% !TeX spellcheck = en_US
% !TeX encoding = UTF-8
\documentclass[8pt,pdf,t]{beamer}
\usepackage{amsmath,amssymb,theorem,mathtools,cancel,hyperref}
%\usepackage[cm]{fullpage}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
%\usepackage[english,russian]{babel}
\usepackage[shortcuts]{extdash}

%\usepackage[notref,notcite]{showkeys}

\usepackage{tikz}
\usetikzlibrary{arrows,shapes}
\usetikzlibrary{arrows.meta}
\usetikzlibrary{decorations.markings}
\usetikzlibrary{intersections}
\tikzset{>={Stealth[length=2mm]}}
\tikzstyle{every picture} += [remember picture]
\tikzstyle{na} = [baseline=-.5ex]
\tikzset{->-/.style={decoration={markings,mark=at position #1 with {\arrow{>}}},postaction={decorate}}}

\usetheme{Warsaw}
\usefonttheme{serif}


\def\smallskip{\vskip\smallskipamount}
\def\medskip{\vskip\medskipamount}
\def\bigskip{\vskip\bigskipamount}

\let\{=\lbrace
\let\}=\rbrace
\let\[=\lbracket
\let\]=\rbracket

\long\def\items#1{\begin{itemize}\parsep=0pt\itemsep=0pt#1\end{itemize}}
\long\def\enum#1{\begin{enumerate}\parsep=0pt\itemsep=0pt#1\end{enumerate}}

\def\eq#1$$#2$${\begin{equation#1}#2\end{equation#1}}
\long\def\subeq#1{\begin{subequations}#1\end{subequations}}
\def\Split$$#1$${\begin{split}#1\end{split}}
\def\Align#1$$#2$${\begin{align#1}#2\end{align#1}}
\def\AlignAt#1$$#2$${\begin{alignat}{#1}#2\end{alignat}}
\def\Aligned#1{\begin{aligned}{}#1\end{aligned}}
\def\Alignedb#1{\begin{aligned}[b]#1\end{aligned}}
\def\Alignedt#1{\begin{aligned}[t]#1\end{aligned}}
\def\Gather#1$$#2$${\begin{gather#1}#2\end{gather#1}}
\def\Gathered#1{\begin{gathered}{}#1\end{gathered}}
\def\Gatheredb#1{\begin{gathered}[b]#1\end{gathered}}
\def\Gatheredt#1{\begin{gathered}[t]#1\end{gathered}}
\def\Multline#1$$#2$${\begin{multline#1}#2\end{multline#1}}
\def\Matrix#1{\begin{matrix}#1\end{matrix}}
\def\pMatrix#1{\begin{pmatrix}#1\end{pmatrix}}
\def\Cases#1{\begin{cases}#1\end{cases}}

\def\d{\partial}
\def\bd{{\bar\d}}
\def\bpsi{{\bar\psi}}
\def\bvarphi{{\bar\varphi}}
\def\Re{\mathop{\rm Re}\nolimits}
\def\Im{\mathop{\rm Im}\nolimits}
\def\sh{\mathop{\rm sh}\nolimits}
\def\ch{\mathop{\rm ch}\nolimits}
\def\tg{\mathop{\rm tg}\nolimits}
\def\ctg{\mathop{\rm ctg}\nolimits}
\def\tr{\mathop{\rm tr}\nolimits}
\def\arctg{\mathop{\rm arctg}\nolimits}
\def\mathspan{\mathop{\rm span}\nolimits}
\def\bn{{\text{\bfseries\itshape n}}}
\def\bJ{{\text{\bfseries\itshape J}}}
\def\Z{{\mathbb Z}}
\def\R{{\mathbb R}}
\def\C{{\mathbb C}}
\def\cH{{\cal H}}
\def\cN{{\cal N}}
\def\bz{{\bar z}}
\def\bw{{\bar w}}
\def\bw{{\bar w}}
\def\tphi{{\tilde\phi}}
\def\vq{{\vec q}}
\def\vx{{\vec x}}
\def\vy{{\vec y}}
\def\vJ{{\vec J}}
\def\va{{\vec a}}
\def\vb{{\vec b}}
\def\i{{\rm i}}
\def\e{{\rm e}}
\def\ve{\varepsilon}
\def\eff{{\rm eff}}
\def\const{\mathinner{\rm const}}

\def\phys{{\text{phys}}}

\def\lcolon{\mathopen:}
\def\rcolon{\mathclose:}

%%%%%%% For figures: %%%%%%%
\def\ar#1 #2, #3 #4 {\arrow <0.3truecm> [0.1,0.3] from #1 #2 to #3 #4 }
\catcode`!=11
\def\ln#1 #2, #3 #4 {\!start(#1,#2)\!ljoin(#3,#4)}
\catcode`!=12
\def\rl#1 #2, #3 #4 {\putrule from #1 #2 to #3 #4 }
\def\trl#1 #2, #3 #4 {\linethickness .8pt
	\putrule from #1 #2 to #3 #4 \linethickness .4pt}
\def\point#1 #2 {\put{\hbox{\kern -1pt .}} [Bl] at #1 #2 }
\def\bfpoint#1 #2 {%
\put{\hbox{\kern -2pt \raise -2.3pt%
\hbox{$\textstyle\bullet$}}} [Bl] at #1 #2 }
\def\figcap#1#2\par{\medskip{\narrower\eightpoint\noindent
	Fig.~#1. \ignorespaces#2\par}\medskip}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\makeatletter
\let\over\@@over
\let\atop\@@atop
\let\above\@@above
\let\overwithdelims\@@overwithdelims
\let\atopwithdelims\@@atopwithdelims
\let\abovewithdelims\@@abovewithdelims
\makeatother

\def\Maketitle{{\def\newpage{}\maketitle}}
\def\emph#1{{\color{blue}#1}}
\def\red{\color{red}}
\def\blue{\color{blue}}
\def\olive{\color{olive}}
\def\black{\color{black}}

\def\unpause{\onslide<+->\relax}

\begin{document}

\title[Lecture 2. Six\-/vertex model]{Lecture 2\\Six\-/vertex model}
\subtitle{A mini-course ``Solvable lattice models and Bethe Ansatz''\\(Ariel University, spring 2021)}
\author{Michael Lashkevich}
\institute{Landau Institute for Theoretical Physics,\\
Kharkevich Institute for Information Transmission Problems}
\date{}

