line of curvature question

A curve α on M is a line of curvature if α (t) is an eigenvector of the shape operator for all t. This is equivalent to saying that the unit tangent vector Tα is a principal vector. Show that α is a line of curvature iff α is parallel to ∇_{(α )}U along α

Re: line of curvature question

Denote $\displaystyle T=\alpha'$ the tangent vector of the curve, U the unit normal vector of the surface, and $\displaystyle D_T U$ the absolute differential of n along $\displaystyle \alpha$ (∇(α )U in your post). We will show that $\displaystyle \alpha$ is a curvature line iff. $\displaystyle D_T U = k T$ where k is the normal curvature.

We have shown the first part in the post line of curvature is developable. Note that there is a "-" in that post. We're choosing a different orientation to calculate the curvature so that we can omit the "-" here.

We need only to show the converse. That is, if $\displaystyle D_T U = k T$, T must be a principal vector.

Still let $\displaystyle B=U \times T$ is the tangent vector to the surface and orthogonal to T. To show T is a principal vector, we show that $\displaystyle \mathrm{I\!I}(T) = kT$, that is, T is the eigenvector of the second fundamental form. This is because:

$\displaystyle \langle \mathrm{I\!I}(T), T \rangle=\mathrm{I\!I}(T, T)=\langle D_T U, T \rangle = \langle kT, T \rangle = k$

$\displaystyle \langle \mathrm{I\!I}(T), B \rangle=\mathrm{I\!I}(T, B)=\langle D_T U, B \rangle = \langle kT, B \rangle = 0$

So $\displaystyle \mathrm{I\!I}(T) = kT$