Rep-tiles

Rep-tiles induce self-similar tilings of Euclidean space. Thus, they can potentially be used to construct non-periodic and aperiodic tilings of the plane and higher-dimensional Euclidean spaces. Self-similar tilings have connections to combinatorial group theory [Con90], propositional logic [Wan60, Ber66, Rob71] (where some of the questions in the field originated), and dynamical systems [Thu89], among others. Our rep-tiles give new self-similar tilings of $\mathbb{R}^{n}$ by tiles with interesting topology. Additionally, in our proof of Theorem 1.2 we show that every compact smooth $n$-manifold with connected boundary in $\mathbb{R}^{n}$ is topologically isotopic to a polycube that tiles a cube. In [Gol66], Golomb developed a hierarchy for polycubes that tile $\mathbb{R}^{n}$, and tiling a cube is the most restrictive level of his hierarchy. Hence all compact smooth $n$-manifolds with connected boundary lie in the most restrictive level of Golomb’s hierarchy, up to isotopy.

A stack of $n$-cubes with stacking direction $x_{n}$ is an $n$-dimensional polycube $S\subset\mathbb{R}^{n}$ such that: (1) All $n$-cubes in $S$ lie above the hyperplane $x_{n}=0$ (that is, every point in $S$ has non-negative $x_{n}$ coordinate), and (2) for every $n$-cube in $S$ that does not have a face contained in $x_{n}=0$, there is another $n$-cube of $S$ directly below it (where height is measured by the $x_{n}$-axis).

Let the subspace of $\mathbb{R}^{n}$ determined by $x_{n}=0$ have the standard tiling by $(n-1)$-cubes induced by the integer lattice in $\mathbb{R}^{n}$. Given $S$, a stack of $n$-cubes with stacking direction $x_{n}$, we consider its projection to the hyperplane $x_{n}=0$, which we call its footprint. By the definition of a stack of cubes, we can think of $S$ as consisting of columns of $n$-cubes lying above each $(n-1)$-cube in its footprint $\mathcal{F}_{S}$, which is itself an $(n-1)$-dimensional polycube. In other words, the homotopy type of $S$ is determined by $\mathcal{F}_{S}$; and $S$ itself is determined by $\mathcal{F}_{S}$, together with integer labels in each $(n-1)$-cube of $\mathcal{F}_{S}$, specifying the height of the column of $n$-cubes which lie above it. Therefore, we can describe $S$ by such a labeled footprint. Figure 2 illustrates a 2-dimensional stack of cubes (left) and its description via a labeling on its 1-dimensional footprint (right).

The image of such a stack of cubes under an isometry of $\mathbb{R}^{n}$ is also called a stack of cubes, with the image of $x_{n}$ under the isometry being the stacking direction.

We use cube-stacking notation as above to describe a rep-tile homeomorphic to $S^{n}\times D^{2}$ for all $n\geq 0$. This description is a simplification, suggested by Richard Schwartz [Sch25], of the construction given in [Bla24]. We define a stack of cubes $S\subset[0,4]^{n+2}$ as follows. The footprint $\mathcal{F}_{S}$ is a polycube in $[0,4]^{n+1}$.

Throughout the following discussion, the reader should refer to Figure 3. Define the core, denoted $\mathcal{C}$, of $[0,4]^{n+1}$ to be the union of unit cubes in the standard integer-lattice tiling of $[0,4]^{n+1}$ containing the point $(2,...,2)$. The shell of $[0,4]^{n+1}$ is $\overline{[0,4]^{n+1}\setminus\mathcal{C}}.$ To create the labeled footprint $\mathcal{F}_{S}$ of our stack of cubes $S$, we first partition $\mathcal{C}$ into two halves: $\mathcal{C}^{+}$, those containing cubes with $x_{n+1}$-coordinate at least 2; and $\mathcal{C}^{-}$, those containing cubes with $x_{n+1}$-coordinate less than 2. Finally, we label each cube in $\mathcal{C}^{+}$ with a 4, and each cube in $\mathcal{C}^{-}$ with a 0. All cubes in the shell are labeled 2. (We recall that the label of each $(n+1)$-cube in the footprint indicates the height of the column of $(n+2)$-cubes stacked on top of it.) Observe that $\mathcal{F}_{S}$, which consists of all unit cubes in $[0,4]^{n+1}$ with nonzero label, is homeomorphic to the shell, which is in turn homeomorphic to $S^{n}\times D^{1}$. Similarly, the stack of cubes $S$ determined by this labeling is homeomorphic to $\mathcal{F}_{S}\times I\cong S^{n}\times D^{2}$.

