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

\centerline{\bf Distributive lattices}
\bigskip

Let~$P$ be a poset.
A {\em greatest lower bound} for~$x,y\in P$ is an element~$v\in P$ that is a
lower bound, i.e., $v\leq x$ and~$v\leq y$, and such that for all~$w\in P$
\[
  w\leq x\  \text{and}\ w\leq y\quad\Longrightarrow w\leq v.
\]
A {\em least upper bound} for elements~$x,y\in P$ is an element~$u\in
P$ that is an upper bound, i.e., $x\leq u$ and~$y\leq u$, and such that for
all~$w\in P$,  
\[
  x\leq w\ \text{and}\ y\leq w\quad\Longrightarrow u\leq w.
\]
A greatest lower bound for~$x$ and~$y$, if it exists, is unique.  It is called
the {\em meet} of~$x$ and~$y$ and denoted~$x\wedge y$
A least upper bound for~$x$ and~$y$, if it exists, is unique.  It is called
the {\em join} of~$x$ and~$y$ and denoted~$x\vee y$.

A {\em lattice} is a poset~$L$ in which every pair of elements~$x,y\in L$ as a
meet and a join.  Here are Hasse diagrams for all lattices with five elements:
\begin{center}
  \begin{tikzpicture}[scale=0.5]
    \node[ball] at (0,0) (1) {};
    \node[ball] at (0,1) (2) {};
    \node[ball] at (0,2) (3) {};
    \node[ball] at (0,3) (4) {};
    \node[ball] at (0,4) (5) {};
    \draw (1)--(2)--(3)--(4)--(5);
    \begin{scope}[xshift=4cm]
      \node[ball] at (0,0) (1) {};
      \node[ball] at (-1,1) (2) {};
      \node[ball] at (1,1) (3) {};
      \node[ball] at (0,2) (4) {};
      \node[ball] at (0,3) (5) {};
      \draw (1)--(2)--(4)--(5);
      \draw (1)--(3)--(4);
    \end{scope}
    \begin{scope}[xshift=8cm]
      \node[ball] at (0,0) (1) {};
      \node[ball] at (0,1) (2) {};
      \node[ball] at (-1,2) (3) {};
      \node[ball] at (1,2) (4) {};
      \node[ball] at (0,3) (5) {};
      \draw (1)--(2)--(3)--(5);
      \draw (2)--(4)--(5);
    \end{scope}
    \begin{scope}[xshift=12cm]
      \node[ball] at (0,0) (1) {};
      \node[ball] at (-1,1) (2) {};
      \node[ball] at (1,1) (3) {};
      \node[ball] at (0,2) (4) {};
      \node[ball] at (0,1) (5) {};
      \draw (1)--(2)--(4)--(3)--(1);
      \draw (1)--(4);
    \end{scope}
    \begin{scope}[xshift=16cm]
      \node[ball] at (0,0) (1) {};
      \node[ball] at (-1,1) (2) {};
      \node[ball] at (1,1.5) (3) {};
      \node[ball] at (-1,2) (4) {};
      \node[ball] at (0,3) (5) {};
      \draw (1)--(2)--(4)--(5);
      \draw (1)--(3)--(5);
    \end{scope}
  \end{tikzpicture}
\end{center}

Here is a poset that is not a lattice:
\begin{center}
  \begin{tikzpicture}[scale=0.5]
    \node[ball] at (0,0) (1) {};
    \node[ball] at (-1,1) (2) {};
    \node[ball] at (1,1) (3) {};
    \node[ball] at (-1,2) (4) {};
    \node[ball] at (1,2) (5) {};
    \draw (1)--(2)--(4);
    \draw (1)--(3)--(4);
    \draw (2)--(5)--(3);
  \end{tikzpicture}
\end{center}
\medskip

