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% \textbf{Chapter 1:} The regularity lemma and applications.
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% \textbf{Chapter 2:} Hypergraph regularity.
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% \textbf{Chapter 3:} The Green--Tao theorem.
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% \textbf{Chapter 4:} Connections with higher-order Fourier analysis.

\newpage
\section{The regularity lemma and applications}

\begin{fcthm}[Szemerédi]
  \label{thm:regularity}
  For all $\eps > 0$, there exists $k = k(\eps)$ such that the vertex set of any
  sufficiently large graph $G = (V, E)$ can be partitioned into $V_1, \ldots,
  V_s$, $s \le k$ such that for all but an $\eps$-proportion of pairs $(V_i,
  V_j)$, $G(V_i, V_j)$ is \emph{$\eps$-regular}.
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/4154728035754969.png}
\end{center}

\begin{remark*}
  \leavevmode
  \begin{itemize}
    \item $\eps$-regular means: looks like a ``random'' graph.
      We will define it more thoroughly at the end of the lecture.
    \item In \cref{thm:regularity}, it is also possible to ensure that all the
      $V_i$ are almost the same size.
    \item $k$ is known to have a bad dependence on $\eps$: we have
      \[
        2^{2^{2^{\iddots^{2}}}}
      ,\]
      where the tower of exponentials is of size $\eps^{-C}$.
  \end{itemize}
\end{remark*}

\begin{fcthm}[Removal]
  For all $\delta > 0$, there exists $\eta > \eta(\delta)$ such that the
  following holds.

  If $G$ is a graph on $n$ vertices with at most $\eta n^3$ triangles, then it
  is possible to remove $\le \delta n^2$ edges to make it triangle-free.
\end{fcthm}

\begin{proof}[Sketch proof]
  Apply \cref{thm:regularity} with $\eps = \frac{\delta}{4}$ to obtain $V_1,
  \ldots, V_s$, $s \le k = k(\delta)$ of almost equal size.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/5491dc19ca7c4edb.png}
  \end{center}
  Remove all edges between $V_i$ and $V_j$ where $G(V_i, V_j)$ fails to be
  $\eps$-regular, or where the density of $G(V_i, V_j) \le 2\eps$.
  The number of edges removed in this way is $\le \eps n^2 + 2\eps n^2 < \delta
  n^2$.

  \textbf{Claim:}
  The ``reduced'' graph is triangle-free.
  
  Indeed, if $x, y, z$ form a triangle in $G'$, then $(x, y, z)$ must lie in
  $V_i \times V_j \times V_k$ with $G(V_i, V_j)$, $G(V_i, V_k)$, $G(V_j, V_k)$
  all regular and dense (density $\ge 2\eps$).
  Hence we in fact have $\ge \eps^3 \left( \frac{n}{k} \right)^3 =
  \frac{\delta^3}{2^6} \frac{n^3}{k(\delta)}$ triangles in $G$.
  This is a contradiction if $\eta < \frac{\delta^3}{2^6k(\delta)}$.
\end{proof}

\begin{fcthm}[Corners]
  \label{thm:corners}
  For all $\alpha > 0$, there exists $N_0 = N_0(\alpha)$ such that for all $N
  \ge N_0$, the following holds.

  Let $A \subseteq [N]^2$ of density $\alpha$ ($|A| / N^2 = \alpha$).
  Then $A$ contains a triple of the form $(x, y)$, $(x + d, y)$, $(x, y + d)$
  with $d > 0$.
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/08116290a08c4a6e.png}
\end{center}

\begin{remark*}
  The theorem as stated can be fairly easily deduced from a version where we
  only ask that $d \neq 0$.
  To do this: note that if $A$ is symmetric, then existence of a triangle with
  $d < 0$ implies existence of one with $d > 0$.
  Then one can ``make $A$ symmetric'' (in exchange for a loss in density) by
  intersecting $A$ with a reflection of $A$ through a suitably chosen point.
\end{remark*}

\begin{proof}
  Let
  \begin{align*}
    X
    &= \{v_u = \{(x, y) : x = u\} : u \in [N]\} \\
    Y
    &= \{h_t = \{(x, y) : y = t\} : t \in [N]\} \\
    Z
    &= \{d_s = \{(x, y) : x + y = s\}\}
  \end{align*}
  We define a tripartite graph on parts $X, Y, Z$, where vertices are joined by
  an edge if and only if the intersection of the corresponding lines lies in
  $A$.
  A triangle in this graph corresponds to three points
  \[
    (u, t), (u, s - u), (s - t, t) \in A
  .\]
  Setting $d = s - u - t$, these points are $(u, t)$, $(u, t + d)$, $(u + d, t)$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/a95074a88c154ba2.png}
  \end{center}
  If $A$ contains no corner with $d \neq 0$, then the only triangles in this
  graph are the degenerate ones ($d = 0$).
  There are $\alpha N^2 = |A|$ many of these, and they are edge disjoint.
  Pick $\delta = \frac{\alpha}{2}$, then by triangle-removal, there exists $\eta
  = \eta(\alpha)$ such that we can destroy $\eta N^3$ triangles by removing at
  most $\delta N^2$ edges.

  If $\alpha N^2 < \eta N^3$, then should be able to remove all triangles.
  But this is a contradiction since all the triangles are edge disjoint.
\end{proof}

\begin{fcthm}[Roth]
  For all $\alpha > 0$, there exists $N_0 = N_0(\alpha)$ such that for all $N
  \ge N_0$, every $A \subseteq [N]$ of density $\alpha$ ($= |A| / N$) contains a
  non-trivial $3$-AP (a triple $x, x + d, x + 2d$).
\end{fcthm}

\begin{proof}[Sketch proof]
  Let $B = \{(x, y) \in [N]^2 : x - y \in A\}$.
  By \cref{thm:corners}, $B$ contains $(x, y), (x, y + d), (x + d, y)$ with $d
  \neq 0$.
  Then $x - y, x - (y + d) = x - y + d, x + d - y = x - y + d \in A$.
\end{proof}