An Analogue of Weil's Converse Theorem for Harmonic Maass Forms of Polynomial Growth
aa r X i v : . [ m a t h . N T ] J a n AN ANALOGUE OF WEIL’S CONVERSE THEOREM FOR HARMONICMAASS FORMS OF POLYNOMIAL GROWTH
KARAM DEO SHANKHADHAR, RANVEER KUMAR SINGH
Abstract.
We construct a family of examples of harmonic Maass forms of polynomialgrowth for any level whose shadows are Eisenstein series of integral weight. We furtherconsider Dirichlet series attached to a harmonic Maass form of polynomial growth, studyits analytic properties, and prove an analogue of Weil’s converse theorem. Introduction
The first glimpse of harmonic Maass forms and mock modular forms goes back to theenigmatic “deathbed” letter that Ramanujan wrote to Hardy in 1920. For many decades,very little was understood about the functions written in this letter and no comprehensivetheory was available to explain them or their role in mathematics. Finally, Zwegers [24] inhis thesis showed that Ramanujan’s mock theta functions can be realized as holomorphicparts of some special families of nonholomorphic modular forms and fit very well with severalimportant theories in mathematics. At almost the same time, Bruinier and Funke [4] wrotea very important paper in which they defined the notion of a harmonic Maass form andthe nonholomorphic modular forms constructed by Zwegers turned out to be weight 1 / §
2. For a systematic developmentand detailed treatment of general theory of harmonic Maass forms we refer the reader to thebook of Bringmann, Folsom, Ono and Rolen [3], which is essentially self-contained.In this article, we study those harmonic Maass forms of manageable growth which haveat most polynomial growth at all the cusps. We denote the space of harmonic Maass formsof polynomial growth of weight k for the group Γ ( N ) with character χ by H k ( N, χ ). Forthe precise definition and other properties of the functions in the space H k ( N, χ ) we referto §
3. We construct a family of examples in H k ( N, χ ) whose shadows are Eisenstein series(see § H k ( N, χ ),study analytic properties, and prove an analogue of Weil’s converse theorem for this space.A converse theorem in the theory of automorphic forms refers to the equivalence of Dirich-let series satisfying certain analytic properties, on the one hand, and automorphic forms oversome group, on the other. Most familiar is the converse theorem due to Hecke [9], whichestablishes an equivalence between modular forms on SL ( Z ) and Dirichlet series satisfyinga certain functional equation as well as meromorphic continuation and a certain bounded-ness property in vertical strips. In our context, the meaning of a converse theorem is bestillustrated by the Weil’s converse theorem for modular forms over congruence subgroups Mathematics Subject Classification.
Primary 11F12, 11F25; Secondary 11F66, 11M36.
Key words and phrases.
Harmonic Maass forms, differential operators, Dirichlet series, converse theorem. ( N ) [23], which is a very significant generalization of the corresponding Hecke’s theoremfor N = 1. Other results of this kind are Maass’ converse theorem for Maass waveformsof level 1 [15], its generalization to Γ ( N ) by Neururer and Oliver [20], converse theoremsfor Jacobi forms [16, 17], Siegel modular forms [14], and Maass Jacobi forms [10]. The con-verse theorem for GL n is a great achievement of several authors through a string of papers[5, 12, 13].1.1. Statement of the main result.
In order to state our converse theorem for the space H k ( N, χ ), we first introduce some basic notations that will be used throughout the article.Let C denote the complex plane. For each z ∈ C , denote the real and imaginary parts of z by Re( z ) and Im( z ), respectively. We also define i := √−
1. Let H := { z ∈ C : Im( z ) > } be the upper half-plane. For τ ∈ H , let τ = u + iv and q = e πiτ .Let us fix two integers k and N with N ≥ χ modulo N suchthat χ ( −
1) = ( − k . We assume that k is a negative integer (see Proposition 1 and Remark3). Let f and g be two functions on H given by the following formal Fourier series f ( τ ) = ∞ X n =0 c + f ( n ) q n + c − f (0) v − k + X n< c − f ( n )Γ(1 − k, − πnv ) q n ,g ( τ ) = ∞ X n =0 c + g ( n ) q n + c − g (0) v − k + X n< c − g ( n )Γ(1 − k, − πnv ) q n , (1)with c ± f ( n ) , c ± g ( n ) bounded by O ( | n | α ) , n ∈ Z , for some α ≥ ν ∈ R , define(2) W ν ( s ) = Z ∞ Γ( ν, x ) e − x x s dxx , Re( s ) > , where Γ( ν, x ) is the incomplete gamma function given by (5). Now, we associate completedDirichlet series Λ N ( f, s ) and Ω N ( f, s ) to the function f as follows.Λ N ( f, s ) = √ N π ! s (cid:2) Γ( s ) L + ( f, s ) + W − k ( s ) L − ( f, s ) (cid:3) , Ξ N ( f, s ) = √ N π ! s (cid:2) Γ( s + 1) L + ( f, s ) − W − k ( s + 1) L − ( f, s ) (cid:3) , Ω N ( f, s ) = − N ( f, s ) + k Λ N ( f, s ) , where L + ( f, s ) = ∞ X n =1 c + f ( n ) n s , L − ( f, s ) = ∞ X n =1 c − f ( − n ) n s are Dirichlet series attached to f . Similarly, we associate Dirichlet series L ± ( g, s ) , Λ N ( g, s )and Ω N ( g, s ) to the function g .As in the case of classical modular forms, we twist the Fourier series f ( τ ) by a Dirichletcharacter ψ with conductor m ψ to get the following twisted Fourier series.(3) f ψ ( τ ) = ∞ X n =0 ψ ( n ) c + f ( n ) q n + ψ (0) c − f (0) v − k + X n< ψ ( − n ) c − f ( n )Γ(1 − k, − πnv ) q n . gain, we consider the Dirichlet series L ± ( f, s, ψ ) = ∞ X n =1 ψ ( n ) c ± f ( ± n ) n s and the completed Dirichlet series Λ N ( f, s, ψ ) , Ω N ( f, s, ψ ) associated to f ψ . Similarly, wetwist the Fourier series g to g ψ and attach the Dirichlet series L ± ( g, s, ψ ) , Λ N ( g, s, ψ ) , Ω N ( g, s, ψ )to g ψ . Note that if ψ is the trivial character with conductor m ψ = 1 then all the twistedDirichlet series will be same as the Dirichlet series attached to f and g .Finally, we denote by P a set of odd prime numbers or 4 which are relatively prime to N and whose intersection with every arithmetic progression of the form a + nb of two coprimeintegers a, b , is non-empty. Also, let ω ( N ) be the standard Fricke involution defined asfollows. f | k ω ( N ) = f | k (cid:18) − N (cid:19) = N k/ ( N τ ) − k f (cid:18) − N τ (cid:19) . Now we state the main theorem of this paper which characterize the Dirichlet seriesassociated to the functions in the space H k ( N, χ ) and hence can be considered as an analogueof Weil’s converse theorem in this case.
Theorem 1.
Let k, N, χ be as above. Let f and g be two functions on H given by the formalFourier series as in (1) with c ± f ( n ) , c ± g ( n ) bounded by O ( | n | α ) for some α ≥ . Then thefollowing two statements are equivalent.(1) The Fourier series f ( τ ) ∈ H k ( N, χ ) , g ( τ ) ∈ H k ( N, χ ) and f | k ω ( N ) = g .(2) (a) Each one of the completed Dirichlet series Λ N ( f, s ) , Λ N ( g, s ) , Ω N ( f, s ) , Ω N ( g, s ) admits a meromorphic continuation to the whole complex plane and satisfies thefunctional equation Λ N ( f, s ) = i k Λ N ( g, k − s ) , Ω N ( f, s ) = − i k Ω N ( g, k − s ) . Moreover, each one of the following functions Λ ∗ N ( f, s ) = Λ N ( f, s ) + c + f (0) s + c + g (0) i k k − s + c − f (0) N − k s − k + 1 + c − g (0) i k N − k − s Λ ∗ N ( g, s ) = Λ N ( g, s ) + c + g (0) s + c + f (0) i − k k − s + c − g (0) N − k s − k + 1 + c − f (0) i − k N − k − s Ω ∗ N ( f, s ) = Ω N ( f, s ) + k c + f (0) s − c + g (0) i k k − s + c − f (0) N − k s − k + 1 − c − g (0) i k N − k − s ! Ω ∗ N ( g, s ) = Ω N ( g, s ) + k c + g (0) s − c + f (0) i − k k − s + c − g (0) N − k s − k + 1 − c − f (0) i − k N − k − s ! is entire and bounded in any vertical strip.(b) For any primitive Dirichlet character ψ whose conductor m ψ ∈ P , each oneof the completed Dirichlet series Λ N ( f, s, ψ ) , Λ N ( g, s, ψ ) , Ω N ( f, s, ψ ) , Ω N ( g, s, ψ ) can be analytically continued to the whole s -plane, bounded on any vertical strip, nd satisfies the functional equation Λ N ( f, s, ψ ) = i k C ψ Λ N ( g, k − s, ¯ ψ ) , Ω N ( f, s, ψ ) = − i k C ψ Ω N ( g, k − s, ¯ ψ ) , where C ψ = C N,ψ = χ ( m ) ψ ( − N ) τ ( ψ ) /τ ( ψ ) = χ ( m ) ψ ( N ) τ ( ψ ) /m. Remark 1. (1) While proving the direct part (1) = ⇒ (2) in § f, g satisfying the condition (1), the analytic properties given in (2)(b) is true forDirichlet series twisted by any primitive Dirichlet character with conductor coprimeto N .(2) In the proof of the converse part (2) = ⇒ (1) presented in § m ψ ∈ P . In fact for a fixed N ≥ , having the condition (2)(b) only for finitely manyprimitive Dirichlet characters are sufficient. The same is true in the case of modularforms as well, evident from the proof of [18, Theorem 4.3.15] . For any fixed N ≥ , byusing a computational platform one can find out finitely many m ψ which are enoughto establish the converse part. For example, for N = 7 , we need to assume 2(b) for m ψ = 11 , , , , , and m ψ = 13 , , , , , , , , , respectively.Unfortunately, we are not able to find such a finite set which works uniformly for all N . This article is organized as follows. In the next section, we recall basic facts about Har-monic Maass forms of manageable growth, study differential operators and Fricke involutionfor these forms, and prove some lemmas which are used throughout the paper. In §
3, wedescribe the space H k ( N, χ ) and construct interesting examples in this space. In §
4, we dis-cuss some intermediate results which will be useful in establishing Theorem 1. In particular,we establish an analogue of Hecke’s converse theorem for the space H k ( N, χ ) which mightbe of independent interest as well in some other context. In § §
6, we prove the directand converse parts of Theorem 1 respectively. To prove Theorem 1, we essentially follow theproof of Weil’s converse theorem presented in [18, Theorem 4.3.15]. The key differences arethe holomorphicity of functions and the shape of Fourier series expansions. Certain differ-ential operators and some crucial properties of incomplete Gamma functions are helpful toovercome the technical difficulties presented due to the nature of harmonic Maass forms.2.
