The paragraph beginning with "Yes" got inadvertently truncated in previous e-mail. The following is, I hope, fixed. —Dan
On Friday/25December/2020, at 4:40 PM, Dan Asimov <dasimov@earthlink.net> wrote:
This is a very interesting question.
Yes, I think this is a proof that any convex polyhedron rolling on its mirror image has trivial holonomy. (Meaning: If they both start by sharing a common face in 3-space, a reflection in whose plane identifies each polyhedron with its mirror image ... then after rolling to any number of adjacent faces so the two polyhedra end up sharing the same face they did initially — reflection in that face's plane associates the same pairs of point, one in each polyhedron, as were associated initially.)
Sketch of proof: Rotating one of them by angle theta about an edge of the latest shared face is equivalent to rotating both of them in opposite directions by an angle theta/2 about the same edge.
As for rolling one sphere on an identical one (with neither slipping nor twirling about the line connecting their centers): Indeed, essentially the same proof sketch as above shows that once each point of a sphere is identified (once and for all) with its mirror image, then (after some rolling) reflection about the tangent plane separating the spheres will always carry a point of one sphere to its (original) mirror image on the other sphere.