Let $M$ be a smooth manifold, and let $S$ be an open smooth submanifold of $N$.

Assume the topological boundary of $S$, $\partial S :=\bar S \setminus S$ is a smooth submanifold of codimension 1 in $N$.

Is it true that $\bar S$ is a submanifold with boundary of $N$?

Somehow, I am not even sure whether $\bar S$ must be a topological manifold with boundary...

I am trying to see if the smoothness of the boundary is the only obstruction for being a submanifold with boundary. (If we do not assume this, we can take $S$ to be the interior of a square).

  • 3
    $\begingroup$ Are $M$ and $N$ supposed to be the same? $\endgroup$ – Jack Lee Aug 13 '16 at 20:30

I'm assuming that your first sentence was meant to say "Let $N$ be a smooth manifold, $\dots$ ."

Yes, it is always the case that $\overline S$ is a smooth submanifold with boundary in $N$. However, the boundary of $\overline S$ need not be equal to the boundary of $S$. See this post for an example in which it's not.

To prove that $\overline S$ is a smooth submanifold with boundary, you just need to show that each point of $\overline S$ has a coordinate neighborhood in $N$ whose image is either an open subset of $\mathbb R^n$ or an open subset of the half-space $\mathbb R^{n-1} \times [0,\infty)$. Certainly each point of $S$ has such a neighborhood. For a point of $\partial S$, you can choose a slice chart $(U,\phi)$ for $\partial S$, such that $\phi(U)$ is a coordinate ball centered at the origin, and $\phi(\partial S\cap U)$ is the subset of $\phi(U)$ where $x^n=0$. Now you have to show that one of the following three possibilities holds:

  1. $\phi(\overline S\cap U) = \phi(U) \cap \{x^n\ge 0\}$,
  2. $\phi(\overline S\cap U) = \phi(U) \cap \{x^n\le 0\}$, or
  3. $\phi(\overline S\cap U) = \phi(U) $.

I'll leave it to you to work out the details.

  • $\begingroup$ Commenting to add that the converse need not be true, and it may be a good exercise for the original questioner to dream up counterexamples. $\endgroup$ – Neal Aug 13 '16 at 21:04
  • $\begingroup$ I am not sure what is the converse statement you refer to: Are you saying that if $S$ is a manifold with boundary (of codimension zero), than it is not always true that $\partial S$ is a smooth submanifold of codimension 1? $\endgroup$ – Asaf Shachar Aug 14 '16 at 9:09

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.