Dan: << I'm pretty sure that the product of any two regular polytopes is regular. >> Consider the product of a square with an equilateral triangle: this is both isocellular and isogonal; but its faces comprise squares and triangles, so it is not regular. WFL On 3/23/15, Dan Asimov <dasimov@earthlink.net> wrote:
There is a standard well-defined concept of (abstract) n-dimensional polytope P, which roughly involves a set of points (the vertices) and sets of vertices that are the vertices of the k-faces for 0 <= k <= n. With natural consistency conditions.
This is a combinatorial definition.
Define a "flag" F as a sequence F = (F_n, F_(n-1),...,F_k,...,F_1,F_0) of faces of P of all dimensions such that j < k implies F_j is a subset of F_k.
Then the n-polytope P is called "regular" if for any two flags F, G there is a permutation of the vertices of P that induces a map from F to G (i.e., taking the k-dimensional face in F to that of G for all k, 0 <= k <= n).
I'm pretty sure that the product of any two regular polytopes is regular.
--Dan
On Mar 21, 2015, at 4:19 PM, Tom Karzes <karzes@sonic.net> wrote:
I was thinking about genus-1 polytopes in 4-space, and I was wondering if some of them would be considered "regular". For example, the cartesian product of two identical regular polygons is a two-dimensional surface (like the surface of a torus), which can be distorted to fit in 3-space but which is much more symmetrical in 4-space.
As a specific example, here's a torus surface in 4-space whose cross sections are squares:
max(|w|, |x|) = 1 max(|y|, |z|) = 1
Unless I'm mistaken, all of its faces are squares (16 of them, the cartesian products of the edges of the squares), and it is face-transitive, edge-transitive, and vertex-transitive. Does that make it regular?
Tom
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