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 the tangent vector of the curve, U the unit normal vector of the surface, and the absolute differential of n along (∇(α )U in your post). We will show that is a curvature line iff. 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 , T must be a principal vector.

Still let is the tangent vector to the surface and orthogonal to T. To show T is a principal vector, we show that , that is, T is the eigenvector of the second fundamental form. This is because:

So