4.1. Preliminaries on multiplication cubes

Our model for the d-dimensional hypercube is the set $[-1,0]^d$ . The bottom and top $(d-1)$ -dimensional hyperfaces orthogonal to $e_i$ are

$$ \begin{align*} {\mathrm{\beta}}_i=\{v\in [-1,0]^d\mid v[i]=-1\}\quad\mbox{and}\quad{\mathrm{\tau}}_i=\{v\in [-1,0]^d\mid v[i]=0\}, \end{align*} $$

respectively.

One can take a collection of hypercubes, color their $(d-1)$ -dimensional hyperfaces ${\mathrm {\beta }}_i$ and ${\mathrm {\tau }}_i$ with various colors, and form tessellations of the space with these cubes in such a way that the adjacent hyperfaces of neighboring hypercubes have matching colors. An example of such a colored set of two-dimensional cubes and a part of a two-dimensional tessellation is presented in Figure 1. More precisely (and more abstractly), this is done as in the following definition.

Figure 1

The multiplication cube set $T_{(2,5)}$ and a part of a tiling from $X_{(2,5)}=X_{T_{(2,5)}}$ (with the upper right corner of the cube at the origin marked by a black dot). Consecutive powers of two (starting with $4,8,16,\ldots $ ) can be found in the tiling along diagonals that go from bottom left to top right: for an explanation of this, see Example 4.21 or Proposition 4.26.

The pair $(X_T,\sigma _z)$ is a dynamical system for all $z\in \mathbb {Z}^d$ , and as a d-dimensional subshift, $X_T$ is in fact an example of a so-called subshift of finite type. Wang hypercubes and their tessellations are the most commonly studied in the case $d=2$ , and particularly, then the elements of T are called Wang tiles. These originate from Wang’s paper [Reference Wang13].

We will proceed to define a particular class of Wang hypercubes which could be called multiplication hypercubes, but which we will call multiplication cubes for simplicity. Two-dimensional multiplication cubes have previously appeared in [Reference Cook, Stérin, Woods, Lakin and Sulc3] as elements of the so-called Collatz tile set. Higher dimensional such cubes will in fact have labels on all of their lower-dimensional hyperfaces, not just the $(d-1)$ -dimensional ones. The set of k-dimensional hyperfaces within the d-dimensional hypercube is denoted by $S_{d,k}$ and the set of all hyperfaces is denoted by $S_d$ . Given $s\in S_d$ , the set of all hyperfaces that are subsets of s is denoted by $S_d(s)$ .

We develop two different sets of notation to refer to individual hyperfaces. For the first notation, whenever $p,u\in \{-1,0\}^d$ are orthogonal (in other words, $p[i]=u[i]=-1$ cannot happen for any $1\leq i\leq d$ ), we say that the hyperface at p along u is

$$ \begin{align*}{\mathrm{faceAt}}(p,u)=\{v\in [-1,0]^d\mid v[i]=p[i]\quad\mbox{when }u[i]=0\}.\end{align*} $$

For the second notation, let

$$ \begin{align*}V_d=\{(v_1,v_2)\in (\{-1,0\}^d)^2\mid v_1\geq v_2\}.\end{align*} $$

For any $(v_1,v_2)\in V_d$ , the minimal k-dimensional hyperface of $[-1,0]^d$ containing both $v_1$ and $v_2$ is ${\mathrm {faceAt}}(v_1,v_2-v_1)$ , where k is the number of ones in $v_1-v_2$ . This gives a one-to-one correspondence between $V_d$ and $S_d$ .

Let $n\in \mathbb {Z}_+^d$ . To any $a\in \Sigma _N$ with $N=\prod _{i=1}^d n[i]$ , we will associate a d-dimensional Wang hypercube, a so-called multiplication cube ${\mathrm {cube}}_{n}(a)$ , whose underlying structure is that of a function:

$$ \begin{align*}{\mathrm{cube}}_{n}(a):V_d\to\Sigma_N,\quad {\mathrm{cube}}_{n}(a)[v_1,v_2]={\mathrm{base}}_{n}(a,(v_1,v_2))[2]\quad\mbox{for }(v_1,v_2)\in V_d.\end{align*} $$

The interpretation is that ${\mathrm {cube}}_{n}(a)[v_1,v_2]$ is the label of the hyperface spanned by the vectors $v_1$ and $v_2$ : this way, we get labels on all lower-dimensional hyperfaces. If $s\in S_d$ is the hyperface spanned by $v_1$ and $v_2$ , we may also define $s({\mathrm {cube}}_{n}(a))=s\ {\mathrm {cube}}_n(a)={\mathrm {cube}}_{n}(a)[v_1,v_2]$ . In particular, the labels of the bottom and top hyperfaces of ${\mathrm {cube}}_{n}(a)$ orthogonal to $e_i$ are

$$ \begin{align*}{\mathrm{\beta}}_i({\mathrm{cube}}_{n}(a))={\mathrm{cube}}_{n}(a)[-e_i,-{\mathbf{1}}], \quad {\mathrm{\tau}}_i({\mathrm{cube}}_{n}(a))={\mathrm{cube}}_{n}(a)[{\mathbf{0}},-{\mathbf{1}}+e_i],\end{align*} $$

respectively. To write this out more explicitly, let $a=a_1n[i]+a_0$ be such that ${\mathrm {base}}(a,(N,n[i]))=(0,a_1,a_0)$ and let $a=a_2'\prod _{j\neq i}n[j]+a_1'$ be such that ${\mathrm {base}}(a, (\prod _{j\neq i}n[i],1))=(a_2',a_1',0)$ . Then we see that

$$ \begin{align*}{\mathrm{\beta}}_i({\mathrm{cube}}_{n}(a))=a_1,\quad {\mathrm{\tau}}_i({\mathrm{cube}}_{n}(a))=a_1'.\end{align*} $$

An example of a three-dimensional multiplication cube together with (some of) the labels of faces and edges is presented in Figure 2.

Figure 2

The multiplication cube ${\mathrm {cube}}_{(2,3,5)}(10)$ with the faces ${\mathrm {\tau }}_1$ (right), ${\mathrm {\beta }}_2$ (front), and ${\mathrm {\tau }}_3$ (top) visible. One can also verify that these faces and their adjacent edges yield the lower-dimensional multiplication cubes ${\mathrm {cube}}_{(3,5)}(10)$ , ${\mathrm {cube}}_{(2,5)}(3)$ , and ${\mathrm {cube}}_{(2,3)}(4)$ .

For $n\in \mathbb {Z}_+^d$ and $N=\prod _{i=1}^d n[i]$ , the set of base-n multiplication cubes and the valuations of cubes are defined as

$$ \begin{align*}T_{n}=\{{\mathrm{cube}}_{n}(a)\mid a\in\Sigma_N\}, \quad {\mathrm{val}}_{n}(t)=t[{\mathbf{0}},-{\mathbf{1}}]\quad\mbox{for }t\in T_n.\end{align*} $$

In other words, ${\mathrm {val}}_{n}(t)$ is the label of the unique d-dimensional hyperface of t. More importantly, this also equals the original value that was used to form the hypercube and the maps ${\mathrm {val}}_{n}$ , ${\mathrm {cube}}_{n}$ are inverses of each other, because $a=0\cdot N+a\cdot 1+0\cdot 1$ and ${\mathrm {base}}(a,(N,1))=(0,a,0)$ for any $a\in \Sigma _N$ , and therefore

$$ \begin{align*}{\mathrm{val}}_{n}({\mathrm{cube}}_{n}(a))={\mathrm{cube}}_{n}(a)[{\mathbf{0}},-{\mathbf{1}}]=a\quad\mbox{and}\quad{\mathrm{cube}}_{n}({\mathrm{val}}_{n}({\mathrm{cube}}_{n}(a)))={\mathrm{cube}}_{n}(a).\end{align*} $$

The set of valid tilings over $T_n$ is denoted by $X_n=X_{T_n}$ .

Later we will see that calling the elements of $T_n$ multiplication cubes is justified. If a tiling contains a base-N representation of a real number on its main diagonal, then shifting the tiling by $\sigma _{e_i}$ corresponds to multiplying this real number by $n[i]$ as will be seen in Proposition 4.26. The tilings will also be connected to so-called multiplication automata in §5.2.

We observe that in the one-dimensional case $n=(N)$ , the labels of top and bottom hyperfaces satisfy ${\mathrm {\beta }}_1({\mathrm {cube}}_{(N)}(a))={\mathrm {\tau }}_1({\mathrm {cube}}_{(N)}(a))=0$ for all $a\in \Sigma _N$ and therefore $X_{(N)}=T_{(N)}^{\mathbb {Z}}$ . For a higher dimensional example, Figure 1 shows the tile set $T_{(2,5)}$ and a part of a tiling from $X_{(2,5)}$ . In the higher dimensional cases, it is not necessarily obvious that there exist any valid tilings except the trivial one consisting only of copies of ${\mathrm {cube}}_n(0)$ , but in Corollary 4.39, we will see that any arrangement of multiplication cubes on the main diagonal of the d-dimensional space can be extended to a valid tessellation of the whole space.

Lemma 4.2. Let $n\in \mathbb {Z}_+^d$ . For a d-directive sequence $(v_j)_{j=1}^k$ and $a\in \mathbb {N}$ , it holds that

$$ \begin{align*}{\mathrm{base}}_{n}(a,(v_1,v_k))[2]=\sum_{i=1}^{k-1}m(n,v_{i}-v_1)\ {\mathrm{base}}_{n}(a,(v_i,v_{i+1}))[2].\end{align*} $$

Proof. The cases $k\leq 2$ are trivial. If $k\geq 3$ , let $(a_0',a_1',a_k')={\mathrm {base}}_{n}(a,(v_1,v_k))$ and $(a_j)_{j=0}^k={\mathrm {base}}_{n}(a,(v_j)_{j=1}^k)$ . It follows that

$$ \begin{align*} a_0'+a_1'm(n,v_1)+a_k'm(n,v_k)=a=\sum_{i=0}^k a_i m(n,v_i) \end{align*} $$

(with the convention that $v_0={\mathbf {0}}$ ). From Lemma 3.4, we see that

$$ \begin{align*} a_i={\mathrm{base}}_{n}(a,(v_j)_{j=1}^k)[i+1]={\mathrm{base}}_{n}(a,(v_i,v_{i+1}))[2]\quad\mbox{for }0\leq i< k. \end{align*} $$

The first part of Lemma 3.4 shows that $a_0=a_0'$ and the last part of Lemma 3.4 shows that $a_k=a_k'$ , so

$$ \begin{align*} a_1'=\sum_{i=1}^{k-1}a_i m(n,v_i)/m(n,v_1)=\sum_{i=1}^{k-1}m(n,v_i-v_1)\ {\mathrm{base}}_{n}(a,(v_i,v_{i+1}))[2]. \end{align*} $$

Lemma 4.3. Let $n\in \mathbb {Z}_+^d$ . For a binary d-directive sequence $(v_i)_{i=1}^k$ and $t\in T_{n}$ , it holds that

$$ \begin{align*}m(n,v_1)t[v_1,v_k]=\sum_{j=1}^{k-1}m(n,v_j)t[v_j,v_{j+1}].\end{align*} $$

Proof. Let $t={\mathrm {cube}}_{n}(a)$ . We rewrite the statement of the lemma

$$ \begin{align*} &m(n,v_1){\mathrm{cube}}_{n}(a)[v_1,v_k]=\sum_{j=1}^{k-1}m(n,v_j)\ {\mathrm{cube}}_{n}(a)[v_j,v_{j+1}] \\ &\quad \iff m(n,v_1)\ {\mathrm{base}}_{n}(a,(v_1,v_k))[2]=\sum_{j=1}^{k-1}m(n,v_j)\ {\mathrm{base}}_{n}(a,(v_j,v_{j+1}))[2] \\ &\quad\iff {\mathrm{base}}_{n}(a,(v_1,v_k))[2]=\sum_{j=1}^{k-1}m(n,v_j-v_1)\ {\mathrm{base}}_{n}(a,(v_j,v_{j+1}))[2], \end{align*} $$

and the last equality holds by Lemma 4.2.

4.2. Lower dimensional matchings in tilings by multiplication cubes

Throughout this subsection, let $n\in \mathbb {Z}_+^d$ . The $(d-1)$ -dimensional hyperfaces match in a tessellation of $X_{n}$ by definition, but there are also lower dimensional matchings as we will see in Theorem 4.6.

Recall that any $s\in S_d$ can be represented in the form $s={\mathrm {faceAt}}(p,u)$ for some orthogonal $p,u\in \{-1,0\}^d$ . If u contains $d'$ non-zeroes and ${\mathrm {{\iota }}}:\{1,\ldots ,d'\}\to \{1,\ldots ,d\}$ is an injection with the image set $\{i\mid u[i]=-1\}$ , then the injective affine map ${\mathrm {I}}_{p,{\mathrm {{\iota }}}}:\mathbb {R}^{d'}\to \mathbb {R}^d$ defined by ${\mathrm {I}}_{p,{\mathrm {{\iota }}}}({\mathbf {0}})=p$ and ${\mathrm {I}}_{p,{\mathrm {{\iota }}}}(e_i)=p+e_{{\mathrm {{\iota }}}(i)}$ for $1\leq i\leq d'$ satisfies ${\mathrm {I}}_{p,{\mathrm {{\iota }}}}([-1,0]^{d'})=s$ .

