Duals of Convex Isohedra are Uniform?

Question

All of the convex isohedra are either Platonic, Catalan, bipyramids, or trapezohedra. Their duals are Platonic, Archimedean, prisms, or antiprisms respectively, all of which are uniform, i.e. having each face a regular polygon.

Why does every convex isohedron have a uniform dual? Is there a good explanation of this, other than exhausting this list?

Answer 1

"iso" sometimes is being mixed up with "equi". Both meaning "same". Wrt. facial shapes (isohedra/equihedra) resp. vertex figures (isogonal/equigonal polyhedron) those clearly ask for congruence of the respective shape, i.e. is more a local comparation of local patches. But "iso" additionally includes some symmetry transitivity too, i.e. kind of is a global comparation. E.g. the Miller polyhedron J37 well is equigonal (all are of type 3,4,4,4) but is not isogonal (polar and tropal vertices cannot be interchanged by global symmetry).

Using this distinction then it becomes evident that the duals of (truely) isohedral polyhedra would indeed be isogonal, i.e. have a symmetry group which acts on the vertex transitive. But the latter description is just one of the requirements for uniform polyhedra. Additionally there is the requirement for unit edges throughout (which thus ensures regular polygonal (or regular polygrammal) faces only. (And for uniform polytopes (of arbitrary dimension) even dimensional hierarchical application is required as well, where (for induction start) the uniform polygons are just the regular polygons.) - Those further additions clearly cannot be ensured by solely isohedral preimages.

--- rk

Answer 2

Not every convex isohedron has a uniform dual. The dual of a bipyramid or trapezohedron is always a prism or antiprism, respectively, but is usually not uniform.