[proofplan] A nowhere-vanishing top-degree form orients each tangent space by declaring a basis positive exactly when the form evaluates positively on it. The smoothness of this orientation is checked in coordinates: the form has a smooth nonzero coefficient, and after possibly reflecting one coordinate, that coefficient is positive near any prescribed point. Conversely, a smooth orientation gives an oriented atlas; using a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to that atlas, we patch the local positive coordinate volume forms. The positivity of oriented transition Jacobians ensures that the resulting global $n$-form is nowhere zero. [/proofplan]

[step:Construct a smooth orientation from a nowhere-vanishing $n$-form]Assume that $\omega \in \Omega^n(M)$ satisfies $\omega_p \neq 0$ for every $p \in M$. For each $p \in M$, define an orientation of the real [vector space](/page/Vector%20Space) $T_pM$ as follows: an ordered basis $(v_1,\dots,v_n)$ of $T_pM$ is declared positive if and only if \begin{align*} \omega_p(v_1,\dots,v_n) > 0. \end{align*} Because $\omega_p$ is a nonzero alternating $n$-linear form on the $n$-dimensional vector space $T_pM$, this definition selects exactly one of the two orientation classes on $T_pM$. We verify that this pointwise orientation is smooth. Let $p \in M$, and let $(U,\varphi)$ be a smooth coordinate chart with \begin{align*} \varphi: U &\to \varphi(U) \subseteq \mathbb{R}^n, \\ q &\mapsto (x_1(q),\dots,x_n(q)). \end{align*} On $U$, the coordinate $n$-form $dx_1 \wedge \cdots \wedge dx_n$ is a local frame for $\Lambda^nT^*M$, so there is a unique smooth function \begin{align*} a: U &\to \mathbb{R} \end{align*} such that \begin{align*} \omega|_U = a\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Since $\omega_q \neq 0$ for every $q \in U$, we have $a(q) \neq 0$ for every $q \in U$. If $a(p) > 0$, shrink $U$ to an open neighbourhood $U_p \subseteq U$ of $p$ on which $a>0$. Then $(U_p,\varphi|_{U_p})$ is positively oriented for the orientation just defined. If $a(p)<0$, let \begin{align*} R: \mathbb{R}^n &\to \mathbb{R}^n, \\ (y_1,\dots,y_n) &\mapsto (-y_1,y_2,\dots,y_n) \end{align*} be the reflection in the first coordinate, and define the chart \begin{align*} \psi: U &\to R(\varphi(U)), \\ q &\mapsto R(\varphi(q)). \end{align*} Writing $\psi(q)=(y_1(q),\dots,y_n(q))$, we have \begin{align*} dy_1 \wedge \cdots \wedge dy_n = - dx_1 \wedge \cdots \wedge dx_n. \end{align*} Thus \begin{align*} \omega|_U = (-a)\, dy_1 \wedge \cdots \wedge dy_n. \end{align*} Since $-a(p)>0$, shrink $U$ to an open neighbourhood $U_p$ of $p$ on which $-a>0$. Then $(U_p,\psi|_{U_p})$ is positively oriented. It remains to check compatibility of these positive charts. Let $(U,\varphi)$ and $(V,\psi)$ be two charts whose coordinate bases are positive for the orientation defined by $\omega$. Write \begin{align*} \varphi(q) &= (x_1(q),\dots,x_n(q)), \\ \psi(q) &= (y_1(q),\dots,y_n(q)). \end{align*} On $U \cap V$, there are smooth positive functions $a_\varphi,a_\psi:U\cap V\to (0,\infty)$ such that \begin{align*} \omega = a_\varphi\, dx_1\wedge\cdots\wedge dx_n = a_\psi\, dy_1\wedge\cdots\wedge dy_n. \end{align*} The coordinate change formula for top-degree forms gives \begin{align*} dy_1\wedge\cdots\wedge dy_n = \det J(\psi\circ\varphi^{-1})\, dx_1\wedge\cdots\wedge dx_n. \end{align*} Therefore \begin{align*} a_\varphi = a_\psi\, \det J(\psi\circ\varphi^{-1}). \end{align*} Since $a_\varphi>0$ and $a_\psi>0$, it follows that \begin{align*} \det J(\psi\circ\varphi^{-1})>0 \end{align*} on $\varphi(U\cap V)$. Hence the positive charts form an oriented atlas, so $M$ is orientable.[/step]

