Dyson's Rank, overpartitions, and weak Maass forms
aa r X i v : . [ m a t h . N T ] A ug DYSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS
KATHRIN BRINGMANN AND JEREMY LOVEJOY
Abstract.
In a series of papers the first author and Ono connected the rank, a partition statisticintroduced by Dyson, to weak Maass forms, a new class of functions which are related to modular forms.Naturally it is of wide interest to find other explicit examples of Maass forms. Here we construct anew infinite family of such forms, arising from overpartitions. As applications we obtain combinatorialdecompositions of Ramanujan-type congruences for overpartitions as well as the modularity of rankdifferences in certain arithmetic progressions. Introduction and Statement of Results A partition of a positive integer n is any non-increasing sequence of positive integers whose sum is n .Let p ( n ) denote the number of partitions of n (with the usual convention that p (0) := 1, and p ( n ) := 0for n N ).Ramanujan proved that for every positive integer n , we have: p (5 n + 4) ≡ ,p (7 n + 5) ≡ ,p (11 n + 6) ≡ . (1.1)In a celebrated paper Ono [27] treated these kinds of congruences systematically (also see [29]). Com-bining Shimura’s theory of modular forms of half-integral weight with results of Serre on modularforms modulo ℓ he showed that for any prime ℓ ≥ An + B such that p ( An + B ) ≡ ℓ ) . In order to explain the congruences in (1.1) with moduli 5 and 7 combinatorially, Dyson [16] intro-duced the rank of a partition. The rank of a partition is defined to be its largest part minus the numberof its parts. Dyson conjectured that the partitions of 5 n + 4 (resp. 7 n + 5) form 5 (resp. 7) groups ofequal size when sorted by their ranks modulo 5 (resp. 7). This conjecture was proven in 1954 by Atkinand Swinnerton-Dyer [2]. In [9] and [6], Ono and the first author showed that Dyson’s rank partitionfunction also satisfies congruences of Ramanujan type. One of the main steps in their proof is to showthat generating functions related to the rank are the “holomorphic parts” of “weak Maass forms”, anotion we will explain later. This new theory has many applications, such as congruences [9, 6] andasymptotics [5] for ranks as well as modularity for rank differences [10].Naturally it is of wide interest to find other explicit examples of weak Maass forms. After partitions,the next place to look is overpartitions. Recall that an overpartition is a partition where the firstoccurrence of a summand may be overlined (see [13]). For example, there are 14 overpartitions of 4:4 , , , , , , , , , , , , , . Overpartitions have arisen in many areas where ordinary partitions play an important role, most notablyin q -series and combinatorics (e.g. [3, 11, 12, 13, 21, 30, 34]), but also in mathematical physics (e.g. Date : October 27, 2018.2000
Mathematics Subject Classification. [17, 18]), symmetric functions (e.g. [4, 15], representation theory (e.g. [19]) and algebraic numbertheory (e.g. [20, 24]). To give a few specific examples, the combinatorial theory of overpartitions leadsto natural and straightforward bijective proofs of q -series identities like Ramanujan’s ψ summation [12,34]; in the theory of symmetric functions in superspace, overpartitions play the role that partitions playin the classical theory of symmetric functions [17, 18]; and certain Dedekind zeta functions associatedto rings of integers of real quadratic fields can be regarded as generating functions for weighted countsof overpartitions [20, 24].Returning to Dyson’s rank, this statistic applies just as well to overpartitions. Indeed, this rankand its generalizations have already proven fundamental in the combinatorial theory of overpartitions[14, 22, 23]. The main result of the present paper will be the construction of an infinite family of weakMaass forms whose holomorphic parts are related to the generating function for Dyson’s rank of anoverpartition. As applications, we discuss congruence properties of overpartitions and the modularityof rank differences in arithmetic progressions.For a positive integer n we denote by p ( n ) the number of overpartions of n . We have the generatingfunction [13] P ( q ) := X n ≥ p ( n ) q n = η (2 z ) η ( z ) = 1 + 2 q + 4 q + 8 q + 14 q + · · · . (1.2)Here η ( z ) := q Q ∞ n =1 (1 − q n ) is Dedekind’s eta function and we write q := e πiz . Moreover we denoteby N ( m, n ) the number of overpartitions of n with rank m . It is shown in [22] that O ( u ; q ) := 1 + ∞ X n =1 N ( m, n ) u m q n = ∞ X n =0 ( − n q n ( n +1) ( uq, q/u ) n = ( − q ) ∞ ( q ) ∞ X n ≥ (1 − u ) (cid:0) − u − (cid:1) ( − n q n + n (1 − uq n ) (1 − u − q n ) . (1.3)Here for a, b ∈ C , n ∈ N ∪ {∞} , we employ the standard q -series notation:( a ) n : = n − Y r =0 (1 − aq r ) , ( a, b ) n : = n − Y r =0 (1 − aq r ) (1 − bq r ) , ( a ) ∞ : = lim n →∞ ( a ) n . It turns out that the function O ( u ; q ) for u a root of unity = 1 is the holomorphic part of a weak Maassform.To make this precise, we recall the notion of a weak Maass form of half-integral weight k ∈ Z \ Z .If z = x + iy with x, y ∈ R , then the weight k hyperbolic Laplacian is given by(1.4) ∆ k := − y (cid:18) ∂ ∂x + ∂ ∂y (cid:19) + iky (cid:18) ∂∂x + i ∂∂y (cid:19) . If v is odd, then define ǫ v by(1.5) ǫ v := ( v ≡ ,i if v ≡ . A (harmonic) weak Maass form of weight k and Nebentypus χ on a subgroup Γ ⊂ Γ (4) is any smoothfunction f : H → C satisfying the following: YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 3 (1) For all A = (cid:0) a bc d (cid:1) ∈ Γ and all z ∈ H , we have f ( Az ) = (cid:18) cd (cid:19) k ǫ − kd χ ( d )( cz + d ) k f ( z ) . (2) We have that ∆ k f = 0.(3) The function f ( z ) has at most linear exponential growth at all the cusps of Γ.Suppose that 0 < a < c are integers, and let ζ c := e πic . Define the theta function of weight (1.6) θ ( α, β ; τ ) := X n ≡ α (mod β ) ne πiτn β , and let Θ a,c ( τ ) := θ (cid:0) a + c, c ; τ c (cid:1) if c is odd , θ (cid:0) a + c , c ; τ c (cid:1) if 2 k c, θ (cid:0) a + c , c ; τc (cid:1) if 4 | c. Using these cuspidal theta functions, we define for c = 2, the non-holomorphic integral(1.7) J (cid:16) ac ; z (cid:17) := πi · tan (cid:0) πac (cid:1) c Z i ∞− ¯ z ( − iτ ) − · Θ a,b (cid:0) − τ (cid:1)p − i ( τ + z ) dτ. Moreover define M (cid:0) ac ; z (cid:1) by(1.8) M (cid:16) ac ; z (cid:17) := O (cid:16) ac ; q (cid:17) − J (cid:16) ac ; z (cid:17) , where O (cid:0) ac ; q (cid:1) := O ( ζ ac ; q ). If u = −
1, we define M ( − z ) := O ( − z ) − I ( − z ) . with I ( − z ) := √ πi Z i ∞− ¯ z η ( τ ) η (2 τ ) · ( − i ( τ + z )) dτ. The main result of this paper is the following theorem which establishes that those real analytic functionsare Maass forms.
Theorem 1.1.
The following statements are true: (1) If < a < c with ( a, c ) = 1 and c = 2 , then M (cid:0) ac ; z (cid:1) is a weak Maass form of weight on Γ (16 c ) . If | c and | c , then it is a weak Maass form on Γ (4 c ) and Γ ( c ) , respectively. (2) The function M ( − z ) is a weak Maass form of weight on Γ (16) .Five remarks.
1) If c is odd, we actually obtain Maass forms for the larger group n (cid:16) α βγ δ (cid:17) ∈ SL ( Z ) (cid:12)(cid:12)(cid:12) α ≡ δ ≡ c ) , γ ≡ c ) o .