Proof. Let $p',u'\in \{-1,0\}^{d'}$ and let $a\in \mathbb {N}$ be such that $s'={\mathrm {faceAt}}(p',u')$ and $t={\mathrm {cube}}_{n}(a)$ (it follows that ${\mathrm {I}}_{p,{\mathrm {{\iota }}}}(p')\leq {\mathrm {I}}_{p,{\mathrm {{\iota }}}}({\mathbf {0}})$ and ${\mathrm {I}}_{p,{\mathrm {{\iota }}}}(p'+u')\geq {\mathrm {I}}_{p,{\mathrm {{\iota }}}}(-{\mathbf {1}})$ ). Let $(a_0,a_1,a_2)\hspace{-1pt}=\hspace{-1pt}{\mathrm {base}}_{n}(a,({\mathrm {I}}_{p,{\mathrm {{\iota }}}}({\mathbf {0}}),{\mathrm {I}}_{p,{\mathrm {{\iota }}}}(-{\mathbf {1}})))$ and $(b_0,b_1,b_2)\hspace{-1pt}=\hspace{-1pt}{\mathrm {base}}_{n}(a_1,({\mathrm {I}}_{p,{\mathrm {{\iota }}}}(p')- {\mathrm {I}}_{p,{\mathrm {{\iota }}}}({\mathbf {0}}), {\mathrm {I}}_{p,{\mathrm {{\iota }}}}(p'+u')-{\mathrm {I}}_{p,{\mathrm {{\iota }}}}({\mathbf {0}}))$ . Then by Lemma 3.3,

$$ \begin{align*}{\mathrm{base}}_{n}(a,({\mathrm{I}}_{p,{\mathrm{{\iota}}}}({\mathbf{0}}),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'+u'),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(-{\mathbf{1}})))=(a_0,b_0,b_1,b_2,a_2).\end{align*} $$

From this, we can see that

$$ \begin{align*} &({\mathrm{I}}_{p,{\mathrm{{\iota}}}}(s'))(t)=t[{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'), {\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'+u')]={\mathrm{base}}_{n}(a,({\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'+u')))[2] \\ &\quad \overset{L.~{3.4}}{=}{\mathrm{base}}_{n} (a,({\mathrm{I}}_{p,{\mathrm{{\iota}}}}({\mathbf{0}}),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(p'+u'),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}({\mathbf{-1}})))[3]=b_1. \end{align*} $$

On the other hand,

$$ \begin{align*}s(t)={\mathrm{base}}_{n}(a,({\mathrm{I}}_{p,{\mathrm{{\iota}}}}({\mathbf{0}}),{\mathrm{I}}_{p,{\mathrm{{\iota}}}}(-{\mathbf{1}})))[2]=a_1\end{align*} $$

and

$$ \begin{align*} s'({\mathrm{cube}}_{n'}(s(t)))&={\mathrm{cube}}_{n'}(s(t))[p',p'+u']={\mathrm{base}}_{n'}(s(t),(p',p'+u'))[2] \\ &={\mathrm{base}}_{n'}(a_1,(p',p'+u'))[2] \\ & \overset{L.~{3.6}}{=}{\mathrm{base}}_{n} (a_1,({\mathrm{I}}_{{\mathbf{0}},{\mathrm{{\iota}}}}(p'),{\mathrm{I}}_{{\mathbf{0}},{\mathrm{{\iota}}}}(p'+u')))[2]=b_1. \end{align*} $$

We conclude that

$$ \begin{align*}({\mathrm{I}}_{p,{\mathrm{{\iota}}}}(s'))(t)=b_1=s'({\mathrm{cube}}_{n'}(s(t))).\end{align*} $$

Proof. Let $p,v\in \{-1,0\}^{d}$ be such that $s_1={\mathrm {faceAt}}(p,v)$ , which means that $s_2={\mathrm {faceAt}}(p+u,v)$ . Assume that the cardinality of $D=\{i\mid v[i]=-1\}$ is $d'$ and let $\iota :\{1,\ldots ,d'\}\to \{1,\ldots ,d\}$ be an injection with image D. Let $n'\in \mathbb {Z}_+^{d'}$ be defined by $n'[i]=n[{\mathrm {{\iota }}}(i)]$ for $1\leq i\leq d'$ . Let $s'\in S_{d'}$ be such that ${\mathrm {I}}_{p,{\mathrm {{\iota }}}}(s')=s_1'$ , which means that ${\mathrm {I}}_{p+u,{\mathrm {{\iota }}}}(s')=s_2'$ . By Lemma 4.4,

$$ \begin{align*} s_1'(t_1)&={\mathrm{I}}_{p,{\mathrm{{\iota}}}}(s')(t_1)=s'({\mathrm{cube}}_{n'}(s_1(t_1))) \\ &=s'({\mathrm{cube}}_{n'}(s_2(t_2)))={\mathrm{I}}_{p+u,{\mathrm{{\iota}}}}(s')(t_2)=s_2'(t_2). \end{align*} $$

The results of this section allow us to speak of values on edges in a tiling without referring to individual cubes. To be more precise, we imagine $\mathbb {Z}^d$ to be the vertex set of an infinite directed graph, the d-dimensional grid, having an edge $(z-e_i,z)$ (the edge from $z-e_i$ to z) for every $z\in \mathbb {Z}^d$ and $1\leq i\leq d$ . When we consider a tiling $f\in X_{n}$ and say that the cube $f[z]$ is at position $z\in \mathbb {Z}^d$ , we visualize this as the unit cube $[-1,0]^d$ whose ‘top-right’ corner, the point ${\mathbf {0}}$ , is positioned at z. Then the edges of the cube $f[z]$ overlap with some of the edges of the d-dimensional grid: an edge of the form $s={\mathrm {faceAt}}(p,-e_i)$ overlaps the directed edge ${D}_z(s)=(z+p-e_i,z+p)$ in the grid. We also define the undirected edge ${E}_z(s)=\{z+p-e_i,z+p\}$ and the (undirected) label of the undirected edge ${E}_z(s)$ in the tessellation by

$$ \begin{align*} \unicode{x3bb}(f,{E}_z(s))=sf[z].\end{align*} $$

To see that this is well defined, let $z_1,z_2\in \mathbb {Z}^d$ and $s_1={\mathrm {faceAt}}(p_1,-e_i)$ , $s_2={\mathrm {faceAt}}(p_2,-e_i)$ be such that ${E}_{z_1}(s_1)={E}_{z_2}(s_2)$ . Then in particular, $z_1+p_1=z_2+p_2$ and $s_1=s_2+(p_1-p_2)$ . Because of this,

$$ \begin{align*} &\unicode{x3bb}(f,{E}_{z_1}(s_1))=s_1 f[z_1]=(s_2+(p_1-p_2))f[z_2-(p_1-p_2)] \\ &\quad\overset{T.~{4.6}}{=}s_2 f[z_2]=\unicode{x3bb}(f,{E}_{z_2}(s_2)). \end{align*} $$

We note that each edge $\{z-e_i,z\}$ has a label, because ${E}_z({\mathrm {faceAt}}({\mathbf {0}},-e_i))= \{z-e_i,z\}$ . For an example of how a tessellation yields labels on the directed edges, see the left and middle parts of Figure 3 (for the drawn part of the grid, the definition can be applied in the form $\unicode{x3bb} (f,{E}_{{\mathbf {0}}}(s))=sf[{\mathbf {0}}]$ ).

Figure 3

Left: A tessellation with ${\mathrm {cube}}_{(2,3,5)}(10)$ positioned at the origin in $\mathbb {Z}^3$ . Middle: $\mathbb {Z}^3$ as a directed graph together with labels given by ${\mathrm {cube}}_{(2,3,5)}(10)$ . Right: The weights ${\mathrm {wgt}}_{(2,3,5)}(v)$ of the points $v\in \mathbb {Z}^3$ have been added to the grid, the point ${\mathbf {0}}$ is at the upper right with ${\mathrm {wgt}}_{(2,3,5)}({\mathbf {0}})=1$ .

4.3. Path integrals over tilings

Throughout this subsection, let $n\in \mathbb {Z}_+^d$ . We assign to every point $v\in \mathbb {Z}^d$ a weight by ${\mathrm {wgt}}_{n}(v)=m(n,v)$ . A path (of length $k-1$ ) is a sequence $P=(P_i)_{i=1}^k\in (\mathbb {Z}^d)^k$ , where for $1\leq i< k$ , we have $P_{i+1}-P_i=ae_j$ for some $a\in \{-1,1\}$ , $1\leq j\leq d$ . For $1\leq i\leq k-1$ , let $E_P(i)=P_{i+1}-P_i$ . Given a valid tiling $f\in X_{n}$ and a pair of vectors $p,p'\in \mathbb {Z}^d$ satisfying $p'-p=ae_j$ for some $a\in \{-1,1\}$ , $1\leq j\leq d$ , we denote ${\mathrm {sign}}(p'-p)=a$ and

$$ \begin{align*}(p,p')f={\mathrm{sign}}(p'-p)\ {\mathrm{wgt}}_{n}(\max\{p,p'\})\unicode{x3bb}(f,\{p,p'\}).\end{align*} $$

Then the path integral of f over P is

$$ \begin{align*}Pf=\sum_{i=1}^{k-1} (P_{i},P_{i+1})f.\end{align*} $$

See the right part of Figure 3 containing labels given by a tessellation together with the weights of the points. Computing the path integral over a path of length $1$ consists of looking at the corresponding edge, multiplying its (undirected) label with the weight at its end point (now considered as a directed edge), and possibly multiplying the result by $-1$ in the case when the direction of the path opposes the direction of the edge.

For $v\in \mathbb {Z}^d$ , we denote $P+v=(P_i+v)_{i=1}^k$ . We prove an equality connecting the integral of a tessellation f over a shifted path $P+v$ to the integral of a shifted tessellation $\sigma _v(f)$ over a path P.

Proof. We denote $a_i={\mathrm {sign}}((P_{i+1}+v)-(P_i+v))={\mathrm {sign}}(P_{i+1}-P_i)$ and directly compute that

$$ \begin{align*} (P+v)f&=\sum_{i=1}^{k-1}a_i\ {\mathrm{wgt}}_{n}(\max\{P_i+v,P_{i+1}+v\})\unicode{x3bb}(f,\{P_i+v,P_{i+1}+v\})\\ &=\sum_{i=1}^{k-1}a_i\ {\mathrm{wgt}}_{n}(v)\ {\mathrm{wgt}}_{n}(\max\{P_i,P_{i+1}\})\unicode{x3bb}(\sigma_v(f),\{P_i,P_{i+1}\}) \\ &={\mathrm{wgt}}(v)P\sigma_v(f). \end{align*} $$

Next we note a simple cancellation property.

Proof. We may compute

$$ \begin{align*} &(p,p')f+(p',p)f \\ &\quad =a\ {\mathrm{wgt}}_{n}(\max\{p,p'\})\unicode{x3bb}(f,\{p,p'\})-a\ {\mathrm{wgt}}_{n}(\max\{p',p\})\unicode{x3bb}(f,\{p',p\})=0.\qquad \end{align*} $$

In the following lemma, we will show that the path integral over two consecutive edges around a square is equal to the path integral over the other two consecutive edges of the same square, see Figure 4. This may also be tested on the concrete examples of labels and weights in Figure 3.

Figure 4

Let P, $P'$ be the two paths between two opposite corners of a square. The path integrals of a tessellation f over P and $P'$ are equal.

Proof. We first note that $p_1'-p_0=p_2-p_1=a_2e_{i_2}$ and $p_2-p_1'=p_1-p_0=a_1e_{i_1}$ . Therefore, $(p_0,p_1')f$ and $(p_1'+p_2)f$ are defined. To prove the lemma, it is sufficient to consider the case $a_1=a_2=1$ (the case in Figure 4). To see this, we show how to reduce the other cases to this one.

Case 1: $a_1=1, a_2=-1$ . Let $q_0=p_1'$ , $q_1=p_2$ , $q_2=p_1$ , and $q_1'=q_0+(q_2-q_1)= p_1'+(p_1-p_2)=p_0$ . Then $q_1-q_0=p_2-p_1'=a_1e_{i_1}$ , $q_2-q_1=p_1-p_2=(-a_2)e_{i_2}$ and

$$ \begin{align*} &(p_0,p_1)f+(p_1,p_2)f=(p_0,p_1')f+(p_1',p_2)f \\ &\quad \iff (q_1',q_2)f+(q_2,q_1)f=(q_1',q_0)f+(q_0,q_1)f \\ &\quad \overset{L.~{4.8}}{\iff}\hspace{-2pt} (q_0,q_1')f+(q_1',q_2)f=(q_0,q_1)f+(q_1,q_2)f. \end{align*} $$

Case 2: $a_1=-1,a_2=1$ . Let $q_0=p_1$ , $q_1=p_0$ , $q_2=p_1'$ , and $q_1'=q_0+(q_2-q_1)= p_1+(p_1'-p_0)=p_2$ . Then $q_1-q_0=p_0-p_1=(-a_1)e_{i_1}$ , $q_2-q_1=p_1'-p_0=a_2e_{i_2}$ and

$$ \begin{align*} &(p_0,p_1)f+(p_1,p_2)f=(p_0,p_1')f+(p_1',p_2)f \\ &\quad \iff (q_1,q_0)f+(q_0,q_1')f=(q_1,q_2)f+(q_2,q_1')f \\ &\quad \overset{L.~{4.8}}{\iff}\hspace{-2pt} (q_0,q_1')f+(q_1',q_2)f=(q_0,q_1)f+(q_1,q_2)f. \end{align*} $$

Case 3: $a_1\hspace{-1pt}=\hspace{-1pt}a_2\hspace{-1pt}=\hspace{-1pt}-1$ . Let $q_0\hspace{-1pt}=\hspace{-1pt}p_2$ , $q_1\hspace{-1pt}=\hspace{-1pt}p_1'$ , $q_2\hspace{-1pt}=\hspace{-1pt}p_0$ , and $q_1'=q_0+(q_2-q_1)= p_2+ (p_0-p_1')=p_1$ . Then $q_1-q_0=p_1'-p_2=(-a_1)e_{i_1}$ , $q_2-q_1=p_0-p_1'=(-a_2)e_{i_2}$ and

$$ \begin{align*} &(p_0,p_1)f+(p_1,p_2)f=(p_0,p_1')f+(p_1',p_2)f \\ &\quad\iff (q_2,q_1')f+(q_1',q_0)f=(q_2,q_1)f+(q_1,q_0)f \\ &\quad\overset{L.~{4.8}}{\iff}\hspace{-2pt} (q_0,q_1)f+(q_1,q_2)f=(q_0,q_1')f+(q_1',q_2)f. \end{align*} $$

