Additive bases with coefficients of newforms

TL;DR

This paper proves that for certain modular forms with integer coefficients, every integer can be expressed as a sum of a bounded number of these coefficients, with the bound depending on the form’s level and weight.

Contribution

It establishes a uniform bound on the number of Fourier coefficients needed to represent any integer as their sum, depending explicitly on the form's level and weight.

Findings

Every integer is a sum of at most C(f) Fourier coefficients.

The bound C(f) depends polynomially on the level N.

Explicit bound C(f) ollows from the form's parameters.

Abstract

Let $f(z)=\sum_{n=1}^{\infty}a(n) e^{2\pi i nz}$ be a normalized Hecke eigenform in $S_{2k}^{\text{new}}(\Gamma_0(N))$ with integer Fourier coefficients. We prove that there exists a constant $C(f)>0$ such that any integer is a sum of at most $C(f)$ coefficients $a(n) $. It holds $C(f)\ll_{\varepsilon,k}N^{\frac{6k-3}{16}+\varepsilon}$.

Equations

$$Z=\sum _{j=1}^{74000}\tau (n_j).$$

$$Z=\sum _{j=1}^{\ell }a(n_j)$$

$$C(f){\ll }_{\epsilon ,k}{N}^{\frac{6k-3}{16}+\epsilon }.$$

$$|a(p)|>(\sqrt{2}-\epsilon )p^{\frac{2k-1}{2}}$$

$$n_f\ll (4k^{2}N{)}^{\frac{3}{8}},$$

$$\gamma =\{\begin{array}{ll}\theta +2,& \text{if }p=2\text{and }\theta >0,\\ \theta +1,& \text{otherwise},\end{array}$$

$$K=\prod _{(p-1)|k}{p}^{\gamma }.$$

$$s\ge \{\begin{array}{ll}{2}^k,& \text{if }k\le 10\\ 2k^2(2\log k+\log \log k+\frac{5}{2}),& \text{if }k>10,\end{array}$$

$$p_1^k+\cdots +p_s^k=Z,\text{for primes }{p}_1,\dots ,p_s,$$

$$I_s(Z)=\mathfrak{G}(Z)\frac{{\Gamma }^s(1/k)}{\Gamma (s/k)}\frac{Z^{\frac{s}{k}-1}}{{\log }^{s}Z}+O_{k,s}\left(\frac{Z^{\frac{s}{k}-1}}{{\log }^{s+1}Z}\log \log Z\right),$$

$$\mathfrak{G}(Z)=\sum _{q=1}^{\infty }\left(\sum _{(h,q)=1}{\left(\sum _{(l,q)=1}{e}^{2\pi ih\frac{l^k}{q}}\right)}^s{e}^{2\pi i\frac{-h}{q}Z}\right),$$

$$0<A\le \mathfrak{G}(Z)<B.$$

$${\mathcal{P}}_0(M)=\mathcal{P}\cap (n_f,M].$$

$$\sum _{i=1}^{k}a(p_i)\ne \sum _{i=k+1}^{2k}a(p_i)$$

$$p_1<\dots <p_k,p_{k+1}<\dots <p_{2k},(p_1,\dots ,p_k)\ne (p_{k+1},\dots ,p_{2k}).$$

$$p^{\frac{2k-1}{2}}<|a(p)|.$$

$${\mathcal{P}}^{\prime }\cap [1,T]=\alpha \pi (T)\left(1+o(1)\right),$$

$$A_i:={\mathcal{P}}^{\prime }\cap \left[2^{{\ell }_{0}i},2^{{\ell }_{0}i+1}\right].$$

$$|A_i|>\frac{\alpha }{2}\frac{2^{{\ell }_{0}i}}{\log 2^{{\ell }_{0}i}},$$

$$|a(p_1)|<\cdots <|a(p_k)|<|a(p_{k+1})|<\cdots <|a(p_{2k})|.$$

$$q_1<\dots <q_k,q_{k+1}<\dots <q_{2k},(q_1,\dots ,q_k)\ne (q_{k+1},\dots ,q_{2k}),$$

