q-series and tails of colored Jones polynomials

0.17 MB·

by  Paul Beirne

q-SERIES AND TAILS OF COLORED JONES POLYNOMIALS
6
1 PAUL BEIRNEAND ROBERTOSBURN
0
2
Abstract. WeextendthetableofGaroufalidis, LˆeandZagierconcerningconjecturalRogers-

g Ramanujan type identities for tails of colored Jones polynomials to all alternating knots up to
u 10 crossings. Wethen provethese new identities using q-series techniques.
A

0
3

] 1. Introduction
T
N The colored Jones polynomial J (K;q) for a knot K is an important quantum invariant of
N
. knots. Here, we use the normalization J (K;q) = 1 for the unknot K, J (K;q) = 1 for all
h N 1
t knots K and J2(K;q) is the Jones polynomial of K. Thetail of JN(K;q) is a power series whose
a
first N coefficients agree (up to a common sign) with the first N coefficients for J (K;q) for
m N
all N ≥ 1. If K is an alternating knot, then the tail exists and equals an explicit q-multisum
[
Φ (q) (see [1], [3], [5]).
K
2 Recently, Garoufalidis and Lˆe (with Zagier) presented a table (see Table 6 in [5]) of 43 conjec-
v
turalRogers-Ramanujan typeidentities between thetails Φ (q)andproductsof theta functions
7 K
7 and/or false theta functions. Thistable consisted of the following knots K: all alternating knots
5 up to 8 , the twist knots K , p > 0 or p < 0, the torus knots T(2,p), p > 0, each of their mirror
4 p
5
knots −K and −8 . For example, if we define for a positive integer b
0 5
.
1
60 hb = hb(q) = Xǫb(n)qbn(n2+1)−n
n∈Z
1
: where
v
i
X
(−1)n if b is odd,
r 
a ǫb(n)=  1 if b is even and n ≥ 0,
−1 if b is even and n < 0

and
n
(a) = (a;q) = (1−aqk−1),
n n Y
k=1
valid for n ∈ N∪{∞}, then
Date: August 31, 2016.
2010 Mathematics Subject Classification. Primary: 33D15; Secondary: 05A30, 57M25.
Key words and phrases. q-series identities, colored Jones polynomial, tails.
1
2 PAULBEIRNEANDROBERTOSBURN
q3a2+2a+b2+bg+ac+ad+ae+af+ag+cd+de+ef+fg+c+d+e+f+g
Φ (q) =(q)7
72 ∞ X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g b+g a+c a+d a+e a+f a+g
a,b,c,d,e,f,g≥0
?
=h .
6
(1.1)
Note that h = 0, h = 1 and h = (q) . In general, h is a theta function if b is odd and a
1 2 3 ∞ b
false theta function if b is even. Using q-series techniques, Keilthy and the second author [10]
proved not only (1.1), but all of the remaining conjectural identities in [5].
The purpose of this paper is to extend the table of Garoufalidis, Lˆe and Zagier to include all
alternating knots up to 10 crossings. This is done in Tables 1 and 2 below. One immediately
observes that their table is not “complete” in the sense that there exist knots K such that
ΦK(q) 6= ΦK′(q) for any knot K′ in Table 6 of [5]. For example, Φ87(q) = h3h5. Our main result
is the following.
Theorem 1.1. The identities in Tables 1 and 2 are true.
K Φ (q) Φ (q) K Φ (q) Φ (q) K Φ (q) Φ (q)
K −K K −K K −K
8 h h h 9 h h h 9 ? ?
6 3 4 5 6 3 6 4 24
8 h h h2 9 h h h 9 h3 ?
7 3 5 3 7 3 4 6 25 3
8 h h h2 9 h h h2 9 h2h h3
8 3 5 3 8 3 6 3 26 3 4 3
8 h h h h 9 h h h 9 h3 h2h
9 3 4 3 4 9 4 5 4 27 3 3 4
8 ? h2 9 h2 h 9 ? ?
10 3 10 4 5 28
8 h h h h 9 h h h2 9 ? ?
11 3 4 3 4 11 4 5 3 29
8 h h h h 9 h h h h 9 h3 ?
12 3 4 3 4 12 3 4 3 5 30 3
8 h2h h2 9 h2 h h 9 h4 h3
13 3 4 3 13 4 3 4 31 3 3
8 h h h3 9 h2h h2 9 ? ?
