RANK-CRANK TYPE PDES AND
GENERALIZED LAMBERT SERIES IDENTITIES
arXiv:1201.1663v1 [math.NT] 8 Jan 2012
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
Dedicated to our friends, Mourad Ismail and Dennis Stanton
Abstract. We show how Rank-Crank type PDEs for higher order Appell functions
due to Zwegers may be obtained from a generalized Lambert series identity due to
the first author. Special cases are the Rank-Crank PDE due to Atkin and the third
author and a PDE for a level 5 Appell function also found by the third author. These
two special PDEs are related to generalized Lambert series identities due to Watson,
and Jackson respectively. The first author’s Lambert series identities are common
generalizations. We also show how Atkin and Swinnerton-Dyer’s proof using elliptic
functions can be extended to prove these generalized Lambert series identities.
1. Introduction
F. J. Dyson [9], [10, p. 52] defined the rank of a partition as the largest part minus
the number of parts. Dyson conjectured that the residue of the rank mod 5 divides
the partitions of 5n + 4 into 5 equal clases thereby providing a combinatorial interpretation of Ramanujan’s famous partition congruences p(5n + 4) ≡ 0 (mod 5). He
also conjectured that the rank mod 7 likewise gives Ramanujan’s partition congruence p(7n + 5) ≡ 0 (mod 7). Dyson’s rank conjectures were proved by A. O. L. Atkin
and H. P. F. Swinnerton-Dyer [3]. The following was the crucial identity that Atkin
and Swinnerton-Dyer needed for the proof of the Dyson rank conjectures. It was first
proved by G.N. Watson [18].
∞
[ζ 2 ]∞ X (−1)n q 3n(n+1)/2
[ζ]∞ [ζ 2]∞ (q)2∞
ζ
+
[ζ]∞ n=−∞
1 − zq n
[z/ζ]∞ [z]∞ [ζz]∞
∞
X
ζ −3n
ζ 3n+3
n 3n(n+1)/2
=
(−1) q
.
+
n /ζ
n
1
−
zq
1
−
zζq
n=−∞
(1.1)
Date: January 8, 2012.
2010 Mathematics Subject Classification. 11F11, 11P82, 11P83, 33D15.
Key words and phrases. Rank-Crank PDE, higher level Appell function, Lambert series, partition,
q-series, basic hypergeometric function, quasimodular form.
The third author was supported in part by NSA Grant H98230-09-1-0051.
1
2
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
Throughout we use the standard q-notation
(x)0 := (x; q)0 := 1,
(x)n := (x; q)n :=
n−1
Y
m=0
(1 − xq m ),
(x1 , · · · , xm )n := (x1 , · · · , xm ; q)n := (x1 ; q)n · · · (xm ; q)n ,
[x1 , · · · , xm ]n := [x1 , · · · , xm ; q]n := (x1 , q/x1 , · · · , xm , q/xm ; q)n .
when n is a nonnegative integer. Assuming |q| < 1 we also use this notation when
n = ∞ by interpreting its meaning as the limit as n → ∞. Later M. Jackson [14]
proved an analogue of the above identity,
∞
[ζ]∞ [ζ 2 ]∞ [xζ]∞ [ζ/x]∞ (q)2∞
ζ 2[ζ 2 ]∞ [xζ]∞ [x/ζ]∞ X (−1)n q 5n(n+1)/2
+
[ζ]∞ [x]2∞
1 − zq n
[z/x]∞ [z/ζ]∞ [z]∞ [zζ]∞ [zx]∞
n=−∞
∞
ζ [ζ]∞ [ζ 2]∞ X
x−5n
x5n+5
n 5n(n+1)/2
+
(−1) q
(1.2)
+
x [x]∞ [x2 ]∞ n=−∞
1 − zq n /x 1 − zxq n
∞
X
ζ 5n+5
ζ −5n
n 5n(n+1)/2
.
+
=
(−1) q
−1 q n
n
1
−
zζ
1
−
zζq
n=−∞
Recently, the first author [8, p.603] found a generalization of the above two identities,
namely,
∞
xm [x2 /x1 , · · · , xm /x1 , x1 xm , · · · , x1 x2 , x2 ]∞ X (−1)n q (2m+1)n(n+1)/2
1
1
[x1 ]∞ [x2 , · · · , xm ]2∞
n=−∞
2
, x1 x2 , x1 ]∞ (q)2∞
1 − zq n
[x1 /x2 , · · · , x1 /xm , x1 , x1 xm , · · ·
[z/x1 , z/x2 , · · · , z/xm , z, zxm , · · · , zx1 ]∞
x1 [x1 /x3 , · · · , x1 /xm , x1 , x1 xm , · · · , x1 x3 , x21 ]∞
+
(1.3)
x2 [x2 /x3 , · · · , x2 /xm , x2 , x2 xm , · · · , x22 ]∞
!
∞
(2m+1)(n+1)
−(2m+1)n
X
x
x
2
2
n (2m+1)n(n+1)/2
×
(−1) q
+
+ idem(x2 ; x3 , · · · , xm )
1 − zq n /x2
1 − zx2 q n
n=−∞
!
∞
−(2m+1)n
(2m+1)(n+1)
X
x
x
1
(−1)n q (2m+1)n(n+1)/2
=
+ 1
,
n /x
n
1
−
zq
1
−
zx
q
1
1
n=−∞
+
where g(a1 , a2 , · · · , am )+ idem(a1 ; a2 , · · · , an ) denotes the sum
n
X
i=1
g(ai , a2 , · · · , ai−1 , a1 , ai+1 , · · · , am ),
in which the i-th term of the sum is obtained from the first by interchanging a1 and ai .
Equation (1.3) was proved using partial fractions. Indeed, the m = 1 case of (1.3)
is equivalent to (1.1), while the m = 2 case is equivalent to (1.2). The fact that the
right-hand side of (1.2) is independent of x, and that the right-hand side of (1.3) is
RANK-CRANK TYPE PDES
3
independent of x2 , x3 , · · · , xm seems to be intriguing at first. Indeed, one purpose of
this article is to show that the left-hand sides of (1.2) and (1.3) are really elliptic
functions of order less than 2, in fact entire functions as we show, in the respective
variables (x for (1.2) and x2 for (1.3) while holding x3 , · · · , xm fixed) and therefore
that they must be constants which are nothing but the right-hand sides of (1.2) and
(1.3) respectively. Since (1.2) follows from (1.3), we show this only for (1.3). This is
done in Section 2.
Let N(m, n) denote the number of partitions of n with rank m. Then the rank
generating function R(z, q) is given by
2
∞
∞ X
n
X
X
qn
m n
N(m, n)z q =
R(z, q) =
.
(1.4)
(zq)n (z −1 q)n
n=0
n=0 m=−n
In [1], G. E. Andrews and the third author defined the crank of a partition, a partition
statistic hypothesized by Dyson in [9]. It is the largest part if the partition contains
no ones, and otherwise is the number of parts larger than the number of ones minus
the number of ones. For n > 1, we let M(m, n) denote the number of partitions of n
with crank m. If we amend the definition of M(m, n) for n = 1, then the generating
function can be given as an infinite product. Accordingly, we assume
M(0, 1) = −1, M(−1, 1) = M(1, 1) = 1, and M(m, 1) = 0 otherwise.
Then the crank generating function C(z, q) is given by
∞ X
n
X
(q)∞
.
