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\newcommand{\theo}[1]{{\bf \color{olive} \# #1 \#}}
\newcommand{\alex}[1]{{\bf \color{blue} \# #1 \#}}

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%Rundom first attempt, more influenced by the property of fitting in a single line
\title[Laplacian tilings]{Generalized Laplacian tilings and root zonotopes}



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%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
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\author[Theo Douvropoulos and Alex McDonough]{{
\texorpdfstring{Theo Douvropoulos \thanks{\href{mailto:tdouvropoulos@brandeis.edu}{tdouvropoulos@brandeis.edu}.}\addressmark{1} \and Alex McDonough\thanks{ \href{mailto:alexmcd@uoregon.edu}{alexmcd@uoregon.edu}. Alex McDonough was partially supported by NSF grant DMS-2039316.}\addressmark{2}}{Theo Douvropoulos and Alex McDonough} }
}

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\address{\addressmark{1}Department of Mathematics, Brandeis University, Waltham, MA, USA.\\ \addressmark{2} Department of Mathematics, University of Oregon, Eugene, OR, USA.}

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\abstract{Given any two equi-oriented collections $A$ and $B$ of vectors in $\mathbb{R}^r$, we construct a family of tilings of $\mathbb{R}^r$. The tiles themselves are related to the bases of $A$, while their relative frequency are related to bases of $B$. Our construction gives a common generalization to Penrose tilings as well as zonotopal space tilings associated to regular matroids. }

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%\resume{We pr{\`o}ve a v{\`e}ry co{\`o}l theor{\`e}m about tilings}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{periodic tilings, aperiodic tilings, Laplacian, root systems, zonotopes}

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\begin{document}

\maketitle
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\section{Introduction}

% \alex{We should start with an example in the introduction. The code is totally functional now, so I can cook something up very fast. A few desirable properties for this example (that don't all have to be met, but are generally from most to least important).}
% \begin{enumerate}
% \item We should have $r=2$ or else we can't really show it. 
% \item $k$ shouldn't be too big so that it isn't too complicated. Probably at least 2 and at most 5 is best (and likely closer to 2). 
% \item It would be nice for it to come from a root zonotope (maybe a specific set of roots and coroots). 
% \item It would be nice to do an example where the matrices aren't the same.
% \item It would be nice to do an example where the tiling is aperiodic (but we could also save this for later if you'd prefer the first example to have a fundamental domain). 
% \end{enumerate}


A well known theorem by Shephard \cite{shephard} states that the zonotope associated with a real matrix $M$ tiles its affine span if and only if the matroid represented by $M$ is regular; this happens in particular if $M$ is unimodular. A less well known generalization by Crapo and Senechal \cite{Crapo_Senechal} states that for general $M$, one can still tile space using the bases of $M$, but some tiles will appear with higher frequency than others. In particular, their tilings consist of translates of the zonotope associated with $M$ along with some additional copies of certain tiles to fill in the gaps. Crapo and Senechal showed further that when the entries of $M$ are rational, the tiling is periodic.

Their work utilizes the cut-and-project method, which uses a lattice $\ZZ^n$ and a lower dimensional affine space $X$ (that "cuts" $\mathbb{Z}^n$). The tiling is constructed by projecting certain points of $\ZZ^n$ on to $X$ when these points are sufficiently close to $X$. This construction goes back to DeBruijn who used it to give an algebraic proof of the properties of the Penrose tilings of the plane with thin and thick rhombi~\cite{deBruijn}. In the Crapo--Senechal construction, every tile appears with frequency proportional to its volume. In particular, Penrose tilings are aperiodic because the ratio of the volumes of the two types of rhombi is the irrational number $\phi=\frac{1+\sqrt{5}}{2}$.

In cases like the Penrose tiling, where the frequency of each tile is proportional to its volume, it follows that the total volume covered by all tiles of a given type is proportional to the \emph{square} of its volume. When the tilings are periodic, this property essentially implies (via the Cauchy--Binet formula) that the volume of the fundamental domain of the tiling must agree with the determinant of the Laplacian matrix $L:=M\cdot M^T$. This was already observed in particular when $M$ is unimodular: the zonotope $Z(M)$ has volume equal to $\det(L)$ (see for instance \cite[Proposition~2.52]{DeConcini_Procesi_book}). 

The second author first explored tilings as a way to understand a Laplacian result in \emph{high-dimensional chip-firing}, where a matrix is multiplied by its own transpose~\cite{McDonough_multijections}. However, later work suggested that related constructions can be used to give a geometric interpretation for more general applications of the Cauchy--Binet formula, where one is interested in $A\cdot B^T$ for two possibly distinct matrices $A$ and $B$~\cite{Doolittle_McDonough_Signed}. This idea ties in with works of the first author, which suggest that such generalizations often encode nontrivial combinatorics \cite{Chapuy_Douvropoulos_WL,Chapuy_Douvropoulos_MT}. In this project we unite the two approaches and prove the following main result (see \Cref{thm:maintiling,thm:mainproportions} and \Cref{cor:rational} for more details): 

\begin{thm}
For a pair $A,B$ of equi-oriented $r\times n$ matrices, there exist tilings of $\mathbb{R}^r$ whose tiles are given by the bases $A_I$ of $A$, $I\subset [n]$, and appear with frequencies equal to the volumes $\op{Vol}(B_I)$. When all frequencies are integers, the tiling is periodic and its fundamental domain has volume equal to $\det(L)$ for the Laplacian $L:=A\cdot B^T$.   
\end{thm}

\begin{wrapfigure}{R}{0.45\textwidth}
    \centering
    \includegraphics[width=.95\linewidth]{tiling_I2_6.PNG}
    \caption{Root-Coroot tiling for $I_2(6)$}
    \label{fig:I2(6)}
\end{wrapfigure}


As an application to (irreducible) Weyl groups $W$, we construct periodic tilings $\tile(W)$, whose combinatorics are described by root-theoretic invariants: the Coxeter number $h$, the rank $n$, and the connection index $I(W)$ of $W$. In particular, their fundamental domains have volume $h^n$, which explains the following well observed \cite{root_zonotopes} phenomenon. The volumes of the root zonotopes in type-A are given by a product formula $\op{vol}\big(Z(A_{n-1})\big)=n^{n-2}$, while for the other root systems they are not, even though the associated Laplacian determinants do factor. Indeed, it is the fundamental domain of $\tile(W)$ that behaves well and is related to the Laplacian, as the previous theorem suggests, not the root zonotope $Z(W)$ which is often a strict subset of the fundamental domain.

