Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

My primary question is: given a cellular decomposition of a sphere is there any way to check if it can be embedded as the boundary of a polytope?

My question is motivated by the following problem. I began with a polyhedral cell complex $P$ homeomorphic to a ball. Then I had an unbounded polyhedral cone $C$ of the same dimension as $P$. I made the cone homeomorphic to a ball by taking its intersection with a ball sufficiently large so that all the bounded portions of the cone were not changed. Call this new polyhedral ball $C^*$. Then I defined a continuous mapping from $\partial C^*$ to $\partial P$ and used the pasting lemma to define my cellular decomposition of the resulting sphere.

Since I am beginning with two polyhedral objects I would like to end up with something polyhedral, in this case a polytope. My method seems equivalent to "blocking" the unbounded portion of the cone with $P$, which seems should remain polyhedral.

share|improve this question

3 Answers 3

up vote 5 down vote accepted

The "primary question" is a well-known hard problem, and I think this reference:

MR0889977 (89b:52009) Reviewed Bokowski, Jürgen(D-DARM); Sturmfels, Bernd(D-DARM) Polytopal and nonpolytopal spheres: an algorithmic approach. Israel J. Math. 57 (1987), no. 3, 257–271. 52A25 (05B35) PDF Clipboard Journal Article Make Link

Is still close to the last word 25 years later (maybe @GilKalai can correct me).

share|improve this answer

Indeed the primary question is a well-known hard problem. To have a chance to be a polytope, the cellular decomposition should be regular (the closure of an open cell must be a closed cell) and satisfies the lattice property (The intersection of two cells must be a cell - possibly empty). It also needed to be a polyhedral complex - each cell by itself should be combinatorially equivalent to a polytope. These conditions are sufficient for 2-spheres by a fundamental theorem of Steinitz (and also when the number of vertices is at most the dimension plus 4) but not for d-spheres for $d>2$. The smallest examples of polyhedral spheres which are non-polytopal are 3-spheres with 8 vertices.

share|improve this answer

The question reminds me of the idea behind David W. Jones's 1984 Bangor thesis on "Poly-T-complexes" available here, as follows: we have globular sets, simplicial sets, cubical sets, but what is wrong with pentagons, and why is there a prejudice against rhombic dodecahedra? Also the part of geometric group theory known as "van Kampen diagrams" uses complicated 2-dimensional diagrams to deduce consequences of relations. There are also families of polyhedra such as Stasheff polyhedra.

David defined quite general "cone complexes" with a shelling criterion to avoid some wild examples. In order to model some group theory, and to define polyhedral sets, he also defined "marked cone complexes"; this allows the modelling of the relations $ x^3=1 $, $ xyx=yxy $, for example.

It turned out recently that the notion of marked complex is equivalent to that of the later defined "discrete vector field" on a complex, for which a web search shows many publications.

See the published version

Jones, D.W. A general theory of polyhedral sets and the corresponding $T$-complexes. Dissertationes Math. (Rozprawy Mat.) 266 (1988) 110.

share|improve this answer

Your Answer

 
discard

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.