Differentiable submanifold

**AlexanderW** · April 29th 2010, 12:35 PM

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 -

**Laurent** · April 29th 2010, 01:08 PM

Originally Posted by AlexanderW

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 $T_pM$ .

**AlexanderW** · April 29th 2010, 01:57 PM

Hello
and thank you for your answer.

Originally Posted by Laurent

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 $T_pM$ .