# Resonance between the Representation Function and Exponential Functions over Arithemetic Progression

Authors: Li Ma; Xiaofei Yan
Journal: Journal of Mathematics
 (2021)

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## Abstract

Letrn denote the number of representations of a positive integer n as a sum of two squares, i.e., n=x12+x22, where x1 and x2 are integers. We study the behavior of the exponential sum twisted by rn over the arithmetic progressions ∑n∼Xn≡lmodqrneαnβ, where 0≠α∈ℝ, 0<β<1, ex=e2πix, and n∼X means X<n≤2X. Here, X>1 is a large parameter, 1≤l≤q are integers, and l,q=1. We obtain the upper bounds in different situations.


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## Body

## 1. Introduction

In analytic number theory, the problems concerning nonlinear exponential twisting arithmetic functions arise naturally in investigating equidistribution theory, zero-distribution ofL-functions, and so on. Let an be some arithmetic number-theoretic function. We usually consider the general nonlinear exponential sum of the form(1)SX,α=∑n∼Xaneαnβ,0≠α∈ℝ,0≤β≤1.Here,n∼X means X≤n≤2X, and ez=e2πiz. When β=1/2 and an=Λn is the von Mangoldt function, the sum SX,α was studied by Vinogradov [1]. For an=Λn and an=μn (μ is the Möbius function), the sums SX,α were studied by Iwaniec, Luo, and Sarnak, and they showed these sums are intimately related to L-functions of GL2. If f is a holomorphic cusp form of even weight on the upper half plane, they also proved that a good upper bound of SX,α implies a quasi-Riemann hypothesis for Ls,f [2]. In addition, studying the behavior of the Fourier coefficients of automorphic forms has great significance in modern number theory. Analytic number theorists always estimate the mean value or the twisted sums such as SX,α mentioned above to obtain some information about the Fourier coefficients (for examples, see [3–10]).Ifβ is a variable and an are the Fourier coefficients of automorphic forms, these sums were studied by Ren and Ye [7] and Sun and Wu [11]. They proved that the resonance phenomenon occurs only when β=1/2 and α is close to 2k,k∈ℤ+. Let rn denote the number of representations of a positive integer n as a sum of two squares, i.e., n=x12+x22, where x1 and x2 are integers. Sun and Wu [11] also studied the case that an=rn and obtained the resonance phenomenon. Yan [12] studied the nonlinear exponential sum twisting the Fourier coefficients of Maass forms over the arithmetic progress and obtained an asymptotic formula for the sum(2)∑n∼Xn≡lmodqλgneαnβ,0≠α∈ℝ,0≤β≤1,where g is a Maass cusp form for SL2,Z and λgn is the n-th Fourier coefficient of g. These analogues to the arithmetic progression are the main motivation of this paper.In this paper, we study the nonlinear exponential sum(3)∑n∼Xn≡lmodqrneαnβ,where 0≠α∈ℝ, 0<β<1. Here, X>1 is a large parameter. 1≤l≤q are integers, and l,q=1. We consider the case that q tends to infinity as X⟶∞ and obtain analogues of the result of Sun and Wu [11].The principal aim of this paper is to prove the following result.Theorem 1.
LetX>1, 0<β<1, and 0≠α∈ℝ. Let l,q∈ℕ and l≤q≤X1/2. Let δ=q,4.(i)
IfαβqXβ<1/4δX, then we have(4)∑n∼Xn≡lmodqrneαnβ≪q1/3+εXαβXβ−1.(ii)
IfαβqXβ≥1/4δX and β≠1/2, then we have(5)∑n∼Xn≡lmodqrneαnβ≪q1/2+ε2β−1−1/2αβXβ1+ε.(iii)
IfαβqXβ≥1/4δX and β=1/2, then for α<1/q or α>X/q, we have(6)∑n∼Xn≡lmodqrneαnβ≪q1/2+εX1/4+εα1/2+ε.For1/q≤α≤X/q, we have(7)∑n∼Xn≡lmodqrneαnβ=1q∑c|qεa,ncc−3/2X3/4rnc,Q∗S−l,−Dc¯nc;cnc−1/4GQc,a+Oq1/2+εX1/4+εα1/2+ε,where(8)εa,nc=δcεαa12∫12u−1/4esgnαα−2δnc2cXudu,andδc=1 or 0 according to if there exists a positive integer nc for c|q satisfying(9)cα−δnc≤X−1/2,or not.(iv)
In particular, ifα=δk/q with 1≤k≤X/δ, then we have(10)∑n∼Xn≡lmodqrne±δknq=22323/4−1δ−1/41±iS−l,−Dc¯k;qq−5/2X3/4rk,Q∗k−1/4GQq,a+Ok−1/4X1/4qXkε.
To prove Theorem1, we shall follow the steps in [7, 11, 12] first. Then, we will use a new Voronoi-type summation formula generalized by Hu et al. [13] to get the asymptotic formula, and this is the key to success. Thus, we can get the Kloosterman sum, use Weil’s bound to get the saving in the q-aspect, and then obtain a similar main term as that in [12].