Notations and Preliminaries
Throughout this section k, N ∈ Z with N ≥
1. Let GL +2 ( Q ) be the group of 2 × ( Z ) be the subgroup of GL +2 ( Q )having integer entries and determinant 1. For any positive integer N , letΓ ( N ) = (cid:26)(cid:18) a bc d (cid:19) ∈ SL ( Z ) : c ≡ N ) (cid:27) . For any γ = ( a bc d ) ∈ GL +2 ( Q ) and k ∈ Z , we define the weight k slash operator as follows.If f : H −→ C is a function, define f | k γ ( τ ) = (det γ ) k/ ( cτ + d ) − k f ( γτ ) here γτ = ( aτ + b )( cτ + d ) − .A holomorphic function f : H −→ C is called a weakly holomorphic modular form (respec-tively modular form and cusp form) of weight k , level N and character χ if f | k γ = χ ( d ) f forevery γ ∈ Γ ( N ) and f is meromorphic (respectively holomorphic and vanishes) at all cuspsof Γ ( N ). We denote the C -vector space of weakly holomorphic modular forms (respectivelymodular and cusp forms) by M ! k ( N, χ ) (respectively M k ( N, χ ) and S k ( N, χ )).Following Bruiner and Funke [4], we define harmonic Maass forms. First define the weight- k hyperbolic Laplacian operator ∆ k on H by∆ k = − v (cid:18) ∂ ∂u + ∂ ∂v (cid:19) + ikv (cid:18) ∂∂u + i ∂∂v (cid:19) = − v ∂∂τ ∂∂ ¯ τ + 2 ikv ∂∂ ¯ τ . Definition 2.
A smooth function f : H → C is called a harmonic Maass form of weight k ,level N and character χ if(i) f | k γ = χ ( d ) f for every γ ∈ Γ ( N ) .(ii) ∆ k ( f ) = 0 .(iii) There exists a polynomial P f ( τ ) ∈ C [ q − ] such that f ( τ ) − P f ( τ ) = O ( e − ǫv ) as v → ∞ for some ǫ > . Analogous conditions are required at all other cusps of Γ ( N ) .If the third condition in the above definition is replaced by f ( τ ) = O ( e ǫv ) , then f is said tobe a harmonic Maass form of manageable growth . The space of harmonic Maass forms ofweight k , level N and character χ is denoted by H k ( N, χ ) and that of harmonic Maass formsof manageable growth by H ! k ( N, χ ) . Any f ∈ H ! k ( N, χ )( k = 1) admits the following Fourier series expansion at the cusp i ∞ [3, Lemma 4.3].(4) f ( τ ) = f ( u + iv ) = X n>> −∞ c + f ( n ) q n + c − f (0) v − k + X n<< ∞ n =0 c − f ( n )Γ(1 − k, − πnv ) q n , where Γ(1 − k, − πnv ) is the incomplete gamma function defined by (5). The notation P n>> −∞ means that P ∞ n = α f for some α f ∈ Z . The other notation P n<< ∞ is analogous. Inparticular, if f ∈ H k ( N, χ ) then the above expansion takes the following shape. f ( τ ) = f ( u + iv ) = X n>> −∞ c + f ( n ) q n + X n< c − f ( n )Γ(1 − k, − πnv ) q n . The incomplete gamma function Γ( s, z ) [3, Page 63] is given by(5) Γ( s, z ) := ∞ Z z e − t t s dtt . This integral converges absolutely for Re( s ) > z ∈ C (or s ∈ C and z ∈ H ). Thefunction Γ( s, z ) satisfies the following asymptotic behaviour [3, Eq. 4.6].(6) Γ( s, x ) ∼ x s − e − x as x ∈ R and | x | → ∞ . Let ρ be any cusp of Γ ( N ) and Γ ρ = { g ∈ Γ ( N ) | gρ = ρ } . Let γ ρ ∈ SL ( Z ) such that γ ρ ( i ∞ ) = ρ . Then γ − ρ Γ ρ γ ρ fixes i ∞ and hence generated by − I and (cid:18) t ρ (cid:19) for some ositive integer t ρ . The integer t ρ is called the width of the cusp ρ . Let g ρ ∈ Γ ρ such that γ − ρ g ρ γ ρ = (cid:18) t ρ (cid:19) . For any f ∈ H ! k ( N, χ ), we have( f | k γ ρ ) | k (cid:18) t ρ (cid:19) = ( f | k g ρ ) | k γ ρ = χ ( d ρ ) f | k γ ρ , g ρ = (cid:18) a ρ b ρ c ρ d ρ (cid:19) . Let κ ρ ∈ [0 ,
1) such that χ ( d ρ ) = e πiκ ρ . The real number κ ρ is called the cusp parameter (cf. [7, § Lemma 1.
Let f ∈ H ! k ( N, χ )( k = 1) and ρ, γ ρ , t ρ , κ ρ be as above. Then the Fourier seriesexpansion of f at the cusp ρ will have the following shape. ( f | γ ρ )( τ ) = X n>> −∞ c + f ( n ) q n + κρtρ + c − f (0) v − k q κρtρ + X n<< ∞ c − f ( n )Γ(1 − k, − πnv/t ρ ) q n + κρtρ . Proof.
From the discussion before the lemma, we have (cid:0) e − πiκ ρ τ/t ρ f | k γ ρ (cid:1) | k (cid:0) t ρ (cid:1) = e − πiκ ρ τ/t ρ f | k γ ρ . Now by following the proof of [3, Lemma 4.3] (for detailed calculation, see [22, Proof ofTheorem 4.6]), we see that e − πiκ ρ τ/t ρ f | γ ρ has Fourier expansion of the form X n>> −∞ c + f ( n ) q n/t ρ + c − f (0) v − k + X n<< ∞ c − f ( n )Γ(1 − k, − πnv/t ρ ) q n/t ρ . The lemma now follows. (cid:3)
Remark 2.
