives the essential definitions and fixesnotation concerning the Alexander module and the Blanchfield pairing. Sections 3and 4 contain the 4–dimensional and 3–dimensional part of the proof of Theorem 1.1,respectively. In Section 5, Theorem 1.10 is proven. Section 6 is devoted to link-ing pairings of branched covers and calculations of g Z . It contains the proof ofTheorem 1.2. Appendix A provides background on Hermitian pairings. Acknowledgments.
The first author thanks Matthias Nagel, Patrick Orson, andMark Powell for a fun Blanchfield pairing calculation session. Thanks to Filip Misevfor drawing Figure 1 together with the second author. Thanks to Maciej Borodzik,Anthony Conway, Jim Davis, Chuck Livingston, Duncan McCoy, and MatthiasNagel for comments on a first version of this paper. The first author gratefullyacknowledges support by the Swiss National Science Foundation Grant 181199. Thesecond author was supported by the Emmy Noether Programme of the DFG.2.
Preliminaries
In this section, we collect definitions of and known facts about links and Seifertsurfaces, Hermitian pairings and their presentation matrices, the Alexander module,and the Blanchfield pairing.2.1.
Matrix presentations of pairings on torsion modules.
Let R denote acommutative unital ring with an involution, denoted by r (cid:55)→ r . Some examplesare the integers Z with the identity as involution, the ring Λ := Z [ t ± ] of Laurentpolynomials with integer coefficients and involution given by f ( t ) (cid:55)→ f ( t − ), and,for all integers n ≥
2, the rings Λ / ( t n −
1) and Λ / (Φ n ) (where Φ n denotes the n –th cyclotomic polynomial) with the involution induced from Λ. We denote by Q ( R ) the total quotient ring —the localization S − R of R with respect to S := R \ { zero-divisors of R } . Of course, for integral domains, Q ( R ) is simply the fieldof fractions; e.g. Q and Q ( t ) for Z and Λ, respectively. An R –module M is called torsion if it is annihilated by some non-zero-divisorof R . A Hermitian pairing on such an M is an R –sesquilinear map (anti-linear withrespect to · in the first factor) (cid:96) : M × M → Q ( R ) /R such that (cid:96) ( y, x ) = (cid:96) ( x, y ) forall x and y in M . Such a pairing is called non-degenerate if for all x ∈ M there is a y ∈ M with (cid:96) ( x, y ) (cid:54) = 0.A Hermitian square matrix A ∈ Mat n × n ( R ) whose determinant is not a zerodivisor defines a non-degenerate Hermitian pairing on the cokernel of A as follows: (cid:96) A : R n /AR n × R n /AR n → Q ( R ) /R, ( x, y ) (cid:55)→ x (cid:62) A − y + R. A Hermitian square matrix A ∈ Mat n × n ( R ) with non-zero-divisor determinantis said to present a Hermitian pairing (cid:96) : M × M → Q ( R ) /R if (cid:96) is isometricto (cid:96) A , i.e. there exists an R –module isomorphism φ : M → R n /AR n such that (cid:96) A ( φ ( x ) , φ ( y )) = (cid:96) ( x, y ) for all x and y in M .See the appendix for a different perspective on Hermitian pairings, and a basechange proposition.2.2. Links, Seifert surfaces, and 3D–cobordism.
Fix a 3–manifold M . A link is a smooth oriented non-empty closed 1–submanifold of M ; a connected link isa knot . A Seifert surface is a smooth oriented connected compact 2–submanifoldwith non-empty boundary. A Seifert surface for a link L is a Seifert surface withboundary L . A between two knots K and J is a Seifert surface withboundary a 2–component link, one component of which is isotopic to K , and theother to J with reversed orientation. In previous work, the authors considered3D–cobordisms between multi-component links [FL18], but this will not be neededhere.2.3. Twisted homology.
Let X be a space admitting a universal cover. A sur-jective group homomorphism φ : π ( X ) → G to some group G (we consider onlyAbelian G ) yields a notion of twisted homology. For this, take the ker( φ )–coverof X , use the deck transformation group action by G to endow the singular Z –chaincomplex with a Z [ G ]–module structure, and define H ∗ ( X ; Z [ G ]) to be the homologyof this Z [ G ]–chain complex.2.4. Alexander module.
Let K be a knot in S . The Abelianization π ( S \ K ) → Z induces a covering space, the infinite cyclic cover , which we denoteby S \ K cyc . The first homology H ∗ ( S \ K cyc ; Z ) becomes a Z [ Z ]-module usingthe deck transformation group action and can be canonically identified with thetwisted homology with respect to the Abelianization H ∗ ( S \ K ; Λ) (where weidentify the group ring Z [ Z ] with Λ). The Alexander polynomial ∆ K ∈ Λ of K isusually defined (as we also did in the introduction) as the order of H ( S \ K ; Λ),which is well-defined up to multiplication with a unit in Λ. Here, the order ideal ofa finitely presented torsion-module is the ideal generated by the determinants of n × n minors of an n × m presentation matrix with n ≤ m . Since the Alexandermodule can be presented by a square matrix, the order ideal is a principal ideal.The order is a generator of the order ideal.Alternative (and equivalent) definitions of the Alexander module and polynomialuse the zero-surgery M K of K instead of the complement S \ K . Recall thatthe map π ( S \ K ) → π ( M K ) induced by inclusion is surjective, and its kernel—normally generated by the class of a zero-framed longitude—is contained in thesecond derived commutator subgroup. Since H ( · ; Z ) is canonically isomorphic to ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 11 the Abelianization of the commutator subgroup of π ( · ), the inclusion also inducesa Λ–module isomorphisms between H ( S \ K ; Λ) and H ( M K ; Λ). Thus, it isconsistent with the usual definitions to see the Alexander module as H ( M K ; Λ),and the Alexander polynomial as its order.Also, we normalize the Alexander polynomial ∆ K to be symmetric and satisfy∆ K (1) = 1.2.5. The Blanchfield pairing.
The Alexander module being a torsion module,one can define the
Blanchfield pairing as the following non-degenerate Hermitianpairing on the Alexander module:Bl( K ) : H ( S \ K ; Λ) × H ( S \ K ; Λ) → Q ( t ) / Λ , ( x, y ) (cid:55)→ (Ψ( x ))( y ) , where Ψ is the composition of the following maps H ( S \ K ; Λ) → H ( S \ K, ∂ ( S \ K ); Λ) ∼ = −→ H ( S \ K ; Λ) ∼ = −→ H ( S \ K ; Q ( t ) / Λ) ev −→ Hom Λ ( H ( S \ K ; Λ) , Q ( t ) / Λ) . The first map is given by canonical inclusion on the chain complex level, thesecond map is the inverse of Poincar´e duality for the twisted homology of the3–manifold S \ K [Wal99], the third map is the inverse of the Bockstein map—theconnecting homomorphism in the long exact sequence of cohomology induced bythe short exact sequence of coefficients0 → Λ → Q ( t ) → Q ( t ) / Λ → , and the fourth map is the so-called Kronecker evaluation map. We only give thisbrief treatment since we will not make use of the definition of the Blanchfield pairing,and refer the reader to [FP17] for a detailed treatment.2.6. The Blanchfield pairing via twisted intersection forms on –manifolds. Above we recalled the definition of the Blanchfield pairing using (twisted) Poincar´eDuality of 3–manifolds. Much like linking numbers in 3–manifolds can be calculatedby intersecting surfaces in 4–manifolds with boundary the 3–manifold in question(most classically, the linking number of two disjoint oriented curves in S equalsthe oriented intersection of generic surfaces bounding them in B ), the Blanchfieldpairing has a presentation via the twisted homology of 4–manifolds W with bound-ary M K . Borodzik and Friedl established the following rather general statement,which only asks for natural homological assumptions on W . We also note that theresult holds in the topological category. Theorem 2.1 ([BF15, Theorem 2.6]) . Let K be a knot and W a connected compactoriented topological 4–manifold with infinite cyclic fundamental group and boundary M K such that the inclusion of M K into W descends to an isomorphism on H ( · ; Z ) .Then the twisted homology H ( W ; Z [ π ( W )]) is free of rank b ( W ) . Furthermore,if B is an integral matrix for the ordinary intersection pairing of W , then thereexists a basis B for H ( W ; Z [ π ( W )]) such that the matrix A ( t ) of the twistedintersection pairing with respect to B presents the Blanchfield pairing Bl( K ) of K ,and A (1) = B . Although we only apply Theorem 2.1 to a rather special manifold W in the nextsection, we do not know of a faster proof that the above holds for this manifold W than the one by Borodzik and Friedl, which goes through a quite general argumentemploying the universal coefficient spectral sequence. The four-dimensional part of the proof— (1) ⇒ (4)Let F be a Z –slice surface of genus g in B with boundary K . We calculatethe Blanchfield pairing of K by using F to define a 4–manifold W with boundarythe zero-surgery of K , denoted by M K , such that the intersection pairing twistedby π ( W ) on W is a Hermitian presentation of the Blanchfield pairing. For thispurpose, we construct W such that π ( W ) ∼ = Z , b ( W ) = 2 g, σ ( W ) = 0 and theinclusion of M K into W descends to an isomorphism on integral first homologygroups.Given such a 4–manifold W , Theorem 2.1 yields a Hermitian 2 g × g –matrix A ( t ) over the ring Λ that presents the Blanchfield pairing of K such that A (1) isa unimodular matrix with signature 0. Unimodularity follows since A (1) presentsthe ordinary intersection form on W , which is unimodular because the fact thatthe inclusion of M K = ∂W into W induces an isomorphism on H ( · ; Z ) impliesthat the long exact sequence of the pair ( W, ∂W ) induces an isomorphism on H ( W, ∂W ; Z ) → H ( W ; Z ). And, by definition, the signature of W is the signaturethe ordinary intersection form on W , thus σ ( A (1)) = σ ( W ) = 0.Therefore, to prove (1) ⇒ (4), it only remains to actually construct the 4–manifold W with the desired properties. Our construction is modeled on what oneoften sees in the literature when F is a pushed-in Seifert surface; see e.g. [COT04,Proof of Lemma 5.4]. See also [Pow17], where this construction is considered forstrong slice surfaces of links. We build W in two steps.3.1. Step I.
We set W (cid:48) := B \ νF , where νF denotes an open tubular neighborhood.Concretely, νF may be taken as open disk subbundle of the normal bundle of F in the sense of [FQ90, Section 9.3]. In particular, the boundary ∂νF of νF as asubspace of B is a locally flat 3–manifold with boundary, properly embedded in B and homeomorphic to F × S . We note that ∂W (cid:48) = ∂νF ∪ ( S \ νK ), where νK denotes an open tubular neighborhood of K in S . The two pieces, ∂νF and S \ νK , intersect in a torus, which we denote by Σ. It can be understood as unitnormal bundle of K in S . Claim 3.1.
We have (i) π ( W (cid:48) ) ∼ = Z , (ii) b ( W (cid:48) ) = 2 g , and (iii) σ ( W (cid:48) ) = 0 . Before proving the claim, let us consider the following special case of Novikov-Wallnon-additivity, which will be needed for the proof of (iii).
Lemma 3.2.
Let Z be a closed surface, let X ± , X be 3–manifolds with boundary Z , let Y ± be topological 4–manifolds with boundaries ∂Y ± = X ± ∪ Z X , and let Y bethe topological 4–manifold given as Y + ∪ X Y − (see Figure 2(i)). Consider the threemaps on H ( Z ; Q ) induced by the inclusions of Z into X ± and X . If the kernelsof two of these maps agree, then σ ( Y ) = σ ( Y + ) + σ ( Y − ) .Proof. Novikov-Wall non-additivity [Wal69] holds in the topological category [Kir89].It states that σ ( Y ) = σ ( Y + ) + σ ( Y − ) − σ ( N ), where N is a certain space with abilinear form. To prove the lemma, it suffices to prove that dim N = 0. Let A, B, C be the three kernels mentioned in the statement of the lemma, then N is defined as N = A ∩ ( B + C )( A ∩ B ) + ( A ∩ C ) ⊆ H ( Z ; Q ) . Permuting
A, B, C does not change N . If there are two among A, B, C that agree,then clearly dim N = 0. (cid:3) ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 13 Y − Y + X X + X − Z νFW (cid:48) ∂νFS \ νKνK Σ H × S W (cid:48) ∂νF ∼ = F (cid:48) × S S \ νKH \ F (cid:48) × S Σ(i) Y (ii) B (iii) W Figure 2.
