Let $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary.$^1$

We know that there is a countable family $((\Omega_i,\phi_i))_{i\in I}$ of $k$-dimensional $C^1$-charts$^2$ of $M$ with $$M\subseteq\bigcup_{i\in I}\Omega_i.$$

We know that there is a $k$-dimensional boundary $C^1$-atlas$^1$ $((\Omega_i,\phi_i))_{i\in I}$ of $M$ for some $I\subseteq\mathbb N$.

Let $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$. Note that $(\mathbb H^k)^\circ=\mathbb R^{k-1}\times(0,\infty)$ and $\partial\mathbb H^k=\mathbb R^{k-1}\times\{0\}$.

Let $B$ denote the closed unit ball in $\mathbb R^k$, $B_+:=B\cap(\mathbb H^k)^\circ$ and $B_0:=B\cap\partial\mathbb H^k$.

Why can we choose $((\Omega_i,\phi_i))_{i\in I}$ such that the manifold interior$^3$ $\Omega_i^\circ$ is equal to $\phi_i^{-1}(B_+)$ and the manifold boundary $\partial\Omega_i$ is equal to $\phi_i^{-1}(B_0)$?

$^1$ i.e. each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^k$.

If $E_i$ is a $\mathbb R$-Banach space and $B_i\subseteq E_i$, then $f:B_1\to E_2$ is called $C^1$-differentiable if $f=\left.\tilde f\right|_{B_1}$ for some $E_1$-open neighborhood $\Omega_1$ of $B_1$ and some $\tilde f\in C^1(\Omega_1,E_2)$ and $g:B_1\to B_2$ is called $C^1$-diffeomorphism if $g$ is a homeomorphism from $B_1$ onto $B_2$ and $g$ and $g^{-1}$ are $C^1$-differentiable.

$^2$ A $k$-dimensional $C^1$-chart of $M$ is a $C^1$-diffeomorphism from an open subset of $M$ onto an open subset of $\mathbb H^k$.

$^3$ $x\in M^\circ$ if and only if there is a $k$-dimensional $C^1$-chart $(\Omega,\phi)$ of $M$ such that $x\in\Omega$ and $\phi(\Omega)$ is $\mathbb R^k$-open.

$x\in\partial M$ if and only if there is a $k$-dimensional $C^1$-chart $(\Omega,\phi)$ of $M$ such that $x\in\Omega$ and $\phi(x)\in\partial\mathbb H^k$.

  • $\begingroup$ The whole manifold interior $\Omega_i^\circ$ equal to $\phi_i^{-1}(B_+)$? It can't be so if they are not homeomorphic. For example, take a torus in $\mathbb{R}^3$. Do you mean the charts' interior or the manifold's? $\endgroup$ Jul 12, 2020 at 5:11
  • $\begingroup$ @Chrystomath What do you mean by the "interior" of the chart? I mean the manifold interior of $\Omega_i$. $\endgroup$
    – 0xbadf00d
    Jul 12, 2020 at 6:11
  • $\begingroup$ The question is what the definition of what a manifold with boundary is. You may be interested in mathoverflow.net/questions/38503/… $\endgroup$ Jul 12, 2020 at 12:15
  • $\begingroup$ @Chrystomath The definition is: Each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^k$. $\endgroup$
    – 0xbadf00d
    Jul 12, 2020 at 13:38
  • $\begingroup$ @Chrystomath Maybe we need to assume that $M$ is compact? $\endgroup$
    – 0xbadf00d
    Jul 13, 2020 at 7:37

1 Answer 1


When working with a (smooth or $C^k$) manifold $M$, implicitly you are saying that a (smooth or $C^k$) structure exists, that is, there exists some associated atlas $\{ \phi_i : V_i \to \mathbb R^k \}_{i \in I}$ corresponding to the (smooth or $C^k$) structure of interest. The key word is exists; there really isn't a canonical choice of atlas. Enter the notion of compatible atlases. We say another atlas $\{\psi_j : U_j \to \mathbb R^k\}_{j \in J}$ is compatible if $$\psi_j \circ \phi_i^{-1} : \phi_i (V_i) \to \mathbb R^k$$ is smooth for every $i \in I$ and $j \in J$. Philosophically speaking, this means both atlases are essentially the same. Mathematically, compatibility forms an equivalence relation on the collection of all atlases of a manifold, which each equivalence class being what is called a (smooth or $C^k$) structure.

