[proofplan] We first differentiate the $G$-invariance of $B$ to obtain the infinitesimal skew-adjointness identity for the Lie algebra action $\rho_*: \mathfrak g \to \mathfrak{gl}(V)$. Then we compute the desired identity in a local trivialization determined by a local section of the principal bundle $\pi:P\to M$ and the principal connection form $\omega\in\Omega^1(P;\mathfrak g)$. In the associated-bundle convention $[pg,v]=[p,\rho(g)v]$, the induced connection is the ordinary derivative corrected by the local connection form with a plus sign, and the infinitesimal invariance identity cancels exactly the two correction terms. Since equality of $1$-forms can be checked locally, the local computation proves the global statement. [/proofplan]

[step:Differentiate the $G$-invariance of $B$ to obtain infinitesimal skew-adjointness] Let \begin{align*} \rho_* := d\rho_e: \mathfrak g \to \mathfrak{gl}(V) \end{align*} denote the differential of $\rho$ at the identity element $e \in G$. We claim that for every $\xi \in \mathfrak g$ and every $v,w \in V$, \begin{align*} B(\rho_*(\xi)v,w) + B(v,\rho_*(\xi)w) = 0. \end{align*} To prove this, fix $\xi \in \mathfrak g$ and let \begin{align*} \gamma: (-\varepsilon,\varepsilon) \to G \end{align*} be a smooth curve satisfying $\gamma(0)=e$ and $\gamma'(0)=\xi$. For fixed $v,w \in V$, define $F: (-\varepsilon,\varepsilon) \to \mathbb R$ by \begin{align*} F(r) = B(\rho(\gamma(r))v,\rho(\gamma(r))w). \end{align*} By the $G$-invariance of $B$, the function $F$ is constant and equal to $B(v,w)$. Hence $F'(0)=0$. Differentiating at $r=0$ using bilinearity of $B$ gives \begin{align*} 0 = B(\rho_*(\xi)v,w) + B(v,\rho_*(\xi)w). \end{align*} This proves the infinitesimal identity. [/step]

[step:Write the associated bilinear form and connection in a local trivialization] Let $M$ denote the smooth base manifold, let $\pi:P\to M$ denote the principal $G$-bundle projection, and let $\omega\in\Omega^1(P;\mathfrak g)$ denote the fixed principal connection form. Fix an [open set](/page/Open%20Set) $U \subset M$ and a smooth local section \begin{align*} \sigma: U \to P \end{align*} of $\pi: P \to M$. Let \begin{align*} A := \sigma^*\omega \in \Omega^1(U;\mathfrak g) \end{align*} be the local connection form determined by $\sigma$. Let $s,t \in \Gamma(E)$. Over $U$, there are unique smooth maps $u: U \to V$ and $v: U \to V$ such that \begin{align*} s(x) = [\sigma(x),u(x)] \end{align*} and \begin{align*} t(x) = [\sigma(x),v(x)] \end{align*} for every $x \in U$. In this local trivialization, the definition of $B_E$ gives \begin{align*} B_E(s,t)(x) = B(u(x),v(x)). \end{align*} Let $\mathfrak X(U) := \Gamma(TU)$ denote the space of smooth vector fields on $U$. For a smooth vector field \begin{align*} X \in \mathfrak X(U), \end{align*} using the convention $[\sigma(x)g,a]=[\sigma(x),\rho(g)a]$ for the associated bundle, the induced connection on the associated bundle is represented by \begin{align*} (\nabla_X s)(x) = [\sigma(x), X(u)(x) + \rho_*(A(X)_x)u(x)] \end{align*} and \begin{align*} (\nabla_X t)(x) = [\sigma(x), X(v)(x) + \rho_*(A(X)_x)v(x)]. \end{align*} Here $A(X): U \to \mathfrak g$ is the smooth map obtained by evaluating the $\mathfrak g$-valued $1$-form $A$ on $X$, and $X(u): U \to V$ denotes the directional derivative of the smooth map $u: U \to V$ along $X$. [/step]

[step:Compute the derivative of $B_E(s,t)$ and cancel the connection terms]Fix $x \in U$ and $X \in \mathfrak X(U)$. Using the local expression \begin{align*} B_E(s,t)(x) = B(u(x),v(x)), \end{align*} the ordinary product rule for the fixed bilinear map $B: V \times V \to \mathbb R$ gives \begin{align*} d(B_E(s,t))_x(X_x) = B(X(u)(x),v(x)) + B(u(x),X(v)(x)). \end{align*} On the other hand, using the local formula for the induced connection, \begin{align*} B_E(\nabla_X s,t)(x) = B(X(u)(x) + \rho_*(A(X)_x)u(x),v(x)) \end{align*} and \begin{align*} B_E(s,\nabla_X t)(x) = B(u(x),X(v)(x) + \rho_*(A(X)_x)v(x)). \end{align*} Adding these two identities and using bilinearity of $B$ gives \begin{align*} B_E(\nabla_X s,t)(x) + B_E(s,\nabla_X t)(x) = B(X(u)(x),v(x)) + B(u(x),X(v)(x)) + B(\rho_*(A(X)_x)u(x),v(x)) + B(u(x),\rho_*(A(X)_x)v(x)). \end{align*} Apply the infinitesimal identity from the first step with \begin{align*} \xi := A(X)_x \in \mathfrak g, \end{align*} with $u(x) \in V$ in place of $v$, and with $v(x) \in V$ in place of $w$. It gives \begin{align*} B(\rho_*(A(X)_x)u(x),v(x)) + B(u(x),\rho_*(A(X)_x)v(x)) = 0. \end{align*} Therefore \begin{align*} B_E(\nabla_X s,t)(x) + B_E(s,\nabla_X t)(x) = B(X(u)(x),v(x)) + B(u(x),X(v)(x)). \end{align*} Comparing with the derivative computed above yields \begin{align*} d(B_E(s,t))_x(X_x) = B_E(\nabla_X s,t)(x) + B_E(s,\nabla_X t)(x). \end{align*}[/step]

