MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 is '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|cite|improve this question
$d=3$ 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
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|cite|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

For any even $d$, the answer is yes; take the cyclic polytope $C_d(v)$, consisting of $v$ points on the moment curve $(t, t^2, \dotsc, t^d)$. Any choice of points gives a combinatorially identical polytope, which is combinatorially vertex-transitive in even dimension; and it is always possible to realize such a polytope so that all its combinatorial automorphisms are Euclidean isometries. Both results are from this chapter "Automorphism Groups of Cyclic Polytopes" by V Kaibel and A Waßmer.

For odd $d$, you can get a vertex-transitive polytope for any even $v \geq 2d$ by taking a prism over the $(d-1)$-dimensional cyclic polytope $C_{d-1}(v/2)$.

For odd $d$ and odd $v$ it is hard to find any examples of vertex-transitive $d$-polytopes with $v$ vertices. The only example I know if is the rectified 5-simplex, with 15 vertices. This seems to be because most symmetry groups in odd dimension include central inversion, which pairs up the vertices. For odd $d \geq 7$, taking the Cartesian product of a cyclic $(d-5)$-polytope with the rectified 5-simplex, you get a vertex-transitive $d$-polytope with $15k$ vertices for any $k \geq d-4$.

share|cite|improve this answer

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|cite|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

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.