A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS

Transcription

1 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS K. L. PETERSEN Abstract. The Jk, l knots, often called the double twist knots, are a subclass of two-bridge knots which contains the twist knots. We show that the A-polynomial of these knots can be determined by an explicit resultant. We present this resultant in two different ways. We determine a recursive definition for the A-polynomials of the J4, 2n and J5, 2n knots, and for the canonical component of the A-polynomials of the J2n, 2n knots. Our work also recovers the A-polynomials of the J1, 2n knots, and the recursive formulas for the A-polynomials of the A2, 2n and A3, 2n knots as computed by Hoste and Shanahan. Contents 1. Introduction 2 2. The A-polynomial 4 3. The Jk, l knots 5 4. Representations Reducible Representations Irreducible Representations The Longitude Proof of Theorem Resultants A Special Resultant Proof of Theorem Explicit computations for Jk, 2n knots for small k J1, 2n, torus knots J2, 2n, the twist knots J3, 2n J4, 2n J5, 2n J2m, 2m Appendix A4, 2n terms A5, 2n terms A2m, 2m terms Acknowledgements 32 References 32 1

2 2 K. L. PETERSEN 1. Introduction The A-polynomial of a 3-manifold M 3 with a single torus cusp was introduced in [4]. It is a two variable polynomial, usually written in terms of the variables M and L, which encodes how eigenvalues of a fixed meridian and longitude are related under representations from π 1 M 3 into SL 2 C. This polynomial is closely related to the SL 2 C character variety of M 3, and for hyperbolic manifolds it encodes information about the deformation of the hyperbolic structure of M 3, the existence of non-hyberbolic fillings of M 3, and can be used to determine boundary slopes of essential surfaces in M 3. Specifically, the boundary slopes of the Newton polygon of the A-polynomial are the boundary slopes detected by the SL 2 C character variety [4], and the Newton polygon is dual to the fundamental polygon of the Culler-Shalen seminorm [7]. This seminorm can be used to classify finite and exceptional surgeries of M 3 [1]. k l Figure 1. The knot Jk, l This polynomial has proven difficult to compute; data has been collected for most knots up to nine crossings see Knot Info [3] and calculations by Marc Culler [6], and a few families of knot complements in S 3 have proven amenable to computations. Tamura and Yokota [13] computed the A- polynomials of the 2, 3, 3 + 2n pretzel knots. Garoufalidis and Mattman [9] studied these A-polynomials, showing that they satisfy a specific type of linear recurrance relation. Hoste and Shanahan [10] established a recursively defined formula for the A-polynomials of the twist knots and the J3, 2n knots. This paper shows that the A-polynomial of the Jk, l knots can be computed as an explicit resultant, and recovers the work of Hoste and Shanahan. We also recursively determine the A-polynomial for the J4, 2n and J5, 2n knots, and the canonical component of the A-polynomial for the J2n, 2n knots.

3 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 3 We consider the two-bridge knots Jk, l as described in Figure 1 where k and l are integers denoting the number of half twists in the labeled boxes; positive numbers correspond to right-handed twists and negative numbers correspond to left-handed twists. Such a projection determines a knot if kl is even, and we can reduce to considering the Jk, 2n knots as Jk, l = Jl, k. The knot J k, l is the mirror image of the knot Jk, l, and as a result the A-polynomial of J k, l is the A-polynomial of Jk, l with each M replaced by a M 1 [5]. Therefore we may assume k or l is positive. As discussed in 3 the Jk, l knots are a particularly attractive family as the fundamental groups of their complements have a relatively simple form. In Section 5.1 we show that the contribution to the A-polynomial from reducible representations is the term L 1. Therefore our main theorems focus on the term of the A-polynomials corresponding to irreducible representations. We write Ak, 2n to denote the contribution of factors of the A-polynomial corresponding to irreducible representations. This is welldefined up to multiplication by elements in Q and by powers of M and L. Our first main theorem is the following, where the polynomials F k,n and G k,n are defined in Definition 4.6 and Theorem 1.1. Assume k 1, 0, 1 and n 0. common vanishing set of F k,n r and G k,n r. Then Ak, 2n is the The A-polynomial can be determined by the resultant of these two polynomials, eliminating the variable r. This can be done, for example, using the Sylvester matrix and has been implemented in many computer algebra programs. The degree of F k,n as function of r is roughly 1 4 n k2 and the degree of G k,n is roughly 1 2 k. The next theorem demonstrates that the resultant can be computed using the polynomial H k,n, of degree about 2 n, which is defined in Definition 6.8, in place of F k,n. This resultant will differ from the resultant of F k,n and G k,n by factors of β = M 2 +l and γ = M 2 1l 1. As in Definition 3.3, l = L if k is even and l = LM 4n if k is odd. Theorem 1.2. Let ɛ = 1 if k is positive and 0 if k is negative, let ɛ = 1 if k is even and 0 if k is odd, and let ɛ = 1 ɛ1 ɛ. For k 1, 0, 1 and n 0 we have Ak, 2nγ m +ɛ n ɛ nk < 0 ResH k,n r, G k,n r = Ak, 2nβγ m +ɛ n ɛ n, k > 0 Ak, 2nδγ m +ɛ n ɛ n, k < 0 Theorem 1.1 is proven in Section 4.4 and Theorem 1.2 is proven in Section 6. In addition, we perform explicit computations in some special cases. If k = 0, then Jk, l is the unknot and the A-polynomial is L 1. The knots J±1, 2n are the torus knots whose A-polynomials are well known [4]. We present their A-polynomials in Theorem 7.1 for completeness. In Theorem 7.2 and Theorem 7.3 we recover the recursive formulas of Hoste

4 4 K. L. PETERSEN and Shanahan [10] for the A-polynomials of the twist knots the J±2, 2n knots and the J±3, 2n knots. We also compute recursive formulas for the A-polynomials of the J±4, 2n knots and the J±5, 2n knots. These are given in Theorem 7.5 and Theorem 7.7, respectively. Finally, we consider the J2n, 2n knots. These knots have an additional symmetry that the other Jk, l knots do not have, seen by flipping the corresponding four-plat upside down. This symmetry effectively factors the representation variety see [11] and this factorization can be seen on the level of the A-polynomial as well. A canonical component of the A-polynomial is an irreducible polynomial which contains the image of the discrete and faithful representation. We determine a recursive formula for the canonical component for these knots, given in Theorem The A-polynomial We follow the construction of the A-polynomial given in [5]. We consider a knot K in S 3, and let µ be a natural oriented meridian of K and λ a natural oriented longitude. Let Γ be the fundamental group of the complement of the knot K in S 3. For an oriented loop α S 3 K we write [α] Γ to be a base pointed homotopy class. We define R U to be the subset of the affine algebraic variety R = HomΓ, SL 2 C consisting of all representations ρ such that ρ[µ] and ρ[λ] are upper triangular. The set R U is an affine algebraic variety as well, since it simply has two additional equations specifying that these 2, 1 entries equal zero. Define the eigenvalue map ξ = ξ µ ξ λ : R U C 2 by setting ξρ = e µ, e λ where e µ is an eigenvalue of ρ[µ], and e λ is an eigenvalue of p[λ], both chosen consistently. We let M be the 1, 1 entry of ρ[µ] and L be the 1, 1 entry of ρ[λ]. We will use an alternate longitude, denoted by l, which we define below. If C is an algebraic component of R U, then the Zariski closure of ξc, ξc, is an algebraic subset of C 2. If ξc is a curve, then there is a polynomial that defines ξc which is unique up to constant multiples. The A-polynomial is defined as the product of all such polynomials. This polynomial may be taken to have integral coefficients with content zero, so it is well-defined up to sign. See [4]. There are an infinite number of abelian representations of Γ into SL 2 C. In such a representation every element of the commutator subgroup, including the longitude, is sent to the identity matrix. Therefore, L 1 is always a factor of the A-polynomial. This factor is often ignored in the definition of the A-polynomial. For the Jk, l knots, we will show that all reducible representations correspond only to the factor L 1, and we will usually omit this factor except when referring to the unknot.

