# Some Density Results on Sets of Primes for Hecke Eigenvalues

Authors: Aiyue Zou; Huixue Lao; Shu Luo
Journal: Journal of Mathematics
 (2021)

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## Abstract

Letf and g be two distinct holomorphic cusp forms for SL2ℤ, and we writeλfn and λgn for their corresponding Hecke eigenvalues. Firstly, we study the behavior of the signs of the sequences λfpλfpj for any even positive integer j. Moreover, we obtain the analytic density for the set of primes where the product λfpiλfpj is strictly less than λgpiλgpj. Finally, we investigate the distribution of linear combinations of λfpj and λgpj in a given interval. These results generalize previous ones.


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## Body

## 1. Introduction

LetHk∗ be the set of all normalized Hecke primitive cusp forms of even integral weight k≥2 for the full modular group SL2ℤ denoted by λfn the n-th Hecke eigenvalue of f∈Hk∗. The Hecke eigenvalues of cusp forms have been extensively studied (see, e.g., [1–7]). From the theory of Hecke operators, we know that λfn satisfies the standard Hecke relation as follows: for any integers m,n≥1,(1)λfmλfn=∑d|m,nλfmnd2.In particular,λfn is a real and multiplicative function and(2)λf1=1.Furthermore, it is also known thatλfn satisfies the Ramanujan conjecture [8]:(3)λfn≤dn,where dn is the Dirichlet divisor function.There are many papers that focus on the sign changes of the Hecke eigenvaluesλfn. It is well known that λfnn≥1 changes sign infinitely often. Meher et al. [9] studied the distribution of the signs of λfpj as p varies over the prime numbers. They calculated the natural density of the sets explicitly in terms of the celebrated Sato–Tate conjecture (see Theorem B in [10]). In [11], a joint version of the pair-Sato–Tate conjecture (as outlined in Proposition 2.2 in [12]) gives the result that the set p|λfpjλgpj<0 has natural density 1/2 for any odd positive integer j.In this paper, based on the now-proven Sato–Tate conjecture, we first study the behavior of the signs ofλfpλfpj for any even positive integer j.Theorem 1.
Letf∈Hk∗ be a cusp form. Then, for any even positive integer j, the sets p|λfpλfpj>0 and p|λfpλfpj<0 both have natural density 1/2.In [13], Kowalski et al. first proved that if the signs of λfp and λgp coincide for all primes up to the exceptional set of analytic density at most 1/32, then f=g. Subsequently, Matomäki [14] improved the above result by utilizing linear programming to take full advantage of all the available information.Inspired by [13], Chiriac [15] started to compare Hecke eigenvalues over prime numbers and simultaneously showed that the sets of primes for λfp<λgp and λf2p<λg2p both have analytic density at least 1/16. Notice that the pair-Sato–Tate conjecture yields a stronger result for the former set in [15] with natural density 1/2 in replace of at least 1/16 (see Proposition 2.1 (iii) in [16]). Of course, this result is also valid for the analytic density since the existence of the natural density implies that of the analytic density, and they are equal.Most recently, Lao [17] further studied Chiriac’s questions [15] by considering the Hecke eigenvalues at prime powers. She established that the sets p|λfpj<λgpj with 1≤j≤8 and p|λf2pj<λg2pj with 1≤j≤4 have analytic density at least 1/16j+1/22 and 1/4jj+12, respectively.In this paper, our second aim is to obtain the analytic density for the setp|λfpiλfpj<λgpiλgpj with j≥1 and 0≤i≤j.Theorem 2.
Letf,g∈Hk∗ be two distinct cusp forms. Then, for j≥1 and 0≤i≤j, the set p|λfpiλfpj<λgpiλgpj has analytic density at least 1+i−δ1/4ij+j+δ22, where(4)δ1=0,ifi≠j,1,ifi=j,(5)δ2=i+1,if2∤i+j,0,if2|i+j.Remark 1.
By symmetry, forj≥1 and 0≤i≤j, Theorem 2 implies that the set p|λfpiλfpj>λgpiλgpj has analytic density at least 1+i−δ1/4ij+j+δ22. Thus, for j≥1 and 0≤i≤j, the set p|λfpiλfpj=λgpiλgpj has analytic density at most 1−1+i−δ1/2ij+j+δ22.Remark 2.
One can easily find that fori=0,j=1, or i=j=1, we have 1+i−δ1/4ij+j+δ22=1/16, which are consistent with the numerical values of Chiriac’s results (see Theorems 1 and 1.3 in [15]). By comparison, 1+i−δ1/4ij+j+δ22=1/16j+1/22 with i=0,1≤j≤8, and 1+i−δ1/4ij+j+δ22=1/4jj+12 with 1≤i=j≤4. Hence, Theorem 2 also generalizes previous results of Lao (see Theorems 1 and 2 in [17]).It is natural to ask whether Theorem2 can be refined by the pair-Sato–Tate conjecture. In fact, with the help of this conjecture, some specific cases can be dealt with by calculating the corresponding double integral.Finally, we concern the distribution of linear combinations ofλfpj and λgpj in a specified interval. Chiriac and Jorza (see Proposition 5.9 in [18]) obtained the density bound for the set v|a<c1avπ1+c2avπ2<b in the context of unitary cuspidal representations π that satisfy the Ramanujan conjecture. For holomorphic cusp forms, we establish the following theorem.Theorem 3.
Letf,g∈Hk∗ be two distinct cusp forms. Let a,b,c1,c2∈ℝ with a<b. Then, for j≥1, the set p|a<c1λfpj+c2λgpj<b has analytic density at least(6)j+1c14+c24+δ3c13+c23+6c12c22+a+b2+2abc12+c22+a2b22j+12c1+c22+j+1a+bc1+c2+ab2−c12+c22+ab2j+12c1+c22+j+1a+bc1+c2+ab,where(7)δ3=0,ifjis odd,1,ifjis even.The proofs of Theorems2 and 3 rely on Lemma 4 involving the analytic density of a particular set of primes. Recently, there was a big breakthrough on the automorphy of all symmetric powers for cuspidal Hecke eigenforms (Theorem A in [19]), which implies that the L-function Lsymjf,s is automorphic for j≥1 and f∈Hk∗. Then, with the help of the properties of symmetric power L-functions and their Rankin–Selberg L-functions, we obtain the desired results.