\frame{\titlepage}


\begin{frame}
\frametitle{Ice model: configurations}
\unpause

The `ice model' (\tikz[baseline=-2pt,scale=.5]{\draw[fill=black] (0,0) circle (2mm);} is Oxygen, \tikz[baseline=-2pt,scale=.5]{\draw[-,fill=white] (0,0) circle (1mm);} is Hydrogen):
$$
\begin{tikzpicture}[scale=.5,baseline=.5*1.5cm-2pt]
\draw[-] (0,-.5) -- (0,3.5);
\draw[-] (1,-.5) -- (1,3.5);
\draw[-] (2,-.5) -- (2,3.5);
\draw[-] (3,-.5) -- (3,3.5);
\draw[-] (-.5,0) -- (3.5,0);
\draw[-] (-.5,1) -- (3.5,1);
\draw[-] (-.5,2) -- (3.5,2);
\draw[-] (-.5,3) -- (3.5,3);
\foreach \x in {0,...,3} {
  \foreach \y in {0,...,3} {
    \draw[fill=black] (\x,\y) circle (2mm);
  }
}
\draw[-,fill=white,yshift=0cm] (0,.35) circle (1mm);
\draw[-,fill=white,yshift=1cm] (0,.65) circle (1mm);
\draw[-,fill=white,yshift=2cm] (0,.65) circle (1mm);
\draw[-,fill=white,yshift=3cm] (0,.35) circle (1mm);
%
\begin{scope}[xshift=1cm]
  \draw[-,fill=white,yshift=-1cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=0cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=1cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=2cm] (0,.65) circle (1mm);
\end{scope}
%
\begin{scope}[xshift=2cm]
  \draw[-,fill=white,yshift=-1cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=0cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=1cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=2cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=3cm] (0,.35) circle (1mm);
\end{scope}
%
\begin{scope}[xshift=3cm]
  \draw[-,fill=white,yshift=0cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=1cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=2cm] (0,.35) circle (1mm);
\end{scope}
%
\draw[-,fill=white,xshift=0cm] (.35,0) circle (1mm);
\draw[-,fill=white,xshift=1cm] (.35,0) circle (1mm);
\draw[-,fill=white,xshift=2cm] (.65,0) circle (1mm);
%
\begin{scope}[yshift=1cm]
  \draw[-,fill=white,xshift=-1cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=0cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=1cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=2cm] (.65,0) circle (1mm);
\end{scope}
%
\begin{scope}[yshift=2cm]
  \draw[-,fill=white,xshift=0cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=1cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=2cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=3cm] (.35,0) circle (1mm);
\end{scope}
%
\begin{scope}[yshift=3cm]
  \draw[-,fill=white,xshift=0cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=1cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=2cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=3cm] (.35,0) circle (1mm);
\end{scope}
\end{tikzpicture}
\uncover<.(1)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.5*1.5cm-2pt]
% vertical lines
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=0cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=0cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=2cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=3cm] (0,1) -- (0,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=1cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=0cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=2cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=3cm] (0,0) -- (0,1);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=2cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=0cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=2cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=3cm] (0,1) -- (0,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=3cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=0cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=2cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=3cm] (0,0) -- (0,1);
\end{scope}
%
% horizontal lines
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=0cm]
  \draw[-,postaction=decorate,xshift=-1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=0cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=2cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=3cm] (0,0) -- (1,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=1cm]
  \draw[-,postaction=decorate,xshift=-1cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=0cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=2cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=3cm] (0,0) -- (1,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=2cm]
  \draw[-,postaction=decorate,xshift=-1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=0cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=2cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=3cm] (1,0) -- (0,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=3cm]
  \draw[-,postaction=decorate,xshift=-1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=0cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=1cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=2cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=3cm] (1,0) -- (0,0);
\end{scope}
%
\end{tikzpicture}
}
\uncover<.(2)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]
% vertical lines
\begin{scope}[xshift=0cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$-$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$-$};
\end{scope}
%
\begin{scope}[xshift=1cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$+$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[xshift=2cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$+$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$-$};
\end{scope}
%
\begin{scope}[xshift=3cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$-$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$+$};
\end{scope}
%
% horisontal lines
\begin{scope}[yshift=0cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$-$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[yshift=1cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$+$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[yshift=2cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$-$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$-$};
\end{scope}
%
\begin{scope}[yshift=3cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$-$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$-$};
\end{scope}
\end{tikzpicture}
}
$$
\uncover<.->{Each oxygen atom has two hydrogen atom next to it.}
\uncover<.(2)->{Small arrows on the right figure define the orientation of the lattice lines and vertices, which will be important later.}

\Align*$$
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (0,0) -- (1,0); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (1,0) circle (2mm); \draw[-,fill=white] (.65,0) circle (1mm);
\end{tikzpicture}
&\uncover<.(1)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (0,0) -- (1,0);
\end{tikzpicture}
}
\uncover<.(2)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
\end{tikzpicture}
}
\\
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (0,0) -- (1,0); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (1,0) circle (2mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
&\uncover<.(1)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (1,0) -- (0,0);
\end{tikzpicture}
}
\uncover<.(2)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
\end{tikzpicture}
}
\\
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-] (0,0) -- (0,1); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (0,1) circle (2mm); \draw[-,fill=white] (0,.65) circle (1mm);
\end{tikzpicture}
&\uncover<.(1)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (0,0) -- (0,1);
\end{tikzpicture}
}
\uncover<.(2)->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
\end{tikzpicture}
}
\\
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-] (0,0) -- (0,1); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (0,1) circle (2mm); \draw[-,fill=white] (0,.35) circle (1mm);
\end{tikzpicture}
&\uncover<+->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (0,1) -- (0,0);
\end{tikzpicture}
}
\uncover<+->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
\end{tikzpicture}
}
$$

