# Differentiable Manifolds/Orientation

Definition (orientation):

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

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

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

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

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

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

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

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

Indeed, let ${\displaystyle \alpha ,\beta \in A}$. Then

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

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

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

contrary to the definition of charts that contain a piece of the boundary. Hence, upon carrying out a Leibniz expansion of ${\displaystyle D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})}$ along the first row, we obtain that the matrix obtained from ${\displaystyle D(\varphi _{\alpha }\circ \varphi _{\beta }^{-1})}$ 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

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