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Show that if a curve $C ⊂ S$ is both a line of curvature and a geodesic, then $C$ is a plane curve.

Give an example of a line of curvature which is a plane curve and not a geodesic. (My thoughts: Take a sphere for $S$ and let the curve $C$ be any of the latitudes of $C$ that are not the equator or the poles. Then $C$ will be a plane curve. Additionally, since the curvature is constant on the whole of $S$, every direction is a principle direction (of sorts; this might be the flaw in my argument), so any curve is a line of curvature. But, $C$ is not a segment of a great circle on $S$, so it is not a geodesic.)

Prove that a curve $C ⊂ S$ is both an asymptotic curve and a geodesic if and only if $C$ is a (segment of a) straight line.


Can I have some hints? I am not very good at covariant derivatives yet, so hand-holding with detailed examples there would be nice.

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Let $\alpha$ be an arc length parametrization of $C$. If $C$ is both a line of curvature and a geodesic then we must have $n$ parallel to $N$ where $n$ is the unit normal of $C$ and $N$ is the unit normal of $S$. Why? This tells us that $b$ the binormal vector is constant. Why? Thus $\tau=0$ where $\tau$ is the torsion of $C$ so $C$ must be a plane curve.

Suppose that $C$ is an asymptotic curve and a geodesic. Then $k_{n}=0$ and $k_{g}=0$. What does this tell us about the curvature of $C$?

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  • $\begingroup$ I'm also a beginner trying to solve this problem. Here is what I though: Let $\alpha$ be a parametrization by arc lenght of $C$. In particular, $<\alpha'',\alpha'>=0$. Since $C$ is a geodesic, $<\alpha'', \alpha'\wedge N>=0$, therefore $\alpha''\parallel N$. But the fact that $C$ is assymptotic means $<\alpha'',N>=0$, which leaves us with $\alpha''=0\Rightarrow \alpha'=\text{constant}\Rightarrow C$ is a line. Is this correct? $\endgroup$ – rmdmc89 May 4 '17 at 14:14

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