5 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 5 3. The Jk, l knots There are many relations amongst the Jk, l knots. The knot Jk, l is ambient isotopic to the knot Jl, k, so we will consider the knots Jk, 2n as if kl is odd then Jk, l is a two component link. The twist knots are the knots J±2, l. The figure-eight and the trefoil are J2, 2 and J2, 2, respectively. Also, J k, l is the mirror image of the knot Jk, l. The A-polynomial of a knot and its reflection differ only by replacing M with M 1 [5], so we may assume that k or l is positive. The Jk, l knots are hyperbolic unless k or l is less than 2 or k = l = ±2. We turn to the fundamental group of S 3 Jk, l. See [10, 11]. Proposition 3.1. The fundamental group of the complement of the knot Jk, 2n in S 3 is isomorphic to the group Γk, 2n = a, b : awk n = wn k b where { ab 1 m a 1 b m if k = 2m w k = ab 1 m aba 1 b m if k = 2m + 1. Among the relations mentioned above, the knot Jk, l is ambient isotopic to Jl, k, so the corresponding groups are isomorphic, but the above presentations are different. For a word v Γk, 2n written in powers of a and b, let v refer to the word obtained by reading v backwards. Definition 3.2. Let ɛv to be the exponent sum of v, written as a word in a and b, and let ɛk, n = ɛwk n. So that ɛ2m, n = 0 and ɛ2m+1, n = 2n. A natural meridian, µ, of the knot corresponds to a and a natural longitude, λ, to wk nwn k a 2ɛk,n. That is, [µ] = a and [λ] = wk nwn k a 2ɛk,n. We will also make use of an alternative longitude, which we now define. See [10]. Definition 3.3. Let λ 1 be the longitude corresponding to the word w n k wn k, so that [λ 1 ] = w n k wn k, and let l be a preferred eigenvalue of [λ 1 ]. This is chosen so that if k = 2m we have l = L and if k = 2m + 1 then l = LM 4n. 4. Representations First, we define a few polynomials we will use throughout. Definition 4.1. The j th Fibonacci polynomial, f j, is the Chebyshev polynomial defined recursively by the relation f j+1 x + f j 1 x = xf j x and initial conditions f 0 x = 0, and f 1 x = 1. With the substitution x = y + y 1 we have f j y + y 1 = y j y j /y y 1. Additionally, define the polynomials g j x = f j x f j 1 x. We will use the following several times. Lemma 4.2. If j 0, 1 then the polynomials f j x and f j 1 x share no common factors.

6 6 K. L. PETERSEN Proof. Let x = y + y 1 so that f j x = y j y j /y y 1. A root of f j x determines a solution to y 2j = 1. Similarly, a root of f j 1 f determines a solution to y 2j 2 = 1. Since gcd2j, 2j 2 = 2 the only simultaneous solutions are y = ±1. It follows that if f j x and f j 1 x share a root, it must be x = ±2. The Fibonacci recursion implies that f j 2 = j and f j 2 = ±j. Therefore, as j ±1, x = ±2 is not a root and the polynomials are relatively prime. Furthermore, we define certain terms in M and L to shorten some expressions. Definition 4.3. Let α = M 2 +1l+1, β = M 2 +l, γ = M 2 1l 1, δ = M 2 l+1, σ = β 2 +δ 2 and τ = l 1 2 M 2 + 4l + 2l M 2 + 4M 4 l + l 1 2 M 6. In many of the calculations to follow, we will use the Cayley-Hamilton theorem, which we now explicitly state for matrices in SL 2 C. Theorem 4.4 Cayley-Hamilton. Let X SL 2 C with trace x. For any integer j we have X j = f j xx f j 1 xi where I is the 2 2 identity matrix. For a matrix X, we will use X ij to denote the i, j th entry of X. As the eigenvalue map is invariant under conjugation, we consider representations ρ : Γk, 2n SL 2 C up to conjugation. A representation ρ : Γ SL 2 C is called reducible if all images share a one-dimensional eigenspace. Otherwise, a representation is called irreducible Reducible Representations. A reducible representation of Γk, 2n can be conjugated so that it is upper triangular with M s M t ρa = A = 0 M 1 and ρb = B = 0 M 1. We include the calculation of these terms of the A-polynomial for completeness. Proposition 4.5. The contribution to the A-polynomial from the reducible representations is the factor L 1. Proof. We use the presentation of the fundamental group from Proposition 3.1. Let W k = ρw k, Wk n = ρwn k, and let 0 denote the 2 2 zero matrix. First, consider k = 2m. The alternate longitude corresponds to ρ[λ 1 ] = Wk nw k n and is the identity in this case. That is, l = 1 for these representations. The defining word, in terms of matrices is AW2m n W 2m n B = 0. The matrix on the left is identically zero except for the 1, 2 entry which

7 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 7 is nmm nm + nmm 2 s t. This equals zero for infinitely many representations, and M can be any value independent of l. Therefore, these representations contribute the factor l 1, which is L 1 to the A-polynomial. Similarly, consider k = 2m+1. The relation gives AWk n W k n B, a matrix whose only non-zero entry is the 1, 2 entry which is s t M 2n 1 + M 2n+1 mm M 1 M 2n M 2n /M + M 1 We conclude that there are infinitely many representations, and that M can be any value, independent of l. The 1, 1 entry of ρ[λ 1 ] is always M 4n, so l = M 4n. This is L = 1, and gives the factor L Irreducible Representations. To compute the A-polynomial, we now restrict attention to the irreducible representations ρ : Γ SL 2 C which map the meridian and longitude to upper triangular matrices. For such a representation, there are M and L in C such that M A = ρ[µ] = ρa = 0 M 1 and Λ = ρ[λ] = ρw n k wn k a 2ɛk,n = L 0 L 1. The alternative longitude, λ 1, corresponds to a matrix with 1, 1 entry l = LM 2ɛk,n. Let W k = ρw k, W = ρwk n and W = ρwk n. Up to conjugation all irreducible representations ρ : Γ SL 2 C are given by A = M 1 0 M 1 and B = ρ[b] = M 0 2 r M 1 where r 2. The 2,1 entry of W = ρwk n, W 21, equals 2 rw 12 for these groups. Therefore, the words W and W can be written as W W = 11 W 12 and W 2 rw 12 W W = 22 W rw 12 W 11 where W ij is W ij with all M s exchanged with M 1 s. This follows from [10]. The relation awk n = wn k b implies that, on the level of matrices, AW = W B. It follows by direct computation that the variables M and r define a valid representation if M M 1 W 12 + W 22 = 0. In fact, Riley [12] shows that a similar, more general statement is true for all two-bridge knots. Definition 4.6. Let F k,n r, t be defined by F k,n r = f n t k rf k,1 r f n 1 t k r

8 8 K. L. PETERSEN with F k,1 r defined by and t = t k r defined by F 2m,1 r = f m rg m rm 2 + M 2 r + 1 F 2m+1,1 r = f m rg m+1 rm 2 + M 2 r + 1 t 2m r = f m rg m+1 r g m rm 2 + M 2 r + 2 t 2m+1 r = g m+1 r 2 M 2 + M 2 r + 2. By [11] Proposition 3.7 the condition for ρ to be a representation can be encoded by these polynomials. Specifically, we have the following. Proposition 4.7. The map ρ as above determines a representation of Γk, 2n into SL 2 C if and only if F k,n r = The Longitude. The equation F k,n r = 0 with r 2 holds if and only if ρ is an irreducible representation of Γk, 2n. We now turn to the condition that the boundary subgroup is upper triangular. A natural meridian corresponds to the group element a, which was already taken to map to an upper triangular matrix A. Therefore, the remaining condition is to ensure that an element of Γk, 2n corresponding to a longitude, either λ or λ 1, also maps to an upper triangular matrix. The image ρ[λ] can be explicitly computed, as stated in the following lemma. We will let Λ = ρ[λ] henceforth. Lemma 4.8. With W = ρwk n, Λ = ρ[λ], Λ 11 = W 11 W 22 + W 12 W 122 r M 2ɛk,n Λ 12 = W 11 W 22 + W 12 W 122 r f 2ɛk,n M + M 1 + W 11 W 12 + W 12 W 11 M 2ɛk,n Λ 21 = 2 r W 12 W 22 + W 22 W 12 M 2ɛk,n Λ 22 = 2 r W 12 W 22 + W 22 W 12 f 2ɛk,n M + M 1 + W 22 W 11 + W 12 W 122 r M 2ɛk,n. Proof. Since [λ] = wk nwn k a 2ɛk,n one can compute Λ = W W A 2ɛk,n explicitly. A direct computation using the Cayley-Hamilton theorem confirms the above. From the lemma above, we now deduce an algebraic condition for Λ to be upper triangular. Lemma 4.9. For an irreducible representation ρ, Λ = ρ[λ] is upper triangular if and only if W 12 l + W 12 = 0, where l = LM 2ɛk,n. Proof. By Lemma 4.8, Λ 21 = 0 reduces to W 12 W 22 + W 22 W 12 = 0.

