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Let $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, i.e. $M$ is locally $\mathcal C^1$-diffeomorphic$^1$ to $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$, $$T_xM:=\left\{v\in\mathbb R^d\mid\exists\varepsilon>0,\gamma\in C^1((-\varepsilon,\varepsilon),M):\gamma(0)=x,\gamma'(0)=v\right\}$$ denote the tangent space of $M$ and $M^\circ$ and $\partial M$ denote the manifold interior and boundary, respectively.

Let $x\in M$, $(\Omega,\phi)$ be a $k$-dimensional $C^1$-chart of $M$ around $x$, i.e. $\Omega$ is an $M$-open neighborhood of $x$ and $\phi$ is a $C^1$-diffeomorphism from $\Omega$ onto an open subset of $\mathbb R^k$ or $\mathbb H^k$ and $u:=\phi(x)$.

Question 1: Can we generally show that $$T_xM={\rm D}\phi^{-1}(u)\mathbb R^k\tag1?$$ This is easy to show if $x\in M^\circ$ and $(\Omega,\phi)$ is an interior chart, i.e. $\phi$ is a $C^1$-diffeomorphism from $\Omega$ onto an open subset of $\mathbb R^k$. It should hold in the general case as well, but I'm unsure whether there is some subtlety I'm missing.

Question 2: We know that $\partial M$ is a $(k-1)$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary. If $x\in\partial M$ and $(\Omega,\phi)$ is a boundary chart, i.e. $\phi$ is a $C^1$-diffeomorphism from $\Omega$ onto an open subset of $\mathbb H^k$ with $u=\phi(x)\in\partial\mathbb H^k$, then$^2$ $(\tilde\Omega,\tilde\phi):=(\Omega\cap\partial M,\pi\circ\left.\phi\right|_{\Omega\:\cap\:\partial M}$ is a $(k-1)$-dimensional $C^1$-chart of $\partial M$ around $x$. From $(1)$ and this question, it should follow that $$T_x\partial M={\rm D}\tilde\phi^{-1}(\tilde\phi(x))\mathbb R^{k-1}={\rm D}\phi^{-1}(u)\partial\mathbb H^k\tag2.$$ Is this correct? And is it possible to construct a (unique) unit normal field on $\partial M$ from that?

In order to compute the normal space $N_x\partial M$, I've tried the following: By $(2)$ we know that each $v\in T_x\partial M$ is of the form $v=Bh$ for some $h\in\partial\mathbb H^k$, where $B:={\rm D}\phi^{-1}(u)$. If $A:={\rm D}\phi(x)$, we should obtain $AB=\operatorname{id}_{\mathbb R^k}$ and $BA=\operatorname{id}_{\mathbb R^d}$. If $(e_1,\ldots,e_k)$ denotes the standard basis of $\mathbb R^k$, then $$\langle Bh,A^Te_k\rangle=\langle ABh,e_k\rangle=\langle h,e_k\rangle=0\tag3.$$ So, $A^Te_k\in N_x\partial M$. Can we prove that and maybe argue by dimensionality that $N_x\partial M=\mathbb RA^Te_d$?

$^1$ 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 at $x_1\in B_1$ if there is an $E_1$-open neighborhood $\Omega_1$ of $x_1$ and a $\tilde f\in\mathcal C^\alpha(\Omega_1,E_2)$ with $\left.f\right|_{B_1\:\cap\:\Omega_1}=\left.\tilde f\right|_{B_1\:\cap\:\Omega_1}$. $f$ is called $\mathcal C^1$-differentiable if $f$ is $C^\alpha$-differentiable at $x_1$ for all $x_1\in B_1$.

$g$ is called $C^1$-diffeomorphism from $B_1$ onto $B_2$ if $g$ is a homeomorphism from $B_1$ onto $B_2$ and $g$ and $g^{-1}$ are $C^1$-differentiable.

