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\section{Euclidean Ramsey}

Launched in 1970 by Erd\H{o}s, Graham, Montgomery,
Rothchild, Spencer and Straus.

Here we want actual copies of some objects.

Colourings of $\Rbb^n$.

Let us $2$-colour $\Rbb^2$. Then have 2 points of
distance $1$ of same colour (consider equilateral
triangle of side length $1$).

If we $3$-colour $\Rbb^3$, we can also get 2
points of distance $1$, by considering a regular
tetrahedron of side length $1$.

In general, if we $k$-colour $\Rbb^k$, then by
looking at the unit regular simplex $(k + 1)$,
then any $2$ have distance $1$ between them, so we
get 2 points having the same colour and being unit
distance apart.

\begin{definition*}[Isometric copy]
  We say $X'$ is an isometric copy of $X$ if there
  exists $\phi : X \to X'$ bijection such that
  $d(x, y) = d(\phi(x), \phi(y))$.
\end{definition*}

\begin{definition*}
  \glsadjdefn{eram}{Ramsey}{finite set}%
  We say that a set $X$ (finite) is \emph{Ramsey}
  (Euclidean Ramsey) if for all $k$ there exists a
  finite set $S \subseteq \Rbb^n$ ($n$ could be
  very big) such that whenever $S$ is
  $k$-coloured, there exists a monochromatic copy
  of $X$.
\end{definition*}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $\{0, 1\}$ is \gls{eram}: for $k$-colours,
      take a $k$-dimensional unit simplex.
    \item Unit equilateral triangle is \gls{eram}: for
      $k$-colours, can take $2k$-dimensional unit
      simplex.
    \item Similarly have that any $\{0, a\}$ is
      \gls{eram}.
    \item Similarly any regular simplex is
      \gls{eram}.
  \end{enumerate}
\end{example*}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $X$ is infinite, then can build a
      $2$-colouring of $\Rbb^n$ with no
      monochromatic $X$ (exercise).
    \item We took for $k$-colours $S$ to be in
      $\Rbb^k$. Can we do better? For $\{0, 1\}$,
      can we do it in $\Rbb$? Colour $x \mapsto
      \left\lfloor x \right\rfloor \bmod 2$. Need
      $2$ dimensions or higher.

      What about $\{0, 1\}$? Can we
      do it in $\Rbb^2$? Yes we can:
      \begin{center}
        \includegraphics[width=0.3\linewidth]{images/3aa811923cc343ee.png}
      \end{center}
      This shows $\chi(\Rbb^2) \ge 4$ (max
      $\chi(G)$ where $G$ is a graph on $\Rbb^2$
      iwth $x \sim y$ if and only if $d(x, y) =
      1$).

      Up until 1990, $4 \le \chi(\Rbb^2) \le 7$
      (by using hexagonal colouring idea).

      De Grey -- 2018: showed $\chi(\Rbb^2) \le 5$
      using a graph on $\approx 1500$ vertices.
      Uses nice ideas and computer assistance.

      In general, $1.1^n \le \chi(\Rbb^n) \le
      3^n$. Lower bound is hard -- by
      Frankl-Wilson. Upper bound is by a type of
      hexagonal colouring, by Posy.
  \end{enumerate}
\end{remark*}

\begin{proposition}
  \label{prop:3.1}
  % Proposition 3.1
  $X$ is \glsref[eram]{Euclidean Ramsey} if and
  only if for all $k$, there exists $n$ such that
  whenever $\Rbb^n$ is $k$-coloured, there exists
  a monochromatic copy of $X$.
\end{proposition}

\begin{iffproof}
  \rightimpl
  If $X$ is \glsref[eram]{Euclidean Ramsey} then
  take $S$ finite in $\Rbb^n$ (for $k$-colours).

  \leftimpl
  We know that for all $k$-colourings of $\Rbb^n$,
  there exists a monochromatic copy of $X$ (by
  compactness). Suppose not. Therefore, for any
  set $S$ finite in $\Rbb^n$, there exists a bad
  colouring (i.e. not a copy of $X$
  monochromatic). Space of all $k$-colourings is
  $[k]^{\Rbb^n}$, which is compact by Tychonoff.
  Let
  \[
    C_{X'} = \{\text{colourings that do not make
    $X$ monochromatic}\}
  .\]
  This is a closed set. Let
  $\{C_{X'}\}_{\text{$X'$ a copy of $X$}}$. It has
  the finite intersection property, because any
  finite $S$ has a bad $k$-colouring. Hence the
  intersection of all $C_{X'}$ is non-empty. Hence
  there exists a colouring of $\Rbb^m$ with no
  monochromatic $X$, contradiction.
\end{iffproof}

How can we generate Ramsey sets?

\begin{fclemma}[]
  \label{lemma:3.2}
  % Lemma 3.2
  Assuming:
  - $X \subset \Rbb^n$ \gls{eram}
  - $Y \subset \Rbb^m$ \gls{eram}
  Then:
  $X \times Y \subset \Rbb^{n + m}$ is also
  \gls{eram}.
\end{fclemma}

\begin{remark*}
  \phantom{}
  \begin{itemize}
    \item $\{0, a\}$ is \gls{eram}, and so is
      $\{0, b\}$. So $a \times b$ rectangles are
      \gls{eram}. In particular, any right angled
      triangle is \gls{eram}.

      By considering $\{0, a\} \times \{0, b\}
      \times \{0, c\}$, we can acute angled
      triangles:
      \begin{center}
        \includegraphics[width=0.3\linewidth]{images/53bc088825854e6f.png}
      \end{center}
  \end{itemize}
\end{remark*}

\begin{proof}
  Let $k$ be a colouring of $S \times T$, where
  $S$ is $k$-\gls{eram} for $X$ and $T$ is
  $k^{|S|}$-\gls{eram} for $T$.

  We $k^{|S|}$-colour $T$ as follows:
  \[
    c''(t) = (c(s_1, t), c(s_2, t), \ldots,
    c(s_{|S|}, t))
  .\]
  By choice of $T$, there exists a monochromatic
  $Y'$ (copy of $Y$) with respect to $c'$, i.e.
  $c(s, y) = c(s, y')$ for all $y, y' \in Y$ and
  any $s \in S$.

  Now $k$-colour $S$ via $c''(s) = c''(s, y)$ for
  some $y \in Y'$ (which is well-defined by the
  above). By the choice of $X$, there exists
  monochromatic $X'$ with respect to $c''$, and
  hence $X' \times Y'$ is monochromatic with
  respect to $c$.
\end{proof}

\textbf{Homework:} Convince yourself that this is
a \emph{very} standard product argument.

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item In general to prove sets are \gls{eram}
      we will first embed them into other sets
      (with `cool' symmetry groups) and show that
      those sets are \gls{eram}.
    \item \textbf{Spoiler:}
      \begin{itemize}
        \item Obtuse triangles are
          \gls{eram} (twisted prism).
        \item Any regular $n$-gon is \gls{eram}.
        \item 3D Platonic solids.
      \end{itemize}
  \end{enumerate}
\end{remark*}

Next time: non-\gls{eram}. (think about $\{0, 1,
2\}$?)