$$\sum _{i=1}^{k}a(q_i)=\sum _{i=k+1}^{2k}a(q_i).$$

$$\sum _{i=1}^{t}a(q_i)=\sum _{i=k+1}^{k+t}a(q_i),$$

$$a(q_{k+t})=\sum _{i=1}^{t}a(q_i)-\sum _{i=k+1}^{k+t-1}a(q_i).$$

$$|a(q_1)|<\cdots <|a(q_t)|,|a(q_{k+1})|<\cdots <|a(q_{k+t})|,$$

$$|a(q_{k+t})|$$

$$\le k(|a(q_t)|+|a(q_{k+t-1})|).$$

$$k(|a(q_{)}|+|a(q_{k+t-1})|)\le 2k|a(p_{s-1})|\le 4k{2}^{({\ell }_0(s-1)+1)\frac{2k-1}{2}},$$

$$|a(q_{k+t})|=|a(p_s)|\ge 2^{{\ell }_{0}s\frac{2k-1}{2}}.$$

$$2^{{\ell }_{0}s(2k-1)/2}\le |a(q_{k+t})|\le k(|a(q_t)|+|a(q_{k+t-1})|)\le 4k{2}^{({\ell }_0(s-1)+1)\frac{2k-1}{2}},$$

$${\ell }_0\le 1+\frac{\log 4k}{(2k-1)\log 2},$$

Full text

Key words: newform; Fourier coefficients; additive basis

MSC: 11F30 11P05

1 Introduction

The set $\{\tau(n)\mid n\geq 1\}$ of values of Ramanujan’s function is an additive basis of the integers: any integer $Z$ can be written as

[TABLE]

See [2], [11]. Here we prove a similar property for the Fourier coefficients of normalized Hecke eigenforms. For integers $k\geq 1$, $N\geq 1$ denote by $S_{2k}^{\text{new}}(\Gamma_{0}(N))$ the space of newforms of weight $2k$ on $\Gamma_{0}(N)$.

Theorem 1**.**

Lef $f(z)=\sum_{n=1}^{\infty}a(n)e^{2\pi inz}$ be a normalized Hecke eigenform in $S_{2k}^{\text{new}}(\Gamma_{0}(N))$ with integer Fourier coefficients. There exists a constant $C(f)>0$ such that any integer $Z$ is a sum

[TABLE]

for some $\ell\leq C(f)$ and integers $n_{j}\ll|Z|^{\frac{2}{2k-1}}+1.$ It holds

[TABLE]

Our method follows the idea of  [2] to connect the solubility of (1) with the Waring–Goldbach problem. We use results of Ram Murty on oscillations of Fourier coefficients of newforms, of Matomäki on signs of the coefficients, and of Hua on the Waring-Goldbach problem, which are stated in the second section. The third section contains the proof of the theorem.

2 Lemmas

We apply the following facts.

Lemma 2** (Ram Murty).**

Lef $f(z)=\sum_{n=1}^{\infty}a(n)e^{2\pi inz}$ be a normalized Hecke eigenform in $S_{2k}^{\text{new}}(\Gamma_{0}(N))$. For any $\varepsilon>0$ we have

[TABLE]

for a positive density of primes $p$.

Proof.

This was proved in [9, Corollary 2] for forms on the full modular group, but the statement is true also for forms on $\Gamma_{0}(N)$ [10, Chapter 4, Theorem 8.6 (ii) with $m=1$, page 89]. ∎

Lemma 3**.**

Lef $f(z)=\sum_{n=1}^{\infty}a(n)e^{2\pi inz}$ be a normalized Hecke eigenform in $S_{2k}^{\text{new}}(\Gamma_{0}(N))$. Let $n_{f}$ be the smallest integer such that $a(n_{f})<0$ and $(n_{f},N)=1$. Then

[TABLE]

where the implied constant is absolute.

Proof.

