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Math Help - characteristic with shape operator

  1. #1
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    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.
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  2. #2
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    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.
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