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In general, a smooth curve $C$ in $R^3$ that lies entirely on the boundary of its convex hull, $\mathcal{H}(C)$, is defined to be convex.

  1. Does any one know of a characterization of a curve in space as convex depending on the sign of its torsion $\tau$?
  2. Is the projection of a non-singular "short" curve in space to any plane always convex?

Where a "short" curve in space is such that there is a 1-1 correspondence between the space curve and the corresponding projected curve in a plane.


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Any set $A$ lies entirely on its convex hull by definition: $A\subset CH(A)$. Or did I missunderstood something in your question? – Luc Jul 30 '13 at 9:22
I think it means "on the boundary of its convex hull". – alvarezpaiva Jul 30 '13 at 9:40
Sorry about not being precise. I am referring to the definition of a convex curve as provided by Sedykh in – Sunayana Ghosh Jul 30 '13 at 9:42
I cannot make sense of the 2nd question. Perhaps this is what is meant: If some curve $C$ in space has the property that every projection of $C$ to a plane is simple, i.e., has no self-intersections, must then $C$ lie on its convex hull in 3D? – Joseph O'Rourke Jul 30 '13 at 11:53
In any case it is impossible that all projections be simple: join two points of the curve by a segment and project in the direction perpendicular to it. The most you can ask is that almost all projections be simple. I guess this only says the curve is planar. – alvarezpaiva Jul 30 '13 at 12:38

The first question is not tractable, as one can glean by reading this paper by Bates and Melko (whence one learns that it took a century to produce a curve of constant torsion and non-vanishing curvature.) However, this paper has a lot of interesting stuff if you are interested in this circle of questions.

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That's an interesting reference! I like their CONJECTURE 3.3. There are no closed curves of nonzero constant torsion on an ovaloid. – alvarezpaiva Jul 30 '13 at 17:19
Thanks for the link to the article. The notion of signed curvature for space curves seems interesting. – Sunayana Ghosh Aug 2 '13 at 11:57

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