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\begin{document}
\Large
\vskip 1in
\bc
{\bf \LARGE  Recent progress on the Kakeya conjecture }\\
\bigskip
\bigskip
\bigskip
\bigskip
\bigskip
Nets Katz (University of Illinois Chicago)\\
\bigskip
Izabella Laba (Princeton University) \\
\bigskip
Terence Tao (UCLA)\\
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\ 
\bigskip
{\tt www.math.ucla.edu/$\sim$tao/preprints } 
\ec

\page

\noindent
{\LARGE
The purpose of this talk is to:
}
\bigskip
\bi
 \item State the Kakeya conjecture in all dimensions, 
and discuss recent results

 \item Present some new progress for the three-dimensional case

\ei

\page

\noindent
What is the Kakeya conjecture?

\bi
\item Define a Besicovitch set in $\R^n$ to be a set which contains
a unit line segment in every direction.  A 1918 construction of Besicovitch
shows that such sets can have measure zero.

\item The Kakeya conjecture states
that every Besicovitch set has Hausdorff and Minkowski dimension $n$.

\item This conjecture is known for $n=2$, but only partial results are known
in higher dimensions.

\item The Kakeya conjecture is related to many other problems in harmonic
analysis and PDE, such as the restriction conjecture and the Bochner-Riesz
conjecture, and also to the Montgomery conjecture in number theory.

\item We won't discuss these ramifications in this talk, and concentrate
purely on geometric and combinatorial issues.

\ei
\page

\bigskip
\noindent
Minkowski dimension

\bi
\item We will work with the Minkowski dimension in this talk, as it is
simpler than the Hausdorff dimension.  Our approach has technical
problems in moving from the Minkowski dimension to the Hausdorff dimension.

\item The upper Minkowski dimension is defined as
$$ \dimsup E = \limsup_{\delta \to 0} \log_{1/\delta} |\E_\delta(E)|$$
where $\E_\delta(E)$ is the smallest number of $\delta$-balls needed
to cover $E$.

\item The Minkowski dimension of a set is always at least as large
as the Hausdorff dimension.  Thus it is easier to prove lower bounds
for the Minkowski dimension than for the Hausdorff dimension.

\item In order to show that the Minkowski dimension of $E$ exceeds $d$,
one has to show that $\E_\delta(E) \gtrsim \delta^{-d}$ for
arbitrarily small $\delta$.  To show that the Hausdorff
dimension of $E$ exceeds $d$, one has to show this bound for
\emph{all} sufficiently small $\delta$.  Actually we need to show
something slightly stronger, allowing the possibility that $E$ can be
covered by balls of different sizes.

\ei

\page

\bigskip
\noindent
Previous results

\bi
\item In all previous results, there was no distinction between
Hausdorff and Minkowski dimension.

\item It is quite simple to show that a Besicovitch set $E$ in $\R^n$ has
dimension at least $(n+1)/2$.  Consider the map from $E \times E \to
\R^{n-1} \times \R \times \R$ defined by
$$ ((\underline{x},x_n), (\underline{y},y_n)) \mapsto
(\frac{\underline{y}-\underline{x}}{y_n-x_n}, x_n, y_n).$$
Since $E$ contains a unit line segment in every direction, we see that
the image of this map contains an open set, i.e. is $n+1$-dimensional.
Thus $E \times E$ must be at least $n+1$ dimensional.

\item In two dimensions there is a superior argument of C\'ordoba, which solves
the full conjecture.  In other words, it is known that every Besicovitch
set in $\R^2$ has dimension 2.  The argument is based on the Cauchy-Schwarz
inequality, and the fact that any two lines intersect in at most one point.

\item We discretize the problem, and look at the $\delta$-neighbourhood
$N_\delta(E)$ of $E$.  This set contains a $\delta \times 1$ rectangle
in every direction.  We discretize further, and restrict ourselves to a
$\delta$-separated set of directions, so that we have $\delta^{-1}$
$\delta \times 1$ rectangles $R$ all pointing in genuinely different directions.