## 2. Preliminaries

### 2.1. Primitive AutomorphicL-Functions

In this section, we will briefly recall some fundamental facts about primitive automorphicL-functions and give the main tools and definitions. For more details to learn automorphic L-functions, refer [17, 20–28].Letf,g∈Hk∗ be two cusp forms. The j-th symmetric power L-function attached to f is defined by(8)Lsymjf,s=∏p∏m=0j1−αfpj−mβfpmps−1,Res>1,where αfp and βfp are two complex numbers with(9)λfp=αfp+βfp,αfp=βfp=αfpβfp=1.One can write it as a Dirichlet series: forRes>1,(10)Lsymjf,s=∑n=1∞λsymjfnns=∏p1+∑v≥1λsymjfpvpvs,where λsymjfn is a real multiplicative function, and(11)λsymjfp=∑m=0jαfpj−2m=λfpj.The Rankin–SelbergL-function attached to symif and symjg is defined by(12)Lsymif×symjg,s=∏p∏m=0i∏m′=0j1−αfpi−2mαgpj−2m′ps−1=∑n=1∞λsymif×symjgnns,Res>1,where λsymif×symjgn is a real multiplicative function, and(13)λsymif×symjgp=∑m=0i∑m′=0jαfpi−2mαgpj−2m′=λsymifpλsymjgp.We make the convention that(14)Lsym0f,s=ζs,Lsym1f,s=Lf,s,Lsym0f×symjf,s=Lsymjf,s.A key ingredient of proving Theorems2 and 3 is the analytic properties of various automorphic L-functions. By a series of deep works [29–36], we learn that for 1≤j≤8, Lsymjf,s is an automorphic L-function. Recently, Newton and Thorne (see Theorem A in [19]) proved the automorphy of the symmetric power lifting symjf for j≥1 and f∈Hk∗. Hence, by standard arguments, we have the following.Lemma 1.
Letf∈Hk∗ be a cusp form and Lsymjf,s be defined as in (8). For j≥1, Lsymjf,s has an analytic continuation as an entire function in the whole complex plane ℂ.Combining Lemma1 with (10), we deduce that for j≥1,(15)∑pλsymjfpps=O1,ass⟶1+.Moreover, based on the automorphy ofsymjf for j≥1 and the work [37–42] on the Rankin–Selberg theory, we have the following.Lemma 2.
Letf,g∈Hk∗ be two cusp forms and Lsymif×symjg,s be defined as in (12). For i,j≥1, Lsymif×symjg,s has an analytic continuation as an entire function in the whole complex plane ℂ (except possibly for simple poles at s=0,1 when symjf≅symjg).Thus, whenf=g, s=1 is a simple pole of Lsymjf×symjg,s. By (12), we have(16)∑pλsymjf×symjgpps=∑p1ps+O1,ass⟶1+.In other cases,Lsymif×symjg,s is an entire function and does not vanish at s=1. Thus,(17)∑pλsymif×symjgpps=O1,ass⟶1+.


