# Differentiable submanifold

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

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

Originally Posted by AlexanderW
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 $\displaystyle p \in M$.

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

Quote:

Originally Posted by Laurent
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?

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.