\end{frame}

\begin{frame}
\frametitle{Ice model: Boltzmann weights}
\unpause

\emph{Six\-/vertex model}: the Boltzmann weights are associated with vertices:
$$
Z=\sum_{\substack{\text{configu-}\\\text{rations}}}\prod_\text{vertices}R^{\ve'_1\ve'_2}_{\ve_1\ve_2},
\quad
R^{\ve'_1\ve'_2}_{\ve_1\ve_2}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$\ve_2$} -- (.5,0) node[right] {\tiny$\ve'_2$};
\draw[<-] (0,-.6) node[below] {\tiny$\ve_1$} -- (0,.5) node[above] {\tiny$\ve'_1$};
\end{tikzpicture},
\quad
{\red\mathop{\boxed{\black\ve'_1+\ve'_2=\ve_1+\ve_2}}\limits_\text{\normalsize\strut Ice condition}}.
$$
\unpause
We have six vertex configurations
\Align*$$
R^{++}_{++}=a
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,-.35) circle (1mm); \draw[-,fill=white] (-.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[>->] (-.5,0) -- (.5,0);
\draw[>->] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-,red] (-.6,0) node[left] {\tiny$+$} -- (.5,0) node[right] {\tiny$+$};
\draw[<-,red] (0,-.6) node[below] {\tiny$+$} -- (0,.5) node[above] {\tiny$+$};
\end{tikzpicture},
&R^{--}_{--}=a'
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,.35) circle (1mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[<-<] (-.5,0) -- (.5,0);
\draw[<-<] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-,blue] (-.6,0) node[left] {\tiny$-$} -- (.5,0) node[right] {\tiny$-$};
\draw[<-,blue] (0,-.6) node[below] {\tiny$-$} -- (0,.5) node[above] {\tiny$-$};
\end{tikzpicture}
\\
R^{+-}_{+-}=b
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,-.35) circle (1mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[<-<] (-.5,0) -- (.5,0);
\draw[>->] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-,blue] (-.6,0) node[left] {\tiny$-$} -- (.5,0) node[right] {\tiny$-$};
\draw[<-,red] (0,-.6) node[below] {\tiny$+$} -- (0,.5) node[above] {\tiny$+$};
\end{tikzpicture},
&R^{-+}_{-+}=b'
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,.35) circle (1mm); \draw[-,fill=white] (-.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[>->] (-.5,0) -- (.5,0);
\draw[<-<] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-,red] (-.6,0) node[left] {\tiny$+$} -- (.5,0) node[right] {\tiny$+$};
\draw[<-,blue] (0,-.6) node[below] {\tiny$-$} -- (0,.5) node[above] {\tiny$-$};
\end{tikzpicture}
\\
R^{-+}_{+-}=c
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,-.35) circle (1mm); \draw[-,fill=white] (0,.35) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[<->] (-.5,0) -- (.5,0);
\draw[>-<] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-,blue] (-.6,0) node[left] {\tiny$-$} -- (0,0) -- (0,.5) node[above] {\tiny$-$};
\draw[<-,red] (0,-.6) node[below] {\tiny$+$} -- (0,0) -- (.5,0) node[right] {\tiny$+$};
\end{tikzpicture},
&R^{+-}_{-+}=c'
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (-.35,0) circle (1mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[>-<] (-.5,0) -- (.5,0);
\draw[<->] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-,red] (-.6,0) node[left] {\tiny$+$} -- (0,0) -- (0,.5) node[above] {\tiny$+$};
\draw[<-,blue] (0,-.6) node[below] {\tiny$-$} -- (0,0) -- (.5,0) node[right] {\tiny$-$};
\end{tikzpicture}
$$
\unpause
$$
R=\pMatrix{a&&&\\&b&c&\\&c'&b'&\\&&&a'}\
\text{in the basis $(++),(+-),(-+),(--)$.}
$$

\end{frame}

\begin{frame}
\frametitle{Six-vertex models: toroidal boundary conditions and exact solvability}
\unpause

We will consider the six\-/vertex models with \emph{toroidal} boundary conditions: any upper line connects to the lower one and any lower line connects to the upper one.
\unpause

With these conditions the ratio $c'/c$ is not essential. Indeed, consider a configuration
$$
\begin{tikzpicture}[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]
% vertical lines
\begin{scope}[xshift=0cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$+$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[xshift=1cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$+$} -- (0,1);
  \draw[-,yshift=0cm,blue,very thick] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[xshift=2cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$+$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=1cm,blue,very thick] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=2cm,blue,very thick] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[xshift=3cm]
  \draw[<-,yshift=-1cm,blue,very thick] (0,.5) node[left=-2pt] {\tiny$-$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=3cm,blue,very thick] (0,0) -- (0,.5) node[left=-2pt] {\tiny$-$};
\end{scope}
%
% horisontal lines
\begin{scope}[yshift=0cm]
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[<-,xshift=-1cm,blue,very thick] (.5,0) node[above=-2pt] {\tiny$-$} -- (1,0);
  \draw[-,xshift=0cm,blue,very thick,shorten >=-.5pt] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0) node[below right=-1pt] {$c$};
  \draw[-,xshift=3cm,blue,very thick,shorten <=-.5pt] (0,0) node[below right=-1pt] {$c'$} -- (.5,0) node[above=-2pt] {\tiny$-$};
\end{scope}
%
\begin{scope}[yshift=1cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$+$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$+$};
  \draw[-,xshift=1cm,blue,very thick,shorten <=-.5pt,shorten >=-.5pt] (0,0) node[below right=-1pt] {$c'$}
   -- node[above=-2pt] {\tiny$-$} (1,0) node[below right=-1pt] {$c$};
\end{scope}
%
\begin{scope}[yshift=2cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$+$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[yshift=3cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$+$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$+$};
  \draw[-,xshift=2cm,blue,very thick,shorten <=-.5pt,shorten >=-.5pt] (0,0) node[below right=-1pt] {$c'$}
    -- node[above=-2pt] {\tiny$-$} (1,0) node[below right=-1pt] {$c$};
\end{scope}
\end{tikzpicture}
$$
\unpause
You see that the number of $c$ and $c'$ is equal. Since the signs ``$-$'' can be organized in such paths and these paths must be closed on the torus, this will be valid for all configurations.
\unpause
Thus we may assume $c'=c$ without loss of generality.
\unpause

The difference between $a$ and $a'$ is more essential, but it can be interpreted as an homogeneous external field.
\unpause
Indeed, $2\#(a)-2\#(a')=\#(+)-\#(-)$.
\unpause

The difference between $b$ and $b'$ cannot be reduced to an ``anisotropic'' external field, which has opposite signs for horizontal and vertical lines.
\unpause

It turns out that the model is \emph{exactly solvable}, if $b'=b$.
\unpause
If also $a'=a$ it admits spontaneous symmetry breaking. Below we will just consider this case.
\unpause
We will see that the solution for $a'\ne a$ is also based on the symmetric case.