### 2.2. Sato–Tate Conjecture

Firstly, let us introduce the definition of natural density.Definition 1.
For a subsetA⊆ℙ which denotes the set of all primes, the natural density of A in ℙ is defined as(18)limx⟶∞#p≤x|p∈A#p≤x|p∈ℙ,provided the limit exists.Secondly, let us define the Sato–Tate measure and state the Sato–Tate conjecture (see Theorem 2.3 in [9]), which will be used to prove Theorem 1.Definition 2.
The Sato–Tate measureμST is the probability measure on 0,π given by 2/πsin2θdθ.For any subintervalI⊆0,π, one has(19)πIx=#p≤xθp∈I.Lemma 3.
Letf∈Hk∗ be a cusp form. The sequence θp is equidistributed in 0,π with respect to the Sato–Tate measure μST. In particular, for any subinterval I⊆0,π, we have(20)limx⟶∞πIxπx=μSTI=2π∫Isin2θdθ.Lemma3 implies that if A is a finite set, then the natural density of the set p|θp∈A is 0.


### 2.3. An Analytic Density Lemma

We also recall the definition of analytic density.Definition 3.
A setℬ of primes is said to have analytic density (or Dirichlet density) δ>0 if and only if,(21)∑p∈ℬ1ps∼δ∑p1ps,ass⟶1+.In order to prove Theorems2 and 3, we need the following lemma, which is inspired by the ideas outlined in Section 3 of [5].Lemma 4.
Lethp be real numbers only determined by prime p and satisfy hp≤B with a bound B that does not depend on p. There exist real constants m and M such that as s⟶1+, the following equalities hold:(22)∑php2ps=m∑p1ps+O1,(23)∑phpps=M∑p1ps+O1.Then, the setℬ=p|hp<0 has analytic density at least m−MB/2B2.Proof.
Ifp∉ℬ, hp≥0, and therefore hp2≤Bhp, whereas if p∈ℬ, then −hp≤B and −Bhp≤B2. We see that(24)∑php2ps=∑p∈ℬhp2ps+∑p∉ℬhp2ps≤B2∑p∈ℬ1ps+B∑phpps−∑p∈ℬhpps+O1≤2B2∑p∈ℬ1ps+B∑phpps+O1,ass⟶1+.
Inserting (23) into (24), we get(25)∑php2ps≤2B2∑p∈ℬ1ps+MB∑p1ps+O1,ass⟶1+.
Comparing (22) with (25) leads to(26)m−MB2B2∑p1ps+O1≤∑p∈ℬ1ps,ass⟶1+,i.e., the set ℬ has analytic density at least m−MB/2B2.


## 2.1. Primitive AutomorphicL-Functions

In this section, we will briefly recall some fundamental facts about primitive automorphicL-functions and give the main tools and definitions. For more details to learn automorphic L-functions, refer [17, 20–28].Letf,g∈Hk∗ be two cusp forms. The j-th symmetric power L-function attached to f is defined by(8)Lsymjf,s=∏p∏m=0j1−αfpj−mβfpmps−1,Res>1,where αfp and βfp are two complex numbers with(9)λfp=αfp+βfp,αfp=βfp=αfpβfp=1.One can write it as a Dirichlet series: forRes>1,(10)Lsymjf,s=∑n=1∞λsymjfnns=∏p1+∑v≥1λsymjfpvpvs,where λsymjfn is a real multiplicative function, and(11)λsymjfp=∑m=0jαfpj−2m=λfpj.The Rankin–SelbergL-function attached to symif and symjg is defined by(12)Lsymif×symjg,s=∏p∏m=0i∏m′=0j1−αfpi−2mαgpj−2m′ps−1=∑n=1∞λsymif×symjgnns,Res>1,where λsymif×symjgn is a real multiplicative function, and(13)λsymif×symjgp=∑m=0i∑m′=0jαfpi−2mαgpj−2m′=λsymifpλsymjgp.We make the convention that(14)Lsym0f,s=ζs,Lsym1f,s=Lf,s,Lsym0f×symjf,s=Lsymjf,s.A key ingredient of proving Theorems2 and 3 is the analytic properties of various automorphic L-functions. By a series of deep works [29–36], we learn that for 1≤j≤8, Lsymjf,s is an automorphic L-function. Recently, Newton and Thorne (see Theorem A in [19]) proved the automorphy of the symmetric power lifting symjf for j≥1 and f∈Hk∗. Hence, by standard arguments, we have the following.Lemma 1.
Letf∈Hk∗ be a cusp form and Lsymjf,s be defined as in (8). For j≥1, Lsymjf,s has an analytic continuation as an entire function in the whole complex plane ℂ.Combining Lemma1 with (10), we deduce that for j≥1,(15)∑pλsymjfpps=O1,ass⟶1+.Moreover, based on the automorphy ofsymjf for j≥1 and the work [37–42] on the Rankin–Selberg theory, we have the following.Lemma 2.
Letf,g∈Hk∗ be two cusp forms and Lsymif×symjg,s be defined as in (12). For i,j≥1, Lsymif×symjg,s has an analytic continuation as an entire function in the whole complex plane ℂ (except possibly for simple poles at s=0,1 when symjf≅symjg).Thus, whenf=g, s=1 is a simple pole of Lsymjf×symjg,s. By (12), we have(16)∑pλsymjf×symjgpps=∑p1ps+O1,ass⟶1+.In other cases,Lsymif×symjg,s is an entire function and does not vanish at s=1. Thus,(17)∑pλsymif×symjgpps=O1,ass⟶1+.