Let f ∈ H ! k ( N, χ ) and α ∈ GL +2 ( Q ) . Then f | k α satisfies similar growth conditionas f at every cusp. Let s ∈ Q ∪ { i ∞} be a cusp and s = β ( i ∞ ) for some β ∈ SL ( Z ) . Weneed to show that ( f | k α ) | k β = f | k αβ has similar Fourier expansion as f . We can assumethat αβ has integer entries by multiplying αβ by an integer without affecting f | k αβ . Thenthere exists a matrix γ ∈ SL ( Z ) and a, b, d ∈ Z with a, d ≥ (cf. [19, Problem 7.1.5] ) suchthat αβ = γ (cid:18) a b d (cid:19) . Since f has a Fourier series expansion of the form (4) at every cusp, therefore we have f | k αβ = f | k γ | k (cid:18) a b d (cid:19) = X n>> −∞ f c + f ( n ) q n + f c − f (0) v − k + X n<< ∞ n =0 f c − f ( n )Γ(1 − k, − πnv ) q n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k (cid:18) a b d (cid:19) = e πib/dh ( a/d ) k/ " X n>> −∞ f c + f ( n ) q an/d + ( a/d ) − k f c − f (0) v − k + X n<< ∞ n =0 f c − f ( n )Γ(1 − k, − πanv/d ) q an/d . Since a, d > , we see that f | k αβ also has similar Fourier expansion as f . ext, we discuss some useful differential operators acting on the space H ! k ( N, χ ). Following[3, Chapter 5], we define the
Maass raising operator R k and the Maass lowering operator L k . R k = 2 i ∂∂τ + kv = i (cid:18) ∂∂u − i ∂∂v (cid:19) + kv .L k = − iv ∂∂ ¯ τ = − iv (cid:18) ∂∂u + i ∂∂v (cid:19) . Let f be any smooth function on H . From [3, Lemma 5.2], we have the following identity.(7) − ∆ k = L k +2 R k + k = R k − L k . We define the shadow operator ξ k (cf. [3, § ξ k = 2 iv k ∂∂ ¯ τ . It is related to the lowering operator L k [3, Page 74] and the Laplacian operator ∆ k [11, Eq.2, Page 1] as follows.(8) ξ k = v k − L k , ∆ k = − ξ − k ◦ ξ k . Suppose k = 1. Then ξ k : H ! k ( N, χ ) → M !2 − k ( N, χ ) is a surjective linear map [3, Theorem5.10], [2, Lemma 2.2]. Moreover, for any f ∈ H ! k ( N, χ ) with Fourier series expansion as in(4), we have(9) ξ k ( f ( τ )) = ( k − c − f (0) − (4 π ) k − X n>> −∞ c − f ( − n ) n k − q n . For k ≤
0, we have the differential operator D − k : H ! k ( N, χ ) → M !2 − k ( N, χ ) , D = πi ∂∂τ ,called the Bol operator [3, Theorem 5.5]. We have [3, Lemma 5.3](10) D − k = 1( − π ) k − R − kk . From [3, Theorem 5.5], for any f ∈ H ! k ( N, χ ) with Fourier series expansion as in (4), we have(11) D − k ( f ( τ )) = − (4 π ) k − (1 − k )! c − f (0) + X n ≫−∞ c + f ( n ) n − k q n . Lemma 2.
Let f be a smooth function on H and α ∈ GL +2 ( Q ) . Then we have R k ( f | k α ) = R k ( f ) | k +2 α, L k ( f | k α ) = L k ( f ) | k − α. ∆ k ( f | k α ) = ∆ k ( f ) | k α, ξ k ( f | k α ) = ξ k ( f ) | − k α. Moreover, if k ≤ then D − k ( f | k α ) = D − k ( f ) | − k α .Proof. In view of (7), (8) and (10), it is sufficient to have the claimed commutation relationof the slash operator with the differential operators R k and L k . To prove this we followthe proof of [3, Lemma 5.2 (i)] (for detailed calculation, see [22, Theorem 4.12 (i)]). Theonly difference is that α ∈ GL +2 ( Q ) but the power of det α will get balanced from both thesides. (cid:3) Lemma 3. If f ∈ H ! k ( N, χ ) then f | k ω ( N ) ∈ H ! k ( N, χ ) . roof. For any γ = (cid:18) a bcN d (cid:19) ∈ Γ ( N ), we have( f | k ω ( N )) | k γ = f | k (cid:18) d − c − bN a (cid:19) | k ω ( N ) = χ ( a ) f | k ω ( N ) = ¯ χ ( γ ) f | k ω ( N ) . Next, by using Lemma 2 we have∆ k ( f | k ω ( N )) = ∆ k ( f ) | k ω ( N ) = 0 . Finally, by using Remark 2 we get the required cusp conditions. (cid:3) The space of harmonic Maass forms of polynomial growth H k ( N, χ )Throughout this section k, N ∈ Z with N ≥
1. In order to attach Dirichlet series toharmonic Maass forms of manageable growth, we consider a further subspace of H ! k ( N, χ )consisting of those forms f ∈ H ! k ( N, χ ) which have at most polynomial growth at any cuspof Γ ( N ), that is, the Fourier expansion have the shape(12) f ( τ ) = ∞ X n =0 c + f ( n ) q n + c − f (0) v − k + X n< c − f ( n )Γ(1 − k, − πnv ) q n , at the cusp i ∞ and analogous expansions at all other cusps of Γ ( N ). We denote the subspaceof all such forms by H k ( N, χ ) (note that the same notation has been used in [3, Eq. 4.9] fora slightly different subspace) and call them harmonic Maass forms of polynomial growth . Inthis section, our aim is to study the space H k ( N, χ ) and construct examples in this space.
Proposition 1.
Suppose k = 1 . Then we have the following observations.(i) If f ∈ H k ( N, χ ) then f | k ω ( N ) ∈ H k ( N, χ ) .(ii) If f ∈ H k ( N, χ ) then ξ k ( f ) ∈ M − k ( N, χ ) .(iii) Suppose k ≤ . If f ∈ H k ( N, χ ) then D − k ( f ) ∈ M − k ( N, χ ) .(iv) Suppose k ≤ . If f ∈ H k ( N, χ ) with Fourier series expansion (12) then c ± f ( n ) = O (1) , n ∈ Z .(v) For k > , H k ( N, χ ) = M k ( N, χ ) .Proof. By using Lemma 2 and Lemma 3, we have ( i ). We have ξ k ( f ) ∈ M !2 − k ( N, χ ) and D − k ( f ) ∈ M !2 − k ( N, χ ). Since ξ k and D − k commutes with the slash operator, by using (9)and (11) we have ( ii ) and ( iii ). By using ( ii ) , ( iii ) together with (9), (11) and the bound forcoefficients of modular forms, we have ( iv ). Since M − k ( N, χ ) = { } for k >
2, by using (9)we have ( v ). (cid:3) Remark 3. (i) In view of the above proposition, the only interesting case for our purposewill be k ≤ and k = 2 .(ii) Suppose k ≤ . If f ∈ H k ( N, χ ) with Fourier series expansion (12) then the Dirichletseries L ± ( f, s ) attached to f are absolutely convergent in the half-plane Re ( s ) > . .1. Examples.
Now we construct examples in the space H k ( N, χ ) for k negative integer.We begin by recalling the basic facts about Eisenstein series of integral weight. We referthe reader to [6, Chapter 8] for details. Let k be a negative integer . Let ρ be a cusp ofΓ ( N ) and γ ρ ∈ SL ( Z ) such that γ ρ ( i ∞ ) = ρ . Put Γ ρ = Γ ( N ) ∩ γ ρ Γ ∞ γ − ρ where Γ ∞ is thestabiliser of i ∞ into SL ( Z ). Assume that χ is trivial on Γ ρ . We define the Eisenstein seriescorresponding to the cusp ρ as follows. E − k,ρ ( N, χ ; τ ) = X g ∈ Γ ρ \ Γ ( N ) χ ( g ) j (cid:0) γ − ρ g, τ (cid:1) k − , where j ( γ, τ ) = cτ + d for γ = ( a bc d ). Then we have the following result. Theorem 3. (Theorem 8.2.3, [6])
The Eisenstein series E − k,ρ ( N, χ ; τ ) ∈ M − k ( N, χ ) .Moreover, E − k,ρ ( N, χ ; τ ) vanishes at every other cusp of Γ ( N ) except at ρ where it is . We construct the preimage of this Eisenstein series under the shadow operator ξ k . Considerthe following function.(13) F k,ρ ( N, χ ; τ ) = v − k − k X g ∈ Γ ρ \ Γ ( N ) χ ( g ) j (cid:0) γ − ρ g, τ (cid:1) − k | j ( γ − ρ g, τ ) | k − , τ = u + iv ∈ H . This series converges absolutely for any negative integer k . Moreover, we prove the followingtheorem. Theorem 4.