See [8, Theorem 1]. See also [5], [4], [6]. ∎

Let $k\geq 1$ be an integer, $p$ a prime number, $\theta\geq 0$ the integer with $p^{\theta}\mid k$ and $p^{\theta+1}\nmid k,$

[TABLE]

[TABLE]

Lemma 4** (Hua).**

Let $k\geq 1$ be an integer and $K$ as in (3). If

[TABLE]

then for any $Z\equiv s\pmod{K}$ the number $I_{s}(Z)$ of solutions of the equation

[TABLE]

satisfies the following asymptotic formula

[TABLE]

where $\mathfrak{G}(Z)$ is the singular series

[TABLE]

which is absolutely convergent and there exist positive constants $A,B$ independent of $Z$ such that

[TABLE]

Proof.

See [3, Theorem 11 and Theorem 12, pages 78 and 100 respectively]. ∎

Kumchev and Wooley proved in [7, Theorem 1] that equation (4) has solution if $k$ is large and $s\geq(4k-2)\log k+k-7$.

3 Proof of the theorem

We denote by ${\mathcal{P}}$ the set of prime numbers which do not divide $N$ and ${\mathcal{P}}(X)={\mathcal{P}}\cap[1,X]$. By Lemma 3 let $n_{f}>1$ be the smallest integer such that $a(n_{f})<0$ and $(n_{f},N)=1$. Let $M$ be a large parameter and set

[TABLE]

We will say that $D_{M}\subset{\mathcal{P}}_{0}(M)$ is an admissible subset of ${\mathcal{P}}_{0}(M)$ if

[TABLE]

for any $p_{1},\ldots,p_{2k}\in D_{M}$ such that

[TABLE]

We prove that admissible sets exist. The Ramanujan–Petersson conjecture, proved by Deligne [1], states that $|a(p)|\leq 2p^{\frac{2k-1}{2}}$ for any prime number $p$. Let ${\mathcal{P}}^{\prime}\subset{\mathcal{P}}$ be the set of prime numbers such that

[TABLE]

From Lemma 2 it follows that there exists a constant $0<\alpha\leq 1,$ which depends on $f,$ such that for $T$ large enough we have

[TABLE]

where $\pi(T)$ is the prime counting function. Let $\ell_{0}>10\log k$ be an integer. For any $1\leq i\leq 2k$ let

[TABLE]

From  (7) it follows that

[TABLE]

so we can choose M and $\ell_{0}$ sufficiently large such that $A_{i}\subset{\mathcal{P}}_{0}(M)$ and $|A_{i}|>1$ for any $1\leq i\leq 2k.$ Let $p_{i}\in A_{i},\,1\leq i\leq 2k$. From  (6) and the Ramanujan-Petersson-Deligne estimate $|a(p_{i})|\leq 2p_{i}^{\frac{2k-1}{2}}$ it follows that

[TABLE]

The set ${\mathcal{Q}}:=\{p_{1},\ldots,p_{2k}\}$ is an admissible subset of ${\mathcal{P}}_{0}(M)$. Indeed, let $q_{1},\ldots,q_{2k}\in{\mathcal{Q}}$ be such that

[TABLE]

and

[TABLE]

Let $t$ be the largest index $1\leq t\leq k$ such that $q_{t}\neq q_{k+t}$. From  (11) it follows that

[TABLE]

[TABLE]

From  (9) and  (10) it follows that

[TABLE]

hence

[TABLE]

Without loss of generality we can suppose that $q_{t}<q_{k+t}$. Let $1\leq s\leq 2k$ be such that $q_{k+t}=p_{s}$. It holds that $q_{t},q_{k+t-1}\leq p_{s-1},$ and from  (6), (8) and the Ramanujan-Petersson-Deligne estimate $|a(p_{s-1})|\leq 2p_{s-1}^{(2k-1)/2}$ it follows that

[TABLE]

[TABLE]

From the above estimates and (3) it follows that

[TABLE]

hence

[TABLE]

which contradicts the assumption $\ell_{0}>10\log k$. So ${\mathcal{Q}}$ is an admissible subset of ${\mathcal{P}}_{0}(M)$.

