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\begin{document}

\title{Projection theory notes}

\maketitle

\setcounter{tocdepth}{1}
\tableofcontents

\section{Introduction and overview}

In 18.156 this spring,  we will study projection theory.   Projection theory studies how a set $X$ behaves under different orthogonal projections.    Questions of this type aren't usually emphaisized in the graduate analysis curriculum,  but they come up in many areas of math,  including harmonic analysis,  analytic number theory,  additive combinatorics,  and homogeneous dynamics.   It is an especially good time to study projection theory,  because there have been some striking recent applications,  and because one of the central problems of the field was very recently solved.   At the same time,  there are many interesting open problems which I am excited to discuss and reflect on.

The goals of the course are:

\begin{itemize}

\item Learn the classical techniques and results of projection theory (with full details).

\item Learn about applications in several areas.

\item Learn about open questions.

\item Learn some of the main ideas in the recent work in the field.   Level of detail will depend on everyone's interest.

\end{itemize}


\subsection{What is projection theory?}

Suppose that we have a set $X \subset \RR^n$.   For any subspace $V \subset \RR^n$,  let $\pi_V: \RR^n \rightarrow V$ denote the orthogonal projection.   Projection theory studies the relationship between the properties of the set $X$ and the properties of the projections $\pi_V(X)$ as $V$ varies among $k$-dimensional subspaces.  Informally, we are looking at $X$ from many different points of view and trying to coordinate the different information.

The most basic question concerns the cardinality of $X$ and the cardinality of $\pi_V(X)$ for different sets $V$.   Suppose that $X$ is a finite subset of $\RR^2$, and write $|X|$ for the cardinality of a finite set.   For almost every line $L$,  $|\pi_L(X)| = |X|$,  but there could be some special lines $L$ where $|\pi_L(X)| < |X|$.   For any number $S < |X|$,  let $E_S(X)$ be the set of lines $L$ with $|\pi_L(X)| \le S$.   The first question of projection theory is: 

\begin{ques} \label{q1} Suppose $X \subset \RR^2$ is a finite set and $S < |X|$.  Given $|X|$ and $S$,  what is the maximum possible size of $E_S(X)$?  
\end{ques}

A key example,  suggested by Erd{\H o}s,  is when $X$ is an integer grid.   In this case,  when the slope of $L$ is a rational number of small height,  $|\pi_L(X)|$ is small.   Erdos conjectured that this example is the worst possible up to a constant factor,  and in the early 1980s,  Szemeredi and Trotter proved this conjecture.

\begin{theorem} \label{st} (Szemeredi-Trotter 1982) If $X$ is a finite subset of $\RR^2$,  and $S < \frac{1}{2} |X|$,  then

$$|E_S(X)| \le C S^2 |X|^{-1} + 1. $$

\end{theorem}

The proof of the Szemeredi-Trotter theorem uses topology,  and it started an interesting interaction between combinatorial geometry questions and topology.  

There are many variations of this question.  For finite sets $X$,  we can consider higher dimensions $\RR^n$.  Or we can consider other fields, like $X \subset \FF_q^n$ where $\FF_q$ is a finite field with $q$ elements.   Many of these questions are open.  

We can also consider infinite sets $X$.   This angle was taken in geometric measure theory,  where the size of an infinite set is measured using Hausdorff dimension.   We write $\HD(X)$ for the Hausdorff dimension of $X$.   Suppose that $X \subset \RR^2$.   We now write $E_s(X)$ for the set of lines $L$ where $\HD(\pi_L(X)) < s$.   Our second main question is the geometric measure theory analogue of the first question above.

\begin{ques} \label{q2} Suppose $X \subset \RR^2$ and $s < \HD(X)$.  
Given $\HD(X)$ and $s$,  what is the maximum possible Hausdorff dimension of $E_s(X)$?
\end{ques}

This question was studied by Marstrand,  Kaufman,  and Falconer in the 60s and 70s.  Kaufman,  building on Marstrand,  proved some results using a double counting argument.   And Falconer proved other results using Fourier analysis.   Their results are interesting but they don't give the full answer to Question 2.

This question was also studied by Furstenberg and Wolff.   Furstenberg was motivated by a question related to ergodic theory and Wolff was motivated by the Kakeya conjecture and by other problems in geometric measure theory.   Wolff studied the proof of Theorem \ref{st} and tried to adapt the topological methods there to Question \ref{q2}.   He was able to prove some interesting estimates and he even applied them to prove some new estimates for the wave equation,  but he was not able to answer Question 2.   Wolff identified a key obstacle to addressing this question: the answer is different over $\CC^2$ compared to $\RR^2$,  but most methods do not distinguish these two problems.   Similarly,  the projection problem in $\FF_q^2$ is different depending on whether $q$ is prime or not prime.

Around 2000,  Bourgain proved the first estimates in projection theory that distinguish between $\RR^2$ and $\CC^2$.   However,  Bourgain's proof improves the previous exponents only by a tiny number $\epsilon$.   For twenty years,  the bounds in Question 2 were only tiny improvments of the old bounds of Kaufman and Falconer.    But very recently,  Question 2 was answered completely by Orponen,  Shmerkin,  Ren, and Wang.

\begin{theorem} \label{furst} (Orponen-Shmerkin-Ren-Wang) If $X \subset \RR^2$,  and $s < \HD(X)$,  then

$$ \HD(E_s(X)) \le \max( 2s - \HD(X), 0). $$

\end{theorem}

The bound here is the natural analogue of the Szemeredi-Trotter theorem in the setting of Hausdorff dimension.   There are many variations on this question too,  and many of them are open.  The field is developing rapidly.

\subsection{Applications of projection theory}

We will survey several applications of projection theory.   For each topic,  we will introduce and motivate the topic and see how it connects with projection theory.   We will prove something about each topic but not necessarily the strongest results.

\vskip10pt

{\bf Sieve theory.} Projection theory is closely parallel to some topics in sieve theory.   Suppose now that $X \subset \ZZ$.   For any integer $q$,  let $\pi_q: \ZZ \rightarrow \ZZ / q \ZZ$ be the quotient map,  which takes an integer $n$ and outputs $n$ mod $q$.   Sieve theory studies the relationship between the properties of the set $X$ and properties of $\pi_q(X)$ for different $q$.   

One interesting result in sieve theory concerns the set of square numbers $S$.   For every prime $p$,  $|\pi_p(S)| = \frac{p+1}{2} \approx \frac{p}{2}$.   Linnik proved that if $X \subset \{1, ... , N\}$ and $|\pi_p(X)| \le \frac{p+1}{2}$ for every prime $p$,  then $|X| \lessapprox N^{1/2}$.   The set of square numbers up to $N$ shows that Linnik's theorem is tight.   The only known tight examples are close cousins of the square numbers,  and it is an important open problem to understand whether there are other examples. 

Another important direction in sieve theory is to understand how prime numbers are distributed modulo $q$ for different $q$.   Let $P_x$ denote the set of prime numbers up to $x$.   Dirichlet proved in the early 1800s that if $q$ is fixed and $x \rightarrow \infty$,  then $P_x$ is evenly distributed modulo $q$ among the residue classes that are relatively prime to $q$.  Dirichlet's method only works when $q$ is far smaller than $x$ -- the exact statement is messy but $q$ needs to be smaller than $x^\epsilon$ for any $\epsilon > 0$.   On the other hand,  it is conjectured that for every $q \le x^{1 - \epsilon}$,  the prime numbers are evenly distributed modulo $q$.   The generalized Riemann hypothesis would imply that the prime numbers are evenly distributed modulo $q$ for every $q \le x^{1/2 - \epsilon}$.

Sieve theory leads to equidistribution results that hold for most $q$.   In particular,  Bombieri-Vinogradov proved that for almost all $q \le x^{1/2 - \epsilon}$,  the primes are evenly distributed modulo $q$.   The point of sieve theory here is that we consider $\pi_q(P_x)$ for many different $q$ and how these different ``projections'' are related to each other.

There is a close analogy between classical methods in projection theory and classical methods in sieve theory.   In particular,  Falconer's work in projection theory (based on Fourier analysis) is closely analagous to the `large sieve' method developed by Linnik and used by Bombieri-Vinogradov.  And Kaufman's work in projection theory (based on double counting) is closely analogous to the `larger sieve' method developed by Gallagher.

One important problem in this area is to try to understand the distribution of $P_x$ mod $q$ for most $q$ when $q > x^{1/2}$.   Yitang Zhang proved the first results of this kind in his proof of bounded gaps between primes.   While we probably won't do the full proof,  we will discuss this problem and some of the issues that make it difficult. 

\vskip10pt

{\bf Sum-product problems.} Suppose that $A$ is a finite set of a field $\FF$,  such as $\RR$ or $\FF_p$.   We write $A + A$ for the set of sums $\{a_1 + a_2: a_1, a_2 \in A \}$ and we write $A \cdot A$ for the set of products $\{ a_1 a_2 : a_1, a_2 \in A \}$.   Erdos raised the question whether $\max ( |A + A |,  |A \cdot A|)$ must be much bigger than $|A|$.   He conjectured that $\max( |A+A|,  |A \cdot A|) \gtrsim |A|^{2 - \epsilon}$,  and Erdos and Szemeredi proved that there is some $c > 0$ so that $\max( |A+A|,  |A \cdot A|) \gtrsim |A|^{1+c}$.   Elekes connected the sum product problem to the Szemeredi-Trotter theorem and used the latter to prove a bound with a much better exponent: $\max( |A+A|,  |A \cdot A|) \gtrsim |A|^{5/4}$.  

Ever since Elekes's work,  there has been a close connection between sum product problems and projection theory.   This connection has been a two way street.   Initially,  Elekes used ideas from projection theory to prove new bounds in sum product theory.   But the work of Bourgain and the recent work of Orponen-Shmerkin-Ren-Wang goes in the other direction,  proving results in sum product theory first and then applying the results to projection theory in general.

Bourgain and Gamburd went on to apply these ideas in sum product theory to questions about random walks on finite groups such as $SL_2(\FF_p)$.   Suppose that $g_1,  ..., g_k$ are a set of generators of $SL_2(\FF_p)$ where we imagine that $k = O(1)$ and $p$ is large.   This set of generators determines a random walk on the group $SL_2(\FF_p)$.   Bourgain and Gamburd showed that this random walk mixes very fast for any choice of generators. 

\vskip10pt

{\bf Homogeneous dynamics.} The setting of homogeneous dynamics is a homogenous space such as $SL_n(\RR) / SL_n(ZZ)$.   This homogeneous space can be viewed as the space of lattices in $\RR^n$.   It comes up in many problems in number theory.  If $H \subset SL_n(\RR)$ is a Lie subgroup,  and $x \in SL_n(\RR) / SL_n(\ZZ)$,  then we can consider the orbit $Hx \subset SL_n(\RR) / SL_n(\ZZ)$,  and we can ask how this orbit is distributed.   If $H$ is a unipotent subgroup,  then there is a very rigid classification theorem due to Ratner,  building on special cases proven by Dani and Margulis.  Ratner's theorem says that either the orbit $Hx$ is dense and evenly distributed,  or else there is a very specific algebraic structure that describes the orbit.   

Recently,  Lindenstrauss and Mohammadi returned to this question and worked on proving good quantitative bounds in Ratner's theorem.   So far,  they were able to do so in some special cases.   One of their key new ideas is to connect these problems in homogeneous dynamics with projection theory.   Lindenstrauss discussed this work in his first Simons lecture at MIT last year.

We will introduce this area,  motivate the question,  and learn how projection theory enters the story.

\vskip10pt

{\bf Imaging.} Projection theory also comes up in different imaging technologies,  from CAT scans to Cryo-electron-microscopy.   In these settings,  one tries to reconstruct a set $X$ or function $f$ from some information about its projections.   We will discuss this briefly and see that some of the math involved is related to the math involved in projection theory in pure math.  In particular, there is a close connection between projection theory and Fourier analysis.  Some of the algorithms behind imaging technologies build on this connection.  And the same connection plays an important role in the pure math of projection theory.

\vskip10pt

{\bf Fourier analysis.}  Projection theory has a close connection with Fourier analysis.  Philosophically, projection theory is closely related to additive structure: the key feature of a projection $\pi_V: \RR^n \rightarrow V$ is that it is a group homomorphism of abelian groups.  Fourier analysis is also closely related to the additive structure of $\RR^n$: in Fourier analysis we study the characters of an abelian group.  This leads to nice formulas relating projections and Fourier transforms.  We will use Fourier analysis in our study of projection theory.

Recent work in Fourier analysis, especially related to decoupling theory, is closely related to projection theory, and ideas have gone in both directions.  Depending on interest we may introduce this direction as well.

\section{Fundamental methods in projection theory}

Thursday Feb 6

In this lecture, we introduce two fundamental methods for proving estimates in projection theory: the double counting method and the Fourier method.

These methods are cleanest in the setting of finite fields, so we begin with that case.  

We write $\FF_q$ for the finite field with $q$ elements.  Our projections will be a set of linear maps $\FF_q^2 \rightarrow \FF_q$.  For each $\theta \in \FF_q$, we define $\pi_\theta: \FF_q^2 \rightarrow \FF_q$ by

\begin{equation}
\pi_\theta(x_1, x_2) = x_1 + \theta x_2    
\end{equation}

Consider the following setup.

{\bf Setup.}

$$ X \subset \FF_q^2$$

$$ D \subset \FF_q \textrm{ (set of directions)}$$

$$ S = S(X,D) := \max_{\theta \in D} | \pi_\theta(X) |. $$

\vskip10pt

The first example of a set which has many small directions is an integer grid.

\vskip10pt

{\bf Example 1.} (Integer grid example) For simplicity suppose that $q=p$ is prime.  Write $[N]$ for $\{1, ..., N\}$.  For some $N \le p$, define

$$ X = \{ (x_1,x_2): x_1, x_2 \in [N] \}$$

\noindent For some $A \le p$, define

$$ D = \{ a_1/a_2: a_1, a_2 \in [A] \}$$

If $\theta \in D$, and $(x_1, x_2) \in X$, we have 

$$\pi_\theta(x_1, x_2) = \frac{ a_2 x_1 + a_1 x_2}{a_2}. $$

Therefore, $|\pi_\theta(X)| \lesssim A N$.  So we get

$$ S(X,D) \sim \max(AN, p).$$  The configuration is interesting when $S \le p/2$.  In this case, we have $S \sim AN$ and so

\begin{equation} \label{gridbound}
|D| \sim \frac{S^2}{|X|}
\end{equation}

This example generalizes to any finite field $\FF_q$ (or any field).  But when $q = p^r$ with $r>1$, there is also a more dramatic example based on the subfields of $\FF_q$.  We illustrate this in the case $q = p^2$.

\vskip10pt

{\bf Example 2.} (Subfield example) Suppose that $q=p^2$ with $p$ prime.  Define

$$ X = \FF_p^2 \subset \FF_q^2$$

$$ D = \FF_p \subset \FF_q$$

\noindent If $\theta \in D$, and $(x_1, x_2) \in X$, then we have $\pi_\theta(x_1,x_2) = x_1 + \theta x_2 \in \FF_p$.  So $|\pi_\theta(X)| \le p$.

So $|X| = p^2 = q$, $|D| = p = q^{1/2}$, and $S = S(X,D) = p = q^{1/2}$.

Comparing with Example 1, we see that $|D|$ is much larger than $\frac{S^2}{|X|}$.

\vskip10pt

Over $\FF_p$, there is no known example more dramatic than the integer grid example.  In fact, all known examples with many small projections are small variations of the integer grid example.  This leads to the following conjecture.

\vskip10pt

\begin{conj} Suppose $X \subset \FF_p^2$, $D \subset \FF_p$, and $S = \max_{\theta \in D} |\pi_\theta(X)|$.  If $S \le p/2$, then

$$ |D| \lesssim \frac{S^2}{|X|}$$
    
\end{conj}

Here we need $S \le p/2$ because for any sets $X$, $D$, we always have $S \le p$.  If $S=p$, then we cannot get any information about $|D|, |X|$.  For fields $\FF_q$, I have not seen a conjecture written down anywhere, but informally it is expected that the extreme examples are minor variations on Examples 1 and 2.

We will prove two fundamental estimates about projection theory in $\FF_q^2$.  The proofs of these results introduce two main techniques that we will use repeatedly: double counting and the orthogonality / Fourier method.

\begin{theorem} \label{kaufmanfq} (Double counting) Suppose $X \subset \FF_q^2$, $D \subset \FF_q$, and $S = \max_{\theta \in D} |\pi_\theta(X)|$.  If $S \le |X|/2$, then

$$ |D| \lesssim S$$
    
\end{theorem}


\begin{theorem} \label{falconerfq}(Orthogonality/ Fourier) Suppose $X \subset \FF_q^2$, $D \subset \FF_q$, and $S = \max_{\theta \in D} |\pi_\theta(X)|$.  If $S \le q/2$, then

$$ |D| \lesssim \frac{S q}{|X|}. $$
    
\end{theorem}

Remark. When $S =q/2$, or when $S \sim q$, Theorem \ref{falconerfq} matches the grid example and it is sharp.  Theorem \ref{falconerfq} is also sharp for the subfield example.  If $q=p$, then whenever $S$ is much less than $q$, Theorem \ref{falconerfq} does not appear to be sharp.  And even if $q=p^2$, there are many values of $S$, $|X|$ where Theorem \ref{falconerfq} does not appear to be sharp.

These theorems give interesting bounds but they don't give a complete picture of projection theory over $\FF_q^2$.  In part, this is because the techniques that we study today don't distinguish prime fields from non-prime fields, but the optimal projection estimates do depend on whether the field is prime.  It is fairly difficult to prove bounds going beyond these two theorems, and we will return to that later in the course.

\subsection{Double Counting}

\begin{proof} [Proof of Theorem 1.]  We will apply double counting to the set

$$(*):= \{ \theta \in D, x_1 \not= x_2 \in X: \pi_\theta(x_1) = \pi_\theta(x_2) \}$$

(Note on notation: here $x_1, x_2$ are points in $X$, not components of a vector.)

We call $(*)$ the set of coincidences.  The idea of the proof is as follows.  If there are many directions $\theta$ where $\pi_\theta(X)$ is small, then there must be a lot of coincidences.  But for any $x_1 \not= x_2 \in X$, there is only one direction $\theta$ so that $\pi_\theta(x_1) =\pi_\theta(x_2)$, and so there can't be that many coincidences.

If $\theta \in D$, then we have $|\pi_\theta(X)| \le S \le |X|/2 $.  Therefore, using Cauchy-Schwarz, we get

$$ \# \{x_1 \not= x_2 \in X: \pi_\theta(x_1) = \pi_\theta(x_2) \} \gtrsim S \left( \frac{|X|}{S} \right)^2. $$

(Details of this argument are on the first problem set.). And so

$$ (*) \gtrsim |X|^2 S^{-1} |D|.$$

On the other hand, for each $x_1 \not= x_2 \in X$, there is only one direction $\theta$ so that $\pi_\theta(x_1) =\pi_\theta(x_2)$, and so

$$ (*) \le |X|^2.$$

All together we have

$$ |X|^2 S^{-1} |D| \lesssim (*) \lesssim |X|^2, $$

and so $|D| \lesssim S$.
\end{proof}

\subsection{Orthogonality / Fourier method}

\begin{proof} [Proof of Theorem \ref{falconerfq}.]
    
The fibers of the map $\pi_\theta$ are parallel lines in $\FF_q^2$.  So if $|\pi_\theta(X)| \le S$, then we can cover $X$ using at most $L$ lines coming from fibers of $\pi_\theta$.

Recall that for each $\theta \in D$, $| \pi_\theta(X)| \le S$.  Let $\LL_\theta$ be a set of $S$ fibers of $\pi_\theta$ which covers $X$.  Let $\LL = \cup_{\theta \in D} \LL_\theta$.  Note that

$$ | \LL | = |D| S.$$

If $L$ is a line in $\FF_q^2$, we write $L(x)$ for the characteristic function of $L$.  We define

$$ f(x) = \sum_{L \in \LL} L(x)$$

Notice that for each $x \in X$,

$$ f(x) = |D|. $$

We will estimate the function $f$ using orthogonality.  To do that, we first break up each function $L$ as a constant function plus a mean zero part:

\begin{equation}
L(x) = \underbrace{ \frac{1}{q} }_{L_0(x)} + \underbrace{ L(x) - \frac{1}{q}}_{L_h(x)}
\end{equation}

Here $L_0(x) = 1/q$ is the mean value of $L(x)$, and so $L_h(x)$ has mean zero.  (The mean value of a function $g: \FF_q^d \rightarrow \CC$ is $\frac{1}{q^d} \sum_{x \in \FF_q^d} g(x)$.)  We can break up $f$ in a similar way:

\begin{equation}
f(x) = \sum_{L \in \LL} L(x) = \underbrace{ \frac{|\LL|}{q} }_{f_0(x)} + \underbrace{ \sum_{L \in \LL} L_h(x)}_{f_h(x)}
\end{equation}

The constant function $f_0$ is very simple to understand.  Since $|\LL| = S D$, and since we assumed $S \le q/2$, we have $f_0(x) \le |D|/2$.  Now for every $x \in X$, $f(x) = |D|$, and so

$$ |f_h(x)| \ge |D|/2 \textrm{ for all } x \in X$$

The key point is that the functions $L_h(x)$ are essentially orthogonal, and we can use this to estimate the function $f_h$.  We state the orthogonality as a lemma.

\begin{lemma} \label{orth} If $L_1, L_2$ are two different lines in $\FF_q^2$, then

$$ \sum_{x \in \FF_q^2} L_{1,h}(x) L_{2,h}(x) \le 0.$$
\end{lemma}

Using Lemma \ref{orth}, we can bound the $L^2$ norm of $f_h$:

$$\sum_{x \in \FF_q^2} |f_h(x)|^2 = \sum_{L_1, L_2 \in \LL} \sum_{x \in \FF_q^2} L_{1,h}(x) L_{2,h}(x) \le \sum_{L \in \LL} \sum_{x \in \FF_q^2} |L_h(x)|^2 .$$

For each line $L$, we can compute $\sum_{x \in \FF_q^2} |L_h(x)|^2$ by hand.  It is slightly smaller than $\sum_{x \in \FF_q^2} L(x)= q$.  So all together we get the $L^2$ bound

\begin{equation} \label{L2high}
   \sum_{x \in \FF_q^2} |f_h(x)|^2 \le | \LL | q 
\end{equation}

Combining everything we have done so far, we see that

$$ |X| |D|^2 \lesssim  \sum_{x \in \FF_q^2} |f_h(x)|^2 \le | \LL | q = |D| S q$$

Rearranging gives $|D| \lesssim \frac{S q}{|X|}$.

\end{proof}
    
Before we prove Lemma \ref{orth}, we make some comments about the proof.  Our bounds here are interesting when $| \LL |$ is much larger than $q$.  The key input is the $L^2$ estimate for $f_h$ in (\ref{L2high}.  When $| \LL |$ is much bigger than $q$, then this estimate shows that $\sum_{x \in \FF_q^2} |f_0(x)|^2$ is much bigger than  $\sum_{x \in \FF_q^2} |f_h(x)|^2$.  So $f(x)$ is equal to a constant function $f_0$ plus a perturbation $f_h$, and for most $x$, $|f_h(x)|$ is much smaller than $|f_0(x)|$.  Informally, we could say that the function $f(x)$ is almost constant.

Looking back at the proof of our $L^2$ estimate (\ref{L2high}), the argument applies to any set of lines $\LL$.  The crux of the matter is that if $| \LL |$ is much bigger than $q$, and if $f(x) = \sum_{L \in \LL} L(x)$, then $f = f_0 + f_h$ where $f_0$ is a constant function, and $f_h$ has small $L^2$ norm.

The key to the $L^2$ estimate is the orthogonality in Lemma \ref{orth}.
Now we discuss the proof of Lemma \ref{orth}.  One simple proof is just to compute $\sum_{x \in \FF_q^2} L_{1,h}(x) L_{2,h}(x)$.  Recall that

$$ L_{1,h}(x) = \begin{cases} 1 - 1/q & x \in L_1 \\ - 1/q & x \notin L_1 \\
    \end{cases}
$$

We can now compute $\sum_{x \in \FF_q^2} L_{1,h}(x) L_{2,h}(x)$.  With a little algebra, we find

$$\sum_{x \in \FF_q^2} L_{1,h}(x) L_{2,h}(x) \begin{cases} =0 & \textrm{ if $L_1, L_2$ are not parallel} \\ < 0 & \textrm{if $L_1, L_2$ are parallel}  \\
    \end{cases}
$$

The main case is when $L_1, L_2$ are not parallel.  In this case something interesting is happening that causes the sum to be zero, and we should look for a conceptual explanation.  One explanation comes from independence.  After a change of coordinates, we can assume that $L_1$ is the vertical axis and $L_2$ is the horizontal axis.  In these coordinates, $L_{1,h}$ only depends on $x_1$ and $L_{2,h}$ only depends on $x_2$, and so $L_{1,h}$ and $L_{2,h}$ are independent.  Therefore,

$$ \sum_{x \in \FF_q}^2 L_{1,h}(x) L_{2,h}(x) = \left(  \sum_{x \in \FF_q}^2 L_{1,h}(x)\right) \left(  \sum_{x \in \FF_q}^2 L_{2,h}(x)\right) = 0 \cdot 0 = 0.  $$

Another conceptual explanation comes from Fourier analysis.  We now pause to review the Fourier transform over finite fields, and then we use Fourier analysis to explain why $L_{1,h}$ and $L_{2,h}$ are orthogonal when $L_1, L_2$ are not parallel.

Suppose that $e: \FF_q \rightarrow \CC^*$ is a non-trivial homomorphism from the group $\FF_q^+$ to the group $\CC^*$.  If $q = p$ is prime, then we can take $e(x) = e^{2 \pi i \frac{x}{p}}$.

If $x, \xi \in \FF_q^d$, we define the dot product $x \cdot \xi$ by

$$ x \cdot \xi = x_1 \xi_1 + ... + x_d \xi_d.$$

If $f: \FF_q^d \rightarrow \CC$, then we define its Fourier transform $\hat f: \FF_q^d \rightarrow \CC$ by

\begin{equation}
    \hat f(\xi) := \sum_{x \in \FF_q^d} f(x) e(- x \cdot \xi)
\end{equation}

With this setup, we can write down the two fundamental theorems in Fourier analysis: Fourier inversion and Plancherel.

\begin{theorem} \label{fourinv} If $f: \FF_q^d \rightarrow \CC$, then

$$ f(x) = \frac{1}{q^d} \sum_{\xi \in \FF_q^d} \hat f(\xi) e(x \cdot \xi)$$
\end{theorem}


\begin{theorem} \label{plancherel} If $f, g: \FF_q^d \rightarrow \CC$, then

$$ \sum_{x \in \FF_q^d} f(x) \overline{g(x)} = \frac{1}{q^d} \sum_{\xi \in \FF_q^d} \hat f(\xi) \overline{ \hat g(\xi)}$$
\end{theorem}

Let us now revisit how we broke up a function $f$ as $f_0 + f_h$.  Starting with Fourier inversion, we can write $f$ as

$$ f(x) = \frac{1}{q^d} \sum_{\xi \in \FF_q^d} \hat f(\xi) e(x \cdot \xi) = \underbrace{\frac{1}{q^d} \hat f(0)}_{f_0(x)} +  \underbrace{\frac{1}{q^d} \sum_{\xi \not= 0} \hat f(\xi) e(x \cdot \xi)}_{f_h(x)}$$

Since $\hat f(0)= \sum_{q \in \FF_q^d} f(x)$, we see that $f_0$ is just the mean value of $f(x)$.  So this decomposition is the same one we used above in the proof of Theorem \ref{falconerfq}.  We can think of $f_0$ as the contribution of the zero frequency, and we think of $f_h$ as the contribution of the non-zero frequencies.  The letter $h$ stands for `high', and we think of $f_h$ as the `high-frequency' part of $f$.  In general, for any function $f$, we can define $f_h$ as above, and we have

$$ \hat f_h(\xi) = \begin{cases} \hat f(\xi) & \xi \not= 0 \\ 0 & \xi=0 \\
    \end{cases}$$

The Fourier transform interacts in a nice way with lines, and more generally with affine subspaces.  
Suppose that $P \subset \FF_q^d$ is an affine $k$-plane.  We write $P(x)$ for the characteristic function of $P$.
We define $P^\perp$ as

$$ P^\perp = \{ \xi \in \FF_q^d: (x_1 - x_2) \cdot \xi = 0 \textrm{ for all } x_1, x_2 \in P\}.$$

(Here the vector $x_1 -x_2$ is tangent to $P$, and so $P^\perp$ is the set of vectors perpendicular to $P$.  Note that $P$ is affine, so it may not contain 0, whereas $P^\perp$ is a subspace, and it does contain 0.)

\begin{lemma} \label{Phat} If $P(x)$ is the characteristic function of an affine $k$-plane in $\FF_q^d$, then

$$ | \hat P(\xi)| = \begin{cases} q^k & \xi \in P^\perp \\ 0 & \xi \notin P^\perp\\ \end{cases}$$

\end{lemma}

The proof of Lemma \ref{Phat} is on the first problem set.  The main point is that when $P$ is an affine plane, then $\hat P(\xi) = \sum_{x \in P} e(- x \cdot \xi)$ is a geometric series, and so we can sum it exactly.  For most $\xi$, the geometric series sums to zero because of symmetry.

Using Fourier analysis, we can now give another proof that when $L_1, L_2$ are not parallel, then $L_{1,h}$ and $L_{2,h}$ are orthogonal.  By Plancherel, we have

$$ \sum_{x \in \FF_q^2} L_{1,h}(x) L_{2,h}(x) = \frac{1}{q^d} \sum_{\xi \in \FF_q^2} \hat L_{1,h}(\xi) \overline{ \hat L_{2,h}(\xi)} =\frac{1}{q^d} \sum_{\xi \not= 0} \hat L_{1}(\xi) \overline{ \hat L_{2}(\xi)} $$

But by Lemma \ref{Phat}, the support of $\hat L_1$ is $L_1^\perp$ and the support of $\hat L_2$ is $L_2^\perp$.  Since the two supports intersect only at $\xi=0$, our last sum is zero.

To summarize, $\hat L_{1,h}$ and $\hat L_{2,h}$ have disjoint supports, and so $L_{1,h}$ and $L_{2,h}$ are orthogonal.

Remark. We don't necessarily need Fourier analysis to prove Theorem \ref{falconerfq}, but in some further developments the Fourier analysis is helpful.  For instance, if we want to generalize Theorem \ref{falconerfq} to higher dimensions, the Fourier analysis point of view is important.  You will explore this on the first problem set.

\subsection{Projection theory for balls in Euclidean space}

Next we will start to study projection theory in Euclidean space.  We will consider the projections of a set of unit balls in Euclidean space, and we will adapt our two fundamental methods to that setting.  There is a new issue that appears for balls in Euclidean space, which has to do with how the balls are clustered.  In this lecture, we start to set up our problems in the context of balls in Euclidean space, and we see how the clustering comes into play.

\vskip10pt

In this section, for a set $X \subset \RR^d$, we write $|X|$ for the $d$-dimensional measure of $X$.

\vskip10pt

{\bf Setup}

Suppose that $X$ is a set of disjoint unit balls in $B_R \subset \RR^2$.

Suppose that $D$ is a finite set in $S^1$, which is $1/R$-separated.

Define $S(X,D) = \max_{\theta \in D} | \pi_\theta(X) |$, the maximal 1-dimensional measure of $\pi_\theta(X)$.

\vskip10pt

Here we suppose that the directions in $D$ are $1/R$-separated because otherwise the projections would be essentially equivalent.

\vskip10pt

Next we can consider some examples.  There is an integer grid example which is analogous to the one we mentioned in finite fields.

\vskip10pt

{\bf Example 1.} (Widely spaced integer grid example) 

We let $X$ be an $N \times N$ grid of unit balls in $B_2^R$, spaced as widely as possible.  The centers of the balls lie on the lattice $\frac{R}{N} \ZZ \times \frac{R}{N}$.  

We choose a parameter $A \le R$, and we let $D$ be the set of directions with slope in the set $\{ a_1/a_2: a_1, a_2 \in [A] \}$.  

By a similar analysis to the one in finite fields, we see that 
$$ S(X,D) \sim \max(AN, R).$$  

The configuration is interesting when $S \le R/2$.  In this case, we have $S \sim AN$ and so

\begin{equation} \label{gridbound}
|D| \sim \frac{S^2}{|X|}
\end{equation}

\vskip10pt

Notice that the numerology of Example 1 in the setting of balls exactly matches the numerology for the integer grid in $\FF_p^2$.

We recall that for projection theory in $\FF_q^2$, there were interesting examples related to subfields of $\FF_q$.  The field $\RR$ does have subfields, such as the field of rational numbers.  However, these subfields do not lead to interesting sets of unit balls in $B_R$.  I think that the issue is that $\QQ$ is not closed.  To get a set of unit balls, we might take the 1-neighborhood of $\QQ \times \QQ$, but that is all of $\RR^2$.

But there is a new phenomenon for projection theory of balls in Euclidean space which has to do with clustering.  As a second example, we consider a tightly clustered set of balls.

\vskip10pt

{\bf Example 2.} (Clustered example) 

For some $N \le R$, we let $X$ be a set of $\sim N^2$ disjoint unit balls in $B_N \subset B_R$.  We have $|X| \sim N^2$

Now for every direction $\theta$, we have $|\pi_\theta(X)| \lesssim N$.  

So we can let $D$ be a maximal set of $1/R$ separated directions, so $|D| \sim R$, and we can take  $S = 2N$.  

Plugging in, we find that $|D|$ is much larger than $\frac{S^2}{|X|} \sim 1$.  And so this example is more extreme than Example 1.

