Difference between revisions of "Kakeya problem"

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Define a '''Kakeya set''' to be a subset A of <math>[3]^n \equiv ({\Bbb Z}/3{\Bbb Z})^n</math> that contains an [[algebraic line]] in every direction.  Thus, for every <math>r \in ({\Bbb Z}/3{\Bbb Z})^n</math>, there exists <math>a \in ({\Bbb Z}/3{\Bbb Z})^n</math> such that a, a+r, a+2r all lie in A.  Let <math>k_n</math> be the smallest size of a Kakeya set in n dimensions.
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A '''Kakeya set''' in <math>{\mathbb F}_3^n</math> is a subset <math>E\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>e\in{\mathbb F}_3^n</math> such that <math>e,e+d,e+2d</math> all lie in <math>E</math>.  Let <math>k_n</math> be the smallest size of a Kakeya set in <math>{\mathbb F}_3^n</math>.
  
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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, 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 26 elements), and I am very curious to know whether <math>k_5=53</math>: notice the pattern in
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== Basic Estimates ==
  
:<math>3,7,13,27,53,\ldots</math>
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Trivially, we have
  
As to the general estimates, we have
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:<math>k_n\le k_{n+1}\le 3k_n</math>.
  
:<math>k_r(k_r-1)\ge 3(3^r-1)</math>
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Since the Cartesian product of two Kakeya sets is another Kakeya set, the upper bound can be extended to
  
and, on the other hand,
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:<math>k_{n+m} \leq k_m k_n</math>;
  
:<math>k_r\le 2^{r+1}-1</math>:
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this implies that <math>k_n^{1/n}</math> converges to a limit as <math>n</math> goes to infinity.
  
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 d, 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 1 and 2 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
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== Lower Bounds ==
  
:<math>k_r\le k_{r+1}\le 3k_r</math>;
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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
  
can the upper bound be strengthened to <math>k_{r+1}\le 2k_r+1</math>?
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:<math>k_n\ge 3^{(n+1)/2}.</math>
  
[http://arxiv.org/abs/0901.2529 Dvir, Kopparty, Saraf, and Sudan] showed that <math>k_r \geq 3^r / 2^r</math>.
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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
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<math>|E|\gtrsim\sqrt{6N} \approx 3^{(n+1)/2}</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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The better estimate
 +
 
 +
:<math>k_n\ge (9/5)^n</math>
 +
 
 +
is obtained in [http://arxiv.org/abs/0901.2529 a paper of Dvir, Kopparty, Saraf, and Sudan]. (In general, they show that a Kakeya set in the <math>n</math>-dimensional vector space over the <math>q</math>-element field has at least <math>(q/(2-1/q))^n</math> elements).
 +
 
 +
A still 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 [http://front.math.ucdavis.edu/math.CO/9906097 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,
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 +
:<math>k_n \ge 3^{6(n-1)/11}.</math>
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 +
== 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>
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or
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:<math>(1.8207+o(1))^n \le k_n \le (1.8899+o(1))^n.</math>

Latest revision as of 00:35, 5 June 2009

A Kakeya set in [math]{\mathbb F}_3^n[/math] is a subset [math]E\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]e\in{\mathbb F}_3^n[/math] such that [math]e,e+d,e+2d[/math] all lie in [math]E[/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.

Basic Estimates

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, the upper bound can be extended to

[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.

Lower Bounds

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\ge 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].

The better estimate

[math]k_n\ge (9/5)^n[/math]

is obtained in a paper of Dvir, Kopparty, Saraf, and Sudan. (In general, they show that a Kakeya set in the [math]n[/math]-dimensional vector space over the [math]q[/math]-element field has at least [math](q/(2-1/q))^n[/math] elements).

A still 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]

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]