The classical finite field Kakeya conjecture state as following (for conveinent, all version of kakeya conjecture is state in hasdorff dimension version):
$\mathrm{Finite\ Field\ Kakeya\ Conjecture}$: Let $\mathrm{F}=\mathrm{F}/\mathrm{qF}$ be a finite field, let $K \subseteq \mathrm{F}^{\mathrm{n}}$ be a Kakeya set, i.e. for each vector $y \in \mathbf{F}^{n}$ there exists $x \in \mathbf{F}^{n}$ such that $K$ contains a line $\{x+t y: t \in \mathbf{F}\}$. Then the set $K$ has size at least $c_{n}q^{n}$ where $c_{n}>0$ is a constant that only depends on $n$.

.....................(*)

The Drir's proof to finite field version of Kakeya conjecture is elegant and influential. On the other hand, it seems the corresponding version of (*) in $\mathrm{R^n}$ is that,

a set $K$ if satisfied, for every $\theta \in S^{n-1}$ $K$ contain a line $l_{\theta, a_{\theta}}$ and $l$ has direction $\theta$ , then the haussdorff dimension of $K$ is $n$.

But this version in Euclidean space is not essential difficult, in particuler because every line in $R^n$ is not compact and a suitable infinite sequence of line shall group a set which has hausdroff dimension $n$. The original version of Kakeya conjecture in $R^n$ change line to a segement with length 1. So, is there a possibility that a more subtle version of finite field analoge of Kakeya conjecture maybe true? i.e.:

$\mathrm{Version \ 1}$: Fix a sequence $\{a_m\}_{m=1}^{\infty}$, $a_{m} \to \infty$ as $m \to \infty$ and we define a set $K$ in $F_{p_m}^n= (Z/Z_{p_m})^n $ to be a Kakeya set iff for every vector $v\in F_{p_m}^n, v\neq 0 $, such that $K$ contain consecutively(or not) at least $a_n$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_m}^n$, i.e. contain every direction line with at least $a_m$ elements, then exist $c>0$ only depend on $m$ such that $|K|\geq cp_m^{n-1}$?\

or more easier problem(compare with version 1) but the directly use of polynominal method Invalidation is a following:

$\mathrm{Version \ 2}$: we define a set $K$ in $F_{p_m}^n= (Z/Z_{p_m})^n $ to be a Kakeya set iff for every vector $v\in F_{p_m}^n, v\neq 0 $, such that $K$ contain consecutive(or not) at least $p_m^{\frac{1}{2}}$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_m}^n$, i.e. contain every direction line with at least $p_m^{\frac{1}{2}}$ elements, then exist $c>0$ only depend on $n$ such that $|K|\geq cp_m^{n-1}$?

for $\mathrm{Version \ 2}$, the polynomial argument gain a bound $O(p^{\frac{m}{2}})$ which coincide with the combinatorics argument two point depending a line. And it is esay to show if take $a_m=2, \forall m \in N^*$ in $\mathrm{Version \ 1}$, then this set $K $ in general do not have full dimension, this corresponde to the situation in $R^n$ when we have a scale $\delta$ kakeya set but instead of $T^{\delta}_{a_{\theta},\theta}$ is a $1\times \delta^{n-1}$ tube, we only have a $O(\delta) \times\delta^{n-1}$ tube.

The classical finite field Kakeya conjecture state as following (for conveinent, all version of kakeya conjecture is state in hasdorff dimension version):
$\mathrm{Finite\ Field\ Kakeya\ Conjecture}$: Let $\mathrm{F}=\mathrm{F}/\mathrm{qF}$ be a finite field, let $K \subseteq \mathrm{F}^{\mathrm{n}}$ be a Kakeya set, i.e. for each vector $y \in \mathbf{F}^{n}$ there exists $x \in \mathbf{F}^{n}$ such that $K$ contains a line $\{x+t y: t \in \mathbf{F}\}$. Then the set $K$ has size at least $c_{n}q^{n}$ where $c_{n}>0$ is a constant that only depends on $n$.

.....................(*)

Dvir's proof of the finite field version of Kakeya conjecture is elegant and influential. On the other hand, it seems the corresponding version of (*) in $\mathrm{R^n}$ is that,

a set $K$ if satisfied, for every $\theta \in S^{n-1}$ $K$ contain a line $l_{\theta, a_{\theta}}$ and $l$ has direction $\theta$ , then the haussdorff dimension of $K$ is $n$.