{\bf Distributive lattices.}  Let~$L$ be a lattice.  Then~$L$ is {\em
distributive} if for all~$x,y,z\in L$,
\[
  x\vee(y\wedge z)=(x\vee y)\wedge(x\vee z).
\]
One may show that this condition is equivalent to the condition that
\[
  x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z).
\]
Examples of distributive lattices include the Boolean posets~$B_n$, the
nonnegative integers~$\N$, and Young's lattice (of partitions of integers). Here
are two examples of lattices that are not distributive:
\begin{center}
  \begin{tikzpicture}
    \node[ball,label={right:{$x$}}] at (-1,0) (x) {};
    \node[ball,label={right:{$y$}}] at (0,0) (y) {};
    \node[ball,label={right:{$z$}}] at (1,0) (z) {};
    \node[ball,label={right:{$\hat{1}$}}] at (0,1.4) (1) {};
    \node[ball,label={right:{$\hat{0}$}}] at (0,-1.4) (0) {};
    \draw (0)--(x)--(1);
    \draw (0)--(y)--(1);
    \draw (0)--(z)--(1);
    \node at (0,-2.2) {Diamond lattice, $M_{3}$};
    \begin{scope}[xshift=6cm]
      \node[ball,label={left:{$y$}}] at (-1,0.5) (y) {};
      \node[ball,label={left:{$x$}}] at (-1,-0.5) (x) {};
      \node[ball,label={right:{$z$}}] at (1,0) (z) {};
      \node[ball,label={right:{$\hat{1}$}}] at (0,1.4) (1) {};
      \node[ball,label={right:{$\hat{0}$}}] at (0,-1.4) (0) {};
      \draw (0)--(x)--(y)--(1);
      \draw (0)--(z)--(1);
      \node at (0,-2.2) {Pentagon lattice, $N_{5}$};
    \end{scope}
  \end{tikzpicture}
\end{center}
These examples characterize non-distributive lattices.
\medskip

\textbf{Proposition.} A lattice is distributive if and only if none of its
sublattices is isomorphic to $M_{3}$ or $N_{5}$.
\smallskip

A \emph{sublattice} of a lattice is a subset that is closed under the meet and
join operations of the original lattice.  
\medskip

\textbf{Example.} Here is an example of a lattice that contains $N_{5}$ as a
subset by not a sublattice:
\begin{center}
  \begin{tikzpicture}
      \node[ball,label={left:{$y$}}] at (-1,0.5) (y) {};
      \node[ball,label={left:{$x$}}] at (-1,-0.5) (x) {};
      \node[ball,label={left:{$a$}}] at (0,-0.5) (a) {};
      \node[ball,label={right:{$z$}}] at (1,0) (z) {};
      \node[ball,label={right:{$\hat{1}$}}] at (0,1.4) (1) {};
      \node[ball,label={right:{$\hat{0}$}}] at (0,-1.4) (0) {};
      \node[ball,label={right:{$\hat{0}$}}] at (0,-1.4) (0) {};
      \draw (0)--(x)--(y)--(1);
      \draw (0)--(z)--(1);
      \draw[dashed] (0)--(a)--(y);
      \draw[dashed] (a)--(z);
      \node at (0,-2.2) {A distributive lattice.};
  \end{tikzpicture}
\end{center}

A lattice is {\em finitary} if it is locally finite (meaning that each of its
intervals is finite) and if it has a unique smallest element~$\widehat{0}$.

Given any poset~$P$, we now describe a way of constructing an distributive
lattice.  An {\em order ideal} or {\em down-set} in~$P$ is a subset~$I$ of~$P$
such that if~$x\in I$ and~$y\leq x$, then~$y\in P$.  A {\em principal order
ideal} is an order ideal generated by a single element, i.e., of the
form~$\left\{ y:y\leq x \right\}$ for some~$x\in P$.  For each~$x\in P$ we
denote the corresponding principal order ideal by
\[
  \Lambda_x:=\left\{ y\in P:y\leq x \right\}.
\]
Let~$J(P)$ be the set of all order ideals of~$P$ and give it a poset structure
by inclusion of subsets of~$P$.  Then~$J(P)$ is a lattice in which the meet
operation is intersection and the join operation is union (as subsets of~$P$).

