[proofplan] We construct the correspondence in both directions. A nowhere-vanishing $n$-form selects exactly those frames on which it has value $1$, and the determinant transformation law shows that these frames form a principal $SL_n(\mathbb{R})$-subbundle. Conversely, an $SL_n(\mathbb{R})$-reduction lets us define an $n$-form by declaring its value on every reduced frame to be $1$; determinant one makes this independent of the chosen reduced frame. Finally, the two constructions are inverse to each other and the sign of the resulting form records the associated orientation. [/proofplan]

[step:Build an $SL_n(\mathbb{R})$-subbundle from a nowhere-vanishing $n$-form]Let $\omega \in \Omega^n(M)$ be nowhere vanishing. Define \begin{align*} Q_\omega = \{u \in \operatorname{Fr}(TM) : \omega_{\pi(u)}(u(e_1), \dots, u(e_n)) = 1\}. \end{align*} For $A \in GL_n(\mathbb{R})$, the right action on $\operatorname{Fr}(TM)$ is given by $u \cdot A = u \circ A$. If $A \in SL_n(\mathbb{R})$, then multilinearity and alternation of $\omega_{\pi(u)}$ give \begin{align*} \omega_{\pi(u)}((u \circ A)(e_1), \dots, (u \circ A)(e_n)) = \det(A)\,\omega_{\pi(u)}(u(e_1), \dots, u(e_n)) = 1. \end{align*} Thus $Q_\omega$ is preserved by the right action of $SL_n(\mathbb{R})$. For each $x \in M$, the fibre $(Q_\omega)_x$ is nonempty. Indeed, since $\omega_x \neq 0$, choose any frame $v: \mathbb{R}^n \to T_xM$. Let \begin{align*} c = \omega_x(v(e_1), \dots, v(e_n)) \in \mathbb{R}\setminus\{0\}. \end{align*} Replacing $v$ by the frame $u = v \circ A$, where $A \in GL_n(\mathbb{R})$ is any matrix with $\det(A) = c^{-1}$, gives \begin{align*} \omega_x(u(e_1), \dots, u(e_n)) = \det(A)c = 1. \end{align*} The action of $SL_n(\mathbb{R})$ on each fibre $(Q_\omega)_x$ is free because the right action of $GL_n(\mathbb{R})$ on frames is free. It is transitive as follows. If $u, v \in (Q_\omega)_x$, then there is a unique $A \in GL_n(\mathbb{R})$ such that $v = u \circ A$. Applying $\omega_x$ to the two frames gives \begin{align*} 1 = \omega_x(v(e_1), \dots, v(e_n)) = \det(A)\,\omega_x(u(e_1), \dots, u(e_n)) = \det(A), \end{align*} so $A \in SL_n(\mathbb{R})$. Hence $(Q_\omega)_x$ is a principal homogeneous space for $SL_n(\mathbb{R})$. Smoothness is local. On an [open set](/page/Open%20Set) $U \subset M$ with a smooth local frame \begin{align*} s: U \to \operatorname{Fr}(TM), \end{align*} define the smooth function \begin{align*} f: U \to \mathbb{R}\setminus\{0\}, \qquad x \mapsto \omega_x(s(x)(e_1), \dots, s(x)(e_n)). \end{align*} After possibly replacing $s$ by $s \circ A_0$ for a fixed matrix $A_0 \in GL_n(\mathbb{R})$ with negative determinant, we may assume $f > 0$ on a smaller neighbourhood. Define \begin{align*} r: U \to \operatorname{Fr}(TM), \qquad x \mapsto s(x) \circ \operatorname{diag}(f(x)^{-1}, 1, \dots, 1). \end{align*} Then $r$ is a smooth local section of $Q_\omega$. These local sections give $Q_\omega$ the structure of a smooth principal $SL_n(\mathbb{R})$-subbundle of $\operatorname{Fr}(TM)$.[/step]

