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 $\pi(\mathbb R^{k-1}\times\{0\})=\mathbb R^{k-1}$.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    Jul 20, 2020 at 14:04
  • $\begingroup$ 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). $\endgroup$
    – Jan Bohr
    Jul 20, 2020 at 15:04
  • $\begingroup$ 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 $\mathbb R^d=T_x\partial M\oplus N_x\partial M$. $\endgroup$
    – 0xbadf00d
    Jul 20, 2020 at 17:09
  • 1
    $\begingroup$ 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. $\endgroup$
    – Jan Bohr
    Aug 5, 2020 at 14:50
  • 1
    $\begingroup$ Yes, I have fixed it. $\endgroup$
    – Jan Bohr
    Sep 2, 2020 at 8:57

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