Kakeya problem

2009-03-14T19:45:58Z

217.132.211.250:

Define a '''Kakeya set''' to be a subset <math>A</math> of <math>[3]^r\equiv{\mathbb F}_3^r</math> that contains an [[algebraic line]] in every direction; that is, for every <math>d\in{\mathbb F}_3^r</math>, there exists <math>a\in{\mathbb F}_3^r</math> such that <math>a,a+d,a+2d</math> all lie in <math>A</math>. Let <math>k_r</math> be the smallest size of a Kakeya set in <math>{\mathbb F}_3^r</math>.

Clearly, we have <math>k_1=3</math>, and it is easy to see that <math>k_2=7</math>. Using a computer, I also found <math>k_3=13</math> and <math>k_4\le 27</math>. I suspect that, indeed, <math>k_4=27</math> holds (meaning that in <math>{\mathbb F}_3^4</math> one cannot get away with just <math>26</math> elements), and I am very curious to know whether <math>k_5=53</math>: notice the pattern in

:<math>3,7,13,27,53,\ldots</math>

As to the general estimates, we have

:<math>k_r(k_r-1)\ge 3(3^r-1)</math>

and, on the other hand,

:<math>k_r\le 2^{r+1}-1</math>:

the former since for each <math>d\in {\mathbb F}_3^r\setminus\{0\}</math> there are at least three ordered pairs of elements of a Kakeya set with difference <math>d</math>, the latter due to the fact that the set of all vectors in <math>{\mathbb F}_3^r</math> such that at least one of the numbers <math>1</math> and <math>2</math> is missing among their coordinates is a Kakeya set. (I actually can improve the lower bound to something like <math>k_r\gg 3^{0.51r}</math>.) Also, we have the trivial inequalities

:<math>k_r\le k_{r+1}\le 3k_r</math>;

can the upper bound be strengthened to <math>k_{r+1}\le 2k_r+1</math>?

[http://arxiv.org/abs/0901.2529 Dvir, Kopparty, Saraf, and Sudan] showed that <math>k_r \geq 3^r / 2^r</math>.

The Cartesian product of two Kakeya sets is another Kakeya set; this implies that <math>k_{r+s} \leq k_r k_s</math>.

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Kakeya problem