# A straight line is an asymptotic line

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

I am using the following definitions.

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 $$\kappa_1=\kappa_2=0$$?

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 $$\kappa_n = \langle \tilde N{}',\tilde T\rangle = -\langle \tilde N,\tilde T{}' \rangle = -\langle \tilde N,0\rangle = 0.$$