10
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ 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 $\mathbb{R}^{100}$? $\endgroup$
    – Boris Bukh
    Feb 14, 2017 at 2:04
  • 2
    $\begingroup$ @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. $\endgroup$ Feb 14, 2017 at 2:13
  • 2
    $\begingroup$ @AnthonyQuas: You guess correctly. And there is a lot more that I am not assuming. $\endgroup$ Feb 14, 2017 at 4:18
  • 3
    $\begingroup$ Are there non-convex counterexamples? $\endgroup$ Apr 5, 2017 at 17:26
  • 3
    $\begingroup$ 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. $\endgroup$ Apr 5, 2017 at 19:20

1 Answer 1

10
+50
$\begingroup$

[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.)

$\endgroup$
4
  • $\begingroup$ May I ask: What does it mean for a set in $\mathbb{R}^n$ to "tile" $\mathbb{R}^d$ when $d > n$? $\endgroup$ Apr 6, 2017 at 12:33
  • $\begingroup$ 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. $\endgroup$ Apr 6, 2017 at 14:28
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 12, 2017 at 17:24
  • 1
    $\begingroup$ 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). $\endgroup$ Apr 12, 2017 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.