An element~$x$ in a lattice~$L$ is {\em join irreducible} if~$x\neq\widehat{0}$
and if it is not possible to write~$x=y\vee z$ with~$y<x$ and~$z<x$.  In other
words,~$x$ is join irreducible if it covers exactly one element.
\medskip

{\bf Example.} Hasse diagrams for a poset~$P$ and its lattice of join
irreducibles~$J(P)$ appear below.  Subsets are listed as words, e.g., $\left\{
a,b,d \right\}=abd$.  The join irreducibles in~$J(P)$ are circled.
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) (a) {a};
    \node at (2,0) (b) {b};
    \node at (1,1) (c) {c};
    \node at (3,1) (d) {d};
    \node at (1.5,-1) {$P$};
    \draw (a)--(c)--(b)--(d);
    \begin{scope}[xshift=5cm,yshift=1cm]
      \node at (1,-1) (e) {$\emptyset$};
      \node[draw,circle] at (0,0) (a) {a};
      \node[draw,circle] at (2,0) (b) {b};
      \node at (1,1) (ab) {ab};
      \node at (1,1) (ab) {ab};
      \node at (1,3) (abcd) {abcd};
      \node[draw,circle] at (3,1) (bd) {bd};
      \node at (2,2) (abd) {abd};
      \node[draw,circle] at (0,2) (abc) {abc};
      \node at (1.5,-2) {$J(P)$};
      \draw (e)--(a);
      \draw (e)--(b)--(bd)--(abd);
      \draw (a)--(ab)--(b);
      \draw (ab)--(abd)--(abcd);
      \draw (ab)--(abc)--(abcd);
    \end{scope}
  \end{tikzpicture}
\end{center}
\medskip
  The following
theorem is quoted from Stanley's {\em Enumerative Combinatorics, Volume 1}:
\medskip