But this version in Euclidean space is not essential difficult, in particuler because every line in $R^n$ is not compact and a suitable infinite sequence of line shall group a set which has hausdroff dimension $n$. The original version of Kakeya conjecture in $R^n$ change line to a segement with length 1. So, is there a possibility that a more subtle version of finite field analoge of Kakeya conjecture maybe true? i.e.:

$\mathrm{Version \ 1}$: Fix a sequence $\{a_m\}_{m=1}^{\infty}$, $a_{m} \to \infty$ as $m \to \infty$ and we define a set $K$ in $F_{p_m}^n= (Z/Z_{p_m})^n $ to be a Kakeya set iff for every vector $v\in F_{p_m}^n, v\neq 0 $, such that $K$ contain consecutively(or not) at least $a_n$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_m}^n$, i.e. contain every direction line with at least $a_m$ elements, then exist $c>0$ only depend on $m$ such that $|K|\geq cp_m^{n-1}$?\

or more easier problem(compare with version 1) but the directly use of polynominal method Invalidation is a following:

$\mathrm{Version \ 2}$: we define a set $K$ in $F_{p_m}^n= (Z/Z_{p_m})^n $ to be a Kakeya set iff for every vector $v\in F_{p_m}^n, v\neq 0 $, such that $K$ contain consecutive(or not) at least $p_m^{\frac{1}{2}}$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_m}^n$, i.e. contain every direction line with at least $p_m^{\frac{1}{2}}$ elements, then exist $c>0$ only depend on $n$ such that $|K|\geq cp_m^{n-1}$?

for $\mathrm{Version \ 2}$, the polynomial argument gain a bound $O(p^{\frac{m}{2}})$ which coincide with the combinatorics argument two point depending a line. And it is esay to show if take $a_m=2, \forall m \in N^*$ in $\mathrm{Version \ 1}$, then this set $K $ in general do not have full dimension, this corresponde to the situation in $R^n$ when we have a scale $\delta$ kakeya set but instead of $T^{\delta}_{a_{\theta},\theta}$ is a $1\times \delta^{n-1}$ tube, we only have a $O(\delta) \times\delta^{n-1}$ tube.

The classical finite field Kakeya conjecture state as following (for conveinent, all version of kakeya conjecture is state in hasdorff dimension version):
$\mathrm{Finite\ Field\ Kakeya\ Conjecture}$: Let $\mathrm{F}=\mathrm{F}/\mathrm{qF}$ be a finite field, let $K \subseteq \mathrm{F}^{\mathrm{n}}$ be a Kakeya set, i.e. for each vector $y \in \mathbf{F}^{n}$ there exists $x \in \mathbf{F}^{n}$ such that $K$ contains a line $\{x+t y: t \in \mathbf{F}\}$. Then the set $K$ has size at least $c_{n}q^{n}$ where $c_{n}>0$ is a constant that only depends on $n$.

.....................(*)

The Drir's proof to finite field version of Kakeya conjecture is elegant and influential. On the other hand, it seems the corresponding version of (*) in $\mathrm{R^n}$ is that,

a set $K$ if satisfied, for every $\theta \in S^{n-1}$ $K$ contain a line $l_{\theta, a_{\theta}}$ and $l$ has direction $\theta$ , then the haussdorff dimension of $K$ is $n$.

But this version in Euclidean space is not essential difficult, in particuler because every line in $R^n$ is not compact and a suitable infinite sequence of line shall group a set which has hausdroff dimension $n$. The original version of Kakeya conjecture in $R^n$ change line to a segement with length 1. So, is there a possibility that a more subtle version of finite field analoge of Kalaya conjecture maybe true? i.e.:

$\mathrm{Version \ 1}$: Fix a sequence $\{a_n\}_{n=1}^{\infty}$, $a_{n} \to \infty$ as $n \to \infty$ and we define a set $K$ in $F_{p_n}^m= (Z/Z_{p_n})^m $ to be a Kakeya set iff for every vector $v\in F_{p_n}^m, v\neq 0 $, such that $K$ contain consecutively(or not) at least $a_n$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_n}^m$, i.e. contain every direction line with at least $a_n$ elements, then exist $c>0$ only depend on $m$ such that $|K|\geq cp_n^m$?\

or more easier problem(compare with version 1) but the directly use of polynominal method Invalidation is a following:

