arXiv:math/0508138v4 [math.GT] 4 Mar 2006 where ipiilvlm ftecmlmn of complement the of volume simplicial ioaih nain n h yeblcvlm ftecmlmn of complement the of volume hyperbolic the and invariant dilogarithm tetrahedron. yeblcpee nteJJdcmoiino h opeetof complement the the of form, JSJ-decomposition original the in pieces hyperbolic ojcue11(ahe 2,Mrkm-uaai[5]). Murakami-Murakami [2], (Kashaev 1.1 Conjecture reformulate follows. and as Kashaev Murakami by and Murakami proposed was conjecture volume The Introduction 1 ealthat Recall ahmtc ujc lsicto 00 72,57N10 57M25, 2000: Classification knot Subject torus Mathematics double, Whitehead conjecture, volume Keywords: ro ftevlm ojcuefrWhitehead for conjecture volume the of Proof h ouecnetr o htha obe fafml fto of family a observations. of interesting doubles some presented Whitehead show is for , conjecture of volume doubles the Whitehead the by illustrated J K,N ehiu ocluaeteclrdJnsplnmaso sa of polynomials Jones colored the calculate to technique A ste(omlzd ooe oe oyoilof polynomial Jones colored (normalized) the is obe fafml ftrsknots torus of family a of doubles v 3 eateto ahmtc,ZoghnUniversity Zhongshan Mathematics, of Department k S 3 2 π \ K N K -al [email protected] E-mail: lim →∞ shproi n h qaini ntrso h quantum the of terms in is equation the and hyperbolic is k sntigbttesmo h yeblcvlmsof volumes hyperbolic the of sum the but nothing is unzo 125 China 510275, Guangzhou log

J a Zheng Hao K,N Abstract N ( e K 2 1 π √ N and − 1 )

v = 3 stevlm fteielregular ideal the of volume the is v 3 k S 3 \ K k ,Wieedlink Whitehead s, hnw prove we Then . K u nt and knots rus , o n knot any For elt knots, tellite K n endby refined and d k S nKashaev’s In . 3 \ K K . k sthe is (1.1) K , The conjecture is marvellous in the sense that it reveals the topological meaning of the quantum invariants of knots which is quite unobvious from definition. How- ever, it also turns out to be rather hard to be proved. Till now, besides positive numerical evidences (ref. [1, 6]) for some hyperbolic knots, only the cases of torus knots (Kashaev-Tirkkonen [3]) and the simplest hyperbolic knot, the figure 8 knot (ref. [2]) have been verified. In view of the compatible behavior of both sides of the conjectured equation (1.1) under connect sum

J = J J , (1.2) K1♯K2,N K1,N · K2,N S3 K ♯K = S3 K + S3 K , (1.3) k \ 1 2k k \ 1k k \ 2k the volume conjecture, in fact, may be reduced to the consideration of prime knots. By Thurston’s Hyperbolization Theorem (ref. [7]), the prime knots further fall into three families: torus knots, hyperbolic knots and satellite knots. In this article, we deal with the conjecture by examining a special case of the third family, the Whitehead doubles of torus knots. The approach is emphasized on the relation between the colored of a and those of the associated companion knot and pattern . In particular, we show a technique to calculate the colored Jones polynomial of satellite knots by cutting and gluing method.

A Whitehead double of a knot K is a knot obtained as follows. Remove the regular neighborhood of one component of the Whitehead link from S3 thus get a knot inside a torus, then knot the torus in the shape of a knot K. Note that, when K is nontrivial, a Whitehead double K′ of K is a satellite knot whose complement contains an obvious essential torus T 2. Cutting along the torus,

2 we get 3 2 3 3 (S K′) T = (S Whitehead link) (S K) (1.4) \ \ ∼ \ ∪ \ thus 3 3 3 S K′ = S Whitehead link + S K . (1.5) k \ k k \ k k \ k In particular, if K is a nontrivial , the complement of K is Seifert fibred and the complement of the Whitehead link is hyperbolic, hence 3 3 v S K′ = vol(S Whitehead link). (1.6) 3k \ k \ The article proceeds as follows. First, we calculate the colored Jones polynomi- als of the twisted Whitehead links and the Whitehead doubles of knots in section 2. Next, as a warming-up we prove in the next two consecutive sections the following two theorems, of which the former one is, in fact, the volume conjecture for twisted Whitehead links and both extends the estimation (1.1) to the second order. Theorem 1.2. For every twisted Whitehead link L, we have