The phrase "we can choose" translates to "we can pick a compatible atlas". Again, philosophically speaking, there is no difference in working with either of two compatible atlases. However, mathematically one atlas may be more convenient to work with than another. For instance, we might want the images $\psi_j (U_j)$ to be balls to (what I like to call) "abuse" the local convex structure of Euclidean space. You see this a lot if you're working with Riemann surfaces.

So how does one arrive at such an atlas? Well, we know from Euclidean topology that open sets, for example $\phi_i (V_i)^\circ \subseteq \mathbb R^k$, can be written as countable unions of open balls $B_{i, j}$. This lets us define bijections $\psi_{i, j} : \phi_i^{-1} (B_{i, j}) \to B_{i, j}$ by $\psi_{i, j} := \phi_i$.

So what about the boundary points? Well, we know that $\phi_i (V_i)$ is open in $\mathbb H^k$, so there exists a set $O \subseteq \mathbb R^k$ open in the Euclidean topology such that $$\phi_i (V_i) = O \cap \mathbb H^k.$$ In particular, suppose $x \in \partial M \cap V_i$, then $\phi_i (x) \in \partial \mathbb H^k$ and is an interior point of $O$. So we draw a ball around $\phi_i (x)$ which is contained in $O$. Then we can draw a ball which is contained in this ball and $\mathbb H^k$ with boundary point $\phi_i (x)$, call it $B_{i, x}$. Then as with before we have a well-defined bijection $\psi_{i, x} : \phi_i^{-1} (B_{i, x}) \to B_{i, x}$ given by $\psi_{i, x} := \phi_i$.

It's useful to draw a picture: enter image description here

And now we're done! Since the $\psi_{i, j}$ and $\psi_{i, x}$ maps were defined using $\phi_i$, collecting all of them together gives a (smooth or $C^k$) atlas. Of course this won't be countable since there are uncountably many boundary points $x \in \partial M$, however you can adapt the argument, instead of using balls tangent to $\partial \mathbb H$ you can use parallelepipeds (with rounded corners) and take a diffeomorphism sending it to a ball in $\mathbb R^k$.

  • $\begingroup$ Thank you for your amazing answer. It was really helpful! I've got two remaining questions: (a) The claim in the question was with respect to the unit open ball $B$. The atlas in your answer maps to different balls with different centers and radii. (b) I'm mostly interested in the case, where $\Omega$ is bounded and open and $k=d$. Does this solve the uncountability problem on the boundary? $\endgroup$
    – 0xbadf00d
    Jul 31, 2020 at 15:06
  • 1
    $\begingroup$ @0xbadf00d (a) if $\phi: \Omega_i \to {\mathbb R}^k$ is a chart and $D: {\mathbb R}^k \to {\mathbb R}^k$ is a diffeomorphism, then $D \circ \phi$ is also a chart (think chain rule). In particular, if the image $\phi_i (\Omega_i)$ is any ball, scaling and translating can ensure $(D \circ \phi_i)(\Omega_i)$ is the unit ball centered at the origin. With charts, we don't really care what the image looks like in Euclidean space, so we have a lot of freedom in that sense. $\endgroup$
    – Reavered
    Jul 31, 2020 at 16:49
  • $\begingroup$ @0xbadf00d (b) not really, you still need to rely on the same trick with "smooth" rectangles to make everything countable. To make the point clear, the dimension of the ambient Euclidean space where the manifold is embedded doesn't really matter in this problem. It's useful in some scenarios if you want to use certain global properties, like inheriting Euclidean distance, but the problem here is entirely local. $\endgroup$
    – Reavered
    Jul 31, 2020 at 16:54
  • $\begingroup$ Thank you for your reply. My point in (b) is that then $\overline M$ is compact and we know that any compact submanifold has a countable atlas (yes, I know that $\overline M$ does not need not be a submanifold if $M$ is bounded and open, but it is at least compact and hence has a finite cover of open sets). $\endgroup$
    – 0xbadf00d
    Jul 31, 2020 at 17:01

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