\begin{example}
In \Cref{fig:I2(6)} we see a tiling $\tile\big(I_2(6)\big)$; five translates  of the fundamental domain are indicated by thick black boundaries, while the central copy contains in orange dashed boundary the root zonotope $Z(\Phi^+)$. There is a total of $9$ types of tiles, which form three classes: small rhombi, large rhombi, and parallelograms. Moreover, the little rhombi have volume $1$ and relative frequency $3$, while the opposite is true for the large rhombi; the parallelograms have volume $2$ and relative frequency $2$. \Cref{thm:root_zonotopes} expresses the probability that a point lies in a translate of some tile $\Phi^+_{\sigma}$ as $I(W_{\sigma})/h^n$, where $I(W_{\sigma})$ is the connection index of the reflection subgroup $W_{\sigma}\leq W$ associated to the tile. Indeed, a point belongs to the translate of a fixed (big or little) rhombus with probability $3/36$ (note that $3=I(A_2)$) and it belongs to the translate of a fixed parallelogram with probability $4/36$ (note that $4=I(A_1\times A_1)$).
\end{example}



\section{The Main Construction}\label{sec:construction}

Throughout this article, fix positive integers $r$, $k$, and $n$ such that $n = r+k$. We will use the notation $\binom{[n]}{r}$ for the $r$ element subsets of the numbers $1$ through $n$. We often use the variable $\sigma$ for an element of $\binom{[n]}{r}$, and will write $\widehat \sigma$ for the set $[n] \setminus \sigma$. For any matrix $M$ with $n$ columns, and any $\sigma \in \binom{[n]}{r}$, we write $M_\sigma$ for the submatrix of $M$ restricted to columns indexed by $\sigma$ and $M_{\hsigma}$ for the submatrix of $M$ restricted to the remaining columns. Similarly, if $\mbf v$ is an $n$-dimensional vector, we will write $\mbf v_{\sigma}$ for the restriction of $\mbf v$ to entries indexed by $\sigma$ and $\mbf v_{\hsigma}$ for the restriction of $\mbf v$ to the remaining entries. 

Let $A$ and $B$ be two full rank $r\times n$ matrices with real valued entries. Note that many of our results depend only on the subspaces of $\RR^n$ generated by the rows of these matrices, but our construction is easier to describe when specific values are provided. More precisely, row operations on $A$ do not change the combinatorics of the tiling, while row operations on $B$ have no impact at all. 

For our construction, we also require that $A$ and $B$ are \emph{equi-oriented}. We say that $A$ and $B$ are \defn{equi-oriented} if for every $\sigma \in \binom{[n]}{r}$, we have \[\det(A_\sigma)\det(B_\sigma) \ge 0.\]
%A few important examples of equi-oriented matrices are two copies of the same matrix, or two \emph{totally nonnegative} matrices (see \Cref{sec:penrose}).

In addition to equi-oriented matrices $A$ and $B$, our tiling also depends on a choice of vector $\gamma \in \RR^n$ which must satisfy a certain genericity condition. 

For a fixed $r\times n$ matrix $A$ and a choice of $\sigma\in\binom{[n]}{r}$, we will write $\mathcal T_\sigma$ for the \defn{fundamental parallelepiped} of $A_\sigma$. In particular, $\mathcal T_\sigma$ is the polytope whose vertices are the sums of all $2^r$ subsets of columns of $A_\sigma$. Equivalently,  
 \[\mathcal T_\sigma := \left\{A_\sigma \mbf x~\mid ~ \mbf x \in [0,1]^r\right\}.\]

The set of parallelepipeds of the form $\mathcal T_\sigma$ for $\sigma\in\binom{[n]}{r}$ are the \emph{prototiles} of our tiling (with possible duplicates and volume 0 tiles). The translates of the prototiles correspond to the vectors in $\ZZ^r$. More precisely, for every $\sigma\in\binom{[n]}{r}$ and $\mbf z \in \ZZ^r$, we will construct a translation vector $\mbf s_{(\sigma,\mbf z)}\in \RR^r$ and then place a copy of $\mathcal T_\sigma$ that is translated by this amount. The formula for $\mbf s_{(\sigma,\mbf z)}$ is given by~\eqref{eq:translate}, but it will take some work to define the necessary terminology. 

Instead of working directly with $B$, most of our calculations involve a matrix $\widetilde B$ that is in some sense dual to $B$. To be precise, $\widetilde B$ is any full rank $k \times r$ matrix whose row space is dual to the row space of $B$. In other words, 
\begin{equation}\label{eq:Btilde}\widetilde BB^T = \mbf 0 \text{~ and $\widetilde B$ is full rank.}\end{equation}

There is a simple procedure to compute $\widetilde B$ from $B$. First, permute the columns of $B$ until the first $r$ columns are linearly independent (which must be possible since $B$ is full rank). Second, use row operations to put this matrix into reduced row echelon form. This produces a matrix of the form $\begin{pmatrix} I_r & D \end{pmatrix}$, where $I_r$ is the $r\times r$ identity matrix and $D$ is any $r\times k$ real matrix. Next, define a new $k \times n$ matrix $\begin{pmatrix} -D^T & I_k \end{pmatrix}$. Finally, undo the initial row permutation. For straightforward linear algebra reasons, this final matrix must satisfy the conditions for $\widetilde B$. Furthermore, there is a nice relationship between the minors of $B$ and $\widetilde B$. 

\begin{lem}\label{lem:detmatch}
    The matrix $\widetilde B$ can be chosen such that for every $\sigma\in \binom{[n]}r$, we have $\det(B_\sigma) = \det(\wtilde B_{\hsigma})$. 
\end{lem}

Note that \Cref{lem:detmatch} can be generalized to any $\widetilde B$ satisfying~\eqref{eq:Btilde} to say that the ratio $\det(B_\sigma)/\det(\wtilde B_\hsigma)$ is always a fixed value, but is not necessarily 1. 

