A note on reduction of tiling problems

We prove the claim only for $d\geq 2$, since the proof for the case $d=1$ requires slightly different considerations but is not more difficult.

Let $d,s\in\mathbb{N}$ and $L<\mathbb{Z}^{d}$ be given. Since $L$ is a finitely generated abelian group of rank at most $d$, there exists $r\leq d$ and $w_{1},\ldots,w_{r}\in L$ that generate $L$ as a free abelian group in the sense that $L=\mathbb{Z}w_{1}\oplus\ldots\oplus\mathbb{Z}w_{r}$. Also, let $e_{1},\ldots,e_{d}\in\mathbb{Z}^{d}$ be a set of free generators for $\mathbb{Z}^{d}$ (say, the standard generators), and for $n\geq 0$, let

$$B_{n}:=\{-n,\ldots,n\}^{d}\subset\mathbb{Z}^{d}$$

and

$$S_{n}:=B_{n}\setminus B_{n-1}\mbox{ for }n\geq 1.$$

The $S_{i}$’s play the role of the bumps and dents in our tiles, and the properties that we need from them are that they are all bounded and that for every $i\neq j$ the set $S_{i}$ does not contain a translated copy of the set $S_{j}$.

Choose $m\in\mathbb{N}$ large enough so that $B_{m}$ contains disjoint translated copies of all the boxes $B_{l}$ for $1\leq l\leq d+rs$ (e.g. $m\geq(d+rs)^{2}$), and let $N=2m+1$.

Choose translation vectors $v_{1},\ldots,v_{d+rs}\in B_{m}$ so that

	$\displaystyle\forall 1\leq i\leq d+rs:\quad(v_{i}+B_{d+rs})\subseteq B_{m},$
	and
	$\displaystyle\forall 1\leq i_{1}<i_{2}\leq d+rs:\quad(v_{i_{1}}+B_{d+rs})\cap(v_{i_{2}}+B_{d+rs})=\emptyset.$

We first define the tile $T\subset\mathbb{Z}^{d}$ as follows:

$$T:=\left(B_{m}\setminus\biguplus_{i=1}^{d}\left(v_{i}+S_{i}\right)\right)\uplus\biguplus_{i=1}^{d}\left(Ne_{i}+v_{i}+S_{i}\right).$$

(3)

We refer to the missing translates of $S_{i}$’s in $T$ as dents, and to the disconnected translates of the $S_{i}$’s as bumps. We claim that $\mathit{Tile}_{0}(T;\mathbb{Z}^{d})=\{N\mathbb{Z}^{d}\}$. The claim is fairly evident from Figure 1. We provide some details below.

The properties of $T$ that we use, which are straightforward from (3), are the following: Firstly, $T$ is a fundamental domain for $N\mathbb{Z}^{d}$. Secondly, for every $1\leq i\leq d$ the only $v\in T$ such that $v+S_{i}\cap T=\emptyset$ is $v_{i}$. Thus, $Ne_{i}+T$ is the unique translate of $T$ that can cover $Ne_{i}$ without intersecting the corresponding bump $Ne_{i}+v_{i}+S_{i}$. See Figure 1. Since $T$ is a fundamental domain for $N\mathbb{Z}^{d}$, we have $N\mathbb{Z}^{d}\in\mathit{Tile}_{0}(T;\mathbb{Z}^{d})$. Now suppose $C\in\mathit{Tile}_{0}(T;\mathbb{Z}^{d})$, then for every $v\in C$ and $1\leq i\leq d$ we also have $v\pm Ne_{i}\in C$, because each bump of $v+T$ must fit into a corresponding dent and vice versa. So $C$ must be a union of cosets of $N\mathbb{Z}^{d}$. But since $T$ is a fundamental domain for $N\mathbb{Z}^{d}$, it follows that $C$ must be a coset of $N\mathbb{Z}^{d}$. Since $0\in C$ we must have $C=N\mathbb{Z}^{d}$.

