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\begin{document}
\centerline{Math 372 lecture for Friday, Week 2}
\bigskip

The goal for the next two class periods is to develop general tools which we
will apply to see that the boolean poset~$B_n$ is Sperner.
\medskip

Let~$P$ be a graded poset of rank~$n$.  For each~$0\leq i\leq n$, let~$P_i$ be
the~$i$-th level of~$P$, and let~$p_i=|P_i|$.  
An \emph{order matching from~$P_i$ to~$P_{i+1}$} is a function
\[
  \mu\colon P_i\hookrightarrow P_{i+1}
\]
such that~$x<\mu(x)$ for all~$x\in P_i$. An   
\emph{order matching from ~$P_{i+1}$ to~$P_i$} is a function
\[
  \mu\colon P_{i+1}\hookrightarrow P_{i}
\]
such that~$\mu(x)<x$ for all~$x\in P_{i+1}$. 
\medskip

{\bf Proposition 4.4.} 
Suppose there exists~$0\leq j\leq n$ and order matchings
\[
  P_0\xhookrightarrow{\mu_0}
  P_i\xhookrightarrow{\mu_2}\cdots\xhookrightarrow{\mu_{j-1}}P_j
  \xhookleftarrow{\mu_{j+1}}P_{j+1}\xhookleftarrow{\mu_{j+2}}\cdots\xhookleftarrow{\mu_n}P_n.
\]
Then~$P$ is unimodal and Sperner.
\smallskip

{\bf Proof.} 
If an order matching exists, since the~$\mu_i$ are injections, then
\[
  p_0\leq p_1\leq\cdots\leq p_j\geq p_{j+1}\geq\cdots\geq p_n,
\]
i.e.,~$P$ is unimodal.  Recall the Sperner property:
\[
  \max\left\{ |A|: A\text{ an antichain of~$P$} \right\}
  =\max\left\{ p_i:0\leq i\leq n \right\}.
\]
To see that~$P$ is Sperner:
\begin{enumerate}[noitemsep]
  \item In the Hasse diagram for~$P$, for each~$i\neq j$, color the edge
    connecting~$x$ to~$\mu_{i}(x)$ red for all~$x\in P_i$.
  \item Let~$G$ be the graph whose vertices are the elements of~$P$, and whose
    edges are the edges in the Hasse diagram that are colored red.  Some of the
    vertices in~$P_j$ may be isolated.  Color these isolated vertices red.
  \item The components of~$G$ partition~$P$ into disjoint chains (considering
    each isolated (red) vertex as a chain of length~$0$). Each of these chains
    meets~$P_j$ in a unique point.  So the number of chains is~$|P_j|$.
  \item Let~$A$ be an antichain in~$P$.  Then: (i) each~$a\in A$ is in one of
    the above chains, and (ii) each of the above chains contains at most one
    element of~$A$.  Thus, the number of chains above is at least~$|A|$.
  \item Done.
\end{enumerate}
See Figure~\ref{fig: 02friHasse} for an example.
\hfill$\Box$
\begin{figure}[ht] 
  \begin{center}
    \begin{tikzpicture}
      \node[ball] at (1*5/3,0) (01) {};
      \node[ball] at (2*5/3,0) (02) {};
      \node[ball] at (1*5/3,1) (11) {};
      \node[ball] at (2*5/3,1) (12) {};
      \node[ball] at (1*5/4,2) (21) {};
      \node[ball] at (2*5/4,2) (22) {};
      \node[ball] at (3*5/4,2) (23) {};
      \node[ball] at (0,3) (31) {};
      \node[ball] at (1,3) (32) {};
      \node[ball] at (2,3) (33) {};
      \node[ball] at (3,3) (34) {};
      \node[ball] at (4,3) (35) {};
      \node[ball] at (5,3) (36) {};
      \node[ball] at (1*5/6,4) (41) {};
      \node[ball] at (2*5/6,4) (42) {};
      \node[ball] at (3*5/6,4) (43) {};
      \node[ball] at (4*5/6,4) (44) {};
      \node[ball] at (5*5/6,4) (45) {};
      \node[ball] at (1*5/3,5) (51) {};
      \node[ball] at (2*5/3,5) (52) {};
      \draw (51)--(41);
      \draw[->,red, ultra thick] (51)--(42);
      \draw (51)--(43);
      \draw (52)--(43);
      \draw (52)--(44);
      \draw[->,red, ultra thick] (52)--(45);