\vskip10pt

The new theme in this setting is that projection estimates depend on how much $X$ is clustered.  It turns out that it is important to consider both how $X$ is clustered and how $D$ is clustered.  We can quantify the clustering of $X$ and $D$ with the following definitions.

We write $B(c,r)$ for the ball with center $c$ and radius $r$.  For any $1 \le r \le R$, we define

\begin{equation}
N_X(r) = \max_{c \in B_R} | X \cap B(c,r)|
\end{equation}

We write $\gamma$ for an arc of $S^1$, and $|\gamma|$ for its length.  For any $\rho \in [1/R, 1]$, we define

\begin{equation}
N_D(\rho) = \max_{|\gamma| = \rho} \# (D \cap \gamma)
\end{equation}

Our goal will be to prove projection estimates that depend on the functions $N_X(r)$ and $N_D(\rho)$.

In the next lecture we will work out analogues of Theorem \ref{kaufmanfq} and Theorem \ref{falconerfq} for balls in $\RR^2$.  The main idea will be to adapt the methods we used today in order to take account of clustering information from $N_X(r)$ and $N_D(\rho)$.


\section{Projection estimates for balls in the plane}

Tuesday February 11 

\newcommand{\R}{\mathbb{R}}

\textbf{SETUP}\\
Let $X$ be a set of disjoint unit balls in $B_R \subseteq \RR^2$. Let $D \subset S^1$ be a set of $1/R$ separated directions. 
$$S=S(X,D)=\max_{\theta \in D}|\pi_\theta(X)|.$$
$$N_X(r)=\max_{c \in \R^2}|X \cap B(c,r)|.$$
$$N_D(\rho)=\max_{\substack{\sigma \in S' \text{arc}\\ |\sigma|=\rho}}|D \cap \sigma|.$$
\textbf{Double Counting}\\
\begin{theorem}(Double Counting Real Version)\label{doubling couting}\\
If \textbf{SETUP}, then
$$|D|\lesssim \frac{S}{|X|}\sum_{1\leq r \leq R}N_X(r)N_D(1/r).$$
\end{theorem}
\begin{proof}
    $$*=\#\{B_1,B_2 \text{ unit balls }\in X, \theta \in D: \pi_\theta(B_1) \cap \pi_\theta(B_2)\not=\emptyset\}.$$
    \underline{Lower bound}: $*\gtrsim |D|\left(\frac{|X|}{S}\right)^2S$. It basically follows from the same argument as in the finite field setting.\\
    \underline{Upper bound}: Fix $B_1,B_2$ with $dist(B_1,B_2)\sim r$, let $c(B_1)$, $c(B_2)$ be the center of $B_1$ and $B_2$. Write 
    $$v=\frac{c(B_2)-c(B_1)}{|c(B_2)-c(B_1)|}$$
    to be the angle from $B_1$ to $B_2$. (see figure 1). If $\pi_\theta(B_1)\cap \pi_\theta(B_2)\not=\emptyset$, then $angle(\theta,\gamma)\lesssim 1/r$.  
    Thus, 
    $$\#\{\theta: \pi_\theta(B_1)\cap \pi_\theta(B_2)\}\lesssim N_D(1/r).$$
    $$\#\{B_1,B_2\in X: dist(B_1,B_2)\lesssim r\}\lesssim |X|N_X(r).$$
    Thus, $$* \lesssim \sum_{\substack{r \text{ dyadic }\\ 1 \leq r \leq R}} |X|N_X(r)N_D(1/r)$$
    so $$|X|^2S^{-1}D\lesssim * \lesssim \sum_{\substack{r \text{ dyadic }\\ 1 \leq r \leq R}}|X|N_X(r)N_D(1/r) .$$
\end{proof}
\begin{Example}
    For $N_X(r)$,
    \begin{enumerate}
        \item $X$ neighborhood of a curve. (see figure 2)
        \item Well-spaced $N \times B$ grid. 
        $$N_X(r)=\begin{cases}
            1  & r \leq R/N \\
            r^2\frac{N^2}{R^2} & r> R/N
        \end{cases}$$ (see figure 3)
        \item A cluster of $N^2$ unit balls 
        $$N_X(r)\sim \begin{cases}
            r^2 & r \leq N\\
            N^2 & r>N
        \end{cases}$$
        (see figure 4)
    \end{enumerate}
    \underline{Pictures of How $N_r(X)$ depends on $r$} (see figure 5)\\
    Normalize $N=R^{1/2}$. 
\end{Example}
\underline{Straight Line Case }\\
$$|X|=R^\alpha, N_X(r)\sim r^\alpha$$
We call this regular $\alpha$ dim spacing. \\
\underline{Below Straight Line Case }\\
$$|X|=R^\alpha, N_X(r)\lesssim r^\alpha$$
We call this $\alpha$ dim spacing. \\
\begin{Definition}
We say that $X$ has Hausdorff spacing if it has $\alpha$ dimension spacing for some $\alpha$. Another way to say this is that 
$$N_{R^\beta}(X)\lesssim |X|^\beta$$
for any $0\leq \beta \leq 1$. 
\end{Definition}
\begin{Corollary}(Double Counting Real Version)\\
    If \textbf{SETUP} $X$, $D$ has Hausdorff spacing then 
    $$|D|\lesssim \log R (|S|+\frac{|S|}{|X|}|D|) \Rightarrow (S \sim X \text{ or }|D| \lesssim S).$$
\end{Corollary}
\begin{proof}
    Let's calculate $N_X(r)N_D(1/r)$. Suppose $r=R^\beta$, the Hausdorff condition implies
    $$N_X(R^\beta)N_D(R^{-\beta})\lesssim |X|^\beta |D|^{1-\beta}$$
    Thus, by theorem \ref{doubling couting}, 
    $$|D| \lesssim \log R\frac{|S|}{|X|}\max_{0\leq \beta \leq 1}|X|^\beta |D|^{1-\beta}\lesssim \log R(|S|+\frac{|S|}{|X|})|D|.$$
\end{proof}
Recall the theorem in the finite field case. 
\begin{theorem}
 If $X \subseteq \F_q^2$, $D \subseteq \F_q, S=\max_{\theta\in D}|\pi_\theta(X)|$ then $S \sim |X|$ or $|D|\lesssim S$.    
\end{theorem}
Note that in the $\R$ setting if we impose the Hausdorff spacing condition, then we get basically the same result as in the finite field case. \\
Now let's compare result in projection theory in $\F_q^2$ vs unit balls in $B_R^2$ with Hausdorff spacing. 
\begin{theorem}(Fourier Method Finite Field)
    If $\F_q$-SETUP and $S \leq q/2$, then 
    $$|D|\lesssim \frac{Sq}{|X|}. $$
\end{theorem}
\begin{Corollary}
    If $\R$-SETUP and $X,D$ has Hausdorff spacing. Then, $|D|\lesssim \frac{SR}{|X|}$
\end{Corollary}
\begin{conj}
    If $p $ primes, $\F_p$ SETUP and $S \leq \frac{1}{2}\min(q,|X|)$ then $$|D| \lesssim \frac{|S|^2}{X}$$
\end{conj}
\begin{conj}(Furstenberg)
   If SETUP, $X$ and $D$ has Hausdorff spacing  and 
   $$S \leq R^{-\epsilon}\min(R,|x|),$$
   then $$|D|\lessapprox \frac{|S|^2}{|R|}.$$
\end{conj}
The above conjecture is proven in 2024 by (Orponen, Shmerkin, Ren and Wang) 
\subsection{Fourier Method}
\begin{lemma}(Main lemma in finite field)\\
If $\L$ is a set of lines in $\F_q^2$. Write $f=\sum_{L \in \L}1_L(x)$. Then, 
$f=f_0+f_1$ so $\spp \hat{f}_0=\{0\}$, $\spp \hat{f}_h=\{0\}^c$.  Then, $f_0$ is a constant function. Then, $\|f_0\|_2^2=|\L|^2$, $\|f_h\|_2^2=|\L|q$. 
\end{lemma}
Now, let's look at the $\R$ setting. Let $\T$ be a set of $1 \times R$ in $\R^2$. Let $\phi_T$ be a smooth approximation of $1_T$.  
\begin{lemma}(Main lemma in real)\\
Let $\T$ be a set of $1 \times R$ rectangles in $\R^2$. Let $f=\sum_{T \in \T} \phi_T(x)$. Then, $$f=\sum_{\substack{1\leq r \leq R\\ dyadic}}f_r(x)$$
such that $\spp \hat{f}_r \subseteq B(1/r)$ and $\|f_r\|_2^2\lesssim N_{\T}(r)|\T|r^{-1}R$ where

$$N_\T(r):=\max_{\Tilde{T}: 2r \times 2R \text{rect}} \#\{T \in \T: T \subset \Tilde{T}\}. $$
\end{lemma}
\begin{proof}(proof sketch of main lemma)\\
$\spp \hat{\phi}_T\subset T^*$ where
$T^*:=\{\xi \in \R^2: |(x_1-x_2)\cdot \xi|\leq 1, \text{ any }x_1,x_2\in T\}$. (see figure 6)
\end{proof}
\textbf{Littlewood-Paley decomposition}\\
Write $1=\sum_{\substack{1\leq r \leq R\\ dyadic}} \eta_r(\xi)$ with $\eta_r \geq 0$ such that 
$$\spp \eta_r \subseteq Ann(\frac{1}{10r} \leq |\xi| \leq \frac{1}{r}), 1 < r <R$$
and $\spp \eta_R \subseteq B(1/R)$ and $\spp \eta_1 \subseteq \{\xi: |\xi|>1/10\}$. 
Define $f_r=(\eta_r \hat f)^\vee$ so $\spp \hat{f_r} \subseteq B(1/r)$. 
In particular, we can write $\phi_{T,r}={(\eta_r \hat{\phi}_T)}^{\vee}$.\\
\textbf{Visual of $\hat{\eta_r}$ and $\phi_{T,r}$}\\
We have $\eta_r(\xi)\sim 1 $ on $Ann(1/r)$ and 
$$|\check{\eta_r}(x)|\sim \begin{cases}
    1/r^2 \text{ on }|x|\lesssim r\\
    \text{rapidly decay if } |x|>r. 
\end{cases}$$
(see figure 7)
where $\check{\eta_r}(x)=\int e^{i x \cdot \xi }\eta_r(\xi)d \xi$. Note that $\int \check{\eta_r}(x) dx =\eta_r(0)$. 
As $\int |\check{\eta_r}(x)|dx=\eta_r(0)$, we have that $f*|\check{\eta_r}(x)|\sim $ Average of $f$ on $B(x,r)$. As $\phi_{T,r}=\phi_T*\check{\eta_1}$, we have 
$|\phi_{T,r}(X)|\sim r^{-1} 1_{r \text{ neighborhood of } T}$. (see figure 8)
\begin{lemma}(Orthogonality)\\
If $T_1,T_2$ are $1 \times R$ tubes then $|\langle \phi_{T_1,r}, \phi_{T_2,r}\rangle|\lesssim R^{-1000}$ unless there exists $\Tilde{T}$, a $R^\epsilon r \times R^{1+\epsilon}$ rectangle such that $T_1, T_2 \in \Tilde{T}$.  
\end{lemma}
\begin{proof}(proof sketch)
If $angle(T_1,T_2)\gtrsim R^\epsilon \frac{r}{R}$, then $\spp \hat{\phi_{T_1,R}}\cap \spp \hat{\phi_{T_2,R}}=\emptyset$.  If $N_r(T_1)$ and $N_r(T_2)$ are disjoint, then 
$$\int \phi_{T_1,r}\phi_{T_2,r}=\int \phi_{T_1}*\hat{\eta_r}\phi_{T_1}*\hat{\eta_r} \lesssim R^{-1000}$$
as $\phi_{T_1}*\hat{\eta_r}$ and $\phi_{T_1}*\hat{\eta_r}$ have essentially disjoint support. 
\end{proof}
$\mathbf{L^2}$ \text{estimates} (see figure 9)\\ 
\begin{eqnarray}
    \|f_r\|_2^2&=& \|\sum_{T \in \T}\phi_{T,r}\|_{L^2}^2\\
    &=& \sum_{T_1,T_2} \langle \phi_{T_1,r}, \phi_{T_2,r}\rangle\\
    &=& \sum_{T_1 \sim_r T_2} \langle \phi_{T_1,r}, \phi_{T_2,r}\rangle + neglible \\
    &\leq& \sum_{T_1\sim_r T_2} \|\phi_{T,r}\|_2^2+\|\phi_{T_2,r}\|^2_{2}\\
    & \leq & N_r(\T) \sum_{T \in \T} \|\phi_{T,r}\|_2^2\\
    &=& N_{\T}(r) \sum_{T \in \T}\|\phi_{T,r}\|_2^2\\
    &=& N_{\T}(r)|\T|r^{-2}rR
\end{eqnarray}
where $r^{-2}$ is the amplitude and $rR$ is the area. 

\section{Fourier Method Continued}

Thursday February 13.

Before we dive into the Fourier method in Euclidean space, let us overview the result in the case of finite fields.
The main lemma used in the finite case is the following.
\begin{lemma} [Main Lemma 2F]
    If $\LL$ is a collection of lines in $\FF_q^2$, and $L(x)$ is the characteristic function for $L \in \LL$, then we can decompose
    \begin{align*}
        f(x) = \sum_{L \in \LL} L(x)
    \end{align*}
    as $f = f_0 + f_h$, where $f_0 = \frac{|\LL|}{q}$, $f_0$ is orthogonal to $f_h$, and 
    \begin{align*}
        ||f_0||^2_{L^2} \lesssim |\LL|^2 \, , \quad ||f_h||^2_{L^2} \lesssim |\LL| q\, .
    \end{align*}
\end{lemma}
Now one can use this lemma to get $L^2$ bounds on $f$ quite easily, we immediately get $||f||^2_{L^2} \lesssim ||f_0||_{L^2} + ||f_h||_{L^2}$, however, there are easier ways to get this same bound.
\begin{lemma}[Elementary $L^2$ bounds on $f$]
    We have $||f||^2_{L^2} \lesssim |\LL| q + |\LL|^2$.
\end{lemma}
\begin{proof}
    We can directly compute
    \begin{align*}
        ||f||^2_{L^2} =& \sum_{x \in \FF_q^2} \left[\sum_{L \in \LL} L(x)\right]^2
        = \sum_{x \in \FF_q^2} \left[\sum_{L_1,L_2 \in \LL} L_1(x)L_2(x)\right]
        \\
        =& \sum_{x \in \FF_q^2} \left(\left[\sum_{L_1=L_2 \in \LL} L_1(x)L_2(x)\right]
        +
        \left[\sum_{L_1 \neq L_2 \in \LL} L_1(x)L_2(x)\right]
        \right)
    \end{align*}
    Now different lines always meet at exactly one point, so $\sum_{x \in \FF_q^2} L_1 (x) L_2 (x) = 1$ for $L_1 \neq L_2$. Thus we have
    \begin{align*}
        ||f||^2_{L^2} \leq \sum_{x \in \FF_q^2} \left(\left[\sum_{L \in \LL} L^2(x)\right]\right)
        +
        \sum_{L_1 \neq L_2 \in \LL} 1 \leq |\LL| q + |\LL|^2
    \end{align*}
\end{proof}

One could then ask, isn't the Fourier method then useless if we can arrive at the same norm bound in an easier way? And in some regimes, it is, if $|\LL| \sim q$ then the Main Lemma does not give us any extra information. However, in the case where $|\LL| \gg q$ we not only get the $L^2$ bounds, but we also get the extra piece of information the constant part, the zeroth frequency, of $f$, dominates the contributions to the norm. We can interpret this information as asserting that $f$ is in some sense 'almost constant'. The usefulness of this will become clear in the Euclidean case.

\begin{wrapfigure}{r}{0.25\textwidth}
\input{lec4diagram1.tex}
\end{wrapfigure}

We now recall the setup for the Fourier Method in Euclidean Space.

{\bf Setup}

Suppose that $\TT$ is a set of $1 \times R$ rectangles.

Suppose that for each rectangle $T \in \TT$, $\psi_t$ is a smooth approximation for $1_T$. 

Let $f = \sum_{T \in \TT} \psi_T$ and $N_\TT (r) = \max_{\tilde{T}} |\{ T \in \TT : T \subset \tilde{T}\}|$ where $\tilde{T}$ ranges across all $2 r \times 2 R$ rectangles, as can be seen in the diagram on the right.

\begin{lemma} [Main Lemma 2R]
    If the setup holds then we can decompose $f$ as 
    \begin{align*}
        f = \sum_{\substack{1 \leq r \leq R\\ r \text{ dyadic}}} f_r
    \end{align*}
    with $f_r$ (nearly) orthogonal to each other, and for each $r$,
    \begin{align*}
        \hat{f_r} \subset B(1/r) \quad \text{and} \quad ||f_r||^2_{L^2} \lesssim |\TT| N_\TT (r) \frac{R}{r}
    \end{align*}
\end{lemma}
Now again we can use this lemma to arrive at a quick $L^2$ bound, simply adding up over $r$ we get
$
    ||f_r||^2_{L^2} \lesssim \sum_{r \text{ dyadic}} |\TT| N_\TT (r) \frac{R}{r}.
$
But once again, there are easier ways to get this bound, which we will now show.

\begin{minipage}{0.6 \textwidth}
For two tubes $T_1, T_2$ we will write $r(T_1, T_2)$ to be the minimal $r$ such that $T_1$ and $T_2$ are both contained in a $2r \times 2R$ rectangle.

A simple look at the geometry of the rectangles gives us the following lemma
\begin{lemma}
    For any two tubes $T_1,T_2$ we have
    \begin{align*}
        \int T_1 (x) T_2 (x) dx \sim \frac{R}{r(T_1,T_2)}
    \end{align*}
\end{lemma}
\end{minipage}
\hspace{0.02\textwidth}
\begin{minipage}{0.38 \textwidth} 
\input{lec4diagram2}
\end{minipage}
\newline

In a similar way to the elementary bound in the finite case we can compute directly, we will use the previous lemma, and group the terms in the sum by $r$
\begin{align*}
    \int f^2 = \sum_{T_1, T_2 \in T} \int T_1 (x) T_2 (x) d x = \sum_{r \text{ dyadic}} \sum_{\substack{T_1, T_2 \in T\\ r \sim r(T_1,T_2) }} \frac{R}{r}
\end{align*}
Now fix $r$, for the first tube we have $|\TT|$ choices and for the second we have at most $N_\TT (r)$ choices. This gives us 
\begin{align*}
    \sum_{r \text{ dyadic}} \sum_{\substack{T_1, T_2 \in T\\ r \sim r(T_1,T_2) }} \frac{R}{r} =  \sum_{r \text{ dyadic}} |\TT| N_\TT (r) \frac{R}{r}\, .
\end{align*}
Once again we get the same $L^2$ bound as from the Main Lemma. 

Thus we again find that the important part of the Lemma, isn't just the $L^2$ bound, its the extra information we get about the frequency structure of the function. We will want to think about this information in a particular way, which we will call the 'locally constant intuition'. 

\textbf{Intuition} If $\spp \hat{g} \subset B_{1/r}$ then $g \approx$ constant on each $B_r$. 
This intuitively should make sense, if $\spp \hat{g} \subset B_{1/r}$ then $g$ is a combination of waves with frequency at most $1/r$, since each wave is then approximately constant on any given $B_r$ then it is plausible that their combination is as well.

Now to use this intuition in our setup let us consider the following diagrams
\begin{center}
\input{lec4diagram3}
\end{center}
The left diagram shows us what happens in a setup where our $f$ is dominated by some $f_r$ with $r$ large, our function then is dominated by the scale $r$ and we can see as expected by our intuition, that for most balls of radius $r$, our function is relatively constant. Furthermore, the locations where $f$ is large will all look like the blob we have drawn in red, and they will have more geometric structure to exploit there.

On the other hand when $f$ is dominated by $f_1$, it is dominated by high frequencies and it might look like the diagram on the right, here we have less points where $f$ is large but they are more scattered and have less structure.

Now let us formalize this intuition before using it with our main lemma. Consider a function $g$ with $\spp \hat{g} \subset B_{1/r}$, what can we say about it? Well in analysis there is often a specific way we deal with supports we know, and that is using a bump function. That is, let $\eta$ be a compactly supported smooth function with $\eta = 1$ on $B(1,r)$, then we have $\hat{g} = \hat{g} \cdot \eta$ and so applying inverse Fourier to this equation we get
$
    g = g * \check{\eta}.
$
We will need three important properties of of $\check{\eta}$.
\begin{itemize}
    \item $|\check{\eta}(x)| \lesssim r^{-2}$, which comes from simple triangle inequality applied to the integral defining $\check{\eta}$.
    \item $|\check{\eta}(x)| \lesssim r^{-2} (\frac{|x|}{r})^{-1000}$, which comes from integration by parts.
    \item If $\eta$ is radial, then $\check{\eta}$ is also radial, which we will assume to be the case henceforth.
\end{itemize}
Now we can use these two facts to get information about $g$. We define $\psi_r := |\check{\eta}|$ and derive the following.
\begin{lemma}
    If $\spp \hat{g} \subset B_{1/r}$, then $|g(x)| \leq |g| * \psi_r$.
\end{lemma}
\begin{proof}
    We compute
    \begin{align*}
        |g(x)| = |(g * \check{\eta}) (x)| = \left|\int g(y) \check{\eta}(x - y)\right| \leq \int  |g(y)| |\check{\eta}(x - y)| = |g| * \psi_r
    \end{align*}
\end{proof}
\begin{lemma}
    If $\spp \hat{g} \subset B_{1/r}$, then $|g(x)|^2 \lesssim |g|^2 * \psi_r$.
\end{lemma}
\begin{proof}
    We again compute
    \begin{align*}
        |g(x)| = |(g * \check{\eta}) (x)|^2 = \left|\int g(y) \check{\eta}(x - y)\right|^2
    \end{align*}
    Now we write $g(y) \check{\eta}(x - y) = (g(y) \check{\eta}(x - y)^{1/2}) \cdot (\check{\eta}(x - y)^{1/2})$ and apply Cauchy Schwarz to get
    \begin{align*}
        \left|\int g(y) \check{\eta}(x - y)\right|^2
        \leq&
        \int (g(y) \check{\eta}(x - y)^{1/2})^2
        \int (\check{\eta}(x - y)^{1/2})^2
        \\ =&
        \int g(y)^2 \check{\eta}(x - y)
        \int \check{\eta}(x - y)
        \\ \lesssim& \,
        (|g|^2 * \psi_r) (1)
    \end{align*}
\end{proof}
Back to our setup, we can now apply all these computations to improve our $L^2$ bound and derive the Euclidean version of theorem 2F. We recall our setup.

\textbf{Setup}. 
$X$ is a set of unit balls in $B_R \subset \RR^2$. 

$D \subset S^1$ is a set of directions, which is $1/R$-separated.

$S = \max_{\theta \in D} |\pi_\theta (X)|$.

$N_X (r) = \max_{c \in \RR^2} |X \cap B(c,r)|$ and $N_D (\rho) = \max_{\substack{\sigma \subset S^1\\ |\sigma| = \rho}} |D \cap \sigma|$.

We will use $\lessapprox$ to mean $g(R,x) \leq C \log (R) f(R,x)$ for some constant $C$.
\begin{theorem}
    If our setup holds then \[
    |D| \lessapprox \frac{S R}{|X|} \max_{r} \frac{N_X (r) N_D (r/R)}{r^2}.
    \]
\end{theorem}
\begin{proof}
    First for all $\theta \in D$ we define $\TT_\theta$ to be the set of $S$ different $1 \times R$ tubes $T$ at angle $\theta$ that cover $X$.
    We then set
    \[
        \TT = \bigcup_{\theta \in D} \TT_\theta \quad f(x) = \sum_{T \in \TT}\psi_T (x).
    \]
    Then for any $x \in X$ we have $|f(x)| \geq |D|$ so we get the simple lower bound
    \[
        |D|^2 |X| \leq \int_{X} |f|^2
    \]

    
    Now the upper bound will be a bit trickier, let us think again about the picture we had before, and notice that if our $X$ set is quite spread apart, that is when $N_\TT (r)$ is small, then estimating $\int_{X} |f_r|^2$ by $||f_r||^2_{L^2}$ will be quite a lossy comparison, we can do better.
    \begin{center}
    \input{lec4diagram4}
    \end{center}

    First we will use the fact that $\spp \hat{f_r} \subset B_{1/r}$ to get
    \begin{align*}
        \int_X |f_r|^2 = \int 1_X |f_r|^2 dx
        \leq \int 1_X \cdot (|f_r|^2 * \psi_r) dx
        = \int \int 1_X (x) |f_r|^2 (y) \psi_r (x - y) dy dx
    \end{align*}
    Now let us assume that $\eta$ and hence $\psi_r$ are radial, then they are also symmetric, so this entire expression is symmetric with respect to swapping $x$ and $y$. Hence we have
    \begin{align*}
        \int_X |f_r|^2 = \int |f_r|^2 (x) \int 1_X (y) \psi_r (x - y) dy dx
        = \int |f_r|^2 (x) (1_X * \psi_r) dx
    \end{align*}
    Now morally $\psi_r$ is approximately $r^{-2}1_{B_{1/r}}$ so we have that $1_X * \psi_r \lesssim r^{-2} N_{X} (r)$. This then gives us
    \[
        \int_X |f_r|^2 \lesssim r^{-2} N_{X} (r) \int |f_r|^2 \lesssim \frac{R |\TT| N_{\TT} (r) N_{X} (r)}{r^3}\, .
    \]
    Now let us estimate $N_\TT (r)$, for any fixed $\theta$ we know that the number of rectangles of size $1 \times R$ that can fit inside a rectangle of size $2r \times 2R$ is $\lesssim r$ since no more can fit. The maximum angle (with respect to the large rectangle) that can fit is going to be $\lesssim r/R$, so as many as $N_D (r/R)$ different $\theta$ can count, hence we have a bound of $N_\TT (r) \lesssim r N_D (r/R)$. We thus have
    \[
        \int_X |f_r|^2 \lesssim \frac{R |\TT| N_{D}(r/R) N_{X} (r)}{r^2}.
    \]
    We also have $|\TT| = S |D|$ so putting it all together we have
    \begin{align*}
        |X| |D|^2 \leq& \sum_{\substack{1 \leq r \leq R\\ r \text{ dyadic}}} \int_X |f_r|^2 \lesssim S |D|R \sum_{\substack{1 \leq r \leq R\\ r \text{ dyadic}}} \frac{ N_{D}(r/R) N_{X} (r)}{r^2}
        \\ \leq& \, S |D|R \log {R}\max_{\substack{1 \leq r \leq R\\ r \text{ dyadic}}} \frac{ N_{D}(r/R) N_{X} (r)}{r^2}
    \end{align*}
    which we can rewrite into
    \[
        |D| \lessapprox \frac{S R}{|X|} \max_{\substack{1 \leq r \leq R\\ r \text{ dyadic}}} \frac{ N_{D}(r/R) N_{X} (r)}{r^2}
    \]
\end{proof}

Now this result looks a little ugly, so let us see what it looks like with the Hausdorff assumption we discussed last class.
Recall that we say $X$ has Hausdorff spacing if $N_X (R^\beta) \lesssim |X|^\beta$ for all $0 \leq \beta \leq 1$. If then $X$ and $D$ both have Hausdorff spacing then we have
\[
    \max_{\substack{1 \leq r \leq R\\ r \text{ dyadic}}} \frac{ N_{D}(r/R) N_{X} (r)}{r^2} \sim 1 + \frac{|X||D|}{R^2}
\]
\begin{Corollary}
    If the setup holds and $X,D$ both have Hausdorff spacing then
    \[
        |D| \lessapprox \frac{S R}{|X|} + \frac{S |D|}{|X|}.
    \]
    In particular either $R \lessapprox S$ or $|D| \lessapprox \frac{S R}{|X|}$.
\end{Corollary}

We will end off this section with a little bit of history about the Fourier and double counting method. 

\textbf{Fourier Method History}
\begin{itemize}
    \item 1940s - First use of Fourier method by Linnik in Sieve Theory.
    \item 1970s - Fourier method use by Rot for the Heilbronn triangle problem.
    \item 1980s - Falconer uses the method for geometric measure theory (what we are currently doing).
    \item Recently - Vinh used the Fourier method in the finite field setting.
\end{itemize}

\textbf{Double Counting Method History}
\begin{itemize}
    \item 60s - Kaufmann uses double counting method for geometric measure theory.
    \item 60s - Gallagher uses double counting method for Sieve theory.
\end{itemize}
\subsection{Sieve Theory}
We will now move on to the study of Sieve theory, which as we will see is very similar to what we have done so far.

We will be interested in studying the maps $\pi_q : \ZZ \to \ZZ_q := \ZZ / q \ZZ$ given by $\pi_q (x) = x \mod{q}$. These will play the role of our projections, in the sense that they are also group homomorphisms of Abelian groups. 

We will use $[N]$ to denote the set $\{1,...,N\}$ and we will study the projections of subsets of $[N]$. 

\textbf{Example} Consider the set $X = \{ n^2 : 1 \leq n \leq N^{1/2}\} \subset [N]$, we know from basic algebra that $|\pi_p (X)| = \frac{p + 1}{2}$ for all primes $p$. This should seem unusual since we could have an extremely large set and yet all of its projections miss half of their co-domain. The natural next question is, how large can a set $S$ be and still have this property?

\begin{theorem}[Linnik]
    If $X \subset [N]$ and $\pi_{p} (X) \leq \frac{p+1}{2}$ for all prime $p$ then $|X| \lessapprox N^{1/2}$.
\end{theorem} 
The only known sharp families for this theorem are square numbers and their close relatives, namely images of specific quadratic polynomials.

Let us now begin analyzing this problem using the double counting method.
\begin{theorem}[1S]
    If $X \subset [N]$, $D$ a set of primes less than $N$ and for all $p \in D$ we have that $|\pi_p(X)| \leq S$, then either $|X| \leq 2 S$ or $|D| \lessapprox S$.
\end{theorem}
\begin{proof}
    We start as usual by considering the set of coincidences
    \[
        (*) = \{x_1,x_2 \in X, p \in D : \pi_p (x) = \pi_p (x_2) \}
    \]
    by the same argument as usual we have the lower bound
    \[
        |(*)| \geq |D| \left(\frac{|X|}{S}\right)^2 = |X|^2 |D| S^{-1}.
    \]

    For the upper bound fix $x_1$ and $x_2$ and count the number of $p$'s for which the condition can hold, if $\pi_p (x_1) = \pi_p (x_2)$ then we have $p | x_2 - x_1$. We now have two cases

    If $x_1 = x_2$ then any $p$ works, this gives us a $|X||D|$ term.

    If $x_1 \neq x_2$ then only the prime divisors of $x_1 - x_2$ work of which there are at most $\log N$, so this gives us an $|X|^2 \log N$ term.

    Together we get
    \[
        |X|^2 |D| S^{-1} \leq |(*)| \leq |X||D| + |X|^2 \log N
    \]
    which we can rewrite as
    \[
        |D| \leq \frac{S |D|}{|X|} + S \log N
    \]
    so either the first term dominates and we have $S \leq 2 |X|$ or the second term dominates and we get $|D| \lessapprox S$.
\end{proof}
As an example if $|\pi_p (X)| \leq N^{2/3}$ for any $p \in D$ with $|D| = O(\log N) N^{2/3}$ then $|X| \lessapprox N^{2/3}$. % Why is this true? I don't think this setup gives us any bounds on |X|.

\newpage

\section{The large sieve}

Notes coming.

Thur Feb 20

% Hanzul



\newpage



\section{Projections and smoothing}

%Note taker Hang Du

Before starting the main subject of the class,  we review the difficult problem at the end of Problem Set 1 about the projection theory of $\FF_q^3$.

\subsection{Projection theory in $\FF_q^3$}
\ \\
\textbf{Setup}\\
Let $\FF_q$ be a finite field and $\operatorname{Gr}_q(1,3)$ be the set of one-dimensional subspaces of $\FF_q^3$. \\For $W\in D$, let $\pi_W:\FF_q^3\to \FF_q$ be a linear map with kernel $W$. \\
For $X\subset \FF_q^3$ and $D\subset \operatorname{Gr}_q(1,3)$, define $S=S(X,D)=\max_{W\in D}|\pi_WX|$. 
\begin{theorem}
    If $|S|\le \frac{q^2}{2}$, then $|D|\lesssim \frac{Sq^2}{|X|}$. 
\end{theorem}
\begin{proof}
    As before, we consider the set of lines 
    \[
    \mathcal L=\{x+W:W\in D,x\in \pi_WX\}\,,
    \]
    and consider the function $f:\FF_q^3\to \CC$ defined by
    \[
    f(x)=\sum_{L\in \mathcal L}\mathbf{1}_L(x)\,.
    \]
    By definition, every vertex in $X$ is covered by at least $|D|$ lines in $\mathcal L$. Hence $f(x)\ge |D|$ for all $x\in X$. 
    
