TRANSFORMATION PROPERTIES FOR DYSON’S RANK FUNCTION
F. G. GARVAN
ABSTRACT. At the 1987 Ramanujan Centenary meeting Dyson asked for a coherent group- theoretical structure for Ramanujan’s mock theta functions analogous to Hecke’s theory of modular forms. Many of Ramanujan’s mock theta functions can be written in terms of R(ζ, q), where R(z, q) is the two-variable generating function of Dyson’s rank function and ζ is a root of unity. Building on earlier work of Watson, Zwegers, Gordon and McIntosh, and motivated by Dyson’s question, Bringmann, Ono and Rhoades studied transformation prop- erties of R(ζ, q). In this paper we strengthen and extend the results of Bringmann, Rhoades and Ono, and the later work of Ahlgren and Treneer. As an application we give a new proof of Dyson’s rank conjecture and show that Ramanujan’s Dyson rank identity modulo 5 from the Lost Notebook has an analogue for all primes greater than 3. The proof of this analogue was inspired by recent work of Jennings-Shaffer on overpartition rank differences mod 7.
1. INTRODUCTION Let p(n) denote the number of partitions of n. There are many known congruences for the partition function. The simplest and most famous were found and proved by Ramanujan: p(5n + 4) ≡ 0 (mod 5), p(7n + 5) ≡ 0 (mod 7), p(11n + 6) ≡ 0 (mod 11). In 1944, Dyson [19] conjectured striking combinatorial interpretations of the first two con- gruences. He defined the rank of a partition as the largest part minus the number of parts and conjectured that the rank mod 5 divided the partitions of 5n + 4 into 5 equal classes and that the rank 7 divided the partitions of 7n + 5 into 7 equal classes. He conjectured the existence a partition statistic he called the crank which would likewise divide the partitions of 11n + 6 into 11 equal classes. Dyson’s mod 5 and 7 rank conjectures were proved by Atkin and Swinnerton-Dyer [9]. The mod 11 crank conjecture was solved by the author and Andrews [5].
Date: June 17, 2016. Revised February 7, 2017. Revised February 27, 2017. 2010 Mathematics Subject Classification. Primary 11F37, 11P82; Secondary 05A19, 11B65, 11F11, 11P83, 11P84, 33D15. Key words and phrases. Dyson’s rank function, Maass forms, mock theta functions, partitions, Mordell integral. The author was supported in part by a grant from the Simon’s Foundation (#318714). A preliminary version of this paper was first given on May 10, 2015 at the International Conference on Orthogonal Polynomials and q-series at University of Central Florida, Orlando, in honour of Mourad Ismail’s 70th birthday. 1 2 F. G. GARVAN
Let N(m, n) denote the number of partitions of n with rank m. We let R(z, q) denote the two-variable generating function for the Dyson rank function so that ∞ X X R(z, q) = N(m, n) zm qn. n=0 m P Here and throughout the paper we use the convention that any sum m is a sum over all integers m ∈ Z, unless otherwise stated. Also throughout this paper we will use the standard q-notation: ∞ Y k (a; q)∞ = (1 − aq ), k=0
(a; q)∞ (a; q)n = n , (aq ; q)∞
(a1, a2, . . . , aj; q)∞ = (a1; q)∞(a2; q)∞ ... (aj; q)∞,
(a1, a2, . . . , aj; q)n = (a1; q)n(a2; q)n ... (aj; q)n. We have the following identities for the rank generating function R(z, q):
∞ 2 X qn (1.1) R(z, q) = 1 + (zq, z−1q; q) n=1 n ∞ n n −1 ! 1 X (−1) (1 + q )(1 − z)(1 − z ) 1 n(3n+1) (1.2) = 1 + q 2 . (q; q) (1 − zqn)(1 − z−1qn) ∞ n=1 See [21, Eqs (7.2), (7.6)]. Let N(r, t, n) denote the number of partitions of n with rank congruent to r mod t, and let ζp = exp(2πi/p). Then
∞ p−1 ! X X k n (1.3) R(ζp, q) = N(k, p, n) ζp q . n=0 k=0 We restate Dyson’s Rank Conjecture 1.1 (1944). For all nonnegative integers n, 1 (1.4) N(0, 5, 5n + 4) = N(1, 5, 5n + 4) = ··· = N(4, 5, 5n + 4) = 5 p(5n + 4), 1 (1.5) N(0, 7, 7n + 5) = N(1, 7, 7n + 5) = ··· = N(6, 7, 7n + 5) = 7 p(7n + 5). Dyson’s rank conjecture was first proved by Atkin and Swinnerton-Dyer [9]. As noted in [21], [22], it can be shown that Dyson’s mod 5 rank conjecture (1.4) follows from an identity in Ramanujan’s Lost Notebook [41, p.20], [4, Eq. (2.1.17)]. We let ζ5 be a primitive 5th root of unity. Then 5 −1 5 5 −1 2 5 (1.6) R(ζ5, q) = A(q ) + (ζ5 + ζ5 − 2) φ(q ) + q B(q ) + (ζ5 + ζ5 ) q C(q ) ψ(q5) − (ζ + ζ−1) q3 D(q5) − (ζ2 + ζ−2 − 2) , 5 5 5 5 q5 DYSON’S RANK FUNCTION 3 where 2 3 5 5 5 5 5 5 4 5 5 (q , q , q ; q )∞ (q ; q )∞ (q ; q )∞ (q, q , q ; q )∞ A(q) = 4 5 2 ,B(q) = 4 5 ,C(q) = 2 3 5 ,D(q) = 2 3 5 2 , (q, q ; q )∞ (q, q ; q )∞ (q , q ; q )∞ (q , q ; q )∞ and ∞ 2 ∞ 2 X q5n X q5n φ(q) = −1 + , ψ(q) = −1 + . (q; q5) (q4; q5) (q2; q5) (q3; q5) n=0 n+1 n n=0 n+1 n By multiplying by an appropriate power of q and substituting q = exp(2πiz), we recognize the functions A(q), B(q), C(q), D(q) as being modular forms. In fact, we can rewrite Ramanujan’s identity (1.6) in terms of generalized eta-products: ∞ 1 Y n η(z) := q 24 (1 − q ), q = exp(2πiz), n=1 and
N P2(k/N) Y m (1.7) ηN,k(z) = q 2 (1 − q ), m>0 m≡±k (mod N)
2 1 where z ∈ h, P2(t) = {t} − {t} + 6 is the second periodic Bernoulli polynomial, and {t} = t − [t] is the fractional part of t. Here as in Robins [42], 1 ≤ N - k. We have
1 − 24 4 5 4 −2 5 (1.8) q R(ζ5, q) − (ζ5 + ζ5 − 2) φ(q ) + (1 + 2ζ5 + 2ζ5 ) q ψ(q )
η(25z) η5,2(5z) η(25z) 4 η(25z) 4 η(25z) η5,1(5z) = 2 + + (ζ5 + ζ5 ) − (ζ5 + ζ5 ) 2 . η5,1(5z) η5,1(5z) η5,2(5z) η5,2(5z) Equation (1.6), or equivalently (1.8), give the 5-dissection of the q-series expansion of R(ζ5, q). We observe that the function on the right side of (1.8) is a weakly holomorphic 1 modular form (with multiplier) of weight 2 on the group Γ0(25) ∩ Γ1(5). In Theorem 1.2 below we generalize the analogue of Ramanujan’s result (1.8) to all primes 1 p > 3. For p > 3 prime and 1 ≤ a ≤ 2 (p − 1) define ∞ pn2 X q , if 0 < 6a < p, (qa; qp) (qp−a; qp) n=0 n+1 n (1.9) Φp,a(q) := ∞ 2 X qpn −1 + , if p < 6a < 3p, (qa; qp) (qp−a; qp) n=0 n+1 n and (1.10) 1 2 (p−1) 1 1 1 − X a 3a+ 2 (p+1) −3a− 2 (p+1) Rp (z) := q 24 R(ζp, q) − χ12(p) (−1) ζp + ζp a=1 1 1 a p2 3a+ 2 (p−1) −3a− 2 (p−1) (p−3a)− p −ζp − ζp q 2 24 Φp,a(q ), 4 F. G. GARVAN where 1 if n ≡ ±1 (mod 12), 12 (1.11) χ (n) := = −1 if n ≡ ±5 (mod 12), 12 n 0 otherwise, and as usual q = exp(2πiz) with =(z) > 0. One of our main results is Theorem 1.2. Let p > 3 be prime. Then the function 2 η(p z) Rp (z) 2 is a weakly holomorphic modular form of weight 1 on the group Γ0(p ) ∩ Γ1(p). Remark. The form of this result is suggested by Ramanujan’s identity (1.8). The proof of this result uses the theory of weak harmonic Maass forms and was inspired by a recent result of Jennings-Shaffer [31] on overpartition rank differences mod 7. Jennings-Shaffer was the first to prove a result of this type using the theory of weak harmonic Maass forms. As a consequence we have
1 2 Corollary 1.3. Let p > 3 be prime and sp = 24 (p − 1). Then the function ∞ ∞ p−1 ! Y pn X X k n (1 − q ) N(k, p, pn − sp) ζp q n=1 1 k=0 n=d p (sp)e is a weakly holomorphic modular form of weight 1 on the group Γ1(p). Remark. Corollary 1.3 is the case m = 0 of Proposition 6.2(i) below. In Ramanujan’s identity (1.6) we see that the coefficient of q5n+4 is zero, and this implies Dyson’s rank 5 conjecture (1.4) in view of (1.3). The analog of (1.6) for the prime 7 does not appear in Ramanujan’s lost notebook although Ramanujan wrote the left side [41, p.19] and wrote some of the functions involved in coded form on [41, p.71]. The complete identity is given by Andrews and Berndt [4, Eq. (2.1.42)]. As noted by the author [22, Theorem 4, p.20] Ramanujan’s identity (1.6) is actually equivalent to Atkin and Swinnerton-Dyer’s [9, Theorem 4, p.101]. In Section 6 we give a new proof of Dyson’s Conjecture 1.1 as well as some of Atkin and Hussain’s [8] results on the rank mod 11 and O’Brien’s [37] results on the rank mod 13. Until this paper, Atkin and Swinnerton-Dyer’s proof has remained the only known proof of Dyson’s Rank Conjectures 1.1. We now explain the connection between Ramanujan’s mock theta functions and Dyson’s rank functions. On [41, p.20] Ramanujan gives four identities for some mock theta functions of order 5. For example, χ0(q) = 2 + 3φ(q) − A(q), where ∞ ∞ X qn X q2n+1 χ (q) = = 1 + . 0 (qn+1; q) (qn+1; q) n=0 n n=0 n+1 The other three identities correspond to [6, Eqs (3.2), (3.6), (3.7)]. These identities are the Mock Theta Conjectures which were later proved by Hickerson [26]. Thus Ramanujan’s DYSON’S RANK FUNCTION 5 mock theta functions of order 5 are related to R(z, q) when z = ζ5. All of Ramanujan’s third order mock theta functions can also be written in terms of R(z, q). For example,