We will therefore assume that $a_1=a_2=1$ and thus $p_0<p_1<p_2$ . We will now represent all terms in the claimed equality in terms of a single cube $t=f[p_2]$ in the tessellation f as follows:

$$ \begin{align*} &(p_0,p_1)f/{\mathrm{wgt}}_{n}(p_2)={\mathrm{wgt}}_{n}(p_1-p_2)\unicode{x3bb}(f,E_{p_2}({\mathrm{faceAt}}(-e_{i_2},-e_{i_1}))) \\ &\quad=n[i_2]{\mathrm{faceAt}}(-e_{i_2},-e_{i_1})t=n[i_2]t[-e_{i_2},-e_{i_2}-e_{i_1}], \\ &(p_1,p_2)f/{\mathrm{wgt}}_{n}(p_2)={\mathrm{wgt}}_{n}(p_2-p_2)\unicode{x3bb}(f,E_{p_2}({\mathrm{faceAt}}({\mathbf{0}},-e_{i_2}))) \\ &\quad={\mathrm{faceAt}}({\mathbf{0}},-e_{i_2})t=t[{\mathbf{0}},-e_{i_2}], \\&(p_0,p_1')f/{\mathrm{wgt}}_{n}(p_2)={\mathrm{wgt}}_{n}(p_1'-p_2)\unicode{x3bb}(f,E_{p_2}({\mathrm{faceAt}}(-e_{i_1},-e_{i_2})))\\ &\quad=n[i_1]{\mathrm{faceAt}}(-e_{i_1},-e_{i_2})t=n[i_1]t[-e_{i_1},-e_{i_1}-e_{i_2}]\quad\mbox{and } \\ &(p_1',p_2)f/{\mathrm{wgt}}_{n}(p_2)={\mathrm{wgt}}_{n}(p_2-p_2)\unicode{x3bb}(f,E_{p_2}({\mathrm{faceAt}}({\mathbf{0}},-e_{i_1})))\\ &\quad={\mathrm{faceAt}}(0,-e_{i_1})t=t[{\mathbf{0}},-e_1]. \end{align*} $$

From this, we see that

$$ \begin{align*} &((p_0,p_1)f+(p_1,p_2)f)/{\mathrm{wgt}}_{n}(p_2) \\ &\quad =n[i_2]t[-e_{i_2},-e_{i_2}-e_{i_1}]+t[{\mathbf{0}},-e_{i_2}]\overset{L.~{4.3}}{=}t[{\mathbf{0}},-e_{i_1}-e_{i_2}]\quad\mbox{and}\\ &((p_0,p_1')f+(p_1',p_2)f)/{\mathrm{wgt}}_{n}(p_2) \\ &\quad =n[i_1]t[-e_{i_1},-e_{i_1}-e_{i_2}]+t[{\mathbf{0}},-e_{i_1}]\overset{L.~{4.3}}{=}t[{\mathbf{0}},-e_{i_1}-e_{i_2}]. \end{align*} $$

For a path $P=(P_i)_{i=1}^k$ and for $1\leq i<k$ , let $a_i\in \{-1,1\}$ and let $1\leq m(i)\leq d$ be such that $a_i e_{m(i)}=E_P(i)$ . We say that an integer i with $1\leq i<k-1$ is a return if $m(i)=m(i+1)$ and $a_i\neq a_{i+1}$ . We say that i is an inversion if $m(i)>m(i+i)$ . The abelianization of P is the unique path ${\mathrm {Ab}}(P)=(P_i')_{i=1}^{k'}$ with $a_i' e_{m'(i)}=E_{P'}(i)$ for $1\leq i<k'$ that satisfies:

• • $P_1'=P_1$ and $P_k'=P_k$ ( $P'$ and P have the same start and end points);

• • if $1\leq i<k'-1$ , then $m'(i)\leq m'(i+1)$ ( $P'$ has no inversions);

• • if $1\leq i<k'-1$ and $m'(i)=m'(i+1)$ , then $a_{i}'=a_{i+1}'$ ( $P'$ has no returns).

We will use the abelianization to show that the integral of a tiling over a path depends only on the choice of the start and end points of the path. For the proof of the following lemma, let $\ell (P)$ be the natural number whose base-d expansion is $m(1)\cdots m(k)$ .

Proof. We will manipulate any given path $P\neq {\mathrm {Ab}}(P)$ into another path $P'$ which satisfies $\ell (P')<\ell (P)$ . We also make sure that $P'$ has the same start and end points as P (so ${\mathrm {Ab}}(P')={\mathrm {Ab}}(P)$ ) and that $P'f=Pf$ . Since $\ell (P)$ is a natural number, this argument can be repeated only finitely many times and it will eventually yield the path ${\mathrm {Ab}}(P)$ . Assume therefore that $P\neq {\mathrm {Ab}}(P)$ and let $a_i e_{m(i)}=E_P(i)$ for $1\leq i<k$ . There are two possibilities.

Case 1, a return: $m(i)=m(i+1)$ and $a_i\neq a_{i+1}$ for some $1\leq i<k-1$ . Let ${P'=(P_1,\ldots ,P_i,P_{i+3},\ldots ,P_k)}$ . Then, using the fact that $P_{i+2}=P_i$ , we see from Lemma 4.8 that $(P_i,P_{i+1})f+(P_{i+1},P_{i+2})f=0$ and therefore

$$ \begin{align*} Pf=&\sum_{j=1}^{i-1}(P_j,P_{j+1})f+(P_{i+2},P_{i+3})f+\sum_{j=i+3}^{k-1}(P_j,P_{j+1})f \\ =&\sum_{j=1}^{i-1}(P_j,P_{j+1})f+(P_i,P_{i+3})f+\sum_{j=i+3}^{k-1}(P_j,P_{j+1})=P'f. \end{align*} $$

The path $P'$ has the same start and end points as P, and it is shorter than P, so $\ell (P')<\ell (P)$ .

Case 2, an inversion: $m(i)>m(i+1)$ for some $1\leq i<k-1$ . Let $P_{i+1}'=P_i +(P_{i+2} -P_{i+1})$ and let $P'=(P_1,\ldots ,P_{i},P_{i+1}',P_{i+2},\ldots ,P_k)$ . It follows from Lemma 4.9 that $(P_i,P_{i+1})f+(P_{i+1},P_{i+2})f=(P_i,P_{i+1}')f=(P_{i+1}',P_{i+2})f$ and therefore $Pf=P'f$ . Letting $b_j e_{m'(i)}=E_{P'}(j)$ for $1\leq j<k$ , we check that

$$ \begin{align*}b_i e_{m'(i)}=P_{i+1}'-P_i=P_{i+2}-P_{i+1}=a_{i+1}e_{m(i+1)}\end{align*} $$

and thus $m'(i)=m(i+1)<m(i)$ , so $\ell (P')<\ell (P)$ .

Remark 4.12. The inspiration for the terminology of ‘path integral’ comes from line integrals of holomorphic functions, and Theorem 4.11 is analogous to the fact that the line integral of a holomorphic function over a closed curve is equal to zero. In the case of meromorphic functions, if the line integral over a closed curve is not zero, its interior necessarily contains a non-removable singularity. Similarly, if we are given only a partial tiling of the plane and the path integral around a non-tiled part is not zero, then the interior of the path cannot be completed into a valid tiling: see Figure 5 for an example.

Figure 5

A partial valid tiling f using $T_{(2,5)}$ . There is no way to complete the tiling, because the path integral around the non-tiled part is not zero.

The situation here is reminiscent of the tiling group approach of [Reference Thurston12] for determining whether a collection of polygonal tiles can tile a finite region of the plane. In that approach, the tiles can be interpreted as describing relators in a group, and if the group element describing the boundary of the region is not the identity element, then tiling the region is not possible.

Now, given a tessellation $f\in X_{n}$ and a pair of vectors $p,p'\in \mathbb {Z}^d$ , define

$$ \begin{align*}(p,p')f=Pf\end{align*} $$

for any path P going from p to $p'$ : by Theorem 4.11, the choice of the path P does not matter. For the directed pair $(p,p')$ , we also define the (directed) label

$$ \begin{align*}\unicode{x3bb}(f,(p,p'))=(p,p')f/{\mathrm{wgt}}_{n}(p').\end{align*} $$

The directed and undirected labels agree in the sense that $\unicode{x3bb} (f,(z-e_i,z))=\unicode{x3bb} (f, \{z-e_i,z\})$ for $z\in \mathbb {Z}^d$ and $1\leq i\leq d$ . As one would hope, the labels do not change when the tessellation is shifted.

Lemma 4.13. Let $f\in X_{n}$ and let $p,p'\in \mathbb {Z}^d$ . For any $v\in \mathbb {Z}^d$ , it holds that

$$ \begin{align*}\unicode{x3bb}(f,(p+v,p'+v))=\unicode{x3bb}(\sigma_v(f),(p,p')).\end{align*} $$

Proof. We have

$$ \begin{align*} &\unicode{x3bb}(f,(p+v,p'+v))=(p+v,p'+v)f/{\mathrm{wgt}}_{n}(p'+v) \\ &\quad\overset{L.~{4.7}}{=}{\mathrm{wgt}}_{n}(v)(p,p')\sigma_v(f)/{\mathrm{wgt}}_{n}(p'+v)=(p,p')\sigma_v(f)/{\mathrm{wgt}}_{n}(p') \\ &\quad =\unicode{x3bb}(\sigma_v(f),(p,p')). \end{align*} $$

The labeling map inherits a summation property from the summation property of paths. The next lemma associates to any label in a tiling a mixed base representation of sorts, where the digits also come from labels in the tiling. In a special case, the base consists of powers of a single number $m(n,v)\in \mathbb {Q}$ , which is however not necessarily an integer.

Lemma 4.14. Let $p_0,p_1,\ldots ,p_k\in \mathbb {Z}^d$ and let $f\in X_{n}$ . Then

$$ \begin{align*}\unicode{x3bb}(f,(p_k,p_0))=\sum_{i=0}^{k-1} m(n,p_{i}-p_0)\unicode{x3bb}(f,(p_{i+1},p_{i})).\end{align*} $$

In particular, for $v\in \mathbb {Z}^d$ and $k\in \mathbb {N}$ , it holds that

$$ \begin{align*}\unicode{x3bb}(f,(kv+p_0,p_0))=\sum_{i=0}^{k-1} m(n,v)^i\unicode{x3bb}(f,((i+1)v+p_0,iv+p_0)).\end{align*} $$

Proof. We have

$$ \begin{align*} &\unicode{x3bb}(f,(p_k,p_0))=(p_k,p_0)f/{\mathrm{wgt}}_{n}(p_0)=\sum_{i=0}^{k-1}(p_{i+1},p_{i})f/{\mathrm{wgt}}_{n}(p_0) \\ &\quad=\sum_{i=0}^{k-1} {\mathrm{wgt}}_{n}(p_i)\unicode{x3bb}(f,(p_{i+1},p_{i}))/{\mathrm{wgt}}_{n}(p_0)=\sum_{i=0}^{k-1} {\mathrm{wgt}}_{n}(p_i-p_0)\unicode{x3bb}(f,(p_{i+1},p_{i})). \end{align*} $$

For the latter part, choose $p_i=iv+p_0$ and note that

$$ \begin{align*}m(n,p_i-p_0)=m(n,iv)=\prod_{j=1}^d n[j]^{-iv[j]}=m(n,v)^i.\end{align*} $$

The base $m(n,v)$ of Lemma 4.14 is a natural number at least when $v\leq {\mathbf {0}}$ . In that case, the ‘digits’ $\unicode{x3bb} (f,((i+1)v+p_0,iv+p_0))$ are actually just usual base- $m(n,v)$ digits, which follows from the next lemma.

Proof. By Lemma 4.13, it is sufficient to show that if $p\leq {\mathbf {0}}$ , then $\unicode{x3bb} (f,(p,{\mathbf {0}}))\in \mathbb {N}$ and $\unicode{x3bb} (f,(p,{\mathbf {0}}))<m(n,p)$ . The proof is by induction on the distance of p from the origin, the base case $p={\mathbf {0}}$ is simple. Assume then that the claim holds for some $p\leq {\mathbf {0}}$ and consider the vector $p-e_i$ for some $1\leq i\leq d$ . Then

$$ \begin{align*} &\unicode{x3bb}(f,(p-e_i,{\mathbf{0}}))=(p-e_i,{\mathbf{0}})f=(p-e_i,p)f+(p,{\mathbf{0}})f \\ &\quad=\underbrace{{\mathrm{wgt}}_{n}(p)}_{\in\mathbb{N}}\underbrace{\unicode{x3bb}(f,\{p-e_i,p\})}_{\in\mathbb{N}}+\underbrace{\unicode{x3bb}(f,(p,{\mathbf{0}}))}_{\in\mathbb{N}}<m(n,p)\ {\mathrm{faceAt}}({\mathbf{0}},-e_i)f[p]+m(n,p) \\ &\quad=m(n,p)({\mathrm{base}}_{n}(f[p],({\mathbf{0}},-e_i))[2]+1)\leq m(n,p) n[i]=m(n,p-e_i), \end{align*} $$

where we use the fact that ${\mathrm {base}}_{n}(f[p],({\mathbf {0}},-e_i))[2]<m(n,-e_i)/m(n,{\mathbf {0}})=n[i]$ .

Lemma 4.16. Let $p_k\leq \cdots \leq p_0$ and let $f\in X_{n}$ . Then for $1\leq i\leq k'\leq k$ ,

$$ \begin{align*}{\mathrm{base}}_{n}(\unicode{x3bb}(f,(p_k,p_0)),(p_j-p_0)_{j=1}^{k'})[i]=\unicode{x3bb}(f,(p_i,p_{i-1})).\end{align*} $$

Proof. By Lemma 4.14,

$$ \begin{align*} \unicode{x3bb}(f,(p_k,p_0))=\sum_{i=0}^{k-1} m(n,p_{i}-p_0)\unicode{x3bb}(f,(p_{i+1},p_{i})). \end{align*} $$

From this, the claim follows for $k'=k$ by the definition of a mixed base representation, because by Lemma 4.15, $0\leq \unicode{x3bb} (f,(p_{i+1},p_{i}))<m(n,p_{i+1}-p_{i})= m(n,p_{i+1}-p_0)/m(n,p_{i}-p_0)$ for $1\leq i\leq k$ . The claim for general $k'$ follows from the case $k'=k$ by Lemma 3.4.

