# 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

## 1 Answer

[The following is not quite an answer, but it refutes a natural generalization suggested in the Comments, and is too long to be a comment itself.]

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

(I learned about this from Francisco Santos's accepted answer to Timothy Chow's Mathoverflow question 49915, which the MO algorithm helpfully put at the top of its list of questions "Related" to this one.)

• 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
• The question has been floating around for quite some time; this is the best answer so far. The 3-dim. case will have to wait a bit longer. – Wlodek Kuperberg Apr 12 '17 at 17:24
• Thanks. I thought about some way to get down to 3 dimensions but didn't find anything. I guess one thing to ask is what's the smallest dimension to which this construction applies (the GLT paper must yield some bound but maybe in a specific case one can do significantly better). – Noam D. Elkies Apr 12 '17 at 18:05