2) The proof of the second part of Theorem 1.1 is harder than the first since the generating functionhas double poles. To overcome this problem, we introduce new functions O r ( q ) having an additionalparameter r but only simple poles such that one can obtain O ( − z ) by a process of differentiation.This differentiation accounts for the augmentation of the weight by 1 in this case. It is worth mentioningthat for the case of the classical Dyson’s rank generating functions in [9], the weak Maass forms haveweight 1 / = 1.3) The authors [7] show that in the context of overpartition pairs, the analogous generating functionsassociated to the appropriate generalization of Dyson’s rank are not weak Maass forms, but classicalmodular forms. KATHRIN BRINGMANN AND JEREMY LOVEJOY
4) We should stress that the analysis of the transformation behavior of O ( u ; q ) is much more involvedthan in the case of the Dyson’s rank generating functions in [9]. One of the reasons is that the half-integer weight modular form ( − q ) ∞ ( q ) ∞ that shows up in (1.3) is not mapped to itself as in the case of theusual ranks. This prohibits “guessing” images under M¨obius transformations as in [9]. There the firstauthor and Ono started with part of images of the generating function that they were able to guess.Thus the idea of proof in [9] which builds on old results of Watson, cannot be employed here. Insteadwe have to determine explicitly the images under all M¨obius transformations with different techniques.In view of Theorem 1.1 one can obtain results on overpartitions by arguing as in work of Ono andthe first author [5, 6, 8, 9, 10]. In this direction we exhibit congruences for N ( r, t ; n ), the numberof overpartitions of n whose rank is congruent to r (mod t ), and provide a theoretical framework forproving identities for rank differences in arithmetic progressions. Other possible applications, which wedo not address here, would be to asymptotics or inequalities for ranks, exact formulas or distributionquestions. We first consider congruences satisfied by N ( r, t ; n ). For ease of notation we restrict to thecase that t is odd, the case t even can be considered similarly. Theorem 1.2.
Let t be a positive odd integer, and let ℓ ∤ t be a prime. If j is a positive integer, thenthere are infinitely many non-nested arithmetic progressions An + B such that for every ≤ r < t wehave N ( r, t ; An + B ) ≡ ℓ j ) . Theorem 1.3.
Suppose that ℓ ≥ is a prime, m, u, β ∈ N with (cid:0) − βℓ (cid:1) = − . Then a positive proportionof primes p ≡ − ℓ ) have the property that for every ≤ r ≤ ℓ m − N (cid:0) r, ℓ m ; p n (cid:1) ≡ ℓ u ) for all n ≡ β (mod ℓ ) that are not divisible by p . This directly implies.
Corollary 1.4. If ℓ ≥ is a prime, m, u ∈ N , then there are infinitely many non-nested arithmeticprogressions An + B such that N ( r, ℓ m ; An + B ) ≡ ℓ u ) for all ≤ r ≤ ℓ m − .Remark. The congruences in Theorems 1.2 and 1.3 may be viewed as a combinatorial decomposition of theoverpartition function congruence p ( An + B ) ≡ ℓ u ) . (1.9)That (1.9) holds for infinitely many non-nested arithmetic progressions An + B was first observed byTreneer [32].We next put identities involving rank differences for overpartitions in the framework of weak Maassforms (see also [10]). For this define for a prime ℓ and integers s and s the function R s ,s ( d ) := ∞ X n =0 (cid:0) N ( s , ℓ, ℓn + d ) − N ( s , ℓ, ℓn + d ) (cid:1) q ℓn + d . We provide a framework that could be used to show an infinite family of identities (see also [10] forrelated results for usual ranks).
Theorem 1.5. If (cid:0) dℓ (cid:1) = − (cid:0) − ℓ (cid:1) , then the function R s ,s ( d ) is a weakly holomorphic modular form on Γ (cid:0) ℓ (cid:1) . YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 5
Using Theorem 1.5, we could prove concrete identities using the valence formula. Since the compu-tations are straightforward but lengthy (coming from the fact that Γ (cid:0) ℓ (cid:1) has a lot of cusps), wechose not to prove individual identities. Instead we just list some identities, and their truth followsfrom work of the second author and Osburn [25].The paper is organized as follows. In Section 2, we prove a transformation law for the rank generatingfunctions in the case c = 2. In Section 3, we show the first part of Theorem 1.1. The main step is torecognize the Mordell type integrals occurring the transformation law of the rank generating functionsas integrals of theta functions. In Section 4 we treat the case c = 2 which is more complicated due todouble poles of the generating function. In Sections 5 and 6 we show congruences for N ( r, t, n ). Section7 is dedicated the proof of Theorem 1.5. Acknowledgements
The authors thank the referee for many helpful suggestions which improved the exposition of thepaper. 2.
A transformation law
Here we consider modularity properties for O (cid:0) ac ; q (cid:1) . For this we need some notation. Let c > k be positive integers. Let e k be either 0 or 1 depending on whether k is even or odd. Moreoverlet k := k ( k,c ) , c = c ( c,k ) , and define the integer 0 ≤ l < c by the congruence l ≡ ak (mod c ). If bc ∈ (0 , s ( b, c ) and t ( b, c ) (for bc = ) by s ( b, c ) := < bc ≤ , < bc ≤ , < bc < , t ( b, c ) := ( < bc < , < bc < . In particular let s := s ( l, c ) and t := t ( l, c ). Let h ′ be defined by hh ′ ≡ − k ). Moreover let ω h,k be given by ω h,k := exp πi X µ (mod k ) (cid:16)(cid:16) µk (cid:17)(cid:17) (cid:18)(cid:18) hµk (cid:19)(cid:19) , (2.1)where (( x )) := (cid:26) x − ⌊ x ⌋ − if x ∈ R \ Z , x ∈ Z . KATHRIN BRINGMANN AND JEREMY LOVEJOY
Define for q = e πiz the following functions. U (cid:16) ac ; q (cid:17) := U (cid:16) ac ; z (cid:17) := sin (cid:16) πac (cid:17) η (cid:0) z (cid:1) η ( z ) X n ∈ Z (1 + q n ) q n + n − q n cos (cid:0) πac (cid:1) + q n , U ( a, b, c ; q ) := U ( a, b, c ; z ) := η (cid:0) z (cid:1) η ( z ) e πiac ( bc − − s ( b,c ) ) q s ( b,c ) c + b c − b c X m ∈ Z q m (2 m +1)+ ms ( b,c ) − e − πiac q m + bc , V ( a, b, c ; q ) := V ( a, b, c ; z ) := η (cid:0) z (cid:1) η ( z ) e πiac ( bc − − s ( b,c ) ) · q s ( b,c ) bc + b c − b c X m ∈ Z q m (2 m +1)+ ms ( b,c ) − e − πiac · q m + bc , O ( a, b, c ; q ) := O ( a, b, c ; z ) := η (2 z ) η ( z ) e πiac ( bc − − t ( b,c ) ) · q t ( b,c ) b c + b c − b c X m ∈ Z ( − m q m (2 m +1)+ mt ( b,c )2 − e − πiac · q m + bc , V (cid:16) ac ; q (cid:17) := V (cid:16) ac ; z (cid:17) := η (2 z ) η ( z ) q X m ∈ Z q m + m · (cid:16) e − πiac · q m + (cid:17) − e − πiac · q m + . Moreover let H a,c ( x ) := e x − (cid:0) πac (cid:1) e x + e x . Then H a,c ( − x ) = H a,c ( x ) ,H a,c ( x + 2 πi ) = H a,c ( x ) . Moreover define for an integer ν the Mordell type integral I a,c,k,ν ( w ) := Z R e − πwx k H a,c πiνk − πwxk − e kπi k ! dx. For k even we have to take the principal part of the integral. We are now ready to show the transfor-mation law of O (cid:0) ac ; q (cid:1) . Theorem 2.1.
