6
$\begingroup$

Consider this definition:

A space $M$ is a manifold with boundary if each point $x\in M$ has a neighborhood $U_x$ that is homeomorphic to $\mathbb R^n$ or to $\mathbb R^n_+=\{(x_1,\cdots,x_n)\;|\; x_n \ge 0\}$; the points which have neighborhoods homeomorphic to $\mathbb R^n_+$ form the boundary $\partial M$ of $M$.

I want to understand how precise this definition is. Indeed, a point that has an open neighborhood homeomorphic to $\mathbb R^n$ can also have another open neighborhood that is homeomorphic to $\mathbb R^n_+$ and in that case is this point on the boundary $\partial M$ or in $M\setminus \partial M$ ? How to choose? Is my definition incomplete so that I have to say: "In a manifold with boundary, a point in $M$ has an open neighborhood homeomorphic to $\mathbb R^n$ but if it doesn't have such neighborhood then it has to have an open neighborhood homeomorphic to $\mathbb R^n_+$"

Thank you for your help!

$\endgroup$
5
  • $\begingroup$ The center $(0,0)$ of the disk $D=\{(x,y)|x^2+y^2\le 1\}$ can be put in a little disk or in a little half disk $\endgroup$
    – palio
    Dec 1, 2017 at 18:11
  • 4
    $\begingroup$ This definition is not stated correctly. The way you've stated it, every point in a neighborhood homeomorphic to $\mathbb R^n_+$ would be considered a boundary point, including many points in the interior. You have to say something like "the boundary consists of those points that have no neighborhood homeomorphic to $\mathbb R^n_+$." $\endgroup$
    – Jack Lee
    Dec 1, 2017 at 18:36
  • 1
    $\begingroup$ @JackLee you mean no neighborhood homeomorphic to $\mathbb R^n$ $\endgroup$
    – palio
    Dec 1, 2017 at 18:47
  • $\begingroup$ Oops, you're right. $\endgroup$
    – Jack Lee
    Dec 1, 2017 at 18:48
  • $\begingroup$ What text is the source of the quoted definition? All quotations should be attributed. $\endgroup$
    – bof
    Dec 8, 2017 at 1:43

3 Answers 3

3
$\begingroup$

As mentioned in answers and comments, the definition given for a boundary point is not the correct one. A possible one should read:

A point $p \in M$ is a boundary point if it is such that $\pi_n(\phi(p))=\phi_n(p)=0$ for some chart $\phi:U \to \mathbb{R}^n_{+}$.

Then, I think, you would be willing to prove that if this is true for one chart $\phi: U \to \mathbb{R}^n_+$ as per the definition above, then it is true for any such chart. As one other answer suggests and references, this is not a big deal in the smooth setting.

This is true in the topological setting too, but due to a more involved reason: namely, the invariance of domain theorem. I don't know a reference which manages to avoid this.

Now, moving forward: Indeed, let $\psi: V \to \mathbb{R}^n_+$ be another chart. What we must prove then is that $\psi_n(p)=0$.

Instead of doing that, let's prove that if $\phi_n(p)>0$, then $\psi_n(p)>0$. It is easy to see (exchange places of $\phi$ and $\psi$) that this will prove that $\phi_n(p)>0$ if and only if $\psi_n(p)>0$, thus concluding that $\phi_n(p)=0$ if and only if $\psi_n(p)=0$.

Thus, suppose $\phi_n(p)>0$. We have that $\psi \circ \phi^{-1}$ is a continuous injection from $U \cap\mathbb{R}^n_+$ to $\mathbb{R}^n_+$, where $U$ is an open subset of $\mathbb{R}^n$. Restricting, $\psi \circ \phi^{-1}|_{U \cap \mathbb{R}^n_{>0}}: U \cap \mathbb{R}^n_{>0} \to \mathbb{R}^n$ is then injective and continuous. By invariance of domain, we know that such an image is open in $\mathbb{R}^n$. But this image is inside $\mathbb{R}^n_+$, therefore it can't intersect the set $\{x \mid x_n=0\}$. Since $\psi(\phi^{-1}(\phi(p)))=\psi(p)$ is in such image, it follows that it can't be such that $\psi_n(p)=0$.

As a sidenote which may also be of interest, a similar use of invariance of domain also yields that if $p$ is a boundary point, then there is no chart $\phi: U \to \mathbb{R}^n$ around $p$ (which is a homeomorphism with $\mathbb{R}^n$).

$\endgroup$
2
$\begingroup$

If $M$ is a manifold with boundary, $U$ is an open set in $M$ and $\varphi\colon U \to \Bbb \varphi[U]\subseteq \Bbb R^n_+$ is a chart, you can prove that if $p \in U$ is such that $\varphi(p) = (x^1(p),\ldots,x^{n-1}(p),0)$, then if $\widetilde{\varphi}\colon \widetilde{U} \to \widetilde{\varphi}[\widetilde{U}]\subseteq \Bbb R^n_+$ is another chart, then $\widetilde{\varphi}(p)$ has also the form $(\widetilde{x}^1(p),\ldots,\widetilde{x}^{n-1}(p),0)$.

See Lemma 24.2 in page 205 of Munkres' Analysis on Manifolds.

$\endgroup$
3
  • $\begingroup$ I'm sorry Ivo, I'm not sure I understand your point here ! $\endgroup$
    – palio
    Dec 1, 2017 at 18:56
  • 2
    $\begingroup$ Let's say that a point $p$ is a boundary point "with respect" to $\varphi$ if the last component of $\varphi(p)$ is zero. What I am trying to say is that if $p$ is a boundary point with respect to one chart, then it must be with respect to all charts. This is stated (and proven) along with the correct definition in Munkres' book, for example. $\endgroup$
    – Ivo Terek
    Dec 1, 2017 at 19:39
  • $\begingroup$ I'm working purely in the topological setting, I think you are speaking in differential terms as does Munkres in the book you mentioned.. $\endgroup$
    – palio
    Dec 2, 2017 at 18:08
0
$\begingroup$

This is about existence of atleast one neighborhood of a point $p\in M$ that is homeomorphic to $\mathbb{R}^n$. Obviously it might have some other neighborhood which intersect with boundary and will be homeomorphic to $\mathbb{R}_{+}^{n}$ but we don't have to worry about those. If a point $p \in M$ has no neighborhood homeomorphic to $\mathbb{R}^n$, then $p \in \partial M$.

$\endgroup$

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.