\documentstyle[amssymb,amsfonts,psfig,12pt]{article} \input MathMacro.tex \newcommand{\eqn}[1]{(\ref{#1})} \newcommand{\is}{$I^{s}\;$} \newcommand{\mbf}[1]{\mbox{\boldmath ${#1}$}} \newcommand{\bc}{\begin{center}} \newcommand{\ec}{ \end{center}} \newcommand{\page}{\vfil \break} \newcommand{\bi}{\begin{itemize}} \newcommand{\ei}{ \end{itemize}} \newcommand{\benum}{\begin{enumerate}} \newcommand{\eenum}{ \end{enumerate}} \newcommand{\bd}{\begin{description}} \newcommand{\ed}{ \end{description}} \def\rp{{r^\prime}} \def\rpt{{\tilde r^\prime}} \def\rtil{{\tilde r}} \def\qtil{{\tilde q}} \def\kp{{k^\prime}} \def\kt{{\tilde{k}}} \def\qp{{q^\prime}} \def\pp{{p^\prime}} \def\qpt{{\tilde q^\prime}} \def\R{{\hbox{\bf R}}} \def\C{{\hbox{\bf C}}} \def\E{{\cal E}} \def\rn{{\R^n}} \def\Z{{\hbox{\bf Z}}} \def\dimsup{{\overline{\rm dim}}} \def\eps{{\varepsilon}} \def\implies{{\Rightarrow}} \def\RR{{\frak R}} \def\emph#1{{\it #1}} % \newcommand{\lesssim}{\mbox{$\,\stackrel{\textstyle{<}}{\sim}\,$}} \renewcommand {\theequation}{\thesection.\arabic{equation}} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{remark}{Remark}[section] \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)\\ \bigskip \ \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}