The Satellite crossing number conjecture for cables of knots

Transcription

1 The Satellite crossing number conjecture for cables of knots Alexander Stoimenow Department of Mathematical Sciences, KAIST April 25, 2009 KMS Meeting Aju University

2 Contents Crossing number Satellites and cables Satellite crossing number conjecture Lower bounds for asymptotic satellite crossing number of alternating knots Lower bounds for adequate knots 1

3 Crossing number of knots and links knot K S 1 S 3 link L S 1... S 1 }{{} n components S 3 K L represented by diagrams D; c(d) = number of crossings of D. If K is a knot or link, c(k) = min { c(d) : D is a diagram of K } is the crossing number of K. This is in general very hard to compute! 2

4 For example, consider the connected sum # = Conjecture 1 (Additivity conjecture) c(k 1 #K 2 ) = c(k 1 ) + c(k 2 ). This conjecture is open for 100+ years! 3

5 Satellites and cables Satellites are constructed from a companion knot and pattern. K S 3 companion knot or link solid torus L T 2 = D 2 S 1 pattern 4

6 The satellite K L with companion K and pattern L is given by: = = (We ignore framing issues for simplicity.) Patterns L T come with two types of degree: the geometric degree d g (L) (sometimes called wrapping number), counting the total number of windings (or minimal number of transverse intersections with a meridional disk of T ), and the algebraic degree d a (L) (absolute homology class [L] H 1 (T 2, Z) = Z), which counts the windings with signs according to the direction they run along the torus. 5

7 If both degrees coincide, we call a satellite K L also a cable of K. Remark 2 If d g (L) = 1, then L = L K L = K#L. Thus the sattellite operation generalizes the connected sum. 6

8 Satellite crossing number conjecture Conjecture 3 (Satellite crossing number conjecture; most general form) Let L be a pattern with n = d g (L). Then for every knot K, we have intuition: c(k L) n 2 c(k). (1) one crossing n 2 crossings Remark 4 If K = is the trivial knot (unknot), then Conjecture 3 is trivial. But for no other K has the conjecture been proved for general pattern L! 7

9 Remark 5 If n = 1 then Conjecture 3 says in particular that for two knots K and L, c(k#l) max ( c(k), c(l) ). This is a weaker version of the Additivity conjecture. (But even this inequality is open in general.) Conjecture 3 was proposed by Freedman-He, who prove: Theorem 6 (Freedman-He) c(k L) d a (L) 2 ( 2g(K) 1 }{{} (*) Here g(k) is the genus of K, given by g(k) = min { g(s) : S is a Seifert surface of K }, ). 8

10 where a Seifert surface S of K is as = S S 3, S = K, and its genus is g(s) = # holes of S. Theorem 7 (He) c(k L) d a (L) 2 ( some hyperbolic invariants of K }{{} (**) ). Often ( ), ( ) c(k). 9

11 Lower bounds for asymptotic satellite crossing number of alternating knots Motivation: Consider special patterns, but want a more realistic estimate. Consider a torus pattern L = T n,p T 2. n = 4, p =

12 It is determined by two integers, n = d g (L) (= d a (L)) > 0 and p = # ( = [L] H1 (S 3 \ T T L) ) (counted with sign). Let K n,p = K T n,p. Our goal will be to provide estimates on the minimal c m (K) and asymptotic cable crossing number c k (K) of K, defined by c m (K) := inf n 2 min p Z c(k n,p ) n 2 and c k (K) := lim inf n min p Z c(k n,p ) n 2. (We assume n > 1 but pose no restriction on p Z.) Clearly, c m (K) c k (K) sup n 2 min p Z c(k n,p ) n 2 c(k), and the suggestive problem is Conjecture 8 c m (K) = c k (K) = c(k). 11

13 Theorem 9 If a knot K is prime alternating, then c k (K) = c(k) and c m (K) 7c(K)/8. If K is in addition achiral, then also c m (K) = c(k), i.e. conjecture 8 holds for K. prime : alternating : K K 1 #K 2 with K 1, K 2 has alternating diagram, where strand goes over-under achiral : isotopic to its mirror image 12

14 Lower bounds for adequate knots Our theorem is a special case of estimates that apply for adequate and semiadequate knots K. A B B A A A B B Each crossing of a diagram has A- and B-corners. The corner A (resp. B) is the one passed by the overcrossing strand when rotated counterclockwise (resp. clockwise) towards the undercrossing strand. A type A (resp. B) splitting is obtained by connecting the A (resp. B) corners of the crossing. We leave a trace of the crossing (indicated by a dashed line). 13

15 If all crossings of D are A-split, we call the resulting set of (solid line) loops, together with the (dashed line) traces connecting them the A-state A(D) of D. Similarly, one has the B-state B(D). A(D) D B(D) Definition 10 (Lickorish and Thistlethwaite) The property that D is A-adequate means that in A(D) each crossing trace connects two different loops, i.e. there 14

16 is no self-trace in A(D): Replacing A(D) by B(D), we have the property B-adequate. A diagram is adequate it is A-adequate and B-adequate, semiadequate it is A-adequate or B-adequate. (semi)adequate link link with a (semi)adequate diagram The trefoil diagram above is adequate. Exercise 11 A (reduced) alternating diagram is adequate. Thus alternating links are adequate. 15

17 We need two quantities from A(D), B(D). The first is: #A(D) := number of loops of A(D), #B(D) := number of loops of B(D). The other one is the atom number a(a(d)) and a(b(d)). To obtain this, first decompose A(D) resp. B(D) along separating loops: + then decompose these parts under connected sum: γ 16

18 + Here the curve γ must be chosen so that γ =, ( ) # γ = 2, and both interior and exterior of γ contain more than an arc. Then count the number of parts (atoms). This is a(a(d)) or a(b(d)). 17

19 Theorem 12 Let K be an adequate knot with an adequate diagram D. Then 1. c k (K) = c(k) = c(k n,p ) n 2 c(k) holds for n 5 (and all p Z). 2. Either c m (K) = c(k), or c(k) 7 and the following estimate holds: c m (K) 9 8 c(k) max( #A(D) + a(a(d)), #B(D) + a(b(d)) ) + { 5 / 8 c(d) odd 3/ 4 c(d) even }. 4 + Corollary 13 Assume K is an adequate knot. Then either c m (K) = c(k), or the following estimates apply: (a) If K is prime alternating, then c m (K) 7c(K) 8 + { 3 / 4 c(k) even 5/ 8 c(k) odd. (2) 18

20 If K is in addition achiral, then always c m (K) = c(k). (b) If K is general adequate, then c m (K) 5c(K) 8 + { 2 c(k) even 15/ 8 c(k) odd }. This is proved using the Kauffman polynomial and work of Stong and Thistlethwaite. There are other estimates using the skein and Alexander polynomial. Here is one of the simpler inequalities obtained this way: Theorem 14 If K is an adequate knot and its adequate diagram D is positive (all crossings look like c m (K) 9 8 ), then c m (K) = c(k), or c(k) s(d) b(k) 2 + { 1 / 4 c(d) even 1/ 8 c(d) odd }. (3) Here s(d) is the number of Seifert circles of D, and b(k) the braid index of K. 19

21 Thank you! 20

22 KMS Meeting April 25, 2009 Aju University, Korea