Let ${\mathcal{P}}_{0}^{\prime}(M)\subset{\mathcal{P}}_{0}(M)$ be some addmisible subset with largest cardinality. We prove that

[TABLE]

Since the admissible subset ${\mathcal{Q}}$ constructed above has $2k$ elements it follows that

[TABLE]

Let

[TABLE]

Given $\lambda\in S_{k},$ let $T(\lambda)$ be the number of solutions $(p_{1},\ldots,p_{k})$ of the equation

[TABLE]

It holds that

[TABLE]

The Cauchy–Schwarz inequality implies that

[TABLE]

Note that $\sum_{\lambda\in S_{k}}T^{2}(\lambda)$ is the number of solutions of the equation

[TABLE]

with

[TABLE]

Since ${\mathcal{P}}_{0}^{\prime}(M)$ is admissible  (15) holds only if $(p_{1},\ldots,p_{k})=(p_{k+1},\ldots,p_{2k})$. From this and  (14) it follows that

[TABLE]

The estimate $|a(p)|\leq 2p^{\frac{2k-1}{2}}$ implies

[TABLE]

so from  (16) we have

[TABLE]

and (13) is proved.

Let $p\in{\mathcal{P}}_{0}(M)\setminus{\mathcal{P}}_{0}^{\prime}(M)$. We proceed as in [2, Page 39] to prove that there exist $p_{1},\ldots,p_{2k-1}$ in ${\mathcal{P}}^{\prime}_{0}(M)$ such that

[TABLE]

Indeed, the maximality of ${\mathcal{P}}^{\prime}_{0}(M)$ implies that there exist

[TABLE]

such that

[TABLE]

[TABLE]

Moreover,

[TABLE]

and, by  (17), $p$ occurs at most twice in the sequece $q_{1},\ldots,q_{2k}$. If $p$ occurs twice, then it appears in $q_{1},\ldots,q_{k}$ and in $q_{k+1},\ldots,q_{2k}$, thus

[TABLE]

for some $q^{\prime}_{1},\ldots,q^{\prime}_{2k}$ in ${\mathcal{P}}^{\prime}_{0}(M)$ with

[TABLE]

This is impossible, since ${\mathcal{P}}_{0}^{\prime}(M)$ is admissible with at least $2k$ elements.

Therefore, for any $p\in{\mathcal{P}}_{0}(M)\backslash{\mathcal{P}}^{\prime}_{0}(M)$ there exist $p_{1},\ldots,p_{2k-1}$ in ${\mathcal{P}}^{\prime}_{0}(M)$

 such that

[TABLE]

Multiplying by $a(p)$ and taking into account that $(p,p_{i})=1$ we get

[TABLE]

since the coefficients of $f$ are multiplicative. Subtracting $a(p^{2})$ and applying the identity $p^{2k-1}=a(p)^{2}-a(p^{2})$ which is satisfied by the coefficients of $f$ it follows that

[TABLE]

Let

[TABLE]

We prove that for $Z$ large there exist $p_{1},\ldots,p_{s_{0}}\in{\mathcal{P}}_{0}(Z^{1/(2k-1)})\setminus{\mathcal{P}}^{\prime}_{0}(Z^{1/(2k-1)})$ such that

[TABLE]

Let $Z_{k}:=Z^{\frac{1}{2k-1}}$ and

[TABLE]

with $\gamma$ defined as in (2). Since $2k-1$ is odd, the only prime number $p$ with $p-1\lvert 2k-1$ is $p=2$, and for $p=2$ we have

[TABLE]

hence

[TABLE]

By Lemma 4 there exists a positive constant $c_{1}=c_{1}(k)$ such that for any $Z\equiv s_{0}\pmod{2},$ with $Z$ large the number of solutions $I_{s_{0}}(Z)$ of

[TABLE]

with $p_{1},\ldots,p_{s_{0}}\in{\mathcal{P}}_{0}(Z_{k}),$ satisfies

[TABLE]

