Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Basic definitions: a tiling of d-dimensional Euclidean space is a decomposition of that space into polyhedra such that there is no overlap between their interiors, and every point in the space is contained in some one of the polyhedra.

A vertex-uniform tiling is a tiling such that each vertex figure is the same: each vertex is contained in the same number of k-faces, etc: the view of the tiling is the same from every vertex.

A vertex-transitive tiling is one such that for every two vertices in the tiling, there exists an element of the symmetry group taking one to the other.

Clearly all vertex-transitive tilings are vertex-uniform. For n=2, these notions coincide. However, Grunbaum, in his book on tilings, mentions but does not explain that for n >= 3, there exist vertex uniform tilings that are not vertex transitive. Can someone provide an example of such a tiling, or a reference that explains this?

share|improve this question
Could you clarify vertix-transitive a bit more? –  Casebash Jul 21 '10 at 6:15
I'm not sure I understand your definition of vertex-uniform. Could you clarify? –  Qiaochu Yuan Jul 28 '10 at 7:45
ok, sorry guys; closed this question because I was being stupid and, as asked, it's not a real question. –  Jamie Banks Aug 5 '10 at 23:08

3 Answers 3

up vote 0 down vote accepted

Transitive action on the vertices is usually the definition of "looking the same from each vertex". So maybe you have two different groups in mind:

  1. The automorphism group of the combinatorics of the tiling; the abstract structure of vertices, faces, edges, etc.

  2. The group of Euclidean motions that leave the tiling invariant. Motions meaning isometries of space, where one should also specify whether orientation-reversal is allowed or not. Any instance of this is also an automorphism of the tiling combinatorics.

If tilings that are vertex-transitive under group #1 are "vertex-uniform" and those that are vertex-transitive in the stricter sense of group #2 are "vertex transitive", then it is easy to give examples where some of the polyhedra are deformations of others (so not isometrically transitive in sense #2). For example, a one-dimensional periodic tiling with several different sizes of interval will have all vertices equivalent combinatorially but a finite number of distinct vertex types under geometric equivalence. This is maybe too simple to be what Grunbaum had in mind, can you quote the book?

share|improve this answer

The Wikipedia article on honeycombs has several examples of 3d tilings by a single polyhedron which aren't vertex-transitive. I am not positive if they are vertex-uniform, though, since (again) I don't really understand your definition and can't find a good one online.

share|improve this answer

Let me discuss an analogous situation with regard to convex polyhedra.

Archimedes, generalizing what are often called the Platonic Solids or the convex regular polyhedra (there are five of them), seems to have discovered 13 polyhedra with the property that the pattern of faces around each vertex was the same for every vertex of the polyhedron, and all of the faces of the polyhedron were regular polygons. I say seems to have done this because the manuscript that describes what he did is lost. Pappus who lived many years after Archimedes describes these polyhedra explicitly and mentions 13 convex solids. Years latter many artists and mathematicians talked about these polyhedra. Kepler explicitly mentions two infinite families of polyhedra which have this property: the prisms (consisting of two regular n-gons and n squares) and the anti-prisms (consisting of two regular n-gons and 2n equilateral triangles). Kepler purports to give a proof a proof that there 13 such solids (with at least two types of faces). I say purports because there are in fact 14 convex polyhedra which meet the local symmetry condition described above. The one Archimedes, Pappus, Kepler and others missed is often called the pseudo-rhombicuboctahedron. (Kepler created confusion in referring to 14 solids in a place other than where he gave his "proof.")


Many books to this day continue to talk about 13 Archimedean convex solids (other than the prisms and antiprisms). With the definition that Archimedes almost certainly had in mind this is wrong - there are 14 such solids. However, if one considers convex polyhedra with at least two regular polygons as faces, and where the symmetry group of the solid is transitive on the vertices (that is, the symmetry group can take any vertex to any other vertex) then there are only 13 such solids. However, almost certainly Archimedes had no knowledge of the idea of a symmetry group in the modern sense. Depending on whether one uses a local symmetry notion or a global symmetry notion there are either 14 or 13 such convex solids (aside from the prisms and antiprisms).

Branko Grünbaum has a nice article about this called "An enduring error."


share|improve this answer
interesting! Thanks. –  Jamie Banks Aug 28 '10 at 21:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.