For any negative integer k , the function F k,ρ ( N, χ ; τ ) ∈ H ! k ( N, χ ) which has atmost polynomial growth at any cusp of Γ ( N ) . In particular, F k,ρ ( N, χ ; τ ) ∈ H k ( N, χ ) withthe shadow ξ k ( F k,ρ ( N, χ ; τ )) = E − k,ρ ( N, χ ; τ ) .Proof. Since the series considered in (13) is absolutely convergent, it is routine to check thatthe function F k,ρ ( N, χ ; τ ) satisfies modularity with character χ for the group Γ ( N ). A quickcalculation shows that ξ k (cid:16) χ ( g ) v − k j (cid:0) γ − ρ g, τ (cid:1) − k | j ( γ − ρ g, τ ) | k − (cid:17) = χ ( g )(1 − k ) j ( γ − ρ g, τ ) k − . Since the series P g ∈ Γ ρ \ Γ ( N ) j ( γ − ρ g, τ ) k − is uniformly convergent on any compact subset of H , we have ξ k ( F k,ρ ( N, χ ; τ )) = 11 − k X g ∈ G ρ \ Γ ( N ) ξ k (cid:16) χ ( g ) v − k j (cid:0) γ − ρ g, τ (cid:1) − k | j ( γ − ρ g, τ ) | k − (cid:17) . Thus we see that ξ k ( F k,ρ ( N, χ ; τ )) = E − k,ρ ( N, χ ; τ ). Next, since ξ k ( F k,ρ ( N, χ ; τ )) is holo-morphic, it follows that ∆ k ( F k,ρ ( N, χ ; τ )) = ξ − k ( ξ k ( F k,ρ ( N, χ ; τ ))) = 0. Finally, we verifythe cusp conditions. Let ν be any cusp of Γ ( N ) with γ ν ( i ∞ ) = ν . We have F k,ρ ( N, χ ; τ ) | k γ ν = v − k − k X g ∈ Γ ν \ Γ ( N ) χ ( g ) j ( γ − ρ gγ ν , τ ) − k | j ( γ − ρ gγ ν , τ ) | k − . herefore we have |F k,ρ ( N, χ ; τ ) | k γ ν ( τ ) | ≤ v − k − k X g ∈ Γ ρ \ Γ ( N ) | j ( γ − ρ gγ ν , τ ) | k − = v − k − k X h = ( a bc d ) ∈ γ − ρ Γ ρ γ ρ \ γ − ρ Γ ( N ) γ ν | cτ + d | k − . In the last line we used the fact that the correspondence g γ − ρ gγ ν is a bijection betweenΓ ρ \ Γ ( N ) and γ − ρ Γ ρ γ ρ \ γ − ρ Γ ( N ) γ ν . We now split the sum in two sums, one with c = 0 andthe other one with c = 0. By definition γ − ρ Γ ρ γ ρ = Γ ∞ ∩ γ − ρ Γ ( N ) γ ρ . We have |F k,ρ ( N, χ ; , τ ) | k γ ν ( τ ) | ≤ v − k − k X ( a bc d ) ∈ γ − ρ Γ ρ γ ρ \ γ − ρ Γ ( N ) γ ν c =0 X ( a bc d ) ∈ γ − ρ Γ ρ γ ρ \ γ − ρ Γ ( N ) γ ν c =0 | cτ + d | k − Now, if ν is not equivalent to ρ modulo Γ ( N ) then gγ ν ( i ∞ ) = γ ρ ( i ∞ ) for any g ∈ Γ ( N ),which implies that γ − ρ gγ ν / ∈ Γ ∞ , that is, Γ ∞ ∩ γ − ρ Γ ( N ) γ ν = ∅ and hence the first termin the above sum is zero. On the other hand if ν is Γ ( N )-equivalent to ρ then we choose γ ν = γ ρ and hence the first sum will have only one term equal to 1. Thus we have |F k,ρ ( N, χ ; τ ) | k γ ν ( τ ) | ≤ v − k − k δ ρ,ν + X ( a bc d ) ∈ γ − ρ Γ ρ γ ρ \ γ − ρ Γ ( N ) γ ν c =0 | cτ + d | k − , where δ ρ,ν is equal to 1 if ν is Γ ( N )-equivalent to ρ and 0 otherwise. Now by using thebound given in [21, Theorem 5.1.1], we get a constant C such that |F k,ρ ( N, χ ; τ ) | k γ ν ( τ ) | ≤ v − k − k h δ ρ,ν + C ( | τ | k − + | τ | k − ) i . This implies that F k,ρ ( N, χ ; τ ) | γ ν ( τ ) = O ( δ ρ,ν v − k + v − k/ ) for v → ∞ . This implies that F k,ρ ( N, χ ; τ ) ∈ H ! k ( N, χ ) which has at most polynomial growth at any cuspof Γ ( N ). Also we have already seen that the shadow of F k,ρ ( N, χ ; τ ) is E − k,ρ ( N, χ ; τ ). Thiscompletes the proof. (cid:3) Remark 4.
The harmonic Maass forms constructed in [11] whose shadows are more generalEisenstein series than ours at the cusp i ∞ are also examples of the harmonic Maass formsof polynomial growth. In fact, our example corresponding to the cusp ρ = i ∞ is a specialcase of the examples constructed in [11] . Note that the approach adopted in Theorem 4 canbe easily generalized to any congruence subgroup with a multiplier system whereas we restrictourselves to remain in the context of the paper. Let us denote the C -span of the set of all Eisenstein series corresponding to all cusps ofΓ ( N ) by E − k ( N, χ ). We have the following result. roposition 2. (Proposition 8.5.15, [6]) For any negative integer k , the dimension of thespace E − k ( N, χ ) is equal to e ( N, χ ) = X C | N gcd ( C,N/C ) | N/m χ ϕ (gcd( C, N/C )) where m χ is the conductor of χ and ϕ denotes the Euler’s phi function. If χ is trivial thenthe dimension of E − k ( N, χ ) is equal to the number of Γ ( N ) -inequivalent cusps. Denote by E k ( N, χ ) ⊆ H k ( N, χ ), the C -span of the set of functions F k,ρ ( N, χ ; τ ) as ρ varies over the Γ ( N )-inequivalent cusps. We have the following corollary. Corollary 1.
The restriction of the shadow operator ξ k to E k ( N, χ ) is an isomorphism from E k ( N, χ ) to E − k ( N, χ ) . Moreover, we havedim E k ( N, χ ) = X C | N gcd ( C,N/C ) | N/m χ φ (gcd( C, N/C )) , where m χ is the conductor of χ .Proof. Indeed, Theorem 4 shows that the above restriction of the shadow map is surjective.By using (9) and Lemma 2, we see that the kernel of the restricted shadow map is M k ( N, χ )(see [3, Eq. 5.12]). For k negative, M k ( N, χ ) = { } . (cid:3) Intermediate results
The aim of this section is to prepare ourselves for the next two sections by establishingsome lemmas and Theorem 5 which can be considered as an analogue of the Hecke’s conversetheorem for the space H k ( N, χ ). Some of these intermediate results might be of independentinterest as well.
Lemma 4.
Let k ≤ . Suppose f ( τ ) = ∞ X n =0 c + f ( n ) q n + c − f (0) v − k + X n< c − f ( n )Γ(1 − k, − πnv ) q n with c ± f ( n ) bounded by O ( | n | σ ) for some σ ≥ . Then the series in the right-hand side isconvergent absolutely and uniformly on any compact subset of H , and defines a real analyticfunction on H . Moreover, we have f ( τ ) = O ( v − σ − ) as v → ,f ( τ ) − c + f (0) − c − f (0) v − k = O (cid:0) e − πv (cid:1) as v → ∞ , uniformly in Re( τ ) .Proof. To prove this lemma we make appropriate changes in the proof of [18, Lemma 4.3.3]as we deal with the real analytic functions in place of holomorphic functions. By usingEuler-Gauss formula [18, Eq. 3.2.9], we have(14) lim n →∞ n σ ( − n (cid:0) − σ − n (cid:1) = Γ( σ + 1) , here (cid:18) − σ − n (cid:19) = ( − σ − − σ − − − σ − − . . . (( − σ − − ( n − n ! . Since c + f ( n ) = O ( n σ ) , n ≥
1, there exists a constant
L > | c + f ( n ) | ≤ L ( − n (cid:18) − σ − n (cid:19) , n ≥ . Since k ≤
0, 1 − k ≥ n ≥ − k, πnv ) = Γ(1 − k ) e − πnv − k X l =0 (4 πnv ) l l != (cid:26) O ( e − πnv ) if v → ,O ( v − k e − πnv ) if v → ∞ . Since c − f ( − n ) = O ( n σ ), by using (14) once again, we have | c − f ( − n )Γ(1 − k, πnv ) | ≤ L ( − n e − πnv − k X l =0 (4 πv ) l l ! (cid:18) − σ − l − n (cid:19) . Thus we have | f ( τ ) | ≤ L ∞ X n =0 ( − n (cid:18) − σ − n (cid:19) e − πnv + O ( v − k ) + L − k X l =0 (4 πv ) l l ! ∞ X n =1 ( − n (cid:18) − σ − l − n (cid:19) e − πnv ≤ L (1 − e − πv ) − σ − + O ( v − k ) + L − k X l =0 (4 πv ) l l ! ((1 − e − πv ) − σ − − . (15)This estimate implies that the given series f ( τ ) is convergent absolutely and uniformly onany compact subset of H . Also, from (9) we have ξ k ( c + f ( n ) q n ) = 0 , ξ k ( c − f (0) v − k ) = ( k − c − f (0) ,ξ k ( c − f ( − n )Γ(1 − k, πnv ) q − n ) = − (4 π ) k − c − f ( − n ) n k − q n , n ≥ . By using the bound c − f ( − n ) = O ( n σ ), we get that the series( k − c − f (0) − (4 π ) k − X n ≥ c − f ( − n ) n k − q n is uniformly convergent on any compact subset of H and defines a holomorphic function.Moreover, by using (8) and (9) we have∆ k ( f ( τ )) = − ξ − k ( ξ k ( f ( τ ))) = 0 . Since solutions of elliptic partial differential equations with real analytic coefficients, such as∆ k ( F ) = 0, are real-analytic (cf. [8, Page 57]), therefore f ( τ ) defines a real-analytic functionon H .Since 1 − k ≥ − e − πv ) − σ − = O ( v − σ − ) as v →
0, by using (15) we have | f ( τ ) | = v − σ − as v → ext, we have f ( τ ) − c + f (0) − c − f (0) v − k = e πiτ g ( τ ) , where g ( τ ) = ∞ X n =0 c + f ( n + 1) q n + X n ≤− c − f ( n + 1)Γ(1 − k, − π ( n + 1) v ) q n . By using the asymptotic formula given in (6), for any n ≤ − | Γ(1 − k, − π ( n + 1) v ) q n | ≤ (4 π ( − n − v ) − k e π ( n +2) v as v → ∞ . Now by following the calculations done to obtain (15), we get that g ( τ ) is bounded in aneighborhood of i ∞ . Thus we have f ( τ ) − c + f (0) − c − f (0) v − k = O ( e − πv ) as v → ∞ . (cid:3) Lemma 5.
