# Tangent Bundle of a smooth manifold with boundary

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.

• 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 Nov 20 at 19:38
• @Didier Is this because of the invariance of the boundary theorem? I wrote my idea above. Nov 21 at 3:50
• 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. Nov 21 at 8:22