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\begin{document}
\centerline{Math 372 lecture for Wednesday, Week 14}
\bigskip

\centerline{\bf Firing posets II.}
\bigskip

Last time, we discussed the following two results:
\medskip

{\bf Theorem 1.} (Fundamental theorem for finitary distributive lattices)
Let~$P$ be a poset in which every principal order ideal is finite.  Then the
poset~$J_f(P)$ of {\em finite} order ideals of~$P$, ordered by inclusion, is a
finitary distributive lattice.  Conversely, if~$L$ is a finitary distributive
lattice and~$P$ is its subposet of join-irreducibles, then every principal order
ideal of~$P$ is finite, and~$L\simeq J_f(P)$.
\medskip

{\bf Theorem 2.} Let~$P$ be an  ideal in which every principal order ideal is
finite. Define a poset~$\widetilde{P}$ by adding elements~$\alpha$ and~$\beta$
to~$P$, then defining
\[
\alpha<\beta< x
\]
for all~$x\in P$.  Let~$G=G(P)$ be the Hasse diagram for~$\widetilde{P}$.  For each~$x\in P$,
let~$n(x)$ be the number of covers of~$x$ in~$P$.  Define~$D\in\Div(G)$ by
\[
  D(x)=
  \begin{cases}
    0&\text{if~$x\in\left\{ \alpha,\beta \right\}$},\\
    \deg_G(x)&\text{if~$x$ is a minimal element of~$P$},\\
    n(x)&\text{otherwise}.
  \end{cases}
\]
Then the firing graph~$\F(D)$ has no cycles, and when considered as a poset,~$\F(D)\simeq J_f(P)$.
\medskip

Combining these, we get the immediate corollary:
\medskip

{\bf Corollary.} Every finitary distributive lattice is the lattice associated
with the firing graph of a divisor on some graph.
\medskip

{\bf Example.} We saw previously that Young's lattice is given by the firing
lattice of the divisor:
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={above:$2$}] at (0,0) (0) {};
    \node[ball,label={above:$1$}] at (1,0) (1) {};
    \node[ball,label={above:$1$}] at (2,0) (2) {};
    \node[ball,label={above:$1$}] at (3,0) (3) {};
    \node at (4,0) (4) {};
    \node[ball,label={above:$1$}] at (-1,0) (-1) {};
    \node[ball,label={above:$1$}] at (-2,0) (-2) {};
    \node[ball,label={above:$1$}] at (-3,0) (-3) {};
    \node at (-4,0) (-4) {};
    \draw (0)--(1)--(2)--(3);
    \draw (0)--(-1)--(-2)--(-3);
    \draw[->] (3)--(4);
    \draw[->] (-3)--(-4);
  \end{tikzpicture}
\end{center}
However, this is not the divisor that would be given by Theorem~$2$.  To see
this, first note that the join irreducibles of Young's lattice correspond to the
Young diagrams that are~$(m,n)$ rectangles for any~$m,n\in\mathbb{P}:=\left\{
1,2,\dots \right\}$.  (We use $\mathcal{P}$ here rather than $\N$ since the
minimal element of a poset is not join irreducible.)
These are the only Young diagrams from which there is
exactly one choice for a rectangle to remove and still have a Young diagram.
Thus, according to Theorem~1, Young's diagram is isomorphic
to~$J_f(\mathbb{P}\times\mathbb{P})\simeq J_f(\N\times\N)$.  

Applying Theorem~2 to~$\N\times \N$, we get the much more complicated divisor
whose firing graph also gives Young's diagram:
\begin{center}
  \begin{tikzpicture}[scale=0.6]
    \def\h{1.732}
    \node[ball,label={left:$0$}] at (0,-1) (beta) {};
    \node[ball,label={left:$0$}] at (0,-2) (alpha) {};
    \node[ball,label={above:$3$}] at (0,0) (00) {};
    \draw (alpha)--(beta)--(00);
    \foreach \x in {1,2,3,4} {
      \node[ball,label={above:$2$}] at (-\x,\x) (0\x) {};
      \node[ball,label={above:$2$}] at (\x,\x) (\x0) {};
    }
    \node at (-5,5) (05) {};
    \node at (5,5) (50) {};
    \draw (00)--(40);
    \draw (00)--(04);
    \draw[->] (40)--(50);
    \draw[->] (04)--(05);
    \node[ball,label={above:$2$}] at (0,2) (11) {};
    \node[ball,label={above:$2$}] at (-1,3) (11) {};
    \node[ball,label={above:$2$}] at (1,3) (11) {};
    \node[ball,label={above:$2$}] at (-2,4) (11) {};
    \node[ball,label={above:$2$}] at (0,4) (11) {};
    \node[ball,label={above:$2$}] at (2,4) (11) {};
    \node[ball,label={above:$2$}] at (-3,5) (11) {};
    \node[ball,label={above:$2$}] at (-1,5) (11) {};
    \node[ball,label={above:$2$}] at (1,5) (11) {};
    \node[ball,label={above:$2$}] at (3,5) (11) {};
    \node[ball,label={above:$2$}] at (-2,6) (11) {};
    \node[ball,label={above:$2$}] at (0,6) (11) {};
    \node[ball,label={above:$2$}] at (2,6) (11) {};
    \node[ball,label={above:$2$}] at (-1,7) (11) {};
    \node[ball,label={above:$2$}] at (1,7) (11) {};
    \draw[->] (-1,1)--(4,6);
    \draw[->] (-2,2)--(3,7);
    \draw[->] (-3,3)--(2,8);
    \draw[->] (-4,4)--(1,9);
    \draw[->] (1,1)--(-4,6);
    \draw[->] (2,2)--(-3,7);
    \draw[->] (3,3)--(-2,8);
    \draw[->] (4,4)--(-1,9);
    \node[ball,label={above:$2$}] at (0,8) (11) {};
  \end{tikzpicture}
\end{center}
{\bf Question.}  Is there some way to look at the above divisor and infer our
earlier, simpler, solution?
\medskip

