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	<title>Comments on: Dvir&#8217;s proof of the finite field Kakeya conjecture</title>
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	<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/</link>
	<description>Updates on my research and expository papers, discussion of open problems, and other maths-related topics.  By Terence Tao</description>
	<pubDate>Thu, 07 Aug 2008 21:34:12 +0000</pubDate>
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		<title>By: Francesco Mazzocca</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-31317</link>
		<dc:creator>Francesco Mazzocca</dc:creator>
		<pubDate>Sun, 20 Jul 2008 17:12:05 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-31317</guid>
		<description>The results by A.Blokhuis and F.Mazzocca about the finite Kakeya problem in two dimensional case, cited by Simeon Ball, will appear on the last volume of Bolyai Society Mathematical Studies dedicated to L.Lovász’s 60th birthday ( http://www.springer.com/math/numbers/book/978-3-540-85218-6?detailsPage=toc )</description>
		<content:encoded><![CDATA[<p>The results by A.Blokhuis and F.Mazzocca about the finite Kakeya problem in two dimensional case, cited by Simeon Ball, will appear on the last volume of Bolyai Society Mathematical Studies dedicated to L.Lovász’s 60th birthday ( <a href="http://www.springer.com/math/numbers/book/978-3-540-85218-6?detailsPage=toc" rel="nofollow">http://www.springer.com/math/numbers/book/978-3-540-85218-6?detailsPage=toc</a> )</p>
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		<title>By: Simeon Ball</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-29723</link>
		<dc:creator>Simeon Ball</dc:creator>
		<pubDate>Tue, 13 May 2008 08:19:42 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-29723</guid>
		<description>I forgot to mention in the previous post that L should be a tangent to C at a point P and that m is incident with P.</description>
		<content:encoded><![CDATA[<p>I forgot to mention in the previous post that L should be a tangent to C at a point P and that m is incident with P.</p>
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		<title>By: Simeon Ball</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-29638</link>
		<dc:creator>Simeon Ball</dc:creator>
		<pubDate>Fri, 09 May 2008 14:23:02 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-29638</guid>
		<description>The bound in the two-dimensional case for q odd was recently improved to q(q+1)/2+(q-1)/2 by Blokhuis and Mazzocca. 

This bound is tight as seen by the conic construction. Let C be a conic in PG(2,q), L the line at infinity, and m a bisecant to C (a line is incident with two points of the conic). Let E be the external points to C (which are the points that are incident with two tangents of C). Then the set
(E U C U m) \ L is a Kakeya (Besikovitch) set with q(q+1)/2+(q-1)/2 points.

Moreover, Blokhuis and Mazzocca have shown that all such sets that attain the bound come from this construction. The bound follows from an old result of Jamison. I have detailed this in
http://www-ma4.upc.es/~simeon/kakeya.pdf
To classify the sets meeting the bound, you use similar arguments to those used by Segre in the fifties to prove that a set of q+1 points in PG(2,q), q odd, with the property that every line is incident with at most two points of the set, is a conic. The proof of this has been done by Blokhuis and Mazzocca, although I'm not sure this preprint is available yet.</description>
		<content:encoded><![CDATA[<p>The bound in the two-dimensional case for q odd was recently improved to q(q+1)/2+(q-1)/2 by Blokhuis and Mazzocca. </p>
<p>This bound is tight as seen by the conic construction. Let C be a conic in PG(2,q), L the line at infinity, and m a bisecant to C (a line is incident with two points of the conic). Let E be the external points to C (which are the points that are incident with two tangents of C). Then the set<br />
(E U C U m) \ L is a Kakeya (Besikovitch) set with q(q+1)/2+(q-1)/2 points.</p>
<p>Moreover, Blokhuis and Mazzocca have shown that all such sets that attain the bound come from this construction. The bound follows from an old result of Jamison. I have detailed this in<br />
<a href="http://www-ma4.upc.es/~simeon/kakeya.pdf" rel="nofollow">http://www-ma4.upc.es/~simeon/kakeya.pdf</a><br />
To classify the sets meeting the bound, you use similar arguments to those used by Segre in the fifties to prove that a set of q+1 points in PG(2,q), q odd, with the property that every line is incident with at most two points of the set, is a conic. The proof of this has been done by Blokhuis and Mazzocca, although I&#8217;m not sure this preprint is available yet.</p>
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		<title>By: Terence Tao</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-29605</link>
		<dc:creator>Terence Tao</dc:creator>
		<pubDate>Thu, 08 May 2008 14:49:08 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-29605</guid>
		<description>Dear Seva: You do have a point.  Previous literature on the Kakeya problem did not seek to control the dependence of constants on the dimension, and so there were basically no interesting non-trivial results in the asymptotically high dimensional regime, but it is certainly possible to pose these questions in this regime.  Very different techniques would probably be needed; in particular, in analogy with the Hales-Jewett theory, one might expect methods from ergodic theory to start playing some role.  (In the F_3 case, it's also possible, by analogy with the theory surrounding Roth's theorem, that Fourier-analytic and additive-combinatorial techniques could also be effective.)