      \draw[->,red, ultra thick] (41)--(31);
      \draw (41)--(32);
      \draw[->,red, ultra thick] (42)--(32);
      \draw (42)--(33);
      \draw[->,red, ultra thick] (43)--(33);
      \draw (44)--(35);
      \draw[->,red, ultra thick] (44)--(34);
      \draw (45)--(35);
      \draw[->,red, ultra thick] (45)--(36);
      \draw (01)--(12);
      \draw[->,red, ultra thick] (01)--(11);
      \draw[->,red, ultra thick] (02)--(12);
      \draw[->,red, ultra thick] (11)--(21);
      \draw (12)--(23);
      \draw[->,red, ultra thick] (12)--(22);
      \draw (21)--(32);
      \draw[->,red, ultra thick] (21)--(31);
      \draw (22)--(32);
      \draw (22)--(33);
      \draw[->,red, ultra thick] (22)--(34);
      \draw (23)--(34);
      \draw (23)--(35);
      \draw[->,red, ultra thick] (23)--(36);
      \begin{scope}[xshift=8cm]
      \node[ball] at (1*5/3,0) (01) {};
      \node[ball] at (2*5/3,0) (02) {};
      \node[ball] at (1*5/3,1) (11) {};
      \node[ball] at (2*5/3,1) (12) {};
      \node[ball] at (1*5/4,2) (21) {};
      \node[ball] at (2*5/4,2) (22) {};
      \node[ball] at (3*5/4,2) (23) {};
      \node[ball] at (0,3) (31) {};
      \node[ball] at (1,3) (32) {};
      \node[ball] at (2,3) (33) {};
      \node[ball] at (3,3) (34) {};
      \node[ball] at (4,3) (35) {};
      \node[ball] at (5,3) (36) {};
      \node[ball] at (1*5/6,4) (41) {};
      \node[ball] at (2*5/6,4) (42) {};
      \node[ball] at (3*5/6,4) (43) {};
      \node[ball] at (4*5/6,4) (44) {};
      \node[ball] at (5*5/6,4) (45) {};
      \node[ball] at (1*5/3,5) (51) {};
      \node[ball] at (2*5/3,5) (52) {};
      \draw[red, ultra thick] (51)--(42);
      \draw[red, ultra thick] (52)--(45);
      \draw[red, ultra thick] (41)--(31);
      \draw[red, ultra thick] (42)--(32);
      \draw[red, ultra thick] (43)--(33);
      \draw[red, ultra thick] (44)--(34);
      \draw[red, ultra thick] (45)--(36);
      \draw[red, ultra thick] (01)--(11);
      \draw[red, ultra thick] (02)--(12);
      \draw[red, ultra thick] (11)--(21);
      \draw[red, ultra thick] (12)--(22);
      \draw[red, ultra thick] (21)--(31);
      \draw[red, ultra thick] (22)--(34);
      \draw[red, ultra thick] (23)--(36);
      \draw[red,fill] (4,3) circle (0.8mm);
      \end{scope}
    \end{tikzpicture}
  \end{center}
  \caption{Order matchings and chains}\label{fig: 02friHasse} 
\end{figure}
\bigskip

{\bf Question.} How should we go about finding order relations as in
Proposition~4.4?
\bigskip

And \emph{order-raising operator} is a linear function $U\colon \R P_i\to \R P_{i+1}$ such that for all $x\in P_i$, we have that~$U(x)$ is a linear
combination of elements~$y\in P_{i+1}$ such that~$x\leq y$.  
\bigskip

{\bf Lemma 4.5.}
  Suppose $U\colon \R P_i\to\R P_{i+1}$ is an injective order-raising operator.
  Then there exists an order matching~$P_i\hookrightarrow P_{i+1}$.
  Similarly, suppose $U\colon \R P_{i}\to\R P_{i+1}$
  is a surjective order-raising operator.
  Then there exists order matching~$P_{i+1}\hookrightarrow P_{i}$.
\smallskip

