Vassiliev knot invariants derived from cable -polynomials - PowerPoint PPT Presentation

Vassiliev knot invariants derived from cable Γ -polynomials 滝岡 英雄 京都大学 日本学術振興会特別研究員 結び目の数理 II 日本大学 2019 年 12 月 20 日

Outline of my presentation The HOMFLYPT and Kauffman polynomials P , F and their coefficient polynomials P 2 i , F i . We focus on P 0 (= F 0 = Γ) . ⇓ ( P 0 ) p/q is the ( p, q ) -cabling of P 0 for coprime integers p, q . ⇓ ( P 0 ) ( d ) p/q (1) is the d th derivative of ( P 0 ) p/q at t = 1 , which is a Vassiliev knot invariant of order ≤ d . ⇓ In particular, we will show some results about ( P 0 ) ( d ) n/ 1 (1) .

HOMFLYPT ( P ) and Kauffman ( F ) polynomials P ( L ; t, z ) ∈ Z [ t ± 1 , z ± 1 ] and F ( L ; a, z ) ∈ Z [ a ± 1 , z ± 1 ] are invariants for oriented links in S 3 . For the trivial knot ⃝ , we have P ( ⃝ ) = F ( ⃝ ) = 1 . For a skein triple ( L + , L − , L 0 ) of oriented links, we have t − 1 P ( L + ) − tP ( L − ) = zP ( L 0 ) . For a skein quadruple ( D + , D − , D 0 , D ∞ ) of oriented link diagrams, we have aF ( D + ) + a − 1 F ( D − ) = z ( F ( D 0 ) + a − 2 ν F ( D ∞ )) , where 2 ν = w ( D + ) − w ( D ∞ ) − 1 and w ( D + ) , w ( D ∞ ) are the writhes of D + , D ∞ , respectively.

Coefficient polynomials of P and F L : an oriented r -component link. P ( L ) = ( − t − 1 z ) − r +1 � � � P 2 i ( L ; t ) z 2 i , i ≥ 0 F ( L ) = ( az ) − r +1 � � � F i ( L ; a ) z i , i ≥ 0 where P 2 i ( L ; t ) ∈ Z [ t ± 1 ] and F i ( L ; a ) ∈ Z [ a ± 1 ] are called the 2 i th coefficient polynomial of P ( L ) and the i th coefficient polynomial of F ( L ) , respectively. Fact [Lickorish 1988] P 0 ( L ; t ) = F 0 ( L ; √− 1 t − 1 ) . P 0 ( L ; t ) is a Laurent polynomial in t − 2 . Putting t − 2 = x , we call it the Γ -polynomial Γ( L ; x ) ∈ Z [ x ± 1 ] , that is, Γ( L ; t − 2 ) = P 0 ( L ; t ) = F 0 ( L ; √− 1 t − 1 ) . In this talk, we use P 0 not Γ to apply Kanenobu’s results.

Cable knot p ( > 0) , q : coprime integers. K : a knot. N ( K ) : a tubular neighborhood of K . K ( p,q ) : the ( p, q ) -cable knot of K , that is, an essential loop in ∂N ( K ) with [ K ( p,q ) ] = p [ l ] + q [ m ] in H 1 ( ∂N ( K ); Z ) , where ( m, l ) is a meridian-longitude pair of K with lk( K ∪ l ) = 0 and lk( K ∪ m ) = +1 . D : a diagram of K . w ( D ) : the writhe of D . K ( p,q ) has a diagram D ( p,q ) which consists of the p -parallel of D and the p -braid ( σ 1 · · · σ p − 1 ) q − pw ( D ) .

Cabling for knot invariants I : a knot invariant. The map sending a knot K to the value I ( K ( p,q ) ) is also a knot invariant, which is called the ( p, q ) -cabling of I denoted by I p/q . In this talk, we focus on ( P 0 ) p/q .

Singular knot invariant from a knot invariant A singular knot is an oriented immersed circle in S 3 whose singularities are only transverse double points. We assume that each double point on a singular knot is a rigid vertex. Let v be an invariant of an oriented knot in S 3 , which takes values in Q . Then v can be uniquely extended to a singular knot invariant by the Vassiliev skein relation: v ( K × ) = v ( K + ) − v ( K − ) , where K × is a singular knot with a double point × and K + , K − are singular knots obtained from K × by replacing × by a positive crossing and a negative crossing, respectively.

Vassiliev knot invariants We call v a Vassiliev knot invariant of order d if there exists an integer d such that v ( K >d ) = 0 for any singular knot K >d with more than d double points and v ( K d ) ̸ = 0 for a singular knot K d with d double points. The set of all Vassiliev knot invariants of order ≤ d forms a vector space over Q , which is denoted by V d . There is a filtration V 0 ⊂ V 1 ⊂ V 2 ⊂ · · · ⊂ V d ⊂ · · · in the entire space of Vassiliev knot invariants. Each V d is finite-dimensional. In particular, we have V 0 = V 1 = < 1 > .