Dear Cliff,

In two dimensions, there are reasonably sharp results (giving sets close to q^2/2) in

http://arxiv.org/abs/math/0510356
http://arxiv.org/abs/math/0607734

(and see also a comment in

http://quomodocumque.wordpress.com/2008/03/25/give-the-people-who-like-the-kakeya-problem-what-they-want/

)

In higher dimensions, the best construction I know of uses the two-dimensional construction, combined with the trivial observation that the Cartesian product of two Kakeya sets is still a Kakeya set (in a higher dimensional space, of course).  This gives about q^d / 2^{d/2} in the case when d is even, and slightly worse than this when d is odd.  This is a fair distance away from Dvir's bound, which is about q^d/d!.  So there is still a little bit of room to close the gap here, and in particular to settle the question of whether the density of Kakeya sets needs to decay faster than exponentially (i.e. it is asymptotically less than c^d q^d as d -&#62; infinity for any c &#62; 0).</description>
		<content:encoded><![CDATA[<p>Dear Seva: You do have a point.  Previous literature on the Kakeya problem did not seek to control the dependence of constants on the dimension, and so there were basically no interesting non-trivial results in the asymptotically high dimensional regime, but it is certainly possible to pose these questions in this regime.  Very different techniques would probably be needed; in particular, in analogy with the Hales-Jewett theory, one might expect methods from ergodic theory to start playing some role.  (In the F_3 case, it&#8217;s also possible, by analogy with the theory surrounding Roth&#8217;s theorem, that Fourier-analytic and additive-combinatorial techniques could also be effective.)</p>
<p>Dear Cliff,</p>
<p>In two dimensions, there are reasonably sharp results (giving sets close to q^2/2) in</p>
<p><a href="http://arxiv.org/abs/math/0510356" rel="nofollow">http://arxiv.org/abs/math/0510356</a><br />
<a href="http://arxiv.org/abs/math/0607734" rel="nofollow">http://arxiv.org/abs/math/0607734</a></p>
<p>(and see also a comment in</p>
<p><a href="http://quomodocumque.wordpress.com/2008/03/25/give-the-people-who-like-the-kakeya-problem-what-they-want/" rel="nofollow">http://quomodocumque.wordpress.com/2008/03/25/give-the-people-who-like-the-kakeya-problem-what-they-want/</a></p>
<p>)</p>
<p>In higher dimensions, the best construction I know of uses the two-dimensional construction, combined with the trivial observation that the Cartesian product of two Kakeya sets is still a Kakeya set (in a higher dimensional space, of course).  This gives about q^d / 2^{d/2} in the case when d is even, and slightly worse than this when d is odd.  This is a fair distance away from Dvir&#8217;s bound, which is about q^d/d!.  So there is still a little bit of room to close the gap here, and in particular to settle the question of whether the density of Kakeya sets needs to decay faster than exponentially (i.e. it is asymptotically less than c^d q^d as d -&gt; infinity for any c &gt; 0).</p>
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		<title>By: Clifford Smyth</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-29528</link>
		<dc:creator>Clifford Smyth</dc:creator>
		<pubDate>Mon, 05 May 2008 00:16:57 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-29528</guid>
		<description>Hi,

What are some upper bounds on the minimum size of a Kakeya set in F_q^n?

