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formerly Central European Journal of Mathematics

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Volume 16, Issue 1

Issues

Volume 13 (2015)

On the algebraicity of coefficients of half-integral weight mock modular forms

SoYoung Choi
  • Department of Mathematics Education and RINS, Gyeongsang National University, 501 Jinjudae-ro, Jinju, 52828, South Korea
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/ Chang Heon Kim
Published Online: 2018-11-15 | DOI: https://doi.org/10.1515/math-2018-0112

Abstract

Extending works of Ono and Boylan to the half-integral weight case, we relate the algebraicity of Fourier coefficients of half-integral weight mock modular forms to the vanishing of Fourier coefficients of their shadows.

Keywords: Weakly holomorphic modular forms

MSC 2010: 11F11; 11F67; 11F37

1 Introduction and statement of results

Let k be an integer greater than 1 and let N be a positive integer. The space of cusp forms of weight 2k for Γ0(N) is denoted by S2k(N). Throughout this paper let p =1 or a prime number. For κZ+12 we denote by Mκ!(Γ0(4p)) the space of weakly holomorphic modular forms of weight κ on Γ0(4p) . As usual, Mκ0 (4p)) (resp. Sκ(Γ0(4p))) stands for the space of weight κ modular forms (resp. cusp forms) on Γ0(4p) . Let Hκ0(4p)) be the space of weight κ harmonic weak Maass forms on Γ0(4p) . Let Mκ!(p) (resp. ℍ2−κ(p) ) denote the subspace of Mκ!(Γ0(4p)) (resp. H2κ(Γ0(4p))), in which each form satisfies Kohnen’s plus space condition, that is, its Fourier expansion is supported only on those n∈ ℤ for which (1)κ12n(mod4p). Let κ=k+12 and 𝕄κ(p) (resp. 𝕊κ(p) ) denote the subspace of Mκ0 (4p)) (resp. Sκ0 (4 p)) ), in which each form satisfies Kohnen’s plus space condition.

Let Δ(τ) ∈ S 12 (1) be the Ramanujan’s Delta function. The famous Lehmer’s conjecture states that the Fourier coefficients of Δ(τ) never vanish. Concerning this conjecture, Ono [1] related the algebraicity of Fourier coefficients of weight −10 mock modular form whose shadow is Δ(τ) to the vanishing of Fourier coefficients of Δ(τ) . Generalizing Ono’s results, Boylan [2] related the algebraicity of Fourier coefficients of weight 2 − 2k mock modular forms to the vanishing of Fourier coefficients of their shadows when dim S2k (1) =1 . In this paper we will extend their works to the half-integral weight case. In the following we recall some known facts.

Fact 1

By Shimura correspondence [3, Proposition 1] we have

dimSk+12(1)=dimS2k(1),

which implies that

dimSk+12(1)=1k=6,8,9,10,11,131k=5,7,8,9,10,12.

Now we assume that k ∈ {6,8,9,10,11,13} . For κ > 2 there is an antilinear differential operator ξ2κ:H2κ(Γ0(4p))Sκ(Γ0(4p)) defined by

ξ2κ(f)(τ):=2iy2κfτ¯¯.

Fact 2

For k ∈ {6,8,9,10,11,13}, it follows from [4, Theorem 1.1-(iii) and Lemma 4.2-(c)] that

ξ32k:H32k(1)Sk+12(1)issurjective.

For any κZ+12, the Duke-Jenkins basis [5] for Mκ!:=Mκ!(1) is constructed as follows. Let 2κ − 1 = 12κ + k′ with uniquely determined κ ∈ ℤ and k′∈ {0, 4,6,8,10,14}. If Aκ denotes the maximal order of a non-zero fMκ! at i∞ , then by the Shimura correspondence [3] one has

Aκ=2κ(1)κ1/2ifκis odd,2κotherwise.(1)

A basis for Mκ! then consists of functions of the form

fκ,m(τ)=qm+n>Aκaκ(m,n)qn,(2)

where mAκ satisfies (1)κ3/2m0,1(mod4). Using (1) and (2) we deduce the following facts.

Fact 3

For κ=32k with k ∈ {6,8,9,10,11,13}, the maximal order Aκ of a non-zero fMκ! at iis given by Aκ = − 4 . Thus for each m ≥ 4 satisfying (−1)km ≡ 0,1(mod 4) , there exist unique modular forms f32k,m(τ)M32k! with Fourier development

f32k,m(τ)=qm+n3a32k(m,n)qn,

which form a basis for the space M32k!.