{\bf Theorem.} (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 Young's lattice.} Let~$Y$ be the infinite path graph with vertex
set~$\Z$ and edges $\left\{ i,i+1 \right\}$ for all~$i\in\Z$.  Let~$D\in\Div(Y)$
be given by
\[
  D(i)=
  \begin{cases}
    2&\text{if~$i= 0$}\\
    1&\text{otherwise}.
  \end{cases}
\]
Then~$\F(D)$ is isomorphic to Young's lattice:
\begin{center}
  \begin{tikzpicture}[scale=0.3]
    \def\ad{1.0}
    \def\yad{5.8cm}
    \node at (33cm,14cm) {\ }; 
    \node at (33cm,-11cm) {\ }; 
    \begin{scope}
      \node at (0,-\ad) {$2$};
      \node at (1,1-\ad) {$1$};
      \node at (2,2-\ad) {$1$};
      \node at (3,3-\ad) {$1$};
      \node at (-1,1-\ad) {$1$};
      \node at (-2,2-\ad) {$1$};
      \node at (-3,3-\ad) {$1$};
      \node[mminiball] at (0,0) {};
      \node[mminiball] at (1,1) {};
      \node[mminiball] at (-1,1) {};
      \node[mminiball] at (2,2) {};
      \node[mminiball] at (-2,2) {};
      \node[mminiball] at (3,3) {};
      \node[mminiball] at (-3,3) {};
      \draw (0,0)--(3.4,3.4);
      \draw (0,0)--(-3.4,3.4);
    \end{scope}
    \begin{scope}[yshift=8cm]
      \node at (0,-\ad) {$0$};
      \node at (1,1-\ad) {$2$};
      \node at (2,2-\ad) {$1$};
      \node at (3,3-\ad) {$1$};
      \node at (-1,1-\ad) {$2$};
      \node at (-2,2-\ad) {$1$};
      \node at (-3,3-\ad) {$1$};
      \node[mminiball] at (0,0) {};
      \node[mminiball] at (1,1) {};
      \node[mminiball] at (-1,1) {};
      \node[mminiball] at (2,2) {};
      \node[mminiball] at (-2,2) {};
      \node[mminiball] at (3,3) {};
      \node[mminiball] at (-3,3) {};
      \draw (0,0)--(3.4,3.4);
      \draw (0,0)--(-3.4,3.4);
      \draw (-1,1)--(0,2)--(1,1);
      \draw[green,thick] (0,-2.5)--(0,-4); % Hasse line
    \end{scope}
    \begin{scope}[xshift=-26cm,yshift=14cm]
      \begin{scope}[yshift=-5cm,color=blue]
	\draw[thick] (1,0)--(2,1)--(3,0);
	\node at (6.5,0.4) {$= 0$};
      \end{scope}
      \begin{scope}[yshift=-8cm,color=blue]
	\draw[thick] (0,0)--(-1,1);
	\node at (2,0.4) {and};
	\draw[thick] (3.5,0)--(4.5,1);
	\node at (6.5,0.4) {$= 1$};
      \end{scope}
      \begin{scope}[yshift=-11cm,color=blue]
	\draw[thick] (1,1)--(2,0)--(3,1);
	\node at (6.5,0.4) {$= 2$};
      \end{scope}
    \end{scope}
      \begin{scope}[xshift=-7.5cm,yshift=2.6*\yad]
      \node at (0,-\ad) {$1$};
      \node at (1,1-\ad) {$2$};
      \node at (2,2-\ad) {$1$};
      \node at (3,3-\ad) {$1$};
      \node at (-1,1-\ad) {$0$};
      \node at (-2,2-\ad) {$2$};
      \node at (-3,3-\ad) {$1$};
      \node[mminiball] at (0,0) {};
      \node[mminiball] at (1,1) {};
      \node[mminiball] at (-1,1) {};
      \node[mminiball] at (2,2) {};
      \node[mminiball] at (-2,2) {};
      \node[mminiball] at (3,3) {};
      \node[mminiball] at (-3,3) {};
      \draw (0,0)--(3.4,3.4);
      \draw (0,0)--(-3.4,3.4);
      \draw (-1,1)--(0,2)--(1,1);
      \draw (-2,2)--(-1,3)--(0,2);
      \draw[green,thick] (1.8,-1.3)--(3.5,-2.9);  % Hasse line (2)
  \end{scope}
    \begin{scope}[xshift=7.5cm,yshift=2.6*\yad]
      \node at (0,-\ad) {$1$};
      \node at (1,1-\ad) {$0$};
      \node at (2,2-\ad) {$2$};
      \node at (3,3-\ad) {$1$};
      \node at (-1,1-\ad) {$2$};
      \node at (-2,2-\ad) {$1$};
      \node at (-3,3-\ad) {$1$};
      \node[mminiball] at (0,0) {};
      \node[mminiball] at (1,1) {};
      \node[mminiball] at (-1,1) {};
      \node[mminiball] at (2,2) {};
      \node[mminiball] at (-2,2) {};
      \node[mminiball] at (3,3) {};
      \node[mminiball] at (-3,3) {};
      \draw (0,0)--(3.4,3.4);
      \draw (0,0)--(-3.4,3.4);
      \draw (-1,1)--(0,2)--(1,1);
      \draw (2,2)--(1,3)--(0,2);
      \draw[green,thick] (-1.8,-1.3)--(-3.5,-2.9);  % Hasse line (1+1)
  \end{scope}
  \begin{scope}[xshift=0cm,yshift=3.8*\yad]
      \node at (0,-\ad) {$2$};
      \node at (1,1-\ad) {$0$};
      \node at (2,2-\ad) {$2$};
      \node at (3,3-\ad) {$1$};
      \node at (-1,1-\ad) {$0$};
      \node at (-2,2-\ad) {$2$};
      \node at (-3,3-\ad) {$1$};
      \node[mminiball] at (0,0) {};
      \node[mminiball] at (1,1) {};
      \node[mminiball] at (-1,1) {};
      \node[mminiball] at (2,2) {};
      \node[mminiball] at (-2,2) {};
      \node[mminiball] at (3,3) {};
      \node[mminiball] at (-3,3) {};
      \draw (0,0)--(3.4,3.4);
      \draw (0,0)--(-3.4,3.4);
      \draw (-1,1)--(0,2)--(1,1);
      \draw (-2,2)--(-1,3)--(0,2);
      \draw (2,2)--(1,3)--(0,2);
      \draw[green,thick] (-1.7,-1.3)--(-3.3,-2.8);    % Hasse line (2+1)
      \draw[green,thick] (1.7,-1.3)--(3.3,-2.8);    % Hasse line (2+1)
      \draw[green,densely dotted, very thick] (0,5)--(0,7);    % Hasse line (2+1)
    \end{scope}
    \begin{scope}[xshift=-14cm,yshift=3.8*\yad]
      \node at (0,-\ad) {$1$};
      \node at (1,1-\ad) {$2$};
      \node at (2,2-\ad) {$1$};
      \node at (3,3-\ad) {$1$};
      \node at (-1,1-\ad) {$1$};
      \node at (-2,2-\ad) {$0$};
      \node at (-3,3-\ad) {$2$};
      \node[mminiball] at (0,0) {};
      \node[mminiball] at (1,1) {};
      \node[mminiball] at (-1,1) {};
      \node[mminiball] at (2,2) {};
      \node[mminiball] at (-2,2) {};
      \node[mminiball] at (3,3) {};
      \node[mminiball] at (-3,3) {};
      \draw (0,0)--(3.4,3.4);
      \draw (0,0)--(-3.4,3.4);
      \draw (-1,1)--(0,2)--(1,1);
      \draw (-2,2)--(-1,3)--(0,2);
      \draw (-3,3)--(-2,4)--(-1,3);
      \draw[green,thick] (1.1,-1.3)--(2.7,-2.9);  % Hasse line (3)
    \end{scope}
    \begin{scope}[xshift=14cm,yshift=3.8*\yad]
      \node at (0,-\ad) {$1$};
      \node at (1,1-\ad) {$1$};
      \node at (2,2-\ad) {$0$};
      \node at (3,3-\ad) {$2$};
      \node at (-1,1-\ad) {$2$};
      \node at (-2,2-\ad) {$1$};
      \node at (-3,3-\ad) {$1$};
      \node[mminiball] at (0,0) {};
      \node[mminiball] at (1,1) {};
      \node[mminiball] at (-1,1) {};
      \node[mminiball] at (2,2) {};
      \node[mminiball] at (-2,2) {};
      \node[mminiball] at (3,3) {};
      \node[mminiball] at (-3,3) {};
      \draw (0,0)--(3.4,3.4);
      \draw (0,0)--(-3.4,3.4);
      \draw (-1,1)--(0,2)--(1,1);
      \draw (2,2)--(1,3)--(0,2);
      \draw (3,3)--(2,4)--(1,3);
      \draw[green,thick] (-2.7,-2.7)--(-1.5,-1.5); % Hasse line (1+1+1)
  \end{scope}
  \end{tikzpicture}
\end{center}