Proof. Up to shifting f, we may assume that $z={\mathbf {0}}$ . Let $t=f[{\mathbf {0}}]$ . For $0\leq i\leq d$ , let $e_i'=\sum _{j=1}^i e_j$ . Then $e_i'$ form a binary d-directive sequence. Using the definition of the undirected label of a single edge, we see that

$$ \begin{align*}\unicode{x3bb}(f,\{e_{i+1}',e_i'\})={\mathrm{faceAt}}(e_i',e_{i+1})f[{\mathbf{0}}]=t[e_i',e_{i+1}'].\end{align*} $$

Using this, we compute that

$$ \begin{align*} \unicode{x3bb}(f,(-{\mathbf{1}},{\mathbf{0}}))&\overset{L.~{4.14}}{=}\sum_{i=0}^{d-1}m(n,e_i')\unicode{x3bb}(f,\{e_{i+1}',e_i'\}) \\ &=\sum_{i=0}^{d-1}m(n,e_i')t[e_i',e_{i+1}']\overset{L.~{4.3}}{=}t[{\mathbf{0}},{\mathbf{1}}] ={\mathrm{val}}_{n}(t). \end{align*} $$

For a face $s={\mathrm {faceAt}}(p,v)$ and $z\in \mathbb {Z}^d$ , denote ${D}_z(s)=(z+p+v,z+p)$ (this generalizes an earlier definition where s was an edge). We generalize the defining formula for the label of an edge in the following lemma.

Lemma 4.18. Let $f\in X_{n}$ . For $s={\mathrm {faceAt}}(p,v)$ and $z\in \mathbb {Z}^d$ , it holds that

$$ \begin{align*}\unicode{x3bb}(f,{D}_z(s))=sf[z].\end{align*} $$

Proof. Up to shifting f, we may assume that $z={\mathbf {0}}$ . Let $a\in \mathbb {N}$ be such that ${\mathrm {cube}}_{n}(a)=f[{\mathbf {0}}]$ . Then, by applying Lemma 4.16 with the choice $(p_0,p_1,p_2,p_3)=({\mathbf {0}},p,p+v,{\mathbf {1}})$ ,

$$ \begin{align*} \unicode{x3bb}(f,({D}_{{\mathbf{0}}}(s)))&=\unicode{x3bb}(f,(p+v,p))\overset{L.~{4.16}}{=}{\mathrm{base}}_{n}(\unicode{x3bb}(f,(-{\mathbf{1}},{\mathbf{0}})),(p,p+v))[2] \\ &\overset{L.~{4.17}}{=}{\mathrm{base}}_{n}(a,(p,p+v))[2]={\mathrm{cube}}_n(a)[p,p+v]=sf[{\mathbf{0}}]. \end{align*} $$

As we shall see, by Lemma 4.17, labels along the direction of ${\mathbf {1}}$ can be found by reading the values of a sequence of cubes and interpreting this sequence of values as the representation of a number in base $N=\prod _{i=1}^d n[i]$ . We actually prove a more general statement by using Lemma 4.18.

Lemma 4.19. Let $p_k\leq \cdots \leq p_0$ with $p_{i+1}-p_i\geq -{\mathbf {1}}$ for $0\leq i<k$ and let $f\in X_{n}$ . Then

$$ \begin{align*}\unicode{x3bb}(f,(p_k,p_0))=\sum_{i=0}^{k-1} m(n,p_{i}-p_0)\ {\mathrm{faceAt}}({\mathbf{0}},p_{i+1}-p_i)f[p_i].\end{align*} $$

In particular, for any $k\in \mathbb {N}$ and $v\in \mathbb {Z}^d$ ,

$$ \begin{align*}\unicode{x3bb}(f,(v-{\mathbf{k}},v))=\sum_{i=0}^{k-1} N^i\ {\mathrm{val}}_n(f[v-{\mathbf{i}}]).\end{align*} $$

We conclude this subsection with some justification for why the elements of $T_n$ are called multiplication cubes.

Proposition 4.20. Let $p, p'\in \mathbb {Z}^d$ , $v\in \mathbb {Z}^d$ , and let $f\in X_n$ . Then

$$ \begin{align*} \unicode{x3bb}(f,(p+v,p'+v)) &= m(n,-v)\unicode{x3bb}(f,(p,p'))\\ &\quad + m(n,-p'-v)((p+v,p)f+(p',p'+v)f).\end{align*} $$

Proof. By Theorem 4.11, $(p+v,p'+v)f=(p+v,p)f+(p,p')f+(p',p'+v)f$ , and by replacing path integrals with labels, we find

$$ \begin{align*} m(n,p'+v)\ \unicode{x3bb}(f,(p+v,p'+v))&=(p+v,p)f\\ &\quad+m(n,p')\ \unicode{x3bb}(f,(p,p'))+(p',p'+v)f.\end{align*} $$

The claim follows by dividing both sides of the equation with $m(n,p'+v)$ .

The message of this proposition is that moving in $\mathbb {Z}^d$ by a vector v changes the values of the labeling map by the factor of $m(n,-v)$ , possibly up to an error term. In particular, moving to the direction of a basis vector $e_i$ corresponds to multiplication by $n[i]$ in this sense. Sometimes, the error term can be made equal to zero. We also remark that another type of tile set performing multiplication in a similar sense has appeared in [Reference Kari4].

Example 4.21. We may now explain the powers of two in Figure 1. Observe that in the figure, $(-{\mathbf {3}}+e_1,-{\mathbf {3}})f=({\mathbf {0}},e_1)f=0$ . Then by applying the previous proposition for $p=-{\mathbf {3}}$ , $p'={\mathbf {0}}$ , and $v=e_1$ , it follows that

$$ \begin{align*}\unicode{x3bb}(f,(-{\mathbf{3}}+e_1,e_1))=m((2,5),-e_1)\unicode{x3bb}(f,(-{\mathbf{3}},{\mathbf{0}}))=2\unicode{x3bb}(f,(-{\mathbf{3}},{\mathbf{0}})).\end{align*} $$

Alternatively, by using the latter part of Lemma 4.19, the values $\unicode{x3bb} (f,(-{\mathbf {3}},{\mathbf {0}}))=64$ and $\unicode{x3bb} (f,(-{\mathbf {3}}+e_1,e_1))=128$ can be read directly from the figure. Similar arguments connect the other powers of two appearing in the figure with each other.

Next we show that Proposition 4.20 can be extended to path integrals over some infinite paths so that the error terms vanish under very natural assumptions, thus connecting multiplication operations to moving around tessellations in an even more satisfactory sense. Given vectors $p,v\in \mathbb {Z}^d$ with $v\leq {\mathbf {0}}$ , we consider the infinite path going through p along v, and given $f\in X_n$ , we denote

$$ \begin{align*}{\mathrm{frac}}_{p,v}(f)=\lim_{i\to\infty}(p,p-iv)f\mbox{ and }{\mathrm{int}}_{p,v}(f)=\lim_{i\to\infty}(p+iv,p)f.\end{align*} $$

Alternatively, we can rewrite

$$ \begin{align*} {\mathrm{frac}}_{p,v}(f)&=\sum_{i=0}^\infty(p-i v,p-(i+1)v)f \\& ={\mathrm{wgt}}_n(p)\sum_{i=0}^\infty{\mathrm{wgt}}_n(v)^{-(i+1)}\unicode{x3bb}(f,(p-i v,p-(i+1)v))\quad\mbox{and} \\{\mathrm{int}}_{p,v}(f)&=\sum_{i=0}^\infty(p+(i+1)v,p+i v)f \\&={\mathrm{wgt}}_n(p)\sum_{i=0}^\infty{\mathrm{wgt}}_n(v)^{i}\unicode{x3bb}(f,(p+(i+1)v,p+iv)), \end{align*} $$

where $\unicode{x3bb} (f,(p-i v,p-(i+1)v))$ and $\unicode{x3bb} (f,(p+(i+1)v,p+iv))$ are natural numbers less than ${\mathrm {wgt}}_n(v)$ by Lemma 4.15. The names of these quantities come from the fact that these are essentially (up to ignoring the factor ${\mathrm {wgt}}_n(p)$ ) the fractional parts and integral parts of some quantity represented in base ${\mathrm {wgt}}_n(v)$ : ${\mathrm {frac}}_{p,v}(f)/{\mathrm {wgt}}_n(p)\in [0,1]$ and ${\mathrm {int}}_{p,v}(f)/{\mathrm {wgt}}_n(p)\in \mathbb {N}$ if it is finite.

We write

$$ \begin{align*} &{\mathrm{real}}_{p,v}(f)={\mathrm{int}}_{p,v}(f)+{\mathrm{frac}}_{p,v}(f)=\lim_{i\to\infty}(p+iv,p-iv)f \\ &\quad ={\mathrm{wgt}}_n(p)\sum_{i=-\infty}^\infty{\mathrm{wgt}}_n(v)^{i}\unicode{x3bb}(f,(p+(i+1)v,p+iv)). \end{align*} $$

In the special case $p={\mathbf {0}}$ , we may omit the subscript p, and in this case, the sum above yields a base- ${\mathrm {wgt}}_n(v)$ representation for ${\mathrm {real}}_v(f)$ . If additionally $v=-{\mathbf {1}}$ , we may also omit the subscript v and say that the tessellation f represents the number ${\mathrm {real}}(f)$ , but in Proposition 4.25, it will turn out that the choices of p and v do not matter much. By Lemma 4.17, the number ${\mathrm {real}}(f)$ can be read directly from the cubes on the main diagonal of the tessellation f:

$$ \begin{align*}{\mathrm{real}}(f)=\sum_{i=-\infty}^\infty{\mathrm{wgt}}_n(-{\mathbf{1}})^{i}\ {\mathrm{val}}_n(f[-{\mathbf{i}}])=\sum_{i=-\infty}^\infty N^i\ {\mathrm{val}}_n(f[-{\mathbf{i}}]).\end{align*} $$

If ${\mathrm {wgt}}_n(v)>1$ , then in the sum defining ${\mathrm {int}}_{p,v}(f)$ , the part ${\mathrm {wgt}}_n(v)^{i}$ tends to infinity and so the finiteness of the sum is equivalent to the equality $\unicode{x3bb} (f,(p+(i+1)v,p+iv))=0$ (or equivalently $(p+(i+1)v,p+i v)f=0$ ) holding for all sufficiently large i. In fact, finiteness of ${\mathrm {int}}_{p,v}(f)$ also implies that many other labels are equal to zero.

Proof. To show the implication (1) $\implies $ (2), assume that ${\mathrm {int}}_{p,v}(f)$ is finite for some $v\ll {\mathbf {0}}$ and ${\mathrm {wgt}}_n(v)>1$ , and consider any $p_2\leq p_1$ such that ${\mathrm {wgt}}_n(p_1)>{\mathrm {real}}_{p,v}(f)$ . Since ${v\ll {\mathbf {0}}}$ , we may fix some $I\in \mathbb {N}$ such that $p_1\leq p-I v$ and $p+I v\leq p_2$ . If it were the case that $\unicode{x3bb} (f,(p_2,p_1))>0$ , then

$$ \begin{align*} &{\mathrm{real}}_{p,v}(f)\geq(p+Iv,p-I v)f=(p+Iv,p_2)f+(p_2,p_1)f+(p_1,p-Iv)f \\ &\quad \geq (p_2,p_1)f={\mathrm{wgt}}_n(p_1)\unicode{x3bb}(f,(p_2,p_1))>{\mathrm{real}}_{p,v}(f)\unicode{x3bb}(f,(p_2,p_1)) \end{align*} $$

and $\unicode{x3bb} (f,(p_2,p_1))>0$ implies that ${\mathrm {real}}_{p,v}(f)>0$ , but dividing by ${\mathrm {real}}_{p,v}(f)$ in the inequalities yields $1>\unicode{x3bb} (f,(p_2,p_1))$ , which is a contradiction.

To show the implication (2) $\implies $ (3), assume that $\unicode{x3bb} (f,(p_2,p_1))=0$ whenever $p_2\leq p_1$ and ${\mathrm {wgt}}_n(p_1)$ is sufficiently large. The number ${\mathrm {wgt}}_n(p+i v)$ tends to infinity as i tends to infinity, so by the assumption, $\unicode{x3bb} (f,(p+(i+1)v,p+iv))=0$ for sufficiently large i and thus ${\mathrm {int}}_v(f)$ is finite.

