# Regular tessellations of Euclidean spaces

May 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

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)?

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

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

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:

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

*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]

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

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

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]

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

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

posted by madcaptenor at 7:03 AM on May 3, 2010

This thread is closed to new comments.

posted by alligatorman at 9:53 AM on May 2, 2010 [2 favorites]