I found one of size (q-1)^n + 2^n - 1. Surely there are smaller ones?

Cliff</description>
		<content:encoded><![CDATA[<p>Hi,</p>
<p>What are some upper bounds on the minimum size of a Kakeya set in F_q^n?</p>
<p>I found one of size (q-1)^n + 2^n - 1. Surely there are smaller ones?</p>
<p>Cliff</p>
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		<title>By: Seva</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-28325</link>
		<dc:creator>Seva</dc:creator>
		<pubDate>Tue, 08 Apr 2008 02:00:45 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-28325</guid>
		<description>Actually, why is "Kakeya only non-trivial in the regime when $latex d$ is fixed and $latex k$ goes to infinity"? What is the smallest size of a subset of $latex F_3^r$, containing a line in every direction? (The field $latex F_3$ seems to be particularly interesting in this context as in this case the notion of a line is identical to that of a three-term arithmetic progression, not to mention that this is the first non-trivial case.) The trivial lower bound for the smallest size of such a set is $latex 3^{r/2}$, and it seems that a random subset of $F_3^r$ of size at least $latex 3^{(2/3+\epsilon)r}$ possesses the property in question with positive probability. Isn't the gap between $latex 3^{r/2}$ and $latex 3^{(2/3+\epsilon)r}$ worth investigation?</description>
		<content:encoded><![CDATA[<p>Actually, why is &#8220;Kakeya only non-trivial in the regime when <img src='http://l.wordpress.com/latex.php?latex=d&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='d' title='d' class='latex' /> is fixed and <img src='http://l.wordpress.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='k' title='k' class='latex' /> goes to infinity&#8221;? What is the smallest size of a subset of <img src='http://l.wordpress.com/latex.php?latex=F_3%5Er&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='F_3^r' title='F_3^r' class='latex' />, containing a line in every direction? (The field <img src='http://l.wordpress.com/latex.php?latex=F_3&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='F_3' title='F_3' class='latex' /> seems to be particularly interesting in this context as in this case the notion of a line is identical to that of a three-term arithmetic progression, not to mention that this is the first non-trivial case.) The trivial lower bound for the smallest size of such a set is <img src='http://l.wordpress.com/latex.php?latex=3%5E%7Br%2F2%7D&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='3^{r/2}' title='3^{r/2}' class='latex' />, and it seems that a random subset of $F_3^r$ of size at least <img src='http://l.wordpress.com/latex.php?latex=3%5E%7B%282%2F3%2B%5Cepsilon%29r%7D&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='3^{(2/3+\epsilon)r}' title='3^{(2/3+\epsilon)r}' class='latex' /> possesses the property in question with positive probability. Isn&#8217;t the gap between <img src='http://l.wordpress.com/latex.php?latex=3%5E%7Br%2F2%7D&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='3^{r/2}' title='3^{r/2}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=3%5E%7B%282%2F3%2B%5Cepsilon%29r%7D&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='3^{(2/3+\epsilon)r}' title='3^{(2/3+\epsilon)r}' class='latex' /> worth investigation?</p>
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		<title>By: Gil</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-27851</link>
		<dc:creator>Gil</dc:creator>
		<pubDate>Wed, 02 Apr 2008 14:11:12 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-27851</guid>
		<description>Dear Terry, 

One nice problem for practicing discretization+polynomial method or other  methods is this: What is the largest measure of a subset $latex X$ of the $latex d$-sphere which does not contain two orthogonal vectors. a natural guess is the twice the measure of a spherical cup of radius $latex \pi/4$. When $latex d$ is large this is roughly a fraction $latex (1/\sqrt 2) ^d$  of the total measure of the sphere. 

