> [!definition] > > Let $X$ be an [[Orientation on Manifold|oriented]] $n$-manifold [[Manifold with Boundary|with boundary]], then > 1. $\partial X$ is orientable. > 2. All outward-pointing vector fields on $\partial X$ determine the same orientation on $\partial X$. > > known as the **Stokes orientation** on $\partial X$. > > *Proof*. Let $\omega \in \Omega^n$ be an orientation form for $X$, and $\xi \in \vf(\partial X)$ be an outward pointing vector field. Define > $ > \omega^*_p(v_1, \cdots, v_{n-1}) = \iota^*_{\partial X}\omega(\xi_p, v_1, \cdots, v_{n-1}) > $ > then $\omega^* \in \Omega^{n - 1}(\partial X)$ gives an orientation $\partial X$. > [!theoremb] Theorem > > Let $X$ be an oriented $n$-manifold with boundary, and $\omega \in \Omega^{n - 1}$ be a [[Compactly Supported|compactly supported]] [[Differential Form|(n-1)-form]], then > $ > \int_X d\omega = \int_{\partial X}\omega > $ > where > 1. $d\omega$ is the [[Exterior Derivative|exterior derivative]]. > 2. $\omega = \iota_{\partial X}^*\omega$ is the [[Pullback of Sections|pullback]] onto $\partial X$. > 3. $\partial X$ is given the Stokes orientation. > > *Proof*. Suppose that $X = \mathbb{H}^n$ or $X = \real^n$. In the first case, extend $\omega$ to $\real^n$ by filling in $0$ on $\real^n \setminus \mathbb{H}^n$. Write > $ > \omega = \sum_{j = 1}^n \omega_j dx^1 \wedge \cdots \wedge \wh{dx^j} \wedge \cdots \wedge dx^{n} > $ > where $\wh{dx^j}$ corresponds to omitting $dx^j$. This allows writing > $ > \begin{align*} > d\omega &= \sum_{j = 1}^n d\omega_j \wedge dx^1 \wedge \cdots \wedge \wh{dx^j} \wedge \cdots \wedge dx^{n} \\ > &= \sum_{j, k = 1}^n \part{\omega_j}{x_k}dx^k \wedge dx^1 \wedge \cdots \wedge \wh{dx^j} \wedge \cdots \wedge dx^{n} \\ > &= \sum_{j = 1}^n (-1)^{j - 1}\part{\omega_j}{x^j} dx^1 \wedge \cdots \wedge dx^n > \end{align*} > $ > Since $\omega$ has compact support, there exists $R \ge 0$ such that $\supp{\omega} \subset [-R, R]^n$. > $ > \begin{align*} > \int_{\mathbb{H}^n}d\omega &= \sum_{j = 1}^n(-1)^{j - 1}\int_{[-R, R]^{n - 1}}\int_{[-R, R]}\part{\omega_j}{x^j}dx_jd(x_1,, \cdots, \wh{x_j} , \cdots, x_{n}) \\ > &= (-1)^{n-1}\int_{[-R, R]^{n - 1}}\int_{[-R, R]}\part{\omega_n}{x^n}d(x_1, \cdots, x_{n-1})dx_n > \end{align*} > $ > where all other coordinates cancel out due to compact support and FTC. If $X = \real^n$, then > $ > \int_{[-R, R]^{n - 1}}\part{\omega_n}{x^n}d(x_1, \cdots, x_{n-1}) = 0 > $ > which corresponds to the lack of boundary. If $X = \mathbb{H}^n$, then > $ > \begin{align*} > \int_{\mathbb{H}^n}d\omega &= (-1)^{n-1}\int_{[-R, R]^{n - 1}}\int_{[-R, R]}\part{\omega_n}{x^n}d(x_1, \cdots, x_{n-1})dx_n \\ > &= (-1)^n\int_{[-R, R]^{n-1}}\omega_n(x_1, \cdots, x_{n - 1})d(x_1, \cdots, x_{n - 1}) \\ > &= \int_{\partial \mathbb{H}^n}\omega_n dx^{1} \wedge \cdots \wedge dx^{n - 1} \\ > &= \sum_{j = 1}^n \int_{\partial \mathbb{H}^n}\omega_n dx^{1} \wedge \cdots \wedge \wh{dx^j}\wedge \cdots \wedge dx^{n} \\ > &= \int_{\partial \mathbb{H}^n}\omega > \end{align*} > $ > where the induced Stokes orientation on $\partial \mathbb{H}^n$ is $(-1)^n dx^1 \wedge \cdots \wedge dx^{n - 1}$. > > Now suppose that $X$ is an arbitrary $n$-manifold, and there exists a positively oriented chart $(U, \varphi)$ such that $\supp{\omega} \subset U$. In this case, > $ > \int_X d\omega = \int_{\mathbb{H}^n}(\varphi^{-1})^*d\omega = \int_{\mathbb{H}^n}d\braks{(\varphi^{-1})^*\omega} > $ > Since positively oriented charts preserve the inwards/outwards pointing tangent vectors, they also preserve the Stokes orientation. Therefore > $ > \int_{\partial \mathbb{H}^n}\paren{\varphi^{-1}}^*\omega = \int_{\partial X}\omega > $ > Lastly, let $\omega \in \Omega^k$ be arbitrary and compactly supported. Let $\{(U_j, \varphi_j)\}_1^k$ be a finite family of positively oriented charts covering its support, and $\{\psi_j\}_1^k$ be a corresponding [[Partition of Unity|partition of unity]]. In this case, > $ > \begin{align*} > \int_{\partial X}\omega &= \sum_{j = 1}^k \int_{\partial X}\psi_i \omega = \sum_{j = 1}^k \int_X d(\psi_i\omega) = \sum_{j = 1}^k\int_X d\psi_j \wedge \omega + \psi_jd\omega \\ > &= \int_X d\paren{\sum_{j = 1}^k \psi_j}\wedge \omega + \int_X \paren{\sum_{j = 1}^k \psi_j} d\omega = 0 + \int_X d\omega > \end{align*} > $