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Reading the Tu.'s book "introduction to manifolds" it first defines manifolds as topological spaces which are locally euclidean (which means locally they are homeomorphic to the euclidean space $R^n$). Then, when it introduces manifolds with boundaries, it takes as a model space the upper half space $H^n$ (defined as all the points of $R^n$ having the n-th coordinate $x_n \ge 0$). But the space $H^n$ is not homeomorphic to $R^n$ (as you can see from this post). So I don't understand, how we are allowed to substitute the model space like this (in the sense that the manifolds using $R^n$ as model space woud still be manifolds under this substitution)? Thank you.

Andrea
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  • Each point of $\Bbb R^n$ is contained in a subset of $\Bbb R^n$ that is homeomorphic to $H^n$. – Karl Jan 19 '24 at 09:38

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This is because $H^n$ contains the open half-space $\mathbb R^n_+ = \{(x_i) \in \mathbb R^n \mid x_n > 0 \}$ which is open in $\mathbb R^n$. It is homeomorphic (even diffeomorphic) to $\mathbb R^n$, but this is irrelevant here.

A chart $(U,\phi)$ with model $H^n$ is a homeomorphism $\phi : U \to V$ from an open subset of $M$ to an open subset of $H^n$. There are two types of charts around points $p \in M$:

  1. Charts $(U,\phi)$ such that $\phi(p) \in \partial H^n$

  2. Charts $(U,\phi)$ such that $\phi(p) \notin \partial H^n$

A subtype of 2. is

  1. Charts $(U,\phi)$ such that $\phi(U) \subset \mathbb R^n_+$

All charts of type 3. are also "ordinary" charts with model $\mathbb R^n$.

In the second case you can always shrink $U$ to an open subset $U'$ containing $p$ such that $V' := \phi(U') \subset \mathbb R^n_+$ which gives you the chart $(U',\phi \mid_{U'} : U' \to V')$ of type 3.

This partitions $M$ into interior points (having an "ordinary" chart around it) and boundary points (having no "ordinary" chart around it).

It is clear that the set of interior points is a manifold without boundary.

The essential point is this: For a manifold $M$ without boundary all charts with model $H^n$ have type 3. This is a non-trivial fact; Tu proves it for charts in a smooth atlas (but it is true for any chart on a topological manifold). Thus the charts on $M$ with model $H^n$ are also charts with model $\mathbb R^n$. There are less charts with model $H^n$ because the range of such charts is contained in $\mathbb R^n_+$. The collection of all these charts is still an atlas, but is not the maximal atlas in the sense of the standard definition with model $\mathbb R^n$.

Paul Frost
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  • Just to be clear, this is one of those things that requires a maximal atlas. Otherwise you can take, say, $\Bbb R^1$, a well-known manifold without boundary, cover it $(-\infty, 0]$ and $[0, \infty)$, and then you have the interior point $0$ being only on the boundary on any of its charts. – Arthur Jan 19 '24 at 10:11
  • @Arthur $(-\infty,0]$ and $[0,\infty)$ are not open in $\mathbb R$, thus cannot be domains of charts. – Paul Frost Jan 19 '24 at 10:29
  • Wasn't the point here that we are using only charts homeomorphic to $H^n$ in the manifold-with-boundary definition? – Arthur Jan 19 '24 at 10:34
  • @Arthur See my update. – Paul Frost Jan 19 '24 at 10:37
  • Ok, that makes sense. I see now. Charts are open subsets homeomorphic to an open subset of $H^n$, rather than subsets homeomorphic to $H^n$. – Arthur Jan 19 '24 at 10:38
  • Thank you for your asnwer, maybe I'm totally mistaken but for a function to be a homeomorphism doesn't it need to be surjective (and how can it be between $R^n_+$ and $R^n$)? – Andrea Jan 19 '24 at 11:12
  • @Andrea The map $h : \mathbb R^n \to \mathbb R^n_+, h(x_1,\ldots,x_n) = (x_1,\ldots, x_{n-1},e^{x_n})$, is a homeomorphism. But actually this is irrelevant here and thus was misleading. See my update. – Paul Frost Jan 19 '24 at 11:18