## 2. Some Lemmas

To prove Theorem1, we need to quote some lemmas. First, we consider the Kloosterman sum, which is defined as(11)Sm,n;c=∑dmodce∗md+nd¯c,where d¯ denotes the inverse of d modulo c. The famous Weil’s bound of the Kloosterman sum is(12)Sm,n;c≤m,n,c1/2c1/2τc,where dc denotes the divisor function.LetJv denote the standard J-Bessel function. Let rn,Q denote the number of representations of n by the quadratic form Q, namely,(13)rn,Q=♯m∈Zl:Qm=n.IfQx=x12+x22+⋯+xl2, we denote rn,Q:=rln. Let A be a symmetric positive definite integral matrix associated to Q, and let D=detA denote the discriminant of A. Let GQc,a=∑xmodceaQx/c, δ=c,D, D=δDc, and Q∗ be a positive definite integral quadratic form, which is defined in terms of the Smith normal form of Q† (see [14]), and Q† the adjoint form of Q.Then, we have the following Voronoi summation formula [13].Lemma 1.
Letc,a=1 and F∈C0∞R+ be a smooth compact function. We have(14)∑n=1∞rn,QeancFn=2πm/2c−mΓm/2DGQc,aΦm2+∑n=1∞rn,Q∗e−aDc¯ncδn1−m/2/2Gn,where(15)Gn=4πDm/4−1cm/2+1GQc,a∫0∞Fxxm/2−1/2Jm/2−14πδnxcDdx,and Φs is the Mellin transform of Fx, which is given by(16)Φs=∫0∞Fxxs−1dx.Remark 1.
In our situation,D=4, but we still want to compute the dependence of D in our proof. If one can obtain the asymptotic formula for general rn,Q, then our result can be applied directly to get the analogues for rn,Q.
For asymptotic expansions of the Bessel functions, we quote the following lemma.Lemma 2.
Forz>0 large, we have(17)J±vz=2πzcosz∓π2v−π4−v2−1/42zsinz∓π2v−π4+Oz−2.
For the mean value ofrn, we have the classical result [16].Lemma 3.
(18)∑n≤xrn=πx+Ox1/3.
We also need the following result [17].Lemma 4.
LetGx and Fx be real functions in a,b with Gx/F′x being monotonic. Suppose that Gx≤M.(a)
IfF′x≥u>0 or F′x≤−u<0, then(19)∫abGxeFxdx≪Mu.(b)
IfF″x≥v>0 or F″x≤−v<0, then(20)∫abGxeFxdx≪Mv.