Next we show that $S$ is a rep-tile. Let $r_{\pi}:\mathbb{R}^{n+2}\rightarrow\mathbb{R}^{n+2}$ denote rotation by $\pi$ about the $n$-plane which is the intersection of $x_{n+2}=2$ and $x_{n+1}=2$. Observe that the closure of the complement of $S$ in $[0,4]^{n+2}$ is also a stack of cubes, with stacking direction $-x_{n+2}$, is isometric to $S$, and in particular, is the image of $S$ under $r_{\pi}.$ As $S$ and $r_{\pi}(S)$ tile the cube $[0,4]^{n+2}$, and since $S$ is a union of cubes, $S$ is a rep-tile.

We note that there is a lot of flexibility regarding the heights of columns in the construction of a rep-tile $S\cong S^{n}\times D^{2}$ given above. Consider any column $H$ in $S$ of height $h\in\{1,2,3\}$. Let $H^{\prime}$ denote the column of $S$ which shares a footprint with $r_{\pi}(H)$. Since $H^{\prime}\cup r_{\pi}(H)$ form a column of height 4, the heights of $H$ and $H^{\prime}$ add up to 4. Moreover, unit cubes can be traded between $H$ and $H^{\prime}$ while preserving the property that the resulting polycube and its image under $r_{\pi}$ tile $[0,4]^{n+2}$. As long as both columns remain of height strictly between 0 and 4 and their heights add up to 4, this swap preserves both the homeomorphism type of $S$ and the property that two copies of $S$ tile a cube.

More generally, let $R$ be any non-empty $n$-dimensional polycube in $\mathbb{R}^{n}$. Let $G$ be a group of isometries of $\mathbb{R}^{n}$ such that the orbit of $R$ under $G$ tiles a cube $\mathcal{C}$. (As before, this implies that $R$ is a rep-tile.) Let $u$ denote any unit cube contained in $R$ and let $g$ be an arbitrary element of $G$. Denote by $gu$ the image of $u$ under $g$. We note that $R^{\prime}:=\overline{R\backslash u}\cup gu$ is also a polycube whose orbit under $G$ is $\mathcal{C}$. Hence, $R^{\prime}$ is also a rep-tile. We will refer to this move as a ball swap. (Aside: the fact that performing a ball swap on a polycube that tiles an $n$-cube produces another polycube that tiles an $n$-cube does not depend in an essential way on the fact that $u$ is a $n$-cube. More complicated pieces could be swapped as well, preserving the tiling property.) Ball swapping, which was inspired by work of Adams [Ada95, Ada97], turns out to be a powerful tool for building rep-tiles, as we will see in Section 3. To be precise, a version of the ball swap – one which involves an action of a group of order $2^{m}$ on $[-1,1]^{m}$ and trading multiple balls simultaneously across their individual orbits – is the key idea in the Proof of Theorem 1.2. The second main ingredient in the proof is this: a priori, $R^{\prime}$ might not have a clear relationship to $R$; to guarantee that $R^{\prime}$ is homeomorphic to or isotopic to $R$, care must be taken in the choice of group action and the choice of $u$.