## 2.2. Sato–Tate Conjecture

Firstly, let us introduce the definition of natural density.Definition 1.
For a subsetA⊆ℙ which denotes the set of all primes, the natural density of A in ℙ is defined as(18)limx⟶∞#p≤x|p∈A#p≤x|p∈ℙ,provided the limit exists.Secondly, let us define the Sato–Tate measure and state the Sato–Tate conjecture (see Theorem 2.3 in [9]), which will be used to prove Theorem 1.Definition 2.
The Sato–Tate measureμST is the probability measure on 0,π given by 2/πsin2θdθ.For any subintervalI⊆0,π, one has(19)πIx=#p≤xθp∈I.Lemma 3.
Letf∈Hk∗ be a cusp form. The sequence θp is equidistributed in 0,π with respect to the Sato–Tate measure μST. In particular, for any subinterval I⊆0,π, we have(20)limx⟶∞πIxπx=μSTI=2π∫Isin2θdθ.Lemma3 implies that if A is a finite set, then the natural density of the set p|θp∈A is 0.


## 2.3. An Analytic Density Lemma

We also recall the definition of analytic density.Definition 3.
A setℬ of primes is said to have analytic density (or Dirichlet density) δ>0 if and only if,(21)∑p∈ℬ1ps∼δ∑p1ps,ass⟶1+.In order to prove Theorems2 and 3, we need the following lemma, which is inspired by the ideas outlined in Section 3 of [5].Lemma 4.
Lethp be real numbers only determined by prime p and satisfy hp≤B with a bound B that does not depend on p. There exist real constants m and M such that as s⟶1+, the following equalities hold:(22)∑php2ps=m∑p1ps+O1,(23)∑phpps=M∑p1ps+O1.Then, the setℬ=p|hp<0 has analytic density at least m−MB/2B2.Proof.
Ifp∉ℬ, hp≥0, and therefore hp2≤Bhp, whereas if p∈ℬ, then −hp≤B and −Bhp≤B2. We see that(24)∑php2ps=∑p∈ℬhp2ps+∑p∉ℬhp2ps≤B2∑p∈ℬ1ps+B∑phpps−∑p∈ℬhpps+O1≤2B2∑p∈ℬ1ps+B∑phpps+O1,ass⟶1+.
Inserting (23) into (24), we get(25)∑php2ps≤2B2∑p∈ℬ1ps+MB∑p1ps+O1,ass⟶1+.
Comparing (22) with (25) leads to(26)m−MB2B2∑p1ps+O1≤∑p∈ℬ1ps,ass⟶1+,i.e., the set ℬ has analytic density at least m−MB/2B2.


## 3. Proof of Theorem 1

By (9), for any prime p, we can write,(27)λfp=2cosθp,for some θp∈0,π. And, λfpj is expressible by the following elementary trigonometric formula:(28)λfpj=sinj+1θpsinθp.When the values ofθp are 0 or π, the values of λfpj are j+1 or −1jj+1.Then, we have(29)λfpλfpj=sin2θpsinθp·sinj+1θpsinθp=2cosθp⋅sinj+1θpsinθp,θp∈0,π.Since the setp|θp=0orπ has natural density 0 which only has finitely many primes, we may assume that θp∈0,π. For θp∈0,π, we know that sinθp>0. Hence, the sign of λfpλfpj is the same as the sign of cosθpsinj+1θp.The proof of Theorem1 can be divided into two cases when j≡2mod4 and j≡0mod4.