[guided]Assume that a smooth nowhere-vanishing top-degree form $\omega \in \Omega^n(M)$ is given. At each point $p \in M$, the value \begin{align*} \omega_p: T_pM \times \cdots \times T_pM \to \mathbb{R} \end{align*} is an alternating $n$-[linear map](/page/Linear%20Map). Because $T_pM$ has dimension $n$ and $\omega_p \neq 0$, evaluating $\omega_p$ on an ordered basis never gives zero. We therefore define an ordered basis $(v_1,\dots,v_n)$ of $T_pM$ to be positive exactly when \begin{align*} \omega_p(v_1,\dots,v_n)>0. \end{align*} This gives one of the two possible orientations of $T_pM$. The remaining issue is smooth dependence on $p$. Fix $p \in M$ and choose a coordinate chart \begin{align*} \varphi: U &\to \varphi(U)\subseteq \mathbb{R}^n, \\ q &\mapsto (x_1(q),\dots,x_n(q)). \end{align*} On $U$, the form $dx_1\wedge\cdots\wedge dx_n$ spans the one-dimensional vector bundle $\Lambda^nT^*M$. Hence there is a unique smooth coefficient function \begin{align*} a: U &\to \mathbb{R} \end{align*} with \begin{align*} \omega|_U = a\,dx_1\wedge\cdots\wedge dx_n. \end{align*} The hypothesis that $\omega$ is nowhere vanishing implies $a(q)\neq 0$ for every $q\in U$. If $a(p)>0$, continuity of $a$ gives an open neighbourhood $U_p\subseteq U$ of $p$ on which $a>0$. In this chart, the coordinate basis is positive according to the orientation defined by $\omega$. If $a(p)<0$, we reverse exactly one coordinate. Define \begin{align*} R: \mathbb{R}^n &\to \mathbb{R}^n, \\ (y_1,\dots,y_n) &\mapsto (-y_1,y_2,\dots,y_n), \end{align*} and set $\psi=R\circ\varphi$. If $\psi(q)=(y_1(q),\dots,y_n(q))$, then the Jacobian determinant of $R$ is $-1$, so \begin{align*} dy_1\wedge\cdots\wedge dy_n = - dx_1\wedge\cdots\wedge dx_n. \end{align*} Thus \begin{align*} \omega|_U = (-a)\,dy_1\wedge\cdots\wedge dy_n. \end{align*} Because $-a(p)>0$, after shrinking to an open neighbourhood $U_p$ of $p$, the coefficient $-a$ is positive throughout $U_p$. Therefore every point has a coordinate neighbourhood whose coordinate basis agrees with the orientation defined by $\omega$. Finally we check that these locally positive charts are compatible. Let $(U,\varphi)$ and $(V,\psi)$ be two such charts, with coordinate functions $x_1,\dots,x_n$ and $y_1,\dots,y_n$. Since both charts are positive, there are smooth functions \begin{align*} a_\varphi,a_\psi:U\cap V\to (0,\infty) \end{align*} such that \begin{align*} \omega = a_\varphi\,dx_1\wedge\cdots\wedge dx_n = a_\psi\,dy_1\wedge\cdots\wedge dy_n. \end{align*} The change-of-coordinates formula for top-degree forms gives \begin{align*} dy_1\wedge\cdots\wedge dy_n = \det J(\psi\circ\varphi^{-1})\, dx_1\wedge\cdots\wedge dx_n. \end{align*} Substituting this into the expression for $\omega$ yields \begin{align*} a_\varphi = a_\psi\,\det J(\psi\circ\varphi^{-1}). \end{align*} Both coefficients are positive, so the determinant of the transition map is positive. Therefore all transition functions between these charts preserve orientation, and the charts form an oriented atlas. Hence $M$ is orientable.[/guided]