We now define the tiles $T_{1},\ldots,T_{s}$. For $1\leq j\leq s$ the tile $T_{j}$ is defined by adding additional $r$ bumps and $r$ dents to the tile $T$. The bumps and dents are of the form $S_{l}$, for $d+r(j-1)<l\leq d+jr$:

$$T_{j}:=\left(T\setminus\biguplus_{l=1}^{r}\left(v_{d+r(j-1)+l}+S_{d+r(j-1)+l}\right)\right)\uplus\biguplus_{l=1}^{r}\left(Nw_{l}+v_{d+r(j-1)+l}+S_{d+r(j-1)+l}\right).$$

(4)

Direct inspection of (4) shows that for any $1\leq j\leq s$ we have

$$T_{j}\oplus NL=T\oplus NL$$

(5)

From (5) and $T\oplus N\mathbb{Z}^{d}=\mathbb{Z}^{d}$ it follows that whenever $\{\tilde{C}_{1},\ldots,\tilde{C}_{s}\}$ is a partition of $N\mathbb{Z}^{d}$ into $NL$-periodic sets, we have $(\tilde{C}_{1},\ldots,\tilde{C}_{s})\in\mathit{Tile}_{0}(T_{1},\ldots,T_{s};\mathbb{Z}^{d})$.

Now we show that for any $(\tilde{C}_{1},\ldots,\tilde{C}_{s})\in\mathit{Tile}_{0}(T_{1},\ldots,T_{s};\mathbb{Z}^{d})$ we have that $\{\tilde{C}_{1},\ldots,\tilde{C}_{s}\}$ is a partition of $N\mathbb{Z}^{d}$ into $NL$-periodic sets. As soon as we show that each $\tilde{C}_{j}$ is $NL$-periodic, it will follow from (5) that $\tilde{C}_{j}\oplus T_{j}=\tilde{C}_{j}\oplus T$ and so $\biguplus_{j=1}^{s}\tilde{C}_{j}\in\mathit{Tile}_{0}(T;\mathbb{Z}^{d})$, thus $\biguplus_{j=1}^{s}\tilde{C}_{j}=N\mathbb{Z}^{d}$. So it is left to explain why each $\tilde{C}_{j}$ is $NL$-periodic. The argument is essentially the same as the argument showing that each $C\in\mathit{Tile}_{0}(T;\mathbb{Z}^{d})$ is $N\mathbb{Z}^{d}$-periodic, and should be fairly clear from Figure 2. In view of (4), the additional property of the collection $T_{1},\ldots,T_{s}$ that is needed here is the following: For every $1\leq l\leq r$ and every $1\leq j,j^{\prime}\leq s$, if $v+T_{j^{\prime}}$ covers $v_{d+r(j-1)+l}$ without intersecting the bump $Nw_{l}+v_{d+r(j-1)+l}+S_{d+r(j-1)+l}$ of $T_{j}$, then we must have $j=j^{\prime}$ and $v=Nw_{l}$. See Figure 2. ∎

1. (a)

   The equality $\pi(A)+\pi(B)=\pi(C)$ follows directly because $\pi$ is a group homomorphism. We need to show that the sum is direct, namely that the representation $\pi(c)=\pi(a)+\pi(b)$ with $a\in A$, $b\in B$ is unique for every $c\in C$. Suppose $\pi(a_{1})+\pi(b_{1})=\pi(a_{2})+\pi(b_{2})$. Since every $c\in C$ has a unique representation $c=a+b$ with $a\in A$ and $b\in B$, we conclude that

   $$v:=(a_{1}+b_{1})-(a_{2}+b_{2})\in L.$$

   Set $b_{3}:=b_{2}+v$. By the assumption that $B$ is $L$-periodic, it follows that $b_{3}\in B$. Then $a_{2}+b_{3}=a_{1}+b_{1}$ are two representations of the same element as a sum of an element of $A$ and an element of $B$. It follows that $a_{1}=a_{2}$ and $b_{1}=b_{3}$, so $b_{1}=b_{2}+v$. This means that $\pi(a_{1})=\pi(a_{2})$ and $\pi(b_{1})=\pi(b_{2})$. We have thus proved that the representation $\pi(c)=\pi(a)+\pi(b)$ with $a\in A$, $b\in B$ is unique for every $c\in C$.