Let $P_{n}=\{\textbf{x}\in\mathbb{R}^{n+2}|x_{n+2}=x_{n+1}=2\}$. The construction in Section 2.2 has produced stacks of $(n+2)$-dimensional cubes in $[0,4]^{n+2}$ with the following useful properties:

1. (1)

   Each polycube intersects $x_{1}=0$ in an $(n+1)$-ball equal to $\{0\}\times[0,4]^{n}\times[0,2]$ and intersects $x_{1}=4$ in an $(n+1)$-ball equal to $\{4\}\times[0,4]^{n}\times[0,2]$;

2. (2)

   the polycube and its image under rotation by $\pi$ about $P_{n}$ tile $[0,4]^{n+2}$.

Note that any two such polycubes of the same dimension $R_{1}$ and $R_{2}$ can be placed side-by-side in the $x_{1}$ direction so that $R_{1}$ is contained in $0\leq x_{1}\leq 4$ and $R_{2}$ is contained in $4\leq x_{1}\leq 8$. For example, place two copies of the stack of cubes in the top of Figure 3 back-to-back. In this configuration $R_{1}\cap R_{2}=\{4\}\times[0,4]^{n}\times[0,2]\cong B^{n+1}$. Thus, $R_{1}\cup R_{2}$ has the homotopy type of the wedge $R_{1}\vee R_{2}$; and, after rescaling in the $x_{1}$ direction and subdividing the integer lattice, it too satisfies the conditions (1) and (2) above.

Now consider $S_{m}$ and $S_{k}$, two of the rep-tiles constructed in Section 2.2 of dimension $m$ and $k$ respectively. If $m\leq k$, then $S_{m}\times D^{k-m}$ can be embedded in $[0,4]^{k+2}$ so that conditions (1) and (2) hold. By stacking $S_{k}$ and this embedding of $S_{m}\times D^{k-m}$ as in the previous paragraph, we construct a rep-tile in the homotopy type of $S^{m}\vee S^{k}$, itself capable of becoming part of a further rep-tilean wedge. By iterating this process, rep-tiles in the homotopy type of any finite wedge of spheres can be constructed.

Let $r_{\pi}$ be an order 2 rotation about some $(n-2)$-subspace in $\mathbb{R}^{n}$. We note that if $R$ is any connected $n$-dimensional stack of cubes such that two copies of $R$, related by $r_{\pi}$, tile an $n$-cube, then $R$ can be used to construct an $(n+1)$-dimensional rep-tile in the homotopy type of the suspension of $R$. We sketch this construction with a specific choice of coordinates below. For clarity, we assume that $R\cup r_{\pi}(R)$ tile the cube $[0,4]^{n}$.

Let $R$ denote any $n$-dimensional stack of unit cubes which has the property that $R$ and its image under under $r_{\pi}$ tile $[0,4]^{n}$. (For instance, $R$ could be one of the rep-tiles in the homotopy type of a wedge of spheres that we previously constructed.) Because $R\cup r_{\pi}(R)=[0,4]^{n},$ we know that $r_{\pi}$ takes cubes at height 4 (with respect to the stacking direction) to holes at height zero; and vice-versa. In particular, $R$ contains as many cubes at height 4 as it has unit-cube-sized holes at height 0. Therefore, we may suspend $R$ as by the following steps

1. (1)

   embed $R\times[0,4]$ into $\mathbb{R}^{n+1}$;

2. (2)

   cubify in the natural way, writing $R\times[0,4]$ as a union of unit $(n+1)$-cubes of the form $(n\text{-cube in }R)\times[i,i+1]$;

3. (3)

   move all height-4 cubes in $R\times[0,1]$ to fill all holes at height zero in that slice;

4. (4)

   repeat the last step in $R\times[3,4]$.

Crucially, steps 3 and 4 constitute ball swaps (see Section 2.3). This guarantees that the resulting polycube is still a rep-tile. Moreover, since the slice $R\times[i,i+1]$ is a stack of cubes, filling all cubes that correspond to “height-0 holes” in $R\times[i,i+1]$ (that is, those holes in $R\times[i,i+1]$ which are height-0 holes in $R$ crossed with $[i,i+1]$) turns $R\times[i,i+1]$ into a ball. Therefore, as before, ball swapping in the first and last slices of $R\times[0,4]$ has the effect, up to homotopy, of contracting each of the ends of $R\times[0,4]$ to a point. This completes the suspension of $R$. Figures 4 and 5 illustrate the suspensions of rep-tiles homeomorphic to $S^{0}\times D^{2}$ and $S^{1}\times D^{2}$, respectively.