    We decompose $f$ into low and high frequencies, 
    \[
    f=f_c+f_h\,,\quad f_c=\frac{|\mathcal L|}{q^2}\mathbf{1}\,.
    \]
    Since $|\mathcal L|=\sum_{W\in D}|\pi_WX|\le |D|S$, we have for $|S|\le \frac{q^2}{2}$, $f_c\le \frac{|D|}{2}\mathbf{1}$. Therefore, $f_h(x)\ge |D|-\frac{|D|}{2}=\frac{|D|}{2}$ for all $x\in X$, yielding that 
    \[
    \|f_h\|_{L^2}^2=\sum_{x\in \FF_q^3}|f_h(x)|^2\ge \frac{|D|^2|X|}{4}\,.
    \]

    We now proceed to derive an upper-bound of $\|f_h\|_{L^2}^2$ (and the difference between $\FF_q^3$ and $\FF_q^2$ comes out). For each $W\in \operatorname{Gr}_q(1,3)$, let 
    \[
    f_W=\sum_{L\in \mathcal L,L\parallel W}\mathbf{1}_L\,,
    \]
    the crucial observation is that $\operatorname{supp}(\hat{f}_W)\subset W^\perp$ (a fact shown in problem set 1), and hence for any $\xi\in \FF_q^3\setminus\{0\}$, $f_W(\xi)\neq 0$ only if $W\subset \xi^\perp$. Using this fact and Plancherel's theorem, we have 
    \begin{align*}
    \|f_h\|_{L^2}^2 =&\ \frac{1}{q^3}\sum_{\xi\in \mathbb{F}^3_q\setminus \{0\}}|\hat{f}(\xi)|^2=\frac{1}{q^3}\sum_{\xi\in \FF_q^3\setminus\{0\}}\Bigg|\sum_{W\subset \xi^\perp}\hat{f}_W(\xi)\Bigg|^2\,.
    \end{align*}
Since for each $\xi\in\FF_q^3\setminus\{0\}$, there are at most $q+1$ many $W\in \operatorname{Gr}_q(1,3)$ such that $W\in \xi^\perp$, using Cauchy-Schwartz we get
\[
\|f_h\|_{L^2}^2\le \frac{q+1}{q^3}\sum_{\xi\in\FF_q^3}\sum_{W\subset\xi^\perp}|\hat{f}_W(\xi)|^2\le (q+1)\sum_{W\in \operatorname{Gr}_q(1,3)}\|f_{W,h}\|_{L^2}^2\,.
\]
For any $L_1\neq L_2,L_1\parallel L_2$, it is easy to check that $\langle \mathbf{1}_{L_1}-\frac{1}{q^2},\mathbf{1}_{L_2}-\frac{1}{q^2}\rangle<0$, so for any $W\in \operatorname{Gr}_q(1,3)$,
\[
\|f_{W,h}\|_{L^2}^2\le \sum_{L\in \mathcal L,L\parallel W}\|\mathbf{1}_L-\frac{1}{q^2}\|_{L^2}^2\le q\#\{L\in \mathcal L,L\parallel W\}\,.
\]
Consequently, we have
\[
\|f_h\|_{L^2}^2\le (q+1)\sum_{W}q\#\{L\in \mathcal L,L\parallel W\}=(q+1)q|\mathcal L|\le 2q^2|D|S\,.
\]

Combining the lower and upper bounds, we obtain 
\[
\frac{|D|^2|X|}{4}\le \|f_h\|_{L^2}^2\le 2q^2|D|S\quad\Rightarrow\quad |D|\le \frac{8Sq^2}{|X|}\,,
\]
as desired.
\end{proof}

The above proof suggests that, given a set $\mathcal L$ of lines in $\FF_q^3$, for a typical vertex in $\FF_q^3$, the number of lines containing it deviates from the average number $\frac{|\mathcal L|}{q^2}$ by $$O\left(\sqrt{\frac{1}{q^3}\|f_h\|_{L^2}^2}\right)\le O\left(\sqrt{\frac{|\mathcal L|}{q}}\right)\,.$$
When $\sqrt{\frac{|\mathcal L|}{q}}\ll \frac{|\mathcal L|}{q^2}\iff |\mathcal L|\gg q^3$, almost all vertices are covered by roughly the same number of lines in $\mathcal L$. This bound is also tight in the following sense: there exists $\mathcal L$ with $|\mathcal L=\Theta(q^3)$ such that half vertices in $\FF_q^3$ does not contain in any line in $\mathcal L$. Here is an example: consider a two-dimensional subspace of $\FF_q^3$ and all its cosets, pick half of them and let $\mathcal L$ be the set of lines in these selected cosets. 

\subsection{Projections and smoothing}
\ \\
\textbf{Setup}\\
Let $f:\mathbb{R}^d\to \CC$ be a $L^2$ function. For $V\subset \RR^d$ a subspace, we define
\[
\pi_Vf(y)=\int_{V^\perp}f(y+z)\operatorname{d}\!\operatorname{vol}_{V^\perp}(z)\,,\forall y\in V\,.
\]
In particular, for $\theta\in \mathbb{S}^{d-1}$ we write $\pi_\theta f=\pi_{\operatorname{span}(\theta)}f$.

The following theorem states that the projection of a high dimensional function onto a typical direction is fairly smooth. 

\begin{theorem}\label{thm-smoothing-by-projecting-to-typical-direction}
    If $f\in L^2(\RR^d)$, $\operatorname{supp}(f)\subset B_1$, then provided that $\frac{d-1}{2}>\frac12+{k}$, it holds
    \[
    \int_{\mathbb{S}^{d-1}}\|\pi_\theta f\|_{{C}^k}^2\operatorname{d}\!\theta\lesssim \|f\|_{L^2}^2\,.
    \]
\end{theorem}

The following lemma builds a connection between the Fourier transform of $f$ and that of its projection.
\begin{lemma}[Dictionary]
    For any subspace $V\subset \RR^d$ and any $\xi\in V$, $\widehat{\pi_Vf}(\xi)=\hat{f}(\xi)$. 
\end{lemma}
\begin{proof}
    By definition we have
    \begin{align*}
    \widehat{\pi_Vf}(\xi)=&\ \int_Vf(x)e^{-ix\cdot \xi}\operatorname{d}\!x=\int_V\int_{V^\perp}f(x+y)\operatorname{d}\!\operatorname{vol}_{V^\perp}(y)e^{-ix\cdot \xi}\operatorname{d}\!x\\
    =&\ \int_V\int_{V^\perp}f(x+y)e^{-i(x+y)\cdot \xi}\operatorname{d}\!\operatorname{vol}_{V^\perp}(y)\operatorname{d}\!x\\
    =&\ \int_{\RR^d}f(x)e^{-ix\cdot \xi}\operatorname{d}\!x=\hat{f}(x)\,.\qedhere
    \end{align*}
\end{proof}

Now we prove Theorem~\ref{thm-smoothing-by-projecting-to-typical-direction}.
\begin{proof}
Using the dictionary lemma, we now relate $\|f\|_{L^2}$ with $\pi_\theta f$. By Plancherel's theorem and the polar coordinate transform, we have
\begin{align*}
\|f\|_{L^2}^2=&\ \|\hat{f}\|_{L^2}^2=\int_{\RR^d}|\hat{f}(\xi)|^2\operatorname{d}\!\xi\\
=&\ \int_{\mathbb{S}^{d-1}}\int_0^\infty|\hat{f}(r\theta)|^2r^{d-1}\operatorname{d}\!r\operatorname{d}\!\theta\\
=&\ \frac{1}{2}\int_{\mathbb{S}^{d-1}}\int_{-\infty}^\infty|\widehat{\pi_\theta f}(r)|^2r^{d-1}\operatorname{d}\!r\operatorname{d}\!\theta\,.
\end{align*}
Recall the Sobolev norms $\|\cdot\|_{\dot{H}_s}$ and $\|\cdot\|_{H_s}$: for $f:V\to \CC$, 
\[
\|f\|_{\dot{H}^s}^2=\int_V|\hat{f}(\xi)|^2|\xi|^{2s}\operatorname{d}\!\xi\,,\quad \|f\|_{{H}^s}^2=\int_V|\hat{f}(\xi)|^2(1+|\xi|)^{2s}\operatorname{d}\!\xi\,.
\]
The above estimate shows that
\[
\int_{\mathbb{S}^{d-1}}\|\pi_\theta f\|_{\dot{H}^{\frac{d-1}{2}}}^2\operatorname{d}\!\theta\lesssim \|f\|_{L^2}^2\,.
\]

By the Sobolev embedding theorem, if $s>\frac{1}{2}+k$, then $\|\pi_\theta f\|_{C^k}\lesssim \|\pi_\theta f\|_{H^{\frac{d-1}{2}}}$. We have
\begin{align*}
    \int_{\mathbb{S}^{d-1}}\|\pi_\theta f\|_{C^k}^2\operatorname{d}\!\theta \lesssim&\ \int_{\mathbb{S}^{d-1}}\|\pi_\theta f\|_{H^{\frac{d-1}{2}}}^2\operatorname{d}\!\theta\lesssim\int_{\mathbb{S}^{d-1}}\int_{-\infty}^\infty|\widehat{\pi_\theta f}(r)|^2(1+r^{d-1})\operatorname{d}\!r\operatorname{d}\!\theta\\
    \lesssim&\ \|f\|_{L^2}^2+\int_{\mathbb S^{d-1}}\int_{|r|\le 1}|\hat{f}(r\theta)|^2\operatorname{d}\!r\operatorname{d}\!\theta\lesssim \|f\|_{L^2}^2+\|f\|_{L^1}^2\,.
\end{align*}
Finally, since $f$ supports on $B_1$, we have $\|f\|_{L^1}\lesssim \|f\|_{L^2}$, the proof is completed.
\end{proof}

\textbf{Connection to probability theory}. Let $X_1,\dots,X_N$ be independent random variables $\operatorname{Uni}([-1/2,1/2])$. The joint density of $X_1,\dots,X_N$ is given by $f(\mathbf{x})=\mathbf{1}_{[-1/2,1/2]^N}(\mathbf x)$. According to central limit theorem, 
\[
\frac{1}{\sqrt N}(X_1+X_2+\cdots+X_N)\Rightarrow \mathcal N(0,1/6)\,.
\]
This suggests that the projection of $f$ onto the direction $(\frac{1}{\sqrt N},\dots,\frac{1}{\sqrt N})$ approximates a Gaussian, which is a much smoother function than $\mathbf{1}_{[-1/2,1/2]}$. 

Indeed, the central limit theorem says something more: provided the direction is not ``close'' to any coordinate axis, then the projection will approximate Gaussian. Similar results can be generalized from high dimensional cube to high dimensional convex bodies. This shows again the point that projecting onto a typical direction smoothens high dimensional functions. 




\newpage


\section{Applications of the Large Sieve to Number Theory}

Linnik initially used the large sieve to study the distribution of quadratic residues.   We will see that work on Problem set 4.

Perhaps the most important application of the large sieve in number theory concerns the distribution of primes mod $q$.

\subsection{Distribution of primes mod $q$}

Let $\pi(N)$ denote the number of primes less than or equal to $N$. Let $\pi(N, q, a)$ be the number of primes $p$ satisfying $p \leq N$ and $p = a \mod{q}$. We want to focus on $a \in \mathbb{Z}_q^*$, since if $a$ and $q$ are not relatively prime, $\pi(N, q, a)$ is at most one. So let $\phi(q) = |\mathbb{Z}_q^*|$. If the primes were evenly distributed mod $q$, then $\pi(N, q, a)$ would be close to $\frac{\pi(N)}{\phi(q)}$. To quantify how badly this fails, we introduce the function
\[
    \Delta_q(N) := \max_{a \in \mathbb{Z}_q^*} \left| \pi(N, q, a) - \frac{\pi(N)}{\phi(q)} \right|.
\]
Here are some results on $\Delta_q(N)$:
\begin{theorem}[Dirichlet]
    For all $q$,
    \[
        \lim_{N \to \infty} \frac{\Delta_q(N)}{N/q} = 0.
    \]
\end{theorem}
\begin{theorem}[Siegel-Walfisz]
    For any $A$, there is some $c_A$ such that
    \[
        \Delta_q(N) \leq c_A N (\log N)^{-A}.
    \]
\end{theorem}
This is the best result that applies to all $q$. If one assumes the generalized Riemann hypothesis, then it is true that
\[
    \Delta_q(N) \leq (C_\epsilon N^\epsilon) N^{1/2}
\]
for any $\epsilon > 0$. Montgomery conjectured that for any $\epsilon > 0$, there is a constant $C_\epsilon$ such that $\Delta_q(N) \leq (C_\epsilon N^\epsilon) \left( \frac{N}{q} \right)^{1/2}$.

Instead of trying to understand what happens for all $q$, we will be concerned with the typical behavior of $\Delta_q(N)$. The theorem we will discuss is
\begin{theorem}[Renyi, Bombiere--Vinogradov]
    For all $\epsilon > 0$ and all $A$,
    \[
        \sum_{q \leq N^{1/2 - \epsilon}} \Delta_q(N) \leq C(\epsilon, A) N (\log N)^{-A}.
    \]
\end{theorem}
This says that for most $q \leq N^{1/2 - \epsilon}$, $\Delta_q(N) \leq \frac{N}{q} (\log N)^{-A} \ll \frac{N}{q}$. So the primes are close to equidistributed mod $q$ for most $q$ up to $N^{1/2 - \epsilon}$.

We will not give the complete proof,  which is somewhat messy, but we will discuss most of the main ideas.   In particular,  we will explain how the large sieve and projection theory enter the story.

\subsection{Multiplicative Convolution and Primes}
To prove this, we will use the \textbf{multiplicative convolution}, which interacts nicely with prime numbers and projections.
\begin{Definition}
    If $f, g : \mathbb{N} \to \mathbb{C}$, then their \textbf{multiplicative convolution} is the function
    \[
        f *_M g(n) = \sum_{n_1, n_2, n_1n_2 = n} f(n_1) g(n_2).
    \]
\end{Definition}
This is related to the prime numbers through the \textbf{sieve}. Sieving is the process of obtaining prime numbers by crossing off all the multiples of $2$, then all the multiples of $3$, and so on, until only the primes are left. If you try to write this down with a formula, the multiplicative convolution will appear. Let $1 = 1_{\mathbb{N}}$ and define
\[
    D_p(n) =
    \begin{cases}
        1 & n = 1, \\
        -1 & n = p, \\
        0 & n \neq 1, p.
    \end{cases}
\]
Then we can calculate
\[
    1_{\mathbb{N}} *_M D_2 = 1_{\mathbb{N}} - 1_{2\mathbb{N}} = 1_{\mathrm{odd}}.
\]
Similarly, $1_{\mathbb{N}} *_M D_2 *_M D_3$ is the indicator function for $n$ relatively prime to $2$ and $3$.

For a set of primes $S$, define
\[
    RP_S(n) =
    \begin{cases}
        1 & (p, n) = 1 \ \forall p \in S, \\
        0 & \text{else}.
    \end{cases}
\]
Note that if $S = P_{N^{1/2}}$ and $N^{1/2} < n \leq N$, then $RP_S(n) = P(n)$.

\begin{lemma}
    If $S = \{ p_1, \ldots, p_r \}$, then
    \[
        RP_S(n) = 1 *_M D_{p_1} *_M \ldots *_M D_{p_r}.
    \]
\end{lemma}

\subsection{Multiplicative Convolution and Projections}
Now we will examine the relationship between multiplicative convolution and projection. Multiplicative convolution interacts nicely with the projection $\mathbb{Z} \to \mathbb{Z}_q$ because this projection is a ring homomorphism.

\begin{lemma}[Lemma 1]
    If $f, g : \mathbb{N} \to \mathbb{C}$, then $\pi_q(f *_M g) = \pi_q f *_M \pi_q g$.
\end{lemma}
To be extra careful, we should say what we mean by multiplicative convolution in $\mathbb{Z}_q$:
\[
    F *_M G(a) = \sum_{a_1, a_2 \in \mathbb{Z}_q, a_1 a_2 = a} F(a_1) G(a_2)
\]
for functions $F, G : \mathbb{Z}_q \to \mathbb{C}$.
\begin{proof}
    Write
    \[
        f = \sum_{n_1} \delta_{n_1} f(n_1), \ g = \sum_{n_2} \delta_{n_2} g(n_2).
    \]
    Then
    \[
        f *_M g = \sum_{n_1, n_2} \delta_{n_1n_2} f(n_1) g(n_2).
    \]
    Here $\delta_n$ is the \textbf{delta function} $\delta_n(m) = \begin{cases}
        1 & n = m \\
        0 & \text{else}
    \end{cases}$. Then
    \begin{align*}
        \pi_q f(a) & = \sum_{n_1} \delta_{n_1 \mod q}(a) f(n_1), \\
        \pi_q f *_M \pi_q g(a) & = \sum_{n_1, n_2} \delta_{n_1 n_2 \mod q} f(n_1) g(n_2) \\
        & = \pi_q(f *_M g)(a).
    \end{align*}
\end{proof}

For our final result, we want $L^\infty$ bounds, but our theory is geared toward $L^2$ bounds. Here's how we can get $L^\infty$ bounds:
\begin{lemma}[Lemma 2]
    If $f, g : \mathbb{N} \to \mathbb{C}$, then
    \[
        \lVert f *_M g \rVert_{L^\infty(\mathbb{Z}_q^*)} \le \lVert f \rVert_{L^2} \lVert g \rVert_{L^2}.
    \]
\end{lemma}
\begin{proof}
    For $a \in \mathbb{Z}_q^*$, $f *_M g(a) = \sum_{b \in \mathbb{Z}_q^*} f(b) g(ab^{-1}) \leq \lVert f \rVert_{L^2} \rVert g \rVert_{L^2}$ by Cauchy-Schwarz.
\end{proof}
There is also the minor technical annoyance of switching between $\mathbb{Z}_q$ and $\mathbb{Z}_q^*$. If $f : \mathbb{Z}_q \to \mathbb{C}$, let $f^* : \mathbb{Z}_q^* \to \mathbb{C}$ be the restriction. Then we can write $f = f_0 + f_h$ and $f^* = f_0^* + f_h^*$, where the starred functions are defined on $\mathbb{Z}_q^*$ and the unstarred functions are defined on $\mathbb{Z}_q$, the subscript zero indicates a constant function, and the subscript $h$ indicates an average zero function.
\begin{lemma}[Lemma 3]
    \[
        \lVert f_h^* \rVert_{L^2(\mathbb{Z}_q^*)} \leq \lVert f_h \rVert_{L^2(\mathbb{Z}_q)}.
    \]
\end{lemma}
Finally, taking the high frequency part commutes with multiplicative convolution:
\begin{lemma}[Lemma 4]
    If $f^*, g^* : \mathbb{Z}_q^* \to \mathbb{C}$, then
    \[
        (f^* *_M g^*)_h = f^*_h *_M g_h^*.
    \]
\end{lemma}
If we combine all of these, we get the following proposition:
\begin{prop}
    \[
        \lVert (\pi_q(f*_M g))^*_h \rVert_{L^\infty} \leq \lVert (\pi_q f)_h \rVert_{L^2} \lVert (\pi_q g)_h \rVert_{L^2}.
    \]
\end{prop}
\begin{proof}
    By Lemma 1 then Lemma 4,
    \[
        (\pi_q(f *_M g))^*_h = ((\pi_q f *_M \pi_q g))^*_h = (\pi_q f)^*_h *_M (\pi_q g)^*_h.
    \]
    Then using Lemma 2 and Lemma 3, we get
    \begin{align*}
        \lVert (\pi_q(f *_M g))_h^* \rVert_{L^\infty} & \leq \lVert (\pi_q f)^*_h \rVert_{L^2} \lVert (\pi_q g)^*_h \rVert_{L^2} \\
        & \leq \lVert (\pi_q f)_h \rVert_{L^2} \lVert (\pi_q g)_h \rVert_{L^2}.
    \end{align*}
\end{proof}

\subsection{Large Sieve and Multiplicative Convolution}

Our goal is to prove that $P(n)$ is evenly distributed mod $q$ for most $q$ of a given size.   We will focus on the case that $q$ is prime,  which avoids technical issues but still shows the main proof ideas.

We have seen that for a large range of $n$, $P(n)$ is equal to $RP_S(n)$,  where $S = P_{N^{1/2}}$.   The key property of $RP_S(n)$ is that it is a multiplicative convolution.   Our next theorem shows that most projections of a multiplicative convolution are nearly constant -- it is the main analytic ingredient in the proof of Bombieri-Vinogradov.


\begin{theorem}
    If $f : [N_1] \to \mathbb{C}$ and $g : [N_2] \to \mathbb{C}$, then $f *_M g : [N] \to \mathbb{C}$, where $N = N_1 N_2$, and
    \[
        \sum_{p \in P_M} \lVert (\pi_q(f *_M g))_h^* \rVert_{L^\infty}^2 \lesssim \left( \left( \frac{N_1}{M} + M \right) \left( \frac{N_2}{M} + M \right) \right)^{1/2} \lVert f \rVert_{L^2} \lVert g \rVert_{L^2}.
    \]
\end{theorem}
\begin{proof}
    We apply the proposition, Cauchy-Schwarz, and then the large sieve:
    \begin{align*}
        \sum_{p \in P_M} \lVert (\pi_p(f *_M g))_h^* \rVert_{L^\infty}^2 & \leq \sum_{p \in P_M} \lVert (\pi_p f)_h \rVert_{L^2} \lVert (\pi_p g)_h \rVert_{L^2} \\
        & \leq \left( \sum_{p \in P_M} \lVert (\pi_p f)_h \rVert_{L^2}^2 \right)^{1/2} \left( \sum_{p \in P_M} \lVert (\pi_p g)_h \rVert_{L^2}^2 \right)^{1/2} \\
        & \lesssim \left( \left( \frac{N_1}{M} + M \right) \left( \frac{N_2}{M} + M \right) \right)^{1/2} \lVert f \rVert_{L^2} \lVert g \rVert_{L^2}.
    \end{align*}
\end{proof}
For the main theorem, we have $|f(n)|, |g(n)| \lessapprox 1$, so $\lVert f \rVert_{L^2}^2 \lessapprox N_1$ and $\lVert g \rVert_{L^2}^2 \lessapprox N_2$, so
\[
    \sum_{p \in P_M} \lVert (\pi_q(f *_M g))_h^* \rVert_{L^\infty}^2 \lesssim \frac{N}{M} + \sqrt{N_1 N} + \sqrt{N_2 N} + M \sqrt{N}.
\]
This will be good if $M \leq N^{1/2 - \epsilon}$ and $N_1, N_2 \ll N$. We cannot have $N_1$ or $N_2$ close to $N$, because in that case the other factor will be close to $1$ and the multiplicative convolution will not result in a more evenly spread function. And the first condition must be true for the projection theory methods to be able to say anything.

Finally,  we give a rough outline the proof of the Bombieri-Vinogradov theorem for $q$ prime. 

I am actually not sure whether the full BV theorem can be proven following this outline.   The proof in books is based on a different way of finding multiplicative convolution structure in the primes,  which is called Vaughn's identity.   Vaughn's identity is more efficient and leads to fewer terms, but I found it a little harder to motivate.
    
    Let $S = P_{< N^{1/2}}$. If $N^{1/2} < n < N$, $RP_S(n) = P(n)$. Also
    \begin{align*}
        RP_S(N) & = [ 1 *_M D_{p_1} ] *_M [\ldots *_M D_{p_R}] \\
        & = f *_M g \\
        & = \left( \sum_{I_1} f 1_{I_1} \right) *_M \left( \sum_{I_2} g 1_{I_2} \right) \\
        & = \sum_{I_1, I_2} f 1_{I_1} *_M g 1_{I_2}.
    \end{align*}
    Here $I_1$ and $I_2$ are intervals that are narrower than dyadic intervals. Let $N_1 = \min I_1$ and $N_2 = \min I_2$. For $n \leq N$,
    \[
        RP_S(n) = \sum_{I_1, I_2, N_1 \cdot N_2 \leq N} f 1_{I_1} *_M g 1_{I_2}.
    \]
    We can then apply the theorem above for each pair of intervals. This works when $1 \ll N_1, N_2 \ll N$. Otherwise, we must group the convolutions for $RP_S(N)$ differently.  It is a possible course project to think this through carefully and see what bounds it gives.

\newpage

\section{The Szemeredi-Trotter theorem}

Tues March 4

\section{Szemerédi–Trotter}

\subsection{Szemerédi–Trotter projection theorem} 

\begin{theorem}
    Let $X$ be a set of points in $\mathbb{R}^2$ and $D$ a set of directions in $S^1$. Then we define 

    \begin{equation}
    S(X,D) = \max_{\theta \in D} |\pi_\theta(X)|
    \end{equation}

    Then 

    \begin{equation}
    |D| \leq \frac{S^2}{|X|} + 1
    \end{equation}
\end{theorem}


Now for the general theorem, let $X$ be a set of points in $\mathbb{R}^2$ and $L$ a set of lines in $\mathbb{R}^2$. Then we define 

\[I(X,L) := \#\{ x\in X, \ell \in L, x \in \ell \} \]  

Note that 

\[I(X,L) = \sum_{\ell \in L} |\ell \cap X| \] 


Then the Szemerédi–Trotter (ST) theorem states that 

\begin{theorem}
  \label{st-inc}
    \begin{equation}
        I(X,L) \leq |X| + |L| + |X|^{2/3} |L|^{2/3}
    \end{equation}
\end{theorem}



\begin{example}[Example 1 for ST Theorem]
    


The ST theorem is sharp with the $|X|$ bound when the number of lines is small and each point lies on a single line.

\hspace{5cm}

\begin{figure}[h]
    \centering
\begin{tikzpicture}
    % \draw[gray, thick] (-1,2) -- (2,-4);
    \draw[gray, thick] (-0.5,-0.5) -- (2,2);
    \filldraw[black] (0,0) circle (2pt);
    \filldraw[black] (1,1) circle (2pt);
    \filldraw[black] (0.5,0.5) circle (2pt);
    \filldraw[black] (1.5,1.5) circle (2pt);
\end{tikzpicture}
\caption{Example of setup where $|X|$ term dominates and $I \sim |X|$}
\end{figure}
\end{example}

\begin{example}[Example 2 for ST Theorem]
    The ST theorem is sharp with the $|L|$ bound when the number of lines is large and each line lies on a single point.


\begin{figure}[h]
    \centering
\begin{tikzpicture}
    % \draw[gray, thick] (-1,2) -- (2,-4);
    \draw[gray, thick] (-1,-1) -- (1,1);
    \draw[gray, thick] (1.4,0) -- (-1.4,0);
    \draw[gray, thick] (0,1.4) -- (0,-1.4);
    \draw[gray, thick] (1,-1) -- (-1,1);
    \filldraw[black] (0,0) circle (2pt);
\end{tikzpicture}
\caption{Example of setup where $|L|$ term dominates and $I \sim |L|$. }
\end{figure}
\end{example}

\begin{example}[Example 3 for ST Theorem]
    


We let $X$ be an $N\times N$ grid, and define $Q_M := \{\frac{a}{b} : a,b \in  [M]\}$. Define $L$ to be the set of lines with slopes in $Q_M$ that pass through points in $X$. Then $|Q_M| \sim M^2$ (the double counting when $\gcd(a,b) > 1$ only affects the magnitude of $|Q_M|$ up to a constant factor). Every point in $X$ has a line passing through it for each slope in $Q_M$. Then 


\begin{equation}
I(X,L) =|X| |Q_M| \sim N^2 M^2
\end{equation}

We now define projection operators for each $s\in Q_M$ as 

\begin{equation}
\pi_s(x_1, x_2) = x_2 = -sx_1
\end{equation} 

The fibers of $\pi_s$ are lines of slope $s$, and the number of lines in $L$ with direction $s$ is $|\pi_s(X)|$. We now prove the following lemma: 

\begin{lemma}
    For all $s\in Q_M$, $|\pi_s(X)| \lesssim MN$
\end{lemma}

\begin{proof}
Take $x_1, x_2 \in [N]$ and $s = \frac{a}{b} $. Then 

\begin{equation}
\pi_s(x_1, x_2) = x_2 - \frac{a}{b} x_2 = \frac{bx_2 - ax_1}{b} 
\end{equation}

Since $a,b \leq M$ and $x_1, x_2 \leq N$, $|bx_2 - ax_1| \lesssim MN$. Since $bx_2 - ax_1$ must be an integer, there are at most $MN$ distinct values in $\pi_s(X)$. 
\end{proof}

Then since $L$ has at most $NM$ lines for every element of $Q_M$. $|L \lesssim |Q_M|NM \sim M^3 N$. Then 

\begin{equation}
I(X,L) \sim N^2 M^2 \gtrsim (M^3 N \cdot N^2)^{2/3} \gtrsim |X|^{2/3} |L|^{2/3}
\end{equation}

Therefore the grid is a sharp example of the Szemerédi–Trotter theorem where the $|X|^{2/3}|L|^{2/3}$ term dominates. Note that the Szemerédi–Trotter theorem implies the ST projection theorem, which is a special case when $L$ is the set of lines with directions in $D$ passing through points in $X$. 


\end{example}

\subsection{Question: Are there other sharp examples for the Szemerédi–Trotter theorem?}

Another example is grids over number fields. Let $R$ be a number field, (for example $\mathbb{Z}[\sqrt{2 }]$). Then define 

\[R_N = \{a_1 + a_2 \sqrt{2}: a_1, a_2 \in \mathbb{Z}, |a_1|, |a_2| \leq N\}\]

\[QR_M = \{\frac{a}{b}: a,b \in R_M\}\]

Then define $X:= R_N \times R_N$ and $L$ as the set of lines with slopes in $QR_M$ that pass through a point in $X$. This is similar to the grid example. 

\subsection{Proof of Szemerédi–Trotter theorem}

We begin the proof the Szemerédi–Trotter theorem with a lemma. 

\begin{lemma}
\begin{displaymath}
I(X,L) \leq |X| |L|^{1/2} + |L|
\end{displaymath}
\label{weak_bound_ST_lemma}
\end{lemma}
\begin{proof}
We start with expanding $I(X,L)$ and applying Cauchy Schwartz to get 

\[I(X,L) = \sum_{\ell \in l} |\ell \cap X| \leq \left(|L|\sum_{\ell \in L} |\ell \cap X|^2 \right)^{1/2}\]


This is advantageous because $|\ell \cap X|^2 \lesssim \binom{|\ell \cap X|}{ 2} + 1$. Then 

\[\sum_{\ell \in L} |\ell \cap X|^2  \lesssim |L| + \sum_{\ell \in L} \binom{|\ell \cap X|}{2}  \] 

Since for every pair of points $x_1, x_2 \in X$, there is at most one line $\ell$ that contains $x_1$ and $x_2$, every pair of points in $X$ can be counted at most once. Then 

\[ \sum_{\ell \in L} \binom{|\ell \cap X|}{2} \leq \binom{|X|}{2} \lesssim |X|^2\]

This gives the final conclusion 

\[I(X,L) \lesssim \left(|L| (|X|^2 + |L|)\right)^{1/2} \leq |X||L|^{1/2} + |L|\]

\end{proof}

Note that this proof uses only the very general fact that any two points define a line. Therefore it holds over spaces such as finite fields. However, the Szemerédi–Trotter theorem does not hold over finite fields. To see this take $X = \mathbb{F}_q^2$ (as the whole space) and $L$ as all lines in $\mathbb{F}_q^2$. Then for every $\ell$, $|\ell \cap X| = q$, so $I(X,L) = q^3$. However, $|X|^{2/3} |L|^{2/3} = q^{8/3} \leq I(X,L)$. Therefore, the Szemerédi–Trotter theorem requires properties of the topology of $\mathbb{R}^2$ to work. In particular, it uses a cell decomposition lemma, which allows cutting the plane into pieces. 

\begin{lemma}[Cell decomposition lemma]
Let $X$ be a set of points in $\mathbb{R}^2$ and pick an integer $s\geq 1$. Then the plane can be disjointly partitioned into a set of open sets $O_i$ and a closed set $W$ such that 

\[\mathbb{R}^2 = W \cup \bigcup_i O_i\] 

and additionally, 

\[|\ell \cap W| \leq s\] 

and for every $i$, 

\[|X| \cap O_i| \lesssim \frac{|X|}{s^2} \]
\end{lemma}

This lemma essentially states that the plane can be split into cells that each contain only a small subset of $X$, and that the walls don't intersect any line too many times. As an example of this theorem, let $X$ be a "roughly" square grid. That is $\subset [N]^2$ and fir every ball $B_1(c)$ of radius 1 (where $c$ is an arbitrary point in the plane), $|X\cap B_1(c)| \lesssim 1$. The below example shows the grid for $s = 2$. 

\begin{figure}[h]
    \centering
    \begin{tikzpicture}
        % \draw[gray, thick] (-1,2) -- (2,-4);
        \foreach \i in {0,1,2,3,4,5} {
        \draw[gray, dashed] (0,\i) -- (5,\i);
        \draw[gray, dashed] (\i,0) -- (\i,5);
        }
        \foreach \i in {5/3, 10/3} {
            \draw[black, thick] (0,\i) -- (5,\i);
            \draw[black, thick] (\i,0) -- (\i,5);
        }
        \filldraw[black] (0.25955310498674833, 0.5630661519871801) circle (1pt);
        \filldraw[black] (0.4137916302144814, 1.2720619913477607) circle (1pt);
        \filldraw[black] (0.9732231939725711, 2.5427334886772703) circle (1pt);
        \filldraw[black] (0.3098666106054, 3.7871072206930636) circle (1pt);
        \filldraw[black] (0.01419593145904452, 4.734899395195148) circle (1pt);
        \filldraw[black] (1.786606807737972, 0.79422295279698) circle (1pt);
        \filldraw[black] (1.67931813829752, 1.8571052177749485) circle (1pt);
        \filldraw[black] (1.1956960247777007, 2.0912689126469677) circle (1pt);
        \filldraw[black] (1.0815919550119522, 3.847137097298062) circle (1pt);
        \filldraw[black] (1.112216822269262, 4.190692648040198) circle (1pt);
        \filldraw[black] (2.9605994936072006, 0.23571202532548774) circle (1pt);
        \filldraw[black] (2.6718713017136357, 1.865754141181469) circle (1pt);
        \filldraw[black] (2.593013450706421, 2.582698705322111) circle (1pt);
        \filldraw[black] (2.6562500412019583, 3.523426303538553) circle (1pt);
        \filldraw[black] (2.4334135790739353, 4.696110600325106) circle (1pt);
        \filldraw[black] (3.546029544303023, 0.9573850322061107) circle (1pt);
        \filldraw[black] (3.1047794604515966, 1.5257811772946877) circle (1pt);
        \filldraw[black] (3.888160864908276, 2.7785089574468333) circle (1pt);
        \filldraw[black] (3.2442635963634507, 3.249918857886983) circle (1pt);
        \filldraw[black] (3.1892303954436376, 4.65910033789462) circle (1pt);
        \filldraw[black] (4.360650110813351, 0.43213244516574956) circle (1pt);
        \filldraw[black] (4.905246089511404, 1.6192226713487996) circle (1pt);
        \filldraw[black] (4.687574700923156, 2.122352916188249) circle (1pt);
        \filldraw[black] (4.912275419613345, 3.480674850602295) circle (1pt);
        \filldraw[black] (4.300124921435576, 4.907912984137379) circle (1pt);
    \end{tikzpicture}
    \caption{The points show the points in $X$. The dashed lines indicate the "rough grid" shape of $X$. The solid lines show $W$, which is the set dividing $X$}
\end{figure}

Each line can only intersect $2s$ lines in $W$, so $|\ell \cap W| \leq 2s$. Since $X$ is roughly grid shaped, and each cell is a square of side length $N/s$, $|X\cap O_i| \lesssim |X|/s^2$, which satisfies the requirements. 

