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Prove or disaprove the following statment:

Let $q$ the power of a prime number $p$ and $n,m\in \mathbb{Z}^{+}$ such that $n<m$. If $K$ is a Kakeya set in $F_q^m$ then $K\times F_q^{n-m}$ is a Kakeya set in $F_q^n$.

Attempt

Actually this is true.

Suppose that $K\times F_q^{n-m}$ isn´t a Kakeya set in $F_q^n$ then we must have $$|K\times F_q^{n-m}|<\frac{q^n}{n!}$$ But notice that $$|K\times F_q^{n-m}|=|K||F_q^{n-m}|=|K|q^{n-m}$$

Since $K$ is a Kakeya set in $F_q^m$ then $|K|\geq \frac{q^m}{m!}$ and $m!<n!\rightarrow \frac{1}{n!}<\frac{1}{m!}$ Therefore

$$|K||F_q^{n-m}|=|K|q^{n-m}\geq \frac{q^m}{m!}q^{n-m}\geq \frac{q^n}{n!}$$ a contradiction.

Then $K$ is a Kakeya set.

Is the proof right? if it had a mistake where and how can i prove that.

Note: Let $q$ the power of a prime number $p$, consider the finite field $F_q^n$, A set $K\subset F_q^n$ is a Kakeya set if for every direction $v\in F_q^n\setminus \lbrace 0 \rbrace$ the line $L_{v,a}\subset K$

Alan Jr
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    Your proof is wrong since you don´t get a contradiction with any hypothesis – Brien Navarro Sep 09 '21 at 01:09
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    Your "proof" does not discuss those lines for every $v\in\Bbb{F}_q^n\setminus{0}$ so it does not even look like a proof. I'm not an expert, but it sure sounds like the cardinality result is just a lower bound for the size of a Kakeya set. So the step where you "deduce" that because a set is not Kakeya set, it must be small, is non-sense. – Jyrki Lahtonen Sep 09 '21 at 04:01
  • So why don't you pick an arbitrary vector $v\in\Bbb{F}_q^n\setminus{0}$, and then split the argument into two cases according to whether the projection of $v$ to the subspace $\Bbb{F}_q^m$ is non-zero or not? – Jyrki Lahtonen Sep 09 '21 at 04:22

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I haven't checked any sources, but presumably the correct definition of a Kakeya set is the following.

$K\subseteq \Bbb{F}_q^m$ is a Kakeya set if to each vector $v$ in $\Bbb{F}_q^m\setminus\{0\}$ (= a direction) there exists a vector $a\in\Bbb{F}_q^m$ such that the line $$L_{v,a}=\{a+tv\mid t\in\Bbb{F}_q\}\subseteq K.$$

Apparently $|K|\ge q^m/m!$ is a known lower bound to the size of a Kakeya set $K$. A fundamental mistake in your attempt is that you treat this lower bound (a necessary condition) as a sufficient condition. After such a logical mistake no recovery is possible.

Assuming I guessed the correct defition of a Kakeya set (please provide one!), the claim is straight forward using

The observation: A vector $(x,y)\in \Bbb{F}_q^m\times \Bbb{F}_q^{n-m}$ is an element of $K\times \Bbb{F}_q^{n-m}$ if and only if $x\in K$.

The main claim follows from this immediately. Let us fix a direction $v=(v_1,v_2)\neq0$, where $v_1\in\Bbb{F}_q^m, v_2\in\Bbb{F}_q^{n-m}$.

  • If $v_1\neq0$ then by the definition of a Kakeya set there exists a vector $a_1\in\Bbb{F}_q^m$ such that $a_1+tv_1\in K$ for all $t\in\Bbb{F}_q$. Let us choose $a=(a_1,0)\in \Bbb{F}_q^m\times\Bbb{F}_q^{n-m}$. By the observation all the vectors $a+tv=(a_1+tv_1,tv_2)$ are elements of $K\times\Bbb{F}_q^{n-m}$. Hence this line $L_{a,v}$ works.
  • If $v_1=0$ then we can select any point $a_1\in K$, and the line $L_{a,v}$ with $a=(a_1,0)$ works.

Hence $K\times \Bbb{F}_q^{n-m}$ is a Kakeya set.

Jyrki Lahtonen
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