Three 4–manifolds, each as union of two others.
Proof of Claim 3.1. (i) is satisfied by the definition of Z –slice surface.(ii) follows quickly from the Mayer-Vietoris sequence of B = W (cid:48) ∪ ∂νF νF (see Fig-ure 2(ii)), which implies that H ( ∂νF ; Q ) ∼ = H ( W (cid:48) ; Q ) ⊕ H ( νF ; Q ). Alternatively,one could use an appropriate variant of Alexander duality in the ball.For (iii), we wish to apply the Lemma 3.2 to B = W (cid:48) ∪ νF (see Figure 2(ii)).The hypothesis is satisfied because the two maps induced by the inclusions of Σinto S \ νK and ∂νF , respectively, have the same kernel, which is generated by theclass of a zero-framed longitude. Thus, σ ( B ) = σ ( W (cid:48) ) + σ ( νF ) . Clearly, we have σ ( B ) = 0 (since B has no H ) and σ ( νF ) = 0 (since νF ishomeomorphic the product of F with a closed 2–disk, so it has no H either). Itfollows that σ ( W (cid:48) ) = 0. (cid:3) Step II.
The 4–manifold W (cid:48) satisfies the desired properties, with the exceptionthat ∂W (cid:48) is not M K . In this step, we fix that.Let H be a genus g handlebody. In the boundary of H , a closed surface ofgenus g , we pick a subsurface F (cid:48) of genus g with one boundary component (i.e. thecomplement of a small open neighborhood of a point in ∂H ). Let us glue H × S to W (cid:48) to obtain W via a certain homeomorphism φ : F (cid:48) × S ∂νF.H × S W (cid:48) ⊆ ⊆ It turns out that the choice of homeomorphism matters. Before we explain ourchoice, let us discuss properties that hold for any such homeomorphism φ .The homeomorphism φ restricts on the boundary to a homeomorphism ( ∂F (cid:48) ) × S → Σ of tori. Let us write (cid:96), m respectively for a zero-framed longitude and ameridian of K on Σ. The curve φ ( ∂F (cid:48) × { } ) is homotopic to (cid:96) , since this curveand K bound the disjoint surfaces φ ( F (cid:48) × { } ) and F in B . Next, fix a base point f ∈ ∂F (cid:48) and consider the curve φ ( { f } × S ). Its homology class and the class of (cid:96) form a basis of H (Σ; Z ). Therefore, its homology class is equal to ± [ m ] + λ [ (cid:96) ] forsome λ ∈ Z . Since H ( W (cid:48) ; Z ) is generated by [ m ], and (cid:96) is null-homologous in W (cid:48) ,it follows that the homology class of φ ( { f } × S ) generates H ( W (cid:48) ; Z ) ∼ = π ( W (cid:48) ).This can be rephrased as follows. Let us denote by i and i and j the homomor-phisms of fundamental groups induced by the inclusions F (cid:48) → F (cid:48) × S , f (cid:55)→ ( f, and S → F (cid:48) × S , z (cid:55)→ ( f , z ) and ∂νF (cid:44) → W (cid:48) . Then the composition j ◦ φ ∗ ◦ i isan isomorphism π ( S ) → π ( W (cid:48) ), which we denote by k .Let us now discuss our choice of φ . We wish to choose φ such that the compositionof the following two maps is the zero map: π ( F (cid:48) ) π ( ∂νF ) π ( W (cid:48) ) φ ∗ ◦ i j To construct φ , start by picking any homeomorphism φ (cid:48) : F (cid:48) × S → ∂νF . Let g : π ( F (cid:48) ) → π ( S ) denote the composition k − ◦ j ◦ φ (cid:48)∗ ◦ i . Let ψ : F (cid:48) → S be a continuous map such that ψ ∗ is − g (we write the group action additively in π ( S ), and multiplicatively in S and π ( F (cid:48) × S )). For this, recall that all grouphomomorphisms from π ( F (cid:48) ) to Z are induced by a continuous map as a consequenceof the following chain of canonical identificationsHom( π ( F (cid:48) ) , Z ) ∼ = Hom( H ( F (cid:48) ; Z ) , Z ) ∼ = H ( F (cid:48) ; Z ) ∼ = [ F (cid:48) , S ] . We define a homeomorphism ω : F (cid:48) × S → F (cid:48) × S by ( f, z ) (cid:55)→ ( f, z · ψ ( f )) and set φ := φ (cid:48) ◦ ω .Let us check that the composition j ◦ ( φ ∗ ◦ i ) is indeed zero for this choice of φ .It is sufficient that k − ◦ j ◦ φ ∗ ◦ i = 0, which one finds as follows: k − ◦ j ◦ φ ∗ ◦ i = k − ◦ j ◦ φ (cid:48)∗ ◦ ω ∗ ◦ i = k − ◦ j ◦ φ (cid:48)∗ ◦ ( i · ( i ◦ ψ ∗ ))= ( k − ◦ j ◦ φ (cid:48)∗ ◦ i ) + ( k − ◦ j ◦ φ (cid:48)∗ ◦ i ◦ ψ ∗ )= g + ( k − ◦ k ◦ ( − g )) = 0 . Remark . The careful reader will notice that we in fact will only use that thekernel of j ◦ φ ∗ contains the kernel of the map induced by the inclusion of F (cid:48) × S into H × S . If g >
0, not all homeomorphisms φ : F (cid:48) × S → ∂νF satisfy this,and choosing one which does not would make the fundamental group of W a finiteinstead of infinite cyclic group.We define W := W (cid:48) ∪ φ H × S and note that ∂W = ∂M K (see Figure 2(iii)). Wewill establish the following properties W , thereby concluding the proof of (1) ⇒ (4). Claim 3.4.
We have that (i) π ( W ) ∼ = Z , (ii) b ( W ) = 2 g , (iii) σ ( W ) = 0 and (iv)the inclusion of M K into W descends to an isomorphism on H ( · ; Z ) .Proof of Claim 3.4. (i) A simple application of the Seifert-van Kampen theoremimplies that the inclusion of W (cid:48) into W induces an isomorphism on fundamentalgroups. For this, recall that W is obtained by gluing W (cid:48) and H × S along F (cid:48) × S via φ . Consider the following commutative diagram of groups: π ( F (cid:48) × S ) π ( H × S ) π ( W (cid:48) ) π ( W ) ⊆ ∗ j ◦ φ ∗ ⊆ ∗ ⊆ ∗ By our assumption on φ , j ◦ φ ∗ factors through π ( F (cid:48) × S ) ⊆ ∗ −−→ π ( H × S ) (thatis, the map π ( H × S ) → π ( W (cid:48) ) given as composition of k with the canonicalprojection π ( H × S ) → π ( S ) commutes with the other maps). This implies that ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 15 the bottom arrow is a group isomorphism since the diagram is a push-out diagramby the Seifert-van Kampen theorem.(iv) This follows because the inclusions S \ νK into W (cid:48) and M K and the inclusionof W (cid:48) into W descend to isomorphism on H ( · ; Z ) (all of these first homology groupsare generated by the class of a meridian of K on Σ).(ii) To show that b ( W ) = 2 g , we consider the long exact Mayer-Vietoris sequencefor homology with rational coefficients: H ( W ) H ( F (cid:48) × S ) H ( H × S ) ⊕ H ( W (cid:48) ) H ( W ) H ( F (cid:48) × S ) H ( H × S ) ⊕ H ( W (cid:48) ) H ( W ) . . . ∂∂ We note that H ( W ; Q ) = 0. Indeed, H ( W ; Q ) ∼ = H ( W, M K ; Q ) ∼ = H ( W, M K ; Q )by Poincar´e duality and the universal coefficient theorem, and H ( W, M K ; Q ) = 0by the long exact sequence for the pair ( W, M K ) and the fact that H ( M K ; Q ) → H ( W ; Q ) is an isomorphism (see (iv)). Therefore, the dimensions of the homologyspaces in the Mayer-Vietoris sequence are (from left to right) 0 , g, g, b ( W ) , g +1 , g + 2 ,
1. Since the alternating sum of dimensions is zero, we have b ( W ) = 2 g .(iii) We prove σ ( W ) = 0 by applying Lemma 3.2 to W = W (cid:48) ∪ H × S ; compareFigure 2(iii). The hypothesis of the lemma is satisfied since the inclusion of Σ in S \ νK and ∂νF have the same kernel (compare the proof of Claim 3.1(iii)). Wehave σ ( W (cid:48) ) = 0 (see Claim 3.1(iii)) and σ ( H × S ) = 0 (since, for example, anypair of classes can be represented by two disjoint closed surfaces). It follows that σ ( W ) = 0. (cid:3) Remark . If the Z –slice surface F of K is obtained as a pushed-in 3D–cobordismbetween K and a knot with Alexander polynomial 1, then a presentation matrixof the Blanchfield pairing can be given more explicitly using Ko’s formula [Ko89].This yields a direct proof of (2) ⇒ (4), the details of which we omit.4. The three-dimensional part of the proof
This section completes the proof of Theorem 1.1. We will show (4) ⇒ (3) and(3) ⇒ (2) in the next two subsections, respectively.4.1. Unknotting information from the Blanchfield pairing— (4) ⇒ (3) . The main result of [BF14] can be phrased as follows:
Theorem 4.1 ([BF14, Thm. 5.1]) . Let A ( t ) be a Hermitian presentation matrix ofthe Blanchfield pairing of a knot K . Assume that the symmetric bilinear form A (1) isdiagonalizable, and denote the number of its positive and negative eigenvalues countedwith multiplicity by p, n ∈ Z +0 , respectively. Then K can be turned into a knot withAlexander polynomial by changing p positive and n negative crossings. (cid:3) We show that in case A (1) is indefinite, the diagonalization assumption is un-necessary. More precisely, we prove the following proposition, which might be ofindependent interest. Proposition 4.2.
Let A ( t ) be a Hermitian presentation matrix of the Blanchfieldpairing of a knot K . Assume that the symmetric bilinear form A (1) is indefinite, anddenote the number of its positive and negative eigenvalues counted with multiplicity by p, n ∈ N , respectively. Then K can be turned into a knot with Alexander polynomial by changing p positive and n negative crossings. The case p = n of this proposition is exactly the desired implication (4) ⇒ (3).The remainder of this section is devoted to the proof of the proposition. Rather thanunpacking Borodzik and Friedl’s intricate 3–dimensional argument turning algebraicinformation into unknotting information, we use a purely algebraic argument aboutHermitian pairings over Λ := Z [ t, t − ] and Λ := Z [ t, t − , ( t − − ] to reduce Propo-sition 4.2 to Theorem 4.1. As a first step, we recall that multiplication by ( t − rather than Λ. More precisely, wehave the following. Lemma 4.3.
Let A ( t ) be a Hermitian Λ –matrix presenting the Blanchfield pairingof K . If T ( t ) is a Λ –matrix such that det T ( t ) is a unit in Λ , and B ( t ) = T ( t ) (cid:62) A ( t ) T ( t ) is a Λ –matrix, then B ( t ) also presents the Blanchfield pairing of K . (cid:3) This statement is implicit in [BF15, Proof of Prop. 2.1] and [COT04, Proof ofLemma 5.4], but we thought it beneficial to make the statement explicit. We usethe occasion to formulate a general principle for arbitrary rings, which we prove indetail; see appendix. The above lemma follows as a special case of Corollary A.4.Now, Proposition 4.2 is a direct consequence of Theorem 4.1 and the next lemma.
