Maybe this mistake will lead to some insight(s). The symmetric group S_n does act by permutations on the standard basis vectors in R^n, and each of these has expression as an integer matrix. S_n certainly has elements of order n, for instance (12...n). Thus this element of S_n corresponds to a rotation of R^n that is of order n. So, (12...n) generates the cyclic group, isomorphic to Z/n, of integer-matrix rotations of R^n. Let (K,L,...,M) be an integer lattice point such that the iterates of this point under the group {(12...n)^k | 0 <= k <= n-1} are all distinct. Then: What is the orbit by this group of an integer lattice point (K,L,...,M) ??? Especially in the case n = p prime, if we assume the orbit {(12...n)^k (KL...M) | 0 <= k <= n-1} has cardinality = n but does not lie in any 2-plane of R^n, then . . . . . . what is it??? Simple case: n = p = 5. Let g denote the rotation of R^5 corresponding to (12...n). Since 5 is odd there is always one eigenvalue = 1, so we consider the rotation on the R^4 perpendicular to an eigenvector of this value. But a rotation g of R^4 of order 5 can always be resolved into two rotations, one on each of two perpendicular 2-planes, say E_1 and E_2. if we let tau = exp(2pi*i/5) then the rotations on E_1 and E_2 of order 5, thought of as complex numbers, are given by (z, w) |—> (tau^j z, tau^k w) where 1 <= i, j <= 4. Suppose we take as generator g the one with j = 1, k = 2: (z, w) |—> (tau z, tau^2 w) . Then the five rotations generated by g, as viewed on the torus {(exp(2pi*si), exp(2pi*ti)) | (s,t) in [0,1] x [0,1]} form the vertices of the (self-dual) tiling of the square torus by 5 squares. —Dan Jim Propp wrote: ----- I remember making the same mistake (confusing an n-simplex with a regular n-gon), and drawing the same sort of incorrect conclusions (a regular heptagon can be constructed with straightedge and compass in seven dimensions) when I was in college. If there's a compendium of commonly made mistakes in higher mathematics, this one should probably be added to the list! On Sat, Aug 12, 2017 at 1:51 PM, Dan Asimov <dasimov@earthlink.net> wrote:

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

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From: Eugene Salamin Sent: Aug 11, 2017 5:16 PM

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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.

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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.