2 X qn f(q) = = R(−1, q), (−q; q)2 n≥0 n X q2n(n+1) ω(q) = = g(q; q2), (q; q2)2 n≥0 n+1 where 1 (1.12) g(x, q) = x−1 −1 + R(x, q) . 1 − x Hickerson and Mortenson [28, Section 5] give a nice catalogue of these and other known analogous identities for all of Ramanujan’s mock theta functions. These known results are orginally due to Watson [44], Andrews [3], Hickerson [27], Andrews and Hickerson [7], Choi [16], [17] Berndt and Chan [10], McIntosh [36], and Gordon and McIntosh [23]. The main emphasis of this paper is transformations for Dyson’s rank function R(ζ, q) and thus for Ramanujan’s mock theta functions. We begin with a quote from Freeman Dyson. The mock theta-functions give us tantalizing hints of a grand synthesis still to be discovered. Somehow it should be possible to build them into a coher- ent group-theoretical structure, analogous to the structure of modular forms which Hecke built around the old theta-functions of Jacobi. This remains a challenge for the future. Freeman Dyson, 1987 Ramanujan Centenary Conference In this paper we continue previous work on Dyson’s challenge. First we describe the genesis of this work. Watson [44] found transformation formulas for the third order functions in terms of Mordell integrals. For example, r r Z ∞ −1/24 2π 4/3 2 3α −3αx2/2 sinh αx q f(q) = 2 q1 ω(q1) + 4 e dx, α 2π 0 sinh 3αx/2 where 2 q = exp(−α), q1 = exp(−β), αβ = π . The derivation of transformation formulas for the other Ramanujan mock theta functions was carried out in a series of papers by Gordon and McIntosh. A summary of these results can be found in [25]. The big breakthrough came when Zwegers [47], [48] realised how Ramanujan’s mock theta functions occurred as the holomorphic part of certain real analytic modular forms. An example for the third order functions f(q), ω(q), ω(−q) is given in
T Theorem 1.4 (Zwegers [47]). Define F (z) = (f0, f1, f2) by
−1/24 1/3 1/2 1/3 1/2 f0(z) = q f(q), f1(z) = 2q ω(q ), f1(z) = 2q ω(−q ), 6 F. G. GARVAN where q = exp(2πiz), z ∈ h. Define Z i∞ T √ (g1(τ), g0(τ), −g2(τ)) G(z) = 2i 3 p dτ, −z −i(τ + z) where X 1 2 g (z) = (−1)n n + q3(n+1/3) /2, 0 3 n 1 X 1 3(n+ )2/2 g (z) = − n + q 6 , 1 6 n X 1 2 g (z) = n + q3(n+1/3) /2. 2 3 n Then H(z) = F (z) − G(z) is a (vector-valued) real analytic modular form of weight 1/2 satisfying −1 ζ24 0 0 H(z + 1) = 0 0 ζ3 H(z), 0 ζ3 0 0 1 0 1 −1 √ H = 1 0 0 H(z). −iz z 0 0 −1 Bringmann and Ono [13], [12], [38] extended this theorem to R(ζ, q) when ζ is a more general root of unity. Theorem 1.5 (Bringmann and Ono, Theorem 3.4 [13]). Let c be an odd positive integer and −1/24 a let 0 < a < c. Then q R(ζc , q) is the holomorphic part of a component of a vector valued weak Maass form of weight 1/2 for the full modular group SL2(Z). Remark. By component of a vector valued weak Maass form we mean an element of the set Vc defined on [13, p.441]. We give this result explicitly below in Corollary 3.2. The definition of a vector valued Maass form of weight k for SL2(Z) is given on [13, p.440]. Bringmann and Ono used this theorem to obtain a subgroup of the full modular group on which q−1/24R(ζ, q) is the holomorphic part a weak Maass form of weight 1/2. We state their theorem in the case where ζ is a pth root of unity. Theorem 1.6 (Bringmann and Ono [13]). Let p > 3 be prime, and 0 < a < p. Define X 2 θ(α, β; z) := nqn q = exp(2πiz), n≡α (mod β)
a X aπ(6m + 1) z Θ ; z := (−1)m sin θ 6m + 1, 12p, , p p 24 m (mod 2p) DYSON’S RANK FUNCTION 7