For any choice of $\mbf z \in \ZZ^r$ and $\sigma \in \binom{[n]}{r}$, the formula for $\mbf s_{(\sigma,\mbf z)}$ is given by~\eqref{eq:translate}. In this formula, we use $\lceil{\cdot}\rceil$ for the element-wise ceiling function. In other words, we round up each entry to the nearest larger integer. Recall that $\gamma\in \RR^n$ is a generic vector, and $\gamma_\sigma$ and $\gamma_\hsigma$ are its restrictions to entries in and out of $\sigma$ respectively. The precise genericity condition for $\gamma$ is that for all $\mbf z \in \ZZ^r$ and $\sigma \in \binom{[n]}r$, the vector given by $\widetilde B_{\widehat\sigma}^{-1}\widetilde B_\sigma(\gamma_\sigma- \mathbf z) + \gamma_{\widehat\sigma}$ does not have any integer entries. 

\begin{equation}
    \label{eq:translate}\mbf s_{(\sigma,\mbf z)} := A_\sigma (\mathbf z-\gamma_\sigma) + A_{\widehat\sigma} \left(\left\lceil\widetilde B_{\widehat\sigma}^{-1}\widetilde B_\sigma(\gamma_\sigma- \mathbf z) + \gamma_{\widehat\sigma}\right\rceil - \gamma_{\widehat \sigma}\right).
\end{equation}

We are now ready to state the two main theorems of this article. 

\begin{thm}\label{thm:maintiling}
    For each $\sigma \in\binom{[n]}{r}$ and $\mbf z \in \ZZ^r$, place a copy of $\mathcal T_\sigma$ translated by $\mbf s_{(\sigma,\mbf z)}$. This gives a tiling of $\RR^r$ with no gaps or overlaps (other than the volume 0 overlap at the boundaries of the tiles). 
\end{thm}

In \Cref{sec:proofs}, we show how to obtain this tiling by taking a slice of a periodic tiling of $\RR^n$. One appealing property of the tiling is the following probabilistic result. 

\begin{thm}\label{thm:mainproportions}Consider the tiling described in \Cref{thm:maintiling}. 
The probability that a randomly chosen point of the tiling lies in a translate of $\mathcal T_\sigma$ is equal to \begin{equation}\label{eq:probability}\det(A_\sigma)\det(B_\sigma)/\det(AB^T).\end{equation}
\end{thm}

\begin{cor}\label{cor:rational}
    If $\det(B_\sigma)$ is an integer for all $\sigma\in \binom{[n]}r$, then the tiling is periodic. If the determinants do not all have a common factor, then the fundamental domain of the tiling contains $\det(B_\sigma)$ copies of the tile $\mathcal T_\sigma$ for each $\sigma \in \binom{[n]}r$. 
\end{cor}

\begin{rem}
    To make the statement of \Cref{thm:mainproportions} precise, fix any convex polytope $P$ in $\RR^r$ and consider the uniform probability density when restricting to a dilation of $P$. \Cref{thm:mainproportions} says that as the dilations increase in volume, the limiting probability of choosing a point in a translate of $\mathcal T_\sigma$ is given by~\eqref{eq:probability}. 
\end{rem}

\begin{example}\label{ex:typeB}
Consider the matrices
    \[A = \begin{pmatrix} 1 & \matminus1 & 0 & 1 \\ 0 & 1 & 1 & 1 \end{pmatrix}\hspace{ 1 cm}\text{ and }\hspace{ 1 cm}
B = \begin{pmatrix} 1 & \matminus1 & \matminus1 & 0 \\ 2 & \matminus1 & 0 & 1 \end{pmatrix}.\]
It is straightforward to show that these are full rank and equi-oriented matrices. Thus, the construction from \Cref{thm:maintiling} will produce a tiling that satisfies the conditions of \Cref{thm:mainproportions}. This tiling is given in \Cref{fig:TypeB}. 

The first step is to choose a matrix $\wtilde B$ which satisfies~\eqref{eq:Btilde} and a sufficiently generic vector $\gamma$. In particular, we can choose \[\wtilde B = \begin{pmatrix} 1 & 0 & 1 & \matminus2 \\ 0 & 1 & \matminus1 & 1 \end{pmatrix}\hspace{ 1 cm}\text{ and }\hspace{ 1 cm} \gamma = (\sqrt2,\sqrt3,\sqrt5,\sqrt7)^T.\]


\begin{figure}
\centering
    \includegraphics[width=0.2\textwidth]{TypeBNew.png}
    \caption{This is the tiling obtained in \Cref{ex:typeB}. Note that 5 of the 6 prototiles have the same area, while the sixth has double the area. This corresponds to the fact that $|\det(A_{\{2,4\}})| = 2$, but $|\det(A_\sigma)| = 1$ for every other $\sigma \in \binom{[n]}r$. Furthermore, $|\det(B_{\{1,3\}})| = 2$, but $|\det(B_\sigma)| = 1$ for every other $\sigma \in \binom{[n]}r$. This means that $\mathcal T_{\{1,3\}}$ (the brown square) appears twice as often as every other prototile.}
    \label{fig:TypeB}
\end{figure}

Applying the construction will give a tiling of $\RR^2$ with $\binom{4}{2} = 6$ different tiles. For example, the tile $\mathcal T_{\{1,3\}}$ a unit cube such that the proportion of the plane covered is
\[\frac{\det(A_{\{1,3\}})\det(B_{\{1,3\}})}{\det(AB^T)} = \frac{\det\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \det\begin{pmatrix}1 & -1 \\ 2 & 0\end{pmatrix}}{\det\begin{pmatrix}2 & 4 \\ -2 & 0\end{pmatrix}} = \frac 28.\]
This means that the brown squares in \Cref{fig:TypeB} cover a quarter of $\RR^2$ when the tiling is repeated. 

To calculate the position where each copy of $\mathcal T_\sigma$ is placed, we use the translation vector from~\eqref{eq:translate}. For example, the translation vector corresponding to $\mbf z = (1,2)^T$ is given by
\begin{align*}\mbf s_{(\{1,3\},\mbf z)} &= \scriptstyle{\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 \matminus\sqrt 2 \\ 2 \matminus \sqrt 5\end{pmatrix} + \begin{pmatrix}\matminus1 & 1 \\ 1 & 1\end{pmatrix}\left(\left\lceil\begin{pmatrix}0 & \matminus 2 \\ 1 & 1\end{pmatrix}^{-1} \begin{pmatrix}1 & 1 \\ 0 & \matminus 1\end{pmatrix}\begin{pmatrix}\sqrt 2 \matminus1 \\ \sqrt 5 \matminus 2\end{pmatrix}+ \begin{pmatrix} \sqrt 3 \\ \sqrt 7\end{pmatrix}\right\rceil- \begin{pmatrix} \sqrt 3 \\ \sqrt 7\end{pmatrix}\right)}\\
&= \begin{pmatrix}1 \matminus\sqrt 2 \\ 2 \matminus \sqrt 5\end{pmatrix} + \begin{pmatrix}\matminus1 & 1 \\ 1 & 1\end{pmatrix}\left(\left\lceil\frac12\begin{pmatrix}1 + \sqrt2 +2\sqrt3 - \sqrt 5 \\ 3 -\sqrt2 - \sqrt 5 + 2\sqrt 7\end{pmatrix}\right\rceil- \begin{pmatrix} \sqrt 3 \\ \sqrt 7\end{pmatrix}\right)\\
&= \begin{pmatrix}1 \matminus\sqrt 2 \\ 2 \matminus \sqrt 5\end{pmatrix} + \begin{pmatrix}\matminus1 & 1 \\ 1 & 1\end{pmatrix}\begin{pmatrix}2 \matminus \sqrt 3 \\ 3 \matminus\sqrt 7\end{pmatrix}\hspace{ .5 cm} = \hspace{ .5 cm}\begin{pmatrix}2 -\sqrt2 + \sqrt 3 - \sqrt 7 \\ 7 - \sqrt 3 - \sqrt 5 - \sqrt 7\end{pmatrix}\end{align*}
This calculation shows that one of the squares in the tiling must have its bottom left vertex at this coordinate. 
\end{example}

\section{Why the tiling works}\label{sec:proofs}

Our construction is based on a periodic tiling of $\RR^n$ that was introduced by the second author and then generalized by Doolittle and the second author~\cite{alexthesis,McDonough_multijections,Doolittle_McDonough_Signed}. In the original construction, the tiles correspond to \emph{bases} of an \emph{oriented arithmetic matroid}, while the lattice points in the fundamental domain correspond to elements of the matroid's \emph{sandpile group}. When this tiling is intersected with a plane that is perpendicular to the last $k$ coordinates, this produces the tiling of $\RR^r$ that is defined in \Cref{sec:construction}. 

As before, we begin with a pair of equi-oriented matrices $A$ and $B$, and then find a matrix $\wtilde B$ whose row space is orthogonal to the row space of $A$. We also need a new $n\times n$ matrix $M$ which is given by 
\[M = \begin{pmatrix} A \\ \matminus\widetilde B\end{pmatrix}.\]

For each $\sigma \in \binom{[n]}{r}$, let $S_\sigma(M)$ be the $n\times n$ matrix whose $i^{th}$ column is given by:
\begin{enumerate}
    \item If $i \in \sigma$, then the first $r$ entries of the $i^{th}$ column of $S_\sigma(M)$ match the $i^{th}$ column of $A$ and the last $k$ entries are all $0$. 
    \item If $i \not\in \sigma$, then the first $r$ entries of the $i^{th}$ column of $S_\sigma(M)$ are all $0$ and the last $k$ entries match the $i^{th}$ column of $\widetilde B$.
\end{enumerate}

For example, if $A$ and $\wtilde B$ are the same as in \Cref{ex:typeB}, then

\[ M = \begin{pmatrix} 1 & \matminus1 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ \matminus1 & 0 & \matminus1 & 2 \\ 0 & \matminus1 & 1 & \matminus1 \end{pmatrix} \hspace{ 1 cm}\text{ and } \hspace{ 1 cm}S_{\{1,3\}}(M) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \matminus 2 \\ 0 & 1 & 0  & 1 \end{pmatrix}.\]

\begin{thm}[{\cite[Corollary 3.10]{Doolittle_McDonough_Signed}}, {\cite[Corollary 9.2.8]{alexthesis}}]\label{thm:ntiling}
Suppose that $\det(S_{\sigma}(M))$ is non-negative for all $\sigma \in \binom{[n]}{r}$ or non-positive for all $\sigma \in \binom{[n]}{r}$. Then, every point in $\RR^n$ can be expressed as a sum of the form $M\mbf z + S_\sigma(M)\mbf x$ for some $\mbf z \in \ZZ^n$, $\mbf x \in [0,1]^n$, and $\sigma \in \binom{[n]}{r}$. For all but a measure 0 set of points, such an expression is unique. 
\end{thm}

Another way to state \Cref{thm:ntiling} is to say that the fundamental parallelepipeds of each matrix $S_\sigma(M)$ form a periodic tiling of $\RR^n$ with the columns of $M$ as translation vectors. Note that the ``measure 0 set overlap'' caveat can be removed by working with \emph{half-open parallelepipeds}. See~\cite{Doolittle_McDonough_Signed} for more details. 

The condition required by the first sentence of \Cref{thm:ntiling} can be shown to be equivalent to our requirement for $\wtilde B$ in \Cref{sec:construction}. In particular, $\det(S_{\sigma}(M))$ is always non-negative or always non-positive if and only if the row space of $\wtilde B$ is dual to the row space of a full rank matrix $B$ where $A$ and $B$ are equi-oriented.  

\subsection{Proof of \texorpdfstring{\Cref{thm:maintiling}}{Theorem 2.2}}
To obtain the construction from \Cref{sec:construction}, we intersect the tiling of $\RR^n$ from \Cref{thm:ntiling} with a hyperplane that is perpendicular to the last $k$ coordinates. This idea was partially explored in~\cite[Section 7]{McDonough_multijections}, but the technique used is only applicable for integer matrices, and cannot be used to produce an aperiodic tiling. 

Before intersecting the tiling with a hyperplane, we first translate all of the tiles by $-M\gamma$ for some vector $\gamma$. Note that since $M$ is the fundamental domain of the $n$-dimensional tiling, the entries of $\gamma$ only matter mod 1. In particular, we can think of $\gamma$ as a vector in $[0,1]^n$.

