Yes, that's exactly (one way that) I solved this, and agrees with what Fred Lunnon wrote — and also Allan Wechsler had another solution that works in Hilbert space. Note that the Voronoi regions* of the integer points in R^3 are unit cubes. The way I came to this originally was from realizing that if we define the "3D plus sign" as the union of the 7 Voronoi cubes Plus = V(0) ∪ V(e_1) ∪ V(-e_1) ∪ V(e_2) ∪ V(-e_2) ∪ V(e_3) ∪ V(-e_3) (where {e_1,e_2,e_3} is the standard basis of R^3), then appropriate translates of Plus will tessellate R^3. One lattice of translations that works is generated by the vectors {(2,-1,0), (0,2,-1), (-1,0,2)}. I think the Voronoi region of this lattice is a (non-uniform) truncated octahedron, but I'm not sure yet. —Dan ————— * For a discrete point set X in R^n, the Voronoi region V(x) of x ∊ X is V(x) = {v ∊ R^n | ‖x-v‖ ≤ ‖v-y‖ for all y ∊ X} — the points of R^n at least as close to x as to any other y ∊ X. Scott Huddleston wrote: ----- Yes. Let L = { (a_1, a_2, a_3) | a_1 + 2a_2 + 3a_3 = 0 mod 7 } This also generalizes to d dimensions, using L = { (a_1, a_2, ..., a_d) | a_1 + 2a_2 + ... + d*a_d = 0 mod 2d+1 } I wonder what the Voronoi regions look like. On Tue, Sep 22, 2020 at 11:01 AM Dan Asimov <dasimov@earthlink.net> wrote:

Let L be any discrete subgroup of R^3 isomorphic to Z^3.

The the quotient R^3/L is a "flat 3-torus", and it carries a well-defined metric. Its metric can be be described as the result of identifying opposite faces of a parallelepiped (the bounded intersection of three slabs* in R^3) in the obvious way.

Of course the cubical 3-torus R^3/Z^3 is one example, but a flat 3-torus can have surprising properties. Note that the cubical 3-torus can be tessellated by 8 cubes each touching three of the other seven along two common square faces.

Puzzle: ------- Does there exist a flat 3-torus that can be tessellated by seven cubes each touching the other six along a common face?

—Dan

————— * A "slab" in R^3 is the closed region between two parallel planes.