A Periodicity Result for Tilings of $\mathbb{Z}^{3}$ by Clusters of Prime-Squared Cardinality

Note that if $T|_{[a,b]}$ is known for some interval $[a,b]$ of length exceeding $\text{diam}(F)$, then $T$ is determined entirely. Now the assertion follows by a pigeonhole argument. ∎

We may assume that $g$ is primitive. Using Lemma A.1, after applying a suitable $GL_{3}(\mathbb{Z})$ transformation, we may assume that $g=(0,0,1)$. Let $S$ be the image of $F$ under the map $f:\mathbf{R}^{3}\to\mathbf{R}^{2}$ defined by

for all $(x,y,z)\in\mathbb{Z}^{3}$, that is, $f$ is the orthogonal projection on the $x,y$-plane. Let $(a,b)$ be an extreme point of the convex hull of $S$ in $\mathbf{R}^{2}$ and let $\ell$ be a line in $\mathbf{R}^{2}$ such that $\ell\cap S=\{(a,b)\}$. Let $\pi$ be a plane parallel to $g$ in $\mathbf{R}^{3}$ such that the image of $\pi$ under $f$ is $\ell$. Then $\pi\cap F=\{(a,b,z)\in\mathbb{Z}^{3}:\ (a,b,z)\in F\}$. By hypothesis, $|\pi\cap F|$ has size divisible by $p$. Now the set $F\setminus\pi\cap F$ satisfies the same hypothesis as $F$ and now we can finish inductively. ∎

Using Lemma A.2, after a suitable $GL_{3}(\mathbb{Z})$ transformation, we may assume that $\pi^{\prime}$ is the plane $\Lambda=\mathbb{Z}^{2}\times\{0\}$. Since $\pi^{\prime}$ is not parallel to $g$ by hypothesis, we see that $\mathbb{Z}g+\Lambda$ is a subgroup of rank $3$ in $\mathbb{Z}^{3}$. Also, without loss of generality assume that $g$ is a primitive vector.³³3See Appendix A for the definition of a primitive vector. By translating $F$ if necessary, we may assume that $\Lambda$ intersects $F$ but $\mathbb{Z}^{2}\times\{k\}$ does not intersect $F$ whenever $k$ is a negative integer. Define $B=\{n\geq 0:\ (\mathbb{Z}^{2}\times\{n\})\cap F\neq\emptyset\}$ and say $B=\{0=b_{0},b_{1},\ldots,b_{k}\}$. Since, by hypothesis, each translate of $\Lambda$ intersects $F$ in a set of size divisible by $p$, we see that $(k+1)p=|B|p\leq|F|=p^{2}$ and hence $k+1\leq p$. By Lemma 3.1 we know that whenever $\ell$ is a line in $\mathbf{R}^{3}$ parallel to $g$, we have that $\ell\cap F$ has size divisible by $p$. Thus there are at least $p$ translates of $\Lambda$ in the direction of $g$ which intersect $F$. Thus $k+1=|B|\geq p$. So we have $k+1=p$. This forces that $(\mathbb{Z}^{2}\times\{b_{i}\})\cap F$ has size $p$ for each $i\in\{0,1,\ldots,k\}$.

It follows that any translate $\ell$ of $\mathbb{Z}g$ must intersect each $\mathbb{Z}^{2}\times\{b_{i}\}$ non-trivially. Let $p_{0},p_{1},\ldots,p_{k}\in\mathbb{Z}^{2}$ be such that $(p_{i},b_{i})-(p_{0},b_{0})$ are parallel to $g$ for each $i\in\{1,\ldots,k\}$. Since $g$ is a primitive vector, we can find integers $n_{1},\ldots,n_{k}$ such that $n_{i}g=(p_{i},b_{i})-(p_{0},b_{0})$. It follows that

and hence $F$ is prism with base $\Lambda$ and axis $g$, and foundation of size $p$. ∎

If the affine span of $F$ has dimension smaller than $3$ then the result follows by Theorem 2.3. Thus we may assume that the dimension of the affine span of $F$ is $3$.

