Thread: characteristic with shape operator

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.

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.