[proofplan] We show the rule "$(n_{\rm out}, e_1, \dots, e_{n-1})$ positive for $M$" defines a well-posed orientation on $\partial M$. In a boundary chart $\varphi: U \hookrightarrow \mathbb{R}^n_+ := \{x_1 \ge 0\}$, we identify the outward normal at $p \in \partial M$ with $-\partial_{x_1}\big|_p$, reduce the orientation of $\partial M$ to a sign condition on $(n-1)$-frames in $\{x_1 = 0\}$, and verify chart-independence using the block structure of boundary transition maps. We then confirm the rule produces a nowhere-vanishing smooth volume form on $\partial M$ via the interior product $\iota_{n_{\rm out}} \omega$. [/proofplan]

[step:Set up boundary charts and identify the outward normal] Let $p \in \partial M$. By definition of a manifold with boundary, there exists a chart $(U, \varphi)$ around $p$ with \begin{align*} \varphi : U &\to \mathbb{R}^n_+ = \{(x_1, \dots, x_n) \in \mathbb{R}^n : x_1 \ge 0\}, \\ \varphi(p) &= 0, \end{align*} and $\varphi(U \cap \partial M) = \varphi(U) \cap \{x_1 = 0\}$. The coordinate vector field $\partial_{x_1}\big|_p$ points from $\{x_1 = 0\}$ into $\{x_1 > 0\} = \varphi(U \setminus \partial M)$, hence into the interior of $M$. The [outward-pointing normal](/page/Outward%20Normal%20Vector) at $p$ is any $n_{\rm out} \in T_p M \setminus T_p(\partial M)$ pointing out of $M$; in the chart $\varphi$ we fix the representative $n_{\rm out} = -\partial_{x_1}\big|_p$. [/step]

[step:Reduce the orientation rule on $\partial M$ to a sign condition on $(n-1)$-frames] Let $\omega_M$ denote the orientation on $M$, given by a nowhere-vanishing $\omega \in \Omega^n(M)$. Define the interior product \begin{align*} \omega_{\partial M} := \iota_{n_{\rm out}}\omega \big|_{T(\partial M)} \in \Omega^{n-1}(\partial M), \end{align*} that is, for $v_1, \dots, v_{n-1} \in T_p(\partial M)$, set $(\omega_{\partial M})_p(v_1, \dots, v_{n-1}) := \omega_p(n_{\rm out}, v_1, \dots, v_{n-1})$. Smoothness of $\omega_{\partial M}$ follows from smoothness of $\omega$ and smoothness of any collar extension of $n_{\rm out}$ to a neighbourhood of $\partial M$. By the rule in the statement, a basis $\{e_1, \dots, e_{n-1}\}$ of $T_p(\partial M)$ is positively oriented on $\partial M$ iff $\omega_p(n_{\rm out}, e_1, \dots, e_{n-1}) > 0$ iff $(\omega_{\partial M})_p(e_1, \dots, e_{n-1}) > 0$. [/step]

[step:Verify that $\omega_{\partial M}$ is nowhere vanishing on $\partial M$] Fix $p \in \partial M$ and pick a basis $\{e_1, \dots, e_{n-1}\}$ of $T_p(\partial M)$. Since $n_{\rm out} \notin T_p(\partial M)$ and $T_p(\partial M)$ is a codimension-one subspace, $(n_{\rm out}, e_1, \dots, e_{n-1})$ is a basis of $T_p M$. Because $\omega_p$ is a non-zero top form on $T_p M$, it evaluates non-zero on every basis: \begin{align*} (\omega_{\partial M})_p(e_1, \dots, e_{n-1}) = \omega_p(n_{\rm out}, e_1, \dots, e_{n-1}) \ne 0. \end{align*} Thus $\omega_{\partial M}$ is a nowhere-vanishing top form on $\partial M$, so it defines an orientation of $\partial M$ via the theorem used in [Orientability via Transition Maps](/theorems/1527) (volume form $\Leftrightarrow$ orientation). [/step]

[step:Check independence of the choice of outward normal representative]Let $n_{\rm out}, \tilde{n}_{\rm out}$ be two outward-pointing normals at $p$. Both lie in $T_p M \setminus T_p(\partial M)$ and point outward, so they lie on the same side of $T_p(\partial M)$. Writing $\tilde{n}_{\rm out} = c \cdot n_{\rm out} + v$ with $v \in T_p(\partial M)$ and $c \in \mathbb{R}$, the condition that $\tilde{n}_{\rm out}$ points outward forces $c > 0$. Then \begin{align*} \omega_p(\tilde{n}_{\rm out}, e_1, \dots, e_{n-1}) &= \omega_p(c \cdot n_{\rm out} + v, e_1, \dots, e_{n-1}) \\ &= c \cdot \omega_p(n_{\rm out}, e_1, \dots, e_{n-1}) + \omega_p(v, e_1, \dots, e_{n-1}). \end{align*} Since $v, e_1, \dots, e_{n-1} \in T_p(\partial M)$ and $\dim T_p(\partial M) = n - 1$, the vectors $v, e_1, \dots, e_{n-1}$ are linearly dependent, so the alternating property gives $\omega_p(v, e_1, \dots, e_{n-1}) = 0$. Hence \begin{align*} \omega_p(\tilde{n}_{\rm out}, e_1, \dots, e_{n-1}) = c \cdot \omega_p(n_{\rm out}, e_1, \dots, e_{n-1}), \end{align*} and since $c > 0$, the two expressions have the same sign. The orientation rule is independent of the chosen outward normal.[/step]