9 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 9 Moreover, as L = Λ 11 by Lemma 4.8, L = W 11 W rw 12 W 12 M 2ɛk,n from which it follows that M 2ɛk,n L = W 11 W rw 12W 12. The alternate longitude is related by [λ 1 ] = a 2ɛk,n [λ]. By the Cayley-Hamilton theorem M 2ɛk,n = A 2ɛk,n 11, so that as ρa and ρ[λ] are upper triangular Therefore, l = ρ[λ 1 ] 11 = M 2ɛk,n L. 1 l = W 11 W rw 12 W 12. We consider this as the defining equation for l. Now we explore the condition for Λ to be upper triangular. We follow [10]. We multiply equation 1 by W 12, and have W 12 l = W 11 W 12 W rw 2 12W 12. Since W 12 W 22 = W 12 W 22 we rewrite this as W 12 l = W 11 W 12W rw 2 12W 12. Next, since detw = W 11 W 22 2 rw 2 12 = 1 W 12 l = W 11 W 12W 22 + W 11 W 22 1W 12 = W 12. Combining this, W 12 l + W 12 = 0. As l, M C this condition is equivalent to Λ 21 = 0. Now we show that the condition for Λ to be upper triangular can be expressed in terms of the entries of W k instead of W = W k n. Lemma For an irreducible representation ρ, Λ = ρ[λ] is upper triangular if and only if W k 12 l + W k 12 = 0, where l = LM 2ɛk,n. Proof. Using the Cayley-Hamilton theorem, W = W k n = f n t k rw k f n 1 t k ri. Therefore, W 12 = f n t k rw k 12 and W 12 = f nt k rw n 12 since f nt k r is symmetric in M and M 1. As a result, W 12 l + W 12 = f n t k r W k 12 l + W k 12. It suffices to show that f n t k r is not zero. If f n t k r = 0 then by Definition 4.6 F k,n r = f n 1 t k r = 0 and so f n 1 t k r = 0. This cannot occur if n 1, 0, 1 as by Lemma 4.2 f n and f n 1 are relatively prime. If n = 1, 0, 1 then either f n t k r or f n 1 t k r equals ±1 and cannot be zero.

10 10 K. L. PETERSEN 4.4. Proof of Theorem 1.1. We now use the Cayley-Hamilton theorem to determine explicit equations for the polynomials above. First, we determine explicit equations for the entries of W k. Lemma With W k = ρw k then W k 21 = 2 rw k 12. If k = 2m then W k 11 = f m r 2 2 rm 2 + [f m rr 1 f m 1 r] 2 W k 12 = f m r 2 [M + M 1 rm 1 ] + f m rf m 1 r[m 1 M] W k 22 = f m r 2 M 2 2 r + [f m r f m 1 r] 2. If k = 2m + 1 then W k 11 = f m r 2 [M 2 r 1 2 r 2r 2 ] + 2f m rf m 1 r[1 rm 2 + rr 2] + f m 1 r 2 [2 + M 2 r] W k 12 = f m r 2 [Mr 1 + r r 2 M 1 ] f m rf m 1 r[2r 1M 1 M] + f m 1 r 2 M 1 W k 22 = f m r 2 [r 1 2 M r] 2f m rf m 1 rr 1M 2 + f m 1 r 2 M 2. Proof. We calculate AB 1 r 1 M = r 2M 1 1 A 1 r 1 M 1 B = 2 rm 1 both of which have trace r, so that by the Cayley-Hamilton theorem AB 1 m r 1fm r f = m 1 r Mf m r r 2M 1 f m r f m r f m 1 r A 1 B m r 1fm r f = m 1 r M 1 f m r. 2 rmf m r f m r f m 1 r As W k = ρw k, using Proposition 3.1 for k = 2m we have and for k = 2m + 1, W k = AB 1 m A 1 B m W k = AB 1 m ABA 1 B m. Upon multiplying we obtain the stated expressions. In light of Lemma 4.10 we now compute W k 12 l + W k 12 using these equations. Definition Let G k,n r be defined by G 2m,n r = f m rrδ α γf m+1 r G 2m+1,n r = βf m+1 r δf m r.

11 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 11 Using the Fibonacci identities, one can write these polynomials in different ways. For example, when k = 2m we also have G k,n r = f m rα rβ γf m 1 r. Lemma For an irreducible representation ρ, the condition for Λ = ρ[λ] to be upper triangular is equivalent to G k,n r = 0. Proof. By Lemma 4.10 it suffices to consider W k 12 l + W k 12. With Lemma 4.11 we see that for k = 2m, W k 12 l+w k 12 is exactly f mrm 1 times the expression for G k,n. When k = 2m + 1, by Lemma 4.11 similar to the above, W k 12 l + W k 12 is g m+1rm 1 times the expression for G k,n. It suffices to show that f m r 0 when k = 2m and that g m+1 r 0 when k = 2m+1. First, consider the even case. If f m r = 0 then F k,n r can be explicitly computed using Proposition 4.7 and is the A-polynomial. This implies that F k,1 r = 1 and t k r = 2. Therefore, F k,n r = f n 2 f n 1 2. As f j 2 = j for all j, we conclude that F k,n r = 1. Therefore the A- polynomial is L 1, with the inclusion of the reducible factor. By [2, 8] the only knot in S 3 with A-polynomial equal to L 1 is the unknot. Similarly, if k is odd we conclude that F k,1 r = 1 and t = 2 so that F k,n r = 1. Lemma For k 1, 0, 1 the polynomial G k,n r is an irreducible non-constant polynomial in Q[M ±1, l ±1 ]. Proof. Let R = Q[M ±1, l ±1 ]. Consider the case when k = 2m, so that G 2m,n r = f m rm 2 + 1l + 1 M 2 + lr + f m 1 rm 2 1l 1. Modulo the ideal generated by M 2 1, R can be identified with Q[l] and G 2m,n r = l + 1f m rr 2. As a result, any factorization of G 2m,n r in R[r] modulo M 2 1, gives a factorization of f m rr 2 in Q[r]. Modulo the ideal generated by M 2 + l we have G 2m,n r = l + 1l 1 f m+1 r f m r. Similarly, a factorization modulo M 2 + l gives a factorization of f m r f m+1 r in Q[r]. As f m rr 2 and f m r f m+1 r share no common roots by Lemma 4.2 we see that there is no factorization. We conclude that G 2m,n r is irreducible. Next consider k = 2m + 1, so that G 2m+1,n r = M 2 l + 1f m r + f m+1 rm 2 + l. First, assume that 2m+1 > 0, reducing modulo M 2 l+1 we have G 2m+1,n r = M 2 + lf m+1 r, and reducing modulo M 2 + 1l + 1 G 2m+1,n r = M 2 +l f m r+f m+1 r. Again, a factorization modulo either M 2 l+1 or M 2 + 1l + 1 gives a factorization of either f m+1 r or f m r + f m+1 r in Q[r]. But these are relatively prime by Lemma 4.2. Therefore G 2m+1,n r is irreducible in this case as well. Now consider the case when 2m + 1 < 0. Reducing modulo M 2 + l, G 2m+1,n r = M 2 l + 1f m r and reducing modulo M 2 1l 1, G 2m+1,n r = M 2 + l f m r f m+1 r. As before, we conclude that G 2m+1,n r is irreducible.