Lemma 4.4.
Let A ( t ) be a Hermitian matrix over Λ such that A (1) is unimodularand indefinite. Then there is a transformation matrix over Λ with determinant aunit in Λ , transforming A ( t ) into a Hermitian matrix B ( t ) over Λ such that B (1) isdiagonal with ± diagonal entries and σ ( B (1)) = σ ( A (1)) .Proof. Indefinite symmetric unimodular forms over the integers have been classified,see e.g. [MH73]. The classification differentiates between even forms (forms θ suchthat θ ( v, v ) is even for all vectors v ) and odd forms (all other forms). Odd formsadmit diagonal matrices, whereas even forms are isometric to a sum of a positivenumber of copies of the hyperbolic plane H , which has the matrix (cid:18) (cid:19) , and anon-negative number of copies of E .Thus, if A (1) is odd, there is a base change over Z that diagonalizes A (1). Thesame base change transforms A ( t ) into a matrix B ( t ) that satisfies the desiredproperties. So let us consider the case that A (1) is even. Then there is a basechange over Z transforming A (1) into H ⊕ R for some R . Assume w.l.o.g. that A (1)is already of this form. Then all the entries of the first two rows of A (1) − ( H ⊕ N )(where N is a zero matrix of size 2 g −
2) evaluate to 0 at t = 1 and are thus divisibleby ( t − A ( t ) is Hermitian, we see that its top-left 2 × (cid:18) xb − t ) b − t − ) b xb (cid:19) , where we write x = (1 − t ) + (1 − t − ) = (1 − t )(1 − t − ). Furthermore A ij = (1 − t ) b ij , A ji = (1 − t − ) b ij for i ∈ { , } and j > b ij ∈ Λ. We will now considerthe parity of b (1), and in each case give a transformation matrix over Λ withdeterminant a unit in Λ, which transforms A ( t ) into a matrix C ( t ) with odd C (1). ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 17
In this way the case that A (1) is even is reduced to the case that A (1) is odd, whichhas already been discussed.If b (1) is even, add 1 / (1 − t ) times the first row to the second, and then1 / (1 − t − ) times the first column to the second. This is a base change over Λ coming from a transformation matrix with determinant 1. It yields a Hermitianmatrix C ( t ) over Λ with top-left 2 × (cid:18) xb − t )( b + b )1 + (1 − t − )( b + b ) xb + 1 + b + b + b (cid:19) . One finds C (1) = 0 + 1 + b (1) + 2 b (1) to be odd. This concludes the case that b (1) is even.If b (1) is odd, one proceeds similarly: one may divide the first row by (1 − t ),and multiply the second row by (1 − t − ), and apply the corresponding changes tothe columns. This is a base change over Λ coming from a transformation matrixwith determinant − t − . It yields a Hermitian matrix C ( t ) over Λ with top-left 2 × (cid:18) b − t ) b − t − ) b x b , (cid:19) . Clearly C (1) = b (1) is odd, which concludes the case of odd b (1). (cid:3) We have thus completed the proof of the implication (4) ⇒ (3), and turn to thenext part of the proof.4.2. (3) ⇒ (2) . This section is de-voted to the proof of the following proposition, from which (3) ⇒ (2) follows sinceit corresponds to the special case that K (cid:48) is a knot with Alexander polynomial 1. Proposition 4.5.
Let g be a non-negative integer. If a knot K (cid:48) can be obtainedfrom a knot K by changing g positive and g negative crossings (in any order), thenthere exists an oriented connected surface Σ of genus g in S with oriented boundarya two-component link whose components are isotopic to K (cid:48) with reversed orientationand K , respectively. (a) (b) Figure 3. (a) A balanced crossing change, diagrammatically. (b)Two isotopic drawings of the intersection of ball with a genus 1Seifert surface realizing a balanced crossing change.
Figure 4.
Handle slides changing the order of two intersectionpoints of crossing disks with a knot; one handle slide works for+1–framed Dehn surgery on the boundary of the blue (darker) disk,the other for − Proof.
Let us first consider two simple cases. If g = 0, just take Σ as a knottedband. Next, consider the case that g = 1 and that the two crossing changes happeninside of a small ball as shown in Figure 3(a). Then, one may construct Σ by gluinga knotted band (as in the case g = 0) outside of the ball, and the surface shownin Figure 3(b) inside of the ball. In fact, a similar construction—gluing g copiesof the surface shown in Figure 3(b) inside of g balls, and a knotted band in thecomplement of the g balls—gives the desired surface in the general case, due to thefollowing lemma. (cid:3) Lemma 4.6.
Let K and K (cid:48) be knots as in Proposition 4.5. Then there exist pairwisedisjoint balls B , . . . , B g , such that K and K (cid:48) agree outside of (cid:83) i B i , and for all i , B i ∩ K and B i ∩ K (cid:48) look like the top and bottom of Figure 3(a), respectively.Proof. By the assumption on K and K (cid:48) , one may choose 2 g crossing disks such thatthe corresponding surgeries (in the right order) transform K into K (cid:48) , and there are g surgeries of each sign. Since the crossings are not changed simultaneously, thecrossing disks may a priori intersect. However, by a general position argument thedisks may be chosen to be disjoint (see e.g. [Sch98, Prop. 1.5] for details); in otherwords, crossing changes may be assumed to happen simultaneously.Finally, by the handle slides shown in Figure 4, one may arbitrarily change theorder in which the 4 g intersection points of crossing disks with K occur on K . Wechange this order such that the 2 g disks can be arranged in g pairs D, D (cid:48) with thefollowing properties. The surgeries corresponding to D and D (cid:48) are of opposite sign;and there is a closed interval I ⊆ K with endpoints on D and D (cid:48) , such that I ◦ doesnot intersect any other crossing disks. Now, for each such pair take a ball that is aneighborhood of D ∪ D (cid:48) ∪ I , and make these g balls small enough so that no two ofthem intersect. These balls form the desired collection B , . . . , B g . (cid:3) ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 19 More general ambient 3– and 4–manifolds (proof of Theorem 1.10)
The proof of Theorem 1.10 follows the same structure as the proof of Theorem 1.1.We note some preliminary lemmas. The first one explains why Z –slice surfaces areautomatically null-homologous. Lemma 5.1.
Let F be a properly locally-flatly embedded surface in a four-manifold V = B C P C P ) c S × S ) c , where c , c ≥ and B is a compact orientable contractible topological four-manifoldwith boundary an integral homology sphere M . If H ( V \ F ; Z ) ∼ = Z , then F isnull-homologous in V .Proof. For F to be null-homologous in V means that the inclusion i : ( F, K ) → ( V, M = ∂V ) induces the zero-map on second homology groups, i.e. i ∗ [ F ] = 0 ∈ H ( V, M ; Z ) for [ F ] ∈ H ( F, K ; Z ) the fundamental class of F . Since H ( ∂V ; Z ) ∼ = H ( ∂V ; Z ) ∼ = 0, the inclusion j : ( V, ∅ ) → ( V, M ) induces an isomorphism on secondhomology groups. So to establish the lemma it will be sufficient to show that( j ∗ ) − ( i ∗ ([ F ])) =: e = 0 ∈ H ( V ; Z ) . Since F and V \ F are disjoint, the intersection form H ( V ; Z ) × H ( V ; Z ) → Z of V evaluates to 0 on ( e, e (cid:48) ) for all e (cid:48) ∈ im k ∗ ⊂ H ( V ; Z ) for k the inclusion map V \ F → V . Since M is an integral homology sphere, the intersection form isnon-degenerate. So to show e = 0 and thus conclude the proof, it only remains toshow that k ∗ is surjective.For this, let νF ⊂ V be a tubular neighborhood of F , and consider the Mayer-Vietoris sequence of νF ∪ ( V \ F ) of integral homology groups:0 → H ( V \ F ) k ∗ −→ H ( V ) → H ( νF \ F ) α −→ H ( νF ) ⊕ H ( V \ F ) → . Both domain and target of α are free abelian of rank 2 g ( F ) + 1, and α is surjective.It follows that α is an isomorphism, and hence so is k ∗ . (cid:3) Let F be a Seifert surface of genus g of a link in some Z HS . We denotethe intersection form on H ( F ; Z ) by (cid:104) · , · (cid:105) . A symplectic basis of F is a tuple( a , b , a , b , . . . , a g , b g ) of homology classes in H ( F ; Z ) such that (cid:104) a i , b k (cid:105) = δ ik and (cid:104) a i , a k (cid:105) = (cid:104) b i , b k (cid:105) = 0 for all i, k . Such a symplectic basis generates a summandof rank 2 g of H (Σ; Z ), and descends to a basis of H (Σ; Z ) /ι ∗ H ( ∂ Σ; Z ), where ι denotes the inclusion ∂ Σ → Σ. A half basis of F is a tuple ( a , . . . , a g ) of homologyclasses in H ( F ; Z ) that may be extended to a symplectic basis ( a , b , . . . , a g , b g )of F .A link L is called proper if each of its components K has even linking numberwith L \ K (for example, all knots are proper links). The Arf invariant of a properlink L may be defined as follows (see [Lic97, Chapter 10]). Let F be a genus g Seifert surface of L , and pick a symplectic basis ( a , b , a , b , . . . , a g , b g ) for F . For v ∈ H ( F ; Z ), let q ( v ) ∈ Z / Z denote the reduction mod 2 of the Seifert formevaluated at ( v, v ). Then Arf( L ) := g (cid:88) i =1 q ( a i ) q ( b i ) . This definition is independent of the choice of F and of the symplectic basis. Lemma 5.2.
Let F be Seifert surface of a proper link L in some Z HS . (i) There exists a half basis a , . . . , a g of F such that the framing induced by F on a i is even for ≤ i ≤ g and odd for a .(ii) There exists a half basis a , . . . , a g of F such that the framings induced by F of a i are even for ≤ i ≤ g if and only if Arf( L ) = 0 .Proof. (i) Let us first show that there exists a symplectic basis ( a , b , . . . , a g , b g ) of F at least one element of which has odd framing. Indeed, if all basis elements haveeven framing, then replacing a by a + b yields such a symplectic basis. Thus,there exists a half basis a , . . . , a g of F and k ∈ { , . . . , g } such that a i has oddframing iff i ≤ k . Then, ( a , a + a , . . . , a k + a , a k +1 , . . . , a g ) is a half basis inwhich only the first element has odd framing.(ii) The ‘only if’ direction follows directly from the definition of the Arf invariant.For the ‘if’ direction, assume that Arf( L ) = 0 and let a , . . . , a g be a half basis asin (i). We have 0 = Arf( L ) = q ( a ) q ( b ), so q ( b ) = 0. Thus a + b , a , . . . , a g is adesired half basis. (cid:3) Let us say a Seifert surface F is H –null (or H –null mod 2 ) if every simple closedcurve on F has zero (or even) linking number with every boundary component of F .Note that the boundary of an H –null mod 2 surface is a proper link. Equivalently, F is H –null (or H –null mod 2) if the inclusion of F ◦ into the complement of ∂F induces the zero map on the first homology group with integer coefficients (or with Z / Lemma 5.3.
