[proofplan] We work in a local gauge determined by a section $\sigma:U\to P$, where sections of the associated vector bundle $E|_U$ become $V$-valued functions and the induced covariant derivative is the first-order operator $X+d\rho(A(X))$, with $A:=\sigma^*\omega$. We restrict the global vector fields $X,Y\in\mathfrak X(M)$ to $U$, write these restrictions again as $X,Y$, and expand the curvature commutator on an arbitrary $V$-valued function $u:U\to V$. The terms involving second derivatives and the Lie bracket $[X,Y]$ cancel, leaving exactly the [exterior derivative](/theorems/1525) contribution $dA(X,Y)$ and the commutator contribution $[A(X),A(Y)]$. After defining the local curvature form $F_A:=dA+\frac{1}{2}[A\wedge A]$, the result follows from the identity $F_A=\sigma^*F_\omega$ and from the fact that $d\rho$ is a Lie algebra representation. [/proofplan] [step:Write the induced covariant derivative in a local gauge] Fix an [open set](/page/Open%20Set) $U\subset M$ and a smooth local section $\sigma:U\to P$. In this local computation, replace the global vector fields $X,Y\in\mathfrak X(M)$ by their restrictions to $U$ and denote the restrictions again by $X,Y$. Define the local connection form \begin{align*} A:=\sigma^*\omega\in\Omega^1(U;\mathfrak g). \end{align*} We use the associated-bundle convention \begin{align*} [p,v]=[pg,\rho(g)^{-1}v] \end{align*} for $p\in P$, $v\in V$, and $g\in G$. With this convention, the section $\sigma$ determines the vector bundle trivialization $E|_U\cong U\times V$ in which a section has the form $s(x)=[\sigma(x),u(x)]$ for a smooth map \begin{align*} u:U\to V. \end{align*} Let $\pi:P|_U\to U$ denote the restricted principal bundle projection. Let $f_s:P|_U\to V$ be the smooth equivariant representative of $s$, defined by the condition $s(\pi(p))=[p,f_s(p)]$. The displayed associated-bundle convention implies \begin{align*} f_s(pg)=\rho(g)^{-1}f_s(p) \end{align*} for $p\in P|_U$ and $g\in G$. For $a\in\mathfrak g$, define the fundamental vertical vector field $a^\#\in\mathfrak X(P)$ by \begin{align*} a^\#_p=\frac{d}{dt}\bigg|_{t=0}p\exp(ta) \end{align*} for $p\in P$. Hence \begin{align*} a^\# f_s=-d\rho(a)f_s. \end{align*} For $Z\in\mathfrak X(U)$, the horizontal lift of $Z_x$ at $\sigma(x)$ is \begin{align*} d\sigma_x(Z_x)-A(Z)(x)^\#_{\sigma(x)}, \end{align*} because $\omega(d\sigma_x(Z_x))=A(Z)(x)$ and $\omega(A(Z)(x)^\#)=A(Z)(x)$. The induced connection on $E=P\times_\rho V$ is defined by differentiating the equivariant representative $f_s$ along the horizontal lift determined by $\omega$. Therefore the induced covariant derivative $\nabla_Zs$ is represented by the smooth map $D_Zu:U\to V$ defined by \begin{align*} D_Zu=Z(u)+d\rho(A(Z))u. \end{align*} Here $A(Z):U\to\mathfrak g$ is the smooth function obtained by evaluating the $\mathfrak g$-valued $1$-form $A$ on $Z$, and $d\rho(A(Z)):U\to\mathfrak{gl}(V)$ is the smooth endomorphism-valued function acting pointwise on $u$. The sign in this formula is fixed by the displayed [equivalence relation](/page/Equivalence%20Relation) for $P\times_\rho V$. [guided] The point of choosing the local section $\sigma:U\to P$ is that it turns the associated bundle into a product over $U$. Define the local connection form \begin{align*} A:=\sigma^*\omega\in\Omega^1(U;\mathfrak g). \end{align*} Using the convention \begin{align*} [p,v]=[pg,\rho(g)^{-1}v], \end{align*} a section of $E|_U$ is written uniquely in this gauge as $s(x)=[\sigma(x),u(x)]$ for a smooth map $u:U\to V$. Let $\pi:P|_U\to U$ denote the restricted principal bundle projection. Let $f_s:P|_U\to V$ be the equivariant representative of $s$, defined by $s(\pi(p))=[p,f_s(p)]$. The equivalence relation gives \begin{align*} f_s(pg)=\rho(g)^{-1}f_s(p). \end{align*} For $a\in\mathfrak g$, define the fundamental vertical