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I am trying to prove that a curve $C\subset S$ is both an asymptotic curve and a geodesic if and only if $C$ is a (segment of) a straight line.

$\Rightarrow$ NTS: If $C$ is both asymptotic and a geodesic, then it is a straight line.

By definition asymptotic curves have normal curvature of zero. Hence $k_n = 0$. The relationship of normal, geodesic, and space curvature of $C$ gives: $(k_n)^2 + (k_g)^2=k^2$. Hence $\mid k_g \mid = k.$ A curve is geodesic if and only if $k_g = 0$ at each point of the curve. So $k=0$. Since $k=\vert \alpha ''(x) \vert \equiv 0$, then by integration $\alpha(s) = cs + d$, so the curve is a (segment of) a straight line.

$\Leftarrow$ NTS: If $C$ is a straight line, then it is both an asymptotic curve and a geodesic. If $C$ is a straight line, the curvature of $C$ is zero, $k=0$, but from the relation $(k_n)^2+(k_g)^2=k^2$, $(k_n)^2 \geq 0$, and $(k_g)^2 \geq 0$, so $k_n$ and $k_g$ must both be zero.

$\therefore$ $C \subset S$ is asymptotic and geodesic curve if and only if $C$ is a (segment of) a straight line.

Is this correct? Thanks

[Edited]

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    $\begingroup$ This was delightful to read, i have no idea about this subject, but assuming your very clear definitions are correct, this looks like a valid proof. Just wait for someone to fact check your definitions here. $\endgroup$ – frogeyedpeas May 6 '18 at 3:16
  • $\begingroup$ @frogeyedpeas Thanks you! $\endgroup$ – T J. Kim May 6 '18 at 3:59
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Yes, this is correct.

However, I think you mean to write "$C \subset S$" to mean that the curve $C$ is contained in the surface $S$, rather that "$C \in S$."

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  • $\begingroup$ Thanks for verifying. You're right, I really meant to write $C \subset S$. $\endgroup$ – T J. Kim May 6 '18 at 4:00

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