[proofplan] A bundle metric determines the subbundle of frames that are isometries from Euclidean space onto the fibers of $E$. We first verify directly that this orthonormal frame bundle is a smooth principal $O(r)$-subbundle of $\operatorname{Fr}(E)$, using local Gram-Schmidt orthonormalization. Conversely, an $O(r)$-reduction determines a fiberwise [inner product](/page/Inner%20Product) by declaring every frame in the reduction to be orthonormal; the orthogonal transition ambiguity makes this definition independent of choices. Finally, the two constructions recover each other because the frames declared orthonormal by one construction are exactly the frames used in the other. [/proofplan]

[step:Construct the orthonormal frame reduction from a bundle metric] Let $h$ be a smooth bundle metric on $E$. For each $p \in M$, define \begin{align*} Q_h(p) := \{\nu:\mathbb{R}^r \to E_p \mid \nu \text{ is a linear isomorphism and } h_p(\nu(a),\nu(b)) = a \cdot b \text{ for all } a,b \in \mathbb{R}^r\}. \end{align*} Define \begin{align*} Q_h := \bigcup_{p \in M} Q_h(p) \subset \operatorname{Fr}(E). \end{align*} The right action of $O(r)$ preserves $Q_h$. Indeed, if $\nu \in Q_h(p)$ and $A \in O(r)$, then for all $a,b \in \mathbb{R}^r$, \begin{align*} h_p((\nu \cdot A)(a),(\nu \cdot A)(b)) = h_p(\nu(Aa),\nu(Ab)) = Aa \cdot Ab = a \cdot b. \end{align*} Thus $\nu \cdot A \in Q_h(p)$. The action is free because the right action on $\operatorname{Fr}(E)$ is free. It is transitive on each fiber $Q_h(p)$: if $\nu,\mu \in Q_h(p)$, define the [linear map](/page/Linear%20Map) \begin{align*} A:\mathbb{R}^r &\to \mathbb{R}^r \end{align*} by $A := \nu^{-1} \circ \mu$. Then $\mu = \nu \cdot A$, and for all $a,b \in \mathbb{R}^r$, \begin{align*} Aa \cdot Ab = h_p(\nu(Aa),\nu(Ab)) = h_p(\mu(a),\mu(b)) = a \cdot b. \end{align*} Hence $A \in O(r)$. It remains to verify smooth local triviality. Let $U \subset M$ be an [open set](/page/Open%20Set) over which $E$ admits a smooth local frame $e_1,\dots,e_r:U \to E$. Apply the Gram-Schmidt construction fiberwise using the metric $h$. Explicitly, define smooth local sections $u_1,\dots,u_r:U \to E$ recursively by \begin{align*} u_1(p) := \frac{e_1(p)}{\sqrt{h_p(e_1(p),e_1(p))}}. \end{align*} For $k \ge 2$, define first \begin{align*} \widetilde u_k(p) := e_k(p) - \sum_{j=1}^{k-1} h_p(e_k(p),u_j(p))u_j(p), \end{align*} and then \begin{align*} u_k(p) := \frac{\widetilde u_k(p)}{\sqrt{h_p(\widetilde u_k(p),\widetilde u_k(p))}}. \end{align*} The denominators are positive because $e_1(p),\dots,e_r(p)$ are linearly independent and $h_p$ is positive definite. Since $h$ and the $e_i$ are smooth, each $u_k$ is smooth. For each $p \in U$, the vectors $u_1(p),\dots,u_r(p)$ form an $h_p$-[orthonormal basis](/page/Orthonormal%20Basis) of $E_p$. Define a smooth map \begin{align*} \sigma:U &\to Q_h \end{align*} by requiring \begin{align*} \sigma(p)(a_1,\dots,a_r) := \sum_{i=1}^r a_i u_i(p). \end{align*} Then $\sigma$ is a smooth local section of $Q_h \to M$. The map \begin{align*} U \times O(r) &\to Q_h|_U \end{align*} given by \begin{align*} (p,A) &\mapsto \sigma(p)\cdot A \end{align*} is a bijection, and it is a diffeomorphism because it is the restriction of the standard frame-bundle trivialization determined by the smooth frame $u_1,\dots,u_r$. Therefore $Q_h$ is a smooth principal $O(r)$-subbundle of $\operatorname{Fr}(E)$. [/step]

[step:Construct a bundle metric from an orthogonal reduction]Let $Q \subset \operatorname{Fr}(E)$ be a principal $O(r)$-reduction. For each $p \in M$, each pair $v,w \in E_p$, and each frame $\nu \in Q_p$, choose the unique vectors $a,b \in \mathbb{R}^r$ such that $\nu(a)=v$ and $\nu(b)=w$. Define \begin{align*} h^Q_p(v,w) := a \cdot b. \end{align*} This definition is independent of the chosen frame $\nu \in Q_p$. If $\mu \in Q_p$ is another frame, then the principal $O(r)$-action is transitive on $Q_p$, so there exists $A \in O(r)$ with $\mu=\nu \cdot A$. If $\mu(a')=v$ and $\mu(b')=w$, then $\nu(Aa')=v$ and $\nu(Ab')=w$, hence $Aa'=a$ and $Ab'=b$. Since $A$ is orthogonal, \begin{align*} a' \cdot b' = Aa' \cdot Ab' = a \cdot b. \end{align*} Thus $h^Q_p(v,w)$ is well-defined. For each $p \in M$, the map \begin{align*} h^Q_p:E_p \times E_p &\to \mathbb{R} \end{align*} is bilinear and symmetric because the Euclidean inner product on $\mathbb{R}^r$ is bilinear and symmetric. It is positive definite because if $v \in E_p$ and $v \ne 0$, then the unique $a \in \mathbb{R}^r$ satisfying $\nu(a)=v$ is nonzero, so \begin{align*} h^Q_p(v,v)=a \cdot a > 0. \end{align*}[/step]

