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\begin{document}

\title{$\SL_2$-Tilings Do Not Exist in Higher Dimensions (mostly)}

\author[Demonet]{Laurent Demonet}
\address[Laurent Demonet]{Graduate School of Mathematics, Nagoya University, Furocho, Chiku\-saku, 464-8602 Nagoya, Japan}
\email{Laurent.Demonet@normalesup.org}
\thanks{The first author was partially supported by JSPS Grant-in-Aid for Young Scientist (B) 26800008.}

\author[Plamondon]{Pierre-Guy Plamondon}
\address[Pierre-Guy Plamondon]{Laboratoire de Math\'ematiques d'Orsay, Univ. Paris-Sud, CNRS, Univ.
Paris-Saclay, 91405 Orsay, France.}
\email{pierre-guy.plamondon@math.u-psud.fr}
\thanks{The second author was partially supported by the French ANR grant SC3A (ANR-15-CE40-0004-01).}

\author[Rupel]{Dylan Rupel}
\address[Dylan Rupel]{Department of Mathematics, University of Notre Dame, Notre Dame, Indiana 46556, USA.}
\email{drupel@nd.edu}

\author[Stella]{Salvatore Stella}
\address[Salvatore Stella]{IN$d$AM - Marie Curie Actions fellow, Universit\`a ``La Sapienza'', Roma, Italy.}
\email{stella@mat.uniroma1.it}

\author[Tumarkin]{Pavel Tumarkin}
\address[Pavel Tumarkin]{Department of Mathematical Sciences, Durham University, South Road, Durham, DH1 3LE, UK.}
\email{pavel.tumarkin@durham.ac.uk}

\begin{abstract}
  We define a family of generalizations of $\SL_2$-tilings to higher dimensions called $\bepsilon$-$\SL_2$-tilings.
  We show that, in each dimension 3 or greater, $\bepsilon$-$\SL_2$-tilings exist only for certain choices of $\bepsilon$.
  In the case that they exist, we show that they are essentially unique and have a concrete description in terms of odd Fibonacci numbers.
\end{abstract}

\maketitle

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\markright{\its S\'eminaire Lotharingien de Combinatoire \bfs 76 \rms (2018), Article~B76e\hfill}
\def\thepage{}

\section{Introduction}
  An $\SL_2$-tiling of the plane (see Definition \ref{defi::sl2plane}) is, loosely speaking, obtained by assigning a positive integer to each integral point of the plane in such a way that all the $2\times 2$ matrices formed by squares of adjacent entries have determinant $1$.
  This definition was introduced by I.~Assem, C.~Reutenauer and D.~Smith in \cite{AssemReutenauerSmith}, and was used by these authors to obtain explicit formulas describing cluster variables in cluster algebras of type $A$.
  $\SL_2$-tilings can also be viewed as $T$-systems of type $A_1$ (e.g. see \cite[Remark 2.1]{DiFrancescoKedem}), where all variables are evaluated in positive integers.

  The fact that $\SL_2$-tilings of the plane even exist 
may be a little surprising in itself.
  On the contrary, already in \cite{AssemReutenauerSmith} it was shown that there are infinitely many of them.
  More recently C.~Bessenrodt, T.~Holm and P.~J{\o}rgensen \cite{BessenrodtHolmJorgensen} classified all such tilings using triangulations of some suitable infinity-gon.

  In this note, we introduce a higher-dimensional analogue of $\SL_2$-tilings: integers will be assigned to points in $\mathbb{Z}^n$, for $n\geq 2$, and $2\times 2$ matrices of adjacent entries in every slice will be
required to satisfy a determinantal identity.

  Although our definition seems like a natural generalization of the notion of $\SL_2$-tilings, our main result (Theorem~\ref{thm:classification}) states that, if $n\geq 3$, then the objects it defines almost never exist.

