%! TEX root = HOU.tex
% vim: tw=80 ft=tex
% 03/03/2026 10AM

$X = X_1 \cup X_2 \cup X_3 \cup X_4$.

$X_{-i}$ means $X_1 \cup \cdots \cup X_{i - 1} \cup X_{i + 1} \cup \cdots \cup
X_4$.
$x_{-i} = (x_1, \ldots, x_{i - 1}, x_{i + 1}, x_4) \in X_{-i}$.
(for all $i \in [4]$)

Work with $4$-partite $3$-uniform weighted hypergraphs.
\[
  g = (g_{-i})_{i \in [4]}
\]
with each $g_{-i} : X_{-i} \to \Rbb$.
For two such hypergraphs, $g$ and $\nu$, write $g \le \nu$ whenever
$g_{-i}(x_{-i}) \le \nu_{-i}(x_{-i})$ for all $i \in [4]$ and all $x_{-i} \in
X_{-i}$.

Given a weighted $3$-uniform hypergraph $h$ on $X_{-i}$, define
\[
  \|h\|_{\square, i}
  = \sup_{A_j \subseteq X_{-j}} \left|
    \Ebb_{x_{-i} \in X_{-i}} h(x_{-i})
    \prod_{j \in [4] \setminus \{i\}} \indicator{A_j} (x_{-i, -j})
  \right|
.\]
Given a weighted $4$-partite $3$-uniform hypergraph on $X$, define
\[
  \|g\|_{\square}
  = \max_{i \in [4]} \{\|g_{-i}\|_{\square, i}\}
.\]

\begin{fcdefn}[]
  % Definition 3.12
  \label{defn:3.12}
  Say that a $4$-partite $3$-uniform weighted hypergraph $\nu$ satisfies the
  \emph{$3$-LFC} if
  \[
    \Ebb_{\substack{x_1^{(0)}, x_1^{(1)} \in X_1 \\ \vdots x_4^{(0)}, x_4^{(1)}
    \in X_4}}
    \prod_{j = 1}^{4} \prod_{\omega \in \{0, 1\}^{[4] \setminus \{j\}}}
    \nu(x_{-j}^{(\omega)})^{n_{j, w}}
    = 1 + o(1)
  .\]
\end{fcdefn}

Our definition can be recovered by making the substitution
\[
  (x^{(\omega)})_{-j}
  = \sum_{i = 1}^{4} (j - i) x_i^{(\omega_i)}
.\]
\[
  \begin{tikzcd}
    x_2 + 2 x_3 + 3 x_4
    \ar[d, "-(x_1 + x_2 + x_3 + x_4)"] \\
    -x_1 + x_3 + 2 x_4
    \ar[d, "-(x_1 + x_2 + x_3 + x_4)"] \\
    -2 x_1 - x_2 + x_4
    \ar[d, "-(x_1 + x_2 + x_3 + x_4)"] \\
    -3 x_1 - 2 x_2 - x_3
  \end{tikzcd}
\]

\begin{fcthm}[Sparse counting lemma]
  % Theorem 3.13
  \label{thm:3.13}
  For all $\gamma > 0$, there exists $\eps > 0$ such that the following holds:

  Let $\nu, g, \tilde{g}$ be weighted hypergraphs on $X_1 \cup X_2 \cup X_3 \cup
  X_4$.
  Suppose that $\nu$ satisfies the $3$-LFC (as in \cref{defn:3.12}), $0 \le g
  \le \nu$, $0 \le \tilde{g} \le 1$, and $\|g - \tilde{g}\|_{\square} < \eps$.

  Then
  \[
    \left|
      \Ebb_{x_1, \ldots, x_4} \prod_{j = 1}^{4} g_{-j}(x_{-j})
      - \Ebb_{x_1, \ldots, x_4} \prod_{j = 1}^{4} \tilde{g}_{-j} (x_{-j})
    \right|
    \le \gamma
  .\]
\end{fcthm}

The proof consists of the following steps:

