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\begin{document}
\centerline{Math 372 lecture for Monday, Week 3}
\bigskip

$P=B_n$, boolean poset
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {order-raising operators:};
    \node at (5,-0.05) {
      \begin{minipage}{0.9\linewidth} 
      \begin{align*}
	U_i\colon\R P_i&\to\R P_{i+1}\\
	x&\mapsto \textstyle\sum_{y\in P_{i+1}:x<y} y
      \end{align*}
    \end{minipage}
  };
  \begin{scope}[yshift=-2cm]
    \node at (0,0) {order-lowering operators:};
    \node at (5,-0.05) {
      \begin{minipage}{0.9\linewidth} 
      \begin{align*}
	D_i\colon\R P_{i}&\to\R P_{i-1}\\
	y&\mapsto \textstyle\sum_{x\in P_{i-1}:x<y} x
      \end{align*}
    \end{minipage}
    };
  \end{scope}
  \end{tikzpicture}
\end{center}

{\color{blue} Goal:}
\medskip

{\bf Theorem 4.7.} $U_i$ is injective for~$i<\frac{n}{2}$ and surjective
for~$n\geq\frac{n}{2}$.
\medskip

It then follows from last time that~$B_n$ is Sperner.
\medskip

{\bf Example.} $n=4$
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) (12) {12};
    \node at (1,0) (13) {13};
    \node at (2,0) (14) {14};
    \node at (3,0) (23) {23};
    \node at (4,0) (24) {24};
    \node at (5,0) (34) {34};
    \node at (5/5,-1.3) (1) {1};
    \node at (2*5/5,-1.3) (2) {2};
    \node at (3*5/5,-1.3) (3) {3};
    \node at (4*5/5,-1.3) (4) {4};
    \draw (1)--(12);
    \draw (1)--(13);
    \draw (1)--(14);
    \draw (2)--(12);
    \draw (2)--(23);
    \draw (2)--(24);
    \draw (3)--(13);
    \draw (3)--(23);
    \draw (3)--(34);
    \draw (4)--(14);
    \draw (4)--(24);
    \draw (4)--(34);
  \end{tikzpicture}
\end{center}
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) (dom) {$U_1\colon\R^4$};
    \node at (4,0) (codom) {$\R^6$};
    \draw (dom)--(codom);
    \node at (2.2,1.8) {$
      \left(\begin{array}{cccc}
	  1&1&0&0\\
	  1&0&1&0\\
	  1&0&0&1\\
	  0&1&1&0\\
	  0&1&0&1\\
	  0&0&1&1
      \end{array} \right)
    $};
    \begin{scope}[xshift=6.5cm]
    \node at (0,0) (dom) {$D_2\colon\R^6$};
    \node at (5,0) (codom) {$\R^4$};
    \draw (dom)--(codom);
    \node at (2.8,1.3) {$
      \left(\begin{array}{cccccc}
	  1&1&1&0&0&0\\
	  1&0&0&1&1&0\\
	  0&1&0&1&0&1\\
	  0&0&1&0&1&1
      \end{array} \right)
    $};
  \end{scope}
  \end{tikzpicture}
\end{center}
In general,~$D_{i+1}=U_i^t$.
\medskip

{\bf Lemma 4.6.} $D_{i+1}U_i-U_{i-1}D_i=(n-2i)I$.
\smallskip

{\bf Proof.} See text.
\hfill$\Box$
\smallskip

{\bf Proof of Theorem 4.7.}  If~$A$ is any real matrix, then~$AA^t$ is real and
symmetric, hence, diagonalizable.  We also have that for all~$v$
\[
  v^tAA^tv=(A^tv)^t(A^tv)\geq0,
\]
i.e.,~$AA^t$ is positive semidefinite, which implies all of its eigenvalues are nonnegative
real numbers. By Lemma~4.6, 
\[
  D_{i+1}U_i-U_{i-1}D_i=(n-2i)I.
\]
Say~$v_1,\dots,v_{p_i}$ are the eigenvectors of~$U_{i-1}D_i=U_{i-1}U_{i-1}^t$
with corresponding eigenvectors~$\lambda_k\geq0$. Then
\begin{align*}
  (D_{i+1}U_i)v_k
  &= 
  (U_{i-1}D_i)v_k+(n-2i)v_k\\
  &= \lambda_kv_k+(n-2i)v_k\\
  &= (\lambda_k+n-2i)v_k.
\end{align*}
Hence, the eigenvalues of~$D_{i+1}U_i$ are~$\left\{ \lambda_k+n-2i \right\}$
for~$k=1,\dots,p_i$.  If~$i<\frac{n}{2}$, the $\lambda_k+n-2i>0$ for
all~$k$, and since~$D_{i+1}U_i$ thus has no zero eigenvalues, it is invertible.
That implies~$U_i$ is injective for~$i<\frac{n}{2}$.

For the case~$i\geq\frac{n}{2}$, use Lemma~4.6 again but in the form
\[
  U_iD_{i+1}=D_{i+2}U_{i+1}+(2i+2-n)I.
\]
If~$\mu_1,\dots,\mu_{p_{i+1}}$ are the eigenvalues of~$D_{i+2}U_{i+1}$
then~$\mu_k\geq0$ for all~$k$, and the eigenvalues of~$U_iD_{i+1}$ are
$\left\{ \mu_k+2i+2-n \right\}$, which are all positive
for~$i\geq\frac{n}{2}$.  Hence,~$U_iD_{i+1}$ is invertible.  It follows
that~$U_i$ is surjective.
\hfill$\Box$



\end{document}


