This is exercise 3.19 from John Lee's Introduction to Smooth Manifolds that I can't quite prove.

Suppose $M$ is a smooth manifold with boundary. Show that $TM$ has a natural topology and smooth structure making it into a smooth manifold with boundary, such that if $(U,(x^i))$ is any choundary chart for $M$, then rearranging the coordinates in the natural chart $(\pi^{-1}(U),(x^i,v^i))$ for $TM$ yields a boundary chart $(\pi^{-1}(U),(v^i,x^i))$.

For interior charts we define the chart as given an interior chart $(U,\phi)$ for $M$, since $\pi^{-1}(U) \subset TM$ is the set of all tangent vectors to $M$ at all points of $U$, and letting $(x^1, \dots, x^n)$ denote the coordinate functions of $\phi$, we can define a map $\tilde{\phi}:\pi^{-1}(U)\to \mathbb{R}^{2n}$ by $$\tilde{\phi}(v^i\frac{\partial}{\partial x^i}|_p)=(x^1(p),\dots, x^n(p), v^1, \dots, v^n).$$

Its image set is $\phi(U)\times \mathbb{R}^n$ which is an open subset of $\mathbb{R}^{2n}$, and is a bijection onto its image where its inverse is $$\tilde{\phi}^{-1}(x^1,\dots, x^n, v^1, \dots, v^n)=v^i \frac{\partial}{\partial x^i}|_{\phi^{-1}(x)}.$$

Then smooth compatibility given two smooth charts $(U,\varphi)$ and $(V,\psi)$ for $M$, we have the transition map $\tilde{\psi}\circ \tilde{\varphi}^{-1}:\varphi(U \cap V)\times \mathbb{R}^n \to \psi(U\cap V) \times \mathbb{R}^n$ written as

$$\tilde{\psi} \circ \tilde{\varphi}^{-1}(x^1, \dots, x^n, v^1, \dots, v^n)=(\tilde{x}^1(x),\dots \tilde{x}^n(x), \frac{\partial \tilde{x}^1}{\partial x^j}(x)v^j,\frac{\partial \tilde{x}^n}{\partial x^j}(x)v^j)$$ which is clearly smooth.

The problem I have is establishing smooth compatibility between interior charts and boundary charts. By the Smooth Manifold Chart Lemma (Lemma 1.35), we need to show that whenever $\pi^{-1}(U)$ and $\pi^{-1}(V)$ intersect, the map $\tilde{\psi} \circ \tilde{\varphi}^{-1}$ is smooth.

If we take the transition of two boundary charts as suggested by rearranging the order of $v^i$ and $x^i$, we would get the same kind of equation as above with just a rearrangement and the final component will be in the half plane $\mathbb{H}^n=\{x: (x)_n \ge 0\}$. However, how do we ensure smooth compatibility for interior and boundary charts, i.e. the case where $\psi$ is a boundary chart and $\varphi$ is an interior chart and vice versa?

I think this can be resolved by the invariance theorem of boundary. So $U\cap V$ can only be interior or boundary domain but not both. So in the case where $(U,\phi)$ is an interior chart and $(V,\psi)$ is a boundary chart, $U\cap V$ cannot contain boundary points, so the transition map $\tilde{\psi} \circ \tilde{\phi}^{-1}(x^1, \dots, x^n, v^1, \dots , v^n)=(\frac{\partial \tilde{x}^1}{\partial x^j}(x)v^j,\dots, \frac{\partial \tilde{x}^n}{\partial x^j}(x)v^j, \tilde{x}^1(x),\dots, \tilde{x}^n(x))$, and $\tilde{\phi} \circ \tilde{\psi}^{-1}(v^1, \dots , v^n, \tilde{x}^1, \dots, \tilde{x}^n,)=(x^1(\tilde{x}),\dots, x^n(\tilde{x}), \frac{\partial x^1}{\partial \tilde{x}^j}(\tilde{x})v^j,\dots, \frac{\partial x^n}{\partial \tilde{x}^j}(\tilde{x})v^j)$ which are smooth as maps with domain and codomain contained in $\mathbb{R}^{2n}$ and the first transition is just a rearrangement of the case with two interior charts.

  • 2
    $\begingroup$ An interior chart always intersect a boundary chart in its interior. Hence, this case boils down to the first you mentioned, i.e two interior charts $\endgroup$
    – Didier
    Nov 20 at 19:38
  • $\begingroup$ @Didier Is this because of the invariance of the boundary theorem? I wrote my idea above. $\endgroup$ Nov 21 at 3:50
  • $\begingroup$ Presumably (I don't know any specific name for this result, but it sounds like it is it). This is because an open subset of $\Bbb R^n$ cannot be homeomorphic to an open subset of $\Bbb R^{n+1}\times [0,+\infty)$ that meets the boundary. $\endgroup$
    – Didier
    Nov 21 at 8:22


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