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I'm really struggling to understand the definition of a "submanifold with boundary". Until now, I'm only familiar with the notion of a "submanifold of $\mathbb R^d$. I've defined this notion in the form I'm aware of below$^1$, but meanwhile I guess it should better be called "$C^1$-submanifold" and what I call "chart" should be called "$C^1$-chart", but in particular the latter might be totally wrong.

Let $M\subseteq\mathbb R^d$ for some $d\in\mathbb N$, $k\in\{1,\ldots,d\}$ and $\mathbb H^k:=\{x\in\mathbb R^k:x_k\ge0\}$.

Maybe it's easier to motivate the definition of a "submanifold with boundary" using the following equivalent characterization of a submanifold: Let $$\mathcal D_d:=\{(\Omega,\psi)\mid\Omega\subseteq\mathbb R^d\text{ is open and }\psi\text{ is a diffeomorphism from }\Omega\text{ onto }\psi(\Omega)\}.$$ Then $M$ is a $k$-dimensional embedded ($C^1$-)submanifold of $\mathbb R^d$ if and only if $$\forall x\in M:\exists(\Omega,\psi)\in\mathcal D_d:x\in\Omega\text{ and }\psi(M\cap\Omega)=\psi(\Omega)\cap(\mathbb R^k\times\{0\}).\tag2$$ Now let $\iota_k$ denote the canonical embedding of $\mathbb R^k$ into $\mathbb R^d$ with $\iota\mathbb R^k=\mathbb R^k\times\{0\}$. Then, it's easy to see that if $(\Omega_1,\psi)\in\mathcal D_d$ with $\psi(M\cap\Omega)=\psi(\Omega)\cap(\mathbb R^k\times\{0\})$ and $\Omega_2:=\psi(\Omega_1)$, then $U:=\iota_k^{-1}(\Omega_2)$ is open, $\psi^{-1}\in C^1(\Omega_2,\mathbb R^d)$, $$\phi:=\psi^{-1}\circ\left.\iota_k\right|_U\in C^1(U,\mathbb R^d)$$ and $$\phi(U)=M\cap\Omega_1$$ and hence $(U,\phi)$ is a $k$-dimensional chart of $M$.

Now, and hopefully I got it right, the definition of a "submanifold with boundary" should be as follows: $M$ is called $k$-dimensional embedded ($C^1$-)submanifold with boundary of $\mathbb R^d$ if for all $x\in M$ there is a $(\Omega,\psi)\in\mathcal D_d$ with $x\in\Omega$ and either

  1. $\psi(M\cap\Omega)=\psi(\Omega)\cap(\mathbb R^k\times\{0\})$; or
  2. $\psi(M\cap\Omega)=\psi(\Omega)\cap(\mathbb H^k\times\{0\})$ and $\psi_k(x)=0$.

I'm not sure, but maybe we can show that for each fixed $x$, either all choices of $(\Omega,\psi)$ satisfy (1.) or all choices satisfy (2.). I've asked for that separetegly: Show that these two diffeomorphisms cannot exist simultaneously.

How does the corresponding "chart definition" of a submanifold with boundary look like? Am I right that we can construct such a chart in the same way as before? Assuming $x\in M$ and $(\Omega,\psi)$ is as in the definition above satisfyin (2.). Then, if again $U:=\iota_k^{-1}(\Omega_2)$ and $\phi:=\psi^{1-}\circ\left.\iota_k\right|_U$, we should have $\psi(M\cap\Omega_1)=\iota(U\cap\mathbb H^k)$ and hence $M\cap\Omega_1=\phi(\mathbb H^k\cap U)$. $\phi$ is still an immersion from $U$ to $\mathbb R^d$ and a topological embedding of $U$ into $\phi(U)$. And $U\cap\mathbb H^k$ is open in $\mathbb H^k$; maybe that's what we need to replace.

