Let $M$ be a compact riemannian manifold with boundary $\partial M\neq \varnothing$. I would like to show that there is some neighborhood $U$ of $\partial M$ which is diffeomorphic to $[0,a)\times \partial M$, for some (and thus for all) $a>0$. I would like also this to be true for a differentiable manifold (without having to consider a riemannian metric but since any differentiable manifold admits a riemannian metric, this is not a problem).

What I've tried: first, note that $\partial M$ is also compact, since it's closed in $M$. Also, since $M$ is compact, the injectivity radius $inj(M)>0$. Start with any real number $0<b<inj(M)$. Then define

$$\begin{array}{cccc}F:&[0,b]\times\partial M&\to&M\\ &(t,p)&\mapsto&\exp_p(t\nu(p)) \end{array},$$

where $\nu(p)$ is the inward unit vector normal to $\partial M$ through $p$.

Then I verified that $d F_{(0,p)}$ is an isomorphism, for all $p\in \partial M$ and then tried to use the inverse function theorem.

Using the compactness of $\partial M$ this gave me a finite collection of open sets $[0,a_i)\times U_i$, $U_i\subset \partial M$, such that $U_i$ covers $\partial M$ and the restrictions on any of these sets is a diffeomorphism.

Then I chose $a<\min\{a_i\}$.

I wished $F$ restricted to $[0,a)\times \partial M$ to be a diffeomorphism on its image. But this seems not to work...

For example, how do I avoid such "transversal" intersections? (the injectivity radius condition avoid this?)

Or, if $a$ is too big it seems that something like this can happen:

  • 1
    $\begingroup$ What you want to show here follows from the continuity of the injectivity radius, which (in a compact manifold) implies a lower bound for it. See here for some thoughts on this and for some references, too: mathoverflow.net/questions/283467/… $\endgroup$ – Thomas Oct 20 '18 at 15:46

Since you have that the map $F$ restricted to $[0,a) \times \partial M$ is a local diffeomorphism, it suffices to see that it is injective order to conclude what you want.

Suppose there is no such $a>0$ that makes it injective. It follows that there exists sequences $x_n,y_n \in \partial M$, together with sequences $a_n, b_n$ which go to $0$, for which $F(a_n,x_n)=F(b_n,y_n)$ and $(a_n,x_n) \neq (b_n,y_n)$. Since $\partial M$ is compact, passing to subsequences if necessary, we have that $x_n \to x$ and $y_n \to y$.

Since $F(a_n,x_n) \to F(0,x)=x$ and $F(b_n,y_n) \to F(0,y)=y$, it follows that $x=y$. But since $F$ is locally a diffeomorphism (hence, injective) near $x$, we have a contradiction, because for $n$ sufficiently big $(a_n,x_n),(b_n,y_n)$ are inside a neighbourhood of $(x,0)$ on which $F$ was supposed to be one-to-one, but $F(a_n,x_n)=F(b_n,y_n)$ by construction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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