We now proceed to the proof of the Szemerédi–Trotter theorem. It hinges on the fact that lemma \ref{weak_bound_ST_lemma} is sharp when the number of lines is either small (bounded by a constant) or much larger than the number of points $(|L| > |X|^2)$. We can use this by using the cell decomposition lemma to pick cells where one of these conditions holds. 

\begin{proof}
Given $X$ and $L$, and arbitrary $s$. Then using the cell decomposition lemma, define $X_i = X \cap O_i$. Then $|X_i| \leq |X|/s^2$ and 

\[\sum_i |X_i| \leq |X|\] 

Define 

\[L_i = \{\ell \in L: \ell \cap O_i \neq \emptyset\}\]


From the cell decomposition lemma

\[\sum_i |L_i| \leq s |L|\]

Then as every intersection of a point and a line is either on a cell boundary or within a cell. Then 

\[I(X,L) \leq \sum_i I(X_i, L_i) + I(X\cap W, L)\] 

Applying lemma \ref{weak_bound_ST_lemma} to the first term, and the $|\ell \cap W| \lesssim s$ bound to the second term, 

\begin{align*}
    I(X,L) \lesssim & \sum_i (|X_i| |L_i|^{1/2} + |L_i|) + s|L|\\
    \lesssim & \left(\sum_i |X_i|^2 \sum_i |L_i|\right)^{1/2}+ 2s|L|\\
    \lesssim & \left(\sum_i \frac{|X|}{s^2} |X_i| \right)^{1/2}(s|L|)^{1/2} + 2s|L|\\
    \lesssim & s^{-1/2} |X| |L|^{1/2} + s|L|\\
\end{align*}

We then choose $s$ to minimize this quantity. This is effectively choosing $s$ so that both terms are equal. Then 

\begin{align*}
    s^{-1/2} |X| |L|^{1/2} = & s|L|\\
    |X||L|^{-1/2} = & s^{3/2}\\
    s = & |X|^{2/3} |L|^{-1/3}
\end{align*}

 Plugging $s$ back into the inequality gives 


\[I(X,L) \lesssim |X|^{2/3}|L|^{2/3}\]  

which gives the desired bound.
\end{proof}

 Note that in the above argument $s$ must be an integer, so this can only be done when $|X|^2 > |L|$. When $|X|^2 < |L|$ then setting $s = 1$ gives the the bound $I(X,L) \lesssim |L|$. Additionally, $s^2$ can be at most $|X|$. Then when $|X| <  |X|^{4/3} |L|^{-2/3}$, $|X| > |L|^2$, so setting $s =|X|^{1/2}$ gives the bound $I(X,L) \lesssim |X|$


We now prove the cell decomposition lemma. However, several prelimary theorems must be shown first. 

\begin{theorem}[Borsuk Ulam Theorem]
Let $f: S^n \to \mathbb{R}^n$ be a continuous function that is antipodal, ie for every $\theta \in S^n$, $f(\theta) = -f(- \theta)$. Then 0 is in the image of $f$. 
\end{theorem}

\begin{Corollary}[Ham Sandwich Theorem]
    Let $O_1, O_2, ..., O_n \subset \mathbb{R}^n$ be bounded open subsets. Then there exists a hyperplane $H$ that bisects every $O_i$. 
\end{Corollary}

\begin{proof}
An upper half (hyper)plane can be described as the set $\{x: a \cdot x > b\}$, for some vector $a$ and $b$ a real number. As scaling $a$ and $b$ by a positive real number preserves this hyperplane, the tuple $(a,b)$, can be identified with an element of $S^n$. Then for an element $\theta \in S_n$ corresponding to $(a,b)$, we let $c_\theta$ be the affine operator defined by $c_\theta(x) = a\cdot x + b$. We then define the vector valued function $f$ by

\[f_i(\theta) = \text{Vol}(O_i \cap \{x: c_\theta(x) > 0\}) - \text{Vol}(O_i \cap \{x: c_\theta(x) < 0\})\] 

$f$ is antipodal, so it has a zero. This zero corresponds to each set being bisected, which proves the theorem
\end{proof} 

The Ham Sandwich theorem works when there are up to $n$ subsets of $\mathbb{R}^n$. This is roughly because $n$ degrees of freedom are needed to bisect the $n$ sets. We can then use polynomials to increase the number of degrees of freedom, and so the number of sets that can be bisected. 

\begin{theorem}[Polynomial Ham Sandwich Theorem]
    We use the same setup as the ham sandwich theorem, except that there can be up to $N$ sets $O_i$. Then there exists a polynomial zero set that bisects every $O_i$
\end{theorem} 

\begin{proof}

    First define the space 

    \[\text{Poly}_D(\mathbb{R}^n) = \{p \in \mathbb{R}[x_1, ..., x_n]: \deg  p \leq D\}\]

    $\text{Poly}_D(\mathbb{R}^n)$ is then a vector space of degree $D^n$. We claim that if $N < D^n$, then there is a nonzero element of $\text{Poly}_D(\mathbb{R}^n)$ that satisfies the claim. We define the vector valued function $f$ by 

    \[f_i(p) = \text{Vol}(O_i \cap \{x: p(x) > 0\}) - \text{Vol}(O_i \cap \{x: p(x) < 0\})\] 

    Since scaling each nonzero $p$ by a positive real does not change $f$, $f$ is a function from $S^{D^{N}-1}$ to $\mathbb{R}^n$. Additionally, $f$ is antipodal. Then by the Borsuk Ulam theorem the conclusion follows. 
\end{proof}

The Ham Sandwich theorems allow open subsets of Euclidean space to be subdivided, but the cell decomposition lemma requires dividing sets of points. This is a technical detail that follows from the Polynomial Ham Sandwich Theorem. 

\begin{lemma}[Ham Sandwich theorem for finite sets]
Let $s_1,s_2,... ,s_N$ be a set of finite sets in $\mathbb{R}^n$. Then there exists a polynomial level set such that for every $s_i$, 

\[|s_i \cap \{x: p(x) > 0\}| \leq \frac{|s_i|}{2} \]
\[|s_i \cap \{x: p(x) < 0\}| \leq \frac{|s_i|}{2} \]
\end{lemma}

As individual points cannot be bisected, this lemma instead guarantees that excess points will lie on the level set. 

\begin{proof}
Take some $\epsilon > 0$ and for each $i$ define $N_\epsilon(s_i)$ to be the set of balls of radius $\epsilon$ centered at the points on $s_i$. Then by the Polynomial Ham Sandwich theorem the $N_\epsilon(s_i)$ can all bisected by the zero set of a polynomial of degree $D^n > n$. Then taking $\epsilon$ to 0 we get a sequence of polynomials $p_\epsilon$ that each bisect the $N_\epsilon(s_i)$. Since the sphere $S^n$ is compact, there must a convergent subsequence to some polynomial $p$. To show that this polynomial $p$ satisfies the conclusion, for contradiction assume that there exists $i$ such that

\[|s_i \cap \{x: p(x) > 0\}| > \frac{1}{2} |s_i|\] 

Since every point in $s_i \cap \{x: p(x) > 0\}$ is some nonzero distance from the set $\{x: p(x) = 0\}$, there is some $\epsilon > 0$ such that modifying $p$ by $\epsilon$ and enlarging $s_i$ by $\epsilon$ gives 

\[|N_\epsilon(s_i) \cap \{x: p_\epsilon(x) > 0\}| > \frac{1}{2} |N_\epsilon(s_i)| = \] 

a contradiction. Note that this step requires the boundedness of $s_i$ to take a perturbation of $p$ continuous. 
\end{proof}

We will now prove the cell decomposition lemma. 

\begin{proof}
    We begin with step $k=1$. Then define $p_1$ to be the degree 1 polynomial that splits $X$ into two parts. Then $X_{1,1} := \{x \in X:  p_1(x) > 0\}$ and $X_{1,2} := \{x \in X:  p_1(x) < 0\}$

    Then at stek $k+1$, define $p_k$ to be the polynomial of degree $D_k$ with $D^2 \sim 2^{k}$ such that $p_k$ bisects all $X_{k,1},..., X_{k, 2^k}$. Then $D_k \sim 2^{k/2}$. 

    Then pick $k_{final} $ such that $2^{k_{final}} \sim s^2$. Then let $O_i$ be the sets defined by $\{x: \pm p_1(x) > 0\}  \cap ... \cap \{x: \pm p_{k_{final}}(x) > 0\}$ for all choices of $\pm$. Define $W = \{x: p_1(x) = 0\} \cup ... \cup \{x: p_{k_{final}}(x) = 0\}$
    
    $X$ has been bisected $k_{final}$ times, so then 

    \[|X_{k_{final}, i}| \lesssim |X|/s^2\] 

    A line can intersect a polynomial of degree $D$ at most $D$ times, so then 

    \[|\ell \cap W| \leq 1 + 2 + 4 + ... + 2^{k_{final}/2} \sim s \]

    Then each line can intersect $W$ at most $\sim s$ times. 
\end{proof}

\newpage


\section{Reflections on the Szemeredi-Trotter theorem}

Thur March 6

First let's recall the statement of Szemer\'edi-Trotter theorem. Let $X$ be a set of points, $L$ be a set of lines (both in $\mathbb{R}^2$), we use $I(X,L)$ to denote the set of incidences between them:
\begin{equation*}
    I(X,L) = \{(p,l) \in X \times L: p \in L\}.
\end{equation*}
Szemer\'edi-Trotter claims that 
\begin{equation*}
    |I(X,L)| \lesssim |X| + |L| + |X|^{2/3} |L|^{2/3}.
\end{equation*}

All the current proofs of this theorem, like the cell decomposition method we discussed in the previous lectures, used the topology of Euclidean plane. This is not surprising, as the conclusion of this theorem is indeed related to the structure of the base field. If we replace $\mathbb{R}^2$ by $\mathbb{F}_p^2$, Szemer\'edi-Trotter bound will fail as one can see by taking $L$ to be all the lines in $\mathbb{F}_p^2$. 

On the other hand, the current methods provide little information on some closely related problems, such as:

1. Projection theory over finite fields. 

2. Structure of sharp examples for Szemer\'edi-Trotter. 

3. Projection theory of unit balls, instead of points, in $\mathbb{R}^2$ with spacing conditions (lots of attempts by Wolff). 

\subsection*{Structure of Sharp Examples}

Let's stare at the Szemer\'edi-Trotter bound:
\begin{equation*}
    |I(X,L)| \lesssim |X| + |L| + |X|^{2/3} |L|^{2/3}.
\end{equation*}
There are three terms on the RHS. The first two terms are given by double counting which generalizes to other fields. They dominate when there are too many points or lines, in which case the structure of the sharp examples example is not very rigid, giving us many degrees of freedom.  To be more specific, when the first dominates we have $|X| \gg |L|^2$, which means that the number of points has already exceeded the total number of intersections among the lines. In this case, the upper bound is tight if each point has a line passing through it. The typical sharp example looks like some chains of beads. Similarly, when the second term dominates the sharp examples look like a bunch of stars, where each line doesn't have much chance to pass through too many points.

The case where the third term dominates is the most interesting one. The known sharp examples are integer grids and their variation $R$-grids, where $R$ is the integer rings of number fields. We expect that the sharp examples in this case are highly structured. To see what information about the sharp examples the proof of Szemer\'edi-Trotter theorem tells us, let's briefly review the cell decomposition proof: Divide $\mathbb{R}^2$ into $s^2$ cells. In each cell there are $|X|/s^2$ points and (in average) $|L|/s$ lines. By choosing $s$ to be large enough we will have $|L|/s \gtrsim (|X|/s^2)^2$ and then apply the double counting bound. The proof doesn't tell us much information on the structure of the sharp example unless we can figure out the way our cells interact. Unfortunately the proof of cell decomposition is not very constructive and based on existence theorems from topology.

\begin{remark}
    In the projective plane $\mathbb{P} \mathbb{R}^2$ there is something called point-line duality. It preserves the incidence relationship between points and lines. In fact, the statement that a point with coordinates $[a_0, a_1 , a_2]$ lies on a line with coefficients $[b_0, b_1, b_2]$ simply means $a_0 b_0 + a_1 b_1 +  a_2 b_2 = 0$, where the roles of $a_i$ and $b_i$ are interchangeable. The chain example and the star examples are mapped to each other via the point-line duality, while  the grid example will be mapped to something different. 
\end{remark}

There are also some interesting variations of this problem. For example, one may ask about the structure of $X$ which maximizes the projections for some particular $D$. Define $S_D(N) = \min_{|X|=N} S(X,D)$. For an arbitrary $D$, what can we tell about the structure of $X$ achieving this maximum? All the known examples are for direction sets with special structures.  

Denote the directions in $\mathbb{R}^2$ by elements of $\mathbb{R} \cup \{\infty\}$ with corresponding projections $\pi_t(x) = x_1+tx_2$ for $t \in \mathbb{R}$, $\pi_\infty(x) = x_2$. Since we can use a projective transformation to map any three directions to any three specified directions without changing the incidence structure, let's begin with $|D|=4$. Without of loss of generality we may assume $D = \{0,1,t,\infty\}$. When $t$ is rational with small denominator the grid example still works. Things become more interesting when $t$ is transcendental. For example, we may take $P_{k,s} = \{a_0 + a_1 t + \cdots + a_{k-1} t^{k-1}: a_i \in \mathbb{Z}, 0 \le a_i \le s-1\}$ be a set of polynomials in $t$ and let $X = P_{k,s} \times P_{k,s}$. Then $|X| = s^{2k}$. We have $\pi_0(X) = \pi_\infty(X) = P_{k,s}$, $\pi_1(X) \subset P_{k,2s}$, $\pi_t(X) \subset P_{k+1,2s}$. Since $t$ is transcendental, $S(X,D) \sim |\pi_t(X)| \sim  2^k |X|^{1/2k} |X|^{1/2}$. Choose $k = (\log_2(|X|)/2^{1/2}$ to maximize the RHS, we obtain that is this case $S_D(X) \sim e^{c(\log|X|)^{1/2}} |X|^{1/2}$. It would also be interesting to analyze $S(X,D)$ for other $D$'s, like $D=\{0,1,\infty,t_1,\ldots,t_k\}$ where $t_j$'s are algebraic independent over $\mathbb{Q}$.

\subsection*{Projection Theory over Finite Fields}

We have seen that projection theory over finite fields may be different from that over the reals. Let $X$ be a set of points, $D$ be a set of directions. Let $S = \max_{\theta \in D} \pi_\theta(X)$. Our conjecture is that for $|S| \le p/2$, 
\begin{equation*}
    |S| \gtrsim |D|^{1/2} |X|^{1/2}.
\end{equation*}
It would be attempting to investigate the structure of sharp examples for this bound, and one may conjecture they are essentially grids. 

\begin{remark}
    Let's give an example showing that the original version of Szemer\'edi-Trotter bound fails over complex field. Again, let $X$ be a set of unit balls in $B_R^{\mathbb{C}^2} \subset \mathbb{C}^2$, $D \subset B_1^\mathbb{C} \subset  \mathbb{C}$ be an $R^{-1}$-separated set of directions. For $t \in D$, let $\pi_t: \mathbb{C}^2 \to \mathbb{C}$ be the map $(z_1, z_2) \mapsto z_1 + t z_2$. For our example, choose $X$ to be a maximal set of $R^{-1}$-separated unit balls with centers in $\mathbb{R}^2$, and $D$ to be a maximal $R^{-1}$-separated subset of $\mathbb{R} \cap B_R^\mathbb{C}$. Then $|X| \sim R^2$, $|D| \sim R$ satisfy the Hausdorff spacing condition, while $S(X,D) \sim R \ll |X|^{1/2} |D|^{1/2}$.
\end{remark}

\subsection*{Projection Theory of Unit Balls}

Let $X$ be a set of unit balls in $B_R$, $D$ be a set of $1/R$-separated directions. Define $N_X(r) = \max_{c \in B_R} |X \cap B(c,r)|$, $N_D(\rho) = \max_{\gamma \subset \mathbb{S}^1, |\gamma| = \rho} |D \cap \gamma|$. We will assume that $X$ has Hausdorff spacing, which means there exists $0 \le \alpha \le 1$ such that $|X| \sim R^\alpha$, $N_X(r) \lesssim r^\alpha$. Similarly we will also assume that $|D| \sim R^\beta$, $N_D(r/R) \lesssim r^\beta$. The following conjecture by Furstenberg was recently proved by Orponen, Shmerkin, Ren and Wang:

\begin{theorem}
    Under the above assumption, we have 
    \begin{equation*}
        |D| \stackrel{<}{\approx} |S|^2/|X|
    \end{equation*}
    if $|D| \lesssim R^{-\epsilon}\min(R,|X|)$.
\end{theorem}

We will discuss briefly why cell decomposition doesn't work in this case. Suppose that we have divided $B_R$ into $s^2$ cells. There is no guarantee on the shape of each cell $O_i$, but in one important scenario, most cells   are roughly balls of some radius $r$ so that it is possible for us to apply induction hypothesis. (At first one might think $r = R/s$, but this may not be the case.  It may be that most of the balls of radius $r$ cover only a fractal subset of $B_R$ which contains our set $X$.)  
The problem is, the $R^{-1}$-separated directions may look indistinguishable at smaller scales. In each cell we have to choose an $r$-separated subset $D_i \subset D$. By the Hausdorff assumption, in a typical cell we will have $|X_i| \sim r^\alpha$, $|D_i| \sim r^\beta$. So we can not force $|X_i|^2$ to be smaller than $|L_i|$ by simply passing to smaller balls.

\newpage


\section{Sum-product theory over characteristic $p$}

Projection theory works differently over characteristic $p$ than in characteristic 0. Recall this example:

\begin{example}\label{eg:1.1}
    Let $p$ be a prime, $q=p^2$, $X=\F_p^2\subset\F_q^2$, $D=\F_p\subset\F_q$. For $\theta\in\F_q$, let $\pi_\theta:\F_q^2\to\F_q$ be defined by $\pi_\theta(x_1,x_2)=x_1+\theta x_2$. Then set $S=\max_{\theta\in D}|\pi_\theta(X)|$. We have $\pi_\theta(X)=\F_p$ for all $\theta\in D$, so $S=p$. Then we have $|X|=p^2=q,S=|D|=p=q^{1/2}$. 
\end{example}

So the sizes of the projections of $X$ can be small even when $X$ is large. However, the same cannot happen over $\F_p$:

\begin{theorem}[Bourgain-Katz-Tao]\label{thm:bkt}
    Let $X\subset\F_p^2$ with $|X|=p^s$ for $0\leq s<2$. Let $D\subset\F_p$ with $|D|=p^t$ for $0<t\leq 1$. Then $S=\max_{\theta\in D}|\pi_\theta(X)|\geq p^{s/2+\eps(s,t)}$ where $\eps(s,t)>0$.
\end{theorem}

The Bourgain projection theorem is an analogous theorem for balls in $\R^2$.

Note that the key difference between $\F_p$ and $\F_q$ that allows an example like Example \ref{eg:1.1} to exist while no such example exists over $\F_p$ is the existence of a subfield $\F_q$. A way to quantify the properties of a subfield is a set $X$ with small sum and product sets. As such, we should study the sizes of such sets.

\subsection{Sum-product theory}

\begin{notation}
    For $A\subset\F_p$, let
    $$A+A=\{a_1+a_2:a_i\in A\},\quad A\cdot A=\{a_1a_2:a_i\in A\}.$$
    Also, let $A^{\oplus n}=\underbrace{A+A+\dots+A}_n$.
\end{notation}

\begin{lemma}\label{lem:1}
    If $A\subset\F_p$, then either
    \begin{enumerate}
        \item $\frac{A-A}{A-A}=\F_p$, or
        \item $\left|\frac{(A\cdot A)^{\oplus 3}-(A\cdot A)^{\oplus 3}}{A-A}\right|\geq|A|^2$.\label{2}
    \end{enumerate}
\end{lemma}
\begin{proof}
    First, note that if $c\not\in\frac{A-A}{A-A}$, then $|A+cA|=|A|^2$. Indeed, if this was not the case then there would be some $a_1,a_2,a_1',a_2'\in A$ with $a_1+ca_2=a_1'+ca_2'$. But this implies $c=\frac{a_1'-a_1}{a_2-a_2'}\in\frac{A-A}{A-A}$.

    Next, note that if $\frac{A-A}{A-A}\neq\F_p$ then there is some $b\in\frac{A-A}{A-A}$ with $b+1\not\in\frac{A-A}{A-A}$. Indeed, we can set $b+1$ to be the smallest element of $\F_p\setminus\frac{A-A}{A-A}$, which would imply $b\in\frac{A-A}{A-A}$.

    Now, if $\frac{A-A}{A-A}\neq\F_p$, then we have
    $$\left|A+\left(\frac{A-A}{A-A}+1\right)A\right|\geq|A|^2,$$
    which implies \ref{2} after putting the LHS over a common denominator.
\end{proof}

\subsection{Freiman-Ruzsa theorem}

One question to ask in sum-product theory is when the set $A+A$ is small.

\begin{example}
    \begin{enumerate}
        \item If $A=[L]$, then $|A+A|\leq 2|A|$.
        \item More generally, if $A$ is an arithmetic progression $A=\{a+nd\}_{n\in[L]}$, then $|A+A|\leq 2|A|$.
        \item Even more generally, we can consider $A=\{a+n_1d_1+\dots+n_rd_r\}_{n_i\in[L_i]}$. Then $A+A\subset\{2a+n_1d_1+\dots+n_rd_r\}_{2\leq n_i\leq 2L_i}$, so $|A+A|\leq 2^r|A|$. In this case we call $A$ a \textbf{generalized arithmetic progression (GAP)} of dimension $r$ and volume $L_1\cdots L_r$.
    \end{enumerate}
\end{example}

\begin{theorem}[Freiman-Ruzsa]
    If $A\subset\Z$ and $|A+A|\leq K|A|$, then $A$ is contained in a GAP of dimension $r(K)$ and $\vol\leq V(K)\cdot|A|$.
\end{theorem}

This is a deep theorem that we will not prove, and the quantitative bounds on $r(K)$ and $V(K)$ are weak. In the original paper, the bounds were of the form $r(K)=\exp(K^c),V(K)=\exp(\exp(K^c))$, so the theorem is only meaningful if $K$ is small.

\begin{conj}
    There is a meaningful bound if $K=|A|^\delta$ for some $\delta>0$.
\end{conj}

\subsection{Ruzsa triangle inequality}

\begin{theorem}[Ruzsa]
    Let $Z$ be an abelian group and $A,B,C\subset Z$. Then $|A||B-C|\leq|A-B||A-C|$.
\end{theorem}

\begin{corollary}
    If $|A+A|\leq K|A|$, then $|A-A|\leq K^2|A|$.
\end{corollary}
\begin{proof}
    Use Ruzsa's triangle inequality with $A=A,B,C=-A$. Then we have
    $$|B-C|=|(-A)-(-A)|=|A-A|,\quad|A-B|=|A-C|=|A-(-A)|=|A+A|.$$
    So Ruzsa's triangle inequality tells us that $|A||A-A|\leq|A+A|^2$, which implies the corollary.
\end{proof}

\begin{proof}[Proof of Ruzsa triangle inequality]
    We will construct an injective map $\phi:A\times (B-C)\to(A-B)\times(A-C)$. For all $d\in B-C$, fix some $b(d)\in B,c(d)\in C$ with $d=b(d)-c(d)$. Then set $\phi(a,d)=(a-b(d),a-c(d))$. We need to show that $\phi$ is injective. Suppose $\phi(a,d)=(x,y)$. Then we will recover $a,d$ from $x,y$ and the choices of $b(d),c(d)$. Note that we have $y-x=b(d)-c(d)=d$, so we can recover $d$. Then from $d$ we know $b(d)$, so we can recover $a=x+b(d)$.
\end{proof}

\subsection{Plunnecke inequality}

\begin{theorem}[Plunnecke]
    Let $Z$ be an abelian group and $A,B\subset Z$ with $|A+B|\leq K|A|$. Then $|B^{\oplus m}-B^{\oplus n}|\leq K^{m+n}|A|$.
\end{theorem}

\begin{corollary}
    If $|A+A|\leq K|A|$ then $|A-A|\leq K^2|A|$, $|A+A+A|\leq K^3|A|$.
\end{corollary}

\begin{corollary}
    If $|A-A|\leq K|A|$ then $|A+A|\leq K^2|A|$.
\end{corollary}
\begin{proof}
    Use Plunnecke's inequality with $B=-A$.
\end{proof}

\begin{lemma}\label{lem:2}
    If $A\subset\F_p,|A|=p^s$ for $0\leq s<1$, then $|A^3-A^3|\geq p^{s+\eps(s)}$ for some $\eps(s)>0$.
\end{lemma}
\begin{proof}
    Let $B=(A^2)^{\oplus 3}-(A^2)^{\oplus 3},C=A-A$. Then by Lemma \ref{lem:1} we have $\left|\frac{B}{C}\right|\geq p^{s+\gamma(s)}$ for some $\gamma>0$. Now, assume for contradiction that $|A^3-A^3|\leq K|A|$ where $K\lessapprox 1$. Then we have $|A^3|\leq K|A|$, and since $|A|\leq|A^3|$, we have $|A^3-A^3|\leq K|A^3|$. Then Plunnecke's inequality implies
    $$|(A^3)^{\oplus m}-(A^3)^{\oplus n}|\leq K^{m+n}|A^3|\leq K^{m+n+1}|A|.$$
    In particular, this implies $|B\cdot C|,|A\cdot B|,|A\cdot C|\leq K^{O(1)}|A|$. Then the Ruzsa triangle inequality (on $\F_p$ as a multiplicative set) implies
    $$|A|\left|\frac{B}{C}\right|\leq|A\cdot B||A\cdot C|\leq K^{O(1)}|A|^2,$$
    so we have $p^{s+\gamma}\leq\left|\frac{B}{C}\right|\leq K^{O(1)}|A|=K^{O(1)}p^s$, which contradicts $K\lessapprox 1$.
\end{proof}

In fact, there is actually a stronger statement:

\begin{theorem}[Bourgain-Katz-Tao]
    If $A\subset\F_p$ with $|A|=p^s$, then $\max(|A\cdot A|,|A+A|)\geq p^{s+\eps(s)}$.
\end{theorem}

\begin{notation}
    We define $\poly_K(A)=(A^K)^{\oplus K}-(A^K)^{\oplus K}$.
\end{notation}

\begin{corollary}
    If $0<s<t<1$, then there exists a $K=K(s,t)$ such that for all $A\subset\F_p$ with $|A|=p^s$, we have $|\poly_K(A)|\geq p^t$.
\end{corollary}
\begin{proof}
    Apply Lemma \ref{lem:2} many times.
\end{proof}

The following proof is due to Petridis.

\begin{proof}[Proof of Plunnecke's inequality]
    The proof depends on a key lemma.

    \begin{lemma}
        If $|A+B|\leq K|A|$, then there exists a $X\subset A$ such that for all $C\subset Z$ we have
        $$\frac{|X+C+B|}{|X+C|}\leq K.$$
    \end{lemma}
    \begin{proof}
        Choose $X\subset A$ to minimize the value $\frac{|X+B|}{|X|}$. Then set $\frac{|X+B|}{|X|}=\underline{K}\leq K$. We will show by induction on $|C|$ that $\frac{|X+C+B|}{|X+C|}\leq\underline{K}$ for all $C\subset Z$.

        For the base case, when $|C|=1$ we have $\frac{|X+C+B|}{|X+C|}=\frac{|X+B|}{|X|}=\underline{K}$. For the inductive step, let $C'=C\cup\{c\}$, and assume that $\frac{|X+C+B|}{|X+B|}\leq\underline{K}$. Then set
        $$Y=\{x\in X:x+c+B\subset X+C+B\}.$$
        Note that by construction we have $Y+\{c\}+B\subset X+C+B$. Now, let us bound $|X+C'+B|$ and $|X+C'|$. First, we have
        \begin{align*}
            |X+C'+B|&=|X+C+B|+|(X+\{c\}+B)\setminus(X+C+B)|\\
            &\leq|X+C+B|+|(X+\{c\}+B\setminus(Y+\{c\}+B|\\
            &=|X+C+B|+|X+B|-|Y+B|.
        \end{align*}
        Next, we have
        \begin{align*}
            |X+C'|&=|X+C|+|\{x\in X:x+c\not\in X+C\}|\\
            &=|X+C|+|X|-|\{x\in X:x+c\in X+C\}|\\
            &\geq|X+C|+|X|-|Y|.
        \end{align*}
        Recall that we have $|X+C+B|\leq\underline{K}|X+C|$ and $|X+B|=\underline{K}|X|$, and we also have $|Y+B|\geq\underline{K}|Y|$ by the definition of $X$. So we have
        $$|X+C'+B|\leq|X+C+B|+|X+B|-|Y+B|\leq\underline{K}|X+C|+\underline{K}|X|-\underline{K}|Y|\leq\underline{K}|X+C'|,$$
        completing the proof.
    \end{proof}
    Now, let us return to the proof of Plunnecke's inequality. By the key lemma, there is some $X\subset A$ such that $|X+C+B|\leq K|X+C|$. Plugging in $C=\{c\}$ yields $|X+B|\leq K|X|$. Then plugging in $C=B$ gives $|X+B+B|\leq K|X+B|\leq K^2|X|$. Continuing in this fashion, we get $|X+B^{\oplus m}|\leq K^m|X|$.