Let k ∈ Z . Let f, g be any two real analytic functions on H such that g = f | k ω ( N ) . Let H ( τ ) = 2 iv ∂f∂u ( τ ) + kf ( τ ) , I ( τ ) = 2 iv ∂g∂u ( τ ) + kg ( τ ) , τ = u + iv ∈ H . Then for any t > we have H | k ω ( N )( it ) = N − k/ ( it ) − k H (cid:18) − N ( it ) (cid:19) = − I ( it ) . Proof.
We have g ( τ ) = N − k/ τ − k f (cid:18) − N τ (cid:19) . Differentiating both sides with respect to u , we get ∂g∂u ( τ ) = − kN − k/ τ − k τ f (cid:18) − N τ (cid:19) + N − k/ τ − k ∂f∂u (cid:18) − N τ (cid:19) (cid:18) N τ (cid:19) . Multiplying both sides by 2 iv and evaluating at τ = it , we get2 it ∂g∂u ( it ) + kN − k/ i − k t − k f (cid:18) iN t (cid:19) = − kN − k/ i − k t − k f (cid:18) iN t (cid:19) − itN − k/ i − k t − k (cid:18) N t (cid:19) ∂f∂u (cid:18) iN t (cid:19) = ⇒ I ( it ) = − N − k/ i − k t − k (cid:18) i Im (cid:18) iN t (cid:19) ∂f∂u (cid:18) iN t (cid:19) + kf (cid:18) iN t (cid:19)(cid:19) = ⇒ I ( it ) = − N − k/ i − k t − k H (cid:18) iN t (cid:19) . (cid:3) emma 6. For any ν > , the function W ν ( s ) ( defined by (2)) is an analytic function inthe half-plane Re( s ) > and | W ν ( s ) | ≤ Γ( ν )Γ( Re ( s )) . Moreover, for any x, σ > we have (16) Γ( ν, x ) e − x = 12 πi Z σ + i ∞ σ − i ∞ x − s W ν ( s ) ds. Proof.
Since | Γ( ν, x ) | ≤ Γ( ν )( x ≥ s ∈ C with Re( s ) >
0, we have | W ν ( s ) | ≤ Γ( ν ) Z ∞ x Re( s ) − e − x dx = Γ( ν )Γ(Re( s )) . Therefore W ν ( s ) defines an analytic function in the half-pane Re( s ) >
0. Since W ν ( s ) is theMellin transform of f ν ( x ) = Γ( ν, x ) e − x , to prove (16) it is sufficient to check the hypothesisof Mellin inversion theorem (see [6, Proposition 3.1.22]) for f ν ( x ). Indeed this is the case as f ν ( x ) is a continuous function. (cid:3) Lemma 7. If ν is a positive integer then for any µ > , we have W ν ( s ) = O (Im( s ) − µ ) onany line Re ( s ) = σ, σ > , as | Im( s ) | → ∞ .Proof. Suppose ν is a positive integer. From [1, § ν, x ) = Γ( ν ) e − x ν − X l =0 x l l ! . Therefore we have W ν ( s ) = Γ( ν ) ν − X l =0 l l ! Z ∞ e − x x s + l dxx = Γ( ν ) ν − X l =0 l l ! 13 s + l Γ( s + l ) . Now by using Stirling’s estimate for the gamma function [18, Eq. (3.2.8)], the lemma follows. (cid:3)
Theorem 5.
Let k be a negative integer and N be a positive integer. Let f and g be twofunctions defined on H given by the formal Fourier series (1) with c ± f ( n ) , c ± g ( n ) bounded by O ( | n | α ) , n ∈ Z , for some α ≥ . Then the following two statement are equivalent.(1) g = f | k ω ( N ) .(2) The completed Dirichlet series Λ N ( f, s ) , Λ N ( g, s ) , Ω N ( f, s ) , Ω N ( g, s ) satisfy condition(2) (a) of Theorem 1.Proof. (1) = ⇒ (2) . We have c − f ( − n ) Z ∞ Γ(1 − k, πnt/ √ N ) e − πnt/ √ N t s dtt = (cid:18) πn √ N (cid:19) − s c − f ( − n ) Z ∞ Γ(1 − k, x ) e − x x s dxx = (cid:18) π √ N (cid:19) − s W − k ( s ) c − f ( − n ) n s , nd c + f ( n ) Z ∞ e − πnt/ √ N t s dtt = (cid:18) π √ N (cid:19) − s Γ( s ) c + f ( n ) n s . Now for Re( s ) > α + 1, we have Z ∞ (cid:18) f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k (cid:19) t s dtt = Z ∞ ∞ X n =1 c + f ( n ) e − πnt/ √ N ! t s dtt + Z ∞ ∞ X n =1 c − f ( − n )Γ(1 − k, πnt/ √ N ) e − πnt/ √ N ! t s dtt = (cid:18) π √ N (cid:19) − s (cid:2) Γ( s ) L + ( f, s ) + W − k ( s ) L − ( f, s ) (cid:3) = Λ N ( f, s ) . Therefore for Re( s ) > α + 1, we haveΛ N ( f, s ) = Z ∞ (cid:18) f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k (cid:19) t s dtt = Z + Z ∞ f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k ! t s dtt = Z ∞ f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k ! t s dtt + Z ∞ f (cid:18) i √ N t (cid:19) t − s dtt − c + f (0) s − c − f (0) N − k s − k + 1= Z ∞ f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k ! t s dtt + i k Z ∞ g (cid:18) it √ N (cid:19) t k − s dtt − c + f (0) s − c − f (0) N − k s − k + 1 . For Re( s ) > k , we have Z ∞ c + g (0) t k − s − dt = − c + g (0) k − s and Z ∞ c − g (0) N − k t − k t k − s − dt = − c − g (0) N − k − s . herefore for Re( s ) > max( k, α + 1), we haveΛ ∗ N ( f, s ) = Λ N ( f, s ) + c + f (0) s + c + g (0) i k k − s + c − f (0) N − k s − k + 1 + c − g (0) i k N − k − s = Z ∞ f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k ! t s − dt + i k Z ∞ (cid:18) g (cid:18) it √ N (cid:19) − c − g (0) − c − g (0) N − k t − k (cid:19) t k − s − dt (18)Since g | k ω ( N ) = ( − k f , we haveΛ ∗ N ( g, s ) = Λ N ( g, s ) + c + g (0) s + c + f (0) i − k k − s + c − g (0) N − k s − k + 1 + c − f (0) i − k N − k − s = Z ∞ (cid:18) g (cid:18) it √ N (cid:19) − c + g (0) − c − g (0) N − k t − k (cid:19) t s − dt + i − k Z ∞ f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k ! t k − s − dt (19)Now by using Lemma 4 we see that the integral representations of Λ ∗ N ( f, s ) and Λ ∗ N ( g, s )define analytic functions and are bounded in any vertical strip in the whole complex plane.Replacing s with k − s in the integral expression (19) of Λ N ( g, s ) and then comparing with(18) we obtain Λ N ( f, s ) = i k Λ N ( g, k − s ) . Next, we have ∂f∂u ( τ ) = 2 πi ∞ X n =1 nc + f ( n ) q n + X n< nc − f ( n )Γ(1 − k, − πnv ) q n ! ∂g∂u ( τ ) = 2 πi ∞ X n =1 nc + g ( n ) q n + X n< nc − g ( n )Γ(1 − k, − πnv ) q n ! . For any n ≥
1, we have2 πinc − f ( − n ) Z ∞ Γ(1 − k, πnt/ √ N ) e − πnt/ √ N t s +1 dtt = 2 πinc − f ( − n ) (cid:18) πn √ N (cid:19) − s − Z ∞ Γ(1 − k, x ) e − x t s +1 dtt = W − k ( s + 1) i √ N (cid:18) π √ N (cid:19) − s c − f ( − n ) n s . πinc + f ( n ) Z ∞ e − πnt/ √ N t s +1 dtt = Γ( s + 1) i √ N (cid:18) π √ N (cid:19) − s c + f ( n ) n s . o we get Z ∞ H (cid:18) it √ N (cid:19) − kc + f (0) − k c − f (0) N − k t − k ! t s dtt = 2 i √ N Z ∞ ∂f∂u (cid:18) it √ N (cid:19) t s +1 dtt + k Z ∞ f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k ! t s dtt = − N ( f, s ) + k Λ N ( f, s ) = Ω N ( f, s ) . By proceeding similar to the case of Λ N ( f, s ) , Λ( g, s ) and by using Lemma 5, we haveΩ ∗ N ( f, s ) = Ω N ( f, s ) + k c + f (0) s − c + g (0) i k k − s + c − f (0) N − k s − k + 1 − c − g (0) i k N − k − s ! = Z ∞ H (cid:18) it √ N (cid:19) − kc + f (0) − k c − f (0) N − k t − k ! t s − dt − i k Z ∞ (cid:18) I (cid:18) it √ N (cid:19) − kc + g (0) − k c − g (0) N − k t − k (cid:19) t k − s − dt. Ω ∗ N ( g, s ) = Ω N ( g, s ) + k c + g (0) s − c + f (0) i − k k − s + c − g (0) N − k s − k + 1 − c − f (0) i − k N − k − s ! = Z ∞ (cid:18) I (cid:18) it √ N (cid:19) − kc + g (0) − k c − g (0) N − k t − k (cid:19) t s − dt − i − k Z ∞ H (cid:18) it √ N (cid:19) − kc + f (0) − k c − f (0) N − k t − k ! t k − s − dt. Again by using Lemma 4 we see that the right hand side integral representations of Ω ∗ N ( f, s )and Ω ∗ N ( g, s ) are analytic and bounded in any vertical strip in the whole complex plane.By replacing s with k − s in the integral expression of Ω N ( g, s ) and then comparing withΩ N ( f, s ), we have Ω N ( f, s ) = − i k Ω N ( g, k − s ) . (2) = ⇒ (1) . By using