C(z, q) =
M(m, n)z m q n =
−1 q)
(zq)
(z
∞
∞
n=0 m=−n
(1.5)
Atkin and the third author [2] found the so-called Rank-Crank PDE, a partial differential equation (PDE) which relates R(z, q) and C(z, q). To state this PDE in its original
form, we first define the differential operators
∂
∂
(1.6)
δz = z , δq = q .
∂z
∂q
Then the Rank-Crank PDE can be written as
1 2
1
2
∗
3
(1.7)
z(q)∞ [C (z, q)] = 3δq + δz + δz R∗ (z, q),
2
2
where
R(z, q)
R∗ (z, q) :=
,
1−z
C(z, q)
C ∗ (z, q) :=
.
(1.8)
1−z
In [2], it was shown how the Rank-Crank PDE and certain results for the derivatives
of Eisenstein series lead to exact relations between rank and crank moments. As in
[13], define Nk (m, n) by
∞
X
1 X
n
Nk (m, n)q =
(−1)n−1 q n((2k−1)n−1)/2+|m|n (1 − q n ),
(1.9)
(q)
∞
n=1
n≥0
4
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
of any positive integer k. When k = 1 this is the generating function for the crank, and
when k = 2 it is the generating function for the rank. When k ≥ 2, Nk (m, n) can be
interpreted combinatorially as the number of partitions of n into k −1 successive Durfee
squares with k-rank equal to m. See [13, Eq.(1.11)] for a definition of the k-rank. We
define
n
X X
Rk (z, q) :=
Nk (m, n)z m q n .
(1.10)
n≥0 m=−n
From [13, Eq.(4.5)], this generating function can be written as
2
Rk (z, q) =
X
nk−1 ≥nk−2 ≥···≥n1
2
2
q n1 +n2 +···+nk−1
,
(q)nk−1 −nk−2 · · · (q)n2 −n1 (zq)n1 (z −1 q)n1
≥1
(1.11)
when k ≥ 2. In Section 3, we show that Rk (z, q) is related to the level 2k − 1 Appell
function
∞
X
(−1)n q (2k−1)n(n+1)/2
(2k−1)
Σ
(z, q) :=
.
(1.12)
n
1
−
zq
n=−∞
We obtain the following
Theorem 1.1. For k ≥ 1,
Rk (z, q)
1
=
(q)∞
z k−1 (1 − z)Σ(2k−1) (z, q) − zθ1,2k−1 (q) + z(1 − z)
where
θj,2k−1(q) =
∞
X
k−3
X
m=0
(−1)n q n((2k−1)n+j)/2 ,
!
z m θ2m+3,2k−1 (q) ,
(1.13)
(1.14)
n=−∞
for j = 1, 3, . . . , 2k − 3.
This theorem generalizes Lemma 7.9 in [12] which gives a relation between the rank
generating function R(z, q) and a level 3 Appell function. The k = 1 case of the theorem
gives the familiar partial fraction expansion
of Jacobi’s theta product
Pfor the reciprocal
n n(n+1)/2
(z)∞ (z −1 q)∞ [16, p. 1], [17, p. 136], since ∞
(−1)
q
= 0.
n=−∞
th
A few years ago the third author found a 4 order PDE, which is an analogue of the
Rank-Crank PDE and is related to the 3-rank [13]. To state this PDE we define
1
G(5) (z, q) :=
Σ(5) (z, q).
(1.15)
(q)3∞
Then
24(q)2∞ [C ∗ (z, q)]5
= 24(1 − 10Φ3 (q))G(5) (z, q)
+ 100δq + 50δz + 100δq δz + 35δz2 + 20δq δz2 + 100δq2 + 10δz3 + δz4 G(5) (z, q), (1.16)
RANK-CRANK TYPE PDES
5
where
∞
X
n3 q n
Φ3 (q) :=
.
1 − qn
n=1
(1.17)
This PDE can be written more compactly as
24(q)2∞ [C ∗ (z, q)]5 = (H2∗ − E4 ) G(5) (z, q),
where H∗ is the operator
(1.18)
H∗ := 5 + 10δq + 5δz + δz2 ,
and
E4 := E4 (q) := 1 + 240Φ3 (q),
is the usual Eisenstein series of weight 4. The PDE (1.16) was first conjectured by the
third author and then subsequently proved and generalized by Zwegers [21]. It was also
Zwegers who first observed that (1.16) could be written in a more compact form. We
now describe Zwegers’s generalization. Define for l ∈ Z>0 , the level l Appell function
as
∞
X
(−1)ln q ln(n+1)/2 w n
l/2
,
(1.19)
Al (u, v) := Al (u, v; τ ) := z
1 − zq n
n=−∞
where z = e2πiu , w = e2πiv , q = e2πiτ , and define the modified rank and crank generating
functions as follows.
z 1/2 q −1/24
R(z, q),
1−z
z 1/2 q −1/24
C := C(u; τ ) :=
C(z, q).
1−z
R := R(u; τ ) :=
(1.20)
(1.21)
Here and throughout we assume Im τ > 0 so that |q| < 1. Then the following theorem
due to Zwegers gives for odd l, the (l − 1)th order analogue of the Rank-Crank PDE.
Theorem 1.2 (Zwegers[21]). Let l ≥ 3 be an odd integer. Define
Hk :=
1
l(2k − 1)
l ∂
∂2
+
−
E2 ,
2
2
πi ∂τ
(2πi) ∂u
12
H[k] := H2k−1 H2k−3 · · · H3 H1 ,
(1.22)
P∞
where E2P
(τ ) = 1 − 24 n=1 σ1 (n)q n is the usual Eisenstein series of weight 2 with
σα (n) = d|n dα . Then there exist holomorphic modular forms fj (which can be constructed explicitly), with j = 4, 6, 8, · · · , l − 1, on SL2 (Z) of weight j, such that
(l−5)/2
X
H[(l−1)/2] +
fl−2k−1 H[k] Al (u, 0) = (l − 1)!η l C l ,
(1.23)
k=0
where η is the Dedekind η-function, given by η(τ ) = q 1/24 (q)∞ .
6
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
Zwegers proved this theorem using the formulas and methods motivated from the
theory of Jacobi forms. In contrast to this, the proof of the Rank-Crank PDE by Atkin
and the third author, which corresponds to the l = 3 case of Zwegers’s PDE, depends
upon simply taking the second derivative with respect to ζ of both sides of (1.1). The
main goal of this paper is to show how a generalized Rank-Crank PDE of any odd order
follows from the Lambert series identity (1.3) in a similar fashion. We will obtain
Zwegers’s result in a different form. In our form the coefficients are quasimodular
forms rather than holomorphic modular forms, but in contrast, our coefficients are
given recursively. See Theorem 4.4 and Corollary 4.5.
This paper is organized as follows. In Section 2, we prove (1.3) using the theory
of elliptic functions. Then in Section 3 we prove Theorem 1.1, which is the theorem
that relates Rk (z, q) with the level 2k − 1 Appell function Σ(2k−1) (z, q). In Section 4
we prove our main result that shows how (1.3) can be used to derive the higher order
Rank-Crank-type PDEs of Zwegers.
In the light of (1.19), it should be observed that the identities (1.1) and (1.2) are really
the identities involving certain combinations of level 3 and level 5 Appell functions
respectively while (1.3) is an identity involving a combination of level (2m + 1) Appell
functions. However, the analogue for level 1 Appell function which cannot be derived
from (1.3) was found by R. Lewis [15, Equation 11] and is as follows.