2π√ 1 3 2π log J (e N− ) = vol(S L) N +3π log N + O(1) (1.7) L,N \ · as N . →∞ Theorem 1.3. For every nontrivial torus knot T (p, q) with q =2, we have

2π√ 1 − 2π log JT (p,q),N (e N ) =3π log N + O(1) (1.8) as N . →∞ Then we prove the main theorem in Section 5 and show some observations in the final section. Theorem 1.4. If K is a Whitehead double of a nontrivial torus knot T (p, q) with q =2, then

2π√ 1 − 3 2π log J (e N ) = v S K N +4π log N + O(1) (1.9) K,N 3k \ k· as N . In particular, the volume conjecture is true for K. →∞ Remark 1.5. In their proof of the volume conjecture for torus knots, Kashaev and Tirkkonen [3] derived the following estimation

2π√ 1 − 2π log JT (p,q),N (e N ) = O(log N). (1.10)

But improving the estimation to (1.8) requires a nonvanishing proposition on num- ber theory (see Proposition 4.1) to which both Theorem 1.3 and Theorem 1.4 are reduced in this article. With a technical condition q = 2 we proved the nonvanish- ing proposition in section 4. A complete proof has been beyond the scope of the article. We only mention here that our technique can be sharpened to prove the nonvanishing proposition, hence both theorems, at least for the cases that both p, q are odd or one of them is a power of 2.

3 Remark 1.6. It is noteworthy that the coefficient “4π” of the second term in the asymptotic expansion (1.9) disagrees with the observation due to Hikami [1]

2π√ 1 3 2π log J (e N− ) = v S K N +3π log N + O(1) (1.11) K,N 3k \ k·

for many prime knots K.

2 Calculation of colored Jones polynomial

In this section, we calculate the colored Jones polynomials of the twisted Whitehead link W L(r) and the Whitehead double WD(K,r) of a knot K.

double(K) r twists r twists PSfrag replacements

WL(r) WD(K,r) In the figure, double(K) denotes the (2,2)-tangle obtained by doubling the knot K to a link with zero and then removing a pair of parallel segments. Our trick is cutting the link diagrams into (2,2)-tangles and gluing the tangle invariants together.

PSfrag replacements

twist belt double(K) clasp Colored Jones polynomial is also defined for tangles, but, instead of a Lau- rent polynomial of t, it is in general a module homomorphism of Uq(sl2) (choose t = q2). Especially, the colored Jones polynomial of a (2,2)-tangle is a module homomorphism V V V V (2.1) N ⊗ N → N ⊗ N 4 where VN is the N dimensional irreducible representation of Uq(sl2). Note that the tensor product admits the decomposition

N 1 − V V = V . (2.2) N ⊗ N 2n+1 Mn=0 A straightforward calculation shows that the (framing independent, unnormalized) colored Jones polynomials of the tangles are

N 1 − J˜ = tn(n+1) id , (2.3) twist,N · V2n+1 Mn=0 N 1 N(2n+1)/2 N(2n+1)/2 − t t− J˜ = − id , (2.4) belt,N t(2n+1)/2 t (2n+1)/2 · V2n+1 Mn=0 − − N 1 − J˜ = J id , (2.5) double(K),N K,2n+1 · V2n+1 Mn=0 N 1 − J˜ = ξ id , (2.6) clasp,N N,n · V2n+1 Mn=0 where

N 1 n n − − N i j i+j (N 2 1)/2+N(N 1)/2 N(i+n) (1 t − − )(1 t ) ξ = t − − t− − − . (2.7) N,n 1 tj Xi=0 Yj=1 − Combining the tangle invariants together, one has

