Given a smooth closed connected curve $\gamma$ in $\mathbb R^3$, is there an immersed surface $S$ with boundary, such that its Gaussian curvature is equal to zero and $\partial S=\gamma$?
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1$\begingroup$ 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. $\endgroup$– Benoît KloecknerCommented Oct 2 at 8:07
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4$\begingroup$ 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. $\endgroup$– Robert BryantCommented Oct 2 at 10:43
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$\begingroup$ 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. $\endgroup$– Dmitrii KorshunovCommented Oct 2 at 15:22
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1$\begingroup$ 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.) $\endgroup$– Daniel AsimovCommented Oct 2 at 17:03
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2$\begingroup$ 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, ∞)}. $\endgroup$– Daniel AsimovCommented Oct 3 at 3:04
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