Proof. It suffices to prove that ${\mathrm {frac}}_{-{\mathbf {k}}}(f)=({\mathbf {0}},p)f+{\mathrm {frac}}_{p,v}(f)$ for sufficiently large $k\in \mathbb {N}$ , because then,

$$ \begin{align*}({\mathbf{0}},p)f+{\mathrm{frac}}_{p,v}(f)={\mathrm{frac}}_{-{\mathbf{k}}}(f)=({\mathbf{0}},q)f+{\mathrm{frac}}_{q,w}(f).\end{align*} $$

Let therefore $k\in \mathbb {N}$ be so large that $-{\mathbf {k}}=v+v'$ for some $v'\ll {\mathbf {0}}$ . For sufficiently large ${i\in \mathbb {N}}$ , it holds that $p-iv\leq -iv-iv'$ , and so an application of Lemma 4.15 at the position indicated below shows that the following expression is non-negative and simultaneously bounds it from above:

$$ \begin{align*} &{\mathrm{frac}}_{-{\mathbf{k}}}(f)-{\mathrm{frac}}_{p,v}(f)-({\mathbf{0}},p)f=\lim_{i\to\infty}({\mathbf{0}},i{\mathbf{k}})f-(p,p-iv)f-({\mathbf{0}},p)f \\ &\quad =\lim_{i\to\infty}({\mathbf{0}},i{\mathbf{k}})f-({\mathbf{0}},p-iv) f=\lim_{i\to\infty}(p-iv,i{\mathbf{k}})f \\ &\quad =\lim_{i\to\infty}(p-iv,-iv-iv')f=\lim_{i\to\infty}\unicode{x3bb}(f,(p-iv,-iv-iv'))\ {\mathrm{wgt}}_n(v+v')^{-i} \\ &\quad \overset{L.~{4.15}}{\leq}\lim_{i\to\infty}{\mathrm{wgt}}_n(p+iv'){\mathrm{wgt}}_n(v+v')^{-i}=\lim_{i\to\infty}{\mathrm{wgt}}_n(p)\ {\mathrm{wgt}}_n(v)^{-i}=0. \end{align*} $$

Proof. From Lemma 4.22, it follows that ${\mathrm {int}}_{q,w}(f)$ is also finite. By using Lemma 4.22, choose a sufficiently large $I\in \mathbb {N}$ so that $(p',p+Iv)f=0$ and $(q',q+Iw)f=0$ whenever $p'\leq p+Iv$ and $q'\leq q+Iw$ . In particular, ${\mathrm {int}}_{p,v}(f)=(p+Iv,p)f$ and ${\mathrm {int}}_{q,w}(f)= (q+Iw,q)f$ . Fix some $r\in \mathbb {Z}^d$ such that $r\leq p+Iv$ and $r\leq p+Iw$ . Then

$$ \begin{align*} &{\mathrm{int}}_{p,v}(f)+(p,q)f=(r,p+Iv)f+(p+Iv,p)f+(p,q)f=(r,q)f \\ &\quad =(r,q+Iw)f+(q+Iw,q)f={\mathrm{int}}_{q,w}(f). \end{align*} $$

Proof. Combine the two previous lemmas to see that

$$ \begin{align*}{\mathrm{frac}}_{p,v}(f)+{\mathrm{int}}_{p,v}(f)+(p,q)f=(p,q)f+{\mathrm{frac}}_{q,w}(f)+{\mathrm{int}}_{q,w}(f)\end{align*} $$

and subtract $(p,q)f$ from both sides of the equality.

We are now ready to present the analogue of Proposition 4.20 for infinite paths. The following proposition shows that shifting a tessellation by $\sigma _v$ multiplies the real number it represents by ${\mathrm {wgt}}_n(-v)$ without additional error terms.

Proof. We compute

$$ \begin{align*} &{\mathrm{real}}(\sigma_v(f))\overset{P.~{4.25}}{=}{\mathrm{real}}_{-v,-{\mathbf{1}}}(\sigma_v(f))=\lim_{i\to\infty}(-v-{\mathbf{i}},-v+{\mathbf{i}})\sigma_v(f)\\ &\quad \overset{L.~{4.7}}{=}{\mathrm{wgt}}_n(-v)\lim_{i\to\infty}(-{\mathbf{i}},{\mathbf{i}})f={\mathrm{wgt}}_n(-v)\ {\mathrm{real}}(f). \end{align*} $$

4.4. Macrotiles and microtiles

In Figure 6, tiles of $T_{(2,5)}$ (left) are grouped into $2\times 2$ squares (middle), values in the centers of the new squares are given by computing the labels from the bottom left to the top right in the original $2\times 2$ squares, and labels for the edges of the new squares are given by computing the labels of the boundaries of the original $2\times 2$ squares. For example, assuming that the top right corner of the top right square is at the origin of a tiling f, the value in the center of the new top right square comes from computing

$$ \begin{align*}\unicode{x3bb}(f,(-{\mathbf{2}},{\mathbf{0}}))\overset{L.~{4.19}}{=}\sum_{i=0}^{1} (2\cdot 5)^i\ {\mathrm{val}}_n(f[-{\mathbf{i}}]) = 8 + 10\cdot 3 = 38\end{align*} $$

and the label for the right edge of the new top right square comes from computing (with the choices $p_0={\mathbf {0}}$ , $p_1=-e_2$ , and $p_2=-2e_2$ in Lemma 4.19)

$$ \begin{align*} &\unicode{x3bb}(f,(-2e_2,{\mathbf{0}}))\overset{L.~{4.19}}{=}\sum_{i=0}^{1} m((2,5),p_{i}-p_0)\ {\mathrm{faceAt}}({\mathbf{0}},p_{i+1}-p_i)f[p_i]\\ &\quad ={\mathrm{faceAt}}({\mathbf{0}},-e_2)f[{\mathbf{0}}] + 5\cdot {\mathrm{faceAt}}({\mathbf{0}},-e_2)f[-e_2] = 3 + 5\cdot 2 = 13. \end{align*} $$

It turns out that the resulting squares are tiles of $T_{(4,25)}$ . Grouping these new tiles again into $2\times 2$ squares yields tiles of $T_{(16,625)}$ . In this subsection, we show how partitioning a valid tessellation into larger squares, and even into more general (multidimensional) parallelepipeds, yields new multiplication cubes.

Figure 6

Tiles from $T_{(2,5)}$ (left) grouped into larger macrotiles of $T_{(4,25)}$ (middle) and $T_{(16,625)}$ (right).

Throughout this subsection, let $n\in \mathbb {Z}_+^d$ and let A be a $d\times d'$ natural number matrix. We interpret the columns $Ae_i$ of A as the generating vectors of the parallelepiped which will yield the new multiplication cubes, e.g., in the case of Figure 6, we would choose $A={\mathrm {diag}}(2,2)$ . We denote $n^A=m(n,(-Ae_i)_{i=1}^{d'})$ , in particular, $n^I=n$ for the identity matrix I. We define the A-macrotile map ${\mathrm {M}}_{A,n}:X_{n}\to X_{n^{A}}$ (usually written just as ${\mathrm {M}}_A$ ) by

$$ \begin{align*}{\mathrm{M}}_{A,n}(f)[v]={\mathrm{cube}}_{n^A}((-A{\mathbf{1}},{\mathbf{0}})\sigma_{Av}(f))={\mathrm{cube}}_{n^A}(\unicode{x3bb}(f,(A(v-{\mathbf{1}}),Av)))\end{align*} $$

for $f\in X_{n}$ , $v\in \mathbb {Z}^{d'}$ . One needs to check that ${\mathrm {M}}_A(f)$ really belongs to $X_{n^A}$ . It is simple to verify that $\unicode{x3bb} (f,(A(v-{\mathbf {1}}),Av))\in \Sigma _{N'}$ with

$$ \begin{align*}N'=\prod_{i=1}^{d'} n^A[i]=\prod_{i=1}^{d'} m(n,-Ae_i)=m(n,-A{\mathbf{1}}),\end{align*} $$

because $\unicode{x3bb} (f,(A(v-{\mathbf {1}}),Av))<m(n,-A{\mathbf {1}})=N'$ by Lemma 4.15.

In Theorem 4.27, it will turn out that labels in the original tiling correspond to the labels of faces in the macrotiling. From this, it follows that the hyperfaces of neighboring cubes match in ${\mathrm {M}}_A(f)$ and that ${\mathrm {M}}_A(f)\in X_{n^A}$ .

Let us observe that

$$ \begin{align*}m(n^A,v_i)=\prod_{j=1}^{d'}m(n,-Ae_j)^{-v_i[j]}=\prod_{j=1}^{d'}m(n,A(v_i[j]e_j))=m(n,Av_i)\end{align*} $$

and

$$ \begin{align*}{\mathrm{base}}_{n^A}(a,(v_i))={\mathrm{base}}(a,m(n^A,(v_i)))={\mathrm{base}}(a,m(n,(Av_i)))={\mathrm{base}}_{n}(a,(Av_i)).\end{align*} $$

Proof. By applying Lemma 4.16 with the choice

$$ \begin{align*}(p_0,p_1,p_2,p_3)=(Az,A(z+p),A(z+p+u),A(z-{\mathbf{1}})),\end{align*} $$

we have

$$ \begin{align*} st&=s\ {\mathrm{cube}}_{n^A}(\unicode{x3bb}(f,(A(z-{\mathbf{1}}),Az))) \\ &={\mathrm{base}}_{n^A}(\unicode{x3bb}(f,(A(z-{\mathbf{1}}),Az),(p,p+u))[2] \\ &={\mathrm{base}}_{n}(\unicode{x3bb}(f,(A(z-{\mathbf{1}}),Az)),(Ap,A(p+u)))[2] \\ &\overset{L.~{4.16}}{=}\unicode{x3bb}(f,(A(z+p+u),A(z+p))), \end{align*} $$

which proves the first claim. To prove the second claim, we apply the first claim twice:

$$ \begin{align*} &{\mathrm{\tau}}_i({\mathrm{M}}_A(f)[z])={\mathrm{faceAt}}({\mathbf{0}},-{\mathbf{1}}+e_i)\ {\mathrm{M}}_A(f)[z]=\unicode{x3bb}(f,(A(z-{\mathbf{1}}+e_i),Az)) \\ &\quad ={\mathrm{faceAt}}(-e_i,-{\mathbf{1}}+e_i)\ {\mathrm{M}}_k(f)[z+e_i]={\mathrm{\beta}}_i({\mathrm{M}}_A(f)[z+e_i]). \end{align*} $$

Remark 4.28. Consider the special case where the columns of A are the first $d'$ elementary basis vectors of $\mathbb {Z}^d$ . Then for every $z'\in \mathbb {Z}^{d'}$ , we have $Az'=z'{\mathrm {\vee }}_{d'}{\mathbf {0}}$ , where ${\mathbf {0}}\in \mathbb {Z}^{d-d'}$ . Fix $z'\in \mathbb {Z}^{d'}$ and let $z=Az'$ . Let $s'={\mathrm {faceAt}}(p',u')$ for $p',u'\in \mathbb {Z}^{d'}$ and let $p=Ap'$ , $u=Au'$ . Then $p,u\in \{-1,0\}^d$ are orthogonal, so we may define $s={\mathrm {faceAt}}(p,u)$ and it follows that

$$ \begin{align*} s'({\mathrm{M}}_A(f)[z'])&\overset{T.~{4.27}}{=}\unicode{x3bb}(f,(A(z'+p'+u'),A(z'+p'))) \\ &=\unicode{x3bb}(f,{D}_z(s))\overset{L.~{4.18}}{=}sf[z]. \end{align*} $$

The meaning of this equality is that the tessellation ${\mathrm {M}}_A(f)$ is formed by looking at a cut of the tessellation f along the first $d'$ coordinate axes.

As a more concrete example, consider $f\in X_{(2,3,5)}$ , and let the columns of A be the basis vectors $e_1=(1,0,0)$ and $e_2=(0,1,0)$ . For any $z'=(z_1,z_2)\in \mathbb {Z}^2$ and any face of the two-dimensional cube $s={\mathrm {faceAt}}((p_1,p_2),(u_1,u_2))$ , one finds that

$$ \begin{align*} &{\mathrm{faceAt}}((p_1,p_2),(u_1,u_2))\ {\mathrm{M}}_A(f)[(z_1,z_2)] \\ &\quad ={\mathrm{faceAt}}((p_1,p_2,0),(u_1,u_2,0))f[(z_1,z_2,0)]. \end{align*} $$

Assuming that the cube in Figure 2 is equal to $f[(z_1,z_2,0)]$ , then the values of all the hyperfaces of ${\mathrm {M}}_A(f)[(z_1,z_2)]$ are equal to the values on the corresponding hyperfaces of $f[(z_1,z_2,0)]$ on the plane $z=0$ . This explains why the top face and its adjacent edges in Figure 2 yield a multiplication cube from $T_{(2,3)}$ .

Lemma 4.29. Let $f\in X_{n}$ and let $p,p'\in \mathbb {Z}^{d'}$ . Then

$$ \begin{align*}\unicode{x3bb}({\mathrm{M}}_A(f),(p,p'))=\unicode{x3bb}(f,(Ap,Ap'))\quad\mbox{and}\quad(p,p')\ {\mathrm{M}}_A(f)=(Ap,Ap')f.\end{align*} $$

Proof. The two equalities are equivalent, so it suffices to prove the first one. We first show for $z,v\in \mathbb {Z}^d$ , $v\geq {\mathbf {0}}$ , that

$$ \begin{align*}\unicode{x3bb}({\mathrm{M}}_A(f),(z-v,z))=\unicode{x3bb}(f,(A(z-v),Az)).\end{align*} $$

In the special case $v=e_i$ for some $1\leq i\leq d'$ , let $t={\mathrm {M}}_A(f)[z]$ and $s={\mathrm {faceAt}}({\mathbf {0}},-e_i)$ . Then by Theorem 4.27,

$$ \begin{align*} \unicode{x3bb}({\mathrm{M}}_A(f),(z-e_i,z))=\unicode{x3bb}({\mathrm{M}}_A(f),{E}_z(s))=st=\unicode{x3bb}(f,(A(z-e_i),Az)). \end{align*} $$

In general, $v=\sum _{k=1}^m e_{i_k}$ for some $m\in \mathbb {N}$ . Denote $v_j=\sum _{k=1}^j e_{i_k}$ for $0\leq j\leq m$ . Then

$$ \begin{align*} \unicode{x3bb}({\mathrm{M}}_A(f),(z-v,z))&\overset{L.~{4.14}}{=}\sum_{i=0}^{m-1}m(n^A,-v_i)\ \unicode{x3bb}({\mathrm{M}}_A(f),(z-v_{i}-e_{i+1},z-v_i)) \\ &=\sum_{i=0}^{m-1}m(n,-Av_i)\ \unicode{x3bb}(f,(A(z-v_{i}-e_{i+1}),A(z-v_i))) \\ &\overset{L.~{4.14}}{=}\unicode{x3bb}(f,(A(z-v),Az). \end{align*} $$

To prove the statement of the lemma, let $p"\in \mathbb {Z}^d$ such that $p"\geq p$ and $p"\geq p'$ . The previous part is applied in the following with the choices $z=p"$ and $v=p"-p$ or $v=p"-p'$ , thus yielding

$$ \begin{align*} &\unicode{x3bb}({\mathrm{M}}_A(f),(p,p')) \\ &\quad \overset{L.~{4.14}}{=}m(n^A,p"-p')\unicode{x3bb}({\mathrm{M}}_A(f),(p,p"))+\unicode{x3bb}({\mathrm{M}}_A(f),(p",p')) \\ &\quad =m(n,A(p"-p'))\unicode{x3bb}(f,(Ap,Ap"))+\unicode{x3bb}(f,(Ap",Ap'))\\ &\quad\overset{L.~{4.14}}{=}\unicode{x3bb}(f,(Ap,Ap')).\end{align*} $$

Now we can show that the real number represented by a tessellation is preserved by the macrotile map ${\mathrm {M}}_A$ .