Each tile of the $n$-dimensional tiling corresponds to a choice of $\sigma \in \binom{[n]}{r}$ and $\mbf z \in \ZZ^n$. In particular, this tile is made up of all points of the form $M(\mbf z-\gamma) + S_\sigma(M)\mbf x$, where $\mbf x \in [0,1]^n$. From the structures of $M$ and $S_\sigma(M)$, it is straightforward to show that the first $r$ values of this point are given by 
\[ A_\sigma(\mbf z_\sigma- \gamma_\sigma +\mbf x_\sigma) + A_{\widehat\sigma}(\mbf z_{\widehat\sigma} - \gamma_{\widehat\sigma}),\]
while the last $k$ values are given by 
\[ \widetilde B_\sigma(\gamma_\sigma-\mbf z_\sigma) + \widetilde B_{\widehat\sigma}(\gamma_{\widehat\sigma}-\mbf z_{\widehat\sigma} + \mbf x_{\widehat\sigma}).\]
Observe that for every choice of $\mbf z_\sigma$, there is a unique choice of $\mbf z_{\widehat\sigma}$ and $\mbf x_{\widehat\sigma}$ such that these last $k$ entries are zero. Routine algebraic manipulations show that 
\[\mbf z_{\widehat\sigma} = \widetilde B_{\widehat\sigma}^{-1}\left(\widetilde B_\sigma(\gamma_\sigma- \mathbf z_\sigma)\right) + \gamma_{\widehat\sigma} +\mbf x_{\widehat\sigma}.\] Additionally, since all of the values of $\mbf x_{\widehat\sigma}$ are between 0 and 1, and we choose $\gamma_{\widehat\sigma}$ such that they cannot be exactly 0 or 1, we can conclude that 
\[\mbf z_{\widehat\sigma} = \left\lceil\widetilde B_{\widehat\sigma}^{-1}\left(\widetilde B_\sigma(\gamma_\sigma- \mathbf z_\sigma)\right) + \gamma_{\widehat\sigma}\right\rceil,\]
where $\lceil\cdot\rceil$ is the entry-wise ceiling function. Plugging back into the formula for the first $r$ coordinates gives~\eqref{eq:translate}. 

\subsection{Proof sketch of \texorpdfstring{\Cref{thm:mainproportions}}{Theorem 2.3}}
Let $P$ be a convex polytope with volume $\vol(P)$. If $P$ is sufficiently large relative to the tiles, we get a good approximation for the number of copies of $\mathcal T_\sigma$ that lie inside of $P$ by considering the number of $\mbf z \in \ZZ^r$ such that the translation vector $\mbf s_{(\sigma,\mbf z)}$ lies inside of $P$. We can also ignore tiles that lie partially inside of $P$ since these cases will be sparse in the limit. 

For a given $\mbf z$ and $\sigma$, let $\epsilon_{\sigma,\mbf z}$ be the vector added when the ceiling function is applied in~\eqref{eq:translate}. This lets us rewrite~\eqref{eq:translate} as
\begin{align*}
    A_\sigma (\mathbf z-\gamma_\sigma) + A_{\widehat\sigma} \left(\widetilde B_{\widehat\sigma}^{-1}\left(\widetilde B_\sigma(\gamma_\sigma- \mathbf z)\right) + \epsilon_{\sigma,\mbf z}\right)&=\\
    \left(A_\sigma - A_{\widehat\sigma}\widetilde B_{\widehat\sigma}^{-1}\widetilde B_\sigma\right)\mathbf z + A_{\widehat\sigma} \epsilon_{\sigma,\mbf z} + A_{\widehat\sigma}\widetilde B_{\widehat\sigma}^{-1}\widetilde B_\sigma \gamma_\sigma.
\end{align*}

The vectors $\epsilon_{\sigma,\mbf z}$ and $\gamma_\sigma$ always have entries between 0 and 1. Thus, the first term is the only relevant term in the limit. 

The number of integer points in a region is approximately equal to the volume of the region, and the determinant of a linear transformation describes how this volume changes. In particular, the number of $\mbf s_{(\sigma,\mbf z)}$ that lie within $P$ is approximately $\vol(P)/|\det(A_\sigma - A_{\widehat\sigma}\widetilde B_{\widehat\sigma}^{-1}\widetilde B_\sigma)|$. Furthermore, by the Schur complement formula, 
\[\det(\wtilde B_{\widehat\sigma})\det(A_\sigma - A_{\widehat\sigma}\widetilde B_{\widehat\sigma}^{-1}\widetilde B_\sigma) = \det\begin{pmatrix}A_\sigma & A _\hsigma \\ \wtilde B_\sigma & \wtilde B_\hsigma\end{pmatrix}\]
Notice that the matrix on the right has the same determinant up to sign for every $\sigma\in \binom{[n]}{r}$ (the columns will just be rearranged). This implies that the number of translates of $\mathcal T_\sigma$ contained in $P$ is (in the limit) proportional to $\det(\wtilde B_\hsigma)$. Since each tile has volume $\det(A_\sigma)$, the total volume of points contained in a translate of $\mathcal T_\sigma$ is proportional to $\det(A_\sigma)\det(\wtilde B_\hsigma)$. 

By \Cref{lem:detmatch}, we can choose $\wtilde B_\hsigma$ so that $\det(B_\sigma) = \det(\wtilde B_\hsigma)$ for all $\sigma \in \binom{[n]}r$. Thus, the total volume of points contained in a translate of $\mathcal T_\sigma$ is proportional to $\det(A_\sigma)\det(B_\sigma)$. Finally, it follows from the Cauchy--Binet formula that the sum of $\det(A_\sigma)\det(B_\sigma)$ over all $\sigma\in \binom{[n]}r$ is equal to $\det(AB^T)$. 

\section{The Penrose Tiling and other Special cases}\label{sec:penrose} 
%One special case where $(A,B)$ will always be equi-oriented is if $A$ and $B$ are \emph{totally nonnegative} matrices, i.e., all of the minors are positive.

%When $r=2$, we can think of the columns of $A$ as vectors in $\RR^2$. The determinant of the matrix whose columns are $v_1$ and $v_2$ is nonnegative if and only if the angle from $v_1$ to $v_2$ (in the counterclockwise direction) is at most $\pi$. In particular, if $A$ has columns $v_1,v_2,\dots,v_n$, then $A$ is a totally nonnegative matrix if and only if the vectors are ordered in a counterclockwise direction and the angle between $v_1$ and $v_n$ is at most $\pi$. A simple way to guarantee the second property is to require that all of the vectors are in the upper half plane. Consequently, given two sets of $n$ vectors in  the upper half plane of $\RR^2$, we get an equi-oriented pair of $r\times n$ matrices by ordering the vectors from right to left. 

One simple condition to guarantee that $A$ and $B$ are equi-oriented is to simply choose $A = B$. This condition is used to construct one of the most famous tilings in mathematics: the Penrose tiling. The first step of the tiling is to set the columns of $A$ and $B$ to be the 5th roots of unity. In other words, 
\[A = B = \begin{pmatrix} 1 & \cos(\zeta) & \cos(2 \zeta) & \cos(3\zeta) & \cos(4\zeta) \\ 0 & \sin(\zeta) & \sin(2\zeta) & \sin(3\zeta) & \sin(4\zeta)\end{pmatrix}, \]
where $\zeta = 2\pi/5$. 