\end{frame}

\begin{frame}
\frametitle{Six\-/vertex models: symmetric case}
\unpause

We will consider the \emph{symmetric} six\-/vertex model:
$$
R^{-\ve'_1\>-\ve'_2}_{-\ve_1\>-\ve_2}=R^{\ve'_1\ve'_2}_{\ve_1\ve_2}
$$
or
$$
a'=a,
\quad
b'=b
\quad
c'=c.
$$
\unpause
The transfer matrix
\eq$$
T^{\ve'_1\ldots\ve'_N}_{\ve_1\ldots\ve_N}
=\sum_{\mu_1\ldots\mu_N}R^{\mu_2\ve'_1}_{\mu_1\ve_1}
R^{\mu_3\ve'_2}_{\mu_2\ve_2}\ldots R^{\mu_1\ve'_N}_{\mu_N\ve_N}.
\label{transfermatrixdef}
$$
\unpause
Let us consider the matrix $R$ as an operator in the tensor product of two two\-/dimensional spaces:
$$
R:\C^2\otimes\C^2\to\C^2\otimes\C^2,
\qquad
v_{\ve_1}\otimes v_{\ve_2}\mapsto
R^{\ve_1\ve_2}_{\ve'_1\ve'_2}v_{\ve'_1}\otimes v_{\ve'_2}.
$$
Here $v_\ve$ is the natural basis in $V=\C^2$.
\unpause
Consider the tensor product $V_1\otimes V_2\otimes\cdots\otimes V_k$ of identical spaces $V_i\simeq V$. Let $R_{ij}$ is the $R$ matrix acting on $V_i\otimes V_j$.
\unpause

Then the transfer matrix can be written as
\eq$$
T=\tr_{V_0}(R_{0N}\ldots R_{02}R_{01}):\
V_1\otimes V_2\otimes\cdots\otimes V_N\to V_1\otimes V_2\otimes\cdots\otimes V_N.
\label{transfermatrixtr}
$$
The space $V_1\otimes\cdots\otimes V_N$ is called \emph{quantum space}, while the space $V_0$ is called \emph{auxiliary space}.

\end{frame}

\begin{frame}
\frametitle{Six-vertex models: $L$ operator}
\unpause

The operator under the trace is
\eq$$
L=R_{0N}\ldots R_{02}R_{01}:\ V_0\otimes V_1\otimes\cdots\otimes V_N\to V_0\otimes V_1\otimes\cdots\otimes V_N.
\label{Ldef}
$$
\unpause
We will consider it as an operator in the quantum space and a matrix in the auxiliary space
$$
L=\pMatrix{A&B\\ C&D},
\quad
A,B,C,D:V_1\otimes V_2\otimes\cdots\otimes V_N\to V_1\otimes V_2\otimes\cdots\otimes V_N.
$$
\unpause
Then
\eq$$
T=\tr_{V_0}L=A+D.
\label{transfermatrixAplusD}
$$

\end{frame}

\begin{frame}
\frametitle{Commuting transfer matrices and Yang\--Baxter equation}
\unpause

Integrability demands the existence of extra commuting integrals of motion $I_n$:
$$
[T,I_n]=0,
\quad
[I_m,I_n]=0.
$$
How to construct them?
\unpause

Let use search for the operators $T'=\tr_{V_0}L'$, $L'=R'_{0N}\ldots R'_{02}R'_{01}$ with some matrix $R'$.
\unpause

\begin{block}{Theorem}
If there exist nondegenerate matrices $R'$, $R''$ such that
\eq$$
R''_{12}R'_{13}R_{23}=R_{23}R'_{13}R''_{12},
\label{YBE}
$$
or, graphically
\eq$$
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.25,1) -- (.75,-1);
\draw[->,name path=l2] (.25,1) -- (-.75,-1);
\draw[->,name path=l3] (1,-.5) -- (-1,-.5);
\begin{scope}[name intersections={of=l1 and l2,by=R11},name intersections={of=l1 and l3,by=R1},name intersections={of=l2 and l3,by=R}]
\draw[-] (R) node[below] {$\quad R$};
\draw[-] (R1) node[below] {$R'\quad$};
\draw[-] (R11) node[left] {$R''$};
\end{scope}
\end{tikzpicture}
\quad=\quad
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.75,1) -- (.25,-1);
\draw[->,name path=l2] (.75,1) -- (-.25,-1);
\draw[->,name path=l3] (1,.5) -- (-1,.5);
\begin{scope}[name intersections={of=l1 and l2,by=R11},name intersections={of=l1 and l3,by=R1},name intersections={of=l2 and l3,by=R}]
\draw[-] (R) node[above] {$R\quad$};
\draw[-] (R1) node[above] {$\quad R'$};
\draw[-] (R11) node[left] {$R''$};
\end{scope}
\end{tikzpicture}
\tag{$\ref{YBE}'$}
$$
then
\eq$$
[T,T']=0
\label{TT'-commut}
$$
\end{block}

