A lower bound for the size of Kakeya sets with respect to hyperplanes in $\mathbb{F}_q^n$

TL;DR

This paper establishes a lower bound on the size of Kakeya sets in finite fields, showing such sets must be nearly as large as the entire space, with only a small quadratic deficit.

Contribution

It provides a new lower bound for Kakeya sets with respect to hyperplanes in finite fields, advancing understanding of their minimal size.

Findings

Kakeya sets contain a hyperplane in every direction

Size of Kakeya sets is at least q^n - O(q^2)

Sets are nearly as large as the entire space

Abstract

We prove that a subset of $\mathbb{F}_q^n$ that contains a hyperplane in any direction has size at least $q^{n}-O(q^2)$.

Equations

$$\dim E=n$$

$$\dim E:=n-\underset{\delta \to 0}{\lim }\frac{\log \text{vol}(E_{\delta })}{\log \delta }.$$

$$\text{vol}(S_{\delta })$$

$$=\frac{4\pi \delta }{3}(6r^2+2{\delta }^2)$$

$$\dim S$$

$$=2.$$

$$\text{vol}(B_{\delta })=\frac{4\pi }{3}(r+\delta {)}^3$$

$$\dim B$$

$$=3.$$

$$|E|⩾c{q}^n.$$

$$|E|⩾\frac{q^{2n}-q^n}{q^n+q^2-2q}.$$

$$|E|⩾\frac{q(q+1)}{2}$$

$$|S|=\frac{q^n-1}{q-1}.$$

$$K=\{(k_1,\dots ,k_{n-1})\in \text{mathds}{F}_q^n\times \cdots \times \text{mathds}{F}_q^n|\dim \text{span}(k_1,\dots ,k_{n-1})=n-1\}.$$

$$|K|=\prod _{h=0}^{n-2}(q^n-q^h).$$

$$|K|=\sum _{w\in S}|{\text{fiber}}_{\varphi }(w)|.$$

$$|{\text{fiber}}_{\varphi }(w)|=\prod _{h=0}^{n-2}(q^{n-1}-q^h).$$

$$|S|$$

$$=\prod _{h=1}^{n-1}\frac{q^{h+1}-1}{q^h-1}$$

$$=\frac{q^n-1}{q-1}.$$

$$|B|⩾\frac{|A{|}^2}{|C|}.$$

$$|C|=\sum _{b\in B}|C_b|.$$

$$\Big{(}\sum_{b\in B}|\text{fiber}_{\phi}(b)|\cdot 1\Big{)}^{2}\leqslant\sum_{b\in B}|\text{fiber}_{\phi}(b)|^{2}\sum_{b\in B}1$$

$$I=\{(w,v)\in S\times E|v\in w^{\prime }\}.$$

$$W=\{(w_1,w_2,v)\in S\times S\times E|v\in w_1^{\prime }\cap w_2^{\prime }\}.$$

$$|E|⩾\frac{|I{|}^2}{|W|}.$$

$$|W|=|I|+|S|(|S|-1)q^{n-2}.$$

$$|E|$$

$$=\frac{q^{2n}-q^n}{q^n+q^2-2q}$$

Full text

1 Introduction

The classical Euclidean Kakeya problem is an important problem in harmonic analysis: let $E\subset\mathds{R}^{n}$ be a compact subset. We say $E$ is Kakeya if it contains a unit line segment in every direction. The problem claims that if $E$ is Kakeya then

[TABLE]

where $\dim E$ denotes for either the Hausdorff or the Minkowski dimension (see [Tao08]).

To give a better understanding of the problem we explain the Minkowski dimension. Let $E\subset\mathds{R}^{n}$ and $\delta>0$. Then we define $E_{\delta}=\{x\in\mathds{R}^{n}|\text{dist}(x,E)<\delta\}$ with $\text{dist}(x,E)=\inf\{\|x-a\|\hskip 1.25pt|a\in E\}$. One may think of $E_{\delta}$ as the union of all open balls with radius $r$ around every point in $E$.