\section{$\SL_2$-Tilings of the Plane}
  The aim of this note is to study higher-dimensional analogues of the following object.
  \begin{definition}[\cite{AssemReutenauerSmith}]\label{defi::sl2plane}
    A bi-infinite array $(a_{ij})_{i,j\in\ZZ}$ with $a_{ij}\in\ZZ_{>0}$ is called an \emph{$\SL_2$-tiling of $\ZZ^2$} if the entries satisfy the relation
    \begin{equation}\label{eq:sl2 recursion}
      a_{i,j+1}a_{i+1,j}-a_{ij}a_{i+1,j+1}=1.
    \end{equation}
    A bi-infinite array $(b_{ij})_{i,j\in\ZZ}$ with $b_{ij}\in\ZZ_{>0}$  is called an \emph{anti-$\SL_2$-tiling of $\ZZ^2$} if the entries satisfy the relation
    \begin{equation}\label{eq:anti-sl2 recursion}
      b_{i,j+1}b_{i+1,j}-b_{ij}b_{i+1,j+1}=-1.
    \end{equation}
  \end{definition}
  The notion of an anti-$\SL_2$-tiling is not actually giving anything new as shown by the following lemma, however this notion will be useful for our considerations in higher dimensions.
  \begin{lemma}
    \label{lem:n=2 equivalent}
    If $(a_{ij})_{i,j\in\ZZ}$ is an $\SL_2$-tiling, 
%CK: then taking $b_{ij}=a_{i,-j}$ gives an anti-$\SL_2$-tiling.
then, by taking $b_{ij}=a_{i,-j}$, one obtains an anti-$\SL_2$-tiling.
  \end{lemma}
  One should think of the difference between $\SL_2$-tilings and anti-$\SL_2$-tilings as viewing the lattice $\ZZ^2$ ``from above'' or ``from below.''
  The following result from \cite{AssemReutenauerSmith} was our starting point.
  \begin{theorem}[\cite{AssemReutenauerSmith}]
    There exist infinitely many $\SL_2$-tilings of $\ZZ^2$.
  \end{theorem}

%First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire
%--restoring the headers and pagenumbering
\pagenumbering{arabic}
\addtocounter{page}{1}
\markboth{\SMALL DEMONET, PLAMONDON, RUPEL, STELLA, AND TUMARKIN}{\SMALL $\SL_2$-TILINGS DO NOT EXIST IN HIGHER DIMENSIONS (MOSTLY)}

  In fact, it is shown in \cite{AssemReutenauerSmith} that any admissible frontier of $1$'s in the 
lattice can be completed into a unique $\SL_2$-tiling.
  An interpretation of all possible $\SL_2$-tilings was later given in \cite{BessenrodtHolmJorgensen} in terms of triangulations of a polygon with infinitely many vertices.