{\bf Example.} Young's lattice belongs the family of lattices~$J_f(\N^k)$
for~$k=1,2,\dots$  The case $k=3$ is known as the lattice of plane partitions.
Just as Young's lattice can be thought of as shoving squares into a corner, plane
partitions correspond to pushing~$3$-dimensional boxes into the corner of a
room:
\begin{center}
  \begin{tikzpicture}[scale=0.5]
    \node at (0,0) {\includegraphics[height=2in]{13wed.jpg}};
    \node at (0,-6) {A plane partition};
  \end{tikzpicture}
\end{center}
The plane partition lattice comes from the firing graph of the following
divisor:
\begin{center}
  \begin{tikzpicture}
    \def\xo{3mm}
    \begin{scope}[rotate=30]
      \foreach \angle in {0,60,...,300}{
	\begin{scope}[rotate=\angle]
	  \draw (0,0)--(3,0);
	  \draw[dashed] (3,0)--(1.5,3*0.866);
	  \draw (1.5,3*0.866)--(0,0);
	  \draw (1,0)--(0.5,0.866)--(2.5,0.866);
	  \draw (2,0)--(1,2*0.866)--(2,2*0.866);
	  \draw (1,0)--(2,2*0.866);
	  \draw (2,0)--(2.5,0.866);
	\end{scope}
      }
    \end{scope}
    \draw[ultra thick] (0,0)--(0,3);
    \draw[ultra thick] (30:-3)--(30:0);
    \draw[ultra thick] (-30:0)--(-30:3);
    \node at (0+\xo,0) {\tiny 6};
    \node[xshift=3mm] at (-30:-1) {\tiny 3};
    \node[xshift=3mm] at (-30:-2) {\tiny 3};
    \node[xshift=3mm] at (-30:-3) {\tiny 3};
    \node[xshift=3mm] at (-30:1) {\tiny 5};
    \node[xshift=3mm] at (-30:2) {\tiny 5};
    \node[xshift=3mm] at (-30:3) {\tiny 5};
    \node[xshift=3mm] at (30:1) {\tiny 3};
    \node[xshift=3mm] at (30:2) {\tiny 3};
    \node[xshift=3mm] at (30:3) {\tiny 3};
    \node[xshift=3mm,yshift=1cm] at (30:2) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (30:2) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (30:3) {\tiny 3};
    \node[xshift=3mm,yshift=-2cm] at (30:3) {\tiny 3};
    \node[xshift=3mm] at (30:3) {\tiny 3};
    \node[xshift=3mm] at (30:-1) {\tiny 5};
    \node[xshift=3mm] at (30:-2) {\tiny 5};
    \node[xshift=3mm] at (30:-3) {\tiny 5};
    \node[xshift=3mm] at (90:1) {\tiny 5};
    \node[xshift=3mm] at (90:2) {\tiny 5};
    \node[xshift=3mm] at (90:3) {\tiny 5};
    \node[xshift=3mm,yshift=1cm] at (30:1) {\tiny 3};
    \node[xshift=3mm,yshift=1cm] at (30:2) {\tiny 3};
    \node[xshift=3mm,yshift=2cm] at (30:1) {\tiny 3};
    \node[xshift=3mm] at (90:-1) {\tiny 3};
    \node[xshift=3mm] at (90:-2) {\tiny 3};
    \node[xshift=3mm] at (90:-3) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (-30:1) {\tiny 3};
    \node[xshift=3mm,yshift=-2cm] at (-30:1) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (-30:2) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (30:-1) {\tiny 3};
    \node[xshift=3mm,yshift=-2cm] at (30:-1) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (30:-2) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (-30:-2) {\tiny 3};
    \node[xshift=3mm,yshift=-1cm] at (-30:-3) {\tiny 3};
    \node[xshift=3mm,yshift=-2cm] at (-30:-3) {\tiny 3};
    \node[xshift=3mm,yshift=1cm] at (-30:-1) {\tiny 3};
    \node[xshift=3mm,yshift=1cm] at (-30:-2) {\tiny 3};
    \node[xshift=3mm,yshift=2cm] at (-30:-1) {\tiny 3};
  \end{tikzpicture}
\end{center}
There is also a more complicated divisor arising through an application of
Theorem~2.
\medskip