Let $\Lambda$ and $g$ be the base and axis of $F$ respectively. Let $h$ be a vector in $\mathbb{Z}^{3}$ and $\{n_{0}=0,\ldots,n_{k}\}$ be a set of integers such that $F=\bigcup_{i=0}^{k}(n_{i}g+(h+\Lambda)\cap F)$. By translating $F$ is necessary we may assume that $h=0$. Also by Lemma A.2 we may without loss of generality assume that $\Lambda=\mathbb{Z}^{2}\times\{0\}$.

Thus the affine span of $A:=\Lambda\cap F$ has dimension $2$. Let $T$ be an $F$-tiling of $\mathbb{Z}^{3}$. It is easy to see that for any integer $c$ the set $T_{c}:=T\cap(\mathbb{Z}^{2}\times\{c\})$ is a tiling of $\mathbb{Z}^{2}\times\{c\}\cong\mathbb{Z}^{2}$ by translates of $A\times\{c\}\cong A$. By Theorem 2.2 we know that there is a rank-$2$ subgroup $\Gamma$ of $\mathbb{Z}^{3}$ such that $T_{c}$ is $\Gamma$-invariant for each $c$. Thus $T$ is $\Gamma$-invariant and hence $2$-periodic. ∎

Let $m$ be the size of a maximal rationally independent subset of $\{\gamma_{1},\ldots,\gamma_{n}\}$. By renumbering the $\gamma_{i}$’s and $x_{i}$’s if required, we may assume that $\{\gamma_{1},\ldots,\gamma_{m}\}$ is a maximal rationally independent subset of $\{\gamma_{1},\ldots,\gamma_{n}\}$. Thus we can find vectors $v_{1},\ldots,v_{n}$ in $\mathbb{Z}^{m}$ such that

for $i=1,\ldots,n$, where $v(j)$ denotes the $j$-th coordinate of any $v\in\mathbb{Z}^{m}$. By the algebraic independence of $\gamma_{1},\ldots,\gamma_{m}$, it follows that for $i\in\{1,\ldots,m\}$ we have

Define the Laurent polynomial $f(U)$ in $\mathbf{C}[u_{1}^{\pm},\ldots,u_{m}^{\pm}]$ as

Then we have

By the rational independence of $\gamma_{1},\ldots,\gamma_{m}$, we know that the set $\{(\gamma_{1}^{k},\ldots,\gamma_{m}^{k}):\ k\geq 0\}$ is dense in $(S^{1})^{m}$.⁵⁵5See Theorem 4.14 in [3]. Thus the image of $f$ on a dense set of $(S^{1})^{m}$ is contained in $\mathbb{Z}$. Since $f$ is continuous, this implies that the image of $(S^{1})^{m}$ under $f$ is contained in $\mathbb{Z}$. The connectedness of $(S^{1})^{m}$ now yields that the image of $(S^{1})^{m}$ under $f$ is a singleton. However, for any $\varepsilon>0$, we can find a $k\geq 0$ such that $|\gamma_{i}^{k}-1|<\varepsilon$ for each $1\leq i\leq n$, and thus we must have that $f$ is identically zero on $(S^{1})^{m}$, finishing the proof. ∎

Let $g=(g_{1},g_{2},g_{3})$ and $h=(h_{1},h_{2},h_{3})$. Consider the $2\times 3$-matrix $M$ over $\mathbb{Q}$ defined as

Note that an element $z$ of $\mathbb{T}^{3}=\mathbf{R}^{3}/\mathbb{Z}^{3}$ is in $\ker\chi_{g}\cap\ker\chi_{h}$ if and only if some (any) representative $z^{\prime}$ of $z$ in $\mathbf{R}^{3}$ satisfies $Mz^{\prime}\in\mathbb{Z}^{2}$. Since $g$ and $h$ are linearly independent, the rank of $M$ is $2$, and its nullity is therefore $1$. Thus, since $M$ is a matrix over $\mathbb{Q}$, we can find a nonzero vector $v$ in $\mathbb{Q}^{3}$ such that $Mv=0$. Clearly, any such vector $v$ also spans the null-space of $M$ since the null-space of $M$ is $1$-dimensional. By scaling $v$ if necessary, we may assume that $v\in\mathbb{Z}^{3}$.