[guided]The condition defining $Q_\omega$ says that we keep exactly the frames whose ordered basis has $\omega$-volume equal to $1$. We first check that this condition is stable under changing the frame by an element of $SL_n(\mathbb{R})$. The right action of $GL_n(\mathbb{R})$ on $\operatorname{Fr}(TM)$ is \begin{align*} u \cdot A = u \circ A. \end{align*} Thus the vectors of the new frame are $(u \circ A)(e_1), \dots, (u \circ A)(e_n)$. Since $\omega_x$ is an alternating $n$-linear form, changing the ordered basis by $A$ multiplies its value by $\det(A)$: \begin{align*} \omega_x((u \circ A)(e_1), \dots, (u \circ A)(e_n)) = \det(A)\,\omega_x(u(e_1), \dots, u(e_n)). \end{align*} If $A \in SL_n(\mathbb{R})$, then $\det(A)=1$, so a frame satisfying the defining condition remains satisfying it. Next we verify that every fibre is nonempty. Fix $x \in M$. Since $\omega_x$ is a nonzero alternating $n$-form on the $n$-dimensional [vector space](/page/Vector%20Space) $T_xM$, there is some frame $v: \mathbb{R}^n \to T_xM$ with \begin{align*} c = \omega_x(v(e_1), \dots, v(e_n)) \neq 0. \end{align*} Choose a matrix $A \in GL_n(\mathbb{R})$ with $\det(A)=c^{-1}$. Then $u = v \circ A$ is again a frame, and the determinant transformation law gives \begin{align*} \omega_x(u(e_1), \dots, u(e_n)) = \det(A)c = 1. \end{align*} So $u \in (Q_\omega)_x$. Now take two frames $u, v \in (Q_\omega)_x$. Because $u$ is a linear isomorphism $\mathbb{R}^n \to T_xM$, there is a unique $A \in GL_n(\mathbb{R})$ with $v = u \circ A$. Applying $\omega_x$ gives \begin{align*} 1 = \omega_x(v(e_1), \dots, v(e_n)) = \det(A)\,\omega_x(u(e_1), \dots, u(e_n)) = \det(A). \end{align*} Therefore $A \in SL_n(\mathbb{R})$. This proves transitivity, and freeness follows from freeness of the frame-bundle action. Finally, we check smoothness. Let \begin{align*} s: U \to \operatorname{Fr}(TM) \end{align*} be a smooth local frame over an open set $U \subset M$. The function \begin{align*} f: U \to \mathbb{R}\setminus\{0\}, \qquad x \mapsto \omega_x(s(x)(e_1), \dots, s(x)(e_n)) \end{align*} is smooth because $\omega$ and $s$ are smooth. After restricting $U$ if necessary and, if needed, replacing $s$ by $s \circ A_0$ for a fixed orientation-reversing matrix $A_0$, we may assume $f>0$ on $U$. Then \begin{align*} r: U \to \operatorname{Fr}(TM), \qquad x \mapsto s(x) \circ \operatorname{diag}(f(x)^{-1}, 1, \dots, 1) \end{align*} is smooth, and its determinant correction has determinant $f(x)^{-1}$. Therefore \begin{align*} \omega_x(r(x)(e_1), \dots, r(x)(e_n)) = f(x)^{-1} f(x) = 1. \end{align*} Thus $r$ is a smooth local section of $Q_\omega$, proving that $Q_\omega$ is a smooth principal $SL_n(\mathbb{R})$-reduction.[/guided]

[step:Construct a nowhere-vanishing $n$-form from an $SL_n(\mathbb{R})$-reduction] Let $Q \subset \operatorname{Fr}(TM)$ be a principal $SL_n(\mathbb{R})$-reduction. For each $x \in M$ and each $q \in Q_x$, define an alternating $n$-[linear map](/page/Linear%20Map) \begin{align*} \omega_x: T_xM \times \cdots \times T_xM \to \mathbb{R} \end{align*} as follows. Given $v_1,\dots,v_n \in T_xM$, let $a_1,\dots,a_n \in \mathbb{R}^n$ be the unique vectors satisfying $v_i = q(a_i)$ for every $i \in \{1,\dots,n\}$. Define \begin{align*} \omega_x(v_1,\dots,v_n) = \det(a_1,\dots,a_n), \end{align*} where $\det(a_1,\dots,a_n)$ denotes the determinant of the matrix whose $i$th column is $a_i$. This definition is independent of the chosen reduced frame $q \in Q_x$. If $q' \in Q_x$, then there is a unique $A \in SL_n(\mathbb{R})$ such that $q' = q \circ A$. If $v_i = q(a_i) = q'(b_i)$, then $a_i = A b_i$ for every $i$, hence \begin{align*} \det(a_1,\dots,a_n) = \det(A)\det(b_1,\dots,b_n) = \det(b_1,\dots,b_n). \end{align*} Thus $\omega_x$ is well-defined. The form is smooth because $Q$ has smooth local sections. If \begin{align*} q: U \to Q \end{align*} is a smooth local section over an open set $U \subset M$, let $\theta_1,\dots,\theta_n \in \Omega^1(U)$ be the smooth coframe dual to the smooth frame $q(e_1),\dots,q(e_n)$. Then on $U$, \begin{align*} \omega = \theta_1 \wedge \cdots \wedge \theta_n. \end{align*} This local expression proves smoothness. Also, \begin{align*} \omega_x(q(e_1),\dots,q(e_n)) = \det(e_1,\dots,e_n) = 1, \end{align*} so $\omega_x \neq 0$ for every $x \in M$. Therefore $\omega$ is a nowhere-vanishing smooth $n$-form. [/step]