Vassiliev knot invariants up to order six Fact [Kanenobu 2001] a 2 i : the 2 i th coefficient of the Alexander-Conway polynomial. P ( d ) 2 i (1) : the d th derivative of P 2 i at t = 1 . ( √− 1) : the d th derivative of F i at a = √− 1 . F ( d ) i V 2 = < 1 , a 2 > . V 3 = < 1 , a 2 , P (3) (1) > . 0 V 4 = < 1 , a 2 , P (3) 2 , a 4 , P (4) (1) , a 2 (1) > . 0 0 V 5 = < 1 , a 2 , P (3) 2 , a 4 , P (4) (1) , a 2 P (3) (1) , P (5) (1) , a 2 (1) , 0 0 0 0 ( √− 1) / √− 1 > . P (1) (1) , F (1) 4 4 V 6 = < 1 , a 2 , P (3) 2 , a 4 , P (4) (1) , a 2 P (3) (1) , P (5) (1) , a 2 (1) , 0 0 0 0 ( √− 1) / √− 1 , a 3 P (1) (1) , F (1) 2 , a 2 a 4 , a 2 P (4) (1) , P (3) (1) 2 , 4 4 0 0 ( √− 1) , F (1) ( √− 1) > . P (6) (1) , P (2) (1) , a 6 , F (2) 0 4 4 5

Cabling for Vassiliev knot invariants Fact [Bar-Natan 1995, Stanford 1994] If v is a Vassiliev knot invariant of order d , then the ( p, q ) -cabling v p/q is also a Vassiliev knot invariant of order ≤ d . Since P ( d ) (1) is a Vassiliev knot invariant of order d , 0 ( P 0 ) ( d ) p/q (1) is a Vassiliev knot invariant of order ≤ d . In this talk, we consider ( P 0 ) ( d ) n/ 1 (1) for 1 ≤ d ≤ 6 and 1 ≤ n ≤ 7 .

Results By using Kodama’s program “KNOT”, we can calculate ( P 0 ) n/ 1 ( K ) with 1 ≤ n ≤ 7 for a knot K with small crossings. Therefore, we obtain the following results. Order ≤ 2 V 2 = < 1 , a 2 > . P (2) (1) = − 8 a 2 0 ( P 0 ) (2) ( P 0 ) (2) 2 / 1 (1) = 4 P (2) 2 / 1 (1) = − 32 a 2 (1) 0 ( P 0 ) (2) ( P 0 ) (2) 3 / 1 (1) = 9 P (2) 3 / 1 (1) = − 72 a 2 (1) 0 ( P 0 ) (2) ( P 0 ) (2) 4 / 1 (1) = 16 P (2) 4 / 1 (1) = − 128 a 2 (1) 0 ( P 0 ) (2) ( P 0 ) (2) 5 / 1 (1) = 25 P (2) 5 / 1 (1) = − 200 a 2 (1) 0 ( P 0 ) (2) ( P 0 ) (2) 6 / 1 (1) = 36 P (2) 6 / 1 (1) = − 288 a 2 (1) 0 ( P 0 ) (2) ( P 0 ) (2) 7 / 1 (1) = 49 P (2) 7 / 1 (1) = − 392 a 2 (1) 0

Proposition ( P 0 ) (2) n/ 1 (1) = n 2 P (2) (1) for any n ( ≥ 1) . 0 Proof. Let ∆ K ( t ) be the normalized Alexander polynomial of a knot K , which satisfies ∆ K ( t ) = ∇ K ( t 1 / 2 − t − 1 / 2 ) . Then we have ∆ (1) K (1) = 0 and ∆ (2) K (1) = 2 a 2 ( K ) . By using satellite formula, we have ∆ K ( n, 1) ( t ) = ∆ K ( t n ) . Therefore, we have � a 2 ( K ( n, 1) ) = (1 / 2)∆ (2) K ( n, 1) (1) = (1 / 2)∆ (2) K ( t n ) � � t =1 K ( t n ) t 2 n − 2 + n ( n − 1)∆ (1) � = (1 / 2)( n 2 ∆ (2) K ( t n ) t n − 2 ) � � t =1 = n 2 a 2 ( K ) . By P (2) ( K ; 1) = − 8 a 2 ( K ) , we have 0 ( P 0 ) (2) n/ 1 ( K ; 1) = P (2) ( K ( n, 1) ; 1) = n 2 P (2) ( K ; 1) . 0 0 □

