A new upper bound for the size of $s$-distance sets in boxes

TL;DR

This paper establishes a new upper bound on the size of sets with limited distances within multi-dimensional boxes, utilizing Tao's slice rank method to improve understanding of geometric combinatorics in Euclidean spaces.

Contribution

The paper introduces a novel upper bound for the size of s-distance sets in boxes, extending previous bounds with a new analytical approach.

Findings

Provides a new upper bound formula involving J(q,d)

Applies Tao's slice rank method to geometric combinatorics

Enhances bounds for s-distance sets in product sets of real numbers

Abstract

Let $q,d\geq 2$ be integers. Define $$ J(q,d):=\frac 1q \Big( \min_{0<x<1} \frac{1-x^q}{1-x} x^{-\frac{q-1}{d}}\Big). $$ Let $\mbox{$\cal G$}\subseteq {\mathbb R}^n$ be an arbitrary subset. We denote by $d(\mbox{$\cal G$})$ the set of (non-zero) distances among points of $\mbox{$\cal G$}$: $$ d(\mbox{$\cal G$}):=\{d( p_1, p_2):~ p_1, p_2\in \mbox{$\cal G$}, p_1\ne p_2\}. $$ Our main result is a new upper bound for the size of $s$-distance sets in boxes. More concretely, let $A_i\subseteq \mathbb R$, $|A_i|=q\geq 2$ be subsets for each $1\leq i\leq n$. Consider the box $\mbox{$\cal B$}:=\prod_{i=1}^n A_i\subseteq {\mathbb R}^n$. Suppose that $\mbox{$\cal G$}\subseteq \mbox{$\cal B$}$ is a set such that $|d(\mbox{$\cal G$})|\leq s$. Let $d:=\frac{n(q-1)}{s}$. Then $$|\mbox{$\cal G$}|\leq 2(qJ(q,d))^n.$$ We use Tao's slice rank bounding method in our proof.

Equations

$$J(q,d):=\frac{1}{q}\Big{(}\min_{0<x<1}\frac{1-x^{q}}{1-x}x^{-\frac{q-1}{d}}\Big{)}.$$

$$d(\mbox{$\cal G$}):=\{d(p_{1},p_{2}):~{}p_{1},p_{2}\in\mbox{$\cal G$},p_{1}\neq p_{2}\}.$$

$$|\mbox{$\cal G$}|\leq 2(qJ(q,d))^{n}.$$

$$(n-\frac{3}{4}{)}^{1/2}-\frac{1}{2}\le f(n)\le \frac{cn}{(\log n{)}^{1/2}}.$$

$$d(\mbox{$\cal G$}):=\{d(\mathbf{p}_{1},\mathbf{p}_{2}):~{}\mathbf{p}_{1},\mathbf{p}_{2}\in\mbox{$\cal G$},\mathbf{p}_{1}\neq\mathbf{p}_{2}\}.$$

$$s(\mbox{$\cal G$}):=\{(\mathbf{p}_{1},\mathbf{p}_{2}):~{}\mathbf{p}_{1},\mathbf{p}_{2}\in\mbox{$\cal G$},\mathbf{p}_{1}\neq\mathbf{p}_{2}\}.$$

$$|\mbox{$\cal F$}|\leq{n+s\choose s}.$$

$$|\mbox{$\cal F$}|\leq{n+s-1\choose s}+{n+s-2\choose s-1}.$$

$$|P|\le 2\cdot |\{x_1^{{\alpha }_1}\cdot \dots \cdot x_n^{{\alpha }_n}:0\le {\alpha }_i\le q-1\text{mbox}foreachi,\sum _i{\alpha }_i\le s\}|.$$

$$J(t,d):=\frac{1}{t}\Big{(}\min_{0<x<1}\frac{1-x^{t}}{1-x}x^{-\frac{t-1}{d}}\Big{)}.$$

$$\underset{q\to \infty }{\lim }J(q)=\underset{z>3}{\inf }\frac{z-z^{-2}}{3\log (z)}=0.8414\dots .$$