N 1 (2n+1)/2 (2n+1)/2 N(2n+1)/2 N(2n+1)/2 − t t− t t− J = − trn(n+1) ξ − , (2.8) WL(r),N tN/2 t N/2 · · N,n · t(2n+1)/2 t (2n+1)/2 Xn=0 − − − − and N 1 (2n+1)/2 (2n+1)/2 − t t− J = − trn(n+1) ξ J . (2.9) WD(K,r),N tN/2 t N/2 · · N,n · K,2n+1 Xn=0 − − Note that, in the expression of JWD(K,r),N , the factor JK,2n+1 is contributed by the companion knot K and the other part is precisely obtained from the expression of JWL(r),N by removing the factor contributed by the belt tangle.

5 3 Proof of Theorem 1.2

2π√ 1 Let L denote the twisted Whitehead link W L(r). Setting t = e N− , we have

N 1 N 1 n n i j i+j 2π√ 1 − − − − 1/2 rn(n+1) (1 t− − )(1 t ) J (e N )= t− (2n + 1)t − − L,N − 1 tj Xn=0 Xi=0 Yj=1 − (3.1) N 1 N 1 n π√ 1 − − − − 4r 1 = e− N (2n + 1)a − S − n n,i Xn=0 Xi=0 where n(n+1) n n(n+1 N) π√ 1 π√ 1 − π√ 1 an = e 2N − − 2 − = e 2N − (3.2) and n 2 (i+j)π 4 sin N Sn,i = . (3.3) 2 sin jπ Yj=1 N

First, we prepare a lemma to estimate the norm factor Sn,i. Put

n jπ sn = log 2 sin (3.4) − N Xj=1 and let x L(x)= log 2 sin u du (3.5) − Z0 | | be the Lobachevsky function.

Lemma 3.1. For 0 <α< 1 we have uniform estimations

N mπ N nπ 1 s s = L( ) L( )+ O(N − )(m n) (3.6) m − n π N − π N − on α N

6 Proof. We have

jπ jπ N N log 2 sin + log 2 sin u du (j 1)π − N π Z − | | N π jπ N N jπ = log 2 sin + log 2 sin( u) du (3.9) − N π Z0 N − π jπ N N sin( N u) = log jπ− du. π Z0 sin N

Since sin(x u) log − = O(u) (3.10) sin x α α uniformly on x [ π, (1 )π] as u 0, the first estimation follows as ∈ 2 − 2 → N mπ N nπ s s L( )+ L( ) m − n − π N π N m π jπ N sin( u) N N (3.11) = log jπ− du π Z0 sin j=Xn+1 N 1 = O(N − )(m n). − Note that sin(x u) x − =1+ O(u) (3.12) x u · sin x − thus sin(x u) x u log − = log − + O(u) (3.13) sin x x uniformly on x [ απ,απ ] 0,u as u 0. It follows that ∈ − \{ } → n π jπ N N nπ N sin( N u) sn L( )= log jπ− du − π N π Z0 sin Xj=1 N n π jπ N N N u 1 = log jπ− du + nO(N − ) π Z0 Xj=1 N (3.14) n j 1 = (j 1) log − 1 + O(1) − − j − Xj=1   n! = log n + O(1) − nn − uniformly on 0

L(x)+ L(π x) = 0 (3.16) − and sn 1 + sN n = sN 1. (3.17) − − − In particular, we have π L( ) = 0 (3.18) 2 and, by the second estimation,

sN 1 = s[ N 1 ] + s[ N ] = log N + O(1). (3.19) − 2− 2 − Therefore, nπ sn = sN 1 sN n log 2 sin − − − − N N nπ 1 2(N n)π = log N + L( )+ log( N n) log − + O(1) (3.20) − π N 2 − − N N nπ 1 = L( ) log(N n)+ O(1) π N − 2 − uniformly on (1 α)N