Proof. Finiteness of ${\mathrm {real}}(f)$ will be required for applying Lemma 4.25. We compute

$$ \begin{align*} {\mathrm{real}}({\mathrm{M}}_A(f))&=\lim_{i\to\infty}(-{\mathbf{i}},{\mathbf{i}})\ {\mathrm{M}}_A(f) \\ &\overset{L.~{4.29}}{=}\lim_{i\to\infty}(-iA{\mathbf{1}},iA{\mathbf{1}})f={\mathrm{real}}_{-A{\mathbf{1}}}(f)\overset{P.~{4.25}}{=}{\mathrm{real}}(f). \end{align*} $$

In the following lemmas, we show that the composition of macrotile maps corresponds to matrix multiplication.

Proof.

$$ \begin{align*}(n^A)^B=m(n^A,(-Be_i)_{i=1}^{d"})=m(n,(-ABe_i)_{i=1}^{d"})=n^{AB}.\end{align*} $$

Proof. For $f\in X_{n}$ and $v\in \mathbb {Z}^{d"}$ , we compute

$$ \begin{align*} {\mathrm{M}}_{B,n^A}({\mathrm{M}}_{A,n}(f))[v]&\overset{L.~{4.31}}{=} {\mathrm{cube}}_{n^{AB}}(\unicode{x3bb}({\mathrm{M}}_{A,n}(f),(B(v-{\mathbf{1}}),Bv))) \\ &\overset{L.~{4.29}}{=}{\mathrm{cube}}_{n^{AB}} (\unicode{x3bb}(f,(AB(v-{\mathbf{1}}),ABv)))={\mathrm{M}}_{AB,n}(f)[v].\quad \end{align*} $$

Now let A be a $d\times d'$ natural number matrix all of whose rows contain a positive integer. Then the A-macrotile map is going to have an inverse, the A-microtile map ${{\mathrm {\mu }}_{A,n}:X_{n^{A}}\to X_{n}}$ . By the assumption on A for every element $x\in \mathbb {Z}^d$ , there is a $z_1\in \mathbb {Z}^{d'}$ such that $Az_1\geq x$ and thus x can be written in the form $x=Az_1-v$ with $v\in \mathbb {Z}^d$ , $v\geq {\mathbf {0}}$ . Let $z_2\in \mathbb {Z}^{d'}$ satisfy $z_1\geq z_2$ , $x-{\mathbf {1}}\geq Az_2$ , let $a=\unicode{x3bb} (f,(z_2,z_1))$ for $f\in X_{n^A}$ and define the map ${\mathrm {\mu }}_{A,n}$ (usually written just as ${\mathrm {\mu }}_A$ ) by

$$ \begin{align*}{\mathrm{\mu}}_{A,n}(f)[x]={\mathrm{cube}}_{n}({\mathrm{base}}_{n}(a,(-v,-v-{\mathbf{1}}))[2]).\end{align*} $$

We need to verify that ${\mathrm {\mu }}_A(f)[x]$ is well defined (which we will do in Lemma 4.33) and that ${\mathrm {\mu }}_A(f)$ is indeed in $X_{n}$ (which we will do in Theorem 4.35). It is simple that ${\mathrm {base}}_{n}(a,(-v,-v-{\mathbf {1}}))[2])\in \Sigma _N$ with $N=\prod _{i=1}^d n[i]$ , since

$$ \begin{align*}{\mathrm{base}}_{n}(a,(-v,-v-{\mathbf{1}}))[2]<m(n,-v-{\mathbf{1}})/m(n,-v)=m(n,-{\mathbf{1}})=N.\end{align*} $$

The assumption that all the rows of A contain a positive integer will hold until Theorem 4.40. Frequent appearances of the map $\mu _A$ (which is not defined for other kinds of matrices) are also reminders of this.

Lemma 4.33. Let $f\in X_{n^A}$ and let $x\in \mathbb {Z}^d$ . Let $z_1,z_1'\in \mathbb {Z}^d$ , $v,v'\in \mathbb {Z}^{d}$ , $v,v'\geq {\mathbf {0}}$ be such that $x=Az_1-v=Az_1'-v'$ . Let $z_2,z_2'\in \mathbb {Z}^{d'}$ satisfy $z_1\geq z_2$ , $z_1'\geq z_2'$ , $x-{\mathbf {1}}\geq Az_2, Az_2'$ . Denote $a=\unicode{x3bb} (f,(z_2,z_1))$ and $a'=\unicode{x3bb} (f,(z_2',z_1'))$ . Then

$$ \begin{align*}{\mathrm{base}}_{n}(a,(-v,-v-{\mathbf{1}}))[2]={\mathrm{base}}_{n}(a',(-v',-v'-{\mathbf{1}}))[2].\end{align*} $$

Proof. It is possible to choose $z_1",z_2"\in \mathbb {Z}^{d'}$ and $v"\in \mathbb {Z}^d$ , $v"\geq {\mathbf {0}}$ so that $x=Az_1"-v"$ , $z_1"\geq z_2"$ , $x-{\mathbf {1}}\geq Az_2"$ , and furthermore, $z_1"\geq z_1,z_1'$ and $z_2"\leq z_2,z_2'$ , that is, $z_1"$ and $z_2"$ are common upper and lower bounds for $z_1,z_1'$ and $z_2,z_2'$ . Therefore, to prove the lemma, it is sufficient to consider the case of $z_1'\geq z_1$ and $z_2'\leq z_2$ .

Observe that

$$ \begin{align*} \unicode{x3bb}(f,(z_2',z_1'))&\overset{L.~{4.14}}{=}m(n^A,z_2-z_1')\unicode{x3bb}(f,(z_2',z_2)) \\ &\quad +m(n^A,z_1-z_1')\unicode{x3bb}(f,(z_2,z_1))+\unicode{x3bb}(f,(z_1,z_1')) \\ &=m(n,A(z_2-z_1'))\underbrace{\unicode{x3bb}(f,(z_2',z_2))}_{a_2} \\ &\quad +m(n,A(z_1-z_1'))\underbrace{\unicode{x3bb}(f,(z_2,z_1))}_{a_1}+\underbrace{\unicode{x3bb}(f,(z_1,z_1'))}_{a_0}. \end{align*} $$

By Lemma 4.15,

$$ \begin{align*} a_0&<m(n^A,z_1-z_1')=m(n,A(z_1-z_1')), \\ a_1&<m(n^A,z_2-z_1')/m(n^A,z_1-z_1')=m(n,A(z_2-z_1'))/m(n,A(z_1-z_1')), \end{align*} $$

and therefore,

$$ \begin{align*}{\mathrm{base}}_{n}(\unicode{x3bb}(f,(z_2',z_1')),(A(z_1-z_1'),A(z_2-z_1')))=(a_0,a_1,a_2).\end{align*} $$

Let

$$ \begin{align*} (b_0,b_1,b_2)&={\mathrm{base}}_{n}(\unicode{x3bb}(f,(z_2,z_1))),(-v,-v-{\mathbf{1}})) \\ &={\mathrm{base}}_{n}(a_1,(-v'-A(z_1-z_1'),-v'-A(z_1-z_1')-{\mathbf{1}})). \end{align*} $$

To see that the mixed base expression after this paragraph contains a valid d-directive sequence, we need to check that $A(z_1-z_1')\geq -v'$ and $-v'-{\mathbf {1}}\geq A(z_2-z_1')$ . By definition, $Az_1-v=Az_1'-v'$ and thus $-v'=A(z_1-z_1')-v\leq A(z_1-z_1')$ . The other inequality is shown by

$$ \begin{align*}-v'-{\mathbf{1}}=-Az_1'+x-{\mathbf{1}}\geq-Az_1'+Az_2=A(z_2-z_1').\end{align*} $$

By Lemma 3.3,

$$ \begin{align*}{\mathrm{base}}_{n}(\unicode{x3bb}(f,(z_2',z_1')),(A(z_1-z_1'),-v',-v'-{\mathbf{1}},A(z_2-z_1')))=(a_0,b_0,b_1,b_2,a_2).\end{align*} $$

We conclude by computing

$$ \begin{align*} &{\mathrm{base}}_{n}(a,(-v,-v-{\mathbf{1}}))[2]={\mathrm{base}}_{n}(\unicode{x3bb}(f,(z_1,z_2))),(-v,-v-{\mathbf{1}}))[2]=b_1 \\ &\quad ={\mathrm{base}}_{n}(\unicode{x3bb}(f,(z_2',z_1')),(A(z_1-z_1'),-v',-v'-{\mathbf{1}},A(z_2-z_1')))[3] \\ &\quad\overset{L.~{3.4}}{=}{\mathrm{base}}_{n}(\unicode{x3bb}(f,(z_2',z_1')),(-v',-v'-{\mathbf{1}}))[2]={\mathrm{base}}_{n}(a',(-v',-v'-{\mathbf{1}}))[2].\qquad \end{align*} $$

Lemma 4.34. Let $f\in X_{n^A}$ and let $x\in \mathbb {Z}^d$ . Let $z_1\in \mathbb {Z}^{d'}$ , $v\in \mathbb {Z}^{d}$ , $v\geq {\mathbf {0}}$ be such that $x=Az_1-v$ . Let $z_2\in \mathbb {Z}^{d'}$ satisfy $z_1\geq z_2$ and $x-{\mathbf {1}}\geq Az_2$ . Denote $a=\unicode{x3bb} (f,(z_2,z_1))$ . For any face $s={\mathrm {faceAt}}(p,u)$ , it holds that

$$ \begin{align*}s({\mathrm{\mu}}_A(f)[x])={\mathrm{base}}_{n}(a,(-v+p,-v+p+u))[2].\end{align*} $$

Proof. Let $(a_0,a_1,a_2)={\mathrm {base}}_{n}(a,(-v,-v-{\mathbf {1}}))$ . By Lemma 3.5,

$$ \begin{align*}{\mathrm{base}}_{n}(a,(-v+p,-v+p+u))[2]={\mathrm{base}}_{n}(a_1,(p,p+u))[2],\end{align*} $$

so we can compute

$$ \begin{align*} &{\mathrm{faceAt}}(p,u)({\mathrm{\mu}}_A(f)[Az_1-v])={\mathrm{base}}_{n}({\mathrm{base}}_{n}(a,(-v,-v-{\mathbf{1}}))[2],(p,p+u))[2] \\ &\quad ={\mathrm{base}}_{n}(a_1,(p,p+u))[2]={\mathrm{base}}_{n}(a,(-v+p,-v+p+u))[2]. \end{align*} $$

Proof. We need to show that ${\mathrm {\tau }}_i({\mathrm {\mu }}_A(f)[w])={\mathrm {\beta }}_i({\mathrm {\mu }}_A(f)[w+e_i])$ for any $w\in \mathbb {Z}^d$ and $1\leq i\leq d$ . Let $z_1,z_2\in \mathbb {Z}^{d'}$ be such that Lemma 4.34 can be applied at coordinates w and $w+e_i$ . Let $w=Az_1-v$ and $w+e_i=Az_1-u$ , from which it follows that ${-v+e_i=-u}$ . Denote $a=\unicode{x3bb} (f,(z_2,z_1))$ . By applying Lemma 4.34 two times, we see that

$$ \begin{align*} {\mathrm{\tau}}_i({\mathrm{\mu}}_A(f)[w])&={\mathrm{base}}_{n}(a,(-v,-v-{\mathbf{1}}+e_i))[2] \\ &={\mathrm{base}}_{n}(a,((-v+e_i)-e_i,(-v+e_i)-e_i+(-{\mathbf{1}}+e_i)))[2] \\ &={\mathrm{base}}_{n}(a,(-u-e_i,-u-e_i+(-{\mathbf{1}}+e_i)))[2] \\ &={\mathrm{\beta}}_i({\mathrm{\mu}}_A(f)[w+e_i]). \end{align*} $$

Lemma 4.36. Let $f\in X_{n^A}$ and let $p,p'\in \mathbb {Z}^{d'}$ . Then

$$ \begin{align*} \unicode{x3bb}({\mathrm{\mu}}_A(f),(Ap,Ap'))=\unicode{x3bb}(f,(p,p'))\quad\mbox{and}\quad(Ap,Ap') {\mathrm{\mu}}_A(f)=(p,p')f.\end{align*} $$

Proof. The two equalities are equivalent, so it suffices to prove the first one. We first show for $z_1\in \mathbb {Z}^{d'},v\in \mathbb {Z}^d$ , $v\geq {\mathbf {0}}$ that

$$ \begin{align*}\unicode{x3bb}({\mathrm{\mu}}_A(f),(A(z_1-v),Az))=\unicode{x3bb}(f,(z_1-v,z_1)).\end{align*} $$

We can represent $Av=\sum _{k=1}^m e_{i_k}$ for some $m\in \mathbb {N}$ . Denote $v^{\prime }_j=\sum _{k=1}^j e_{i_k}$ for $0\leq j\leq m$ , so in particular, $Av=v^{\prime }_m$ . Denote $s_i={\mathrm {faceAt}}({\mathbf {0}},-e_i)$ for any $1\leq i\leq m$ . Choose ${z_2\in \mathbb {Z}^{d'}}$ so that Lemma 4.34 can be applied to faces $s_{i_{j+1}}$ of the microtiles at coordinates of the form $Az_1-v^{\prime }_j$ . Let $a=\unicode{x3bb} (f,(z_2,z_1))$ . Then

$$ \begin{align*} \unicode{x3bb}({\mathrm{\mu}}_A(f),(A(z-v),Az))&=\unicode{x3bb}({\mathrm{\mu}}_A(f),(Az-v^{\prime}_m,Az-v^{\prime}_0)) \\ &\overset{L.~{4.14}}{=}\sum_{j=0}^{m-1}m(n,-v^{\prime}_j)\ \unicode{x3bb}({\mathrm{\mu}}_A(f),(Az-v^{\prime}_{j+1},Az-v^{\prime}_j)) \\ &=\sum_{j=0}^{m-1}m(n,-v^{\prime}_j)\ s_{i_{j+1}}({\mathrm{\mu}}_A(f)[Az-v^{\prime}_j]) \\ &\overset{L.~{4.34}}{=}\sum_{j=0}^{m-1}m(n,-v^{\prime}_j)\ {\mathrm{base}}_{n}(a,(-v^{\prime}_j,\underbrace{-v^{\prime}_j-e_{i_{j+1}}}_{-v^{\prime}_{j+1}}))[2] \\ &\overset{L.~{4.2}}{=}{\mathrm{base}}_{n}(a,(-v^{\prime}_0,-v^{\prime}_m))[2]={\mathrm{base}}_{n}(a,({\mathbf{0}},-Av))[2] \\ &={\mathrm{base}}_{n^A}(\unicode{x3bb}(f,(z_2,z_1)),({\mathbf{0}},-v))[2]\overset{L.~{4.16}}{=}\unicode{x3bb}(f,(z_1-v,z_1)). \end{align*} $$

On the last line, we need that $(z_1,z_1-v,z_2)$ is a $d'$ -directive sequence, but this can be achieved by choosing $z_2$ suitably.

