My earlier confusion was between "non-singular" and "non-self-intersecting". I still think a non-self-intersecting example is impossible. On Fri, Sep 4, 2009 at 1:15 PM, Veit Elser <ve10@cornell.edu> wrote:
Dan's construction is exactly the one I came up with. If you don't trust your 3D visualization skills, then you can play with curves expressed as trigonometric polynomials like this:
Let x'(s) = u(s), y'(t) = v(t) be the velocities of the curves and don't require that they have unit speed. All we need to do is bound
d(s,t) = |u x v|^2 = |u|^2 |v|^2 - (u.v)^2
For example, here is something that works:
u[t_] := {18 Cos[t] - 10 Cos[2 t] - 6 Cos[4 t] + 3 Cos[5 t], 18 Sin[t] + 10 Sin[2 t] - 6 Sin[4 t] - 3 Sin[5 t], 10 Cos[3 t] - 4 Cos[6 t]}
v[s_] := {10 Cos[s], 10 Sin[s], 7 Cos[3 s]}
The domains of s and t are angles on two circles (0 to 2Pi). The integrated velocities are also trigonometric polynomials, so x and y are closed curves. It's possible to express d(s,t) as a 2D trigonometric polynomial and show, with a finite set of samples, that it is always positive. It would be nice if there were a strictly algebraic demonstration of d(s,t) > 0 for some very elegant choice of curves. It took a lot of tweaking to get the forms above, which has a reasonably large d. But I'm sure the members of this forum can do a lot better, and reduce the complexity of my example. The inversion symmetric v corresponds to Dan's C, while u is his perturbation D.
These Minkowski-sum tori can never be embedded because their signed volumes are always zero.
There is a physics context for all this where Minkowski makes two appearances. It turns out that you get the "world-sheet" of a cosmic string in Minkowski space by Minkowski summing two null curves!
Veit
On Sep 3, 2009, at 11:18 PM, Dan Asimov wrote:
As a torus fan, I've been working on this.
The answer to the question is Yes.
Proof: We seek two C^oo closed curves x,y: R/Z -> R^3 such that
F(s,t) := x(s) + y(t)
is everywhere non-singular.
F is non-singular at (s,t) precisely when x'(s) x y'(t) is not the 0 vector.
WLOG we assume that x and y are each parametrized by arclength. (So in fact the domain of F will be R/(cZ) x R/(dZ), which will not affect anything.)
Since the only way unit vectors a, b can satisfy a x b = 0 is that a = +-b. So we seek curves x, y such that both x'(s) + y'(t) and x'(s) - y'(t) are nonzero for all s,t.
A known theorem* about the *curve of tangent vectors* to a unit-speed closed curve in R^3 states that C is such a curve on S^2 if [either C is planar or else C contains points on both sides of every great circle].
So consider a monkey saddle, the graph of z = x^3 - 3xy^2 in R^3. Now consider its intersection C with the unit sphere x^2 + y^2 + z^2 = 1.
It's easy to verify that C intersects both sides of every great circle (and so is the curve of unit tangents to a closed space curve, in fact many of them). Further, note that the antipodal map a: S^2 -> S^2 carries C into itself, so any curve D disjoint from C is also disjoint from a(C).
By perturbing C slightly off itself, we get another curve on S^2 that clearly intersects both sides of every great circle, and so it, too, is the curve of unit tangents to a closed space curve.
Thus setting {x'(s)} = C and {y'(t)} = D, we can let x(0) and y(0) be arbitrary points of R^3 and recover the closed curves x and y from x' and y', resp., by integration. Hence (s,t) -> x(s) + y(t) is locally non-singular for all (s,t), or technically speaking, an immersion of T^2 into R^3. ⟡
Note that this does *not* prove that the image {x(s)+y(t)} is non-self-intersecting, no less whether it can be, so I suppose that remains as an unsolved problem (unless Veit know the answer).
--Dan ___________________________________________________________ * See Werner Fenchel, On the differential geometry of closed space curves, Bull. Amer. Math. Soc. 57, (1951). 44–54. This theorem is due to Vigodsky. (For the converse, note that for every plane in R^3, the two planes parallel to it that intersect the space curve at the greatest distance apart must both be tangent to the curve.)
But here's something for torus fans:
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 ?
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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