[guided]We now prove the key local identity in full detail. The point of choosing the local section $\sigma: U \to P$ is that it turns sections of the associated bundle into ordinary $V$-valued functions. Thus the sections $s,t \in \Gamma(E)$ are represented over $U$ by smooth maps \begin{align*} u: U \to V \end{align*} and \begin{align*} v: U \to V \end{align*} through the formulas \begin{align*} s(x) = [\sigma(x),u(x)] \end{align*} and \begin{align*} t(x) = [\sigma(x),v(x)]. \end{align*} In this trivialization the associated [bilinear form](/page/Bilinear%20Form) is exactly the original bilinear form applied fibrewise: \begin{align*} B_E(s,t)(x) = B(u(x),v(x)). \end{align*} Therefore, for a smooth vector field $X \in \mathfrak X(U)$ and a point $x \in U$, the product rule for the fixed bilinear map $B: V \times V \to \mathbb R$ gives \begin{align*} d(B_E(s,t))_x(X_x) = B(X(u)(x),v(x)) + B(u(x),X(v)(x)). \end{align*} The connection introduces correction terms. The principal connection form is the fixed form $\omega\in\Omega^1(P;\mathfrak g)$, and the local connection form determined by the section $\sigma$ is \begin{align*} A := \sigma^*\omega \in \Omega^1(U;\mathfrak g). \end{align*} Evaluating $A$ on the vector field $X$ gives a smooth map \begin{align*} A(X): U \to \mathfrak g. \end{align*} With the associated-bundle convention $[\sigma(x)g,a]=[\sigma(x),\rho(g)a]$, the associated connection is represented locally by \begin{align*} (\nabla_X s)(x) = [\sigma(x), X(u)(x) + \rho_*(A(X)_x)u(x)] \end{align*} and \begin{align*} (\nabla_X t)(x) = [\sigma(x), X(v)(x) + \rho_*(A(X)_x)v(x)]. \end{align*} Applying $B_E$ to these expressions gives \begin{align*} B_E(\nabla_X s,t)(x) = B(X(u)(x) + \rho_*(A(X)_x)u(x),v(x)) \end{align*} and \begin{align*} B_E(s,\nabla_X t)(x) = B(u(x),X(v)(x) + \rho_*(A(X)_x)v(x)). \end{align*} Now expand both terms using bilinearity: \begin{align*} B_E(\nabla_X s,t)(x) + B_E(s,\nabla_X t)(x) = B(X(u)(x),v(x)) + B(u(x),X(v)(x)) + B(\rho_*(A(X)_x)u(x),v(x)) + B(u(x),\rho_*(A(X)_x)v(x)). \end{align*} The last two terms are exactly where the preservation of $B$ by the structure group is used, so we derive the needed infinitesimal identity inside this guided argument. Fix $\xi \in \mathfrak g$ and $a,b \in V$. Choose a smooth curve \begin{align*} \gamma: (-\varepsilon,\varepsilon) \to G \end{align*} satisfying $\gamma(0)=e$ and $\gamma'(0)=\xi$. Define \begin{align*} F: (-\varepsilon,\varepsilon) \to \mathbb R \end{align*} by \begin{align*} F(r) = B(\rho(\gamma(r))a,\rho(\gamma(r))b). \end{align*} Because $B$ is preserved by $\rho(G)$, we have $F(r)=B(a,b)$ for every $r \in (-\varepsilon,\varepsilon)$, so $F'(0)=0$. Differentiating at $r=0$ using the ordinary product rule for the fixed bilinear map $B$ and the definition $\rho_* = d\rho_e$ gives \begin{align*} 0 = B(\rho_*(\xi)a,b) + B(a,\rho_*(\xi)b). \end{align*} Thus every $\xi \in \mathfrak g$ satisfies \begin{align*} B(\rho_*(\xi)a,b) + B(a,\rho_*(\xi)b) = 0 \end{align*} for every $a,b \in V$. We apply this with \begin{align*} \xi := A(X)_x, \end{align*} with $a := u(x)$, and with $b := v(x)$. This gives \begin{align*} B(\rho_*(A(X)_x)u(x),v(x)) + B(u(x),\rho_*(A(X)_x)v(x)) = 0. \end{align*} Thus the two connection correction terms cancel, leaving \begin{align*} B_E(\nabla_X s,t)(x) + B_E(s,\nabla_X t)(x) = B(X(u)(x),v(x)) + B(u(x),X(v)(x)). \end{align*} This is precisely the expression obtained for $d(B_E(s,t))_x(X_x)$, so \begin{align*} d(B_E(s,t))_x(X_x) = B_E(\nabla_X s,t)(x) + B_E(s,\nabla_X t)(x). \end{align*}[/guided]

[step:Pass from the local computation to the global identity] The preceding computation holds on every open set $U \subset M$ admitting a smooth local section $\sigma: U \to P$, for every $x \in U$, and for every tangent vector $X_x \in T_xM$ represented by a local vector field $X \in \mathfrak X(U)$. Principal bundles admit such local sections around every point of $M$, and on each such trivializing open set the preceding formulas are exactly the standard local expressions for associated bundles and their connections under the stated convention. Both sides of \begin{align*} d(B_E(s,t)) = B_E(\nabla s,t) + B_E(s,\nabla t) \end{align*} are globally defined smooth $1$-forms on $M$. Since the two $1$-forms agree on each member of a local trivializing cover, they agree on all of $M$. This proves the stated metric compatibility identity for every $s,t \in \Gamma(E)$. [/step]