[step:Patch local positive coordinate volume forms using an oriented atlas]Conversely, assume that $M$ is orientable. Choose an oriented smooth atlas $\{(U_i,\varphi_i)\}_{i\in I}$ for $M$, where \begin{align*} \varphi_i:U_i &\to \varphi_i(U_i)\subseteq \mathbb{R}^n, \\ q &\mapsto (x_{i,1}(q),\dots,x_{i,n}(q)). \end{align*} For each $i\in I$, define the local coordinate volume form \begin{align*} \theta_i := dx_{i,1}\wedge\cdots\wedge dx_{i,n} \in \Omega^n(U_i). \end{align*} Because a smooth manifold is second-countable and Hausdorff, $M$ is paracompact, and the oriented coordinate atlas $\{U_i\}_{i\in I}$ is an open cover of $M$. Hence the [Smooth Partition of Unity Theorem](/theorems/???) applies to this cover. Choose a locally finite family of smooth functions \begin{align*} \rho_i:M &\to [0,1], \end{align*} indexed by $I$, such that $\operatorname{supp}\rho_i\subseteq U_i$ for every $i\in I$ and \begin{align*} \sum_{i\in I}\rho_i(p)=1 \end{align*} for every $p\in M$. Define a global smooth $n$-form $\beta_i\in \Omega^n(M)$ by \begin{align*} (\beta_i)_q = \begin{cases} \rho_i(q)(\theta_i)_q, & q\in U_i,\\ 0, & q\in M\setminus \operatorname{supp}\rho_i. \end{cases} \end{align*} This is smooth because $\operatorname{supp}\rho_i\subseteq U_i$. Since the family is locally finite, the pointwise sum \begin{align*} \omega := \sum_{i\in I}\beta_i \end{align*} defines a smooth global section of $\Lambda^nT^*M$.[/step]

[guided]Assume now that $M$ is orientable. This means that there is an atlas whose transition maps have positive Jacobian determinant. Choose such an oriented atlas $\{(U_i,\varphi_i)\}_{i\in I}$, where each chart is a smooth map \begin{align*} \varphi_i:U_i &\to \varphi_i(U_i)\subseteq \mathbb{R}^n, \\ q &\mapsto (x_{i,1}(q),\dots,x_{i,n}(q)). \end{align*} Each chart provides a local positive top-degree form \begin{align*} \theta_i := dx_{i,1}\wedge\cdots\wedge dx_{i,n}\in \Omega^n(U_i). \end{align*} The local forms $\theta_i$ cannot simply be glued by declaring them equal on overlaps, because on an overlap they may differ by a positive smooth factor. The standard way to patch local data is a partition of unity. We verify the hypotheses for the partition of unity theorem: by the standing convention for smooth manifolds, $M$ is second-countable and Hausdorff, hence paracompact, and the family $\{U_i\}_{i\in I}$ is an open cover by coordinate domains. Therefore the [Smooth Partition of Unity Theorem](/theorems/???) applies to this cover, and we may choose smooth functions \begin{align*} \rho_i:M &\to [0,1] \end{align*} such that the family is locally finite, $\operatorname{supp}\rho_i\subseteq U_i$, and \begin{align*} \sum_{i\in I}\rho_i(p)=1 \end{align*} for every $p\in M$. For each $i\in I$, define a global $n$-form $\beta_i\in\Omega^n(M)$ by multiplying the local coordinate volume form by the cutoff $\rho_i$: \begin{align*} (\beta_i)_q = \begin{cases} \rho_i(q)(\theta_i)_q, & q\in U_i,\\ 0, & q\in M\setminus \operatorname{supp}\rho_i. \end{cases} \end{align*} The condition $\operatorname{supp}\rho_i\subseteq U_i$ is what makes this extension smooth across the boundary of $U_i$: near every point outside $\operatorname{supp}\rho_i$, the form is identically zero. Since the family $(\rho_i)_{i\in I}$ is locally finite, only finitely many $\beta_i$ are nonzero near any given point. Therefore the sum \begin{align*} \omega := \sum_{i\in I}\beta_i \end{align*} is a well-defined smooth global section of $\Lambda^nT^*M$.[/guided]

