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\begin{document}




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\frame{
\frametitle{\bf BILL, RECORD LECTURE!!!!}
BILL RECORD LECTURE!!!
}


\title{\bf\Huge Ramsey on $\omega^2$}
\author{
{\bf \Large  Results by Joanna Boyland,  William Gasarch, Nathan Hurtig, Robert Rust}\\
}

\maketitle



\begin{frame}\frametitle{\bf Notation}

\pause
\bigskip


\isblue{Def}
Let $\COL \colon \binom{A}{2} \into [1,000,000]$. 
Let $c\in\N$. 

\pause
\smallskip
A set $H\subseteq A$ is \isred{$c$-homogenous} if $\COL$ restricted to 
$\binom{H}{2}$ takes on at most $c$ values. (From now on \isred{$c$-homog}.) 

\pause
\bigskip

\isred{Def} If $L_1,L_2$ are linearly ordered sets then $L_1\equiv L_2$
means there is an order preserving bijection between $L_1$ and $L_2$. 

\end{frame} 

\begin{frame}\frametitle{\bf We Will Prove The Following Two Theorems} 

\pause
\isblue{Thm} $\forall \COL\colon \binom{\omega^2}{2}\into [1,000,000]$
$\exists$ a 4-homog $H\equiv \omega^2$. 

\pause
\bigskip
\isblue{Thm} $\exists \COL\colon \binom{\omega^2}{2}\into [4]$
Such that there is no 3-homog $H\equiv \omega^2$.
\end{frame}

\begin{frame}{ }

\centerline{\isred{\Huge \bf There is Always a  }}

\centerline{\isred{\Huge \bf 4-Homog Set}}

\end{frame} 




\begin{frame}\frametitle{\bf Theorem About Coloring $\omega^2$} 

\isblue{Thm} $\forall \COL\colon \binom{\omega^2}{2}\into [1,000,000]$
$\exists$ a 4-homog $H\equiv \omega^2$. 

\pause
\smallskip
We represent $\omega^2$ as follows, which is standard: 

\pause
\smallskip

\begin{tabular}{r r r r r}
$\om\cdot 1 + 1,$ & $\om\cdot 1 + 2,$ & $\om\cdot 1 + 3,$ & $\om\cdot 1 + 4,$ & \ldots\\
$\om\cdot 2 + 1,$ & $\om\cdot 2 + 2,$ & $\om\cdot 2 + 3,$ & $\om\cdot 2 + 4,$ & \ldots\\
$\om\cdot 3 + 1,$ & $\om\cdot 3 + 2,$ & $\om\cdot 3 + 3,$ & $\om\cdot 3 + 4,$ & \ldots\\
$\vdots$          & $\vdots$          & $\vdots$          & $\vdots$ &   $\vdots$ $\ddots$ \\ 
\end{tabular}

\pause
\smallskip
Edges within a copy of $\omega$ are called \isblue{internal}. 

\pause 
Edges between copies of $\omega$ are called \isblue{external}.

\end{frame} 

\begin{frame}\frametitle{\bf The Internal Edges}  

\pause
\smallskip
For each $i\in\N$ let

$$A_i=\{\om\cdot i + 1,\om\cdot i+ 2, \om\cdot i + 3, \ldots \}.$$

\pause
\smallskip
Apply infinite Ramsey to $\binom{A_i}{2}$ to get infinite $H_i\subseteq A_i$ 
and color $c_i$ such that $\COL$ on $\binom{H_i}{2}$ is always $c_i$. 
Replace $A_i$ with $H_i$. 

\pause
\smallskip
Some color $c$ appears as $c_i$ infinitely often. 

\pause 
\smallskip
\isred{Kill all the $A_j$'s that disagree}

\pause
\smallskip
We assume all Internal edges are $\R$. 

\end{frame}

\begin{frame}\frametitle{\bf Where We Are Now?} 

\pause 
We relabeled so that the $\omega^2$ is represented by 

\pause
\smallskip

\begin{tabular}{r r r r r}
$\om\cdot 1 + 1,$ & $\om\cdot 1 + 2,$ & $\om\cdot 1 + 3,$ & $\om\cdot 1 + 4,$ & \ldots\\
$\om\cdot 2 + 1,$ & $\om\cdot 2 + 2,$ & $\om\cdot 2 + 3,$ & $\om\cdot 2 + 4,$ & \ldots\\
$\om\cdot 3 + 1,$ & $\om\cdot 3 + 2,$ & $\om\cdot 3 + 3,$ & $\om\cdot 3 + 4,$ & \ldots\\
$\vdots$          & $\vdots$          & $\vdots$          & $\vdots$ &   $\vdots$ $\ddots$ \\ 
\end{tabular}

\pause
Within any copy of $\omega$ all of the edges are $\R$. 

