# what abstract regular polyhedra exist?

It is well-known since Plato that there are only 5 regular polyhedra which live in 3D Euclidean space. However abstract polytopes are defined solely by their incidences, and are not confined by the geometry of 3 dimensional Euclidean space, so there may be more of them. Wikipedia mentions one non-Euclidean example, the hemicube. Are there more? Is there an algorithm to enumerate them?

Or to be less ambitious, it seems to me that for any Schläfli symbol of rank 2 you can write down, $\{p,q\}$, there is a regular polyhedron of that type. For example a polyhedron with 13 heptogonal faces. Does that exist?

Thanks.

• I see that Wikipedia also has an extensive article on abstract polytopes, which might discuss this, or whose references might discuss it.
– MJD
Nov 6 '16 at 19:12
• @MJD Wikipedia is where I got the example about the hemicube, and I found no others. The cited reference by McMullen I found a sample PDF of online with a few dozen pages. Judging by the TOC it would definitely discuss the question, and seems like a useful reference. Nov 6 '16 at 19:15
• I don't understand why you say you found no other examples in that article. What about the digon, the hemi-octahedron, the 11-cell, or the hexagonal hosohedron?
– MJD
Nov 6 '16 at 19:18
• @MJD ok yes I see (some?) of those examples in that article, although not in the section on regular polytopes Nov 6 '16 at 19:30

There are infinitely many abstract regular polyhedra. Several are discussed at https://en.wikipedia.org/wiki/Regular_polyhedron#Abstract_regular_polyhedra, including Petrials, spherical polyhedra, hosohedra, and dihedra.

Many abstract regular polyhedra are infinite; any tilings of the Euclidean or hyperbolic planes (as well as tilings of the 2-sphere) are considered abstract polyhedra. These include tilings of Schläfli type $\{p,q\}$ for any $p, q \geq 2$. You can take quotients of these abstract polyhedra by subgroups of their automorphism group to get more abstract regular polyhedra. In fact, every abstract regular polyhedron arises in this way, since any such polyhedron has a Schläfli type $\{p,q\}$, and is a quotient of the tiling of type $\{p,q\}$, which is called the "universal polytope $\{p,q\}$" for this reason. Sometimes these quotients are finite.

You may find data about finite abstract regular polyhedra, with at most 2000 flags (not including 1024 or 1536, which are apparently too numerous), at http://www.abstract-polytopes.com/atlas/r3. This includes 5946 nondegenerate and 993 degenerate polytopes, with 2250 different Schlafli types.

There is no polytope with 13 heptagonal faces, since the number of odd-length faces must be even, but there are three polytopes listed of type $\{7,13\}$. There is one with 14 heptagonal faces.

• Why must number of odd-length faces be even? Is this a purely combinatorial constraint? Apr 26 '17 at 18:13
• @ziggurism: The number of edges in the polytope is the sum of the lengths of each face divided by two (since every edge is in exactly two faces, even for abstract polytopes). So the sum of the lengths of each face must be even. Or, if you like, consider the graph whose nodes are the faces of $P$, with an edge between any two faces that share an edge in $P$, and use the fact that any graph has evenly many odd nodes. Apr 26 '17 at 19:50
• What does "length" mean in this context? Number of rank $n-1$ faces? Apr 27 '17 at 13:38
• @ziggurism: Since we're discussing polyhedra (i.e. 3-polytopes), by "face" I mean 2-face, and by "length" of a face I mean the number of edges (equivalently, the number of adjacent faces). Apr 27 '17 at 15:01
• So there is a tiling of type ${p,q}$ for any $p,q$, and for some $p,q$ subject to some graph theoretic constraints (such as number of odd length faces must be even), there may exist finite quotients. So the general thrust is: many abstract polyhedra exist. The tight geometric constraints of Euclidean space are not present, but some much weaker combinatorial ones are. Would that be a valid summary? Nov 8 '18 at 16:44