For the general case, let $p"\in \mathbb {Z}^d$ such that $p"\geq p$ and $p"\geq p'$ . The previous part is applied in the following with the choices $z=p"$ and $v=p"-p$ or $v=p"-p'$ , thus yielding

$$ \begin{align*} \unicode{x3bb}({\mathrm{\mu}}_A(f),(Ap,Ap')) &\overset{L.~{4.14}}{=}m(n,A(p"-p'))\unicode{x3bb}({\mathrm{\mu}}_A(f),(Ap,Ap"))+\unicode{x3bb}({\mathrm{\mu}}_A(f),(Ap",Ap')) \\ &=m(n^A,p"-p')\unicode{x3bb}(f,(p,p"))+\unicode{x3bb}(f,(p",p'))\overset{L.~{4.14}}{=}\unicode{x3bb}(f,(p,p')). \end{align*} $$

Now we can show that the real number represented by a tessellation is preserved also by the microtile map ${\mathrm {\mu }}_A$ .

Proof. We observe that ${\mathrm {wgt}}_n(-{\mathbf {1}})>1$ if and only if ${\mathrm {wgt}}_n(-A{\mathbf {1}})>1$ . The case ${\mathrm {wgt}}_n(-{\mathbf {1}})={\mathrm {wgt}}_n(-A{\mathbf {1}})=1$ is simple because then ${\mathrm {real}}({\mathrm {\mu }}_A(f))={\mathrm {real}}(f)=0$ .

Assume then that ${\mathrm {wgt}}_n(-{\mathbf {1}})>1$ and ${\mathrm {wgt}}_n(-A{\mathbf {1}})>1$ . Because $-{\mathbf {1}}\ll {\mathbf {0}}$ and $-A{\mathbf {1}}\ll {\mathbf {0}}$ , it follows from Proposition 4.25 that the equality ${\mathrm {real}}({\mathrm {\mu }}_A(f))={\mathrm {real}}_{-A{\mathbf {1}}}({\mathrm {\mu }}_A(f))$ holds if one these quantities is finite (and it trivially holds if both of them are infinite). Therefore,

$$ \begin{align*} {\mathrm{real}}({\mathrm{\mu}}_A(f))&={\mathrm{real}}_{-A{\mathbf{1}}}({\mathrm{\mu}}_A(f))=\lim_{i\to\infty}(-iA{\mathbf{1}},iA{\mathbf{1}}) {\mathrm{\mu}}_A(f) \\ &\overset{L.~{4.36}}{=}\lim_{i\to\infty}(-{\mathbf{i}},{\mathbf{i}})f={\mathrm{real}}(f). \end{align*} $$

Proof. Let $f\in X_{n}$ and consider $x=Az_1-v$ with $z_1\in \mathbb {Z}^{d'}$ , $v\in \mathbb {Z}^{d}$ , $v\geq {\mathbf {0}}$ . Let $z_2\in \mathbb {Z}^{d'}$ satisfy $z_1\geq z_2$ and $x-{\mathbf {1}}\geq Az_2$ . Then

$$ \begin{align*} {\mathrm{\mu}}_A({\mathrm{M}}_A(f))[Az_1-v]&={\mathrm{cube}}_{n}({\mathrm{base}}_{n}(\unicode{x3bb}({\mathrm{M}}_A(f),(z_2,z_1)),(-v,-v-{\mathbf{1}}))[2]) \\ &\overset{L.~{4.29}}{=}{\mathrm{cube}}_{n}({\mathrm{base}}_{n}(\unicode{x3bb}(f,(Az_2,Az_1)),(-v,-v-{\mathbf{1}}))[2]) \\ &\overset{L.~{4.16}}{=}{\mathrm{cube}}_{n}(\unicode{x3bb}(f,(Az_1-v-{\mathbf{1}},Az_1-v))) \\ &\overset{L.~{4.17}}{=}{\mathrm{cube}}_{n}({\mathrm{val}}_{n}(f[Az_1-v]))=f[Az_1-v]. \end{align*} $$

For the other direction, let $f\in X_{n^A}$ and $z\in \mathbb {Z}^{d'}$ . Then

$$ \begin{align*} {\mathrm{M}}_A({\mathrm{\mu}}_A(f))[z]&={\mathrm{cube}}_{n^A}(\unicode{x3bb}({\mathrm{\mu}}_A(f),(A(z-{\mathbf{1}}),Az))) \\ &\overset{L.~{4.36}}{=}{\mathrm{cube}}_{n^A}(\unicode{x3bb}(f,(z-{\mathbf{1}},z)))\overset{L.~{4.17}}{=}{\mathrm{cube}}_{n^A}({\mathrm{val}}_{n^A}(f[z]))=f[z].\qquad \end{align*} $$

As a corollary, we can show that any bi-infinite sequence of cubes occurs on the main diagonal of precisely one valid tessellation.

Proof. Let A be the $d\times 1$ matrix with all entries equal to $1$ , so $n^A=(N)$ and there is a $g\in X_{n^A}=X_{(N)}=T_{(N)}^{\mathbb {Z}}$ such that ${\mathrm {val}}_{(N)}(g[i])=x[i]$ for all $i\in \mathbb {Z}$ . For $f\in X_n$ , the equality ${\mathrm {M}}_A(f)=g$ holds if and only if

$$ \begin{align*} {\mathrm{val}}_n(f)[{\mathbf{i}}]&=\unicode{x3bb}(f,({\mathbf{i-1}},{\mathbf{i}}))=\unicode{x3bb}(f,(A(i-1),Ai))\overset{L.~{4.29}}{=}\unicode{x3bb}({\mathrm{M}}_A(f),(i-1,i)) \\ &={\mathrm{val}}_{(N)}({\mathrm{M}}_A(f)[i])={\mathrm{val}}_{(N)}(g[i])=x[i]\quad\mbox{for all }i\in\mathbb{Z}, \end{align*} $$

so we need to show that there is a unique $f\in X_n$ such that ${\mathrm {M}}_A(f)=g$ . This follows because by Theorem 4.38, the map ${\mathrm {M}}_A$ is bijective.

We will show that ${\mathrm {M}}_A$ is surjective for a $d\times d'$ natural number matrix A even without the assumption that all its rows contain a positive integer.

Proof. By Theorem 4.40, the map ${\mathrm {M}}_A$ is surjective, so it remains to show that $\sigma _z\circ {\mathrm {M}}_A={\mathrm {M}}_A \circ \sigma _{Az}$ . Let $f\in X_{n}$ and $v\in \mathbb {Z}^{d'}$ be arbitrary. Then

$$ \begin{align*} \sigma_z({\mathrm{M}}_A(f))[v]&={\mathrm{M}}_A(f)[z+v]={\mathrm{cube}}_{n^A}(\unicode{x3bb}(f,(A(z+v-{\mathbf{1}}),A(z+v)))) \\ &\overset{L.~{4.13}}{=}{\mathrm{cube}}_{n^A} (\unicode{x3bb}(\sigma_{Az}(f),(A(v-{\mathbf{1}}),Av)))={\mathrm{M}}_A(\sigma_{Az}(f))[v]. \end{align*} $$

In the case when every row of A contains a positive number, the map ${\mathrm {M}}_A$ is bijective by Theorem 4.38.

5.1. Preliminaries on multiplication automata

Let $\Sigma $ be an alphabet. Concatenations of symbols from $\Sigma $ are called words, and for $k\in \mathbb {Z}_+$ , the set of words of length k is denoted by $\Sigma ^k$ . Elements of the one-dimensional subshift $\Sigma ^{\mathbb {Z}}$ are bi-infinite sequences that are called configurations. As previously in this paper, the value of $x\in \Sigma ^{\mathbb {Z}}$ at a coordinate $i\in \mathbb {Z}$ is denoted by $x[i]$ . More generally, given an interval $I=[i,j]$ with $i\leq j\in \mathbb {Z}$ , x has a finite subword denoted by $x[I]=x[i,j]=x[i]x[i+1]\cdots x[j]\in \Sigma ^{j-i+1}$ .

If $F:\Sigma ^{\mathbb {Z}}\to \Sigma ^{\mathbb {Z}}$ is a cellular automaton, then $(\Sigma ^{\mathbb {Z}},F)$ is a dynamical system. Bijective cellular automata are often called reversible, because their inverse maps are also cellular automata (see e.g., [Reference Kari6, Theorem 2]).

For a CA $F:\Sigma ^{\mathbb {Z}}\to \Sigma ^{\mathbb {Z}}$ , a configuration $x\in X$ , and an interval $I=[i,j]$ with ${i\leq j\in \mathbb {Z}}$ , the I-trace of x (with respect to F) is the sequence ${\mathrm {Tr}}_{F,I}(x)$ over the alphabet $\Sigma ^{j-i+1}$ (that is, the symbols of the alphabet are words over $\Sigma $ ) defined by ${\mathrm {Tr}}_{F,I}(x)[t]=F^t(x)[I]$ for $t\in \mathbb {N}$ . If $I=\{i\}$ is the degenerate interval, we may write ${\mathrm {Tr}}_{F,i}(x)$ and if $i=0$ , we may write ${\mathrm {Tr}}_F(x)$ . If the CA F is clear from the context, we may write ${\mathrm {Tr}}_I(x)$ . The I-trace subshift of F is $(\Xi _I(F),\sigma )$ , where

$$ \begin{align*}\Xi_I(F)={\mathrm{Tr}}_{F,I}(\Sigma^{\mathbb{Z}})\subseteq (\Sigma^{j-i+1})^{\mathbb{N}}.\end{align*} $$

It is easy to see that $\Xi _{[i,j]}(F)=\Xi _{[i-j,0]}(F)$ and, therefore, the set of all trace subshifts satisfies

$$ \begin{align*}\{(\Xi_{[i,j]}(F),\sigma)\mid i\leq j\in\mathbb{Z}\}=\{(\Xi_{[-(k-1),0]},\sigma)\mid k\in\mathbb{Z}_+\}.\end{align*} $$

We call $(\Xi _{[-(k-1),0]},\sigma )$ the width k trace subshift of F.

For examining multiplication CA, it is convenient to define

$$ \begin{align*} {\mathcal{N}(\Sigma_N)}&=\{x\in\Sigma_N^{\mathbb{Z}}\mid \text{there exists } j\in\mathbb{Z}: x[i]=0\mbox{ for }i<j\}\quad\mbox{and} \\ {\mathcal{F}(\Sigma_N)}&=\{x\in\Sigma_N^{\mathbb{Z}}\mid x[i]\neq 0\mbox{ for finitely many }i\in\mathbb{Z}\}\subseteq {\mathcal{N}(\Sigma_N)}. \end{align*} $$

The elements of ${\mathcal {N}(\Sigma _N)}$ are analogous to the usual base-N representations of positive numbers, where there may be infinitely many digits to the right of the decimal point but always a finite number of digits to the left of the decimal point (and then the representation can be extended to a bi-infinite sequence by adding an infinite sequence of zeroes to the left end). The elements of ${\mathcal {F}(\Sigma _N)}$ correspond to numbers whose representations are finite to both directions.

Definition 5.2. Let $N>1$ be an integer. If $\xi \geq 0$ is a real number and $\xi =\sum _{i=-\infty }^{\infty }{\xi _i N^{i}}$ is the unique base-N expansion of $\xi $ such that $\xi _i\neq N-1$ for infinitely many $i<0$ , we define ${\mathrm {config}}_N(\xi )\in {\mathcal {N}(\Sigma _N)}$ by

$$ \begin{align*}{\mathrm{config}}_N(\xi)[i]=\xi_{-i}\end{align*} $$

for all $i\in \mathbb {Z}$ . In reverse, for $x\in \Sigma _N^{\mathbb {Z}}$ , we define

$$ \begin{align*}{\mathrm{real}}_N(x)=\sum_{i=-\infty}^{\infty}{x[-i] N^{i}}.\end{align*} $$

Clearly, ${\mathrm {real}}_N({\mathrm {config}}_N(\xi ))=\xi $ and ${\mathrm {config}}_N({\mathrm {real}}_N(x))=x$ for every $\xi \geq 0$ and every ${x\in {\mathcal {N}(\Sigma _N)}}$ such that $x[i]\neq N-1$ for infinitely many $i>0$ .