Let F be a Seifert surface with boundary components L , . . . , L n insome Z HS .(i) There exists an H –null Seifert surface F (cid:48) with a homeomorphism f : F → F (cid:48) such that f ( L j ) is isotopic to L j .(ii) If F is H –null mod 2, then there exist f and F (cid:48) as in (i) such that f preserves the framing of curves mod 2.Proof. We only sketch the proof, and refer the reader to [FL18, Proof of Lemma 18]for details. The Seifert surface F has a handle decomposition into one 0–handle,1–handles h , . . . , h g ( F ) whose belt spheres lie in L n , and 1–handles h (cid:48) , . . . , h (cid:48) n − ,such that the belt sphere of h (cid:48) j has one point in L j and one point in L n . Now weconstruct F (cid:48) and f in three steps:1. Insert twists into the 1–handle h (cid:48) j for all j to make the framing of its corecurve (which is parallel to L j ) zero.2. For all pairs j, k , twist the 1–handles h (cid:48) j and h (cid:48) k around each other to makethe linking number of the cores of the handles h (cid:48) j and h (cid:48) k zero. This can bedone without changing the framing of the cores and the isotopy classes ofthe boundary components.3. For all pairs i, j , twist the 1–handle h i around h (cid:48) j to make the linkingnumber of the cores of the handles h i and h (cid:48) j zero. This can be done withoutchanging the framing of the core of h (cid:48) j , and without changing the isotopyclasses of the boundary components. It may however change the framing ofthe core curve of h i .To prove (ii), observe that if F is H –null mod 2, then every L j have even framing,so the first step only changes its framing by an even number; and similarly, thelinking number of the core curves of h i and h (cid:48) j is even, so the framing of the corecurve of h i is only changed by an even number. (cid:3) ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 21
Lemma 5.4.
Let F be a Seifert surface of a link L in some Z HS . Assume F is H –null mod 2, and has two boundary components L , L . Then Arf( L ) =Arf( L ) + Arf( L ) .Proof. By Lemma 5.3(ii), there exists an H –null Seifert surface F (cid:48) with a home-omorphism f : F → F (cid:48) such that f ( L j ) is isotopic to L j for j ∈ { , } , and f preserves framing mod 2. Pick a Seifert surface Σ of L . There is a stabilization Σ (cid:48) of Σ such that Σ (cid:48) ∩ F (cid:48) = L (see [FL18, Proof of Lemma 18] for details). It is clearfrom the definition of the Arf invariant thatArf( ∂ (Σ (cid:48) ∪ F (cid:48) )) = Arf( ∂ Σ (cid:48) ) + Arf( ∂F (cid:48) ) ⇒ Arf( f ( L )) = Arf( f ( L )) + Arf( f ( L )) , and this implies the statement of the lemma since f preserves framing mod 2. (cid:3) Proof of Theorem 1.10. (1’) ⇒ (4’): The proof of (1) ⇒ (4) in Section 3 can beadapted with minimal changes, using that σ ( V ) = 0 and π ( V ) = 1 and thatBorodzik-Friedl’s Theorem 2.1 holds verbatim for knots in integer homology three-spheres.(4’) ⇒ (3’): This part of the proof is purely 3–dimensional and makes no referenceto the four-manifold V . So all we need for our proof of (4) ⇒ (3) in Section 4.1to adapt is that Borodzik-Friedl’s Theorem 4.1 holds for knots in any Z HS ; see[BF14, Rmk. 5.2].(3’) ⇒ (2’): The geometric construction of the surface in Section 4.2, whichproves (3) ⇒ (2) works in any Z HS , without changes. In fact, the statement ofProposition 4.5 and its proof carry over to knots K (cid:48) and K in any 3–manifold M rather than S (even without any assumptions about the homology of M ).(2’) ⇒ (1’): This is the only step that goes beyond a straight-forward generaliza-tion of the corresponding step in Theorem 1.1. Let Σ be a genus g = h + c = h + c + c K and a knot J with Alexander polynomial 1. Note thatArf( J ) = 0. By Lemma 5.3, we may and do choose Σ to be H –null. K ×{ } K ×{ } J ×{ } J ×{ } DJ × [ , ] K × [ , ×{ } M ×{ } M ×{ } M ×{ } B B Figure 5.
Schematic drawing of F (cid:48) (dimensions reduced by one).1–, 2–, 3– and 4–manifolds are drawn in blue, purple, brown andblack, respectively. We start as in (2) ⇒ (1). See Figure 5 for the following construction. Constructa Z –slice surface F (cid:48) by taking the union of Σ ⊂ M = ∂B with a Z –slice disc in B for J , using that Freedman’s Theorem 1.3 holds in any Z HS , and then pushingthe interior of this union into B . To be more precise, we identify B with a copyof B (we denote it by B ) union M × [ ,
1] obtained by canonically gluing ∂B to M × { } . Now we construct F (cid:48) as follows. We take a Z –slice disk D in B ⊂ B = B ∪ M × [ , , with boundary J × { } ⊂ M × [ ,
1] and the surface Σ (cid:48) ⊂ M × [ ,
1] given asΣ × { } ∪ J × [ , ] ∪ K × [ , , and set F (cid:48) := D ∪ Σ (cid:48) .The argument to show π ( B \ F (cid:48) ) ∼ = Z is similar as in the case of M = S and B = B , which is needed in the proof of (2) ⇒ (1) and for which a detailed argumentis available in [FL18, Proof of Claim 20]. Here is a brief version. By the ‘risingwater principle’ (see [GS99, Proof of Proposition 6.2.1]), ( M × [ , \ ν (Σ (cid:48) ) , M ×{ } \ ν ( J × { } ) has a relative handle decomposition with only 2–handles. As aconsequence, the inclusion M × { } \ ( J × { } ) ⊂ M × [ , \ Σ (cid:48) induces a surjection on π , and, thus, F (cid:48) = D ∪ Σ (cid:48) is a Z –slice surface in B .Next, we want to specify c many simple closed framed curves γ (cid:48) i on Σ (cid:48) ⊂ F (cid:48) suchthat surgery along them corresponds to connected sum with S × S or C P C P and such that Σ (cid:48) may be be compressed c times in the resulting manifold.One picks disjoint simple closed curves γ , . . . , γ c in Σ ◦ with the property thattheir homology classes may be extended to a symplectic basis of Σ. If c = 1,arrange for exactly one of them to have odd framing induced by Σ. If instead c = 0,choose all of them to have even framing. This is possible by Lemma 5.2. Note thatwe need the assumption Arf( K ) = 0(= Arf( J )) exactly in case h = c = 0. Indeed,in this case the homology classes of the γ i form a half basis of Σ, so choosing all ofthem to have even framing is possible since Arf( K ∪ J ) = Arf( K ) + Arf( J ) = 0 byLemma 5.4We denote by γ (cid:48) i the curves in Σ (cid:48) that map to γ i under the projection to the firstfactor M × [ , → M . Note that γ (cid:48) i ⊂ M × { } . Let I be the interval [ − ε, ε ], with ε chosen in (0 , ) for the explicit formulas below to work. Let A = A ( γ ) ∪ · · · ∪ A ( γ c )denote a closed regular neighborhood of the union of the γ i , where A ( γ i ) ∼ = S × I denotes a regular neighborhood of γ i in Σ. We fix a trivialization Φ : A × I (cid:44) → M of a regular neighborhood of A ⊂ M with Φ( a,
0) = a for all a ∈ A (e.g. take therestriction of a trivialization of an embedded normal bundle of Σ in M ). We furtherlet Ψ : A × I × I (cid:44) → M × [ , , ( s, t, h ) (cid:55)→ (Φ( s, t ) , + h ) , which gives a trivialization of a regular neighborhood of A (cid:48) := A ×{ } in M × [ , ⊂ B . For each γ (cid:48) i , we use Ψ to identify the regular neighborhood N i = Ψ( A ( γ i ) × I × I )with A ( γ i ) × I × I ∼ = S × I × I × I . We now perform surgery on M × [ ,
1] and thus B = B ∪ M × [ ,
1] by removing the interior of N i and gluing D × ∂ ( I × I × I ) ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 23 along S × ∂ ( I × I × I ) to the boundary of N i via Ψ. We modify Σ (cid:48) by removingΣ (cid:48) ∩ N ◦ i and adding D × ∂I × { } × { } ⊂ D × ∂ ( I × I × I ). In other, words wehave ambiently compressed γ (cid:48) i in Σ (cid:48) . Performing such a surgery for all the γ (cid:48) i ⊂ B together with the corresponding modification on Σ (cid:48) ⊂ F (cid:48) yields a 4–manifold V andan embedded surface F of genus g − c = h .It remains to check that π ( V \ F ) ∼ = Z and that V is homeomorphic to B C P C P ) c S × S ) c . First, we calculate π ( V \ F ). Writing V \ F = (cid:32) B \ ( F (cid:48) ∪ b (cid:91) i =1 N i ) (cid:33) ∪ (cid:32) b (cid:91) i =1 R i (cid:33) , where R i ∼ = D × ( S \{ N, S } ) is the copy of D × ∂ ( I × I × I ) \ ( D × ∂I ×{ }×{ } ))in V corresponding to γ i , the Seifert-van Kampen theorem shows that π ( V \ F )is a quotient of π ( B \ ( F (cid:48) ∪ (cid:83) bi =1 N i )) ∼ = π ( B \ F (cid:48) ) ∼ = Z . In fact, this quotient isgiven by killing the free conjugation classes of Ψ( γ i × {− ε } × {− ε } ) in B \ F (cid:48) , whichis homotopy equivalent to B \ (cid:16) F (cid:48) ∪ (cid:83) bi =1 N ◦ i ) (cid:17) . However, Ψ( γ i × {− ε } × {− ε } ) isnull-homologous in B \ F (cid:48) (and thus null-homotopic) for all i . For this, note thatΨ( γ i × {− ε } × {− ε } ) is homotopic toΨ( γ i × { } × {− ε } ) ⊂ Σ ◦ × { − ε } ⊂ ( M \ J ) × { − ε } ⊂ B \ Σ (cid:48) ⊂ B \ F (cid:48) , and all the γ i ⊂ Σ ◦ are null-homologous in M \ J by our choice of Σ being H –null.Secondly, we describe V . Recall that embeddings of a finite union of circles ina simply-connected 4–manifold are unknotted, i.e. there exist disjoint embedded4–balls, one for each component, such that each curve lies in standard positionin one of these 4–balls. Therefore, surgery along each component corresponds tochanging the manifold by connect summing with the result of performing surgeryalong a simple closed curve in S . Surgery along a simple closed curve in S yieldseither S × S or C P C P . We check that the surgeries along γ (cid:48) i we performed tofind V yielded S × S when the framing of γ i induced by Σ is even and C P C P when the framing induced by Σ is odd. To check this, we consider, for each i , thesecond homology classes a i and b i in H ( V ; Z ) represented by S i := { } × ∂ ( I × I × I ) T i := ( D × { } × { } × { ε } ) ∪ (Σ i × { + ε } ) , respectively. Here, S i and ( D × { } × { } × { ε } ) are viewed as subsets of the D × ∂ ( I × I × I ) subset of V corresponding to i , and Σ i ⊂ M is a Seifert surfacefor γ i . We have that ( a , b , . . . , a c , b c ) is a basis of H ( V ; Z ) ∼ = Z c .One checks that a i is also represented by a sphere disjoint of S i , S i and T i intersect transversely in one point, and b i can be represented by a surface T (cid:48) i thatalgebraically intersects T i n i times, where n i denotes the framing of γ i inducedby Σ. Hence, the intersection form on H ( V ; Z ) restricted to (cid:104) a i , b i (cid:105) ⊂ H ( V ; Z ) isgiven by (cid:18) ± ± n i (cid:19) . This in turn is equivalent to (cid:18) ± ± (cid:19) if n i is even, and to (cid:18) ± ± (cid:19) and thus (cid:18) − (cid:19) if n i is odd. The former is the intersection form of S × S while the later is the intersection form of C P C P , which proves thatthe surgery corresponding to γ i is as claimed. (cid:3) Linking pairings of cyclic branched coverings of prime order
Theorem 1.1 (4) expresses the Z –slice genus in terms of the existence of certainpresentation matrices for the Blanchfield pairing. In spite of the algebraic nature ofthis characterization, no algorithm to compute g Z of a given knot is evident fromit. One faces the same hurdle as when trying to compute the Nakanishi index (theminimum number of generators of the Alexander module), which bounds g Z frombelow, namely the complexity of the ring Λ = Z [ t ± ] underlying the Blanchfieldpairing. In particular, one lacks a classification of finitely generated modules over Λ,as it is available over rings such as PIDs or Dedekind domains.To obtain lower bounds for g Z , however, one may consider the Hermitian pairinginduced by Bl when taking the quotient of Λ by a suitable ideal. In this section,we pursue this idea for the principal ideals generated by the n –th cyclotomicpolynomials Φ n for n prime. There are two reasons for this choice. Firstly, takingthe quotient by Φ n yields the sesquilinear linking pairing on the first homology groupof the n –fold cyclic branched covering of S along the knot, which is of geometricinterest. The details of this relationship are well known (see e.g. [Dav95]), and willbe explained in Section 6.1.Secondly, the algebraic situation is particularly simple. The conjugation onΛ / (Φ n ) is just complex conjugation, the ring Λ / (Φ n ) is Dedekind, and isometryclasses of Hermitian linking pairings can be fully classified. For n = 2, one hasΦ = t + 1 and Λ / (Φ ) ∼ = Z , hence the Hermitian pairings are in fact simply thesymmetric ones. Isometry classes of symmetric integral pairings on finite Abeliangroups A with odd order have been classified by Wall [Wal63]. We obtain anobstruction for g Z ( K ) ≤ Z –slice genus for knots in the table of primeknots with crossing number 12 and less (see Section 6.3).One might expect the cases n ≥ n ,the isometry class of the linking pairing of the n –fold branched cover is alreadydetermined by the action of the deck transformation group.This section is inspired by Borodzik and Friedl’s pursuit of the analogous lowerbounds for the algebraic unknotting number [BF15]. They show that the problemof finding minimal presentation matrices after quotienting by Φ n is essentially afinite problem, and can thus be solved by a computer using brute force. In contrast,the lower bounds we obtain from the double branched cover in Section 6.2 can for agiven knot be checked by hand. Moreover, our Theorem 1.2 partially answers thequestion implicitly asked in [BF15]: why n –fold branched covers for prime odd n donot supply efficient lower bounds.6.1. Bilinear and sesquilinear linking pairings of finite branched coverings.