\item  From Cauchy-Schwarz we have
$$ \int \sum_R \chi_R(x)\ dx \leq |N_\delta(E)|^{1/2} 
(\int (\sum_R \chi_R(x))^2\ dx)^{1/2}.$$

\item Since there are $\delta^{-1}$ rectangles, each of area $\delta$,
we have
$$ \int \sum_R \chi_R(x)\ dx = 1.$$
We expand the $L^2$ integral as
$$ \int (\sum_R \chi_R(x))^2\ dx = \sum_R \sum_{R'} |R \cap R'|.$$
If we assume that the average angle of intersection
between $R$ and $R'$ is comparable to 1, we get that $|R \cap R'|$
is at most $\delta^2$ on the average.  Hence this integral is $O(1)$,
whence we get $|N_\delta(E)| \gtrsim 1$.  This shows that the Minkowski
dimension of $E$ is 2.

\item  Actually we cheated a little bit; if we count the smaller
angles we pick up a factor of $\log(1/\delta)$, but this
is harmless.

\item Heuristically, this argument shows that a collection of
$\delta \times 1$ rectangles which point in different directions in the
plane must essentially be disjoint.

\item One can adapt this argument to higher dimensions, but it only
gives $\dimsup(E) \geq 2$ regardless of the dimension.

\ei

\page

Wolff's bound

\bi 
\item  In 1991 Bourgain showed that the trivial $(n+1)/2$ bound,
which until then was the best known bound for $n \geq 3$, could be
improved slightly.  In 1995 Wolff carried this further, and improved
the $(n+1)/2$ bound to $(n+2)/2$ in all dimensions.  This became the
best known bound for any $n \geq 2$.

\item  We sketch the proof as follows.  Let $E$ be a Besicovitch set,
and let $N_\delta(E)$ be the $\delta$-neighbourhood of $E$.  Since $E$
is a Besicovitch set, $N_\delta(E)$ contains a $\delta \times 1$ tube
in every direction.  If we discretize the directions to a $\delta$-separated
set, we thus obtain about $\delta^{1-n}$ essentially distinct
$\delta \times 1$ tubes.

\item  Let $\mu$ denote the median multiplicity of these tubes; i.e. $\mu$
is the number of tubes which contain a ``typical'' point $x$.  From general
combinatorics we have (heuristically at least) the relationship
$$ N_\delta(E) \approx 1 / \mu.$$

\item  Consider a single $\delta$-tube $T$, and count the number of
other $\delta$-tubes which intersect $T$.  We may split $T$ up into
$\delta^{-1}$ $\delta$-balls.  Each ball is contained in roughly
$\mu$ tubes, so we expect about $\delta^{-1} \mu$ tubes overall which
intersect $T$.  Call the collection of these tubes a ``hairbrush''.

\item We divide each tube in the hairbrush into an inner part (within
1/10 of the stem $T$) and an outer part (everything else).  The inner
part of a hairbrush can have a lot of overlap.  However, the outer part
cannot.  In fact, if we consider all the 2-planes passing through the
axis of $T$, and let $\Pi$ be a $\delta$-separated net of these planes,
then the outer part of any tube in the hairbrush lies in the 
$\delta$-neighbourhood of exactly one of these planes.  Thus the outer
parts of tubes associated to different planes are completely disjoint.

\item Also, even the tubes that are associated to the same plane
are still essentially disjoint.  This comes from Cordoba's argument.

\item Since there are $\delta^{-1} \mu$ tubes and the outer part
of each tube has volume about $\delta^{n-1}$, we thus see that
the hairbrush has volume at least $\delta^{n-2} \mu$.  In particular
we have
$$ | N_\delta(E) | \gtrsim \delta^{n-2} \mu.$$
Combining this with our previous equation $|N_\delta(E)| \approx 1/\mu$
we obtain
$$ | N_\delta(E) | \gtrsim \delta^{(n-2)/2},$$
which corresponds to the $(n+2)/2$ bound on the Minkowski dimension.

\item There were some small lies in the above argument (we implicitly
assumed that the average angle of intersection was about 1, and
ignored the problem that $\mu$ could vary from point to point).  Correcting
for these errors gives some extra $\log(1/\delta)$ errors to the estimate,
which we will ignore.
\ei

\page

X-ray estimates

\bi

\item Wolff's estimate can be phrased as follows: if $\Omega$ is a
maximal $\delta$-separated set of directions, and for every direction in
$\Omega$ we associate a $\delta$-tube $T$ in that direction, then
$$ | \bigcup_T T | \gtrsim \delta^{(n-2)/2}.$$
To solve the Kakeya conjecture we would need to improve this bound to
$\delta^\eps$ for any $\eps > 0$.

