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Math Help - Differentiable submanifold

  1. #1
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    Differentiable submanifold

    Hello.

    Let M \subset \mathbb R^n be a differentiable submanifold (with the induced metrics from \mathbb R^n),  q \in \mathbb R^n\setminus M,

    and g:M\to \mathbb R, \quad x \mapsto \left \| q-x \right \| has a minimum p \in M.

    Proof, that the vector q-p is orthogonal to T_p M.

    T_p M is the tangent space at p \in M.

    Thanks in advance for help!
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    - Alexander -
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  2. #2
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    Quote Originally Posted by AlexanderW View Post
    Hello.

    Let M \subset \mathbb R^n be a differentiable submanifold (with the induced metrics from \mathbb R^n),  q \in \mathbb R^n\setminus M,

    and g:M\to \mathbb R, \quad x \mapsto \left \| q-x \right \| has a minimum p \in M.

    Proof, that the vector q-p is orthogonal to T_p M.

    T_p M is the tangent space at p \in M.

    Thanks in advance for help!
    Buy,
    - Alexander -
    For any differentiable curve \gamma on M such that \gamma(0)=p, the function s\mapsto \|q-\gamma(s)\|^2 has a minimum at 0, hence its derivative at this point is 0, which gives 2(q-\gamma(0),-\gamma'(0))=0, i.e. (q,\gamma'(0))=0: q is orthogonal to \gamma'(0)\in T_pM. Since T_pM is spanned by the vectors of the form \gamma'(0) (this is even one possible definition), we deduce that q is orthogonal to T_pM.
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  3. #3
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    Hello
    and thank you for your answer.

    Quote Originally Posted by Laurent View Post
    For any differentiable curve \gamma on M such that \gamma(0)=p, the function s\mapsto \|q-\gamma(s)\|^2 has a minimum at 0, hence its derivative at this point is 0, which gives 2(q-\gamma(0),-\gamma'(0))=0, i.e. (q,\gamma'(0))=0: q is orthogonal to \gamma'(0)\in T_pM. Since T_pM is spanned by the vectors of the form \gamma'(0) (this is even one possible definition), we deduce that q is orthogonal to T_pM.
    I don't understand why do you consider the function s\mapsto \|q-\gamma(s)\|^2 and not the function s\mapsto \|q-\gamma(s)\|.
    And why is the derivation at 0 of s\mapsto \|q-\gamma(s)\|^2 the same as 2(q-\gamma(0),-\gamma'(0)) ? Is (.,.) the dot product?

    Thanks in advance!
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  4. #4
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    Quote Originally Posted by AlexanderW View Post
    I don't understand why do you consider the function s\mapsto \|q-\gamma(s)\|^2 and not the function s\mapsto \|q-\gamma(s)\|.
    The minima of x\mapsto\|q-x\| and x\mapsto\|q-x\|^2 are the same, and the second one is easier to differentiate, that's why.

    And why is the derivation at 0 of s\mapsto \|q-\gamma(s)\|^2 the same as 2(q-\gamma(0),-\gamma'(0)) ? Is (.,.) the dot product?
    Yes, it is dot product. The differential of N:x\mapsto \|x\|^2 is dN_x(h)=2(x,h). And I used the chain-rule.
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