# Regular tessellations of Euclidean spacesMay 2, 2010 9:39 AM   Subscribe

MathFilter: Is there a relationship between the number of dimensions in a Euclidean space and the number of regular tessellations of that space?

For example, there are 3 regular tessellations of the Euclidean plane, only 1 regular tessellation of Euclidean 3-space, and 3 regular tessellations of Euclidean 4-space. Is there a general formula for the number of regular tessellations of Euclidean n-space? Can anything else be said about the relationship (e.g. the types of tessellations)?
posted by jedicus to Science & Nature (10 answers total) 2 users marked this as a favorite

Best answer: This suggests (at the end of the first complete paragraph) that for n>=5, there is only 1 regular tesselation.
posted by alligatorman at 9:53 AM on May 2, 2010 [2 favorites]

This person has been working on some applications related to this (I'm not familiar with the mathematical principles I confess).
posted by a womble is an active kind of sloth at 9:55 AM on May 2, 2010

To my knowledge you have at least two regular tessellations of Euclidian 3-space, one with cubes and one with regular tetrahedrons.

The knowledge field where you are looking at is topology... IIRC the answer to your question is related to the Euler-Poincaré, and it entails a number of invariants that constraints the way you build tessellations. Unfortunately I can't go further myself, just an advice to look towards topology.
posted by knz at 10:34 AM on May 2, 2010

Response by poster: To my knowledge you have at least two regular tessellations of Euclidean 3-space, one with cubes and one with regular tetrahedrons.

Ah, no, tetrahedrons do not tessellate Euclidean 3-space. "It is well known that three-dimensional Euclidean space cannot be tiled by regular tetrahedra." Source. I wish I could've found a proper proof, though.
posted by jedicus at 10:47 AM on May 2, 2010

This might be a good question to also post over at http://mathoverflow.net/
posted by artlung at 10:58 AM on May 2, 2010 [3 favorites]

knz is correct. Regular tetrahedrons do not fit together to fill 3-space.
However, rhombic dodecahedrons do fit together to fill 3-space.
posted by hexatron at 12:30 PM on May 2, 2010

Best answer: How are you defining a regular tessellation? A tessellation using a regular polytope as a tile? My guess is that for dimensions 5 and higher, there's only one regular tessellation of that sort, since for dimenstions 5 and higher there are only 3 regular polytopes (simplex, cube, crosspolytope), and I'm pretty sure neither the simplex nor the crosspolytope tiles.

Coxeter's Regular polytopes would be a possible source for some of this.

I wouldn't consider the tiling of three space by rhombic dodecahedra to be a regular tiling, just as a tiling of the plane by parallelograms is not regular, although it is monohedral.
posted by leahwrenn at 1:37 PM on May 2, 2010 [1 favorite]

I agree that this is a perfect mathoverflow question.
posted by madcaptenor at 5:11 PM on May 2, 2010

I wish I could've found a proper proof, though.
There's enough information in the paper you linked to sketch one. Look at the first figure. (1) The angle between adjacent faces of a regular tetrahedron is about 70.5°, which does not divide 360°. Therefore no space-filling tessellation involves regular tetrahedra packed around their edges. (2) The solid angle subtended by a tetrahedron's vertex is about 0.55 steradians, which does not divide 4π. Therefore no space-filling tessellation involves regular tetrahedra packed around a vertex. (Lemma:) Any space-filling tessellation of polyhedra involves packing the polyhedra around their edges, their vertices, or both. So we can't do this with regular tetrahedra.
posted by fantabulous timewaster at 6:45 AM on May 3, 2010 [1 favorite]

Regular tetrahedra don't tile 3-space. But right tetrahedra -- that is, tetrahedra with their corners at (0, 0, 0), (0, 0, 1), (0, 1, 0), and (1, 0, 0) -- do. Six of them fill a unit cube, and we know how to fill space with unit cubes. So this tells you that whether copies of a given tetrahedron fill space or not is a geometric fact, not a graph-theoretic one.
posted by madcaptenor at 7:03 AM on May 3, 2010

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