[guided]The key cancellation is that the "tangential component" $v$ of the altered normal contributes zero when paired with an $(n-1)$-tuple that already spans $T_p(\partial M)$: we are feeding $n$ vectors into $\omega$, but the last $n$ lie in an $(n-1)$-dimensional subspace, so they are linearly dependent and the alternating $n$-form vanishes. Only the scalar $c > 0$ along the true outward direction survives, and since $c$ is positive, signs are preserved. This is the algebraic content of "outward-ness determines the sign".[/guided]

[step:Verify chart-independence of the induced orientation]Let $(U, \varphi)$ and $(V, \psi)$ be two boundary charts at $p$ compatible with the orientation of $M$, with $\varphi(U), \psi(V) \subset \mathbb{R}^n_+$. Write coordinates $x$ for $\varphi$ and $y$ for $\psi$. The transition $T := \psi \circ \varphi^{-1}$ maps $\{x_1 = 0\}$ to $\{y_1 = 0\}$ and $\{x_1 \ge 0\}$ to $\{y_1 \ge 0\}$. Therefore $T_1(0, x_2, \dots, x_n) = 0$ and $T_1(x_1, x_2, \dots, x_n) \ge 0$ when $x_1 \ge 0$. Differentiating $T_1$ at a boundary point and using the one-sided non-negativity of $T_1$: \begin{align*} \frac{\partial T_1}{\partial x_1}(0, x_2, \dots, x_n) \ge 0, \qquad \frac{\partial T_1}{\partial x_j}(0, x_2, \dots, x_n) = 0 \text{ for } j \ge 2. \end{align*} Since $T$ is a diffeomorphism, $\partial T_1/\partial x_1 > 0$ on the boundary. The Jacobian of $T$ at a boundary point thus has the block form \begin{align*} JT = \begin{pmatrix} a & 0 \\ * & B \end{pmatrix}, \qquad a = \frac{\partial T_1}{\partial x_1} > 0, \quad B = \left(\frac{\partial T_i}{\partial x_j}\right)_{i,j \ge 2}. \end{align*} Orientation-compatibility of the atlas on $M$ gives $\det(JT) = a \cdot \det(B) > 0$, hence $\det(B) > 0$. The matrix $B$ is the Jacobian of the induced boundary transition $T\big|_{\{x_1 = 0\}} : \{x_1 = 0\} \to \{y_1 = 0\}$ in the coordinates $(x_2, \dots, x_n), (y_2, \dots, y_n)$. Thus the boundary charts form an atlas with positive transition determinants, and by [Orientability via Transition Maps](/theorems/1527) this atlas orients $\partial M$ consistently with the interior-product construction from Step 2.[/step]

[guided]The block-triangular structure of $JT$ at a boundary point is a geometric fact about how transitions behave: the boundary is preserved, so $T_1$ vanishes on $\{x_1 = 0\}$, which forces $\partial T_1 / \partial x_j = 0$ for tangential indices $j \ge 2$. The $x_1$-derivative $a = \partial T_1 / \partial x_1$ measures how the "outward" coordinate transforms; it must be positive because $T$ maps $\{x_1 > 0\}$ into $\{y_1 > 0\}$ and is a local diffeomorphism. Given $\det(JT) > 0$ and $a > 0$, the lower-right block $B$ — which is the Jacobian of the restricted boundary transition — has $\det B > 0$. This matches the block decomposition $\det(JT) = a \cdot \det(B)$ and propagates the orientation of $M$ to a canonical orientation on $\partial M$.[/guided]

[step:Confirm the sign convention yields Stokes' theorem] In a boundary chart as above, the orientation of $M$ is given by $dx_1 \wedge \cdots \wedge dx_n$. The interior product with $n_{\rm out} = -\partial_{x_1}$ is \begin{align*} \iota_{-\partial_{x_1}}(dx_1 \wedge \cdots \wedge dx_n) = -\iota_{\partial_{x_1}}(dx_1 \wedge \cdots \wedge dx_n) = -dx_2 \wedge \cdots \wedge dx_n. \end{align*} Thus on $\partial M = \{x_1 = 0\}$, the induced orientation is represented by $-dx_2 \wedge \cdots \wedge dx_n$. This is the sign convention that makes [Stokes' Theorem](/theorems/1530) hold without corrective signs, as verified directly in the proof of that theorem. This completes the construction of the canonical orientation on $\partial M$. [/step]