2
$\begingroup$

Suppose $M$ is a smooth manifold with boundary, show that there exists a smooth function $f: M \rightarrow [0, \infty)$ such that $\partial M = f^{-1}(0)$.

My attempt is that given a chart $(U_\alpha, \varphi _\alpha)$ of $M$, $U_\alpha$ is diffeomorphic to some open subset of $\mathbb H ^m = \mathbb R^{m-1} \times [0,\infty)$. Then the m-th coordinate function is a smooth function from $U_ \alpha$ to $[0, \infty)$. But I don't know how to do this globally.

$\endgroup$
5
  • 2
    $\begingroup$ Have you seen partitions of unity? $\endgroup$
    – Jose27
    Nov 7 '14 at 2:00
  • $\begingroup$ I tried but still don't know how partition of unity could work here. Would you please elaborate a bit more? Thanks! @Jose27 $\endgroup$
    – lyx
    Nov 7 '14 at 2:20
  • $\begingroup$ If $f_n$ is the function you obtained in a coordinate patch $U_n$, and $\psi_n$ is a partition of unity subordinate to this cover by closed sets then consider $\sum_n f_n\psi_n$ (Notice that you still have to do something about points in $M\setminus \cup U_n$, but this is not a problem because...) $\endgroup$
    – Jose27
    Nov 7 '14 at 3:08
  • $\begingroup$ Do you already know that $\partial M$ has a neighborhood diffeomorphic to $\partial M \times [0,1)$? $\endgroup$ Nov 7 '14 at 3:53
  • $\begingroup$ @Jose27, thank you so much! The explanation is pretty clear and straightforward $\endgroup$
    – lyx
    Nov 7 '14 at 14:50
0
$\begingroup$

Take a collar neighborhood, ie a diffeomorphism $\partial M \times [0,1) \to U \subset M$ where $U$ is a neighborhood of $\partial M$ and this diffeomorphism takes $\partial M \times 0$ identical to the boundary of $M$. Hence after collapsing $M/M-U$ you get actually the cone $C$ over $\partial M$, which has the projection and you define $M \to M/M-U \to [0,1]$ which is onto.

$\endgroup$

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.