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% 28/11/2024 09AM

Remarkably, cannot beat linear.

\begin{fcprop}[]
  \label{prop_3_3}
  % Proposition 3.3
  Assuming:
  - $r$ is odd
  - $A \subset X\rsubs$ such that $|x \cap y|$
  even for each distinct $x, y \in A$
  Then:
  $|A| \le n$.
\end{fcprop}

\textbf{Idea:} Find $|A|$ linearly independent
vectors in a vector space of dimension $n$, namely
$\Q_n$.

\begin{proof}
  View $\powset X$ as $\Zbb_2^n$, the
  $n$-dimensional space over $\Zbb_2$ (the field
  of order $2$). By identifying $x$ with $\ol{x}$,
  its characteristic sequence (e.g. $1011000
  \ldots$ for $\{1, 3, 4\}$).

  We have $(\ol{x}, \ol{x}) \neq 0$ for each $x$,
  as $r$ is odd ($(\blank, \blank)$ is the usual
  dot-product).

  Also, $(\ol{x}, \ol{y}) = 0$ for distinct $x, y
  \in A$ (as $|x \cap y|$ even).

  Hence the $\ol{x}$, $x \in A$ are linearly
  independent (if $\sum \lambda_i \ol{x_i} = 0$,
  dot with $\ol{x_j}$ to get $\lambda_j = 0$).
\end{proof}

\begin{remark*}
  Hence also if $A \subset X\rsubs$, $r$ even,
  with $|x \cap y|$ odd for all distinct $x, y \in
  A$, then $|A| \le n + 1$ -- just add $n + 1$ to
  each $x \in A$ and apply \cref{prop_3_3} with $X
  = [n + 1]$.
\end{remark*}

Does this modulo $2$ behaviour generalise?

Now show: $s$ allowed values for $|x \cap y|$
modulo $p$ implies $|A| \le$ polynomial of degree
$s$.

\begin{fcthm}[Frankl-Wilson Theorem]
  \label{thm_3_4}
  % Theorem 3.4
  Assuming:
  - $p$ is prime
  - $\lambda_1, \ldots, \lambda_s$ ($s \le r$)
  - $\lambda_i \not\equiv r \pmod p$ for each $i$
  - $A \subset X\rsubs$ such that for all distinct
  $x, y \in A$ have $|x \cap y| \equiv \lambda_i
  \pmod p$ for some $i$
  Then:
  $|A| \le {n \choose s}$.
\end{fcthm}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item This bound is a \emph{polynomial} in $S$
      (as $r$ vares)!
    \item Bound is essentially best possible: can
      achieve $|A| = {n \choose n - r + s} \sim {n
      \choose s}$ (see picture).
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/2d9c58afc5a04284.png}
      \end{center}
    \item Do need no $\lambda_i \equiv r \pmod p$.
      Indeed, if $n = a + \lambda p$ ($0 \le a \le
      p - 1$) then can have $A \subset\rsubs[a +
      kp]$ with $|A| = {\lambda \choose k}$ (not a
      polynomial in $n$, as we can choose any $k$) and
      all $|x \cap y| \equiv a \pmod p$.
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/502742f6a72f4cd7.png}
      \end{center}
  \end{enumerate}
\end{remark*}

\textbf{Idea:} Try to find $|A|$ linearly
independent points in a vector space of dimension
${n \choose s}$, by somehow ``applying the
polynomial $(t - \lambda_1) \cdots (t -
\lambda_s)$ to $|x \cap y|$''.

\begin{proof}
  For each $i \le j$, let $M(i, j)$ be the ${n
  \choose i} \times {n \choose j}$ matrix, with
  rows indexed by $X\rsubs[i]$, columns indexed by
  $X\rsubs[j]$, with
  \[
    M(i, j)_{xy} = \begin{cases}
      1 & \text{if $x \subset y$} \\
      0 & \text{otherwise}
    \end{cases}
  \]
  for each $x \in X\rsubs[i]$, $y \in X\rsubs[j]$.
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/9b1dc4629c6740aa.png}
  \end{center}
  Let $V$ be the vector space (over $\Rbb$)
  spanned by the rows of $M(s, r)$. So $\dim V \le
  {n \choose s}$.

  For $i \le s$, consider $M(i, s) M(s, r)$ (note
  each row belongs to $V$, as we
  \emph{premultiplied} $M(s, r)$ by a matrix). For
  $x \in X\rsubs[i]$, $y \in X\rsubs[r]$:
  \begin{align*}
    (M(i, s) M(s, r))_{xy}
    &= \text{\# of $s$-sets $z$ with $x \subset z$
    and $z \subset y$} \\
    &= \begin{cases}
      0 & \text{if $x \not \subset y$} \\
      {r - i \choose s - i} & \text{if $x \subset
      y$}
    \end{cases}
  \end{align*}
  So
  \[
    M(i, s) M(s, r)
    = {r - i \choose s - i} M(i, r)
  \]
  so all rows of $M(i, r)$ belong to $V$.

  Let $M(i) = M(i, r)^\top M(i, r)$ (note each row
  is in $V$).

  For $x, y \in X\rsubs$, have
  \begin{align*}
    M(i)_{xy}
    &= \#\text{$i$-sets $z$ with $z \subset x$, $z
    \subset y$} \\
    &= {|x \cap y| \choose i}
  \end{align*}
  Write the integer polynomial $(t - \lambda_1)
  \cdots (t - \lambda_s)$ as $\sum_{i = 0}^{s} a_i
  {t \choose i}$, with $a_i \in \Zbb$ -- possible because $t(t - 1)
  \cdots (t - i + 1) = i! {t \choose i}$.
  
  Let $M = \sum_{i = 0}^{s} a_i M(i)$ (each
  row is in $V$).

  Then for all $x, y \in X\rsubs$:
  \[
    M_{xy} = \sum_{i} a_i {|x \cap y| \choose i}
    = (|x \cap y| - \lambda_i) \cdots (|x \cap y|
    - \lambda_s)
  .\]
  So the submatrix of $M$ spanned by the rows and
  columns corresponding to the elements of $A$ is
  \[
    \begin{pmatrix}
      \not\equiv 0
      & 0
      & \cdots
      & 0 \\
      0
      & \not\equiv 0
      & \cdots
      & 0 \\
      \vdots
      & \vdots
      & \ddots
      & \vdots \\
      0
      & 0
      & \cdots
      & \not\equiv 0
    \end{pmatrix}
  .\]
  Hence the rows of $M$ corresponding to $A$ are
  linearly independent over $\Zbb_p$, so also over
  $\Zbb$, so also over $\Qbb$, so also over
  $\Rbb$.

  So $|A| \le \dim V \le {n \choose s}$.
\end{proof}