inverse Mellin transform of Γ( s ) and Lemma 6, for any t > > α + 1, we have12 πi β + i ∞ Z β − i ∞ t − s Λ N ( f, s ) ds = 12 πi β + i ∞ Z β − i ∞ (cid:18) πt √ N (cid:19) − s Γ( s ) L + ( f, s ) ds + β + i ∞ Z β − i ∞ (cid:18) πt √ N (cid:19) − s W − k ( s ) L − ( f, s ) ds = ∞ X n =1 πi β + i ∞ Z β − i ∞ c + f ( n ) (cid:18) πnt √ N (cid:19) − s Γ( s ) ds + 12 πi β + i ∞ Z β − i ∞ c − f ( − n ) (cid:18) πnt √ N (cid:19) − s W − k ( s ) ds = ∞ X n =1 h c + f ( n ) + c − f ( − n )Γ(1 − k, πnt/ √ N ) i e − πnt/ √ N = f (cid:18) it √ N (cid:19) − c + f (0) − c − f (0) N − k t − k . Since L + ( f, s ) and L − ( f, s ) is bounded on Re( s ) = β , by using Stirling’s estimate for Γ( s )and Lemma 7 for W − k ( s ), for any µ > | Λ N ( f, s ) | = O ( | Im( s ) | − µ ) , as | Im( s ) | → ∞ , on Re( s ) = β . Now choose any real number β such that k − β > α + 1. Then by usingthe functional equation for Λ N ( f, s ) and by following the arguments similar to above forΛ N ( g, s ), for any µ > | Λ( f, s ) | = | Λ N ( g, k − s ) | = O ( | Im( s ) | − µ ) , as | Im( s ) | → ∞ , on Re( s ) = β . Since by assumption the function Λ ∗ N ( f, s ) = Λ N ( f, s ) + c + f (0) s + c + g (0) i k k − s + c − f (0) N − k s − k +1 + c − g (0) i k N − k − s is holomorphic and bounded on the vertical strip β ≤ Re( s ) ≤ β ,by applying Phragm´en–Lindel¨of theorem to Λ ∗ N ( f, s ), we get | Λ ∗ N ( f, s ) | = | Λ N ( f, s ) + c + f (0) s + i k c + g (0) k − s + c − f (0) N − k s − k + 1 + i k c − g (0) N − k − s | = O ( | Im( s ) | − µ )for any 0 < µ ≤ | Im( s ) | → ∞ , on the domain β ≤ Re( s ) ≤ β . From this, we concludethat Λ( f, s ) = O ( | Im( s ) | − µ ), for 0 < µ ≤ | Im( s ) | → ∞ . Without loss of generality weassume that β < k −
1. The function t − s Λ( f, s ) has possibly simple poles at s = 0 , k, , k − − c + f (0) , i k c + g (0) t − k , i k c − g (0) N − k t − , − c − f (0) N − k t − k +1 respectively. Now consider arectangular contour with vertical sides Re( s ) = β , β and horizontal sides Im( s ) = T, − T .The integral of the function t − s Λ N ( f, s ) along the horizontal lines | Im( s ) | = T will go to 0as T → ∞ . This will allow us to change the integration path from Re( s ) = β to Re( s ) = β to get f (cid:18) it √ N (cid:19) = 12 πi Z β + i ∞ β − i ∞ Λ N ( f, s ) t − s ds + i k c + g (0) t − k + i k c − g (0) N − k t − . y using functional equation of Λ N ( f, s ), we get f (cid:18) it √ N (cid:19) = 12 πi Z β + i ∞ β − i ∞ i k Λ N ( g, k − s ) t − s ds + i k c + g (0) t − k + i k c − g (0) N − k t − = i k t − k πi Z k − β + i ∞ k − β −∞ Λ N ( g, s ) t s ds + i k c + g (0) t − k + i k c − g (0) N − k t − = i k t − k g (cid:18) i √ N t (cid:19) . Since the above equality is true for any t >
0, by replacing t by 1 / √ N t we get(20) f | k ω ( N )( it ) = g ( it ) . By following the calculations done in the beginning, for any t > β > α + 1 we have12 πi β + i ∞ Z β − i ∞ t − s Ξ N ( f, s ) ds = 12 πi " β + i ∞ Z β − i ∞ (cid:18) πt √ N (cid:19) − s Γ( s + 1) L + ( f, s ) ds − β + i ∞ Z β − i ∞ (cid:18) πt √ N (cid:19) − s W − k ( s + 1) L − ( f, s ) ds = ∞ X n =1 " πi β + i ∞ Z β − i ∞ c + f ( n ) (cid:18) πnt √ N (cid:19) (cid:18) πnt √ N (cid:19) − s − Γ( s + 1) ds − πi β + i ∞ Z β − i ∞ c − f ( − n ) (cid:18) πnt √ N (cid:19) (cid:18) πnt √ N (cid:19) − s − W − k ( s + 1) ds = ti √ N ∞ X n =1 πin h c + f ( n ) − c − f ( − n )Γ( k − , πnt/ √ N ) i e − πnt/ √ N = ti √ N ∂f∂u (cid:18) it √ N (cid:19) . Therefore we have12 πi β + i ∞ Z β − i ∞ t − s Ω N ( f, s ) ds = − ti √ N ∂f∂u (cid:18) it √ N (cid:19) + kf (cid:18) it √ N (cid:19) − kc + f (0) − k c − f (0) N − k t − k . By following all the above calculations done in Λ N ( f, s ) case and by using the functionalequation for Ω( f, s ) we get2 iN t ∂f∂u (cid:18) − N ( it ) (cid:19) + kf (cid:18) − N ( it ) (cid:19) = − N k/ i k t k (cid:20) it ∂g∂u ( it ) + kg ( it ) (cid:21) , that is,(21) H | k ω ( N )( it ) = − I ( it ) . ince ∆ k = − ξ − k ξ k , by applying ∆ k on the given Fourier series expansion of f and g weget that ∆ k ( f ) = 0 = ∆ k ( g ). Put G := f | k ω ( N ) − g . Then we have∆ k ( G ) = 0 . From the theory of differential equations, it is well known (cf. [8, Page 57]) that solutions ofelliptic partial differential equations with real analytic coefficients, such as ∆ k ( F ) = 0, arereal-analytic. So G ( τ ) is real analytic and we write G ( τ ) = G ( u + iv ) = ∞ X n =0 G n ( v ) u n . Since G ( τ ) satisfies ∆ k ( F ) = 0, we have − v (( n + 2)( n + 1) G n +2 ( v ) + G ′′ n ( v )) + ikv (( n + 1) G n +1 ( v ) + iG ′ n ( v )) = 0 , for n ≥ . Therefore we get the following recursive formula for G n .(22) G n +2 ( v ) = ikv ( n + 1) G n +1 ( v ) − kvG ′ n ( v ) − v G ′′ n ( v ) v ( n + 1)( n + 2) , for n ≥ . Also, we have G ( t ) = G ( it ) = N − k/ i − k t − k f (cid:18) − N it (cid:19) − g ( it ) ,G ( t ) = ∂G∂u ( it ) = ∂∂u (cid:18) N − k/ i k τ − k f (cid:18) iN τ (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) τ = it − ∂g∂u ( it ) . By using (20) we get G ( t ) = 0. Next, by following the steps in the proof of Lemma 5 andby using (21) we get G ( t ) = 0. Therefore by using (22) we have G n ( v ) = 0 for all n ≥ G ( τ ) = 0, which implies that f | k ω ( N )( τ ) = g ( τ ). (cid:3) Proof of Theorem 1
Throughout this section, let k be a negative integer and N be a positive integer. Let f and g be two functions defined on H given by the formal Fourier series (1) with c ± f ( n ) , c ± g ( n )bounded by O ( | n | α ) , n ∈ Z , for some α ≥
0. Let ψ be a non-trivial primitive Dirichletcharacter with conductor m ψ ( >
1) and f ψ be the twisted Fourier series (3) of f by ψ . Wehave the following twisted Dirichlet series attached to f . L + ( f, s, ψ ) = L + ( f ψ , s ) = ∞ X n =1 ψ ( n ) c + f ( n ) n s , L − ( f, s, ψ ) = L − ( f ψ , s ) = ∞ X n =1 ψ ( n ) c − f ( − n ) n s . Λ N ( f, s, ψ ) = Λ N ( f ψ , s ) = m ψ √ N π ! s (cid:2) Γ( s ) L + ( f, s, ψ ) + W − k ( s ) L − ( f, s, ψ ) (cid:3) . Ξ N ( f, s, ψ ) = Ξ n ( f ψ , s ) = m ψ √ N π ! s (cid:2) Γ( s + 1) L + ( f, s, ψ ) − W − k ( s + 1) L − ( f, s, ψ ) (cid:3) . Ω N ( f, s, ψ ) = Ω N ( f ψ , s ) = − N ( f, s, ψ ) + k Λ N ( f, s, ψ ) . Similarly, we have the twisted Fourier series g ψ and we attach the Dirichlet series Λ N ( g, s, ψ ),Ξ N ( g, s, ψ ) and Ω N ( g, s, ψ ) to g . roposition 3. Let all the notations be as above. Then the following two statements areequivalent.(1) f ψ | k ω ( N m ψ ) = C ψ g ¯ ψ . (2) Λ N ( f, s, ψ ) , Λ N ( g, s, ψ ) and Ω N ( f, s, ψ ) , Ω N ( g, s, ψ ) can be analytically continued tothe whole s-plane, bounded on any vertical strip, and satisfies the following functionalequations. Λ N ( f, s, ψ ) = i k C ψ Λ N ( g, k − s, ¯ ψ ) , Ω N ( f, s, ψ ) = − i k C ψ Ω N ( g, k − s, ¯ ψ ) , for a constant C ψ .Proof. This proposition follows as an application of Theorem 5 by putting f = f ψ , g = C ψ g ψ and N = N m ψ in the theorem. (cid:3) Lemma 8.
