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# Vassiliev knot invariants derived from cable Γ-polynomials

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### Outline of my presentation

The HOMFLYPT and Kauﬀman polynomials P, F and their coeﬃcient polynomials P2i, Fi. We focus on P0(=F0= Γ).

(P0)p/q is the (p, q)-cabling of P0 for coprime integers p, q.

(P0)(d)p/q(1) is the dth derivative of (P0)p/q at t= 1, which is a Vassiliev knot invariant of order d.

In particular, we will show some results about(P0)(d)n/1(1).

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### HOMFLYPT(P)and Kauﬀman(F)polynomials

P(L;t, z)Z[t±1, z±1]andF(L;a, z)Z[a±1, z±1]are invariants for oriented links inS3.

For the trivial knot, we haveP() =F() = 1.

For a skein triple(L+, L, L0)of oriented links, we have t1P(L+)tP(L) =zP(L0).

For a skein quadruple(D+, D, D0, D)of oriented link diagrams, we have

aF(D+) +a1F(D) =z(F(D0) +aF(D)), where=w(D+)w(D)1andw(D+),w(D)are the writhes ofD+,D, respectively.

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### Coeﬃcient polynomials ofPandF

P(L) = (t1z)r+1XXX

i0

P2i(L;t)z2i,

F(L) = (az)r+1XXX

i0

Fi(L;a)zi,

where P2i(L;t)Z[t±1] and Fi(L;a)Z[a±1] are called the 2ith coeﬃcient polynomial of P(L) and

the ith coeﬃcient polynomial of F(L), respectively.

Fact [Lickorish 1988]P0(L;t) =F0(L;

1t1).

P0(L;t) is a Laurent polynomial in t2. Putting t2=x, we call it theΓ-polynomial Γ(L;x)Z[x±1], that is,

Γ(L;t2) =P0(L;t) =F0(L;

−1t1).

In this talk, we use P0 not Γ to apply Kanenobu’s results.

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### 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 H1(∂N(K);Z), where (m, l) is a meridian-longitude pair of K with

lk(Kl) = 0 and lk(Km) = +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 1· · ·σp1)qpw(D).

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### 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 ofI denoted by Ip/q.

In this talk, we focus on(P0)p/q.

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### Singular knot invariant from a knot invariant

A singular knot is an oriented immersed circle in S3 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 S3, which takes values inQ. 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 fromK× by replacing × by a positive crossing and a negative crossing, respectively.

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### Vassiliev knot invariants

We callv 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 ddouble points and v(Kd)̸= 0 for a singular knot Kd with d double points.

The set of all Vassiliev knot invariants of order d forms a vector space over Q, which is denoted byVd.

There is a ﬁltration

V0 ⊂ V1 ⊂ V2⊂ · · · ⊂ Vd⊂ · · ·

in the entire space of Vassiliev knot invariants. EachVd is ﬁnite-dimensional. In particular, we haveV0=V1 =<1> .

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### Vassiliev knot invariants up to order six

Fact [Kanenobu 2001]

a2i: the 2ith coeﬃcient of the Alexander-Conway polynomial.

P2i(d)(1): the dth derivative ofP2i at t= 1.

Fi(d)(

1): the dth derivative of Fi ata=

1.

V2=<1, a2> .

V3=<1, a2, P0(3)(1)> .

V4=<1, a2, P0(3)(1), a22, a4, P0(4)(1)> .

V5=<1, a2, P0(3)(1), a22, a4, P0(4)(1), a2P0(3)(1), P0(5)(1), P4(1)(1), F4(1)(

1)/

1> .

V6=<1, a2, P0(3)(1), a22, a4, P0(4)(1), a2P0(3)(1), P0(5)(1), P4(1)(1), F4(1)(

1)/

1, a32, a2a4, a2P0(4)(1), P0(3)(1)2, P0(6)(1), P4(2)(1), a6, F4(2)(

1), F5(1)(

1)> .

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### Cabling for Vassiliev knot invariants

Fact [Bar-Natan 1995, Stanford 1994]

Ifv is a Vassiliev knot invariant of order d,

then the (p, q)-cabling vp/q is also a Vassiliev knot invariant of orderd.

SinceP0(d)(1) is a Vassiliev knot invariant of order d, (P0)(d)p/q(1) is a Vassiliev knot invariant of order d.