Fix a prime power n ≥ n ( K ) denote the n –fold cyclic branched cover of S along K . Then H (Σ n ( K ); Z ) is an Abelian group of finite order d , carrying abilinear symmetric non-degenerate linking pairing lk : H (Σ n ( K ); Z ) × H (Σ n ( K ); Z ) → Q / Z , ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 25 which can be defined as lk( x, y ) = x ∩ PD( Y ) d + Z ∈ Q / Z , where Y ∈ H (Σ n ( K ); Z ) is a class with boundary dy , PD denotes the Poincar´edual, and ∩ denotes the cap product.For non-prime powers n , lk may be defined on the torsion part of the firsthomology of Σ n , which need not be a rational homology sphere. But since theresults we prove in the following sections apply only to prime n anyway, we do notfollow that more general setup.The deck transformation group of the branched covering Σ n ( K ) → S is cyclic oforder n . Identifying it with the multiplicative group of units (cid:104) [ t ] (cid:105) ⊂ Λ n := Λ / ( t n − H (Σ n ( K ); Z ) with the structure of a Λ n –module. Clearly, lk is equivariant with respect to the action of the unit group of Λ n , i.e. lk( tx, ty ) = lk( x, y ) for all x, y ∈ H (Σ n ( K ); Z ).One may now define another linking pairing (cid:96) (cid:48) on the Λ n -module H (Σ n ( K ); Z )that is sesquilinear, Hermitian (with respect to the conjugation on Λ n given by t k (cid:55)→ t − k ) and also non-degenerate: (cid:96) (cid:48) ( K ) : H (Σ n ( K ); Z ) × H (Σ n ( K ); Z ) → Q (Λ n ) / Λ n , ( † ) ( x, y ) (cid:55)→ n (cid:88) k =1 [ t ] k lk( t k x, y ) . Note that lk may be recovered from (cid:96) (cid:48) simply by( ‡ ) lk( x, y ) = θ ( (cid:96) (cid:48) ( x, y )) , where θ : Q (Λ n ) = Q [ t ] / ( t n − → Q is the Q –linear map sending t k to 1 if n | k andto 0 otherwise. So we have the following (a variation of [Dav95, Prop. 1.3]). Proposition 6.1.
On a fixed isomorphism type of torsion Λ n –module, the isometrytypes of bilinear symmetric equivariant pairings and sesquilinear Hermitian pairingsare in one-to-one correspondence via ( † ) and ( ‡ ) . This correspondence preservesnon-degeneracy. (cid:3) Next, let us change the ground ring from Λ n to its quotient Λ /ρ n , where ρ n =( t n − / ( t −
1) = 1+ t + . . . + t n − . For this, consider the covering map π : S \ K ∞ → S \ K n and the inclusion ι : S \ K n → Σ n ( K ), where S \ K n denotes the n –foldcyclic cover of S \ K . The deck transformation group endows H ( S \ K n ; Z ) with aΛ n –module structure and as a Λ n –module H ( S \ K n ; Z ) is canonically isomorphicto the twisted homology H ( S \ K ; Λ n ) (twisted with respect to the Abelianization π ( S \ K ) → Z composed with Z → Z /n Z ). Here, we identify the group ring Z [ Z /n Z ] with Λ n . Observe that ρ n H ( S \ K n ; Z ) is contained in the image of thetransfer homomorphism H ( S \ K ; Z ) → H ( S \ K n ; Z ), which is generated bythe class of the meridian of the boundary torus of S \ K n . But since this class iskilled by ι ∗ we find that ρ n annihilates H (Σ n ; Z ). So H (Σ n ; Z ) has the structureof a Λ /ρ n –module, on which (cid:96) (cid:48) induces a sesquilinear Hermitian non-degeneratepairing (cid:96) . By Corollary A.3 the isometry classes of (cid:96) (cid:48) and (cid:96) determine one another.In summary, we have seen three variations of linking pairings on the n –foldbranched covering, all of which contain the same information and can thus be usedinterchangeably. In the following, we will stick to (cid:96) and refer to it as the sesquilinearlinking pairing of Σ n . Let us now discuss the relationship between (cid:96) and Bl. The map( ι ◦ π ) ∗ : H ( M K ; Λ) → H (Σ n ( K ); Z )is a surjection with kernel ( t n − H ( M K ; Λ), which equals ρ n H ( M K ; Λ) because(1 − t ) acts invertibly on H ( M K ; Λ). This provides a canonical identification ofthe Λ /ρ n –modules H ( M K ; Λ) /ρ n H ( M K ; Λ) and H (Σ n ( K ); Z ). Moreover, theBlanchfield pairing Bl( K ) : H ( M K ; Λ) × H ( M K ; Λ) → Q ( t ) / Λinduces a sesquilinear Hermitian pairing on the quotient H ( M K ; Λ) /ρ n H ( M K ; Λ).Under the canonical identification with H (Σ n ( K ); Z ), this pairing corresponds tothe pairing (cid:96) discussed above. This can either be seen using Ko’s presentationmatrix [Ko89] (see [BF15, Sec. 5.1]), or by using the alternative version Bl (cid:48) of theBlanchfield pairing discussed in Appendix A. We omit the details.As a consequence, a presentation matrix of the Blanchfield pairing descends toa presentation matrix of (cid:96) when taking the quotient by ρ n . Taken together withTheorem 1.1 (4), this shows how the isometry class of the sesquilinear linking pairingof the n –fold branched covering of a knot provides a lower bound for g Z . This isconsidered in more detail in the following subsections.6.2. The double branched covering.
For n = 2, we have ρ = Φ = 1 + t ,and Λ / ( ρ ) is simply isomorphic to Z via t (cid:55)→ −
1. Under this identification,the sesquilinear pairing (cid:96) simply equals twice the bilinear pairing lk, i.e. (cid:96) ( x, y ) =2 lk( x, y ) for all x, y ∈ H (Σ ; Z ). Although lk is the pairing that is usually consideredin the literature and that is e.g. presented by the symmetrization of a Seifert matrix,we prefer to stick with (cid:96) for consistency’s sake. We ask the reader to keep thissubtlety in mind when working with results from this section. Let us start byproving the following. Proposition 6.2.
There is a symmetric integral presentation matrix C of size g Z ( K ) for the pairing (cid:96) on H (Σ ; Z ) with det C ≡ ( − g Z ( K ) (mod 4) .Proof. The presentation matrix A ( t ) of the Blanchfield pairing provided by The-orem 1.1 (4) descends to a presentation matrix C of (cid:96) when taking the quotientby Φ (see e.g. [BF15, Lemma 3.3]). It remains to check the condition on det C . Thematrix A ( t ) satisfies σ ( A (1)) = 0 and det A ( t ) = ± ∆ K ( t ) where ∆ K ( t ) = ∆ K ( t )and ∆ K (1) = 1. Combined, this impliesdet A ( t ) = ( − g Z ( K ) ∆ K ( t ) = ( − g Z ( K ) (1 + (1 − t )(1 − t − )∆ (cid:48) ( t ))for some ∆ (cid:48) ( t ) ∈ Λ. So[det A ( t )] = [( − g Z ( K ) ] ∈ Λ / ((1 − t )(1 − t − )) ⇒ [det C ] = [( − g Z ( K ) ] ∈ Λ / (Φ , (1 − t )(1 − t − )) ⇒ det C ≡ ( − g Z ( K ) (mod 4) . (cid:3) In light of this lower bound for g Z it would be useful—and of somewhat inde-pendent interest—to determine for every isometry class of symmetric pairings (cid:96) onfinite Abelian groups A of odd order (as classified by Wall [Wal63]) the minimalsize of a presentation matrix with prescribed determinant modulo 4. We make astart by giving a necessary and sufficient condition for such a symmetric pairing toadmit a 2 × ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 27
It turns out to be more practical to phrase the condition not in terms of Wall’sclassification. Let us instead fix a decomposition of A as orthogonal sum of cyclicgroups of order q , . . . , q n with q i | q i +1 . For each i , let g i be a generator of the i –thsummand. Then (cid:96) ( g i , g i ) ∈ Q / Z can be written as a i /q i for some a i ∈ Z , which arecoprime to q i since (cid:96) is non-degenerate. Clearly, the tuple (cid:18) a q , . . . , a n q n (cid:19) determines the isometry class of (cid:96) . Note that n is a lower bound for the size ofa presentation matrix of (cid:96) . As a warm-up consider that for n = 1, (cid:96) admits a1 × a or − a is a square residue modulo q .To analyze the existence of 2 × Jacobisymbol (cid:16) xy (cid:17) ∈ {± , } defined for all x, y ∈ Z with y odd and positive. It is 0 iffgcd( x, y ) (cid:54) = 1. If y is prime, then it equals the Legendre symbol, i.e. it is 1 if x is asquare residue modulo y , and − y = p · . . . · p k is a product of primenumbers, then (cid:16) xy (cid:17) is the product of the (cid:16) xp i (cid:17) . The Jacobi symbol is multiplicativein both x and y , and (if x, y are positive and odd) satisfies quadratic reciprocity: (cid:16) xy (cid:17) · (cid:16) yx (cid:17) = ( − x − · y − . Moreover, we have (cid:16) − y (cid:17) = ( − y − . Note that for non-prime y , (cid:16) xy (cid:17) = 1 is anecessary, but not a sufficient condition for x being a quadratic residue modulo y —this is the reason that Dirichlet’s prime number theorem is needed in the proof ofthe following (which consists, aside from that, of elementary manipulations). Proposition 1.12.
