# [Differential Geometry] What does an Asymptotic Geodesic Line of Curvature look like?

How could we show that a curve $C \subset S$ on a smooth, orientable surface 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 curve of a regular surface is such that, the tangent line of C at each point of the curve is an asymptotic direction, and along asymptotic directions the normal curvature is zero. Hence $k_n = 0$. The relationship of normal, geodesic, and usual curvature of $C$ gives: $(k_n)^2 + (k_g)^2=k^2.$ And since we have $k_n=0$, $\mid k_g \mid = k.$

Additionally, a curve is geodesic if and only if $k_g = 0$ at each point of the curve. So $k=0$. Since the usual curvature of curve C is zero at all points, if we let $\alpha(s)$ be a regular parametrisation of C, we have that $\vert \alpha ''(s) \vert \equiv 0$. Then by integration, $\alpha(s) = bs + c$, 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 usual curvature of $C$ is zero, hence $k=0$. From the relation $(k_n)^2+(k_g)^2=k^2$, we see that $(k_n)^2 \geq 0$, and $(k_g)^2 \geq 0$, so $k_n$ and $k_g$ must both be zero. Therefore, C is both asymptotic and a geodesic line of curvature.

Therefore, we conclude that $C \subset S$ is asymptotic and geodesic curve if and only if (C) is a (segment of) a straight line.

Note by Bright Glow
3 years, 2 months ago

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