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Coefficients of Homfly Polynomial and Kauffman Polynomial Are Not Finite Type Invariants

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Coefficients of Homfly Polynomial and Kauffman Polynomial Are Not Finite Type Invariants COEFFICIENTS OF HOMFLY POLYNOMIAL AND KAUFFMAN POLYNOMIAL ARE NOT FINITE TYPE INVARIANTS GYO TAEK JIN AND JUNG HOON LEE Abstract. We show that the integer-valued knot invariants appearing as the nontrivial coe±cients of the HOMFLY polynomial, the Kau®man polynomial and the Q-polynomial are not of ¯nite type. 1. Introduction A numerical knot invariant V can be extended to have values on singular knots via the recurrence relation V (K£) = V (K+) ¡ V (K¡) where K£, K+ and K¡ are singular knots which are identical outside a small ball in which they di®er as shown in Figure 1. V is said to be of ¯nite type or a ¯nite type invariant if there is an integer m such that V vanishes for all singular knots with more than m singular double points. If m is the smallest such integer, V is said to be an invariant of order m. q - - - ¡@- @- ¡- K£ K+ K¡ Figure 1 As the following proposition states, every nontrivial coe±cient of the Alexander- Conway polynomial is a ¯nite type invariant [1, 6]. Theorem 1 (Bar-Natan). Let K be a knot and let 2 4 2m rK (z) = 1 + a2(K)z + a4(K)z + ¢ ¢ ¢ + a2m(K)z + ¢ ¢ ¢ be the Alexander-Conway polynomial of K. Then a2m is a ¯nite type invariant of order 2m for any positive integer m. The coe±cients of the Taylor expansion of any quantum polynomial invariant of knots after a suitable change of variable are all ¯nite type invariants [2]. For the Jones polynomial we have Date: October 17, 2000 (561). 2000 Mathematics Subject Classi¯cation. 57M27. Key words and phrases. knot, ¯nite type invariant, homfly polynomial, Kau®man polynomial. This work was supported by Brain Korea 21 Project. 1 2 G.T. JIN AND J.H. LEE Theorem 2 (Birman-Lin). Let K be a knot and let JK (t) be its Jones polynomial. x Let UK (x) be obtained from JK (t) by replacing the variable t by e . Express UK (x) as a power series in x: X1 n UK (x) = un(K)x : n=0 Then u0 ´ 1 and, for each n ¸ 1, un is an invariant of order (at most) n. But the coe±cients of the original Jones polynomial are all non-¯nite type in- variants. Theorem 3 (Zhu). Let K be a knot and let X1 n JK (t) = bn(K)t n=¡1 be the Jones polynomial of K. Then bn is not a ¯nite type invariant for any n. 2. HOMFLY polynomial The HOMFLY polynomial PK (v; z) of an oriented knot or link K is determined by the following two conditions: (H1) PO(v; z) = 1, where O is the trivial knot. (H2) For three knots or links K+, K¡ and K0 which are identical outside a small ball in which they di®er as shown in Figure 2, we have the relation: ¡1 v PK+ (v; z) ¡ vPK¡ (v; z) = zPK0 (v; z) - - - @- ¡- - K+ K¡ K0 Figure 2 Theorem 4. Let K be a knot and let X i j PK (v; z) = cij (K)v z : Then, cij is not a ¯nite type invarianat if i, j are even integers with j ¸ 0, and cij = 0 for any knot otherwise. Let Tn and Kn denote the torus knot of type (2; 2n + 1) and the torus knot of type (2; 2n + 1) together with 2n singular crossings, respectively, as shown in Figure 3. Let E denote the ¯gure eight knot. For a knot K, let Kl denote the iterated connected sum K] ¢ ¢ ¢ ]K of l copies of K. For a knot or a singular knot K, let K¹ denote the mirror image of K. To simplify notations, let P (K) denote the HOMFLY polynomial PK (v; z) of K. Lemma 5. Let n ¸ 1. Then l (a) c2l;2m(T1]Tn) = 0, for l ¸ 0. m m¡l (b) c2l;2m(T1 ]E ]Tn) = 0, for m > l ¸ 0. COEFFICIENTS OF HOMFLY POLYNOMIAL AND KAUFFMAN POLYNOMIAL 3 - 2n + 1 - 2n + 1 2n - - q q Tn Kn Figure 3 Proof. We have 2 2 2 P (T1) = v (2 + z ¡ v ) P (E) = v¡2(1 ¡ (1 + z2)v2 + v4) and inductively, we may show that the minimum v-degree of P (Tn) is 2n. Therefore l l m m¡l the minimum v-degrees of P (T1]Tn) = P (T1) ¢ P (Tn) and P (T1 ]E ]Tn) = m m¡l P (T1) ¢ P (E) ¢ P (Tn) are both 2l + 2n > 2l. This completes the proof. Proof of Theorem 4. It is an elementary result that cij(K) = 0 for any knot K when j < 0 or at least one of i and j is an odd integer. Now we show that there exist singular knots with arbitrarily many singular double points on which c2l;2m with m ¸ 0 does not vanish. When l ¸ 0, we have two cases: Case 1. l ¸ m ¸ 0 : If we resolve any p singular crossings of Kn into negative crossings and the rest 2n ¡ p into positive ones, we obtain T2n¡p. By Lemma 5(a), l 2l 2 2 l and since P (T1) = v (2 + z ¡ v ) , we have µ ¶ X2n 2n c (T l]K ) = (¡1)p c (T l]T ) 2l;2m 1 n p 2l;2m 1 2n¡p p=0 µ ¶ l = c (T l) = 2l¡m 6= 0 2l;2m 1 m for any n ¸ 1: Case 2. m > l ¸ 0 : By Lemma 5(b), and since m m¡l 2l 2 2 m 2 2 4 m¡l P (T1 ]E ) = v (2 ¡ v + z ) (1 ¡ (1 + z )v + v ) we have µ ¶ X2n 2n c (T m]Em¡l]K ) = (¡1)p c (T m]Em¡l]T ) 2l;2m 1 n p 2l;2m 1 2n¡p p=0 m m¡l = c2l;2m(T1 ]E ) = 1 6= 0 for any n ¸ 1: ¡1 Let l < 0. Combining the fact PK¹ (v; z) = PK (¡v ; z) with the above two cases, we obtain ¹jlj ¹ jlj c2l;2m(T1 ]Kn) = cj2lj;2m(T1 ]Kn) 6= 0 if jlj ¸ m ¸ 0, and ¹m m¡jlj ¹ m m¡jlj c2l;2m(T1 ]E ]Kn) = cj2lj;2m(T1 ]E ]Kn) 6= 0 if m > jlj. Consequently, c2l;2m is not a ¯nite type invariant for any l and for any m ¸ 0. 4 G.T. JIN AND J.H. LEE 3. Kauffman polynomial The Kau®man polynomial FK (a; x) of an oriented knot or link K is de¯ned by ¡w(D) FK (a; x) = a ¤D(a; x) where D is a diagram of K, w(D) its writhe and ¤D(a; x) the polynomial determined by the following conditions: (K1) ¤O(a; x) = 1 where O is the trivial knot diagram. (K2) For any four diagrams D+, D¡, D0 and D1 which are identical outside a small disk in which they di®er as shown in Figure 4, we have the relation: ¤D+ (a; x) + ¤D¡ (a; x) = x(¤D0 (a; x) + ¤D1 (a; x)) @ ¡ D+ D¡ D0 D1 Figure 4 (K3) For any three diagrams D+, D and D¡ which are identical outside a small disk in which they di®er as shown in Figure 5, we have the relation: ¡1 a ¤D+ (a; x) = ¤D(a; x) = a ¤D¡ (a; x) @ ¡ D+ D D¡ Figure 5 Theorem 6. Let K be a knot and let X i j FK (a; x) = dij (K)a x : Then, dij is not a ¯nite type invariant if i, j are integers with i + j even and j ¸ 0, and dij = for any knot otherwise. Lemma 7. Let n ¸ 1. Then ¹l ¹ ¹l ¹ (a) d2l;2m(T1]Tn) = d2l+1;2m+1(T1]Tn) = 0, for l ¸ 0. ¹l+1 ¹ (b) d2l+1;2l+1(T1 ]E]Tn) = 0, for l ¸ 0. ¹m m¡l ¹ ¹m m¡l ¹ (c) d2l;2m(T1 ]E ]Tn) = d2l+1;2m+1(T1 ]E ]Tn) = 0, for m > l ¸ 0. Proof. To simplify notations, let F (K) denote the Kau®man polynomial FK (a; x) of K. We have 2 2 2 2 3 F (T¹1) = a (¡2 + x + ax + a (¡1 + x ) + a x) F (E) = a¡2((1 + ax)(¡1 + x2) + a2(¡1 + 2x2) + a3(¡x + x3) + a4(¡1 + x2)) COEFFICIENTS OF HOMFLY POLYNOMIAL AND KAUFFMAN POLYNOMIAL 5 and inductively, we may show that the minimum a-degree of F (T¹n) is 2n. The ¹l ¹ ¹ l ¹ minimum a-degree of F (T1]Tn) = F (T1) ¢ F (Tn) is 2l + 2n > 2l + 1. This proves ¹l+1 ¹ ¹ l+1 ¹ the part (a). The minimum a-degrees of F (T1 ]E]Tn) = F (T1) ¢ F (E) ¢ F (Tn) ¹m m¡l ¹ ¹ m m¡l ¹ and F (T1 ]E ]Tn) = F (T1) ¢ F (E) ¢ F (Tn) are both 2l + 2n > 2l + 1. This proves the parts (b) and (c). Proof of Theorem 6. It is an elementary result that dij(K) = 0 for any knot K when j < 0 or i + j is odd. Part 1. We show that d2l;2m is not of ¯nite type by ¯nding singular knots with arbitrarily many double points making d2l;2m 6= 0. ¹l ¹ If l ¸ m ¸ 0, we consider the singular knot T1]Kn. Since the coe±cient of 2l 2m ¹l ¹ l 2m 2 l a x in F (T1) = F (T1) is equal to that of x in (¡2+x ) and by Lemma 7(a), we have µ ¶ X2n 2n d (T¹l]K¹ ) = (¡1)p d (T¹l]T¹ ) 2l;2m 1 n p 2l;2m 1 2n¡p p=0 µ ¶ l = d (T¹l) = (¡2)l¡m 6= 0 2l;2m 1 m for any n ¸ 1.
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