# Differentiable submanifold

• Apr 29th 2010, 12:35 PM
AlexanderW
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 $p \in M$.

- Alexander -
• Apr 29th 2010, 01:08 PM
Laurent
Quote:

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 $p \in M$.

- 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 $T_pM$.
• Apr 29th 2010, 01:57 PM
AlexanderW
Hello

Quote:

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 $T_pM$.

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?

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.