    Now, Ruzsa's triangle inequality implies
    $$|X||B^{\oplus m}-B^{\oplus n}|\leq|X+B^{\oplus m}||X+B^{\oplus n}|\leq K^{m+n}|X|^2,$$
    so we get
    $$|B^{\oplus m}-B^{\oplus n}|\leq K^{m+n}|X|\leq K^{m+n}|A|.$$
\end{proof}


\section{Sum Product theory over characteristics $p$ continued}
% Jiahui Yu,  Thur March 20

\subsection{Contagious Structure Lemma}
\begin{lemma}
    If $Z$ is an abelian group and $A \subset Z$, and  
    $$|A-tA|\leq K|A| \text{ and }|A-t_2A|\leq K|A|,$$
    then $|A-(t_1\cdot t_2 )A|\leq K^2A$. 
\end{lemma}
\begin{proof}
    Note that $|A-t_2A|\leq K|A|$ implies that $|t_1A-t_1t_2A|\leq K|A|$. Let $\bar{B}=A$, $\bar{C}=t_1t_2A$, $\bar{A}=A$ in Rusza's inequality, so 
    $$|t_1A||A-t_1t_2A|\leq |t_1A-A||t_1-t_1t_2A|.$$
    Thus, $|A||A-t_1t_2A|\leq K^2|A|$.
\end{proof}
\begin{lemma}
    If $|A+tA|\leq K|A|$ then $|A-tA| \leq K^2|A|$.
\end{lemma}
\begin{proof}
    By Rusza's inequality, $|A||A-tA|\leq |A+A||A+tA|$. By Plunnecke's inequality, $|A+A|\leq K^2|A|$. Thus, $|A||A-tA|\leq K^3|A|$. 
\end{proof}
\begin{lemma}
    If $|A+t_1A| \leq |A|$ and $|A+t_2A| \leq K|A|$, then 
    $$|A+(t_1+t_2)A|\leq K^5|A|. $$
\end{lemma}
\begin{proof}
   Note that  $|A+(t_1+t_2)A| \leq |A+t_1A+t_2A|$. By the main lemma, there exists $X_1\subset A$ so for any $C$, we have $|X_1+C+t_1A|\leq |X_1+C|$ and there exists $X_2\subset A$ so for any $C$, we have $|X_2+C+t_1A|\leq |X_2+C|$. 
   \begin{eqnarray*}
       |A+t_1A+t_2A| & \leq & |x_1+x_2+A+t_1A+t_2A|\\
       &\leq& K|x_1+x_2+A+t_2A|\\
       &\leq& K^2|x_1+x_2+A|\\
       &\leq& K^2|A+A+A|\\
    &\leq& K^5|A|\\
   \end{eqnarray*}
\end{proof}
Our goal today is to prove the following theorem. 
\begin{theorem}\label{mainthm}
    If $A\subset \F_p$, $|A|=p^{s_A}$, $D\subset \F_p$, $|D|=p^{s_D}$, $0 <s_A,s_p<1$. Then, there exists $\epsilon(s_A,s_D)>0$, $\max(s_A,s_D)>0$, $\max(|A+tA|)\geq p^{s_A+\epsilon(s_A,s_D)}$.  
\end{theorem}
\begin{corollary}
    $|A+A\cdot A|\geq p^{s_A+\epsilon}$. 
\end{corollary}
Now, let's recall double counting result. 
\begin{lemma}(Double Counting)\\
  Suppose $X \subseteq \F_p^2$, and $D \subseteq \F_p$, then 
  $$\max_{t \in D}|\pi_t(X)| \gtrsim \min(|X|,|D|).$$
\end{lemma}
Note that if $s_D >s_A$, then double counting implies theorem \ref{mainthm}, so the hard cases are the cases in which $0<s_D<s_A$.  
Let's also recall a corollary from the previous section. 
\begin{lemma}\label{sumprod}
    If $0<s<t<1$, then there exists $k=k(s,t)$ so if $A \subseteq \F_p$, $|A|=p^s$ then $|poly_k(A)|\geq p^t$. 
\end{lemma}
The proof idea is to use lemma \ref{sumprod} to increase $s_D$ to be bigger than $s_A$ by taking sums and products and then use the contagious structure.  
\begin{proof}
    By lemma \ref{sumprod}, there exists $K(s_A,s_D)$ so $|poly_k(D)|>p^{s_A+r}$. By double counting there exists $u \in poly_k(D)$ so $|A+uA|>p^{s_A+r}$. But if $\max_{t\in D}|A+tA|\leq K|A|$, then the contagious structure says that 
    $$\max_{u \in poly_k(D)}|A+uA|\leq K^{c(k)}|A|=K^{c(s_A,s_D)}|A|=K^cp^s.$$
    However, this would imply that $K^c\geq p^r$ which would imply that $p^{r/c}=p^\epsilon$ a contradiction. 
\end{proof}
The above theorem \ref{mainthm} is a special case of the following theorem when we put $X=A\times A$. 
\begin{theorem}(BKT)\\
If $X \subseteq \F_p^2$, $|X|=p^{s_X}$ with $0<s_X<2$, and $D \subseteq \F_p$ with $|D|=p^{s_D}$ such that $0<s_D$. 
$$\max_{t\in D}|\pi_t(X)|\geq p^\epsilon |X|$$
\end{theorem}

\section{Possible Topics for the Rest of the Semester}
% Still Jiahui Yu, Thur March 20

\begin{enumerate}
    \item Mixing for random walks on discrete groups. (Bourgain-Ganbund)
    \item Homogenous dynamics mixing orbit (Margulis, Dani, Rutner)
    \item Furstenberg
    \item Kakeya
\end{enumerate}
A typical example of homogeneous space is $X=SL(n;\R)/SL(n;\Z)$. It is homogeneous in the sense that $SL(n;\R)$ acts on the space on the left and the action is transitive.
That is to say for any $x \in X$, $SL(n;\R)\cdot x =X$. 
We can also think of $X$ as the space of lattices of covolume $1$. One can take a subgroup $H \subset SL(n;R)$ and ask how the orbit is distributed. 
\subsubsection{Number theory applications}
Let $Q(x_1,\cdots, x_n)$ be a quadratic form. \\
\textbf{Question}: How is $Q(\Z^n)$ distributed? 
\begin{conj}(Oppenheim) If $n\geq 3$, the signature of $Q$ is mixed and coeffience of $Q$ are not contained in $\Z\alpha$ for any $\alpha$, then $Q(\Z^n)$ is dense in $\R$. 
\end{conj}
This conjecture was proven by Margulis and his proof involves connecting this problem with the set up of an orbit in the space of lattices. After a change of variable we can turn $Q$ into a standard quadratic form and $\Z^n$ into some lattice. 
For example: after a change of variables. 
$Q \Rightarrow x_1^2+x_2^2-x_3^2$, $\Z^3 \Rightarrow \Lambda$ a lattice. Note that the quadratic form has many symmetries. In particular, $SO(2;1) \subset SL(3;\R)$ preserves the quadratic form. Thus, we can use any lattice in $SO(2;1)\Lambda$ instead of $\Lambda$, so we turned this problem into the problem of studying the orbit of $\Lambda$ under the action of $SO(2;1)$. 
\subsubsection{Relation to projection theory}
For different subgroups, the problem behaves very differently. We shall focus on unipotent groups as there are strong theorems on them.\\ 
For example, in $SL(n,\R)$, we define 
$$U=\left\{\begin{bmatrix}
    1& & & t\\
     &1& &  \\
     & &1& \\
     & & & 1\\
\end{bmatrix}: t\in \R \right\}.$$
Note that it is a subgroup of $SL(n,\R)$. Let's focus on $SL(2,\R)$ and $$U=\left\{\begin{bmatrix}
    1& t\\
    0 & 1
\end{bmatrix}:t\in \R\right\}.$$ We are interested in the asymptotic behavior of the orbit so we would like to study $$\{u_tx: 0 \leq t \leq T\}$$
as $T$ tends to infinity. What's important about this group is that it interplays nicely with a different matrix. 
Define $$a_r=\begin{bmatrix}
    e^r& 0\\
    0  & e^{-r}
\end{bmatrix}\subset SL(2;\R).$$
Then, $a_ru_ta_{-r}=u_{e^{rt}}$ and 
$$a_1\cdots a_1 \{u_tx:0 \leq t \leq 1\}a_1^{-1}\cdots a_1^{-1}=\{u_tx: 0 \leq t \leq e^N\}.$$ 
Thus, it is enough to understand what happens when conjugating by $a$. If we look at a little box in the homogeneous space, some direction of it will be stretched and some direction will be contracted. It's almost like that the box is completely flatten in the direction that is contracted so it's sort of like projecting into a lower dimensional space. If we go through a bunch of different boxes, the corresponding direction that gets contracted changes so it's like projecting into different directions. 


\section{Proof of Bourgain-Katz-Tao projection theorem}

% Siqi Wu,  Tues Apr 1

In this section,  we introduce the Balog-Szemeredi-Gowers theorem and use it to finish the proof of the Bourgain-Katz-Tao projection theorem.

\subsection{Proof of BKT}

Recall the Bourgain-Katz-Tao theorem for $\mathbb{F}_p$: 

\begin{theorem}[BKT]
    Let $X \subset \mathbb{F}_p^2$ be a subset with size $|X| = p^{s_X}$, $0 < s_X < 2$ and $D \subset \mathbb{F}_p$ be a set of direction with $|D| = p^{s_D}$, $s_D > 0$. Then 
    \begin{equation*}
        \max_{t \in D} |\pi_t(X)| \gtrsim p^{\epsilon} |X|^{1/2} 
    \end{equation*}
    for $\epsilon = \epsilon(s_X, s_D) > 0$. 
\end{theorem}

The same statement fails for nonprime field $\mathbb{F}_q$, as one can see by taking $(X,D)$ to be $(\mathbb{F}_p^2, \mathbb{F}_p)$ where $q = p^a$. 

Our proof will be based on Theorem 12.1 in previous lectures, which we state here for reader's convenience. 

\begin{theorem}
    Let $A$, $D$ be subsets of $\mathbb{F}_p$ with $|A| = p^{s_A}$, $0 < s_A< 1$ and $|D| = p^{s_D}$, $s_D > 0$. Then
    \begin{equation*}
        \max_D |A+tA| \ge p^{\epsilon_1}|A|
    \end{equation*}
    for $\epsilon_1 = \epsilon_1(s_A, s_D)$.
\end{theorem}

It can be viewed as a special case of BKT where $X$ takes the special form  $A \times A$. Let's try to prove BKT by contradiction using this theorem. Assume that $S_D(X) \ll p^{\epsilon} |X|^{1/2}$ for $\epsilon > 0$ to be determined. Since the size of projections are invariant under projective transformations, we may assume $0,\infty \in D$ without loss of generality. 

Let $A = \pi_0X \cup \pi_\infty X$. Then $X \subset A \times A$. By Theorem 12.1, we have $\max_{t \in D} |\pi_t (A \times A)| \gtrsim p^{\epsilon_1}|A \times A| \ge p^{\epsilon_1} |X|^{1/2}$ with $\epsilon_1 = \epsilon_1(s_X/2, s_D)$. We will win if the size of projections of $A \times A$ does not differ too much from that of $X$. But this is not always the case. 

Consider the following enemy scenario. Here the red plots are points of $X$ and the blue plots are some random elements we added to form $A \times A$. One can easily see that even if the size of $X$ is comparable to the size of $A \times A$, some of their projections may still be quite different.   
\begin{figure}
\centering 
\includegraphics[width=0.7\textwidth]{image.jpg} 
\caption{Enemy scenario} 
\end{figure}

To be more precise, let $X$ be the grid example $X = [N] \times [N]$. Then it has small projections along rational directions. Now choose $A = [N] \cup \Tilde{A}$ to be the projection of $X$ plus some unstructured "garbage" $\Tilde{A}$ with $|\Tilde{A}| = N$. For a generic choice of $\Tilde{A}$ it will cause large projections along most directions. To avoid this kind of difficulties, it is necessary to remove the "garbage" part of $A$. Fortunately, the Balog–Szemer\'edi–Gowers theorem says this is always possible.

\begin{theorem}[BSG]
     Let $X,A,B$ be subsets of an abelian group $(G,+)$. For $t \in G$, let $\pi_t: G \times G \to G$ denote the projection operator $(g_1, g_2) \mapsto g_1 + t g_2$. Assume that $|A|, |B| \le N$, $X \subset A \times B$ and $|X| \ge K^{-1} N^2$, $|\pi_t X| \le KN$ for some $t \in G$. Then exists $A' \subset A$, $B' \subset B$ such that $|X'| \gtrsim  K^{-O(1)}N^2$, $\pi_t(A' \times B') \le K^{O(1)}N$ where $X' = X \cap (A' \times B')$.
\end{theorem}

Assumeing this theorem, it is easy to prove BKT:

\begin{proof}{(of BKT assuming BSG)}
    Assume $S_D(X) \le p^{\epsilon} |X|$ with $\epsilon > 0$ t.b.d., $\{0,\infty\} \subset D$. Let $A = \pi_0 X \cup \pi_\infty X$. By our previous discussion, $\max_{t \in D} |\pi_t (A \times A)| \gtrsim p^{\epsilon_1}|A \times A| \ge p^{\epsilon_1} |X|^{1/2}$ with $\epsilon_1$ depending only on $s_A$, $s_D$. Write $|A| = N \ge |X|^{1/2}$. Then by assumption we have $N^2 \le p^{2\epsilon}|X|$. Fix some $t_1 \in D \setminus \{0,\infty\}$ and apply BSG, we obtain $A', B' \subset A$, $X' = X \cap (A' \times B')$ such that $|X'| \ge p^{-O(\epsilon)} N^2$, $|A' + t_1 B'| \le p^{O(\epsilon)} N$. 

    Since $X' \subset A' \times B'$ has small projection along one direction, we expect it to be highly structured and hence has small projections along many other directions. This is done by the following argument. Let $t \in G$. Consider the map
    \begin{equation*}
        \pi_t \left(A' \times \frac{-1}{t_1} A'\right) \times X' \to (A'-A') \times (A'-t_1 B') \times \pi_t(Y)
    \end{equation*}
    given by 
    \begin{equation*}
        \left(a_1 - \frac{1}{t_1} a_2, (a,b)\right) \mapsto (a_1-a, a_2 + t_1 b, a+tb).
    \end{equation*}
    This map is clearly injective. Hence
    \begin{equation*}
        \left|\pi_t \left(A' \times \frac{-1}{t_1} A'\right)\right| \le \frac{|A'-A'||A'+t_1B'||\pi_t(X')|}{|X'|} \le p^{O(\epsilon)} \frac{N^2}{|X'|} |\pi_t(X')|,
    \end{equation*}
    where we used Pl\"unnecke-Ruzsa to bound $|A'-A'|$. Therefore, 
    \begin{equation*}
        \max_{t \in D} |\pi_t(X')| \ge p^{-O(\epsilon)} \max_{t \in D} \frac{|X'|}{N^2} |\pi_{-t/t_1} (A' \times A')| \ge p^{\epsilon_1 - O(\epsilon)} |X|^{1/2} 
    \end{equation*}
    by Theorem 12.1. A contradiction if $\epsilon$ is sufficiently small w.r.t. $\epsilon_1$. 
\end{proof}

\subsection{Additive energy}

Le $A$, $B$ be finite subsets of an abelian group $(G,+)$. Define the energy 
\begin{equation*}
    E(A,B) = |\{a_1, a_2 \in A, b_1, b_2 \in B: a_1 + b_1 = a_2 + b_2\}|.
\end{equation*}
There is a close relation between $E(A,B)$ and the size of the sumset $A+B$. Basically the energy counts the number of additive relations between $A$ and $B$. Thus $E(A,B)$ must be large if $|A+B|$ is small. One may see this from the following proposition.

\begin{prop}
    We have 
    \begin{equation*}
        |A|^2 |B|^2 \le |A+B| E(A,B)
    \end{equation*}
\end{prop}

\begin{proof}
    For $z \in G$, write $r_{A,B}(z) = |\{(a,b) \in A\times B: a + b = z\}|$. Then $E(A,B) = \sum_{A+B} r_{A,B}(z)^2$. By Cauchy-Schwarz,
    \begin{equation*}
        |A+B|E(A,B) \ge \left( \sum_{A+B} r_{A,B}(z)\right)^2 = (|A||B|)^2.
    \end{equation*}
\end{proof}

However, the converse if not true. Even if there are many additive relations between the elements of $A,B$, these subsets may still contain garbage with size comparable to the size of themselves which forces $|A+B|$ to be very large. One may exhibit this by taking $A=B \subset \mathbb{Z}$ to be $[N] \cup$ (some random subset of $\mathbb{Z}$ with size $N$). Instead, we have the BSG theorem for energy below.

Our previous results like Theorem 12.1 and BSG, BKT can also be formulated in terms of energy instead of cardinality. Let's record some results here without proof (which are similar to what we did). 

\begin{theorem}
    Let $A$, $D$ be subsets of $\mathbb{F}_p$ with $|A| = p^{s_A}$, $0 < s_A< 1$ and $|D| = p^{s_D}$, $s_D > 0$. Then
    \begin{equation*}
        \min_{t \in D} |A+tA| \ge p^{3\epsilon_A - \epsilon_2}|A|,
    \end{equation*}
    for $\epsilon_2 = \epsilon_2(s_A, s_D) > 0$.
\end{theorem}

\begin{theorem}[BSG var]
     Let $A,B$ be subsets of $(G,+)$ with $|A|, |B| = N$ and $E(A,B) \ge K^{-1} N^3$. There exists $A' \subset A$, $B' \subset B$ with $|A'|, |B'|\ge K^{-O(1)} N$ such that $|A'+B'| \le K^{-O(1)}N$.
\end{theorem}

\begin{theorem}[BKT 2]
    With the same setting of BKT, there exists $t \in D$ such that for any subset $Y \subset X$ with $|Y| \ge p^{-\epsilon} |X|$, we have $|\pi_t(Y)| \ge p^\epsilon |X|$.
\end{theorem}

Let's remark that there are both advantages and disadvantages of working with energy. It makes the problem behaves better when passing to large subsets. But there is also a major drawback: Recall that P-R inequality yields the  contagious structure of $|A+tA|$ (see Lemma 12.$x$ with $1 \le x \le 3$). This is no longer the case for energy. Intuitively, to say that $E(A,B)$ is large is equivalent to say that a large part of $A$ is "friendly" (has a lot of nontrivial additive relations) with a large part of $B$. Even if for each $i$ there is a piece of $A$ being friendly with $t_i A$, they are not necessarily the same for each $i$. Consider the example $A = [N] \cup t_1[N] \cup t_2[N]$. Then both $E(A,t_1 A)$ and $E(A,t_2A)$ are large, but $E(A, (t_1+t_2)A)$ doesn't need to be large in general.



\section{Proof of the BSG Theorem}
% Jonathan Buchanan,  Thur Apr 3.

In this lecture we'll prove the BSG theorem that we used in the proof of the BKT theorem. Here's the statement again:
\begin{theorem}[Balog--Szemer\'edi--Gowers]
    Let $A$ and $B$ be subsets of an abelian group and suppose $X \subset A \times B$. If $|A|, |B| \leq N$, $|X| \geq K^{-1} N^2$, and $|\pi_1(X)| \leq KN$, then there are $A' \subset A$ and $B' \subset B$ such that $|A' + B'| \leq K^{O(1)} N$ and $|X'| \geq K^{-O(1)} N^2$, where $X' = X \cap (A' \times B')$.
\end{theorem}
Here $\pi_1(X) = \{ a + b \ : \ (a, b) \in X \}$.
\begin{example}
    As subsets of $\Z$, let $A = B$ be the union of $[N]$ and some garbage. Let $X$ be the union of any subset of $[N] \times [N]$ and a little garbage. Then $\pi_1(X)$ is small ($\lesssim N$), while $|A + B| \gtrsim N^2$ is large. We can take $A' = B' = [N]$.
\end{example}
The theorem was originally proved by Balog and Szemer\'edi, but in the bounds $K^{O(1)}$ was instead $F(k)$ and $K^{-O(1)}$ was $\frac{1}{F(K)}$, and $F(K)$ was some function with crazy growth. The bounds in the version stated above are due to Gowers.

We will proceed by thinking of $X \subset A \times B$ as a bipartite graph with $A$ on the left, $B$ on the right, and an element $(a, b) \in X$ representing an edge from $a$ to $b$. Let
\[
    P_K(a, b) = \# \{ \text{paths of length } K \text{ in the graph } X \text{ from } a \text{ to } b \}.
\]
\begin{lemma}
    \label{lem:bsg_additive_lemma}
    If $A' \subset A$ and $B' \subset B$, and for any $a \in A'$, $b \in B'$, $P_3(a, b) \geq P$, then
    \[
        |A' + B'| \leq \frac{\pi_1(X)^3}{P}.
    \]
\end{lemma}
\begin{proof}
    A path of length $3$ from $a$ to $b$ goes from $a$ to some $b_1 \in B$, then to some $a_1 \in A$, then to $b$. So $(a, b_1), (a_1, b_1), (a_1, b) \in X$ and hence
    \[
        \underbrace{a + b_1}_{z_1}, \underbrace{a_1 + b_1}_{z_2}, \underbrace{a_1 + b}_{z_2} \in \pi_1(X).
    \]
    We can write $a + b = z_1 - z_2 + z_3$. Therefore
    \[
        \# \{ (z_1, z_2, z_3) \in \pi_1(X)^3 \ : \ a + b = z_1 - z_2 + z_3 \} \geq P_3(a, b) \geq P.
    \]
    Summing over $A' + B'$ we get
    \[
        |A' + B'| \cdot P \leq |\pi_1(X)|^3.
    \]
\end{proof}
From now on, everything we prove will be a statement about bipartite graphs, i.e. we don't need the addition law for anything that follows.
\begin{lemma}[Key Lemma]
    If $X \subseteq A \times B$ and $|X| \geq K^{-1} |A| |B|$, then there are $A' \subset A$ and $B' \subset B$ such that $|X'| \geq K^{-O(1)} |A||B|$ where $X' = X \cap (A' \times B')$ and for any $a \in A'$, $b \in B'$,
    \[
        P_3(a, b) \geq K^{-O(1)} |A||B|.
    \]
\end{lemma}
The BSG theorem is proved by combining Lemma \ref{lem:bsg_additive_lemma} and the Key Lemma.

\subsection{Simple Bounds About $P_K(a, b)$}
In this section, we have
\begin{align*}
    \# \text{ edges } & = |X| = \geq K^{-1} |A||B|, \\
    P_\ell & := \# \text{paths of length $\ell$ starting in $A$}, \\
    P_1 & = |X| \geq K^{-1} |A||B|.
\end{align*}
\begin{Definition}
    For $a \in A$, the \textbf{neighborhood} of $a$ is the set $N(a)$ of points that share an edge with $a$.
\end{Definition}
We can average over $|A|$ to get
\[
    \operatorname{Avg}_{a \in A} P_1(a, \cdot) = \frac{|P_1|}{|A|} \geq K^{-1} |B|.
\]
To get an estimate for the average of $P_2$, we use Cauchy-Schwarz to get
\begin{align*}
    P_2 & = \sum_b |N(b)|^2 \\
    & \geq \frac{\left( \sum_b |N(b)| \right)^2}{|B|} \\
    & \geq \frac{(K^{-1}|A||B|)^2}{|B|} \\
    & = K^{-2} |A|^2 |B|.
\end{align*}
Averaging this get us
\[
    \operatorname{Avg}_{a_1, a_2} P_2(a_1, a_2) \geq K^{-2} |B|.
\]
\emph{As an exercise}, use similar methods to prove $|P_3| \geq K^{-3} |A|^2 |B|^2$ and
\[
    \operatorname{Avg}_{a, b} P_3(a, b) \geq K^{-3} |A| |B|.
\]
The Key Lemma says that $P_3(a, b)$ is at least a small fraction of the average for \emph{all} $a \in A'$, $b \in B'$.
\begin{lemma}[Length 2]
    If $X \subset A \times B$, $|X| \geq K^{-1} |A||B|$, $\epsilon > 0$, then there is a subset $A' \subset A$ such that $|A'| \geq \frac{1}{2} K^{-1} |A|$ and $P_2(a_1, a_2) \geq \epsilon K^{-2} |B|$ for $(1 - 2\epsilon) |A'|^2$ choices of $(a_1, a_2) \in (A')^2$.
\end{lemma}
Note that we cannot always take $A' = A$, because there are graphs $X$ where only $\frac{1}{K} |A|$ vertices in $A$ have an edge and there are also graphs with multiple connected components. What we will do is let $A' = N(b)$ for some $b \in B$.
\begin{Definition}
    A pair $(a_1, a_2)$ is \textbf{$\epsilon$-bad} if $P_2(a_1, a_2) < \epsilon K^{-2} |B|$. Let
    \[
        BP_\epsilon(b) = \# \{ (a_1, a_2) \in N(b)^2 \ : \ (a_1, a_2) \text{ is $\epsilon$-bad} \}.
    \]
\end{Definition}
\begin{lemma}[P1]
    \[
        \mathbb{E}_b |BP_\epsilon(b)| \leq \epsilon K^{-2} |A|^2.
    \]
\end{lemma}
\begin{lemma}[P2]
    \[
        \mathbb{E}_b |N(b)|^2 \geq K^{-2} |A|^2.
    \]
\end{lemma}
This says there's only about an $\epsilon$-fraction of bad pairs.
\begin{proof}[Proof of P2]
    By Cauchy-Schwarz,
    \begin{align*}
        \sum_b |N(b)|^2 & \geq \frac{\left( \sum_b |N(b)| \right)^2}{|B|} \\
        & \geq \frac{(K^{-1} |A||B|)^2}{|B|} \\
        & = K^{-2} |A|^2 |B|.
    \end{align*}
    Divide by $|B|$.
\end{proof}
\begin{proof}[Proof of P1]
    \begin{align*}
        \sum_b |BP_\epsilon(b)| & = \# \{ \text{$a_1, a_2, b$ such that $(a_1, b), (a_2, b) \in X$ and $P(a_1, a_2) \leq \epsilon K^{-2} |B|$} \} \\
        & \leq |A|^2 \epsilon K^{-2} |B|.
    \end{align*}
    Divide by $|B|$.
\end{proof}
\begin{proof}[Proof of Length 2]
    Let $A' = N(b)$. Then by the previous two lemmas,
    \[
        \mathbb{E} \left( |N(b)|^2 - \frac{1}{2\epsilon} |BP_\epsilon(b)| \right) \geq \frac{1}{2} K^{-2} |A|^2.
    \]
    So we can pick $b$ to satisfy
    \[
        |N(b)|^2 - \frac{1}{2\epsilon} |BP_\epsilon(b)| \geq \frac{1}{2} K^{-2} |A|^2
    \]
    and let $A' = N(b)$. Then $|BP_\epsilon(b)| \leq 2\epsilon |N(b)|^2$.
\end{proof}
By discarding some $a_1$'s, we can upgrade this.
\begin{lemma}[2]
    If $X \subset A \times B$, $|X| \geq K^{-1} |A||B|$, and $\epsilon > 0$, then there exists $A_2 \subset A$ such that $|A_2| \geq \frac{1}{4} K^{-1} |A|$ and for every $a \in A_2$, there are at most $10\epsilon |A_2|$ choices for $a_2$ such that $(a, a_2)$ is $\epsilon$-bad.
\end{lemma}
We won't prove this, but the idea is to let
\[
    A_2 = A' \setminus \{ a \in A' \ : \ \text{$(a, a_2)$ is $\epsilon$-bad for many $a_2 \in A'$} \}.
\]
The second part of the conclusion can be written as $A = B(a) \cup G(a)$, where $|B(a)| \leq 10\epsilon |A_2|$ and for any $a_2 \in G(a)$, $P_2(a, a_2) \geq \epsilon K^{-2} |B|$.
\begin{proof}[Proof of Key Lemma]
    First, let
    \[
        A_1 = \{ a \in A \ : \ |N(a)| \geq \frac{1}{10} K^{-1} |A| \}.
    \]
    Let
    \[
        X(A', B') := \{ (a, b) \in (A' \times B') \cap X \} = (A' \times B') \cap X.
    \]
    Choose $A' \subset A$ be the $A_2$ of Lemma 2. Let
    \[
        B' = \{ b \in B \ : \ |N(b) \cap A'| > 20\epsilon |A'| \}
    \]
    so
    \[
        |B(a)| \leq 10\epsilon |A'|.
    \]
    For any $a \in A'$, $b \in B'$, we have
    \begin{align*}
        P_3(a, b) & \geq \epsilon K^{-2} |B| (|N(b) \cap G(a)|) \\
        & \geq \epsilon K^{-2} |B| (|N(b) \cap A'| - |B(a)|) \\
        & \gtrsim \epsilon^2 K^{-3} |A||B|
    \end{align*}
    using $|A'| \gtrsim K^{-1}|A|$. Now we just need to check $|X(A', B')| \geq K^{-O(1)}|X|$. Since $|A'| \gtrsim K^{-1} |A|$, $A' \subset A$, $N(a) \geq \frac{1}{10} K^{-1} |B|$ for $a \in A'$, and $|X(A', B)| \gtrsim K^{-2} |A||B|$, so
    \begin{align*}
        |X(A', B \setminus B')| & \leq 20\epsilon |A'||B| \\
        & \leq 20\epsilon |A||B|.
    \end{align*}
    Let $\epsilon = \frac{1}{10^6} K^{-2}$, so $|X(A', B \setminus B')| \ll |X(A', B)|$. Hence
    \[
        |X(A', B')| \sim |X(A', B)| \geq K^{-O(1)} |A||B|.
    \]
\end{proof}





\section{Bourgain's projection theorem over $\RR$}

% April 8 Frank Wang

\subsection{Finite field case}


Let us first recall Bourgain's projection theorem over $\F_p$.

\begin{theorem}[Bourgain]
    Let $0<t<2,0<s\leq 1$ and $p$ be a prime. Then there exists some $\eps=\eps(s,t)>0$ such that for all $X\subset\F_p^2$ with $|X|=p^t$ and all $D\subset\F_p$ with $|D|=p^s$, we have
    $$\max_{\theta\in D}|\pi_\theta X|\geq p^{t/2+\eps}$$
    and
    $$\max_{\theta\in D}\min_{Y\subset X,|Y|\geq p^{-\eps}|X|}|\pi_\theta Y|\geq p^{t/2+\eps}.$$
\end{theorem}

We proved the first part of this theorem in a previous lecture.

\begin{remark}\label{rem:1}
    Note that if we instead consider $\eps=0$ and $|D|\geq 2$, then the bound becomes trivial. Indeed, for any $\theta_1\neq\theta_2$, we have an injective map $X\to\pi_{\theta_1}X\times\pi_{\theta_2}X$, which implies $\max_{\theta\in D}|\pi_\theta X|\geq |X|^{1/2}$.
\end{remark}

\subsection{Real case}

Now, let us consider the analogous theorem for unit balls in $\R^2$. Let $R$ be some positive real number and let $X\subset B_R$ be a (not necessarily disjoint) union of unit balls. Let $D\subset[0,1]$ be a $\frac{1}{R}$-separated set, and set $\pi_\theta(x_1,x_2)=x_1+\theta x_2$ like in the $\F_p$ case.

Note that without any additional assumptions, the trivial bound in Remark \ref{rem:1} does not hold in the real case. So to state Bourgain's projection theorem we will need additional assumptions on $X,D$.

\begin{example}
    \begin{enumerate}
        \item Consider when $X$ is a $1\times R$ rectangle packed with unit balls. Then if we set $D=[0,R^s]$ then we get $\max|\pi_\theta X|\sim R^s$, so if $s<\frac{t}{2}$ then we get $\max|\pi_\theta X|<R^{t/2}$.
        \item Let $X=B(0,R^{1/2})$. Then $|X|\sim R$ and $|\pi_\theta X|\sim R^{1/2}$ for all $\theta$. So in this case we do not get $\max|\pi_\theta X|\geq R^{t/2+\eps}$.
    \end{enumerate}
\end{example}

\begin{theorem}[Bourgain]
    Let $0<t<2,0<s\leq 1$. Then there exist $\eps,\eta>0$, both functions of $s,t$, such that for all $X$ with $|X|=R^t$, $D$ with $|D|=R^s$, if for all $x\in B_R,r\leq R,\theta\in[0,1],\rho\in[0,1]$ we have
    $$|X\cap B(x,r)|\leq R^\eta\left(\frac{r}{R}\right)^t|X|,\quad|D\cap B(\theta,\rho)|\leq R^\eta\rho^s|D|,$$
    then there exists some $\theta\in D$ such that
    $$\inf_{Y\subset X,|Y|\geq R^{-\eta}|X|}|\pi_\theta Y|\geq R^{t/2+\eps}.$$
\end{theorem}

Note that this theorem does not hold over $\CC$. Indeed, if we take $X=B_R\cap\R^2$ and $D$ the set of real directions, then we get a similar counterexample to the $\F_{p^2}$ case.

We would like to adopt the various inequalities we used in the $\F_p$ case (Ruzsa triangle inequality, Plunnecke inequality, Balog-Szemeredi-Gowers) to the real case. To do this, we introduce a new notion of size of a set.

\begin{Definition}
    Let $X\subset\R^d$. Then for any $\delta>0$, the \textbf{$\delta$-covering number} $|X|_\delta$ is the smallest number of $\delta$-balls needed to cover $X$.
\end{Definition}

We make a few observations about delta covering numbers:

\begin{itemize}
    \item If $X$ is $2\delta$-separated, then $|X|_\delta=|X|$.
    \item If $X$ is a union of $\delta$-balls, then $|X|_\delta\sim_d\delta^{-d}|X|$.
    \item Let $\mathcal{D}_\delta=\{\delta k+[0,\delta)^d,k\in\Z^d\}$. Then $|X|_\delta\sim_d|\{Q\in\mathcal{D}_\delta,Q\cap X\neq\emptyset\}|$.
\end{itemize}

In lieu of this last observation, we define
$$X^{(\delta)}=\{k:(\delta k+[0,\delta)^d)\cap X\neq\emptyset\},$$
so we have $|X|_\delta\sim|X^{(\delta)}|$. 

Now, the Ruzsa triangle inequality, Plunnecke inequality, and Balog-Szemeredi-Gowers all hold for $\delta$-covering numbers. For example, for the Ruzsa triangle inequality the statement is now
$$|B|_\delta|A-C|_\delta\lesssim|A-B|_\delta|B-C|_\delta$$
for all $A,B,C\subset\R^d$. 

Recall the key idea for expanding sets over $\F_p$:

\begin{lemma}
    There exists a polynomial $Q$ such that given $s\in(0,1)$, there exists some $\eps(s)>0$ such that for all $A\subset\F_p$ with $|A|=p^s$, we have $|Q(A)|\geq p^{s+\eps}$.
\end{lemma}

Iterating this lemma, we could obtain all of $\F_p$ within some polynomial of $A$ (that depends on $s$). In the proof of this lemma, the key idea was to consider the set $B=\frac{A-A}{A-A}$. If $B=\F_p$, we could run an argument to imply the lemma, and if $B\neq\F_p$, then there would be some $x\in B$ such that $x+1\not\in B$, and we could use this $x$ to prove the lemma. We would like to extend these ideas to the real case.

