Non-slice 3-stranded pretzel knots
NNon-slice 3-stranded pretzel knots
MIN HOON KIM, CHANGHEE LEE, AND MINKYOUNG SONG
Abstract.
Greene-Jabuka and Lecuona confirmed the slice-ribbon conjecture for 3-stranded pretzel knotsexcept for an infinite family P ( a, − a − , − ( a +1) ) where a is an odd integer greater than 1. Lecuona andMiller showed that P ( a, − a − , − ( a +1) ) are not slice unless a ≡ , , , ,
59 (mod 60). In this note, weshow that four-fifths of the remaining knots in the family are not slice.
1. Introduction
A knot K in S is slice if K bounds a smoothly embedded disk in the 4-ball D . A knot K in S is ribbon if K bounds an immersed disk in S having only ribbon singularities. It is easy to see thatribbon knots are slice since we can push such an immersed disk in S around the ribbon singularitiesto get an embedded disk in D . A famous open question posed by Fox [Fox62] asks whether all sliceknots are ribbon or not. This question is usually called the slice-ribbon conjecture . The slice-ribbon conjecture ([Fox62]) . Every slice knot is ribbon.In [Lis07], Lisca confirmed the slice-ribbon conjecture for 2-bridge knots. The slice-ribbon con-jecture has been extensively studied for pretzel knots. In [GJ11], Greene and Jabuka confirmed theslice-ribbon conjecture for 3-stranded pretzel knots P ( p, q, r ) with odd p, q, r . (Figure 1 depicts adiagram of a pretzel knot P ( p, q, r ) where p, q, r in the boxes represent the numbers of half twists.) p q r Figure 1. P ( p, q, r ).Lecuona [Lec15] confirmed the slice-ribbon conjecture for all 3-stranded pretzel knots P ( p, q, r )except for infinitely many 3-stranded pretzel knots of the form P ( a, − a − , − ( a +1) ) for a ≥ P ( a, − a − , − ( a +1) ) with odd a ≥
3, Lecuonaconjectured that none of them are slice even topologically. (Recall that a knot K in S is topologicallyslice if K bounds a locally flat, embedded disk in D .) Conjecture 1 ([Lec15, Remark 2.3]) . For every odd integer a ≥
3, the 3-stranded pretzel knot P ( a, − a − , − ( a +1) ) is not topologically slice. Mathematics Subject Classification. a r X i v : . [ m a t h . G T ] J a n MIN HOON KIM, CHANGHEE LEE, AND MINKYOUNG SONG
Conjecture 1 is of interest since it implies that the slice-ribbon conjecture is true for all 3-strandedpretzel knots. Lecuona showed that Conjecture 1 is true unless a ≡ , , , , ,
59 (mod 60)and a ≥ a ≡ K which is not of the form P ( a, − a − , − ( a +1) ) with a ≥ K istopologically slice if and only if either ∆ K ( t ) = 1 or K is ribbon.Conjecture 1 has remained open when a ≡ , , , ,
59 (mod 60). Our main result is to showthat four-fifths of the remaining knots of Conjecture 1 are not slice.
Theorem 1.1.
Let a ≥ be an odd integer. If a ≡ , ,
59 (mod 60) or a ≡ ,
61 (mod 120) ,then the -stranded pretzel knot P ( a, − a − , − ( a +1) ) is not topologically slice.Proof of Theorem 1.1. In Theorems 3.3 and 4.2 given below, we show that P ( a, − a − , − ( a +1) ) isnot topologically slice if a ≡ a ≡ (cid:3) Hence Conjecture 1 remains open for a ≡ ,
97 (mod 120). By the aforementioned results ofLecuona [Lec15] and Miller [Mil17], we have the following corollary.
Corollary 1.2.
Let K be a -stranded pretzel knot which is not of the form P ( a, − a − , − ( a +1) ) for a ≡ ,
97 (mod 120) . Then K is smoothly slice if and only if K is ribbon. By [Mil17, Theorems 1.2 and 1.6], Theorem 1.1 immediately gives the following corollary.