  The following anti-$\SL_2$-tiling will be relevant in our higher dimensional analysis.
  We will call it the \emph{staircase} anti-$\SL_2$-tiling of $\ZZ^2$.
  \begin{example}\label{ex:Fibonacci}
    Consider the anti-$\SL_2$-tiling $(a_{ij})_{i,j\in\ZZ}$ of $\ZZ^2$ with $a_{ij}=1$ if $i+j\in\{0,1\}$.
    Using \eqref{eq:anti-sl2 recursion} and the well-known recursion $F_{2r-1}F_{2r+3}=F_{2r+1}^2+1$ $(r\ge1)$ for the odd Fibonacci numbers, it is easy to see that
      \[
        a_{ij}
        =
        \begin{cases}
          F_{2r-1} & \text{if $i+j=r\ge1$;}\\
          F_{-2r+1} & \text{if $i+j=r\le0$;}
        \end{cases}
      \]
      where we number the Fibonacci numbers as:
      \[
        \begin{tabular}{|c|c|c|c|c|c|c|c}
          $F_1$ & $F_2$ & $F_3$ & $F_4$ & $F_5$ & $F_6$ & $F_7$ & $\cdots$\\
          \hline $1$ & $1$ & $2$ & $3$ & $5$ & $8$ & $13$ & $\cdots$
        \end{tabular}
      \]
      The following figure is a portion of this tiling.
      Note the bold frontier of $1$'s; it is an ``infinite staircase''.
      \begin{center}
        \begin{tikzpicture}
          \draw (0,1) node{${\bf 1}$} (.75,1) node{${\bf 1}$} (1.5,1) node{$2$} (2.25,1) node{$5$} (3,1) node{$13$} (3.75,1) node{$34$} (4.5,1) node{$89$} (5.25,1) node{$233$};
          \draw (0,.5) node{$2$} (.75,.5) node{${\bf 1}$} (1.5,0.5) node{${\bf 1}$} (2.25,0.5) node{$2$} (3,0.5) node{$5$} (3.75,0.5) node{$13$} (4.5,0.5) node{$34$} (5.25,0.5) node{$89$};
          \draw (0,0) node{$5$} (.75,0) node{$2$} (1.5,0) node{${\bf 1}$} (2.25,0) node{${\bf 1}$} (3,0) node{$2$} (3.75,0) node{$5$} (4.5,0) node{$13$} (5.25,0) node{$34$};
          \draw (0,-.5) node{$13$} (.75,-.5) node{$5$} (1.5,-.5) node{$2$} (2.25,-.5) node{${\bf 1}$} (3,-.5) node{${\bf 1}$} (3.75,-.5) node{$2$} (4.5,-.5) node{$5$} (5.25,-.5) node{$13$};
          \draw (0,-1) node{$34$} (.75,-1) node{$13$} (1.5,-1) node{$5$} (2.25,-1) node{$2$} (3,-1) node{${\bf 1}$} (3.75,-1) node{${\bf 1}$} (4.5,-1) node{$2$} (5.25,-1) node{$5$};
          \draw (0,-1.5) node{$89$} (.75,-1.5) node{$34$} (1.5,-1.5) node{$13$} (2.25,-1.5) node{$5$} (3,-1.5) node{$2$} (3.75,-1.5) node{${\bf 1}$} (4.5,-1.5) node{${\bf 1}$} (5.25,-1.5) node{$2$};
          \draw (0,-2) node{$233$} (.75,-2) node{$89$} (1.5,-2) node{$34$} (2.25,-2) node{$13$} (3,-2) node{$5$} (3.75,-2) node{$2$} (4.5,-2) node{${\bf 1}$} (5.25,-2) node{${\bf 1}$};
          \draw (0,-2.5) node{$610$} (.75,-2.5) node{$233$} (1.5,-2.5) node{$89$} (2.25,-2.5) node{$34$} (3,-2.5) node{$13$} (3.75,-2.5) node{$5$} (4.5,-2.5) node{$2$} (5.25,-2.5) node{${\bf 1}$};
        \end{tikzpicture}
      \end{center}
    \end{example}

\section{$\SL_2$-Tilings in Higher Dimensions}
  For a fixed integer $n\geq 2$, denote vectors in $\ZZ^n$ by $\bi=(i_1\dots,i_n)$ and let $\be_k$ be the $k$-th unit vector in the same lattice.
  A \emph{signature matrix} is a symmetric $n\times n$ matrix $\bepsilon=(\epsilon_{k\ell})$ with $\epsilon_{k\ell}=\pm 1$ whenever $k\neq \ell$ and $\epsilon_{kk}=-1$.
  \begin{definition}
    Fix a signature matrix $\bepsilon$.
    An array $(a_{\bi})_{\bi\in\ZZ^n}$ with $a_{\bi}\in\ZZ_{>0}$ is called an \emph{$\bepsilon$-$\SL_2$-tiling of $\ZZ^n$} if for 
all $k$ and $\ell$ with $k \neq\ell$ we have
    \begin{equation}\label{eq:higher sl2 recursion}
      a_{\bi+\be_\ell}a_{\bi+\be_k}-a_{\bi}a_{\bi+\be_k+\be_\ell}=\epsilon_{k\ell}.
    \end{equation}
  \end{definition}
  The requirement on the diagonal entries of signature matrices might seem arbitrary right now because they do not play any role in the above definition; we will see later on that it is indeed a consistent choice.

  In the case $n=2$ there are only two possible signature matrices recovering the notions of $\SL_2$-tilings and anti-$\SL_2$-tilings of $\ZZ^2$.