This allows us to define the Minkowski dimension for some $E\subset\mathds{R}^{n}$ as

[TABLE]

A good example to see how the definition works is to compute the dimension of a sphere $S\subset\mathds{R}^{3}$ of radius $r$. Then $S_{\delta}=\{x\in\mathds{R}^{3}|r-\delta<||x||<r+\delta\}$. This denotes for a set that contains the open ball with radius $r+\delta$ where the ball with radius $r-\delta$ has been cut out. Thus

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which implies by (1) that

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If $B\subset\mathds{R}^{3}$ is a ball with radius $r$ instead, then

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which means by (1) that

[TABLE]

It is however very hard to solve the Euclidean Kakeya problem which is why the problem is still open for $n>2$. The case $n=2$ was solved by Davies in [Dav71].

In 1999, T. Wolff proposed a disrete analogue of the Kakeya problem. Let $\mathds{F}_{q}$ be the field with $q$ elements. We define a line to be the translate of a one-dimensional linear subspace. The direction of a given line $w^{\prime}$ is the unique one-dimensional subspace $w$ such that $w^{\prime}$ is a translate of $w$. A set $E\subset\mathds{F}_{q}^{n}$ is Kakeya if $E$ contains a line in every direction, i. e. for all $v\in\mathds{F}_{q}^{n}\backslash\{0\}$ there is a $x_{0}\in\mathds{F}^{n}_{q}$ such that $\{x_{0}+tv|t\in\mathds{F}_{q}^{n}\}\subset E$. The problem then claims that for all $n$ there is a $c>0$ (only depending on $n$) such that for all Kakeya sets $E\subset\mathds{F}_{q}^{n}$

[TABLE]

Z. Dvir was able to solve the problem with $c=\frac{1}{n!}$ using the so-called polynomial method.

One generalization of the Kakeya problem is the $k$-plane Furstenberg set problem in $\mathds{F}_{q}^{n}$ (see [EE15], Question 1.3). Let $S_{k}$ be the set of all subspaces $w\subset\mathds{F}_{q}^{n}$ such that $\dim(w)=k$. We define a $k$-plane $w^{\prime}\subset\mathds{F}_{q}^{n}$ to be a translated $k$-dimensional subspace. Given a $k$-plane $w^{\prime}$, its direction is defined as the unique subspace $w\in\mathds{F}_{q}^{n}$ such that $w^{\prime}$ is a translate of $w$. Fix some $c>0$ and let $E\subset\mathds{F}_{q}^{n}$ be such that for every direction $w\in S_{k}$ there is at least one $k$-plane $w^{\prime}$ with $|w^{\prime}\cap E|\geqslant q^{c}$. The problem asks for a lower bound for $|E|$.

This problem implies the finite field Kakeya problem namely when $c=k=1$. Note that $c=k$ means $w^{\prime}\subset E$. We therefore consider this case as the generalization of the Kakeya problem.

Definition 1** (Kakeya set with respect to $k$-planes).**

A set $E\subset\mathds{F}_{q}^{n}$ is Kakeya with respect to $k$-planes if for every $w\in S_{k}$ there is a $k$-plane $w^{\prime}\subset\mathds{F}_{q}^{n}$ in direction $w$ such that $w^{\prime}\subset E$.

We will only discuss the size of Kakeya sets w.r.t. to $(n-1)$-planes also known as hyperplanes and will therefore call a set Kakeya if it is Kakeya w.r.t. hyperplanes. From the result from Dvir one would expect that for every $n$ there is a constant $c>0$ such that for every Kakeya set $E\subset\mathds{F}_{q}^{n}$ the inequality $|E|\geqslant cq^{n}$ from the original problem holds. In fact, we are able to show the following theorem.

Theorem 1**.**

Every set $E\subset\mathds{F}^{n}_{q}$ that is Kakeya fulfills

[TABLE]

A known proof, independent of Dvirs method, solves the problem when $n=2$ and gives

[TABLE]

as a bound for every set $E\subset\mathds{F}_{q}^{2}$ that is Kakeya. This proof is a discrete version of the previously mentioned proof in [Dav71]. We were not able to trace it back in the literature, but to read an exposition of this proof see [Sla14].

We modify and generalize the proof so that it gives Theorem 1. What is surprising is that while for $n=2$ the asymptotic bound is $|E|\geqslant\frac{q^{2}}{2}-O(q)$, for $n>2$ it is $|E|\geqslant q^{n}-O(q^{2})$ rather than the expected $|E|\geqslant cq^{n}-O(q^{n-1})$ with some $c\in(0,1)$.

2 The proof

Before being able to prove the theorem we need to prove two lemmas. Let $S=S_{n-1}$.

