Since there has been only one response to this (via direct e-mail to me), let me try to rephrase it so it doesn't sound so technical.* Because it really isn't. Let's define a "regular map" on an orientable topological surface M to be a tiling of the surface by n-gons for some fixed n, such that for any two such polygons P and Q, and for any of the n rotational positions that P can be mapped to Q (without flipping it over), there is a homeomorphism of the surface carrying P to Q that way. Note that the tiling here means only some graph G (edges and nodes) on the surface M such that the complement M \ G is a disjoint union of topological (open) n-gons. As long as the open polygons satisfy the symmetry condition above. So that, e.g., representing a torus as a single square with its opposite edges identified is fine. PUZZLE: Let M be homeomorphic to the torus T. What is the smallest positive integer K such that there does not exist a regular map on T having K polygons? --Dan ___________________________________________________ * Okay, maybe it doesn't sound less technical, but I hope all the details are clear. On Mar 21, 2014, at 10:23 AM, Dan Asimov <dasimov@earthlink.net> wrote:
For our purposes, a “regular map” on an metric orientable surface M (assumed compact and without boundary) is a tiling of the surface by congruent regular n-gons, such that for any two n-gons P and Q and any orientation-preserving isometry P -> Q (there are n of these), there is an automorphism f: M -> M of the tiling such that f restricted to P is this isometry P -> Q.
(There must be an easier way to say this. But a regular map defined as just a tiling of M by congruent polygons need not have the isometry property.)
PUZZLE: If M = the torus T, what is the least positive integer L such that there exists no regular map on T having exactly L polygons?
(Note: We allow the boundary of any tile to overlap any part of itself. So even one lone square can be a regular map of T.)
—Dan