Timeline for Spanning curves by flat surfaces
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 25 at 15:51 | comment | added | Dmitrii Korshunov | @DanielAsimov it seems that you argument shows that we can always slightly perturb our curve in a neighborhood of a point $P$ to obtain a curve that can be "spanned" (just push the curve into the surface away from $P$ where there can be a problem with smoothness). to ensure that the surface is immersed it is enough to chose $P$ in such a way that no line connecting $P$ with a point on the curve is tangent to the curve, which can be done by a transversality argument. | |
| Oct 3 at 3:04 | comment | added | Daniel Asimov | The question does not require that the surface be compact, so maybe the tangent developable of the curve would suffice. That is, if A : [0, 1] —> R^3 is the curve (with A'(s) nonvanishing and all kth derivatives for k ≥ 0 equal at the endpoints), this is {A(s) + t A'(s) | (s,t) ∈ [0, 1] × [0, ∞)}. | |
| Oct 2 at 17:03 | comment | added | Daniel Asimov | Taking the union of all the line segments from some given point P on the curve to each of the other points on the curve almost gives such a surface. Perhaps that idea can be refined to always work? (Possibly by a judicious choice of P.) | |
| Oct 2 at 15:22 | history | edited | Dmitrii Korshunov | CC BY-SA 4.0 |
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| Oct 2 at 15:22 | comment | added | Dmitrii Korshunov | Dear @RobertBryant, thanks for the comment. I meant an orientable surface (although I believe there can be a punctured flat torus), but both questions are interesting to me. I'll remove this remark. | |
| Oct 2 at 10:43 | comment | added | Robert Bryant | What do you mean by "necessarily a disk"? There are well-known examples of Möbius strips with Gauss curvature equal to zero. One can easily make one with a strip of paper. | |
| Oct 2 at 8:07 | comment | added | Benoît Kloeckner | I would be surprized if that would be the case, but I have no definitive argument. The first thing I would try would be deformations of the unit disk in the plane. Write down the constraint for a deformation along a vector field on the disk to be by curvature-0 surfaces, and look at that constraint on the boundary. You might be able to prove that some vector fields along the boundary cannot extend to the correct class of vector field along the disk. | |
| Oct 1 at 22:25 | history | asked | Dmitrii Korshunov | CC BY-SA 4.0 |