Let $H$ and $H^{\prime}$ denote any pair of columns in Figures 4 or 5 which trade a cube during the ball-swapping operations. Specifically, say $H^{\prime}$ is height 0 and the top cube of $H$ is moved to $H^{\prime}$ during the ball swapping. Now suppose next highest cube of $H$ (now at height 3) is also moved to column $H^{\prime}$. This would constitute a ball swap, since the unit cube remains within its orbit under the rotation. Executing this additional swap between all such pairs has the effect that all columns of heights 3 and 1 become columns of height 2. The result would be the $S^{n}\times D^{2}$ rep-tile constructed in Section 2.2. Put differently, rep-tiles homeomorphic to $S^{n}\times D^{2}$ can also be obtained from the $S^{0}\times D^{2}$ rep-tile in Figure 2 inductively, via a sequence of suspensions and ball swaps. For details on this approach, see Section 2 of [Bla24].

The following was observed by Richard Schwartz [Sch25] while perusing the first version of our article.

This result, together with the existence of cubifications for smooth codimension-0 submanifolds of $\mathbb{R}^{n}$ (see Section 3.5) can be used to prove a version of Corollary 1.2.1. Specifically, we see that it is possible to realize the homotopy type of any compact $n$-dimensional CW complex as a $(2n+2)$-dimensional rep-tile $R$, without appealing to Theorem 1.2. The present approach uses an extra dimension; but it is rather explicit (given a polycube footprint to start with) and has the advantage that just 2 copies of $R$ can tile the $(2n+2)$-cube.

The main ingredient is Theorem  3.3, which we prove using a strategy we refer to as a ball swap. To start, $R$ is smoothly embedded in $C^{n}=[0,1]^{n}$ so that $R\cap\partial C^{n}=\emptyset$. In turn, the unit cube $C^{n}$ sits inside the cube $\boxplus=[-1,1]^{n}$. Since $R$ is disjoint from $\partial C^{n}$ and has a single boundary component, $C^{n}\setminus R$ is connected. By Theorem  3.4, we may decompose $\overline{C^{n}\setminus R}$ into $n$ $n$-dimensional balls $B_{1},\dots,B_{n}$.¹¹1If $b(C^{n}\setminus R)<n$, one could use fewer balls here and tile the cube with fewer copies of $R$, but we use $n$ balls for simplicity in the proof of the main theorem. After a homotopy of $C^{n}$ which restricts to an isotopy on each piece of the decomposition $\{R,B_{1},\dots,B_{n}\}$ of $C^{n}$, we ensure that the pieces of this decomposition intersect an $(n-1)$-disk on $\partial{C}^{n}$ as shown in Figure 8, in what we call a taloned pattern. The defining features of taloned patterns include: there is an $(n-1)$-disk on $\partial C^{n}$ such that $R$ and each of $B_{1},\dots B_{n}$ intersect that disk in an $(n-1)$-ball and intersect the boundary of the disk in an $(n-2)$-ball; the $(n-1)$-balls $B_{i}$ are disjoint inside this disk; and $R$ is adjacent to each ball $B_{i}$ in this disk. (See Section 3.2 for the formal definition.) The homotopy used to create the taloned pattern is achieved in Lemmas 3.5 and 3.6 below. We then isotope $C^{n}$ so that the $(n-1)$-disk which constitutes the taloned pattern of Figure 8 is identified with the union of faces of $C^{n}=[0,1]^{n}$ whose interiors lie in the interior of $\boxplus=[-1,1]^{n}$, with certain additional restrictions. These restrictions guarantee that certain rotated copies of the $B_{i}$ contained in cubes adjacent to $[0,1]^{n}$ in $\boxplus=[-1,1]^{n}$ are disjoint, allowing us to form the boundary connected sum of $R$ with these balls without changing the isotopy class of $R$. Indeed, we give a family of rotations $r_{k}$, $1\leq k\leq\lfloor{n/2}\rfloor$, together with one additional rotation $f$ if $n$ is odd, such that the orbit of $C^{n}$ under these rotations tiles $\boxplus$. By taking the boundary sum of $R\subset C^{n}$ with the image of each $B_{i}$ under an appropriate choice of rotation above, we obtain the desired manifold $R^{*}$. By construction, $R^{*}$ is isotopic to $R$ and, moreover, the orbit of $R^{*}$ under the above set of rotations gives a tiling of $\boxplus$. A 2-dimensional tile $R^{*}$ created via ball swapping is shown in Figure 7. A sketch of $R^{*}$ in dimension $n=3$ is shown in Figure 14. Finally, we show that this construction can be “cubified”, so that $R^{*}$ is a polycube tiling $\boxplus$, completing the proof of Theorem 3.3. Once this is established, Theorem 1.2 follows from Lemma 3.1.

