We show that the Conway polynomials of Fibonacci links are Fibonacci
polynomials modulo 2. We deduce that, when $ n \not\equiv 0 \Mod 4$ and $(n,j)
\neq (3,3),$ the Fibonacci knot $ \cF_j^{(n)} $ is not a Lissajous knot.Comment: 7p. Sumitte

Koseleff, Pierre-Vincent

Pecker, Daniel

English

arXiv

International audienceWe show that the Conway polynomials of Fibonacci links are Fibonacci polynomials modulo 2. We deduce that, when $ n \not\equiv 0 \Mod 4$ and $(n,j) \neq (3,3),$ the Fibonacci knot $ \cF _j^{(n)} $ is not a Lissajous knot

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HAL Id: hal-00408736https://hal.archives-ouvertes.fr/hal-00408736Submitted on 2 Aug 2009HAL is a multi-disciplinary open accessarchive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come fromteaching and research institutions in France orabroad, or from public or private research centers.L’archive ouverte pluridisciplinaire HAL, estdestinée au dépôt et à la diffusion de documentsscientifiques de niveau recherche, publiés ou non,émanant des établissements d’enseignement et derecherche français ou étrangers, des laboratoirespublics ou privés.On Fibonacci KnotsPierre-Vincent Koseleff, Daniel PeckerTo cite this version:Pierre-Vincent Koseleff, Daniel Pecker. On Fibonacci Knots. The Fibonacci Quarterly, DalhousieUniversity, 2010, 48 (2), pp.137-143. ￿hal-00408736￿On Fibonacci knotsP. -V. Koseleff, D. PeckerAugust 2, 2009AbstractWe show that the Conway polynomials of Fibonacci links are Fibonacci polynomials modulo2. We deduce that, when n 6≡ 0 (mod 4) and (n, j) 6= (3, 3), the Fibonacci knot F(n)j is not aLissajous knot.keywords: Fibonacci polynomials, Fibonacci knots, continued fractions1 IntroductionFibonacci knots (or links) were defined by J. C. Turner ([11]) as rational knots with Conway notationC(1, 1, . . . , 1). He also considered the generalized Fibonacci knots F(n)j = C(n, n, . . . , n), where n isa fixed integer and the sequence (n, . . . , n) has length j.In this paper we determine the Conway and Alexander polynomials modulo 2 of Fibonacci knots.We show that the Conway polynomial of a generalized Fibonacci knot is a Fibonacci polynomialmodulo 2.As an application, we show that if n 6≡ 0 (mod 4) and (n, j) 6= (3, 3) the Fibonacci knot F(n)j isnot a Lissajous knot.Our results are obtained by continued fraction expansions.2 Conway notation and Fibonacci knotsThe Conway notation (J. H. Conway, [3]) is particularly convenient for the important class of rational(or two-bridge) knots. The Conway normal form C(a1, a2, ..., am) of a rational knot (or link), is bestexplained by the following figure. The number of twists is denoted by the integer |ai|, and the signof ai is defined as follows: if i is odd, then the right twist is positive, if i is even, then the right twistis negative. On Fig. 1 the ai are positive (the a1 first twists are right twists).The rational links are classified by their Schubert fractionsαβ= a1 +1a2 +1a3 +1· · · +1am= [a1, . . . , am], α > 0. (1)12 P. -V. Koseleff, D. Peckera1a2 am−1ama1a2am−1amFigure 1: Conway’s