  The following is a portion of an $\bepsilon$-$\SL_2$-tiling of $\ZZ^3$ with all entries of $\bepsilon$ equal to $-1$.
  \begin{center}
    \begin{tikzpicture}[x=1.7cm, y=1.3cm, z={(-1.3cm,-0.73cm)}]
      \pgfmathsetmacro{\d}{0.18}
      \foreach \i in {0,1,2}
      {
        \foreach \j in {0,1,2}
        {
          \foreach \k in {0,1}
          {
            \draw[lightgray] (\i,\j,\k+\d) -- (\i,\j,\k+1-\d);
            \draw[lightgray] (\i,\k+\d,\j) -- (\i,\k+1-\d,\j);
            \draw[lightgray] (\k+\d,\i,\j) -- (\k+1-\d,\i,\j);
          }
        }
      }
      \node at (0,0,0) {$1$};

      \node at (1,0,0) {$1$};
      \node at (0,1,0) {$1$};
      \node at (0,0,1) {$1$};

      \node at (2,0,0) {$2$};
      \node at (1,1,0) {$2$};
      \node at (0,2,0) {$2$};
      \node at (0,1,1) {$2$};
      \node at (0,0,2) {$2$};
      \node at (1,0,1) {$2$};

      \node at (1,1,1) {$5$};
      \node at (2,1,0) {$5$};
      \node at (2,0,1) {$5$};
      \node at (1,2,0) {$5$};
      \node at (0,2,1) {$5$};
      \node at (0,1,2) {$5$};
      \node at (1,0,2) {$5$};

      \node at (2,1,1) {$13$};
      \node at (1,2,1) {$13$};
      \node at (1,1,2) {$13$};
      \node at (2,2,0) {$13$};
      \node at (2,0,2) {$13$};
      \node at (0,2,2) {$13$};

      \node at (2,2,1) {$34$};
      \node at (1,2,2) {$34$};
      \node at (2,1,2) {$34$};

      \node at (2,2,2) {$89$};
    \end{tikzpicture}
  \end{center}

  We will say that two $n\times n$ signature matrices $\bepsilon$ and $\bepsilon'$ are \emph{equivalent} if any $\bepsilon$-$\SL_2$-tiling of $\ZZ^n$ can be made into an $\bepsilon'$-$\SL_2$-tiling of $\ZZ^n$ by applying some linear transformation to its indices 
(i.e., elements of $\ZZ^n$).
  By Lemma~\ref{lem:n=2 equivalent}, all signature matrices with $n=2$ are equivalent.
  The situation is more complicated when $n\geq 3$ since not all 
signature matrices are equivalent.
  On the other hand, there are easy ways of constructing signature matrices equivalent to a given signature matrix $\bepsilon$.
  \begin{lemma}\label{le:flip}
    Let $\bepsilon=(\epsilon_{k\ell})$ be any signature matrix and write $\bepsilon^{(r)}$ for the matrix obtained from $\bepsilon$ by changing the sign of all the entries in row $r$ and column $r$, leaving the diagonal entries fixed.
    That is, $\bepsilon^{(r)} =(\epsilon'_{k\ell})$ where $\epsilon'_{k\ell}=-\epsilon_{k\ell}$ if exactly one of $k$ and $\ell$ equals $r$ and $\epsilon'_{k\ell}=\epsilon_{k\ell}$ otherwise.
    If $(a_{\bi})_{\bi\in\ZZ^n}$ is an $\bepsilon$-$\SL_2$-tiling, 
then, by taking $b_{\bi}=a_{\bi-2i_r\be_r}$, one obtains an $\bepsilon^{(r)}-\SL_2$-tiling.
  \end{lemma}
  \begin{proof}
Indeed, all the relations (\ref{eq:higher sl2 recursion}) not involving the index 
$r$ are satisfied for $(b_{\bi})_{\bi\in\ZZ^n}$ because they are satisfied for $(a_{\bi})_{\bi\in\ZZ^n}$.
    However, the relations involving the index 
$r$ pick up a sign when passing from $(a_{\bi})_{\bi\in\ZZ^n}$ to $(b_{\bi})_{\bi\in\ZZ^n}$, as desired.
  \end{proof}
  \begin{definition}
    If $\bepsilon$ is a signature matrix such that $\epsilon_{k\ell}=1$ 
{\em(}respectively $\epsilon_{k\ell}=-1${\em} whenever $k\neq\ell$, we refer to an $\bepsilon$-$\SL_2$-tiling as an \emph{$\SL_2$-tiling} 
{\em(}respectively \emph{anti-$\SL_2$-tiling}{\em)} of $\ZZ^n$.
  \end{definition}

  \begin{lemma}\label{le:constant slices}
    Let $n\ge 3$ and assume $(a_{\bi})_{\bi\in\ZZ^n}$ is either an $\SL_2$-tiling or an anti-$\SL_2$-tiling of $\ZZ^n$.
    Then for any $r\in\ZZ$ the set $\{a_{\bi}: \sum_{j=1}^n i_j=r\}$ consists of a single element.
  \end{lemma}
  \begin{proof}
    We will show that the value of $a_{\bi+\be_k}$ depends only on $a_{\bi}$.
    This is enough to conclude our claim because any vector in $\ZZ^n$ whose entries sum up to a given integer $r$ can be obtained from any other with the same property via ``zig-zagging'' up and down by simultaneously adding and subtracting unit vectors.

