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An orthogonal structure on a real vector bundle $E$ of rank $r$ is a choice of local frames whose transition matrices lie in the orthogonal group $O(r)$. In such frames declareOrthogonal transitions preserve this expression, so it defines a smooth positive-definite fiber metric. Conversely, a fiber metric turns any local frame into a smooth orthonormal frame by the Gram-Schmidt process; the resulting transition matrices are orthogonal. The two constructions are inverse, giving the required equivalence.

An orthogonal local trivialization $E{|}_U\cong U\times {\mathbb{R}}^r$ identifies each fiber isometrically with standard Euclidean space. Equivalently, its coordinate frame satisfies $h(e_i,e_j)={\delta }_{ij}$. This is local and does not assert the existence of a global frame.

Every real vector bundle over a smooth manifold admits a fiber metric. Choose a trivializing open cover and local Euclidean metrics $h_i$. The partition of unity theorem for a Hausdorff second-countable smooth manifold supplies nonnegative smooth functions ${\rho }_i$, with locally finite supports contained in the respective cover members, such that ${\sum }_i{\rho }_i=1$. Setextending each weighted metric by zero off its chart. Local finiteness makes this smooth. For every nonzero fiber vector $v$, at least one positive weight gives $h(v,v)={\sum }_i{\rho }_i{h}_i(v,v)>0$. Thus $h$ is a fiber metric and provides an orthogonal structure.

One definition of an orthogonal connection is that its covariant derivative preserves the fiber metric:for all vector fields $X$ and smooth sections of a vector bundle $s,t$. Another is that its parallel transport along every smooth curve is an isometry between the endpoint fibers.

For the first implication, take parallel sections $s,t$ along a curve. Their covariant derivatives vanish, so the displayed identity makes $h(s,t)$ constant: parallel transport preserves a fibre metric. Conversely, if parallel transport preserves the inner product, choose parallel extensions of arbitrary initial fiber vectors along a curve with prescribed initial tangent $X$. Differentiating their constant inner product gives $({\nabla }_{X}h)(s,t)=0$ at the initial point. Since $X,s,t$ were arbitrary, the connection is metric-compatible.

The equivalent local description is that the connection matrix in an orthonormal frame is skew-symmetric. Writing $\nabla e_j={\sum }_i{A}^i{}_j{e}_i$ in an orthogonal trivialization, differentiate $h(e_i,e_j)={\delta }_{ij}$ to obtainConversely this matrix identity gives the metric-preservation formula by expanding arbitrary sections. It is the orthogonal group connection description of the same condition.

Let $E\to M$ have rank $r$. A vector bundle covariant derivative is a real-linear operatorPairing its one-form slot with a vector field $X$ gives ${\nabla }_{X}s$, with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$.

In the horizontal description, a connection on a vector bundle $A$ is a smooth splitting $T_{a}E=V_{a}E\oplus H_{a}E$, where $V_{a}E=\ker (d\pi {)}_a$, and the horizontal choice is compatible with fibre addition and scalar multiplication. This linearity is essential; an arbitrary nonlinear horizontal splitting is not a vector-bundle connection. Identify a vertical vector at $a\in E_x$ with the element of $E_x$ obtained by differentiating the curve $a+tv$. Let $K_A:T_{a}E\to E_x$ be vertical projection followed by this identification. The induced vector bundle covariant derivative isIn a local frame of a vector bundle $e=(e_1,\dots ,e_r)$, a linear horizontal splitting has the formfor a matrix-valued connection one-form $\Gamma$. Thus for $s=ev$, ${\nabla }^{A}s=e(dv+\Gamma v)$, which directly verifies the Leibniz rule.

Conversely, the Leibniz rule makes $\nabla$ local: if a section vanishes near a point, multiply it by a smooth cutoff function supported in that neighborhood and equal to one near the point to see that its derivative vanishes there too. Local sections can therefore be differentiated by any extension agreeing near the point. Given $\nabla$, define its local connection one-form by $\nabla e_b={\sum }_a{e}_a{\Gamma }_b^a$. The Leibniz rule gives $\nabla (ev)=e(dv+\Gamma v)$. Under $e^{\prime }=eg$, $v^{\prime }=g^{-1}v$, one hasThe tangent fibre coordinate transforms by $w^{\prime }=g^{-1}w-g^{-1}(dg)(\xi )g^{-1}v$, so $w=-\Gamma (\xi )v$ transforms precisely to $w^{\prime }=-{\Gamma }^{\prime }(\xi )v^{\prime }$. Consequently these local horizontal subspaces of a vector bundle connection glue to a smooth linear splitting inducing the given $\nabla$. Its local formula also shows uniqueness. This proves the horizontal connection associated to a covariant derivative construction.

A horizontal lift $\tilde{\gamma }$ of $\gamma$ is a smooth curve in $E$ with $\pi \circ \tilde{\gamma }=\gamma$ and $\dot{\tilde{\gamma }}(t)\in H_{\tilde{\gamma }(t)}E$. In a local frame write $\tilde{\gamma }(t)=e(\gamma (t))v(t)$. Horizontality is exactly the linear ordinary differential equationWe use existence and uniqueness for an ODE with smooth coefficients, and the continuation theorem that a solution extends across a finite interval when it remains bounded inside the coordinate domain. For a linear system $\dot{v}=-C(t)v$ with continuous coefficients bounded by $L$ on a compact time interval, the Gronwall inequality bounds $|v(t)|$ by its initial norm times $e^{L|t-t_0|}$, preventing finite-time fibre escape. Cover a compact subinterval of $I$ by finitely many bundle charts and partition it so each piece lies in one chart. Solve successively, changing fibre frames at the partition points. Coordinate invariance of the horizontal equation and uniqueness make the pieces agree. Exhausting $I$ by compact intervals containing $0$ produces a unique horizontal lift on all of $I$ for every prescribed $a_0$. This is global continuation of linear parallel transport.

The lift equation is linear in its initial value. Hence the endpoint map $P_{\gamma }:E_{\gamma (0)}\to E_{\gamma (1)}$ is linear. Solving the same equation backwards, or transporting along $t\mapsto \gamma (1-t)$, gives its inverse, soThis is parallel transport. For two parallel lifts $s_1(t),s_2(t)$, the assumed metric connection identity extends along the embedded curve and becomesHere $D_t$ is the covariant derivative along a curve. The permitted extension of sections and the tangent vector along the embedded curve justifies applying the bundle identity; equivalently it follows in each local frame. Thereforeso parallel transport is an isometry. This is the principle that parallel transport preserves a fibre metric.