Lemma 1**.**

The set $S$ fulfills

[TABLE]

Proof.

Let

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We may choose any element of $\mathds{F}^{n}_{q}\backslash\{0\}$ for $k_{1}$. If $k_{2}\in\text{span}(k_{1})$ then $\text{span}(k_{1},k_{2})=\text{span}(k_{1})$. If however $k_{2}\notin\text{span}(k_{1})$ then $\text{span}(k_{1})\subsetneq\text{span}(k_{1},k_{2})$ and hence $\dim\text{span}(k_{1},k_{2})=2$. Thus $k_{2}\in\mathds{F}_{q}^{n}\backslash\text{span}(k_{1})$. One can easily generalize this argument and conclude that $k_{h}\in\mathds{F}^{n}_{q}\backslash\text{span}(k_{1},\dots,k_{h-1}).$ Therefore

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Define $\phi:K\mapsto S$ to be $\phi(k_{1},\dots,k_{n-1})=\text{span}(k_{1},\dots,k_{n-1})$. Thus

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We count the number of $(k_{1},\dots,k_{n-1})\in\text{fiber}_{\phi}(w)$. We may choose any element from $w\backslash\{0\}$ for $k_{1}$. Using a similar argument as above we see that $k_{h}\in w\backslash\text{span}(k_{1},\dots,k_{h-1})$ which shows that

[TABLE]

Therefore

[TABLE]

∎

Lemma 2**.**

Let $A$ and $B$ be finite sets and let $\phi:A\rightarrow B$ be a map. If $C=\{(a_{1},a_{2})\in A\times A|\phi(a_{1})=\phi(a_{2})\}$ then

[TABLE]

Proof.

Let $b\in B$ and define $C_{b}=\{(a_{1},a_{2})\in C|\phi(a_{1})=b\}$. Note that

[TABLE]

For a fixed $b$ one may choose any $a_{1}\in\text{fiber}_{\phi}(b)$ and any $a_{2}\in\text{fiber}_{\phi}(b)$. Thus $|C_{b}|=|\text{fiber}_{\phi}(b)|^{2}$. By applying the Cauchy-Schwarz inequality we obtain

[TABLE]

which is by the previous observations equivalent to the lemma. ∎

Proof of Theorem 1.

For every $w\in S$ take a $v_{w}\in\mathds{F}_{q}^{n}$ such that $w^{\prime}=v_{w}+w$ is contained in $E$. Let

[TABLE]

There are $|S|$ possibilities for $w$ while every $w$ contributes $q^{n-1}$ points since $|w^{\prime}|=q^{n-1}$. Thus $|I|=|S|q^{n-1}$.

Let $\gamma:I\rightarrow E$ be $\gamma(w,v)=v$ and define

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It is easy to see that there is a bijection between $W$ and $K=\{((w_{1},v_{1}),(w_{2},v_{2}))\in I\times I|v_{1}=v_{2}\}$ given by $(w_{1},w_{2},v)\mapsto((w_{1},v),(w_{2},v))$. This implies by Lemma 2 that

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The last step before being able to obtain the final inequality is to compute $|W|$. To do so, we differ two cases:

$w_{1}=w_{2}$: This case contributes $|I|$ elements since for every $(w_{1},v)\in I$ there is a $(w_{1},w_{1},v)\in W$ and vice versa. 2. 2.

$w_{1}\neq w_{2}$: Note first that by assumption $w_{1}\nparallel w_{2}$. By the dimension formula for subscpaces $w_{1}^{\prime}\cap w_{2}^{\prime}$ has cardinality $q^{n-2}$. There are $|S|(|S|-1)$ ordered pairs of not parallel hyperplanes. Hence this case contributes $|S|(|S|-1)q^{n-2}$ elements to $W$.

This leaves us with

[TABLE]

By using (2) and Lemma 1 we deduce that

[TABLE]

∎

Note how our final bound for $|E|$ has the form $|E|\geqslant q^{n}-O(q^{2})$ for $n>2$ and $|E|\geqslant\frac{q^{2}}{2}-O(q)$ for $n=2$. Namely, when $n>2$ the leading term in the denominator is $q^{n}$ while it is $2q^{2}$ when $n=2$.

3 Acknowledgements

I want to thank SwissMAP for providing me with the opportunity to participate in this program. I also want to thank Kaloyan Slavov for his support throughout the whole project and his proposal to do this research project.