Additionally, in the special case $N=1$ , we define ${\mathrm {config}}_1:\{0\}\to \Sigma _1^{\mathbb {Z}}$ and ${\mathrm {real}}_1:\Sigma _1^{\mathbb {Z}}\to \{0\}$ in the only possible way.

Definition 5.3. (Multiplication CA)

Let $N\in \mathbb {Z}_+$ . For $p\in \mathbb {Z}_+$ dividing N, we define a $(0,1)$ local rule $g_{p,N}:\Sigma _{N}\times \Sigma _{N}\to \Sigma _{N}$ for a CA $\Pi _{p,N}$ as follows. Let $q\in \mathbb {Z}_+$ be such that $N=pq$ . The numbers $a,b\in \Sigma _{N}$ are represented as $a=a_1q+a_0$ and $b=b_1q+b_0$ , where $a_0,b_0\in \Sigma _q$ and $a_1,b_1\in \Sigma _p$ : such representations always exist and they are unique. Then

$$ \begin{align*}g_{p,N}(a,b)=g_{p,pq}(a_1q+a_0,b_1q+b_0)=a_0p+b_1.\end{align*} $$

It can be shown that the map $g_{p,N}$ performs multiplication by p in base N in the sense that $\Pi _{p,n}({\mathcal {F}(\Sigma _N)})\subseteq {\mathcal {F}(\Sigma _N)}$ , $\Pi _{p,N}({\mathcal {N}(\Sigma _N)})\subseteq {\mathcal {N}(\Sigma _N)}$ , and ${\mathrm {real}}_N(\Pi _{p,N}(x))=p{\mathrm {real}}_N(x)$ for $x\in {\mathcal {N}(\Sigma _N)}$ . To put it shortly, this is because $g_{p,N}$ encodes the usual algorithm for long multiplication by p in base N (for more details, see e.g., [Reference Kari, Yen and Ibarra5, Reference Kopra7]). All these maps commute because

$$ \begin{align*}\Pi_{p_1,N}(\Pi_{p_2,N}(x))={\mathrm{config}}_N(p_1p_2\ {\mathrm{real}}_N(x))=\Pi_{p_2,N}(\Pi_{p_1,N}(x))\end{align*} $$

for $p_1,p_2$ dividing N and for x in the dense set ${\mathcal {F}(\Sigma _N)}$ . They are reversible, because for $pq=N$ ,

$$ \begin{align*}\Pi_{p,N}(\Pi_{q,N}(x))={\mathrm{config}}_N(N\ {\mathrm{real}}_N(x))=\sigma(x)\end{align*} $$

for x in the dense set ${\mathcal {F}(\Sigma _N)}$ and $\sigma $ is reversible. Whenever $\alpha =p/q$ , where p and q are products of factors $p_i$ and $q_i$ of N, we may define $\Pi _{\alpha ,N}$ as the composition of corresponding $\Pi _{p_i,N}$ and $\Pi _{q_i,N}^{-1}$ . The map $\Pi _{\alpha ,N}$ does not depend on the choice of the composition into $p_i$ and $q_i$ by arguments similar to those above.

5.2. The connection between multiplication cube tessellations and multiplication automata

Throughout this subsection, let $n\in \mathbb {Z}_+^d$ and let $N=\prod _{i=1}^d n[i]$ . From any configuration $x\in \Sigma _N^{\mathbb {Z}}$ , one can form a tessellation ${\mathrm {tess}}_{n}(x)\in X_{n}$ as follows: for any $z\in \mathbb {Z}^d$ , we define $\alpha (z,n)=\prod _{i=1}^d n[i]^{z[i]}$ and define

$$ \begin{align*}{\mathrm{tess}}_{n}(x)[z]={\mathrm{cube}}_{n}(\Pi_{\alpha(z,n),N}(x)[0])\end{align*} $$

(one still has to verify that this is indeed a valid tiling, which we will do in Lemma 5.4). On the other hand, from any tessellation $f\in X_{n}$ , one can extract a configuration ${{\mathrm {diag}}_{n}(f)\in \Sigma _N^{\mathbb {Z}}}$ by looking at the tiles on the main diagonal: let

$$ \begin{align*}{\mathrm{diag}}_{n}(f)[i]={\mathrm{val}}_{n}(f[{\mathbf{i}}])\mbox{ for }i\in\mathbb{Z}.\end{align*} $$

Proof. Let $z\in \mathbb {Z}^d$ and $1\leq i\leq d$ be arbitrary. We need to show that ${\mathrm {\tau }}_i({\mathrm {tess}}_{n}(x)[z])={\mathrm {\beta }}_i({\mathrm {tess}}_{n}(x)[z+e_i])$ . Let $y=\Pi _{\alpha (z,n),N}(x)$ . By writing $y[0]=a_1\prod _{j\neq i}n[j]+a_0$ for ${0\leq a_0<\prod _{j\neq i}n[i]}$ and $0\leq a_1<n[i]$ , we see that $\Pi _{n[i],N}(y)[0]=a_0n[i]+a_1'$ for some $0\leq a_1'<n[i]$ . Then

$$ \begin{align*} &{\mathrm{\tau}}_i({\mathrm{tess}}_{n}(x)[z])={\mathrm{\tau}}_i({\mathrm{cube}}_{n}(\Pi_{\alpha(z,n),N}(x)[0])) \\ &\quad ={\mathrm{\tau}}_i({\mathrm{cube}}_{n}(y[0]))={\mathrm{\tau}}_i({\mathrm{cube}}_{n}(a_1\prod_{j\neq i}n[j]+a_0))=a_0\quad\mbox{and} \\ &{\mathrm{\beta}}_i({\mathrm{tess}}_{n}(x)[z+e_i])={\mathrm{\beta}}_i({\mathrm{cube}}_{n}(\Pi_{\alpha(z+e_i,n),N}(x)[0])) \\ &\quad ={\mathrm{\beta}}_i({\mathrm{cube}}_{n}(\Pi_{n[i],N}(y)[0]))={\mathrm{\beta}}_i({\mathrm{cube}}_{n}(a_0n[i]+a_1'))=a_0. \end{align*} $$

Proof. Let $x\in \Sigma _N^{\mathbb {Z}}$ and $z'\in \mathbb {Z}^d$ be arbitrary. Then

$$ \begin{align*} &\sigma_z({\mathrm{tess}}_{n}(x))[z']={\mathrm{tess}}_{n}(x)[z'+z]={\mathrm{cube}}_{n}(\Pi_{\alpha(z'+z,n),N}(x)[0]) \\ &\quad ={\mathrm{cube}}_{n}(\Pi_{\alpha(z',n),N}(\Pi_{\alpha(z,n),N}(x))[0])={\mathrm{tess}}_{n}(\Pi_{\alpha(z,n),N}(x))[z']. \end{align*} $$

We define ‘partial’ shifts on macrotilings by conjugating to a microtiling. For the rest of this subsection (unless specified otherwise), let A be a $d\times d'$ natural number matrix and let $N'=\prod _{i=1}^{d'} n^A[i]$ . If additionally all the rows of A contain a positive integer, we define $\sigma _{A,z}:X_{n^A}\to X_{n^A}$ by $\sigma _{A,z}={\mathrm {M}}_A\circ \;\sigma _z\circ {\mathrm {\mu }}_A$ for any $z\in \mathbb {Z}^d$ . The assumption that all the rows of A contain a positive integer will hold until Theorem 5.12. Appearances of the maps $\sigma _{A,z}$ and $\mu _A$ (which are not defined for other kinds of matrices) are also reminders of this.

Proof. First observe that $\Pi _{\alpha (z,n),N'}$ always exists, because every factor of N is a factor of $N'$ . It is sufficient to show for any $1\leq k\leq d$ that $\Pi _{n[k],N'}\circ {\mathrm {diag}}_{n^A}={\mathrm {diag}}_{n^A}\circ \;\sigma _{A,e_k}$ . After composing by ${\mathrm {M}}_A$ on the right, we need to show for all $f\in X_{n}$ and $i\in \mathbb {Z}$ that $\Pi _{n[k],N'}({\mathrm {diag}}_{n^A}({\mathrm {M}}_A(f)))[i]={\mathrm {diag}}_{n^A}({\mathrm {M}}_A(\sigma _{e_k}(f)))[i]$ . We compute

$$ \begin{align*} &\Pi_{n[k],N'}({\mathrm{diag}}_{n^A}({\mathrm{M}}_A(f)))[i]=g_{n[k],N'}({\mathrm{diag}}_{n^A}({\mathrm{M}}_A(f))[i],\ {\mathrm{diag}}_{n^A}({\mathrm{M}}_A(f))[i+1]) \\ &\quad =g_{n[k],N'}(\underbrace{\unicode{x3bb}(f,(A({\mathbf{i-1}}),A{\mathbf{i}}))}_{a},\underbrace{\unicode{x3bb}(f,(A{\mathbf{i}},A({\mathbf{i+1}})))}_{b}) \end{align*} $$

and

$$ \begin{align*} &{\mathrm{diag}}_{n^A}({\mathrm{M}}_A(\sigma_{e_k}(f)))[i]={\mathrm{val}}_{n^A}({\mathrm{M}}_A(\sigma_{e_k}(f)))[{\mathbf{i}}] \\ &\quad =\unicode{x3bb}(\sigma_{e_i}(f),(A({\mathbf{i-1}}),A{\mathbf{i}}))=\underbrace{\unicode{x3bb}(f,(A({\mathbf{i-1}})+e_i,A{\mathbf{i}}+e_i))}_{c}, \end{align*} $$

where $a,b,c<N'=m(n,-A{\mathbf {1}})$ . We need to show that $g_{n[k],N'}(a,b)=c$ . After defining $a_0,a_1,b_0,b_1$ by

$$ \begin{align*} (a_0,a_1)&={\mathrm{base}}(a,(N'/n[k]))={\mathrm{base}}_{n}(a,(-A{\mathbf{1}}+e_k,-A{\mathbf{1}}))[0,1] \\ &={\mathrm{base}}_{n}(a,((A({\mathbf{i-1}})+e_k)-A{\mathbf{i}},A({\mathbf{i-1}})-A{\mathbf{i}}))[0,1]\quad\mbox{and} \\ (b_0,b_1)&={\mathrm{base}}(b,(N'/n[k]))={\mathrm{base}}_{n}(b,(-A{\mathbf{1}}+e_k,-A{\mathbf{1}}))[0,1] \\ &={\mathrm{base}}_{n}(b,((A{\mathbf{i}}+e_k)-A({\mathbf{i+1}}),A{\mathbf{i}}-A({\mathbf{i+1}})))[0,1], \end{align*} $$

it amounts to showing that $c=a_0n[k]+b_1$ . By Lemma 4.16, we see that

$$ \begin{align*}a_1=\unicode{x3bb}(f,(A({\mathbf{i-1}}),A({\mathbf{i-1}})+e_i))\quad\mbox{and}\quad b_1=\unicode{x3bb}(f,(A{\mathbf{i}},A{\mathbf{i}}+e_i)).\end{align*} $$

By Lemma 4.14, we see that

$$ \begin{align*} &a_1N'+c \\ &\quad =\unicode{x3bb}(f,(A({\mathbf{i-1}}),A({\mathbf{i-1}})+e_i))m(n,-A{\mathbf{1}})+\unicode{x3bb}(f,(A({\mathbf{i-1}})+e_i,A{\mathbf{i}}+e_i)) \\ &\quad =\unicode{x3bb}(f,(A({\mathbf{i-1}}),A{\mathbf{i}}+e_i)) \\ &\quad =\unicode{x3bb}(f,(A({\mathbf{i-1}}),A{\mathbf{i}}))m(n,-e_k)+\unicode{x3bb}(f,(A{\mathbf{i}},A{\mathbf{i}}+e_i))=an[k]+b_1. \end{align*} $$

From this, one can solve $c=(a-a_1N'/n[k])n[k]+b_1=a_0n[k]+b_1$ , and we are done.

Proof. First, let $x\in \Sigma _N^{\mathbb {Z}}$ and $i\in \mathbb {Z}$ be arbitrary. It holds that

$$ \begin{align*}{\mathrm{tess}}_{n}(x)[{\mathbf{i}}]={\mathrm{cube}}_{n}(\Pi_{N^i,N}(x)[0])={\mathrm{cube}}_{n}(\sigma^i(x)[0])={\mathrm{cube}}_{n}(x[i])\end{align*} $$

and therefore ${\mathrm {diag}}_{n}({\mathrm {tess}}_{n}(x))[i]={\mathrm {val}}_{n}({\mathrm {cube}}_{n}(x[i]))=x[i]$ .

Next let $f\in X_{n}$ and $z\in \mathbb {Z}^d$ be arbitrary, and denote $f'=\sigma _z(f)$ . It holds that

$$ \begin{align*} {\mathrm{tess}}_{n}({\mathrm{diag}}_{n}(f))[z]&=\sigma_z({\mathrm{tess}}_{n}({\mathrm{diag}}_{n}(f)))[{\mathbf{0}}] \\ &\overset{T. {5.5}}{=}{\mathrm{tess}}_{n}(\Pi_{\alpha(z,n),N}({\mathrm{diag}}_{n}(f)))[{\mathbf{0}}] \overset{L. {5.6}}{=}{\mathrm{tess}}_{n}({\mathrm{diag}}_{n}(f'))[{\mathbf{0}}] \\ &={\mathrm{cube}}_{n}({\mathrm{diag}}_{n}(f')[0])={\mathrm{cube}}_{n}({\mathrm{val}}_{n}(f'[{\mathbf{0}}]))=f'[{\mathbf{0}}]=f[z]. \end{align*} $$

Since ${\mathrm {tess}}_{n}$ is the inverse of ${\mathrm {diag}}_n$ , the last claim follows from Lemma 5.6.

It is a simple observation that the maps we have defined for mapping between tessellations and configurations are such that the represented real numbers are preserved.