a i∞ 2 a πa √ Z Θ p ; 24p τ dτ S1 ; z := −i sin 2 2p p . p p −z −i(τ + z) Then a 2 2 a D ; z := q−p R ζa; q24p − S ; z p p 1 p 4 is a weak Maass form of weight 1/2 on Γ1(576 · p ). Bringmann, Ono and Rhoades [14] applied this theorem to prove Theorem 1.7 (Bringmann, Ono and Rhoades; Theorem 1.1 [14]). Suppose t ≥ 5 is prime, 0 ≤ r1, r2 < t and 0 ≤ d < t. Then the following are true: 1−24d (i) If t = −1, then ∞ X 24(tn+d)−1 (N(r1, t, tn + d) − N(r2, t, tn + d))q n=0 6 is a weight 1/2 weakly holomorphic modular form on Γ1(576 · t ). 1−24d 1 (ii) Suppose that t = 1. If r1, r2 6≡ 2 (±1 ± α) (mod t), where α is any integer for which 0 ≤ α < 2t and 1 − 24d ≡ α2 (mod 2t), then ∞ X 24(tn+d)−1 (N(r1, t, tn + d) − N(r2, t, tn + d))q n=0 6 is a weight 1/2 weakly holomorphic modular form on Γ1(576 · t ). Ahlgren and Treneer [1] strengthened this theorem to include the case 24d ≡ 1 (mod t). Theorem 1.8 (Ahlgren and Treneer [1]). Suppose p ≥ 5 is prime and that 0 ≤ r < t. Then ∞ X pn + 1 1 pn + 1 N r, p, − p qn 24 t 24 n=1 4 is a weight 1/2 weakly holomorphic modular form on Γ1(576 · p ). Remark. Ahlgren and Trenneer [1] also derived many analogous results when t is not prime. In this paper we strengthen many of the results of Ahlgren, Bringmann, Ono, Rhoades and Trenneer [1], [13], [14]. In particular (i) We make Theorem 1.5 (Bringmann and Ono [13, Theorem 3.4] more explicit. See Theorem 3.1 below. In this theorem we have corrected Bringmann and Ono’s defini- a tion of the functions G2 c ; z and G2(a, b, c; z). (ii) For the case p > 3 prime we strengthen Theorem 1.6 (Bringmann and Ono [13, 4 2 Theorem 1.2]). We show that the group Γ1(576·p ) can be enlarged to Γ0(p )∩Γ1(p) but with a simple multiplier. See Corollary 4.2. We prove more. In Theorem 4.1 we a z basically determine the image of Bringmann and Ono’s function D p ; 24p2 under the group Γ0(p). 8 F. G. GARVAN
(iii) We strengthen Theorem 1.7(i) (Bringmann, Ono and Rhoades [14, Theorem 1.1(i)]), and Theorem 1.8 (Ahlgren and Treneer [1, Theorem 1.6, p.271]). In both cases we 4 enlarge the group Γ1(576 · p ) to the group Γ1(p) but with a simple multiplier. See Corollary 6.5 below. (iv) We take a different approach to Bringmann, Ono and Rhoades Theorem 1.7(ii). We 1−24d handle the residue classes t = 1 in a different way. For these residue classes we show the definition of the functions involved may be adjusted to make them holomor- phic modular forms. In particular see equation (6.5), Theorem 6.3 and its Corollary 6.5. The paper is organized as follows. In Section 2 we go over Bringmann and Ono’s transfor- mation results for various Lambert series and Mordell integrals. We make all results explicit and correct some errors. In Section 3 we describe Bringmann and Ono’s vector-valued Maass 1 form of weight 2 , correcting some definitions and making all results explicit. In Section 4 we ` derive a Maass form multiplier for Bringmann and Ono’s function G1 p ; z for the group Γ0(p) when p > 3 is prime. In Section 5 we prove Theorem 1.2 which extends Ramanujan’s Dyson rank identity (1.6) or (1.8) to all primes p > 3. In Section 6 we give a new proof of the Dyson Rank Conjecture (1.4)–(1.5). We extend Ahlgren, Bringmann, Ono, Rhoades and Treneer’s results, mentioned in (ii)–(iv) above, for all primes p > 3. We also use our method to derive Atkin and Hussain’s [8] results for the rank mod 11 and O’Brien’s [37] results for rank mod 13. In Section 7 we mention other approaches and methods of proof of Bringmann and Ono’s work, particularly Zagier’s [46] simple proof and Hickerson and Mortenson’s [30] more elementary approach.
2. PRELIMINARIES Following [13] we define a number of functions. Suppose 0 < a < c are integers, and assume throughout that q := exp(2πiz). We define
∞ n n+ a a 1 X (−1) q c 3 2 n(n+1) M ; z := n+ a q c (q; q) 1 − q c ∞ n=−∞ ∞ n n 2πa ! a 1 X (−1) (1 + q ) 2 − 2 cos c 1 n(3n+1) N ; z := 1 + q 2 . c (q; q) 1 − 2 cos 2πa qn + q2n ∞ n=1 c For integers 0 ≤ a < c, 0 < b < c define
∞ n n+ a 1 X (−1) q c 3 2 n(n+1) M(a, b, c; z) := n+ a q , (q; q) 1 − ζbq c ∞ n=−∞ c b 1 1 5 where ζc := exp(2πi/c). In addition for c 6∈ {0, 6 , 2 , 6 } define 0 if 0 < b < 1 , c 6 1 if 1 < b < 1 , k(b, c) := 6 c 2 2 if 1 < b < 5 , 2 c 6 5 b 1 3 if 6 < c < 6 , DYSON’S RANK FUNCTION 9 and −a b ∞ 1 iζ2c q 2c X m(3m+1) N(a, b, c; z) := + K(a, b, c, m; z)q 2 , b (q; q)∞ −a 2 1 − ζc q c m=1 where
K(a, b, c, m; z) sin πa − b + 2k(b, c)m πz + sin πa − b − 2k(b, c)m πz qm := (−1)m c c c c . 2πa 2πbz m 2m 1 − 2 cos c − c q + q We need the following identities.
