COEFFICIENTS OF HOMFLY AND ARE NOT FINITE TYPE INVARIANTS

GYO TAEK JIN AND JUNG HOON LEE

Abstract. We show that the integer-valued invariants appearing as the nontrivial coefficients of the HOMFLY polynomial, the Kauffman polynomial and the Q-polynomial are not of finite type.

1. Introduction A numerical V can be extended to have values on singular 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 differ as shown in Figure 1. V is said to be of finite type or a finite 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 coefficient of the Alexander- Conway polynomial is a finite type invariant [1, 6]. Theorem 1 (Bar-Natan). Let K be a knot and let

2 4 2m ∇K (z) = 1 + a2(K)z + a4(K)z + ··· + a2m(K)z + ··· be the Alexander-Conway polynomial of K. Then a2m is a finite type invariant of order 2m for any positive integer m. The coefficients of the Taylor expansion of any quantum polynomial invariant of knots after a suitable change of variable are all finite type invariants [2]. For the we have

Date: October 17, 2000 (561). 2000 Mathematics Subject Classification. 57M27. Key words and phrases. knot, finite type invariant, homfly polynomial, Kauffman 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: X∞ 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 coefficients of the original Jones polynomial are all non-finite type in- variants. Theorem 3 (Zhu). Let K be a knot and let X∞ n JK (t) = bn(K)t n=−∞ be the Jones polynomial of K. Then bn is not a finite type invariant for any n.

2. HOMFLY polynomial

The HOMFLY polynomial PK (v, z) of an oriented knot or 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 differ 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 finite type invarianat if i, j are even integers with j ≥ 0, and cij = 0 for any knot otherwise.

Let Tn and Kn denote the knot of type (2, 2n + 1) and the of type (2, 2n + 1) together with 2n singular crossings, respectively, as shown in Figure 3. Let E denote the figure 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 K¯ 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 PK¯ (v, z) = PK (−v , z) with the above two cases, we obtain ¯|l| ¯ |l| c2l,2m(T1 ]Kn) = c|2l|,2m(T1 ]Kn) 6= 0 if |l| ≥ m ≥ 0, and ¯m m−|l| ¯ m m−|l| c2l,2m(T1 ]E ]Kn) = c|2l|,2m(T1 ]E ]Kn) 6= 0 if m > |l|. Consequently, c2l,2m is not a finite type invariant for any l and for any m ≥ 0. 4 G.T. JIN AND J.H. LEE

3. Kauffman polynomial

The Kauffman polynomial FK (a, x) of an oriented knot or link K is defined by −w(D) FK (a, x) = a ΛD(a, x) where D is a diagram of K, w(D) its and ΛD(a, x) the polynomial determined by the following conditions:

(K1) ΛO(a, x) = 1 where O is the trivial knot diagram. (K2) For any four diagrams D+, D−, D0 and D∞ which are identical outside a small disk in which they differ as shown in Figure 4, we have the relation:

ΛD+ (a, x) + ΛD− (a, x) = x(ΛD0 (a, x) + ΛD∞ (a, x))