8 1 Lemma 3.2. For any 2 <δ< 1 there exist ǫ> 0 and C > 0 such that

2δ 1 ǫN − Sn,i < Ce− S N N (3.25) [ 2 ],[ 4 ] for n N + i N N δ | − 2 | | − 4 | ≥ Proof. Since f has a unique critical point ( π , π ) in the region 0 x, y, x + y π, 2 4 ≤ ≤ we have f(x, y) max f(x′,y′) (3.26) ≤ x π + y π =πN δ 1 | ′− 2 | | ′− 4 | − π π δ 1 for x + y πN − . By (3.24), there exist ǫ> 0 and C′ > 0 such that | − 2 | | − 4 | ≥ π π δ 1 2 max f(x′,y′) < f( , ) 2πǫ(N − ) + C′ (3.27) x π + y π =πN δ 1 2 4 − | ′− 2 | | ′− 4 | −

Therefore, by (3.21) there exists C′′ > 0 such that

2δ 1 log Sn,i < log S[ N ],[ N ] ǫN − + C′′ (3.28) 2 4 − for n N + i N N δ. | − 2 | | − 4 | ≥ 1 2 Lemma 3.3. For any α 0, β R and 2 <δ< 3 there exists a nonzero constant C C such that ≥ ∈ ∈ α β α+1 βN π√ 1 3δ 2 (2n + 1) a Sn,i = CN e− 8 − S N N (1 + O(N − )). (3.29) n [ 2 ],[ 4 ] n N +Xi N

N N Proof. For simplicity, we use the notation n′ = n , i′ = i in the proof. Note − 2 − 4 that

2 α π (n 2+2n i +2i 2)+ βn′ π√ 1 (2n + 1) e− N ′ ′ ′ ′ 2N − n N +Xi N

N nπ iπ N π π δ 1 log Sn,i log S[ N ],[ N ] = f( , ) f( , )+ O(N − ) − 2 4 π N N − π 2 4 (3.31) π 2 2 3δ 2 = (n′ +2n′i′ +2i′ )+ O(N − ) −N

9 N N δ uniformly on n 2 + i 4 < N . Moreover, on the same region we have the uniform estimation| − | | − |

n(n+1 N) n 2 N n − π√ 1 ( ′ + )π√ 1 an = e 2N − = e 2N − 8 2N − 2 (3.32) ( n′ N + 1 )π√ 1 δ 1 = e 2N − 8 4 − (1 + O(N − )).

Therefore, by (3.30),

2 2 2 βx √ α β π(x +2xy+2y )+ 2 π 1 (2n + 1) anSn,i = e− − dxdy ZR2 n N +Xi N

2 β π√ 1 π(x2+2xy+2y2)+ βx π√ 1 C = e 4 − e− 2 − dxdy. (3.34) ZR2

It follows from Lemma 3.2 and Lemma 3.3 that, in the same notations as the lemmas,

2δ 1 4r 1 3 ǫN − (2n + 1)a − Sn,i = N e− S N N O(1), n [ 2 ],[ 4 ] n N +Xi N N δ | − 2 | | − 4 |≥ (3.35) 4r 1 2 O(1) (2n + 1)a − Sn,i = N S N N e , n [ 2 ],[ 4 ] n N +Xi N

From (3.7) we also have 4N π 1 log S[ N ],[ N ] = L( ) log N + O(1). (3.37) 2 4 π 4 − 2 Therefore,

2π√ 1 π 2π log J (e N− ) =8L( ) N +3π log N + O(1) L,N 4 · (3.38) = vol(S3 L) N +3π log N + O(1) \ · as N . In the last row, we used the fact →∞ π vol(S3 L) = vol(S3 Whitehead link) = 8L( ). (3.39) \ \ 4

10 4 Proof of Theorem 1.3

The colored Jones polynomial of the torus knot T (p, q) was calculated in [4] as

2 (n 1)/2 pq(n 1)/4 − t− − pk(qk+1) qk+1/2 qk 1/2 JT (p,q),n = n/2 n/2 t (t t− − ). (4.1) t t− − k= X(n 1)/2 − − − We put pq 1 − 2 j Nj π√ 1 2k jπ jπ A± (N,k)= ( 1) e− 2pq − j sin sin . (4.2) p,q ± p q Xj=1 Note that

Ap,q± (N,k)= Ap,q∓ (N +2pq,k)= Ap,q± (N +4pq,k), (4.3) so ( 1)n + Ap,q− (N,k)= Ap,q(N +2npq,k), + ( 1)n (4.4) Ap,q(N,k)= Ap,q− (N +2npq,k).

