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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. $\endgroup$
    – Yoav Kallus
    Commented Feb 21, 2012 at 19:22

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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.

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  • $\begingroup$ 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
    Commented Feb 22, 2012 at 22:49
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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.

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    $\begingroup$ Very nice answer. Just wanted to add that the "rectified 5-simplex" is sometimes called the "2nd hypersimplex of dimension 5". Hypersimplices of dim $d$ are the slices of the $(d+1)$-cube obtained by intersecting with the hyperplane $\sum x_i=k$. In our case $k=2$, hence the name "second". ($k= 1$ gives the standard simplex). Note: Some days ago I inadvertently placed this as a comment to my answer. It probably looked weird, to say the least, that I called my own answer very nice... $\endgroup$
    – Francisco Santos
    Commented Oct 18, 2022 at 13:27
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Gale transforms allow to settle the case of odd $d$ and even $v$ completely, in the positive (Matteo's answer in this case needed $v\ge 2d$).

Remember that the Gale transform of a $d$-polytope with $v$ vertices is (or can be thought as) a configuration of $v$ points in the sphere of dimension $v-d-2$. If $v$ is odd and $d$ even, $v-d-2$ is odd so our sphere lives in an even dimensional space. If we take as Gale transform the vertices of a cyclic polytope in their Caratheodory embedding (see e.g. Exercise 2.21 in Ziegler's "Lectures on polytopes"), the Gale dual will be vertex-transitive.

Thus: vertex-transitive polytopes exist for all values of $d\ge 2$ and $v\ge d+1$ if at least one of $v$ or $d$ is even. If $d$ is even you have cyclic polytopes, if $v$ is even you have their "Gale duals".

So, only the odd-odd cases are difficult to settle (as already said by Matteo).

In the smallest odd-odd case, with $v=d+2$, Gale transforms also tell that vertex-transitive polytopes do not exist: you need to distribute $v$ points in a $0$-sphere, which consists of two points. In Gale transforms you are allowed to count the same point several times, but in order for the Gale dual to be vertex-transitive you need the same number on both points of the $0$-sphere, so $v$ needs to be even. I presume a similar argument implies a negative answer for $v=d+4$, taking into account that there is no vertex-transitive way of placing an odd number of points in a $2$-sphere.

So, the smallest possible odd-odd vertex-transitive polytope has $d\ge 5$ and $v\ge d+6 \ge 11$. This shows that the rectified $5$-simplex mentioned by Matteo (a.k.a. the second hypersimplex) is probably the smallest one. It would be strange to have a vertex transitive $5$-polytope with $11$ or $13$ vertices; since they are prime, the symmetry group would need to contain the cyclic group of that order...

Edited to mention the comment by M.Winter that no odd-dimensional vertex-transitive polytope can have an odd prime number of vertices. Hence the rectified $5$-simplex is indeed the smallest possible odd-odd example, both with respect to number of vertices ($15$) and dimension ($5$).

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    $\begingroup$ As you say, the symmetry group of a 5-dimensional 11/13-vertex example would need to contain the cyclic group of order 11/13. If I am not mistaken, there is no real representation of these cyclic groups in 5 dimensions (with a full-dimensional orbit polytope). Likewise, for any odd prime $p$ there is no odd-dimensional representation of the cyclic group $C_p$ and so no odd-dimensional $p$-vertex example. $\endgroup$
    – M. Winter
    Commented Oct 18, 2022 at 17:16
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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

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  • $\begingroup$ 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 $\endgroup$
    – Gerhard Paseman
    Commented Feb 21, 2012 at 21:30

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