$\mathrm{Version \ 2}$: we define a set $K$ in $F_{p_n}^m= (Z/Z_{p_n})^m $ to be a Kakeya set iff for every vector $v\in F_{p_n}^m, v\neq 0 $, such that $K$ contain consecutive(or not) at least $p_n^{\frac{1}{2}}$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_n}^m$, i.e. contain every direction line with at least $p_n^{\frac{1}{2}}$ elements, then exist $c>0$ only depend on $m$ such that $|K|\geq cp_n^m$?

for $\mathrm{Version \ 2}$, the polynomial argument gain a bound $O(p^{\frac{n}{2}})$ which coincide with the combinatorics argument two point depending a line. And it is esay to show if take $a_n=2, \forall n \in N^*$ in $\mathrm{Version \ 1}$, then this set $K $ in general do not have full dimension, this corresponde to the situation in $R^n$ when we have a scale $\delta$ kakeya set but instead of $T^{\delta}_{a_{\theta},\theta}$ is a $1\times \delta^{n-1}$ tube, we only have a $O(\delta) \times\delta^{n-1}$ tube.

The classical finite field Kakeya conjecture state as following (for conveinent, all version of kakeya conjecture is state in hasdorff dimension version):
$\mathrm{Finite\ Field\ Kakeya\ Conjecture}$: Let $\mathrm{F}=\mathrm{F}/\mathrm{qF}$ be a finite field, let $K \subseteq \mathrm{F}^{\mathrm{n}}$ be a Kakeya set, i.e. for each vector $y \in \mathbf{F}^{n}$ there exists $x \in \mathbf{F}^{n}$ such that $K$ contains a line $\{x+t y: t \in \mathbf{F}\}$. Then the set $K$ has size at least $c_{n}q^{n}$ where $c_{n}>0$ is a constant that only depends on $n$.

.....................(*)

The Drir's proof to finite field version of Kakeya conjecture is elegant and influential. On the other hand, it seems the corresponding version of (*) in $\mathrm{R^n}$ is that,

a set $K$ if satisfied, for every $\theta \in S^{n-1}$ $K$ contain a line $l_{\theta, a_{\theta}}$ and $l$ has direction $\theta$ , then the haussdorff dimension of $K$ is $n$.

But this version in Euclidean space is not essential difficult, in particuler because every line in $R^n$ is not compact and a suitable infinite sequence of line shall group a set which has hausdroff dimension $n$. The original version of Kakeya conjecture in $R^n$ change line to a segement with length 1. So, is there a possibility that a more subtle version of finite field analoge of Kakeya conjecture maybe true? i.e.:

$\mathrm{Version \ 1}$: Fix a sequence $\{a_m\}_{m=1}^{\infty}$, $a_{m} \to \infty$ as $m \to \infty$ and we define a set $K$ in $F_{p_m}^n= (Z/Z_{p_m})^n $ to be a Kakeya set iff for every vector $v\in F_{p_m}^n, v\neq 0 $, such that $K$ contain consecutively(or not) at least $a_n$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_m}^n$, i.e. contain every direction line with at least $a_m$ elements, then exist $c>0$ only depend on $m$ such that $|K|\geq cp_m^{n-1}$?\

or more easier problem(compare with version 1) but the directly use of polynominal method Invalidation is a following:

$\mathrm{Version \ 2}$: we define a set $K$ in $F_{p_m}^n= (Z/Z_{p_m})^n $ to be a Kakeya set iff for every vector $v\in F_{p_m}^n, v\neq 0 $, such that $K$ contain consecutive(or not) at least $p_m^{\frac{1}{2}}$ point in a line $\{x+t v: t \in \mathbf{F}\}$ for some $x\in F_{p_m}^n$, i.e. contain every direction line with at least $p_m^{\frac{1}{2}}$ elements, then exist $c>0$ only depend on $n$ such that $|K|\geq cp_m^{n-1}$?

for $\mathrm{Version \ 2}$, the polynomial argument gain a bound $O(p^{\frac{m}{2}})$ which coincide with the combinatorics argument two point depending a line. And it is esay to show if take $a_m=2, \forall m \in N^*$ in $\mathrm{Version \ 1}$, then this set $K $ in general do not have full dimension, this corresponde to the situation in $R^n$ when we have a scale $\delta$ kakeya set but instead of $T^{\delta}_{a_{\theta},\theta}$ is a $1\times \delta^{n-1}$ tube, we only have a $O(\delta) \times\delta^{n-1}$ tube.