\vspace{-2cm}
The proof is obtained by looking at the path that runs along the top of each
diagram.  It extends infinitely in the~$(-1,1)$ and in the~$(1,1)$ directions
and has bumps along the top of the Young diagram.  The method of decoding the
lines and bumps into the numbers~$0$,~$1$, and~$2$ is listed in blue on the
left.
\medskip

{\bf Theorem.} Let~$P$ be a poset 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,~$\F(D)\simeq J_f(P)$.
\smallskip

{\bf Proof.} Homework.
\hfill$\Box$
\medskip

{\bf Example.}  Apply the construction in the theorem to the zigzag poset
example, above:
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) (a) {a};
    \node at (2,0) (b) {b};
    \node at (1,1) (c) {c};
    \node at (3,1) (d) {d};
    \node at (1.5,-1) {$P$};
    \draw (a)--(c)--(b)--(d);
    \draw[->] (4,0.5)--(5,0.5);
    \begin{scope}[xshift=6.5cm,yshift=0.7cm]
      \node at (0,0) (a) {a};
      \node at (2,0) (b) {b};
      \node at (1,1) (c) {c};
      \node at (3,1) (d) {d};
      \node at (1,-1) (beta) {$\beta$};
      \node at (1,-2) (alpha) {$\alpha$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \node at (-0.2,-1.7) {$G$};
    \end{scope}
  \end{tikzpicture}
\end{center}