## 3. Proof of Theorem1

In this section, we will finish the proof of Theorem1. By the formula of the Ramanujan sum(21)∑d|q∑a=1a,d=1deand=0,ifq∤n,q,ifq|n,we get(22)∑n∼Xn≡lmodqrneαnβ=1q∑c|q∑amodce∗−alc∑n∼Xrneanceαnβ,where ∑∗ means the summation is restricted by a,c=1.LetΔ>1, and let 0≤ϕx≤1 be a C∞ function supported on [1, 2], which is identically 1 on 1+Δ−1,2−Δ−1 and satisfies ϕrx≪Δr for r≥0. Using the bound in Lemma 3, we get(23)∑n∼Xrneanceαnβ=∑n=1∞rneancWn+R1N,X,Δ,where(24)Wx=ϕxXeαxβ,(25)R1N,X,Δ≪∑X<n≤X+X/Δrn≪X+XΔ+X+XΔ1/3−X−X1/3≪XΔ1+ε.Applying Lemma1 with m=2 and Fn=Wn, we have(26)∑n=1∞rneancWn=2πc2DGQc,a∫0∞Wxdx+∑n=1∞rn,Q∗e−aDc¯ncGn,where(27)Gn=4πc2DGQc,a∫0∞WxJ04πδnxcDdx.For the first term, changing variablesx=Xt and applying Lemma 4(a), we get(28)∫0∞Wxdx=X∫12ϕteαXtβdt≪XαβXβ−1.By [16], we have GQc,a≪c. Thus, the contribution of the first term to (22) is(29)1q∑c|q∑amodce∗−alc2πc2DGQc,aXαβXβ−1≪1q∑c|q∑amodce∗−alc1cXαβXβ−1≪q−1+εXαβXβ−1.Next, we turn to estimate the contribution from the term involvingGn. Using Lemma 2, we have(30)J0z=1−i2πzeiz+ie−iz−1+i16πz3eiz−ie−iz+Oz−5/2.Takingz=4πδnx/cD, we obtain(31)J04πδxycD=1−ic1/2D1/44πδ1/4xy−1/4e2δxycD+ie−2δxycD−1+ic3/2D3/4128π2δ3/4xy−3/4e2δxycD−ie−2δxycD+Oc5/2xy−5/4.Putting this inGy, we get(32)Gy=1−ic3/2D1/4δ1/4GQc,a∫0∞xy−1/4ϕxXeαxβe2δxycD+ie−2δxycDdx−1+iD1/432πc1/2δ3/4GQc,a∫0∞xy−3/4ϕxXeαxβe2δxycD−ie−2δxycDdx+Oc3/2∫0∞xy−5/4ϕxXdx.Changing variablex=Xt2, we obtain(33)Gy=21−ic3/2D1/4δ1/4GQc,aX3/4y−1/4∫0∞t−1/2ϕt2eαXβt2βe2δXycDt+ie−2δXycDtdt−1+iD1/416πc1/2δ3/4GQc,aX1/4y−3/4∫0∞t−1/2ϕt2eαXβt2βe2δXycDt−ie−2δXycDtdt+Oc3/2∫0∞xy−5/4ϕxXdx=G1y+Oc3/2X−1/4y−5/4,where(34)G1y=a1c−3/2X3/4y−1/4GQc,aP+2δXycD+iP+−2δXycD+a2c−1/2X1/4y−3/4GQc,aP−2δXycD−iP−−2δXycD,with(35)a1=21−iD1/4δ1/4,a2=−1+iD1/416πδ3/4,(36)P±w=∫0∞t±1/2ϕt2eαXβt2β+wtdt.TheO-term contributes(37)1q∑c|q∑n=1∞rn,Q∗S−l,−Dc¯n;cc3/2X−1/4n−5/4≪q1+εX−1/4+ε.The integralP±w defined in (36) was studied by Ren and Ye [7], Sun and Wu [11], and Yan [12]. Here, we follow their steps and choose the parameters with a few differences