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.