%! TEX root = Ramsey.tex
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% 03/12/2024 11AM

  Let's prove that there is a colouring of
  $\Rbb$ such that $\sum_{i = 1}^{n} c_i y_i = b$
  does not have a monochromatic solution.
  Let $\sum_{i = 1}^{n - 1}
  c_i(y_i - y_n) = b$. By rescaling, we may assume
  $b = \half$. Now we split $[0, 1)$ into intervals
  $[0, \delta), [\delta, 2\delta), \ldots$ where
  $\delta$ is very small. Let
  \[
    c(y) = (\text{interval where $\{c_1 y\}$ is},
    \text{interval where $\{c_2 y\}$ is}, \ldots)
  .\]
  A $\left( \left\lfloor \frac{1}{\delta}
  \right\rfloor \right)^{n - 1}$-colouring.

  Assume $y_1, \ldots, y_{n - 1}$ monochromatic
  under $c$ and such that $\sum c_i (y_i - y_n) =
  \half$.

  Hence the sum is within $(n - 1)\delta$ of an
  integer, so not $\half$, if $\delta$ small
  enough.
\end{proof}

We have showed \gls{eram} implies \gls{spher}.

What about \gls{spher} implies \gls{eram}?

Still an open question (1975).

What is known:
\begin{enumerate}[(1)]
  \item Triangles are \gls{eram}, simplices are
    \gls{eram}, (old stuff). In 1991, Kriz showed
    that a regular pentagon is \gls{eram} and that
    any regular $m$-gon is \gls{eram}. His proof
    is unbelievably \emph{clever}!
\end{enumerate}

\textbf{Aim:} To show tht $X = \{1, 2, \ldots,
m\} =$ a regular $m$-gon is \gls{eram}.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/048b0300d21c4f22.png}
\end{center}
Roughly speaking:
\begin{enumerate}[(1)]
  \item First find a copy of $X$ such
    that $1$ and $2$ are monochromatic.
  \item Use a product argument to get a copy of
    $X^N$ (with $N$ very large), such that the
    colouring is invariant under $1, 2$. e.g.
    \begin{center}
      \includegraphics[width=0.6\linewidth]{images/16a72d450f7343d1.png}
    \end{center}
  \item The above plus some clever stuff to find a
    copy of $X$ on which $1, 2, 3$ is
    monochromatic.
\end{enumerate}

\begin{definition*}[$A$-invariant]
  For a finite $A \subseteq X$, we say that a
  colouring of $X^n$ is
  $A$-invariant if
  it is invariant under chaning the coordinates
  within $A$. i.e. for $x = (x_1, \ldots, x_n)$,
  $x' = (x_1', \ldots, x_n')$ if $\forall i$
  either $x_i = x_i'$ or $x_i, x_i' \in A$ implies
  $c(x') = c(x)$.
\end{definition*}

\begin{fcprop}[Our product argument]
  \label{prop:3.7}
  % Proposition 3.7
  Assuming:
  - $X \subseteq \Rbb^d$
  - $A \subseteq X$
  - $\forall k, \exists$ a finite $S \subseteq
  \Rbb^e$ such that whenever $S$ is $k$-coloured,
  there exists a copy of $X$ that is constant on
  $A$
  Then:
  for all $n, k$, there exists $S'$ finite such
  that whenever $S'$ is $k$-coloured there exists
  a copy of $X^n$ that is $A$-invariant.
\end{fcprop}

``boosting from $A$-constant to $A$-invariant.''

\begin{proof}
  (Yawn, product argument\ldots)

  We will by induction on $n$ (and all $k$ at
  once).

  $n = 1$ is just the assumption.

  Suppose true for $n - 1$. Fix $k$.
  Let $S$ be a finite
  set such that whenever $S$ is $k^{|X|}$
  coloured, there exists an $A$-invariant copy of
  $X^{n - 1}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/db0468dee806420b.png}
  \end{center}
  $T$ a finite set such that whenever
  $T$ is $k^{|S|}$-coloured, there exists a copy
  of $X$ with $A$ monochromatic.

  \textbf{Claim:} $S \times T$ works for $X^n$.

  By definition of $T$, if we look at $c'(s, t) =
  (c(s_1, t), c(s_2, t), \ldots, c(s_{|S|}, t))$,
  a $k^{|S|}$-colouring. Thus there exists a copy
  of $X$ with $A$ monochromatic.

  This induces a colouring of $S$ as follows:
  $c''(s) = (c(s, a), c(s, x_1), \ldots, c(s,
  x_{|X| - |A|}))$ for some $a \in A$
  (note $c(s, a)$ is well-defined). This is a
  $k^{|X| - |A| + 1}$-colouring.

  Therefore by the choice of $n$, there exists a
  copy of $X^{n - 1}$ that is $A$-invariant.

  Then we are done as this copy of $X$ in $T$ is
  $A$-invariant.
\end{proof}

Next time:

\begin{theorem}[Kriz Theorem]
  \label{thm:3.8}
  % Theorem 3.8
  Every regular $m$-gon is \gls{eram}.
\end{theorem}

\begin{note*}
  We will show that we can find $(1, 2, \ldots,
  m), (2, 3, \ldots, m, 1), (3, 4, \ldots, m, 1,
  2), \ldots, (m, 1, 2, \ldots, m - 1)$
  monochromatic, which is a copy of $X$, but
  scaled by $\sqrt{m}$ (which is fine as
  $\sqrt{m}$ is constant).
\end{note*}