Corollary 1.3.
Let K be a -stranded pretzel knot which is not of the form P ( a, − a − , − ( a +1) ) for a ≡ ,
97 (mod 120) . Then K is topologically slice if and only if either K is ribbon or ∆ K ( t ) = 1 .
2. Alexander polynomial of P ( a, − a − , − ( a +1) ) In this section, we recall elementary facts on the Alexander polynomial of a slice knot and collect ob-servations of Lecuona on the Alexander polynomials of 3-stranded pretzel knots P ( a, − a − , − ( a +1) )following [Lec15, pp. 2151–2152].Fox and Milnor observed in [FM66] that the Alexander polynomial of a slice knot can be factoredin a special form. Theorem 2.1 ([FM66]) . The Alexander polynomial ∆ K ( t ) of a topologically slice knot K satisfies ∆ K ( t ) . = f ( t ) f ( t − ) for some f ( t ) ∈ Z [ t ± ] and . = means equality up to multiplication by a unit of Z [ t ± ] . Let R be either Z or Z p for some prime p . We say that a Laurent polynomial ∆( t ) ∈ R [ t ± ] admitsa Fox-Milnor factorization if ∆( t ) . = f ( t ) f ( t − )for some f ( t ) ∈ R [ t ± ] and . = means equality up to multiplication by a unit of R [ t ± ]. By Theorem 2.1,for a topologically slice knot K , the Alexander polynomial ∆ K ( t ) and its mod p reduction ∆ pK ( t )admit Fox-Milnor factorizations. Remark 2.2.
A polynomial f ( t ) ∈ R [ t ± ] is self-reciprocal if f ( t ) = t deg f f ( t − ). The Alexanderpolynomials of knots and the cyclotomic polynomials are self-reciprocal. Irreducible factors of a self-reciprocal polynomial are all self-reciprocal or come in reciprocal pairs , that is, if g ( t ) | f ( t ) and g ( t ) isnot self-reciprocal then t deg g g ( t − ) is also a factor of f ( t ). It follows that a self-reciprocal polynomialadmits a Fox-Milnor factorization if and only if the multiplicities of its irreducible self-reciprocalfactors are all even. For an odd integer a ≥
3, let P a be the 3-stranded pretzel knot P ( a, − a − , − ( a +1) ). In [Lec15,Appendix], Lecuona computed the Alexander polynomial ∆ P a ( t ):∆ P a ( t ) . = ( t a +2 +1)( t a +1)( t +1) − ( a +1) t a − ( t − = (cid:81) d | ad (cid:54) =1 Φ d ( − t ) (cid:81) d | a +2 d (cid:54) =1 Φ d ( − t ) − ( a +1) t a − ( t − where Φ d ( t ) is the d th cyclotomic polynomial. For a prime p dividing a +12 , the mod p reduction∆ pP a ( t ) of ∆ P a ( t ) is a product of the mod p reductions of some cyclotomic polynomials:∆ pP a ( t ) . = (cid:81) d | ad (cid:54) =1 Φ pd ( − t ) (cid:81) d | a +2 d (cid:54) =1 Φ pd ( − t ) ∈ Z p [ t ± ] . (1)Suppose that P a is topologically slice. Then both ∆ P a ( t ) and ∆ pP a ( t ) admit Fox-Milnor factoriza-tions by Theorem 2.1. Since gcd( p, a ) = gcd( p, a + 2) = gcd( a, a + 2) = 1, all the irreducible factorsof Φ pd ( − t ) appearing in (1) are distinct and have multiplicity 1. Let N pd be the number of irreduciblefactors of Φ pd . By Remark 2.2, N pd is even for any integer d > a or a + 2. Hence wehave the following proposition which is due to Lecuona (see [Lec15, pp. 2151–2152]). Proposition 2.3 ([Lec15, pp. 2151–2152]) . For an odd integer a ≥ , let P a be the -stranded pretzelknot P ( a, − a − , − ( a +1) ) . Let p be a prime dividing a +12 and d > be an integer dividing either a or a + 2 . If P a is topologically slice, then the following holds. (1) ∆ P a ( t ) admits a Fox-Milnor factorization. (2) ∆ pP a ( t ) admits a Fox-Milnor factorization. (3) Φ pd ( − t ) has no irreducible self-reciprocal factor. (4) N pd is even. Lecuona showed that N pd is odd for some such p and d when a ≡ , , a ≡ , , ,
35 (mod 60) to obtain the non-sliceness of P a . So did Miller [Mil17] when a ≡
49 (mod 60).For the case a ≡ t + t + 1 of ¯∆ P a ( t ),which has multiplicity 1.The remainder of this paper is organized as follows. In Section 3, we interpret the condition (4)in terms of a Legendre symbol and show that (4) is not satisfied if either a ≡ , a ≡ a ≡ P a satisfying (2), (3) and (4).
