Tilings of Convex Polyhedral Cones and Topological Properties of Self-Affine Tiles

Abstract

Let ${\varvec{a}}_1,\ldots , {\varvec{a}}_r$ be vectors in a half-space of ${\mathbb {R}}^n$. We call $C={\varvec{a}}_1{\mathbb {R}}^{+}+\cdots +{\varvec{a}}_r {\mathbb {R}}^{+}$ a convex polyhedral cone and $\{{\varvec{a}}_1,\ldots , {\varvec{a}}_r\}$ a generator set of C. A generator set with the minimal cardinality is called a frame. We investigate the translation tilings of convex polyhedral cones. Let $T\subset {\mathbb {R}}^n$ be a compact set such that T is the closure of its interior, and ${{\mathcal {J}}}\subset {\mathbb {R}}^n$ be a discrete set. We say $(T,{{\mathcal {J}}})$ is a translation tiling of C if $T+{{\mathcal {J}}}=C$ and any two translations of T in $T+{{\mathcal {J}}}$ are disjoint in Lebesgue measure. We show that if the cardinality of a frame of C is larger than the dimension of C, then C does not admit any translation tiling; if the cardinality of a frame of C equals the dimension of C, then the translation tilings of C can be reduced to the translation tilings of $({\mathbb {Z}}^+)^n$. As an application, we characterize all the self-affine tiles possessing polyhedral corners (that is, there exists a point of the tile such that a neighborhood of the point is congruent to a neighborhood of the vertex of a convex polyhedral cone), which generalizes a result of Odlyzko (Proc. Lond. Math. Soc. 37, 213–229 (1978)).

Appendix A: Proof of Lemma 7.3

Proof

For simplicity, we identify ${\mathbb {R}}^2$ with the complex plane ${\mathbb {C}}$. Let $L_0=\{x+h {\mathbf {i}}:x\in [a,b]\}$. Assume that $t_1, t_2,\ldots , t_p$ are points in L from left to right. Suppose on the contrary that $(T, \{t_0,t_1,\ldots , t_p\})$ is a tiling of A. Let

$$\begin{aligned} I=\{x+y(a+h{\mathbf {i}}):x\in [0,t_1],\,y\in [0,1]\} \end{aligned}$$

be a parallelogram on the left part of A. Clearly $I\subset T$. Let M be the largest integer such that $Mt_1<1$. We claim that $T\cap \bigcup _{m=0}^{M-1}(I+mt_1)$ is a union of translations of I. To prove this, we need only prove the following two statements: for each integer m, $0\le m\le M-1$, we have:

• (i) ${{\mathcal {J}}}\cap [0, mt_1]\subset t_1{\mathbb {Z}}^+$.

• (ii) For every integer $0\le u\le m$, $I+ut_1$ belongs to one tile except a measure zero set.

We prove (i) and (ii) by induction on m. Clearly, (i) and (ii) hold for $m=0$. Now we assume that (i) and (ii) hold for $m-1$ with $m\ge 1$.

First, we prove (i). If $t\in {{\mathcal {J}}}\cap [0, mt_1)$, then $t+I$ and $\bigcup _{j=0}^{m-1}(I+jt_1)$ overlap, we have $t\notin ((m-1)t_1,mt_1)$ by the induction hypothesis of (ii). Therefore, by the induction hypothesis of (i), no matter $mt_1\in {{\mathcal {J}}}$ or not, (i) holds for m.

Now we prove (ii). Suppose on the contrary that (ii) is false. Then $I+mt_1$ does not belong to one tile. This first implies that $mt_1\notin {{\mathcal {J}}}$. Secondly, if there exists $1\le m'\le m-1$, such that $m't_1\in {{\mathcal {J}}}$ and $0<\mu ((T+m't_1)\cap (I+mt_1))<\mu (I)$, then $0<\mu (T\cap (I+(m-m')t_1))<\mu (I)$, which contradicts the assumption (ii). Therefore, if a tile $T+t$ satisfies that $0<\mu ((T+t)\cap (I+mt_1))<\mu (I)$, then either $t=0$, or $mt_1<t<(m+1)t_1$. In the latter case, there is only one t satisfying this property, and we denote it by $t^*$. Then

$$\begin{aligned} I+mt_1\subset T\cup (T+t^*). \end{aligned}$$

Denote $U=\{x+y(a+h{\mathbf {i}}):x\in [mt_1,t^*),\,y\in [0,1]\}$. By $U\cap (T+t^*)=\emptyset$ and the above equation, we have $U\subset T$. Then the intersection of $T+t_1$ and $T+t^*$ contains $U+t_1$ as a subset, which is a contradiction. So (ii) holds for m.

Since $t_p$ is the rightmost point of ${{\mathcal {J}}}$, $T+t_p$ must contain a relative neighborhood $B(1,r)\cap A$ of 1, for all small enough r ($<1-Mt_1$). Moreover, we have $B(1,r)\cap A=(T+t_p)\cap B(1,r)$, thus

$$\begin{aligned} (B(1,r)\cap A)-t_p=T\cap B(1-t_p,r). \end{aligned}$$

(A.1)

On the other hand, since $0\le 1-t_p\le Mt_1$ and $T\cap \bigcup _{m=0}^{M-1}(I+mt_1)$ is a union of translations of I, then for small enough r, $T\cap B(1-t_p,r)$ is a half-ball or a translation of $I\cap B(0,r)$, or a translation of $I\cap B(t_1,r)$, which contradicts with the shape of $T\cap B(1-t_p,r)$ in (A.1). The lemma is proven. $\square$