Classification of vertex-transitive zonotopes Zonotopes are convex polytopes that can be defined in several equivalent ways:


*

*parallel projections of cubes,

*Minkowsi sums of line segments,

*only centrally symmetric faces,

*...


I wonder whether there exists a calssification of all vertex-transitive zonotopes. I know only of the following examples:


*

*omnitruncations of uniform polytopes (this is probably the same as $W$-permutahedra, see comments). This already includes the interval $[0,1]$, all regular $2n$-gons, and, e.g. the following polyhedra in $\smash{\Bbb R^3}$:


$\qquad\qquad\qquad\qquad\qquad$




*

*cartesian products of any of these above. This includes $d$-cubes, prisms, duo-prisms, ...


Are there any more? For that matter, are there even any more zonotopes for which all vertices are on a common sphere?
 A: Update
I recently uploaded a preprint in which I work out the details that are missing below.
So in fact, vertex-transitive zonotopes are $\Gamma$-permutahedra.

I believe to have (at least a roadmap to) a proof of the following:

Theorem. If $P\subset\Bbb R^d$ is a vertex-transitive zonotope, then $P$ is a $\Gamma$-permutahedron. That is, $P$ is the convex hull of the orbit of an appropriately chosen point $\smash{v\in\Bbb R^d}$ under a finite reflection group $\smash{\Gamma\subset\mathrm{GL}(\Bbb R^d)}$.

In other words, $P$ is the omnitruncation of some uniform polytope (when considered with a certain subgroup of its symmetries).

I will give some thoughts about my proof, since I have not thought through every detail:


*

*Every zonotope can be uniquely written as the Minkowski sum of line segments with pair-wise trivial intersection.

*Let's call $r\in\Bbb R$ a root of $P$ if $\mathrm{conv}\{-r,r\}$ is one of these line segments.

*One then shows that the set of roots of $P$ forms a root system (without integrality condition).1

*One further shows, that the zonotope $P$ has the same symmetries as its set of roots, hence that its symmetry group is a reflection group.


(until here, I think, David had another approach using the normal fan of $P$).


*

*Let $\tilde \Gamma$ be the symmetry group of $P$. Since $P$ is vertex-transitive, $P$ is the orbit polytope of some point $\smash{v\in\Bbb R^d}$ w.r.t $\smash{\tilde \Gamma}$. As David observed, this group might be too large to call $P$ a $\smash{\tilde\Gamma}$-permutahedron.

*Consider the subgroup $\Gamma\subseteq\tilde\Gamma$ generated by all reflections in $\tilde\Gamma$ that fix no vertex of $P$. Then $\Gamma$ is a reflection group.

*Show that $P$ is the orbit polytope of $v$ under $\Gamma$. Then $\Gamma$ acts vertex-transitively and -regularly on $P$, hence $P$ is a $\Gamma$-permutahedron.



Some notes on 1
Let $R$ be the set of roots of $P$. How to show that $R$ is a root system:


*

*Choose any two (linearly independent) $r,r'\in R$ and consider the 2-dimensional set $R':=\mathrm{span}\{r,r'\}\cap R$.

*Let $P'$ be the zonotope generated by $R'$. This zonotope is a 2-face of $P$, and by using the argument that $\mathrm{Aut}(P)=\mathrm{Aut}(R)$ one can conclude that from the vertex-transitivity of $P$ follows the vertex-transitivity of $P'$. (This part is sketchy right now, and makes some trouble. How to fix this? I think that that the faces of a vertex-transitive polytope do not necessarily have to be vertex transitive! Update: yes they are vertex-transitive, see the preprint)

*It follows that $P'$ is a $2n$-gon with possibly alternating edge lengths.

*One convinces oneself that the roots of $P'$ are a root system ($2n$ roots equally spaces by $\pi/n$, maybe of alternating lengths), that is, $R'$ (and hence $R$) contains the reflection of $r'$ on the hyperplane defined by $r$.

*Since $r$ and $r'$ were chosen arbitrarily, this shows that $R$ is a root system.