∞
X
[z]∞ [ζ 2 ]∞ (q)2∞
ζ −n
ζ n+1
n n(n+1)/2
(−1) q
=
+
.
(1.24)
[zζ]∞ [ζ]∞ [zζ −1 ]∞ n=−∞
1 − zq n /ζ 1 − zζq n
2. General Lambert series identity through elliptic function theory
Atkin and Swinnerton-Dyer’s proof of (1.1) depends in essence on the theory of
elliptic functions. In this section, we show how this method of proof can be used to
prove (1.3). Let
z = e2πiu , x1 = e2πiv , x2 = e2πiw ,
(2.1)
and let
xj = e2πiaj ,
j = 3, · · · , m,
(2.2)
where u, v, w, a3, . . . , am are all complex numbers. Also recall that q = e2πiτ , where Im
τ > 0.
Let
[b]∞ =: J(a, q) =: J(a), where b = e2πia , a ∈ C.
(2.3)
Then using the Jacobi triple product identity [4, p. 10, Theorem 1.3.3], we easily find
that,
ieπia q −1/8
J(a, q) =
θ(a),
(2.4)
(q)∞
where
∞
X
1
1
1 2
τ +2πi n+
z+
πi n+
2
2
2
.
(2.5)
θ(z) = θ(z; τ ) :=
e
n=−∞
RANK-CRANK TYPE PDES
7
Comparing this with the classical definition of θ1 (z) [11, p. 355, Section 13.19, Equation
10], we find that upon replacing q by q 1/2 in this classical definition, θ(z) = −θ1 (z).
From [20, p. 8], we see that
θ(z + 1) = −θ(z),
θ(z + τ ) = −e−πiτ −2πiz θ(z),
θ(−z) = −θ(z),
′
θ (0; τ ) = −2πq 1/8 (q)3∞ .
(2.6)
(2.7)
(2.8)
(2.9)
Using (2.6) and (2.7), we have
J(a + 1, q) = J(a, q),
(2.10)
J(a + τ, q) = −e−2πia J(a, q),
(2.11)
J(−a, q) = −e−2πia J(a, q).
(2.13)
J(a − τ, q) = −q −1 e2πia J(a, q),
(2.12)
Using (2.3), we rephrase (1.3) as follows:
∞
e2πimv J(w − v)J(w + v)J(2v) Y J(aj − v)J(aj + v) X (−1)n q (2m+1)n(n+1)/2
J(v)(J(w))2
(J(aj ))2
1 − e2πiu q n
n=−∞
3≤j≤m
Y J(v − aj )J(v + aj )
J(v − w)J(v)J(v + w)J(2v)(q)2∞
J(u − v)J(u − w)J(u)J(u + w)J(u + v) 3≤j≤m J(u − aj )J(u + aj )
J(v)J(2v) Y J(v − aj )J(v + aj )
+ e2πi(v−w)
J(w)J(2w) 3≤j≤m J(w − aj )J(w + aj )
−2πiw(2m+1)n
∞
X
e
e2πiw(2m+1)(n+1)
n (2m+1)n(n+1)/2
×
(−1) q
+
+ idem(w; a3, · · · , am )
1 − e2πi(u−w) q n
1 − e2πi(u+w) q n
n=−∞
−2πiv(2m+1)n
∞
X
e
e2πiv(2m+1)(n+1)
n (2m+1)n(n+1)/2
(−1) q
=
+
.
(2.14)
1 − e2πi(u−v) q n
1 − e2πi(u+v) q n
n=−∞
+
Fix a3 , · · · , am , consider the left-hand side of (2.14) as a function of w only and denote
it by g(w). Let f1 (w) denote the expression in line 1 of (2.14), f2 (w) the expression in
line 2 of (2.14) and f3 (w) the expression in lines 3 and 4 of (2.14). Then, using (2.10),
(2.11) and (2.12), we see
f1 (w + 1) = f1 (w) = f1 (w + τ ),
f2 (w + 1) = f2 (w) = f2 (w + τ ),
f3 (w + 1) = f3 (w).
(2.15)
8
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
Another application of (2.11) and (2.12) gives
f3 (w + τ )
J(v − aj )J(v + aj )
e2πi(v−w−τ ) J(v)J(2v) Y
=
J(w + τ )J(2w + 2τ ) 3≤j≤m J(w + τ − aj )J(w + τ + aj )
−2πi(w+τ )(2m+1)n
∞
X
e
e2πi(w+τ )(2m+1)(n+1)
n (2m+1)n(n+1)/2
×
(−1) q
+
1 − e2πi(u−w−τ ) q n
1 − e2πi(u+w+τ ) q n
n=−∞
+
m
X
e2πi(v−ak )
k=3
∞
X
J(v)J(2v)J(v − w − τ )J(v + w + τ )
J(ak )J(2ak )J(ak − w − τ )J(ak + w + τ )
Y
2<j<m+1
j6=k
J(v − aj )J(v + aj )
J(ak − aj )J(ak + aj )
e−2πiak (2m+1)n
e2πiak (2m+1)(n+1)
+
(−1) q
×
2πi(u−ak ) q n
1
−
e
1 − e2πi(u+ak ) q n
n=−∞
∞
Y J(v − aj )J(v + aj ) X
2πiv+4πimw J(v)J(2v)
=e
(−1)n q (2m+1)(n+1)(n+2)/2
J(w)J(2w) 3≤j≤m J(w − aj )J(w + aj ) n=−∞
∞
2πiw(2m+1)n (2m+1)n
X
q
e−2πiw(2m+1)(n+1) q −(2m+1)(n+1)
n (2m+1)n(n−1)/2 e
(−1) q
+
×
1 − e2πi(u−w) q n
1 − e2πi(u+w) q n
n=−∞
+
m
X
n (2m+1)n(n+1)/2
e2πi(v−ak )
k=3
×
= f3 (w).
∞
X
J(v)J(2v)J(v − w)J(v + w)
J(ak )J(2ak )J(ak − w)J(ak + w)
n (2m+1)n(n+1)/2
(−1) q
n=−∞
Y
2<j<m+1
j6=k
J(v − aj )J(v + aj )
J(ak − aj )J(ak + aj )
e2πiak (2m+1)(n+1)
e−2πiak (2m+1)n
+
1 − e2πi(u−ak ) q n
1 − e2πi(u+ak ) q n
(2.16)
Thus from (2.15) and (2.16), we deduce that g is a doubly periodic function in w with
periods 1 and τ . Our next task is to show that g is an entire function of w and hence
a constant (with respect to w). We show that the poles of g at w = u and w = −u are
actually removable singularities by proving that limw→±u (w ∓ u) (f2 (w) + f3 (w)) = 0
which readily implies that limw→±u (w ∓ u)g(w) = 0. Let
Y
A := A(v, a3 , · · · , am ; q) := J(v)J(2v)
J(v − aj )J(v + aj ).