2. (b)

   Using induction, it is enough to prove the case $k=2$. Clearly, $f(C_{1}\cup C_{2})=f(C_{1})\cup f(C_{2})$ for every function $f$ and sets $C_{1},C_{2}$ in its domain, so we only need to show that $\pi(C_{1})\cap\pi(C_{2})=\emptyset$ under the assumptions that $C_{1}\cap C_{2}=\emptyset$ and that each $C_{i}$ is $L$-periodic. Otherwise, there are $c_{1}\in C_{1}$ and $c_{2}\in C_{2}$ such that $\pi(c_{1})=\pi(c_{2})$. This means that $c_{1}-c_{2}\in L$. But $C_{1}$ is $L$-periodic and $c_{1}\in C_{1}$ so $c_{2}\in C_{1}$, contradicting $C_{1}\cap C_{2}=\emptyset$.

∎

Given $d,k\in\mathbb{N}$ and a surjective homomorphism $\pi:\mathbb{Z}^{d}\to\Gamma$ let $L=\ker(\pi)$, let $D$ be a fundamental domain for $L$ that contains $0$, and let $T,T_{1},\ldots,T_{2k}\subseteq\mathbb{Z}^{d}$ and $N\in\mathbb{N}$ satisfy the conclusion of Lemma 2.1 with respect to $d,s=2k$ and $L$.

Given finite sets $F_{1},\ldots,F_{k}\subset\Gamma$, let $\tilde{F}_{1},\ldots,\tilde{F}_{k}\subset\mathbb{Z}^{d}$ be as in (2). For the first inclusion, suppose $(A_{1},\ldots,A_{k})\in\mathit{Tile}_{0}(F_{1},\ldots,F_{k};\Gamma)$, then by definition $\biguplus_{j=1}^{k}(F_{j}\oplus A_{j})=\Gamma$. Applying $\pi^{-1}$, we obtain that

$$\biguplus_{j=1}^{k}\pi^{-1}(F_{j}\oplus A_{j})=\mathbb{Z}^{d}.$$

(6)

For $1\leq j\leq k$ we define

$$\tilde{C}_{j}:=\left((N\pi^{-1}(F_{j}\setminus\{0\})\cap D)\oplus N\pi^{-1}(A_{j})\right)\mbox{ and }\tilde{C}_{j+k}:=N\pi^{-1}(A_{j}).$$

(7)

By the identity

$$\pi^{-1}(F_{j}\oplus A_{j})=(\pi^{-1}(F_{j})\cap D)\oplus\pi^{-1}(A_{j})=(\pi^{-1}(F_{j}\setminus\{0\})\cap D)\oplus\pi^{-1}(A_{j})\uplus\pi^{-1}(A_{j}),$$

and in view of (7), we see that for every $1\leq j\leq k$ we have $\pi^{-1}(F_{j}\oplus A_{j})=\frac{1}{N}\left[\tilde{C}_{j}\uplus\tilde{C}_{j+k}\right]$. Thus multiplying the identity in (6) by $N$ yields $\biguplus_{j=1}^{2k}\tilde{C}_{j}=N\mathbb{Z}^{d}$.

Note that $N\pi^{-1}(A_{j})$ is $NL$-periodic, thus $\{\tilde{C}_{1},\ldots,\tilde{C}_{2k}\}$ is a partition of $N\mathbb{Z}^{d}$ into $NL$-periodic sets. By Lemma 2.1 it follows that

$$\left(\tilde{C}_{1},\ldots,\tilde{C}_{2k}\right)\in\mathit{Tile}_{0}(T_{1},\ldots,T_{2k};\mathbb{Z}^{d}).$$

Using (7), the expression $\biguplus_{j=1}^{2k}T_{j}\oplus\tilde{C}_{j}=\mathbb{Z}^{d}$ can be rearranged to

$$\biguplus_{j=1}^{k}\left[\left((N\pi^{-1}(F_{j}\setminus\{0\})\cap D)\oplus T_{j}\right)\uplus T_{j+k}\right]\oplus N\pi^{-1}(A_{j})=\mathbb{Z}^{d}.$$

But in view of the definition of the $\tilde{F}_{j}$ in (2), this means that $(N\pi^{-1}(A_{1}),\ldots,N\pi^{-1}(A_{k}))\in\mathit{Tile}_{0}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})$. We conclude that

$$N\Big{(}(\pi^{\otimes k})^{-1}\mathit{Tile}_{0}\left(F_{1},\ldots,F_{k};\Gamma\right)\Big{)}\subseteq\mathit{Tile}_{0}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d}).$$

To see the other inclusion, let $(\tilde{A}_{1},\ldots,\tilde{A}_{k})\in\mathit{Tile}_{0}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})$. Then the following statements hold:

1. (1)

   $0\in\biguplus_{j=1}^{k}\tilde{A}_{j}$.

