Dan, 1. By the product of polygons P1 and P2, I just mean the Cartesian product: {p1 + p2 | p1 \in P1 && p2 \in P2} Or if you prefer a route through 4D space, you can take their 4D duoprism and project down to 3D by any (full rank) linear transformation to get an affine family of equivalent surfaces. The necessity of their having self-intersections in 3D is a consequence of the linearity of the transformation, but I don't see an instant proof of that at the moment. 2. When watching the video, I expect you'll see there is a natural notion of a "homogeneous" zonohedral surface that you get by starting from any one zigzag path in space and applying the algorithm. Perhaps "homogeneous" is not the best term, as it already has many other specific uses, but I intended the term to generally characterize that there is a certain uniformity to the structure that entails long-distance constraints. I'm open to suggestions for better a term. George http://georgehart.com On 11/16/2020 12:04 PM, Dan Asimov wrote:
Thanks, George! I look forward to watching the video, which unfortunately I don't have time to do this morning.
But I'm wondering a couple of things from this post:
1. How do you define the product of two polygons as a polyhedron in 3-space?
Abstractly, it makes perfect sense: The product of two vertices is a vertex; of a vertex and an edge (in either order) is an edge; and of two edges is a quadrilateral. So far this is just a combinatorial object, but it could even be an abstract geometrical object, as if each polygon were in a distinct plane and the product becomes a specific subset of 2+2 = 4-space.
But that seems to allow a lot of possibilities for it to end up in 3-space.
How do you see that any way of doing this results in a polyhedron that's self-intersecting?
2. What does "homogeneous" mean in this situation? It could refer just to the result being combinatorially homogeneous, in that the result has maps taking vertices to vertices, edges to edges, and faces to faces while preserving incidence, that will carry any vertex to any other (transitivity on vertices). Is that it?
Thanks,
Dan
----- The talk Jim had referred to is now online here in a more completed form. Polyhedra fans might enjoy it:
https://www.youtube.com/watch?v=_PSdVX02Vbs
The question which I thought might have been intended (during the CoM discussion, or initially by Jim) was whether one could make a *homogeneous* zonohedral surface in 3D which is topologically toroidal without self-intersection. By homogeneous, I mean following the algorithm described in the video from a given initial state. (One with all vertices of degree four would be homogeneous, but that isn't a necessary condition; homogeneous examples might have vertices of any order, e.g., polar zonohedra.) One can take a product of two polygons to create a toroidal surface composed of parallelograms and having all vertices of degree four, but it will be self-intersecting. And one can approximate any continuous surface by joining homogeneous patches, e.g., gluing together small cubes. (The video shows how to join larger homogeneous patches seamlessly.) But I would conjecture that a homogeneous zonohedral surface can only be toroidal if it is self-intersecting. -----
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