Now, the corresponding problem for the discrete cube (when the dimension is 4 times a prime number) can be solved (up to a factor 2 or so) by the polynomial method based on calculations over the field of 2 elements. This result translates to a result for the Euclidean question which is the best known but it is far from the conjecture. (It gives 
measure $latex 2^{ (H(1/4)-1) d}$. ) This is essentially Frankl-Wilson's theorem. 

One can try to discretize differently and use the polynomial methods for other fields and finding an appropriate "weighting" is indeed part of the difficulty but so far nobody managed to do it. There are promising competing methods which at present gives weaker bounds.</description>
		<content:encoded><![CDATA[<p>Dear Terry, </p>
<p>One nice problem for practicing discretization+polynomial method or other  methods is this: What is the largest measure of a subset <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='X' title='X' class='latex' /> of the <img src='http://l.wordpress.com/latex.php?latex=d&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='d' title='d' class='latex' />-sphere which does not contain two orthogonal vectors. a natural guess is the twice the measure of a spherical cup of radius <img src='http://l.wordpress.com/latex.php?latex=%5Cpi%2F4&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='\pi/4' title='\pi/4' class='latex' />. When <img src='http://l.wordpress.com/latex.php?latex=d&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='d' title='d' class='latex' /> is large this is roughly a fraction <img src='http://l.wordpress.com/latex.php?latex=%281%2F%5Csqrt+2%29+%5Ed&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='(1/\sqrt 2) ^d' title='(1/\sqrt 2) ^d' class='latex' />  of the total measure of the sphere. </p>
<p>Now, the corresponding problem for the discrete cube (when the dimension is 4 times a prime number) can be solved (up to a factor 2 or so) by the polynomial method based on calculations over the field of 2 elements. This result translates to a result for the Euclidean question which is the best known but it is far from the conjecture. (It gives<br />
measure <img src='http://l.wordpress.com/latex.php?latex=2%5E%7B+%28H%281%2F4%29-1%29+d%7D&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='2^{ (H(1/4)-1) d}' title='2^{ (H(1/4)-1) d}' class='latex' />. ) This is essentially Frankl-Wilson&#8217;s theorem. </p>
<p>One can try to discretize differently and use the polynomial methods for other fields and finding an appropriate &#8220;weighting&#8221; is indeed part of the difficulty but so far nobody managed to do it. There are promising competing methods which at present gives weaker bounds.</p>
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		<title>By: Terence Tao</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-27831</link>
		<dc:creator>Terence Tao</dc:creator>
		<pubDate>Wed, 02 Apr 2008 01:39:26 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-27831</guid>
		<description>Hmm, now that I think about it, there is a significant difference between finite field Kakeya and HJ.  Both take place on a discrete cube $latex {}[k]^d$ (in the finite field case we identify [k] with a finite field).  But Kakeya is only non-trivial in the regime when d is fixed and k goes to infinity, while in contrast Hales-Jewett is only interesting in the regime when k was fixed and d goes to infinity.  For bounded d the combinatorial lines split into easily understood families and there is not much of interest to say.

