[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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