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\title{18.156 Notes}
\author{Frank Wang}
\date{March 18, 2025}

\begin{document}

\maketitle

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

\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}
\end{document}