### 3.1.j≡2mod4

Assume thatj≡2mod4, we get(30)j=4j4+2.Then, we observe that(31)cosθpsinj+1θp>0⟺θp∈∪m=1j/4+12m−2πj+1,2m−1πj+1∪π2,j+2π2j+1∪m=j/4+22j/4+12m−1πj+1,2mπj+1,(32)cosθpsinj+1θp<0⟺θp∈∪m=1j/42m−1πj+1,2mπj+1∪jπ2j+1,π2∪m=j/4+12j/4+12mπj+1,2m+1πj+1.Especially, we rewrite the third term of the right-hand side of (31) and (32) by using (30).(33)∪m=j/4+22j/4+12m−1πj+1,2mπj+1=∪m=1j/44j/4−2m+3πj+1,4j/4−2m+4πj+1=∪m=1j/4j−2m+1πj+1,j−2m+2πj+1,∪m=j/4+12j/4+12mπj+1,2m+1πj+1=∪m=1j/4+14j/4−2m+4πj+1,4j/4−2m+5πj+1=∪m=1j/4+1j−2m+2πj+1,j−2m+3πj+1.Next, we consider the setsA and B consisting of the following forms:(34)A=∪m=1j/4+12m−2πj+1,2m−1πj+1∪π2,j+2π2j+1∪m=1j/4j−2m+1πj+1,j−2m+2πj+1=∪m=1j/4+1Im∪J∪m=1j/4Km,B=∪m=1j/42m−1πj+1,2mπj+1∪jπ2j+1,π2∪m=1j/4+1j−2m+2πj+1,j−2m+3πj+1=∪m=1j/4Km′∪J′∪m=1j/4+1Im′.We will prove the results in Theorem1 by showing that the Sato–Tate measure of the two sets A and B is equal, i.e., μSTA=μSTB. Since the sets A and B are divided into disjoint intervals given in (34), it is sufficient to prove that μSTIm=μSTIm′ for each m with 1≤m≤j/4+1 and μSTKm=μSTKm′ for each m with 1≤m≤j/4 and μSTJ=μSTJ′.Now, we restrict our attention to proveμSTJ=μSTJ′. Using the fact(35)∫sin2xdx=x2−sin2x4+C,we obtain(36)∫Jsin2xdx=π4j+1−14sinj+2πj+1.On the other hand, we have(37)∫J′sin2xdx=π4j+1+14sinjπj+1.We know that(38)sinj+2πj+1=sinπ+πj+1=−sinjπj+1.Thus, the second term on the right-hand side of equations (36) and (37) is same. Therefore, we see that(39)∫Jsin2xdx=∫J′sin2xdx.Let us proveμSTKm=μSTKm′. Applying (35) again yields the following equations:(40)∫Kmsin2xdx=π2j+1−14sin2j−2m+2πj+1−sin2j−2m+1πj+1,∫Km′sin2xdx=π2j+1−14sin4mπj+1−sin22m−1πj+1.It is easy to find that(41)sin2j−2m+2πj+1=sin2π+21−2mπj+1=−sin22m−1πj+1,sin2j−2m+1πj+1=sin2π−4mπj+1=−sin4mπj+1.From the above discussion, we deduce that(42)∫Kmsin2xdx=∫Km′sin2xdx.Similarly, we can have(43)∫Imsin2xdx=∫Im′sin2xdx.


### 3.2.j≡0mod4

Assumej≡0mod4. We see that(44)cosθpsinj+1θp>0⟺θp∈∪m=1j/42m−2πj+1,2m−1πj+1∪jπ2j+1,π2∪m=j/4+1j/22m−1πj+1,2mπj+1,(45)cosθpsinj+1θp<0⟺θp∈∪m=1j/42m−1πj+1,2mπj+1∪π2,j+2π2j+1∪m=j/4+1j/22mπj+1,2m+1πj+1.Obviously,(46)∪m=j/4+1j/22m−1πj+1,2mπj+1=∪m=1j/4j−2m+1πj+1,j−2m+2πj+1,∪m=j/4+1j/22mπj+1,2m+1πj+1=∪m=1j/4j−2m+2πj+1,j−2m+3πj+1.Next, we define the right-hand side of (44) and (45) as(47)C=∪m=1j/42m−2πj+1,2m−1πj+1∪jπ2j+1,π2∪m=1j/4j−2m+1πj+1,j−2m+2πj+1=∪m=1j/4Im∪J′∪m=1j/4Km,D=∪m=1j/42m−1πj+1,2mπj+1∪π2,j+2π2j+1∪m=1j/4j−2m+2πj+1,j−2m+3πj+1=∪m=1j/4Km′∪J∪m=1j/4Im′.The rest of the proof runs as before.