Assume the notation above. Moreover, let w ∈ C with Re( w ) > , q := e πik ( h + iw ) , and q := e πik ( h ′ + iw ) . (1) If c | k and k is even, then we have O (cid:16) ac ; q (cid:17) = ( − k i · e − πia h ′ k c · tan (cid:16) πac (cid:17) · cot (cid:18) πah ′ c (cid:19) ω h,k ω h,k/ w − · O (cid:18) ah ′ c ; q (cid:19) + 4 · sin (cid:0) πac (cid:1) · ω h,k ω h,k/ · k · w X ν (mod k ) ( − ν e − πih ′ ν k · I a,c,k,ν ( w ) . (2) If c | k and k is odd, then we have O (cid:16) ac ; q (cid:17) = √ i · e πih ′ k − πia h ′ k c tan (cid:16) πac (cid:17) ω h,k ω h,k · w − · U (cid:18) ah ′ c ; q (cid:19) + 4 √ · sin (cid:0) πac (cid:1) · ω h,k ω h,k · k · w X ν (mod k ) e − πih ′ k (2 ν − ν ) · I a,c,k,ν ( w ) . YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 7 (3) If c ∤ k , | k , and c = 2 , then we have O (cid:16) ac ; q (cid:17) = − e − πia h ′ k c c tan (cid:16) πac (cid:17) ω h,k ω h,k/ w − · ( − c ( l + k ) O (cid:18) ah ′ , lcc , c ; q (cid:19) + 4 sin (cid:0) πac (cid:1) · ω h,k ω h,k/ · k · w X ν (mod k ) ( − ν e − πih ′ ν k · I a,c,k,ν ( w ) . (4) If c ∤ k , | k , and c = 2 , then we have O (cid:16) ac ; q (cid:17) = − e − πia h ′ k c · tan (cid:16) πac (cid:17) ω h,k ω h,k/ · w − · V (cid:18) ah ′ c ; q (cid:19) + 4 sin (cid:0) πac (cid:1) · ω h,k ω h,k/ · k · w X ν (mod k ) ( − ν e − πih ′ ν k · I a,c,k,ν ( w ) . (5) If c ∤ k , ∤ k , and c = 4 , then we have O (cid:16) ac ; q (cid:17) = −√ e πih ′ k − πih ′ a k cc · tan (cid:16) πac (cid:17) ω h,k ω h,k w − · U (cid:18) ah ′ , lcc , c ; q (cid:19) + 4 √ · sin (cid:0) πac (cid:1) · ω h,k ω h,k · k · w X ν (mod k ) e − πih ′ k (2 ν − ν ) · I a,c,k,ν ( w ) . (6) If c ∤ k , ∤ k , and c = 4 , then we have O (cid:16) ac ; q (cid:17) = − e πih ′ k − πih ′ a k cc · tan (cid:16) πac (cid:17) ω h,k √ · ω h,k · w − · V (cid:18) ah ′ , lcc , c ; q (cid:19) + 4 √ · sin (cid:0) πac (cid:1) · ω h,k ω h,k · k · w X ν (mod k ) e − πih ′ k (2 ν − ν ) · I a,c,k,ν ( w ) . Corollary 2.2.
Assume that z ∈ H , < a < c with c = 2 . (1) If c = 4 , then we have O (cid:18) ac ; − z (cid:19) = −√ (cid:16) πac (cid:17) · ( − iz ) · U (0 , a, c ; z ) + 4 √ · sin (cid:16) πac (cid:17) · ( − iz ) − · I a,c, , (cid:18) iz (cid:19) . (2) If c = 4 , then we have O (cid:18) ac ; − z (cid:19) = − tan (cid:0) πac (cid:1) √ − iz ) · V (0 , a, c ; z ) + 4 √ · sin (cid:16) πac (cid:17) · ( − iz ) − · I a,c, , (cid:18) iz (cid:19) . Corollary 2.3.
Assume that (cid:16) α βγ δ (cid:17) ∈ Γ ( c ) with c odd, z ∈ H , and let γ := γ ( c,γ ) . (1) If | γ , then the holomorphic part of O (cid:16) ac ; αz + βγz + δ (cid:17) is given by − i · e πia δγ c · tan (cid:16) πac (cid:17) · cot (cid:18) πaδc (cid:19) ω α,γ ω α,γ/ ( − i ( γz + δ )) · O (cid:18) aδc ; z (cid:19) . (2) If γ is odd, then the holomorphic part of O (cid:16) ac ; αz + βγz + δ (cid:17) is given by √ i · e − πiδ γ + πia δγ c tan (cid:16) πac (cid:17) ω α,γ ω α,γ · ( − i ( γz + δ )) · U (cid:18) − aδc ; z (cid:19) . KATHRIN BRINGMANN AND JEREMY LOVEJOY
Proof of Theorem 2.1.
We proceed similarly as in [1, 5]. First we rewrite (1.3) as(2.2) O (cid:16) ac ; q (cid:17) = 4 sin (cid:16) πac (cid:17) η (cid:0) k ( h + iw ) (cid:1) η (cid:0) k ( h + iw ) (cid:1) X n ∈ Z ( − n e πin h + iw ) k · H a,c (cid:18) πink ( h + iw ) (cid:19) . Now let e O (cid:16) ac ; q (cid:17) := η (cid:0) k ( h + iw ) (cid:1) (cid:0) πac (cid:1) · η (cid:0) k ( h + iw ) (cid:1) · O (cid:16) ac ; q (cid:17) . (2.3)Writing n = km + ν with 0 ≤ ν < k , m ∈ Z gives that e O (cid:0) ac ; q (cid:1) equals k − X ν =0 ( − ν e πihν k X m ∈ Z ( − km H a,c (cid:18) πihνk − πwk ( km + ν ) (cid:19) · e − πwk ( km + ν ) . (2.4)Using Poisson summation and substituting x kx + ν gives that the inner sum equals1 k X n ∈ Z Z R H a,c (cid:18) πihνk − πxwk (cid:19) · e πik (2 n + e k )( x − ν ) − πwx k dx. (2.5)Strictly speaking for c | k there may lie a pole at x = 0. In this case we take the principal part of theintegral. Inserting (2.5) into (2.4) we see that the summation only depends on ν (mod k ). Moreover,by changing ν into − ν , x into − x , and n into − ( n + e k ), we see that the part of the sum over n with n ≤ − n ≥
0. Thus (2.4) equals(2.6) 2 k X ν (mod k ) ( − ν e πihν k X n ∈ N Z R H a,c (cid:18) πihνk − πxwk (cid:19) · e πik (2 n + e k )( x − ν ) − πwx k dx. Next we introduce the function S a,c,k ( x ) := sinh( c x )sinh (cid:0) xk + πiac (cid:1) · sinh (cid:0) xk − πiac (cid:1) which is entire as a function of x . Here we need that c = 2. We rewrite the integrand in (2.6) as( − hc ν e πi (2 n + e k )( x − ν ) k − πwx k · S a,c,k ( πxw − πihν )4 sinh( πc xw ) . From this we see that the only poles can lie in the points x m := imc w ( m ∈ Z ) . If c | k , then c = 1; thus poles can only lie in points of the form x m = imw . One can easily compute thateach choice ± leads at most for one ν (mod k ) to a non-zero residue, and that this ν can be chosen as ν ± m := − h ′ ( m ∓ ak ) . If c ∤ k , then we can only have a nontrivial residue if m ≡ ± ak (mod c ). We write c m ± l insteadof m with m ≥ (1 ∓ − there corresponds exactly one ν (mod k )and we can choose ν as ν ± m := − h ′ (cid:18) m ± c ( l − ak ) (cid:19) . Now shift the path of integration through the points ω n := (cid:16) n + e k (cid:17) i w . YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 9
Which points x m ( m ≥
0) we have to take into account when we use the Residue Theorem depends onwhether c | k or not and on whether k is even or odd. The cases that c = 2 , • If c | k and k is even, then we have to take those x m into account for which 2 m ≤ n . The poleson the path of integration are x and ω n . • If c | k and k is odd, then we have to take those x m into account for which 2 m ≤ n . The point x is the only pole on the path of integration . • If c ∤ k , k is even, and c = 2, then there is no pole on the path of integration. Moreover in thiscase we have to take those x m into account for which n ≥ m + (1 ± t ). • If c ∤ k , k is even, and c = 2, then the only pole on the path of integartion lies in ω n . We haveto take those x m into account for which n ≥ m ± • If c ∤ k , k is odd, and c = 4, then there is no pole on the path of integration. Moreover wehave to take those x m into account for which n ≥ m ± s . • If c ∤ k , k is