# Space-tiling convex prisms

A convex prism is a subset of $\mathbb{R}^3$ congruent to the Cartesian product of a convex polygon (the prism's base) with the interval $[0,1]$.

Question. If a family of congruent convex prisms tiles space (not necessarily in a face-to-face manner), must there exist a tiling of the plane with polygons congruent to the prism's base?

• Is there anything special about $\mathbb{R}^3=\mathbb{R}^2\times \mathbb{R}^1$? Do you have counterexamples for products of $2$-dimensional base and $[0,1]^{98}$ in $\mathbb{R}^{100}$ for example? Or proofs for a base in $\mathbb{R}^{99}$ and $[0,1]$ in $\mathbb{R}^{100}$? – Boris Bukh Feb 14 '17 at 2:04
• @BorisBukh: No, I do not. But I think $\mathbb{R}^2\times\mathbb{R}^1$ is an interesting case, easy to visualize. Also, a convex polygon with more than six sides cannot tile the plane. – Wlodek Kuperberg Feb 14 '17 at 2:13
• @AnthonyQuas: You guess correctly. And there is a lot more that I am not assuming. – Wlodek Kuperberg Feb 14 '17 at 4:18
• Are there non-convex counterexamples? – Noam D. Elkies Apr 5 '17 at 17:26
• Yes, Noam. A $3\times4$ rectangle with a $1\times2$ rectangular hole. The prism of height 1 tiles space. The example can be cut into two congruent pieces to make a simply-connected example. Also, if we allow the prism to be an affine image of the product, i.e., a slant prism, then convex counterexamples exist. – Wlodek Kuperberg Apr 5 '17 at 19:20

Counterexample in ${\bf R}^N \times {\bf R}$ for some $N>2$: any lattice hexagon $H$ with angles $90^\circ$, $90^\circ$, $135^\circ$, $135^\circ$, $135^\circ$, $135^\circ$ in that order. (That is, $H$ is obtained from a lattice rectangle by truncating two adjacent vertices by two isosceles right lattice triangles, not necessarily congruent. Alternatively, remove two congruent lattice triangles, not necessarily isoceles, related by a 90-degree rotation.) Then $H$ does not tile the plane, but four copies do tile a (non-simply-connected) polyomino. But it was recently shown that any polyomino in some ${\bf Z}^n$ (which need not be simply connected, or even connected at all!) tiles ${\bf Z}^d$ for some $d$:
Vytautas Gruslys, Imre Leader, and Ta Sheng Tan: Tiling with arbitrary tiles. Proc. London Math. Soc. (2016) 112 (6): 1019-1039. https://doi.org/10.1112/plms/pdw017 $\cong$ http://arxiv.org/abs/1505.03697
• May I ask: What does it mean for a set in $\mathbb{R}^n$ to "tile" $\mathbb{R}^d$ when $d > n$? – Joseph O'Rourke Apr 6 '17 at 12:33
• These are polyominos, i.e. finite subsets of the tiling of ${\bf R}^n$ by ${\bf Z}^n$ translates of $[0,1]^n$. If $d>n$, a polyomino in ${\bf Z}^n$ can be regarded as a polyomino in ${\bf Z}^d$ by taking its product with $[0,1]^{d-n}$. This is consistent with the prisms in Wlodek Kuperberg's question. – Noam D. Elkies Apr 6 '17 at 14:28