31, 41, and 63 knots of Garoufalidis, Le and Zagier. Indeed a trio of ..... [4] George
Gasper and Mizan Rahman, Basic hypergeometric series, Encyclopedia of.
Knots and q-Series George E. Andrews∗
Abstract Generalizations are given for the q-series identities arising from the 31 , 41 , and 63 knots of Garoufalidis, Le and Zagier. Indeed a trio of new parameters can be added in each case while preserving the infinite product side of the identities.
1
Introduction
In [3, p. 7], Garoufalidis and Le prove two surprising q-series/infinite product identities related to the 31 and 41 respectively: 3a2
−2 (q)∞
a
q 2 +ab+ac+bc+ 2 +b+c (−1) = . (q) a (q)b (q)c (q)a+b (q)a+c a,b,c≥0 X
a
b2
−3 (q)∞
=
X
b+d
(−1)
a,b,c,d,e≥0 a+b=d+e
d2
(1.1) a
e
q 2 + 2 +bc+ac+ad+be+ 2 +c+ 2 . (q)b+c (q)a (q)b (q)c (q)d (q)e (q)c+d
(1.2)
They then state: “... for the amphicheiral knot 63 , we conjecture that a
−4 (q)∞
=
X
a,b,c,d,e,f ≥0 a+e≥b,b+f ≥a
3a2
b
b2
e
3e2
2
(−1)a−b+e q 2 + 2 + 2 + 2 +c+ac+d+ad+cd+ 2 +2ae−2be+de+ 2 −af +bf +f . (q)a (q)b (q)c (q)a+c (q)d (q)a+d (q)e (q)a−b+e (q)a−b+d+e (q)f (q)−a+b+f (1.3)
It is indicated in [2, p. 13] that a knot-theoretic proof of (1.1), (1.2), and (1.3) may be deduced from results in [1]. ∗
Partially supported by National Security Agency Grant H98230-12-1-0205.
1
In these identities, (A)n = (A : q)n =
∞ Y (1 − Aq m ) . m+n ) (1 − Aq m=0
(1.4)
Note that now (A)n is defined for all integers n and 1/(q)n = 0 if n < 0. Our object in this paper is to prove the following identities which reduce to (1.1), (1.2), and (1.3) respectively when x = y = z = 1. Theorem 1. 2
X (−1)a q 3a2 + a2 +ab+ac+bc+b+c xa y a+b z c 1 = (yq)∞ (zq)∞ a,b,c≥0 (q)a (q)b (q)c (xq)a+b (yq)a+c b2
d2
(1.5) a
e
X (−1)b+d q 2 + 2 +bc+ac+ad+be+ 2 +c+ 2 xq y b z c 1 = (xq)∞ (yq)∞ (zq)∞ a,b,c,d,e≥0 (yq)b+c (q)a (q)b (q)c (q)d (q)e (xq)c+d
(1.6)
a+b=d+e
and 1 (wq)∞ (yq)∞ (zq)∞ (z −1 q)∞ 2
a
3a2
b
b2
e
3e2
q f +(b−a)f + 2 + 2 + 2 + 2 +c+ac+d+ad+cd+ 2 +2ae−2be+ 2 y e z c−d wf = (q)f (wq)b+f −a (q)a (q)b (q)c (xq)a+c (q)d (zq)a+d (q)e (q)a−b+e (yq)a−b+d+e a,b,c,d,e,f ≥0 X
(1.7) In Section 2, we provide the necessary results from the literature. Sections 3, 4, and 5 are devoted to the proofs of (1.5), (1.6), and (1.7) respectively. Finally in Section 6 we examine the q-series related to the 85 knot.
2
Background
We begin with some classical identities. The first two are due to Euler [?Gasper1990, (II.1) and (II.2), p. 236]. ∞ X 1 tn = , (q)n (t)∞ n=0
2
(2.1)
and ∞ X (−1)n tn q n(n−1)/2
(q)n
n=0
= (t)∞ .
(2.2)
Next the q-binomial theorem [?Gasper1990, (II.4), p. 236] j N X (−x)j q (2)
j=0
(q)j (q)N −j
=
(x)N . (q)N
(2.3)
The following lemma is well-known although perhaps not in the following generality [?Gasper1990, eq. (1.6.3), p. 12]. Lemma 2. For any integer A, −∞ < A < ∞, 2 ∞ X q n +An z n 1 = . (q) (zq) (zq) n n+A ∞ n=0
(2.4)
Proof. If A ≥ 0, this follows immediately from the corrected [?Gasper1990, eq. (1.6.3), p. 12, z → zq A ]. If A < 0, set A = −B, and 2 2 ∞ ∞ X X q n (zq −B )n q n −Bn z n 1−B = (zq )B (q)n (zq)n−B (q)n (zq −B q)n n=0 n=0
= (zq 1−B )B =
1 (zq −B )∞
(by the A = 0 case)
1 . (zq)∞
Lemma 3. For integers M and N with M > 0, ∞ X j=0
2 ∞ X q M j λj 1 q j +(M +N )j λj µj = . (q)j (µq)j+N (λq M )∞ j=0 (q)j (µq)j+N
Proof. ∞ X j=0
∞
X (τ )j (τ )j q M j λj q M j λi 1 = lim (q)j (µq)j+N (µq)N τ →0 j=0 (q)j (µq N +1 )j 3
(2.5)
2 M −N −1
=
−1
(τ q λµ )∞ 1 lim (µq)N τ →0 (λq M )∞
∞ X j=0
µq N +1 τ
2
τ 2 q M −N −1 λµ−1
j
j
(q)j (µq N +1 )j
(by [?Gasper1990, eq. (1.4.6), p. 107]) =
2 ∞ X q j +(M +N )j λj µj
1 (λq M )∞
j=0
(q)j (µq)j+N
.