∞ n2 ∞ n 3 n(n+1) 1 X q z X (−1) q 2 (2.1) − 1 + = , 1 − z (zq, z−1q; q) (q; q) 1 − zqn n=0 n ∞ n=−∞ and
∞ n 3 n(n+1) ∞ n n 1 X (−1) q 2 z X (−1) q 3 n(n+1) (2.2) = q 2 . (q; q) 1 − zqn (q; q) 1 − zqn ∞ n=−∞ ∞ n=−∞
Equation (2.1) is [21, Eq.(7.10), p.68] and (2.2) is an easy exercise. Replacing q by qc and z by qa in (2.1), (2.2) we find that
∞ cn2 a ∞ n 3c n(n+1) X q q X (−1) q 2 (2.3) = 1 + (qa; ac) (qc−a; qc) (qc; qc) 1 − qcn+a n=0 n+1 n ∞ n=−∞ a = 1 + qa M ; cz . c By (1.2) we have a (2.4) R(ζa, q) = N ; z . c c Extending earlier work of Watson [44], Gordon and McIntosh [24], Bringmann and Ono [13] found transformation formula for all these functions in terms of Mordell integrals. We define the following Mordell integrals
Z ∞ 3a 3a a − 3 αx2 cosh c − 2 αx + cosh c − 1 αx J ; α := e 2 dx, c 0 cosh(3αx/2) and Z ∞ b −αx 2b −2αx 3 2 a ζc e + ζc e − 2 αx +3αx c J(a, b, c; α) := e x b dx. −∞ cosh 3α 2 − 3πi c 10 F. G. GARVAN
Following [13] we adjust the definitions of the N- and M-functions so that the transformation formulas are tidy. We define (2.5)
a aπ − 1 a N ; z := csc q 24 N ; z , c c c (2.6) a 3a 1− a − 1 a M ; z := 2q 2c ( c ) 24 M ; z , c c (2.7) 3a 1− a − 1 M(a, b, c; z) := 2q 2c ( c ) 24 M(a, b, c; z), (2.8) 2 a b 2a −b b k(b,c)− 3b − 1 N (a, b, c; z) := 4 exp −2πi k(b, c) + 3πi − 1 ζ q c 2c2 24 N(a, b, c; z). c c c c We now restate Bringmann and Ono’s [13, Theorem 2.3, p.435] more explicitly. Theorem 2.1. Suppose that c is a positive odd integer, and that a and b are integers for which 0 ≤ a < c and 0 < b < c. (1) For z ∈ h we have a a (2.9) N ; z + 1 = ζ−1 N ; z , c 24 c ( 3b2 −1 ζ2c2 ζ24 N (a − b, b, c; z) if a ≥ b, (2.10) N (a, b, c; z + 1) = 3b2 −3b −1 −ζ2c2 ζc ζ24 N (a − b + c, b, c; z) otherwise, a 2 (2.11) M ; z + 1 = ζ5a ζ−3a ζ−1 M(a, a, c; z), c 2c 2c2 24 M(a, a + b, c; z) if a + b < c, 2 5a −3a −1 a (2.12) M(a, b, c; z + 1) = ζ2c ζ2c2 ζ24 M c ; z if a + b = c, M(a, a + b − c, c; z) otherwise, where a is assumed to be positive in the first and third formula. (2) For z ∈ h we have 1 a 1 a √ √ a (2.13) √ N ; − = M ; z + 2 3 −iz J ; −2πiz , −iz c z c c 1 1 √ √ (2.14) √ N a, b, c; − = M(a, b, c; z) + ζ−5b 3 −iz J(a, b, c; −2πiz), −iz z 2c √ 1 a 1 a 2 3i a 2πi (2.15) √ M ; − = N ; z − J ; , −iz c z c z c z √ 1 1 3i 2πi (2.16) √ M a, b, c; − = N (a, b, c; z) − ζ−5b J a, b, c; , −iz z 2c z z where again a is assumed to be positive in the first and third formula. DYSON’S RANK FUNCTION 11
We will write each of the Mordell integrals as a period integral of a theta-function. Before we can do this we need some results of Shimura [43]. For integers 0 ≤ k < N we define ∞ ∼ X πiz θ(k, N; z) := (Nm + k) exp (Nm + k)2 . N m=−∞ We note that this corresponds to θ(z; k, N, N, P ) in Shimura’s notation [43, Eq.(2.0), p.454] (with n = 1, ν = 1, and P (x) = x). For integers 0 ≤ a, b < c we define
6c−1 ∼ 3ab −a X m π −2πima 2 Θ (a, b, c; z) := ζ 2 ζ (−1) sin (2m + 1) exp θ(2mc−6b+c, 12c ; z), 1 c 2c 3 c m=0 and 2c−1 X −πib ∼ Θ (a, b, c; z) := (−1)` exp (6` + 1) θ(6c` + 6a + c, 12c2; z) 2 c `=0 −πib ∼ +(−1)` exp (6` − 1) θ(6c` + 6a − c, 12c2; z) . c An easy calculation gives (2.17) ∞ 3ab −a X n n 1 b π −2πina Θ (a, b, c; z) = 6c ζ 2 ζ (−1) + − sin (2n + 1) exp 1 c 2c 3 6 c 3 c n=−∞ ! n 1 b2 × exp 3πiz + − . 3 6 c
We calculate the action of SL2(Z) on each of these theta-functions. Proposition 2.2. For integers 0 ≤ a, b < c and z ∈ h we have