\item Suppose we added more tubes.  Specifically, suppose for every
direction in $\Omega$ we associated $m$ disjoint $\delta$-tubes $T$ in that
direction.  One would expect the size of the union to increase.  Although
the above argument does not directly give such an improved bound, in 1997 Wolff
modified his argument in the three-dimensional case $n=3$ to show that
$$ | \bigcup_T T | \gtrsim m^{1/4} \delta^{1/2}$$
in this setting.  These estimates can also be recast as mixed-norm smoothing
estimates for the x-ray transform, and are therefore known as x-ray estimates.

\ei

\page

A different approach: additive combinatorics

\bi

\item In 1998 Bourgain gave a completely different argument to that of Wolff,
which used almost no geometry but instead used the combinatorics of sum sets
and difference sets.  In fact, he showed that the dimension of a Besicovitch
set in $\R^n$ was at least $13n/25 + 12/25$.  This is inferior to Wolff's bound
for $n < 26$ but superior for $n > 26$.

\item The key combinatorial lemma is the following.  Suppose that $A$, $B$
are finite subsets of a torsion-free abelian group (such as $\Z^n$), with cardinality
$\# A, \# B \leq N$.  Suppose that $G$ is a subset of $A \times B$ such that
$$ \# \{ a+b: (a,b) \in G \} \leq N.$$
Then $\# \{ a-b: (a,b) \in G \} \leq N^{2 - 1/13}$.

Roughly speaking, this says that if the sumset of $A$ and $B$ is extremely small,
then the difference set cannot be extremely huge.

\item The proof of this lemma is almost purely combinatorial, and relies on
some recent ideas of Gowers.  The only non-combinatorial facts used are the
trivial observations $a+b = c+d \iff a-c = d-b$ and $a-b = (a-b') - (a'-b') + (a'-b)$.

\item The application to the Kakeya problem is as follows.
If $E$ is a Besicovitch set of Minkowski dimension $d$, 
consider the discretized set $D = N_\delta(E) \cap \delta \Z^n$.
let $A$, $B$, $C$ be the slices of $D$ at $x_n = 0$, $x_n = 1/2$, and $x_n = 1$, respectively.
Generically one expects $A$, $B$, $C$ to consist of about $\delta^{1-d}$ points.

\item There are about $\delta^{1-n}$ distinct $\delta$-tubes in $N_\delta(E)$, each
pointing in different directions.  Each such tube intersects $A$, $B$, $C$ in a single
point.  We can thus create a subset $G$ of $A \times B$ such that
$$ \# \{ a-b: (a,b) \in G \} \sim \delta^{1-n}$$
$$ \# \{ a+b: (a,b) \in G \} \lesssim \delta^{1-d}.$$
$$ \# A, \# B \lesssim \delta^{1-d}.$$
If one applied the above lemma without the gain of 1/13, one would obtain the trivial
bound $d \geq (n+1)/2$.  The gain of 1/13 is what gives the improved bound of $13n/25 + 12/25$.

\ei

\page

New results in three dimensions

\bi

\item  In three dimensions, the trivial bound on the dimension is
$\dimsup(E) \geq 2$, which incidentally is all one can obtain from Cordoba's
argument alone.

\item  Wolff's argument improves this to 5/2, whereas Bourgain's argument
improves this to 2 + 1/25.

\item  By combining the two methods, and introducing several new ideas, we have managed
to get the following improvement:

\begin{theorem}[Katz,Laba,T.]  There exists an $\eps > 0$ such that $\dimsup E > 5/2 + \eps$
for all Besicovitch sets $E$ in $\R^3$.
\end{theorem}

\item  Unfortunately our $\eps$ is very small; we know that $\eps = 10^{-10}$ works.
The best value of $\eps$ from our argument is likely to be in the $10^{-5}$ range.

\item  Also, for reasons we shall see shortly, our method is currently restricted
to the Minkowski dimension alone.

