Difference between revisions of "Kakeya problem"

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== General lower bounds ==
 
== General lower bounds ==
  
Trivially, we have
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Trivially,  
  
 
:<math>k_n\le k_{n+1}\le 3k_n</math>.
 
:<math>k_n\le k_{n+1}\le 3k_n</math>.
  
Next, the Cartesian product of two Kakeya sets is another Kakeya set; hence,  
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Since the Cartesian product of two Kakeya sets is another Kakeya set, we have
  
:<math>k_{n+m} \leq k_m k_n</math>,
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:<math>k_{n+m} \leq k_m k_n</math>;
  
implying that <math>k_n^{1/n}</math> converges to a limit as <math>n</math> goes to infinity.
+
this implies that <math>k_n^{1/n}</math> converges to a limit as <math>n</math> goes to infinity.
  
 
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, but this is superseded by the estimates given below.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, but this is superseded by the estimates given below.
  
We have
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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, etermining this direction. Therefore, <math>\binom{k_n}2\ge 3(3^n-1)/2<math>, and hence
  
:<math>k_n(k_n-1)\ge 3(3^n-1)</math>
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:<math>k_n\gtsym \sqrt{3^{(n+1)/2}}</math>
  
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>.
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One can get essentially the same conclusion using the "bush" argument.  There are <math>N := (3^n-1)/2</math> different directions.  Take a line in every direction, let E be the union of these lines, and let <math>\mu</math> be the maximum multiplicity of these lines (i.e. the largest number of lines that are concurrent at a point).  On the one hand, from double counting we see that E has cardinality at least <math>3N/\mu</math>.  On the other hand, by considering the "bush" of lines emanating from a point with multiplicity <math>\mu</math>, we see that E has cardinality at least <math>2\mu+1</math>.  If we minimise <math>\max(3N/\mu, 2\mu+1)</math> over all possible values of <math>\mu</math> one obtains approximately <math>\sqrt{6N} \approx 3^{(n+1)/2}</math> as a lower bound of |E|, which is asymptotically better than <math>(3/2)^n</math>.
 
+
For instance, we can use the "bush" argument.  There are <math>N := (3^n-1)/2</math> different directions.  Take a line in every direction, let E be the union of these lines, and let <math>\mu</math> be the maximum multiplicity of these lines (i.e. the largest number of lines that are concurrent at a point).  On the one hand, from double counting we see that E has cardinality at least <math>3N/\mu</math>.  On the other hand, by considering the "bush" of lines emanating from a point with multiplicity <math>\mu</math>, we see that E has cardinality at least <math>2\mu+1</math>.  If we minimise <math>\max(3N/\mu, 2\mu+1)</math> over all possible values of <math>\mu</math> one obtains approximately <math>\sqrt{6N} \approx 3^{(n+1)/2}</math> as a lower bound of |E|, which is asymptotically better than <math>(3/2)^n</math>.
+
  
 
Or, we can use the "slices" argument.  Let <math>A, B, C \subset ({\Bbb Z}/3{\Bbb Z})^{n-1}</math> be the three slices of a Kakeya set E.  We can form a graph G between A and B by connecting A and B by an edge if there is a line in E joining A and B.  The restricted sumset <math>\{a+b: (a,b) \in G \}</math> is essentially C, while the difference set <math>\{a-b: (a-b) \in G \}</math> is all of <math>({\Bbb Z}/3{\Bbb Z})^{n-1}</math>.  Using an estimate from [http://front.math.ucdavis.edu/math.CO/9906097 this paper of Katz-Tao], we conclude that <math>3^{n-1} \leq \max(|A|,|B|,|C|)^{11/6}</math>, leading to the bound <math>|E| \geq 3^{6(n-1)/11}</math>, which is asymptotically better still.
 