odd, and c = 4, then there lies a pole in ω n and we have to take those x m intoaccount for which n ≥ m ± s + ( − ± λ ± n,m . It is not hard to compute λ ± n,m = ± ik πw sin (cid:0) πac (cid:1) e πik (2 n + e k ) ( x m − ν ± m ) − πwx mk . ¿From this one directly sees that λ ± n +1 ,m = exp (cid:18) πik (cid:0) x m − ν ± m (cid:1)(cid:19) · λ ± n,m . Shifting the path of integration through ω n , we obtain by the Residue Theorem e O (cid:16) ac ; q (cid:17) = X + X , where X := 2 k X ν (mod k ) ( − ν e πihν k X n ∈ N Z ∞ + ω n −∞ + ω n H a,c (cid:18) πihνk − πxwk (cid:19) · e πi (2 n + e k )( x − ν ) k − πx wk dx. For the definition of P we have to distinguish several cases. We set r := and r m := 1 for m ∈ N .If c | k and k is even, then X := 4 πik X m ≥ ǫ ∈{±} r m ( − ν ǫm e πih ( νǫm )2 k λ ǫ m +1 ,m − exp (cid:0) πik ( x m − ν ǫm ) (cid:1) + 12 λ ǫ m,m ! . An easy calculation shows that this equals( − k ie − πih ′ a k c sin (cid:16) πh ′ ac (cid:17) w · sin (cid:0) πac (cid:1) X n ∈ Z ( − n q n + n − q n cos (cid:0) πah ′ c (cid:1) + q n . (2.7)If c | k and 2 ∤ k , then X := 4 πik X m ≥ ǫ ∈{±} r m ( − ν ǫm e πih ( νǫm ) k λ ǫ m,m − exp (cid:0) πik ( x m − ν ǫm ) (cid:1) . (2.8)We assume without loss of generality that h ′ is even. Then we can show that (2.8) equals i sin (cid:16) πh ′ ac (cid:17) w · sin (cid:0) πac (cid:1) e − πih ′ a k c X n ∈ Z (1 + q n ) q n + n − q n cos (cid:0) πah ′ c (cid:1) + q n . (2.9) If c ∤ k , 2 | k , and c = 2, then X := 4 πik X m ≥ ǫ ∈{±} ( − ν ǫm e πih ( νǫm )2 k λ ǫ m + (1 ± t ) ,m − exp (cid:0) πik ( x m − ν ǫm ) (cid:1) . It can be calculated that this equals − w sin (cid:0) πac (cid:1) q tl c − l c + l c · e πih ′ ac “ lc − ak c − t − ” ( − c ( l + k ) X m ∈ Z ( − m q m (2 m +1)+ mt − e − πiah ′ c · q m + lc . If c ∤ k , 2 | k , and c = 2, then X := 4 πik X m ≥ ( − ν + m e πih ( ν + m )2 k λ +2 m +2 ,m − exp (cid:0) πik (cid:0) x m − ν + m (cid:1)(cid:1) + 12 λ +2 m +1 ,m ! + X m ≥ ( − ν − m e πih ( ν − m )2 k λ − m,m − exp (cid:0) πik (cid:0) x m − ν − m (cid:1)(cid:1) + 12 λ − m − ,m ! . One can show that this equals. − q w sin (cid:0) πac (cid:1) e − πia h ′ k c X m ∈ Z q m + m (cid:18) e − πih ′ ac q m + (cid:19) − e − πiah ′ c · q m + . (2.10)If c ∤ k , k is odd, and c = 4, then X := 4 πik X m ≥ ( − ν + m e πih ( ν + m )2 k λ +2 m + s,m − exp (cid:0) πik (cid:0) x m − ν + m (cid:1)(cid:1) + X m ≥ ( − ν − m e πih ( ν − m )2 k λ − m − s,m − exp (cid:0) πik (cid:0) x m − ν − m (cid:1)(cid:1) . Without loss of generality we may assume that h ′ is even. With this assumption we can compute that P equals − w sin (cid:0) πac (cid:1) q slc − l c + l c · e πih ′ ac “ lc − ak c − s − ” X m ∈ Z q m (2 m +1)+ ms − e − πiah ′ c · q m + lc . If c ∤ k , k is odd, and c = 4, then X := 4 πik ∞ X m =0 ( − ν + m e πih ( ν + m )2 k λ +2 m + s +1 ,m − exp (cid:0) πik (cid:0) x m − ν + m (cid:1)(cid:1) + 12 λ m + s,m ! + ∞ X m =1 ( − ν − m e πih ( ν − m )2 k λ − m − s,m − exp (cid:0) πik (cid:0) x m − ν − m (cid:1)(cid:1) + 12 λ m − s − ,m !! . One can show that this equals(2.11) − w sin (cid:0) πac (cid:1) q slc − l c + l c e πih ′ ac “ lc − ak c − s ” X m ∈ Z q m (2 m +1)+ ms (cid:18) e − πiah ′ c q m + lc (cid:19) − e − πiah ′ c · q m + lc . YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 11
We next turn to the computation of P . If there is a pole in ω n we take the principal part of the integral.With the same argument as before we can change the sum over N into a sum over Z . Moreover wemake the translation x x + ω n and write n = 2 p + δ with p ∈ Z and δ ∈ { , − } . This gives X = 1 k X ν (mod k ) ( − ν e πihν k X p ∈ Z δ ∈{ , − } e − πik (4 p +2 δ + e k ) ν − π kw (4 p +2 δ + e k ) Z ∞−∞ H a,c (cid:18) πihνk − πwxk − πi k (cid:16) p + 2 δ + e k (cid:17)(cid:19) · e − πwx k dx. Next we change ν into − h ′ ( ν + p ) and distinguish whether k is even or odd.If k is even, then we have, since h ′ is odd, X = 1 k X ν (mod k ) δ ∈{ , − } ( − ν e πih ′ k ( − δ +4 ν ( δ − ν ) ) X p ∈ Z ( − p q (4 p +2 δ )216 Z ∞−∞ H a,c (cid:18) πiνk − πwxk − πiδk (cid:19) · e − πwx k dx. Now the integral is independent of p . Moreover the sum over p vanishes for δ = − p th andthe ( − p + 1)th term cancel. For δ = 0 the sum over p equals X p ∈ Z ( − p · q p = η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) . (2.12)Thus X = η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) k · η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) X ν (mod k ) ( − ν e − πih ′ ν k · I a,c,k,ν ( w ) . (2.13)If k is odd, then we may without loss of generality assume that h ′ is even. In this case we obtain X = 1 k e − πih ′ k X ν (mod k ) δ ∈{ , − } e πih ′ k ( − ν + ν (2 δ +1) ) X p ∈ Z q (4 p +2 δ +1)216 Z ∞−∞ H a,c (cid:18) πiνk − πwxk − πi (2 δ + 1)2 k (cid:19) · e − πwx k dx. Making the substitutions p
7→ − p , x
7→ − x , and ν
7→ − ν , one can easily see that the contribution for δ = 0 and for δ = − p equals X p ∈ Z q (4 p +1)216 = η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) . (2.14)Thus X = 2 k · e − πih ′ k η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) η (cid:0) k (cid:0) h ′ + iw (cid:1)(cid:1) X ν (mod k ) e πih ′ k ( − ν + ν ) · I a,c,k,ν ( w ) . (2.15)To finish the proof of Theorem 2.1, we require the well-known transformation law of Dedekind’s η -function. η (cid:18) k ( h + iw ) (cid:19) = e πi k ( h − h ′ ) · ω − h,k · w − · η (cid:18) k (cid:18) h ′ + iw (cid:19)(cid:19) . (2.16) This implies that for k even, we have η (cid:18) k ( h + iw ) (cid:19) = e πi k ( h − e h ) · ω − h,k/ · w − · η (cid:18) k (cid:18)e h + iw (cid:19)(cid:19) , (2.17)where h e h ≡ − k/ k is odd, (2.16) implies that η (cid:18) k ( h + iw ) (cid:19) = e πi k (2 h − (2 h ) ′ ) · ω − h,k · (2 w ) − · η (cid:18) k (cid:18) (2 h ) ′ + i w (cid:19)(cid:19) . (2.18)Combining (2.7), (2.9), (2.10), (2.11), (2.13), (2.15), (2.16), (2.17), and (2.18) gives (after a lengthy butstraightforward calculation) the theorem. (cid:3) Construction of the weak Maass forms
In this section we prove the first part of Theorem 1.1. First we interpret the Mordell type integraloccurring in Corollary 2.2 as an integral of theta functions. For this let I z := 4 √ (cid:0) πac (cid:1) ( − iz ) Z R e − πix z · H a,c (cid:18) πixz + πi (cid:19) dx. (3.1) Lemma 3.1.