Let $\Lambda=M(\mathbb{Z}^{3})$, that is, $\Lambda\subseteq\mathbb{Z}^{2}$ is the image of $\mathbb{Z}^{3}$ under $M$. Since $M$ has rank $2$, we see that $\Lambda$ is a finite index subgroup of $\mathbb{Z}^{2}$. Let $n$ be the smallest positive integer such that $(n,0)$ and $(0,n)$ are both in $\Lambda$. Say $u_{1}$ and $u_{2}$ in $\mathbb{Z}^{3}$ be such that $Mu_{1}=(n,0)$ and $Mu_{2}=(0,n)$. We will show that

Let $w\in\ker\chi_{g}\cap\ker\chi_{h}$ be arbitrary. Let $w^{\prime}$ be a representative of $w$ in $\mathbf{R}^{3}$. Then $Mw^{\prime}=(a,b)$ for some $(a,b)\in\mathbb{Z}^{2}$. Therefore

giving

Thus $w^{\prime}-(a/n)u_{1}-(b/n)u_{2}$ is in the null-space of $M$, and hence there is $t\in\mathbf{R}$ such that $w^{\prime}-(a/n)u_{1}+(b/n)u_{2}=tv$. Therefore $w^{\prime}=(au_{1}+bu_{2})/n+tv$. We can find $0\leq i,j,k<n$ such that $(au_{1}+bu_{2})/n\equiv(i/n,j/n,k/n)\pmod{\mathbb{Z}^{3}}$. Therefore

showing the desired containment. ∎

Without loss of generality we may assume that the $a_{i}$’s are all non-negative. Define the polynomial $Q(z)=z^{a_{1}}+\cdots+z^{a_{m}}$. Then since $\zeta$ is a root of $Q(z)$, we have $Q(z)$ is divisible by the $p^{k}$-th cyclotomic polynomial $\Phi_{p^{k}}(z)$. Let $\Psi(z)$ be an integer polynomial such that $Q(z)=\Phi_{p^{k}}(z)\Psi(z)$. Now using the fact that $\Phi_{p^{k}}(z)=\Phi_{p}(z^{p^{k-1}})$ we have $Q(z)=\Phi_{p}(z^{p^{k-1}})\Psi(z)$. Now substituting $z=1$ gives $m=p\Psi(1)$ and hence $p$ divides $m$. ∎

Let $n$ be the product of all primes which divide $|F|$. Then for each non-negative integer $k$ we have $nk+1$ is relatively prime to $|F|$. Fix $(a,b,c)\in Z$. Then we have, for all non-negative integers $k$ that

giving

Therefore

for all $N\geq 1$. Taking the limit $N\to\infty$ we see that there must exist $g\in F\setminus{\{0\}}$ such that $a^{ng_{1}}b^{ng_{2}}c^{ng_{3}}$ is $1$,⁶⁶6 This is because if $z\in S^{1}$ then the average $(z+z^{2}+\cdots+z^{N})/N$ does not converge to zero if and only if $z=1$. and thus $(a,b,c)\in\ker(\chi_{ng})$. Therefore $Z\subseteq\bigcup_{g\in F\setminus\{0\}}\ker\chi_{ng}$.