\ei

\page

Preliminaries: the sticky reduction

\bi

\item Henceforth we shall always be working in $\R^3$.  I'll be somewhat lax about the 
distinction between  such words as "all" and "most", and will pretend certain quantities
are essentially constant when in reality they could vary over a reasonably wide range.
These technicalities can be patched over by routine
combinatorics such as the pigeonhole principle.

\item We prove by contradiction.  We assume that there exists a Besicovitch set $E$ with
Minkowski dimension exactly $5/2$, and seek to obtain a contradiction.  By pushing
our arguments a bit we can get a contradiction even if the Minkowski dimension is $5/2 + \eps$,
hence our result.

\item  Since $E$ has dimension $5/2$, the set $N_\delta(E)$ has volume roughly $\delta^{1/2}$.
However, because we are working with the upper Minkowski dimension, we also know (among other 
things) that $N_{\sqrt{\delta}}$ has volume roughly $\delta^{1/4}$.  We will need \emph{both}
of these estimates to obtain our contradiction, which is why we are restricted to the Minkowski
dimension.

\item  The set $N_\delta(E)$ consists of about $\delta^{-2}$ $\delta$-tubes, each
pointing in a different direction.  Consider all the tubes which point within $\sqrt{\delta}$
of a given angle $\omega$; there are about $\delta^{-1}$ of these.  Ordinarily one has
no control over the location of this subcollection of tubes.  However, from our bound
on $N_{\sqrt{\delta}}(E)$ we can say that this subcollection of tubes is entirely contained
a single $\sqrt{\delta}$-tube $T_\omega$, for arbitrary $\omega$.  This property we call
"stickiness".

\item The above observation is due to Wolff and is derived as follows.  Suppose for 
contradiction that for most directions $\omega$ the tubes pointing within $\sqrt{\delta}$
of $\omega$ could not be contained inside a single $\sqrt{\delta}$-tube, but instead
could only be covered by $m$ $\sqrt{\delta}$-tubes for some $m \gg 1$.  Then 
$N_{\sqrt{\delta}}(E)$ would consist of $m$ $\sqrt{\delta}$-tubes in every direction.
Wolff's x-ray estimate then gives
$$ | N_{\sqrt{\delta}}(E)| \gtrsim m^{1/4} \delta^{1/4},$$
contradicting our assumption that $N_{\sqrt{\delta}}(E) \sim \delta^{1/4}$.

\item Because of stickiness, our original collection of $\delta$-tubes, or "thin tubes"
can be organized into about $\delta^{-1}$ bunches of $\delta^{-1}$ tubes each,
such that each bunch of thin tubes is contained in a $\sqrt{\delta}$-tube, or "fat tube".
In other words, we have a certain self-similar structure on our Besicovitch set.

\centering\ \psfig{figure=rough.eps,height=2.5in,width=3.3in}
\centerline{Local well-posedness results for $n=4$.  }

\begin{figure}[htbp] \centering
\ \psfig{figure=nonsticky.eps,height=3in,width=3.6in}
\caption{An example of a non-sticky direction.}
        \label{fig:nonsticky}
        \end{figure}

\begin{figure}[htbp] \centering
\ \psfig{figure=sticky.eps,height=3in,width=1.2in}
\caption{An example of a sticky direction $\omega$.}
        \label{fig:sticky}
        \end{figure}

\ei

\page

Local structure of the bunches

\bi

\item Inside every fat tube we have a bunch of about $\delta^{-1}$ thin tubes.  What
can we say about the bunches?

\item One important observation is that these bunches are actually affinely rescaled 
versions of a $\sqrt{\delta}$-neighbourhood of a Besicovitch set.  The idea is to
apply a linear transformation to convert the fat tube into a unit cylinder.  This allows 
us to control the size of the bunches using (for instance) Wolff's estimates.

\item Another thing we can say is that the thin tubes in each bunch are very close to being
parallel, being only $\sqrt{\delta}$ separated in angle.  Thus when restricted to a
$\sqrt{\delta}$-ball, the bunch actually looks like a collection of parallel
$\delta \times \sqrt{\delta}$-tubes, or "short tubes".

\begin{figure}[htbp] \centering
\ \psfig{figure=adef.eps,height=3in,width=2.8in}
\caption{A fat tube, its bunch of thin tubes, and its intersection with
a ball, which is a union of short tubes.
 }
        \label{fig:atsdi}
        \end{figure}

\ei

\page

Overlap of short tubes.