(2.17)
3≤j≤m
Next, applying (2.4), (2.13) and (2.9), we see that
lim (w − u) (f2 (w) + f3 (w))
Y
1
J(v − w)J(v + w)(q)2∞
= A lim (w − u)
w→u
J(u − w)J(u + w)J(u − v)J(u)J(u + v) 3≤j≤m J(u − aj )J(u + aj )
∞
X
1
1
e2πi(v−w) Y
+
(−1)n q (2m+1)n(n+1)/2
+
2πi(u−w)
J(w)J(2w) 3≤j≤m J(w − aj )J(w + aj ) 1 − e
n=−∞
w→u
n6=0
RANK-CRANK TYPE PDES
9
∞
2πiw(2m+1)(n+1)
X
e−2πiw(2m+1)n
n (2m+1)n(n+1)/2 e
+
(−1) q
×
1 − e2πi(u−w) q n n=−∞
1 − e2πi(u+w) q n
e2πi(v−u) Y
1
−iq 1/8 (q)3∞
1
=A
+
J(u)J(2u) 3≤j≤m J(u − aj )J(u + aj )
θ′ (0)
2πi
= 0.
(2.18)
Similarly, limw→−u (w +u) (f2 (w) + f3 (w)) = 0. Now the only other possibility of a pole
of g is at 0, which arises from f1 and f3 each having a pole at 0. Again, to show that
this is a removable singularity, it suffices to show that limw→0 w (f1 (w) + f3 (w)) = 0.
To show this, we need Jacobi’s duplication formula for theta functions [19, p. 488, Ex.
5]
θ(2w)θ2 θ3 θ4 = 2θ(w)θ2 (w)θ3 (w)θ4 (w).
(2.19)
Let
∞
Y J(aj − v)J(aj + v) X
(−1)n q (2m+1)n(n+1)/2
.
J(v) 3≤j≤m
(J(aj ))2
1 − e2πiu q n
n=−∞
2πimv J(2v)
B := B(u, v, a3, · · · , am ; q) := e
(2.20)
Then from (2.14) and (2.20),
lim w (f1 (w) + f3 (w))
1
J(w − v)J(w + v)
e2πi(v−w) Y
= lim w B
+
A
w→0
(J(w))2
J(w)J(2w) 3≤j≤m J(w − aj )J(w + aj )
∞
X
e2πiw(2m+1)(n+1)
e−2πiw(2m+1)n
n (2m+1)n(n+1)/2
+
×
(−1) q
1 − e2πi(u−w) q n
1 − e2πi(u+w) q n
n=−∞
w→0
+ lim w
w→0
m
X
e2πi(v−ak )
k=3
J(v)J(2v)J(v − w)J(v + w)
J(ak )J(2ak )J(ak − w)J(ak + w)
Y
2<j<m+1
j6=k
J(v − aj )J(v + aj )
J(ak − aj )J(ak + aj )
−5πiw Y
w
1
θ(w − v)θ(w + v)
2πiv 1/4
2 e
= lim
B
− Ae q (q)∞
w→0 θ(w)
θ(w)
θ(2w) 3≤j≤m J(w − aj )J(w + aj )
−2πiw(2m+1)n
∞
X
e2πiw(2m+1)(n+1)
e
n (2m+1)n(n+1)/2
+
×
(−1) q
1 − e2πi(u−w) q n
1 − e2πi(u+w) q n
n=−∞
=
1
D(w)
lim
,
θ (0) w→0 θ(w)
′
(2.21)
10
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
where
D(w) := D(w, v, a3, · · · , am ; q) := Bθ(w − v)θ(w + v) − Ae2πiv q 1/4 (q)2∞ E(w)
1
e−5πiw θ(w) Y
E(w) := E(w; u, a3 · · · , am ; q) :=
θ(2w) 3≤j≤m J(w − aj )J(w + aj )
−2πiw(2m+1)n
∞
X
e2πiw(2m+1)(n+1)
e
n (2m+1)n(n+1)/2
.
+
×
(−1) q
2πi(u−w) q n
2πi(u+w) q n
1
−
e
1
−
e
n=−∞
(2.22)
Now using (2.19), (2.4) and (2.13), we find that as w → 0,
∞
X
(−1)n q (2m+1)n(n+1)/2
e2πiv Aq 1/4 (q)2∞
D(w) → −Bθ (v) −
(−1)m−2 e−2πi(a3 +···+am ) (J(a3 ) · · · J(am ))2 n=−∞
1 − e2πiu q n
2
= 0,
(2.23)
which is observed by putting the expressions for B and C back in the first expression
on the right side in (2.23). Thus,
′
D (0)
lim w (f1 (w) + f3 (w)) = ′ 2 .
w→0
θ (0)
(2.24)
′
Now we calculate D (0).
′
′
′
′
D (w) = B θ (w − v)θ(w + v) + θ(w − v)θ (w + v) − e2πiv Aq 1/4 (q)2∞ E (w). (2.25)
Using (2.4) and (2.19), we have
1
e−πiw(2m+1) θ2 θ3 θ4 Y
2θ2 (w)θ3 (w)θ4 (w) 3≤j≤m θ(w − aj )θ(w + aj )
−2πiw(2m+1)n
∞
X
e2πiw(2m+1)(n+1)
e
n (2m+1)n(n+1)/2
.
+
×
(−1) q
2πi(u−w) q n
2πi(u+w) q n
1
−
e
1
−
e
n=−∞
E(w) = (−1)m−2 q
m−2
4
2(m−2)
(q)∞
(2.26)
Differentiating both sides with respect to w and simplifying, we obtain
m−2
1
′
2(m−2) −πiw(2m+1)
E (w) = (−1)m−2 q 4 (q)∞
e
θ2 θ3 θ4
2
′
′
′
X θ′ (w − aj ) θ′ (w + aj )
θ2 (w) θ3 (w) θ3 (w)
+
πi(2m + 1) + θ2 (w) + θ3 (w) + θ3 (w) +
θ(w − aj )
θ(w + aj )
3≤j≤m
Y
× −
θ2 (w)θ3 (w)θ4 (w)
θ(w − aj )θ(w + aj )
3≤j≤m
∞
X
e2πiw(2m+1)(n+1)
e
+
×
(−1) q
1 − e2πi(u−w) q n
1 − e2πi(u+w) q n
n=−∞
Y
1
1
′
+
F (w) ,
θ2 (w)θ3 (w)θ4 (w) 3≤j≤m θ(w − aj )θ(w + aj )
n (2m+1)n(n+1)/2
−2πiw(2m+1)n
(2.27)
RANK-CRANK TYPE PDES
11
where
F (w) := F (w, u, m; q) :=
∞
X
n (2m+1)n(n+1)/2
(−1) q
n=−∞
e2πiw(2m+1)(n+1)
e−2πiw(2m+1)n
.
+
1 − e2πi(u−w) q n
1 − e2πi(u+w) q n
(2.28)
It is straightforward to see that
′
F (0) = 2πi(2m + 1)
From (2.8), we have
∞
X
(−1)n q (2m+1)n(n+1)/2
.
2πiu q n
1
−
e
n=−∞
′
(2.29)
′
θ (−z) = θ (z).
(2.30)
′
Then letting w → 0 in (2.27), and using (2.8), (2.30), (2.29) and the fact that θk (0) = 0
for 2 ≤ k ≤ 4, we find that
′
E (0) = 0.
(2.31)
′
Using (2.30) and (2.31) in (2.25), we finally deduce that D (0) = 0.
