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\begin{document}
\centerline{Math 372 lecture for Friday, Week 9}
\bigskip

\centerline{\bf Order complexes}
\bigskip

Let~$P$ be a finite poset.  Recall that a {\em chain} of length~$\ell$ is a
sequence of elements of~$P$ of the form~$x_0<x_1<\dots<x_{\ell}$.  Create a
simplicial complex~$\Delta(P)$ corresponding to~$P$ by taking the vertices to be
the elements of~$P$ and the~$i$-faces to be the chains of length~$i$.  Note that
a subset of a chain is again a chain.  The facets of~$\Delta(P)$ are the maximal
chains of~$P$.

To create the boundary mapping, we need to linearly order the vertices, i.e.,
the elements of~$P$.  We can do this arbitrarily, although we will always do
this by refining the given ordering on~$P$.  Denote the linear ordering
by~$\prec$, as opposed to the ordering~$<$ on~$P$. Then we will choose~$\prec$
so that~$x<y$ implies~$x\prec y$.  
\medskip

{\bf Example.} (Bruhat order on~$\mathfrak{S_n}$).  For this example, we will
  represent each permutation~$\pi$ by the word consisting of the
  list~$\pi(1)\pi(2)\dots\pi(n)$.  For example, if~$n=3$ and~$\pi(1)=2$,
  $\pi(2)=3$, and~$\pi(3)=1$, we represent~$\pi$ by~$231$.
  An {\em inversion} of a
  permutation~$\pi\in\mathfrak{S}_n$ is a pair~$1\leq i<j\leq n$ such
  that~$\pi(i)>\pi(j)$.  Define~$\ell(\pi)$ to be the number of inversions
of~$\pi$.   So for example,~$\ell(123)=0$ and~$\ell(321)=3$, the latter
  since~$3>2$, $3>1$, and~$2>1$.  The {\em Bruhat order} on~$\mathfrak{S}_n$ has
  covering relations~$\pi\lessdot\tau$ if~$\tau$ can be obtained from~$\pi$ by
  a transposition (swapping two elements) and~$\ell(\tau)=\ell(\pi)+1$.  

  The Hasse diagram for Bruhat order on~$\mathfrak{S}_n$:
  \begin{center}
    \begin{tikzpicture}[scale=1.3]
      \node at (0,0) (123) {$123$};
      \node at (-1,1) (213) {$213$};
      \node at (1,1) (132) {$132$};
      \node at (1,2) (312) {$312$};
      \node at (-1,2) (231) {$231$};
      \node at (0,3) (321) {$321$};
      \draw (123)--(213);
      \draw (123)--(132);
      \draw (213)--(231);
      \draw (132)--(231);
      \draw (213)--(312);
      \draw (312)--(321);
      \draw (231)--(321);
      \draw (132)--(312);
      \node at (2,1.5) {=};
      \begin{scope}[xshift=4cm]
	\node at (0,0) (123) {$a$};
	\node at (-1,1) (213) {$b$};
	\node at (1,1) (132) {$c$};
	\node at (1,2) (312) {$e$};
	\node at (-1,2) (231) {$d$};
	\node at (0,3) (321) {$f$};
	\draw (123)--(213);
	\draw (123)--(132);
	\draw (213)--(231);
	\draw (132)--(231);
	\draw (213)--(312);
	\draw (312)--(321);
	\draw (231)--(321);
	\draw (132)--(312);
	\end{scope}
    \end{tikzpicture}
  \end{center}
  The order complex for this poset has four facets:
  \[
    \overline{abdf},\ \overline{abef},\ \overline{acdf},\ \overline{acef}.
  \]
  These represent four solid tetrahedra sharing the edge~$\overline{af}$ and
  forming a three dimensional triangulation of a solid sphere. It thus has no
  homology (i.e., all of the homology groups are~$0$). 
\medskip

\medskip

{\bf Definition.} The {\em (reduced) Euler characterisic} of a $d$-dimensional simplicial
complex~$\Delta$ is 
\[
  \widetilde{\chi}(\Delta):=\sum_{i\in\Z}(-1)^i\dim
  \widetilde{H}_i(\Delta)=\sum_{i=0}^d(-1)^i\tilde{\beta}_i.
\]

{\bf Exercise.} Show that
\[
  \widetilde{\chi}(\Delta)=-f_{-1}+f_0-f_1+\dots+f_d,
\]
where~$f_i=|F_i|$, the number of faces of dimension~$i$.
\medskip

\medskip

{\bf Proposition.} (Philip Hall's theorem) Give an finite poset~$P$, let~$\widehat{P}$ be the
poset formed from~$P$ by adjoining~$\hat{0}$ and~$\hat{1}$ where~$\hat{0}<p<\hat{1}$ for
all~$p\in P$.  Let~$c_i$ be the number of chains of the
form~$\hat{0}=p_0<\dots<p_i=\hat{1}$.  (In particular,~$c_0=0$ and~$c_1=1$.)
Then
\[
  \mu_{\hat{P}}(\hat{0},\hat{1})=
  \widetilde{\chi}(\Delta(P))=c_0-c_1+c_2-\cdots.
\]
\smallskip

{\bf Proof.} Using results from the homework, we have
\begin{align*}
  \mu_{\hat{P}}(\hat{0},\hat{1})
  &= \frac{1}{\zeta}(\hat{0},\hat{1})\\
  &= \frac{1}{\delta+(\zeta-\delta)}(\hat{0},\hat{1})\\[5pt]
&= (\delta-(\zeta-\delta)+(\zeta-\delta)^2-\cdots)(\hat{0},\hat{1})\\
&=\delta(\hat{0},\hat{1})-(\zeta-\delta)(\hat{0},\hat{1})+(\zeta-\delta)^2(\hat{0},\hat{1})-\cdots\\[5pt]
&= c_0-c_1+c_2-\cdots\\
&= 0 -f_{-1}+f_0-\cdots\\
&= \widetilde{\chi}(\Delta(P)).
\end{align*}
\hfill$\Box$
\medskip

{\bf Corollary.} Let~$P$ be a locally finite poset, and~$P^{*}$ be its {\em dual
poset}, equal to~$P$ as a set but with~$s<t$ in~$P^{*}$ if~$t<s$ in~$P$.  (So
  the Hasse diagram for~$P^*$ is the flip of that for~$P$.)  Then for
  all~$s<t$ in~$P$, we have
  \[
    \mu_P(s,t)=\mu_{P^*}(t,s).
  \]
\smallskip

{\bf Proof.} Let~$Q$ be the subposet of~$P$ formed from the interval~$[s,t]$ by
removing~$s$ and~$t$, and let~$Q^*$ be its dual.  Then~$\widehat{Q}\simeq[s,t]$,
and
\[
  \mu_{\widehat{Q}}(\hat{0},\hat{1})=\mu_P(s,t).
\] 
Similarly
\[
  \mu_{\widehat{Q^*}}(\hat{0},\hat{1})=\mu_{P^*}(s,t).
\]
However, the~$c_i$'s for~$Q$ and for~$Q^*$ are equal.
\hfill$\Box$

\end{document}
