# Thread: characteristic with shape operator

1. ## characteristic with shape operator

Let $v = v_1 \boldsymbol{x_u} +v_2\boldsymbol{x_v}$ show that v is principal if and only if $S(v) \times v = 0$. (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 $(\boldsymbol{x_u},\ \boldsymbol{x_v})$ will be the basis for the entire set. So what I have so far is

$S(v) \times v = S(v_1 \boldsymbol{x_u} +v_2\boldsymbol{x_v}) \times (v_1 \boldsymbol{x_u} +v_2\boldsymbol{x_v})$ $= dN(u'(0) \boldsymbol{x_u} +v'(0)\boldsymbol{x_v}) \times (v_1 \boldsymbol{x_u} +v_2\boldsymbol{x_v})$ (where N is the unit normal vector)

Now I don't know how I would expand it further.

2. 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.