Is finite field version kakeya conjecture still true when changing the line of every direction with only 2(or several but not the full line)element?

Question

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.

Answer 1

In his paper "On the size of Kakeya sets in finite fields" (where the proof of the finite field Kakeya conjecture has appeared), Dvir also introduces the notion of a $(\delta,\gamma)$-Kakeya set, which is, essentially, a set $K\subset\mathbb F_q^n$ with the following property: there are at least $\delta(q^n-1)/(q-1)$ directions in $\mathbb F_q^n$ such that in each of these directions there is a line containing at least $\gamma q$ points of $K$. He then proves the following result:

Theorem. If $K\subset\mathbb F_q^n$ is a $(\gamma,\delta)$-Kakeya set, then $$ |K| \ge \binom{n+d-1}{n-1}, $$ where $d=\lfloor q\cdot\min\{\delta,\gamma\}\rfloor-2$.

Applying this theorem with $\delta=1$ and $\gamma=\varepsilon q^{-1+c}$, we conclude that if $K\subset\mathbb F_q^n$ contains at least $\varepsilon q^c$ collinear points in every direction, then $$ |K| \ge \binom{n+d-1}{n-1},\quad d=\lfloor \varepsilon q^c\rfloor-2. $$

Does this answer the question?