With the help of (2.24), this then implies that limw→0 w (f1 (w) + f3 (w)) = 0 and
thus limw→0 wg(w) = 0. Thus w = 0 is also a removable singularity, which implies that
g is an doubly periodic entire function and hence a constant, say K (which may very
well depend on v). Finally, since J(0) = 0, we have
−2πiv(2m+1)n
∞
X
e2πiv(2m+1)(n+1)
e
n (2m+1)n(n+1)/2
.
+
K = g(v) =
(−1) q
1 − e2πi(u−v) q n
1 − e2πi(u+v) q n
n=−∞
This completes the proof.
3. Proof of Theorem 1.1
From [13, Eq.(4.3)], we see that
∞
1 X
1
z −1 q n
n−1 n((2k−1)n−1)/2
n
Rk (z, q) =
(−1) q
(1 − q )
+
(q)∞ n=1
1 − zq n 1 − z −1 q n
∞
z −1 X
1 − qn
=
(−1)n−1 q n((2k−1)n+1)/2
.
(q)∞ n=−∞
1 − z −1 q n
n6=0
(3.1)
Replacing z by z −1 in (1.11) and (3.1), we see that
Rk (z, q)
∞
z X
1 − qn
=
(−1)n−1 q n((2k−1)n+1)/2
(q)∞ n=−∞
1 − zq n
n6=0
∞
(1 − z)q n
z X
n−1 n((2k−1)n+1)/2
1−
(−1) q
=
(q)∞ n=−∞
1 − zq n
n6=0
12
=
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
∞
X
(−1)n q n((2k−1)n+3)/2
z
(−1)n−1 q n((2k−1)n+1)/2 + (1 − z)
(q)∞ n=−∞
1 − zq n
n=−∞
n6=0
=
∞
X
n6=0
∞
X
∞
X
z
(−1)n q n((2k−1)n+3)/2
(−1)n q n((2k−1)n+1)/2 + (1 − z)
1 −
(q)∞
1 − zq n
n=−∞
n=−∞
n6=0
∞
X
z
−zθ1,2k−1 (q)
(−1)n q n((2k−1)n+3)/2
+
1 + (1 − z)
=
(q)∞
(q)∞
1 − zq n
n=−∞
n6=0
k−2 (k−2)n
∞
X
z q
z
1 − (zq n )k−2
−zθ1,2k−1 (q)
n n((2k−1)n+3)/2
1 + (1 − z)
(−1) q
+
+
=
(q)∞
(q)∞
1 − zq n
1 − zq n
n=−∞
n6=0
∞
X
−zθ1,2k−1 (q)
z
=
+
1 + (1 − z)
(−1)n q n((2k−1)n+3)/2
(q)∞
(q)∞
n=−∞
z
k−2 (k−2)n
n6=0
q
1 − zq n
+
k−3
X
z m q mn
m=0
!
∞
X
z
(−1)n q (2k−1)n(n+1)/2
−zθ1,2k−1 (q)
k−2
1 + z (1 − z)
+
=
(q)∞
(q)∞
1 − zq n
n=−∞
n6=0
+ (1 − z)
k−3
X
z
m
m=0
∞
X
n n((2k−1)n+2m+3)/2
(−1) q
n=−∞
n6=0
∞
X
z
(−1)n q (2k−1)n(n+1)/2
−zθ1,2k−1 (q)
z k−2 (1 − z)
+
=
(q)∞
(q)∞
1 − zq n
n=−∞
+ (1 − z)
1
=
(q)∞
k−3
X
z
m
m=0
∞
X
n n((2k−1)n+2m+3)/2
(−1) q
n=−∞
−zθ1,2k−1 (q) + z k−1 (1 − z)Σ(k) (z, q) + z(1 − z)
This completes the proof of Theorem 1.1.
k−3
X
!
z m θ2m+3,2k−1 (q) .
m=0
4. Higher order Rank-Crank-type PDEs
In this section we show how the generalized Lambert series identity (1.3) can be used
to derive general Rank-Crank PDEs of the type found by Zwegers.
4.1. The idea. First we let xi = ζ i , 1 ≤ i ≤ m in (1.3) to obtain
Ym (ζ, z, q) (q)2∞ = S2m+1 (ζ, z, q)+
m−1
X
Fj,m (ζ, q) S2m+1(ζ j+1, z, q)−F0,m (ζ, q) Σ(2m+1) (z, q),
j=1
(4.1)
RANK-CRANK TYPE PDES
where
Sk (ζ, z, q) :=
∞
X
n kn(n+1)/2
(−1) q
n=−∞
for k odd and
13
ζ −kn
ζ k(n+1)
+
1 − zζ −1 q n 1 − zζq n
,
(4.2)
[ζ m+1 ]
F0,m (ζ, q) := ζ m m ∞ ,
[ζ ]
−(m−1)∞ −(m−2)
ζ
,ζ
, · · · , ζ −(m−j) ∞
Fj,m (ζ, q) :=
(for 1 ≤ j ≤ m − 1),
[ζ m+2 , · · · , ζ m+j+1]∞
−(m−1) −(m−2)
ζ
,ζ
, · · · , ζ −2, ζ −1, ζ, ζ 2, · · · , ζ m, ζ m+1 ∞
.
Ym (ζ, z, q) :=
[zζ −m , zζ −(m−1) , · · · , zζ −2 , zζ −1 , z, zζ, zζ 2 , · · · , zζ m−1 , zζ m ]∞
(4.3)
(4.4)
(4.5)
The basic idea is to apply the operator D2m to both sides of (4.1) where
ℓ
∂
Dℓ := ζ
= δζℓ ζ=1 .
∂ζ
(4.6)
ζ=1
We will also need the differential operator
Hk∗ := kδz + 2kδq + δz2 .
(4.7)
We note that the operator Hk∗ differs from Zwegers’s Hk although they are similar. First
we need to write the functions Σ(k) (z, q) and Sk (z, q) as double series. Throughout we
assume that 0 < |q| < 1, z 6∈ {q n : n ∈ Z} ∪ {0} and ζ 6∈ {z ±1 q n : n ∈ Z} ∪ {0}. We
obtain
!
∞
∞
∞
X
X
X
z m ζ m q mn
z m ζ −mq mn + ζ k(n+1)
Sk (ζ, z, q) =
(−1)n q kn(n+1)/2 ζ −kn
−
=
n=0
∞
X
(−1)n q kn(n−1)/2
n=1
∞ X
∞
X
ζ kn
m=0
∞
X
z −m ζ m q mn + ζ k(−n+1)
−
and
(k)
(−1)n z m q kn(n+1)/2+mn ζ −kn−m + ζ k(n+1)+m
Σ (z, q) =
n m kn(n+1)/2+mn
(−1) z q
n=0 m=0
We have
(−1)n z −m q kn(n−1)/2+mn ζ kn+m + ζ −kn+k−m
−
∞ X
∞
X
!
(4.8)
(−1)n z −m q kn(n−1)/2+mn .
(4.9)
n=1 m=1
∞ X
∞
X
z −m ζ −mq mn
m=1
m=1
n=0 m=0
∞
∞ X
X
m=0
∞
X
n=1 m=1
Theorem 4.1. Suppose k is odd and 1 ≤ ℓ ≤ k − 1. Then
Dℓ Sk (ζ, z, q) = Pk,ℓ(Hk∗ )Σ(k) (z, q),
(4.10)
14
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
where
Pk,ℓ (x) =
⌊ℓ/2⌋
X ℓ(ℓ − m − 1)!
xm k ℓ−2m .