    Pick any three distinct indices $j,k,\ell\in[1,n]$.
    To prove our claim we compute $a_{\bi+\be_j+\be_k+\be_\ell}$ in terms of $a_{\bi}, a_{\bi+\be_j}, a_{\bi+\be_k}, a_{\bi+\be_\ell}$ in three different ways.
    For simplicity of notation we 
set
    \[
      \epsilon_{jk}=\epsilon_{j\ell}=\epsilon_{k\ell}=\epsilon,
      \quad\quad
      a_{\bi}=a,
      \quad\quad
      a_{\bi+\be_j}=x,
      \quad\quad
      a_{\bi+\be_k}=y,
      \quad\quad
      a_{\bi+\be_\ell}=z.
    \]
    The following picture will be useful.
    \begin{center}
      \begin{tikzpicture}
        \draw (0,0) -- ++(1.5,1)  node[above] {$z$} ;
        \draw (0,0) node[left] {$\frac{xz-\epsilon}{a}$} ;
        \draw (0,0) -- ++(0,-1.6) node[left] {$x$} ;
        \draw [dashed] (1.5,1) -- ++(0,-1.6) node[above right] {$a$};
        \draw [dashed] (0,-1.6) -- ++(1.5,1) ;
        \draw (0,0) -- ++(1.5,-1);
        \draw (1.5,1) -- ++(1.5,-1) node[right] {$\frac{yz-\epsilon}{a}$};
        \draw (0,-1.6) -- ++(1.5,-1) node[below]{$\frac{xy-\epsilon}{a}$};
        \draw [dashed] (1.5,-.6) -- ++(1.5,-1);
        \draw (1.5,-1) -- ++(0,-1.6) ;
        \draw (1.5,-1) -- ++(1.5,1) ;
        \draw (1.5,-2.6) -- ++(1.5,1) ;
        \draw (3,0) -- ++(0,-1.6) node[right]{$y$};
      \end{tikzpicture}
    \end{center}
    Using \eqref{eq:higher sl2 recursion} three 
times, we get
    \[
      a_{\bi+\be_j+\be_k}=\frac{xy-\epsilon}{a},
      \quad\quad
      a_{\bi+\be_k+\be_\ell}=\frac{yz-\epsilon}{a},
      \quad\quad
      a_{\bi+\be_j+\be_\ell}=\frac{xz-\epsilon}{a}.
    \]
    Three more applications of \eqref{eq:higher sl2 recursion} then
lead to
    \begin{equation*}
      a_{\bi+\be_j+\be_k+\be_\ell}=\left\{
        \begin{array}{l}
          \frac{a_{\bi+\be_j+\be_k}a_{\bi+\be_j+\be_\ell}-\epsilon}{a_{\bi+\be_j}}=\frac{xyz}{a^2}-\epsilon\frac{y+z}{a^2}-\epsilon\frac{a^2-\epsilon}{a^2x}\\
          \frac{a_{\bi+\be_j+\be_k}a_{\bi+\be_k+\be_\ell}-\epsilon}{a_{\bi+\be_k}}=\frac{xyz}{a^2}-\epsilon\frac{x+z}{a^2}-\epsilon\frac{a^2-\epsilon}{a^2y}\\
          \frac{a_{\bi+\be_j+\be_\ell}a_{\bi+\be_k+\be_\ell}-\epsilon}{a_{\bi+\be_\ell}}=\frac{xyz}{a^2}-\epsilon\frac{x+y}{a^2}-\epsilon\frac{a^2-\epsilon}{a^2z}
        \end{array}\right.
    \end{equation*}
    It follows that $\frac{x-y}{a^2}=\frac{a^2-\epsilon}{a^2x}-\frac{a^2-\epsilon}{a^2y}$ or $(xy+a^2-\epsilon)(x-y)=0$.
    But $xy+a^2-\epsilon\ge1$ since $a,x,y\ge1$, hence $x=y$.
    Similarly $y=z$.
    The result then follows by iterating on all possible triples of distinct indices $j,k,\ell$.
  \end{proof}

