# Orientation in the proof of Stokes Theorem

I'm reading the proof of Stokes theorem at page 83 of "Godinho, Natàrio, An introduction to Riemannian geometry" and I can't understand a passage in it, probably because the definition of orientability is not very clear to me. The definition of orientability and orientation used in the book are:

Definition (Oriented manifold) Let $$M$$ be a $$n$$-differentiable manifold. An orientation for $$M$$ consists in a choice of orientation for all the $$n$$-dimensional real vector spaces $$T_pM$$ s.t. for every $$p \in M$$ there exist a parametrization $$(U, \varphi)$$ s.t. $$$$\text{the linear map } \, \, d\varphi_x : T_xU \cong \mathbb{R}^n \to T_{\varphi(x)}M \, \, \text{ is orientation preserving} \quad \quad \quad (*)$$$$ where $$\mathbb{R}^{n}$$ is equipped with the standard orientation (i.e. the one that assigns a positive sign to the standard basis). A manifold admitting an orientation is called oriented manifold. A parametrization $$(U, \phi)$$ is said to be compatible with the orientation if it satisfies (*).

I also know that

Lemma A $$n$$-differentiable manifold is orientable iff there exists an atlas s.t. the (matrices representing the) differentials of the overlapping maps have positive determinant.

And then I have this definition of induced orientation on $$\partial M$$:

Definition (Induced orientation) Let $$M$$ be a $$n$$-differentiable manifold with boundary $$\partial M$$ (which is a $$(n-1)$$-differential manifold). Chosen coordinates $$(x^1, \dots, x^n)$$ in $$M$$ around $$p \in \partial M$$ then $$(x^1, \dots, x^{n-1})$$ are coordinates in $$\partial M$$ around $$p$$. $$\{ \partial_1, \dots, \partial_n \}$$ is a basis for $$T_p(M)$$ and $$\{ \partial_1, \dots, \partial_{n-1} \}$$ is a basis for $$T_p (\partial M)$$. We call induced orientation by $$M$$ on $$\partial M$$ the one for which a basis $$\mathcal{B}$$ of $$T_p(\partial M)$$ has positive sign if it does the basis $$\{ -\partial_n, \mathcal{B} \}$$ of $$T_pM$$.

Then in the proof of Stokes theorem I have a parametrization of $$M$$ $$(U, \varphi)$$ with $$U \subset \mathbb{H}^n$$ s.t. $$\varphi(U) \cap \partial M \ne \emptyset$$. Then the book defines $$\tilde{U} : = \{ (x^1, \dots, x^{n-1} ) \mid ((-1)^n x^1, \dots, x^{n-1}, 0) \in U \}$$ and the parametrization $$\tilde{\varphi} : \tilde{U} \to \partial M \quad \quad (x^1, \dots, x^{n-1}) \mapsto \varphi ((-1)^n x^1, \dots, x^{n-1}, 0))$$

My question : why is $$(\tilde{U}, \tilde{\varphi})$$ a parametrization of $$\partial M$$ preserving the induced orientation of $$M$$ on $$\partial M$$?

Edit (Just to see if I understood the suggestions of Balloon) Define $$\overline{U} := \{ (x^1, \dots, x^{n-1}) \in \mathbb{R}^{n-1} \mid (x^1, \dots, x^{n-1}, 0) \in U \} \\ \overline{\varphi} := \varphi \biggr |_{\overline{U}} : \overline{U} \to \partial M \\ \tilde{U}:= \{ (x^1, \dots, x^{n-1} \in \mathbb{R}^{n-1} \mid ((-1)^n x^1, \dots, x^{n-1}) \in \overline{U} \} \\ \tau : \tilde{U} \to \overline{U} \quad \quad (x^1, \dots, x^{n-1}) \mapsto ((-1)^n x^1, \dots, x^{n-1})\\ \tilde{\varphi} := \overline{\varphi} \circ \tau : \tilde{U} \to \partial M$$ I want to prove $$(\tilde{U}, \tilde{\varphi})$$ is a parametrization of $$\partial M$$ preserving the standard orientation of $$\mathbb{R}^{n-1}$$ w.r.t. the orientation induced by $$M$$ on $$\partial M$$.

It is enough to prove that the determinant of the Jacobian of $$\tilde{\varphi}$$ (written w.r.t. the standard basis of $$\mathbb{R}^{n-1}$$ and a positive basis in $$\partial M$$) is positive at each point $$x \in \tilde{U}$$ . I observe that $$d\tilde{\varphi}_x = d \overline{\varphi}_{\tau(x)} \circ d\tau_x$$

1. $$d \tau_x = (-1)^n I_{n-1}$$ when I write it w.r.t. the standard basis (both) of $$\mathbb{R}^{n-1}$$ and then $$\det(d\tau_x) = (-1)^n$$ for any $$x \in \tilde{U}$$.