However, there are some problems with the real case. First, $B$ can be unbounded, as the denominator $A-A$ could be very small. Also, if $A$ is a segment, then $A+A,A\cdot A$ are segments with $|A+A|,|A\cdot A|\sim |A|$, so we have no real growth when we take a polynomial of $A$. It is also not immediately clear what the equivalent of adding 1 to get from $x\in B$ to $x+1\not\in B$ is in the real case. Finally, $\R$ has subgroups of uncountable size, so we need to be able to ``escape" such a subgroup.

\begin{Definition}
    Let $X\subset B^d(0,1),\delta\in(0,1),s\in[0,d],C\geq 1$. Then $X$ is a \textbf{$(\delta,s,C)_d$-set} if $|X\cap B(x,r)|_\delta\leq Cr^s|X|_\delta$ for all $x$ and all $\delta\leq r\leq 1$.
\end{Definition}

For Bourgain's projection theorem, we will take $C=\delta^{-\eta}$.

\begin{lemma}
    There exists a polynomial $Q$ such that given $s\in(0,1)$, there exists some $\eps(s)>0$ such that for all $A\subset[0,1]$ with $|A|_\delta=\delta^{-s}$, if $A$ a $(\delta,s,\delta^{-\eps})$-set, then $|Q(A)|_\delta\geq\delta^{-s-\eps}$.
\end{lemma}

\begin{proof}[Proof idea]
    Pick some $\gamma\in(0,1)$. Then set 
    $$B=\{\frac{a_1-a_2}{a_3-a_4}:a_i\in A,|a_3-a_4|>\delta^\gamma\}\cap[0,1].$$
    This $\gamma$ will have to be chosen carefully to make the rest of the proof work, but we omit the details here.

    \begin{lemma}
        Let $B\subset[0,1]$ be closed with $0,1\in B$, and let $\rho$ be the supremum of the lengths of the segments in $[0,1]\setminus B$. Then there exists a $b\in B$ such that either $d(\frac{b}{2},B)\geq\frac{\rho}{5}$ or $d(\frac{b+1}{2},B)\geq\frac{\rho}{5}$.
    \end{lemma}
    \begin{proof}
        Let $B'=\frac{B}{2}\cup\frac{B+1}{2}\subset[0,1]$. Then it suffices to show there is an element of $B'$ that is a distance $\frac{\rho}{5}$ away from $B$. Note that $\frac{1}{2}\in B'$ since $0\in B$, so the longest segment in $[0,1]\setminus B'$ has length at most $\frac{\rho}{2}$. Now, consider an interval of length $\rho-\eps$ in $[0,1]\setminus B$, and consider the middle $\frac{\rho}{2}$ interval inside it. By the above this middle interval contains some point in $B'$. But by construction this middle interval has distance at least $\frac{\rho}{5}$ from $B$, which completes the proof.
    \end{proof}

    Now, for $\rho\in(0,1)$, we have two cases. First, if $B$ is $\rho$-dense in $[0,1]$, then we have an argument similar to the $\frac{A-A}{A-A}=\F_p$ case in the finite field version of this lemma. Otherwise, by the above lemma there is some $b\in B$ such that either $\frac{b}{2},\frac{b+1}{2}$ are far from $B$, in which case we can run an argument similar to the case in the $\F_p$ version where we have $x\in B,x+1\not\in B$.
\end{proof}

Now, given this lemma, we would like to iterate it like in the $\F_p$ case to prove Bourgain's projection theorem. However, there is an issue with this iteration, which is that we do not know whether $Q(A)$ is a $(\delta,s+\eps,\delta^{-\eps})$-set, and in fact this is likely not true in general. Instead, we will use that $Q(A)$ contains a $(\delta,s+\eps,\delta^{-\eps})$-set, which we will prove next time.




\section{Bourgain's Projection Theorem}

%Jacob Reznikov April 10

\begin{Definition}
    A $(\delta, s, C)_d$-set is a set $X \subset B^d(0,1)$ such that 
    $$|X \cap B(x,r)|_\delta \leq C r^s 
    |x|_\delta.$$
\end{Definition}
\begin{remark}
    We think of a $(\delta,s,C)$ set as a set which is 'non-concentrated' on the scale $\delta$ with degree $s$.
\end{remark}
Using this language we can rewrite the Bourgain projection theorem as.
\begin{theorem}
    Given $0 < t < 2$, $0 < s \leq 1$, there exist $\varepsilon, \eta > 0$ such that
    
    If $X \subset B^2 (0,1)$ is a $(\delta, t, \delta^{-\eta})_2$-set with $|X|_\delta = \delta^{-t}$ and $D \subset [0,1]$ is a $(\delta, s, \delta^{-\eta})_1$-set. Then there exists some $\theta \in D$ such that
    $$
        \min_{\substack{X' \subset X\\ |X'|_\delta \geq \delta^\eta |X|_\delta}} |\pi_\theta X'| \geq \delta^{-\frac{t}{2} - \varepsilon}
    $$
\end{theorem}
Last time we saw that there exists a polynomial $Q$ such that for every $0 < s < 1$ there exists $\varepsilon, \eta > 0$ such that if $A$ is a $(\delta, s, \delta^{-\eta})_1$-subset of $[0,1]$ and $|A|_\delta = \delta^{-s}$ then $|Q(A)|_\delta \geq \delta^{-s-\varepsilon}$. Now we cannot yet iterate this because we do not know that $Q(A)$ is a non-concentrated, in fact this is not true, but we can ask for $Q(A)$ to contain a $(\delta, s + \varepsilon, \delta^{-\eta})$ set (though with different $\varepsilon, \eta$).

Let us quickly confirm some properties of non-concentrated sets.
\begin{lemma}
    If $X$ is a $(\delta, s, C)_d$-set then:
    \begin{enumerate}
        \item $|X|_\rho \geq C^{-1} \rho^{-s}$ for all $\rho \in [\delta,1]$. 
        \item If $Y \subset X$ and $|Y|_\delta \geq \frac{1}{K} |X|_\delta$ then $Y$ is a $(\delta, s, CK)_1$-set.
    \end{enumerate}
\end{lemma}
Intuitively (i) tells us that if $X$ is non-concentrated on scale $\delta$ then it is large on all scales at least $\delta$, (ii) tells us that this concept is preserved under taking 'dense' subsets.
\begin{proof}
    \begin{enumerate}
        \item If $X \subset \bigcup_{i=1}^m B(x_i, \rho)$ then 
        $$
            |X|_\delta \leq \sum_{i=1}^m |X \cap B(x_i, \rho)|_\delta  
        $$
        but we know that $|X \cap B(x_i, \rho)|_\delta \leq C \rho^s |X|_\delta$
        so
        $$
            |X|_\delta \leq m C \rho^s |X|_\delta \implies m \geq C^{-1} \rho^{-s}.
        $$
        \item This is even simpler since
        $$
            |Y \cap B(x, \rho)|_\delta \leq |X \cap B(x,\rho)|_\delta \leq C \rho^s |X|_\delta \leq (C K) \rho^s |Y|_\delta
        $$
    \end{enumerate}
\end{proof}

Now due to this lemma if we want $Q(A)$ to contains a $(\delta, s + \varepsilon, \delta^{-\eta})$ set then we must have. We need $|Q(A)|_\rho \geq \rho^{-s-\varepsilon}$ $\forall \delta \in [\delta,1]$.

Now we notice two important things about the above property.
\begin{itemize}
    \item We don't get this for free because $A$ need not be a $(\rho, s, \delta^{-\eta})$-set for $\rho in [\delta,1]$.
    \item This property is necessary but not sufficient.
\end{itemize}
We can fix both of these problems with one framework, that of the 'uniform set', which is very useful even outside of this theory.

Assume that $\delta$ is some negative power of $2$, we will denote by $\mathcal{D}_\delta$ the set of $\delta$-mesh cubes tiling $\RR^d$. For any given set $X$ we denote by $\mathcal{D}_\delta (X)$ the set of those cubes that intersect $X$. We then define $|X|_\delta^* := |\mathcal{D}_\delta (X)|$ and notice that $|X|_\delta^* \sim |X|_\delta$ as we saw in the last lecture.

\begin{Definition}
    Given $\Delta \in 1/\mathbb{N}$ and $m \in \mathbb{N}$, A set $X \subset [0,1]^d$ is $(\Delta, m)$-uniform if for any $j \in \{0,\dots,m-1\}$ and for any cube $Q \in \mathcal{D}_\delta (X)$ we have
    $$
      |Q \cap X|^*_{\Delta^{j+1}} = R_j
    $$
    where $R_j$ is independent of $Q$.

    The numbers $R_j$ are called 'branching factors' of $X$.
\end{Definition}

\begin{figure}[h]
    \centering
    \input{uniformsetsdiagram.tex}
    \caption{A $(1/2, 2)$-uniform set with its 3 branching factors}
    \label{fig:uniform-set}
\end{figure}

We can see why these uniform sets are useful with the following lemma.
\begin{lemma}
    Let $X \subset [0,1]^d$ be a $(\Delta, m)$-uniform set and let $\delta = \Delta^m$.

    (a) If $|X|_\rho \geq C^{-1} \rho^{-s}$ for all $\rho \in \{1, \Delta, \Delta^2, \dots, \Delta^m\}$, then $X$ is a $(\delta, s,O_\Delta (C))_\delta$ set.

    (b) If $X$ is a $(\delta,s,C)$ set then $X$ is also a $(\rho, s, O_\Delta (C))$ for all $\rho \in [\delta, 1]$.
\end{lemma}
If we believe this, and we know that $A$ and $Q(A)$ are both uniform, then that immediately solves both our problems and lets us continue the proof. Before we explain how to make $A$ and $Q(A)$ uniform let us prove this lemma.
\begin{proof}
    (a) Let $\rho = \Delta^j$ and $Q$ some cube in $\mathcal{D}_{\Delta^j} (X)$. We clearly have the recursive relation $|X \cap Q|_{\Delta^{i+1}} = R_{i} |X \cap Q|_{\Delta^{i}}^*$ which when iterated gives us
    $$
        |X \cap Q|_{\Delta^m}^* = R_{j} R_{j+1} \cdots R_{m-1} |X \cap Q|_{\Delta^j}^*
    $$
    but we know that $|X \cap Q|_{\Delta^j}^* = 1$ precisely because $Q$ is a $\Delta^j$ cube. We thus have
    $$
    |X \cap Q|_{\Delta^m}^* = \frac{R_{0} R_{1} \cdots R_{m-1} }{R_{0} R_{1} \cdots R_{j-1}}\, ,
    $$
    Now the numerator here is precisely $|X|^*_{\Delta^m}$ and the denominator is $|X|^*_{\Delta^j}$ so by assumption we have $|X|^*_{\Delta^j} \gtrsim C^{-1} \rho^{-s}$ which gives us 
    $$
    |X \cap Q|_{\Delta^m}^* \lesssim C \rho^{s} |X|^*_{\Delta^m}.
    $$
    This shows that $X$ is a $(\delta,s,C)_\delta$ set at scales $1,\Delta,\dots,\Delta^m$. For the scales in between we can sandwich them between two powers of $\Delta$, this loses us an extra factor of at most $O_\Delta (C)$.

    (b) Again let $\rho = \Delta^{j_0}$, then for any $j$ with $0 \leq j \leq j_0$, let $Q$ be some square in $\mathcal{D}_{\Delta^j} (X)$ then we have again
    $$
        |X \cap Q|_{\Delta^{j_0}} = \frac{R_0 \cdots R_{j_0 - 1}}{R_0 \cdots R_{j-1}} = \frac{|X|^*_{\rho}}{|X|^*_{\Delta^j}}
        \lesssim C \rho^s |X|_\rho
    $$
    where in the last step we applied the previous lemma for $(\delta, s, C)$ sets.
    Again the sandwiching gives us an extra factor of $O_\Delta (C)$.
\end{proof}

Now we learn an important tool, which is the method to make any set uniform.
\begin{lemma}[Uniformization]
    Let $\delta = \Delta^m$, $X \subset [0,1]^d$, and let $\mu$ be an arbitrary sub-additive set function (eg. $\mu(B) = |B|_\delta$). Then there exists a subset $Y \subset X$ such that $Y$ is $(\Delta,m)$-uniform and
    $$
    \mu(Y) \geq \left[2 d \ln\left(\frac{1}{\Delta}\right)\right]^{-m} \mu(X) = \delta^{- \sigma} \mu(X)
    $$
    where $\sigma = \frac{\ln(2 \ln \frac{1}{\Delta})}{\ln \frac{1}{\Delta}}$.
\end{lemma}
Note that as $\Delta \to 0$ we have $\sigma \to 0$ so we can make this power arbitrarily small by picking $\Delta$ at the end.
\begin{proof}
    We will construct a uniform subset by thinking of $X$ as a tree, and pruning it from the leaves to make it uniform. We will do this step by step, first we set $X_m = X$, then at each step, from $X_j$ we construct $X_{j-1}$ by removing enough mass from level $j$ to make it uniform. 

    To do this let $X_{j-1, \ell} = \{Q \in \mathcal{D}_{\Delta^{j-1}}(X) : |Q \cap X|_{\Delta^j}^* \in [2^\ell, 2^{\ell+1}] \}$ where $\ell$ ranges between $0$ and $d \ln \frac{1}{\Delta}$. This splits $X_j$ into $d \ln \frac{1}{\Delta}$ different pieces across which we have similar magnitude branching on level $j$. 
    Then because we have a sub-additive function
    $$
    \mu(X_j) \leq \sum_{\ell=0}^{d \ln \frac{1}{\Delta}} \mu(X_{j-1, \ell}).
    $$
    so we can pick the 'largest' piece and lost at most a factor of $d \ln{1}{\Delta}$. Assume that $X_{j-1,\ell}$ is that piece, we set $X_{j-1}$ to be the $X_{j-1,\ell}$ where at level $j$ we removed enough of the set to get the branching factor to be exactly $2^\ell$. Since the branching factors are all within a factor of 2 away from $2^\ell$ this loses us at most half of the 'measure' of $X_{j-1,\ell}$ so that 
    $$
        2 d \ln \frac{1}{d} \mu(X_{j-1}) \geq d \ln \frac{1}{d} \mu(X_{j-1, \ell}) \geq \mu(X_{j})
    $$
    iterating this process $m$ times gives us exactly the lemma.
\end{proof}
% Pattern Info
\begin{figure}[h]
    \centering
    \input{uniformizationdiagram}
    \caption{Applying uniformization with $\Delta = 1/2$ and $m = 2$ to a set.}
    \label{fig:uniformization}
\end{figure}

Now back to our original goal, we recall that $A$ is a $(\delta,s,C)$-set and $|Q(A)|_\delta \geq \delta^{-s-\varepsilon}$.
We make progress using the following steps.
\begin{enumerate}
    \item We may assume that $A$ is uniform by uniformization, we pick $\Delta$ very small and use $\mu(B) = |B|_\delta$. We then get a subset $A'$ which is uniform, and from previous lemmas we get that $A'$ is a $(\delta, s, C \delta^{-\sigma})$ but we can make $\sigma$ arbitrarily small.
    \item Next we recall from the previous lecture and the finite field settings that there is some $a \in A$ such that $|A + a A|_\delta \geq \delta^{-s - \varepsilon_1}$. Since $A$ is uniform we in fact also have $|A + a A|_\rho \geq \rho^{-s-\varepsilon_1}$ for all $\rho \in [\delta,1]$.
    \item This is in fact enough to show that this holds for almost all $a in A$. To see this we define $A_{bad} = \{ a \in A : (2) \,\text{is false}\}$, now if $A_{bad}$ is too large then we can apply steps 1 and 2 to $A_{bad}$ and arrive at a contradiction.
    \item If $X \subset A \times A$ and $|X|_\delta \geq \delta^{\eta^2} |A \times A|_{\delta}$ then we have for some $a \in A$ we have $|\pi_a (X)|_\rho \geq \rho^{-s - \varepsilon/2}$. This is the key step and is non-trivial, to get some intuition before we jump into the proof assume for the moment that $1 \in A$. Now we have two cases, either $\pi_1 X$ works in which we are happy, or it doesn't. If it does not then there must exist some subset $X \subset A \times A$ which is dense in the sense that $|X|_\delta \geq \delta^{\eta^2} |A \times A|_\delta$ and its projection is small in the sense that $|\pi_1 (X)|_\delta \leq \delta^{-s - \varepsilon/2}$. But now this is exactly the setting in which we can use BSG.
\end{enumerate}

\section{Homogenous Dynamics}
 

We here apply projection theory to studying the behavior of random walks on a finite group. Let $G$ be a finite group and $\mu$ be a probability measure on $G$. A random walk starting at $g_0$ is defined as a sequence of random variables $(g_n)_{n\geq 0}$ such that $g_{n+1} = g_n g$ with probability $\mu(g)$. Essentially, at every step, a random element is chosen from $G$ using $\mu$, and then the current state is right multiplied by the chosen element. The guiding question is how evenly distributed the random walk is after $K$ steps. We now develop several formal definitions to phrase this question more precisely. First, define a convolution of functions $f_1, f_2: G \to \mathbb{C}$ in the standard way: 

\begin{equation}
f_1 * f_2(g) = \sum_{g_1, g_2 \in G: g_1g_2 = g} f_1(g_1)f_2(g_2)
\end{equation} 

We now view the random walk as a Markov chain with transitions given by a linear operator $T_\mu$ defined as 


\begin{equation}
T_\mu f = f * \mu 
\end{equation}

When $f$ is viewed as a probabilty distribution of a state $g_n$, $T_\mu f$ gives the probability distribution of $g_{n+1}$. When the random walk starts at a state $g_0$, that is equivalent to starting with initial probability distribution $\delta_{g_0}$. Then after one step the probability distribution is $T_\mu \delta_{g_0}$, so the probability of state $g_0 h$ is 

\begin{align}
    T_\mu \delta_{g_0} (g_0 h)  = \sum_{g_1 g_2 = g_0 h} \delta_{g_0}(g_1) \mu(g_2)\\
    = \delta_{g_0}(g_0) \mu(h) = \mu(h)
\end{align} 

After $K$ steps, the probability distribution of the random walk position $g_K$ is $T_\mu^K \delta_{g_0}$. This leads to the first question, which is to estimate the $L^2$ norm 

\begin{equation}
|| T_\mu^K \delta_{g_0} - \frac{1}{|G|} ||_{L^2}
\end{equation}

or alternatively, other $L^p$ norms. The $1/|G|$ term is the average value of the distribution over all of $G$, so the $L^p$ norms are measures of the regularity of the distribution. Since $T_\mu$ is a linear operator, we can approach this by examining the singular values of $T_\mu$. The squares of the singular values are the eigenvalues of the matrix $T_\mu^T T_\mu$ where $T_\mu^T$ is the transpose. When $T$ is symmetric, the singular values are the same as the eigenvalues, but in general they are different. 

We first show the following lemma:

\begin{lemma}
\begin{equation}
||T_\mu f||_{L^2} \leq ||f||_{L^2}
\end{equation}

\end{lemma}
\begin{proof}
    \begin{align}
        T_\mu f(g) = & f * \mu(g) \\
        =& \sum_{g_1g_2= g} f(g_1)\mu(g_2)       \\
         =& \sum_{g_2} f(g g_2^{-1})\mu(g_2)
        % ||T_\mu f||_{L^2}^2 = & \sum_g \sum_{g_1g_2= g} f(g_1)\mu(g_2)
    \end{align}

    We then define the right multiplication operator $R_g$ so that $R_g f(h) = f(hg^{-1})$
Then applying the triangle inequality and the translation invariance of the $L^2$ norm,

\begin{align}
    || T_\mu f||_{L^2} = &|| \sum_{g_2} \mu(g_2) R_{g_2} f ||_{L^2}\\
    \leq & \sum_{g_2} \mu(g_2) ||   R_{g_2} f ||_{L^2}\\
    \leq & ||f||_{L^2}
\end{align}
\end{proof}

Then since $T_\mu 1 = 1$, $1$ is the largest singular value of $T_\mu$. We now define the subspace 

\[L^2(G)_0 = \{ f\in L^2(G): \langle f, 1\rangle = 0\}\] 

where $\langle ,\rangle$ is the standard inner product with the counting measure on $G$. We can then analyze the restriction of $T_\mu$ 

\[ T_\mu: L^2(G)_0 \to L^2(G)_0\] 

This restriction quotients out the trivial singular value 1 and allows us to examine the next singular value, which governs the decay rate of the $L^p$ norms. Denote $\sigma_1(T_\mu)$ as the largest singular value of $T_\mu$ restricted to $L^2(G)_0$. Then we can quantitatively express the decay of the $L^2$ norm in terms of the following proposition: 

\begin{prop}
\[ ||T_\mu^K \delta_{g_0} - \frac{1}{|G|} ||_{L^2} \leq |\sigma_1(T_\mu)|^K \]
\label{l2_sigma_1}
\end{prop}
\begin{proof}
Note that $$\langle \delta_{g_0} - \frac{1}{|G|}, 1\rangle = 0$$ 

so 

$$\delta_{g_0} - \frac{1}{|G|} \in L^2(G)_0$$ 

Then $T_\mu$ maps $\delta_{g_0} - \frac{1}{|G|} $ to $L^2(G)_0$, so the claim follows from the fact that the largest singular value of a linear operator is also its operator norm. 
\end{proof} 

This proposition leads to the second guiding question, which is to estimate $\sigma_1(T_\mu)$. The proposition shows that an estimate on $\sigma_1(T_\mu)$ is sufficient to give an estiamte on the decay of the $L^2$ norm. We additionally remark that since we are using the counting measure, the $L^\infty$ norm is bounded by the $L^2$ norm, so this gives an estimate of the $L^\infty$ norm as well. 

We now examine the group $G = SL_2(\mathbb{F}_p)$ where $p$ is prime. The case where $\mu$ is the uniform measure on a subset $A$ of $G$ was studied by Selberg. For convenience, define $T_A \equiv T_{\mu_A}$ to be the operator corresponding to the measure on $A$. In particular, Selberg studied the particular set 

\[A = \left\{ \begin{pmatrix}
    1 & \pm 1\\
    0 & 1
\end{pmatrix}, \begin{pmatrix}
    1 & 0\\
    \pm 1 & 1
\end{pmatrix}\right\}\]

which has four elements. Selberg proved the following theorem about this case: 

\begin{theorem}
There exists a universal constant $c > 0$ so that for every prime $p$, then 

\begin{equation}
    \sigma_1(T_A) \leq 1 - c 
\end{equation}
\end{theorem}

This is not the original theorem that Selberg proved, but it is a corollary of a different theorem about the spectrum of the Laplacian on particular hyperbolic surfaces. However, a discussion of the precise result is beyond the scope of these notes. 

We now examine an approach to random walks using expansion graphs. For a finite group $G$ and a subset $A \subset G$ define a graph $C(G,A) = (V, E)$ with set of vertices $V$ indexed by $G$ and an edge $(g_1, g_2) \in E$ if $g_1^{-1} g_2 \in A$, or equivalently there exists $a\in A$ such that $g_2 = g_1 a$. Therefore the nodes that are connected by edges are the nodes that can be connected by a single step of the random walk. Now for two subsets $S,T\subset V$, define 

\[ E(S,T) \equiv \{ (g_1, g_2) \in S \times T: (g_1, g_2) \in E\}\] 

or equivalently, $E(S,T) = E \cap S \times T$. We now consider the following proposition: 

\begin{prop}
If $S$ is a subset of $G$, then 

\begin{equation}
|E(S, S^c)| \geq (1 - \sigma_1(T_A))\frac{|A| |S| |S^c|}{|G|} 
\end{equation}
\end{prop}
\begin{proof}
We first prove that 

\begin{equation}
    E(S, S^c)| = |A| \langle T_A 1_S, 1_{S^c}\rangle
    \label{E_S_Sc_eq}
\end{equation}


which follows from the following computation: 
\begin{align*}
    T_A 1_S(g) = \frac{1}{|A|} \sum_{a \in A} 1_S(ga^{-1})\\
    \langle T_A 1_S(g), 1_{S^c} \rangle = \sum_{g\in G} \frac{1}{|A|} \sum_{a \in A} 1_S(ga^{-1}) 1_{S^c}(g)
\end{align*} 

Note that $1_S(ga^{-1}) 1_{S^c}(g) = 1$ if $ga^{-1} \in S$ and $g \in S^c$, which is equivalent to the statement $(ga^{-1}, g) \in E(S, S^c)$, which shows equation \ref{E_S_Sc_eq}. We then decompose $1_S$ into a constant and mean zero part as  

\[ 1_S = \frac{|S|}{|G|}  +  1(_S - \frac{|S|}{|G|}) \]

Applying this decomposition to $1_{S^c}$ as well gives 

\begin{align*}
    \langle T_A 1_S, 1_{S^c} \rangle 
    = & \langle T_A \left(\frac{|S|}{|G|}  +  1_S - \frac{|S|}{|G|}\right),  1 - \frac{|S|}{|G|} +  1_{S^c} - (1 - \frac{|S|}{|G|} )\rangle \\
     = &  \langle \frac{|S|}{|G|}  +  T_A \left(1_S - \frac{|S|}{|G|}\right),  1 - \frac{|S|}{|G|} +  1_{S^c} - (1 - \frac{|S|}{|G|} )\rangle 
\end{align*} 

This then decomposes into the inner products of the constant and the non-constant terms. The inner product of the constant terms is 

$$ |G| \frac{|S|}{|G|} (1 - \frac{|S|}{|G|} )  = \frac{|S||S^c|}{|G|}  $$ 

The inner product of the non-constant terms is 

\begin{align*}
     &  \langle  T_A \left(1_S - \frac{|S|}{|G|}\right),  1_{S^c} - (1 - \frac{|S|}{|G|} )\rangle 
\end{align*}

Applying Proposition \ref{l2_sigma_1} and Cauchy Schwartz gives the upper bound 

\[ \sigma_1(T_A) ||1_S - \frac{|S|}{|G|}||_{L^2}||1_{S^c} - (1 - \frac{|S|}{|G|} )||_{L^2} = \sigma_1(T_A) \frac{|S||S^c|}{|G|} \] 

Then combining the terms from the constant and nonconstant parts gives 

\[ \langle T_A 1_S, 1_{S^c} \rangle  \geq (1 - \sigma(T_A))\frac{|S||S^c|}{|G|}\]

Multiplying by $A$ and applying equation \ref{E_S_Sc_eq} then gives the desired result: 



\[E(S, S^c) \geq (1 - \sigma(T_A))\frac{|A||S||S^c|}{|G|}\]


\end{proof}

Without loss of generality $S$ can be chosen so that $|S| \leq |G|/2$. Then if $\sigma(T_A) \leq 1$

\[E(S, S^c) \gtrsim |S| \] 

where the implicit constants depend on $A$. This property of a subset of vertices and its complement sharing a large number of edges is known as an expander graph. Note that when $A$ is a subset of a proper subgroup $H$ of $G$, then the set of elements generated by $A$ is at most $H$. The distribution will therefore never become uniform after repeatedly applying $T_A$, which implies that $\sigma_1(T_A) = 1$. 

We now return to the group $SL_2(\mathbb{F}_p)$ and examine its representation theory. Consider the following proposition:

\begin{prop}
If $\rho: SL_2(\mathbb{F}_p) \to U(d)$ is a nontrivial representation of $SL_2(\mathbb{F}_p)$ mapping to the unitary group with $d$ dimensions, then $d \geq \frac{p+1}{2} $ 
\end{prop}

\begin{proof}
This proof relies on the existence of the elements 

\[ u = \begin{pmatrix} 1 & 1\\
0 & 1 \end{pmatrix}\] 

and 

\[ v = \begin{pmatrix} 1 & 0\\
1 & 1 \end{pmatrix}\] 

These elements generate $SL_2(\mathbb{F}_p)$, and are tranposes of each other, so without loss of generality we assume that $\rho(u) \neq e$. $u$ and $v$ have the property that they are conjugate to powers of themselves. In particular: 

\begin{align*}
    \begin{pmatrix} a & 0\\
    0  & a^{-1}  \end{pmatrix} 
    \begin{pmatrix}
        1 & 1\\
        0 & 1
    \end{pmatrix}
    \begin{pmatrix} a^{-1} & 0\\
    0 & a \end{pmatrix} = \begin{pmatrix} 1 & a^2 \\
    0 & 1 \end{pmatrix} = u^{a^2}
\end{align*}

The conjugates of $v$ similarly are powers of $v$. Then because representations preserve conjugacy classes, $\rho(u)$ must be conjugate to $\rho(u)^{a^2}$. Since conjugate matrices have the same eigenvalues, then $\rho(u)$ and $\rho(u)^{a^2}$ must have the same set of eigenvalues. $\rho(u)$ has order $p$, so its eigenvalues must be roots of unity of order $p$, or equivalently of the form $e^{2\pi i n/p}$ for integer $n$. Then the eigenvalues of $\rho(u)^{a^2}$, and equivalently of $\rho(u)$, are of the form $e^{2\pi i a^2 n/p}$. Since this is true for arbitrary $a$, a single nontrivial eigenvalue $e^{2\pi i n/p}$ generates all eigenvalues corresponding to $a^2 n \mod p$. $\rho(u)$ is by hypothesis not the identity, so must have at least one eigenvalue not equal to 1. Since there are $\frac{p-1}{2} $ distinct nonzero quadratic residues (and 1 is an eigenvalue of $\rho(u)$), then $\rho(u)$ has at least $\frac{p+1}{2} $ distinct eigenvalues, and so has dimension at least $\frac{p+1}{2} $. This completes the proof. 
\end{proof}

We now apply this proposition to prove a further proposition. 

\begin{prop}
Let $\mu$ be a measure on $SL_2(\mathbb{F}_p)$. Then 

\[\sigma_1(T_\mu)^2 \frac{p+1}{2} \leq |SL_2(\mathbb{F}_p)| \ || u||_{L^2}^2\] 

In particular, since $|SL_2(\mathbb{F}_p)| \sim p^3$, this implies 

\[\sigma_1(T_\mu) \lesssim p || u||_{L^2}\] 

\label{sig_one_Tmu_bound}
\end{prop}

\begin{proof}
Note that $\sigma_i(T_\mu)^2$ is the $i$th eigenvalue of $T_\mu T_\mu^*$. Since $T_\mu T_\mu^*$ is a right action, its eigenspaces have a left $G$ action $L_g f(h) = f(g^{-1}h)$, which is nontrivial except for the constant functions. Each action on an eigenspace induces a representation of $SL_2(\mathbb{F}_p)$, which is unitary because 

\begin{align*}
    \langle L_g f, h \rangle =& \sum_{\ell \in G} f(g^{-1} \ell) h(\ell)\\
    =& \sum_{\ell \in G} f( \ell) h(g\ell)\\
    = & \langle f, L_{g^{-1} }h \rangle 
\end{align*}

Therefore the representation must have dimension at least $ \frac{p+1}{2} $, so the singular values must value multiplicity at least $\frac{p+1}{2} $. Then because the Frobenius is invariant under unitary operations, and since $T_\mu T_\mu^*$ is symmetric it is diagonalizable by a unitary transformation:

\begin{align*}
    \frac{p+1}{2} \sigma_1(T_\mu)^2 \leq &  \sum_{i\text{ with multiplicity}}  \sigma_i(T_\mu)^2\\
    = & \sum_{g_1, g_2} |(T_\mu)_{g_1, g_2}|^2 \\
    = & \sum \mu(g_1 g_2^{-1})^2 \\
    =& |SL_2(\mathbb{F}_p)| \sum_g \mu(g)^2 \\
    =& |SL_2(\mathbb{F}_p)| \ ||\mu||_{L^2}^2
\end{align*}


\end{proof}

Then returning to the case that $\mu = \mu_A$ for a subset $A$

\[ ||\mu_A||_{L^2}^2 = \frac{1}{|A|^2} |A| = \frac{1}{|A|} \] 

This together with proposition \ref{sig_one_Tmu_bound} implies the following corollary: 

\begin{corollary}
\[\sigma_1(T_A)^2 \lesssim \frac{p^2}{|A|} \]
\end{corollary}

This bound is only nontrivial when $|A| \gtrsim p^2$. However, it is strong enough to prove one additional corollary. 

\begin{corollary}
If $H$ is a proper subgroup of $SL_2(\mathbb{F}_p)$ then $|H| \lesssim p^2$. 
\end{corollary}
\begin{proof}
If $H$ is a proper subgroup, then $\sigma(T_H) = 1$, which implies that $p^2/|H| \gtrsim 1$. Multiplying both sides by $|H|$ gives the desired result. 
\end{proof}

Note that the order of $SL_2(\mathbb{F})_p$ is $p(p-1)(p+1)$, so this corollary is not a consequence of Lagrange's theorem. 