12 12 K. L. PETERSEN Theorem 1.1 now follows from Proposition 4.7, Lemma 4.13, and Lemma Remark Let K = QM, l. To determine the A-polynomial by eliminating the variable r from F k,n t k r and G k,n r using resultants, we require F k,n t k r and G k,n r to be non-constant polynomials in K[r]. The polynomial G k,n r is a non-constant polynomial in K[r] unless k = 1, 0, 1. For these values we have G 1,n r = δ, G 0,n r = γ, and G 1,n r = β. The polynomial F k,1 x is a non-constant polynomial in K[x] unless k = 0, 1. We have F 0,1 x = F 1,1 x = 1. The defining equation for t k r is nonconstant in K[r] unless k = 0, in which case t 0 r = 2. It is non-linear if k > 1 and if k = ±1 then t = M 2 + M r. It follows that F k,n t k r is constant only when n = 0, k = 0, or when n = k = 1. For these values there are no irreducible representations. These correspond to Jk, 0 = J0, 2n = J1, 2 which are all the unknot. We conclude that one of F k,n t k r and G k,n r is constant when k = 1, 0, 1 or n = 0. These values of k are the excluded values in Theorem Resultants In this section we collect facts about resultants that will be used later. If px and qx are polynomials with leading coefficients P and Q, the resultant of px and qx is Res px, qx = P deg q Q deg p r p r q where the product ranges over all roots r p of px and r q of qx. First, we summarize some basic facts about resultants. Lemma 5.1. Assume that px, qx, and rx are polynomials and the leading coefficient of qx is Q. 1 Resp, q = 1 deg p+deg q Resq, p. 2 Respr, q = Resp, qresr, q. 3 Q degp+rq Res p, q = Q deg p Res p + rq, q. The following polynomial will be useful in our calculations, as rational functions will naturally come out of our calculation. Definition 5.2. For a fixed x and y, and a polynomial ϕ, define the polynomial ϕx, y = y degϕ ϕ x y. We now collect a few useful lemmas. Lemma 5.3. Let ϕz be a polynomial of degree d and assume that tp 1 = p 2 + gp 3. Then p d 1 ϕt = ϕp 2, p 1 + gp 4 for some polynomial p 4.

13 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 13 Proof. Notice that if ϕz = c d z d + c d 1 z d c 1 z + c 0 then since tp 1 = p 2 + gp 3 we have p d 1ϕt = c d p 1 t d + c d 1 p 1 p 1 t d c 1 p d 1 1 p 1 t + c 0 p d 1 = c d p 2 + gp 3 d + c d 1 p 1 p 2 + gp 3 d c 1 p d 1 1 p 2 + gp 3 + c 0 p d 1. For each n > 0 the term p 2 + gp 3 n = p n 2 + gp where p is a polynomial. Therefore, p d 1ϕt = c d p d 2 + c d 1 p 1 p d c 1 p d 1 1 p 2 + c 0 p d 1 + gp 4 = p d 1ϕ p 2 p 1 + gp 4 = ϕp 2, p 1 + gp 4. The following can be verified directly from the Fibonacci recursion. Lemma 5.4. Let A and B be constants. Then p k x, y = Af k x, y + Bf k 1 x, y satisfies the following recursion. We have p 1 x, y = A, p 2 x, y = Ax + B and for k > 2 the polynomial p k x, y = xp k 1 x, y y 2 p k 2 x, y. We have p 0 x, y = B, p 1 x, y = A xb and for k < 1 the polynomial p k x, y = xp k+1 x, y y 2 p k+2 x, y A Special Resultant. In this section we compute the resultant R j = Res f j xx + A + f j 1 xb, px where px = x 2 ax + b and a, b, A and B are constants. This type of resultant will be used several times in our computations. With X j = R j + R j+1 we will first show that X j satisfies a recursion, from which a recursion for R j follows. All of the proofs will rely on the recursive definition of the Fibonacci polynomials, f j+1 x + f j 1 x = xf j x. Let x 1 and x 2 be the two roots of px, so that x 1 x 2 = b and x 1 + x 2 = a. Let F j = f j x 1 f j x 2 = Resf j x, px G j = f j x 1 f j 1 x 2 + f j x 2 f j 1 x 1 H j = x 1 f j x 1 f j 1 x 2 + x 2 f j x 2 f j 1 x 1. In this section, let d 1 = b, d 2 = 2 a 2 + 2b, d 3 = b and d 4 = 1. Lemma 5.5. The following identities hold. 1 H j = F j+1 + bf j + F j 1 2 H j H j 2 = F j 1 a 2 2b 2bF j 2 3 F j = bf j a 2 + 2bF j 2 + bf j 3 F j 4 4 G j = af j 1 G j 1.

14 14 K. L. PETERSEN Proof. For the first identity, F j = x 1 f j 1 x 1 f j 2 x 1 x 2 f j 1 x 2 f j 2 x 2 = x 1 x 2 F j 1 + F j 2 x 1 f j 1 x 1 f j 2 x 2 + x 2 f j 1 x 2 f j 2 x 1 = bf j 1 + F j 2 H j 1. This proves the first assertion. Next, consider H j = x 1 x 1 f j 1 x 1 f j 2 x 1 f j 1 x 2 + x 2 x 2 f j 1 x 2 f j 2 x 2 f j 1 x 1 = F j 1 x x 2 2 x 1 f j 1 x 2 f j 2 x 1 + x 2 f j 1 x 1 f j 2 x 2 = F j 1 a 2 2b x 1 f j 2 x 1 x 2 f j 2 x 2 f j 3 x 2 + x 2 f j 2 x 2 x 1 f j 2 x 1 f j 3 x 1 = F j 1 a 2 2b 2x 1 x 2 F j 2 + x 1 f j 2 x 1 f j 3 x 2 + x 2 f j 3 x 1 f j 2 x 2 = F j 1 a 2 2b 2bF j 2 + H j 2. This shows the second assertion. With the above, we conclude that the third assertion holds. Finally, G j = f j 1 x 2 f j x 1 + f j 1 x 1 f j x 2 = f j 1 x 2 x 1 f j 1 x 1 f j 2 x 1 + f j 1 x 1 x 2 f j 1 x 2 f j 2 x 2 = x 1 + x 2 f j 1 x 2 f j 1 x 1 f j 1 x 2 f j 2 x 1 + f j 1 x 1 f j 2 x 2 = af j 1 G j 1 proving the final identity. Lemma 5.6. The following identity holds X j = BF j+2 + b + Aa + A 2 + bb BF j+1 + F j b + Aa + A 2 + bb + aab + B 2 + B + B 2 + BF j 1. Proof. The resultant R j is R j = f j x 1 x 1 + A + f j 1 x 1 B f j x 2 x 2 + A + f j 1 x 2 B = F j b + Aa + A 2 + B 2 F j 1 + ABG j + BH j. By Lemma 5.5, we can substitute H j and rewrite R j as R j = BF j+1 + F j b + Aa + A 2 + bb + B 2 + BF j 1 + ABG j. By another application of Lemma 5.5 we can substitute G j so that X j is as stated. Lemma 5.7. The term X j satisfies the recursion X j = bx j a 2 + 2bX j 2 + bx j 3 X j 4. Since X j = R j + R j 1 this implies the following recursion for R j.

