On Sep 2, 2015, at 1:33 PM, Dan Asimov <asimov@msri.org> wrote:
I. Consider the vertices V of an equilateral triangle as a metric space with all distances equal to 1.
In the cartesian power V^n, we take the metric is taken to be
D(x=(x_1,...,x_n), y=(y_1,...,y_n)) := sqrt((x_1-y_1)^2+...+(x_n-y_n)^2),
where all x_j and y_k belong to V.
So D(x,y) is the sqrt of the number of dimensions in which the coordinates differ.
Question: Clearly, V^n embeds isometrically in R^2n, since V embeds isometrically in the plane. But is there a smaller dimension for which V embeds isometrically?
Let E(n) = E_V(n) be the least dimension in which V^m embeds.* What is E(n) ??? -----
II. By permuting the coordinates 1,...,n of V^n, and independently permuting the 3 elements of each of the n factors V, it follows that the isometry group of V^n contains the subgroup S_n x S_3 of size 3(n!). Is S_n x S_3 equal to the full isometry group Isom(V^n) ? ----- . . . . . .
Correction: In question II I rashly ignored the fact that n independent S_3's form the group (S_3)^n. So the group that is clearly a subgroup of Isom(V^n) is in fact S_n x (S_3)^n , of size n! x 6^n. That's the group I suspect is the full isometry group of V^n, but am not sure. —Dan