\section{Homogeneous Dynamics Continued}
Today we continue our discussion on homogeneous dynamics. Let $G=SL_2(\R)$, and $\Gamma=SL_2(\Z)$. Let $m$ be a right invariance metric on $G/\Gamma$. The left action of $G$ on $G/\Gamma$ might distort the metric but it preserves the volume. 
Define $U=\{\begin{bmatrix}
  1 &t \\
    & 1
\end{bmatrix}_{t\in \R}\}$ and $u_t=\begin{bmatrix}
  1 &t \\
    & 1
\end{bmatrix}$
Our goal is to study $U \cdot x$ in $G/\Gamma$ for $x \in G/\Gamma$. 
\begin{theorem}(Hedlund 30's)\\
$U\cdot x$ is either periodic or dense. 
\end{theorem}
Define $a_r=\begin{bmatrix}
  e^r & \\
    & e^{-r}
\end{bmatrix}$. After some calculation, we see that 
$a_ru_ta_r^{-1}=u{e^{2r}t}$.
Denote $U_{[0,T]}x=\{u_tx\}_{t \in [0,T]}$. 
Note that 
$$U_{[0,T]}x=a_RU_{[0,1]} a_R^{-1}x$$
where $e^{2R}=T$. Also note that if $R=Jr$, $a_R=a_r^j$. \\
\textbf{Goal}: Understand how $a_r$ acts on unipotent orbits. \\
We shall consider a tube in the fundamental domain for $G/\Gamma$. The key point is that the compressed direction is twisting relative to the unipotent orbits. \\
\includegraphics[scale=0.1,angle=270]{may1p1.jpeg}\\
Now, let's see what's the precise statement about what is twisting. We call $v_{comp}(t)\in T_{u_tg_0}$ the direction that was compressed when we apply $L_{a_r^{-1}}$, i.e. the smallest singular value vector for $dL_{a_r^{-1}}$. We also define an orbit vector $v_{orb(t)}$ such that $u_t(g_0+\epsilon v_0)=u_tg_0+\epsilon v_{orb}(t)$. The moral the the story is that at each point there is an orbit vector and a compression vector and the angle between them is changing. \\
Now let's try to compute $v_{comp}(t)$.
$$R_{(a_rg_0)^{-1}}\circ L_{a_r^{-1}}\circ R_{g_0h}=a_rhg_0(a_rg_0)^{-1}=a_rha_r^{-1}=C_{a_r}h.$$
Here $C_{a_r}:G \rightarrow G$ mean conjugation by $a_r$. Note that $C_{a_r}: e \mapsto e$. 
$d C_{a_r}: T_e G \circlearrowleft $
$$d C_{a_r}\begin{bmatrix}
  a &b \\
c & d
\end{bmatrix}= \begin{bmatrix}
  e^r & \\
 & e^{-r}
\end{bmatrix}\begin{bmatrix}
  a &b \\
c & d
\end{bmatrix}\begin{bmatrix}
  e^{-r} & \\
 & e^r
\end{bmatrix}.$$
\textbf{Recall:} Orthonormal basis for $T_eG:$ 
$$n=\begin{bmatrix}
  0 &1 \\
   0 & 0
\end{bmatrix},\Tilde{n}=\begin{bmatrix}
  0 &0 \\
   1 & 0
\end{bmatrix},\Tilde{d}=\begin{bmatrix}
  \frac{1}{\sqrt{2}} &0 \\
   0 & -\frac{1}{\sqrt{2}}
\end{bmatrix} $$
After some calculation, we see that 
$$dC_{a_r}(n)=e^{2r}n, dC_{a_r}(\Tilde{n})=e^{-2r}\Tilde{n},dC_{a_r}(d)=d$$
Thus, $R_{g_0}(\Tilde{n})$ is the singular vector of $dL_{a_r^{-1}}$ with singular value $e^{-2r}$ so
$v_{comp}(t)=R_{u_tg_0}(\Tilde{n})=\Tilde{n}u_tg_0$.It depends on a different way from $v_{orb}(t)$. 
\textbf{Tracking the spread of an orbit}
$$U_{[0,T]}x=a_RU_{[0,1]}a_R^{-1}x.$$
Put $\Tilde{x}=a_R^{-1}x$ and assume that $\Tilde{x}$ is not deep in the cusp. This implies that $U\cdot x$ is not close to being periodic. Let $R=Jr$ and put $U_j=a_r^jU_{[0,1]}\Tilde{x}$. 
Define $|X|_\delta$ to be the nubmer of $\delta$ balls needed to cover $X$. 
\textbf{Goal:} Estimate $|U_j|_\delta$ in terms of $j, \delta, r$. Define $X_j$ to be the top layer of $U_j$, then $|U_j|_\delta=\delta^{-1}|X_j|_\delta$. We say that we are in the very spread situation if $|U_j|_\delta\sim \delta^{-3}$ and $|X_j|_\delta\sim \delta^{-2}$. \\
\textbf{Using the Key Picture}\\
\begin{lemma}
    If $e^{2r}=\delta$. Then, 
    $$|X_{j+1}|_\delta\sim \sum_{\substack{t \in \delta \Z\\ 0 \leq t \leq 1}} \sim \delta^{-1}Avg_{0 \leq t \leq 1} |\pi_t X_j|_\delta.$$
\end{lemma}
For proof sketch see picture\\ 
\includegraphics[scale=0.08,angle=180]{may1p2.jpeg}\\

Let $f_t$ be $L_{a_r^{-1}}$ restricted to time $t$. $f_t$ looks like the projection map $\pi_t$. $f_t$ is not linear but is smooth. 
\begin{prop}
    If $X \in B_1^2$, then 
    $$Avg_{\theta \in S^1}|\pi_\theta X|_\delta\gtrsim |X|_\delta^{1/2}.$$
\end{prop}
For examples, see pictures\\
% \includegraphics[scale=0.1]{mayp3.jpeg}\\
The proof for $\pi_t$ holds for $f_t$ as well. 
\begin{corollary}
    $$|X_{j+1}|_\delta \gtrsim \delta^{-1}|X_j|^{1/2}.$$
\end{corollary}
\begin{proof}
    $|X_{j+1}|_\delta \gtrsim \delta^{-1}Avg_t|f_t X_j|_\delta \gtrsim \delta^{-1}|X_j|_\delta^{1/2}$. 
\end{proof}
Suppose $|X_0|_\delta=1$, then $|X_1|_\delta\gtrsim \delta^{-1}$, $|X_2|_\delta\gtrsim \delta^{-3/2}$, $|X_3|_\delta\gtrsim \delta^{-7/4}\cdots$
\begin{remark}
    Even this shows that $X_j$ is well spread, this doesn't show that the orbit is dense. 
\end{remark}
\textbf{What about $SL_n(\R)$?}
For example, $G=SL_3(\R)$, $\Gamma=SL_3(\Z)$. Put, $$U=\begin{bmatrix}
    1& t& t^2\\
    0& 1& t\\
    0& 0& 1\\
\end{bmatrix}\text{ and } a_r=\begin{bmatrix}
    e^r& 0& 0\\
    0& 1& 0\\
    0& 0& e^{-r}\\
\end{bmatrix}$$
Then, $a_ru_ta_r^{-1}=U_{e^rt}$. Now, we shall study the action of $L_{a_r^{-1}}$. We can study the singular value of $L_{a_r^{-1}}$ by studying the singular value of $C_{a_r}$. They are 
$$e^{-2r}, e^{-r},e^{-r},1,1,e^r,e^r,e^{2r}.$$
\textbf{Difference from previous time}
\begin{enumerate}
    \item Now the approximate projection goes from $B^7 $ some vector space $V$ of dimension $3$. This approximately a linear map with singular values  
    $$e^{-2r}, e^{-r},e^{-r},1,1,e^r,e^r,e^{2r}.$$
    Our singular values are a little bit more complicated, but we shall not focus on this. 
    \item The more serious problem is that now we only have one parameter family of linear maps, so they are not nearly all the projection. 
\end{enumerate}
Thus, this lead people to consider the restricted projection problem. 
\begin{question}(Fassler-Orponen 2013)\\
Let $\gamma$ be curve in $S^2$. If $X \subseteq B^3$, $X$ a $(\delta, s, C)$ set  estimate $Avg_{\theta\in \gamma}|\pi_\theta(X)|_\delta$. 
Here, for $\theta \in S^2, \pi_\theta: \R^3 \rightarrow \theta^\perp$.  
\end{question}
The answer depends on whether $\gamma$ lies in an equator or not. 
\begin{example}
    Let $\gamma$ be the equator and $X$ a $\delta \times 1 \times 1$ slab. Then, $Avg_{\theta \in \gamma} |\pi_\theta(X)|_\theta \sim \delta^{-1}$. 
\end{example}
\begin{theorem}(G,G,G,H,M,W)\\
    If $X \subseteq B^3$ is a $(\delta, 2, C)$ set and $\gamma$ is non-degenerate. Then, $Avg_{\theta \in \gamma}|\pi_\theta(X)|_\delta \geq C_\epsilon \delta^{-2+\epsilon}$ for any $\epsilon >0$. 
\end{theorem}
The proof is based on decoupling. This theorem was used to prove the following theorem. 
\begin{theorem}(LMMY)
If $G=SL(3,\R)$, $\Gamma=SL(3,\Z)$. $U$ as above and $U \cdot x $ is not close to a proper homogeneous subspace. Then, $U_{[0,T]} x$ is $\delta$ dense in $(G\ \Gamma)$ where $\delta=T^{-c}$ where $c>0$ is a universal constant. 
\end{theorem}
One key step in the proof of this theorem is that 
$$|U_{[0,T]} x|_\delta \geq c_\epsilon \delta^{-(\dim G+\epsilon)}.$$


\newpage
\section{Random walk on groups}

\noindent\textbf{Setup}:
\begin{itemize}
    \item Let $G$ be a finite group.
    \item $\mu:G\to \mathbb R$ is a probability measure on $G$, i.e., $\mu(g)\ge 0,\sum_{g\in G}\mu(g)=1$.
    \item Starting with $g_0\in G$, let $h_t\in G,t=1,2,\dots$ be sampled according to $\mu$, and define the random walk on $G$ by $g_t=g_{t-1}\cdot h_t,t=1,2,\dots$. 
\end{itemize}
\noindent\textbf{Question}: how evenly distributed is $g_K$ on $G$ for large $K$?

To state our question more precisely, we introduce some definitions. For two functions $f_1,f_2:G\to \mathbb C$, define
\[
f_1\ast f_2(g)=\sum_{g_1,\cdot g_2=g}f_1(g_1)f_2(g_2)\,,\quad\forall g\in G\,.
\]
Define the operator $T_\mu:\ell^2(G)\to \ell^2(G)$ by $T_\mu f=f\ast \mu$.  It is straightforward to check that for any $K$, $T_\mu^K\delta_{g_0}$ is the distribution of $g_K$ defined as above. Our main question is to estimate
\[
\|T_\mu^K\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\|_{\ell^2(G)}
\]
for large $K\in \mathbb N$.

We start with some easy observations.
\begin{lemma}
    $T_\mu \mathbf 1=\mathbf 1$, and $\|T_\mu f\|_{\ell^2(G)}\le \|f\|_{\ell^2(G)},\forall f\in \ell^2(G)$. 
\end{lemma}
\begin{proof}
    The first claim can be checked straightforwardly. For the second claim, we define the right shift operator $R_g:\ell^2(G)\to \ell^2(G)$ by
    \[
    R_g f(h)=f(f\cdot g^{-1})\,,\quad\forall f\in \ell^2(G),g,h\in G\,.
    \]
    It is easy to check that $R_g:\ell^2(G)\to \ell^2(G)$ is an isometry, and it holds that
    \[
    T_\mu f=f\ast \mu=\sum_{g\in G}\mu(g)R_gf\,,\quad \forall f\in \ell^2(G)\,.
    \]
    Therefore, it follows from the triangle inequality that
    \[
    \|T_\mu f\|_{\ell^2(G)}\le \sum_{g\in G}\mu(g)\|R_gf\|_{\ell^2(G)}=\|f\|_{\ell^2(G)}\,.\qedhere
    \]
\end{proof}
Denote $\ell^2(G)_0$ as the orthogonal complement of the constant functions in $\ell^2(G)$. One can verify that $T_\mu$ maps $\ell^2(G)_0$ to itself. Denote by $\sigma_1(T_\mu)$ the largest singular value of the operator $T_\mu:\ell^2(G)_0\to \ell^2(G)_0$.
\begin{lemma}\label{lem-mixing}
    For any $K\in \mathbb N$, it holds that
    \[
    \|T_\mu^K\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\|_{\ell^2(G)}\le \sigma_1(T_\mu)^K\,.
    \]
\end{lemma}
\begin{proof}
    Write $\delta_{g_0}=\tfrac{1}{|G|}\mathbf{1}+(\delta_{g_0})_h$, where $(\delta_{g_0})_h=\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\in \ell^2(G)_0$. We have for any $K\in \mathbb N$, $T_\mu^K\delta_{g_0}=\tfrac{1}{|G|}\mathbf{1}+T_\mu^K(\delta_{g_0})_h$, and thus
    \[
     \|T_\mu^K\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\|_{\ell^2(G)}\le \|T_\mu^K(\delta_{g_0})_h\|_{\ell^2(G)}\le \sigma_1(T_\mu)^K\|(\delta_{g_0})_h\|_{\ell^2(G)}\le \sigma_1(T_\mu)^K\,.\qedhere
    \]
\end{proof}

For a subset $A$ of $G$, we denote $\mu_A=\tfrac{1}{|A|}\mathbf{1}_A$ and abbreviate $T_{\mu_A}$ as $T_A$. Of particular interest of us is the following concrete example: let $G=\operatorname{SL}_2(\F_p)$ where $p$ is a large prime and 
\[
A_{\operatorname{sel}}=\left\{\left(\begin{matrix}
    1&\pm 1\\
    0&1
\end{matrix}\right),\left(\begin{matrix}
    1&0\\
    \pm 1&1
\end{matrix}\right)\right\}\subset \operatorname{SL}_2(\F_p)\,.
\]
Our main goal is to prove the following theorem due to Selberg.
\begin{theorem}\label{thm-selb}
    There exists a universal constant $c>0$ such that $\sigma_1(T_{A_{\operatorname{sel}}})\le 1-c$. 
\end{theorem}
In general, for a pair $(G,A)$ where $A$ is a subset of the group $G$, we are interested in $\sigma_1(T_A)$. This is not only because it is related to the mixing of random walks on $G$ with steps in $A$ (see Lemma~\ref{lem-mixing}), but also because the spectral gap $1-\sigma_1(T_A)$ reflects a certain expansion property of the corresponding Cayley graph.

More precisely, for $A \subset G$ that is symmetric and generates $G$, we define $\operatorname{C}(G,A)$ as the graph with vertices corresponding to the elements of $G$, and with edges $(g_1, g_2) \in E$ if there exists $a \in A$ such that $g_2 = g_1 \cdot a$. Note that $A$ generates $G$, which implies that $\operatorname{C}(G,A)$ is connected. Moreover, for a subset $S \subset G$, we denote $E(S, S^c)$ as the set of edges $(g_1, g_2) \in E$ such that $g_1 \in S$ and $g_2 \in S^c$.

\begin{lemma}\label{lem-expansion}
    For any $S\subset G$, it holds that
    \[
    |E(S,S^c)|\ge (1-\sigma_1(T_A))\frac{|A||S||S^c|}{|G|}\,.
    \]
\end{lemma}
\begin{proof}
    It is straightforward to check that
    \begin{align*}
    |E(S,S^c)|=&\ |A|\langle T_A\mathbf{1}_S,\mathbf{1}_{S^c}\rangle\\
    =&\ |A|\langle\tfrac{|S|}{|G|}\mathbf{1},\tfrac{|S^c|}{|G|}\mathbf{1}\rangle+|A|\langle T_A(\mathbf{1}_S)_h,(\mathbf{1}_{S^c})_h\rangle\\
    \ge&\ \frac{|A||S||S^c|}{|G|}-\sigma_1(T_A)\|(\mathbf{1}_S)_h\|_{\ell^2(G)}\|(\mathbf{1}_{S^c})_h\|_{\ell^2(G)}\\
    \ge&\ (1-\sigma_1(T_A))\frac{|A||S||S^c|}{|G|}\,.\qedhere
    \end{align*}
\end{proof}
Combining Lemma~\ref{lem-mixing} with Theorem~\ref{thm-selb}, we obtain that for any $S\subset G=\operatorname{SL}_2(\F_p)$ with $|S|\le \tfrac{|G|}{2}$, in the graph $\operatorname{C}(\operatorname{SL}_2(\F_p),A_{\operatorname{sel}})$,
\[
E(S,S^c)\ge \frac{c|A||S|}{2}\,.
\]
For a large graph $H=(V,E)$, we say $H$ is an expander graph, if there exists a universal constant $c>0$ such that for any $S\subset V$ with $S\le \tfrac{|V|}{2}$, $E(S,S^c)\ge \frac{c|E||S|}{|V|}$. The above result indicates that $\operatorname{C}(\operatorname{SL}_2(\F_p),A_{\operatorname{sel}})$ is a sparse expander graph (here sparse means that the graph has average degree $O(1)$). Indeed, this is the very first proof of the existence of sparse expander graphs.

In what follows we fix a large prime number $p$ and let $G=\operatorname{SL}_2(\F_p)$. We now turn to prove Theorem~\ref{thm-selb}.

\textbf{$\ell^2$-bound}. 
We claim the following $\ell^2$-estimate of $\sigma_1(T_\mu)$. 
\begin{theorem}\label{thm-ell^2}
    There exists a universal constant $C>0$ such that
    \[
    \sigma_1(T_\mu)^2\le Cp^2\|\mu\|^2_{\ell^2(G)}\,.
    \]
\end{theorem}

We begin with a lemma on non-trivial representations of $G=\operatorname{SL}_2(\F_p)$.

\begin{lemma}\label{lem-rep}
    Let $\rho:G\to \operatorname{U}(d)$ be a non-trivial representation of $G$, then $d\ge \tfrac{p-1}{2}$. 
\end{lemma}
\begin{proof}
Consider the following two elements in $G$: 
\[
u=\left(\begin{matrix}
    1&1\\
    0&1
\end{matrix}\right)\,,\quad 
v=\left(\begin{matrix}
    1&0\\
    1&1
\end{matrix}\right)\,.
\]
It is easy to check that $u,v$ generates $G$. Since $\rho$ is non-trivial, without loss of generality we may assume that $\rho(u)\neq I_d$. Note that
\[
\left(\begin{matrix}
    a&0\\
    0&a^{-1}
\end{matrix}\right)\left(\begin{matrix}
    1&1\\
    0&1
\end{matrix}\right)\left(\begin{matrix}
    a^{-1}&0\\
    0&a
\end{matrix}\right)=\left(\begin{matrix}
    1&a^2\\
    0&1
\end{matrix}\right)\,,\quad\forall a\in \F_p^*\,.
\]
This implies that $u$ is conjugate to $u^{a^2}$ for any $a\in \F_p^*$. Let $\Lambda$ be the multi-set of eigenvalues of $\rho(u)$, we have $\Lambda=\Lambda^{a^2},\forall a\in \F_p^*$. On the other hand, since $u^p=1$, we have $\Lambda\subset \{z\in \mathbb C,z^p=1\}$. Moreover, one can check that $\Lambda\neq \{1,\dots,1\}$, as this would imply that $\rho(u)^p\neq I_d$ (unless $\rho(u)=I_d$). Consequently, we can pick $\lambda\in \Lambda$ such that $\lambda\neq 1$. Then, the $\tfrac{p-1}{2}$ distinct elements $\lambda^{a^2},a\in \F_p^*$ all lie in $\Lambda$. We conclude that $d\ge |\Lambda|\ge \frac{p-1}{2}$, as desired.
\end{proof}
The above lemma says that any non-trivial representation of $G=\operatorname{SL}_2(\F_p)$ has dimension at least of order $p$. This lower bound is order tight: consider the subgroup $U$ of $G$:
\[
U=\left\{\left(\begin{matrix}
    a&t\\
    0&a^{-1}
\end{matrix}\right)\,,a\in \F_p^*,t\in \F_p\right\}\,,
\]
which has size of order $p^2$. We have $G$ acts on $G/U$ induces a non-trivial representation with dimension of order $p$.

Before proving Theorem~\ref{thm-ell^2}, we first introduce some notations. For $\mu:G\to \mathbb{R}$, define $\mu^*(g)=\mu(g^{-1}),\forall g\in G$. One can check that $T_\mu^*$, the adjoint of $T_\mu$, equals $T_{\mu^*}$. Moreover, $T_{\mu}T_\mu^*=T_{\mu}T_{\mu^*}=T_{\mu\ast \mu^*}$. Denote $\nu=\mu\ast \mu^*$. 

\begin{proof}[Proof of Theorem~\ref{thm-ell^2}]
    Let $V\subset \ell^2(G)_0$ be the eigenspace of $T_{\nu}=T_\mu T_\mu^*$ that corresponds to the eigenvalue $\lambda_1(T_\nu)=\sigma_1(T_\mu)^2$. Consider the left shift operator $L_g:\ell^2(G)\to \ell^2(G)$ defined by $L_gf(h)=f(g^{-1}h),\forall f\in \ell^2(G),g,h\in G$. It is straightforward to check that $L_g$ commutes with $T_\nu$, and thus $L_g$ maps $V$ to itself. Since $V$ does not contain any constant function, $L_g$ induces a non-trivial representation of $G$ on $V$. By Lemma~\ref{lem-rep}, we have $\operatorname{dim}(V)\ge \frac{p-1}{2}$, and thus $\lambda_1(T_\nu)$ has multiplicity at least $\frac{p-1}{2}$. Therefore,
    by the trace formula we have
    \begin{align*}
    \frac{p-1}{2}\sigma_1(T_\mu)^2&\ =\frac{p-1}{2}\lambda_1(T_\mu)\le \sum_i \lambda_i(T_\nu)=\operatorname{Tr}(T_\nu)\\&\ =\operatorname{Tr}(T_\mu T_\mu^*)=\sum_{g_1,g_2\in G}T_{\mu,g_1,g_2}^2=|G|\sum_{g\in G}\mu(g)^2=|G|\|\mu\|^2_{\ell^2(G)}\,.
    \end{align*}
    Since $|G|\sim p^3$, the desired result follows.
\end{proof}
As a corollary, we see that for any set $A\subset G$ with $|A|\ge 2Cp^2$, it holds that $\sigma_1(T_A)^2\le Cp^2\|\mu_A\|_{\ell^2(G)}^2\le 1/2$. Note that for $U$ the subgroup of $G$ defined as above, we have $|U|\sim p^2$ and $\sigma_1(T_U)=1$. This example also shows that the result of Theorem~\ref{thm-ell^2} is order-tight.

We say $\mu$ is symmetric if $\mu=\mu^*$ (i.e. $\mu(g)=\mu(g^{-1})$). When $\mu$ is symmetric, we have $T_\mu$ is self-adjoint and thus $\sigma_1(T_\mu)^K=\lambda_1(T_\mu)^K=\lambda_1(T_\mu^K)=\lambda_1(T_{\mu^{\ast K}})=\sigma_1(T_{\mu^{\ast K}})$. Applying Theorem~\ref{thm-ell^2}, we obtain that for any $K\in \mathbb{N}$, it holds that
\[
\sigma_1(T_{A_{\operatorname{sel}}^K})^K\le Cp^2\|\mu_{A_{\operatorname{sel}}}^{\ast K}\|^2_{\ell^2(G)}\,.
\]
Our plan is to pick $K=C_0\log p$ for some universal constant $C_0>0$, and show that $\|\mu_{A_{\operatorname{se;}}}^{\ast K}\|^2_{|ell^2(G)}\le p^{-2.1}$. This would imply that $\sigma_1(T_{A_{\operatorname{sel}}})\le 1-c$ for some universal $c=c(C,C_0)>0$. 

\textbf{Lifting to $\operatorname{SL}_2(\ZZ)$}. Consider the projection $\pi_p:\mathbb Z\to \F_p$, which induces a group homomorphism $\Pi_p:\operatorname{SL}_2(\ZZ)\to \operatorname{SL}_2(\F_p)$. Let $M$ be a probability measure on $\operatorname{SL}_2(\mathbb Z)$ and we let $\mu=\Pi_p(M)$ be its push-forward onto $\operatorname{SL}_2(\F_p)$. It holds that $\Pi_p(M^{\ast K})=\mu^{\ast K}$ for any $K\in \mathbb N$. Therefore, to understand $\mu_{A_{\operatorname{sel}}}^{\ast K}$ for large $K\in \mathbb N$, we may try to first understand $M_{A_{\operatorname{sel}}}^{\ast K}$, where $M_{A_{\operatorname{sel}}}=\frac{1}{4}\mathbf{1}_{{A}_{\operatorname{sel}}}$, and then understand how it projects onto $\operatorname{SL}_2(\F_p)$. 

\textbf{Some good features about $\operatorname{SL}_2(\ZZ)$}: 
\begin{itemize}
    \item[(1)] $\operatorname{SL}_2(\ZZ)$ is virtually free, meaning that it has a finite index free subgroup. 
    \item [(2a)] $\operatorname{SL}_2(\ZZ)\subset \operatorname{SL}_2(\RR)$ closely related to Lie groups.
    \item [(2b)] $\operatorname{SL}_2(\ZZ)$ acts nicely on the $\mathbb{H}^2$ hyperbolic plane. 
\end{itemize}

\textbf{Intuition about convolution about $\operatorname{SL}_2(\ZZ)$}: As a warm-up, consider the convolution on $\mathbb Z$. Let $\mu=\frac{1}{2}(\delta_1+\delta_{-1})$. By the central limit theorem, $\mu^K$ can be approximated by Gaussian, which intimately relates to the heat kernel on $\mathbb R$. In light of this, we might hope that for a probability measure $M$ on $\operatorname{SL}_2(\ZZ)$, there is some central limit theorem for matrices, and the convolution $M^{\ast K}$ would be related to the ``heat kernel'' on $\operatorname{SL}_2(\RR)$.  

Consider the ``ball'' in $\operatorname{SL}_2(\RR)$ with radius $T$, defined as follows:
\[
B_T:=\left\{\left(\begin{matrix}
    a&b\\
    c&d
\end{matrix}\right)\in \operatorname{SL}_2(\RR):a^2+b^2+c^2+d^2\le T^2\right\}\,.
\]
Moreover, we denote $B_T(\ZZ):=B_T\cap \operatorname{SL}_2(\ZZ)$. 
\begin{lemma}
   For $T$ large, we have $|B_T(\ZZ)|\approx T^2$.
\end{lemma}
\begin{proof}[Proof sketch]
    We need to count the solutions of $ad-bc=1,a,b,c,d\in \mathbb Z,a^2+b^2+c^2+d^2\le T^2$. For a typical pair $(a,d)\in [-T,T]^2$, the number of pairs $(b,c)\in [-T,T]^2$ such that $bc=ad-1$ is at least $1$, and at most $T^{o(1)}$. This suggests $|B_T(\ZZ)|\approx T^2$. 
\end{proof}

Vague statement: for large $K$, $M^{\ast K}$ is roughly equally distributed on $B_T(\ZZ)$, where $T\sim \exp(c(M)\cdot K)$. 

\begin{lemma}\label{lem-B_T}
    If $\mu$ is symmetric, then $\|\mu^{\ast K}\|_{\ell^2(G)}^2=\mu^{\ast 2K}(1)$, where $1\in G$ is the identity element.  
\end{lemma}
\begin{proof}
    By definition we have
    \[
    \|\mu^{\ast K}\|_{\ell^2(G)}^2=\sum_{g\in G}\mu^{\ast K}(g)^2=\sum_{g\in G}\mu^{\ast K}(g)\mu^{\ast K}(g^{-1})=\mu^{\ast 2K}(1)\,.\qedhere
    \]
\end{proof}

Let $\Gamma_p\subset\operatorname{SL}_2(\ZZ)$ be the pre-image of $I_2\in \operatorname{SL}_2(\F_p)$ under $\Pi_p$, i.e.,
\[
\Gamma_p:=\left\{a,b,c,d\in \ZZ,ad-bc=1,\left(\begin{matrix}
    a&b\\
    c&d
\end{matrix}\right)\equiv \left(\begin{matrix}
    1&0\\
    0&1
\end{matrix}\right)\pmod p\right\}
\]
Following the vague statement, we expect that 
\[
\|\mu_{A_{\operatorname{sel}}}^{\ast K}\|_{\ell^2(G)}^2=\mu_{A_{\operatorname{sel}}}^{\ast 2K}(I_2)=\Pi_p(M_{\operatorname{sel}}^{\ast 2K}(I_2))\approx \frac{|\Gamma_p\cap B_T(\ZZ)}{|B_T(\ZZ)|}\,.
\]
Since $\operatorname{SL}_2(\F_p)$ has size of order $p^3$, it is natural to expect that for large $T$, $\Gamma_p\cap B_T(\ZZ)$ occupies nearly a $p^{-3}$-fraction in $B_T(\ZZ)$. The following lemma shows that this is indeed the case.

\begin{lemma}\label{lem-Gamma_pcap B_T}
    For $T>p^2$, it holds that $|\Gamma_p\cap B_T(\ZZ)|\lessapprox p^{-3}T^2$. 
\end{lemma}
\begin{proof}
    For any $\left(\begin{matrix}
        a&b\\
        c&d
    \end{matrix}\right)\in \Gamma_p$, we have $p\mid b,p\mid c$, and thus $p^2\mid bc=ad-1$. Meanwhile, we have $p\mid a-1,p\mid d-1$, which implies $p^2\mid (a-1)(d-1)=ad-a-d+1$. Altogether we conclude that $p^2\mid a+d-2$. In light of this, we see that for $\left(\begin{matrix}
        a&b\\
        c&d
    \end{matrix}\right)\in \Gamma_p\cap B_T(\ZZ)$, $a\in [-T,T]$ has at most $O(p^{-1}T)$ choices, and given $a$, $d$ satisfies $d\equiv -2-a\pmod{p^2}$ has at most $O(p^{-2}T)$ choices (here we use the fact that $T>p^2$). Finally, given $a,d$, $b,c$ satisfies $bc=ad-1$ has at most $T^{o(1)}$ choices. Combining things together, we obtain the desired bound.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm-selb}]
    We pick $K$ such that $T\sim\exp(c(M_{\operatorname{sel}})K)\sim p^{1.1}$, and thus $K\le C_0\log p$ for a universal constant $C_0>0$. By the vague statement and Lemmas~\ref{lem-B_T}, \ref{lem-Gamma_pcap B_T}, we have 
    \[
    \|\mu_{A_{\operatorname{sel}}}^{\ast K}\|_{\ell^2(G)}^2\lessapprox \frac{|\Gamma_p\cap B_T(\ZZ)|}{|B_T(\ZZ)|}\lessapprox \frac{p^{-3}T^2}{T^2}=p^{-3}\,.
    \]
    Applying Theorem~\ref{thm-ell^2}, we obtain that $\sigma_1(T_{A_{\operatorname{sel}}})^K\lessapprox p^{-1}$. This yields that $\sigma_1(T_{A_{\operatorname{sel}}})\le 1-c$ for some universal constant $c>0$.  
\end{proof}

\textbf{Connection to hyperbolic geometry}

Recall that $\operatorname{SL}_2(\ZZ)$ acts on $\mathbb{H}^2$. Let $X(p)=\mathbb{H}^2/\Gamma_p$. We have $X(p)$ is a cover of $X(1)$, and the group of deck tranform of $X(p)$ is $\operatorname{SL}_2(\F_p)$. 

Consider the spectrum of the Laplacian of $X(p)$. We have $0$ lies in the spectrum, but above $0$ there is a gap. Denote $\lambda_1(X(p))$ the smallest positive eigenvalue of the Laplacian of $X(p)$. Selberg proved that $\lambda_1(X(p))\ge \frac 3{16}$ and conjectured that $\lambda_1(X(p))\ge \frac{1}{4}-o(1)$. 

\textcolor{blue}{The remaining part about the heat kernel is a bit difficult to read.}

\section{Sharp Projection Theorems I: Introduction and Beck's Theorem}


We begin this section by restating the Szemeredi Trotter theorem, an important sharp theorem in projection theory. 

\begin{theorem}[1982]

    Let $E$ be a set of points in $\R^2$. Pick some integer $S > 1$. For every $x$ in $E$, let $L_x$ be a set of $S$ lines passing through $x$. Define $L = \bigcup_{x\in E} L_x$. Then 
    
    \begin{equation}
        |L| \gtrsim \min(|E|\cdot S, |E|^{1/2} S^{3/2})
    \end{equation}
    
\end{theorem}

This theorem is discussed in more detail earlier in these notes. It is one of the earlier examples of a sharp theorem in projection theory. Until recently, sharp theorems for the continuous case of the Szemeredi Trotter theorem did not exist. For the continuous case, recall the $(\delta, s, C)$ spacing condition. 

\begin{Definition}
    A set $E \subset \R^d$ contained in the unit ball centered at the origin is $(\delta, s, C)$ if 

    \begin{equation}
        |E \cap B_x(r)|_{\delta} \leq C r^s |E|_\delta
    \end{equation} 

    for all balls of radius $r$ with $r \geq \delta$ centered at arbitrary points $x$, where $|\cdot |_\delta$ is the $\delta$-ball covering number. 
\end{Definition}

The following theorem by Orponen, Shmerkin, Ren, Wang (OSRW) gives an analogous statement to the Szemeredi Trotter theorem for $(\delta, s, C)$ sets (this statement is also known as the Furstenberg conjecture (or FC)) :

\begin{theorem}[Furstenberg Conjecture, OSRW (2024)]
    Let $E \subset \R^2$ be a $(\delta, t, C)$ set. Define a $\delta$ tube as a $1\times \delta$ rectangle in the plane. For every $x \in E$, let $\mathbb{T}_x$ be a set of $\delta$ tubes in $\R^2$ passing through $x$. For each $x$, let $\Dir \mathbb{T}_x \subset S^1$ be the set of directions of the $\delta$ tubes in $\mathbb{T}_x$. Assume that for all $x\in E$, $\Dir \mathbb{T}_x $ is a $(\delta, s, C)$ subset of $S^1$. Define 

    \[ \mathbb{T}  = \bigcup_x \mathbb{T}_x\] 

    Then for every $\epsilon > 0$

    \begin{equation}
        |\mathbb{T}| \geq c_\epsilon C^{-O(1)}\delta^\epsilon \min (\delta^{-t-s}, \delta^{-t/2 - 3s/2}, \delta^{-1-s})
    \end{equation} 

    where $c_\epsilon$ is a constant depending on $\epsilon$. 

\end{theorem}

Denote the three cases of the minimum value, A, B, and C, in order. The first two cases of the minimum value are analogous to the cases of the Szemeredi Trotter theorem. In case $A$, each point has many lines passing through it, and each line passes through only one point. The second case corresponds to a grid of points with lines corresponding to rational angles. The third case corresponds to a set $E$ and tubes that are both randomly dispersed in the plane. 

\subsection{History of the Furstenberg conjecture}


The first set of methods that were applied are classical methods, due to Kaufman, Faulkner, and Wolff. These consisted of double counting arguments and Fourier methods. Double counting methods are sufficient for the $s\geq t$ case, and Fourier methods can show the lower bounds given by cases A and C. Case B rougly corresponds to a grid, and was the last case to be proved. The classical methods generally are not sharp, and one of their key deficiencies that they cannot distinguish between $\R$ and $\mathbb{C}$. As the Furstenberg conjecture is false in $\mathbb{C}$, it is essential to use methods that distinguish the two spaces. 

The second set of methods, from 2000 to 2022 was $\epsilon$ improvements. These methods began with Bourgain's projection thoerem in 2000, and showed bounds that were $\epsilon$ better than the trivial or classical bounds. For instance, these methods were used to show that if $t=1$ and $s= 1/2$, then 

$$|\mathbb{T}| \gtrsim \delta^{-1-\epsilon} $$ 

Without the $\epsilon$, this bound is trivial, but the $\epsilon$ improvement was a large step forward. In 2021, Orponen and Shmerkin pushed these methods further, and proved that if $t > s > 0$ then 

$$|\mathbb{T} |\gtrsim \delta^{-2s-\epsilon} $$ 

This result was based on Bourgain's projection thoerem, as well as other ideas. It is a very involved result, and the resulting $\epsilon$ is small, but it was one of the first results to distinguish $\R$ from $\mathbb{C}$. 