{\bf Example.}  Consider the firing graph of the following divisor:
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={above:$1$}] at (0,0) (0) {};
    \node[ball,label={above:$1$}] at (1,0) (1) {};
    \node[ball,label={above:$1$}] at (2,0) (2) {};
    \node[ball,label={above:$1$}] at (3,0) (3) {};
    \node at (4,0) (4) {};
    \draw (0)--(1)--(2)--(3);
    \draw[->] (3)--(4);
  \end{tikzpicture}
\end{center}
The corresponding lattice is the lattice of ``shifted shapes''.  ``A shifted
shape'' is a Young diagram for which the corresponding partition has unequal
parts.  For instance, the shifted shape for the partition~$(4,3,1)$ can be
written like this
\begin{center}
  \begin{tikzpicture}[scale=0.5]
    \fill[color=blue!20] (0,0)--(-1,1)--(0,2)
    --(-1,3)--(0,4)--(-1,5)--(0,6)--(1,5)--(2,6) --(4,4)--(0,0);
    \node[ball,label={below:$0$}] at (0,0) {};
    \node[ball,label={below:$2$}] at (1,1) {};
    \node[ball,label={below:$0$}] at (2,2) {};
    \node[ball,label={below:$1$}] at (3,3) {};
    \node[ball,label={below:$2$}] at (4,4) {};
    \node[ball,label={below:$1$}] at (5,5) {};
    \node[ball,label={below:$1$}] at (6,6) {};
    \draw[->] (0,0)--(7,7);
    \draw (0,0)--(-1,1)--(0,2);
    \draw (0,2)--(-1,3)--(0,4);
    \draw (0,4)--(-1,5)--(0,6);
    \draw (0,6)--(-1,7)--(0,8);
    \draw (0,2)--(1,1);
    \draw (0,4)--(2,2);
    \draw (0,6)--(3,3);
    \draw (0,2)--(3,5)--(4,4);
    \draw (0,4)--(2,6)--(4,4);
  \end{tikzpicture}
\end{center}
What does the poset of join irreducibles look like for the lattice of shifted
shapes?
\bigskip

We have seen that every finitary distributive lattice is the lattice associated
to a firing graph for the divisor of some graph.
\begin{quote}
  \textbf{Main question:} Can we characterize the posets arising from firing graphs?
\end{quote}
We will focus on this question now.

A \emph{firing script} is a formal sum of vertices: $\sigma=\sum_{i=1}^{n} a_i
v_i$.  It is {\em legal} for a divisor~$D$ if there is a legal sequence of
vertex firings~$w_1,\dots,w_k$  where $k=\sum_{i=1}^{n} a_i$ such
that~$\sigma=\sum_{i=1}^{k}w_i$.  

A lattice is {\em locally free} if for each element~$x$, the interval from~$x$
to the join of all elements covering~$x$ is a Boolean lattice (i.e., isomorphic
to a lattice of subsets of a given set).  
\medskip

{\bf Theorem.} (Bj\"orner, Lov\`asz, Shor, 1991).  Let~$G$ be a finite graph,
and let~$D\in\Div(G)$ be stabilizable.  Then the firing graph~$\F(D)$ is a
locally free lattice.  Given~$D',D''\in\F(D)$, let~$\sigma'$ and~$\sigma''$ be
legal firing scripts such that~$D\xrightarrow{\sigma'}D'$
and~$D\xrightarrow{\sigma''}D''$.  Then~$D\xrightarrow{\sigma'\vee\sigma''}D''$,
where~$(\sigma'\vee\sigma'')(v):=\max\{\sigma'(v),\sigma''(v)\}$ for all
vertices~$v$ of~$G$.
\medskip

In our case, given~$D'\in\F(D)$ (with~$D$
stabilizable), we look at all the vertices of~$D'$ that are unstable.  Say these
are~$v_1,\dots,v_k$.  Suppose~$D\xrightarrow{\sigma'}D'$ via a legal firing
script~$\sigma'$.  Then let~$\tau=\sigma'+v_1+\cdots+v_k$.  The join discussed
above is~$D'':=D-L\tau$, so~$D\xrightarrow{\tau}D''$.
\end{document}
