[proofplan] Choose a local section $s:U\to P$ and use it to write the principal connection as the $\mathfrak g$-valued local connection form $A_s=s^*A$. The induced connection on the associated bundle has local connection matrix $\rho_*(A_s)$, so its curvature is computed by the usual vector bundle formula. Since $\rho_*:\mathfrak g\to\mathfrak{gl}(V)$ is a [Lie algebra](/page/Lie%20Algebra) homomorphism and commutes with exterior differentiation as a constant [linear map](/page/Linear%20Map), the curvature formula becomes $F_E=\rho_*(F_{A,s})$. Evaluating a $GL(V)$-invariant polynomial then gives the pulled-back invariant polynomial on $\mathfrak g$, and the local identities glue because both sides are the local representatives of globally defined forms. [/proofplan]

[step:Write the induced connection in the local frame determined by the section]Fix an [open set](/page/Open%20Set) $U\subset M$ and a smooth local section $s:U\to P$. Define the local principal connection form \begin{align*} A_s:=s^*A\in\Omega^1(U;\mathfrak g). \end{align*} The section $s$ determines the vector bundle trivialization \begin{align*} \Phi_s:E|_U\to U\times V \end{align*} by sending the equivalence class $[s(x),v]$ to $(x,v)$. This uses the associated-bundle convention $(pg,v)\sim(p,\rho(g)v)$ from the theorem statement. Let \begin{align*} \sigma:U\to E|_U \end{align*} be a smooth local section. In the trivialization $\Phi_s$, write $\sigma(x)=[s(x),f(x)]$ for a smooth map \begin{align*} f:U\to V. \end{align*} By the construction of the associated connection from the horizontal distribution of $A$, equivalently by the local connection formula in [citetheorem:6271] with $Q=P$ and $\omega=A$, the induced covariant derivative has local expression \begin{align*} \nabla^E\sigma=df+\rho_*(A_s)f. \end{align*} Thus the local connection matrix of $\nabla^E$ in the frame determined by $s$ is \begin{align*} B_s:=\rho_*(A_s)\in\Omega^1(U;\mathfrak{gl}(V)). \end{align*}[/step]

[guided]The purpose of choosing $s$ is to compare two objects that naturally live in different places. The principal curvature $F_A$ is a $\mathfrak g$-valued form on $P$, while the vector bundle curvature is locally a $\mathfrak{gl}(V)$-valued form on $U$. Pulling back by \begin{align*} s:U\to P \end{align*} puts the principal connection on the same base open set: \begin{align*} A_s:=s^*A\in\Omega^1(U;\mathfrak g). \end{align*} The same local section gives a trivialization \begin{align*} \Phi_s:E|_U\to U\times V \end{align*} by the rule $\Phi_s([s(x),v])=(x,v)$, using the quotient convention $(pg,v)\sim(p,\rho(g)v)$. Therefore every smooth section \begin{align*} \sigma:U\to E|_U \end{align*} can be written uniquely as $\sigma(x)=[s(x),f(x)]$ for a smooth map \begin{align*} f:U\to V. \end{align*} In this trivialization, the local connection formula for an associated vector bundle, applied as in [citetheorem:6271] with $Q=P$ and $\omega=A$, says that the connection induced from the principal connection differentiates the $V$-valued representative $f$ and adds the infinitesimal action of the local principal connection form. Hence \begin{align*} \nabla^E\sigma=df+\rho_*(A_s)f. \end{align*} This identifies the local connection matrix of $\nabla^E$ as \begin{align*} B_s:=\rho_*(A_s)\in\Omega^1(U;\mathfrak{gl}(V)). \end{align*} This is the only place where the construction of the associated connection is used: after this identification, the proof is a curvature computation.[/guided]

