# gauss map takes geodesics to geodesics

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.

• 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 2 '13 at 4:27

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.
• But what if $S$ it's a surface that is homeomorphic to the sphere, then it's true? – Trafalgar Law Jul 3 '13 at 20:56
• @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 4 '13 at 1:50
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).