# [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 Tasha Kim
3 months, 1 week ago

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