Let A be an Abelian group of odd order with two generators,equipped with a non-degenerate symmetric pairing (cid:96) : A × A → Q / Z . Then A decomposes as orthogonal sum of two subgroups generated cyclically by g , g , whichare of respective order q and q with q | q . Note q , q are odd and may equal 1.Let a i q i = (cid:96) ( g i , g i ) . Note that a i and q i are coprime. Then, for a given u ∈ {− , } ,the two following statements (A) and (B) are equivalent:(A) (cid:96) can be presented by an odd symmetric × integer matrix M with det M ≡ u (mod 4) .(B) a , a , q , q , u satisfy both of the following two conditions.(B1) ( − ( q q − u ) / a a is a square residue modulo q ,(B2) u = 1 , or q q ≡ , or the Jacobi symbol (cid:16) a q /q (cid:17) equals .Proof. The existence of an orthogonal decomposition is part of Wall’s classifica-tion [Wal63]. The proof then proceeds by showing that (A) and (B) are bothequivalent to the following intermediate statement:(C) There exist integers α, β, γ, λ , λ satisfying the following conditions:(C1) α is odd,(C2) αγ − β = ( − ( q q − u ) / q /q ,(C3) λ α ≡ a (mod q ),(C4) λ α ≡ ( − ( q q − u ) / a (mod q ). ‘(C) ⇒ (A)’: We claim that the matrix M = q · (cid:18) α ββ γ (cid:19) has the desired properties. Indeed, by (C1) and (C2), this is an odd symmetric 2 × q ( αγ − β ) = ( − ( q q − u ) / q q ≡ u (mod 4). Itremains to check that M presents (cid:96) , for which we will use M − = ( − q q − u q (cid:18) γ − β − β α (cid:19) . Consider v = ( α, β ) (cid:62) and v = (0 , (cid:62) . Then M − v = (1 , (cid:62) /q is of order q in Q / Z and M − v = ( − ( q q − u ) / ( − β, α ) (cid:62) /q of order q (using (C4)). Theseare equal, respectively, to the orders of [ v ] , [ v ] ∈ coker M , and so the cokernelof M is isomorphic to Z /q ⊕ Z /q , with [ v ] , [ v ] generating the two summands.Since (C3) and (C4) imply gcd( λ i , q i ) = 1, λ i v i are also generators of the summands.Furthermore, one computes (cid:96) M ([ λ v ] , [ λ v ]) = λ λ v (cid:62) M − v = 0 ,(cid:96) M ([ λ v ] , [ λv ]) = λ v (cid:62) M − v = λ α/q ∈ Q / Z ,(cid:96) M ([ λ v ] , [ λ v ]) = λ v (cid:62) M − v = ( − ( q q − u ) / λ α/q ∈ Q / Z . By conditions (C3) and (C4), this implies that (cid:96) M is isometric to (cid:96) . ‘(A) ⇒ (C)‘: Pick a non-trivial vector x (cid:48) ∈ Z with [ x (cid:48) ] = g ∈ coker M = A . Let λ = gcd( x (cid:48) ) and x = x (cid:48) /λ . Then, x may be extended to a basis ( y, x ) of Z , i.e. T = ( y | x ) ∈ GL ( Z ). Let α, β, γ ∈ Z such that N := T − M ( T − ) (cid:62) = q · (cid:18) α ββ γ (cid:19) . We claim that these integers satisfy (C2)–(C4). Indeed, (C2) follows from det N =det M . Next, one computes N − = ( − q q − u q (cid:18) γ − β − β α (cid:19) and a q = (cid:96) ( g , g ) = x (cid:48)(cid:62) M − x (cid:48) = λ x (cid:62) ( T − ) (cid:62) N − T − x = λ (cid:18) (cid:19) (cid:62) N − (cid:18) (cid:19) = ( − q q − u λ αq ∈ Q / Z ⇒ λ α ≡ ( − q q − u a (mod q ) , so (C4) is satisfied. Finally, one checks that v = (cid:18) αβ (cid:19) ⇒ N − v = (cid:18) /q (cid:19) , which implies that [ v ] is of order q in coker N and orthogonal to [(0 , (cid:62) ] withrespect to N − . So, [ T v ] is of order q in coker M and orthogonal to [ x ] with respectto M − . Therefore, [ T v ] and g generate the same subgroup, and thus λ [ T v ] = g for some λ ∈ Z , which implies (C3) by a calculation similar to the one for (C4)above. ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 29
So if α is odd, (C1) is satisfied and we are done. If α is even, then γ must be oddbecause N is an odd matrix. Replacing α and β by α + 2 q β + q γ and β + q γ ,respectively, preserves (C2)–(C4), while also satisfying (C1). ‘(C) ⇒ (B)‘: Condition (B1) follows immediately from (C3) and (C4). To prove(B2), we assume u = − q q ≡ (cid:16) a q /q (cid:17) = 1. By(C2), we have αγ − β = − q /q , so q /q is a square modulo α , which implies (cid:16) q /q | α | (cid:17) = 1. But this concludes the proof since1 = (cid:16) q /q | α | (cid:17) = (cid:16) | α | q /q (cid:17) (by quadratic reciprocity and q /q ≡ (cid:16) ± a q /q (cid:17) (by (C4))= (cid:16) a q /q (cid:17) (because q /q ≡ . ‘(B) ⇒ (C)‘: Dirichlet’s prime number theorem states that for any x, y ∈ Z + ,there exists a positive prime number equivalent to x modulo y . So in our situation,for any σ , σ ∈ {− , } , there exists a positive prime number p satisfying p ≡ σ · ( − ( q q − u ) / · a (mod q ) and p ≡ σ (mod 4). Set α = σ · p . By definition,it satisfies (C1) and (C4) with λ = 1. By (B1), there exists λ ∈ Z such that λ ( − ( q q − u ) / a ≡ a (mod q ) . This implies (C3). Finally, note that the existence of some β, γ ∈ Z solving (C2) isequivalent to − ( − ( q q − u ) / · q /q being a square residue modulo p . Because p isprime, this is equivalent to (cid:16) ( − q q − u q /q p (cid:17) = 1 . One computes (cid:16) ( − q q − u q /q p (cid:17) = (cid:16) ( − q q − u p (cid:17) · (cid:16) q /q p (cid:17) = ( − σ − · q q − u +22 · ( − σ − · q q − · (cid:16) pq /q (cid:17) = ( − σ − · u +12 · (cid:16) σ ( − ( q q − u ) / a q /q (cid:17) = ( − σ − · u +12 · ( − σ − q q − u · q q − · (cid:16) a q /q (cid:17) . Now, if u = 1, switching σ changes the sign of the first factor, and so σ may bechosen to make the whole product 1. Similarly, if q q ≡ σ may bechosen to make the whole product 1. Else we have u = − q q ≡ (cid:3) Now let us spell out the obstructions for u a and g Z obtained by combining thetwo previous propositions. In concrete cases, the obstructions can be checked byhand. Corollary 6.3.
If the symmetric pairing (cid:96) as in the proposition above is the linkingpairing of the double branched covering of a knot K , then we have the followingobstructions.(i) If q ≡ and a a is not a square residue modulo q , then u a ( K ) ≥ and g Z ( K ) ≥ .(ii) If q ≡ q ≡ , a a is not a square residue modulo q and (cid:16) a q /q (cid:17) = − , then u a ( K ) ≥ and g Z ( K ) ≥ .(iii) If q ≡ and q ≡ and a a is not a square residue modulo q ,then g Z ( K ) ≥ .(iv) If q ≡ q ≡ and − a a is not a square residue modulo q , then g Z ( K ) ≥ .(v) If q q ≡ and (cid:16) a q /q (cid:17) = − , then g Z ( K ) ≥ . (cid:3) Obstruction to g Z = 1 Knots to which the obstruction applies
Levine-Tristram-Signatures 1 12n749Hasse-Taylor [LM19] 6 12a787, 12n { } H (Σ ( K ); Z ) needs more thantwo generators 7 12a554, 12a750, 12n { } u a ≥ , 11n148, 12a327, 12a921, 12a1194,12n147Corollary 6.3(iii) 26 9 , 10 , 11a { } , 11n {
71, 75,167 } , 12a { } , 12n { } Corollary 6.3(iv) 12 9 , 11a135, 12a { } ,12n { } Table 1. Z –slice genus calculations for small knots.6.3. Calculations of the Z –slice genus for small knots. Let us try out theknown lower bounds for the Z –slice genus on prime knots with crossing number 12 orless. In [LM19], g Z was determined for all but 54 of these knots. Let us summarizethose calculations, and see how much further we can go using Corollary 6.3.Four of those 2,977 knots have Alexander polynomial 1 and thus g Z = 0. For allothers g Z ≥ g Z from a randomizedSeifert matrix algorithm [BFLL18, LM19]. For 1,998 knots, this upper bound is 1and g Z = 1 follows. For another 901 knots, the upper bound equals | σ | / g Z = | σ | / g Z equals 1 or 2. Table 1 lists those knots for which g Z = 2 can be proven, and the used obstruction. The Z –slice genus was previouslyunknown for the 38 knots listed in the last two rows, for which Corollary 6.3 provesto be an effective tool.There remain the 16 knots with unknown g Z ∈ { , } , all of them alternatingwith crossing number 12. These are their numbers in the knot table: 735, 1013,1047, 1168, 1203, 1211, 1221, 1222, 1225, 1226, 1229, 1230, 1248, 1260, 1283, 1288. ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 31
Branched coverings of odd prime order.
This section is devoted to theproof of the following.
Theorem 1.2.
Let p be an odd prime, Σ p ( K ) the p –fold branched covering of aknot K , and N p ( K ) := H (Σ p ( K ); Z [ t ] / ( t p − the first homology group of Σ p ( K ) ,on which t acts as a generator of the deck transformation group. Then, any twonon-degenerate Hermitian pairings on N p ( K ) are isometric. In particular, if fortwo knots K, K (cid:48) , the Z [ t ] / ( t p − –modules N p ( K ) and N p ( K (cid:48) ) are isomorphic, thenthe linking pairings on these modules are isometric. Let us start with an example, which already contains one crucial ingredient ofthe proof.
Example . Let K be the (6 , N p ( K ) is a cyclic module for all primes p . For example, N ( K ) ∼ = Z [ t ] / (7 , t + 1), which is simply Z / t acts as −
1. The sesquilinear linkingpairing (cid:96) : N ( K ) × N ( K ) → Q / Z is completely characterized by its value (cid:96) ( x, x )on a generator x of N ( K ). We have (cid:96) ( x, x ) = a/ a ∈ Z . Now, for K (cid:48) themirror image of K , one has N ( K (cid:48) ) ∼ = N ( K ) and (cid:96) ( x (cid:48) , x (cid:48) ) = − a/ x (cid:48) of N ( K (cid:48) ). Suppose there exists an isometry between the linkingpairings on N ( K ) and N ( K (cid:48) ). It would send x to λx (cid:48) , which would imply a/ λ · ( − a/ ∈ Q / Z ⇔ λ ≡ − . But − K and K (cid:48) are not distinguished by their N –modules, which are isomorphic; but they are distinguished by their linkingpairings on those N –modules, which are not isometric. Theorem 1.2 tells us that p = 2 is the only prime for which this may happen. Indeed, let us consider thesame pair of knots K, K (cid:48) for p = 3. One finds N ( K ) ∼ = N ( K (cid:48) ) to be of order 7, i.e. N ( K ) ∼ = Z [ t ] / (7 , t + t + 1). Note that the underlying Abelian group of this moduleis isomorphic to Z / × Z /
7. Again, the linking pairing on N is determined by thevalue (cid:96) ( x, x ) ∈ ( Q ( t ) / ( t + t + 1)) / ( Z [ t ] / ( t + t + 1)) for x a generator of N ( K ).Because (cid:96) is Hermitian, (cid:96) ( x, x ) = (cid:96) ( x, x ), where · is the conjugation that linearlyextends t (cid:55)→ t − . Thus (cid:96) ( x, x ) = b/ b ∈ Z . Once more, (cid:96) ( x (cid:48) , x (cid:48) ) = − b/ x (cid:48) of N ( K (cid:48) ). But, in contrast to the case p = 2, for p = 3 thelinking pairings on the modules N p ( K ) and N p ( K (cid:48) ) are isometric. An isometry isgiven by sending x to (3 + 6 t ) x (cid:48) . Indeed, one checks that (cid:96) ((3 + 6 t ) x (cid:48) , (3 + 6 t ) x (cid:48) ) = (3 + 6 t )(3 + 6 t − ) (cid:96) ( x (cid:48) , x (cid:48) )= ( − t − ( t + t + 1) + 28) · ( − b/
7) = b/ (cid:96) ( x, x ) ∈ ( Q ( t ) / ( t + t + 1)) / ( Z [ t ] / ( t + t + 1)) . The key is that modulo (7 , t + t ), one can write − ∈ Z [ t ] as λλ for some λ ∈ Z [ t ]. In fact, every y ∈ Z [ t ] with y = y can be written in that way.The theorem is an instance of a more general statement about Hermitian pairingson Dedekind rings, which we formulate as the following proposition. Proposition 6.5.