(ℓ
−
2m)!m!
m=0
Proof. Suppose k is odd and 1 ≤ ℓ ≤ k − 1 . First we prove that
√
√
Pk,ℓ (x) = ( 12 k − 12 k 2 + 4x)ℓ + ( 12 k + 21 k 2 + 4x)ℓ .
(4.11)
(4.12)
We have
√
1
√
1
⌊ℓ/2⌋
ℓ
( 12 k − 2 k 2 + 4x) + ( 12 k + 2 k 2 + 4x) =
k ℓ−2j 21−ℓ (k 2 + 4x)j
2j
j=0
⌊ℓ/2⌋
⌊ℓ/2⌋ j
⌊ℓ/2⌋
X
XX ℓ
X
j
j m ℓ−2m 2m−ℓ+1
ℓ
x k
2
.
xm k ℓ−2m 22m−ℓ+1 =
=
m
m
2j
2j
m=0
m=0
j=0
j=m
ℓ
ℓ
X
The result (4.12) now follows from the binomial coefficient identity
⌊ℓ/2⌋
X ℓ
ℓ(ℓ − m − 1)!
j
= 2ℓ−2m−1
,
m
2j
(ℓ − 2m)!m!
j=m
(4.13)
which we leave as an exercise.
We observe that if x = km + m2 + k 2 n(n + 1) + 2mnk then
1
k
2
1
k
2
−
+
√
k 2 + 4x = (k + 2m + 2kn)2 ,
1
k2
2
√
1
k2
2
+ 4x = −kn − m,
+ 4x = k(n + 1) + m,
and we see that
Dℓ ζ −kn−m + ζ k(n+1)+m = (−kn−m)ℓ +(k(n+1)+m)ℓ = Pk,ℓ (km+m2 +k 2 n(n+1)+2mnk).
Similarly we find that
Dℓ ζ kn+m + ζ −kn+k−m = (kn+m)ℓ +(−kn+k−m)ℓ = Pk,ℓ (−km+m2 +k 2 n(n−1)+2mnk).
We note that
Hk∗ q kn(n+1)/2+mn z m = (km + m2 + k 2 n(n + 1) + 2mnk) q kn(n+1)/2+mn z m ,
Hk∗ q kn(n−1)/2+mn z −m = (−km + m2 + k 2 n(n − 1) + 2mnk) q kn(n−1)/2+mn z −m .
Thus
and
Dℓ q kn(n+1)/2+mn z m (ζ −kn−m + ζ kn+k+m) = Pk,ℓ (Hk∗ ) q kn(n+1)/2+mn z m ,
Dℓ q kn(n+1)/2+mn z m (ζ kn+m + ζ −kn+k−m) = Pk,ℓ (Hk∗ ) q kn(n−1)/2+mn z −m .
The result (4.10) follows from equations (4.8) and (4.9).
Next we calculate D2m of each term in (4.1).
RANK-CRANK TYPE PDES
15
4.2. The term Ym (ζ, z, q). It is clear that the function Ym (ζ, z, q) has a zero of order
2m at ζ = 1. It is well-known that
D2m (f (ζ)) =
2m
X
S(2m, j)f (j) (1),
i=1
where the numbers S(2m, j) are Stirling numbers of the second kind. Since S(2m, 2m) =
1 it follows that
D2m (Ym (ζ, z, q))) = Ym(2m) (1, z, q) = (−1)m−1 (2m)! (m+1)! (m−1)! [C ∗ (z, q)]2m+1 (q)2m−1
∞
(4.14)
by an easy calculation.
4.3. The term F0,m (ζ, q). By logarithmic differentiation we have
δζ F0,m (ζ, q) = L0,m (ζ, q) F0,m(ζ, q).
(4.15)
where
∞
X
ζ m+1 q i
ζ −m−1 q i
L0,m (ζ, q) = K0,m (ζ) − (m + 1)
−
1 − ζ m+1 q i 1 − ζ −m−1q i
i=1
∞
X
ζ −m q i
ζ mqi
−
+m
1 − ζ mq i 1 − ζ −m q i
i=1
X
X
= K0,m (ζ) − (m + 1)
(ζ m+1q i )n − (ζ −m−1 q i )n + m
(ζ m q i )n − (ζ −mq i )n
i,n≥1
i,n≥1
(4.16)
K0,m (ζ) = m + Jm (ζ) − Jm−1 (ζ),
m
X
nζ n
Jm (ζ) =
n=1
m
X
(4.17)
.
ζ
(4.18)
n
n=0
For any positive integer k we define
G2k
∞
X
1
B2k
n2k−1 q n
:= G2k (q) := ζ(1 − 2k) +
=−
+ Φ2k−1 (q),
n
2
1−q
4k
n=1
where B2n is the (2n)-th Bernoulli number, and
∞
X
Φ2k−1 := Φ2k−1 (q) :=
σ2k−1 (n)q n .
(4.19)
(4.20)
n=1
The function G2k is a normalized Eisenstein series. For k > 1 it is an entire modular
form of weight 2k. We need the following
Lemma 4.2. If m and a are positive integers, then
Ba+1
(m + 1)a+1 − 1 .
Da (Jm (ζ)) =
a+1
16
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
Proof. Suppose m and a are positive integers. It is well-known that
∞
X
x
Bk xk
=
.
ex − 1 k=0 k!
(4.21)
By taking the logarithmic derivative of (ζ m+1 − 1)/(ζ − 1) we find that
1
1
−
.
Jm (ζ) = m + (m + 1) m+1
ζ
−1 ζ −1
Hence by (4.21) we have
∞
X
Bk+1
x
(m + 1)k+1 − 1 xk .
Jm (e ) = m +
(k + 1)!
k=0
The result now follows since
Da (Jm (ζ)) =
d
dx
a
Jm (ex )
.
x=0
Corollary 4.3. Suppose a, m are integers a ≥ 0 and m ≥ 1.
a+1
− (m + 1)a+1 )Ga+1 (q)
2(m
Da (L0,m (ζ, q)) = m + 12
0
Then
if a is odd,
if a = 0,
otherwise.
(4.22)
Proof. The proof of (4.22) when a = 0 is straightforward. Suppose a is even and
positive. Then by Lemma 4.2
Da (L0,m (ζ, q)) = Da (K0,m (ζ))
Ba+1
=
(−ma+1 + (m + 1)a+1 )
a+1
= 0,
since the Bernoulli numbers Bk are zero when k > 1 is odd. Now suppose a is odd.
Then again by Lemma 4.2
Da (L0,m (ζ, q)) = Da (K0,m (ζ)) + 2(ma+1 − (m + 1)a+1 )Φa (q)
Ba+1
=
(−ma+1 + (m + 1)a+1 ) + 2(ma+1 − (m + 1)a+1 )Φa (q)
a+1
= 2(ma+1 − (m + 1)a+1 )Ga+1 (q).
By applying Da−1 to both sides of (4.15) and using (4.22) we obtain the following
recurrence
Da (F0,m (ζ, q)) = (m + 12 )Da−1 (F0,m (ζ, q))
⌊a/2⌋
X
a−1
+
(m2i − (m + 1)2i )G2i (q)Da−2i (F0,m (ζ, q)).