\end{frame} 


\begin{frame}\frametitle{\bf Coloring 4-Sets} 

\pause
\smallskip
\isblue{Plan} 
\begin{itemize}
\item
\pause 
We will create a clever coloring $\COL'\colon \binom{\N}{4}\into [1,000,000]^3$.
\item 
\pause
Let $H'$ be an infinite homog set (relative to $\COL'$).
\item
\pause 
We will use $H'$ to create $H$, such that 

\pause 
(a) $H$ is 4-homog relative to $\COL$, and 

\pause
(b) $H\equiv \om^2$. 
\end{itemize}

\end{frame} 


\begin{frame}\frametitle{\bf Coloring 4-Sets (cont)} 

\pause 
\smallskip
We color $x_1<x_2<x_3<x_4$ with the 3-tuple of colors below:

\pause
$\COL(\om\cdot x_1 + x_2, \om\cdot x_3 + x_4)$

\pause 
$\COL(\om\cdot x_1 + x_3, \om\cdot x_2 + x_4)$

\pause 
$\COL(\om\cdot x_1 + x_4, \om\cdot x_2 + x_3)$

\pause 
By the 4-ary Ramsey Theorem there is a homog set $H'$. 

\pause
\smallskip
Let $(\B,\G,\Y)$ be the color of the homog set. 

\end{frame} 

\begin{frame}\frametitle{\bf The Set We Begin With} 
\pause 
Let $H'=\{h_1 < h_2 < h_3 < h_4< \cdots \}$

\pause
\smallskip
Let $H''$ be 

\pause
\begin{tabular}{r r r r r}
$\om\cdot h_{1}+ h_{1},$ & $\om\cdot h_{1}+ h_{2},$ & $\om\cdot h_{1}+ h_{3},$ & $\om\cdot  h_{1}+ h_{4},$ & \ldots\\
$\om\cdot h_{2}+ h_{1},$ & $\om\cdot h_{2}+ h_{2},$ & $\om\cdot h_{2}+ h_{3},$ & $\om\cdot  h_{2}+ h_{4},$ & \ldots\\
$\om\cdot h_{3}+ h_{1},$ & $\om\cdot h_{3}+ h_{2},$ & $\om\cdot h_{3}+ h_{3},$ & $\om\cdot  h_{3}+ h_{4},$ & \ldots\\
$\vdots$          & $\vdots$          & $\vdots$          & $\vdots$ \\ 
\end{tabular}
\pause

We will thin out $H''$ to get the desired $H$.

\pause
For which external edges $(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)$ 
do we get 

\pause 

$$\COL(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)\in\{\B,\G,\Y\}$$

\pause

from $H'$ being homog of color $(\B,\G,\Y)$?  

\pause
We continue this on the next slide.

\end{frame}

\begin{frame}{\bf Which External Edges Must Be in $\mathbf{\B,\G,\Y}$}
\pause
For which external edges $(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)$ 
do we get 

\pause 

$$\COL(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)\in\{\B,\G,\Y\}.$$

\pause

from $H'$ being homog of color $(\B,\G,\Y)$?  

\pause 
Since the edge is external we assume $h_i<h_k$.

\pause
Recall that $\COL'(x_1<x_2<x_3<x_4)=$
\pause

$\COL(\om\cdot x_1 + x_2, \om\cdot x_3 + x_4)$

$\COL(\om\cdot x_1 + x_3, \om\cdot x_2 + x_4)$

$\COL(\om\cdot x_1 + x_4, \om\cdot x_2 + x_3)$

\pause
\smallskip
If $h_i<h_j<h_k<h_\ell$ then $\COL'(h_i<h_j<h_k<h_\ell)=(\B,\G,\Y)$ so 
$\COL(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)=\B$.

\pause
\smallskip
If $h_i<h_k<h_j<h_\ell$ then $\COL'(h_i<h_k<h_j<h_\ell)=(\B,\G,\Y)$ so 
$\COL(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)=\G$.

\pause
\smallskip
If $h_i<h_\ell<h_j<h_k$ then $\COL'(h_i<h_\ell<h_j<h_k)=(\B,\G,\Y)$ so 
$\COL(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)=\Y$.