\end{frame}

\begin{frame}
\frametitle{Commuting transfer matrices: a proof}
\unpause

A graphical proof:
\Multline$$
T'T=
\\
\uncover<+->{\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,4.5) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L$};
\draw[-] (1,4.5) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L'$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}}
\uncover<+->{\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,6) to[out=-90,in=90] (1,5) to[out=-90,in=90] (0,4); \draw[-] (.5,5.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (1,6) to[out=-90,in=90] (0,5) to[out=-90,in=90] (1,4); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R''$}};
\draw[-] (0,4) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L$};
\draw[-] (1,4) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L'$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}}
\uncover<+->{\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,5) to[out=-90,in=90] (1,4) to[out=-90,in=90] (0,3); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (1,5) to[out=-90,in=90] (0,4) to[out=-90,in=90] (1,3); \draw[-] (.5,3.5) node[right=4pt] {\rlap{$R''$}};
\draw[-] (0,3) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6);
\draw[-] (1,3) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6);
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}}
\uncover<+->{\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,5) to[out=-90,in=90] (1,4); \draw[-] (1,4) -- (1,3) to[out=-90,in=90] (0,2); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (1,5) to[out=-90,in=90] (0,4); \draw[-] (0,4) -- (0,3) to[out=-90,in=90] (1,2); \draw[-] (.5,2.5) node[right=4pt] {\rlap{$R''$}};
\draw[dashed] (0,2) -- (0,.5); \draw[->] (0,.5) -- (0,-.6);
\draw[dashed] (1,2) -- (1,.5); \draw[->] (1,.5) -- (1,-.6);
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}}
\uncover<+->{\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,5) to[out=-90,in=90] (1,4);
\draw[-] (1,5) to[out=-90,in=90] (0,4);
\draw[-] (0,4) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[-] (0,.5) -- (0,0) node[above left] {$L'$};
  \draw[->] (0,0) to[out=-90,in=90] (1,-1) -- (1,-1.3);
\draw[-] (1,4) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[-] (1,.5) -- (1,0) node[above right] {$L$};
  \draw[->] (1,0) to[out=-90,in=90] (0,-1) -- (0,-1.3);
\draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (.5,-.5) node[right=4pt] {\rlap{$R''$}};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}}
\uncover<+->{\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,6) to[out=-90,in=90] (1,5) to[out=-90,in=90] (0,4); \draw[-] (.5,5.5) node[right=4pt] {\rlap{$R''$}};
\draw[-] (1,6) to[out=-90,in=90] (0,5) to[out=-90,in=90] (1,4); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (0,4) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L'$};
\draw[-] (1,4) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}}
\uncover<+->{\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,4.5) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L'$};
\draw[-] (1,4.5) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}}
\\
\uncover<+->{{}=TT'.}
$$
\unpause
A more conventional proof is based on the relation
$$
R''_{12}L'_1L_2=L_2L'_1R''_{12},
$$
which is proved by induction.
\unpause
Then
$$
\Aligned{
T'T
&=\tr_{V_1\otimes V_2}(L'_1L_2)
=\tr_{V_1\otimes V_2}((R''_{12})^{-1}R''_{12}L'_1L_2)
=\tr_{V_1\otimes V_2}((R''_{12})^{-1}L_2L'_1R''_{12})
\\
&=\tr_{V_1\otimes V_2}(R''_{12}(R''_{12})^{-1}L_2L'_1)
=\tr_{V_1\otimes V_2}(L_2L'_1)=TT'.
}
$$

\end{frame}

\begin{frame}
\frametitle{Yang\--Baxter equation: solution for the six\-/vertex model}
\unpause

The solution can be found in the form
\eq$$
\Aligned{
R
&=R(\lambda,u_2-u_3),
\\
R'
&=R(\lambda,u_1-u_3),
\\
R''
&=R(\lambda,u_1-u_2)
}\label{YBEsolution}
$$
with a given matrix\-/valued function $R(\lambda,u)$.
\unpause
Since the common factor of $a,b,c$ is arbitrary, assume $a(\lambda,u)=1$. Trigonometric solution(s):
$$
\Aligned{
b(\lambda,u)
&={\sin u\over\sin(\lambda-u)},
&\uncover<.(1)->{b(\lambda,u)}
&\uncover<.(1)->{{}={\sh u\over\sh(\lambda-u)},}
\\
c(\lambda,u)
&={\sin\lambda\over\sin(\lambda-u)}
&\uncover<.(1)->{c(\lambda,u)}
&\uncover<.(1)->{{}={\sh\lambda\over\sh(\lambda-u)}}
\\
\span(a<b+c,\ b<a+c,\ c<a+b);\quad
&\span\uncover<+->{(c>a+b).}
}
$$
\unpause
The cases $a>b+c$ and $b>a+c$ and not interesting from the thermodynamic point of view and will be discussed later.
\unpause
The parameter $\lambda$ is the same for $R,R',R''$ and can be expressed as
\eq$$
\left.\Matrix{-\cos\lambda\\-\ch\lambda}\right\}=\Delta\equiv{a^2+b^2-c^2\over2ab}.
\label{Deltadef}
$$
\unpause
Thus we will omit the parameter $\lambda$ from now on:
$$
R(u)\equiv R(\lambda,u),\ a(u)\equiv a(\lambda,u)\text{ etc.}
$$

\end{frame}

\begin{frame}
\frametitle{Yang\--Baxter equation: spectral parameter}
\unpause

The spectral parameters can be associated to lines:
$$
R(\lambda,u-v)^{\ve_3\ve_4}_{\ve_1\ve_2}
=\begin{tikzpicture}[baseline=-2pt]
\draw[<-] (-.6,0) node[left] {$\ve_2$} -- node[pos=.25,above] {$v$} (.5,0) node[right] {$\ve_4$};
\draw[<-] (0,-.6) node[below] {$\ve_1$} -- node[pos=.25,right] {$u$} (0,.5) node[above] {$\ve_3$};
\end{tikzpicture}
$$
\unpause
This $R$ matrix is the solution to the Yang\--Baxter equation in the form
\Multline$$
R_{12}(\lambda,u_1-u_2)R_{13}(\lambda,u_1-u_3)R_{23}(\lambda,u_2-u_3)
\\
=R_{23}(\lambda,u_2-u_3)R_{13}(\lambda,u_1-u_3)R_{12}(\lambda,u_1-u_2).
\label{YBrel}
$$
\unpause
Graphically:
\eq$$
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.25,1) -- (.75,-1) node[below] {$\ u_1$};
\draw[->,name path=l2] (.25,1) -- (-.75,-1) node[below] {$u_2$};
\draw[->,name path=l3] (1,-.5) -- (-1,-.5) node[left] {$u_3$};
\end{tikzpicture}
\quad=\quad
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.75,1) -- (.25,-1) node[below] {$\ u_1$};
\draw[->,name path=l2] (.75,1) -- (-.25,-1) node[below] {$u_2$};
\draw[->,name path=l3] (1,.5) -- (-1,.5) node[left] {$u_3$};
\end{tikzpicture}
\tag{$\ref{YBrel}'$}
$$
\unpause
Besides, the $R$ matrix satisfy the relations
\eq$$
b(u)R(\lambda-u)^{\ve_3\ve_4}_{\ve_1\ve_2}=R(u)^{\ve_2\>-\ve_3}_{\ve_4\>-\ve_1},
\qquad
R_{12}(u)R_{21}(-u)=1,
\qquad
R(0)=P=\tikz[baseline=-2pt,>={Stealth[length=1mm]},scale=.5]{
  \draw[->] (0,.5) -- (0,.2) to[out=-90,in=0] (-.2,0) -- (-.5,0);
  \draw[->] (.5,0) -- (.2,0) to[out=180,in=90] (0,-.2) -- (0,-.5);
}\ .
\label{addrels}
$$