Fact 4

For κ=k+12 with k ∈ {6,8,9,10,11,13}, the space 𝕊κ is spanned by

fk:=fκ,α

where α is given by

α=αk=Aκ=Ak+12=1, k even, 3, k odd, 

and fk has the form qα+O(q4).

Fact 5

Let κ=k+12 and f(z)Hκ(Γ0(4p)) with Fourier expansion f(τ)=nZc(y;n)e2πinx where τ = x + iy . For each prime l with gcd(l, 4 p) =1, the l2 -th Hecke operator is defined by

f|κT(l2)(τ)=nZc(y/l2;nl2)+(1)knllk1c(y;n)+l2k1c(l2y;n/l2))e2πinx.

Then for each ℳ∈ ℍ2−κ (p) , we obtain from [6, (2.6)] or [7, (7.2)] that for κ > 2,

ξ2κ(M)|κT(l2)=l2κ2ξ2κ(M|2κT(l2)).

As a corollary of Fact 5, one has that if f(z)=n1(1)kn0,1(4)af(n)qnSk+12, then

f|k+12T(l2)=n1(1)kn0,1(4)af(l2n)+(1)knllk1af(n)+l2k1af(n/l2)qnSk+12.

(or see [3, Theorem 1-(i)].)

Let (V,Q) be a non-degenerate rational quadratic space of signature (b+ ,b) and L an even lattice with dual L′ . Denote the standard basis elements of the group algebra ℂ[L′ / L] by 𝔢γ for γL′ / L . Let Mp 2 (ℤ) denote the integral metaplectic group, which consists of pairs (γ ,ϕ) , where γ = (abcd) ∈ SL2(ℤ) and ϕ:HC is a holomorphic function with ϕ(τ)2 = + d . It is well known that Mp2(ℤ) is generated by S=0110,T and T=1101,1.Then there is a unitary representation ρL of the group Mp 2(ℤ) on ℂ[L′ / L] , the so-called Weil representation, which is defined by

ρL(T)(eγ):=e(Q(γ))eγ,ρL(S)(eγ):=e((bb+)/8)|L/L|δL/Le((γ,δ))eδ,

where e(z) := e2πiz and (X,Y):=Q(X+Y)Q(X)Q(Y) is the associated bilinear form. One has the relations

S2=ST3=ZZ=1001,i

from which we note that

ρL(Z)eγ=ibb+eγ.(3)

We write < ⋅,⋅ > for the standard scalar product on ℂ[L′ / L] , i.e.

<γL/Lλγeγ,γL/Lμγeγ>=γL/Lλγμγ¯.

For γ, δL′ / L and (M,ϕ )∈ Mp 2 (ℤ) the coefficient ργδ(M, ϕ) of the representation ρL is defined by

ργδ(M,ϕ)=<ρL(M,ϕ)eδ,eγ>.

Following [9], for an integer r we denote by Hr+1/2,ρL(resp.Mr+1/2,ρL)!, the space of ℂ[L′ / L] -valued harmonic weak Maass forms (resp. weakly holomorphic modular forms) of weight r +1/ 2 and type ρL .

Let Lr be the lattice 2 pℤ of signature (1, 0) (resp. (0,1) ) when r is even (resp. odd) equipped with the quadratic form Qr(x)=(1)rx2/4p. Then its dual lattice Lr is equal to ℤ . For a vector valued modular form F=γFγeγ, we define a map Φ by

Φ(F)(τ):=γFγ(4pτ).(4)

It then follows from [8, Theorem 1] that the map Φ defines an isomorphism from Hr+1/2,ρLr to Hr+1/2(p) since ρ¯Lr=ρLr+1.

For a ℂ[L′ / L] -valued function f and (M,ϕ) ∈ Mp2 (ℤ) we define the Petersson slash operator by

(f|r+1/2L(M,ϕ))(τ)=ϕ(τ)2r1ρL(M,ϕ)1f(Mτ).

