I want to prove that a straight line on a regular surface is an asymptotic line.

I am using the following definitions.

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Here it is assumed that $M \subset \mathbb{R}^2$.

Let $\gamma: [0,L] \to M$ be an arclenth-parametrization of a straight line in $M$ and consider $\tilde{\gamma}=f \circ \gamma$. Since $\gamma$ is a planar curve we have

$$ \gamma''=T'=\kappa_{\gamma} N_{\gamma} $$

where $\kappa_{\gamma}$ denotes the curvature of $\gamma$ and $N_{\gamma}$ the normal. Since $\gamma$ is a straight line we have $\kappa_{\gamma}=0$ and therefore $\gamma''=0$. Let $A$ denote the shape operator. For the normal curvature we have

$$ \kappa_n =\langle \tilde{N} , \tilde{T} \rangle =\langle A \gamma' , \gamma' \rangle $$

so the normal curvature equals the directional curvature of $f$ in the direction of $\gamma'$. Deriving the expression yields

$$ \kappa'_n=\langle A \gamma'' , \gamma' \rangle+\langle A \gamma' , \gamma'' \rangle=0. $$

This implies that the normal curvature is constant so we have $\kappa_n=0$ or $\kappa_n \neq 0$. Since $\kappa_n$ is a directional curvature we consider the formula

$$ \kappa_n=\kappa_1 \cos^2(\theta)+\kappa_2 \sin^2(\theta) $$

where $\theta \in [0,2 \pi]$ and $\kappa_1, \kappa_2$ denote the principal curvatures. We know that $\kappa_1, \kappa_2$ are constants but how can I conclude that $\kappa_1=\kappa_2=0$?


1 Answer 1


You certainly do not want to conclude that $\kappa_1=\kappa_2=0$. No one said that the surface is planar. You have a typo in your formula for $\kappa_n = \langle \tilde N{}',\tilde T\rangle$.

This is the standard and important "trick" you should never forget: Since $\langle \tilde N,\tilde T \rangle = 0$ for all $s$, we get $$\langle \tilde N{}',\tilde T\rangle + \langle \tilde N,\tilde T{}' \rangle = 0,$$ so in this case — since $\gamma$ is a line — we have $$\kappa_n = \langle \tilde N{}',\tilde T\rangle = -\langle \tilde N,\tilde T{}' \rangle = -\langle \tilde N,0\rangle = 0.$$


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