\end{frame}

\begin{frame}
\frametitle{Integrals of motion}
\unpause

We have
\eq$$
[T(u),T(u')]=0
\quad\forall u,u'.
\label{commfam}
$$
But not all the integrals of motion $T(u)$ are independent.
\unpause

First of all, $T(0)$ is nothing but the shift operator:
\eq$$
T(0)=\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,4.8) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,0.5); \draw[->] (0,.5) -- (0,-.8) node[below] {$u$};
\draw[->] (.8,4) -- (-.8,4) node[left] {$u$};
\draw[->] (.8,3) -- (-.8,3) node[left] {$u$};
\draw[->] (.8,0) -- (-.8,0) node[left] {$u$};
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[->] (0,4.8) -- (0,4.2) to[out=-90,in=0] (-.2,4) -- (-.8,4);
\draw[->] (.8,4) -- (.2,4) to[out=180,in=90] (0,3.8) -- (0,3.2) to[out=-90,in=0] (-.2,3) -- (-.8,3);
\draw[-] (.8,3) -- (.2,3) to[out=180,in=90] (0,2.8) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,0.5);
  \draw[->] (0,.5) -- (0,.2) to[out=-90,in=0] (-.2,0) -- (-.8,0);
\draw[->] (.8,0) -- (.2,0) to[out=180,in=90] (0,-.2) -- (0,-.8);
\end{tikzpicture}
\label{T0-shift}
$$
\unpause

Then decompose the product $T^{-1}(0)T(u)$ in~$u$:
\eq$$
T^{-1}(0)T(u)=1-\sum_{n=1}^\infty{H_nu^n\over n!}.
\label{T-decom}
$$
\unpause
Hamiltonians $H_n$ commute with $T(u)$ and mutually commute:
\eq$$
[T(0),H_n]=[H_m,H_n]=0
\quad\forall m,n.
\label{Hn-commut}
$$
The set $T(0)$, $H_1,\ldots,H_{N-1}$ form a set of independent integrals of motion.
\unpause
Operators $H_n$ are \emph{local} in the sense that each of them is a sum of term, which involves a finite number ($n+1$) of neighboring nodes.

\end{frame}

\begin{frame}
\frametitle{Six\-/vertex model and XXZ Heisenberg chain}
\unpause

Let us find the Hamiltonian $H_1$ explicitly:
\Multline*$$
-H_1=T^{-1}(0)T'(0)=
\\
\uncover<+->{\left.d\over du\right|_{u=0}\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[<-] (1,4.8) node[above] {$0$} -- (1,4.2) to[out=-90,in=180] (1.2,4) -- (1.8,4);
\draw[<-] (-.8,4) node[left] {$0$} -- (.8,4) to[out=0,in=90] (1,3.8) -- (1,3.2) to[out=-90,in=180] (1.2,3) -- (1.8,3);
\draw[<-] (-.8,3) node[left] {$0$} -- (.8,3) to[out=0,in=90] (1,2.8) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,0.5);
  \draw[-] (1,.5) -- (1,.2) to[out=-90,in=180] (1.2,0) -- (1.8,0);
\draw[<-] (-.8,0) node[left] {$0$} -- (.8,0) to[out=0,in=90] (1,-.2) -- (1,-.8);
\draw[-] (0,4.8) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,0.5); \draw[->] (0,.5) -- (0,-.8) node[below] {$u$};
\end{tikzpicture}}
\uncover<+->{{}=\left.d\over du\right|_{u=0}\sum^N_{n=1}
\vbox spread -1cm{\vss\hbox{\begin{tikzpicture}[scale=.5,baseline=1.5cm-2pt]
\draw[<-] (-.8,6) node[above] {\tiny$V_N$} -- (-.2,6) to[out=0,in=-90] (0,6.2) -- (0,6.8) to[out=90,in=180] (.2,7) -- (.8,7) to[out=0,in=90]
  (1,6.8) -- (1,6.2) to[out=-90,in=180] (1.2,6) -- (1.8,6);
\draw[<-] (-.8,4) node[above] {\tiny$V_{n+2}$} -- (-.2,4) to[out=0,in=-90] (0,4.2) -- (0,4.5); \draw[dashed] (0,4.5) -- (0,5.5);
  \draw[-] (0,5.5) -- (0,5.8) to[out=90,in=180] (.2,6) -- (.8,6) to[out=0,in=-90] (1,5.8) -- (1,5.5); \draw[dashed] (1,5.5) -- (1,4.5);
  \draw[-] (1,4.5) -- (1,4.2) to[out=-90,in=180] (1.2,4) -- (1.8,4);
\draw[<-] (-.8,3) node[above] {\tiny$V_{n+1}$} -- (.8,3) to[out=0,in=90] (1,2.8) -- (1,2.2) to[out=-90,in=180] (1.2,2) -- (1.8,2);
\draw[<-] (-.8,2) node[above] {\tiny$V_n$} -- (-.2,2) to[out=0,in=-90] (0,2.2) -- (0,3.8) to[out=90,in=180] (.2,4)
  -- (.8,4) to[out=0,in=90] (1,3.8) -- (1,3.2) to[out=-90,in=180] (1.2,3) -- (1.8,3);
\draw[<-] (-.8,0) node[above] {\tiny$V_1$} -- (-.2,0) to[out=0,in=-90] (0,.2) -- (0,.5); \draw[dashed] (0,.5) -- (0,1.5);
  \draw[-] (0,1.5) -- (0,1.8) to[out=90,in=180] (.2,2) -- (.8,2) to[out=0,in=90] (1,1.8) -- (1,1.5); \draw[dashed] (1,1.5) -- (1,.5);
  \draw[-] (1,0.5) -- (1,0.2) to[out=-90,in=180] (1.2,0) -- (1.8,0);
\end{tikzpicture}}}}
\uncover<+->{{}=\left.d\over du\right|_{u=0}\sum^N_{n=1}
\begin{tikzpicture}[scale=.5,baseline=1.5cm-2pt]
\draw[<-] (-.8,6) node[above] {\tiny$V_N$} -- (1.8,6);
\path (.5,6) -- node[pos=.4] {$\vdots$} (.5,4);
\draw[<-] (-.8,4) node[above] {\tiny$V_{n+2}$} -- (1.8,4);
\draw[<-] (-.8,3) node[above] {\tiny$V_{n+1}$} -- (0,3) to[out=0,in=180] (1,2) -- (1.8,2);
\draw[<-] (-.8,2) node[above] {\tiny$V_n$} -- (0,2) to[out=0,in=180] (1,3) -- (1.8,3);
\draw[-] (.5,2.5)+(240:.3) arc (240:120:.3) node[midway,left=-2pt] {$u$};
\path (.5,2) -- node[pos=.4] {$\vdots$} (.5,0);
\draw[<-] (-.8,0) node[above] {\tiny$V_1$} -- (1.8,0);
\end{tikzpicture}}
\\
\uncover<+->{{}=\sum^N_{n=1}\check R'_{n,n+1}(0),}
$$
\unpause
where
$$
\check R(u)
\arraycolsep=2pt\def\arraystretch{.9}=PR(u)
=\pMatrix{a(u)&&&\\&c(u)&b(u)&\\&b(u)&c(u)&\\&&&a(u)}
=1+{u\over\sin\lambda}\pMatrix{0&&&\\&\cos\lambda&1&\\&1&\cos\lambda&\\&&&0}
+O(u^2)
$$

