On Wed, Sep 2, 2009 at 6:05 AM, Seb Perez-D <sbprzd+mathfun@gmail.com<sbprzd%2Bmathfun@gmail.com>
wrote:
On Wed, Sep 2, 2009 at 05:45, Allan Wechsler<acwacw@gmail.com> wrote:
Oh, man, I cannot rigorize this at all, but I have a *strong* intuition that the answer is "no". It really feels like there is a simple topological fixed-point argument that would make it clear that this is impossible.
Do degenerate solutions count?
If the x(s) and y(t) surfaces are flat and coplanar, the Minkowski sum is also flat. For instance if x(s) and y(t) are unit circles in the (x,y) plane, then z(s,t) is the disk with radius 2.
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.
Cheers,
Seb
On Tue, Sep 1, 2009 at 12:23 PM, Veit Elser <ve10@cornell.edu> wrote:
Can the Minkowski sum of two smooth closed curves in R^3 ever be a smooth torus?
If you don't know what a Minkowski sum is, let x(s) and y(t) be smooth closed curves in R^3 parameterized on the unit interval. Now consider the set
z(s,t) = x(s) + y(t), (s,t) in [0,1]^2.
Question: can you find an x(s) and y(t) so z(s,t) is a smooth surface for all (s,t) in [0,1]^2 ?
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