[step:Verify that the two constructions are inverse] Starting with a nowhere-vanishing smooth $n$-form $\omega$, form $Q_\omega$, and then construct the form $\widetilde{\omega}$ from $Q_\omega$. For $x \in M$, choose $q \in (Q_\omega)_x$. By definition, \begin{align*} \omega_x(q(e_1),\dots,q(e_n)) = 1. \end{align*} If $v_i = q(a_i)$ for $a_i \in \mathbb{R}^n$, then the determinant transformation law gives \begin{align*} \omega_x(v_1,\dots,v_n) = \det(a_1,\dots,a_n)\,\omega_x(q(e_1),\dots,q(e_n)) = \det(a_1,\dots,a_n). \end{align*} This is exactly the defining value of $\widetilde{\omega}_x(v_1,\dots,v_n)$, so $\widetilde{\omega}=\omega$. Conversely, start with an $SL_n(\mathbb{R})$-reduction $Q$ and construct $\omega_Q$. Let $Q_{\omega_Q}$ be the reduction obtained from $\omega_Q$. We prove $Q_{\omega_Q}=Q$. If $q \in Q_x$, then \begin{align*} \omega_{Q,x}(q(e_1),\dots,q(e_n)) = 1, \end{align*} so $q \in (Q_{\omega_Q})_x$. Hence $Q \subset Q_{\omega_Q}$. For the reverse inclusion, let $u \in (Q_{\omega_Q})_x$. Choose $q \in Q_x$. There is a unique $A \in GL_n(\mathbb{R})$ such that $u = q \circ A$. Since $u \in Q_{\omega_Q}$, \begin{align*} 1 = \omega_{Q,x}(u(e_1),\dots,u(e_n)) = \det(A). \end{align*} Thus $A \in SL_n(\mathbb{R})$, and because $Q$ is stable under the right action of $SL_n(\mathbb{R})$, we have $u = q \circ A \in Q_x$. Therefore $Q_{\omega_Q} \subset Q$, and the two reductions are equal. [/step]

[step:Identify the orientation determined by the form] For a nowhere-vanishing smooth $n$-form $\omega$, define an orientation by declaring a frame $u: \mathbb{R}^n \to T_xM$ to be positive exactly when \begin{align*} \omega_x(u(e_1),\dots,u(e_n)) > 0. \end{align*} This agrees with the orientation determined by $Q_\omega$. Indeed, every $q \in (Q_\omega)_x$ satisfies \begin{align*} \omega_x(q(e_1),\dots,q(e_n)) = 1 > 0, \end{align*} so every reduced frame is positive for the orientation induced by $\omega$. If $q,q' \in (Q_\omega)_x$, then $q'=q\circ A$ for some $A \in SL_n(\mathbb{R})$, and $\det(A)=1>0$, so all reduced frames determine the same orientation. Conversely, if $Q$ is an $SL_n(\mathbb{R})$-reduction and $\omega_Q$ is the form constructed from it, then each $q \in Q_x$ satisfies \begin{align*} \omega_{Q,x}(q(e_1),\dots,q(e_n)) = 1 > 0. \end{align*} Thus the sign convention for $\omega_Q$ is precisely the orientation for which the reduced frames in $Q$ are positive. This completes the natural equivalence between nowhere-vanishing smooth $n$-forms and $SL_n(\mathbb{R})$-reductions together with their induced orientations. [/step]