On Wed, Sep 2, 2009 at 8:40 AM, Veit Elser <ve10@cornell.edu> wrote:
On Sep 2, 2009, at 7:53 AM, Andy Latto wrote:
The question was whether the sum could be a smooth torus, not a smooth
manifold with boundary. So this doesn't count.
My intuition is no, but I haven't found a proof yet.
Maybe this way of stating the problem helps:
Consider two smooth maps from S^1 to R^3 given by
s --> x(s) t --> y(t)
Question: can the map S^1 x S^1 to R^3 given by
(s,t) --> x(s) + y(t)
be smooth?
I believe that this is actually a different question, and that the answer to this question is "yes", while the answer to the original question is "no"! That is, the two curves in R^3 give a map from the torus to R^3. I believe that this map can be an immersion (that is, smooth, but not 1-1, so the torus in R3 is smooth, but self-intersects) but not an imbedding (a 1-1 smooth map, so that the image 'looks like' a torus). How to choose x and y to get an immersion: A smooth curve x give us a smooth map s-> x'(s)/|x'(s)| from S^1 to S^2. This map must intersect every closed hemisphere in S^2; otherwise, the original space curve is always going "up" and never "down", so it can't close up. Further, unless it is a great circle, it touches every open hemisphere of S^2, for the same reason. I believe that any smooth map from S1 to S2 that satisfies these constraints comes from a closed space curve (hand-waving argument: if it doesn't close up, see what direction it fails to close up in, and increase the magnitude of x' when it has a component in that direction). The map from the torus to R3 fails to be smooth exactly when x'(s) is a multiple of y'(s). So to immerse the torus smoothly, we need two smooth closed curves on the sphere, each of which touches every open hemisphere. Take a sin wave around the equator, and move it slightly up and slightly down (but by an amount less than the amplitude) to get two such curves. I don't have a proof yet that the immersion has to be an imbedding. But you can think of the process of making the torus as taking the curve x(t), and "moving it around" along the path y(t) to get a surface, which is a closed surface when you finish because y(t) is closed. When you move a circle around in a circle to make a torus in the conventional way, you "turn the circle around" as you go, and this is necessary to avoid self-intersection. The definition of "turn around as you go" involves some induced map to O(3), and the loop involved has to be of the non-identity homotopy class. And the minkowski sum construction will give you something in the identity class. But I don't have the details worked out yet. Andy
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