\end{frame}

\begin{frame}
\frametitle{Six\-/vertex model and XXZ Heisenberg chain}
\unpause

We have
$$
\check R(u)=1-{1\over\sin\lambda}\left(h+{\cos\lambda\over2}\right)u+O(u^2),
$$
\unpause
where
$$
h=-{1\over2}(\sigma^x\otimes\sigma^x+\sigma^y\otimes\sigma^y-\cos\lambda\>\sigma^z\otimes\sigma^z).
$$
\unpause
Hence
$$
H_1\sin\lambda=H_{\rm XXZ}+{N\over2}\cos\lambda,
$$
\unpause
where $H_{\rm XXZ}$ is the Hamiltonian of the \emph{XXZ Heisenberg} chain:
\eq$$
H_{\rm XXZ}=-{1\over2}\sum_{n=1}^N(\sigma^x_n\sigma^x_{n+1}+\sigma^y_n\sigma^y_{n+1}+\Delta\sigma^z_n\sigma^z_{n+1})
\label{XXZdef}
$$
with $\Delta$ given by~(\ref{Deltadef}):
$$
\Delta={a^2+b^2-c^2\over2ab}=\Cases{-\cos\lambda\\-\ch\lambda}.
$$
\unpause
This leads to the identification of the space $\cH_N$ and the quantum space of the six\-/vertex model:
$$
\cH_N=\underbrace{V\otimes\cdots\otimes V},
\quad
v_\pm=|\pm\rangle.
$$

\end{frame}

\begin{frame}
\frametitle{XXZ Heisenberg chain: pseudovacuums}
\unpause

Due to the ice condition the $z$ component of total spin
$$
S^z={1\over2}\sum^N_{i=1}\sigma^z_n
$$
is a conserved charge:
\eq$$
[T(u),S^z]=[H_{\rm XXZ},S^z]=0.
\label{Szcommut}
$$
Thus the space of states is split into the sum over eigenvalues of~$S^z$.
\unpause

Recall that
\Gather*$$
S^z|\Omega_\pm\rangle=\pm{N\over2}|\Omega_\pm\rangle,
\quad
H_{\rm XXZ}|\Omega_\pm\rangle=-{N\Delta\over2}|\Omega_\pm\rangle.
$$
\unpause
The action of the transfer matrix:
\vspace{-5pt}
$$
T(u)|\Omega_+\rangle
\uncover<+->{=\sum_{\ve=\pm}\begin{tikzpicture}[baseline=-2pt,scale=.4,>={Stealth[length=1mm]}]
\draw[<-] (-.5,1.5) node[left] {\tiny$+$} -- (.5,1.5) node[right] {\tiny$+$};
\draw[<-] (-.5,.5) node[left] {\tiny$+$} -- (.5,.5) node[right] {\tiny$+$};
\draw[<-] (-.5,-.5) node[left] {\tiny$+$} -- (.5,-.5) node[right] {\tiny$+$};
\draw[<-] (-.5,-1.5) node[left] {\tiny$+$} -- (.5,-1.5) node[right] {\tiny$+$};
\draw[->] (0,2) node[above] {\tiny$\ve$} -- (0,-2) node[below] {\tiny$\ve$};
\end{tikzpicture}}
\uncover<+->{=(a^N(u)+b^N(u))|\Omega_+\rangle.}
$$
\unpause
Finally,
$$
T(u)|\Omega_\pm\rangle=(a^N(u)+b^N(u))|\Omega_\pm\rangle.
$$