$^2$ For convenience, let $\iota$ denote the canonical embedding of $\mathbb R^{k-1}$ onto $\mathbb R^k$ with $\iota\mathbb R^{k-1}=\mathbb R^{k-1}\times\{0\}$ and $\pi$ denote the canonical projection of $\mathbb R^k$ onto $\mathbb R^{k-1}$ with $\pi(\mathbb R^{k-1}\times\{0\})=\mathbb R^{k-1}$.

0xbadf00d
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1 Answers1

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You've got it all right.

For Q1, the point is that $\phi$ is a diffeomorphism $ V \xrightarrow{\sim} U\subset \mathbb{H}^k$, sending $x\in V$ to $u\in U$, hence $D\phi(x):T_xM\rightarrow T_u\mathbb{H}^k \cong\mathbb{R}^{k}$ is a linear isomorphism (with inverse given by the differential of $\phi^{-1})$. This gives (1) in your question.

For Q2, the same reasoning applies to $\tilde \phi$. However, the notation $T_u \partial \mathbb{H}^k \cong\mathbb{R}^{k-1}$ (emphasis on the linear structure!) is maybe better than $\partial \mathbb{H}^{k}$ on the right hand side of (2). Regarding the normal, your construction works perfectly fine, indeed $N_x\partial M = (A^Te_k) \mathbb{R}$ (note that you misses the transpose in your suggestion): You know that the normal bundle has one-dimensional fibres (because together with the $k-1$-dimensional space $T_x\partial M$ it spans the $k$-dimensional space $T_xM)$, and the only thing you're saying is that this one-dimensional space is spanned by a non-zero element (=basis) in it.

