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

Given positive integers $d$ and $v$ with $v \geq d+1$, does there always exist a (convex) vertex-transitive $d$-polytope with $v$ vertices? It seems that the answer should be "obviously" true, but I don't know of any natural constructions. A couple of points:

1) If the question is changed to the existence of regular polytopes (i.e., polytopes that are transitive with respect to all the faces), then the answer if 'no' at least for arbitrary $v$.

2) For $d = 2$ the answer is trivially 'yes' by taking the convex hull of $v$ equi-spaced points on the circle.

share|improve this question
3  
$d=3$ and $v=5$ is an obvious small counterexample. By symmetry, the vertices must all lie in a plane, hence their convex hull is not a polyhedron. –  Yoav Kallus Feb 21 '12 at 19:22
add comment

2 Answers

up vote 4 down vote accepted

There is no three dimensional vertex transitive polyhedron with 7 vertices. The rotation groups of three dimensional polyhedra are finite quaternion groups. The only finite quaterionion groups whose order is divisible by 7 are associated with polygons with 7 or a multiple of 7 sides. So there have to be another element in the group to move points out of the plane and this result in more than 7 points. There is a complete classification of finite isometric groups that leave one point fixed in dimensions 3 and 4.

There are also no three dimensional vertex transitive polyhedra with 5 vertices but in that case there are more finite quaternion groups whose order is divisible by 5. When these are examined there are no three dimensional vertex transitive polyhedra but there is one in four dimensions namely the four dimensional simplex.

share|improve this answer
    
Thanks, this is very useful. I've been wondering more generally about interesting families of vertex-transitive polytopes (indexed by dimension $d$) that have about $O(d^\alpha)$ vertices for any integer $\alpha$. My question was a first attempt at trying to understand what the issues might be. –  Donald Feb 22 '12 at 22:49
add comment

For v prime, I don't know. For v even, and d=3, one can take a regular polygonal prism. I can think of toriodal versions for odd composite v, but I am unsure they are vertex transitive. I can go to higher dimensions at the cost of adding a multiplicative factor, but for d+1 I don't see how to do it for v with less than d prime factors; without starting with a lower dimensional seed.

Gerhard "Ask Me About System Design" Paseman, 2012.02.21

share|improve this answer
    
Of course, for prime p, there are d-simplices for d+1=p. I know of no other examples. Gerhard "Ask Me About System Design" Paseman, 2012.02.21 –  Gerhard Paseman Feb 21 '12 at 21:30
add comment

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.