We have I z = π tan (cid:0) πac (cid:1) c Z ∞ Θ a,c ( iu ) p − i ( iu + z ) du. Proof.
We modify a proof of [9, 35]. By analytic continuation it is enough to show the claim for z = it with t >
0. Making the change of variables x xt , we find that I it = 4 √ (cid:16) πac (cid:17) Z R e − πtx · H a,c (cid:18) πx + πi (cid:19) dx. (3.2)We next rewrite H a,c (cid:0) πx + πi (cid:1) using the Mittag-Leffler theory of partial fraction decomposition. Thiseasily gives that(3.3) H a,c (cid:18) πx + πi (cid:19) = i π sin (cid:0) πac (cid:1) X m ∈ Z x − i (cid:0) m − ac − (cid:1) − x − i (cid:0) m + ac − (cid:1) ! . We plug (3.3) back into (3.2) and interchange summation and integration. For this we introduce theextra summands i ( m − ac − ) and i ( m + ac − ) which enforce absolute convergence and cancel when weintegrate. This gives I it = i tan (cid:0) πac (cid:1) √ X m ∈ Z Z R e − πtx x − i (cid:0) m − ac − (cid:1) − x − i (cid:0) m + ac − (cid:1) ! dx. Next use that for all s ∈ R \ { } , we have the identity Z ∞−∞ e − πtx x − is dx = πis Z ∞ e − πus √ u + t du (this follows since both sides are solutions of (cid:0) − ∂∂t + πs (cid:1) f ( t ) = πis √ t f ( t ) and have the same limit 0 as t
7→ ∞ and hence are equal). Again interchanging summation and integration and making the change
YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 13 of variables u u gives I it = − π tan (cid:16) πac (cid:17) Z ∞ √ u + t X m ∈ Z (cid:18)(cid:18) m − ac − (cid:19) · e − πu ( m − ac − ) − (cid:18) m + ac − (cid:19) · e − πu ( m + ac − ) (cid:19) du. From this the claim can be easily deduced. (cid:3)
Lemma 3.2.
For z ∈ H , we have J (cid:16) ac ; z + 1 (cid:17) = J (cid:16) ac ; z (cid:17) , √− iz · J (cid:18) ac ; − z (cid:19) = I z + πi tan (cid:0) πac (cid:1) c Z i ∞− ¯ z Θ a,c ( τ ) p − i ( τ + z ) dτ. Proof.
We only show the lemma in the case that c is odd, the case c even is shown similarly. The firstclaim follows from the fact that Θ a,c (cid:0) − τ (cid:1) is invariant under τ τ + 1. Indeed Shimura’s work [31]implies that(3.4) ( − i cτ ) − · Θ a,c (cid:18) − τ (cid:19) = − i (2 c ) − X k (mod 2 c ) exp (cid:18) πik (4 a + c )2 c (cid:19) · Θ( k, c ; 4 cτ ) . To prove the second transformation law we directly compute1 √− iz · J a,c (cid:18) − z (cid:19) = − πi tan (cid:0) πac (cid:1) c · √− iz Z i ∞ z ( − iτ ) − · Θ a,c (cid:0) − τ (cid:1)q − i (cid:0) τ − z (cid:1) dτ. Making the change of variable τ
7→ − τ now easily gives the claim. (cid:3) Proof of Theorem 1.1 (1).
Again we assume that c is odd. By [31] Proposition 2.1, the functionsΘ( k, c ; τ ) are cusp forms for Γ(4 c ). Thus the functions Θ( k, c ; 4 cτ ) are cusp forms for Γ (16 c ).Using Theorem 2.1 one can conclude that also M (cid:0) ac ; z (cid:1) transforms correctly under Γ (16 c ). That itis an eigenfunction under ∆ follows as in [9] page 21. (cid:3) The case u = − u = −
1. We assume the same notation as in Section 2. Equation (1.3)gives O ( − q ) = 4 ( − q ) ∞ ( q ) ∞ X n ∈ Z ( − n q n + n (1 + q n ) . This function is more complicated than O (cid:0) ac ; q (cid:1) with c = 2 since double poles occur. To overcome thisproblem, we first prove a transformation law for the function O r ( q ) := 4 ( − q ) ∞ ( q ) ∞ X n ∈ Z ( − n +1 q n (1 + e πir q n ) . This function is related to O ( − q ) by12 πi (cid:20) ∂∂r O r ( q ) (cid:21) r =0 = O ( − q ) . To state the transformation law for O r ( q ) we additionally need the functions U r ( q ) := e πir η ( z ) η (2 z ) X m ∈ Z m odd q ( m + m ) − e πir q m ,I ± k,ν,r ( w ) := Z R e − πwx k e ± πiνk ∓ πi e k k +2 πir − πwxk dx. Theorem 4.1.
Assume that r ∈ R with | r | sufficiently small. (1) If k is odd, then O r ( q ) = − √ iw − ω h,k ω h,k e πih ′ k + πkr w U irw (cid:18) q (cid:19) − √ w ω h,k ω h,k k X ν (mod k ) ± e πih ′ k ( − ν + ν ) I ± k,ν,r ( w ) . (2) If k is even, then O r ( q ) = − i ω h,k ω h,k/ w − e πkr w O riw ( q ) − ω h,k w kω h,k/ X ν (mod k ) ( − ν e − πih ′ ν k I + k,ν,r ( w ) . Proof.
