18
$\begingroup$

Any (non-self-intersecting) quadrilateral tiles the plane.

QuadTilings
    (MathWorld image.)

Q. What is the strongest known generalization of this statement to higher dimensions? I.e., $\mathbb{R}^d$ filling with combinatorial cuboids? Is Michael Goldberg's 37-yr-old paper the latest in $\mathbb{R}^3$?

Goldberg, Michael. "On the space-filling hexahedra." Geometriae Dedicata 6.1 (1977): 99-108.

GoldbergSnippet
    (Snippet from Goldberg.)

Improve this question
$\endgroup$
7
  • 10
    $\begingroup$ Any polyhedron which tiles must have $0$ Dehn invariant, which doesn't seem easy to force with a combinatorial type, so that should be an additional requirement. $\endgroup$
    – Douglas Zare
    Commented Sep 13, 2014 at 1:01
  • 2
    $\begingroup$ @DouglasZare: I don't doubt you but nor do I see immediately why the Dehn invariant must be zero. Could you elaborate a bit? (And if you put that in an answer, I'll accept it.) $\endgroup$
    – Joseph O'Rourke
    Commented Sep 13, 2014 at 12:21
  • 5
    $\begingroup$ Tile a large roughly round region of radius $r$. The Dehn invariant is additive so the Dehn invariant of the union of tiles is "proportional" to $r^3$. However, the Dehn invariant is also carried by the surface, and only finitely many angles can occur, so it is bounded by $\pm cr^2$. This forces the Dehn invariant to be $0$. You have to be a little careful since the Dehn invariant is not defined to be in $\mathbb R$, but in $\mathbb R/\pi \otimes \mathbb R.$ By the way, the analogue fails for hyperbolic space which is tiled by arbitrarily small tiles with arbitrary Dehn invariant. $\endgroup$
    – Douglas Zare
    Commented Sep 13, 2014 at 12:25
  • $\begingroup$ @DouglasZare: could you please explain how tilings of hyperbolic space with non-trivial Dehn invariant arise? I tried to find some literature, but without success. The only thing I know is that if the tiling is by fundamental domains for a torsion-free subgroup $\Gamma\subseteq SO(n,1)$ such that $\mathbb{H}^n/\Gamma$ is a hyperbolic manifold of finite volume, then the Dehn invariant is zero, by a theorem of Goncharov. $\endgroup$
    – Matthias Wendt
    Commented Oct 30, 2014 at 16:15
  • $\begingroup$ @Matthias Wendt: AFAIK, the construction is mine, but not terribly surprising to some. I mentioned it in a few places but didn't publish it. Start with a "horobrick," something like a fundamental domain for a Baumslag-Solitar group, or higher dimensional generalizations, between two horospheres. For example, take a pentagon with $3$ vertices on one horocircle, and $2$ vertices on another concentric horocircle. You can add a bump on one side, and take away a copy of that bump from the other two sides to get polygons of arbitrary areas which tile. There are $2$-reptiles in higher dimensions. $\endgroup$
    – Douglas Zare
    Commented Oct 30, 2014 at 17:42

1 Answer 1

Reset to default
8
$\begingroup$

I just stumbled over this, maybe you find it useful:

As Douglas Zare already discussed in his comments, a polytope tiling $\mathbb{R}^n$ must have Dehn invariant zero. This statement has appeared at least twice in the literature:

Once here: H. Debrunner: Über Zerlegungsgleichheit von Pflasterpolyedern mit Würfeln. Arch. Math. 35 (1980) 583-587.

And once here: J.C. Lagarias and D. Moews: Polytopes that fill $\mathbb{R}^n$ and scissors congruence. Discrete and computational geometry 13 (1995), 573-583

A similar problem for tilings by translations has been solved by H. Hadwiger: Mittelpunktspolyeder und translative Zerlegungsgleichheit. Math. Nachr. 8 (1952) 53-58. This was also reproduced by Lagarias and Moews in the paper mentioned above.

See also the acknowledgement of priority by Lagarias and Moews, which discusses some of the history, all the literature references are taken from there.

I have not found papers discussing if triviality of Dehn invariants is sufficient for tiling (at least in dimensions 3 and 4, where Dehn invariant and volume completely characterize scissors congruence). (Maybe it is possible to use the scissors congruence to the cube together with the cube tiling to prove the converse and construct a tiling?)

Improve this answer
$\endgroup$
2
  • $\begingroup$ Great trove of relevant references---Thanks! $\endgroup$
    – Joseph O'Rourke
    Commented Sep 20, 2014 at 12:36
  • $\begingroup$ Dehn invariant zero is not sufficient for tiling, I believe: all polycubes have dihedral angles of $\pi/2$ or $3\pi/2$, but there are polycubes homeomorphic to the ball which do not tile space (consider a $3\times 3\times 3$ cube with the center of one of the faces removed). $\endgroup$
    – RavenclawPrefect
    Commented Jun 25, 2021 at 23:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .