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\begin{document}
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{\bf \LARGE  The Kakeya conjecture}

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An example of research in analysis at UCLA
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Terence Tao (UCLA)\\

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\noindent

Geometric combinatorics

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\item The Kakeya conjecture is a question in \emph{geometric combinatorics}.  This branch of mathematics deals with the most efficient way to arrange various configurations of points, lines, circles, tubes, etc. in order to extremize some quantity.

\item A typical example: How can one arrange $n$ points in the plane in order to minimize the total number of distinct differences between them?  (The Erd\"os distance problem).  

\item These problems are studied by combinatorial methods, but they often have application to many fields outside of combinatorics.

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The Kakeya needle problem

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\item In 1917 S. Kakeya posed the {\it Kakeya needle problem}: what is the smallest area required to rotate a unit line segment (a ``needle'') by 180 degrees in the plane?  Rotating around the midpoint requires $\pi/4$ units of area, whereas a ``three-point U-turn'' requires $\pi/8$.  In 1927 the problem was solved by A. Besicovitch, who gave the surprising answer that one could rotate a needle using arbitrarily small area.

\item Besicovitch's solution to the Kakeya needle problem relied on two observations.  The first observation, which is elementary, is that one can translate a needle to any location using arbitrarily small area.

\psfig{figure=needle.eps}

\item The second observation is that one can construct open subsets of $\R^2$ of arbitrarily small area which contain a unit line segment in every direction.

\psfig{figure=besicovitch.eps}

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The Kakeya tube problem

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\item There is a more quantitative version of this problem in which one thickens unit line segments to $1 \times \delta$ rectangles for some $0 < \delta \ll 1$.

\item Suppose we have a set $E$ in the plane which contains a $1 \times \delta$ rectangle in every direction.  What is the smallest area $E$ can have?

\item Handy notation: use $A \lesssim B$ to denote the estimate $A \leq C B$ for some constant $C$ (or equivalently, $A = O(B)$).

\item Besicovitch 1927: $E$ can have area $\lesssim \log\log(1/\delta)/\log(1/\delta)$.  C\'ordoba 1977: $E$ must have area $\gtrsim 1/\log(1/\delta)$.  Keich 1999: $E$ can have area $\lesssim 1/\log(1/\delta)$. 

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What about three dimensions?

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\item One can ask the same problem in three dimensions.  Let $E$ be a subset of $\R^3$ which contains a $1 \times \delta$ tube in every direction.  What is the smallest volume $E$ can have?  (Or more whimsically, what is the least volume needed for a $1 \times \delta$ Hubble space telescope to look at every star in the universe)?

\item Keich's construction (for instance) shows that $E$ can have volume $\lesssim 1/\log(1/\delta)$.  But lower bounds have been much more difficult.  C\'ordoba 1977: $|E| \gtrsim \delta^{1+\eps}$ for all $\eps > 0$.  Bourgain 1991: $|E| \gtrsim \delta^{2/3+\eps}$.  Wolff 1995: $|E| \gtrsim \delta^{1/2+\eps}$.  Katz-Laba-T. 1999: $|E| \gtrsim \delta^{1/2-10^{-10}+\eps}$ (modulo some technicalities about the choice of $\delta$).

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Why do we care about this problem?

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\item The problem of obtaining bounds for these sets in general dimensions is known as the \emph{Kakeya problem}.  This problem has applications to

\item Bounds on the average growth of Dirichlet series with arbitrary bounded co-efficients (Montgomery conjecture);

\item Convergence of Fourier series in higher dimensions;

\item Multiple focussing behaviour of solutions to the wave and Schr\"odinger equations;

\item Differentiation of rough functions along rough vector fields.

\item The point is that progress on the Kakeya problem allows one to control how much tubes in different directions can overlap each other.  One can then use other arguments to replace ``tubes'' with ``arithmetic progressions'' (for the Dirichlet series application) or ``wave packets'' (for the Fourier series or PDE applications).  

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Some results

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\item Let $E$ be a subset of $\R^n$ which contains a $1 \times \delta$ tube in every direction.  The \emph{Kakeya conjecture} asserts that $|E| \gtrsim \delta^\alpha$ for all $\alpha > 0$.  (Informally: it is not possible to compress many $\delta \times 1$ tubes into a really small set).

\item Drury 1983: This estimate is true for $\alpha \geq  (n-1)/2$.  A modern proof (Bourgain 1991) is as follows.  By construction, $E$ contains a collection of about $\delta^{1-n}$ tubes, whose directions are $\delta$-separated.  Let $\mu$ be the largest number of tubes which can intersect at a point.  Since each tube has volume about $\delta^{n-1}$, we thus have $|E| \gtrsim \delta^{1-n} \delta^{n-1}/\mu = 1/\mu$.  On the other hand, if $x_0$ is a point which contains $\mu$ tubes, then the union of these tubes (a ``bush'') has volume at least $\gtrsim \mu \delta^{n-1}$.  Thus
$$ |E| \gtrsim \delta^{n-1} \mu, 1/\mu$$
and hence
$$ |E| \gtrsim \delta^{(n-1)/2}.$$

\item More sophisticated arguments have pushed this result to $\alpha \geq (n-2)/2$ (Wolff 1995).  The idea is to replace the bush (the union of tubes going through a point) with a ``brush'' (the union of all tubes going through another tube), in order to boost the bound $|E|  \gtrsim \delta^{n-1} \mu$ to $|E| \gtrsim \delta^{n-2} \mu$ (although more effort is needed to make this rigorous).  Unfortunately this argument does not seem to push much further.

\item Another ``arithmetic'' approach bas pushed the bound $\alpha \geq (n-1)/2$ to $\alpha \geq \frac{12}{25}(n-1)$ (Bourgain 1998) and then to $\alpha \geq \frac{3}{7}(n-1)$ (Katz-T. 2000).  The current world record is $\alpha \geq .403\ldots (n-1)$.  (Katz-T. 2001).

\item Informally, the idea is to consider three slices of $E$, namely
$$ A := \{ (x_1, \ldots, x_n) \in E: x_n = 0 \};$$
$$ B := \{ (x_1, \ldots, x_n) \in E: x_n = 1 \};$$
$$ C := \{ (x_1, \ldots, x_n) \in E: x_n = 1/2 \}.$$

Generically one expects the ($n-1$-dimensional) area of $A$, $B$, $C$ to be roughly the same as the volume of $E$, on the average.  

\item The set $E$ contains a large number of line segments connecting $A$, $B$, $C$, with each line segment pointing in a different direction.  Let $G \subset A \times B$ denote the set of all pairs in $A \times B$ connected by this manner.  Because all the line segments point in different directions, the differences $\{ a-b: (a,b) \in G\}$ are all distinct; on the other hand, because $C$ lies halfway between $A$ and $B$, the sums $\{ a+b: (a,b) \in G\}$ all lie in $2C$.  Thus the pairs in $G$ have a large set of differences but a small set of sums.  The idea is to then try to play the two facts off against each other using such identities as
$$ a-b = c-d \iff a+d = b+c$$
and
$$ a-b = (a-b') - (a'-b') + (a'-b).$$

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\end{document}