We proceed similarly as in the proof of Theorem 2.1 and therefore we skip most of the details.Moreover, we only show the claim for k odd, the case k even is treated similarly. Define e O r ( q ) := X n ∈ Z ( − n q n (1 + e πir q n ) . Changing n into ν + km with ν running modulo k and m ∈ Z , we obtain, using Poisson summationand making a change of variables, e O r ( q ) = 1 k X ν (mod k ) ( − ν e πihν k X n ∈ N ± Z R e − πwx k + πik (2 n +1)( x − ν ) e πir ± πihνk ∓ πwk x dx. One can show that poles of the integrand only lie in points x m := iw (cid:18) m + 12 ± kr (cid:19) and we have nontrivial residues at most for one ν (mod k ) which can can chose as ν m := − k m + 1) h ′ . Using the Residue Theorem, we shift the path of integration through ω n := 2 n + 14 w . There are no poles on the real axis or in ω n . Moreover we have to take those x m with m ≥ n ≥ m + 1. We denote by λ ± n,m the residue of each summand and compute λ ± n,m = ± k πw e − πwx mk + πik (2 n +1)( x m − ν m ) . YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 15
Thus λ ± n +1 ,m = e πik ( x m − ν m ) λ ± n,m which gives e O w ( q ) = X + X , where X = 2 πik ( − ν m e πihν mk X m ≥ ± λ ± m +1 ,m − e πik ( x m − ν m ) , X = 1 k X ν (mod k ) ( − ν e πihν k X n ∈ N ± Z R + ω n e − πwx k + πik (2 n +1)( x − ν ) e πir ± πihνk ∓ πwk x dx. A lengthy calculation gives that X = iw e πkr w − πrw X m ∈ Z m odd q ( m + m ) − e − πrw q m . To compute P , we change the sum over n back into a sum over Z and change x x + ω n . This gives X = 1 k X ν (mod k ) ( − ν e πihν k X n ∈ Z e − π kw (2 n +1) − πik (2 n +1) ν Z R e − πwx k e − πi k (2 n +1)+ πihνk − πwxk +2 πir dx. Changing n p + δ (with p ∈ Z and δ ∈ { , − } ) and ν
7→ − h ′ ( ν + p ) gives X = 1 k X ν,p,δ e πih ′ k ( − ν − (2 δ +1) + ν (2 δ +1) ) q (4 p +2 δ +1) Z R e − πwx k − e πir + πiνk − πi k (2 δ +1) − πwxk dx. Now the sum over p equals η ( k ( h ′ + iz )) η ( k ( h ′ + iz )) thus X = η (cid:0) k (cid:0) h ′ + iz (cid:1)(cid:1) kη (cid:0) k (cid:0) h ′ + iz (cid:1)(cid:1) e − πih ′ k X ν (mod k ) ± e πih ′ k ( − ν + ν ) Z R e − πwx k e ± πiνk ∓ πi k +2 πir − πwxk dx which gives the claim using (2.16), (2.17), and (2.18). (cid:3) ¿From Theorem 4.1, we can conclude a transformation law for O ( − q ). For this define U ( q ) := 4 η ( z ) η (2 z ) X m ∈ Z q ( m + m ) (1 − q m ) ,I ± k,ν ( w ) := Z R e − πwx k (cid:16) e ± πiνk ∓ πi e k k − πwxk (cid:17) dx. Corollary 4.2.
Assume the notation above. We have. (1) If k is odd, then O ( − q ) = − √ w − ω h,k ω h,k e πih ′ k U (cid:18) q (cid:19) + 2 √ w ω h,k k ω h,k X ν (mod k ) ± e πih ′ k ( − ν + ν ) I ± k,ν ( w ) . (2) If k is even, then O ( − q ) = − w − ω h,k ω h,k/ O ( − q ) + 4 ω h,k kω h,k/ w X ν (mod k ) ( − ν e − πih ′ ν k I + k,ν ( w ) . Proof.
We only prove (i), (ii) can be shown similarly. We have12 πi ∂∂r e πkr w − πrw − e − πrw q m r =0 = (cid:16) q m − (cid:17) iw (cid:16) − q m (cid:17) . Now X m ∈ Z q ( m + m )1 (cid:16) q m − (cid:17)(cid:16) − q m (cid:17) = − X m ∈ Z q ( m + m )1 (cid:16) − q m (cid:17) + 2 X m ∈ Z q ( m + m )1 (cid:16) − q m (cid:17) = 2 X m ∈ Z q ( m + m )1 (cid:16) − q m (cid:17) , since in the first sum the m th and − m th term cancel. Now (i) can be easily concluded by using that12 πi (cid:20) ∂∂r (cid:18)
11 + e ± πiνk ∓ πi k − πwxk +2 πir (cid:19)(cid:21) r =0 = − (cid:16) e ± πiνk ∓ πi k − πwxk (cid:17) . (cid:3) Remark. ¿From Corollary 4.2 we can obtain the transformation law for O ( − z ) under (cid:16) α βγ δ (cid:17) ∈ SL ( Z )by setting h ′ = α , k = γ , h = − δ , and w = − i ( δ + γτ ).We next realize the integral occurring in Corollary 4.2 for k = 1 as a theta integral. For this let I ± r ( τ ) := Z R e − πix τ ± ie πir − πixτ ,I ( τ ) := 12 πi (cid:20) ∂∂r (cid:0) I + r ( τ ) + I − r ( τ ) (cid:1)(cid:21) r =0 . Lemma 4.3.
We have I ( τ ) = − ( − iτ ) √ π Z ∞ η ( iu ) η (cid:0) iu (cid:1) ( − i ( iτ + iu )) du. Proof.
Via analytic continuation it is sufficient to show the claim for τ = it . Making the change ofvariables x
7→ − xt gives I ± r ( it ) = t Z R e − πtx ± ie πir e πx dx. YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 17
We use the theory of Mittag-Leffler to rewrite11 + ie πir e πx + 11 − ie πir e πx = − π X m ∈ Z x − i (cid:0) − m + − r (cid:1) + 1 x − i (cid:0) m − − r (cid:1) . This implies that I + r ( it ) + I − r ( it ) = − t π X m ∈ Z Z R e − πtx x − i (cid:0) − m + − r (cid:1) + e − πtx x − i (cid:0) m − − r (cid:1) ! dx. From this we conclude I ( it ) = t π X m ∈ Z ± Z R e − πtx (cid:0) x − i (cid:0) m + ∓ (cid:1)(cid:1) dx = t π X m ∈ Z m odd Z R e − πtx (cid:0) x − im (cid:1) dx. We have the integral identity Z R e − πu t uu − is du = 12 √ Z ∞ e − πus ( u + t ) du ( s = 0)which follows since both sides are solutions of the differential equation − ∂∂t + 2 πs = √ t and havelimit 0 as t → ∞ . Using integration by parts gives Z R e − πtx ( x − is ) dx = √ πt Z ∞ e − πus ( u + t ) du. Thus I ( it ) = − t √ π Z ∞ X m ∈ Z m odd e − πum ( u + t ) du which easily gives the claim. (cid:3) Combining Theorem 4.1, Corollary 4.1, and Lemma 4.3 gives
Corollary 4.4.
For z ∈ H , we have O (cid:18) − − z (cid:19) = − i √ − iz ) · U (cid:16) − z (cid:17) + 2 ( − iz ) π Z ∞ η ( iu ) η ( iu/ − i ( iz + iu )) du. We next give the transformation law of the non-holomorphic part of M ( − z ) which can be shownas in proof of Theorem 3.2. Lemma 4.5.
For z ∈ H , we have J ( − z + 1) = J ( − z ) , − iz ) J (cid:18) − − z (cid:19) = − iπ Z i ∞− ¯ z η ( τ ) η ( τ /
2) ( − i ( τ + z )) dτ + 2 iπ Z ∞ η ( iu ) η ( iu/
2) ( − i ( iu + z )) du. Using that η ( τ ) η (2 τ ) is a modular form on Γ (16) gives the claim. Congruences for overpartitions
Following the original strategy of Ono [27] and Ono and the first author [6, 8, 9] we can prove thecongruences in Theorem 1.2. We limit ourselves to a sketch of the proof.
Sketch of Proof of Theorem 1.2.