14 3 4 3 14 3 5 3 32
8 h3 ? 9 h h h h 9 ? ?
15 3 15 3 4 3 5 33
8 ? ? 9 h ? 9 ? ?
16 16 4 34
8 ? ? 9 h2 h2h 9 ? h
17 17 3 3 5 35 3
8 ? ? 9 h h h2 9 ? h2
18 18 3 4 4 36 3
9 h 1 9 h h h3 9 h3 ?
1 9 19 3 5 3 37 3
9 h h 9 h2 h h2 9 ? ?
2 8 3 20 3 3 4 38
9 h h 9 h h h2h 9 ? ?
3 7 4 21 3 4 3 4 39
9 h h 9 ? h2 9 ? ?
4 6 5 22 3 40
9 h h h 9 h2 h3 9 ? ?
5 3 4 6 23 4 3 41
Table 1.
Unfortunately, we were unable to find similar identities not only in each case labelled “?” in
Tables 1 and 2, but for any alternating knot (or its mirror) from 10 to 10 . This is also the
79 123
situation for 8 where although one has (after q-theoretic simplification or the methods in [8])
5
qa2+a+b2+b(q)
Φ (q) = (q)2 a+b, (1.2)
85 ∞ X (q)2(q)2
a,b≥0 a b
q-SERIES AND TAILS OF COLORED JONES POLYNOMIALS 3
K Φ (q) Φ (q) K Φ (q) Φ (q) K Φ (q) Φ (q)
K −K K −K K −K
10 h h 10 h h h2h 10 ? h3
1 9 3 27 3 5 3 4 53 3
10 ? h 10 h h h h2 10 ? h2
2 3 28 3 4 5 3 54 3
10 h h 10 h h2 h h 10 ? h3
3 7 5 29 3 4 3 4 55 3
10 ? h 10 h h2 h3 10 ? h h
4 3 30 3 4 3 56 3 4
10 h h h2 10 h h h2h 10 ? h2h
5 3 7 3 31 3 5 3 4 57 3 4
10 h h h 10 ? h3 10 ? h3
6 3 6 5 32 3 58 3
10 h h h h 10 ? h2h 10 ? h3
7 3 6 3 4 33 3 4 59 3
10 h h h 10 h h h2 10 ? h3
8 3 5 6 34 3 7 3 60 3
10 h h h h 10 h h h h 10 ? h
9 3 6 3 4 35 3 6 3 4 61 3
10 h2h h2 10 h h h3 10 ? h2
10 3 6 3 36 3 6 3 62 3
10 h h h 10 h h h h 10 ? h h
11 4 5 5 37 3 5 3 5 63 3 4
10 h h h h 10 ? h3 10 ? h h
12 3 5 3 5 38 3 64 3 4
10 h h h h 10 h h h2h 10 ? h2h
13 4 5 3 4 39 3 4 3 5 65 3 4
10 h2h h h 10 ? h2h 10 ? ?
14 3 5 3 4 40 3 4 66
10 h2 h2 10 h h2 h3 10 ? h3
15 5 3 41 3 4 3 67 3
10 h h h h 10 h2h ? 10 ? h2
16 4 5 3 4 42 3 4 68 3
10 ? h h 10 h2h h2h 10 ? ?
17 3 5 43 3 4 3 4 69
10 h2h h h 10 h3h h4 10 ? h h
18 3 5 3 4 44 3 4 3 70 3 4
10 h h h h2 10 h4 h4 10 ? h2h
19 3 4 5 3 45 3 3 71 3 4
10 h h h 10 ? h 10 h h ?
20 7 3 4 46 3 72 3 4
10 h h h h 10 ? h2 10 ? h2h
21 3 6 3 4 47 3 73 3 4
10 h h h h 10 ? h h 10 ? h h
22 3 4 4 5 48 3 5 74 3 4
10 h h h2h 10 ? h2h 10 ? ?
23 3 5 3 4 49 3 5 75
10 h h h h 10 ? h h 10 ? h
24 4 5 3 4 50 3 4 76 5
10 h h2 h h 10 ? h2h 10 ? h h
25 3 4 3 4 51 3 4 77 3 5
10 h h2 h h 10 ? h3 10 ? ?