15 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 15 Proposition 5.8. The resultant R j of f j xx+a+f j 1 xb and x 2 ax+b satisfies the following recursion. R j = b 1R j 1 +2 a 2 +3bR j 2 +2 a 2 +3bR j 3 +b 1R j 4 R j 5 with initial conditions R 2 = b 2 + 2Bb + 2Bb 2 + B 2 + 2bB 2 + B 2 b 2 + Aab AaB + abba a 2 B a 2 B 2 + ba 2 R 1 = A 2 + Aa + AaB + b + 2Bb + bb 2 R 0 = B 2 R 1 = b + Aa + A 2 R 2 = B 2 2Bb + b 2 + AaB + Aab + a 2 B + ba 2. Proof. A recursion for X j of the form determines a recursion X j = c 1 X j 1 + c 2 X j 2 + c 3 X j 2 + c 4 X j 4 R j+1 = c 1 1R j + c 1 + c 2 R j 1 + c 2 + c 3 R j 2 + c 3 + c 4 R j 3 + c 4 R j 4. The following is immediate. Proposition 5.9. If S j = cd j R j then for j > 5, S j satisfies the recursion S j = db 1S j 1 + d 2 2 a 2 + 3bS j 2 + d 3 2 a 2 + 3bS j 3 + d 4 b 1S j 4 d 5 S j 5 and for j 5, S j satisfies the recursion S j = db 1S j+1 + d 2 2 a 2 + 3bS j+2 + d 3 2 a 2 + 3bS j+3 + d 4 b 1S j+4 d 5 S j+5. The initial conditions are determined by S j = cd j R j for j > 5 and j Proof of Theorem 1.2 By Theorem 1.1 the A-polynomial can be computed as ResF k,n t k r, G k,n r. The degree of F k,n is approximately 1 4 n k2 and the degree of G k,n is about 1 2 k. In this section we undergo a series of reductions replace F k,n with a polynomial of smaller degree, approximately 2 n. The terms α, β, γ and δ are defined in Definition 4.3. Definition 6.1. Let q 1 r = δβr σ and q 2 r = σr τ, and let g 0 be the leading coefficient of G k,n r. Specifically, g 0 = β if k > 0 and δ if k < 0. The following can be verified using the Fibonacci recursion.

16 16 K. L. PETERSEN Lemma 6.2. We have the following β m γ if k = 2m δ G k,n σ δβ = β m+1 δ m if k = 2m + 1 > 0 δ m β m 1 if k = 2m + 1 < 0. In the next few lemmas we rewrite key terms. Essentially we rewrite these terms as a multiple of G k,n r plus a remainder. This will allow us to simplify the resultant of F k,n r and G k,n r using Lemma 5.1. Let F = QM, L. First, we address F k,1 r. Let F k,1 = γ 1 α βt k r and let F k,n r = f nt k rf k,1 f n 1t k r. Lemma 6.3. For a fixed k 1, 0, 1, and n 0 there are polynomials R 1 r and R 2 r in F [r] such that and F k,1 r = F k,1 R 1rG k,n r F k,n r = F k,n r R 2rG k,n r. Proof. If k = 2m, let R 1 r = γ 1 f m rm 2 + M 2 r. If k = 2m + 1 let R 1 r = γ 1 g m+1 rm 2 + M 2 r. In each case, the first identity can be directly verified using the definition of F k,1 r and G k,n r with the substitution r = s + s 1 so that f j r = s j s j /s s 1. The second assertion follows from the definition of F k,n r and F k,n r letting R 2r = f n t k rr 1 r. This allows us to calculate the resultant of F k,n r and G k,n r in terms of F k,n r. Let A 1 = ResF k,n r, G k,n r and A 2 = ResF k,n r, G k,nr. Lemma 6.4. For a fixed k 1, 0, 1, and n 0 then A 1 nk < 0 A 2 = ±βa 1 n, k > 0. 1 k δa 1 n, k < 0 Proof. When n > 0 by Lemma 6.3 and Lemma 5.1 up to sign, we see that A 1 g degf k,n r 0 = g degf k,nr 0 A 2. Since degf k,n r = degt k r n 1 + degf k,1 r and degf k,n r = degt k r n we have A 1 g deg t kr 0 = A 2 g degf k,1r 0. The statement follows from calculating the degrees of t k r and F k,1 r. The other cases are similar.

17 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 17 Now we turn to the defining equation for t k r. Lemma 6.5. Fix k 1, 0, 1 and n 0. R 3 r F [r] such that Proof. When k = 2m let t k rq 1 r = q 2 r G k,n rr 3 r. Then there is a polynomial R 3 r = M 2 + M 2 rf m rβr α + γf m+1 r and when k = 2m + 1 let R 3 r = r 2M 2 + M 2 rβf m r δf m+1 r. The statement follows immediately. Let A 3 = Resq 1 r n γf k,n r, G k,nr. Lemma 6.6. Fix k 1, 0, 1 and n 0. We have n A 1 γ m ɛ 2 δβ m ɛ 2 G k,n σ δβ nk < 0 n A 3 = ±A 1 βγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k > 0 n ±A 1 δγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k < 0 where ɛ 2 = 0 unless k is odd and negative, in which case ɛ 2 = 1. Proof. We use the identities in Lemma 5.1. First, notice that since deg G k,n = m ɛ 2, we have Next, A 2 γ degf k,nr A 2 γ m ɛ 2 = ResγF k,n r, G k,nr. ResγF k,n r, G k,nrresq 1 r, G k,n r n = A 3. Therefore, A 3 = Resq 1 r, G k,n r n γ m ɛ 2 A 2. By definition, q 1 r = δβr σ δβ. It follows that This simplifies to Resq 1 r, G k,n r = δβ m ɛ 2 G k,n σ δβ. n A 2 γ m ɛ 2 δβ m ɛ 2 G k,n σ δβ = A3. The lemma follows from combining this with the identity from Lemma 6.4. The term F k,n r has a factor of M 2 n in the denominator from the M 2 in the defining equation for t k r. The A-polynomial is well-defined up to multiples of M, so we will multiply by powers of M to make the powers of M all positive. To remove this factor of M 2 n, we compute A 0 = ResM 2 n F k,n r, G k,n r.

18 18 K. L. PETERSEN Therefore, A 0 = M 2 n deg Gk,nr A 1. By Lemma 6.6 we have n A 0 γ m ɛ 2 δβ m ɛ 2 G k,n σ δβ nk < 0 M 2 n deg Gk,nr n A 3 = ±A 0 βγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k > 0 n ±A 0 δγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k < 0 Since A 3 = Resq 1 r n γf k,n r, G k,nr, we conclude that M 2 n deg G k,nr A 3 = ResM 2 n q 1 r n γf k,n r, G k,nr. We conclude the following, with A 4 = ResM 2 n q 1 r n γf k,n r, G k,nr. Lemma 6.7. With ɛ 2 defined in Lemma 6.6, n A 0 γ m ɛ 2 δβ m ɛ 2 G k,n σ δβ nk < 0 n A 4 = ±A 0 βγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k > 0 n ±A 0 δγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k < 0 Now we define the polynomials H k,n r which differ from M 2 n q n 1 γf k,n r by a multiple of G k,n r. Definition 6.8. If n is positive let H k,n r = q 2 rm 2 H k,n 1 r q 1 r 2 M 4 H k,n 2 r with initial conditions H k,1 r = M 2 αq 1 r βq 2 r, and H k,2 r = M 4 αq 1 rq 2 r γq 1 r 2 βq 2 r 2. If n is negative let H k,n r = q 2 rm 2 H k,n+1 r q 1 r 2 M 4 H k,n+2 r with initial conditions H k,0 r = γ and H k, 1 r = M 2 βq 2 r αq 1 r + γq 2 r. The following follows directly from Lemma 5.4. Lemma 6.9. If n is positive, then H k,n r = M 2 n f n q 2, q 1 αq 1 r βq 2 r γq 1 r 2 f n 1 q 2, q 1 and if n is negative, then H k,n r = M 2 n f n q 2, q 1 αq 1 r βq 2 r γf n 1 q 2, q 1. Lemma Fix k 1, 0, 1 and n 0. R 4 r F [r] such that Then there is a polynomial M 2 n q 1 r n γf k,n r = H k,nr + G k,n rr 4 r.

