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Thread: Differentiable submanifold

  1. #1
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    Differentiable submanifold

    Hello.

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

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

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

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

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

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

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

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

    Thanks in advance for help!
    Buy,
    - Alexander -
    For any differentiable curve $\displaystyle \gamma$ on $\displaystyle M$ such that $\displaystyle \gamma(0)=p$, the function $\displaystyle s\mapsto \|q-\gamma(s)\|^2$ has a minimum at 0, hence its derivative at this point is 0, which gives $\displaystyle 2(q-\gamma(0),-\gamma'(0))=0$, i.e. $\displaystyle (q,\gamma'(0))=0$: $\displaystyle q$ is orthogonal to $\displaystyle \gamma'(0)\in T_pM$. Since $\displaystyle T_pM$ is spanned by the vectors of the form $\displaystyle \gamma'(0)$ (this is even one possible definition), we deduce that $\displaystyle q$ is orthogonal to $\displaystyle T_pM$.
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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 $\displaystyle \gamma$ on $\displaystyle M$ such that $\displaystyle \gamma(0)=p$, the function $\displaystyle s\mapsto \|q-\gamma(s)\|^2$ has a minimum at 0, hence its derivative at this point is 0, which gives $\displaystyle 2(q-\gamma(0),-\gamma'(0))=0$, i.e. $\displaystyle (q,\gamma'(0))=0$: $\displaystyle q$ is orthogonal to $\displaystyle \gamma'(0)\in T_pM$. Since $\displaystyle T_pM$ is spanned by the vectors of the form $\displaystyle \gamma'(0)$ (this is even one possible definition), we deduce that $\displaystyle q$ is orthogonal to $\displaystyle T_pM$.
    I don't understand why do you consider the function $\displaystyle s\mapsto \|q-\gamma(s)\|^2$ and not the function $\displaystyle s\mapsto \|q-\gamma(s)\|$.
    And why is the derivation at 0 of $\displaystyle s\mapsto \|q-\gamma(s)\|^2$ the same as $\displaystyle 2(q-\gamma(0),-\gamma'(0)) $ ? Is $\displaystyle (.,.)$ 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 $\displaystyle s\mapsto \|q-\gamma(s)\|^2$ and not the function $\displaystyle s\mapsto \|q-\gamma(s)\|$.
    The minima of $\displaystyle x\mapsto\|q-x\|$ and $\displaystyle x\mapsto\|q-x\|^2$ are the same, and the second one is easier to differentiate, that's why.

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