2π√ 1 − In [3], an estimation of JT (p,q),N (e N ) was derived as

2 3/2 2π√ 1 N 1 p q π√ 1 π√ 1 N 1 − pq − π√ 1 N ( + ) − + − ( 1) J (e N )=2e− 2N − e− q p 2N 4 A − − (N, 1)+ O(1). T (p,q),N (2pq)3/2 p,q (4.5)

In view of the periodicity of Ap,q± , to establish the theorem it suffices to show that N 1 ( 1) − Ap,q− (N, 1) never vanishes if q = 2, or equivalently by (4.4), Proposition 4.1. Let p, q 2 be coprime integers with q = 2. Then for every ≥ integer N, pq 1 − 2 + Nj π√ 1 2 jπ jπ A (N, 1) = e− 2pq − j sin sin =0. (4.6) p,q p q 6 Xj=1 The proof of the proposition is purely arguments on elementary algebraic num- 2π√ 1 − ber theory. In the following, we write ζn = e n for each n N. An algebraic number field means a finite extension of Q contained in C. ∈ For any finite extension E/K of field, one has a K-linear map tr : E E/K → K, called the trace function, which values on x E the trace of the K-linear transformation ρ : E E given by ρ (z)= xz. ∈ x → x Lemma 4.2. Let α be a prime, k,l N and K be an algebraic number field such ∈ that K Q(ζ k+l )= Q. Then we have ∩ α k n 0, α ∤ n, tr (ζ k+l )= (4.7) K(ζαk+l )/K(ζαl ) α k n k  α ζ k+l , α n. · α |

11 i k Proof. The field extension K(ζαk+l )/K(ζαl ) has a basis ζαk+l 0 i<α on n { | k≤ }n which the diagonal of the matrix of ζαk+l consists of only 0 if α ∤ n, or ζαk+l otherwise.

Lemma 4.3. Let α be a prime and K be an algebraic number field such that K Q(ζα)= Q. Then ∩ α 1 − c ζj =0, (4.8) j · α Xj=0 for c K if and only if the c ’s are identical. j ∈ j 2 α 2 Proof. On one hand, the field extension K(ζα)/K has a basis 1,ζα,ζα,...,ζα− . α 1 j { } On the other hand, we have j=0− ζα = 0. Therefore, the summation vanishes if and only if the cj’s are identicalP to cα 1. − Thanks to the next lemma, we are able to eliminate the Gaussian exponential appearing in the expression of Ap,q± . Lemma 4.4. Let α be an odd prime, l N and K be an algebraic number field ∈ such that K Q(ζ l )= Q. Assume that ∩ α Nj2+2aj c ζ− = 0 (4.9) j · αl Xj X ∈ where X is a finite subset of Z, c K and α ∤ a. Then we have j ∈ c ζ2aj = 0 (4.10) j · αl j X: j X0 mod αl 1 ∈ ≡ − if α N, or otherwise, |

cj =0. (4.11) j X: Nj aXmod α[(l+1)/2] ∈ ≡

Proof. If α N, taking the trace function of K(ζαl )/K(ζα) on both sides of the equality assumed,| we find from Lemma 4.2 that

c ζ2aj =0. (4.12) j · αl j X: j X0 mod αl 1 ∈ ≡ − Otherwise, choose b Z such that bN 1 mod αl. From the assumption we have ∈ ≡ b(Nj a)2 ba2 Nj2+2aj c ζ− − = ζ− c ζ− =0. (4.13) j · αl αl j · αl Xj X Xj X ∈ ∈

12 Taking the trace function of K(ζαl )/K(ζα) on both sides of the equality, we get

α 1 − c ζk =0. (4.14)  j · α Xk=0 j X: b(Nj aX)2 kαl 1 mod αl ∈ − − ≡ −

2 l 1 l Since α is an odd prime, the congruence equation x kNα − mod α has no ≡ − k solution for some 0

cj =0. (4.15) j X: b(NjXa)2 0 mod αl ∈ − − ≡ Hence the lemma follows.