2. (2)

   $\biguplus_{j=1}^{k}\tilde{F}_{j}\oplus\tilde{A}_{j}=\mathbb{Z}^{d}$.

The second equation can be rewritten as follows:

$$\biguplus_{j=1}^{k}\left[\left((N\pi^{-1}(F_{j}\setminus\{0\})\cap D)\oplus T_{j}\right)\uplus T_{j+k}\right]\oplus\tilde{A}_{j}=\mathbb{Z}^{d}.$$

This can be rearranged as:

$$\biguplus_{j=1}^{k}\left(N(\pi^{-1}(F_{j}\setminus\{0\})\cap D)\oplus T_{j}\oplus\tilde{A}_{j}\right)\uplus\biguplus_{j=1}^{k}(T_{j+k}\oplus\tilde{A}_{j})=\mathbb{Z}^{d}.$$

(8)

Thus $(\tilde{A}_{1},\ldots,\tilde{A}_{k})\in\mathit{Tile}_{0}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})$ if and only if

$$\left(\tilde{C}_{1},\ldots,\tilde{C}_{2k}\right)\in\mathit{Tile}_{0}(T_{1},\ldots,T_{2k};\mathbb{Z}^{d}),$$

where

$$\tilde{C}_{j}:=N(\pi^{-1}(F_{j}\setminus\{0\})\cap D)\oplus\tilde{A}_{j}\mbox{ and }\tilde{C}_{j+k}:=\tilde{A}_{j}\leavevmode\nobreak\ \text{ for }1\leq j\leq k.$$

By Lemma 2.1, this happens if and only if $\{\tilde{C}_{1},\ldots,\tilde{C}_{2k}\}$ is a partition of $N\mathbb{Z}^{d}$ into $NL$-periodic sets. Since $\tilde{C}_{j}\subseteq N\mathbb{Z}^{d}$ it follows that $\tilde{A}_{j}\subseteq N\mathbb{Z}^{d}$. Now because $\tilde{C}_{j+k}$ is $NL$-periodic, it follows that each $\tilde{A}_{j}$ is $NL$-periodic. This means that for every $1\leq j\leq k$ there exists $A_{j}\subseteq\Gamma$ such that

$$\tilde{A}_{j}=N\pi^{-1}(A_{j}).$$

(9)

It remains to show that $(A_{1},\ldots,A_{k})\in\mathit{Tile}_{0}(F_{1},\ldots,F_{k};\Gamma)$. Because $0\in\biguplus_{j=1}^{k}\tilde{A}_{j}$, it follows that $0\in\biguplus_{j=1}^{k}A_{j}$. Using the property $T_{l}\oplus NL=T\oplus NL$ for every $1\leq l\leq 2k$, and the fact that each $\tilde{A}_{j}$ is $NL$-periodic, we may replace every $T_{l}$ in (8) by $T$ to obtain:

$$\biguplus_{j=1}^{k}\left(N(\pi^{-1}(F_{j}\setminus\{0\})\cap D)\oplus T\oplus\tilde{A}_{j}\right)\uplus\biguplus_{j=1}^{k}(T\oplus\tilde{A}_{j})=\mathbb{Z}^{d}.$$

Using the fact that $0\in D$, the left-hand side can be rearranged to obtain the following:

$$\left[\biguplus_{j=1}^{k}\left(N(\pi^{-1}(F_{j})\cap D)\oplus\tilde{A}_{j}\right)\right]\oplus T=\mathbb{Z}^{d}.$$

Recall that by Lemma 2.1 we have $\mathit{Tile}_{0}(T;\mathbb{Z}^{d})=\{N\mathbb{Z}^{d}\}$, thus

$$\biguplus_{j=1}^{k}\left(N(\pi^{-1}(F_{j})\cap D)\oplus\tilde{A}_{j}\right)=N\mathbb{Z}^{d}.$$