Let $\Delta$ be a non-empty subset of $\mathbb{Z}^{3}\setminus\{0\}$ of smallest possible size such that $Z\subseteq\bigcup_{g\in\Delta}\ker\chi_{g}$. Such a $\Delta$ exists by the above paragraph. We claim that the elements of $\Delta$ are pairwise linearly independent. Suppose not. Then there exist distinct $g,h\in\Delta$ such that $g$ and $h$ are linearly dependent. We can thus find a nonzero vector $v$ such that $v\in(\mathbb{Z}g)\cap(\mathbb{Z}h)$. Let $\Delta^{\prime}=(\Delta\setminus\{g,h\})\cup\{v\}$. It is clear that

and hence $Z\subseteq\bigcup_{u\in\Delta^{\prime}}\ker\chi_{u}$. But $\Delta^{\prime}$ has size strictly smaller than the size of $\Delta$ which is a contradiction to the choice of $\Delta$. This finishes the proof. ∎

Using Lemma A.1, after applying a suitable $GL_{3}(\mathbb{Z})$ transformation, we may assume that $h=(0,0,m)$ for some positive integer $m$.⁷⁷7 This is because of the following. For any linear map $T:\mathbf{R}^{3}\to\mathbf{R}^{3}$ such that $T\in GL_{3}(\mathbb{Z})$, and any $F\subseteq\mathbb{Z}^{3}$ finite, we have $Tp\in Z\left(\sum_{g\in F}\chi_{g}\right)$ if and only if $(T^{-1})^{*}p\in Z\left(\sum_{g\in T(F)}\chi_{g}\right)$. Let $(a,b,c)$ be an arbitrary element in $\ker\chi_{h}\cap Z$, where

Let $h_{0}=(0,0,1)$. Then

Thus there is an $m$-th root of unity $\omega$ such that $(a,b,c)\in((1,1,\omega)\cdot\ker\chi_{h_{0}})\cap Z$. Then $c=\omega$ and hence

for all $\alpha$ relatively prime with $|F|$.

Let $d$ be a positive integer such that $\omega$ is a primitive $d$-th root of unity. Let $d=p^{r}\beta$ where $\beta$ is relatively prime to $p$. Let $n$ be the product of all the primes dividing $|F|$. For all non-negative integers $k$, we have $mnk+1$ is relatively prime to $|F|$. Using the fact that $|F|$ is a power of $p$, we have $(mnk+1)\beta$ is also relatively prime with $|F|$. So we have, by substituting $\alpha=(mnk+1)\beta$, that

for all non-negative integers $k$. Let $\zeta=\omega^{\beta}$, and hence $\zeta$ is a primitive $p^{r}$-th root of unity. Let $\sim$ be a relation on $F$ defined as $g\sim g^{\prime}$ for $g,g^{\prime}\in F$ if and only if $(g_{1},g_{2})=(g_{1}^{\prime},g_{2}^{\prime})$. Then $\sim$ is an equivalence relation, and let $\mathcal{E}$ be the set of all the equivalence classes of $\sim$.⁸⁸8Note that the equivalence classes are precisely the sets that are obtained by intersection $F$ with a line parallel to $h$. For each $E\in\mathcal{E}$ choose an element $g^{E}$ of $E$. For any $v\in\mathbb{Z}^{3}$ let $\hat{v}$ denote the vector in $\mathbb{Z}^{2}$ obtained by dropping the last coordinate of $v$. Then Equation 3.24 gives

for all $k$ non-negative. Let $\theta_{E}=\sum_{g\in E}\zeta^{g_{3}}$ and $\gamma_{E}=\theta_{E}\chi_{\beta\hat{g}^{E}}(a,b)$ for all $E$ in $\mathcal{E}$. If $\theta_{E}$ is $0$ for all $E\in\mathcal{E}$, then by Lemma 3.7 we have each $|E|$ is divisible by $p$. This would mean precisely that (1) holds. So we may assume that there is some $E_{0}\in\mathcal{E}$ such that $\theta_{E_{0}}$, or equivalently $\gamma_{E_{0}}$, is nonzero. Then from Equation 3.25 we have