to get the q-aspect saving.Apply the method given in [7], we obtain(38)P±w≪Δh−1maxαβXβ,wh,h=1,2.We will takew=±2δyX/cD here. Then, w∈1/2αβXβ,4αβXβ implies y∈I, where(39)I=116δ−1c2Dαβ2X2β−1,4δ−1c2Dαβ2X2β−1.Thus, fory=n∉I, one obtains by (34) and (38) that(40)G1y≪c−3/2X3/4y−1/4+c−1/2X1/4y−3/4GQc,aR2X,y,where(41)R2X,y≪Δh−1Xy/ch,h=1,2.Thus, the contrition fromG1n with n∉I is(42)1q∑c|q∑amodce∗−alc∑n∉Irn,Q∗e−aDc¯ncG1n≪1q∑c|q∑n=1∞rn,Q∗S−l,−Dc¯n;cc−1/2X3/4n−1/4+c1/2X1/4n−3/4R2X,n.Let(43)Y=Δ2c2X−1.Applying (41) with(44)h=1,ifn≤Y,2,ifn>Y,we get(45)≪1q∑c|q∑n≤Yrn,Q∗S−l,−Dc¯n;cc−1/2X3/4n−1/4+c1/2X1/4n−3/4cX−1/2n−1/2+1q∑c|q∑n>Yrn,Q∗S−l,−Dc¯n;cc−1/2X3/4n−1/4+c1/2X1/4n−3/4c2ΔX−1n−1≕∑1+∑2.By the bound∑n≤xrn=πx+Ox1/3, we obtain(46)∑1=1q∑c|q∑n≤Yrn,Q∗S−l,−Dc¯n;cc1/2X1/4n−3/4+c3/2X−1/4n−5/4≪1q∑c|qdccX1/4∑n≤Yrn,Q∗l,Dc¯n;c1/2n−3/4+1q∑c|qdcc2X−1/4∑n≤Yrn,Q∗l,Dc¯n;c1/2n−5/4≪q1/2+εΔ1/2+ε.Similarly, using the bound∑n=1∞rn,Q∗2/n1+ε≪1, we get(47)∑2=1q∑c|q∑n>Yrn,Q∗S−l,−Dc¯n;cΔc3/2X−1/4n−5/4+c5/2X−3/4n−7/4≪1qΔ∑c|qdcc2+εX−1/4∑n>Yrn,Q∗l,Dc¯n;c1/2n−5/4+1qΔ∑c|qdcc3+εX−3/4∑n>Yrn,Q∗l,Dc¯n;c1/2n−7/4≪1qΔ∑c|qc2+εX−1/4∑n>Yrn,Q∗2n1+ε1/2∑n>Yn−3/21/2+1qΔ∑c|qc3+εX−3/4∑n>Yrn,Q∗2n1+ε1/2∑n>Yn−5/21/2≪1qΔ∑c|qc2+εX−1/4Y−1/4+ε+1qΔ∑c|qc3+εX−3/4Y−3/4+ε≪q1/2+εΔ1/2+ε.Finally, we obtain the contribution of (45) as(48)≪q1/2+εΔ1/2+ε.IfαβqXβ<1/2δX/D, then I∩Z+=∅, by (25), (29), (37), and (48), and choosing(49)Δ=X2/3q−1/3,we get(50)∑n∼Xn≡lmodqrneαnβ≪q1/3+εXαβXβ−1.Next, we will handle the situation thaty=n∈I. By the trivial estimate ∫12t±1/2ϕt2eftdt≪1 and putting it into (22), we get the second term in the right-hand side of (34) which contributes(51)≪1q∑c|q∑n∈Irn,Q∗S−l,−Dc¯n;cc−1/2X1/4n−3/4GQc,a≪q1/2+εαβXβ1/2+ε.Applying (38) to bound the terms with αw>0, we have(52)P+2δXycD+iP+−2δXycD=εα∫0∞t1/2ϕt2ef1tdt+OR2X,y,c,where(53)ϵα=i,α>0,1,α<0,(54)f1t=f1t,y=sgnααXβt2β−2δXycDt.Since the contribution from theO-term is absorbed by (51), we only need to estimate(55)εαa1q∑c|qc−3/2X3/4∑n∈Irn,Q∗S−l,−Dc¯n;cn−1/4GQc,a∫12t1/2ϕt2ef1t,ndt.We distinguish two cases according toβ=1/2 