I'm not sure I get it. I tried to use that description to show that the familiar torus T^2 can exist in R^3, but the resulting surface intersects itself. Placing the initial T^1 (which is a circle) on the X-Y plane, and treating the Z axis as the "time" dimension, and using the X axis as the direction in which the two copies get pushed while time is moving forward, I end up with the surface intersecting itself at various points in the Y-Z plane. I could use a 4th dimension to prevent this self-intersection, but then it's a bit hard to explain how you can twist half of the result around to embed the surface into R^3 in a non-self-intersecting way, or even to explain whether or not it is a Klein bottle. - Robert On Wed, Dec 1, 2010 at 18:40, Dan Asimov <dasimov@earthlink.net> wrote:
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.
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com