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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?

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  • $\begingroup$ How about permutohedra (en.wikipedia.org/wiki/Permutohedron) and more generally $W$-permutohedra (old name: Coxeterhedra) for Weyl groups $W$? $\endgroup$ – Sam Hopkins Sep 24 '19 at 14:55
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    $\begingroup$ @SamHopkins There is indeed one ;) see E8 polytopes. The omnitruncated one is not listed there, but one can write down its Coxeter diagram (all nodes ringed), so it exists. Maybe a point of confusion was that initially my questions stated "regular polytopes" instead of "uniform polytopes", which was indeed too restrictive. I changed that. $\endgroup$ – M. Winter Sep 24 '19 at 15:07
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    $\begingroup$ What about the truncated 24-cell and the omnitruncated grand antiprism (not a W-permutahedra)? $\endgroup$ – LeechLattice Sep 25 '19 at 3:13
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    $\begingroup$ @Bullet51 Both of these are no counterexamples: 1. the truncated 24-cell is identical to the omnitruncated demitesseract (as written on the Wikipedia page of the truncated 24-cell). As David said, it is the $D_4$-permutahedron. 2. omnitruncations are (as far as I know) only defined for Wythoffian uniform polytopes, so I suppose whatever you image by omnitruncated grand antiprism is probably not vertex-transitive (but it probably gives a counterexample for the question for a zonotope with all vertices on a sphere, as already do some $2n$-gons). $\endgroup$ – M. Winter Sep 27 '19 at 13:56
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    $\begingroup$ Having said that, I probably found a proof that all vertex-transitive zonotopes are $\Gamma$-permutahedra for some finite reflection group $\Gamma$. However, all I could give right now as an answer are some hints, since I have not formulated every detail. Maybe I should do that. $\endgroup$ – M. Winter Sep 27 '19 at 13:58
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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.
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  • $\begingroup$ It's bullet point 3 that I want more detail on. Why is reflection over $r^{\perp}$ necessarily one of the symmetries of your polytope? If you get this, I believe you have the rest. $\endgroup$ – David E Speyer Sep 27 '19 at 14:33
  • $\begingroup$ @DavidESpeyer The point that I have not thought through in detail is how to show that $P$ is an orbit polytope of $v$ w.r.t. $\Gamma$. Might be not too hard, but still a gap for me right now. $\endgroup$ – M. Winter Sep 27 '19 at 14:53
  • $\begingroup$ I filled in a detailed part for point 3. However, I still had to be vague on the second point there, and some details are missing. Help is appreciated. But this breaks the proof down to the question of how wild the 2-faces of such a zonotope can be. $\endgroup$ – M. Winter Sep 27 '19 at 15:20
  • $\begingroup$ I agree that it comes down to $2$-faces. A direction I have been thinking about is that Nathan Reading gives a characterization of permutahedra by their $2$-skeletons: Theorem 10-2.21 in "Finite Coxeter Groups and the Weak Order", which is Chapter 10 of "Lattice Theory: Special Topics and Applications". $\endgroup$ – David E Speyer Sep 27 '19 at 15:43
  • $\begingroup$ Here is how I would rephrase what he does: First, the edge graph must be $n$-regular and every pair of adjacent edges must be in a unique $2$-face of even size, so assume these things. Choose a vertex and color the $n$-edges incident to it with $1$, $2$, ..., $n$. Propogate this to an edge coloring of the full graph such that every two face is colored $(i,j,i,j,\dots, i,j)$. If there are two $2$-faces with the same color pair $(i,j)$ and different numbers of edges, it is not a permutahedron. Otherwise, let $2 m_{ij}$ be the number of edges of an $(i,j)$-colored $2$-face. (continued) $\endgroup$ – David E Speyer Sep 27 '19 at 15:48

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