In this talk, we consider (P0)(d)n/1(1) for 1d6 and 1n7.

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### Results

By using Kodama’s program “KNOT”, we can calculate (P0)n/1(K) with 1n7 for a knot K with small crossings. Therefore, we obtain the following results.

Order2 V2 =<1, a2> .

P0(2)(1) =8a2

(P0)(2)2/1(1) =−32a2 (P0)(2)2/1(1) = 4P0(2)(1) (P0)(2)3/1(1) =−72a2 (P0)(2)3/1(1) = 9P0(2)(1) (P0)(2)4/1(1) =128a2 (P0)(2)4/1(1) = 16P0(2)(1) (P0)(2)5/1(1) =200a2 (P0)(2)5/1(1) = 25P0(2)(1) (P0)(2)6/1(1) =288a2 (P0)(2)6/1(1) = 36P0(2)(1) (P0)(2)7/1(1) =−392a2 (P0)(2)7/1(1) = 49P0(2)(1)

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Proposition (P0)(2)n/1(1) =n2P0(2)(1) for any n(1).

Proof. Let K(t) be the normalized Alexander polynomial of a knotK, which satisﬁes K(t) =K(t1/2t1/2).

Then we have (1)K (1) = 0 and (2)K (1) = 2a2(K).

By using satellite formula, we have K(n,1)(t) = ∆K(tn).

Therefore, we have

a2(K(n,1)) = (1/2)∆(2)K(n,1)(1) = (1/2)∆(2)K (tn)

t=1

= (1/2)(n2(2)K (tn)t2n2+n(n1)∆(1)K (tn)tn2)

t=1

=n2a2(K).

By P0(2)(K; 1) =−8a2(K), we have

(P0)(2)n/1(K; 1) =P0(2)(K(n,1); 1) =n2P0(2)(K; 1).

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Order3 V3 =<1, a2, P0(3)(1)> . P0(2)(1) =8a2.

(P0)(3)2/1(1) =−48a2+ 4P0(3)(1) (P0)(3)2/1(1) = 6P0(2)(1) + 4P0(3)(1) (P0)(3)3/1(1) =192a2+ 9P0(3)(1) (P0)(3)3/1(1) = 24P0(2)(1) + 9P0(3)(1) (P0)(3)4/1(1) =480a2+ 16P0(3)(1) (P0)(3)4/1(1) = 60P0(2)(1) + 16P0(3)(1) (P0)(3)5/1(1) =960a2+ 25P0(3)(1) (P0)(3)5/1(1) = 120P0(2)(1) + 25P0(3)(1) (P0)(3)6/1(1) =−1680a2+ 36P0(3)(1) (P0)(3)6/1(1) = 210P0(2)(1) + 36P0(3)(1) (P0)(3)7/1(1) =−2688a2+ 49P0(3)(1) (P0)(3)7/1(1) = 336P0(2)(1) + 49P0(3)(1)

Question

(P0)(3)n/1(1) = (n1)n(n+ 1)P0(2)(1) +n2P0(3)(1)?

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Order4 V4=<1, a2, P0(3)(1), a22, a4, P0(4)(1)> . P0(2)(1) =8a2.

(P0)(4)2/1(1) =96a2+ 8P0(3)(1) + 1920a22768a4+ 4P0(4)(1) (P0)(4)3/1(1) =−768a2+ 32P0(3)(1) + 11520a224608a4+ 9P0(4)(1) (P0)(4)4/1(1) =−2880a2+ 80P0(3)(1) + 38400a2215360a4+ 16P0(4)(1) (P0)(4)5/1(1) =−7680a2+ 160P0(3)(1) + 96000a2238400a4+ 25P0(4)(1) (P0)(4)6/1(1) =16800a2+ 280P0(3)(1) + 201600a2280640a4+ 36P0(4)(1) (P0)(4)2/1(1) = 12P0(2)(1) + 8P0(3)(1) + 30P0(2)(1)2768a4+ 4P0(4)(1) (P0)(4)3/1(1) = 96P0(2)(1) + 32P0(3)(1) + 180P0(2)(1)24608a4+ 9P0(4)(1) (P0)(4)4/1(1) = 360P0(2)(1) + 80P0(3)(1) + 600P0(2)(1)215360a4+ 16P0(4)(1) (P0)(4)5/1(1) = 960P0(2)(1) + 160P0(3)(1) + 1500P0(2)(1)238400a4+ 25P0(4)(1) (P0)(4)6/1(1) = 2100P0(2)(1) + 280P0(3)(1) + 3150P0(2)(1)280640a4+ 36P0(4)(1)