Besides that I've read that we can always (i.e. for any submanifold with boundary, $x\in M$ and $(\Omega,\psi)\in\mathcal D_d$ with $x\in\Omega$) assume that $\psi(M\cap\Omega)=\psi(\Omega)\cap(\mathbb H^k\times\{0\})$; the only difference would be that if $x\in\partial M$ (I'm not sure if the topological boundary is meant), then $\psi_k(x)=0$ and if $x\in M^\circ$ (I'm not sure if the topological boundary is meant), then $\psi_k(x)>0$. Why is that the case?


$^1$ $(U,\phi)$ is called $k$-dimensional chart of $M$ if $U\subseteq\mathbb R^k$ is open, $\phi:U\to\mathbb R^d$ is an immersion and a topological embedding of $U$ into $M$ and $\phi(U)$ is $M$-open. Let $\mathcal C_k(M)$ denote the set of $k$-dimensional charts of $M$.

$M$ is called $k$-dimensional embedded submanifold of $\mathbb R^d$ if $$\forall x\in M:\exists(U,\phi)\in\mathcal C_k(M):x\in\phi(U)\tag1.$$

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  • $\begingroup$ What is $u_d$? To show that the boundary is a manifold (in $\mathbb{R}^d$), a hint is to adapt the atlas on $M$ to an atlas on $\partial M$ - how can you change each chart? $\endgroup$ – Oliver Clarke Jul 5 '20 at 13:35
  • $\begingroup$ @OliverClarke Sorry, $u_d$ was supposed to be $u_k$; the $k$th component of $u$. $\endgroup$ – 0xbadf00d Jul 5 '20 at 13:41
  • $\begingroup$ @OliverClarke Please take note of my edit. $\endgroup$ – 0xbadf00d Jul 5 '20 at 13:56
  • $\begingroup$ By "manifold" I mean a $k'$ dimensional submanifold of $\mathbb{R^n}$, for some $k'$ - if you write down some charts for boundary of $M$ you'll see what the dimension of the boundary should be. $\endgroup$ – Oliver Clarke Jul 5 '20 at 13:56
  • $\begingroup$ According to Do Carmo's definition, a submanifold $N$ of dimension $k$ of a manifold $M$ of dimension $d$ is just considering open sets $W$ that cover $N$ and verify $\sigma \cap W$ is diffeomorphic to $\mathbb{R}^k$ for all charts $\sigma$ of $M$ that intersect $W$. I think in this context you can define it exactly in this way relative to $\mathbb{H}$ which is somewhat easier and inherits topological properties from $M$ in a more straightforward manner. $\endgroup$ – astro Jul 5 '20 at 14:19
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A point $x$ of $M$ is an interior point if and only if there is an open neighbourhood of $x$ which is homeomorphic to an open subset of $\mathbb{R}^k$. So let's pick a chart $(U, \phi)$ where $x = \phi(u)$ for some $u \in U$.

If $u_k = 0$ then a $\mathbb{H}^k$-open neightbourhood of $u$ is not open in $\mathbb{R}^k$. So $x$ is not an interior point and so must lie on the boundary of $M$.

If $u_k > 0$ then we can restrict $U$ to $V = U \cap \{y \in \mathbb{H}^k : y_k > 0 \}$ and we get a chart $(V, \phi |_V)$. So a neightbourhood of $x$ is homeomorphic to $V$ which is an open subset of $\mathbb{R}^k$. So $x$ is an interior point.

The boundary of $M$ is a submanifold of $\mathbb{R}^n$ (without boundary). Hint: write down charts for $\partial M$ by using the charts for $M$. This will tell you what is the dimension of $\partial M$.

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  • $\begingroup$ Thank you for your effort so far. I think I've already got a better understanding of it, but it still is not totally clear me. I've edited the question and hopefully it's clearer now what I'm struggling to understand. $\endgroup$ – 0xbadf00d Jul 5 '20 at 19:21

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