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\newpage
\section{Isoperimetric Inequalities}

``How do we minimise the boundary of a set of
given size?''

\begin{example*}
  Among all subsets of $\Rbb^2$ of given area, the
  disc minimises the perimeter.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/dff7eeedaec947f3.png}
  \end{center}
  Among all subsets of $\Rbb^3$of given volume,
  the solid sphere minimises the surface area.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/b626d72cbaee461d.png}
  \end{center}
  Among all subsets of $S^2$ of given surface
  area, the circular arc has the smallest
  perimeter.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/93edbd0a48a34672.png}
  \end{center}
\end{example*}

\begin{definition*}[Boundary in a graph]
  \glssymboldefn{b}%
  \glsnoundefn{bound}{boundary}{boundaries}%
  For a set $A$ of vertices of a graph $G$, the
  \emph{boundary} of $A$ is
  \[
    b(A) = \{x \in G : x \notin A, xy \in E
    \text{ for some $y \in A$}\}
  .\]
\end{definition*}

\begin{example*}
  Here, if $A = \{1, 2, 4\}$, then $\bound(A) = \{3,
  5\}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/8476c40faa0a45f3.png}
  \end{center}
\end{example*}

\begin{definition*}[Isoperimetric inequality]
  An \emph{isoperimetric inequality} on $G$ is an
  inequality of the form
  \[
    |\bound(A)| \ge f(|A|) \qquad \forall A \subset G
  ,\]
  for some function $f$.
\end{definition*}

\begin{definition*}[Neighbourhood]
  \glssymboldefn{neigh}%
  \glsnoundefn{neigh}{neighbourhood}{neighbourhoods}%
  Often simpler to look at the
  \emph{neighbourhood} of $A$: $N(A) = A \cup
  \bound(A)$. So
  \[
    N(A) = \{x \in G : d(x, A) \le 1\}
  .\]
\end{definition*}

A good example for $A$ might be a ball $B(x, r) =
\{y \in G : d(x, y) \le r\}$. What happens for
$\Q_n$?

\begin{example*}
  $|A| = 4$ in $\Q_3$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/8b2e165291284932.png}
  \end{center}
\end{example*}

Good guess that balls are best, i.e. sets of the
form $B(\emptyset, r) = X\rsubs[\le r] =
X\rsubs[0] \cup X\rsubs[1] \cup \cdots \cup
X\rsubs$.

What if $|X\rsubs[\le r]| < |A| < |X\rsubs[r +
1]|$?

Guess: take $A$ with $X\rsubs[\le r] < A <
X\rsubs[\le r + 1]$. If $A = X\rsubs[\le r] \cup
B$, where $B \subset X\rsubs[r + 1]$, then
$\bound(A) = (X\rsubs[r + 1] - B) \cup \ushadow
B$. So we'd take $B$ to be an initial segment of
\gls{lex} (by \nameref{KK}).

This suggests...

\glsnoundefn{simpord}{simplicial ordering}{NA}%
In the \emph{simplicial ordering} on $\powset X$,
we set $x < y$ if \emph{either} $|x| < |y|$
\emph{or} $|x| = |y|$ and $x < y$ in \gls{lex}.

\textbf{Aim:} initial segments of \gls{simpord}
minimise the \gls{bound}.

\begin{definition*}[$i$-sections]
  \glsnoundefn{nsect}{$n$-section}{$n$-sections}%
  \glssymboldefn{lowerupperlayers}%
  \glssymboldefn{Ci}%
  For $A \subset \powset X$ and $1 \le i \le n$,
  the \emph{$i$-sections} of $A$ are the families
  $A_-^{(i)}, A_+^{(i)} \subset \powset{X - i}$
  given by:
  \begin{align*}
    A_-^{(i)}
    &= \{x \in A : i \notin x\} \\
    A_+^{(i)}
    &= \{x - i : x \in A, i \in x\}
  \end{align*}
  The \emph{$i$-compression} of $A$ in the family
  $C_i(A) \subset \powset X$ given by:
  \begin{itemize}
    \item $(C_i(A))_-^{(i)}$ is the first
      $|A_-^{(i)}|$ elements of \gls{simpord} on
      $\powset{X - i}$
    \item $(C_i(A))_+^{(i)}$ is the first
      $|A_+^{(i)}|$ elements of \gls{simpord} on
      $\powset{X - i}$
  \end{itemize}
\end{definition*}