Question

(P0)(4)n/1(1) = 2(n1)2n(n+ 1)P0(2)(1)

+ (4/3)(n1)n(n+ 1)P0(3)(1) + (5/2)(n1)n2(n+ 1)P0(2)(1)2

64(n1)n2(n+ 1)a4+n2P0(4)(1)?

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We have the following relations:

a4

=(1/768)(12P0(2)(1) + 8P0(3)(1) + 30P0(2)(1)2 + 4P0(4)(1)(P0)(4)2/1(1))

=(1/4608)(96P0(2)(1) + 32P0(3)(1) + 180P0(2)(1)2 + 9P0(4)(1)(P0)(4)3/1(1))

=(1/15360)(360P0(2)(1) + 80P0(3)(1) + 600P0(2)(1)2 + 16P0(4)(1)(P0)(4)4/1(1))

=(1/38400)(960P0(2)(1) + 160P0(3)(1) + 1500P0(2)(1)2 + 25P0(4)(1)(P0)(4)5/1(1))

=(1/80640)(2100P0(2)(1) + 280P0(3)(1) + 3150P0(2)(1)2 + 36P0(4)(1)(P0)(4)6/1(1)).

Therefore, we see that

V4=<1, P0(2)(1), P0(3)(1), P0(2)(1)2, P0(4)(1),(P0)(4)n/1(1)>

for2n6.

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Order5 V5=<1, a2, P0(3)(1), a22, a4, P0(4)(1), a2P0(3)(1), P0(5)(1), P4(1)(1), F4(1)(

1)/

1> .

(P0)(5)2/1(1) =320a2+ (460/9)P0(3)(1) + 10240a22+ 3840a4+ (170/9)P0(4)(1)

(2720/3)a2P0(3)(1) + (44/9)P0(5)(1)(24320/3)P4(1)(1)(5120/3)F4(1)(

1)/

1 (P0)(5)

3/1(1) =3840a2+ (1040/3)P0(3)(1) + 96000a22+ 7680a4+ (280/3)P0(4)(1)

5440a2P0(3)(1) + (43/3)P0(5)(1)48640P4(1)(1)10240F4(1)(

1)/

1

(P0)(5)4/1(1) =−19840a2+ (11000/9)P0(3)(1) + 427520a2223040a4+ (2500/9)P0(4)(1)

(54400/3)a2P0(3)(1) + (304/9)P0(5)(1)(486400/3)P4(1)(1)(102400/3)F4(1)(

1)/

1

(P0)(5)5/1(1) =−67840a2+ (28400/9)P0(3)(1) + 1329920a22176640a4+ (5800/9)P0(4)(1)

(136000/3)a2P0(3)(1) + (625/9)P0(5)(1)(1216000/3)P4(1)(1)(256000/3)F4(1)(

1)/

1

(P0)(5)6/1(1) =181440a2+ (20300/3)P0(3)(1) + 3333120a22618240a4+ (3850/3)P0(4)(1)

−95200a2P0(3)(1) + (388/3)P0(5)(1)851200P4(1)(1)179200F4(1)(

−1)/

−1

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We have the following relations:

51072000P4(1)(1) + 10752000F4(1)(

1)/

1714000P0(2)(1)P0(3)(1)

= 9144576252000P0(2)(1) + 6096384322000P0(3)(1)

+ 22861441008000P0(2)(1)2+ 3048192119000P0(4)(1)762048000000(P0)(4)2/1(1) + 30800P0(5)(1)6300(P0)(5)2/1(1)

= 1354752504000P0(2)(1) + 451584364000P0(3)(1)

+ 2540161575000P0(2)(1)2+ 127008098000P0(4)(1)14112000000(P0)(4) 3/1(1) + 15050P0(5)(1)1050(P0)(5)3/1(1)

= 1234518541200P0(2)(1) + 274337665000P0(3)(1)