vector field $a^\#\in\mathfrak X(P)$ by \begin{align*} a^\#_p=\frac{d}{dt}\bigg|_{t=0}p\exp(ta) \end{align*} for $p\in P$. Differentiating the equivariance identity at $g=\exp(ta)$ shows that \begin{align*} a^\# f_s=-d\rho(a)f_s. \end{align*} The minus sign comes from the inverse representation $\rho(g)^{-1}$ in the chosen associated-bundle convention. Now fix $Z\in\mathfrak X(U)$ and $x\in U$. The induced connection on the associated bundle is defined by taking the derivative of the equivariant representative $f_s$ in the horizontal direction determined by the principal connection $\omega$. Since $A=\sigma^*\omega$, we have $\omega(d\sigma_x(Z_x))=A(Z)(x)$. The vector \begin{align*} d\sigma_x(Z_x)-A(Z)(x)^\#_{\sigma(x)} \end{align*} is horizontal because applying $\omega$ gives $A(Z)(x)-A(Z)(x)=0$, and it projects to $Z_x$ because vertical fundamental vectors project to zero. Therefore this vector is the horizontal lift of $Z_x$ at $\sigma(x)$. The induced covariant derivative is obtained by differentiating the equivariant representative along this horizontal lift. Hence the local representative of $\nabla_Zs$ is \begin{align*} Z(u)-A(Z)^\# f_s. \end{align*} Using the vertical derivative identity $a^\# f_s=-d\rho(a)f_s$ with $a=A(Z)(x)$ gives \begin{align*} D_Zu=Z(u)+d\rho(A(Z))u. \end{align*} Thus, in the gauge determined by $\sigma$, the induced covariant derivative is the ordinary directional derivative corrected by the endomorphism-valued coefficient $d\rho(A(Z))$. [/guided] [/step] [step:Expand the curvature commutator on a local representative] Let \begin{align*} B_Z:U\to\mathfrak{gl}(V) \end{align*} denote the smooth endomorphism-valued function \begin{align*} B_Z:=d\rho(A(Z)). \end{align*} Thus $D_Zu=Z(u)+B_Zu$. For $X,Y\in\mathfrak X(U)$, compute \begin{align*} D_XD_Yu=X(Y(u))+X(B_Yu)+B_XY(u)+B_XB_Yu. \end{align*} Using the product rule for the derivative of the $V$-valued function $B_Yu$, this becomes \begin{align*} D_XD_Yu=X(Y(u))+(X B_Y)u+B_YX(u)+B_XY(u)+B_XB_Yu. \end{align*} Similarly, \begin{align*} D_YD_Xu=Y(X(u))+(Y B_X)u+B_XY(u)+B_YX(u)+B_YB_Xu. \end{align*} Also, \begin{align*} D_{[X,Y]}u=[X,Y](u)+B_{[X,Y]}u. \end{align*} Subtracting the last two displayed expressions from the first gives \begin{align*} (D_XD_Y-D_YD_X-D_{[X,Y]})u=\bigl(XB_Y-YB_X-B_{[X,Y]}+B_XB_Y-B_YB_X\bigr)u. \end{align*} [guided] The local operator $D_Z$ has two parts: the ordinary directional derivative $Z$ and the connection correction $B_Z=d\rho(A(Z))$. To find the curvature, we must compute the commutator of these first-order operators and then subtract the correction coming from the Lie bracket of vector fields. Define \begin{align*} B_Z:U\to\mathfrak{gl}(V) \end{align*} by \begin{align*} B_Z=d\rho(A(Z)). \end{align*} Then the local formula for the induced covariant derivative is \begin{align*} D_Zu=Z(u)+B_Zu. \end{align*} We first apply $D_X$ to $D_Yu$. Since $D_Yu=Y(u)+B_Yu$, we get \begin{align*} D_XD_Yu=X(Y(u)+B_Yu)+B_X(Y(u)+B_Yu). \end{align*} Expanding and using the product rule on $X(B_Yu)$ gives \begin{align*} D_XD_Yu=X(Y(u))+(X B_Y)u+B_YX(u)+B_XY(u)+B_XB_Yu. \end{align*} The same computation with $X$ and $Y$ interchanged gives \begin{align*} D_YD_Xu=Y(X(u))+(Y B_X)u+B_XY(u)+B_YX(u)+B_YB_Xu. \end{align*} Finally, the covariant derivative in the bracket direction is \begin{align*} D_{[X,Y]}u=[X,Y](u)+B_{[X,Y]}u. \end{align*} Now subtract. The pure second-order part cancels because the Lie bracket of vector fields is defined by \begin{align*} [X,Y](u)=X(Y(u))-Y(X(u)). \end{align*} The mixed first-order terms $B_YX(u)$ and $B_XY(u)$ also cancel with their counterparts in the reversed expression. Therefore the only surviving terms are the derivatives of the connection coefficients and the commutator of the endomorphism coefficients: \begin{align*} (D_XD_Y-D_YD_X-D_{[X,Y]})u=\bigl(XB_Y-YB_X-B_{[X,Y]}+B_XB_Y-B_YB_X\bigr)u. \end{align*} This is the algebraic heart of the curvature computation: curvature measures exactly the failure of the connection-corrected directional derivatives to commute after the ordinary vector-field commutator has been removed. [/guided] [/step] [step:Identify the derivative terms with $d\rho(dA(X,Y))$] Because $d\rho:\mathfrak g\to\mathfrak{gl}(V)$ is linear and constant on the base manifold $U$, differentiating $B_Y=d\rho(A(Y))$ along $X$ gives \begin{align*} XB_Y=d\rho(X(A(Y))). \end{align*} Likewise, \begin{align*} YB_X=d\rho(Y(A(X))). \end{align*} Also, \begin{align*} B_{[X,Y]}=d\rho(A([X,Y])). \end{align*} By the definition of the exterior derivative of a $\mathfrak g$-valued $1$-form, \begin{align*} dA(X,Y)=X(A(Y))-Y(A(X))-A([X,Y]). \end{align*} Therefore \begin{align*} XB_Y-YB_X-B_{[X,Y]}=d\rho(dA(X,Y)). \end{align*} [guided] We now identify the first three surviving terms from the commutator calculation. The function $B_Y:U\to\mathfrak{gl}(V)$ was defined by $B_Y=d\rho(A(Y))$, where $A(Y):U\to\mathfrak g$ is the evaluation of the local connection form on $Y$. Since $d\rho:\mathfrak g\to\mathfrak{gl}(V)$ is a fixed [linear map](/page/Linear%20Map) between finite-dimensional vector spaces, differentiating through it along $X$ gives \begin{align*} XB_Y=d\rho(X(A(Y))). \end{align*} The same reasoning with $X$ and $Y$ interchanged gives \begin{align*} YB_X=d\rho(Y(A(X))). \end{align*} The bracket-direction coefficient is also obtained by the same definition: \begin{align*} B_{[X,Y]}=d\rho(A([X,Y])). \end{align*} The exterior derivative of the $\mathfrak g$-valued $1$-form $A$ is characterized on vector fields by \begin{align*} dA(X,Y)=X(A(Y))-Y(A(X))-A([X,Y]). \end{align*} Substituting the three displayed identities into this formula and using linearity of $d\rho$ gives \begin{align*} XB_Y-YB_X-B_{[X,Y]}=d\rho(dA(X,Y)). \end{align*} This explains why the derivative part of the curvature commutator is exactly the exterior-derivative part of the local curvature form. [/guided] [/step] [step:Identify the product terms with the bracket part of the curvature] Since $\rho:G\to GL(V)$ is a smooth representation, its derivative \begin{align*} d\rho:\mathfrak g\to\mathfrak{gl}(V) \end{align*} is a Lie algebra homomorphism. Hence, for every $a,b\in\mathfrak g$, \begin{align*} [d\rho(a),d\rho(b)]_{\mathfrak{gl}(V)}=d\rho([a,b]_{\mathfrak g}). \end{align*} Applying this pointwise with $a=A(X)(x)$ and $b=A(Y)(x)$ gives \begin{align*} B_XB_Y-B_YB_X=d\rho([A(X),A(Y)]_{\mathfrak g}). \end{align*} The bracket wedge product of the $\mathfrak g$-valued $1$-form $A$ satisfies \begin{align*} \frac{1}{2}[A\wedge A](X,Y)=[A(X),A(Y)]_{\mathfrak g}. \end{align*} Combining this identity with the previous step gives \begin{align*} XB_Y-YB_X-B_{[X,Y]}+B_XB_Y-B_YB_X=d\rho\left(dA(X,Y)+\frac{1}{2}[A\wedge A](X,Y)\right). \end{align*} Since \begin{align*} F_A=dA+\frac{1}{2}[A\wedge A], \end{align*} we obtain \begin{align*} XB_Y-YB_X-B_{[X,Y]}+B_XB_Y-B_YB_X=d\rho(F_A(X,Y)). \end{align*} [guided] It remains to identify the product terms $B_XB_Y-B_YB_X$. Because $\rho:G\to GL(V)$ is a smooth representation, its derivative \begin{align*} d\rho:\mathfrak g\to\mathfrak{gl}(V) \end{align*} is a Lie algebra homomorphism. Thus, for every $a,b\in\mathfrak g$, \begin{align*} [d\rho(a),d\rho(b)]_{\mathfrak{gl}(V)}=d\rho([a,b]_{\mathfrak g}). \end{align*} At each $x\in U$, apply this identity with $a=A(X)(x)$ and $b=A(Y)(x)$. Since $B_X=d\rho(A(X))$ and $B_Y=d\rho(A(Y))$, this gives \begin{align*} B_XB_Y-B_YB_X=d\rho([A(X),A(Y)]_{\mathfrak g}). \end{align*} By definition of the bracket wedge product for a $\mathfrak g$-valued $1$-form, \begin{align*} \frac{1}{2}[A\wedge A](X,Y)=[A(X),A(Y)]_{\mathfrak g}. \end{align*} The derivative