−3b2 (2.18) Θ1(a, b, c; z + 1) = ζ2c2 ζ24 Θ1(a + b, b, c; z), √ − 3 1 3i (2.19) (−iz) 2 Θ a, b, c; − = − Θ (a, b, c; z), 1 z 2 2 3a2 (2.20) Θ2(a, b, c; z + 1) = ζ2c2 ζ24 Θ2(a, b − a, c; z), √ − 3 1 2 3i (2.21) (−iz) 2 Θ a, b, c; − = Θ (a, b, c; z). 2 z 3 1 Proof. Transformations (2.18), (2.20) are an easy calculation. By [43, Eq. (2.4), p.454] we have ∼ 1 (−iz)−3/2 θ 2mc − 6b + c, 12c2; − z X πik ∼ = (−i) (12c2)−1/2 exp (2mc + c − 6b) θ(k, 12c2; z). 6c2 k (mod 12c2) 12 F. G. GARVAN
Therefore
1 (−iz)−3/2 Θ a, b, c; − 1 z 2 −1/2 3ab −a X m π −2πima = (−i) (12c ) ζ 2 ζ (−1) sin (2m + 1) exp c 2c 3 c m (mod 6c) X πik ∼ exp (2mc + c − 6b) θ(k, 12c2; z) 6c2 k (mod 12c2) ∼ (−i) 3ab −a X 2 = √ ζ 2 ζ θ(k, 12c ; z) 2c 3 c 2c k (mod 12c2) X k a (−1)m exp 2πi (2mc + c − 6b) − m 12c2 c m (mod 6c) 1 πi πi × exp (2m + 1) − exp − (2m + 1) 2i 3 3 ∼ (−1) 3ab −a X πik 2 = √ ζ 2 ζ exp (c − 6b) θ(k, 12c ; z) 4c 3 c 2c 6c2 k (mod 12c2) πi X 2πi exp exp (k − 6a + 5c)m 3 6c m (mod 6c) πi X 2πi − exp − exp (k − 6a + c)m 3 6c m (mod 6c) √ ∼ − 3 3ab −a X πi πik 2 = ζ 2 ζ exp + (c − 6b) θ(k, 12c ; z) 2 c 2c 3 6c2 k (mod 12c2) k≡6a−5c (mod 6c)
∼ X πi πik 2 − exp − + (c − 6b) θ(k, 12c ; z) . 3 6c2 k (mod 12c2) k≡6a−c (mod 6c)
In the sum above we let k = 6c` + 6a ± c where 0 ≤ ` ≤ 2c − 1, so that
k 1 a 1 b πi (c − 6b) ± = πi (c − 6b) ± + ` − (6` ± 1) , 6c2 3 c2 2 c DYSON’S RANK FUNCTION 13 and we find 1 (−iz)−3/2 Θ a, b, c; − 1 z √ 2c−1 −i 3 X −πib ∼ = (−1)` exp (6` + 1) θ(6c` + 6a + c, 12c2; z) 2 c `=0 −πib ∼ +(−1)` exp (6` − 1) θ(6c` + 6a − c, 12c2; z) . c √ 3i = − Θ (a, b, c; z), 2 2 which is transformation (2.19). Transformation (2.21) follows immediately from (2.19). In addition we need to define ∞ 2! a X πa(6n + 1) 1 (2.22) Θ ; z := (−1)n(6n + 1) sin exp 3πiz n + . 1 c c 6 n=−∞ a This coincides with Bringmann and Ono’s function Θ c ; z which is given in [13, Eq.(1.6), p.423]. An easy calculation gives a i Θ ; z = − Θ (0, −a, c; z). 1 c 2c 2 From Proposition 2.2 we have Corollary 2.3. Let a, b, c and z be as in Proposition 2.2. Then a a (2.23) Θ ; z + 1 = ζ Θ ; z , 1 c 24 1 c √ − 3 a 1 3 (2.24) (−iz) 2 Θ ; − = Θ (0, −a, c; z). 1 c z 3c 1 Next we define 2 √−2 exp 3πi a − 1 if 0 < a < 1 , −iz z c 6 c 6 a 1 a 5 ε1 ; z := 0 if 6 < c < 6 , c 2 √−2 3πi a 5 5 a −iz exp z c − 6 if 6 < c < 1,
ζb 2 √ 2c exp 3πi a − 1 if 0 ≤ a < 1 , −iz z c 6 c 6 1 a 5 ε1(a, b, c; z) := 0 if 6 < c < 6 , ζ5b 2 √ 2c 3πi a 5 5 a −iz exp z c − 6 if 6 < c < 1,
2 2 exp −3πiz a − 1 if 0 < a < 1 , c 6 c 6 a 1 a 5 ε2 ; z := 0 if < < , c 6 c 6 a 5 2 5 a 2 exp −3πiz c − 6 if 6 < c < 1, 14 F. G. GARVAN and 2 2ζ−2b exp −3πiz a − 1 if 0 ≤ a < 1 , c c 6 c 6 1 a 5 ε2(a, b, c; z) := 0 if < < , 6 c 6 a 5 2 5 a 2 exp −3πiz c − 6 if 6 < c < 1. We are now ready to express each of our Mordell integrals as a period integral of a theta- function.
1 5 Theorem 2.4. Let a, b, c be as in Theorem 2.1, and in addition that a/c 6∈ { 6 , 6 }. Then for z ∈ h we have √ Z i∞ a 2 3 a 2πi i Θ1 c ; τ a (2.25) J ; = √ p dτ + ε1 ; z , iz c z 3 0 −i(τ + z) c Z i∞ √ √ a i Θ1(0, −a, c; τ) a (2.26) 2 3 −iz J ; −2πiz = − p dτ + ε2 ; z , c 3c 0 −i(τ + z) c √ Z i∞ 3 2πi 1 Θ1(a, b, c; τ) (2.27) J a, b, c; = p dτ + ε1(a, b, c; z), −2iz z 6c 0 −i(τ + z) (2.28) √ √ √ Z i∞ −5b −5b i 3 Θ2(a, b, c; τ) ζ2c 3 −iz J (a, b, c; −2πiz) = −ζ2c p dτ + ε2(a, b, c; z). 6c 0 −i(τ + z) Remark. We have corrected the results of Bringmann and Ono [13, p.441] by including the necessary correction factors ε1 and ε2.