Let f and ψ be as above. For any r ∈ R , let T r = (cid:18) r (cid:19) . Then we have f ψ = τ ( ψ ) − m ψ X u =1 ψ ( u ) (cid:0) f | T u/m ψ (cid:1) , where τ ( ψ ) = P ma =1 ψ ( a ) e πia/m is the Gauss sum of ψ .Proof. Put m = m ψ . We have (cid:0) f | T u/m (cid:1) ( τ ) = ∞ X n =0 e πinu/m c + f ( n ) q n + c − f (0) v − k + X n< e πinu/m c − f ( n )Γ(1 − k, − πnv ) q n . From [18, Lemma 3.1.1 (1)] we have m X a =1 ψ ( a ) e πian/m = ψ ( n ) τ ( ψ ) , for any integer n . By using this fact we get m X u =1 ψ ( u ) (cid:0) f | T u/m (cid:1) ( τ ) = ∞ X n =0 m X u =1 ψ ( u ) e πinu/m ! c + f ( n ) q n + c − f (0) v − k m X u =1 ψ ( u )+ X n< m X u =1 ψ ( u ) e πinu/m ! e πinu/m c − f ( n )Γ(1 − k, − πnv ) q n = τ ( ψ ) " ∞ X n =0 ψ ( n ) c + f ( n ) q n + X n< ψ ( n ) c − f ( n )Γ(1 − k, − πnv ) q n = τ ( ψ ) f ψ ( τ ) . (cid:3) .1. Proof of Theorem 1: (1) = ⇒ (2) .Proposition 4. Let χ be a Dirichlet character modulo N with conductor m χ and f ∈ H k ( N, χ ) . Let ψ be a primitive Dirichlet character of conductor m ψ ( > as above and M = lcm( N, m ψ , m ψ m χ ) . Then f ψ ∈ H k ( M, χψ ) .Proof. Put m = m ψ . By using Lemma 2, we have ∆ k ( f | k T u/m ) = ∆ k ( f ) | k T u/m = 0 andtherefore Lemma 8 will give us ∆ k ( f ψ ) = 0. In view of Remark 2, we only need to prove therequired transformation property for f ψ . Let γ = (cid:18) a bcM d (cid:19) ∈ Γ ( M ) . Then T u/m γT − d u/m = (cid:18) a ′ b ′ c ′ d ′ (cid:19) ∈ Γ ( M ) ⊆ Γ ( N ) , for any u = 1 , . . . m . We have d ′ = d − cd uM/m ≡ d (mod m χ ). Therefore we get f | k T u/m γ = χ ( d ) f | k T d u/m . Now by using Lemma 8 and the fact that gcd( d, m ) = 1, we get f ψ | k γ = χ ( d ) τ ( ψ ) − m X u =1 ψ ( u ) f | k T d u/m = χ ( d ) ψ ( d ) f ψ . (cid:3) Proposition 5.
Let f and ψ be as in Proposition 4. Let g = f | k ω N . If gcd( m ψ , N ) = 1 then f ψ | k ω ( N m ψ ) = C ψ g ¯ ψ , where C ψ = C N,χ,ψ = χ ( m ψ ) ψ ( − N ) τ ( ψ ) /τ ( ψ ) = χ ( m ψ ) ψ ( N ) τ ( ψ ) /m ψ . (23) Proof.
Put m = m ψ . Let u be an integer such that gcd( m, N u ) = 1. Choose n, v ∈ Z suchthat nm − N uv = 1. Then from [18, Eq. 4.3.22], we have T u/m ω (cid:0) N m (cid:1) = m · ω ( N ) (cid:18) m − v − uN n (cid:19) T v/m . By using Proposition 1 ( i ), we have f | k T u/m ω (cid:0) N m (cid:1) = χ ( m ) g | k T v/m . hus by using Lemma 8 we have τ ( ψ ) f ψ | k ω ( N m ) = m X u =1 ψ ( u ) f | k T u/m ω (cid:0) N m (cid:1) = χ ( m ) m X v =1 ψ ( − N v ) g | k T v/m = χ ( m ) ψ ( − N ) m X v =1 ψ ( v ) g | k T v/m = χ ( m ) ψ ( − N ) τ ( ψ ) g ψ . (cid:3) Theorem 6.
Let f and ψ be as in Proposition 4 and let g = f | k ω ( N ) . Suppose gcd( m ψ , N ) =1 . Then each one of the Dirichlet series Λ N ( f, s, ψ ) , Λ N ( g, s, ψ ) and Ω N ( f, s, ψ ) , Ω N ( g, s, ψ ) can be analytically continued to the whole s-plane, is bounded on any vertical strip, andsatisfies the following functional equation. Λ N ( f, s, ψ ) = i k C ψ Λ N ( g, k − s, ¯ ψ ) , Ω N ( f, s, ψ ) = − i k C ψ Ω N ( g, k − s, ¯ ψ ) , where C ψ is the constant in (23) . Moreover, we have the direct part of Theorem 1.Proof. By using Proposition 5, we have f ψ | ω (cid:0) N m (cid:1) = C ψ g ¯ ψ . The claimed analytic properties now follows from Proposition 3. Combining this with thedirect part of Theorem 5 we have (1) = ⇒ (2) in Theorem 1. (cid:3) Proof of Theorem 1: (2) = ⇒ (1) . Suppose f and g are two formal Fourier seriessatisfying the condition 2 of Theorem 1. From Lemma 4, we get that the Fourier series f and g are absolutely and uniformly convergent on compact subsets of H and define real analyticfunctions on H . As in the proof of the converse part of Theorem 5, by applying ∆ k on thegiven Fourier series expansions of f and g we get that ∆ k ( f ) = 0 = ∆ k ( g ). By using theconverse part of Theorem 5 and Proposition 3, we have f | k ω ( N ) = g, f ψ | k ω ( N m ψ ) = C ψ g ¯ ψ , for any primitive Dirichlet character ψ with conductor m ψ ∈ P . Therefore to establish theconverse part of Theorem 1, we need to verify the required transformation properties withrespect to Γ ( N )-matrices and the cusp conditions. First we prove Proposition 6 which willensure the required cusp conditions by using Lemma 4 if we have the modularity. Lemma 9.
Let f : H → C be a smooth function given by the Fourier series f ( τ ) = X n ≪−∞ c + f ( n ) q n + κt + c − f (0) v − k q κ/t + X n ≫∞ n =0 c − f ( n )Γ(1 − k, − πnv/t ) q n + κt , or some positive real numbers κ, t , converging uniformly on any compact subset of H . Thenfor any τ = u + iv ∈ H , we have t Z τ + tτ f ( τ ) e − πi ( n + κ ) τ/t dτ = ( c + f ( n ) + c − f ( n )Γ(1 − k, − πnv /t ) if n = 0 ,c + f (0) + c − f (0) v − k if n = 0 . Proof.
