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.

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    $\begingroup$ $d=3$ 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. $\endgroup$ Feb 21, 2012 at 19:22

3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$
    – Donald
    Feb 22, 2012 at 22:49

For any even $d$, and any $v \geq d + 1$, 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 $C_{d-1}(v/2)$. There is also the $d$-simplex with $v = d + 1$.

For odd $d$ and odd $v$ it is hard to find any examples of vertex-transitive $d$-polytopes with $v$ vertices. For instance, none exist at all for $d = 3$ (a consequence of Transitive Planar Graphs, Fleischner & Imrich, 1979)).

The only example I know of is the rectified 5-simplex, with 15 vertices in 5-space. The scarcity of examples seems to occur 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 $k \geq d-4$, so you get some examples that way; the smallest being a 45-vertex example in 7-space.


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

  • $\begingroup$ 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 $\endgroup$ Feb 21, 2012 at 21:30

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