## 3.1.j≡2mod4

Assume thatj≡2mod4, we get(30)j=4j4+2.Then, we observe that(31)cosθpsinj+1θp>0⟺θp∈∪m=1j/4+12m−2πj+1,2m−1πj+1∪π2,j+2π2j+1∪m=j/4+22j/4+12m−1πj+1,2mπj+1,(32)cosθpsinj+1θp<0⟺θp∈∪m=1j/42m−1πj+1,2mπj+1∪jπ2j+1,π2∪m=j/4+12j/4+12mπj+1,2m+1πj+1.Especially, we rewrite the third term of the right-hand side of (31) and (32) by using (30).(33)∪m=j/4+22j/4+12m−1πj+1,2mπj+1=∪m=1j/44j/4−2m+3πj+1,4j/4−2m+4πj+1=∪m=1j/4j−2m+1πj+1,j−2m+2πj+1,∪m=j/4+12j/4+12mπj+1,2m+1πj+1=∪m=1j/4+14j/4−2m+4πj+1,4j/4−2m+5πj+1=∪m=1j/4+1j−2m+2πj+1,j−2m+3πj+1.Next, we consider the setsA and B consisting of the following forms:(34)A=∪m=1j/4+12m−2πj+1,2m−1πj+1∪π2,j+2π2j+1∪m=1j/4j−2m+1πj+1,j−2m+2πj+1=∪m=1j/4+1Im∪J∪m=1j/4Km,B=∪m=1j/42m−1πj+1,2mπj+1∪jπ2j+1,π2∪m=1j/4+1j−2m+2πj+1,j−2m+3πj+1=∪m=1j/4Km′∪J′∪m=1j/4+1Im′.We will prove the results in Theorem1 by showing that the Sato–Tate measure of the two sets A and B is equal, i.e., μSTA=μSTB. Since the sets A and B are divided into disjoint intervals given in (34), it is sufficient to prove that μSTIm=μSTIm′ for each m with 1≤m≤j/4+1 and μSTKm=μSTKm′ for each m with 1≤m≤j/4 and μSTJ=μSTJ′.Now, we restrict our attention to proveμSTJ=μSTJ′. Using the fact(35)∫sin2xdx=x2−sin2x4+C,we obtain(36)∫Jsin2xdx=π4j+1−14sinj+2πj+1.On the other hand, we have(37)∫J′sin2xdx=π4j+1+14sinjπj+1.We know that(38)sinj+2πj+1=sinπ+πj+1=−sinjπj+1.Thus, the second term on the right-hand side of equations (36) and (37) is same. Therefore, we see that(39)∫Jsin2xdx=∫J′sin2xdx.Let us proveμSTKm=μSTKm′. Applying (35) again yields the following equations:(40)∫Kmsin2xdx=π2j+1−14sin2j−2m+2πj+1−sin2j−2m+1πj+1,∫Km′sin2xdx=π2j+1−14sin4mπj+1−sin22m−1πj+1.It is easy to find that(41)sin2j−2m+2πj+1=sin2π+21−2mπj+1=−sin22m−1πj+1,sin2j−2m+1πj+1=sin2π−4mπj+1=−sin4mπj+1.From the above discussion, we deduce that(42)∫Kmsin2xdx=∫Km′sin2xdx.Similarly, we can have(43)∫Imsin2xdx=∫Im′sin2xdx.


## 3.2.j≡0mod4

Assumej≡0mod4. We see that(44)cosθpsinj+1θp>0⟺θp∈∪m=1j/42m−2πj+1,2m−1πj+1∪jπ2j+1,π2∪m=j/4+1j/22m−1πj+1,2mπj+1,(45)cosθpsinj+1θp<0⟺θp∈∪m=1j/42m−1πj+1,2mπj+1∪π2,j+2π2j+1∪m=j/4+1j/22mπj+1,2m+1πj+1.Obviously,(46)∪m=j/4+1j/22m−1πj+1,2mπj+1=∪m=1j/4j−2m+1πj+1,j−2m+2πj+1,∪m=j/4+1j/22mπj+1,2m+1πj+1=∪m=1j/4j−2m+2πj+1,j−2m+3πj+1.Next, we define the right-hand side of (44) and (45) as(47)C=∪m=1j/42m−2πj+1,2m−1πj+1∪jπ2j+1,π2∪m=1j/4j−2m+1πj+1,j−2m+2πj+1=∪m=1j/4Im∪J′∪m=1j/4Km,D=∪m=1j/42m−1πj+1,2mπj+1∪π2,j+2π2j+1∪m=1j/4j−2m+2πj+1,j−2m+3πj+1=∪m=1j/4Km′∪J∪m=1j/4Im′.The rest of the proof runs as before.


