This is a less elegant but maybe more transparent way to see that T^n embeds in R^(n+1): ---------- The induction step is typified by showing T^3 embeds in R^4, assuming T^2 embeds in R^3. Start from a T^2 = T_0 embedded in R^3. Letting the next dimension be "time", we make a short movie that starts with the T^2 embedded in R^3, which immediately separates into two T^2's that move apart to a maximum separation. (Just push T_0 off itself in R^3 in both directions.) This is the first half of the film; the rest of it is just the first half in reverse time order. The result can be thought of as two copies of T^2 x [0,1], with both copies of T^2 x {0} identified with each other, and also both copies of T^2 x {1} identified with each other. This is easily seen to be T^2 x S^1 topologically. ---------- As has been mentioned, (S^1)^3 with its natural embedding in (R^2)^3 = R^6 has the natural metric of R^3/Z^3 (at least up to uniform scaling). So in general each "cubical n-torus" Q^n := R^n/Z^n has a natural flat metric with a lot of symmetry. QUESTION: What is the lowest dimension F(n) of a Euclidean space in which this Q^n = R^n/Z^n can be embedded isometrically? Clearly n+1 <= F(n) <= 2n, but what more can be said about F(n) ??? (Obviously F(1) = 2 and F(2) = 4, but what about for n >= 3 ? In particular, is F(n) == 2n for all n ?) --Dan Robert Munafo wrote: << Hey Bill, that makes a lot of sense. If I read you right, you're saying that just as we can expand a circle into a set of epsilon-sized circles which together make up a 2-torus (the surface of a donut whose "donut hole" is almost as big as the original circle) we can do the same thing in 4 dimensions to get a 3-torus surface, and so on. I still like to embed the 2-torus in 4 dimensions, because then nothing gets stretched. Probably just the way my mind works. Bill Thurston wrote: << 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}.

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