# characteristic with shape operator

• Aug 19th 2010, 10:27 PM
lllll
characteristic with shape operator
Let $\displaystyle v = v_1 \boldsymbol{x_u} +v_2\boldsymbol{x_v}$ show that v is principal if and only if $\displaystyle S(v) \times v = 0$. (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 $\displaystyle (\boldsymbol{x_u},\ \boldsymbol{x_v})$ will be the basis for the entire set. So what I have so far is

$\displaystyle 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})$$\displaystyle = 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.
• Aug 20th 2010, 05:49 AM
xxp9
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.