1 5 Proof. First we prove (2.25). Assume 0 < a < c are integers and a/c 6∈ { 6 , 6 }. We proceed as in the proof of [13, Lemma 3.2, pp.436–437]. By analytic continuation we may assume that z = it and t > 0. We find that Z ∞ Z ∞ a 2π −3πtx2 −3πtx2 J ; = t e f(x) dx = t e g11(x) dx, c t 0 −∞ where cosh 3 a − 2 2πz + cosh 3 a − 1 2πz f(z) = c c , cosh(3πz) exp 3 a − 2 2πz + exp 3 a − 1 2πz g (z) = c c . 11 2 cosh(3πz)
1 n We note that the f(z) has poles at z = zn = −i( 6 + 3 ), where n ∈ Z. We find pole residue of f(z) (−1)n sin( πa (−6n+1)) z = −i(− 1 + n) √c 3n−1 6 πi 3 (−1)n+1 sin( πa (6n+1)) z = −i( 1 + n) √ c 3n 6 πi 3 1 z3n+1 = −i( 2 + n) 0 DYSON’S RANK FUNCTION 15
Applying the Mittag-Leffler Theory [45, pp.134–135], we have
n πa X∗ i (−1) sin (6n + 1) 1 1 f(z) = 2 + √c − π 3 z + i(n + 1 ) i(n + 1 ) n∈Z 6 6 n πa ! (−i)(−1) sin c (−6n + 1) 1 1 + √ 1 − 1 π 3 z + i(n − 6 ) i(n − 6 ) for z 6∈ (± 1 + ), and assuming 1 < a < 5 . Here we assume that P ∗ = lim PN . 6 Z 6 c 6 n∈Z N→∞ n=−N 1 We note that the convergence is uniform on any compact subset of C\−i(± 6 +Z). We must consider three cases.
1 a 5 Case 1.1. 6 < c < 6 . We have (−i)X∗ πa 1 1 f(z) = 2 + √ (−1)n sin (6n + 1) + π 3 c z − i(n + 1 ) i(n + 1 ) n∈Z 6 6 1 1 + 1 + 1 . −z − i(n + 6 ) i(n + 6 ) Thus Z ∞ e−3πtx2 f(x) dx 0 Z ∞ = 2e−3πtx2 dx 0 ! Z ∞ ∗ i −3πtx2 X n πa 1 1 − √ e (−1) sin (6n + 1) 1 + 1 dx π 3 −∞ c x − i(n + ) i(n + ) n∈Z 6 6 By absolute convergence on R we have Z ∞ e−3πtx2 f(x) dx 0 Z ∞ = e−3πtx2 dx −∞ ∗ Z ∞ i X n πa −3πtx2 1 1 − √ (−1) sin (6n + 1) e 1 + 1 dx π 3 c −∞ x − i(n + ) i(n + ) n∈Z 6 6 2 ∗ Z ∞ −3πtx i X n πa e = − √ (−1) sin (6n + 1) 1 dx, π 3 c −∞ x − i(n + ) n∈Z 6 since πa X∗ sin (6n + 1) √ (2.29) (−1)n c = π 3 for 1 < a < 5 . i(n + 1 ) 6 c 6 n∈Z 6 16 F. G. GARVAN
We leave (2.29) as an exercise for the reader. It can be proved using [48, Lemma 1.19, p.19]. By using the identity [47, Eq.(3.8), p.274] Z ∞ e−πtx2 Z ∞ e−πus2 (2.30) dx = πis √ du for s ∈ R \{0}, −∞ x − is 0 u + t we find that 1 −πu(n+ )2 a 2π t X πa Z ∞ e 6 J ; = √ (−1)n (6n + 1) sin (6n + 1) √ du. c t 6 3 c 0 u + 3t n∈Z Letting u = −3iτ in the integral we find
1 3πiτ(n+ )2 Z i∞ 6 a 2π −it X n πa e J ; = (−1) (6n + 1) sin (6n + 1) p dτ. c t 6 c 0 −i(τ + it) n∈Z Arguing as in the proof of [47, Lemma 3.3] we may interchange summation and integration to obtain
Z i∞ 1 a 2πi z X n πa 3πiτ(n+ )2 dτ J ; = − (−1) (6n + 1) sin (6n + 1) e 6 p , c z 6 0 c −i(τ + z) n∈Z when z = it, for t > 0. Equation (2.25) follows in this case.
a 1 Case 1.2. 0 < c < 6 . Observe that in this case a −4πx < 3 − 2 2πx < −3πx, (for x > 0), c and the Mittag-Leffler Theory does not directly apply. We simply note that (1 + e2y) = 2ey, cosh y and find that ∞ a Z 2 exp 3 − 2 2πx 2 π a 2 e−3πtx c 1 + e6πx dx = √ exp 6 − 1 . −∞ cosh(3πx) 3t 12t c Thus we have Z ∞ a a Z ∞ −3πtx2 cosh 3 c − 2 2πx + cosh 3 c − 1 2πx −3πtx2 e dx = e g11(x) dx 0 cosh(3πx) −∞ 2 Z ∞ 1 π a −3πtx2 = √ exp 6 − 1 + e g12(x) dx, 3t 12t c −∞ where exp 3 a − 1 2πz − exp 3 a + 1 2πz g (z) = c c . 12 2 cosh(3πz) DYSON’S RANK FUNCTION 17
We observe that the function g11(z) from Case 1.1 and the function g12(z) have the same poles and the same residue at each pole. We note we may apply the Mittag-Leffler Theory to the function g12(z) since in this case a a −2πx < 3 − 1 2πx < −πx, 2πx < 3 + 1 2πx < 3πx, (for x > 0). c c The remainder of the proof is analogous to Case 1.1.