2
2i
−
1
i=1
(4.23)
RANK-CRANK TYPE PDES
17
This together with the initial value
D0 (F0,m (ζ, q)) = F0,m (1, q) =
m+1
m
(4.24)
uniquely determines the coefficients Da (F0,m (ζ, q)). We compute some examples
3
D0 (F0,2 ) = ,
2
15
D1 (F0,2 ) = ,
4
75
− 15G2 = 10 − 15Φ1 ,
D2 (F0,2 ) =
8
365 225
225 225
D3 (F0,2 ) =
−
G2 =
−
Φ1 ,
16
2
8
2
1875 1125
−
G2 + 450G22 − 195G4 = 82 − 600Φ1 + 450Φ21 − 195Φ3 .
D4 (F0,2 ) =
32
2
4.4. The terms Fj,m (ζ, q) (1 ≤ j ≤ m − 1). Suppose 1 ≤ j ≤ m − 1. We may obtain
a similar recurrence for Da (Fj,m (ζ, q)). This time we find that
δζ Fj,m (ζ, q) = Lj,m (ζ, q) Fj,m(ζ, q),
(4.25)
for some function Lj,m (ζ, q) that satisfies
P
j
a+1
− (m − i)a+1 ) Ga+1 (q) if a is odd,
2 i=1 ((m+ i + 1)
Da (Lj,m (ζ, q)) = −j m + 21
(4.26)
if a = 0,
0
otherwise.
The proof of (4.26) is analogous to the proof of Corollary 4.3. By applying Da−1 to
both sides of (4.25) and using (4.26) we obtain the following recurrence
Da (Fj,m (ζ, q)) = −j(m + 12 )Da−1 (Fj,m (ζ, q))
⌊a/2⌋ j
X X a − 1
((m + k + 1)2i − (m − k)2i )G2i (q)Da−2i (Fj,m (ζ, q)).
2
+
2i
−
1
i=1 k=1
(4.27)
This together with the initial value
D0 (Fj,m (ζ, q)) = Fj,m(1, q) = (−1)
j
j
Y
i=1
m−i
m+i+1
(4.28)
18
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
uniquely determines the coefficients Da (Fj,m (ζ, q)). We compute some examples
1
D0 (F1,2 ) = − ,
4
5
D1 (F1,2 ) = ,
8
25 15
5 15
D2 (F1,2 ) = − − G2 = − − Φ1 ,
16
2
4
2
25 225
125 225
+
G2 =
+
Φ1 ,
D3 (F1,2 ) =
32
4
16
4
625 1125
255
1
255
D4 (F1,2 ) = −
−
G2 − 675G22 −
G4 = − 225Φ1 − 675Φ21 −
Φ3 .
64
4
2
4
2
4.5. The terms S2m+1 (ζ j+1, z, q) (1 ≤ j ≤ m−1). Again suppose that 1 ≤ j ≤ m−1.
Consider the operator Tj that operates on a function f (ζ) by Tj (f (ζ)) = f (ζ j ). Then
δζ ◦ Tj = j (Tj ◦ δζ ) .
(4.29)
δζa ◦ Tj = j a Tj ◦ δζa ,
(4.30)
Da ◦ Tj = j a Da .
(4.31)
∗
Dℓ S2m+1 (ζ j+1, z, q) = (j + 1)ℓ P2m+1,ℓ (H2m+1
)Σ(2m+1) (z, q).
(4.32)
A simple induction argument gives
and
Thus by (4.10) we have
4.6. The main theorem. We are now ready to derive our main theorem. Applying
D2m to both sides of (4.1) and using (4.10), (4.14), (4.23), (4.24), (4.27), (4.28), and
(4.32) we have
Theorem 4.4.
(−1)m+1 (2m)! (m + 1)! (m − 1)![C ∗ (z, q)]2m+1 (q)2m+1
∞
∗
P2m+1,2m (H2m+1
)
=
+
m−1
2m
XX
(j + 1)
2m−a
j=1 a=0
2m
∗
Da (Fj,m (ζ, q))P2m+1,2m−a (H2m+1
)
a
!
− D2m (F0,m (ζ, q)) Σ(2m+1) (z, q)
(4.33)
where the coefficient functions Da (Fj,m )(ζ, q)) (0 ≤ j ≤ m − 1) are given recursively
by (4.23) and (4.27), and their initial values (4.24) and (4.28).
For n ≥ 0 let Vn be the Q-vector space spanned by the monomials Φa1 Φb3 Φc5 with
a + 2b + 3c = n. We Define
n
X
Vn ;
(4.34)
Wn =
k=0
RANK-CRANK TYPE PDES
19
i.e., Wn is the Q-vector space spanned by the monomials Φa1 Φb3 Φc5 with 0 ≤ a+2b+3c ≤
n. We call Wn the space of quasi-modular forms of weight less than or equal to 2n.
This agrees with the definition in [2, p.355] except this time we allow monomials of
weight 0.
Corollary 4.5. Suppose m ≥ 1. Then there exist quasi-modular forms fj ∈ Wj for
1 ≤ j ≤ m such that
!
m−1
X
∗m
∗k
Σ(2m+1) (z, q) = (2m)! [C ∗ (z, q)]2m+1 (q)2m+1
.
(4.35)
H2m+1
+
fm−k H2m+1
∞
k=0
Proof. Suppose m ≥ 1. It is well-known that
Φ2n−1 ∈ Wn .
See [2, Eq.(3.25)]. Equation (4.19), the recurrence (4.27) and a simple induction argument imply that
Da (Fj,m (ζ, q)) ∈ W⌊a/2⌋ ,
for 1 ≤ j ≤ m − 1. Similarly using (4.22) and (4.23) we find that
D2m (F0,m (ζ, q)) ∈ Wm .
(4.36)
∗k
H2m+1
Now we calculate the coefficient of
in the right side of (4.33). The degree of
the polynomial P2m+1,2m−a (x) is ⌊(2m − a)/2⌋. Assuming k ≤ ⌊(2m − a)/2⌋ we have
2k ≤ 2m − a and ⌊a/2⌋ ≤ m − k, and in this case Da (Fj,m (ζ, q)) is in Wm−k . This
∗k
together with (4.36) implies that the coefficient of H2m+1
is in Wm−k for 0 ≤ k ≤ m.
∗m
The coefficient of H2m+1 is
f0 = 2 + 2
m−1
X
j
(−1) (j + 1)
2m
j=1
(m − 1)!(m + 1)!
(m − j − 1)!(m + j + 1)!
j=0
m−1
2m
2(m − 1)!(m + 1)! X
j
2m
.
(−1) (j + 1)
=
m
−
j
−
1
(2m)!
j=0
=2
We show that
m−1
X
j
Y
(m − k)
m+k+1
k=1
(−1)j (j + 1)2m
f0 = (−1)m+1 (m + 1)! (m − 1)!.
In view of (4.37) this is equivalent to showing that
m−1
2m
2 X
m+j+1
2m
= 1,
(−1)
(j + 1)
m−j −1
(2m)! j=0
(4.37)
(4.38)
(4.39)
which we can rewrite as
2
m−1
X
j=0
j
(−1) (m − j)
by replacing j by m − j − 1 in the sum.