All of the maximal minors of $A$ are nonzero, which means that there are a total of $\binom{5}{2} = 10$ prototiles. However, the symmetry of the vectors actually means that there are actually just two kinds of tiles up to rotation. In particular, the parallelepiped formed by $v_i$ and $v_j$ will have an area of $\sin(\zeta)$ if $i-j \equiv \pm 1 (\bmod 5)$ and will have an area of $\sin(2\zeta)$ if $i-j \equiv \pm 2 (\bmod 5)$.

By \Cref{thm:maintiling,thm:mainproportions}, any choice of $\gamma\in  [0,1]^5$ will produce a tiling of $\RR^2$ using the 10 prototiles. Furthermore, each of the 5 possible rotations of each tile appears with the same frequency. In particular, the probability that a randomly chosen point is contained in a large tile is 
\[\frac{5\sin^2(\zeta)}{\det(AB^T)} = \frac{5\sin^2(\zeta)}{5(\sin^2(2\zeta) + \sin^2(\zeta))} = \frac{\sin^2(\zeta)}{\sin^2(2\zeta) + \sin^2(\zeta)} = \frac{5 + \sqrt5}{10} \]

Such a tiling is called a \emph{generalized Penrose tiling}. However, the precise tiling depends on $\gamma$. See \Cref{fig:Penroses} for a few possibilities. For an excellent reference on Penrose tilings and other related tilings, we recommend~\cite{Quasicrystals}.

\begin{figure}
    \centering
    \includegraphics[width=0.26\linewidth]{FPSACPenrose.png}\hspace{.5cm}
    \includegraphics[width=0.26\linewidth]{FPSACPenrose2.png}\hspace{.5 cm}\includegraphics[width=0.26\linewidth]{Seventh2.png}
    \caption{The first two tilings are generalized Penrose tilings that come from our construction. On the left is a tiling obtained from $\gamma = (.5,.5,.5,.00001,.5)^T$, which satisfies the matching rules for a traditional Penrose tiling within this region. In the center is the tiling obtained from $\gamma = (.5,.5,.5,.5,.5)^T$, which was studied in~\cite[Section 7.4]{LePenrose}. Finally, the image on the right is from an analogous construction using seventh roots of unity and $\gamma = (.5,.5,.00001,.00001,.00001,.00001,.00001)^T$.}
    \label{fig:Penroses}
\end{figure}

We learned while researching Penrose tilings that our method of generating tilings is closely related to something called an \emph{oblique tiling}~\cite{Oblique}. For our tilings, we begin with a somewhat complicated tiling of $\RR^n$ and then slice along a plane that is perpendicular to the last $k$ coordinates (see \Cref{sec:proofs}). Conversely, the oblique tiling begins with a simple tiling of unit cubes in $\RR^n$ and then slices along a more complicated hyperplane. While the two constructions work similarly for the $A = B$ case, the oblique tiling perspective does not generalize as naturally to the case where $A$ and $B$ are equi-oriented but not equal. In the last two sections, we will explore a few applications where the equi-oriented generality is necessary. 

\section{The Laplacian of two equi-oriented matrices}\ 

Burman et al. \cite{Burman_MT} define a  Laplacian operator $L(A,B)$ that depends on \emph{two} collections of vectors $A:=(a_i)_{i=1}^n$ and $B:=(b_i)_{i=1}^n$,  from some vector space $V\cong \RR^r$ as follows:
\begin{equation}
    {\color{blue}L(A,B)(x):=}\sum_{i=1}^N\langle b_i,x\rangle\cdot a_i\quad\text{or equivalently}\quad L(A,B)(x)=AB^T\cdot x.\label{eq:defn:Lapl(v,w)}
\end{equation}
Notice that $L(A,B)$ is a sum of rank-$1$ operators ($x\rightarrow \langle b_i,x\rangle\cdot a_i$ has its image in the line spanned by $a_i$). The following proposition gives the characteristic polynomial of $L(A,B)$ in terms of the Grammian determinants of associated sets of columns of $A$ and $B$. Recall here that given two equal sized collections of vectors $K=(k_i)_{i\in I}$ and $P=(p_i)_{i\in I}$, the \defn{Grammian determinant} $\GD(K,P)$ is defined as
\begin{equation}
    \defn{\GD (K,P):=}\det\Big(\langle k_i,p_j\rangle\Big)_{i,j\in I},\label{eq:defn:GramDet}
\end{equation}
as the determinant of the (generalized) Grammian matrix formed by $K$ and $P$. The following is essentially \cite[Lemma~8.2]{Chapuy_Douvropoulos_MT}. In this abstract we are only interested in the determinant $\op{det}\big(L(A,B)\big)$ for which case the following proposition is the Cauchy--Binet formula.


\begin{prop}
The characteristic polynomial of the Laplacian operator \eqref{eq:defn:Lapl(v,w)} is given via:
\[
\det\big(t\cdot \mathbf{I}_r+L(A,B)\big)=\sum_{\substack{\sigma\subseteq [n] \\ |\sigma|\leq r}}\GD(A_{\sigma},B_{\sigma})\cdot t^{r-|\sigma|}.
\]
\end{prop}
% \begin{proof}
% The coefficient of $t^{r-k}$ in the characteristic polynomial above is the trace of the operator $\wedge^k\big( L(A,B)\big)$ on the exterior power $\wedge^k(V)$. We proceed by first computing the action of the operator $\wedge^k\big( L(A,B)\big)$. Pick any element $\mathbf{x}:=x_1\wedge x_2\cdots\wedge x_k\in\wedge^k(V)$. Then,
% \begin{align*}
% \wedge^k\big( L(A,B)\big)(\mathbf{x})&=L(A,B)(x_1)\wedge L(A,B)(x_2)\wedge\cdots\wedge L(A,B)(x_k)\\
% &=\left(\sum_{i=1}^n\langle b_i,x_1\rangle\cdot a_i\right)\wedge \left(\sum_{i=1}^n\langle b_i,x_2\rangle\cdot a_i\right) \wedge\cdots \wedge \left(\sum_{i=1}^n\langle b_i,x_k\rangle\cdot a_i\right)\\
% &=\sum_{(i_1,i_2\ldots,i_k)\in [n]^k}\langle b_{i_1},x_1\rangle\cdot \langle b_{i_2},x_2\rangle\cdots\langle b_{i_k},x_k\rangle \cdot \big(a_{i_1}\wedge a_{i_2}\wedge\cdots\wedge a_{i_k}\big).
% \end{align*}
% In the last expression all indices $i_j$ must be distinct or the $k$-vector $\omega_{i_1}\wedge\cdots\wedge\omega_{i_k}$ is $0$. Then, each $k$-subset of indices appears $k!$ times and they are all multiplied with the same $k$-vector up to sign. When we group them together, the coefficient is in fact a determinant and we have (we write $^{\wedge}\!A_{\sigma}$ for the wedge $a_{i_1}\wedge\cdots\wedge a_{i_k}$, $\sigma=\{i_1,\ldots,i_k\}$; similarly for $B$)
% \[
% \wedge^k\big( L(A,B)\big)(\mathbf{x})=\sum_{\sigma\in\binom{[n]}{k}}\GD(B_{\sigma},\mathbf{x})\cdot (^{\wedge}\!A_{\sigma}).
% \]