[step:Compute the associated curvature using that $\rho_*$ preserves brackets] The curvature formula for a vector bundle connection with local connection matrix $B_s$, again the local curvature formula from [citetheorem:6271] in the associated-bundle setting, is \begin{align*} F_E=dB_s+\frac{1}{2}[B_s\wedge B_s]. \end{align*} Substituting $B_s=\rho_*(A_s)$ gives \begin{align*} F_E=d(\rho_*(A_s))+\frac{1}{2}[\rho_*(A_s)\wedge\rho_*(A_s)]. \end{align*} Since $\rho_*:\mathfrak g\to\mathfrak{gl}(V)$ is a constant linear map, exterior differentiation commutes with applying $\rho_*$: \begin{align*} d(\rho_*(A_s))=\rho_*(dA_s). \end{align*} Since $\rho_*$ is a Lie algebra homomorphism, for all $X,Y\in\mathfrak g$ one has \begin{align*} \rho_*([X,Y])=[\rho_*X,\rho_*Y]. \end{align*} Applying this pointwise to the bracket-wedge product gives \begin{align*} [\rho_*(A_s)\wedge\rho_*(A_s)]=\rho_*([A_s\wedge A_s]). \end{align*} Therefore \begin{align*} F_E=\rho_*(dA_s)+\frac{1}{2}\rho_*([A_s\wedge A_s]). \end{align*} By linearity of $\rho_*$, \begin{align*} F_E=\rho_*\left(dA_s+\frac{1}{2}[A_s\wedge A_s]\right). \end{align*} Pullback commutes with exterior differentiation and with the bracket-wedge product of forms, so \begin{align*} dA_s+\frac{1}{2}[A_s\wedge A_s]=s^*\left(dA+\frac{1}{2}[A\wedge A]\right)=s^*F_A=F_{A,s}. \end{align*} Hence \begin{align*} F_E=\rho_*(F_{A,s}). \end{align*} [/step]

[step:Evaluate invariant polynomials after pulling them back along $\rho_*$] Let $q:\mathfrak{gl}(V)^k\to\mathbb K$ be a $GL(V)$-invariant symmetric $k$-linear polynomial. Define \begin{align*} \rho^*q:\mathfrak g^k\to\mathbb K \end{align*} by \begin{align*} (\rho^*q)(X_1,\dots,X_k):=q(\rho_*X_1,\dots,\rho_*X_k). \end{align*} For $g\in G$, let $\operatorname{Ad}_g:\mathfrak g\to\mathfrak g$ denote the adjoint action of $g$ on the Lie algebra $\mathfrak g$. For $T\in GL(V)$, let $\operatorname{Ad}_T:\mathfrak{gl}(V)\to\mathfrak{gl}(V)$ denote conjugation, so $\operatorname{Ad}_T(S)=TST^{-1}$ for $S\in\mathfrak{gl}(V)$. For $g\in G$ and $X_1,\dots,X_k\in\mathfrak g$, the identity \begin{align*} \rho_*(\operatorname{Ad}_g X_i)=\operatorname{Ad}_{\rho(g)}(\rho_*X_i) \end{align*} and the $GL(V)$-invariance of $q$ imply \begin{align*} (\rho^*q)(\operatorname{Ad}_g X_1,\dots,\operatorname{Ad}_g X_k)=(\rho^*q)(X_1,\dots,X_k). \end{align*} Thus $\rho^*q$ is $\operatorname{Ad}(G)$-invariant. Using the curvature identity from the previous step and the definition of evaluation of a symmetric multilinear polynomial on curvature forms, we obtain \begin{align*} q(F_E)=q(\rho_*(F_{A,s}),\dots,\rho_*(F_{A,s})). \end{align*} By the definition of $\rho^*q$, this is exactly \begin{align*} q(F_E)=(\rho^*q)(F_{A,s}). \end{align*} [/step]

[step:Glue the local identities to obtain the global Chern-Weil identity] The form $F_E$ is the curvature of the globally defined connection $\nabla^E$, and $F_A$ is the curvature of the globally defined principal connection $A$. On each local section $s:U\to P$, the preceding steps show that the local representative of $q(F_E)$ is equal to the local representative of $(\rho^*q)(F_A)$: \begin{align*} q(F_E)=(\rho^*q)(F_{A,s}). \end{align*} On overlaps of two such trivializations, both expressions transform by the same change-of-frame rule for Chern-Weil forms, and the invariant polynomial removes the conjugation dependence. Hence the local equalities agree on overlaps. Therefore the globally defined Chern-Weil form obtained from $q$ and the associated vector bundle connection equals the Chern-Weil form obtained from $\rho^*q$ and the principal connection $A$. This proves the claimed compatibility. [/step]