3. Method 1: Using the Legendre symbol (cid:0) pd (cid:1) The goal of this section is to prove Theorem 3.3 which states that P a is not topologically slice if a ≡ , a ≡ N pd is equivalent to computing the Legendre symbol (cid:0) pd (cid:1) .For the reader’s convenience, we recall the definition and properties of Legendre symbols. For aninteger n and an odd prime d , the Legendre symbol (cid:0) nd (cid:1) is defined by (cid:0) nd (cid:1) = − n is not a square modulo d, n is divisible by d, n is a nonzero square modulo d. We need the following standard properties on Legendre symbols (for example, see [Ser73, pp. 6–7]).
Lemma 3.1.
Let d, p be odd primes and a, b be integers. (1) (cid:0) ad (cid:1) = (cid:0) bd (cid:1) if a ≡ b (mod d ) . (2) (cid:0) abd (cid:1) = (cid:0) ad (cid:1)(cid:0) bd (cid:1) . In particular, (cid:0) a d (cid:1) = 1 . (3) (cid:0) pd (cid:1) = ( − ( p − d − (cid:0) dp (cid:1) . MIN HOON KIM, CHANGHEE LEE, AND MINKYOUNG SONG (4) (cid:0) − d (cid:1) = ( − d − . (5) (cid:0) d (cid:1) = ( − d − . Proposition 3.2.
Let p and d be distinct primes and d is odd. Then the number N pd of irreduciblefactors of Φ pd ( t ) is even if and only if the Legendre symbol (cid:0) pd (cid:1) = 1 .Proof. Note that N pd is equal to the quotient of Euler’s totient function ϕ ( d ) divided by the multi-plicative order of p modulo d and hence N pd = | Z × d /p Z × d | .Suppose that N pd = | Z × d /p Z × d | is even. Since d is prime, Z × d is cyclic. Let y be a generator of Z × d and m be an integer such that y m ≡ p (mod d ). Since y m is trivial in Z × d /p Z × d and y also generates Z × d /p Z × d , m is a multiple of N pd . It follows that m is even and ( y m/ ) ≡ p (mod d ) or (cid:0) pd (cid:1) = 1.Suppose that (cid:0) pd (cid:1) = 1. Choose an integer x ∈ Z × d such that x ≡ p (mod d ). Since x is trivial in Z × d /p Z × d , the order of x in Z × d /p Z × d is either 1 or 2. In the latter case, N pd is automatically even since N pd is equal to the order of Z × d /p Z × d . Suppose that x is trivial in Z × d /p Z × d . Then x ≡ p m (mod d )for some m ∈ Z . Since x ≡ p (mod d ) and x ≡ p m (mod d ), p m − is trivial in Z × d . So | p Z × d | is adivisor of 2 m − | p Z × d | is odd. Since d is an odd prime, | Z × d | is even and hence N pd = | Z × d /p Z × d | = | Z × d || p Z × d | is even. (cid:3) We are ready to prove the main result of this section.