Or, we can use the "slices" argument.  Let <math>A, B, C \subset ({\Bbb Z}/3{\Bbb Z})^{n-1}</math> be the three slices of a Kakeya set E.  We can form a graph G between A and B by connecting A and B by an edge if there is a line in E joining A and B.  The restricted sumset <math>\{a+b: (a,b) \in G \}</math> is essentially C, while the difference set <math>\{a-b: (a-b) \in G \}</math> is all of <math>({\Bbb Z}/3{\Bbb Z})^{n-1}</math>.  Using an estimate from [http://front.math.ucdavis.edu/math.CO/9906097 this paper of Katz-Tao], we conclude that <math>3^{n-1} \leq \max(|A|,|B|,|C|)^{11/6}</math>, leading to the bound <math>|E| \geq 3^{6(n-1)/11}</math>, which is asymptotically better still.

Revision as of 10:21, 18 March 2009

Define a Kakeya set to be a subset [math]A[/math] of [math][3]^n\equiv{\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,

[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

[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, etermining this direction. Therefore, [math]\binom{k_n}2\ge 3(3^n-1)/2\ltmath\gt, and hence :\ltmath\gtk_n\gtsym \sqrt{3^{(n+1)/2}}[/math]

One can get essentially the same conclusion using the "bush" argument. There are [math]N := (3^n-1)/2[/math] different directions. Take a line in every direction, let E be the union of these lines, and let [math]\mu[/math] be the maximum multiplicity of these lines (i.e. the largest number of lines that are concurrent at a point). On the one hand, from double counting we see that E has cardinality at least [math]3N/\mu[/math]. On the other hand, by considering the "bush" of lines emanating from a point with multiplicity [math]\mu[/math], we see that E has cardinality at least [math]2\mu+1[/math]. If we minimise [math]\max(3N/\mu, 2\mu+1)[/math] over all possible values of [math]\mu[/math] one obtains approximately [math]\sqrt{6N} \approx 3^{(n+1)/2}[/math] as a lower bound of |E|, which is asymptotically better than [math](3/2)^n[/math].

Or, we can use the "slices" argument. Let [math]A, B, C \subset ({\Bbb Z}/3{\Bbb Z})^{n-1}[/math] be the three slices of a Kakeya set E. We can form a graph G between A and B by connecting A and B by an edge if there is a line in E joining A and B. The restricted sumset [math]\{a+b: (a,b) \in G \}[/math] is essentially C, while the difference set [math]\{a-b: (a-b) \in G \}[/math] is all of [math]({\Bbb Z}/3{\Bbb Z})^{n-1}[/math]. Using an estimate from this paper of Katz-Tao, we conclude that [math]3^{n-1} \leq \max(|A|,|B|,|C|)^{11/6}[/math], leading to the bound [math]|E| \geq 3^{6(n-1)/11}[/math], which is asymptotically better still.

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.

Question: can the upper bound be strengthened to [math]k_{n+1}\le 2k_n+1[/math]?

Another construction uses the "slices" idea and a construction of Imre Ruzsa. Let [math]A, B \subset [3]^n[/math] be the set of strings with [math]n/3+O(\sqrt{n})[/math] 1's, [math]2n/3+O(\sqrt{n})[/math] 0's, and no 2's; let [math]C \subset [3]^n[/math] be the set of strings with [math]2n/3+O(\sqrt{n})[/math] 2's, [math]n/3+O(\sqrt{n})[/math] 0's, and no 1's, and let [math]E = \{0\} \times A \cup \{1\} \times B \cup \{2\} \times C[/math]. From Stirling's formula we have [math]|E| = (27/4 + o(1))^{n/3}[/math]. Now I claim that for most [math]t \in [3]^{n-1}[/math], there exists an algebraic line in the direction (1,t). Indeed, typically t will have [math]n/3+O(\sqrt{n})[/math] 0s, [math]n/3+O(\sqrt{n})[/math] 1s, and [math]n/3+O(\sqrt{n})[/math] 2s, thus [math]t = e + 2f[/math] where e and f are strings with [math]n/3 + O(\sqrt{n})[/math] 1s and no 2s, with the 1-sets of e and f being disjoint. One then checks that the line [math](0,f), (1,e), (2,2e+2f)[/math] lies in E.

This is already a positive fraction of directions in E. One can use the random rotations trick to get the rest of the directions in E (losing a polynomial factor in n).

Putting all this together, I think we have

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

or

[math](1.8207\ldots+o(1))^n \leq k_n \leq (1.88988+o(1))^n[/math]