+ 2057531704200P0(2)(1)2+ 54867543500P0(4)(1)3429216000(P0)(4)4/1(1) + 10640P0(5)(1)315(P0)(5)

4/1(1)

= 12383951670720P0(2)(1) + 2063992164640P0(3)(1)

+ 19349925434280P0(2)(1)2+ 322498794800P0(4)(1)22256640(P0)(4)5/1(1) + 8750P0(5)(1)126(P0)(5)5/1(1)

= 35833191760800P0(2)(1) + 4777759126000P0(3)(1)

+ 53749788724800P0(2)(1)2+ 614283341000P0(4)(1)17063424000(P0)(4) 6/1(1) + 7760P0(5)(1)60(P0)(5)6/1(1).

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We see that V5

=<1, P0(2)(1), P0(3)(1), P0(2)(1)2, P0(4)(1),(P0)(4)l/1(1), P0(2)(1)P0(3)(1), P0(5)(1),P4(1)(1),F4(1)(

1)/

1>

=<1, P0(2)(1), P0(3)(1), P0(2)(1)2, P0(4)(1),(P0)(4)l/1(1), P0(2)(1)P0(3)(1), P0(5)(1),(P0)(5)m/1(1), F4(1)(

1)/

1>

=<1, P0(2)(1), P0(3)(1), P0(2)(1)2, P0(4)(1),(P0)(4)l/1(1), P0(2)(1)P0(3)(1), P0(5)(1), P4(1)(1),(P0)(5)n/1(1)>

for2l, m, n6.

1, P0(2)(1), P0(3)(1), P0(2)(1)2, P0(4)(1),(P0)(4)l/1(1),

P0(2)(1)P0(3)(1), P0(5)(1),(P0)(5)m/1(1),(P0)(5)n/1(1)are not linearly independent for2l, m, n6, m̸=n.

QuestionV5 is determined by(P0)p/q?

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Order6 V6=<1, a2, P0(3)(1), a22, a4, P0(4)(1), a2P0(3)(1), P0(5)(1), P4(1)(1), F4(1)(

1)/

1, a32, a2a4, a2P0(4)(1), P0(3)(1)2, P0(6)(1), P4(2)(1), a6, F4(2)(

1), F5(1)(

1)> .

(P0)(6)2/1(1) =−160a2+ (1360/3)P0(3)(1) + 21760a2269120a4+ (950/3)P0(4)(1)

6080a2P0(3)(1) + (232/3)P0(5)(1) + 25600P4(1)(1) + 23040F4(1)(

1)/

1

15360a2a41520a2P0(4)(1) + (380/3)P0(3)(1)2+ (80/9)P0(6)(1)47360P4(2)(1)

404480a6+ 7680F4(2)(

1)40960F5(1)(

1)

(P0)(6)3/1(1) =11520a2+ (11840/3)P0(3)(1) + 526080a22368640a4+ (6520/3)P0(4)(1)

−56320a2P0(3)(1) + (1384/3)P0(5)(1)40960P4(1)(1) + 97280F4(1)(

−1)/

−11520640a32 +1152000a2a49120a2P0(4)(1) + 760P0(3)(1)2+ (115/3)P0(6)(1)284160P4(2)(1)

−2565120a6+ 46080F4(2)(−1)245760F5(1)(−1)

(P0)(6)4/1(1) =102080a2+ 17280P0(3)(1) + 3668480a221359360a4+ 8020P0(4)(1)

249600a2P0(3)(1) + (4696/3)P0(5)(1)752640P4(1)(1) + 194560F4(1)(

1)/

112165120a32 +9646080a2a430400a2P0(4)(1) + (7600/3)P0(3)(1)2+ (1024/9)P0(6)(1)947200P4(2)(1)

9195520a6+ 153600F4(2)(

1)819200F5(1)(

1) (P0)(6)

5/1(1) =488960a2+ (160640/3)P0(3)(1) + 15430400a224377600a4+ (65560/3)P0(4)(1)

774400a2P0(3)(1) + (12056/3)P0(5)(1)3389440P4(1)(1) + 168960F4(1)(

1)/

153222400a32 +42777600a2a476000a2P0(4)(1) + (19000/3)P0(3)(1)2+ (2425/9)P0(6)(1)2368000P4(2)(1)

25062400a6+ 384000F4(2)(

1)2048000F5(1)(

1)

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