Let m = 0. Then we have1 t Z τ + tτ f ( τ ) e − πi ( m + κ ) τ/t dτ = ∞ X n = −∞ c + f ( n ) e π ( m − n ) v /t t Z u + tu e − πi ( m − n ) u/t du + c − f (0) v − k e πmv /t t Z u + tu e − πimu/t du + ∞ X n = −∞ n =0 c − f ( n )Γ(1 − k, − πnv /t ) e π ( m − n ) v /t t Z u + tu e − πi ( m − n ) u/t du = c + f ( m ) + c − f ( m )Γ(1 − k, − πmv /t ) . The case m = 0 follows similarly. (cid:3) Proposition 6.
Let f : H → C be a smooth function. Assume that f satisfies the first andsecond conditions of Definition 2. If f ( τ ) = O ( v − σ ) as v → for some σ ≥ uniformly in Re( z ) then f has at most polynomial growth at every cusp of Γ ( N ) .Proof. Since the cusp i ∞ is Γ ( N )-equivalent to a real cusp, it is enough to establish thelemma for real cusps. Let ρ ∈ Q be a cusp of Γ ( N ) with width t and parameter κ . Let α = ( a bc d ) ∈ SL ( Z ) , c = 0, such that ρ = α ( i ∞ ). Then by following the proof of Lemma 1,we have( f | k α )( τ ) = ∞ X n = −∞ c + f ( n ) q n + κt + c − f (0) v − k q κ/t + ∞ X n = −∞ n =0 c − f ( n )Γ(1 − k, − πnv/t ) q n + κt which converges uniformly on any compact subset of H . Note that κ ∈ [0 , τ = u + iv ∈ H we have(24) 1 t Z τ + tτ ( f | k α )( τ ) e − πi ( n + κ ) τ/t dτ = ( c + f ( n ) + c − f ( n )Γ(1 − k, − πnv /t ) if n = 0 ,c + f (0) + c − f (0) v − k if n = 0 . Since c = 0, We have Im( ατ ) = v | cτ + d | → v → ∞ uniformly on | Re( τ ) | ≤ t/
2. Therefore we have( f | k α )( τ ) = ( cτ + d ) − k f ( ατ ) = O ( | cτ + d | − k Im( ατ ) − σ ) = O ( v σ − k ) as v → ∞ uniformly on | Re( τ ) | ≤ t/
2. By taking τ = iy − t/
2, we see that the left hand side of (24)is O ( y σ − k e π ( n + κ ) y/t ) as y → ∞ . Therefore for any n ∈ Z , n = 0,(24) gives us(25) lim y →∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c + f ( n ) + c − f ( n )Γ(1 − k, − πny/t ) y σ − k e π ( n + κ ) y/t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ∞ . y using the asymptotic relation (6) for Γ(1 − k, − πny/t ), we haveΓ(1 − k, − πny/t ) ∼ ( − πny/t ) − k e πny/t as y → ∞ . As κ ∈ [0 , n < ∞ as y → ∞ unless c + f ( n ) = 0. We argue similarly in the case of n > c − f ( n ) = 0. (cid:3) We now proceed to establish the modularity of f . Let us begin by introducing someterminology. For two integers m, v with gcd( m, N v ) = 1, take n, u ∈ Z such that mn − N uv =1. Put γ ( m, v ) = (cid:18) m − v − N u n (cid:19) ∈ Γ ( N ) . Though the choice of γ ( m, v ) is not unique but for any such matrix, we have T u/m ω (cid:0) N m (cid:1) = m · ω ( N ) γ ( m, v ) T v/m , as in the proof of Proposition 5. Moreover, u mod m is uniquely determined. Lemma 10.
Let m be an odd prime number or 4 coprime to N . Suppose that f and g satisfy f ψ | k ω ( N m ) = C ψ g ¯ ψ for all primitive Dirichlet characters ψ mod m with the constant C ψ = χ ( m ) ψ ( − N ) τ ( ψ ) /τ ( ψ ) , then g | k ( χ ( m ) − γ ( m, u )) T u/m = g | k ( χ ( m ) − γ ( m, v )) T v/m for any two integers u and v coprime to m .Proof. The proof is about handling the slash operators and therefore it will be exactly onthe lines of [18, Proof of Lemma 4.3.13]. (cid:3)
Lemma 11.
Let m and n be odd prime numbers or 4. Assume that both m and n arecoprime to N . Then the matrix β ( m, n, u, v ) := (cid:18) − v/m uN/n /mn − (cid:19) is an elliptic matrix whose eigenvalues are not roots of unity. Moreover, if f and g satisfy f ψ | k ω ( N m ) = C ψ g ¯ ψ for all primitive Dirichlet characters ψ whose conductor m ψ = m or n, with the constant C ψ = χ ( m ) ψ ( − N ) τ ( ψ ) /τ ( ψ ) , then (cid:18) g (cid:12)(cid:12)(cid:12)(cid:12) k (cid:18) m − v − N u n (cid:19) − χ ( m ) g (cid:19) | k (1 − β ( m, n, u, v )) = 0 . Proof.
We get this lemma by following the first half of the proof of [18, Lemma 4.3.14] andby using Lemma 10 appropriately. (cid:3)
Here we record the following theorem of Neururer and Oliver (cf. [20, Theorem 3.11])which will be useful in proving our converse theorem.
Theorem 7. If h is a continuous function on H that is invariant under two infinite orderelliptic matrices with distinct fixed points in H then it is a constant. ow we prove the following theorem which is the main step in the proof of the conversepart. Theorem 8.
Let m, n, f, g, c ψ be same as in Lemma 11. Then we have g | k γ = χ ( n ) g for all γ ∈ Γ ( N ) of the form γ = (cid:18) m − v − N u n (cid:19) . Proof.
Put h = g (cid:12)(cid:12)(cid:12)(cid:12) k (cid:18) m − v − N u n (cid:19) − χ ( m ) g. Then by using Lemma 11 we get that h is invariant under the elliptic matrix β ( m, n, u, v ) = (cid:18) − v/m uN/n /mn − (cid:19) . Since β ( m, n, u, v ) has eigenvalues which are not roots of unity, it is an infinite order matrix.The fixed point of β ( m, n, u, v ) in H is given by z = i s − n (cid:0) mn − (cid:1) N u + 4 mvmN u − n (cid:0) mn + 1 (cid:1) N u .
Let m ′ = m − rN u be a prime for some non-zero integer r such that m ′ is coprime to N anddistinct from m and n . Put v ′ = v − rn . Then we have g (cid:12)(cid:12)(cid:12)(cid:12) k (cid:18) m ′ − v ′ − N u n (cid:19) = g (cid:12)(cid:12)(cid:12)(cid:12) k (cid:18) r (cid:19) (cid:18) m − v − N u n (cid:19) = g (cid:12)(cid:12)(cid:12)(cid:12) k (cid:18) m − v − N u n (cid:19) . Since χ ( m ′ ) = χ ( m ), by using Lemma 11 we see that h ′ = g (cid:12)(cid:12)(cid:12)(cid:12) k (cid:18) m ′ − v ′ − N u n (cid:19) − χ ( m ′ ) g = h is invariant under the elliptic matrix β ( m ′ , n, u, v ′ ) = (cid:18) − v ′ /m ′ uN/n /m ′ n − (cid:19) . The fixed point of β ( m ′ , n, u, v ′ ) in H is given by z = i s − n (cid:0) m ′ n − (cid:1) N u + 4 mv ′ m ′ N u − n (cid:0) m ′ n + 1 (cid:1) N u .
By comparing the real parts we see that z = z . Thus by Theorem 7 we get that h is aconstant. Put h ( τ ) = c , a constant in C . Let τ be the unique fixed point of β ( m, n, u, v ) in H and put ρ := 1 τ − ¯ τ (cid:18) − τ − ¯ τ (cid:19) ∈ GL(2 , C ) ,p ( w ) := (cid:0) h | k ρ − (cid:1) ( w ) = (1 − w ) − k h (cid:0) ρ − w (cid:1) = (1 − w ) − k c, w ∈ { z ∈ C : | z | < } . We write ρβ ( m, n, u, v ) ρ − = (cid:18) ζ ζ − (cid:19) , here ζ is an eigenvalue of β ( m, n, u, v ). Since h is invariant under β ( m, n, u, v ), we have p ( ζ w ) = ζ − k p ( w ) . In particular, for w = 0 we have c = ζ − k c which implies that c = 0 as ζ is not a root ofunity. We conclude the proof by observing that χ ( m ) = ¯ χ ( n ). (cid:3) Finally, we deduce the modularity of f by proceeding exactly similar to the proof of themodular property in [18, Theorem 4.3.15] and by making use of Theorem 8 appropriately. References [1] G. B. Arfken and H. Weber, Mathematical methods for physicists (english summary), Fifth edition,Harcourt/Academic Press, Burlington, MA (2001).[2] K. Bringmann and K. Ono,
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Department of Mathematics, Indian Institute of Science Educa-tion and Research Bhopal, Bhopal Bypass Road, Bhauri, Bhopal 462 066, Madhya Pradesh,India
Email address : [email protected], [email protected] (Ranveer Kumar Singh) Department of Physics, Indian Institute of Science Education andResearch Bhopal, Bhopal Bypass Road, Bhauri, Bhopal 462 066, Madhya Pradesh, India
Email address : [email protected], [email protected]@iiserb.ac.in, [email protected]