\begin{center}
  \includegraphics[width=0.5\linewidth]{images/1bc1e5e9f2df4554.png}
\end{center}

\begin{example*}
  \phantom{}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/734b7509a6d74a11.png}
  \end{center}
  \glsadjdefn{icomp}{$i$-compressed}{family}%
  Certainly $|\Ci(A)| = |A|$. Say $A$ is
  \emph{$i$-compressed} if $\Ci(A) = A$. Also,
  $\Ci(A)$ ``looks more like'' a Hamming ball than
  $A$ does.

  \glsnoundefn{hball}{Hamming ball}{Hamming balls}
  Here, a \emph{Hamming ball} is a family $A$ with
  $X\rsubs[\le r] \subset A \subset X\rsubs[\le r
  + 1]$, for some $r$.
\end{example*}

\begin{fcthm}[Harper's Theorem]
  \label{harper}
  % Theorem 2 1
  Assuming:
  - $A \subset \Q_n$
  - $C$ the initial segment of \gls{simpord}
  with $|C| = |A|$
  Then:
  $|\neigh(A)| \ge |\neigh(C)|$. In particular, if
  $|A| = \sum_{i = 0}^{r} {n \choose i}$ then
  $|\neigh(A)| \ge \sum_{i = 0}^{r + 1} {n \choose
  i}$.
\end{fcthm}

\begin{center}
  \includegraphics[width=0.4\linewidth]{images/73422fe3cccc4ad8.png}
\end{center}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If we knew $A$ was a \gls{hball}, then
      we would be done by \nameref{KK}.
    \item Conversely, \nameref{harper} implies
      \nameref{KK}: given $B \subset X\rsubs$,
      then apply \nameref{harper} to $A =
      X\rsubs[\le r - 1] \cup B$.
  \end{enumerate}
\end{remark*}

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

  Given $n > 1$, and $A \subset \Q_n$, and $1 \le
  i \le n$.

  \textbf{Claim:} $|\neigh(\Ci(A))| \le
  |\neigh(A)|$.

  Proof of claim: Write $B$ for $\Ci(A)$. We have
  \begin{align*}
    \neigh(A)\lowerl
    &= \neigh(A\lowerl) \cup A\upperl \\
    \neigh(A)\upperl
    &= \neigh(A\upperl) \cup A\lowerl
  \end{align*}
  and of course
  \begin{align*}
    \neigh(B)\lowerl
    &= \neigh(B\lowerl) \cup B\upperl \\
    \neigh(B)\upperl
    &= \neigh(B\upperl) \cup B \lowerl
  \end{align*}
  Now, $|B\upperl| = |A\upperl|$ and
  $|\neigh(B\lowerl)| \le |\neigh(A\lowerl)|$ (by
  the induction hypothesis). But $B\upperl$ is an
  initial segment of \gls{simpord}, and
  $\neigh(B\lowerl)$ is an initial segment of
  \gls{simpord} (as neighbourhood of initial
  segment is an initial segment).

  So then
  $B\upperl$ and $\neigh(B\lowerl)$ are
  \emph{nested} (one conatined in the other).
  Hence $|\neigh(B)\lowerl| \le
  |\neigh(A)\lowerl|$. Similarly,
  $|\neigh(B)\upperl| \le |\neigh(A)\upperl|$.

  Hence $|\neigh(B)| \le |\neigh(A)|$, which
  completes the proof of our claim.
\end{proof}