[guided]The only delicate point in this construction is choice-independence. The reduction $Q$ gives many frames in the same fiber $E_p$, and the metric must not depend on which one we use. Fix $p \in M$ and choose $\nu \in Q_p$. Since $\nu:\mathbb{R}^r \to E_p$ is a linear isomorphism, every $v \in E_p$ has a unique coordinate vector $a \in \mathbb{R}^r$ with $\nu(a)=v$, and every $w \in E_p$ has a unique coordinate vector $b \in \mathbb{R}^r$ with $\nu(b)=w$. The proposed definition is \begin{align*} h^Q_p(v,w) := a \cdot b. \end{align*} Now suppose we chose another frame $\mu \in Q_p$. Because $Q \to M$ is a principal $O(r)$-bundle, the right $O(r)$-action is transitive on the fiber $Q_p$. Hence there exists $A \in O(r)$ such that \begin{align*} \mu = \nu \cdot A. \end{align*} By the convention for the frame-bundle right action, this means $\mu=\nu\circ A$. Let $a',b' \in \mathbb{R}^r$ be the coordinate vectors with respect to $\mu$, so that $\mu(a')=v$ and $\mu(b')=w$. Substituting $\mu=\nu\circ A$ gives \begin{align*} \nu(Aa')=v. \end{align*} Since $a$ is the unique vector satisfying $\nu(a)=v$, we get $Aa'=a$. Similarly, $Ab'=b$. Orthogonality of $A$ means precisely that it preserves the Euclidean inner product on $\mathbb{R}^r$, so \begin{align*} a' \cdot b' = Aa' \cdot Ab' = a \cdot b. \end{align*} Therefore the value assigned to $h^Q_p(v,w)$ is independent of the chosen frame in $Q_p$. The fiberwise bilinearity and symmetry follow because in any chosen frame $\nu \in Q_p$, the formula for $h^Q_p$ is exactly the Euclidean dot product of coordinate vectors. Positive definiteness follows for the same reason: if $v \ne 0$, then its coordinate vector $a$ is nonzero because $\nu$ is injective, and therefore \begin{align*} h^Q_p(v,v)=a \cdot a > 0. \end{align*}[/guided]

[step:Verify smoothness of the recovered metric] Let $U \subset M$ be an open set admitting a smooth local section \begin{align*} \sigma:U &\to Q. \end{align*} Such sections exist because $Q \to M$ is a smooth principal bundle. The section $\sigma$ determines a smooth vector bundle trivialization \begin{align*} \Phi:U \times \mathbb{R}^r &\to E|_U \end{align*} by \begin{align*} \Phi(p,a) := \sigma(p)(a). \end{align*} Let $s,t:U \to E|_U$ be smooth local sections of $E$. Since $\Phi$ is a smooth trivialization, there are unique smooth maps \begin{align*} a:U &\to \mathbb{R}^r \end{align*} and \begin{align*} b:U &\to \mathbb{R}^r \end{align*} such that \begin{align*} s(p)=\sigma(p)(a(p)) \end{align*} and \begin{align*} t(p)=\sigma(p)(b(p)) \end{align*} for all $p \in U$. By construction of $h^Q$, \begin{align*} h^Q_p(s(p),t(p)) = a(p)\cdot b(p). \end{align*} The right-hand side is a smooth real-valued function on $U$, because $a$ and $b$ are smooth and the Euclidean inner product is smooth. Hence $h^Q$ is a smooth bundle metric on $E$. [/step]

[step:Show the two constructions are inverse] Start with a smooth bundle metric $h$ and form $Q_h$. Then construct $h^{Q_h}$ from $Q_h$. For $p \in M$, choose $\nu \in Q_h(p)$ and let $v,w \in E_p$. If $a,b \in \mathbb{R}^r$ are the unique vectors with $\nu(a)=v$ and $\nu(b)=w$, then $\nu$ is an $h_p$-isometry by definition of $Q_h(p)$, so \begin{align*} h^{Q_h}_p(v,w)=a\cdot b=h_p(\nu(a),\nu(b))=h_p(v,w). \end{align*} Thus $h^{Q_h}=h$. Conversely, start with a principal $O(r)$-reduction $Q$ and form the metric $h^Q$. Let $Q_{h^Q}$ be the corresponding orthonormal frame bundle. If $\nu \in Q_p$, then for all $a,b \in \mathbb{R}^r$, \begin{align*} h^Q_p(\nu(a),\nu(b))=a\cdot b \end{align*} by construction, so $\nu \in Q_{h^Q}(p)$. Hence $Q \subset Q_{h^Q}$. For the reverse inclusion, let $\mu \in Q_{h^Q}(p)$. Choose any $\nu \in Q_p$. Define \begin{align*} A:\mathbb{R}^r &\to \mathbb{R}^r \end{align*} by $A:=\nu^{-1}\circ \mu$. Since $\mu$ is $h^Q_p$-orthonormal and $\nu$ is also $h^Q_p$-orthonormal by construction, for all $a,b \in \mathbb{R}^r$ we have \begin{align*} Aa \cdot Ab = h^Q_p(\nu(Aa),\nu(Ab)) = h^Q_p(\mu(a),\mu(b)) = a\cdot b. \end{align*} Therefore $A \in O(r)$, and $\mu=\nu\cdot A \in Q_p$. Hence $Q_{h^Q} \subset Q$, so $Q_{h^Q}=Q$. The assignments $h \mapsto Q_h$ and $Q \mapsto h^Q$ are therefore mutually inverse, giving the desired bijection between smooth bundle metrics on $E$ and principal $O(r)$-reductions of $\operatorname{Fr}(E)$. [/step]