A Lower Bound for the Large Sieve with Square Moduli

TL;DR

This paper establishes a fundamental lower bound for the large sieve inequality when applied to square moduli, advancing understanding of its limitations and potential applications.

Contribution

It introduces a new lower bound for the large sieve with square moduli, providing insights into its theoretical constraints.

Findings

Established a lower bound for the large sieve with square moduli

Enhanced understanding of sieve limitations in number theory

Potential implications for related analytic number theory problems

Abstract

We prove a lower bound for the large sieve with square moduli.

Equations

$$\sum _{q=1}^Q\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q{\left|\sum _{n=M+1}^{M+N}{a}_{n}e\left(\frac{an}{q}\right)\right|}^2\le (Q^2+N-1)\sum _{n=M+1}^{M+N}|a_n{|}^2.$$

$$\sum _{q=1}^Q\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^{q^2}{\left|\sum _{n=M+1}^{M+N}{a}_{n}e\left(\frac{an}{q^2}\right)\right|}^2\ll Q^{\epsilon }(Q^3+N)\sum _{n=M+1}^{M+N}|a_n{|}^2,$$

$$\sum _{q=1}^Q\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^{q^2}{\left|\sum _{n=M+1}^{M+N}{a}_{n}e\left(\frac{an}{q^2}\right)\right|}^2\ge D{Q}^{\frac{\log 2}{(1+\epsilon )\log \log Q}}(Q^3+N)\sum _{n=M+1}^{M+N}|a_n{|}^2$$

$$\ll (QN{)}^{\epsilon }\left(Q^3+N+\min \left\{\sqrt{Q}N,\sqrt{N}{Q}^2\right\}\right)\sum _{n=M+1}^{M+N}|a_n{|}^2$$

$$\mathcal{S}(Q):=\left\{(a,q)\in \mathbb{N}\times \mathbb{N}:Q<q\le 2Q,1\le a\le q^2,(a,q)=1,\left|\frac{a}{q^2}-\frac{1}{Q}\right|\le \frac{1}{Q^3}\right\}.$$

$$♯\mathcal{S}(Q)\ge Q^{\frac{\log 2}{(1+\epsilon )\log \log Q}},$$

$$q^2\equiv 1(\mathrm{mod}Q)$$

$$q^2=1+aQ$$

$$\left|\frac{a}{q^2}-\frac{1}{Q}\right|=\frac{1}{q^{2}Q}\le \frac{1}{Q^3}.$$

$$♯\mathcal{S}(Q)\ge 2^m.$$

$$\log Q=\sum _{i=1}^m\log p_i\le (1+\epsilon )p_m\le (1+2\epsilon )m\log m$$

$$m\ge \frac{\log Q}{(1+\epsilon )\log \log Q}$$

$$M:=0,N:=\frac{Q^3}{9},a_n:=e\left(-\frac{n}{Q}\right).$$

$$\sum _{n=M+1}^{M+N}{a}_{n}e\left(\frac{an}{q^2}\right)=\sum _{n=1}^{N}e\left({\alpha }_n\right)$$

$${\alpha }_n:=n\left(\frac{a}{q^2}-\frac{1}{Q}\right).$$

$$\left|\frac{a}{q^2}-\frac{1}{Q}\right|\le \frac{1}{Q^3},$$

$$\left|\sum _{n=1}^{N}e\left({\alpha }_n\right)\right|\ge CN$$

$$\begin{array}{rl}& \sum _{q=Q+1}^{2Q}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^{q^2}{\left|\sum _{n=M+1}^{M+N}{a}_{n}e\left(\frac{an}{q^2}\right)\right|}^2\\ \ge & \sum _{(a,q)\in \mathcal{S}(Q)}{\left|\sum _{n=M+1}^{M+N}{a}_{n}e\left(\frac{an}{q^2}\right)\right|}^2\\ \ge & ♯\mathcal{S}(Q)\cdot (CN{)}^2\\ =& C^2\cdot ♯\mathcal{S}(Q)\cdot N\sum _{n=M+1}^{M+N}|a_n{|}^2\\ =& \frac{C^2}{10}\cdot ♯\mathcal{S}(Q)\cdot (Q^3+N)\sum _{n=M+1}^{M+N}|a_n{|}^2\\ \ge & \frac{C^2}{10}\cdot Q^{\frac{\log 2}{(1+\epsilon )\log \log Q}}\cdot (Q^3+N)\sum _{n=M+1}^{M+N}|a_n{|}^2,\end{array}$$

Full text

Mathematics Subject Classification (2010): 11B57, 11J25, 11J71, 11L03, 11L07, 11L40.