% Now the Grammian determinant of \eqref{eq:defn:GramDet} essentially encodes the inner product in exterior powers $\wedge^k(V)$ so that we have $\op{GD}(B_{\sigma},\bm{x})=\langle ^{\wedge}\!B_{\sigma},\bm{x}\rangle$ (where we write $\bm{x}$ both for the k-vector and the collection of vectors $(x_i)_{i=1}^k$). We can then rewrite the last equation as
% \[
% \wedge^k\big( L(A,B)\big)(\mathbf{x})=\sum_{\sigma\in\binom{[n]}{k}}\langle\, ^{\wedge}\!B_{\sigma},\mathbf{x}\rangle\cdot (^{\wedge}\!A_{\sigma}).
% \]

% \noindent Notice the similarity of this with the definition of the Laplacian in \eqref{eq:defn:Lapl(v,w)}. Now, in general when we have an inner product and an orthonormal basis $e_i$, we can compute the trace of any matrix $A$ as 
% \[
% \op{Tr}(A)=\sum_i\langle Ae_i,e_i\rangle,
% \]
% and since in our case $e_J:=e_{j_1}\wedge e_{j_2}\wedge\cdots\wedge e_{j_k}$ forms an orthonormal basis of $\wedge^k(V)$ (over all $k$-subsets $J\subseteq [r]$), we have
% \begin{align*}
% \op{Tr}\left(\wedge^k\big( L(A,B)\big)\right)&=\sum_{J\in\binom{[r]}{k}}\Big\langle \wedge^k\big( L(A,B)\big)(e_J)\, ,\, e_J\Big\rangle =\sum_{J\in\binom{[r]}{k},\, \sigma \in\binom{[n]}{k}}\Big\langle \langle ^{\wedge}\!B_{\sigma},e_J \rangle\cdot (^{\wedge}\!A_{\sigma})\,,\,e_J\Big\rangle \\
% &=\sum_{\sigma\in\binom{[n]}{k},\, J\in\binom{[r]}{k}} \langle ^{\wedge}\!B_{\sigma},e_J \rangle\cdot \langle e_J, ^{\wedge}\!A_{\sigma}\rangle=\sum_{\sigma\in\binom{[n]}{k}}\langle  ^{\wedge}\!B_{\sigma}, ^{\wedge}\!A_{\sigma}\rangle.
% \end{align*}
% The last equality above is by the orthonormality of the $e_J$. This completes the proof since $\op{GD}(B_{\sigma},A_{\sigma})=\langle  ^{\wedge}\!B_{\sigma}, ^{\wedge}\!A_{\sigma}\rangle$.
% \end{proof}

When two matrices $A,B$ are equi-oriented, all grammian determinants $\op{GD}(A_{\sigma},B_{\sigma})$ are non-negative and equal the product of the volumes of the fundamental parallele\-pipeds for bases $A_{\sigma}$ and $B_{\sigma}$. After applying \Cref{cor:rational}, we have proven the following.

\begin{cor}
The volume of the fundamental domain for a periodic tiling $\tile(A,B)$ equals the determinant of the Laplacian $L(A,B)$.
\end{cor}



\section{Tiling space with root zonotopes and their relatives}

A particularly nice application of \Cref{thm:maintiling,thm:mainproportions} is for the case that the collections of vectors $A$ and $B$ are respectively the (positive) roots $\Phi^+$ and coroots $\widehat{\Phi}^+$ of a Weyl group $W$ (we will mostly represent the roots in terms of a basis of the ambient space $V$ of $W$ instead of the root or weight basis because it leads to more appealing pictures but this is not necessary).
We will denote such a tiling by $\tile(W):=\tile\big(\Phi^+$, $\widehat{\Phi}^+\big)$.



\begin{thm}\label{thm:root_zonotopes}
Let $W$ be an irreducible Weyl group of rank $n$ and Coxeter number $h$, and $\tile(W)$ its associated periodic tiling described above. Then, any fundamental domain of $\tile(W)$ has volume equal to $h^n$. Moreover, for every tile $\mathcal{T}_{\sigma}$ of $\tile(W)$ the probability that a random point belongs to a translate of $\mathcal{T}_{\sigma}$ is equal to $\dfrac{I(W_{\sigma})}{h^n}$ where $I(W_{\sigma})$ is the connection index of the reflection subgroup $W_{\sigma}$ associated to $\mathcal{T}_{\sigma}$.
\end{thm}

\begin{proof}
Let $N=hn/2=|\Phi^+|$ denote the number of roots of $W$, and $Q$ and $\widehat{Q}$ its root and coroot lattice. Then, any $\mathbb{R}$-basis of roots corresponding to some $\sigma\in\binom{[N]}{n}$ generates a rank-$n$ reflection subgroup $W_{\sigma}\leq W$. Furthermore, the roots $\Phi^+_{\sigma}$ and coroots $\widehat{\Phi}^+_{\sigma}$ generate respectively the root and coroot lattices $Q_{\sigma}$ and $\widehat{Q}_{\sigma}$ of $W_{\sigma}$ (see \cite[Corollary~1.2]{BW_roots}). This means that $\op{Vol}(\Phi^+_{\sigma})=\op{Vol}(Q_{\sigma})$ and $\op{Vol}(\widehat{\Phi}^+_{\sigma})=\op{Vol}(\widehat{Q}_{\sigma})$. Now, the connection index $I(W_{\sigma})$ is defined as the determinant of the Cartan matrix of $W_{\sigma}$, which is the Laplacian of the two collections of simple roots and simple coroots of $W_{\sigma}$, and hence agrees with the product of the volumes of the two lattices:
\[
I(W_{\sigma})=\op{Vol}(Q_{\sigma})\cdot \op{Vol}(\widehat{Q}_{\sigma}).
\]
Finally the determinant of the Laplacian $L(\Phi^+,\widehat{\Phi}^+)$ is equal to $h^n$ after \cite[Proposition~3.3]{Chapuy_Douvropoulos_WL} and the proof is complete.
\end{proof}

\begin{rem}
The theorem above gives a geometric exegesis to the formula \cite[Section 7.3]{DLM}
\[
h^n=\sum_{W'\leq W}\#\op{RGS}(W')\cdot I(W'),
\]
where the sum is over all rank-$n$ reflection subgroups of $W$ and $\op{RGS}(W')$ denotes the set of minimum generating sets of reflections of $W'$. Indeed, it corresponds to the partition of the fundamental domain of $\tile(W)$ with respect to the distinct tiles $\tile_{\sigma}$.
\end{rem}