19 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 19 Proof. By Lemma 5.3 and Lemma 6.5 since tq 1 r = q 2 r G k,n R 3 r we have when n > 0 q 1 r n f n t α βt γf n 1 t = q 1 r n 1 f n t αq 1 r βq 2 r + βg k,n R 3 γq1 r n f n 1 t = H k,n rm 2 n + G k,n R 4. Here R 4 = βr 3 q1 n 1 f n t k r + P 4 αq 1 βq 2 γq1p 2 4 with P 4 and P 4 from Lemma 5.3. If n is negative, q 1 r n f n t α βt γf n 1 t = q 1 r n 1 f n t αq 1 βtq 1 γq1 r n f n 1 t = H k,n rm 2 n + G k,n R 4. Here R 4 = βr 3 q n 1 1 f n t k r + P 4 αq 1 βq 2 γp 4 with P 4 and P 4 as in the previous case. We are now ready to prove Theorem 1.2. Let A 5 = ResH k,n r, G k,n r. It is enough to show that up to sign, A 0 γ m +ɛ n 1 ɛ1 ɛ nk < 0 A 5 = A 0 βγ m +ɛ n 1 ɛ1 ɛ n, k > 0 A 0 δγ m +ɛ n 1 ɛ1 ɛ n, k < 0 where ɛ = 1 if k is even and ɛ = 0 if k is odd, and ɛ = 1 if k is positive and 0 if k is negative. Lemma 6.10 with Lemma 5.1 implies that A 4 g deg H k,nr 0 = g n +degf k,n r 0 A 5. Since the degree of H k,n r is n, we have A 4 = g degf k,n r 0 A 5. The degree of degf k,n r = n degt kr = nk so that A 4 = g nk 0 A 5. Using Lemma 6.7 it follows that n A 0 γ m ɛ 2 δβ m ɛ 2 G k,n σ δβ nk < 0 g nk n 0 A 5 = ±A 0 βγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k > 0 n ±A 0 δγ m ɛ 2 δβ m ɛ 2 G k,n σ δβ n, k < 0. n When k = 2m then ɛ 2 = 0 and δβ m ɛ 2 G k,n σ δβ = ±γ n g nk 0. This reduces to the stated form. When k = 2m + 1, then ɛ 2 depends on the sign

20 20 K. L. PETERSEN n of k. In both cases, δβ m ɛ 2 G k,n σ δβ nk = ±g 0. This reduces to the stated form. This completes the proof of Theorem Explicit computations for Jk, 2n knots for small k When k is small, we can recursively compute these resultants for all n. Here we calculate the A-polynomials for the J1, 2n, J2, 2n, J3, 2n, J4, 2n and J5, 2n knots. Since J k, l is the mirror image of Jk, l, the A-polynomial of J k, l can be determined from the A-polynomial of the Jk, l knot by replacing every M with an M 1. Therefore, these calculations also determine the A-polynomials of the J 1, 2n, J 2, 2n, J 3, 2n, J 4, 2n, and J 5, 2n knots. It is sufficient to determine the contribution from the irreducible representations, as the reducible representations contribute the factor L 1 to the A-polynomial J1, 2n, torus knots. The knot J1, 2n is the torus knot T 2, 2n 1 corresponding to the Schubert pair p, q = 1, 1 2k. By Remark 4.15 when k = ±1, G k,n r is constant. In particular, G 1,n = β and G 1,n = δ. Therefore this determines the irreducible contribution of the A-polynomial. The algebraic condition F 1,k r = 0 merely determines valid r values for ρ to be an irreducible representation. For n = 0, or 1 however, F 1,0 r and F 1,1 r are ±1. This indicates that there are no irreducible representations, which is clear as J1, 0 = J1, 2 are the unknot. Here l = LM 4n and since M 0, G 1,k r = β = 0 determines the factor β = M 2 + l = M 2 + LM 4n. Normalizing, by dividing by M 2, this is equivalent to the factor of 1 + LM 4n 2. We obtain the following, which has been observed before [4]. Theorem 7.1. The A-polynomial of the J1, 2n torus knot is given by { 1 + LM 4n 2 if n > 1 A1, 2n = M 2 4n + L if n J2, 2n, the twist knots. We will directly compute the A-polynomial as the common vanishing set of F 2,n and G 2,n using Theorem 1.1. Here m = 1 and we have the following F 2,n r = f n t 2 rf 2,1 r f n 1 t 2 r F 2,1 r = M 2 + M 2 r + 1 t 2 r = 2 rm 2 + M 2 r + 2 G 2,n r = M 2 + lr M 2 + 1l + 1. Since G 2,n r = 0 we conclude that r = M 2 + 1l + 1/M 2 + l. It follows that F 2,1 r = M 6 + l M 2 M 2 + l

21 and A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 21 t 2 = n 2 d 2 = 1 lm 8 + 2lM 6 + l M 4 + 2lM 2 + l 2 l M 2 + l 2 M 2. Therefore, M 6 + l F 2,n r = f n t 2 M 2 M 2 + l f n 1t 2. One can easily verify the following F 2, 1 = M 8 l + M 6 l + M 4 l M 2 l l /M 2 M 2 + l 2 F 2,0 = 1 F 2,1 = M 6 + l/m 2 M 2 + l F 2,2 = 1 lm lM 12 + l 2 + 2lM 10 M 8 l 2 lm 6 + 2l 2 + lm 4 + 2M 2 l 2 + l 3 l 2 /M 4 M 2 + l 3. The A-polynomial is given by the numerator of F k,n. Using Lemma 5.4 we conclude that the A-polynomials satisfy the recursion given by Hoste and Shanahan [10]. In this case l = L. Theorem 7.2. Let n 2 and d 2 be defined as above. For n positive with initial conditions A2, 2 = M 6 + l A2, 2n = n 2 A2, 2n 2 d 2 2A2, 2n 4 A2, 4 = 1 lm lM 12 + l 2 + 2lM 10 M 8 l 2 lm 6 For negative n with initial conditions + 2l 2 + lm 4 + 2M 2 l 2 + l 3 l 2. A2, 2n = n 2 A2, 2n + 2 d 2 2A2, 2n + 4 A2, 0 = 1 A2, 2 = M 8 l + lm 6 + 2l + l 2 + 1M 4 + M 2 l l J3, 2n. We will directly compute the A-polynomial as the common vanishing set of F 3,n and G 3,n using Theorem 1.1. Here, F 3,n r = f n t 3 rf 3,1 r f n 1 t 3 r F 3,1 r = r 1M 2 + M 2 r + 1 t 3 r = r 1 2 M 2 + M 2 r + 2 G 3,n r = M 2 + lr M 2 l + 1.

22 22 K. L. PETERSEN Since G 3,n r = 0 we have r = M 2 l + 1/M 2 + l. Using this, F 3,1 r = 1 lm 8 + lm 6 + 2M 4 l + M 2 l + l 2 l M 2 + l 2 M 2. Therefore, F 3,n is defined by 1 lm 8 + lm 6 + 2M 4 l + M 2 l + l 2 l F 3,n r = f n t 3 M 2 + l 2 M 2 f n 1 t 3 and t = n 3 /d 3 where n 3 = l 1 2 M l2 lm 8 + l 2 + 4l + 1M 6 + ll 2 + 4l + 1M 4 + 2l2l 1M 2 + ll 1 2 and d 3 = M 2 +l 3 M 2. The A-polynomial is given by the numerator of F k,n. Using Lemma 5.4 we see that the numerators satisfy the recursion given by Hoste and Shanahan [10]. Recall that since k = 3 we have l = LM 4n. The results that helped us determine the recursion for the resultant rely on the recursive definitions of F k,n r and G k,n r as polynomials in l, M ±2 and r. Therefore, to write A3, 2n in terms of the variables L and M, one uses the recursive definition below to determine the polynomial in l and M and then substitutes l = LM 4n. Theorem 7.3. Let n 3 and d 3 be defined as above. For n positive with initial conditions A3, 2n = n 3 A3, 2n 2 d 2 3A3, 2n 4 A3, 2 = 1 lm 8 + lm 6 + 2M 4 l + M 2 l + l 2 l A3, 4 = 1 3l + 3l 2 l 3 M l 2 + 3l 3 + 5lM 16 For n negative with initial conditions A3, 0 = 1 + 3l 2 l 3 + 5lM l 3 2l + 13l 2 l 4 M 12 + l + 2l l 2 3l 3 M l 3 3l 2 + 2l l 4 M l 3 l 5l 2 2l 4 M 6 + 3l 3 l 2 + 5l 4 M 4 + 8l 3 + 5l 4 + 3l 2 M 2 3l 4 + l 5 + 3l 3 l 2. A3, 2n = n 3 A3, 2n + 2 d 2 3A3, 2n + 4 A3, 2 = l 2 lm 10 + l 2 + 2lM lM 6 + 2l 2 + l 3 M 4 + l + 2l 2 M 2 + l l 2.