Let L := Lk and Q := Qk . Following [9] we define the vector valued cuspidal Poincaré series Pβ,nL(τ) as follows: for each β ∈ ℤ / 2 pℤ and n ∈ ℤ + Q(β) with n > 0 ,

Pβ,nL(τ):=12(M,ϕ)Γ~Mp2(Z)eβ(nτ)|κ(M,ϕ).

Then we know from [9] that Pβ,nL(τ) belongs to the space Sκ,ρL. Let Dk denote the set of all integers D such that (−1)k D > 0 and D is congruent to a square modulo 4 p .

Theorem 1.1

([4, Theorem 1.1]). For an integer k > 2 we let κ=k+12 and L := Lk . For each DDk, we define

PD+:=Φ(Pβ,|D|4pL),

where β is an integer such that Dβ2 (mod 4 p) . Then the following assertions are true.

  1. (i) PD+Sκ(p) and the defintion of PD+ does not depend on the choice of β .

  2. (ii) For each f=n1af(n)qnSκ(p) , we have

(f,ck,DPD+)=af(|D|)

where (⋅,⋅) denotes the Petersson inner product and

ck,D=(4π|D|)κ1Γ(κ1)s(D)3, if p = 1 (4π|D|)κ1Γ(κ1)s(D)4, if p = 2 (4π|D|)κ1Γ(κ1)s(D)6, if p > 2

with s(D)=1, if p|D,2, otherwise. 

(iii) The set {PD+DDk} spans the space 𝕊κ(p) . Moreover, if we let t := dim 𝕊κ( p) and { f1 , f2 , , ft }be a basis for 𝕊κ(p) satisfying fi=q|Di|+O(q|Di|+1) for some DiDk(i=1,,t) and 0 <| D1 |<| D2 |< <| Dt | , then the set

{PD1+,PD2+,,PDt+}

forms a basis for 𝕊κ(p) .

(iv) Let I be a nonempty finite subset of ℕ . Then the following two conditions are equivalent.

  1. (a) iIαiPDi+(τ)0 for some αiC and DiDk .

  2. (b) There exists gM2κ!(p) with the principal part iIαi¯|Di|1κq|Di|.

Remark 1.2

Let p =1 and take

Dk=1, if k is even3, if k is odd.

Then in Theorem 1.1 one can choose β =1 and

PDk+:=Φ(P1,|Dk|4pLk).

We let Γ~:=<T>. We define for s∈ ℂ and y ∈ ℝ −{0}:

Ms(y)=y(2κ)/2M(2κ)/2,s1/2(y)(y>0),Ws(y)=|y|(2κ)/2W2κ2sgn(y),s1/2(|y|)

where Mν,μ(z) and Wν,μ(z) denote the usual Whittaker functions. Now we take κ = k +1/ 2 > 2 , L := L1−k , and Q := Q1−k . For each β ∈ ℤ / 2 pℤ and m∈ ℤ +Q(β) with m < 0 , modifying the Poincaré series in [9, $(1.35)$] we define the vector valued Maass Poincaré series Fβ,mL of index (β ,m) by

Fβ,mL(τ,s):=12Γ(2s)(M,ϕ)Γ~Mp2(Z)[Ms(4π|m|y)eβ(mx)]|2κL(M,ϕ)

where τ=x+iyH and s = σ + it∈ ℂ with σ >1. Indeed, since ℳs(4π | m | y) 𝔢 β (mx) is invariant under slash operator |2−κ T , the Maass Poincaré series is well defined. This series has desirable properties as follows. As in Section 1.3 in [9] it converges normally for τ ∈ H and s = σ + itC with σ >1 and hence defines a Mp 2 (ℤ) -invariant function on H under the slash operator |2−κ . Moreover, Fβ,mL(τ,s) is an eigenfunction of Δ2−κ with an eigenvalue s(1− s) +κ(κ − 2) / 4 . Since eβ(τ)|2κZ=eβ by (3), the invariance of Fβ,mL under the action of Z implies Fβ,mL=Fβ,mL.

Let κ=k+12 and L = L1−k with k an integer > 2 . For each β ∈ ℤ / 2 pℤ and m ∈ ℤ + Q(β) with m < 0 , we obtain from [4, Corollary 1.5] that Fβ,mLτ,κ2 belongs to the space H2κ,ρL.