terms were already identified as $d\rho(dA(X,Y))$, so linearity of $d\rho$ gives \begin{align*} XB_Y-YB_X-B_{[X,Y]}+B_XB_Y-B_YB_X=d\rho\left(dA(X,Y)+\frac{1}{2}[A\wedge A](X,Y)\right). \end{align*} The local curvature form of the connection in the gauge $\sigma$ is \begin{align*} F_A=dA+\frac{1}{2}[A\wedge A]. \end{align*} Therefore \begin{align*} XB_Y-YB_X-B_{[X,Y]}+B_XB_Y-B_YB_X=d\rho(F_A(X,Y)). \end{align*} [/guided] [/step] [step:Conclude the local curvature formula and its gauge independence] Substituting the identity from the previous step into the commutator expansion gives \begin{align*} (D_XD_Y-D_YD_X-D_{[X,Y]})u=d\rho(F_A(X,Y))u. \end{align*} Under the trivialization determined by $\sigma$, the left-hand side represents \begin{align*} R^E(X,Y)s=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s. \end{align*} Thus $R^E(X,Y)s$ is represented by the map \begin{align*} U\to V, \end{align*} given by \begin{align*} x\mapsto d\rho(F_A(X,Y)(x))u(x). \end{align*} Let \begin{align*} F_\omega=d\omega+\frac{1}{2}[\omega\wedge\omega]\in\Omega^2(P;\mathfrak g) \end{align*} denote the principal curvature form of $\omega$. Since $A=\sigma^*\omega$, pullback commutes with the exterior derivative, and the bracket-wedge operation is natural under pullback in the explicit sense that \begin{align*} [\sigma^*\omega\wedge\sigma^*\omega]=\sigma^*[\omega\wedge\omega], \end{align*} we have \begin{align*} F_A=dA+\frac{1}{2}[A\wedge A]=\sigma^*d\omega+\frac{1}{2}\sigma^*[\omega\wedge\omega]=\sigma^*F_\omega. \end{align*} Therefore, in the trivialization determined by $\sigma$, the curvature operator $R^E(X,Y)$ is represented by multiplication by \begin{align*} d\rho((\sigma^*F_\omega)(X,Y)). \end{align*} Since the computation was made in an arbitrary local section $\sigma$ and the left-hand side is the globally defined curvature of the induced connection on $E$, this local formula is the gauge-independent assertion that the associated-bundle curvature is induced from the principal curvature form by applying $d\rho$. This proves the theorem. [guided] We now translate the local computation back into the global statement. From the commutator expansion and the identification of the local curvature form, we have \begin{align*} (D_XD_Y-D_YD_X-D_{[X,Y]})u=d\rho(F_A(X,Y))u. \end{align*} The trivialization determined by $\sigma$ identifies $s$ with $u$ and identifies the induced covariant derivative $\nabla_Z$ with the local operator $D_Z$. Hence the left-hand side is precisely the local representative of \begin{align*} R^E(X,Y)s=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s. \end{align*} Thus $R^E(X,Y)s$ is represented by the smooth map \begin{align*} x\mapsto d\rho(F_A(X,Y)(x))u(x). \end{align*} The remaining point is to connect the gauge-dependent symbol $F_A$ with the global curvature form of the principal connection. Define \begin{align*} F_\omega=d\omega+\frac{1}{2}[\omega\wedge\omega]\in\Omega^2(P;\mathfrak g) \end{align*} to be the principal curvature form. Since $A=\sigma^*\omega$, pullback commutes with the exterior derivative, and the bracket-wedge operation is natural under pullback, we compute \begin{align*} F_A=dA+\frac{1}{2}[A\wedge A]=\sigma^*d\omega+\frac{1}{2}\sigma^*[\omega\wedge\omega]=\sigma^*F_\omega. \end{align*} Substituting this identity into the local formula gives that $R^E(X,Y)s$ is represented by the smooth map \begin{align*} U\to V \end{align*} given by \begin{align*} x\mapsto d\rho((\sigma^*F_\omega)(X,Y)(x))u(x). \end{align*} Because the local section $\sigma$ was arbitrary and the expression on the left is the globally defined curvature of the induced connection, this is exactly the gauge-independent statement that the curvature of the associated vector bundle is obtained by applying $d\rho$ to the principal curvature form. [/guided] [/step]