@

D+ D− D0 D∞

Figure 4

(K3) For any three diagrams D+, D and D− which are identical outside a small disk in which they differ as shown in Figure 5, we have the relation: −1 a ΛD+ (a, x) = ΛD(a, x) = a ΛD− (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 finite 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 ¯l ¯ ¯l ¯ (a) d2l,2m(T1]Tn) = d2l+1,2m+1(T1]Tn) = 0, for l ≥ 0. ¯l+1 ¯ (b) d2l+1,2l+1(T1 ]E]Tn) = 0, for l ≥ 0. ¯m m−l ¯ ¯m m−l ¯ (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 Kauffman polynomial FK (a, x) of K. We have 2 2 2 2 3 F (T¯1) = 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 (T¯n) is 2n. The ¯l ¯ ¯ l ¯ minimum a-degree of F (T1]Tn) = F (T1) · F (Tn) is 2l + 2n > 2l + 1. This proves ¯l+1 ¯ ¯ l+1 ¯ the part (a). The minimum a-degrees of F (T1 ]E]Tn) = F (T1) · F (E) · F (Tn) ¯m m−l ¯ ¯ m m−l ¯ 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 finite type by finding singular knots with arbitrarily many double points making d2l,2m 6= 0. ¯l ¯ If l ≥ m ≥ 0, we consider the singular knot T1]Kn. Since the coefficient of 2l 2m ¯l ¯ 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 (T¯l]K¯ ) = (−1)p d (T¯l]T¯ ) 2l,2m 1 n p 2l,2m 1 2n−p p=0 µ ¶ l = d (T¯l) = (−2)l−m 6= 0 2l,2m 1 m for any n ≥ 1. ¯m m−l ¯ If m > l ≥ 0, we consider the singular knot T1 ]E ]Tn. Since the coeffi- 2l 2m ¯m m−l ¯ m m−l 2m cient of a x in F (T1 ]E ) = F (T1) · F (E) is equal to that of x in (−2 + x2)m(−1 + x2)m−l and by Lemma 7(c), we have µ ¶ X2n 2n d (T¯m]Em−l]K¯ ) = (−1)p d (T¯m]Em−l]T¯ ) 2l,2m 1 n p 2l,2m 1 2n−p p=0 ¯m m−l = d2l,2m(T1 ]E ) µ ¶µ ¶ Xm m m − l = (−1)m−l 2m−p 6= 0 p m − p p=0 for any n ≥ 1. −1 Let l < 0. Combining the fact FK¯ (a, x) = FK (a , x) with the above, we obtain |l| ¯|l| ¯ d2l,2m(T1 ]Kn) = d|2l|,2m(T1 ]Kn) 6= 0 if |l| ≥ m ≥ 0 and

d2l,2m(= −1 When l < 0, using the fact FK¯ (a, x) = FK (a , x), we have

d2l,2m(K) = d−2l,2m(K¯ ).

Applying the above arguments to d−2l,2m with the mirror images of the knots used, we complete the proof of Part 1.

Part 2. We show that d2l+1,2m+1 is not of finite type in a similar manner. ¯l ¯ If l > m ≥ 0, we consider the singular knot T1]Kn. Since the coefficient of 2l+1 2m+1 ¯l ¯ l 2m+1 2 l a x in F (T1) = F (T1) is equal to that of ax in (−2 + x + ax) and by Lemma 7(a), we have 6 G.T. JIN AND J.H. LEE

µ ¶ X2n 2n d (T¯l]K¯ ) = (−1)p d (T¯l]T¯ ) 2l+1,2m+1 1 n p 2l+1,2m+1 1 2n−p p=0 µ ¶ l − 1 = d (T¯l) = l (−2)l−m−1 6= 0 2l+1,2m+1 1 m for any n ≥ 1. ¯l+1 ¯ If l = m ≥ 0, we consider the singular knot T1 ]E]Kn. Since the coeffi- 2l+1 2l+1 ¯l+1 ¯ l+1 2l+1 cient of a x in F (T1 ]E) = F (T1) · F (E) is equal to that of ax in (−2 + x2 + ax)l+1(−1 + x2)(1 + ax) and by Lemma 7(b), we have µ ¶ X2n 2n d (T¯l+1]E]K¯ ) = (−1)p d (T¯l+1]E]T¯ ) 2l+1,2l+1 1 n p 2l+1,2l+1 1 2n−p p=0 ¯l+1 = d2l+1,2l+1(T1 ]E) = l + 1 6= 0 for any n ≥ 1. ¯m m−l ¯ If m > l ≥ 0, we consider the singular knot T1 ]E ]Kn. We first compute ¯m m−l 2l+1 d2l+1,2m+1(T1 ]E ). Let A ∼ B mean that the coefficients of a in A and B are the same. Then we have ¯m m−l F (T1 ]E ) ∼ a2l(−2 + x2 + ax)m(−1 + x2)m−l(1 + ax)m−l ∼ a2l(−1 + x2)m−l((−2 + x2)m + m(−2 + x2)m−1ax)(1 + (m − l)ax) ∼ a2l(−1 + x2)m−l(−2 + x2)m−1(−2 + x2 + max)(1 + (m − l)ax) ∼ a2l+1x(−1 + x2)m−l(−2 + x2)m−1(2l − m + (m − l)x2) Therefore the coefficient of x2m in (−1 + x2)m−l(−2 + x2)m−1(2l − m + (m − l)x2) ¯m m−l is equal to d2l+1,2m+1(T1 ]E ), hence ¯m m−l d2l+1,2m+1(T1 ]E ) µ ¶ µ ¶ mX−l m − l m − 1 = (2l − m) (−1)m−l−p (−2)p−1 p m − p p=1 µ ¶ µ ¶ mX−l m − l m − 1 + (m − l) (−1)m−l−p (−2)p p m − p − 1 p=0 µ ¶ · µ ¶ µ ¶¸ mX−l m − l m − 1 m − 1 = (−1)m−l 2p−1 2(m − l) − (2l − m) . p m − p − 1 m − p p=0 It is not hard to see that the quantity in the square bracket of the last formula is always positive. Therefore, by Lemma 7(c), we have µ ¶ X2n 2n d (T¯m]Em−l]K¯ ) = (−1)p d (T¯m]Em−l]T¯ ) 2l+1,2m+1 1 n p 2l+1,2m+1 1 2n−p p=0 ¯m m−l = d2l+1,2m+1(T1 ]E ) 6= 0. If l < 0, we proceed as at the end of Part 1. This completes Part 2. COEFFICIENTS OF HOMFLY POLYNOMIAL AND KAUFFMAN POLYNOMIAL 7