or not.Ifβ≠1/2, we have(56)f1″t=sgnαα2β2β−1Xβt2β−2≫αβ2β−1Xβ,fort∈1,2.By integration by parts and [17], we have(57)∫12t1/2ϕt2ef1t,ndt≪max1≤t≤2∫1tef1udu≪αβ2β−1Xβ−1/2.Hence, (55) is(58)≪1q∑c|qc−3/2X3/4∑n∈Irn,Q∗S−l,−Dc¯n;cn−1/4GQc,aαβ2β−1Xβ−1/2≪q1/2+ε2β−1−1/2αβXβ1+ε.Ifβ=1/2, let(59)Hr=2−rcαX1/2,1≤r≤r0=log2cαX1/2+1,and write(60)AHr=n:HrX−1/2<cα−2δnD≤2HrX−1/2.Then,(61)AHr≤HrcαX−1/2.Moreover, forn∈AHr, we have(62)f1′t=α−2δncDX1/2>Hrc−1.Then, we obtain(63)∫12t1/2ϕt2ef1tdt≪Hr−1c.Hence, in this situation, (55) contributes(64)≪1q∑c|qc−1/2X3/4+ε∑r=1r0Hr−1∑n∈AHrrn,Q∗S−l,−Dc¯n;cn−1/4GQc,a≪1q∑c|qc−1/2X3/4+ε∑r=1r0Hr−1α2c2−1/4+εc3/2+εAHr≪q1/2+εX1/4+εα1/2+ε.It remains to estimate(65)εαa1q∑c|qc−3/2X3/4∑n∈I∩I0rn,Q∗S−l,−Dc¯n;cn−1/4GQc,a∫12t1/2ϕt2ef1t,ndt,where(66)I0=n:cα−2δnD≤X−1/2.Note that(67)n:cα−2δnD≤X−1/2≤cαX−1/2.Ifα≤1/q, then I0=∅, and we have (65) which vanishes. And if α≥X/q, then by trivial estimate, (65) contributes(68)1q∑c|qc−3/2X3/4∑cα−2δn/D≤X−1/2rn,Q∗S−l,−Dc¯n;cn−1/4GQc,a≪1q∑c|qc1/2X3/4I0c2α2−1/4+ε≪qεX1/4+εα1/2+ε.If1/q≤α≤X/q, then there is at most one integer satisfying(69)cα−2δnD≤X−1/2,for every c|q, say n=nc. Hence, (55) becomes(70)εαa1q∑c|qδcc−3/2X3/4rnc,Q∗S−l,−Dc¯nc;cnc−1/4GQc,a∫12t1/2ϕt2ef1t,ncdt=εαa12q∑c|qδcc−3/2X3/4rnc,Q∗S−l,−Dc¯nc;cnc−1/4GQc,a∫12u−1/4ef1u1/2,ncdu+Oc−1/2+εΔ−1X3/4rnc,Q∗nc−1/4=1q∑c|qεa,ncc−3/2X3/4rnc,Q∗S−l,−Dc¯nc;cnc−1/4GQc,a+O1,where(71)εa,nc=δcεαa12∫12u−1/4esgnαα−2δnccDXudu,and δc=1 or 0 according to if there exists such nc or not. Recall(72)a1=21−iD1/4δ1/4,εα=i,α>0,1,α<0.Whenβ=1/2 and 1/q≤α≤X/q and applying (64) and (68), we derive that (55) equals to(73)1q∑c|qεa,ncc−3/2X3/4rnc,Q∗S−l,−Dc¯nc;cnc−1/4GQc,a+Oq1/2+εX1/4+εα1/2+ε.This proves Theorem1(iii).In particular, ifα=2/qδk/D with 1≤k≤DX/4δ, then for c|q, c≠q, we have(74)cα−2δkD=2δkDcq−1≥2qδkD≥2δkDX≥1.Thus,δc=0 for c≠q in this situation.Hence, (73) becomes(75)∑n∼Xn≡lmodqrne±2qδknD=4323/4−1D−1/4δ−1/41±iS−l,−Dc¯k;qq−5/2X3/4rk,Q∗k−1/4GQq,a+Ok−1/4X1/4qXkε.This proves Theorem1(iv).



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