26 3 4 3 4 52 3 78
Table 2.
themodular(or false theta, mock/mixed mock, quantummodular)propertiesof the doublesum
in(1.2)arenotclear. Thedifficultyinfindingniceidentitiesforthesetails isduetothestructure
of their reduced Tait graphs (see [6]). Another approach to Theorem 1.1 is to utilize the skein-
theoretic techniques in [2], [4] and [9]. It would be of considerable interest to investigate the
connection between skein theory and q-series to gain a better understanding of these unknown
cases and of a general framework.
It would also be desirable to study q-series identities in other settings which arise from knot
theory. For example, the q-multisum Φ (q) occurs as the “0-limit” of J (K;q) (see Theorem
K N
2 in [5]). Garoufalidis and Lˆe have also obtained an explicit formula (see Theorem 3 in [5]) for
the “1-limit” of J (K;q). Finally, do tails exist (in some appropriate sense) for generalizations
N
of J (K;q) (see [7], [11]–[13])?
N
The paper is organized as follows. In Section 2, we recall the necessary background from [10].
In Section 3, we prove Theorem 1.1.
4 PAULBEIRNEANDROBERTOSBURN
2. Preliminaries
We first recall six q-series identities (see (2.1)–(2.3), Lemma 2.1, (4.3) and the proof of (4.1)
in [10]). Namely,
∞
tn 1
= , (2.1)
X (q) (t)
n ∞
n=0
∞ (−1)ntnqn(n−1)/2
= (t) , (2.2)
∞
X (q)
n
n=0
∞ qn2+An 1
= (2.3)
X (q) (q) (q)
n n+A ∞
n=0
for any integer A,
(−1)nqm2+m+mn+n(n2+1) = h , (2.4)
X (q) (q) 4
m n
m,n≥0
(−1)l+nq3l(l2+1)+m2+m+n(n2+1)+2lm+ln+mn = h (2.5)
X (q) (q) (q) 5
l m n
l,m,n≥0
and
n−1 n−2 k k n−1k−1
qna(a2+1)−a+akP=1ck 1 nP−2 Pk ijq12kP=1(cid:0)jP=1ij(cid:1)(cid:0)1+jP=1ij(cid:1)+kP=2jP=1ckij
(−1)na = (−1)k=1j=1
X n−1 (q) X n−2 n−2
∞
a≥0 (q)akQ=1(q)a+ck i1,...,in−2≥0 kQ=1(q)ik kQ=1(q)ck+Pk ij
j=1
(2.6)
for any n > 2 and integers c .
k
Let K be an alternating knot with c crossings and T its associated Tait graph. The reduced
K
Tait graph T′ is obtained from T by replacing every set of two edges that connect the same
K K
two vertices by a single edge. The tail Φ (q) is given by
K
Φ (q) = (q)c S (q) (2.7)
K ∞ K
where S (q) is an explicitly constructed q-multisum (see pages 261–264 in [10]). Now, by
K
Theorem2in[2],ifT′ isthesameasT′ fortwoalternatingknotsK andL,thenΦ (q) = Φ (q).
K L K L
Thus, by comparing the reduced Tait graphs for those knots in Table 1 of [10] and Tables 1 and
2 above, it suffices to verify the conjectural identities in the following cases: 8 , 8 , −9 , 9 ,
7 13 5 14
−9 , −9 , −9 , 9 , 10 , −10 , 10 , 10 , 10 , 10 , 10 , 10 . Note that Corollary 2 in [5]
17 20 27 31 5 8 10 15 19 26 28 44
is false as stated since T′ ∼= T′ , but Φ (q)6= Φ (q).
