Differentiable Manifolds/Orientation

Definition (orientation):

Let $M$ be a differentiable manifold. An orientation of $M$ is a family of atlases $(\varphi _{\alpha })_{\alpha \in A}$ on $M$ such that for all $\alpha ,\beta \in A$

$\det \left(D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})\right)>0$.

Proposition (an orientation of a manifold induces an orientation on its boundary):

Let $M$ be a differentiable manifold. If $(\varphi _{\alpha })_{\alpha \in A}$ is an orientation of $M$, then an orientation of $\partial M$ is given by

$(\pi _{2,\ldots ,n}\circ \varphi _{\alpha }\upharpoonright \partial M)_{\alpha \in A}$,

where $\pi _{2,\ldots ,n}$ is defined as follows:

$\pi _{2,\ldots ,n}:\mathbb {R} ^{n}\to \mathbb {R} ^{n-1},~\pi _{2,\ldots ,n}(x_{1},\ldots ,x_{n}):=(x_{2},\ldots ,x_{n})$

Proof: The given family of functions defines an atlas on $\partial M$, so that the proposition will be proven once it is demonstrated that the requirement regarding the positivity of the determinant is satisfied.

Indeed, let $\alpha ,\beta \in A$. Then

$\pi _{2,\ldots ,n}\circ \varphi _{\alpha }\upharpoonright \partial M\circ (\pi _{2,\ldots ,n}\circ \varphi _{\beta }\upharpoonright \partial M)^{-1}=\pi _{2,\ldots ,n}\circ (\varphi _{\alpha }\circ \varphi _{\beta }^{-1})\circ (\pi _{2,\ldots ,n}\upharpoonright \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\})^{-1}$.

Since $\varphi _{\alpha }\circ \varphi _{\beta }^{-1}$ maps the set $\{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\}$ to itself, the first row of $D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})$ is zero so long as $x\in \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\}$, except the very first entry. Yet the very last entry must be non-negative so long as $x\in \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\}$, since otherwise the fundamental theorem of calculus and the continuity of the derivative would imply that for a $y:=(y_{1},\ldots ,y_{n})\in \mathbb {R} ^{n}$ such that $y_{n}>0$ was sufficiently small, $\varphi _{\alpha }\circ \varphi _{\beta }^{-1}(y)$ would be contained within

$\{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{n}<0\}$,

contrary to the definition of charts that contain a piece of the boundary. Hence, upon carrying out a Leibniz expansion of $D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})$ along the first row, we obtain that the matrix obtained from $D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})$ by removing the first row and the first column has positive determinant. Yet the definitions of the respective partial derivatives as limits show that this matrix is exactly the matrix

$D(\pi _{2,\ldots ,n}\circ (\varphi _{\alpha }\circ \varphi _{\beta }^{-1})\circ (\pi _{2,\ldots ,n}\upharpoonright \{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|x_{1}=0\})^{-1})$. $\Box$