  We now come to our first main result: in dimension $n$, an ``infinite staircase'' of $1$'s yields the only possible anti-$\SL_2$-tiling.

  \begin{theorem}
    \label{thm:antitiling}
    For $n\ge3$, there exists a unique {\em(}up to translation{\em)} anti-$\SL_2$-tiling of $\ZZ^n$.
    Any of its ``two-dimensional slices'' obtained by fixing all but two of the entries of $\bi$ is a translation of the staircase anti-$\SL_2$-tiling of $\ZZ^2$ from Example \ref{ex:Fibonacci}.
    In particular, all the integers appearing are odd Fibonacci numbers.
  \end{theorem}
  \begin{proof}
    Assume $(a_{\bi})_{\bi\in\ZZ^n}$ is an anti-$\SL_2$-tiling of $\ZZ^n$.
    Pick $\bi$ with $a_{\bi}$ minimal.
    Applying \eqref{eq:higher sl2 recursion},
we get
    \[
      a_{\bi+\be_1}a_{\bi-\be_2}=a_{\bi}a_{\bi+\be_1-\be_2}+1=a_{\bi}^2+1,
    \]
    where we 
used Lemma~\ref{le:constant slices} in the last equality.
    If $a_{\bi}>1$, this implies $a_{\bi+\be_1}<a_{\bi}$ or $a_{\bi-\be_2}<a_{\bi}$, contradicting minimality, so we must have $a_{\bi}=1$.
    In turn, again leveraging Lemma~\ref{le:constant slices}, this implies $\{a_{\bi+\be_k},a_{\bi-\be_k}\}=\{1,2\}$ for any $k\in[1,n]$.
    Without loss of generality (by replacing $\bi$ by $\bi+\be_1$ if needed) we will assume $a_{\bi+\be_k}=2$ and $\sum_{j=1}^n i_j=1$.
    Then, applying \eqref{eq:higher sl2 recursion} repeatedly, we see that $a_{\bi'}$ with $\sum_{j=1}^n i'_j=r\ge1$ is exactly the $r^{th}$ odd Fibonacci number $F_{2r-1}$ (see Example~\ref{ex:Fibonacci}).
    Similarly one sees that $a_{\bi'}$ with $\sum_{j=1}^n i'_j =r\le0$ is the odd Fibonacci number $F_{-2r+1}$.
  \end{proof}

  \begin{proposition}\label{pr:nonexistence}
    There does not exist any $\SL_2$-tiling of $\ZZ^n$ for $n\geq3$.
  \end{proposition}
  \begin{proof}
    Since any $3$-dimensional slice of an $\SL_2$-tiling of $\ZZ^n$ is an $\SL_2$-tiling of $\ZZ^3$, it suffices to show that there is no $\SL_2$-tiling of $\ZZ^3$.
    Assume $(a_{\bi})_{\bi\in\ZZ^3}$ is an $\SL_2$-tiling of $\ZZ^3$.
    Pick $\bi$ with $a_{\bi}$ minimal.
    Applying \eqref{eq:higher sl2 recursion},
% gives
we get
    \[
      a_{\bi+\be_1}a_{\bi-\be_2}=a_{\bi}a_{\bi+\be_1-\be_2}-1=a_{\bi}^2-1,
    \]
    where we 
used Lemma~\ref{le:constant slices} in the last equality.
    But this implies $a_{\bi+\be_1}<a_{\bi}$ or $a_{\bi-\be_2}<a_{\bi}$, contradicting minimality.
  \end{proof}

  \begin{corollary}\label{co:n=3}
    For $n=3$, there are precisely $4$ signature matrices $\bepsilon$ for which there exists an $\bepsilon$-$\SL_2$-tiling.
    For such $\bepsilon$, this $\bepsilon$-$\SL_2$-tiling is unique {\em(}up to translation{\em)}.
    More precisely, an $\bepsilon$-$\SL_2$-tiling of $\ZZ^3$ exists if and only if $\epsilon_{12}\epsilon_{13}\epsilon_{23}=-1$.
  \end{corollary}
  \begin{proof}
    The claim follows immediately from the observation that any signature matrix for $n=3$ is either one of the two satisfying $\epsilon_{12}=\epsilon_{13}=\epsilon_{23}$ or is obtained from one of these with a single application of Lemma~\ref{le:flip}.
  \end{proof}