normal forms, m odd, m evenTwo rational links of fractionsαβandα′β′are equivalent if and only if α = α′ and β′ ≡ β±1(mod α).The integer α is the determinant of the link, it is odd for a knot, and even for a two-component link.The following result is a useful consequence of the continued fraction description of rational links(see [4] p. 207).Theorem 1. Any rational link has a Conway normal form C(2a1, 2a2, . . . , 2am).The Fibonacci knots (or links) are defined by their Conway notation Fj = C(1, 1, . . . , 1), wherej is the number of crossings. The Schubert fraction of Fj isFj+1Fj, and its determinant is theFibonacci number Fj+1. It is the reason why J. C. Turner named these knots Fibonacci knots. Healso introduced the generalized Fibonacci knots F(n)j = C(n, n, . . . n), where n is a fixed integer.We first observeProposition 2. F(n)j is a knot if and only if n ≡ 0 (mod 2) and j ≡ 0 (mod 2) or n 6≡ 0 (mod2)and j 6≡ 2 (mod3).Proof. Let us consider the Möbius transformation P (z) = [n, z] = n+1z. It is convenient to considerits matrix notation P =(n 11 0). Letαβ= [n, . . . , n] = P j(∞), it is also( αβ)= P j(10).If n ≡ 1 (mod 2) then P ≡(1 11 0)(mod 2), P 2 ≡(0 11 1)(mod 2), P 3 ≡ 1l (mod 2). We deducethat α ≡ β ≡ 1 (mod 2) when j ≡ 1 (mod3), α ≡ 0, β ≡ 1 (mod 2) when j ≡ 2 (mod3) andα ≡ 1, β ≡ 0 (mod 2) when j ≡ 0 (mod3). The case n ≡ 0 (mod 2) is similar. 2Fibonacci Knots 3F(1)3 F(1)4 F(2)1F(2)2 F(2)3 F(3)1F(3)2 F(3)3Figure 2: Some Fibonacci knots and links3 The Conway and Alexander polynomialsThe Alexander polynomial, discovered in 1928, is one of the most famous invariant of knots. J. H.Conway discovered an easy way to calculate it. He introduced the “Skein relations” which relate thepolynomial of a link K to the polynomials of links obtained by changing one crossing of K.The following result is a beautiful application of his algorithm.Theorem 3 ([4]). Let K = C(2a1, 2a2, . . . , 2am) be a rational knot (or link).The Conway polynomial of K is∇K(z) = ( 1 0 )(−a1z 11 0)(a2z 11 0)· · ·((−1)mamz 11 0)(10)The Alexander polynomial of K is∆K(t) = ∇K(t1/2 − t−1/2).Let us consider a simple example.Example 4 (The torus links). The torus link T(2, m) has Conway normal form C(m) = F(m)1 .It is the link of fractionm1orm1 − m. We have the continued fraction (of length m − 1)m1 − m= [−2, 2, . . . , (−1)m−1 · 2].Hence, the Conway polynomial is∇(z) = ( 1 0 )(z 11 0)m−1 ( 10).4 P. -V. Koseleff, D. PeckerIt is well known that(z 11 0)m=(fm+1(z) fm(z)fm(z) fm−1(z))where fm(z) are the Fibonacci polynomials defined by f0(z) = 0, f1(z) = 1, fm+1(z) = zfm(z) +fm−1(z) ([12]). We conclude that the Conway polynomial of T(2, m) is the Fibonacci polynomialfm(z) (see also [6]). If m = 2k + 1 (i.e. T(2, m) is a knot) we obtain the Alexander polynomial∆(t) = f2k+1(t1/2 − t−1/2)= (tk + t−k) − (tk−1 + tk−1) + · · · + (−1)k.The recently introduced Lissajous knots ([2, 5, 10, 4]) are non singular Lissajous space curves. Wewill show that in many cases, Fibonacci knots are not Lissajous knots. Let us first recall the followingTheorem 5 ([5, 10]). If K is a rational