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.