86 924 86 924
The strategy for proving Theorem 1.1 is now as follows. For each of the 16 cases, we first
compute S (q) using the methods from [10]. We then employ (2.1)–(2.6) to reduce this q-
K
multisum to (1.1) or one of the following key identities proven in [10]:
q-SERIES AND TAILS OF COLORED JONES POLYNOMIALS 5
S (q) := (−1)a qa(5a2+3)+ab+ac+ad+ae+bc+cd+de+b+c+d+e = 1 h , (2.8)
51 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)5 5
a b c d e a+b a+c a+d a+e ∞
a,b,c,d,e≥0
S (q) := (−1)e q2f2+f+e(3e2+1)+ab+af+bc+bf+cd+ce+cf+de+a+b+c+d = 1 h ,
62 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)5 4
a b c d e f a+f b+f c+e c+f d+e ∞
a,b,c,d,e,f≥0
(2.9)
S (q) := (−1)a qa(7a2+5)+ab+ac+ad+ae+af+ag+bc+cd+de+ef+fg+b+c+d+e+f+g
71 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g a+b a+c a+d a+e a+f a+g
a,b,c,d,e,f,g≥0
1
= h ,
(q)7 7
∞
(2.10)
q2f2+f+2g2+g+ab+ag+bc+bg+cd+cf+cg+de+df+ef+a+b+c+d+e
S (q):=
74 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g a+g b+g c+f c+g d+f e+f
a,b,c,d,e,f,g≥0 (2.11)
1
= h2,
(q)7 4
∞
q3e2+e+3f2+f+3g2+g+ab+ad+ae+af+bf+cd+cg+de+dg+a+b+c
S (q):= (−1)e+f+g 2 2 2 2 2 2
77 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g a+e d+e a+f b+f c+g
a,b,c,d,e,f,g≥0
qd
×
(q)
d+g
1
= ,
(q)4
∞
(2.12)
S (q):= (−1)bq3a2+2a+b(3b2+1)+ad+ae+af+ag+ah+bc+bd+cd+de+ef+fg+gh+c+d+e+f
82 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g h b+c b+d a+d a+e a+f
a,b,c,d,e,f,g,h≥0
qg+h
×
(q) (q)
a+g a+h
1
= h
(q)7 6
∞
(2.13)
and
6 PAULBEIRNEANDROBERTOSBURN
S (q):= (−1)gqg(5g2+3)+h(2h+1)+ab+ah+bc+bh+cd+cg+ch+de+dg+ef+eg+fg+a+b+c+d
−84 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g h a+h b+h c+g c+h d+g
a,b,c,d,e,f,g,h≥0
qe+f
×
(q) (q)
e+g f+g
1
= h h .
(q)8 4 5
∞
(2.14)
3. Proof of Theorem 1.1
Proof of Theorem 1.1. We give full details for 8 , −9 and −10 . As the remaining cases are
7 5 8
handled similarly, we sketch their proofs. For Φ (q), it suffices to prove
87
S (q) := (−1)h+i qi(5i2+3)+h(3h2+1)+g2+ab+ag+ah+bc+bh+bi+cd+ci+de+di+ei+a+b+c
87 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e g h i a+g a+h b+h b+i c+i
a,b,c,d,e,g,h,i≥0
qd+e
×
(q) (q)
d+i e+i
1
= h .
(q)7 5
∞
(3.1)
We now have
S (q)= 1 (−1)h+i qi(5i2+3)+h(3h2+1)+ab+ah+bc+bh+bi+cd+ci+de+di+ei+a+b+c
87 (q) X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ a b c d e h i a+h b+h b+i c+i d+i
a,b,c,d,e,h,i≥0
qd+e
×
(q)
e+i
(evaluate the g-sum with (2.3))
= 1 (−1)h+i qi(5i2+3)+h(h2+1)+ab+ah+bc+bi+cd+ci+de+di+ei+a+b+c+d+e
(q)2 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ a b c d e h i b+h b+i c+i d+i e+i
a,b,c,d,e,h,i≥0
(apply (2.6) to the h-sum with n = 3)
= 1 (−1)i qi(5i2+3)+bc+bi+cd+ci+de+di+ei+b+c+d+e
(q)2 X (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ b c d e i b+i c+i d+i e+i
b,c,d,e,i≥0
(evaluate the a-sum with (2.1), simplify, then use (2.2) for the h-sum).
Thus, (3.1) then follows from (2.8) after letting i → a.
For Φ (q), it suffices to prove
813
q-SERIES AND TAILS OF COLORED JONES POLYNOMIALS 7
qg(3g+1)+(3h+1)+i(2i+1)+af+ag+ci+cd+de+di+ef+eh+ei
S (q) := (−1)g+h 2 2
813 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a c d e f g h i a+g c+i d+i e+i e+h
a,c,d,e,f,g,h,i≥0
qfh+fg+a+c+d+e+f
×
(q) (q)
f+h f+g
1
= h .
(q)6 4
∞
(3.2)
Apply (2.6) with n = 3 to the g-sum, (2.1) to the a-sum, then simplify and (2.2) to the g-sum
to obtain
S (q) = 1 (−1)h qh(3h2+1)+i(2i+1)+ci+cd+de+di+ef+eh+ei+fh+c+d+e+f .