$$0.8414\le J(q)\le 0.9184$$

$$|\mbox{$\cal F$}|\leq 2(qJ(q,d))^{n}.$$

$$|\mbox{$\cal F$}|\leq{n+s\choose s}.$$

$$|\mbox{$\cal F$}|\leq|\{x_{1}^{\alpha_{1}}\cdot\ldots\cdot x_{n}^{\alpha_{n}}:~{}0\leq\alpha_{i}\leq q-1\mbox{ for each }i,\ \sum_{i}\alpha_{i}\leq s\}|.$$

$$|\mbox{$\cal F$}|>2\sum_{i=0}^{d/2}{n\choose i}.$$

$$P(x_1,\dots ,x_n,y_1,\dots ,y_n,)\in \mathbb{F}[x_1,\dots x_n,y_1,\dots y_n]$$

$$|\mbox{$\cal F$}|\leq 2|\{x_{1}^{\alpha_{1}}\ldots x_{n}^{\alpha_{n}}:~{}0\leq{\alpha_{i}\leq t-1}\mbox{ for each }1\leq i\leq n\mbox{ and}$$

$$\sum _{j=1}^n{\alpha }_j\le \frac{\text{mbox}deg(P)}{2}\}|.$$

$$P(x_1,\dots ,x_n,y_1,\dots ,y_n,)\in \mathbb{F}[x_1,\dots x_n,y_1,\dots y_n]$$

$$|\mbox{$\cal F$}|\leq 2(tJ(t,d))^{n}.$$

$$P(x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}):=\prod_{i=1}^{s}\Big{(}\sum_{j=1}^{n}(x_{j}-y_{j})^{2}-d_{i}^{2}\Big{)}\in\mathbb{R}[x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}].$$

$$P(\mathbf{a},\mathbf{a})=\prod _{i=1}^s(-d_i{)}^2\ne 0$$

$$P(\mathbf{a},\mathbf{b})=\prod_{i=1}^{s}\Big{(}\sum_{j=1}^{n}(a_{j}-b_{j})^{2}-d_{i}^{2}\Big{)}=\prod_{i=1}^{s}\Big{(}d(\mathbf{a},\mathbf{b})^{2}-d_{i}^{2}\Big{)}.$$

$$|\mbox{$\cal F$}|\leq|\{x_{1}^{\alpha_{1}}\cdot\ldots\cdot x_{n}^{\alpha_{n}}:~{}0\leq\alpha_{i}\leq q-1\mbox{ for each }i,\ \sum_{i}\alpha_{i}\leq s\}|.$$

$$\mbox{$\cal F$}:=\{\mathbf{v}_{A}:~{}A\subseteq[n],|A|=s\}\subseteq\{0,1\}^{n}\subseteq{\mathbb{R}}^{n}.$$

Full text

[TABLE]

Let \mbox{\cal G}\subseteq{\mathbb{R}}^{n} be an arbitrary subset. We denote by d(\mbox{\cal G}) the set of (non-zero) distances among points of $\cal G$:

[TABLE]

Our main result is a new upper bound for the size of $s$-distance sets in boxes. More concretely, let $A_{i}\subseteq\mathbb{R}$, $|A_{i}|=q\geq 2$ be subsets for each $1\leq i\leq n$. Consider the box \mbox{\cal B}:=\prod_{i=1}^{n}A_{i}\subseteq{\mathbb{R}}^{n}. Suppose that \mbox{\cal G}\subseteq\mbox{\cal B} is a set such that |d(\mbox{\cal G})|\leq s. Let $d:=\frac{n(q-1)}{s}$. Then

[TABLE]

We use Tao’s slice rank bounding method in our proof.

00footnotetext: **Keywords. $s$-distance sets, distance problem, slice rank bounding method

2010 Mathematics Subject Classification: 52C45, 12Y99, 05D99**

1 Introduction

In this manuscript we give upper bounds for $s$-distance sets in the direct product of finite set of points in the Euclidean space. Two independent research directions motivated our results.

Our first motivation comes from the following question of Erdős: Given $n$ points in the plane, what is the smallest number of distinct distances they can determine?

Erdős proved in [8] the following result.

Theorem 1.1

The minimum number of $f(n)$ of distances determined by $n$ points of the plane satisfies the inequalities

[TABLE]

Erdős conjectured that the square grid is essentially the extremal example, consequently the upper bound for the function $f(n)$ is sharp.

Our second motivation comes from the investigation of different types of $s$-distance sets in the $d$-dimensional Euclidean space.