+ Proof of Proposition. Assume that Ap,q(N, 1)=0 for

p = αl1 αl2 αlr βk1 βk2 βks , (4.16) 1 2 ··· r · 1 2 ··· s where the αi’s and βi’s are distinct odd primes not dividing and dividing N, re- + spectively. Rewrite Ap,q(N, 1)=0 as

1 2 Nj2+2qj j j j ζ− (ζ ζ− ) = 0 (4.17) −4 · 4pq · 2q − 2q pq

σ(ζ )= ζ ζ l1 ζ lr ζ k1 ζ ks . (4.18) 4pq 4q α αr β βs · 1 ··· · 1 ··· Under the Galois action of σ, the equality becomes

1 2 Nj2+2qj pj pj j (ζ ζ l1 ζ lr ζ k1 ζ ks )− (ζ ζ− )=0. (4.19) 4q α αr β βs 2q 2q −4 · · 1 ··· · 1 ··· · − pq

[(l1+1)/2] [(lr+1)/2] k1 1 ks 1 α = α α , β = β − β − , p′ = β β . (4.20) 1 ··· r 1 ··· s 1 ··· s It follows from Lemma 4.4 that

1 2 2qj Nj2+2qj pj pj j (ζ k1 ζ ks ) ζ− (ζ ζ− ) = 0 (4.21) β βs 4q 2q 2q −4 · 1 ··· · · − Xj X ∈ where X = pq

Nj2+4j pj pj 0, 2 j, ζ8− (ζ4 ζ4− )= N (j 1)/2 p p | (4.23) · −  ζ− ( 1) − (ζ ζ− ), 2 ∤ j. − 8 · − 4 − 4 Dropping a nonzero factor, (4.21) becomes

2 4j/β (j 1)/2 j ζ ( 1) − =0. (4.24) · p′ · − j XX:2∤j ∈

Choose 0 j0 < 2α so that Nβj0 2 mod α and j0 1 mod 2. Then the left hand side≤ of above equality, up to a≡ sign, is ≡

8pβζ4j0 2 2 8αj+4j0 j p′ 2α β (2αj + j0) ζp ( 1) = 8α j0 8α . (4.25) · ′ · − 1+ ζ  − 1+ ζ−  p/αβXj

Therefore, we must have p′ = 1 and j0 = α. But from the choice of j0, it follows that α = 1. Hence p = 1, a contradiction.

5 Proof of Theorem 1.4

Let K denote the r-twisted Whitehead double of the torus knot T (p, q). Then

N 1 1 − J = trn(n+1)ξ Jˆ (5.1) K,N tN/2 t N/2 N,n T (p,q),2n+1 − − Xn=0 where n/2 n/2 Jˆ =(t t− )J . (5.2) T (p,q),n − T (p,q),n 2π√ 1 − N/2 N/2 Setting t = e N , one notices that the denominator t t− vanishes. There- fore, one has to apply L’Hospital’s rule, i.e. take derivative− of both the denominator and the numerator. It follows that

N 1 N 1 n 1/2 − − − t− 4r 1 d J = − a − S b Jˆ + t Jˆ (5.3) K,N N n n,i n,i T (p,q),2n+1 dt T (p,q),2n+1 − Xn=0 Xi=0   where

n N i j i + j j b = rn(n + 1) N(i + n)+ − − + (5.4) n,i − − 1 t N+i+j − 1 t i j 1 t j Xj=1  − − − − − − −  and an,Sn,i are same as Section 3.

