# Difference between revisions of "Kakeya problem"

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

Clearly, we have <math>k_1=3</math>, and it is easy to see that <math>k_2=7</math>. Using a computer, it is not difficult to find that <math>k_3=13</math> and <math>k_4\le 27</math>. Indeed, it seems likely that <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. | Clearly, we have <math>k_1=3</math>, and it is easy to see that <math>k_2=7</math>. Using a computer, it is not difficult to find that <math>k_3=13</math> and <math>k_4\le 27</math>. Indeed, it seems likely that <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. | ||

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== General lower bounds == | == General lower bounds == | ||

− | Trivially, | + | Trivially, we have |

:<math>k_n\le k_{n+1}\le 3k_n</math>. | :<math>k_n\le k_{n+1}\le 3k_n</math>. | ||

− | Since the Cartesian product of two Kakeya sets is another Kakeya set, we have | + | Since the Cartesian product of two Kakeya sets is another Kakeya set, we also have |

:<math>k_{n+m} \leq k_m k_n</math>; | :<math>k_{n+m} \leq k_m k_n</math>; | ||

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From a paper of [http://arxiv.org/abs/0901.2529 Dvir, Kopparty, Saraf, and Sudan] it follows that <math>k_n \geq 3^n / 2^n</math>, but this is superseded by the estimates given below. | From a paper of [http://arxiv.org/abs/0901.2529 Dvir, Kopparty, Saraf, and Sudan] it follows that <math>k_n \geq 3^n / 2^n</math>, but this is superseded by the estimates given below. | ||

− | To each of the <math>(3^n-1)/2</math> directions in <math>{\mathbb F}_3^n</math> there correspond at least three pairs of elements in a Kakeya set, | + | To each of the <math>(3^n-1)/2</math> directions in <math>{\mathbb F}_3^n</math> there correspond at least three pairs of elements in a Kakeya set, determining this direction. Therefore, <math>\binom{k_n}{2}\ge 3\cdot(3^n-1)/2</math>, and hence |

:<math>k_n\gtrsim 3^{(n+1)/2}.</math> | :<math>k_n\gtrsim 3^{(n+1)/2}.</math> |

## Revision as of 13:36, 21 March 2009

A **Kakeya set** in [math]{\mathbb F}_3^r$ is a subset \ltmath\gtA\subset{\mathbb F}_3^n[/math] that contains an algebraic line in every direction; that is, for every [math]d\in{\mathbb F}_3^n[/math], there exists [math]a\in{\mathbb F}_3^n[/math] such that [math]a,a+d,a+2d[/math] all lie in [math]A[/math]. Let [math]k_n[/math] be the smallest size of a Kakeya set in [math]{\mathbb F}_3^n[/math].

Clearly, we have [math]k_1=3[/math], and it is easy to see that [math]k_2=7[/math]. Using a computer, it is not difficult to find that [math]k_3=13[/math] and [math]k_4\le 27[/math]. Indeed, it seems likely that [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.

## General lower bounds

Trivially, we have

- [math]k_n\le k_{n+1}\le 3k_n[/math].

Since the Cartesian product of two Kakeya sets is another Kakeya set, we also have

- [math]k_{n+m} \leq k_m k_n[/math];

this implies that [math]k_n^{1/n}[/math] converges to a limit as [math]n[/math] goes to infinity.

From a paper of Dvir, Kopparty, Saraf, and Sudan it follows that [math]k_n \geq 3^n / 2^n[/math], but this is superseded by the estimates given below.

To each of the [math](3^n-1)/2[/math] directions in [math]{\mathbb F}_3^n[/math] there correspond at least three pairs of elements in a Kakeya set, determining this direction. Therefore, [math]\binom{k_n}{2}\ge 3\cdot(3^n-1)/2[/math], and hence

- [math]k_n\gtrsim 3^{(n+1)/2}.[/math]

One can derive essentially the same conclusion using the "bush" argument, as follows. Let [math]E\subset{\mathbb F}_3^n[/math] be a Kakeya set, considered as a union of [math]N := (3^n-1)/2[/math] lines in all different directions. Let [math]\mu[/math] be the largest number of lines that are concurrent at a point of [math]E[/math]. The number of point-line incidences is at most [math]|E|\mu[/math] and at least [math]3N[/math], whence [math]|E|\ge 3N/\mu[/math]. On the other hand, by considering only those points on the "bush" of lines emanating from a point with multiplicity [math]\mu[/math], we see that [math]|E|\ge 2\mu+1[/math]. Comparing the two last bounds one obtains [math]|E|\gtrsim\sqrt{6N} \approx 3^{(n+1)/2}[/math].

A better bound follows by using the "slices argument". Let [math]A,B,C\subset{\mathbb F}_3^{n-1}[/math] be the three slices of a Kakeya set [math]E\subset{\mathbb F}_3^n[/math]. Form a bipartite graph [math]G[/math] with the partite sets [math]A[/math] and [math]B[/math] by connecting [math]a[/math] and [math]b[/math] by an edge if there is a line in [math]E[/math] through [math]a[/math] and [math]b[/math]. The restricted sumset [math]\{a+b\colon (a,b)\in G\}[/math] is contained in the set [math]-C[/math], while the difference set [math]\{a-b\colon (a,b)\in G\}[/math] is all of [math]{\mathbb F}_3^{n-1}[/math]. Using an estimate from a paper of Katz-Tao, we conclude that [math]3^{n-1}\le\max(|A|,|B|,|C|)^{11/6}[/math], leading to [math]|E|\ge 3^{6(n-1)/11}[/math]. Thus,

- [math]k_n \ge 3^{6(n-1)/11}.[/math]

## General upper bounds

We have

- [math]k_n\le 2^{n+1}-1[/math]

since the set of all vectors in [math]{\mathbb F}_3^n[/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.

This estimate can be improved using an idea due to Ruzsa (seems to be unpublished). Namely, let [math]E:=A\cup B[/math], where [math]A[/math] is the set of all those vectors with [math]r/3+O(\sqrt r)[/math] coordinates equal to [math]1[/math] and the rest equal to [math]0[/math], and [math]B[/math] is the set of all those vectors with [math]2r/3+O(\sqrt r)[/math] coordinates equal to [math]2[/math] and the rest equal to [math]0[/math]. Then [math]E[/math], being of size just about [math](27/4)^{r/3}[/math] (which is not difficult to verify using Stirling's formula), contains lines in a positive proportion of directions: for, a typical direction [math]d\in {\mathbb F}_3^n[/math] can be represented as [math]d=d_1+2d_2[/math] with [math]d_1,d_2\in A[/math], and then [math]d_1,d_1+d,d_1+2d\in E[/math]. Now one can use the random rotations trick to get the rest of the directions in [math]E[/math] (losing a polynomial factor in [math]n[/math]).

Putting all this together, we seem to have

- [math](3^{6/11} + o(1))^n \le k_n \le ( (27/4)^{1/3} + o(1))^n[/math]

or

- [math](1.8207+o(1))^n \le k_n \le (1.8899+o(1))^n.[/math]