## 4. Proof of Theorem 2

The next two lemmas are generalizations of Lemmas1 and 2 and Lemmas 3.1 and 3.2 of Lao [17], respectively.Lemma 5.
Letf,g∈Hk∗ be two distinct cusp forms. Then, for j≥1 and 0≤i≤j, we have(48)∑pλfpiλfpj−λgpiλgpj2ps=21+i−δ1∑p1ps+O1,ass⟶1+,where δ1 is defined as in (4).Proof.
By observing (11) and (13), we find that for 0≤i≤j,(49)λfpiλfpj=∑m=0i∑m′=0jαfpi−2mαfpj−2m′=∑l1=0iλsymi+j−2l1fp.
In particular, fori=j,(50)λf2pj=λsymjf×symjfp=1+∑l1=1jλsym2l1fp.
Hence, we have(51)λfpiλfpj×λgpiλgpj=∑l1=0i∑l2=0iλsymi+j−2l1f×symi+j−2l2gp,(52)λf2piλg2pj=1+∑l1=1iλsym2l1fp+∑l2=1jλsym2l2gp+∑l1=1i∑l2=1jλsym2l1f×sym2l2gp.
By (51) and (52), we see that for j≥1 and 0≤i≤j,(53)∑pλfpiλfpj−λgpiλgpj2ps=∑p∑l1=1iλsym2l1fpps+∑p∑l2=1jλsym2l2fpps+∑p∑l1=1iλsym2l1gpps+∑p∑l2=1jλsym2l2gpps+2∑p1ps+∑p∑l1=1i∑l2=1jλsym2l1f×sym2l2fpps+∑p∑l1=1i∑l2=1jλsym2l1g×sym2l2gpps−2∑p∑l1=0i∑l2=0iλsymi+j−2l1f×symi+j−2l2gpps.
Note that using (17), for j≥1 and 0≤i≤j, we have(54)∑p∑l1=0i∑l2=0iλsymi+j−2l1f×symi+j−2l2gpps=δ1∑p1ps+O1,ass⟶1+,where δ1 is defined as in (4). In particular, taking i=j=l1=l2, we apply (2) and get(55)λsymi+j−2l1f×symi+j−2l2gp=λfpi+j−2l1λgpi+j−2l2=1.
Now, using (15)–(17) and (54) to estimate the terms in (53), we obtain Lemma 5 immediately.Lemma 6.
Letf,g∈Hk∗ be two distinct cusp forms. Then, for j≥1 and 0≤i≤j, we have(56)λfpiλfpj−λgpiλgpj≤2ij+j+δ2,where δ2 is defined as in (5).Proof.
By (49), we know that for 0≤i≤j,(57)λfpiλfpj−λgpiλgpj=∑l1=0iλsymi+j−2l1fp−λsymi+j−2l1gp.Case 1.
Ifi+j is odd, then so is i+j−2l1. Then, (11) gives(58)λsymi+j−2l1fp=αfpi+j−2l1+αfpi+j−2l1−2+⋯+αfp−i−j+2l1+2+αfp−i−j+2l1,λsymi+j−2l1gp=αgpi+j−2l1+αgpi+j−2l1−2+⋯+αgp−i−j+2l1+2+αgp−i−j+2l1.
Sinceαfp=1 (see (9)), we have(59)λfpiλfpj−λgpiλgpj≤2ij+j+i+1.Case 2.
Ifi+j is even, then so is i+j−2l1. Similarly, we have(60)λsymi+j−2l1fp=αfpi+j−2l1+αfpi+j−2l1−2+⋯+1+⋯+αfp−i−j+2l1+2+αfp−i−j+2l1,λsymi+j−2l1gp=αgpi+j−2l1+αgpi+j−2l1−2+⋯+1+⋯+αgp−i−j+2l1+2+αgp−i−j+2l1.
Thus,(61)λfpiλfpj−λgpiλgpj≤2ij+j.
By the above estimates, we conclude that forj≥1 and 0≤i≤j,(62)λfpiλfpj−λgpiλgpj≤2ij+j+δ2,where δ2 is defined as in (5).Now, we are ready to prove Theorem2.Proof.
of Theorem2. Let f,g∈Hk∗ be two distinct cusp forms. For j≥1 and 0≤i≤j, define(63)hp=λfpiλfpj−λgpiλgpj.
We infer from (15) and (49) that(64)∑pλfpiλfpj−λgpiλgpjps=O1,ass⟶1+.
Observe thatM=0.
Moreover, we know from Lemmas5 and 6 that m=21+i−δ1 and B=2ij+j+δ2.
Therefore, from the proof of Lemma4, we have(65)1+i−δ14ij+j+δ22∑p1ps+O1≤∑p∈A1ps,ass⟶1+,i.e., the set p|λfpiλfpj<λgpiλgpj has analytic density at least 1+i−δ1/4ij+j+δ22.