The third phase is based on repeatedly applying the $\epsilon$ improvement results to reach a sharp result. The result by OSRW is a key example of these methods. We here show a simpler example where an $\epsilon$ improvement leads to a sharp bound 

\begin{theorem}[Beck]
    Let $E$ be a set of points in $\R^2$ and for any line $\ell$, assume that $\ell$ intersects at most half of the points in $E$. i.e. 

    $$ |\ell \cap E| \leq \frac{1}{2} |E|$$ 

    For every $x\in E$, let $L_{x,E}$ be the set of lines passing through $x$ that also pass through an additional point in $E$, i.e. 

    $$ L_{x,e} = \{\ell: \text{$\ell$ is a line passing through $x$ such that } |\ell \cap E | \geq 2\}$$ 

    Then for every $x$

    $$ |L_{x,E}| \gtrsim |E|$$ 
\end{theorem}
\begin{proof}
    We assume that the lines in each $L_{x,E} $ are uniform, that is each $L_{x,E}$ contains approximately the same number of lines, and each point has approximately the same number of lines passing through it. It is possible to reduce the general statement to this case. This implies that

    $$ |E\cap \ell| \sim \frac{|E|}{|L_{x,E}|} $$ 

    for each $\ell$ and each $x$. Let $S \sim |L_{x,E}|$ be the number of lines per point. Then 

    $$ |E| \cdot S \sim |L| \cdot |E|/S$$ 

    The left hand side is the number of points multiplied by the number of lines per point, so is the total number of lines multiplied by the number of points per line. The right hand side is number of lines multiplied by $|E|/S$, which is the number of points per line. The two sides are therefore equal, so by manipulating the equation:

    $$ |L| \sim S^2$$ 

    The Szemeredi Trotter theorem implies that  

    $$ |L_E| \gtrsim \min (S|E|, S^{3/2} |E|^{1/2})$$ 

    which implies that 

    $$ S \gtrsim |E|$$ 

    the desired conclusion. 
\end{proof}

Note that to obtain this conclusion, a weaker version of Szemeredi Trotter is sufficient, that if 

$$ |E| \gg S$$ 

then 

$$ |L| \gg S^2 $$ 

This matches the OS $\epsilon$-improvement approach. 

\begin{theorem}[Continuum Beck's Theorem, OSW (2023)]
    Choose $\eta > 0$ and let $E$ be a $(\delta, u, C)$ set in the plane such that for all  $\rho \times 1$ rectangle $R$, 

    $$ |E\cap R |_\delta \leq C \rho^\eta |E|_\delta$$ 

    Then for every $x\in E$, 

    $$ |L_{x,E} \gtrsim \delta^\epsilon \min(\delta^{-u}, \delta^{-1})$$ 
\end{theorem}

This theorem is sharp, but is false over $\mathbb{C}$. The proof imitates the proof of Beck's theorem, with adjustments for the continuous setting. Here we provide only a proof sketch, and omit the details of several important lemmas. If $L_{x,E}$ is a $(\delta, s, C)$ set then 

    $$ |L_{x,E}| \gtrsim \delta^{-s}$$ 

    but this lower bound is not necessarily sharp. We therefore go to the uniform case, where all $L_{x,E}$ have the same size. Using double counting, it is possible to show that 

    $$ |L| \sim |L_{x,E}|^2$$ 
    
    An $\epsilon$ improvement theorem by OS shows that if $0 < s < \min(u,1)$ then 

    $$ L\gtrsim \delta^{-2s-\epsilon}$$ 

    for some small $\epsilon$. Then 

    $$ |L_{x,E} \gtrsim \delta^{-s-\epsilon}$$ 

    This gives a corollary: If $L_{x,E}$ is a $(\delta, s, C)$ set and 

    $$ |L_{x,E}| \sim \delta^{-s}$$ 

    then 

    $$ |L_{x,E} | \gtrsim \delta^\epsilon \min (\delta^{-u}, \delta^{-1})$$ 

    However, the $(\delta, s, C)$ assumption is a large assumption. We now state an important lemma 

    \begin{lemma}
        If $0<s< \min(u,1)$, and $L_{x,E}$ is a typical set, and $L_{x,E}$ is $(\delta, s, C)$, then 

        $$ |L_{x,E} | \gtrsim \delta^{-s-\epsilon}$$ 
    \end{lemma}

    We now sketch the bootstrapping argument to obtain the sharp case. Suppose that $L_{x,E}$ is uniform and $(\delta, s, c)$ for $s$, where $0 , s < \min(u,1)$. Then $L_{x,E}$ is $(\rho, s,C)$ for $\rho \geq \delta$. The lemma then implies that 

    $$ |L_{x,E} | \gtrsim \rho^{-s-\epsilon}$$ 

    for every $\rho \geq \delta$. From the assumption tht $L_{x,E}$ is a uniform set, $L_{x,E}$ is therefore a $(\delta, s + \epsilon, C)$ set. By iterating until $s\to \min(u,1)$, we reach the sharp case $s = \min(u,1)$. 

    To begin the argument by showing the necessary non-concentration conditions, note that the non-concentrarion of $E$ in rectangles implies that $L_{x,E}$ is a $(\delta, \eta, C)$ set. This is nontrivial, but is a consequence of other results. 


\subsection{OSRW proof of Furstenberg conjecture}
For the proof of the Furstenberg conjecture of OSRW, the problem is split into cases based on the spacing of the set $E$. The problem is first reduced to the case where $E$ is a uniform set with $\delta = \Delta^m$ for some large $m$. Recall that the uniform condition on $E$ means that for any dyadic $\Delta^j$ cube $Q$ with $j$ an integer between 1 and $m$, then 

$$ |E\cap Q|_{\delta^{j+1}} \sim R_j$$ 

where $1\leq R_j \leq \Delta^{-2}$ is a splitting number that determines the spacing of $E$. The AD regular case is where all the $R_j$ are approximately equal, i.e. within a constant factor of each other. The well spaced case is where $R_j$ is large up to some threshold $J < m$, then is small for $j > J$. 

We now present an outline for the proof of the Furstenberg conjecture, which will be proven over the next several lectures. A theorem by OS in 2024 shows that the FC holds in the AD regular case, and builds on the continuum Beck theorem. A different theorem by OSW in 2019 showed that the FC holds in the well-spaced case, using Fourier methods. A multi-scale argument then shows that the full conjecture reduces to the AD regular and semi-well spaced case (the semi-well spaced case is a generalization of the well-spaced case). The next few lectures will cover these three separate pieces that together provide a proof of the Furstenberg conjecture. 


\section{Sharp Projection Theorems II: AD Regular Case}

Recall that we want to prove the following:
\begin{theorem}[OSRW]
    If $E \subset \R^2$ is a $(\delta, t, C)$-set and for all $x \in E$, $\mathbb{T}_x$ is a set of $\delta$-tubes going through $X$, $\mathrm{Dir}(\mathbb{T}_X)$ is a $(\delta, s, C)$-set, with $\mathbb{T}_X$ uniform, $|\mathbb{T}_x| \sim \delta^{-s}$, and $s > 0$, then
    \[
        |\mathbb{T}| \geq c_\epsilon \delta^\epsilon C^{-O(1)} \min \left( \delta^{-s - t}, \delta^{-\frac{t}{2} - \frac{3s}{2}}, \delta^{-1 - s} \right).
    \]
\end{theorem}
When $\delta^{-s - t}$ is the minimum, call this case A. When $\delta^{-\frac{t}{2} - \frac{3s}{2}}$ is the minimum, call this case B. And if $\delta^{-1 - s}$ is the minimum, call this case C.

In case A, $s \geq t$ and the result follows by double counting. In case C, $s + t \geq 2$, and we can deduce the theorem using the Fourier method. This leaves case B, which is the essentially new content of this theorem.

It will be a little easier to think about things in terms of
\[
    R(E, \mathbb{T}) := \text{``typical number of $\delta$-balls of $E$ on a typical tube of $\mathbb{T}$''}.
\]
More precisely,
\[
    R(E, \mathbb{T}) = \frac{|E| \delta^{-s}}{|\mathbb{T}|}.
\]

We will be interested in the AD-regular case. Suppose $E$ is uniform. Let $\delta = \Delta^m$ ($m$ large). Then
\[
    |E \cap Q_{\Delta^j}|_{\Delta^{j + 1}} \sim B_j,
\]
where $B_j$ is the branching number, for all dyadic cubes $Q_{\Delta^j}$ intersection $E$.

\begin{Definition}
    $E$ is \textbf{$(\delta, t, C)$-AD-regular} if
    \[
        \frac{1}{C} (\Delta^J)^{-t} \leq \left| \prod_{j = 1}^J B_j \right| \leq C(\Delta^J)^{-t}.
    \]
\end{Definition}
Let
\[
    R_{AD}(s, t, \delta, C) = \max_{\text{$E$, $\mathbb{T}$ obey hypotheses of theorem, $E$ is $(\delta, t, C)$-AD-regular}} R(E, \mathbb{T}).
\]
We won't worry about $C$, so we'll just set $C = 1$. The argument works if $C \lessapprox 1$. And we'll abbreviate the above to $R_{AD}(\delta)$. Then in terms of these quantities, the theorem in the AD-regular case is
\begin{theorem}[OS]\label{os}
    \[
        R_{AD}(s, t, \delta) \lessapprox \max \left( 1, \delta^{-\frac{t}{2}} \delta^{\frac{s}{2}}, \delta^{1 - t} \right).
    \]
\end{theorem}

The AD regular case seems like a very special case, but as we'll see, this is a very important case that is crucial to proving the theorem. Breaking into cases was an important step to proving the theorem.

\begin{lemma}[Submultiplicative Lemma]
    If $\delta = \delta_1 \delta_2$, $\delta_1, \delta_2 < 1$, then
    \[
        R_{AD}(\delta) \lessapprox R_{AD}(\delta_1) R_{AD}(\delta_2).
    \]
\end{lemma}
\begin{proof}[Proof Sketch]
    The idea is to take a set $E$ of $\delta$-balls and $\mathbb{T}$ of $\delta$-tubes and thicken it to set $E_1$ of $\delta_1$-balls and a set $\mathbb{T}_1$ of $\delta_1$-tubes. We can also restrict $E$ and $\mathbb{T}$ to a $\delta_1$-ball and magnify it. Then we'll get a set $E_2$ of $\delta_2$-balls and a set $\mathbb{T}_2$ of $\delta_2$-tubes. Then $(E_1, \mathbb{T}_1)$ and $(E_2, \mathbb{T}_2)$ satisfy the hypotheses, and
    \begin{align*}
        R_{AD}(\delta) & \leq \left( \text{number of $\delta_1$-balls in a $\delta_1$-tube} \right) \\
        & \quad \cdot \left( \text{number of $\delta$-balls in a $\delta$-tube within one $\delta_1$-ball} \right) \\
        & \leq R_{AD}(\delta_1) R_{AD}(\delta_2).
    \end{align*}
\end{proof}


\begin{figure}
\centering 
\includegraphics[width=0.7\textwidth]{adreg_0.jpeg} 
\caption{Proof of the submultiplicative lemma} 
\end{figure}


\begin{enumerate}
    \item If $E$ and $\mathbb{T}_x$ are uniform and $E$ is $(\delta, t)$, then $E_1$ is $(\delta_1, t)$. If $\mathbb{T}_x$ is $(\delta, s)$, then $\mathbb{T}_{1, x}$ is $(\delta_1, s)$. If $E$ is AD-regular then so is $E_1$.
    \item Because $E$ is AD-regular, $E \cap B_{\delta_1}$ magnifies to a set that is $(\delta_2, t)$ and AD-regular.
\end{enumerate}
This is why we need to work with AD-regular sets.

This is similiar to a submultiplicative lemma from decoupling theory in Fourier analysis.

\subsection{Applications}
One can give a brute force proof of the AD-regular OS theorem. For some specific $\delta_0$, check by brute force
\[
    R_{AD}(s, t, \delta_0) \leq \max(1, \delta_0^{-\frac{t}{2}} \delta_0^{\frac{s}{2}}, \delta_0^{1 - t}) \delta_0^{-\epsilon}
\]
for some $\epsilon > 0$. There are essentially only finitely (but a very very large number!) many cases for a fixed $\delta_0$, so this can theoretically be check by brute force. Then we can use the submultiplicative lemma many times to get
\[
    R_{AD}(s, t, \delta_0^2) \leq \max(1, (\delta_0^2)^{-\frac{t}{2}} (\delta_0^2)^{\frac{s}{2}}, (\delta_0^2)^{1 - t}) (\delta_0^2)^{-\epsilon}
\]
and so on.

\subsection{General AD vs Projective AD}
This theorem is more general than orthogonal projections
\begin{Definition}
    We say $(E, \mathbb{T})$ is \textbf{projective} if $\mathrm{Dir}(\mathbb{T}_{x_1}) = \mathrm{Dir}(\mathbb{T}_{x_2})$ for any $x_1, x_2 \in E$.
\end{Definition}
Let
\[
    R_{AD, \mathrm{proj}}(\delta) = \max_{\text{$(E, \mathbb{T})$ satisfy hypotheses, $E$ is AD-reg, $(E, \mathbb{T})$ projective}} R(E, \mathbb{T}).
\]
Then clearly $R_{AD, \mathrm{proj}}(\delta) \leq R_{AD}(\delta)$.

Notice that from the proof of the submultiplicative lemma, if we let $\delta_1 = \delta_2 = \sqrt{\delta}$, then the small ball problems are all projective: We can only distinguish the angles of tubes for the small ball up to $\sim \sqrt{\delta}$, so they are the angles of the larger $\sqrt{\delta}$-tubes. So
\begin{align*}
    R_{AD}(\delta) & \lesssim R_{AD}(\delta^{1/2}) R_{AD, \mathrm{proj}}(\delta^{1/2}) \\
    & \lesssim R_{AD}(\delta^{1/4}) R_{AD, \mathrm{proj}}(\delta^{1/4}) R_{AD, \mathrm{proj}}(\delta^{1/2}) \\
    & \lesssim \ldots
\end{align*}
So to prove the theorem, it suffices to check the projective case. The philosophy to prove this is that we need an $\epsilon$-improvement to the submultiplicative lemma.

Then, either $R_{AD}(\delta^{1/2})$ obeys the theorem, or from an improved submultiplicative lemma,
\[
    R_{AD}(\delta) \lesssim \delta^\epsilon R_{AD}(\delta^{1/2})^2.
\]
If $R_{AD}(\delta) \leq \delta^{-\gamma}$, then $R_{AD}(\delta^2) \leq (\delta^2)^{-\gamma + \epsilon}$, $R_{AD}(\delta^4) \leq (\delta^4)^{-\gamma + 2\epsilon}$, and so on.

So what does it mean for the submultiplicative lemma to be sharp? The following pictures show the difference (they show a typical $\delta_1$-tube in the submultiplicative lemma).

\begin{figure}
\centering 
\includegraphics[width=0.7\textwidth]{adreg_1.jpeg} 
\caption{Top: Submultiplicative Lemma is Not Sharp, Bottom: Submultiplicative Lemma is sharp} 
\end{figure}


We already reduced to the projective case. Let $D$ be the set of directions, so $D \subseteq S^1$ is a $(\delta, s)$-set. Let $\theta$ be the $\delta^{1/2}$-arc corresponding to the fat tube. Let $A$ be the projection of $E$ along the direction of the fat tube. Let $B$ be the projection of $E_1$ along the direction perpendicular to the fat tube. If the submultiplicative lemma is sharp, then in the main case, we have subsets $A, B, C \subset \R$ with
\begin{itemize}
    \item $A$ is a $(\rho, a)$-set with $|A| \sim \rho^{-a}$.
    \item $B$ is a $(\rho, b)$-set with $|B| \sim \rho^{-b}$.
    \item $C$ is a $(\rho, c)$-set with $|C| \sim \rho^{-c}$.
    \item For any $c \in C$, $|A + cB| \lessapprox |A|$.
\end{itemize}

\begin{lemma}
    If the above is true, then $a \geq b + c$.
\end{lemma}
We stop iterating the improved submultiplicative lemma when the submultiplicative lemma is essentially sharp, in which case we can apply the above. We have to decide what it means for the submultiplicative lemma to be ``essentially sharp''. We'll say it's sharp if
\[
    R_{AD}(\delta) \geq \delta^{\epsilon_0} R_{AD}(\sqrt{\delta})^2
\]
for some tiny power $\epsilon_0$. We'd replace $|A + cB|$ with $|\pi_c(X)|$ for some subset $X \subset A \times B$ with $|X| \geq \delta^\epsilon |A||B|$.

For the first step of the proof, we use BSG to reduce to $|A' + cB'| \lessapprox |A'|$. Then use Plunnecke-Rusza to get $|A' + c_1 B' + c_2B' + c_3B'| \lessapprox |A'|$.


\section{Sharp Projection Theorems III: Combining different scales}
For the last case we want to prove, we will use two tools:
\begin{itemize}
  \item Fourier method and well-spaced case.
  \item Combining different scales.
\end{itemize}
\subsection{Well spaced case}
For the well-spaced case, we want to have some sort of Geometric Measure Theory version of the Szemerédi–Trotter (theorem \ref{st-inc}). Let us remind ourselves what the classic theorem looks like in our setting. 
\begin{theorem}
  If $E \subset \RR^2$ is a set of $N$ points and $L_{R} (E)$ the set of lines
  $
    |L_{R} (E)| \lesssim \frac{N^2}{R^3} + \frac{N}{R}
  $
\end{theorem}
For the well-spaced case, we do have a version of this theorem.
\begin{theorem}[GSW]
  Let $E \subset \RR^2$ be a set of $N$ $\delta$-balls with $E \subset B_1$ which is well-spaced, in the sense that $|E \cap B_{N^{-1/2}}|_\delta \lesssim 1$. \\
  Let $\TT_{R}(E)$ be a set of $\delta$-tubes which are essentially distinct with $|T \cap E|_\delta \geq R$. \\
  Assume also that $R > \delta^{-\varepsilon} \delta \dot |E|_\delta$. Then
  \[
  |\TT_{R} (E)| \lessapprox \frac{N^2}{R^3}.
  \]
\end{theorem}


When we compare the two theorems two things stand out to us.
\begin{itemize}
  \item First we no longer have a $\frac{N}{R}$ term, this isn't to interesting because that term dominates only when $R > \sqrt{N}$ which isn't possible in the well-spaced case since each line intersects roughly $\sqrt {N}$ squares.
  \item The second difference is that we do need to assume some lower bound on $R$. To see why this is necessary, let us consider a random $\delta$-tube $T$, then the expected number of balls on the line is
    \[
      \mathbb{E}[|T \cap E|_\delta] \sim \delta |E|_\delta.
    \]
    If $R$ is equal to $\delta |E|_\delta$,  then an average tube will be $R$-rich,  and so $|\TT_R(E)|$ can be comparable to the total number of essentially distinct $\delta$-tubes (about $\delta^{-2}$).   In this regime,  the theorem is not true.   But if we increase $R$ slightly,  then we get the sharp bound in the theorem.  
    It is  quite remarkable that there such a sharp phase transition once we increase $R$ past the threshold in the theorem.
\end{itemize}

\begin{proof}
  We will now sketch the proof of this theorem, the tools we will need are the Fourier Method, Double Counting, and Playing with different scales.

  Now the Fourier method implies that under the hypothesis of the theorem, we get that
  \[
    |\TT_{R} (E)| \lessapprox \delta^{-1} |E|_\delta R^{-2} = \delta^{-1} N R^{-2}.
  \]
  Now the specicial case where $R$ is as small as it can be, that is
  $R = \delta^{-\varepsilon} \delta |E|_\delta$ then $\delta^{-1} \approx \frac{N}{R}$ so
  \[
    \delta^{-1} N R^{-2} = \frac{N^2}{R^3},
  \]
  which exactly matches the theorem. Unfortunately, this breaks down when we increase $R$. However, this bound gets better as we increase $\delta$, that is if we increase the width of our tubes. To examine this in more detail, we are going to introduce a scale parameter to nicely parametrize between our bounds.
  Since we know $\delta^{-\varepsilon} \delta |E|_\delta \leq R \leq N^(1/2)$ then we set the scale parameter $\rho$ such that $\rho \cdot N \sim R$. This way
  \[
    \delta < \rho = \frac{R}{N} < N^{-\frac{1}{2}}
  \]
  and then we are going to study $E_\rho$, the $\rho$-neighborhood of $E$.
\begin{figure}[h]
\centering
  \input{wellspacedsetdiagram}
  \caption{An example of a well spaced set with $N$ points, along with its $E_\rho$ neighborhood in red}
\end{figure}
  Now we want to understand $\delta$-tubes that hit a lot of balls, but now that we have thickened our set, it makes sense to study thickened tubes intersecting our set.
  \[
    \TT_{\tilde{R}} (E_\rho) = \{\rho \text{-tubes } T_\rho : |T_\rho \cap E_\rho|_\rho \geq \tilde{R} \}.
  \]
  Now we can again apply the Fourier method, and we get the bounds
  \[
    |\TT_{\tilde{R}} (E_\rho)| \lesssim \rho^{-1} |E|_{\delta} \tilde{R}^{-2} = \frac{N^2}{R \cdot \tilde{R}^2}.
  \]
  where we importantly used the fact that $|E|_\rho = |E|_\delta = N$ because our set is well-spaced. In particular, if we pick $\tilde{R} = R$ then
  \[
    |\TT_{R} (E_\rho) \lesssim \frac{N^2}{R^3}.
  \]
  Now one might think that we are now done, but this isn't quite the case. Recall that originally we want to count thin $\delta$-tubes, where as this rescaling result gives us a bound for thick $\rho$-tubes. While each $\delta$-tube can be expanded to give a single $\rho$-tube, each $\rho$-tube can contain many $\delta$-tubes and so we are not quite done yet. This does, however, reduce our problem to the following case.

\begin{figure}[h]
  \centering
  \input{smalltubelargetubediagram}
\end{figure}
  Here we have one large tube with many smaller roughly aligned tubes within it and we have $|T_\rho \cap E|_\delta = |T_\rho \cap E|_\rho \sim \tilde{R} \geq R$ (???). Induction then allows us to assume that the balls are not concentrated on one side, as if they are we can split our original set $E$ and reduce to a smaller tube size. 

  We now apply double counting, each pair of balls on opposite ends (of which there are around $\tilde{R}^2$ many) gives us a unique small $\delta$-tube inside the larger $\rho$ tube. Applying D.C. to the set
  \[
    \{ (T,x_1,x_2) | T \in \TT_R (E_\delta) \cap \text{ tubes in } T_\rho, x_1,x_2 \in T \text{ near opposite ends }\}
  \]
  on one side we get roughly $R^2$ pairs of balls inside each tube on opposite ends, on the other hand each opposite pair of balls gives us a unique small tube.
  Hence we have
  \[
    |\TT_R (E_\delta) \cap \text{ tubes in } T_\rho| \cdot R^2 \lesssim \tilde{R}^2.
  \]
  Now to solve our original problem, we can dyadically sum over scales $\tilde{R}$. Then for each scale, we will consider all $\rho$ tubes, and apply the double counting to estimate small tubes inside it. This gives us
  \begin{align*}
    |\TT_R (E_\delta)| 
    \leq& \sum_{\tilde{R} > R, \text{ dyadic}} |\TT_{\tilde{R}} (E_\rho)| \cdot |\TT_R (E_\delta) \cap \text{ tubes in } T_\rho| 
    \\ \lessapprox& \, \sum_{\tilde{R} > R, \text{ dyadic}} \frac{N^2}{R \cdot \tilde{R}^2} \cdot \frac{\tilde{R}^2}{R^2} 
    \\ \lessapprox& \, \frac{N^2}{R^3}
  \end{align*}
\end{proof}
It can be instructive to check where we used each hypothesis of the result.
\begin{itemize}
  \item The well-spaced hypothesis was only used to control the rescaled size $|E|_\rho$ of $E$, and a slightly weaker version was used for the Fourier analysis.
  \item The lower bound on $R$ was necessary for the Fourier analysis part. It was necessary to assume because otherwise the lower frequencies of the characteristic functions of the tubes dominate and we get a bad bound.
\end{itemize}

Another thing that is interesting is that it seems oddly coincidental for the lower bound given by simple examples mathces the upper bounds given by ST, not just in this case but for the other cases of ST as well. This is especially interesting because in cousin problems involving other collections of points such as circles or parabolas, the same methods give mismatches upper and lower bounds, and are in general open.

\subsection{Combining scales}
This is the most recently developed part of the theorem, and it is reasonable to assume that this tool could find many uses outside of this domain.
\begin{theorem}[OSRW]
  Let $E$ be a $(\delta, t)$ set in $B_1 \subset \RR^2$ and $|E| = \delta^{-t}$. \\
  For every $x \in E$ let $\TT_x$ be a $(\delta, s)$ set of tubes passing through $x$ with $|\TT_x| = \delta^{-s}$. Set $\TT = \bigcup_{x \in E} \TT_x$. \\
  Let $R = |E \cap T|_\delta$ be the size of a typical intersection between the tubes and $E$. \\
  Then $R \lessapprox \max(\underbrace{1}_{A}, \underbrace{\delta^{-s} \delta^{-\frac{t}{2}}}_{B}, \underbrace{\delta^{1-t}}_{C})$.
\end{theorem}

\begin{figure}[h]
  \centering
  \input{osrwcasesdiagram}
  \caption{The 3 regimes of the OSRW theorem, $A$ - unrelated balls with many tubes through each, $B$ - an integer grid of balls, $C$ - randomly picked tubes}
\end{figure}

We have already used many tools and techniques to prove this theorem for specific cases and regimes. Let us quickly document these.
\begin{enumerate}
  \item In the case where $A$ dominates, i.e. $s \geq t$, this is true by D.C.
  \item In the case where $C$ dominates, i.e. $s + t \geq 2$, this is true by the Fourier method.
  \item IN the case where $B$ dominates we have $s < t < 2 - 2$. In this case we do not yet know if the theorem holds. However, we proved it for two special cases:
    \begin{itemize}
      \item If $E$ is AD-regular, we proved this last class (theorem \ref{os}).
      \item If $E$ is well-spaced, which we just showed.
    \end{itemize}
\end{enumerate}
Now the interesting point here is that AD-regular and well-spaced sets are sort of the two extremes of how a set can be spaced. In fact such a situation happens for many similar problems in this field, where we know a result holds at these two extremes but not if it holds in the middle.

The last idea of the proof which comes from Ren and Wang, comes in two steps.
First they relax the well-spaced condition in the result we proved to a semi-well-spaced set, which we will define in a moment.
Secondly they put together the known regimes by using a multiscale approach and a variant of the submultiplicative lemma we used for the AD-regular case.

Let $\rho$ be a scale parameter with $\delta < \rho < 1$. Let us assume now that $E$ is a general uniform set. We can imagine it like a bunch of $\rho$-balls inside of each we have a bunch of $\delta$-balls.

\begin{figure}[h]
  \centering
  \input{multiscalediagram}
  \caption{Multiscale argument}
\end{figure}

Now $R(E_\delta, \TT_\delta)$ is the number of $\delta$-balls of $E_\delta$ that are in $T_\delta$. From the diagram we can compute this by first calculating the number of $\rho$-balls contained in $T_\rho$, which we will denote $R(E_\rho, \TT_\rho)$. Then if we call one of these balls $B$, then for each such ball we take all the short segments of $\delta$-tubes and see how many $\delta$-balls each of them hits, we will call this $R(E_B,\TT_B)$. We thus have
  \[
  R(E_\delta, \TT_\delta) \lessapprox R(E_\rho, \TT_\rho) \cdot R(E_B, \TT_B)
  \]
But now we can rescale $B$ to $B_1$, so we will assume from now on that $E_B$ is a set of $\frac{\delta}{\rho}$-balls, and $\TT_B$ is a set of $\frac{\delta}{\rho}$-tubes.

Now this is very powerful, we started with one scale $\delta$, and using this multi-scale argument we broke it up into two similar problems with scales $\rho$ and $\frac{\delta}{\rho}$. Because we can choose $\rho$ freely, we can then keep doing this splitting which will eventually reduce us to basic pieces. We then hope these pieces will fit into one of the cases we already solved. At that point we will also hope that we can multiply the bounds together to get a sharp bound at the initial scale.

Now to discuss these scaling argument we will use the language of the branching function of a uniform set. Because we are concerned with scaling we will reparametrize the function by setting $f : \log_\delta (\rho) \mapsto \log_{\frac{1}{\delta}} |E_\rho|$, with domain $[0,1]$. What do we know about $f$?
\begin{enumerate}
  \item $f$ is trivially increasing, since $|E| = \delta^{-t}$ we have that $f(0) = 0$ and $f(1) = t$.
  \item Because $E$ is a $(\delta,t)$ set we know that $f(x) \geq t \cdot x$ for all $x \in [0,1]$.
  \item Because we are in 2 dimensional Euclidean space, we can always cover a $C \rho$ ball with $C^2$ smaller $\rho$ balls and so our function satisfies $f(x + \Delta x) \leq f(x) + 2 \Delta x$, i.e. is 2-Lipschitz.
\end{enumerate}
All of these properties give us a range of 'admissable' branching functions, which we can represent in the following graph.
\begin{figure}[h]
  \centering
  \input{branchingfunctiondiagram}
  \caption{An example of a branching function (blue). The two known cases of an AD-regular set and a Well-spaced set bound an admissible region in which the function can lie (red). The semi-well-spaced case corresponds to any function lying above the orange dashed line.}
\end{figure}

Using this language we can define what a semi-well-spaced set is. A well spaced set is formed on the graph with two lines of slope $2$ and $0$ which meet in the middle. We then slightly weaken this to have two lines of slope $2 - s$ and $s$. Any branching function above this new graph corresponds to a semi-well-spaced set.

Now how does our multiscale argument interact with this branching function? Well $R(E_\rho, \TT_\rho)$ we get by only looking at scales larger than $\rho$, this is equivalent to restricting our branching function to $[0,\log_\delta (\rho)]$. Similarly we can the scales smaller than $\rho$ by restricting to $[\log_\delta (\rho), 1]$. In essence the multiscale argument splits our branching function into two pieces which we can analyze separately.

In our graph this looks like splitting the graph into a left and a right part. The left part corresponds to the branching function of $E_\rho$ and the right describes the branching function of $E_B$. Because the branching function of $E_\delta$ can be recovered from the two pieces by placing them side by side, we will call this the Concatenation method. Let us attempt an explicit example to help reason about this method.

\begin{figure}[h]
  \centering
  \input{concatenationmethoddiagram}
\end{figure}
Consider a branching function as above, by splitting at $\rho$ corresponding to where the two lines meet, we get
\[
  \delta^{-t} = |E_\delta| = |E_\rho| |E_B| = \rho^{-t_1} \cdot \left(\frac{\delta}{\rho}\right)^{-t_2}
\]
Now let us try to estimate $R(E_\delta,\TT_\delta)$ using this splitting. We already know that
\[
  R(E_\delta, \TT_\delta) \leq R(E_\rho, \TT_\rho) R(E_B, \TT_B),
\]
Now we have two scenarios that can happen depending on the values $t_1,t_2$. 
\begin{itemize}
  \item $s < t_1,t_2 < 2 - s$. In this case we can estimate both $R(E_\rho, \TT_\rho)$ and $R(E_B, \TT_B)$ by the $B$ bound in the theorem. This gives us
  \[
    R(E_\rho, \TT_\rho) R(E_B, \TT_B) \leq
    \rho^\frac{s}{2} \rho^{-\frac{t_1}{2}} \left(\frac{\delta}{\rho}\right)^{\frac{s}{2}} \left(\frac{\delta}{\rho}\right)^{-\frac{t_2}{2}}
    =
    \delta^{\frac{s}{2}} \delta^{-\frac{t}{2}},
  \]
    where we used the equation for $\delta^{-t}$ we had above. This is exactly the bound we want when $t$ is in the $B$ regime.
  \item $2 - s < t_1$ and $t_2 < s$. Now when we bound $R(E_\rho, \TT_\rho)$ we get the $C$ bound of the theorem, and when we bound $R(E_B, \TT_B)$ we get the $A$ bound of the theorem. This gives us
    \[
R(E_\rho, \TT_\rho) R(E_B, \TT_B) \leq
    \rho^{1 - t_1} \cdot 1
    \gg \delta^{\frac{s}{2}} \delta^{-\frac{t}{2}}.
    \]
    Unfortunately, in this regime, we do not get the desired bound.
\end{itemize}
What can we learn from this? Well we want to avoid the second scenario, so we want to keep splitting our function into pieces that all fall in the first scenario, but that are simple enough for us to analyze. The only such pieces are the AD-regular branching functions and Semi-well-spaced branching functions, so we want all of our pieces to be one of those.

The last argument of the theorem is then to show that this is actually possible.
\begin{lemma}
  If $f : [0,1] \to \RR$ is 2-Lip, increasing with $f(1) = t$, $f(x) \geq t \cdot x$ and $s < t < 2 - s$.\\
  Then there is a decomposition $[0,1] = \bigsqcup I$ (plus some leftovers) where on each interval $I$ either
  \begin{itemize}
    \item $f$ is almost linear with slope $t_I$, $s < t_I < 2 - s$.
    \item $f$ is semi-well-spaced.
  \end{itemize}
\end{lemma}
We do not show the full proof here, but an interesting tool used here is the Radamacher theorem. Because our function is 2-Lipschitz our function must be differentiable almost everywhere. Thus as we split into smaller and smaller pieces, our pieces will look more and more like constant slope functions, i.e. the AD-regular case. We then use the semi-well-spaced case to get rid of the slopes that are outside our range.

This lemma was the last tool necessary to combine different scales, and so the theorem is at last proven.

\end{document}

