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\begin{document}
\centerline{Math 372 lecture for Friday, Week 13}
\bigskip

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

\textbf{Example 1.}
  Consider dropping equilateral triangles into a V-shape:
  \medskip

  \begin{center}
    \begin{tikzpicture}
      \def\t{0.866cm}
      \draw (0,0)--(60:4);
      \draw (0,0)--(120:4);
      \begin{scope}[shift={(0,0)}]
	%down triangle
	\draw[fill=blue!40,draw] (0,0)--(0.5,\t)--(-0.5,\t)--(0,0);
      \end{scope}
      %level 2
      \begin{scope}[shift={(0,2*\t)}]
	%up triangle
	\draw[fill=blue!40,draw] (0,0)--(0.5,-\t)--(-0.5,-\t)--(0,0);
      \end{scope}
      \begin{scope}[shift={(0.57*\t,\t)}]
	%down triangle
	\draw[fill=blue!40,draw] (0,0)--(0.5,\t)--(-0.5,\t)--(0,0);
      \end{scope}
      \begin{scope}[shift={(-0.57*\t,\t)}]
	%down triangle
	\draw[fill=blue!40,draw] (0,0)--(0.5,\t)--(-0.5,\t)--(0,0);
      \end{scope}
      %level three
      \begin{scope}[shift={(-0.57*\t,3*\t)}]
	%up triangle
	\draw[fill=blue!40,draw] (0,0)--(0.5,-\t)--(-0.5,-\t)--(0,0);
      \end{scope}
      \begin{scope}[shift={(-0.57*2*\t,2*\t)}]
	%down triangle
	\draw[fill=blue!40,draw] (0,0)--(0.5,\t)--(-0.5,\t)--(0,0);
      \end{scope}
      \begin{scope}[shift={(0.57*\t,3*\t)}]
	%up triangle
	\draw[fill=blue!40,draw] (0,0)--(0.5,-\t)--(-0.5,-\t)--(0,0);
      \end{scope}
    \end{tikzpicture}
  \end{center}
Analogous to Young's lattice, we get a distributive lattice~$T$:
\begin{center}
  \includegraphics[height=3in]{triangle.jpg}
\end{center}

For homework, you are asked to find its subposet of join irreducibles.  

Let $t(n)$ be the number of elements of $T$ having rank~$n$, and let
$\tau(x)=\sum_{n\geq0}t(n)x^{n}$ be the rank generating function.  Here are a few
things some students I have shown:

\begin{enumerate}
  \item The rank generating function is
    \begin{align*}
      \tau(x)
  &= \prod_{i\geq 1}\frac{(1+x^{2i-1})}{(1-x^{2i})}\\
  &= 1 + 1x + 1x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 7x^7 + 10x^8 + 13x^9
  + 16x^{10} + 21x^{11} + \cdots
    \end{align*}
  \item The number $t(n)$ is:
    \begin{itemize}
      \item the number of partitions of~$n$ in which each even part occurs with even
	multiplicity and with no restriction on the odd parts;
      \item the number of partitions of~$n$ in which all odd parts occur with
	multiplicity 1 and with no restriction on the even parts;
      \item the number of partitions of $n$ into parts not congruent to $2\bmod4$.
    \end{itemize}
\end{enumerate}

\textbf{Question.} What is the simplest, most elegant, realization of $T$ as a
firing lattice?
\medskip

\textbf{Problem.} Think of nice variations of these ``stacking lattices'' and calculate the generating functions.
\bigskip

\textbf{Example 2.}
Consider the divisor on the integer grid graph pictured below:
\begin{center}
  \begin{tikzpicture}
    \foreach \x in {-3,-2,-1,0,1,2,3} {
      \draw[<->] (\x,-3.8)--(\x,3.8);
      \draw[<->] (-3.8,\x)--(3.8,\x);
    }
    \foreach \x in {-3,-2,-1,1,2,3} {
      \foreach \y in {-3,-2,-1,1,2,3} { 
	\node[circle,fill=white,text=black] at (\x,\y) {2};
      }
    }
    \node[circle,fill=white,text=black] at (0,0) {4};
    \foreach \x in {-3,-2,-1,1,2,3} {
      \node[circle,fill=white,text=black] at (\x,0) {3};
      \node[circle,fill=white,text=black] at (0,\x) {3};
     }
  \end{tikzpicture}
\end{center}

Our goal is to describe the firing lattice~$\F(D)$ for this divisor.  The
minimal element in~$\F(D)$ is the divisor itself.  It has one unstable vertex,
which when fired unstablizes its four neighbors.  Then there are four
possibilities.  So the rank generating function starts out~$1,1,4$.  We will
prove the following:
\medskip

{\bf Theorem.} Imagine a stack of oranges with levels~$1,2,3,\dots$, starting at
the top with level~$1$.  Level~$k$ consists of a~$k\times k$ square of $k^{2}$
oranges, and it sits
on top of the~$(k+1)\times(k+1)$-square of oranges on level~$k+1$.
Let~$o(\ell)$ be the number of ways of removing~$\ell$ oranges from the stack so
that none fall.  (So one may only remove oranges that do not support oranges on
a previous level.)  Then~$\F(D)(x):=\sum_{i\geq0}o(n)x^n$ is the rank generating
function for the divisor~$D$, above.
\smallskip

