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Let $M$ be a smooth $n$-manifold with boundary and let $p \in M$. If $p \in \partial M$, I would like to say that there exists a smooth boundary chart $(U, \varphi)$ such that $\varphi(U) \subseteq \mathbb H^n$ and $\varphi(p) = 0$. However, the boundary $\partial M$ is defined topologically -- it is the set of points contained in some topological chart taking $U$ to an open subset of $\mathbb H^n$ and $p$ to $\partial \mathbb H^n$. If I do such a thing, am I implicitly using the theorem on invariance of the boundary?

If that's the case, it seems like a lot of Lee's results in Introduction to Smooth Manifolds implicitly use the theorem on invariance of the boundary in this sense. This seems wrong to me since the theorem requires some heavy machinery to prove, so ideally its use should be avoided if possible. What gives?

Justin T.
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    Who says that the boundary is defined topologically? – Ted Shifrin Apr 23 '23 at 23:45
  • That is the definition given in Introduction to Smooth Manifolds, at least. (Namely, $p \in \partial M$ iff there exists a topological chart $(U, \varphi)$ containing $p$ such that $\varphi(U) \subseteq \mathbb H^n$ and $\varphi(p) \in \partial \mathbb H^n$.) – Justin T. Apr 23 '23 at 23:53
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    I don’t know Lee’s books, but this strikes me as an unfortunate choice. If we use smooth charts, then the inverse function theorem gives well-definedness of the boundary points. I agree that topological invariance of domain is a sledgehammer to be avoided. – Ted Shifrin Apr 23 '23 at 23:55

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The relevant definition here is that of a smooth manifold with boundary (see page 29 in ISM). This is a topological manifold with boundary equipped with a maximal smoothly compatible atlas of interior charts and boundary charts. In this case, invariance of the boundary is easy to prove using smooth boundary charts (see Theorem 1.46 in ISM); this doesn't depend on the topological invariance of the boundary.

Jack Lee
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    Thanks for your answer Prof. Lee! I guess my confusion is this -- if $M$ is a smooth manifold with boundary and we say that $p \in \partial M$, do we mean that there exists a smooth boundary chart taking $p$ to $\partial \mathbb H^n$, or do we only mean that there exists a topological (i.e., not necessarily smooth) boundary chart taking $p$ to $\partial \mathbb H^n$? The definition of "boundary point" (p. 25) is given in the context of topological manifolds with boundary, and I can't find any clarification on the meaning of $\partial M$ when smooth manifolds with boundary are introduced. – Justin T. Apr 25 '23 at 20:55
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We can infer from the text, that a manifold with boundary, is a Haudsorff, Second-Countable, topological space that is also at least locally Half-Euclidean of dimension $n$.

Interpreting the last of these, we have:

$$\forall p\in M:\text{ }\exists\text{ }\color{red}{\varphi_p}\in Homeo\big(U_{\text{p, open}}\subseteq M,\mathbb{H}^n\text{ or }\mathbb{R}^n\big).$$

Insert the distinction between topological and geometric boundaries: $$\partial M := \overline{M} - Int (M)$$ $$\text{vs.}$$ $$\partial M := \bigg\{p\in M\text{ }\bigg|\text{ }Im(\varphi_p) \cap \partial\mathbb{H}^n\neq \emptyset\text{ and }(\varphi_p(p))^n=0\bigg\}.$$ Recall that $\mathbb{H}^n:= \bigg\{x\in \mathbb{R}^n\text{ }\bigg|\text{ }x^n\geq 0\bigg\}.$ For ex: $\mathbb{H}^2 = \bigg\{(x,y)\in \mathbb{R}^2\text{ }|\text{ }y\geq0\bigg\}$.

From here, we may arrange for the chart (associated to a geometric boundary point) to have: $$\forall i\in\{1,...,n\}:\text{ }\varphi^i_p(p)=0.$$ by declaring: $$\forall q\in U_p:\text{ }\text{ }\color{red}{\widetilde{\varphi}_p}(q):= \varphi_p(q) - \varphi_p(p).$$ (a.k.a. re-centering a chart)

I claim this re-declaration to $\widetilde{\varphi}_p$ preserves smoothness $\color{orange}{\text{[Pf:Ex]}}$. Thus, in the presence of a smooth atlas $\mathscr{A}$ on $M$, we have: $$\forall p\in \partial M:\text{ }\exists \widetilde{\varphi}_{p,\text{ smooth}}: U_p \xrightarrow{\cong} \mathbb{H}^n;$$ $$\widetilde{\varphi}_p:p\mapsto 0$$ which was to be shown. $\square$

Kevin
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  • If $M$ is a topological space, then $\operatorname{Int}(M) = M$. $(\varphi_p(p))^n =0$ will frequently be true when $p \notin \partial M$. – aschepler Apr 24 '23 at 03:41
  • I updated the definition of $\partial M$ to include $Im(\varphi_p) \cap \partial\mathbb{H}^n\neq \emptyset$. Hopefully this suffices! – Kevin Apr 25 '23 at 06:26