2. I know that $$d \overline{\varphi}_y (e^k) = \partial_k$$ for every $$k=1, \dots, n-1$$ and every $$y \in \overline{U}$$. And then $$d \overline{\varphi}_{\tau(x)}=I_{n-1}$$ when I write it w.r.t. the standard basis of $$\mathbb{R}^{n-1}$$ and the basis $$\mathcal{B}=\{\partial_1, \dots, \partial_{n-1} \}$$ of $$T_{\tilde{\varphi}(x)} \partial M$$. Then, since $$\mathcal{B}$$ has $$(-1)^n$$ times the sign of a the positive basis in $$T_{\tilde{\varphi}(x)} \partial M$$, $$d \overline{\varphi}_{\tau(x)}=(-1)^nI_{n-1}$$ when I write it w.r.t. the standard basis of $$\mathbb{R}^{n-1}$$ and a positive basis of $$T_{\tilde{\varphi}(x)} \partial M$$. Then $$\det(d \overline{\varphi}_{\tau(x)})= (-1)^n$$ for any $$x \in \tilde{U}$$.

And then $$\det(d\tilde{\varphi}_x) = (-1)^n (-1)^n = 1 >0$$ for any $$x \in \tilde{U}$$ when I write the jacobian matrix w.r.t. the standard basis of $$\mathbb{R}^{n-1}$$ and a positive basis in $$T_{\tilde{\varphi}(x)} \partial M$$.

You want your basis $$(\varepsilon\partial_1,\dots,\partial_{n-1})$$ of $$T(\partial M)$$ being such that $$(-\partial_n,\varepsilon \partial_1,\dots,\partial_{n-1})$$ is a direct basis of $$TM$$, where $$\varepsilon=\pm1$$ is a sign we will adjust. We will try to write this in term of the standard basis $$(\partial_1,\dots,\partial_n)$$, which is direct by assumption of your parametrization being orientation-preserving. If we want to put the $$-\partial_n$$ in the $$n$$-th position, we have to apply $$(n-1)$$ transpositions, changing each time the sign of the orientation to finally get \begin{align*} (-\partial_n,\varepsilon\partial_1,\dots,\partial_{n-1})\text{ is direct }&\iff (-1)^{n-1}(\varepsilon\partial_1,\dots,\partial_{n-1},-\partial_n)\text{ is direct}\\ &\iff\varepsilon(-1)^n(\partial_1,\dots,\partial_n)\text{ is direct.} \end{align*}
Then precomposing by the map your book describes make appear the correct $$\varepsilon=(-1)^n$$, which guarantees that the parametrization $$\tilde{\varphi}:\tilde{U}\to\partial M$$ is orientation preserving.
For example, the orientation induced by $$\mathbb{H}^2$$ on $$\mathbb{R}=\partial\mathbb{H}^2$$ is the standard one, while the one induced by $$\mathbb{H}^3$$ on $$\mathbb{R}^2=\partial\mathbb{H}^3$$ is the opposite of the standard one.
• Thank you for your answer. The first part is clear to me, I understand why the basis $(\partial_1, \dots, \partial_{n-1})$ has $(-1)^n$ the sign of the positive orientation induced by $M$ on $\partial M$. What I can't understand is why $\tilde{\phi}$ is orientation preserving. I should prove that its differential at any point maps the standard basis of $\mathbb{R}^{n-1}$ is a positive oriented basis in $T_pM$ or something similar... – Bremen000 Oct 25 '18 at 18:38
• As a shortcut to show that, you can note that $\tilde\varphi:\tilde U\to \partial M$ is the composition $\varphi\circ \tau$, where $\tau:\tilde U\to U$ is the the map $(x_1,\dots,x_{n-1})\mapsto((-1)^nx_1,\dots,x_{n-1})$. So $$\det (d\tilde\varphi)=\det(d\varphi)\det(d\tau)=(-1)^n(-1)^n=1,$$ so is orientation-preserving. – Balloon Oct 25 '18 at 18:47
• Ok I thought it was clear to me, but it is not. I understand why $\det(d\varphi) = (-1)^n$ but, in order to consider the composition it must be $\tau: (x_1, \dots, x_{n-1}) \mapsto ((-1)^n, \dots, x_{n-1}, 0 )$ and its jacobian matrix is not squared! I'm missing something, could you please write me the details? – Bremen000 Oct 26 '18 at 12:08