The standard name for an n-cube with opposite sides identified via translation = (S^1)^n is the n-torus, T^n. It's also equivalent to the quotient of the additive group of vectors in R^n by the integer lattice subgroup, T^n = R^n/Z^n. In general, the n-torus can be embedded in R^{n+1}. For the circle, this is immediate. Everything else can be done by induction: if the n-torus is smoothly embedded in R^{n+1}, then the boundary of an epsilon neighborhood in R^{n+2} gives a smooth embedding of T^{n+1} in R^{n+2}. Bill On Dec 1, 2010, at 1:54 PM, Allan Wechsler wrote:
Six dimensions is overdoing it; as Mike pointed out, the object can be embedded in a four dimensional space. I am having a vague memory that this manifold (S1^3, as Robert points out) is surprisingly equivalent to some other 3-manifold, but I forget all details. Perhaps Bill Thurston knows this one?
On Wed, Dec 1, 2010 at 1:33 PM, Robert Munafo <mrob27@gmail.com> wrote:
Just like a normal torus is a circle squared, and a cylinder is a circle multiplied by a line, I think of your torus as a circle raised to the 3rd power, with the 3-d surface being embedded in 6 dimensions (three orthogonal sets of two dimensions, each defining a plane that gives a circle as the cross-section).
On Wed, Dec 1, 2010 at 13:16, Michael Beeler <mikebeeler@verizon.net> wrote:
What's the proper name for the following "4 dimensional donut"?
Identifying opposite edges of a sheet of paper makes a torus, actually just the surface, either 0 or "1 cell" thick. Identifying 2 pairs of opposite sides of a 3-D rectangular solid (brick) makes a torus skin with some thickness, like the glaze of a donut. Now also identify the third pair of sides of the brick. This connects the inner surface of the glaze to the outer surface. That seems to require a fourth dimension, so I call this a "4 dimensional donut". But what is its correct name?
I'm still working on domino-nets. The smallest "4-D donut" that might have solutions is 4x5x5, for donimos with 1 to 25 pips (no double dominos). This is probably too large for me to either find a solution or to prove there are none, unfortunately.
Thanks, -- Mike Beeler
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