\bi

\item Let's look at the Besicovitch set inside a single $\sqrt{\delta}$-ball $B$.  A typical
such ball $B$ will have many fat tubes containing it; each fat tube comes with a bunch of
thin tubes, which when restricted to the $B$, look like a collection of parallel short
tubes.  However, different fat tubes give short tubes which point in different directions.

\item We are assuming the union of the thin tubes, $N_\delta(E)$, to be extremely small.
This forces the thin tubes to overlap a lot.  In particular, the collections of short
tubes in $B$ coming from different fat tubes should overlap very often, almost $100\%$ in fact.


\item Geometric observation \# 1: this $100\%$ overlap is only possible if the fat tubes 
containing $B$ are coplanar.

\item Geometric observation \# 2: Even if the fat tubes containing $B$ are coplanar, one
only gets $100\%$ overlap if the short tubes look like unions of $\delta \times \sqrt{\delta}
\times \sqrt{\delta}$ slabs.

\item We call the first property ``planiness'' and the second property ``graininess''.

\begin{figure}[htbp] \centering
\ \psfig{figure=three.eps,height=3in,width=3.6in}
\caption{ A non-planar and planar triple of collection of short tubes.}
        \label{fig:three}
        \end{figure}

\ei

\page

\bi

\item Roughly speaking, the planiness property states that for any given
point $x$, the fat tubes through $x$ lie in a plane.  By replacing $\delta$
by $\sqrt{\delta}$, one can also assume that the thin tubes through $x$
also lie in a plane.  (This is another instance where we use the Minkowski
dimension hypothesis to exploit several scales simultaneously).

\item Roughly speaking, the graininess property states that the intersection
of $N_\delta(E)$ with any $\sqrt{\delta}$-cube looks like a union of 
parallel $\delta \times \sqrt{\delta} \times \sqrt{\delta}$ ``grains''.

\begin{figure}[htbp] \centering
\ \psfig{figure=grainy.eps,height=3in,width=3.6in}
        \label{fig:grainy}
        \end{figure}

\item It is easy to show a certain consistency property - the plane which
contains the thin tubes through $x$, is parallel to the grain containing
$x$.

\begin{figure}[htbp] \centering
\ \psfig{figure=consistency.eps,height=3in,width=3.6in}
\caption{Consistency of planes and grains.}
        \label{fig:bilinear}
        \end{figure}

\item Now, we restrict our attention to what happens inside a single
fat tube, which we orient to be vertical.  We adapt the argument of Bourgain, 
and consider three equally spaced horizontal slices of this fat tube.
In each 
of these cubes, the set $N_\delta(E)$ looks like a collection of grains
which are parallel to the fat tube by consistency.  By an affine transformation
one can make normalize the orientation of the grains as displayed.

\begin{figure}[htbp] \centering
\ \psfig{figure=additive.eps,height=3in,width=3.6in}
\caption{A typical fat tube, and its intersection with three equally
spaced cubes.
 }
        \label{fig:additive}
        \end{figure}

\ei

\page

\bi

\item  We could apply Bourgain's additive combinatorial arguments to
these three slices directly.  However, this would just reproduce
Bourgain's bound of $\dimsup(E) \geq 13n/25 + 12/25$, which
is inferior to Wolff's bound of $5/2$ and thus insufficient to obtain
a contradication.  To do better, we exploit both graininess and planiness.

\item  The point is that there are a number of symmetries one can
``factor out'', reducing the three-dimensional problem to a two-dimensional 
problem.  A typical example is the following observation.  For every
point on the bottom slice, consider the thin tubes through that point inside
the fat tube.  These tubes are aligned in a plane parallel to the
$x$-axis, and hit the top slice in a number of disks, also parallel to the
$x$-axis.  It turns out that this collection of disks depends only
on the $y$ co-ordinate of the original point, and is independent of the
$x$ co-ordinate.  This is proven by considering the dual
configurations of thin tubes coming from points in the top slice.
These types of independence results allow one to factor out a dimension
in a certain combinatorial sense.  This in turn improves
Bourgain's dimension bound to $5/2 + \eps$, and we get the desired
contradiction.

\ei
\end{document}