From past experience with HJ I would have to say that how one defines the "number" of lines involved is going to be very important; a naive notion of cardinality is probably not the right one (it assigns too much weight to those lines with about half of the coordinates being active).  A randomly formulated problem in this area is likely to be either trivially true, trivially false, or otherwise not terribly interesting.</description>
		<content:encoded><![CDATA[<p>Hmm, now that I think about it, there is a significant difference between finite field Kakeya and HJ.  Both take place on a discrete cube <img src='http://l.wordpress.com/latex.php?latex=%7B%7D%5Bk%5D%5Ed&amp;bg=ffffff&amp;fg=545454&amp;s=0' alt='{}[k]^d' title='{}[k]^d' class='latex' /> (in the finite field case we identify [k] with a finite field).  But Kakeya is only non-trivial in the regime when d is fixed and k goes to infinity, while in contrast Hales-Jewett is only interesting in the regime when k was fixed and d goes to infinity.  For bounded d the combinatorial lines split into easily understood families and there is not much of interest to say.</p>
<p>From past experience with HJ I would have to say that how one defines the &#8220;number&#8221; of lines involved is going to be very important; a naive notion of cardinality is probably not the right one (it assigns too much weight to those lines with about half of the coordinates being active).  A randomly formulated problem in this area is likely to be either trivially true, trivially false, or otherwise not terribly interesting.</p>
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		<title>By: Gil Kalai</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-27830</link>
		<dc:creator>Gil Kalai</dc:creator>
		<pubDate>Wed, 02 Apr 2008 01:31:48 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-27830</guid>
		<description>Right, but density H-J is only a vaguely Kakeya type question. If you insist on having a "combinatorial line" in each "direction" how large should the set be? (There are various combinatorial notions of lines/direction you can take but as the condition is weaker maybe the set need not be so very large. )</description>
		<content:encoded><![CDATA[<p>Right, but density H-J is only a vaguely Kakeya type question. If you insist on having a &#8220;combinatorial line&#8221; in each &#8220;direction&#8221; how large should the set be? (There are various combinatorial notions of lines/direction you can take but as the condition is weaker maybe the set need not be so very large. )</p>
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		<title>By: Terence Tao</title>
		<link>http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/#comment-27826</link>
		<dc:creator>Terence Tao</dc:creator>
		<pubDate>Tue, 01 Apr 2008 23:51:38 +0000</pubDate>
		<guid isPermaLink="false">http://terrytao.wordpress.com/?p=291#comment-27826</guid>
		<description>Dear Harrison,

One can certainly interpret this proof in coding language; the proposition in the post can be viewed as an assertion that a Reed-Solomon code of a certain order can be uniquely decoded by looking at the restriction of that code to a Kakeya set, and the Lemma can be viewed as a lower bound on the capacity of that code.  I believe that this proof of Kakeya was motivated by some computer science considerations that are somewhat related to coding theory, but I do not know the full details.

Dear Gil: Well, there is of course the density Hales-Jewett theorem of Furstenberg and Katznelson, which (viewed contrapositively) is the assertion that if a set contains a point in every combinatorial line, then it has asymptotically full measure, which is an assertion of a vaguely Kakeya-type nature.  Of course, algebraic methods such as the polynomial method are unlikely to apply directly to such a combinatorial situation that has no obvious algebraic structure, but it is still an interesting question as to whether some other proof of the density Hales-Jewett theorem (including, perhaps, a simplified ergodic theory proof) is available.</description>
		<content:encoded><![CDATA[<p>Dear Harrison,</p>
<p>One can certainly interpret this proof in coding language; the proposition in the post can be viewed as an assertion that a Reed-Solomon code of a certain order can be uniquely decoded by looking at the restriction of that code to a Kakeya set, and the Lemma can be viewed as a lower bound on the capacity of that code.  I believe that this proof of Kakeya was motivated by some computer science considerations that are somewhat related to coding theory, but I do not know the full details.</p>
<p>Dear Gil: Well, there is of course the density Hales-Jewett theorem of Furstenberg and Katznelson, which (viewed contrapositively) is the assertion that if a set contains a point in every combinatorial line, then it has asymptotically full measure, which is an assertion of a vaguely Kakeya-type nature.  Of course, algebraic methods such as the polynomial method are unlikely to apply directly to such a combinatorial situation that has no obvious algebraic structure, but it is still an interesting question as to whether some other proof of the density Hales-Jewett theorem (including, perhaps, a simplified ergodic theory proof) is available.</p>
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