First we introduce some notation. Let ${\mathbb{R}}^{d}$ denote the $d$-dimensional Euclidean space. Let \mbox{\cal G}\subseteq{\mathbb{R}}^{n} be an arbitrary set. We denote by d(\mbox{\cal G}) the set of (non-zero) distances among points of $\cal G$:

[TABLE]

Let $(\mathbf{x},\mathbf{y})$ stand for the standard scalar product. Let \mbox{\cal G}\subseteq{\mathbb{R}}^{n} be an arbitrary set. We denote by s(\mbox{\cal G}) the set of scalar products between the dintinct points of $\cal G$.

[TABLE]

Bannai, Bannai and Stanton proved the following result in [1] Theorem 1.

Theorem 1.2

Suppose that \mbox{\cal F}\subseteq{\mathbb{R}}^{n} is a set and that |d(\mbox{\cal F})|\leq s. Then

[TABLE]

Delsarte, Goethals and Seidel investigated the spherical $s$-distance sets. They proved the following Theorem in [7].

Theorem 1.3

Suppose that \mbox{\cal F}\subseteq{\mathbb{S}}^{n-1} is a set and that |d(\mbox{\cal F})|\leq s. Then

[TABLE]

We state here our main results.

Theorem 1.4

Let $A_{i}\subseteq\mathbb{R}$, $|A_{i}|=q\geq 2$ for each $1\leq i\leq n$. Consider the box \mbox{\cal B}:=\prod_{i=1}^{n}A_{i}\subseteq{\mathbb{R}}^{n}. Suppose that \mbox{\cal F}\subseteq\mbox{\cal B} is a set such that |d(\mbox{\cal F})|\leq s. Then

[TABLE]

We use Tao’s slice rank bounding method in our proof (see the blog [11]). Tao developed this method as a proof technique to prove Ellenberg and Gijswijt’s breakthrough about the upper bounds for the size of subsets $A$ in $({\mathbb{Z}_{p}})^{n}$ without three-term arithmetic progressions (see [9]).

Let $t,d\geq 2$ be integers. Define

[TABLE]

Define $J(q):=J(q,3)$ for each $q>1$.

This $J(q)$ constant appeared in the proof of the Ellenberg and Gijswijt’s bound for the size of three-term progression-free sets (see [9]). It was proved in [2] Proposition 4.12 that $J(q)$ is a decreasing function of $q$ and

[TABLE]

We can verify easily that $J(3)=0.9184$, consequently $J(q)$ lies in the range

[TABLE]

for each $q\geq 3$. The following Corollary gives a more concrete upper bound for the size of $s$-distance sets in boxes.

Corollary 1.5

Let $A_{i}\subseteq\mathbb{R}$, $|A_{i}|=q\geq 2$ for each $1\leq i\leq n$. Consider the box \mbox{\cal B}:=\prod_{i=1}^{n}A_{i}\subseteq{\mathbb{R}}^{n}. Suppose that \mbox{\cal F}\subseteq\mbox{\cal B} is a set such that |d(\mbox{\cal F})|\leq s. Let $d:=\frac{n(q-1)}{s}$. Then

[TABLE]

Deza and Frankl proved the following statement in [6] Theorem 4.

Theorem 1.6

Suppose that \mbox{\cal F}\subseteq{\mathbb{R}}^{n} is an arbitrary set such that |s(\mbox{\cal F})|\leq s. Then

[TABLE]

It is easy to prove the following result using a slight modification of Tao’s slice rank bounding method.

Theorem 1.7

Let $A_{i}\subseteq\mathbb{R}$, $|A_{i}|=q\geq 2$ for each $1\leq i\leq n$. Consider the box \mbox{\cal B}:=\prod_{i=1}^{n}A_{i}\subseteq{\mathbb{R}}^{n}. Suppose that \mbox{\cal F}\subseteq\mbox{\cal B} is an arbitrary set which satisfies the following properties:

• (i)

(\mathbf{f},\mathbf{f})\notin s(\mbox{\cal F})* for each \mathbf{f}\in\mbox{\cal F};*

• (ii)

|s(\mbox{\cal F})|\leq s.

Then

[TABLE]

In Section 2 we present our proofs.

2 Proofs

The following simple statement was proved in [5] Lemma 1.