Theorem 3.3.
For an odd integer a ≥ , let P a be the -stranded pretzel knot P ( a, − a − , − ( a +1) ) .If a ≡ , or a ≡ , then the Alexander polynomial ∆ P a ( t ) does not have aFox-Milnor factorization and hence P a is not topologically slice.Proof. Assume that a is an integer with a ≡ a +12 and a has aprime divisor d such that d ≡ d ≡ (cid:0) d (cid:1) as follows: (cid:0) d (cid:1) = ( − ( d − / = − . By Proposition 3.2, N d is odd and ∆ P a ( t ) does not have a Fox-Milnor factorization.Assume that a is an integer such that a ≡ a +12 and a has a prime divisor d such that d ≡ d ≡ (cid:0) d (cid:1) = (cid:0) d (cid:1) ( − ( d − / . When d ≡ (cid:0) d (cid:1) = (cid:0) (cid:1) = − − ( d − / = 1. When d ≡ (cid:0) d (cid:1) = (cid:0) (cid:1) = 1 and ( − ( d − / = −
1. Hence (cid:0) d (cid:1) = (cid:0) d (cid:1) ( − ( d − / = − . By Proposition 3.2, N d is odd and ∆ P a ( t ) does not have a Fox-Milnor factorization.Assume that a is an integer such that a ≡ a +12 ≡ a +12 has a primedivisor p such that p ≡ a = r (cid:81) i =1 p e i i · s (cid:81) j =1 q f i i be the prime factorization of a such that p i ≡ q j ≡ i and j . Since a ≡ f = (cid:80) sj =1 f i is even. Since f is even and a ≡ − p ), we can observe that (cid:81) ri =1 (cid:0) pp i (cid:1) e i (cid:81) sj =1 (cid:0) pq j (cid:1) f j = (cid:81) ri =1 (cid:0) p i p (cid:1) e i (cid:81) sj =1 ( − f j (cid:0) q j p (cid:1) f j = ( − f (cid:0) ap (cid:1) = (cid:0) − p (cid:1) = − d of a such that (cid:0) pd (cid:1) = −
1. By Proposition 3.2, N pd isodd and ∆ P a ( t ) does not have a Fox-Milnor factorization. (cid:3) Remark 3.4.
The case a ≡ a ≡
4. Method 2: Finding a self-reciprocal irreducible factor
The goal of this section is to prove Theorem 4.2 which states that P a is not topologically slice if a ≡ p reduction Φ pd ( t ) of the d th cyclotomic polynomial Φ d ( t ) for some d and p . For this purpose, in Theorem 4.1, we discussequivalent conditions regarding the existence of a self-reciprocal irreducible factor of Φ pd ( t ). Note thatΦ pd ( t ) has an irreducible self-reciprocal factor if and only if Φ pd ( − t ) has an irreducible self-reciprocalfactor. From this, we can understand when Φ pd ( − t ) has an irreducible self-reciprocal factor and applyProposition 2.3(3) to conclude that P a is not slice if a ≡ p and d .To simplify our discussion, for a nonzero integer n and a prime q , we denote v q ( n ) and u q ( n ) bethe integers such that n = q v q ( n ) u q ( n )and u q ( n ) is not divisible by q . The number v q ( n ) is usually called the q -adic valuation of n . Notethat u ( n ) is odd and if m divides n , then u q ( m ) divides u q ( n ). Theorem 4.1 (cf. [WYF20, Theorem 2.3]) . Let p and d are distinct primes. Suppose that d is odd.Then the following are equivalent. (1) Φ pd ( t ) has a self-reciprocal irreducible factor. (2) There exists an integer w such that p w ≡ − d ) . (3) p u (cid:54)≡ d ) where u = u ( ϕ ( d )) and ϕ ( d ) denotes Euler’s totient function.Proof. As we mentioned, the equivalence between (1) and (2) is proved by Wu, Yue and Fan in[WYF20, Theorem 2.3].We first show that (2) implies (3). Let w be an integer such that p w ≡ − d ) and supposethat p u ≡ d ) to get a contradiction. Let s = gcd(2 w, u ) so that there are integers x, y ∈ Z such that s = 2 wx + uy . Observe that p s = p wx + uy ≡ d ). Since s is odd and s divides 2 w , s divides w . It follows that p w ≡ d ) which is a contradiction.It remains to show that (3) implies (2). Let r = ord( p, d ) be the order of p in Z × d . By Euler’stheorem, r divides ϕ ( d ). If r is odd, then r divides u = u ( ϕ ( d )). This implies that p u ≡ d )which is a contradiction. We can conclude that r is even. Observe that p r/ (cid:54)≡ d ) and( p r/ ) ≡ d ). Since d is prime, Z d is a field and hence p r/ ≡ − d ). (cid:3) Theorem 4.2.