5 a Case 1.3. 6 < c < 1. The proof is analogous to Case 1.2. This time we have ∞ a Z 2 exp 3 − 1 2πx 2 π a 2 e−3πtx c 1 + e−6πx dx = √ exp 6 − 5 . −∞ cosh(3πx) 3t 12t c Thus we have Z ∞ a a Z ∞ −3πtx2 cosh 3 c − 2 2πx + cosh 3 c − 1 2πx −3πtx2 e dx = e g11(x) dx 0 cosh(3πx) −∞ 2 Z ∞ 1 π a −3πtx2 = √ exp 6 − 5 + e g13(x) dx, 3t 12t c −∞ where exp 3 a − 2 2πz − exp 3 a − 4 2πz g (z) = c c . 13 2 cosh(3πz)
We observe that the function g11(z) from Case 1.1 and the function g13(z) have the same poles and the same residue at each pole, and we may apply the Mittag-Leffler Theory to the function g13(z) since in this case a a πx < 3 − 2 2πx < 2πx, −3πx < 3 − 4 2πx < −2πx, (for x > 0). c c Equation (2.26) follows easily from equations (2.24) and (2.25). The proof of (2.27) is analogous to that of (2.25). This time we assume 0 ≤ a < c, 0 < b < c are integers and a/c 6∈ {1/6, 5/6}. Again by analytic continuation we may assume z = it and t > 0. We find that Z ∞ 2π −3πtx2 J a, b, c; = t e g21(x) dx, t −∞ where b −2πz 2b −4πz ζc e + ζc e a g21(z) = b exp 6πz c . cosh 3πz − 3πi c 1 n b We note that the g21(z) has poles at z = zn = −i( 6 + 3 − c ), where n ∈ Z. We find that n+1 2(−1) π πia Res g21(z) = sin 3 (2n + 1) exp 2 (6b − c − 2nc) . z=zn 3π c 18 F. G. GARVAN
Applying the Mittag-Leffler Theory [45, pp.134–135], we have
X∗ 2(−1)n+1 πia g (z) = g (0) + sin π (2n + 1) exp (6b − c − 2nc) 21 21 3π 3 c2 n∈Z ! 1 1 × n b 1 − n b 1 z + i( 3 + c − 6 ) i( 3 + c − 6 )
1 a 5 for z 6= zn, n ∈ Z, and assuming 6 < c < 6 . We note that the convergence is uniform on any compact subset of C \{zn : n ∈ Z}. Again we must consider 3 cases.
1 a 5 Case 2.1. 6 < c < 6 . Proceeding as in Case 1.1, using the analog of (2.29) (2.31) X∗ 2(−1)n+1 πia 1 sin π (2n + 1) exp (6b − c − 2nc) = g (0), 3π 3 c2 i( n + b − 1 ) 21 n∈Z 3 c 6 applying (2.30) and using (2.17) we find that 2πi J a, b, c; z 2iz 6πiab πia = −√ exp − 3 c2 c Z i∞ n 1 b X π a 3πiτ( + − )2 dτ n n 1 b 3 6 c × (−1) 3 + 6 − c sin (2n + 1) exp −2πin e p , 0 3 c −i(τ + z) n∈Z Z i∞ 2iz Θ1(a, b, c; τ) = − √ p dτ, 6 3c 0 −i(τ + z) which gives (2.27) for this case.
a 1 Case 2.2. 0 < c < 6 . We proceed as in Case 1.2. This time we need
Z ∞ a b −3πtx2+6πx −4πx (1 + exp 6πx − 6πi ) 2 π 2 ζ2be c c dx = √ ζb exp 6 a − 1 . c b 2c 12t c −∞ cosh 3πx − 3πi c 3t We have √ Z ∞ 2π 2 b π a 2 −3πtx2 J a, b, c; = √ t ζ2c exp 6 c − 1 + t e g22(x) dx, t 3 12t −∞ where ! b −2πz −b 2πz a ζ e − ζ e 6πz c c c g22(z) = b e . cosh 3πz − 3πi c
We observe that the function g21(z) from Case 2.1 and the function g22(z) have the same poles and the same residue at each pole. The result follows. DYSON’S RANK FUNCTION 19
5 a Case 2.3. 6 < c < 1. We proceed as in Case 2.2. This time we need
Z ∞ a b −3πtx2+6πx −2πx (1 + exp −6πx + 6πi ) 2 π 2 ζbe c c dx = √ ζ5b exp 6 a − 5 . c b 2c 12t c −∞ cosh 3πx − 3πi c 3t We have √ Z ∞ 2π 2 5b π a 2 −3πtx2 J a, b, c; = √ t ζ2c exp 6 c − 5 + t e g23(x) dx, t 3 12t −∞ where ! 4b −8πz 2b −4πz a −ζ e + ζ e 6πz c c c g23(z) = b e . cosh 3πz − 3πi c
We observe that the function g21(z) from Case 2.1 and the function g23(z) have the same poles and the same residue at each pole. The result follows. Finally (2.28) follows easily from (2.19) and (2.27).
3. VECTOR VALUED MAASSFORMSOFWEIGHT 1/2 1 We describe Bringmann and Ono’s vector valued Maass forms of weight 2 making all functions and transformations explicit. Suppose 0 ≤ a < c and 0 < b < c are integers where (c, 6) = 1. We define Z i∞ a a i Θ1 c ; τ (3.1) T1 ; z := −√ p dτ, c 3 −z −i(τ + z) Z i∞ a i Θ1(0, −a, c; τ) (3.2) T2 ; z := p dτ, c 3c −z −i(τ + z) −5b Z i∞ ζ2c Θ1(a, b, c; τ) (3.3) T1(a, b, c, ; z) := p dτ, 3c −z −i(τ + z) √ Z i∞ −5b i 3 Θ2(a, b, c; τ) (3.4) T2(a, b, c; z) := ζ2c p dτ 6c −z −i(τ + z)
1 We can now define a family of vector valued Maass forms of weight 2 . a a a (3.5) G ; z := N ; z − T ; z , 1 c c 1 c a a a a (3.6) G ; z := M ; z + ε ; z − T ; z , 2 c c 2 c 2 c (3.7) G1(a, b, c; z) := N (a, b, c, z) − T1(a, b, c; z),
(3.8) G2(a, b, c; z) := M(a, b, c, z) + ε2(a, b, c; z) − T2(a, b, c; z).