Now consider equation (19) with at least one $p_{i}\in{\mathcal{P}}^{\prime}_{0}(Z_{k})$ and denote by $I^{\prime}_{s_{0}}(Z)$ its number of solutions. $I^{\prime}_{s_{0}}(Z)$ should be less than $s_{0}I^{\prime}_{s_{0}-1},$ where $I^{\prime}_{s_{0}-1}$ denotes the number of solutions of the equation

[TABLE]

with $p_{1},\ldots,p_{s_{0}-1}\in{\mathcal{P}}_{0}(Z_{k}),$ and $p_{s_{0}}\in{\mathcal{P}}^{\prime}_{0}(Z_{k}).$ Note that

[TABLE]

where $I^{\prime}_{s_{0}-1}(Z-p^{2k-1}_{s_{0}})$ denotes the number of solutions of (21) for $p_{s_{0}}$ given. Therefore we have

[TABLE]

Afterwards, for some $p^{\prime}_{s_{0}}\in{\mathcal{P}}^{\prime}_{0}(Z_{k})$ we get

[TABLE]

In order to estimate $I^{\prime}_{s_{0}-1}(Z-p^{\prime 2k-1}_{s_{0}})$ we apply Lemma (4) with $s_{0}-1$ variables. Recalling that $Z-p^{\prime 2k-1}_{s_{0}}>n_{f}^{2k-1},$ we obtain

[TABLE]

Combining equations (22), (23) and estimate (13) we get

[TABLE]

The number of solutions for (19) with $p_{i}\in{\mathcal{P}}_{0}(Z^{1/(2k-1)})\backslash{\mathcal{P}}^{\prime}_{0}(Z^{1/(2k-1)})$ is equal to $I_{s_{0}}(Z)-I^{\prime}_{s_{0}}(Z).$ The estimates (20) and (24) imply that

[TABLE]

Therefore equation (19) is solvable for primes in ${\mathcal{P}}_{0}(Z^{1/(2k-1)})\backslash{\mathcal{P}}^{\prime}_{0}(Z^{1/(2k-1)}).$ From this and (15) it follows that any large integer $Z$ with $Z\equiv s_{0}\pmod{2}$ has a representation

[TABLE]

for some integers $n_{i},n_{j}\leq Z^{2/(2k-1)}$ with $(n_{f}!N,n_{i})=(n_{f}!N,n_{j})=1.$ Note that $-Z$ has a similar representation. We also note that any integer $Z_{0}$ can be represented as

[TABLE]

with $Z\equiv s_{0}\pmod{2},\;0\leq r_{0}<2$, thus if $Z_{0}$ is large then

[TABLE]

since $a(1)=1$. Recall that $n_{f}$ satisfies $a(n_{f})<0$. Let $C_{0}:=-a(n_{f})$. We have

[TABLE]

with $0\leq r_{0}<K$. As above, we note that for $Z_{1}$ large enough there exist integers $Z_{0}$ and $0\leq r_{1}<C_{0}$ such that

[TABLE]

Therefore, any integer $Z$ with $|Z|\geq T$ can be expressed as in  (25) with $r_{0}\leq 2,$ $r_{1}\leq C_{0}.$ The number of summands $a(n)$ in (25) is

[TABLE]

For integers $Z$ with $|Z|\leq T$ let $n^{\prime}$ be such that $2Z<a_{n^{\prime}}$. It holds that $|Z-a(n^{\prime})|>T,$ so $Z-a(n^{\prime})$ can be written in the form  (25). Hence any integer $Z$ can be written in the form

[TABLE]

with

[TABLE]

[TABLE]

since $r_{0}\leq 2$ and $r_{1}\leq C_{0}$. The theorem is proved with

[TABLE]

Since $s_{0}$ depends only on $k$ we have

[TABLE]

By the Ramanujan-Petersson-Deligne estimate we have

[TABLE]

where $d(\cdot)$ is the number of divisors function which satisfies

[TABLE]

so

[TABLE]

By Lemma 3 we have

[TABLE]

hence

[TABLE]

$\Box$