{\bf Proof.} 
\begin{enumerate}
  \item Consider the matrix~$A$ representing~$U$.  Its columns are are indexed
    by~$P_i$ and its rows are indexed by~$P_{i+1}$.
    \begin{center}
      \begin{tikzpicture}
	\def\xs{0.1cm}
	\node at (0,0) {$
	  \left[\begin{array}{cccccc}
	      &&&&&\\
	      &\ \ \ \ &&&&\\
	      &&&&&\\
	      &&&&&\\
	      &&&&&\\
	      &&&&&\\
	      &&&&&\\
	  \end{array} \right]
	$};
	\node at (0,0) {\huge $\star$};
	\node at (-10*\xs,1.8) {$x_1$};
	\node at (-4*\xs,1.8) {$x_2$};
	\node at (2*\xs,1.8) {$\cdots$};
	\node at (9*\xs,1.8) {$x_{p_i}$};
	\node at (-19*\xs,1.3) {$y_1$};
	\node at (-19*\xs,0.6) {$y_2$};
	\node at (-19*\xs,0.1) {$\vdots$};
	\node at (-19*\xs,-0.5) {$y_{p_i}$};
	\node at (-19*\xs,-1.0) {$\vdots$};
	\node at (-19*\xs,-1.5) {$y_{p_{i+1}}$};
	\node at (-25*\xs,-2.3) (a) {$\R^{p_i}$};
	\node at (25*\xs,-2.3) (b) {$\R^{p_{i+1}}$};
	\draw[->] (a)--(b);
      \end{tikzpicture}
    \end{center}
  \item  Since~$U$ is injective, the columns of~$A$ are linearly independent.
    So~$A$ has rank~$p_i$.  Since column-rank equals row-rank, this means there
    is a selection of~$p_i$ linearly independent rows of~$A$.  Throwing out the other rows, we arrive
    at a square submatrix~$A'$ of~$A$ with rank~$p_i$ indexed by all of~$P_i$ and a
    subset of~$P_{i+1}$ of size~$p_i$.
    \begin{center}
      \begin{tikzpicture}
	\def\xs{0.1cm}
	\node at (0,0) {$
	  \left[\begin{array}{cccccc}
	      &&&&&\\
	      &\ \ \ \ &&&&\\
	      &&&&&\\
	      &&&&&\\
	  \end{array} \right]
	$};
	\node at (0,0) {\huge $\star$};
	\node at (-10*\xs,1.3) {$x_1$};
	\node at (-4*\xs,1.3) {$x_2$};
	\node at (2*\xs,1.3) {$\cdots$};
	\node at (9*\xs,1.3) {$x_{p_i}$};
	\node at (-19*\xs,0.8) {$y_{j_1}$};
	\node at (-19*\xs,0.1) {$y_{j_2}$};
	\node at (-19*\xs,-0.4) {$\vdots$};
	\node at (-19*\xs,-0.9) {$y_{j_{p_i}}$};
	\node at (-3,0) {$A'=$};
      \end{tikzpicture}
    \end{center}
  \item We have~$0\neq \det(A')=\sum_{\pi}\mathrm{sgn}(\pi)a'_{1\pi(1)}\cdots
    a'_{p_i\pi(p_i)}$, summing over permutations~$\pi$ of~$[p_i]:=\left\{ 1,\cdots,p_i
    \right\}$.  In particular, there  exists a~$\pi$ for which the corresponding
    summand is nonzero.  Think of this~$\pi$ as a {\em rook placement} in the
    matrix~$A'$, i.e., a selection of entries in each row such that no two are
    in the same column.  The entries in this rook placement are nonzero.
    Let~$e$ be one of these entries of~$A'$.  Then~$e$ is in a row indexed by an
    element~$y\in P_{i+1}$ and is in a column indexed by an element~$x$
    of~$P_i$.  We define the order matching by letting~$x\mapsto y$.  For
    example, we might have
    \begin{center}
      \begin{tikzpicture}
	\node at (0,0) {$
	  \left[\begin{array}{ccc}
	      *&*&{\color{blue}\otimes}\\
	      {\color{blue}\otimes}&*&*\\
	      *&{\color{blue}\otimes}&*\\
	  \end{array} \right]
	$};
	\node at (-0.7,1) {$x_1$};
	\node at (0.0,1) {$x_2$};
	\node at (0.7,1) {$x_3$};
	\node at (-1.6,0.5) {$y_{j_1}$};
	\node at (-1.6,0.0) {$y_{j_2}$};
	\node at (-1.6,-0.6) {$y_{j_3}$};
      \end{tikzpicture}
    \end{center}
\end{enumerate}
\[
  x_1\mapsto y_{j_1},\quad x_2\mapsto y_{j_2},\quad x_1\mapsto y_{j_3}.
\]
\hfill$\Box$

 
\end{document}