% \section{Introduction}


% \theo{TODO:

% \begin{itemize}
%  \item Theo does section on root systems by Monday.
%  \item 
% \end{itemize}
% }

% \theo{Here is some idea for how to contextualize or advertise the results for the FPSAC audience:}
% \begin{itemize}
%     \item \theo{We can construct a Laplacian operator given two distinct collections of vectors $A$ and $B$. This is a sum of the grammian matrices for each pair $(A_I,B_I)$ of corresponding bases from them. }

%     \item \alex{Sounds good! Should we have a version in full generality, or do we want to focus entirely on the equi-??? case?}

%     \theo{I propose to focus on the equi-oriented case}
    
%     \item \theo{Given two collections of vectors $A,B$, we can construct a tiling of space with tiles from $A$ whose frequencys are determined by the volumes of the corresponding tiles from $B$, when all pairs of tiles are \emph{equi-directed? equi-oriented?.. some name}}

%     \item \alex{In my thesis, I say that the top matrix and bottom matrix have ``complementary orientation'', but I'm not attached to this term. One related decision is whether we want to work with an $r \times n$ and $(n-r)\times n$ matrix with complementary orientation or a pair of $r \times n$ matrices with compatible orientation (which here means that the minors are either the same sign or one is zero). This advantage of the second version is we can talk about pairing a collection of vectors with itself and how that always works. The downside is that we need to work a little bit harder to make it canonical (I think it's probably worth it though). 

%     If we want to be trendy, we could even focus on the case where both sets of vectors are elements of the totally nonnegative grassmannian.}
    
%     \item \theo{This generalizes constructions of DeBruijn, the cut-and-project method as well as the grid method and we have pretty pictures.}

%     \item \alex{Yeah, the pictures are key! We should definitely also mention the oblique tiling, since this is one most directly related.}
    
%     \item \theo{The tiling has some connection to the Laplacian. For example, when the Laplacian is an integer matrix (not really, some other condition\alex{$\rightarrow$ integrally independent}), the tiling is periodic and the volume of the fundamental domain is the determinant of the Laplacian.

%     Can we say something more about the interpretation when the matrix is not rational?}

%     \item \alex{ I checked the Quasicrystals book, and Senechal uses the term "integrally independent", meaning that you can't get 0 from adding integer linear combinations of the vectors.
    
%     In the general case, we can talk about the limiting frequencies. Essentially, if we take an increasing ball (probably any reasonable convex shape is fine), then the limiting proportion of a given tile is the relative size of the determinant. I know a statement of this form has been used to talk about how the proportion of big diamonds to small diamonds in the Penrose tiling is the golden ratio. } 
    
%     \item \theo{In the case that $A$ and $B$ are the roots and coroots of a Weyl system, we get a fundamental domain of are $h^n$ and the frequencies and volumes of tiles are expressed in terms of root-theoretic data.}

%     \item \alex{I think that this would be a good topic to focus on since it feels the most novel compared to my earlier publications on the tiling.}
    
%     \item \theo{This last point resolves the sort of dissonance between type-A and other types about the volumes of their root zonotopes. In type-A the volume is a product formula and agrees with the determinant of the Laplacian; in other types it does not. The reason is that the correct analogue is the area of the fundamental domain of the tiling which equals the zonotope only when the system is unimodular. }

%     \item \alex{Sounds good! An interesting related concept is how we can find copies of the zonotope with certain shifting vectors, but maybe not with others.}
% \end{itemize}




% \theo{Note immediately after theorem that frequency can be irrational as well as in Penrose tiling}

% \section{Various questions we are working on}

% \begin{enumerate}
%     \item Does our construction contain the Penrose tilings? In \cite{generalised_penrose} the authors indicate that the De Bruijn's pentagrid construction generates Penrose tilings only when the sum of the offsets for each grid line is an integer (I think de Bruijn says just $0$, but that could be equivalent to any integer). They study what happens when the sum of the offsets is not necessarily an integer and their pictures look like ours. 

%     So we would like to find how the sum-of-offsets statistic translates to the the McDonough construction; then we should be able to specify a particular choice of bottom matrix and/or offset vector for which the tiling is one (or any one) of the Penrose tilings.

%     \alex{I think I did get code that gives the Penrose tiling (although I wanted to clean it up before sharing since it's a mess at the moment). One key idea is to work with half integers instead of integers. I still don't fully understand all of the details though, so I'll think some more about it.}
    
%     \item We can explicitly bound the points of the $\mathbb{Z}^r$ lattice that correspond to tiles in a fixed (rectangular?) domain.

%     \alex{Yeah, I do this this is worth mentioning since we can get pretty excellent bounds, and it is useful for actually coding things up. I'll think about generalizing to other domains as well as the half integer case.} 
    
%     \item If the ordered collection $B$ as in the theorem that encodes the frequencies of the tiles is unimodular, this should imply that the zonotope of $A$ tiles space. This may be equivalent to McMullen's (?) characterization of the zonotopes that tiles space as precisely those whose associated matroid is regular.

%     \alex{This definitely feels related. I feel like if the bottom matrix is totally unimodular, and the top matrix has compatible orientation, it does seem reasonable that the top matrix must regular (even if the particular representation isn't totally unimodular).}

%     \item The question of which tile covers a particular point. \theo{In particular, is it NP hard? Finding closest lattice point to given point is NP hard. Is it true that any voronoi tiling of a lattice also a tiling in our sense?}
% \end{enumerate}

\acknowledgements{We would like to thank Spencer Backman who first suggested that we combine our results on the Laplacian decompositions and on tilings.}

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