## 5. Proof of Theorem 3

The following lemmas play an important role in the Proof of Theorem3.Lemma 7.
Letf,g∈Hk∗ be two distinct cusp forms. Let a,b,c1,c2∈ℝ with a<b. Then, for j≥1 and s⟶1+, we have(66)∑pc1λfpj+c2λgpj−a2c1λfpj+c2λgpj−b2ps=j+1c14+c24+δ3c13+c23+6c12c22+a+b2+2abc12+c22+a2b2∑p1ps+O1,where δ3 is given in (7).Proof.
It is easy to verify that the sum on the left side of (66) equals(67)−2aba+b∑pc1λfpj+c2λgpjps+a+b2+2ab∑pc1λfpj+c2λgpj2ps−2a+b∑pc1λfpj+c2λgpj3ps+∑pc1λfpj+c2λgpj4ps+a2b2∑p1ps.
For the first sum in (67), by using (15), we get(68)∑pc1λfpj+c2λgpjps=O1,ass⟶1+.
For the second sum in (67), one can easily obtain from (15), (17), and (50) that(69)∑pc1λfpj+c2λgpj2ps=c12+c22∑p1ps+O1,ass⟶1+.
As for the third sum in (67), we note that(70)c1λfpj+c2λgpj3=c13λf3pj+3c12c2λf2pjλgpj+3c1c22λfpjλg2pj+c23λg3pj.
In view of (50), we have(71)λf2pjλgpj=λsymjgp+∑l1=1jλsym2l1f×symjgp.
Noticing (11), (13), and (49), we clearly deduce that(72)λf3pj=λsymjfp+∑l1=1jλsymjf×sym2l1fp=j+1λsymjfp+∑i=0j−1i+1λsym3j−2ifp+∑i=1j/2j−2i+1λsymj−2ifp.
Ifj is even and j=2i, we apply (2) and get(73)λsymj−2ifp=λfpj−2i=1.
Hence, we deduce from (15) and (17) that as s⟶1+,(74)∑pc1λfpj+c2λgpj3ps=δ3c13+c23∑p1ps+O1,where δ3 is given in (7).
We now calculate the fourth sum in (67). By (50), we have(75)λf4pj=1+2∑l1=1jλsym2l1fp+∑l1=1j∑l2=1jλsym2l1f×sym2l2fp.
Combining the above result with (15) and (16) leads to(76)∑pλf4pjps=j+1∑p1ps+O1,ass⟶1+.
From (15), (17), and (72), it is easy to check that as s⟶1+,(77)∑pλf3pjλgpjps=j+1∑pλsymjf×symjgpps+∑p∑i=0j−1i+1λsym3j−2if×symjgpps+∑p∑i=1j/2j−2i+1λsymj−2if×symjgpps=O1.
Moreover, we infer from (15), (17), and (52) that(78)∑pλf2pjλg2pjps=∑p1ps+O1,ass⟶1+.
By the above results, it is shown that ass⟶1+,(79)∑pc1λfpj+c2λgpj4ps=c14∑pλf4pjps+c24∑pλg4pjps+4c13c2∑pλf3pjλgpjps+4c1c23∑pλfpjλg3pjps+6c12c22∑pλf2pjλg2pjps=j+1c14+c24+6c12c22∑p1ps+O1.
We finish the proof by inserting (68), (69), (74), and (79) into (67).Lemma 8.
Letf,g∈Hk∗ be two distinct cusp forms. Let a,b,c1,c2∈ℝ with a<b. Then, for j≥1, we have(80)c1λfpj+c2λgpj−ac1λfpj+c2λgpj−b≤j+12c1+c22+j+1a+bc1+c2+ab.Proof.
It follows from (9) and (11) that(81)λfpj≤αfpj+αfpj−2+⋯+αfp−j+2+αfp−j≤j+1.
By similar computations, we infer from (13) and (50) that(82)λfpjλgpj≤j+12,λf2pj≤j+12.
Forj≥1, we have(83)c1λfpj+c2λgpj−ac1λfpj+c2λgpj−b=c12λf2pj+c22λg2pj+2c1c2λfpjλgpj−a+bc1λfpj+c2λgpj+ab.
According to the triangle inequality, we draw the conclusion.Proof.
of Theorem3. Let f,g∈Hk∗ be two distinct cusp forms. It is easy to verify from (15), (17), (50), and (83) that(84)∑pc1λfpj+c2λgpj−ac1λfpj+c2λgpj−bps=c12+c22+ab∑p1ps+O1,ass⟶1+.
Combining the above result with Lemmas7 and 8, Theorem 3 follows apparently by applying Lemma 4 with(85)hp=c1λfpj+c2λgpj−ac1λfpj+c2λgpj−b,B=j+12c1+c22+j+1a+bc1+c2+ab,m=j+1c14+c24+δ3c13+c23+6c12c22+a+b2+2abc12+c22+a2b2,M=c12+c22+ab.



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