We define the desired boundary pattern described above. A $k$-claw is a tree which consists of one central vertex $v$ and $k$ leaves, each connected to $v$ by a single edge. See Figure 8.

Our goal is to construct a boundary pattern on $C^{n}$ such that there exists an embedded disk $D^{n-1}\subset\partial C^{n}$ with the following properties:

• •

  $D^{n-1}\cap B_{i}$ is a single $(n-1)$-disk, for all $1\leq i\leq k$;

• •

  $(D^{n-1}\cap B_{i})\cap\partial D^{n-1}$ is an $(n-2)$-disk, for all $1\leq i\leq k$;

• •

  $D^{n-1}\setminus(\cup_{i=1}^{k}B_{i}\cap D^{n-1})\subset R$.

• •

  $B_{i}\cap B_{j}\cap D^{n-1}=\emptyset$ for $i\neq j$.

We regard the boundary pattern as the regular neighborhood of a $k$-claw, with the following decomposition: $R$ contains a neighborhood of the central vertex; and each $B_{i}$ containing a neighborhood of a leaf. See Figure 8. We call this a taloned pattern of intersections.

We begin by proving Lemma 3.5, which ensures that, in the interior of $C^{n}$, the union of the boundaries of the pieces $\{R,B_{1},\dots,B_{n}\}$ in the interior of our decomposition of $C^{n}$ can be assumed to be connected.

We begin by setting up the necessary notation. For each $i=1,\dots,n$, let $F_{i}$ be the $(n-1)$-dimensional face of the $n$-cube $C^{n}$ contained in the hyperplane $x_{i}=0$. For the moment, we will assume that $n$ is even. The case of $n$ odd requires an extra step, which we leave until the end of the proof.

Let $r_{i}:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be the rotation by $\frac{\pi}{2}$ about the $(n-2)$-plane $x_{2i-1}=x_{2i}=0$ that carries the $x_{2i-1}-$axis to the $x_{2i}-$axis. Note that each $r_{i}$ has order four and that these rotations commute, generating a group isomorphic to $(\mathbb{Z}_{4})^{n/2}$. Given a vector $\mathbf{y}=(y_{1},\dots,y_{n/2})\in(\mathbb{Z}_{4})^{n/2}$, we define the rotation ${\bf r_{y}}$ as follows:

We set $C_{\mathbf{y}}:=\mathbf{r_{y}}(C^{n}).$

We claim that the orbit of a unit sub-cube under this group action is the entire $n$-dimensional cube $\boxplus:=[-1,1]^{n}$. In other words, $\boxplus$ is tiled by the $2^{n}$ distinct unit cubes $\{\mathbf{r_{y}}(C^{n})|\mathbf{y}\in(\mathbb{Z}_{4})^{n/2}\}.$