14 Below, we follow the approach used in [3] to derive an estimation of JˆT (p,q),n d ˆ and t dt JT (p,q),n in the form of (4.5). For any complex number h with Im(h) > 0, one has the integral formula

1/2 2 h pq(n2 1) h pq ( p + q ) h pq nz z JˆT (p,q),n(e )= e− − 4 e− q p 4 dze − h τ(z) (5.5) πh ZC 

π√ 1 where the contour C is given by the line e 4− R and

pz pz qz qz (e e− )(e e− ) τ(z)= − − . (5.6) epqz e pqz − − 2π√ 1 Lemma 5.1. For h = N− we have

k 2 2 k d pq nz z 1 N ( 1)n 1 2k 1/2 − h √ − − − k dze τ(z)= 4π 1 Ap,q (N,k)+ O(N ) (5.7) dh ZC  − − pq 4pq  uniformly on n N < N . | − | 2pq Proof. Put z = n h = n π√ 1. We have 0 2 N −

2 z2 2 pq nz z 2k pq 0 pq z 2k 1/2 dze − h z τ(z)= e h dze− h (z + z0) τ(z + z0)= O(N − ) ZC+z0  ZC (5.8) uniformly on n N < N , since the function z2kτ(z) is bounded on the region | − | 2pq

π√ 1 1 z e 4− x + yπ√ 1 x, y R, y 1 < . (5.9) ∈ n − ∈ | − | 2pq o

jπ√ 1 Counting the residues of the integrand at pq− , 0

2 2k+1 pq nz z 2k π√ 1 ( 1)n 1 ( )dze − h z τ(z)= 4 − A − − (N,k). (5.10) pq p,q ZC − ZC+z0  −   Therefore,

k 2 k 2 d pq nz z pq pq nz z 2k − h − h k dze τ(z)= 2 dze z τ(z) dh ZC  h  ZC  2 (5.11) k n 1 1 N ( 1) − 2k 1/2 = 4π√ 1 Ap,q− (N,k)+ O(N − ) − − pq 4pq  uniformly on n N < N . | − | 2pq

15 2π√ 1 Lemma 5.2. For t = e N− we have

JˆT (p,q),n = O(1) (5.12) and

2 5/2 n 1 p q π√ 1 π√ 1 n 1 d pq − π√ 1 N ( + ) − + − ( 1) 2 t Jˆ = 2e− 2N − e− q p 2N 4 A − − (N, 1)+ O(N ) dt T (p,q),n − (2pq)3/2 p,q (5.13) uniformly on n N < N . | − | 2pq Proof. From (5.5) and Lemma 5.1 we have

2 1/2 n 1 p q π√ 1 π√ 1 n 1 pq − π√ 1 N ( + ) − + − ( 1) Jˆ = 4e− 2N − e− q p 2N 4 A − − (N, 0)+ O(1). (5.14) T (p,q),n − (2pq)1/2 p,q

It follows from the periodicity of Ap,q± and the identity

2π√ 1 2π√ 1 Jˆ (e N− )=0 J (e N− ) = 0 (5.15) T (p,q),N · T (p,q),N N 1 ( 1) − that Ap,q− (N, 0), hence by (4.4) Ap,q± (N, 0), identically vanishes. So, the leading term of the right hand side of (5.14) is zero. Then, applying Lemma 5.1 to the derivative of (5.5), one obtains the second estimation. Now we conclude the proof of the theorem. It is clear that d b Jˆ + t Jˆ = O(N 3)O(N)+ O(N 3)= O(N 4) (5.16) n,i T (p,q),2n+1 dt T (p,q),2n+1 uniformly on 0 n, i, n + i < N. Moreover, for 0 <α< 1, since the function ≤ x e√ 1xx = − (5.17) 1 e 2√ 1x 2√ 1 sin x − − − − is bounded on x [0, απ], we have ∈ 2 bn,i = O(N ) (5.18) uniformly on 0 < n, i, n + i<αN. It follows from Lemma 5.2 that

2 bn,iJˆT (p,q),2n+1 = O(N ) (5.19) and

5/2 p q π√ 1 π√ 1 d 4pq N ( + ) − + − + 2 t Jˆ = 2(a )− e− q p 2N 4 A (N, 1)+ O(N ) (5.20) dt T (p,q),2n+1 − n (2pq)3/2 p,q