2m
2m
= (2m)!,
j
(4.40)
20
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
Since
m−1
X
j=0
j
(−1) (m − j)
2m
2m
X
2m
j
2m 2m
,
=
(−1) (m − j)
j
j
j=m+1
it suffices to prove
2m
X
j=0
j
(−1) (m − j)
2m
2m
= (2m)!.
j
(4.41)
Beginning with the elementary identity
2m
X
2m j 2m−j
xy
= (x + y)2m ,
j
j=0
setting x = − √1ζ and y =
√
ζ, we obtain
2m
X
2m
j=0
j
(−1)j ζ m−j = ζ −m(ζ − 1)2m .
(4.42)
We apply D2m to both sides of (4.42) and argue as in Section 4.2 to obtain (4.41) which
completes the proof of (4.38). The final result (4.35) follows from (4.33) by dividing
both sides by f0 and using (4.38).
4.7. Some examples. We illustrate Theorem 4.4 and Corollary 4.5 with some examples. We show details of the calculations for the cases m = 1, 2. In cases m = 3, 4 we
give the quasi-modular forms fj (1 ≤ j ≤ m) in Corollary 4.5, in terms of the functions
Φ2k−1 (q) rather than the G2k (q).
Example m = 1.
!
4 [C ∗ (z, q)]3 (q)3∞ =
=
=
9 + 2 H3 − D2 (F0,1 (ζ, q)) Σ(3) (z, q)
!
9 + 2 H3 − (5 − 12 Φ1 ) Σ(3) (z, q)
!
2 H3 + 4 + 12 Φ1 Σ(3) (z, q),
and
!
H3 + 2 + 6 Φ1 Σ(3) (z, q) = 2 [C ∗ (z, q)]3 (q)3∞ .
This identity implies the Rank-Crank PDE (1.7) as in [2, Section 2].
RANK-CRANK TYPE PDES
21
Example m = 2.
− 144 [C ∗(z, q)]5 (q)5∞
=
625 + 100 H∗5 + 2 H∗ 25 + 16 D0 (F1,2 (ζ, q)) (625 + 100 H∗5 + 2 H∗ 25 )
+ 32 D1 (F1,2 (ζ, q)) (125 + 15 H∗5 ) + 24 D2(F1,2 (ζ, q)) (25 + 2 H∗5 ) + 40 D3 (F1,2 (ζ, q))
!
+ 2 D4 (F1,2 (ζ, q)) − D4 (F0,2 (ζ, q)) Σ(5) (z, q)
=
625 + 100 H∗5 + 2 H∗ 25 − 4 (625 + 100 H∗5 + 2 H∗ 25 )
+ 20 (125 + 15 H∗5 ) − (30 + 180 Φ1 )(25 + 2 H∗ 5 ) + (
125
+ 2250 Φ1 )
2
!
1
+ ( − 450 Φ1 − 1350 Φ21 − 255 Φ3) + (−82 + 600 Φ1 − 450Φ21 + 195 Φ3 ) Σ(5) (z, q),
2
and
!
H∗ 25 + (60Φ1 + 10) H∗ 5 + 300 Φ21 + 10 Φ3 + 350 Φ1 + 24 Σ(5) (z, q) = 24 [C ∗ (z, q)]5 (q)5∞ .
(4.43)
In this case of Corollary 4.5 we see that
f1 = 60Φ1 + 10,
f2 = 300 Φ21 + 10 Φ3 + 350 Φ1 + 24.
We show how this identity implies (1.18). We need the results
δq ((q)∞ ) = −Φ1 (q)∞ ,
δq (Φ1 ) = 61 Φ1 − 2Φ21 + 65 Φ3 .
This implies that
H5∗ (Σ∗ (z, q)) = H5∗ ((q)3∞ G(5) (z, q)) = (q)3∞ (H5∗ − 30Φ1 )G(5) (z, q),
(4.44)
and
H5∗ 2 (Σ∗ (z, q)) = H5∗ 2 ((q)3∞ G(5) (z, q)) = (q)3∞ (H5∗ 2 −60Φ1 H5∗ −50Φ1 +1500Φ21 −250Φ3 )G(5) (z, q).
(4.45)
Substituting (4.44), (4.45) into (4.43) we find
(H5∗ 2 + 10H5∗ + 24 − 240Φ3 )G(5) (z, q) = 24 [C ∗(z, q)]5 (q)5∞ ,
which simplifies to (1.18) since
H∗ = H5∗ + 5.
22
SONG HENG CHAN, ATUL DIXIT AND FRANK G. GARVAN
Example m = 3. We find that
f1 = 210 Φ1 + 28
f2 = 210 Φ3 + 8820 Φ21 + 252 + 4410 Φ1
f3 = 41160 Φ31 + 2450 Φ3 + 14 Φ5 + 720 + 22736 Φ1 + 2940 Φ3 Φ1 + 102900 Φ21
Example m = 4. We find that
f1 = 504 Φ1 + 60
f2 = 1260 Φ3 + 24948 Φ1 + 1308 + 68040 Φ21
f3 = 136080 Φ3 Φ1 + 504 Φ5 + 45360 Φ3 + 2449440 Φ21 + 403704 Φ1 + 2449440 Φ31 + 12176
f4 = 40320 + 2126232 Φ1 + 404082 Φ3 + 9828 Φ5 + 2653560 Φ3 Φ1 + 21820428 Φ21
+ 9072 Φ1 Φ5 + 47764080 Φ31 + 18 Φ7 + 1224720 Φ3 Φ21 + 11340 Φ23 + 11022480 Φ41.
5. Concluding remarks
The main goal of this paper was to show how the generalized Lambert series identity
(1.3) leads to the higher level Rank-Crank-type PDEs of Zwegers. The first author’s
proof [8] of (1.3) only involves a partial fraction argument and this together with the
proof in Section 4 gives an elementary q-series proof of these higher level Rank-Cranktype PDEs. The elliptic function proof of (1.3) in Section 2 is independent of the other
sections. Our form of Zwegers’s result (1.23) was given above in (4.35). In our form
the coefficients are quasimodular forms rather than holomorphic modular forms. The
quasimodular function E2 occurs in Zwegers’s result as part of the definition of his
operator Hk . Our coefficient functions are given recursively. It would be interesting to
find explicit expressions for the coefficients and to derive the form of Zwegers’s result by
our method. The coefficients in the 4th order PDE (1.18) only involve the holomorphic
modular form E4 , and the differential operator H∗ does not involve the quasimodular
E2 . It would be interesting to determine whether there is a renormalization of higher
order Rank-Crank-type PDEs which only involve holomorphic modular forms, either
as coefficients or in the definition of the differential operator. Bringmann, Lovejoy
and Osburn [5], [6] found Rank-Crank-type PDEs for overpartitions. Bringmann and
Zwegers [7] showed how these results fit into the framework of non-holomorphic Jacobi
forms and found an infinite family of these PDEs. However these PDEs only involve
Appell functions of level 1 or 3. It would be interesting to determine whether the
methods of this paper could be extended to find PDEs for higher level analogues.
Acknowledgements
We would like to thank Bruce Berndt and Ken Ono for their comments and suggestions.
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Division of Mathematical Sciences, School of Physical and Mathematical Sciences,
Nanyang Technological University, 21 Nanyang link, Singapore, 637371, Republic of
Singapore
E-mail address: ChanSH@ntu.edu.sg
Department of Mathematics, University of Illinois, 1409 West Green Street, Urbana, IL 61801, USA
E-mail address: aadixit2@illinois.edu
Department of Mathematics, University of Florida, Gainesville, Florida 32611,
USA
E-mail address: fgarvan@ufl.edu