{\bf Proof.} Replace the oranges with cubes, and place them on top the divisor
graph.  So, looking straight down at the stack of cubes, we get a picture like
this:  
\begin{center}
  \begin{tikzpicture}
    \foreach \x in {-3,-2,-1,0,1,2,3} {
      \draw[<->] (\x,-3.8)--(\x,3.8);
      \draw[<->] (-3.8,\x)--(3.8,\x);
    }
    \foreach \x in {-3,-2,-1,1,2,3} {
      \foreach \y in {-3,-2,-1,1,2,3} { 
	\node[circle,fill=white,text=black] at (\x,\y) {2};
      }
    }
    \node[circle,fill=white,text=black] at (0,0) {4};
    \foreach \x in {-3,-2,-1,1,2,3} {
      \node[circle,fill=white,text=black] at (\x,0) {3};
      \node[circle,fill=white,text=black] at (0,\x) {3};
     }
     \foreach \x in {0.5,1.5,2.5} {
       \draw[thick] (0,\x)--(\x,0)--(0,-\x)--(-\x,0)--(0,\x);
     }
  \end{tikzpicture}
\end{center}
Here is another rendition:
\begin{center}
  \begin{tikzpicture}
    \draw[fill=orange] (0,0) rectangle (1,1);
    \draw[fill=orange] (1,0) rectangle (2,1);
    \draw[fill=orange] (2,0) rectangle (3,1);
    \draw[fill=orange] (0,1) rectangle (1,2);
    \draw[fill=orange] (1,1) rectangle (2,2);
    \draw[fill=orange] (2,1) rectangle (3,2);
    \draw[fill=orange] (0,2) rectangle (1,3);
    \draw[fill=orange] (1,2) rectangle (2,3);
    \draw[fill=orange] (2,2) rectangle (3,3);
    
    \draw[fill=orange] (1/2,1/2) rectangle (3/2,3/2);
    \draw[fill=orange] (3/2,1/2) rectangle (5/2,3/2);
    \draw[fill=orange] (1/2,3/2) rectangle (3/2,5/2);
    \draw[fill=orange] (3/2,3/2) rectangle (5/2,5/2);

    \draw[fill=orange] (1,1) rectangle (2,2);
    
    \node  at (3/2,3/2) {4};

    \node  at (3/4,2) {3};
    \node  at (3/4,1) {3};
    \node  at (9/4,2) {3};
    \node  at (9/4,1) {3};

    \node  at (1/4,2.5) {3};
    \node  at (1/4,0.5) {3};
    \node  at (11/4,2.5) {3};
    \node  at (11/4,0.5) {3};

    \node at (3/2,11/4) {2};
    \node at (3/2,1/4) {2};
    \node at (1/4,3/2) {2};
    \node at (11/4,3/2) {2};

    \draw[densely dotted, orange,ultra thick] (-0.7,3/2)--(-1.2,3/2); 
    \draw[densely dotted, orange,ultra thick] (3.7,3/2)--(4.2,3/2); 
    \draw[densely dotted, orange,ultra thick] (1.5,-0.7)--(1.5,-1.2); 
    \draw[densely dotted, orange,ultra thick] (1.5,3.7)--(1.5,4.2); 
  \end{tikzpicture}
\end{center}
Supposing each box has side length~$2$, the numbers now refer to the area of the
top face of the box that is exposed.  Removing a box exposes more area from the
boxes below it, and this corresponds exactly to firing vertices.
\hfill$\Box$
\bigskip

{\bf Remarks.} 
\begin{itemize}
  \item The same argument given above but in one dimension lower gives 
    an isomorphism from Young's lattice to the firing graph for the
    divisor~$\dots,1,1,1,2,1,1,1,\dots$ on the infinite path graph.
  \item \textbf{Open question:} What is the generating function for
    $\mathcal{F}(D)$?  See \href{https://arxiv.org/abs/1607.01890}{On a square-ice analogue of plane partitions}
\end{itemize}
\bigskip

\textbf{Open problem.}  We have seen that Young's lattice is the firing lattice
for
\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}
The generating function for Young's lattice is $\prod_{i>0}\frac{1}{1-x^{i}} $.
We have also seen that the lattice of shifted shapes (integer partitions with unequal parts) is the
firing lattice for
\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}
Its generating function is $\prod_{i>0}(1+x^{i})$.  Notice that the divisor for
Young's lattice comes from gluing together two copies of the divisor for the
shifted shapes lattice.  What if we glue together three copies:
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={below:$3$}] 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);
    \node[ball,label={left:$1$}] at (0,1) (v1) {};
    \node[ball,label={left:$1$}] at (0,2) (v2) {};
    \node[ball,label={left:$1$}] at (0,3) (v3) {};
    \node at (0,4) (v4) {};
    \draw (0)--(v1)--(v2)--(v3);
    \draw[->] (v3)--(v4);
  \end{tikzpicture}
\end{center}

\textbf{Open problem.} What is the generating function for this divisor?
\end{document}
