Gene is right that my argument failed. I was unduly confident that the action of the symmetric group S_n [as rotations of the regular n-simplex centered at the origin in R^n] could be parlayed into a regular lattice n-gon, but this is not to be, thanks to a web page I found this morning: Theorem: -------- (i) The rectangle and octagon are the only two equiangular lattice polygons in Z^2.( Since the equiangular octagon is not equilateral, the square is the only regular lattice polygon in Z^2.) (ii) The square, triangle and hexagon are the only regular lattice polygons in Z^d for any dimension d. The square can be embedded in Z^2 while the triangle and hexagon require Z^3. ----- Proofs and more at <http://dynamicsofpolygons.org/PDFs/LatticePolygons.pdf>. —Dan
From: Eugene Salamin Sent: Aug 11, 2017 5:16 PM
On Friday, August 11, 2017, 5:04:38 PM PDT, Dan Asimov <dasimov@earthlink.net> wrote:
Here is a one way to attack the n-dimensional case of regular polygons whose vertices lie in Z^n.
Suppose there's a regular p-gon P, p in {3,4,5,...} with vertices in Z^n.
If n >= p then R^p \sub R^n, so the standard basis vectors of R^p show that such a p-gon always exists.
-----------------------------------------------------------------
But these standard basis vectors form a (p-1)-simplex, and for p>3, do not lie in a common 2-plane. For example, in Z^4, the basis vectors (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) form a tetrahedron.