Keywords: large sieve, Farey fractions in short intervals, estimates on exponential sums

1. Introduction

The classical large sieve inequality states that for $Q,N\in\mathbb{N}$, $M\in\mathbb{Z}$ and any sequence of complex numbers $\{a_{n}\}$,

[TABLE]

In [Zha], the third author studied the large sieve inequality for square moduli and conjectured that for any $\varepsilon>0$,

[TABLE]

where the implied constant depends only on $\varepsilon$. In his undergraduate thesis, the second author numerically investigated the validity of (1.1). A natural question is whether (1.1) can hold with the factor $Q^{\varepsilon}$ removed. In this note, we answer this question in the negative. More precisely, we prove the following.

Theorem 1**.**

For every $\varepsilon>0$, there are infinitely many natural numbers $Q$ such that for suitable $M\in\mathbb{Z}$, $N\in\mathbb{N}$ and sequences $\{a_{n}\}$ of complex numbers, we have

[TABLE]

for some absolute positive constant $D$.

The above theorem shows that the $Q^{\varepsilon}$ factor in (1.1) cannot be discarded or even replaced by a power of logarithm. We note that the best-known upper bound for the left-hand side of (1.1) is

[TABLE]

due to the first and third authors [SBLZ2].

The large sieve inequality for square (and quadratic) moduli has many applications. For example, it is used in the study of the Bombieri-Vinogradov theorem for square moduli [baier-zhao3], elliptic curves over finite fields [banks-pappalardi-shparlinski, ISLZ], Fermat quotients [bourgain-ford-konyagin-shparlinski], and the representation of primes [baier-zhao3, matomaki].

In [Zha], the third author also studied the large sieve inequality for $k$-power moduli, where $k>2$. The best known result for these $k$-power moduli with $k>2$ is due to K. Halupczok [Hal], who gave a large sieve inequality for $k$-power moduli which is uniform in $k$.

Acknowledgement. The first author would like to thank the University of New South Wales (UNSW) for its financial support and hospitality during his pleasant stay at the School of Mathematics and Statistics. The second author thanks the following organisations for his Research Training Program scholarship: the Department of Education and Training, Australian Government and the School of Mathematics and Statistics, UNSW. The third author was supported by the FRG grant PS43707 and the Faculty Silverstar Fund PS49334 at UNSW during this work.

2. Proof of Theorem 1

We first establish the following lower bound for the number of Farey fractions with square denominators near certain rational points.

Lemma 1**.**

Let $\varepsilon>0$ and $p_{1},...,p_{m}$ be the first $m$ odd primes. Set $Q:=p_{1}\cdots p_{m}$ and

[TABLE]

Then

[TABLE]

provided $m$ is sufficiently large.

Here we note that the expected number of Farey fractions of the form $a/q^{2}$ with $Q<q\leq 2Q$, $1\leq a\leq q^{2}$, $(a,q)=1$ in an interval of length $\Delta$ is, heuristically, of order of magnitude $Q^{3}\Delta$. So the above Lemma 1 shows that under certain circumstances, the true number can exceed the expectation significantly.

Proof of Lemma 1.

Using the Chinese Remainder Theorem, the number of solutions to the congruence

[TABLE]

with $Q<q\leq 2Q$ is exactly $2^{m}$. If $q$ solves the above congruence, then

[TABLE]

for some $a$ with $1\leq a\leq q^{2}$ and $(a,q)=1$, and it follows that

[TABLE]

Hence,

[TABLE]

Moreover, using the prime number theorem, for any given $\varepsilon>0$,

[TABLE]

if $m$ is sufficiently large. Consequently, for any given $\varepsilon>0$,

[TABLE]

if $m$ is large enough. Now the desired inequality (2.2) follows. ∎

Having proved Lemma 1, we are ready to prove Theorem 1. It will suffice to prove (1.2) with the summation range $1\leq q\leq Q$ replaced by $Q<q\leq 2Q$ which we do in the following. Set $Q=p_{1}\cdots p_{m}$ as in Lemma 1. Further, set

[TABLE]

Then

[TABLE]

with

[TABLE]

If

[TABLE]

then $|\alpha_{n}|\leq 1/9$ for $n=1,...,N$ and hence

[TABLE]

for some absolute positive constant $C$.

Define $\mathcal{S}(Q)$ as in (2.1). Then we have

[TABLE]

where the third line follows from (2.3), and the last line follows from Lemma 1. This completes the proof.