Differential Geometry and Its Application - Exercise 2.4.4

Problem:

Solution:

By definition, we have this
$\nabla_{u_1} U = -S_{p}(u_1) = - k_1 u_1$

Suppose $\alpha' = u_1$, then $U' = \nabla_{\alpha'} U = \nabla_{u_1} U = -S_{p}(u_1) = - k_1 u_1  =  - k_1 \alpha'$.

Suppose $\alpha' \ne u_1$ and $\alpha' \ne u_2$ , then $U' = \nabla_{\alpha'} U = -S_{p}(\alpha') \ne c \alpha'$ for constant $c$ as $\alpha'$ is not an eigenvector.

For the second part, as the angle between the surface and the plane is constant, we get

$U_M \cdot U_P = c$.

So we can take the directional derivative for this one.

$0 = \alpha'[U_M \cdot U_P] = \alpha'[U_M] \cdot U_P + \alpha'[U_P] \cdot U_M$

$-\alpha'[U_M] \cdot U_P = \alpha'[U_P] \cdot U_M$

$S_p(\alpha') \cdot U_P =0$

We know $S_p(\alpha')$ is on the tangent plane. We now also know it is on $P$, so it has to be $k\alpha'$. So $\alpha'$ is an eigenvector of the Weingarten map, and the curve is a line of curvature.