3
$\begingroup$

I'm interested in understanding the class of simplicial polytopes in $\mathbb R^n$ whose Euclidean isometry group $G$ acts transitively on the vertices. These are examples that I know of:

  • simplicial regular polytopes: there are not many of these, only the simplex and cross-polytope in $\mathbb R^n$, regular polygons in $\mathbb R^2$, the icosahedron in $\mathbb R^3$, and the 600-cell in $\mathbb R^4$.

  • trigonometric cyclic polytopes: for even $n=2k$, and any positive integer $M$, the convex hull of the $M$ points $$\left(\cos(\frac{2\pi}M\cdot j), \sin (\frac{2\pi}M\cdot j), \cos(\frac{2\pi}M\cdot 2j), \sin (\frac{2\pi}M\cdot 2j), \dots, \cos(\frac{2\pi}M\cdot kj), \sin (\frac{2\pi}M\cdot kj)\right)$$ for $j=0,\dots,M - 1$.

One way of approaching this (for low dimensions $n$) might be to enumerate finite groups $G$ and their irreducible matrix representations. For a given matrix representation, we could select a point and form its orbit, then check if its convex hull is simplicial; however I'm not sure how to compute this check efficiently -- enumerating the faces might not be tractable because there could be exponentially many of them. I wonder if there is some theory that could shed more light on the problem. Aside from the computational difficulty, a couple other potential issues with this approach:

  1. It wouldn't help us find examples where $G$ is reducible (such as for the trigonometric cyclic polytopes). I wouldn't mind restricting attention to the irreducible case as long as there is still a rich set of examples (e.g., more than just the regular polytopes), but I don't know if this is the case or not.

  2. We have to decide how to select which orbit of $G$ to consider. There are infinitely many choices here; for a given matrix representation of $G$, I expect the orbits will fall into finitely many equivalence classes in terms of the abstract polytope that arises, but it is not obvious to me how to enumerate representatives of those classes. And to some extent the geometry matters to me here and not just the abstract polytope; within each class, I would be interested in finding an orbit that makes the vertex distribution on the sphere as "uniform" as possible in some sense, for example by maximizing the distance between a vertex and its nearest neighbor.

Another thought is that it might be useful to restrict to the case where $G$ acts on a lattice, and then a natural choice might be to consider orbits consisting of short vectors of the lattice. The main things I'm wondering specifically are

  • Is there a way to accelerate the check of whether the resulting polytope is simplicial?
  • Are theoretical results that would enable us to narrow down the groups $G$ (and/or its representations) that we would need to consider?
  • Are there even any irreducible examples at all aside from the regular polytopes?
$\endgroup$
3
  • 2
    $\begingroup$ A good question. I asked this myself before. I never had the hope for a classification because my feeling was that if I take just any sufficiently wild finite subgroup of $\mathrm O(\Bbb R^d)$ and just any generic point in $\Bbb R^d$ then the convex hull of the orbit of this point under the group has a good chance of being simplicial (as a polytope with generically chosen vertices is simplicial). But of course, this is just a feeling, and sadly, I have no further examples. I have some candidates in mind but I would need to check whether they are actually simplicial. $\endgroup$
    – M. Winter
    Mar 13, 2021 at 23:57
  • 1
    $\begingroup$ The disphenoidal 30-cell and the disphenoidal 288-cell are further examples. $\endgroup$
    – M. Winter
    Mar 14, 2021 at 0:04
  • $\begingroup$ Another only half-new example is any generic orbit polytope of the orientation preserving symmetries of the tetrahedron (also called the snub tetrahedron, see also here). Combinatorially it is the same as the icosahedron, but it is not regular for most choices of the generating point. And of course, any vertex-transitive (not necessarily regular) polygon counts as well. $\endgroup$
    – M. Winter
    Mar 14, 2021 at 2:30

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.