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Let $S$ be a regular surface, and let's consider $\gamma: I \to S$ be a geodesic. Let $ N: S \to S^2 $ be the gauss map. Then $ \beta(s) = N(\alpha(s))$ is a curve $\beta : I \to S^2$ (where $S^2$ denotes the unit sphere). I want to prove that $\beta$ it's also a geodesic, and also I want to know a more general result considering other kind of maps, and not only the gauss map N.

Trafalgar Law
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    In general, $\beta$ will almost never be a geodesic. Since geodesics on a sphere are great circles, all the normal vectors along $\gamma$ would have to lie in a plane. Of course, it does work on a flat surface, a sphere, or for the meridians on a surface of revolution. – Ted Shifrin Jul 02 '13 at 04:27

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As Ted observes in a comment, if the Gauss map sends geodesics in your surface $S$ to geodesics in the sphere, then geodesics in $S$ are plane curves.

It is a classic result that a (connected!) surface all of whose geodesics are plane curves is in fact (part of) a plane or a sphere.

This last fact is a simple exercise. To start, show that, in general, a geodesic which is a plane curve is a line of curvature. Since in our surface all geodesics are lines of curvature, all of its points are umbilical. Next show that if a surface has all its points umbilical it is a part of a plane or a sphere. These two observations are standard exercises, and should be in pretty much every textbook on the geometry of surfaces.

Mariano Suárez-Álvarez
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  • But what if $S$ it's a surface that is homeomorphic to the sphere, then it's true? – Trafalgar Law Jul 03 '13 at 20:56
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    It has to be isometric to a round sphere for this to be true —homeomorphism is not enough. Already ellipsoids have non-planar geodesics! – Mariano Suárez-Álvarez Jul 03 '13 at 21:00
  • @MarianoSuárez-Alvarez: No, I definitely did not say that the geodesics in $S$ had be planar. Indeed, this is false. Take any helix on a cylinder, for example! – Ted Shifrin Jul 04 '13 at 01:50
  • "Since in our surface all geodesics are lines of curvature, all of its points are umbilical." Can you please explain why this assertion is true? I have no clue how to connect lines of curvatures to umbilical points. – Peanica Jun 24 '19 at 10:12
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Any curve $\gamma$ with this property is either a planar line of curvature (such as a profile curve, or meridian, on a surface of revolution) or a generalized helix — a curve on a generalized cylinder that makes a constant angle with the rulings. These curves are characterized by having $\tau/\kappa$ constant ($\tau$ being torsion and $\kappa$ curvature).

Ted Shifrin
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