Proposition 2.1

Suppose that $n\geq 1$ and $d\geq 0$ are integers, $P$ is a multilinear polynomial in $n$ variables of total degree at most $d$ over a field $\mathbb{F}$, and \mbox{\cal F}\subseteq{\mathbb{F}}^{n} is a subset with

[TABLE]

If $P(\mathbf{a}-\mathbf{b})=0$ for all \mathbf{a},\mathbf{b}\in\mbox{\cal F}, $\mathbf{a}\neq\mathbf{b}$, then $P(\mathbf{0})=0$.

Tao’s slice rank bounding method (see the blog [11] and an other proof in [2] Section 4) gives easily the following generalization of Proposition 2.1, which is a special case of [10] Corollary 1.3.

Theorem 2.2

Let $\mathbb{F}$ be an arbitrary field. Let $A_{i}\subseteq\mathbb{F}$ be fixed subsets such that $|A_{1}|=\ldots=|A_{n}|=t>0$. Let \mbox{\cal F}\subseteq\prod_{i=1}^{n}A_{i} be a finite subset. Suppose that there exists a polynomial

[TABLE]

satisfying the following conditions:

• (i)

$P(\mathbf{a},\mathbf{a})\neq 0$* for each \mathbf{a}\in\mbox{\cal F};*

• (ii)

if \mathbf{a},\mathbf{b}\in\mbox{\cal F}, $\mathbf{a}\neq\mathbf{b}$ are arbitrary vectors, then $P(\mathbf{a},\mathbf{b})=0$.

Then

[TABLE]

[TABLE]

As an easy consequence, we proved the following result in [10] Corollary 1.5.

Theorem 2.3

Let $\mathbb{F}$ be an arbitrary field. Let $A_{i}\subseteq\mathbb{F}$ be fixed subsets such that $|A_{1}|=\ldots=|A_{n}|=t>0$. Let \mbox{\cal F}\subseteq\prod_{i=1}^{n}A_{i} be a finite subset. Suppose that there exists a polynomial

[TABLE]

satisfying the following conditions:

• (i)

$P(\mathbf{a},\mathbf{a})\neq 0$* for each \mathbf{a}\in\mbox{\cal F};*

• (ii)

if \mathbf{a},\mathbf{b}\in\mbox{\cal F}, $\mathbf{a}\neq\mathbf{b}$ are arbitrary vectors, then $P(\mathbf{a},\mathbf{b})=0$.

Let $d:=\frac{2n(t-1)}{deg(P)}$. Then

[TABLE]

Proof of Theorem 1.4:

Consider the set d(\mbox{\cal F})=\{d_{1},\ldots,d_{s}\}. Clearly $d_{i}\neq 0$ for each $i$.

Define the polynomial

[TABLE]

Clearly $deg(P)=2s$. Then

[TABLE]

for each \mathbf{a}\in\mbox{\cal F} On the other hand if \mathbf{a},\mathbf{b}\in\mbox{\cal F}, $\mathbf{a}\neq\mathbf{b}$, then

[TABLE]

But there exists an $i$ such that $d(\mathbf{a},\mathbf{b})=d_{i}$, because d(\mbox{\cal F})=\{d_{1},\ldots,d_{s}\}. Hence $d(\mathbf{a},\mathbf{b})^{2}=d_{i}^{2}$, so $P(\mathbf{a},\mathbf{b})=0$.

Finally we can apply Theorem 2.2 with the choices $\mathbb{F}=\mathbb{R}$ and $t=q$.

3 Concluding remarks

We conjecture that the following upper bound is sharp.

Conjecture 1

Let $A_{i}\subseteq\mathbb{R}$, $|A_{i}|=q\geq 2$ for each $1\leq i\leq n$. Consider the box \mbox{\cal B}:=\prod_{i=1}^{n}A_{i}\subseteq{\mathbb{R}}^{n}. Suppose that \mbox{\cal F}\subseteq\mbox{\cal B} is a set with |d(\mbox{\cal F})|\leq s. Then

[TABLE]

As a lower bound, we give the following construction in the case $q=2$. If $A\subseteq[n]$, then denote by $\mathbf{v}_{A}$ the characteristic vector of $A$. Consider the set

[TABLE]

Then $\cal F$ is an $s$-distance set with |\mbox{\cal F}|={n\choose s}.