Throughout we assume the assumptions of Theorem 1.2. First onecan show, using the results above that the function ∞ X n =0 (cid:18) N ( r, t ; n ) − p ( n ) t (cid:19) q n (5.1)is the holomorphic part of a weak Maass form of weight on Γ (cid:0) c (cid:1) . In order to use results of Serreon p -adic modular forms, we next apply twists to the associated weak Maass forms, which “kill” thenon-holomorphic part. This requires knowing on which arithmetic progressions it is supported. Weprove M (cid:16) ac ; z (cid:17) = 1 + ∞ X n =1 ∞ X m = −∞ N ( m, n ) ζ amc q n + √ πi tan (cid:16) πac (cid:17) X k (mod 2 c ) ( − k e (cid:18) kac (cid:19) X m ≡ k (mod 2 c ) Γ (cid:18)
12 ; 4 πm y (cid:19) q − m , (5.2)where e ( x ) := e πix , and where(5.3) Γ( a ; x ) := Z ∞ x e − t t a − dt is the incomplete Gamma-function. Using this one can show that for a prime p ∤ t the function(5.4) X n ≥ ( np ) = − ( − p ) (cid:18) N ( r, t ; n ) − p ( n ) t (cid:19) q n is a weight weakly holomorphic modular form on Γ (16 c p ). Now the theorem can be concluded asin [9] using a geralization of Serre’s results on p -adic modular forms (Theorem 4.2 of [9]). (cid:3) Proof of Theorem 1.3
Here we prove Theorem 1.3. While we follow the model of Ono [27] and the first author and Ono[6, 9], the proof of the modularity is rather delicate. For brevity we set t := ℓ m . Define the function g r ( z ) := t · η r (2 ℓz ) · η ℓ ( ℓz ) ∞ X n =0 N ( r, t ; n ) q n , where 0 < r <
48 is a solution of r ℓ ≡ − g r ( z ) = η (2 z ) · η r (2 ℓz ) · η ℓ ( ℓz ) η ( z ) + t − X j =1 ζ − rjt (cid:18) M (cid:18) jt ; z (cid:19) + J (cid:18) jt ; z (cid:19)(cid:19) η r (2 ℓz ) · η ℓ ( ℓz ) . We denote the two summands by f ( z ) and f r ( z ), respectively. Define for a function h ( z ) = P n a ( n ) q n the twist e h ( z ) := − (cid:18) − ℓ (cid:19) (cid:18) h ( z ) − (cid:18) − ℓ (cid:19) h ( z ) ℓ (cid:19) ℓ , (6.2) YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 19 where as before(6.3) h ℓ := gℓ X ν (mod ℓ ) (cid:0) νℓ (cid:1) h | (cid:16) − νℓ (cid:17) . Clearly e h ( z ) = X n ( − nℓ ) = − a ( n ) q n . (6.4)The main step in the proof of Theorem 1.3 is the following theorem. Theorem 6.1.
For every u ≥ there exist a character χ , integers λ, λ ′ , N, N ′ , and modular forms h ( z ) ∈ S λ + (Γ ( N ) , χ ) and h r ( z ) ∈ S λ ′ + (Γ ( N ′ )) such that e g r ( z ) η r (2 ℓz ) · η ℓ ( ℓz ) ≡ h ( z ) + h r ( z ) (mod ℓ u ) . The proof of Theorem 6.1 is given later. We first show how Theorem 1.3 follows from Theorem 6.1.
Proof of Theorem 1.3.
We easily see that e g r ( z ) η ℓ ( ℓz ) · η r (2 ℓz ) = t X ( − nℓ ) = − N ( r, t ; n ) q n . Now we let 0 ≤ β ≤ ℓ − (cid:0) − βℓ (cid:1) = − g r,β ( z ) := t X n ≡ β (mod ℓ ) N ( r, t ; n ) q n . Theorem 6.1 gives that g r,β ( z ) ≡ h β ( z ) + h r,β ( z ) (mod ℓ u ) , where h β ( z ) and h r,β ( z ) denote the restrictions of the Fourier expansion of h ( z ) resp. h r ( z ) to thosecoefficients n with n ≡ β (mod ℓ ). Using the theory of Hecke operators, we can show that for all n ≡ β (mod ℓ ) coprime to p we have t N (cid:0) r, t ; p n (cid:1) ≡ ℓ u ) . Dividing by t directly gives the theorem since u is arbitrary. (cid:3) Proof of Theorem 6.1. If a is a positive integer, then define E ℓ,a ( z ) := η ℓ a ( z ) η ( ℓ a z ) ∈ M ℓa − (Γ ( ℓ a ) , χ ℓ,a ) , where χ ℓ,a ( d ) := (cid:0) ( − ( ℓa − / ℓ a d (cid:1) . It is well known that E ℓ,a ( z ) vanishes at those cusps of Γ ( ℓ a ) that arenot equivalent to ∞ and that for all u > E ℓ u − ℓ,a ( z ) ≡ ℓ u ) . (6.5)We now treat the summands in (6.1) separately. We start with f ( z ). Using Theorem 1.64 from [29], itis not hard to see that f ( z ) is a modular form of weight r +2 ℓ − with some character on Γ (2 ℓ ). ¿Fromthis we see that e f ( z ) ∈ M r ℓ − (cid:0) Γ (2 ℓ ) , e χ (cid:1) for some character e χ . For sufficiently large u ′ the function h ( z ) := e f ( z ) · E ℓ u ′ ℓ, ( z ) η r (2 ℓz ) · η ℓ ( ℓz ) is a weakly holomorphic modular form on Γ (8 ℓ ) that vanishes at all cusps with the exception of ∞ and rℓ with r ∈ { , , } , and satisfies h ( z ) ≡ e f ( z ) η r (2 ℓz ) η ℓ ( ℓz ) (mod ℓ u ) . We show that h ( z ) is a cusp form. It is easy to see that in the cusp ∞ the Fourier expansion of e f ( z )starts at least with q ℓ ( r + ℓ )+1 . Since the Fourier expansion of η r (2 ℓz ) · η ℓ ( ℓz ) starts with q ℓ ( r + ℓ ) ,we see that h ( z ) vanishes in ∞ . Since e f ( z ) is a modular form on Γ (cid:0) ℓ (cid:1) and the Fourier expansion of η r (2 ℓz ) · η ℓ ( ℓz ) in rℓ with r even starts with q ℓ ( r + ℓ ) , h ( z ) vanishes also in the cusps ℓ and ℓ .Next consider the cusp ℓ . In (6.3), we can choose a set of respresentatives with ν even. Now foreven ν it is not hard to see that (cid:16) − νℓ (cid:17) (cid:0) ℓ (cid:1) is Γ (2 ℓ )- equivalent to (cid:0) ℓ (cid:1) (cid:16) − νℓ (cid:17) . Thus f ( z ) ℓ = gℓ X ν (mod ℓ ) ν even (cid:18) νℓ (cid:19) f | (cid:0) ℓ (cid:1) (cid:16) − νℓ (cid:17) . It is not hard to see that f ( z ) vanishes in ℓ of order ( − r ℓ + 4 ℓ ). Since the cusp width of ℓ inΓ (2 ℓ ) is 2, the Fourier expansion of f | (cid:0) ℓ (cid:1) starts with q r , where r := ( − r ℓ + 4 ℓ ). Thusthe Fourier expansion of (cid:0) − ℓ (cid:1) f ℓ starts with (cid:18) − ℓ (cid:19) gℓ X ν (mod ℓ ) ν even (cid:18) νℓ (cid:19) e − πir νℓ = (cid:18) − r ℓ (cid:19) = 1 . Since twisting doesn’t decrease the order of vanishing, e f ( z ) has in ℓ a Fourier expansion starting atleast with q r + , whereas η r (2 ℓz ) · η ℓ ( ℓz ) has in ℓ a Fourier expansion starting with q ℓ ( r ℓ )48 . Thus h ( z ) vanishes in all cusps and is therefore a cusp form.We next turn to f r ( z ). Using Theorem 1.1, it is not hard to see that e f r ( z ) is the holomorphic partof a weak Maass form on Γ (16 t ℓ ). Moreover by (5.2) it is easy to see that the corresponding weakMaass form doesn’t have a non-holomorphic part, and thus e f r ( z ) is a weakly holomorphic modularform. Since E ℓ, m ( z ) vanishes at each cusp αγ with t ∤ γ for sufficiently large u ′ , the function f ′ r ( z ) := E ℓ u ′ ℓ, m ( z ) e f r ( z )is a weakly holomorphic modular form on Γ (cid:0) t ℓ (cid:1) that vanishes at all cusps αγ with t ∤ γ andsatisfies f ′ r ( z ) ≡ f r ( z ) (mod ℓ u ) . Therefore to finish the proof it remains to show that e f r ( z ) η r (2 ℓz ) η ℓ ( ℓz ) vanishes also at those cusps αγ with t | γ . Now let (cid:16) α βγ δ (cid:17) ∈ Γ ( t ) . In the following we need the commutation relation for ν ′ ≡ δ ν (mod ℓ ) (cid:18) − νℓ (cid:19) (cid:18) α βγ δ (cid:19) = (cid:18) α ′ β ′ γ ′ δ ′ (cid:19) (cid:18) − ν ′ ℓ (cid:19) (6.6)with (cid:18) α ′ β ′ γ ′ δ ′ (cid:19) = α − γνℓ β − γνν ′ ℓ + αν ′ − δνℓ γ δ + γν ′ ℓ ! ∈ Γ ( t ) . We distinguish the cases whether 2 | γ or not. YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 21
