a principle direction v is defined to be the eigenvector of the Weingarten map S.
So S(v) = kv, where k is the eigenvalue, the principle curvature.
So S(v) x v = kv x v = 0.
And vice versa.
Let show that v is principal if and only if . (where S(v) denotes the shape operator)
I'm having a hard time working out what S(v) \times v will look like once expanded. I figure that will be the basis for the entire set. So what I have so far is
(where N is the unit normal vector)
Now I don't know how I would expand it further.