23 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS J4, 2n. We will show that the resultant defining A4, 2n is of the form considered in Section 5.1. Since k = 4 = 2m we have l = L for all of these knots. By Theorem 1.1 we can compute the A-polynomial as where ResF 4,n r, G 4,n r F 4,n r = f n t 4 rf 4,1 r f n 1 t 4 r F 4,1 r = rr 1M 2 + M 2 r + 1 t 4 r = rr 2 2rM 2 + M 2 r + 2 G 4,n r = M 2 + Lr 2 L + 1M 2 + 1r L 1M 2 1. Moreover, for this section, let a = β 4 M 2 LL 12L 2 L L1 4L + 5L 2 M 2 + L2L 3 + 5L L 1M 4 + 4L1 + 4L + L 2 M L + 10L 2 L 3 M 8 + 2L5 4L + L 2 M L2 L + L 2 M 12 b = β 4 M 4 LL LL 13L 1M 2 + 2LL 3 + L 2 + L 1M 4 + 2LL + 13L + 1M L + 14L 2 + 4L 3 + L 4 M 8 + 2LL + 1L + 3M L + L 2 L 3 M LL 1L 3M L3M 16 c = M 4 β 7 L 2 L 1 2 M M A = α/β C = β/γ B = 1/C. First we show that ResF 4,n r, G 4,n r is of the form considered in Proposition 5.9. Let R n = Res f n xx + A + f n 1 xb, x 2 ax + b and let A 1 = ResF 4,n r, G 4,n r. Lemma 7.4. Let ɛ = 1 if n > 0 and ɛ = 0 if n < 0. We have Proof. First, we define the following P 1 = M 6 + L rm 2 β/β 2 M 2 A 1 L 1 2 M = β 4 n +2 ɛ R n P 2 = M 2 r 2 β 2 M 2 + 1M 4 + Lβr L 1M 2 1M 6 + L/β 3 M 2 t = 2 + L 1 2 M M 6 + L L 1M M Lr /M 2 β 3 F 4,1 = L 1L M 8 + LM 2 M LM r /M 2 β 2.

24 24 K. L. PETERSEN With this, F 4,1 = F 4,1 + G 4,n P 1, and t = t + G 4,n P 2. Moreover, in terms of t we have F 4,1 = Ct +A, and G 4,n = ct 2 at +b. Additionally, let F 4,nr = f n t 4 F 4,1 f n 1 t 4, and F 4,nr = f n t F 4,1 f n 1 t. Let A 2 = ResF 4,n r, G 4,n. By the above, we have that and we conclude that F 4,n = F 4,nr + f n t 4 G 4,n P 1 A 1 g max{4 n 3,4 n 1 4} 0 = A 2 g max{4 n 1 4,4 n 1} 0. Therefore, when n > 0, A 1 = A 2 β 2 and when n < 0, A 1 = A 2. We also have F 4,nr = f n t + G 4,n P 2 F 4,1 f n 1 t + G 4,n P 2 = F 4,nr + G 4,n P 3. Let A 3 = ResF 4,n r, G k,nr as a function of r. The degree of F 4,n r is max{4 n 3, 4 n 1 4} and the degree of F 4,n r is n so we conclude that when n > 0 we have A 2 = β 3n 3 A 3 and when n < 0, A 2 = β 3 n A 3. By definition of A, B, C, a, b, and c, F 4,nr = C[f n t t + A + Bf n 1 t ] and G 4,n r = ct 2 at + b. Therefore, A 3 = A 4 where A 4 = ResC[f n t t + A + Bf n 1 t ], ct 2 at + b, and the resultant is taken with respect to the variable r. Let A 5 = ResC[f n xx + A + Bf n 1 x], cx 2 ax + b with respect to the variable x. The leading coefficient of t is t 0 = L1 LM M /M 2 β 3 and we conclude that these two resultants differ by this leading coefficient to the product of the powers of the two polynomials, A 4 = A 5 t 0 2 n. Explicitly, this is A 4 M 4 n β 6 n = A 5 L 2 n 1 L 2 n M n M n. Finally, since R n = Res f n xx+a+f n 1 xb, x 2 ax+b we conclude that R n = c degfnxx+a+f n 1xB C degx2 ax+b A 5, which reduces to the stated formulas. The M 2 factor in t 4 r introduces a factor of M 4 n into A 1. normalization, A4, 2n = M 4 n A 1 so that by Lemma 7.4 After A4, 2nL 1 2 M = M 4 n β 4 n +2 ɛ R n.

25 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 25 We obtain a recursive formula for A4, 2n using Proposition 5.9. Let c 1 = b 1, and c 2 = 2 a 2 + 3b. Specifically, for n > 5, A4, 2n = c 1 M 4 β 4 A4, 2n 1 + c 2 M 4 β 4 2 A4, 2n 2 + c 2 M 4 β 4 3 A4, 2n 3 + c 1 M 4 β 4 4 A4, 2n 4 M 4 β 4 5 A4, 2n 5. where the initial conditions are A4, 2n = M 4n ResF 4,n r, G 4,n r for 1 n 5. in fact, setting A4, 0 = β 1 we can begin the recursion one step earlier. When n 5, using A4, 0 = 1 the recursion is A4, 2n = c 1 M 4 β 4 A4, 2n c 2 M 4 β 4 2 A4, 2n c 2 M 4 β 4 3 A4, 2n c 1 M 4 β 4 4 A4, 2n + 4 M 4 β 4 5 A4, 2n + 5. where initial terms A4, 2n for 4 n 0 are A4, 2n = M 4 n ResF 4,n r, G 4,n r. Finally, we write this as a self contained theorem, computing these coefficients explicitly. Theorem 7.5. With d 1, d 2, d 3, d 4, and d 5 as in the appendix, we have the following. For n positive A4, 2n = d 1 A4, 2n 1 + d 2 A4, 2n 2 + d 3 A4, 2n 3 + d 4 A4, 2n 4 + d 5 A4, 2n 5. The initial conditions are A4, 2n = ResM 2n F 4,n r, G 4,n r for 0 < n 5. For n negative A4, 2n = d 1 A4, 2n d 2 A4, 2n d 3 A4, 2n d 4 A4, 2n d 5 A4, 2n + 5. The initial conditions are A4, 2n = ResM 2 n F 4,n r, G 4,n r for 4 n J5, 2n. By Theorem 1.1 we can compute the A-polynomial of J5, 2n as ResF 5,n r, G 5,n r where F 5,n r = f n tf 5,1 r f n 1 t F 5,1 r = rr 2 r 1M 2 + M 2 r + 1 t 5 r = r 2 r 1 2 M 2 + M 2 r + 2 G 5,n r = M 2 + lr 2 M 2 l + 1r M 2 l.