If 2 | γ , then one can easily see that the Fourier expansion of η r (2 ℓz ) η ℓ ( ℓz ) in αγ starts with ℓ ( r + ℓ ) =: n . Clearly ℓ | n . Thus we have to prove that the q -expansion of e f r ( z ) in αγ starts with q b with b > n . We may assume that 48 | ν, ν ′ . Then we can write( f r ) ℓ | (cid:16) α βγ δ (cid:17) = X ν (mod ℓ )48 | ν,ν ′ f r | (cid:16) α ′ β ′ γ ′ δ ′ (cid:17) (cid:16) − ν ′ ℓ (cid:17) . Let f r,j ( z ) := M (cid:18) jt ; z (cid:19) · η r (2 ℓz ) · η ℓ ( ℓz )and define f f r,j ( z ) for weak Maass forms as for holomorphic forms. It is enough to show that f f r,j ( z )starts with q b with b > n . As before it follows that f f r,j ( z ) doesn’t have a non-holomorphic part. Weuse twice (6.6), Corollary 2.3, and the transformation law of the η -function. We let e α := ( α ′ ) ′ , e δ := ( δ ′ ) ′ ,and e γ := ( γ ′ ) ′ . The holomorphic part of f r,j | (cid:16) e α e β e γ e δ (cid:17) is given by(6.7) − ie πiℓ e γ ( e α + e δ )( r + ℓ ) e πij e δ e γ t ω e α, e γ ω e α, e γ/ · ω ℓ e α, e γ/ℓ · ω r e α, e γ/ ℓ tan (cid:18) πjt (cid:19) cot πj e δt !(cid:16) − i (cid:16)e γz + e δ (cid:17)(cid:17) (1+ r +2 ℓ ) f r,j e δ ( z ) . Using [26] ω − α,γ · e πi γ ( α + δ ) = ( (cid:0) δγ (cid:1) · i − γ · e πi ( βδ (1 − γ )+ γ ( α + δ )) if γ is odd , (cid:0) γδ (cid:1) e πi ( αγ (1 − δ )+ δ ( β − γ +3)) if δ is odd , (6.8)we can show that in (6.7) we can change e α , e δ , and e γ into α ′ , δ ′ , and γ ′ , respectively if we change z into z + ν ′ ℓ . The Fourier expansion of f r,jδ ′ starts with q n and ℓ | n . Moreover X ν (mod ℓ ) ν ≡ (cid:18) νℓ (cid:19) e − πin ν ′ ℓ = X ν (mod ℓ ) (cid:18) νℓ (cid:19) = 0 . Thus the Fourier expansion of the holomorphic part of ( f r,j ) ℓ | (cid:16) α ′ β ′ γ ′ δ ′ (cid:17) starts with at least q n +1 whichimplies that the Fourier expansion of the holomorphic part of (cid:18) f r,j − (cid:18) − ℓ (cid:19) ( f r,j ) ℓ (cid:19) | (cid:16) α ′ β ′ γ ′ δ ′ (cid:17) starts at least with q n . Arguing in the same way, we obtain that the Fourier expansion of (cid:18) f r,j − (cid:18) − ℓ (cid:19) ( f r,j ) ℓ (cid:19) ℓ | (cid:16) α βγ δ (cid:17) starts with at least q n +1 as desired.Next assume that γ is odd. It is not hard to see that the Fourier expansion of η r (2 ℓz ) η ℓ ( ℓz ) in αγ starts with q ℓ ( r +4 ℓ ) =: q n . Since twisting does not decrease the order of vanishing,it is enough to show that the Fourier expansion of the holomorphic part of (cid:0) f r,j − (cid:0) − ℓ (cid:1) ( f r,j ) ℓ (cid:1) | (cid:16) α βγ δ (cid:17) starts with q b with b > n . This time we use (6.6) once. One can compute that the holomorphic partof f r,j | (cid:16) α ′ β ′ γ ′ δ ′ (cid:17) equals(6.9) √ ie πiℓ γ ′ α ′ ( r + ℓ )+ πiδ ′ ℓ γ ′ ( r +4 ℓ ) − πiδ ′ γ ′ + πij δ ′ γ ′ t · tan (cid:18) πjt (cid:19) ω α ′ ,γ ′ ω α ′ ,γ ′ · ω r α ′ ,γ ′ /ℓ · ω ℓα ′ ,γ ′ /ℓ ( − i ( γ ′ τ + δ ′ )) r ℓ +12 · η ℓ ( ℓz ) · η r (cid:18) ℓz (cid:19) · U (cid:18) − jδt ; z (cid:19) . Again using (6.8), one can show that one can change α ′ , δ ′ , and γ ′ into α , δ , and γ , respectively if onechanges z into z + ν ′ ℓ . The expansion of (6.9) starts with q ( ℓr +4 ℓ − . For r := ( ℓr + 4 ℓ −
3) one clearly has ( r , ℓ ) = 1. Moreover gℓ (cid:18) − ℓ (cid:19) X ν (mod ℓ ) ν,ν ′ ≡ (cid:18) νℓ (cid:19) e − πir ν ′ ℓ = (cid:18) − r ℓ (cid:19) = 1 . Thus the expansion of (cid:0) f r,j − (cid:0) − ℓ (cid:1) ( f r,j ) ℓ (cid:1) | (cid:16) α ′ β ′ γ ′ δ ′ (cid:17) starts at least with q ( ℓr +4 ℓ +45) which implies theclaim. (cid:3) Proof of Theorem 1.5
Proof of Theorem 1.5. ¿From (5.1) it follows that ∞ X n =0 (cid:18) N ( s i , ℓ ; n ) − p ( n ) ℓ (cid:19) q n for i ∈ { , } is the holomorphic part of a weak Maass form on Γ (16 ℓ ). Moreover the non-holomorphicpart is supported on negative squares. The same is true for the function ∞ X n =0 (cid:0) N ( s , ℓ ; n ) − N ( s , ℓ ; n ) (cid:1) q n . The restriction of the associated weak Maass form to those coefficients congruent to d modulo ℓ givesa weak Maass form on Γ (16 ℓ ). Since (cid:0) dℓ (cid:1) = − (cid:0) − ℓ (cid:1) it does not have a non-holomorphic part whichproves Theorem 1.5. (cid:3) Next we state some identities which may be deduced thanks to our theorem.Define for a positive integer N , g, h real numbers that are not simultaneously congruent to 0 (mod N ),the generalized Dedekind eta-function E g,h ( z ) := q B ( gN ) · ∞ Y m =1 (cid:16) − ζ hN · q m − gN (cid:17) (cid:16) − ζ − hN · q m − gN (cid:17) , where B ( x ) := x − x + . We have the following identities.(7.1) ∞ X n =0 (cid:0) N (1 , , n + 2) − N (2 , , n + 2) (cid:1) q n +2 = 2 η (50 z ) E , (25 z ) , (7.2) ∞ X n =0 (cid:0) N (1 , , n + 3) − N (2 , , n + 3) (cid:1) q n +3 = − η (50 z ) E , (25 z ) , YSON’S RANK, OVERPARTITIONS, AND WEAK MAASS FORMS 23 (7.3) ∞ X n =0 (cid:0) N (0 , , n + 3) − N (2 , , n + 3) (cid:1) q n +3 = 2 η (50 z ) E , (25 z ) , (7.4) ∞ X n =0 (cid:0) N (0 , , n + 2) − N (2 , , n + 2) (cid:1) q n +2 = 0 , (7.5) ∞ X n =0 (cid:0) N (0 , , n + 1) − N (1 , , n + 1) (cid:1) q n +1 = 2 η (9 z ) · η (18 z ) η (3 z ) . References [1] G. E. Andrews,
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