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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. \begin{equation} \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 (*)\end{equation} 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 $T_{\tilde{\varphi}(x)} \partial M$.

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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.

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  • $\begingroup$ 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... $\endgroup$ – Bremen000 Oct 25 '18 at 18:38
  • $\begingroup$ 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. $\endgroup$ – Balloon Oct 25 '18 at 18:47
  • $\begingroup$ Oh sure! Thank you! It was really simple! $\endgroup$ – Bremen000 Oct 25 '18 at 18:49
  • $\begingroup$ No problem, you are welcome! $\endgroup$ – Balloon Oct 25 '18 at 18:49
  • $\begingroup$ 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? $\endgroup$ – Bremen000 Oct 26 '18 at 12:08

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