For an odd integer a ≥ , let P a be the -stranded pretzel knot P ( a, − a − , − ( a +1) ) .If a ≡ , then the Alexander polynomial ∆ P a ( t ) does not have a Fox-Milnor factorizationand hence P a is not topologically slice.Proof. Let a ≥ a ≡ p of a +12 and a divisor d of a or a + 2 such that Φ pd ( − t ) has a self-reciprocalirreducible factor. For this purpose, we will use Theorem 4.1. Let u = u ( ϕ ( a ( a + 2))). Note that u is an odd integer.If ( a +12 ) u ≡ a ( a +2)), then 2 u ≡ ( a +1) u (mod a ( a +2)). Since ( a +1) ≡ a ( a +2))and u is odd, 2 u ≡ ( a + 1) u ≡ ( a + 1) (mod a ( a + 2)) . It follows that 2 u ≡ − d ) for any prime factor d of a + 2. Note that 2 is a prime factor of a +12 since a ≡ d ( − t ) has a self-reciprocal irreducible factor for any primedivisor d of a + 2. MIN HOON KIM, CHANGHEE LEE, AND MINKYOUNG SONG
If ( a +12 ) u (cid:54)≡ a ( a + 2)), choose a prime factor p of a +12 such that p u (cid:54)≡ a ( a + 2)). ByChinese remainder theorem, there is a prime factor d of a ( a + 2) such that p u (cid:54)≡ d l ) where l = v d ( a ( a + 2)). Since u ( ϕ ( d l )) = d l − u ( ϕ ( d )) is a divisor of u , ( p u ( ϕ ( d )) ) d l − (cid:54)≡ d l ). ByProposition 4.3 given below, p u ( ϕ ( d )) (cid:54)≡ d ). By Theorem 4.1, Φ pd ( − t ) has a self-reciprocalirreducible factor. (cid:3) It remains to prove the following elementary fact which was used in the above proof.
Proposition 4.3.
For a prime d , if n ≡ d ) , then n d l − ≡ d l ) .Proof. Write n = md + 1 for some integer m . Then n d l − = ( md + 1) d l − = 1 + (cid:80) d l − k =1 m k d k (cid:0) d l − k (cid:1) . Let k be an integer with 1 ≤ k ≤ d l − . Then d l − divides d v d ( k ) (cid:0) d l − k (cid:1) since k (cid:0) d l − k (cid:1) = d l − (cid:0) d l − − k − (cid:1) .Since v d ( k ) < k , d l divies d k (cid:0) d l − k (cid:1) for all such an integer k . Thus n d l − ≡ d l ). (cid:3)
5. Discussion on the remaining cases
The remaining cases of Conjecture 1 are when a ≡ ,
97 (mod 120). In either case, there are examplessatisfying the conditions (2), (3) and (4) of Proposition 2.3: a = 1081 , , , , , , , , . . . . Hence the methods of Sections 3 and 4 do not work for these knots.
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Department of Mathematics, POSTECH, Pohang 37673, Republic of KoreaE-mail address: [email protected]