%! TEX root = Combi.tex
% vim: tw=50
% 12/11/2024 09AM

\subsection{Edge-isoperimetric inequalities}

\glsnoundefn{edgeb}{edge-boundary}{edge-boundaries}%
\glssymboldefn{bounde}%
For a subset $A$ of vertices of a graph $G$, the
\emph{edge-boundary} of $A$ is
\[
  \partial_e A = \partial A = \{xy \in E : x \in
  A, y \notin A\}
.\]
\begin{center}
  \includegraphics[width=0.3\linewidth]{images/9b769489a5c14552.png}
\end{center}

An inequality of the form: $|\bounde A| \ge
f(|A|)$ for all $A \subset G$ is an
\emph{edge-isoperimetric inequality} on $G$.

What happens in $\Q_n$? Given $|A|$, which $A
\subset \Q_n$ should we take, to minimise
$|\bounde A|$?

\begin{example*}
  $|A| = 4$ in $\Q_3$:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e2a9c7bb8538460a.png}
  \end{center}
\end{example*}

This suggests that maybe subcubes are best.

What if $A \subset \Q_n$? with $2^k < |A| < 2^{k +
1}$? Natural to take $A = \powset{[k]} \cup
\{\text{some stuff containing $k + 1$}\}$. Suppose
we are in $\Q_4$, and considering $|A| > 2^3$, eg
$|A| = 12$. We might take the whole of the bottom
layer, and then stuff in the upper layer. Note
that the size of the boundary will be the number
of up edges (which is $12 - 2^3$, a constant), plus the
number of edges in the top layer. So we just want
to minimise the number of edges in the top layer,
i.e. find $A' \subset \Q_3$ with $|A'|$ with
minimal boundary.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/6bff501fb86d425c.png}
\end{center}

So we define: for $x, y \in \Q_n$, $x \neq y$,
say $x < y$ in the \emph{binary ordering} on
$\Q_n$ if $\max x \Delta y \in y$. Equivalently,
$x < y$ if and only if $\sum_{i \in x} 2^i <
\sum_{i \in y} 2^i$. ``Go up in subcubes''.

\begin{example*}
  In $\Q_3$: $\emptyset, 1, 2, 12, 3, 13, 23,
  123$.
\end{example*}

\glssymboldefn{Bi}%
\glsadjdefn{ibced}{$i$-binary compressed}{set}%
For $A \subset \Q_n$, $1 \le i \le n$, we define
the \emph{$i$-binary compression} $B_i(A) \subset
\Q_n$ by giving its $i$-sections:
\begin{align*}
  (B_i(A))_-^{(i)}
  &= \text{initial segment of binary on $\powset{X
  - i}$ of size $|A_-^{(i)}|$} \\
  (B_i(A))_+^{(i)}
  &= \text{initial segment of binary on $\powset{X
  - i}$ of size $|A_+^{(i)}|$}
\end{align*}
so $|B_i(A)| = |A|$. Say $A$ is \emph{$i$-binary
compressed} if $B_i(A) = A$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/f690153b7d4e4bcd.png}
\end{center}

\begin{fcthm}[Edge-isoperimetric inequality in
  $Q_n$]
  \label{thm_2_8}
  % Theorem 2 8
  Assuming:
  - $A \subset \Q_n$
  - let $C$ the initial segment of binary on
  $\Q_n$ with $|C| = |A|$
  Then:
  $|\bounde C| \le |\bounde A|$. In particular: if
  $|A| = 2^k$ then $|\bounde A| \ge 2^k(n - k)$.
\end{fcthm}

\begin{remark*}
  Sometimes called the ``Theorem of Harper,
  Lindsey, Bernstein \& Hart''.
\end{remark*}

\begin{proof}
  Induction on $n$. $n = 1$ trivial.

  For $n > 1$, $A \subset \Q_n$, $1 \le i \le n$:

  \textbf{Claim:} $|\bounde \Bi(A)| \le |\bounde
  A|$.

  Proof of claim: write $B$ for $\Bi(A)$.
  .image
  Have
  \begin{picmath}
    |\bounde A| =
    \ub{|\bounde (A_-)|}_{\text{downstairs}} +
    \ub{|\bounde (A_+)|}_{\text{upstairs}} +
    \ub{|A_+ \Delta A_-|}_{\text{across}}
  \end{picmath} 
  Also
  \[
    |\bounde B| = |\bounde (B_-)| + |\bounde
    (B_+)| + |B_+ \Delta B_-|
  .\]
  Now, $|\bounde (B_-)| \le |\bounde (A_-)|$ and
  $|\bounde(B_+)| \le |\bounde(A_+)|$ (induction
  hypothesis). Also, the sets $B_+$ and $B_-$ are
  nested (one is contained inside the other),
  as each is an initial segment of binary
  on $\powset{X - i}$.

  Whence we certainly have $|B_+ \Delta B_-| \le
  |A_+ \Delta A_-|$. So $|\bounde B| \le |\bounde
  A|$.

  Define a sequence $A_0, A_1, \ldots \subset
  \Q_n$ as follows: set $A_0 = A$. Having defined
  $A_0, \ldots, A_k$, if $A_k$ is \gls{ibced} for
  all $n$ then stop the sequence with $A_k$. If
  not, choose $i$ with $\Bi(A) \neq A$ and put
  $A_{k + 1} = \Bi(A_k)$. Must terminate, as the
  function $k \mapsto \sum_{x \in A_k}
  (\text{position of $x$ in binary})$ is strictly
  decreasing.

  The final family $B = A_k$ satisfies $|B| =
  |A|$, $|\bounde B| \le |\bounde A|$, and $B$ is
  \gls{ibced} for all $i$.

  Note that $B$ need \emph{not} be an initial
  segment of binary, for example $\{\emptyset, 1,
  2, 3\} \subset \Q_3$.

  However:

  \begin{fclemma}[]
    % Lemma 2.9
    Assuming:
    - $B \subset \Q_n$ is \gls{ibced} for all $i$
    - $B$ not an initial segment of binary
    Then:
    $B = \powset{n - 1} - \{1, 2, 3, \ldots, n -
    1\} \cup \{n\}$ (``downstairs minus the last point, plus
    the first upstairs point'').
  \end{fclemma}

  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/5fe1964251d742ba.png}
  \end{center}
  
  (Then done, as clearly $|\bounde B| \ge |\bounde
  C|$, since $C = \powset{n - 1}$).

  \begin{proof}
    As $B$ not an initial segment, there exists $x
    < y$ with $x \notin B$, $y \in B$. Then for
    all $i$: cannot have $i \in x, y$, and cannot
    have $i \notin x, y$ (as $B$ is \gls{ibced}).

    Thus for each $y \in B$, there exists at most
    1 earlier $x \notin B$ (namely $x = y^c$).
    Also for each $x \notin B$ there is at most
    one later $y \in B$ (namely $y = x^c$).

    Then $x$ and $y$ adjacent (since $y$ is the
    unique element in $B$ after $x$, and $x$ is
    the unique element not in $B$ before $y$).

    So $B = \{z : z \le y\} - \{x\}$, where $x$ is
    the predecessor of $y$ and $y = x^c$.

    So must have $y = \{n\}$.
  \end{proof}

  This concludes the proof of \cref{thm_2_8}.
\end{proof}