26 26 K. L. PETERSEN We proceed in the same fashion as the A4, 2n case. For this section, we define the following a = l 4 2l 3 + 3l 2 4l + 2M l l 16l 2 2l 4 M 12 + l 4 12l l 2 + 6l + 2M l 3 + 5l + 30l 2 M l 3 + 5l 2 + 5l 4 M 6 + 6l l 3 12l 2 + l + 2l 5 M l 3 2l + 9l l 4 M 2 + 3l 3 2l 2 + 2l 5 + l 4l 4 M 2 β 5 b = 6l 2 + l 4 4l lM l 4 18l 2 + 8l + 12l 3 M l + 2l 4 6l 3 + 4l 2 M l 2l 4 8l l 2 M l l l 2 + 6lM 10 + l + l l 3 + 6l 4 + 6l 2 M 8 + 8l 2 2l + 8l l 3 M 6 + 2l 5 + 4l 3 6l 2 + 2l 4 + 2lM l 2 2l 18l 3 + 8l 4 M 2 + 6l 3 + l 4l 4 + l 5 4l 2 M 4 β 5 c = β 9 M 4 l 1 4 M M l 2 A = α/β B = γ/β C = β/γ. In addition, let and R n = Resf n xx + A + f m 1 xb, x 2 ax + b A 1 = ResF 5,n r, G 5,n r. Lemma 7.6. Let ɛ equal 1 when n > 0 and 2 when n < 0. Then A 1 γ 2 = β 5 n +ɛ R n. Proof. Let P 1 = r 2 M 2 M 2 + l 2 r1 + M 2 M 4 + lm 2 + l l 1M 2 1M 6 + l /β 3 M 2 P 2 = r 3 M 2 M 2 + l 3 + M 2 + l 2 l + 2M 2 l + 2M 4 + M 6 r 2 + M 2 + lm 6 + lm 2 l 2l 2M 2 + 1r + M 2 l + 1 2M 8 + M 8 l 2lM 6 2M 4 l 2M 2 l 2l 2 + l /M 2 β 4 F 5,1 = l1 + M 2 2 M l 1r M 2 + lm 8 l M 8 lm 6 2M 4 l M 2 l + l l 2 /β 3 M 2 t = ll M 2 2 M r

27 Moreover, let A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS 27 + M 2 + lm 10 2M 10 l + M 10 l 2 2M 8 l 2 + 4M 8 l + 4lM 6 + M 6 + l 2 M 6 + M 4 l 3 + M 4 l + 4M 4 l 2 2M 2 l + 4M 2 l 2 + l 2l 2 + l 3 /M 2 β 4. F 5,nr = f n t 5 F 5,1 f n 1 t 5 and F 5,nr = f n t F 5,1 f n 1 t. A direct calculation shows that F 5,1 = F 5,1 + G 5,n P 1, and t = t + G 5,n P 2. Let A 2 = ResF 5,n r, G 5,n. We conclude that A 1 g max{5 n 4,5 n 1 5} 0 = A 2 g max{5 n 1,5 n 1 5} 0. Therefore when n > 0 we have A 1 = β 3 A 2 and when n 0, A 1 = A 2. Next, we see that F 5,nr = f n t + G 5,n P 2 F 5,1 f n 1 t + G 5,n P 2 = F 5,nr + G 5,n P 3. The degree of F 5,n r is max{5 n 4, 5 n 1 5} and the degree of F 5,n r is n. Let A 3 = ResF 5,n r, G 5,nr in terms of the variable r. Then A 2 = A 3 β 4n 1 when n > 0 and A 2 = A 3 β 4 n when n 0. With the terms as defined in the statement of the lemma, and F G 5,n = c t 2 at + b, 5,nr = C f n t t + A + Bf n 1 t. Let A 4 = Res Cf n t t + A + Bf n 1 t, ct 2 at + b, where the resultant is taken with respect to the variable r. Therefore, A 3 = A 4. Let A 5 = Res Cf n xx + A + Bf n 1 x, cx 2 ax + b. The leading coefficient of t is ll 1 2 M M M 2 β 4. Since the degree of Cf n xx + A + Bf n 1 x is n we conclude that A 4 M 2 β 4 2 n = ll 1 2 M M n A5. With A 6 = Resf n xx + A + Bf n 1 x, x 2 ax + b we see that A 5 = A 6 C 2 c n which is A 5 γ 2 l 1 4 n M n M n l 2 n = A 6 β 9 n +2 M 4 n. The lemma follows. Because of the M 2 in the t 5 r equation, A 1 is a polynomial divided by M 4 n. Therefore, we normalize such that the A-polynomial is A5, 2n = M 4 n A 1. By Lemma 7.6, A5, 2nγ 2 = M 4 n β 5 n +ɛ R n.

28 28 K. L. PETERSEN It now suffices to modify the recursion using Proposition 5.9 to remove the factors of γ 2 and β ɛ β 5 n. Let c 1 = b 1 and c 2 = 2 a 2 + 3b. As a result, for n > 5 A5, 2n = β 5 M 4 c 1 A5, 2n 1 + β 5 M 4 2 c 2 A5, 2n 2 + β 5 M 4 3 c 2 A5, 2n 3 + β 5 M 4 4 c 1 A5, 2n 4 β 5 M 4 5 A5, 2n 5. In fact, setting A5, 0 = β 1 we can begin the recursion one term earlier. For negative n, we have and A5, 2nγ 2 = M 4 n β 5 n +2 R n A5, 2n = β 5 M 4 c 1 A5, 2n β 5 M 4 2 c 2 A5, 2n β 5 M 4 3 c 2 A5, 2n β 5 M 4 4 c 1 A5, 2n + 4 β 5 M 4 5 A5, 2n + 5. As for the positives, setting A5, 0 = β 1 we can begin the recursion one term earlier. Finally, we write this as a self contained theorem. Recall that since k = 5 we have l = LM 4n. Similar to the case when k = 3, the results that helped us determine the recursion for the resultant rely on the recursive definitions of F 5,n r and G 5,n r as polynomials in l, M ±2 and r. Therefore, to write A5, 2n in terms of the variables L and M, one uses the recursive definition below to determine the polynomial in l and M and then substitutes l = LM 4n. Theorem 7.7. With d 1, d 2, d 3, d 4, and d 5 as in the appendix, we have the following. For n positive A5, 2n = d 1 A5, 2n 1 + d 2 A5, 2n 2 + d 3 A5, 2n 3 + d 4 A5, 2n 4 + d 5 A5, 2n 5. The initial conditions are A5, 2n = ResM 2n F 5,n r, G 5,n r for 0 < n 5. For n negative A5, 2n = d 1 A5, 2n d 2 A5, 2n d 3 A5, 2n d 4 A5, 2n d 5 A5, 2n + 5. The initial conditions are A5, 2n = ResM 2 n F 5,n r, G 5,n r for 5 n 0.

29 A-POLYNOMIALS OF A FAMILY OF TWO-BRIDGE KNOTS J2m, 2m. The knots J2m, 2m exhibit an additional symmetry, seen by turning the 4-plat diagram for these two-bridge knots upside down. This effectively factors the character variety, and the canonical component is birational to C see [11]. Moreover, the canonical component of the A- polynomial is determined by the resultant of t m r r and G 2m,m. In this case, l = L. The equations are, up to sign t m r = f m rg m+1 r g m rm 2 + M 2 r + 2 G 2m,m r = f m rα βr γf m 1 r. The polynomial t m r r is reducible. Since t m r r is given by f m rg m+1 r g m rm 2 + M 2 r + 2 r it suffices to see that 2 r is a factor of g m+1 r g m r. Since f j 2 = j for all j, it follows that g j 2 = f j 2 f j 1 2 = j j 1 = 1 and the fact that for all j, g j+1 2 g j 2 = 0 follows. Since r = 2 corresponds to the reducible representations, we do not want to include the contribution of these to the A-polynomial. However, the format of tr r as above is easy to use, so we will compute the resultant of G 2m,m r and tr r and then divide by the extra factor which corresponds to this extra term. The contribution of this extra factor is Resr 2, G 2m,m r = G 2m,m 2 = γ. For small values of m, the resultant can be computed directly using a computer algebra program. We show that this resultant can be computed recursively using Proposition 5.9. Let a = 2σ βδ, b = τ βδ, A = α β, and B = γ β. Let R m = Resr 2 ar+b, f m rr+a+bf m 1 r and let A 1 = Rest m r r, G 2m,m r. Lemma 7.8. For m 0 we have γg 0 Rest m r r, G 2m,m r = β 2 βδ m R m where g 0 is the leading coefficient of G 2m,m r. Proof. Let A 2 = Resδβr σt m r r, G 2m,m. First, notice that A 2 = A 1 Resδβr σ, G 2m,m and since degg 2m,m = m using Lemma 6.2 we have Resδβr σ, G 2m,m = γg 2 m 0. Therefore, A 2 = γg 2 m 0 A 1. By Lemma 6.5 there is a polynomial P 3 such that It follows that t m rδβr σ + τ σr + G 2m,m rp 3 r = 0. δβr σt m r r = βδr 2 + 2σr τ G 2m,m rp 3 r. Let A 3 = Resβδr 2 2σr + τ, G 2m,m r.