(1)
    \begin{fclemma}[Telescoping argument]
      % Lemma 3.14
      \label{lemma:3.14}
      Let $0 \le \tilde{g} \le 1$, and $g_{-j} \le 1 ~\forall j \in [3]$,
      $g_{-4} \le \nu_{-4}$.
      If $\|g - \tilde{g}\|_{\square} \le \eps$, then
      \[
        \left|
          \Ebb_{x_1, \ldots, x_4} \prod_{j = 1}^{4} g_{-j}(x_{-j})
          - \prod_{j = 1}^{4} \tilde{g}_{-j}(x_-j)
        \right|
        \le 4\eps
      .\]
    \end{fclemma}

    \begin{proof}[Sketch proof]
      \begin{align*}
        \Ebb \prod_{j = 1}^{4} g(x_{-j})
        &= \ub{\Ebb \prod_{j = 1}^{3}  g(x_{-j}) (g(x_{-4}) -
        \tilde{g}(x_{-4}))}_{\text{small}}
        + \Ebb \prod_{j = 1}^{3} g(x_{-j}) \tilde{g}(x_{-4})
      \end{align*}
      Now we no longer have to worry about $g_{-4}$.
      Then repeat.
    \end{proof}

(2)
    \begin{fclemma}[Strong linear forms]
      % Lemma 3.15
      \label{lemma:3.15}
      Suppose $\nu$ satisfies the $3$-LFC as defined in \cref{defn:3.12} and
      suppose $0 \le g \le \nu$, $0 \le \tilde{g} \le 1$.
      Then
      \[
        \left|
          \Ebb_{x_1, x_2, x_3, x_4^{(0)}, x_4^{(1)}}
          (\nu(x_{-4}) - 1) \prod_{j = 1}^{3} \prod_{\omega \in \{0, 1\}}
          h_{j, \omega} (x_{-j}^{(\omega)}) 
        \right|
      ,\]
      where $x^{(\omega)} = (x_1, x_2, x_3, x_4^{(\omega)})$ and $h_{j, \omega}$
      is either $g_{-j}$ or $\tilde{g}_{-j}$.
    \end{fclemma}

    \begin{proof}[Sketch proof]
      Use LFC and Cauchy-Schwarz.
    \end{proof}

(3)
    Next (final) lecture: densification strategy

    At each step, if $g_{-j} \le \nu_{-j}$ and $\|g_{-j} -
    \tilde{g}_{-j}\|_{\square, j} \le \eps$, find $0 \le g_{-j}' \le 1$ such
    that $\|g_{-j}' - \tilde{g}_{-j}\|_{\square, j} \le \eps$.

\begin{proof}[Sketch proof of \cref{thm:3.5}]
  Given $f : \Zbb / N\Zbb \to [0, \infty)$ and $\nu : \Zbb / N\Zbb \to [0,
  \infty)$, $0 \le f \le \nu$, satisfying the $3$-LFC:
  \begin{align*}
    \psi : G^4 \to G
    \begin{cases}
      \psi_1(x_1, \ldots, x_4) = x_2 + 2 x_3 + 3 x_4 \\
      \psi_2(x_1, \ldots, x_4) = -x_1 + x_3 + 2 x_4 \\
      \psi_3(x_1, \ldots, x_4) = -2 x_1 - x_2 + x_4 \\
      \psi_4(x_1, \ldots, x_4) = -3 x_1 - 2 x_2 - x_3
    \end{cases}
  \end{align*}
  Define the weighted hypergraphs $g, \nu$ on $(\Zbb / N\Zbb)^4$:
  \begin{align*}
    g_{-j}(x_{-j})
    &= f(\psi_j(x_1, \ldots, x_4)) \\
    \nu_{-j}(x_{-j})
    &= \nu(\psi_j(x_1, \ldots, x_4))
  \end{align*}
  and from the Dense Model Theorem, obtain $\tilde{f} : \Zbb / N\Zbb \to [0, 1]$
  such that $\|f - \tilde{f}\|_{\square} \le \eps$, with corresponding weighted
  hypergraph
  \[
    \tilde{g}_{-j}(x_{-j})
    = \tilde{f}(\psi_j(x_1, \ldots, x_4))
  .\]
  By the sparse counting lemma, we have

  \begin{fccoro}[]
    % Corollary 3.16
    \label{coro:3.16}
    For all $\gamma > 0$, there exists $\eps > 0$ such that the following holds:

    Let $\nu, f, \tilde{f} : \Zbb / N\Zbb \to [0, \infty)$ with $0 \le f \le
    \nu$, $0 \le \tilde{f} \le 1$ and $\|f - \tilde{f}\|_{\square} \le \eps$.
    Then
    \[
      \left|
        \Ebb_{x, d} \prod_{j = 0}^{3} f(x + jd)
        - \Ebb \prod_{j = 0}^{3} \tilde{f}(x + jd)
      \right|
      \le \gamma
    .\]
  \end{fccoro}

  But also, by \cref{coro:3.8}, $\Ebb \tilde{f} = \Ebb f \ge \delta$, so by the
  weighted Szemerédi Theorem we get the result.
\end{proof}