16 N N N uniformly on n 2 + i 4 < 4pq . Therefore,| in− the| case| − that| q = 2, by Lemma 3.2, Lemma 3.3 and Proposition 4.1, in the same notations as the lemmas we have

d 2δ 1 4r 1 ˆ ˆ 6 ǫN − an − Sn,i bn,iJT (p,q),2n+1 + t JT (p,q),2n+1 = N e− S[ N ],[ N ]O(1), dt 2 4 n N +Xi N N δ   | − 2 | | − 4 |≥ 4r 1 ˆ d ˆ 7/2 O(1) an − Sn,i bn,iJT (p,q),2n+1 + t JT (p,q),2n+1 = N S[ N ],[ N ]e , dt 2 4 n N +Xi N

2π√ 1 − 5/2 2π log JK,N (e N ) =2π log(N S N N )+ O(1) [ 2 ],[ 4 ] π =8L( ) N +4π log N + O(1) (5.22) 4 · = v S3 K N +4π log N + O(1) 3k \ k· as N . →∞ 6 Concluding remarks

Although the proof of Theorem 1.4 depends on the simple nature of the Whitehead doubles of torus knots, the approach still works for the satellite knots to which the colored Jones polynomials of the associated companion knot and pattern link satisfy certain mild conditions. Meanwhile, one has to deal with several problems. For example, as shown in expression (2.9), although the volume conjecture itself is only concerned with the value of the colored Jones polynomial JK,N at the N-th root of unity, the values at other roots of unity become crucial once the satellite knots of K are involved. A more challenging problem is due to the estimations (5.19) and (5.20), which have enabled us to neglect the term bn,iJˆT (p,q),2n+1 in summation (5.3). Note that the derivative of the polynomial

N/2 N/2 Jˆ =(t t− )J (6.1) K,N − K,N is related to JK,N via the identity

d 2π√ 1 2π√ 1 t Jˆ (e N− )= NJ (e N− ). (6.2) dt K,N − K,N d ˆ Therefore, the term t dt JT (p,q),2n+1 in summation (5.3), in fact, plays the role of 2π√ 1 − JT (p,q),N (e N ). Hence it is quite natural to see the term bn,iJˆT (p,q),2n+1 be sup- pressed. Following this observation, when the Whitehead doubles of general knots are considered, it is reasonable to expect that a similar suppression also happens.

17 Conjecture 6.1. For every nontrivial knot K we have

2π√ 1 ˆ − JK,2n+1(e N ) 2 2π√ 1 = o(N − ) (6.3) d ˆ N− t dt JK,2n+1(e ) uniformly on n N < N δ for some 1 <δ< 2 . | − 2 | 2 3 Note that the conjecture excludes the case of , for which the statement of the conjecture is obviously false. Indeed, the Whitehead doubles of unknot are no longer satellite knots but the so called twist knots (including unknot, trefoil, figure 8, etc.), whose complements always admit a volume strictly smaller than that of Whitehead link. In the sequel, the conjecture has an interesting implication: the colored Jones polynomial detects unknot.

Acknowledgement

The author is grateful to Xiao-Song Lin for enlightening discussions. He also thanks Lingquan Kong who pointed out to the author the approach to proving the nonva- nishing proposition.

References

[1] K. Hikami, Volume conjecture and asymptotic expansion of q-series, Exper. Math. 12 (2003), 319–337. [2] R. M. Kashaev, The of knots from quantum dilogarithm, Lett. Math. Phys. 39 (1997), 269–275. [3] R. M. Kashaev, O. Tirkkonen, Proof of the volume conjecture for torus knots, J. Math. Sci. 115 (2003), 2033–2036. [4] H. R. Morton, The coloured Jones function and for torus knots, Proc. Cambr. Phil. Soc. 117 (1995), 129–135. [5] H. Murakami, J. Murakami, The colored Jones polynomials and the simplicial volume of a knot, Acta Math. 186 (2001), 85–104. [6] H. Murakami, J. Murakami, M. Okamoto, T. Takata, Y. Yokota, Kashaev’s conjecture and the Chern-Simons invariants of knots and links, Exper. Math. 11 (2002), 427–435 [7] W. P. Thurston, Three-dimensional manifolds, Kleinian groups, and hyper- bolic geometry, Bull. A.M.S. 6 (1982), 357–381.

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