Lemma 4. For integers B and C, ∞ X (−1)n q n(3n+1)/2+(B+C)n λn µn
(q)n (λq)n+B (µq)n+C
n=0
∞ 1 X (−1)n q n(n+1)/2+Bn λn = . (λq)∞ n=0 (q)n (µq)n+C
(2.6)
Proof. ∞ X (−1)n q n(3n+1)/2+(B+C)n λn µn
(q)n (λq)n+B (µq)n+C
n=0
=
∞ X
1 lim (λq)B (µq)C τ →0 n=0
1 3 τ n
3n n n n(2+B+C)
τ λ µ q
(q)n (λq B+1 )(µq C+1 )n
∞ 1 τ λq B+1 ∞ (λµq B+C+2 τ 2 )∞ X (τ µq C+1 )2n (τ λq B+1 )n 1 τ n = lim (λq)B (µq)C τ →0 (λq B+1 )∞ (τ 3 λµq B+C+2 )∞ n=0 (q)n (µq C+1 )n (λµq B+C+2 τ 2 )n (by [?Gasper1990, p. 241, eq. (III.9), d = µq C+1 , e = λq B+1 ]) =
3
∞ 1 X (−1)n q n(n+1)/2+Bn λn (λq)∞ n=0 (q)n (µq)n+C
Proof of (1.5)
Proof. 2
X (−1)a q 3a2 + a2 +ab+ac+bc+b+c xa y a+b z c (q)a (q)b (q)c (xq)a+b (yq)a+c a,b,c≥0 4
∞ 1 X q bc+b+c y b z c X (−1)n q n(n−1)/2+bn xn = (xq)∞ b,c≥0 (q)b (q)c n=0 (q)n (yq)n+c
(by Lemma 4, B = b, C = c, λ = x, µ = y) 1 1 X (−1)n q c+n(n+1)/2 z c xn · = 1+c+n (xq)∞ c,n≥0 (q)c (q)n (yq)n+c (yq )∞ (by (2.1))
4
X (−1)n q n(n+1)/2 xn X q c z c (q)n (q)c n≥0 c≥0
=
1 (xq)∞ (yq)∞
=
1 1 · (xq)∞ (xq)∞ (yq)∞ (zq)∞
=
1 . (yq)∞ (zq)∞
Proof of (1.6)
Proof. b2
d2
a
e
(−1)b+d q 2 + 2 +bc+ac+ad+be+ 2 +c+ 2 xa y b z c (yq)b+c (q)a (q)b (q)c (q)d (q)e (xq)c+d a,b,c,d.e≥0 X
a+b=d+e 3a2
2
e2
a
e
(−1)a+e q 2 +b + 2 +2ab+bc+ac−2ae+ 2 +c+ 2 xa y b z c = c (yq)b+c (q)a (q)b (q)c (q)a+b−e (q)e (xq)a+b+c−e a,b,c,d,e≥0 X
(replacing d by a + b − e) 2 b2 + e2 +bc+c+ 2e
∞ X (−1)e q ybzc 1 X (−1)j xj q j(j+1)/2+(b+c−e)j × = (yq)b+c (q)b (q)c (q)e (xq)∞ j=0 (q)j (q)j+b−e b,c,e≥0
(by (2.6) with B = b + c − e, C = b − e, λ = x, µ = 1) X q b2 +bc+c y b z c (−1)j xj q j(j+1)/2+(b+c)j X (−1)e q e(e+1)/2−ej 1 = × (xq)∞ b,c,j≥0 (yq)b+c (q)b (q)c (q)j (q)e (q)j+b−e e≥0 5
2 1 X z c q c X q b +bc y b = (xq)∞ c≥0 (q)c b≥0 (q)b (yq)b+c
(by (2.3) applied to e-sum, annihilating all j terms except j = 0) =
1 1 1 , (xq)∞ (zq)∞ (yq)∞
by (2.4)
5
Proof of (1.7)
Proof. We start with four additional parameters, x, y, z, and w. This allows us to obtain a more general result than (1.7). Subsequently we obtain (1.7) by setting x = 1/z. a
2
3a2
b
b2
e
3e2
(−1)a−b+e xd y e z c wf q f +(b−a)c+ 2 + 2 + 2 + 2 +c+ac+d+ad+cd+ 2 +2ae−2be+ 2 (q)f (wq)b+f −a (q)a (q)b (q)c (xq)a+c (q)d (zq)a+d (q)e (q)a−b+e (yq)a−b+d+e a,b,c,d,e,f ≥0 X
2
1 = (wq)∞
X (−1)a−b+e q a2 + 3a2 a,b,d,e≥0
2
2
+ 2b + b2 +d+ad+ 2e +2ae−2be+de+ 3e2
xd y e X q c(1+a+d) z c (q)a (q)b (q)d (zq)a+d (q)e (q)a−b+e (yq)a−b+d+e (q)c (xq)a+c c≥0 (by (2.4) applied to the w sum) 2