4. Q-polynomial

The Q-polinomial QK (x) of an unoriented knot or link K is determined by the following two conditions:

(Q1) QO(x) = 1 where O is the trivial knot diagram. (Q2) For any four diagrams D+, D−, D0 and D∞ which are identical outside a small disk in which they differ as shown in Figure 4, we have the relation:

QD+ (x) + QD− (x) = x(QD0 (x) + QD∞ (x)) By an induction on the number of crossings, we know that the lowest term of QL(x) has degree at least 1 − c(L) for any knot or link L, where c(L) is the number of components of L. The highest degree term has degree smaller than the crossing number of L [5]. Theorem 8. Let K be a knot and let X i QK (x) = ei(K)x . i≥0

Then, for any nonnegative integer i, ei is not a finite type invariant.

Lemma 9. Let Qn(x) denote the Q-polynomial of the diagram

k with n crossings and let qn,k denote the coefficient of x in Qn(x). Then, for n ≥ 1, n (a) qn,k 6= 0 if and only if −(1 + (−1) )/2 ≤ k ≤ n − 1. (b) If qn,k > 0 then n − k ≡ 1 or 2 (mod 4). (c) If qn,k < 0 then n − k ≡ 0 or 3 (mod 4). ∞ Proof. The {Qn(x)}n=1 satisfy the recurrence relation: 1. Q1(x) = 1. −1 2. Q2(x) = −2x + 1 + 2x. 3. Qn(x) = x Qn−1(x) − Qn−2 + x, for n ≥ 3. We leave the detail to the reader as an exercise.

Let Tn and Kn be as in Figure 3 with the orientation ignored. Then µ ¶ µ ¶ X2n 2n X2n 2n e (K ) = (−1)p e (T ) = (−1)p q i n p i 2n−p p 4n−2p+1,i p=0 p=0 If 0 ≤ i ≤ 4n, by Lemma 9, we know that there do exist nonzero terms all of which have the same sign. Therefore, for any i ≥ 0, ei(Kn) 6= 0 for all sufficiently large n, and hence ei is not of finite type. This proves Theorem 8.

5. Remarks By Thorem 3 and Theorem 2, we have 1 X∞ u = knb . n n! k k=−∞ 8 G.T. JIN AND J.H. LEE

By Theorem 1 and Theorem 4, we have X∞ a2n = c2k,2n k=−∞ since PK (1, z) = ∇K (z) for any knot K. These are examples of finite type invariants which are infinite linear combinations of non-finite type invariants. Question 1. Are the coefficients of Jones polynomial, HOMFLY polynomial, Kauff- man polynomial and the Q-polynomial linearly independent in the sense that no nontrivial finite linear combination of them is a finite type invariant?

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Department of Mathematics, Korea Advanced Institute of Science and Technology, Taejon 305-701 Korea E-mail address: [email protected]

Department of Mathematics, Korea Advanced Institute of Science and Technology, Taejon 305-701 Korea E-mail address: [email protected]