V 3 = < 1 , a 2 , P (3) (1) > . P (2) Order ≤ 3 (1) = − 8 a 2 . 0 0 ( P 0 ) (3) 2 / 1 (1) = − 48 a 2 + 4 P (3) ( P 0 ) (3) 2 / 1 (1) = 6 P (2) (1) + 4 P (3) (1) (1) 0 0 0 ( P 0 ) (3) 3 / 1 (1) = − 192 a 2 + 9 P (3) ( P 0 ) (3) 3 / 1 (1) = 24 P (2) (1) + 9 P (3) (1) (1) 0 0 0 ( P 0 ) (3) 4 / 1 (1) = − 480 a 2 + 16 P (3) ( P 0 ) (3) 4 / 1 (1) = 60 P (2) (1) + 16 P (3) (1) (1) 0 0 0 ( P 0 ) (3) 5 / 1 (1) = − 960 a 2 + 25 P (3) ( P 0 ) (3) 5 / 1 (1) = 120 P (2) (1) + 25 P (3) (1) (1) 0 0 0 ( P 0 ) (3) 6 / 1 (1) = − 1680 a 2 + 36 P (3) ( P 0 ) (3) 6 / 1 (1) = 210 P (2) (1) + 36 P (3) (1) (1) 0 0 0 ( P 0 ) (3) 7 / 1 (1) = − 2688 a 2 + 49 P (3) ( P 0 ) (3) 7 / 1 (1) = 336 P (2) (1) + 49 P (3) (1) (1) 0 0 0 Question ( P 0 ) (3) n/ 1 (1) = ( n − 1) n ( n + 1) P (2) (1) + n 2 P (3) (1)? 0 0

Order ≤ 4 V 4 = < 1 , a 2 , P (3) 2 , a 4 , P (4) (1) > . P (2) (1) , a 2 (1) = − 8 a 2 . 0 0 0 ( P 0 ) (4) 2 / 1 (1) = − 96 a 2 + 8 P (3) 2 − 768 a 4 + 4 P (4) (1) + 1920 a 2 (1) 0 0 ( P 0 ) (4) 3 / 1 (1) = − 768 a 2 + 32 P (3) (1) + 11520 a 2 2 − 4608 a 4 + 9 P (4) (1) 0 0 ( P 0 ) (4) 4 / 1 (1) = − 2880 a 2 + 80 P (3) 2 − 15360 a 4 + 16 P (4) (1) + 38400 a 2 (1) 0 0 ( P 0 ) (4) 5 / 1 (1) = − 7680 a 2 + 160 P (3) (1) + 96000 a 2 2 − 38400 a 4 + 25 P (4) (1) 0 0 ( P 0 ) (4) 6 / 1 (1) = − 16800 a 2 + 280 P (3) 2 − 80640 a 4 + 36 P (4) (1) + 201600 a 2 (1) 0 0 ( P 0 ) (4) 2 / 1 (1) = 12 P (2) (1) + 8 P (3) (1) + 30 P (2) (1) 2 − 768 a 4 + 4 P (4) (1) 0 0 0 0 ( P 0 ) (4) 3 / 1 (1) = 96 P (2) (1) + 32 P (3) (1) + 180 P (2) (1) 2 − 4608 a 4 + 9 P (4) (1) 0 0 0 0 ( P 0 ) (4) 4 / 1 (1) = 360 P (2) (1) + 80 P (3) (1) + 600 P (2) (1) 2 − 15360 a 4 + 16 P (4) (1) 0 0 0 0 ( P 0 ) (4) 5 / 1 (1) = 960 P (2) (1) + 160 P (3) (1) + 1500 P (2) (1) 2 − 38400 a 4 + 25 P (4) (1) 0 0 0 0 ( P 0 ) (4) 6 / 1 (1) = 2100 P (2) (1) + 280 P (3) (1) + 3150 P (2) (1) 2 − 80640 a 4 + 36 P (4) (1) 0 0 0 0 Question ( P 0 ) (4) n/ 1 (1) = 2( n − 1) 2 n ( n + 1) P (2) (1) 0 + (4 / 3)( n − 1) n ( n + 1) P (3) (1) + (5 / 2)( n − 1) n 2 ( n + 1) P (2) (1) 2 0 0 − 64( n − 1) n 2 ( n + 1) a 4 + n 2 P (4) (1)? 0

We have the following relations: a 4 = (1 / 768)(12 P (2) (1) + 8 P (3) (1) + 30 P (2) (1) 2 0 0 0 + 4 P (4) (1) − ( P 0 ) (4) 2 / 1 (1)) 0 = (1 / 4608)(96 P (2) (1) + 32 P (3) (1) + 180 P (2) (1) 2 0 0 0 + 9 P (4) (1) − ( P 0 ) (4) 3 / 1 (1)) 0 = (1 / 15360)(360 P (2) (1) + 80 P (3) (1) + 600 P (2) (1) 2 0 0 0 + 16 P (4) (1) − ( P 0 ) (4) 4 / 1 (1)) 0 = (1 / 38400)(960 P (2) (1) + 160 P (3) (1) + 1500 P (2) (1) 2 0 0 0 + 25 P (4) (1) − ( P 0 ) (4) 5 / 1 (1)) 0 = (1 / 80640)(2100 P (2) (1) + 280 P (3) (1) + 3150 P (2) (1) 2 0 0 0 + 36 P (4) (1) − ( P 0 ) (4) 6 / 1 (1)) . 0 Therefore, we see that V 4 = < 1 , P (2) (1) , P (3) (1) , P (2) (1) 2 , P (4) (1) , ( P 0 ) (4) n/ 1 (1) > 0 0 0 0 for 2 ≤ n ≤ 6 .