Lissajous knot then ∆K(t) ≡ 1 (mod 2).Consequently, we see that a non trivial torus knot is never a Lissajous knot.Moreover, Theorem 3 provides many examples of knots which are not Lissajous knots.Corollary 6. Let bi ≡ 2 (mod 4), m > 1. The Conway polynomial of C(b1, . . . , bm) is equivalent tofm+1(z) (mod 2).Corollary 7. If n ≡ 2 (mod4), the modulo 2 Conway polynomial of F(n)j is fj+1(z).Hence these knots are not Lissajous knots by Theorem 5.The following result is an immediate consequence of Theorem 3.Corollary 8. If n ≡ 0 (mod4), the modulo 2 Conway polynomial of F(n)j is 0 if j is odd, and 1 ifj is even.It is not known whether the knot F(4)2 = C(4, 4) is Lissajous or not (see [1]).4 The modulo 2 Conway polynomial of Fibonacci knotsWe shall now study the knots F(n)j , where n = 2k + 1 is an odd integer.Lemma 9. Let n = 2k + 1. We have the identities[n, n, x] = [n + 1,−2, 2, . . . ,−2, 2︸ ︷︷ ︸2k,−(1 + x)] (2)[n, n, n, z] = [n + 1,−2, 2, . . . ,−2, 2︸ ︷︷ ︸2k,−(n + 1),−z]. (3)Proof. Let us prove the first formula. We shall use matrix notations for Möbius transformations. LetG(u) = [−2, 2, u] =3u + 2−2u − 1. Its matrix is G =(3 2−2 −1), and consequently we get by inductionGk =(1 + 2k 2k−2k 1 − 2k)=(n n − 11 − n 2 − n). (4)Fibonacci Knots 5LetM(x) = [n + 1,−2, 2, . . . ,−2, 2,−(1 + x)], L(u) = [n + 1, u], T (x) = −x − 1. (5)The corresponding matrices areL =(n + 1 11 0), T =(1 10 −1), M = LGkT. (6)ConsequentlyM =(n + 1 11 0) (n n − 11 − n 2 − n)(1 10 −1)=(n2 + 1 nn 1)=(n 11 0)2, (7)that is M(x) = [n, n, x] which proves the first identity. If we substitute x = [n, z] in Formula (2),we obtain the second identity (3). 2Corollary 10. Let n = 2k + 1. We have the continued fractions[n, n] = [n + 1,−2, 2, . . . ,−2, 2︸ ︷︷ ︸2k], [n, n, n] = [n + 1,−2, 2, . . . ,−2, 2︸ ︷︷ ︸2k,−(n + 1)].Let us denote [n]j = [n, . . . , n︸ ︷︷ ︸j]. If j 6≡ 1 (mod 3), we get the continued fractions[n]j+3 = [n + 1,−2, 2, . . . ,−2, 2︸ ︷︷ ︸2k,−(n + 1),−[n]j ].When j ≡ 1 (mod3), there is no continued fraction expansion of [n]j with even quotients, by Prop.2. In this case, we shall get a continued fraction expansion forαβ − α, which is another fraction of thesame knot. Let s be the Möbius transformation defined by s(x) =x1 − x. We have s(αβ) =αβ − α.Proposition 11. Let n = 2k + 1. We have the continued fractionss([n]1) = s(n) =n1 − n= [−2, 2, . . .− 2, 2︸ ︷︷ ︸2k], n 6= 1, (8)s([n]j+3) = [−2, 2, . . . ,−2, 2︸ ︷︷ ︸2k,−(n + 1),−(n + 1),−s([n]j)] (9).Proof. The first formula has already been proved. Let us prove the second formula. We shall usethe Möbius maps G(x) = [−2, 2, x] =3x + 2−2x − 1, Q(x) = [−(n + 1), x], R(x) = −x corresponding toG =(3 2−2 −1), Q =(−(n + 1) 11 0), R =(1 00 −1).6 P. -V. Koseleff, D. PeckerLet us define the Möbius transformation H = Gk · Q2 · R. We obtain, using Form. (4)H =(n3 + n2 + 2n + 1 n2 + 1−n3 − n n − n2 − 1). (10)Let S be a matrix corresponding to the Möbius map s. We haveS−1HS =(1 01 1)H(1 0−1 1)=(n3 + 2n n2 + 1n2 + 1 n)=(n 11 0)3,and thenS(n 11 0)3= HS.This means that s([n, n, n, x])= h ◦ s(x), which proves our formula. 