You could pucker the tiling out-of-plane. For example, there’s a nice tiling of the plane by 60-degree rhombi, where either 3 obtuse angles meet at every vertex or 6 acute. Now push the vertices out of the plane so the rhombi become squares. Rolling a cube on that polyhedral surface (of squares) has trivial monodromy. You get other examples like this one by using Voronoi cells of lattices (e.g. the rhombic dodecahedron). -Veit
On Jul 17, 2015, at 10:55 AM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Veit.
What if we use re-entrant polyhedra (so that the curvature sum-rule requires modification)?
I think the small stellated dodecahedron (viewed as a re-entrant polyhedron with just 12 vertices) has the property I stated: its faces are pentagrams with angles of pi/5 meeting in groups of 5, so that angle-sum at each of the 12 vertices is pi.
The way I want to "roll" a re-entrant polyhedron on a plane is somewhat peculiar: the faces that are currently rolling along the plane are required to stay on or above it, but other faces may penetrate the plane. Analogy: We can "roll" pentagram ABCDE along a line so that (for instance) as it rests on segment BC, vertices A and D are above the line but vertex E is below the line. I'd like to do something similar with polyhedra, but for simplicity I'd like to start with the "monodromy-free" case. Note that when you roll a cube, there's nontrivial monodromy: when the cube returns to where it started, in need not be in the same orientation. That's because the three 90-degree angles at each corner add up to 3 pi / 2, which is not a submultiple of 2 pi.
But there's more to niceness than trivial monodromy, because if you start rolling a stellated dodecahedron around, its "footprints" will form a re-entrant tiling of the plane by pentagrams, and the set of footprints won't be discrete.
What I'd like is something that I can "re-entrantly roll" on a plane so that the set of footprints is discrete, AND the monodromy is trivial.
Are there any, besides the tetrahedra I mentioned?
Jim Propp
Are there other examples?
Jim Propp
On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
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