Let

Q=Q(k;z):=ΦF1,α4L1kτ,κ2=Q++Q

where Q+ = Q+ (k; z) is the holomorphic part of Q(k; z) and Q = Q (k; z) is the nonholomorphic part of Q(k; z) . Let Q(k; z) have the Fourier development as follows:

Q(k;z)=2qα+cQ+(0)+n1(1)k1n0,1(4)cQ+(n)qn+n1(1)kn0,1(4)cQ(n)Γ(κ1,4πny)qn.

Now we are ready to state our main results.

Theorem 1.3

With the same notations as above the following assertions are true.

(1) Let

fk|k+12T(l2)=λk(l2)fk

for some λk(l2)C. Then one has

λk(l2)=afk(l2α)+(1)kαllk1.

(2) We have

Q|32kT(l2)l12kλk(l2)Q=Q+|32kT(l2)l12kλk(l2)Q+M32k!.

Theorem 1.4

For an odd prime l , the following assertions are true.

(1) We have

cQ+(l2βk)Z[cQ+(βk)]l2k1Q(cQ+(βk)).

(2) Assume that cQ+(βk) is irrational. Then

afk(l2αk)=lk1(1)k1βkl(1)kαkl if and only if cQ+(l2βk)Q.

(3) Assume that βkl=αkl and cQ+(βk) is irrational. Then

afk(l2αk)=0 if and only if cQ+(l2βk)Q.

Remark 1.5

For simplicity, we dealt with the case p =1 in our main results. But we remark that they can be extended to higher level cases whenever dim dimSk+12(p)=1.

2 Proof of Theorem 1.3

First we are in need of two lemmas and one more fact.

Lemma 2.1

([4, Lemma 4.1]). Let κ=k+12 for an integer k > 2 and let DDk.

Then the following assertions are true.

(a) For each GH2κ,ρL1k, we have

(4p)κ1Φξ2κ(G)=ξ2κΦ(G).

(b) For each f=n1cf(n)qnSκ(p),

(f,(4p)κ1Φξ2κ(Fβ,|D|4pL1k(τ,κ2)))=3s(D)cf(|D|), if p = 1 4s(D)cf(|D|), if p = 2 6s(D)cf(|D|), if p > 2.

Lemma 2.2

([4, Lemma 4.2]). With the same notations as in Lemma 2.1, we have the following assertions.

  1. (a) For a vector valued function h=hβ(τ)eβ one has

ξ2κ(h|2κL1k(M,ϕ))=(ξ2κ(h))|κLk(M,ϕ).

  1. (b) ξ2κ(Γ(κ1,4πny))=(4πn)κ1e4πny.

  2. (c) Let m=|D|4p. Then one has

ξ2κ(Fβ,mL1k(τ,κ2))=(4π|m|)κ1Γ(κ1)Pβ,|m|Lk(τ).

Fact 6

Let p =1 and κ=k+12.

  1. (1) It follows from Lemmas 2.1 and 2.2 that

Φξ2κF1,|Dk|4L1kτ,κ2=Φ(π|Dk|)κ1Γ(κ1)P1,|Dk|4Lk(τ)=(π|Dk|)κ1Γ(κ1)PDk+(τ)Sk+12.

  1. (2) We obtain from Theorem 1.1-(ii) that

(fk,ck,DkPDk+)=afk(|Dk|)=1

where ck,Dk=(4π|Dk|)κ13Γ(κ1)R.

For k ∈ {6,8,9,10,11,13}, it follows from Fact 3, Fact 4, and Theorem 1.1-(iii), (iv) that PDk+ does not vanish and

PDk+=ckfk

for some ck ∈ ℂ× . Thus one has from Fact 6 (2) that

1=(fk,ck,Dkckfk)=ck¯ck,Dk||fk||2,

which implies

ck=ck,Dk1||fk||2.