[step:Show the patched form is nowhere vanishing]Fix $p\in M$. Choose an oriented chart $(V,\psi)$ around $p$, with \begin{align*} \psi:V &\to \psi(V)\subseteq \mathbb{R}^n, \\ q &\mapsto (y_1(q),\dots,y_n(q)). \end{align*} For each $i\in I$ with $U_i\cap V\neq\varnothing$, define the smooth map \begin{align*} J_i: U_i\cap V &\to \mathbb{R}^{n\times n}, \\ q &\mapsto J(\varphi_i\circ\psi^{-1})_{\psi(q)} \end{align*} whose value $J_i(q)$ is the Jacobian matrix of the transition map $\varphi_i\circ\psi^{-1}$ at $\psi(q)$. Since the atlas is oriented, for every $q\in U_i\cap V$, \begin{align*} \det J_i(q)>0. \end{align*} On $U_i\cap V$, the coordinate change formula gives \begin{align*} \theta_i = \det J_i\, dy_1\wedge\cdots\wedge dy_n. \end{align*} Hence on a neighbourhood of $p$, \begin{align*} \omega = \left(\sum_{i\in I}\rho_i\,\det J_i\right) dy_1\wedge\cdots\wedge dy_n, \end{align*} where only finitely many terms are nonzero near $p$. Evaluating at $p$, every summand $\rho_i(p)\det J_i(p)$ is nonnegative, and at least one is positive because $\sum_{i\in I}\rho_i(p)=1$. Therefore \begin{align*} \sum_{i\in I}\rho_i(p)\det J_i(p)>0. \end{align*} Since $dy_1\wedge\cdots\wedge dy_n$ is a nonzero element of $\Lambda^nT_p^*M$, it follows that $\omega_p\neq 0$. As $p\in M$ was arbitrary, $\omega$ is nowhere vanishing. Thus an orientable smooth $n$-manifold admits a nowhere-vanishing smooth $n$-form. Together with the first direction, this proves the equivalence.[/step]

[guided]We now prove that the global form constructed from the partition of unity does not vanish anywhere. Fix $p\in M$, and choose an oriented chart \begin{align*} \psi:V &\to \psi(V)\subseteq \mathbb{R}^n, \\ q &\mapsto (y_1(q),\dots,y_n(q)). \end{align*} We will express every local contribution to $\omega$ in this one coordinate frame. For each index $i$ with $U_i\cap V\neq \varnothing$, define the smooth map \begin{align*} J_i: U_i\cap V &\to \mathbb{R}^{n\times n}, \\ q &\mapsto J(\varphi_i\circ\psi^{-1})_{\psi(q)}. \end{align*} For $q\in U_i\cap V$, the matrix $J_i(q)$ is the Jacobian matrix of the transition map from the $\psi$-coordinates to the $\varphi_i$-coordinates at the coordinate point $\psi(q)$. Since both charts belong to the oriented atlas, the transition map is orientation-preserving, so \begin{align*} \det J_i(q)>0 \end{align*} for every $q\in U_i\cap V$. The coordinate change formula for top-degree forms gives \begin{align*} dx_{i,1}\wedge\cdots\wedge dx_{i,n} = \det J_i\,dy_1\wedge\cdots\wedge dy_n. \end{align*} Thus, on $U_i\cap V$, \begin{align*} \beta_i = \rho_i\,\det J_i\,dy_1\wedge\cdots\wedge dy_n. \end{align*} Because the partition of unity is locally finite, there is a neighbourhood of $p$ on which only finitely many of these terms occur. Therefore near $p$ we may write \begin{align*} \omega = \left(\sum_{i\in I}\rho_i\,\det J_i\right) dy_1\wedge\cdots\wedge dy_n. \end{align*} Now evaluate this coefficient at $p$. Each $\rho_i(p)$ is nonnegative, and each relevant determinant $\det J_i(p)$ is positive. Also the partition of unity condition gives \begin{align*} \sum_{i\in I}\rho_i(p)=1, \end{align*} so at least one $\rho_i(p)$ is positive. Hence the coefficient \begin{align*} \sum_{i\in I}\rho_i(p)\det J_i(p) \end{align*} is strictly positive. Since $dy_1\wedge\cdots\wedge dy_n$ is a nonzero basis element of the one-dimensional vector space $\Lambda^nT_p^*M$, the form $\omega_p$ is nonzero. The point $p$ was arbitrary, so $\omega$ is nowhere vanishing on $M$. This proves that every orientable smooth $n$-manifold admits a nowhere-vanishing smooth $n$-form, and the previous direction proved the converse.[/guided]