Plugging in (9) we conclude that $\biguplus_{j=1}^{k}\left(N(\pi^{-1}(F_{j})\cap D)\oplus N\pi^{-1}(A_{j})\right)=N\mathbb{Z}^{d}$. Since $v\mapsto Nv$ induces a group isomorphism between $\mathbb{Z}^{d}$ and $N\mathbb{Z}^{d}$ we get $\biguplus_{j=1}^{k}\left((\pi^{-1}(F_{j})\cap D)\oplus\pi^{-1}(A_{j})\right)=\mathbb{Z}^{d}$. By Lemma 2.3, using the fact that $\pi^{-1}(A_{j})$ is $L$-periodic, it follows that $\biguplus_{j=1}^{k}F_{j}\oplus A_{j}=\Gamma$. We have thus shown that

$$\mathit{Tile}_{0}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})\subseteq N\left((\pi^{\otimes k})^{-1}\mathit{Tile}_{0}\left(F_{1},\ldots,F_{k};\Gamma\right)\right).$$

∎

Suppose $(F_{1},\ldots,F_{k})$ is an aperiodic $k$-tuple of finite sets in $\Gamma$, then $\mathit{Tile}(F_{1},\ldots,F_{k};\Gamma)\neq\emptyset$. By (2), we have that $\mathit{Tile}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})\neq\emptyset$. If we assume that $(\tilde{F}_{1},\ldots,\tilde{F}_{k})$ is not aperiodic, then by (2) there exists $(A_{1},\ldots,A_{k})\in\mathit{Tile}(F_{1},\ldots,F_{k};\Gamma)$ such that $(N\pi^{-1}(A_{1}),\ldots,N\pi^{-1}(A_{k}))$ is a periodic $k$ tuple in $\mathbb{Z}^{d}$. But this implies that $(A_{1},\ldots,A_{k})$ is periodic. Thus, the existence of an aperiodic $k$-tuple in $\mathbb{Z}^{d}$ implies the existence of an aperiodic $k$-tuple in $\Gamma$. ∎

Assume that for a particular $k\in\mathbb{N}$ a tiling by $k$ tiles in $\mathbb{Z}^{d}$ is algorithmically decidable. By definition, this means that there exists a Turing machine $\tilde{M}$ that takes as input a $k$-tuple $\tilde{F}_{1},\ldots,\tilde{F}_{k}$ of finite subsets in $\mathbb{Z}^{d}$ and accepts the input if and only if $\mathit{Tile}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})\neq\emptyset$. The Turing machine $\tilde{M}$ always halts in finite time. Then we can construct a Turing machine $M$ that takes as input a $k$-tuple $F_{1},\ldots,F_{k}$ of finite subsets in $\Gamma$, produces the finite subsets $\tilde{F}_{1},\ldots,\tilde{F}_{k}$ of $\mathbb{Z}^{d}$ described in the statement of Theorem 1.1, then runs $\tilde{M}(\tilde{F}_{1},\ldots,\tilde{F}_{k})$. Clearly $M$ halts in finite time on input $F_{1},\ldots,F_{k}$ if and only if $\tilde{M}$ halts with input $\tilde{F}_{1},\ldots,\tilde{F}_{k}$. The Turing machine $M$ accepts $F_{1},\ldots,F_{k}$ if and only if $\tilde{M}$ accepts $\tilde{F}_{1},\ldots,\tilde{F}_{k}$, which by assumption happens if and only if $\mathit{Tile}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})\neq\emptyset$. By Theorem 1.1 this happens if and only if $\mathit{Tile}(F_{1},\ldots,F_{k};\mathbb{Z}^{d})\neq\emptyset$. ∎

The concatenation of the proof of $\mathit{Tile}(\tilde{F}_{1},\ldots,\tilde{F}_{k};\mathbb{Z}^{d})\neq\emptyset$ in $\mathcal{T}$ with the proof of (2) in $\mathcal{T}$ yields a proof of $\mathit{Tile}(F_{1},\ldots,F_{k};\Gamma)\neq\emptyset$ in $\mathcal{T}$. ∎