\end{frame} 

\begin{frame}\frametitle{\bf What Do We Need the $\mathbf{h_i}$'s To Look Like?}

\isblue{Need}  If $(\om\cdot h_i + h_j, \om\cdot h_k + h_\ell)$ is an external edge with
$h_i<h_k$ then either

\pause 
\smallskip
$h_i<h_j<h_k<h_\ell$, or 

\pause 
\smallskip
$h_i<h_k<h_j<h_\ell$, or 

\pause 
\smallskip
$h_i<h_\ell<h_j<h_k$.

\pause
\bigskip
It suffices to have 
\pause
\smallskip

$h_i<h_j$,

\pause
\smallskip

$h_k<h_\ell$,  and

\pause
\smallskip
$h_i,h_j,h_k,h_\ell$ all different. 

\pause
\smallskip
Next slide has a thinned out version of $H''$ that suffices. 

\end{frame} 


\begin{frame}\frametitle{\bf The Homog Set} 

\pause
\smallskip
Recall  $H'=\{h_1 < h_2 < h_3 < h_4< \cdots \}$

\pause
\smallskip
We define $H$ as: 

\pause 

\begin{tabular}{r r r r r}
$\om\cdot h_{2^1}+ h_{2^2},$ & $\om\cdot h_{2^1}+ h_{2^3},$ & $\om\cdot h_{2^1}+ h_{2^4},$ & $\om\cdot  h_{2^1}+ h_{2^5},$ & \ldots\\
$\om\cdot h_{3^1}+ h_{3^2},$ & $\om\cdot h_{3^1}+ h_{3^3},$ & $\om\cdot h_{3^1}+ h_{3^4},$ & $\om\cdot  h_{3^1}+ h_{3^5},$ & \ldots\\
$\om\cdot h_{5^1}+ h_{5^2},$ & $\om\cdot h_{5^1}+ h_{5^3},$ & $\om\cdot h_{5^1}+ h_{5^4},$ & $\om\cdot  h_{5^1}+ h_{5^5},$ & \ldots\\
$\vdots$          & $\vdots$          & $\vdots$          & $\vdots$ \\ 
\end{tabular}

\pause
\bigskip

$H$ is 4-homog and, 

\pause 
$H\equiv \omega^2$

\pause
So we are done!

\end{frame}

\begin{frame}{ }

\centerline{\isred{\Huge \bf There is a Coloring With  }}

\centerline{\isred{\Huge \bf No 3-Homog Set}}

\end{frame} 

\begin{frame}\frametitle{\bf There is a Coloring With No 3-Homog} 

\pause
\isblue{Thm} $\exists \COL\colon \binom{\omega^2}{2}\into [4]$
such that there is no 3-homog $H\equiv \omega^2$.

\pause
We use this representation

\pause

\begin{tabular}{r r r r r}
$\om\cdot 2^1 + 2^2,$ & $\om\cdot 2^1 + 2^3,$ & $\om\cdot 2^1 + 2^4,$ & $\om\cdot 2^1 + 2^5,$ & \ldots\\
$\om\cdot 3^1 + 3^2,$ & $\om\cdot 3^1 + 3^3,$ & $\om\cdot 3^1 + 3^4,$ & $\om\cdot 3^1 + 3^5,$ & \ldots\\
$\om\cdot 5^1 + 5^2,$ & $\om\cdot 5^1 + 5^3,$ & $\om\cdot 5^1 + 5^4,$ & $\om\cdot 5^1 + 5^5,$ & \ldots\\
$\vdots$          & $\vdots$          & $\vdots$          & $\vdots$ &   $\vdots$ \ldots \\ 
\end{tabular}

\end{frame}

\begin{frame}\frametitle{\bf The Coloring} 

We define $\COL(\om \cdot a + b, \om \cdot c + d)$.

\pause 
1) By the representation, $a<b$, $c<d$, and $b,c,d$ are all different. 

\pause
2) We assume $a\le c$. 

\pause
\smallskip

\begin{equation*}
\COL(\om \cdot a + b, \om \cdot c + d) =
\begin{cases}
\R  & \hbox{If $a=c$}\\
 1  & \hbox{If $a<c$ and $b<c<d$}\\
 2  & \hbox{If $a<c$ and $c<b<d$}\\
 3  & \hbox{If $a<c$ and $c<d<b$}\\
\end{cases}
\end{equation*}

\pause
\smallskip
The proof that there is no 3-homog set is left to the reader.

\end{frame} 

\end{document} 