\end{frame}

\begin{frame}
\frametitle{Six\-/vertex model: three regimes}
\unpause

1.~\emph{Ferroelectric} regime: $\Delta>0$. Let $a>b+c$. \only<1>{Ground configurations:}\only<2>{Excitations?}\only<3->{Excitations:}
$$
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {0,...,3} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$+$} -- (\i,3) -- node[right=-2pt] {\tiny$+$} (\i,2) -- node[right=-2pt] {\tiny$+$} (\i,1)
    -- node[right=-2pt] {\tiny$+$} (\i,0) -- (\i,-.5) node[right] {\tiny$+$};
}
\foreach \i in {0,...,3} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$+$} -- (3,\i) -- node[above=-2pt] {\tiny$+$} (2,\i) -- node[above=-2pt] {\tiny$+$} (1,\i)
    -- node[above=-2pt] {\tiny$+$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$+$};
}
\uncover<2->{
  \draw[fill,white] (1.5,1cm+5pt) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,blue] (1,1) -- node[above=-2pt] {\tiny$-$} (2,1);
}
\uncover<3->{
  \draw[fill,white] (.5,1cm+5pt) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,blue] (0,1) -- node[above=-2pt] {\tiny$-$} (1,1);
  \draw[fill,white] (5pt,.5) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,blue] (0,0) -- node[right=-2pt] {\tiny$-$} (0,1);
  \draw[fill,white] (-.5,5pt) circle (5pt);
  \draw[<-,very thick,shorten >=-.5pt,blue] (-.5,0) node[above=-2pt] {\tiny$-$} -- (0,0);
  \draw[fill,white] (2cm+5pt,1.5) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,blue] (2,1) -- node[right=-2pt] {\tiny$-$} (2,2);
  \draw[fill,white] (2.5,2cm+5pt) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,blue] (2,2) -- node[above=-2pt] {\tiny$-$} (3,2);
  \draw[fill,white] (3cm+5pt,2.5) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,blue] (3,2) -- node[right=-2pt] {\tiny$-$} (3,3);
  \draw[fill,white] (3cm+5pt,3.5) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,blue] (3,3) -- (3,3.5) node[right=-2pt] {\tiny$-$};
}
\end{tikzpicture}
\quad\text{and}\quad
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {0,...,3} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$-$} -- (\i,3) -- node[right=-2pt] {\tiny$-$} (\i,2) -- node[right=-2pt] {\tiny$-$} (\i,1)
    -- node[right=-2pt] {\tiny$-$} (\i,0) -- (\i,-.5) node[right] {\tiny$-$};
}
\foreach \i in {0,...,3} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$-$} -- (3,\i) -- node[above=-2pt] {\tiny$-$} (2,\i) -- node[above=-2pt] {\tiny$-$} (1,\i)
    -- node[above=-2pt] {\tiny$-$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$-$};
}
\end{tikzpicture}
$$
\uncover<4->{On a large lattice any excitations have vanishing weight. $\Rightarrow$ \emph{Frozen order}.}

\uncover<5->{
2.~\emph{Antiferroelectric} regime: $\Delta<-1$, $c>a+b$. \only<5>{Ground configurations:}\only<6>{Excitations?}\only<7->{Excitations:}
$$
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {3,1} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$+$} -- (\i,3) -- node[right=-2pt] {\tiny$-$} (\i,2) -- node[right=-2pt] {\tiny$+$} (\i,1)
    -- node[right=-2pt] {\tiny$-$} (\i,0) -- (\i,-.5) node[right] {\tiny$+$};
}
\foreach \i in {2,0} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$-$} -- (\i,3) -- node[right=-2pt] {\tiny$+$} (\i,2) -- node[right=-2pt] {\tiny$-$} (\i,1)
    -- node[right=-2pt] {\tiny$+$} (\i,0) -- (\i,-.5) node[right] {\tiny$-$};
}
\foreach \i in {3,1} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$-$} -- (3,\i) -- node[above=-2pt] {\tiny$+$} (2,\i) -- node[above=-2pt] {\tiny$-$} (1,\i)
    -- node[above=-2pt] {\tiny$+$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$-$};
}
\foreach \i in {2,0} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$+$} -- (3,\i) -- node[above=-2pt] {\tiny$-$} (2,\i) -- node[above=-2pt] {\tiny$+$} (1,\i)
    -- node[above=-2pt] {\tiny$-$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$+$};
}
\uncover<6->{
  \draw[fill,white] (1.5,1cm+5pt) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,purple] (1,1) -- node[above=-2pt] {\tiny$+$} (2,1);
}
\uncover<7->{
  \draw[fill,white] (1cm+5pt,1.5) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,purple] (1,1) -- node[right=-2pt] {\tiny$-$} (1,2);
  \draw[fill,white] (1.5,2cm+5pt) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,purple] (1,2) -- node[above=-2pt] {\tiny$-$} (2,2);
  \draw[fill,white] (2cm+5pt,1.5) circle (5pt);
  \draw[-,very thick,shorten <=-.5pt,shorten >=-.5pt,purple] (2,1) -- node[right=-2pt] {\tiny$+$} (2,2);
}
\end{tikzpicture}
\quad\text{and}\quad
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {3,1} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$-$} -- (\i,3) -- node[right=-2pt] {\tiny$+$} (\i,2) -- node[right=-2pt] {\tiny$-$} (\i,1)
    -- node[right=-2pt] {\tiny$+$} (\i,0) -- (\i,-.5) node[right] {\tiny$-$};
}
\foreach \i in {2,0} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$+$} -- (\i,3) -- node[right=-2pt] {\tiny$-$} (\i,2) -- node[right=-2pt] {\tiny$+$} (\i,1)
    -- node[right=-2pt] {\tiny$-$} (\i,0) -- (\i,-.5) node[right] {\tiny$+$};
}
\foreach \i in {3,1} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$+$} -- (3,\i) -- node[above=-2pt] {\tiny$-$} (2,\i) -- node[above=-2pt] {\tiny$+$} (1,\i)
    -- node[above=-2pt] {\tiny$-$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$+$};
}
\foreach \i in {2,0} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$-$} -- (3,\i) -- node[above=-2pt] {\tiny$+$} (2,\i) -- node[above=-2pt] {\tiny$-$} (1,\i)
    -- node[above=-2pt] {\tiny$+$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$-$};
}
\end{tikzpicture}
$$
}%
\uncover<8->{The excitations have finite weight. $\Rightarrow$ \emph{Nontrivial thermodynamics}.}

\uncover<9->{
3.~\emph{Disordered} regime: $|\Delta|<1$. No ground configurations. It turns out that this regime is always \emph{critical}.
}

\end{frame}

\end{document}