The divisor described in the theorem is~$D=2a+2b$, and its firing graph is
pictured below:
\begin{center}
  \begin{tikzpicture}[scale=0.7]
    \def\xs{4cm}
    \def\ys{4cm}
    \begin{scope}
      \node at (0,0) (a) {$2$};
      \node at (2,0) (b) {$3$};
      \node at (1,1) (c) {$0$};
      \node at (3,1) (d) {$0$};
      \node at (1,-1) (beta) {$0$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \draw[blue,->] (3.0,1.7)--(3.8,2.5);
      \draw[blue,->] (-0.8,1.7)--(-1.6,2.5);
      \node at (1,0) {$\emptyset$};
    \end{scope}
    \begin{scope}[xshift=-\xs,yshift=\ys]
      \node at (0,0) (a) {$0$};
      \node at (2,0) (b) {$3$};
      \node at (1,1) (c) {$1$};
      \node at (3,1) (d) {$0$};
      \node at (1,-1) (beta) {$1$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \draw[blue,->] (3.0,1.7)--(3.8,2.5);
      \node at (1,0) {a};
    \end{scope}
    \begin{scope}[xshift=\xs,yshift=\ys]
      \node at (0,0) (a) {$2$};
      \node at (2,0) (b) {$0$};
      \node at (1,1) (c) {$1$};
      \node at (3,1) (d) {$1$};
      \node at (1,-1) (beta) {$1$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \draw[blue,->] (3.0,1.7)--(3.8,2.5);
      \draw[blue,->] (-0.8,1.7)--(-1.6,2.5);
      \node at (1,0) {b};
    \end{scope}
    \begin{scope}[xshift=0*\xs,yshift=2*\ys]
      \node at (0,0) (a) {$0$};
      \node at (2,0) (b) {$0$};
      \node at (1,1) (c) {$2$};
      \node at (3,1) (d) {$1$};
      \node at (1,-1) (beta) {$2$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \draw[blue,->] (3.0,1.7)--(3.8,2.5);
      \draw[blue,->] (-0.8,1.7)--(-1.6,2.5);
      \node at (1,0) {ab};
    \end{scope}
    \begin{scope}[xshift=2*\xs,yshift=2*\ys]
      \node at (0,0) (a) {$2$};
      \node at (2,0) (b) {$1$};
      \node at (1,1) (c) {$1$};
      \node at (3,1) (d) {$0$};
      \node at (1,-1) (beta) {$1$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \draw[blue,->] (-0.8,1.7)--(-1.6,2.5);
      \node at (1,0) {bd};
    \end{scope}
    \begin{scope}[xshift=1*\xs,yshift=3*\ys]
      \node at (0,0) (a) {$0$};
      \node at (2,0) (b) {$1$};
      \node at (1,1) (c) {$2$};
      \node at (3,1) (d) {$0$};
      \node at (1,-1) (beta) {$2$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \draw[blue,->] (-0.8,1.7)--(-1.6,2.5);
      \node at (1,0) {abd};
    \end{scope}
    \begin{scope}[xshift=-1*\xs,yshift=3*\ys]
      \node at (0,0) (a) {$1$};
      \node at (2,0) (b) {$1$};
      \node at (1,1) (c) {$0$};
      \node at (3,1) (d) {$1$};
      \node at (1,-1) (beta) {$2$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \draw[blue,->] (3.0,1.7)--(3.8,2.5);
      \node at (1,0) {abc};
    \end{scope}
    \begin{scope}[xshift=0*\xs,yshift=4*\ys]
      \node at (0,0) (a) {$1$};
      \node at (2,0) (b) {$2$};
      \node at (1,1) (c) {$0$};
      \node at (3,1) (d) {$0$};
      \node at (1,-1) (beta) {$2$};
      \node at (1,-2.2) (alpha) {$0$};
      \draw (beta)--(a)--(c)--(b)--(d);
      \draw (alpha)--(beta)--(b);
      \node at (1,0) {abcd};
    \end{scope}
  \end{tikzpicture}
\end{center}
\end{document}