2Remark 12. By considering the case n = 1 where h(x) = [−2,−2,−x], we obtain the followinginteresting continued fractions of length 2m:F3m+2F3m= [2, 2,−2,−2, . . . , (−1)m−1 · 2, (−1)m−1 · 2]Using the corollary 10 we obtain similarlyF3m+1F3m= [2,−2,−2, 2, 2, . . . , (−1)m−1 · 2, (−1)m−1 · 2, (−1)m · 2]of length 2m andF3m+3F3m+2= [2,−F3m+2F3m] of length 2m + 1.Of course, these fractions correspond to Fibonacci knots (or links). They are not Lissajous knotsbecause of corollary 6.It is straightforward to calculate the Conway polynomials of our Fibonacci knots, using Prop. 11.Theorem 13. Let us denote by ∇(n)j (z) the modulo 2 Conway polynomial of the Fibonacci linkF(n)j . We have ∇(n)j (z) = fN (z) where{If n ≡ 1 (mod4), N = ⌊ j+23 ⌋(n − 2) + j + 1,If n ≡ 3 (mod4), N = ⌊ j+23 ⌋(n + 2) − (j + 1).(11)Corollary 14. If n 6≡ 0 (mod4) and (n, j) 6= (3, 3), the Fibonacci link F(n)j is not a Lissajous knot.It is not known whether the knot F(3)3 = C(3, 3, 3) is a Lissajous knot ([1]).Question 15. It would be interesting to study the wider classes of knots defined by their Conwaynotation C(±n,±n, . . .± n).If n = 1 we obtain all the rational knots ([8, 9]).If n = 2 we obtain the important class of rational fibered knots (see [7]).In general, we obtain knots with fractionsαβsuch that (α, β) ≡ (0,±1) or (±1, 0) (modn).Acknowledgements: We would like to thank Pr. C. Lamm for having suggested this problem tous.Fibonacci Knots 7References[1] A. Boocher, J. Daigle, J. Hoste, W. Zheng, Sampling Lissajous and Fourier knots,arXiv:0707.4210, 2007[2] M. G. V. Bogle, J. E. Hearst, V. F .R. Jones, L. Stoilov, Lissajous knots, Journal of KnotTheory and its Ramifications, 3(2): 121-140 (1994)[3] J. H. Conway, An enumeration of knots and links, and some of their algebraic properties,Computational Problems in Abstract Algebra (Proc. Conf., Oxford, 1967), 329-358 Perga-mon, Oxford (1970)[4] P. Cromwell, Knots and links. Cambridge University Press, 2004[5] V. F. R. Jones, J. Przytycki, Lissajous knots and billiard knots, Banach Center Publications,42:145-163 (1998)[6] L.H. Kauffman , On knots, Annals of Mathematics Studies, 115. Princeton University Press,Princeton, NJ, 1987[7] A. Kawauchi A survey of knot theory, Birkhäuser Verlag, Basel, 1996[8] P. -V. Koseleff, D. Pecker, Chebyshev knots, arXiv:0812.1089, 2008[9] P. -V. Koseleff, D. Pecker, Chebyshev diagrams for rational knots,arXiv:0906.4083, 2009[10] C. Lamm, There are infinitely many Lissajous knots, Manuscripta Math., 93: 29-37 (1997)[11] J.C. Turner, On a class of knots with Fibonacci invariant numbers, Fibonacci Quart. 24, 1,61-66 (1986)[12] W. A. Webb, E. A. Parberry, Divisibility of Fibonacci polynomials, Fibonacci Quart. 7, 5,457-463 (1969)Pierre-Vincent Koseleff,Équipe-project INRIA Salsa & Université Pierre et Marie Curie (UPMC-Paris 6)e-mail: koseleff@math.jussieu.frDaniel Pecker,Université Pierre et Marie Curie (UPMC-Paris 6)e-mail: pecker@math.jussieu.fr

On Fibonacci Knots

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On Fibonacci Knots

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