1 = (wq)∞
X (−1)a−b+e q a2 + 3a2 a,b,c,d,e≥0
×
2
2
+ 2b + b2 +d+ad+ 2e +2ae−2be+de+ 3e2
xd y e
(q)a (q)b (q)d (zq)a+d (q)e (q)a−b+e (yq)a−b+d+e 1
(zq 1+a+d )∞
X (−1)j q j 2 +j(1+2a+d) z j xj (q)j (xq)j+a
j≥0
(by (2.4) applied to c-sum) 2
1 = (wq)∞ (zq)∞
2
X (−1)a−b q a2 + 3a2 + 2b + b2 +d+ad+j 2 +j+2aj+dj xd z j xj (q)a (q)b (q)d (q)j (xq)j+a a,b,d,j≥0 2
×
X (−1)e q e2 + 2e +(2a−2b+d)e y e e≥0
(q)e (q)a−b+e (yq)a−b+d+e 6
2
1 = (wq)∞ (zq)∞
2
X (−1)a−b q a2 + 3a2 + 2b + b2 +d+ad+j 2 +j+2aj+dj xd z j xj (q)a (q)b (q)d (q)j (xq)j+a a,b,d,j≥0
n+1 ∞ 1 X (−1)n y n q ( 2 )+(a−b+d)n × (yq)∞ n=0 (q)n (q)n+a−b
(by (2.6), λ = y, µ = 1, B = a − b + d, c = e − b) 1 = (wq)∞ (zq)∞ (yq)∞
×
3a2
a
2
(−1)a+n y n xd+j z j q 2 + 2 +d+ad+j +j+2aj+dj+an+dn+ (q)n (q)d (q)j (xq)j+a (q)n a,d,j,n≥0 X
n2 +n 2 2
−bn X (−1)b q (b+1 2 ) b≥0
(q)b (q)a+n−b 2
X (−1)a q a2 + 3a2 +d+ad+j 2 +j+2aj+dj xd+j z j 1 = (wq)∞ (zq)∞ (yq)∞ a,d,j≥0 (q)a (q)d (q)j (q)j+a (because by (2.3)) the b-sum was 0 unless n = 0 when it was 1) (wq)∞ (zq)∞ (yq)∞ (xq)∞ P =
3a2
a
2
(−1)a q 2 + 2 +j +j+2aj z j xj (q)a (q)j a,j≥0 (by (2.1)) applied to the d-sum 2
X (−1)a q a2 − a2 +j 2 +j (xz)j−a 1 = (wq)∞ (zq)∞ (yq)∞ (xq)∞ a,j≥0 (q)a (q)j − a (shifting j to j − a and noting the comment following (1.4)) 2
X (−1)a q a2 − a2 (xz)−a X 2 1 j +j j P = q (xz) (wq)∞ (zq)∞ (yq)∞ (xq)∞ j≥0 j≥0 (q)a (q)j−a a≥0 2 ∞ X q j +j (xz)j ((xz)−1 )j 1 = , (wq)∞ (zq)∞ (yq)∞ (xq)∞ j=0 (q)j
7
(by (2.3))
To obtain (1.7) we need only set x = z −1 in the above identity and note that the sum on j collapses to 1.
6
The q-series for the 85 knot.
The q-series for the 85 knot [2, p. 13] is given by X S(a, b, c, d, e, f, g, h) a,b,c,d,e,f,g,h≥0
where S = S(a, b, c, d, e, f, g, h) 2
b+f
:= (−1)
3b2
b
q 2a+3a − 2 −2ab+ 2 +c+ac+d+ad (q)a (q)b (q)c (q)d (q)e (q)f (q)g (q)h 5f 2
3f
q cd+e+ae+de+ 2 +4af −4bf +ef + 2 +g+ag−bg+eg+f g+h+ah−bh+f h+gh × . (q)a+c (q)a+d (q)a+e (q)a−b+f (q)a−b+e+f (q)a−b+f +g (q)a−b+f +h Next we note that the sum on f over nonnegative integers may be replaced by −∞ < f < ∞ because of 5f 2 /2 in the numerator of q and the fact that (as noted before) 1/(q)f = 0 if f < 0. Hence it is valid to replace f by f − a + b in the above 8-fold sum, and consequently X S(a, b, c, d, e, f, g, h) a,b,c,d,e,f,g,h≥0
=
X
S(a, b, c, d, e, f − a + b, g, h)
a,b,c,d,e,g,h −∞