  We are finally ready to classify all $\bepsilon$-$\SL_2$-tilings for any $n\geq3$.
  \begin{theorem}
    \label{thm:classification}
    For $n\geq3$, there are precisely $2^{n-1}$ signature matrices $\bepsilon$ for which there exists an $\bepsilon$-$\SL_2$-tiling of $\ZZ^n$.
    They are precisely the signature matrices obtainable from the anti-$\SL_2$-signature matrix by repeated application of Lemma~\ref{le:flip}.
    Whenever an $\bepsilon$-$\SL_2$-tiling exists, it is unique up to translation.
  \end{theorem}
  \begin{proof}
    Let $(a_{\bi})_{\bi\in\ZZ^n}$ be an $\bepsilon$-$\SL_2$-tiling of $\ZZ^n$.
    Fixing all but any three distinct entries of $\bi$ gives an $\bepsilon'$-tiling of $\ZZ^3$ whose signature matrix $\bepsilon'$ is the submatrix of $\bepsilon$ with the corresponding rows and columns.
    Therefore, it follows from Corollary~\ref{co:n=3} that we have an inclusion $E \subset E'$, where $E$ is the set of $n\times n$ signature matrices $\bepsilon$ which admit an $\bepsilon$-$\SL_2$-tiling, and $E'$ is the set of $n\times n$ signature matrices $\bepsilon$ satisfying $\epsilon_{jk} \epsilon_{k\ell} \epsilon_{j\ell} = -1$ for any triple of distinct indices $j,k,\ell$.

    Any row (or equivalently any column) of a matrix $\bepsilon$ in $E'$ uniquely determines all the remaining entries of $\bepsilon$, moreover all possible choices of entries in this fixed row (or equivalently column) are allowed.
    Indeed, assume for the sake of clarity that the matrix $\bepsilon$ has been computed using row $1$, then
    \[
      \epsilon_{jk} \epsilon_{k\ell} \epsilon_{j\ell} = (-\epsilon_{1k}\epsilon_{1j})(-\epsilon_{1k}\epsilon_{1\ell})(-\epsilon_{1j}\epsilon_{1\ell})=-1.
    \]
    Therefore $E'$ is in bijection with $\{\pm 1\}^{n-1}$ and $\#E'= 2^{n-1}$.

    Using Lemma~\ref{le:flip}, there is an action of $(\ZZ/2\ZZ)^{n-1}$ on $E$ given by $\bepsilon \mapsto \bepsilon^{(r)}$ for $1 \leq r \leq n-1$.
    This action is free; indeed the only element of $(\ZZ/2\ZZ)^{n-1}$ leaving invariant the last column of any given matrix of $E$ is the identity.
    
Due to Theorem~\ref{thm:antitiling}, $E$ is not empty, and so we compute $\#E \geq 2^{n-1} = \# E' \geq \#E$ and deduce that $E = E'$.

    The uniqueness claim also follows immediately from Corollary~\ref{co:n=3} by fixing all but any three distinct entries of $\bi$.
  \end{proof}

  \begin{remark}
    It is now clear why we choose the diagonal entries of $\bepsilon$ to be equal to $-1$: any $\bepsilon$-$\SL_2$-tiling consists of odd Fibonacci numbers and \eqref{eq:higher sl2 recursion} is satisfied also for $k=\ell$.
  \end{remark}

\section*{Acknowledgements}
  These results were obtained while we were taking part in the Conference on Cluster Algebras and Representation Theory at the KIAS in Seoul, South Korea.
  We would like to thank the organizers of that meeting for the stimulating environment they provided.

  We are especially grateful to Peter J\o rgensen for his very inspiring talk at the same conference: the results presented here are the content of the exciting discussions he sparked.

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\end{document}