Let R be a Dedekind ring with an involution · . Let R − be thesubring of elements fixed by the involution and assume that R = R − [ ξ ] for some ξ ∈ R \ R − . Let A be a finitely presented R –torsion module with A = A whoseorder is coprime with ξ − ξ − . Then any two non-degenerate R –Hermitian pairings A × A → Q ( R ) /R are isometric. Let us first deduce Theorem 1.2.
Proof of Theorem 1.2.
First note that by Section 6.1 (in particular Proposition 6.1)the various notions of linking pairing are interchangeable; so it suffices to provethat any two non-degenerate Hermitian pairings on the R –module H (Σ p ( K ); Z )are isometric (where R := Λ / Φ p ). Now R is the ring of algebraic integers of the p –th cyclotomic field (cf. [Neu99, Ch. I, § R inherits the involution t (cid:55)→ t − from Λ, and R = R − [ t ], where R − is thefixed point ring. The ideals (∆ K ) and ( t −
1) are coprime over Λ, since ∆ K (1) = 1.Moreover, t + 1 generates all of R , since Φ p ( −
1) = 1 (here, we need p (cid:54) = 2). So (∆ K )is coprime with t ± R , and thus also with t − t − . Thus any two Hermitianforms of A are isometric by Proposition 6.5. (cid:3) The remainder of this section contains the proof of Proposition 6.5, which willrequire some algebraic number theory (see e.g. [Neu99]). In a nutshell, the proof goesas follows. We will first follow Wall’s classification of symmetric pairings of finiteAbelian groups of odd order [Wal63], showing that pairings may be diagonalized,i.e. decomposed as orthogonal sum of pairings on cyclic modules (for this, we will ofcourse need that R is Dedekind). On a cyclic module, there is only one isometryclass of pairings, because every element is a norm modulo a fixed prime ideal (forthis, we will need that R is not fixed by conjugation).First, note that R − is also a Dedekind ring, for the following reasons. The element ξ is a root of the monic polynomial ( x − ξ )( x − ξ ) ∈ R − [ x ], so R is isomorphicto R − × R − as an R − –module. This implies that R − is Noetherian and the ringextension R : R − is integral, whence the Krull-dimension of R − is the same as thatof R . Finally, if u ∈ Q ( R − ) is integral over R − , then u is also integral over R , so u ∈ R . Since R − = Q ( R − ) ∩ R , this implies u ∈ R − , and thus R − is integrallyclosed, completing the proof that it is a Dedekind ring.So we find ourselves in the usual situation of an extension of Dedekind rings, where R − is a Dedekind ring, K : Q ( R − ) is a finite field extension (where K = Q ( R )),and R is the integral closure of R − in K . The relationship of prime ideals of R − and R in this situation is well-understood (cf. [Neu99, Ch. I, § p ⊂ R − be aprime ideal, and P ⊂ R the ideal generated by p . Then there are three scenarios: p may be inert , i.e. P is prime; or p may split , i.e. P is the product of two distinctprime ideals of R that are interchanged by the involution; or p may ramify , i.e. P isthe square of a prime ideal of R . Lemma 6.6.
Let p , P be ideals as above with non-ramifying p . Write E = R/ P and F = R − / p . Define the trace T : E → F and the norm N : E × → F × as T ( x ) = x + x and N ( x ) = x · x , respectively. Then T and N are surjective.Proof. If p is inert, then E is a finite field (cf. [Neu99, Theorem 3.1]) and x = x | F | isthe Frobenius automorphism. So x ∈ ker N if N ( x ) = x | F | +1 = 1. This is satisfiedby at most | F | + 1 elements. On the other hand, | E × | = | F | − | F × | = | F | − | F | + 1 elements. It follows that | ker N | = | F | + 1and | im N | = | F | −
1, so N is surjective. If p splits, then the ring E is isomorphic to F × F , with conjugation interchanging the two components, and F ⊂ E identifiedwith { ( x, x ) | x ∈ F } . Clearly, ( x, x ) = N (( x, N is surjective.The ring E is a two-dimensional F –vector space. Assume that T is not surjective;then it is the zero-map. Since T (1) = 1 + 1 = 0 ∈ F , F must have characteristic 2. ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 33
So for any x ∈ E , T ( x ) = x + x = 0 implies that x = x . However R (cid:54) = R − , so thisis not the case. (cid:3) Now, using the freshly established non-triviality of the trace, let us prove a basechange lemma for homogeneous modules.
Lemma 6.7.
Let p , P be ideals as above with non-ramifying p . Let m, k ≥ , let A = (cid:76) mi =1 R/ P k and (cid:96) : A × A → Q ( R ) /R a non-degenerate Hermitian pairing.Write g i for [1] in the i –th summand. Then there is an automorphism φ : A → A such that P k − (cid:96) ( φ ( g m ) , φ ( g m )) (cid:54) = (0) .Proof. If P k − (cid:96) ( g i , g i ) (cid:54) = (0) for any i , let φ simply exchange the i –th and m –thsummand. If not, we have P k − (cid:96) ( g m , g j ) (cid:54) = (0) for some j because (cid:96) is non-degenerate. By Lemma 6.6, there exists a λ ∈ R such that λ + λ (cid:54)∈ p = R − ∩ P . Set φ ( g m ) = g m + λg j and φ ( g i ) = g i for i < m . One computes P k − (cid:96) ( φ ( g m ) , φ ( g m ))to be equal to P k − ( (cid:96) ( g m , g m ) + λλ(cid:96) ( g j , g j ) + ( λ + λ ) (cid:96) ( g m , g j )) = P k − (cid:96) ( g m , g j ) (cid:54) = (0) . (cid:3) We are now ready to prove diagonalizability for general modules.
Lemma 6.8.
Let A be an R –torsion module as in the Proposition 6.5. Let (cid:96) be anon-degenerate Hermitian pairing on A . Then there is a decomposition A ∼ = m (cid:77) i =1 R/ P k i i , where k i ≥ and the P i ⊂ R are ideals generated by prime ideals p i ⊂ R − . Moreover,if x and y are respective elements of the i –th and j –th summand with i (cid:54) = j , then (cid:96) ( x, y ) = 0 .Proof. Since R is a Dedekind domain, A is isomorphic to a unique sum of termsof the form R/ Q r with Q ⊂ R prime. Let p = Q ∩ R − , and P ⊂ R be the idealgenerated by p . Depending on whether p is inert, split, or ramified, P is equal to Q , QQ , or Q .Let us check that p does not ramify. Ramification is controlled by the differentideal D R : R − ⊂ R (cf. [Neu99, Ch. III, § p ramifies if and only if Q | D R : R − .Since R = R − [ ξ ], the different is the principal ideal generated by f (cid:48) ( ξ ), where f ( x ) = x − ( ξ + ξ − ) x + ξξ ∈ R − [ x ] is the minimal polynomial of ξ . Thus thedifferent is ( ξ − ξ − ). Note Q divides the order of A , which is by assumption coprimewith ( ξ − ξ − ). Thus Q does not divide D R : R − , and so p does not ramify.Since A = A , for every term R/ Q r either it holds that Q = Q , or the term R/ Q r also appears. So for each Q with Q = Q , take P i = Q ; for each pair Q , Q with Q (cid:54) = Q , take P i = QQ . In both cases, let k i = r and p i = Q ∩ R − , so that p i is aprime ideal generating P i over R (here, we use that p i does not ramify). This givesa decomposition of A as desired, which it remains to diagonalize.If x, y are given as above, and P i (cid:54) = P j , then (cid:96) ( x, y ) ∈ Q ( R ) /R is annihilatedby P k i i + P k j j . Note that P k i i + P k j j = R follows by inductively applying that I I + J = R if I + J = I + J = R for ideals I , I and J (a general propertyof commutative rings). Thus, we have (cid:96) ( x, y ) = 0. So, it suffices to solve the case that all P i are equal. Let us write P = P i , order the k i ascendingly, let k be theirmaximum, and let r ∈ { , . . . , m } such that k ≤ k ≤ . . . ≤ k r − < k r = k r +1 = . . . = k m = k. We note that (cid:96) restricts to a non-degenerate Hermitian pairing on the submodule M := (cid:76) mi = r R/ P k . To establish this, assume towards a contradiction that we have x ∈ M \ { } with (cid:96) ( x, y ) = 0 for all y ∈ M . Let n be the maximal integer such that x = ( x r , . . . , x m ) is an element of (cid:76) mi = r P n / P k . By the maximality of n , P k − − n x is a nontrivial submodule of M . Let x (cid:48) be any non-trivial element of it. We canwrite x (cid:48) = px for some p ∈ P k − − n . We note that (cid:96) ( x (cid:48) , y ) = (cid:96) ( x, py ) = 0 for all y in M . Furthermore, since px i ∈ P k − ⊂ P k i for all i ≤ r −
1, we find (cid:96) ( x (cid:48) , y ) = m (cid:88) i = r (cid:96) (( . . . , , px i , , . . . ) , y ) = m (cid:88) i = r (cid:96) (( . . . , , , , . . . ) , px i y )= m (cid:88) i = r (cid:96) (( . . . , , , , . . . ) ,
0) = m (cid:88) i = r y in (cid:76) r − i =1 R/ P k i . Thus, x (cid:48) is in the radical of (cid:96) contradicting the assumptionof (cid:96) being non-degenerate.Let g i be a generator of the i –th summand. By Lemma 6.7, there is a base changeon summands number r to m after which P k r − (cid:96) ( g m , g m ) (cid:54) = (0). This implies that λ i = (cid:96) ( g m , g i ) /(cid:96) ( g m , g m ) lies in R for all i ∈ { , . . . , m − } . Now one may replace g i with g i − λ i g m (as in the Gram-Schmidt algorithm). After this base change, onehas (cid:96) ( g i , g m ) = 0 and may proceed by induction over m . (cid:3) We now conclude the proof of the proposition.
Proof of Proposition 6.5.
