# Definition of a submanifold with boundary

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 $$\forall x\in M:\exists(U,\phi)\in\mathcal C_k(M):x\in\phi(U)\tag1.$$

• 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? – Oliver Clarke Jul 5 '20 at 13:35
• @OliverClarke Sorry, $u_d$ was supposed to be $u_k$; the $k$th component of $u$. – 0xbadf00d Jul 5 '20 at 13:41
• @OliverClarke Please take note of my edit. – 0xbadf00d Jul 5 '20 at 13:56
• 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. – Oliver Clarke Jul 5 '20 at 13:56
• 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. – astro Jul 5 '20 at 14:19

## 1 Answer

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 $$\partial M$$.

• 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. – 0xbadf00d Jul 5 '20 at 19:21