813 (q) X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ c d e f h i c+i d+i e+i e+h f+h
c,d,e,f,h,i≥0
Thus, (3.2) then follows from (2.9) upon (c,d,e,f,h,i) → (a,b,c,d,e,f).
For Φ (q), it suffices to prove
−95
qh(2h+1)+j(3j+2)+ab+ag+ah+aj+bc+bh+ch+de+dj+ef+ej+fg+fj+gj+a+b+c
S (q) :=
−95 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g h j a+h a+j b+h c+h d+j
a,b,c,d,e,f,g,h,j≥0
qd+e+f+g
×
(q) (q) (q)
e+j f+j g+j
1
= h h .
(q)9 4 6
∞
(3.3)
We now have
1 qs2+s+st+t(t2+1)+bs+c(s+t)+j(3j+2)+ab+ag+aj+bc+de+dj+ef+ej+fg
S (q) =
−95 (q) X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ a b c d e f g j s t a+j d+j s+a
a,b,c,d,e,f,g,j,s,t≥0
qfj+gj+a+b+c+d+e+f+g
×
(q) (q) (q) (q)
s+t+b e+j f+j g+j
(apply (2.6) to the h-sum with n = 4)
1 qs2+s+st+t(t2+1)+bs+j(3j+2)+ab+ag+aj+de+dj+ef+ej+fg+fj+gj+a+b+d+e
=
(q)2 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ a b d e f g j s t a+j d+j e+j f+j g+j
a,b,d,e,f,g,j,s,t≥0
qf+g
×
(q)
s+a
(evaluate the c-sum with (2.1) and simplify)
8 PAULBEIRNEANDROBERTOSBURN
1 qj(3j+2)+ag+aj+de+dj+ef+ej+fg+fj+gj+a+d+e+f+g
= h
(q)3 4 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ a d e f f g j a+j d+j e+j f+j g+j
a,d,e,g,j≥0
(evaluate the b-sum with (2.1), simplify, then apply (2.4) to the st-sum).
Now, (3.3) follows from first applying (2.3) the b-sum in (1.1), then letting (a,d,e,f,g,j) →
(c,g,f,e,d,a).
For Φ (q), it suffices to prove
914
S (q):= (−1)h+i+j qh(3h2+1)+i(3i2+1)+j(5j2+3)+ab+ag+ah+ai+bc+bi+bj+cd+cj+de+dj+ej
914 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e g h i j a+h a+i b+i b+j
a,b,c,d,e,g,h,i,j≥0
qgh+a+b+c+d+e+g
×
(q) (q) (q) (q)
c+j d+j e+j g+h
1
= h .
(q)7 5
∞
(3.4)
First, apply (2.6) with n = 3 to the h-sum, (2.1) to the g-sum, simplify and (2.2) to the h-sum,
then (2.6) with n = 3 to the i-sum, (2.1) to the a-sum, simplify and (2.2) to the i-sum to obtain
S (q)= 1 (−1)j qj(5j2+3)+bc+bj+cd+cj+de+dj+ej+b+c+d+e .
914 (q)2 X (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ b c d e j b+j c+j d+j e+j
b,c,d,e,j
Thus, (3.4) follows from (2.8) after j → a.
For Φ (q), it suffices to prove
−917
S (q) := (−1)h+i+j qh(3h2+1)+i(5i2+3)+j(3j2+1)+ab+aj+bc+bi+bj+cd+ci+de+di+ef
−917 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f h i j a+j b+i b+j c+i
a,b,c,d,e,f,h,i,j≥0
qeh+ei+fh+a+b+c+d+e+f
×
(q) (q) (q) (q)
d+i e+h e+i f+h
1
= h .
(q)7 5
∞
(3.5)
First, apply (2.6) with n= 3 to the h-sum, (2.1) to the f-sum, simplify and (2.3) to the h-sum,
then (2.6) with n = 3 to the j-sum, (2.1) to the a-sum, simplify and (2.3) to the j-sum to get
S (q) = 1 (−1)i qi(5i2+3)+bc+bi+cd+ci+de+di+ei+b+c+d+e .
−917 (q)2 X (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ b c d e i b+i c+i d+i e+i
b,c,d,e,i≥0
Thus, (3.5) follows from (2.8) after i → a.
