# Higher dimensional generalization of: Any quadrilateral tiles the plane? (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. (Snippet from Goldberg.)

• 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. Sep 13, 2014 at 1:01
• @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.) Sep 13, 2014 at 12:21
• 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. Sep 13, 2014 at 12:25
• @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. Oct 30, 2014 at 16:15
• @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. Oct 30, 2014 at 17:42

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:
• 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). Jun 25, 2021 at 23:11