Jan Bohr
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  • If I'm not missing anything, we both made a mistake: The conclusion $N_x\partial M=\mathbb R{\rm D}\phi(u)^Te_d$ for all $x\in\Omega\cap M$ and $u=\phi(x)$ is obviously only correct if $k=d$, since generally $N_x\partial M$ is $(d-(k-1))$-dimensional. – 0xbadf00d Jul 20 '20 at 14:04
  • Ah! That depends on what you mean by normal bundle. You can just discard the ambient $\mathbb{R}^d$ and consider the normal bundle as a sub-bundle of $TM$, then it has rank one. The other option is to discard the interior of $M$ and view $\partial M$ as a sub-manifold of $\mathbb{R}^d$ in its own right. Then it has a $d-(k-1)$-dimensional normal bundle. Both viewpoints are valid, but arguably the first one is more natural in many contexts (say Stokes' theorem, e.g. used in the context of electrodynamics where ambient $\mathbb{R}^3$ is lingering around anyways). – Jan Bohr Jul 20 '20 at 15:04
  • Thank you for your comment. I'm definiing $N_x\partial M$ as the orthogonal complement of $T_x\partial M$ in $\mathbb R^d$ so that $\mathbb R^d=T_x\partial M\oplus N_x\partial M$. – 0xbadf00d Jul 20 '20 at 17:09
  • Just another issue I'm worrying about: Is the definition of $T_x:M$ I gave in the question really correct when $M$ is a submanifold with boundary? At other places, I've seen that we need to allow the curves to be defined on intervals of the type $(-\varepsilon,0]$ or $[0,\varepsilon)$ as well. Is this necessary to prevent problems with boundary points? If so? How is $\gamma'(0)$ defined then? – 0xbadf00d Aug 05 '20 at 07:46
  • I think there are two ways: Either as a one-sided limit or by assuming that $\gamma$ is differentiable in the interior of $I$ and its derivative to admit a continuous extension (but this extension is not unique, is it?) to all of $I$. I'm unsure whether these two ways are equivalent or not. – 0xbadf00d Aug 05 '20 at 07:46
  • Two comments about manifolds with boundary: First, given a smooth function, curve, etc. in $M$, you can (say in a chart) always find a smooth extension across $\partial M$. This might not be unique, but any derivatives you take at the boundary are independent of the extension. Second, but in the same spirit, it is always possible to view $M$ as embedded submanifold of a larger manifold $N$ without boundary and of the same dimension (see e.g. Problem 17.14 in Lee's smooth manifolds book). This makes considerations at the boundary quite natural and it is the point of view I typically take. – Jan Bohr Aug 05 '20 at 08:56
  • So, you mean that the extension $\gamma'$ is not unique, but $\gamma'(0)$ is uniquely determined? – 0xbadf00d Aug 05 '20 at 09:13
  • Yes, the reason is that any two smooth curves $\gamma,\tilde \gamma:(-\epsilon,\epsilon)\rightarrow \mathbb{R}^d$ that agree for $-\epsilon<t\le 0$, satisfy $\gamma'(0)=\tilde \gamma'(0)$. – Jan Bohr Aug 05 '20 at 09:21
  • Now I'm totally confused. You are considering two curves $\gamma,\tilde\gamma$ on an open interval $I$ containing $0$. It's then trivial to conclude that $\gamma'(0)=\tilde\gamma'(0)$, since, in total generality, the Fréchet derivative is unique. In this case, no extension is necessary. – 0xbadf00d Aug 05 '20 at 10:42
  • Sorry, I don't know what it is, but there is some subtlety that I'm obviously missing, since this doesn't make sense to me. We can assume that any interval of interest is of the form $I=(-\varepsilon,\varepsilon)$, $I=(-\varepsilon,0]$ or $[0,\varepsilon)$. If $\gamma:I\to M$ is Fréchet differentiable in $I^\circ$, then $\left.\gamma'\right|_{I^\circ}$ is uniquely determined. So, the only crucial question is: If $0\not\in I^\circ$, how do we define $\gamma'(0)$ and in which sense is it unique. – 0xbadf00d Aug 05 '20 at 10:42
  • My point is that $\gamma'$ is not only unique on the interior, but also on the closure of the interval, provided $\gamma$ can be extended smoothly across the closure. In my previous comment, both $\gamma$ and $\tilde \gamma$ are such extensions of $\gamma\vert_{(-\epsilon,0]}:(-\epsilon,0]\rightarrow \mathbb{R}^d$. – Jan Bohr Aug 05 '20 at 10:45
  • Let us continue this discussion in chat. – Jan Bohr Aug 05 '20 at 10:46
  • While I made a mistake in reasoning in the other question, it should work for $f=\gamma$ and $\Omega_1=(-\varepsilon,0]$ or $\Omega_1=[0,\varepsilon)$. Do you agree? – 0xbadf00d Aug 05 '20 at 14:33
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    I do agree. As I said, if $\Omega$ has a smooth boundary (which is trivially satisfied for an half-open interval), then everything is fine. – Jan Bohr Aug 05 '20 at 14:50
  • In your answer to Q1, I guess you mean $T_u\mathbb{H}^{\color{red}k} \cong\mathbb{R}^{k}$, right? – 0xbadf00d Sep 02 '20 at 08:19
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    Yes, I have fixed it. – Jan Bohr Sep 02 '20 at 08:57
  • Thanks for the fix. I think my picture of everything is clearer now. Please tell me if I got this right: Assume $M$ and $(\Omega,\phi)$ are as in the question. (a) $T_x:\Omega=T_x:M$ for all $x\in\Omega$. (b) $\partial\Omega={x\in\Omega:\phi(x)\in\partial\mathbb H^k}\subseteq\partial M$. (c) If $x\in\partial\Omega$, there is a unique outward pointing unit $\nu_{\partial\Omega}(x)\in T_x:\Omega\cap N_x:\partial\Omega$. – 0xbadf00d Sep 02 '20 at 10:05
  • (d) If $k=d$ and $x\in\partial\Omega$, then $$\nu_{\partial\Omega}(x)=\frac{T_x(\phi)^\ast e_d}{\left|T_x(\phi)^\ast e_d\right|},$$ where $T_x(\phi):T_x:\Omega\to T_x:\mathbb H^d=\mathbb R^d$ is the pushforward. (e) The latter formula shows that when $\phi$ is even a $C^\alpha$-diffeomorphism for some $\alpha\in\mathbb N$, then $\nu_{\partial\Omega}$ is $C^{\alpha-1}$-differentiable. Do you agree to those points? – 0xbadf00d Sep 02 '20 at 10:05
  • What do you think? – 0xbadf00d Sep 08 '20 at 09:51