We compute that

ξ2κ(Q(k;z))=ξ2κΦF1,α4L1kτ,κ2=4κ1Φξ2κF1,α4L1kτ,κ2 by Lemma 2.1-(a) =(4π|Dk|)κ1Γ(κ1)PDk+(τ) by Fact 6 (1) =(4π|Dk|)κ1Γ(κ1)ck,Dk1||fk||2fk=3||fk||2fk.(5)

Since

fk|k+12T(l2)=n1(1)kn0,1(4)afk(l2n)+(1)knllk1afk(n)+l2k1afk(n/l2)qn=λk(l2)fk,

one has

λk(l2)=afk(l2α)+(1)kαllk1,

which proves the first assertion. Hence for all n ≥1 with (−1)k n ≡ 0,1(mod 4)

afk(l2α)+(1)kαllk1afk(n)=afk(l2n)+(1)knllk1afk(n)+l2k1afk(n/l2).(6)

It follows from (5) that

3||fk||2fk(z)=ξ2κ(Q(k;z))=nα(1)kn0,1(4)(4πn)κ1cQ(n)¯qn,

which implies that

cQ(n)=3||fk||2(4πn)1κafk(n).(7)

Now we put dk:=3||fk||2(4π)1κ. We obtain that for all positive integers n with (−1)k n ≡ 0,1(mod 4) ,

n1κcQ(nl2)(nl2)κ1+l2k1cQ(n/l2)(n/l2)κ1+(1)knllk1cQ(n)nκ1=n1κdkafk(nl2)+afk(n/l2)l2k1+(1)knllk1afk(n) by (7) =n1κdkafk(l2α)+(1)kαllk1afk(n) by (6) =λk(l2)cQ(n).

Thus we have

Q|T32k(l2)=nZcQ(nl2)+(1)knllkcQ(n)+l12kcQ(n/l2)Γ(κ1,4πny)qn=l12knZcQ(nl2)l2k1+(1)knllk1cQ(n)+cQ(n/l2)Γ(κ1,4πny)qn=l12knZdkn1κcQ(nl2)(nl2)κ1dk1+(1)knllk1nκ1cQ(n)dk1+nκ1cQ(n/l2)dk1Γ(κ1,4πny)qn=l12knZdkn1κafk(nl2)+(1)knllk1afk(n)+l2k1afk(n/l2)Γ(κ1,4πny)qn=l12knZλk(l2)cQ(n)Γ(κ1,4πny)qn since fk|Tk+12(l2)=λk(l2)fk=l12kλk(l2)Q.(8)

We obtain that

l2κ2ξ2κQ|2κTl2=ξ2κQ|κTl2byFact5=2fk2fk|κTl2by5=3fk2λkl2fk=ξ2κλkl2Qsinceλkl2R.

Indeed, we observe that

λk(l2)=afk(l2α)+(1)kαllk1Z.

Thus we have

l2κ2Q|2κT(l2)λk(l2)QM2κ!,

which combined with (8) yields the second assertion.

3 Proof of Theorem 1.4

We observe that

Q|32kT(l2)l12kλk(l2)Q=Q+|32kT(l2)l12kλk(l2)Q+=2qα+cQ+(0)+n1(1)k1n0,1(4)cQ+(n)qn|32kT(l2)l12kλk(l2)2qα+cQ+(0)+n1(1)k1n0,1(4)cQ+(n)qn=2l12kqαl2+2(1)kαllkl12kλk(l2)qα+cQ+(0)(1+l12kl12kλk(l2))+n1(1)k1n0,1(4)cQ+(l2n)+(1)k1nllkcQ+(n)+l12kcQ+(n/l2)l12kλk(l2)cQ+(n)qn=2l12kf32k,αl2sinceα3.

So we find that

2f32k,αl2=l2k1Q+|32kT(l2)λk(l2)Q+

has integral coefficients and for all positive integers n with (−1)k−1n ≡ 0,1(mod 4) ,

l2k1cQ+(l2n)+(1)k1nllk1cQ+(n)+cQ+(n/l2)λk(l2)cQ+(n)=cQ+(n)(1)k1nllk1(1)kαllk1afk(l2α)+l2k1cQ+(l2n)+cQ+(n/l2)2Z.

Then for n = βk with

βk=3, k even, 1, k odd, 

we obtain that

cQ+(βk)(1)k1βkl(1)kαllk1afk(l2α)+l2k1cQ+(l2βk)2Z.

As a consequence of the above identity we get the assertions.

Acknowledgement

We would like to thank KIAS (Korea Institute for Advanced Study) for its hospitality.

Choi was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea goverment (Ministry of Education) (No. 2017R1D1A1A09000691).

Kim was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (NRF-2018R1D1A1B07045618 and 2016R1A5A1008055).

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About the article

Received: 2018-06-11

Accepted: 2018-10-05

Published Online: 2018-11-15


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1335–1343, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0112.

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© 2018 Choi and Kim, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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