To see this, first decompose $\boxplus$ into $2^{n}$ unit sub-cubes of the form $J_{1}\times\dots\times J_{n}$, where each $J_{i}$ is either $[-1,0]$ or $[0,1]$. Fixing $k$, for each choice of $J_{2k-1}$ and $J_{2k}$ from the set $\{[-1,0],[0,1]\}$, the product $J_{2k-1}\times J_{2k}$ is a unit square in the $x_{2k-1}x_{2k}-$ plane, which we denote by $\mathbb{R}^{2}_{k}.$ Let $P_{k}:=C^{n}\cap\mathbb{R}^{2}_{k}$, i.e $P_{k}$ is the unit square in the first quadrant of $\mathbb{R}^{2}_{k}$. Then $J_{2k-1}\times J_{2k}=r^{y_{k}}(P_{k})$ for some $y_{k}\in\{0,1,2,3\}$. Hence, each of the $2^{n}$ unit cubes above can be expressed as

for some $\mathbf{y}=(y_{1},\dots,y_{n/2})\in(\mathbb{Z}_{4})^{n/2}$. Moreover, for each $J_{1}\times\dots\times J_{n}$, the $\mathbf{y}$ such that $J_{1}\times\dots\times J_{n}=C_{\mathbf{y}}$ is unique. To see this, note that each $J_{1}\times\dots\times J_{n}$ has exactly one corner with all nonzero coordinates (and therefore with all coordinates $\pm 1$). On the other hand, the cube $C_{\mathbf{y}}$ also has exactly one corner $(c_{1},\dots,c_{n})$ with all $c_{i}=\pm 1$ (namely, the image of the point $(1,1,1,\dots,1)\in C^{n}$), and its coordinates satisfy the formula $y_{k}=-(c_{2k}-1)-\frac{1}{2}(c_{2k-1}c_{2k}-1).$ In other words, the coordinates $(c_{1},\dots c_{n})$ uniquely determine each component $y_{k},$ and therefore $\mathbf{y}$ itself.

Observe that the cube $r_{k}(C^{n})$ intersects $C^{n}$ along its face $F_{2k-1}$, and the cube $r^{-1}_{k}(C^{n})$ intersects $C^{n}$ along its face $F_{2k}$. Thus, each rotation $r_{k}$ gives a pairing of the faces of $C^{n}$. We use this pairing to carry out a ball swap as previously described. This will allow us to build the rep-tile $R^{*}$.

We will now describe a homotopy of $C^{n}$ which restricts to an isotopy on the interiors of $R$ and the balls $B_{1},\dots,B_{n}$. Our goal is to use Lemma 3.6 to position $R$ and $B_{1},\dots,B_{n}$ so that their intersections with the boundary of $C^{n}$ satisfy:

1. (1)

   For each $1\leq i\leq n$, the only ball meeting the face $F_{i}$ is $B_{i}$ (and thus $F_{i}\setminus(B_{i}\cap F_{i})\subset R$),

2. (2)

   $r_{k}(B_{2k}\cap F_{2k})\subset F_{2k-1}$ is disjoint from $B_{2k-1}$, and

3. (3)

   $r_{k}^{-1}(B_{2k-1}\cap F_{2k-1})\subset F_{2k}$ is disjoint from $B_{2k}$.

In what follows, we refer the reader to a schematic in Figure 12. Figure 13 illustrates this configuration in dimension 4.

For each $k=1,\dots,\frac{n}{2}$, let $\varphi_{2k-1}$ be the $(n-2)$-facet in $C$ equal to the intersection of $C$ with the $(n-2)$-plane given by setting $x_{2k-1}=0$ and $x_{2k}=1$. Likewise, let $\varphi_{2k}$ be the $(n-2)$-facet in $C$ equal to the intersection of $C$ with the $(n-2)$-plane given by setting $x_{2k-1}=1$ and $x_{2k}=0$. Note that this pair of facets are exactly those that are simultaneously parallel to the intersection $F_{2k-1}\cap F_{2k}$ and contained in $F_{2k-1}\cup F_{2k}$.

Now, we fix points $\alpha_{2k-1}\in\varphi_{2k-1}$ and $\alpha_{2k}\in\varphi_{2k}$ by setting