# Characterization of the tangent space of the boundary of an embedded submanifold of $\mathbb R^d$ with boundary

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

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.
• 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
• 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 5 '20 at 14:50