For Φ (q), it suffices to prove
−920
q-SERIES AND TAILS OF COLORED JONES POLYNOMIALS 9
S (q) := (−1)hqh(3h2+1)+i(2i+1)+j(2j+1)+ab+ah+bc+bh+bi+cd+ci+de+di+dj+ef+ej
−920 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f h i j a+h b+h b+i c+i
a,b,c,d,e,f,h,i,j≥0
qfj+a+b+c+d+e+f
×
(q) (q) (q) (q)
d+i d+j e+j f+j
1
= h2.
(q)8 4
∞
(3.6)
Apply (2.6) with n = 3 to the h-sum, (2.1) to the a-sum and simplify, then (2.2) to the h-sum
to obtain
1 qi(2i+1)+j(2j+1)+bc+bi+cd+ci+de+di+dj+ef+ej+fj+b+c+d+e+f
S (q) = .
−920 (q) X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ b c d e f i j b+i c+i d+i d+j e+j f+j
b,c,d,e,f,i,j≥0
Now, (3.6) follows from (2.11) after the substitution (b,c,d,e,f,i,j) → (a,b,c,d,e,g,f).
For Φ (q), it suffices to prove
−927
S (q) := (−1)f+h qf(3f2+1)+g(2g+1)+h(3h2+1)+i2+ab+af+bc+bf+bg+cd+cg+de+dg+dh
−927 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g h i a+f b+f b+g c+g
a,b,c,d,e,f,g,h,i≥0
qeh+ei+a+b+c+d+e
×
(q) (q) (q) (q)
d+g d+h e+h e+i
1
= h .
(q)7 4
∞
(3.7)
Apply (2.3) to the i-sum, (2.6) with n = 3 to the f-sum, (2.1) to the a-sum, simplify and (2.2)
to the f-sum to obtain
S = 1 (−1)h qg(2g+1)+h(3h2+1)+bc+bg+cd+cg+de+dg+dh+eh+b+c+d+e .
−927 (q)2 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ b c d e g h b+g c+g d+g d+h e+h
b,c,d,e,g,h≥0
Now, (3.7) follows from (2.9) after letting (b,c,d,e,g,h) → (a,b,c,d,f,e).
For Φ (q), it suffices to prove
931
10 PAULBEIRNEANDROBERTOSBURN
qg(3g+1)+h(3h+1)+i(3i+1)+j(3j+1)+ab+af+ag+aj+bc+bg+bh
S (q):= (−1)g+h+i+j 2 2 2 2
931 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c e f g h i j a+g a+j b+g
a,b,c,e,f,g,h,i,j≥0
qch+ef+ei+fi+fj+a+b+c+e+f
×
(q) (q) (q) (q) (q)
b+h c+h e+i f+i f+j
1
= .
(q)5
∞
(3.8)
Apply (2.6) with n = 3 to the h-sum, (2.1) to the c-sum, simplify and (2.2) to the h-sum to
obtain
S (q) = 1 (−1)g+i+j qg(3g2+1)+i(3i2+1)+j(3j2+1)+ab+af+ag+aj+bg+ef+ei+fi+fj+a
931 (q) X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ a b e f g i j a+g a+j b+g e+i f+i
a,b,e,f,g,i,j≥0
qb+e+f
× .
(q)
f+j
Now, (3.8) follows from (2.12) after letting (a,b,e,f,g,i,j) → (a,b,c,d,f,g,e).
For Φ (q), it suffices to prove
105
S (q):= (−1)j+kqj(3j2+1)+k(7k2+5)+i2+ab+ai+aj+bc+bj+bk+cd+ck+de+dk+ef+ek+fg
105 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
a b c d e f g i j k a+i a+j b+j
a,b,c,d,e,f,g,i,j,k≥0
qfk+gk+a+b+c+d+e+f+g
×
(q) (q) (q) (q) (q) (q)
b+k c+k d+k e+k f+k g+k
1
= h .
(q)9 7
∞
(3.9)
Apply (2.3) to the i-sum, (2.6) with n = 3 to the j-sum, (2.1) to the a-sum and simplify, then
(2.2) to the j-sum to obtain
S (q) = 1 (−1)k qk(7k2+5)+bc+bk+cd+ck+de+dk+ef+ek+fg+fk+gk+b+c+d+e+f
105 (q)2 X (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q) (q)
∞ b c d e f g k b+k c+k d+k e+k f+k
b,c,d,e,f,g,k≥0
qg
× .
(q)
g+k
Now, (3.9) follows from (2.10) after letting k → a.
For Φ (q), it suffices to prove
−108