Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$(M,g)$ is a Riemannian manifold and $N$ is a submanifold of $M$, then is the function $r(x)=\mathop {\min }\limits_{y \in N} d(x,y)$ smooth near $N$? ($d(x,y)$ is the distance function induced by Riemannian metric $g$ )

I think a possible proof may involve the tubular neighborhood of $N$, that is, the normal bundle of $N$ is diffeomorphic to a neighborhood of $N$ in $M$. But I am not sure how to prove that $r(x)$ is equal to the length of the normal vector $v$ where exp$(y,v)=x$ for some $y$ in $N$.

Note: Here we say $r(x)$ is smooth "near" $N$ means that $r(x)$ is smooth on $U-N$ where $U$ is a neighborhood of $N$. For example, when $M$ is $\mathbb R$, $N$ is $\{0\}$, the function $\left| x \right|$ is smooth near $0$.

share|improve this question
add comment

1 Answer

No the function $r$ is not smooth near $N$.
The simplest counter-example is $M=\mathbb R$ with its usual riemannian metric and $N=\lbrace 0\rbrace$, since $r(x)=\mid x\mid $ which is not differentiable in any neighbourhood of $0$.

share|improve this answer
    
By "near" I mean $r(x)$ is smooth in a small neighborhood of $N$ except on $N$. –  Hezudao Sep 20 '12 at 12:05
1  
Dear Hezudao, I suggest you add this definition of "near" in your question, so as to remove all ambiguity ("near $N$" often means "in a neighbourhood of $N$" and thus does not exclude $N$). –  Georges Elencwajg Sep 20 '12 at 14:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.