By Lemma 6.8, it suffices to prove that any two pairings (cid:96) , (cid:96) on a module of the form A = R/ P k are isometric. A pairing on a cyclicmodule such as A is completely determined by its value (cid:96) i (1 , P k , but not by P k − . Moreover, (cid:96) i (1 ,
1) = (cid:96) i (1 , (cid:96) i (1 ,
1) as x i /y i with x i ∈ R − \ p and y i ∈ p k . The quotient of x /y by x /y is µ = x y x y , which lies in R − \ p . By Lemma 6.6, there exists a λ ∈ R such that[ λ · λ · µ ] = [1] ∈ R − / p . Let φ be the ring endomorphism of A given by multiplicationwith λ . This endomorphism is an automorphism since λ is a unit in A . Indeed, thelatter follows since ( λ )+ P = R implies ( λ )+ P k = R . Then (cid:96) ( φ (1) , φ (1)) = (cid:96) (1 , φ is an isometry between (cid:96) and (cid:96) as desired. (cid:3) Appendix A. Base change for Hermitian pairings
The purpose of this appendix is to establish Proposition A.2, which allows basechanges for Hermitian pairings of R –modules over a different (in the applicationsusually bigger) ring R (cid:48) . As mentioned in Section 4.1, this method is already inimplicit use by experts for the Blanchfield pairing; in Appendix A.2, we make themethod more explicit and generalize it to arbitrary rings R . Along the way, inAppendix A.1, we discuss how one may replace Q ( R ) /R as target of Hermitianpairings by R/I for a suitably chosen ideal I . Once again, this method has beenused before; we provide a general setup and discuss how the ideal I can be chosen,particularly if R is not a PID. ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 35
Let us start with a simple, motivating example to illustrate the results of bothAppendix A.1 and Appendix A.2. Consider the pairing (cid:96) A presented by a matrix A : (cid:96) A : Z /A Z × Z /A Z → Q / Z , (cid:96) A ( x, y ) = x (cid:62) A − y, where A = (cid:18)
12 33 24 (cid:19) . Firstly, note that im (cid:96) A = ( Z ) / Z ⊂ Q / Z , where 93 generates the annihilator ofthe cokernel of A . Since the Abelian groups ( Z ) / Z and Z / Z are isomorphic,one may instead of (cid:96) A equivalently consider the map Z /A Z × Z /A Z → Z / (cid:96) A ( x, y ) = 93 · x (cid:62) A − y .Secondly, a matrix B congruent to A over Z clearly yields an isometric pairing (cid:96) B .But now, consider the matrix B = (cid:18) (cid:19) , which is congruent to A over Z [ ]. One observes that (cid:96) B is isometric to (cid:96) A , eventhough A and B are not congruent over Z (because A is even and B is odd).Let us now formulate general principles based on these two observations.A.1. Change of perspective: Hermitian pairings as maps to
R/I . Let R be a unital commutative rings with involution. Usually, one considers Hermitianpairings of torsion modules M with target Q ( R ) /R (as we did in Section 2.1): (cid:96) : M × M → Q ( R ) /R. However, it will turn out to be more convenient to consider Hermitian pairings on M with target R/I , where I is an ideal contained in Ann( M ): (cid:96) (cid:48) : M × M → R/I.
Note that this makes sense even when M is not torsion.If I is a principal ideal generated by a non-zero-divisor s , then there is thefollowing 1-to-1-correspondence between Hermitian pairings (cid:96) of M with target Q ( R ) /R and Hermitian pairings (cid:96) (cid:48) of M with target R/I . Denote by ι s the injectiongiven by ‘dividing by s ’: ι s : R/I = R/ ( s ) (cid:44) → Q ( R ) /R, r + ( s ) (cid:55)→ rs + R. Let a pairing (cid:96) (cid:48) as above correspond to ι s ◦ (cid:96) (cid:48) . This mapping from pairings withtarget R/I to pairings with target Q ( R ) /R is clearly injective; it is also surjective,because I ⊂ Ann( M ).If M is torsion, but I is not principal, the Hermitian pairings with targets Q ( R ) /R and R/I may not be in such a natural 1-to-1 correspondence; but since we do notencounter such M in this text, we refrain from pursuing this further.Note that the 1-to-1-correspondence depends on the choice of s . However, whenapplying this change of perspective to the Blanchfield pairing of a knot K , where M = H ( M K ; Λ) is the Alexander module, there are two natural choices of I as aprincipal ideal with a canonical generator. One may choose I to be the order ideal,canonically generated by the Alexander polynomial ∆ K ( t ) with ∆ K ( t ) = ∆ K ( t − )and ∆ K (1) = 1. Thus the Blanchfield pairing of a knot K may be considered in acanonical fashion as a Hermitian pairingBl (cid:48) ( K ) : H ( M K ; Λ) × H ( M K ; Λ) → Λ / (∆( t )) . One may also choose I to be the annihilator of H ( M K ; Λ). It follows fromLemma A.1 below that I is principal. Since ∆ K ∈ Ann( H ( M K ; Λ)), any generatorof Ann( H ( M K ; Λ)) divides ∆ K . Let us choose as canonical generator of theannihilator of the Alexander module the unique generator a ( t ) with a ( t ) = a ( t − )and a (1) = 1. So one may equally well consider the Blanchfield pairing of a knot K in a canonical fashion as a Hermitian pairingBl (cid:48)(cid:48) ( K ) : H ( M K ; Λ) × H ( M K ; Λ) → Λ / ( a ( t )) . Note that composing Bl (cid:48)(cid:48) with multiplication by ∆( t ) a ( t ) gives Bl (cid:48) . Lemma A.1.
Let R be a unital commutative unique factorization domain withinvolution, and A an n × n matrix over R with non-zero-divisor determinant. Thenthe annihilator ideal of the cokernel of A is principal.Proof. Write M for the cokernel of A . We have r ∈ Ann( M ) iff for all v ∈ R n it holdsthat rv ∈ AR n , or equivalently rA − v ∈ R n (where A − is a matrix over Q ( R )).Now, A − v is a vector in Q ( R ) n with entries of the form p/q . So Ann( M ) consistsof the intersection of all the principal ideals ( q ) for the q ∈ R that appear in thisway for some v . However, over a unique factorization domain, an intersection ofprincipal ideals is again principal. (cid:3) Let us generalize from the Blanchfield pairing to Hermitian pairings (cid:96) A on an R –torsion module M presented by a square matrix A with non-zero-divisor determinant.Taking I as the order ideal Ord( M ), which is principal and generated by det( A ),gives the following formula for (cid:96) A in terms of A : (cid:96) A : R n /AR n × R n /AR n → R/ (det( A )) , ( x, y ) (cid:55)→ x (cid:62) Adj( A ) y + (det( A )) , where Adj( A ) denotes the adjoint of A . Recall that A Adj( A ) equals det A times theidentity matrix. This explains how this formula and the ‘old’ formula (cf. Section 2.1) (cid:96) A : R n /AR n × R n /AR n → Q ( R ) /R, ( x, y ) (cid:55)→ x (cid:62) A − y + R translate into each other via the above 1-to-1 correspondence coming from thegenerator det A of Ord( M ).A.2. Changing bases over a different ring.
The advantage of the perspectiveon Hermitian pairings given in the previous subsection is that a change of the basering of the module naturally carries over to the target of Hermitian pairings, evenwhen the order is no longer a non-zero-divisor.Let φ : R → R (cid:48) be a homomorphism of unital commutative rings R and R (cid:48) withinvolution and let (cid:96) : M × M → R/I be a Hermitian pairing on an R –module M .Then there is an induced Hermitian pairing (cid:96) φ on the R (cid:48) –module M (cid:48) = M ⊗ R R (cid:48) given by (cid:96) φ : M (cid:48) × M (cid:48) → R (cid:48) /φ ( I ) R (cid:48) , ( x ⊗ r, y ⊗ s ) (cid:55)→ r · s · φ ( (cid:96) ( x, y )) + φ ( I ) R (cid:48) . For well-definedness, observe that φ ( I ) R (cid:48) ⊂ Ann R (cid:48) ( M (cid:48) ). Proposition A.2.
Let φ : R → R (cid:48) be a homomorphism of unital commutative ringswith involution R and R (cid:48) . For i ∈ { , } , let (cid:96) i : M i × M i → R/I i be Hermitianpairings of R –modules M i , where I i is an ideal contained in Ann( M i ) . Assume that φ induces an isomorphism of R –modules between R/I i and R (cid:48) /φ ( I i ) R (cid:48) for i = 1 , .Then (cid:96) φ and (cid:96) φ are isometric over R (cid:48) if and only if (cid:96) and (cid:96) are isometric over R . ALANCED ALG. UNKNOTTING, LINKING FORMS, SURFACES IN 3D & 4D 37
Proof.
It is straightforward that an isometry between (cid:96) and (cid:96) induces an isometrybetween (cid:96) φ and (cid:96) φ . For the other direction, note that the map M i → M i ⊗ R R (cid:48) given by x (cid:55)→ x ⊗ R –module isomorphism, since it can be written as thecomposition of the following R –isomorphisms: M i ∼ = M i ⊗ R R/I i ∼ = M i ⊗ R R (cid:48) /φ ( I i ) R (cid:48) ∼ = M i ⊗ R R (cid:48) . An isometry (cid:96) φ and (cid:96) φ is an R (cid:48) –isomorphism M ⊗ R R (cid:48) → M ⊗ R R (cid:48) . Composing withthe isomorphisms M → M ⊗ R R (cid:48) and M ⊗ R R (cid:48) → M gives an R –isomorphism M → M . It is straight-forward that this isomorphism behaves well with respectto (cid:96) , (cid:96) , and is thus an isometry. (cid:3) Let us consider two corollaries needed in the paper. The first allows one to thinkof the sesquilinear linking pairing of p –fold branched covers of the knot as moduleover the ring Λ / ( g ) rather than Λ / ( t p −
1) where p is a prime and g = 1 + . . . + t p − is the p –th cyclotomic polynomial (see Section 6.1). For this purpose, take f = ∆ K : Corollary A.3.
Let g ∈ Λ and φ : Λ / (( t − g ) → Λ / ( g ) be the canonical projection.For i ∈ { , } , let (cid:96) i : M i × M i → Λ / ( f, ( t − g ) be Hermitian pairings of Λ / (( t − g ) –modules for an f ∈ Λ with f (1) = 1 and f ∈ Ann( M i ) . Then (cid:96) φ and (cid:96) φ are isometricover Λ / ( g ) if and only if (cid:96) and (cid:96) are isometric over Λ / (( t − g ) .Proof. The statement is a direct application of Proposition A.2 with R = Λ / (( t − g )and R (cid:48) = Λ / ( g ); one merely needs to check that the map induced by φ between R/f R and R (cid:48) /f R (cid:48) is an isomorphism—this is the canonical projectionΛ / (( t − g, f ) → Λ / ( g, f ) . Note that f (1) = 1 implies f − t − r for some r ∈ Λ, so g = − ( t − gr + f g .This implies the equality of the ideals (( t − g, f ) = ( g, f ). (cid:3) The following is a version of Proposition A.2 for Hermitian pairings presentedby square matrices, which we use in our proof of (4) ⇒ (3) of Theorem 1.1; seeLemma 4.3. Corollary A.4.
Let φ : R → R (cid:48) be a homomorphism of unital commutative rings R and R (cid:48) with involution. Let A and B be Hermitian n × n R –matrices and denoteby A φ and B φ the R (cid:48) –matrices obtained from applying φ entry-wise to A and B ,respectively. Assume that φ induces isomorphisms R/ (det( A )) ∼ = R (cid:48) / ( φ (det( A ))) and R/ (det( B )) ∼ = R (cid:48) / ( φ (det( B ))) . Then:(i) (cid:96) A is isometric to (cid:96) B if and only if (cid:96) A φ is isometric to (cid:96) B φ .(ii) (cid:96) A is isometric to (cid:96) B if there exists T ∈ GL n ( R (cid:48) ) with B φ = T (cid:62) A φ T .Proof. The existence of T implies that (cid:96) A φ is isometric to (cid:96) B φ , so (ii) follows from (i).Let us show (i). Denote by M be the cokernel of A . The Hermitian pairing ( (cid:96) A ) φ isisometric to (cid:96) A φ via the canonical isomorphism from M ⊗ R (cid:48) to the cokernel of A φ (using right-exactness of the tensor product with R (cid:48) ). Similarly, ( (cid:96) B ) φ is isometricto (cid:96) B φ . Since Ord( M ) = (det( A )), and similarly for the cokernel of B , the statementfollows from Proposition A.2. (cid:3) References [Ale28] J. W. Alexander:
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ETH Zurich, R¨amistrasse 101, 8092 Zurich, Switzerland
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