Manifolds with boundaries: Why map boundary points onto boundary points?

Question

The notion of a manifold with boundary has just been introduced in my script. If $\mathbb{R}^n_+$ is defined as $\mathbb{R}^n_+ := \{ (x^1, \ldots, x^n) \in \mathbb{R}^n \; | \; x^n \geq 0 \}$, then a manifold with boundary $M$ is an atlas whos charts $\phi : M \longrightarrow V \subseteq \mathbb{R}^n_+$.

Now, as far as I understand, it is required that the preimage of each point in $\mathbb{R}^n$ with $x^n = 0$ is a boundary point on $M$. In other words each boundary point of $M$ gets mapped onto the hyperplane defined by $x^n = 0$.

I don't really understand this requirement. I think my intuition of either this definition or of a manifold/coordinate chart is wrong. I always understood the coordinate charts as local coordinates of the manifold in Euclidean space. A ball in 3 dimensions for example can be described by $x,y,z$. But describing its boundary with $z = 0$ is pretty uncommon when parametrising a ball, isn't it?

How is that definition or the coordinate charts of a manifold to be interpreted with respect to this example?

Answer 1

Get rid of that image of a manifold given by an embedding into some $\Bbb R^n$. An atlas describes the manifold without reference of any surrounding space (if such a space exists at all). Think of a classical atlas of the (surface of the) Earth: Nowhere do you work with three-dimensional coordinates, but rather on each page of the atlas is a flatt part of $\Bbb R^2$ (unlike when working with a globe instead of an atlas), you can reference a point on Earth (provided it is in the region depicted in that map) by two coordinates only (e.g., distance from left and from top edge of the page). Thanks to the various maps overlapping (though with some distortions), you can manage to navigate across all of the world.

Now for a manifold with boundary, imagine we cut a hole into the globe, e.g., we remove the continent on Antarctica and deny its existence, or we imagine that its boundary is some unpassable boundary (perhaps an ice wall guarded by armed penguins?). If you look at your standard atlas from your shelf and imagine that you simply erase all parts of Antarctica as terra incognita or more precisely as terra non existens, you will get some maps that extend only to parts of the pages, and only for some directions there are maps that show some extension. For convenience, you may assume that all those maps showing some boundary are distorted to make the a priori wildly curved boundary straight and transform the map further to bring that straight boundary to the $y=0$ line and perhaps write "Here be Dragons" on the other side of that line. (It may be necessary to split up some maps into several, namely when you originally had a map that showed the complete closed boundary of Antarctica). This doesn't change the fact that the atlas can be used to navigate the modified world, it just simplifies our notion of a maps with boundary and thereby simplifies working with such an atlas.

Answer 2

In addition to Hagen von Eitzen's great answer, let me add something basic. $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\B}{\mathbb{B}}$ $\newcommand{\pl}{\partial}$ $\renewcommand{\S}{\mathbb{S}}$

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#1 The definition of a manifold (either with or without boundary) is based on comparison with some model manifold (with or without boundary, respectively). The convention is to take:

• $\R^n$ as the model manifold without boundary,
• $\R^n_+$ as the model manifold with boundary (with flat boundary $\{ x^n = 0 \} \cong \R^{n-1}$).

In principle, you could replace $\R^n$ and/or $\R^n_+$ with your favorite manifold (say, sphere $\S^n$ and closed ball $\B^n$), and the theory would look the same. Oversimplfying a bit, we choose our convention because the spaces $\R^n$ and $\R^n_+$ are easily parametrized and the boundary of $\R^n_+$ is also easily parametrized.

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#2 The reason why the choice of model manifolds doesn't matter is the following: manifolds are locally the same. Of course, they are different globally, but the definition of atlas only considers small parts of the manifold.

And the same goes for manifolds with boundary: the $1$-dimensional manifold with boundary $M = [0,1]$ doesn't have the same boundary as $\R_+$, but near each of the boundary points $0,1$ our $M$ looks the same as $\R_+$. It's also similar for the closed ball $\B^2$, but now one needs to straighten out the boundary (note: this part only makes sense for submanifolds of $\R^n$ and not for general manifolds with boundary). To see that $\pl \B^2$ looks like $\pl \R^2_+$ near the point $(0,-1)$, consider the map $$ \Phi \colon \B^2 \to \R^2_+, \qquad \Phi(x,y) = (x,y+\sqrt{1-x^2}), $$ which maps the interior of $\B^2$ to the interior of $\R^2_+$, and the boundary of $\B^2$ (at least the lower part of it) to the boundary of $\R^2_+$.

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#3 Let me also remark that it's necessary to treat the boundary separately. Consider the following two fake "definitions" that attempt to skip this point:

• $M$ is a manifold with boundary if there's an atlas $\Phi_i \colon U_i \to \R^n$ on open subsets $U_i \subseteq \operatorname{int} M$ of the interior of $M$, covering all of $\operatorname{int} M$.
• $M$ is a manifold with boundary if there's an atlas $\Phi_i \colon U_i \to \R^n$ on open subsets $U_i \subseteq M$ (possibly including the boundary), covering all of $M$, such that each $\Phi_i$ maps $U_i$ homeomorphically to an open subset of $\R^n$.

You can see that the first "definition" accepts more objects than we'd like (as it doesn't say anything about the boundary at all), while the second "definition" effectively accepts only manifolds without boundary (this is a good exercise).