[proofplan] Choose a principal connection $A$ on $Q$ and form the connection $\widetilde A$ induced on the extended principal $G$-bundle $\widetilde Q=Q\times_\varphi G$. The inclusion of the original reduction $j:Q\to \widetilde Q$ pulls $\widetilde A$ back to $d\varphi_{e_H}(A)$, and the corresponding curvature pulls back to $d\varphi_{e_H}(F_A)$. Evaluating an invariant polynomial $P_0$ on this curvature is therefore identical to evaluating the pulled-back polynomial $\varphi^*P_0$ on $F_A$. Since the Chern-Weil homomorphism is defined by the de Rham class of these closed representative forms and is independent of the chosen connection, the equality of representatives gives the desired equality of characteristic classes. [/proofplan] [step:Define the pulled-back invariant polynomial on $\mathfrak h$] First suppose that $P_0\in I^k(G)$ is homogeneous of degree $k$, so that $P_0:\mathfrak g^k\to\mathbb R$ is a symmetric $k$-[linear map](/page/Linear%20Map) invariant under the adjoint action of $G$. Define \begin{align*} \varphi^*P_0:\mathfrak h^k\to\mathbb R \end{align*} by \begin{align*} (\varphi^*P_0)(X_1,\dots,X_k)=P_0(d\varphi_e(X_1),\dots,d\varphi_e(X_k)). \end{align*} This map is symmetric and $k$-linear because $d\varphi_e:\mathfrak h\to\mathfrak g$ is linear and $P_0$ is symmetric and $k$-linear. It remains to verify $H$-invariance. For $h\in H$ and $X_1,\dots,X_k\in\mathfrak h$, the derivative of the identity \begin{align*} \varphi(h\exp_H(tX)h^{-1})=\varphi(h)\exp_G(t\,d\varphi_e(X))\varphi(h)^{-1} \end{align*} at $t=0$ gives \begin{align*} d\varphi_e(\operatorname{Ad}_h X)=\operatorname{Ad}_{\varphi(h)}(d\varphi_e(X)). \end{align*} Therefore \begin{align*} (\varphi^*P_0)(\operatorname{Ad}_hX_1,\dots,\operatorname{Ad}_hX_k)=P_0(\operatorname{Ad}_{\varphi(h)}d\varphi_e(X_1),\dots,\operatorname{Ad}_{\varphi(h)}d\varphi_e(X_k)). \end{align*} Since $P_0$ is $G$-invariant, the right-hand side equals \begin{align*} P_0(d\varphi_e(X_1),\dots,d\varphi_e(X_k))=(\varphi^*P_0)(X_1,\dots,X_k). \end{align*} Thus $\varphi^*P_0\in I^k(H)$. For an inhomogeneous $P_0\in I(G)$, this construction is applied degree by degree. [guided] We first make precise the notation $P_0\circ d\varphi_e$, because $P_0$ is not merely a function of one variable when it is viewed through its polarised Chern-Weil form. Assume first that $P_0$ is homogeneous of degree $k$. This means that $P_0$ is represented by a symmetric $k$-linear map \begin{align*} P_0:\mathfrak g^k\to\mathbb R. \end{align*} The correct pullback along $d\varphi_e:\mathfrak h\to\mathfrak g$ is the map \begin{align*} \varphi^*P_0:\mathfrak h^k\to\mathbb R \end{align*} defined by \begin{align*} (\varphi^*P_0)(X_1,\dots,X_k)=P_0(d\varphi_e(X_1),\dots,d\varphi_e(X_k)). \end{align*} Because $d\varphi_e$ is linear and $P_0$ is symmetric and $k$-linear, the map $\varphi^*P_0$ is also symmetric and $k$-linear. The only non-formal point is invariance under $H$. Fix $h\in H$ and $X\in\mathfrak h$. The Lie [group homomorphism](/page/Group%20Homomorphism) $\varphi$ intertwines conjugation by $h$ with conjugation by $\varphi(h)$: for all sufficiently small $t\in\mathbb R$, \begin{align*} \varphi(h\exp_H(tX)h^{-1})=\varphi(h)\exp_G(t\,d\varphi_e(X))\varphi(h)^{-1}. \end{align*} Differentiating this identity at $t=0$ gives \begin{align*} d\varphi_e(\operatorname{Ad}_h X)=\operatorname{Ad}_{\varphi(h)}(d\varphi_e(X)). \end{align*} Applying this identity to each argument of $\varphi^*P_0$, we get \begin{align*} (\varphi^*P_0)(\operatorname{Ad}_hX_1,\dots,\operatorname{Ad}_hX_k)=P_0(\operatorname{Ad}_{\varphi(h)}d\varphi_e(X_1),\dots,\operatorname{Ad}_{\varphi(h)}d\varphi_e(X_k)). \end{align*} Since $P_0$ is invariant under the adjoint action of $G$, this last expression equals \begin{align*} P_0(d\varphi_e(X_1),\dots,d\varphi_e(X_k))=(\varphi^*P_0)(X_1,\dots,X_k). \end{align*} Thus $\varphi^*P_0$ is an $H$-invariant polynomial. If $P_0$ has several homogeneous pieces, the same construction is performed on each homogeneous component and then summed. [/guided] [/step] [step:Relate the induced connection and curvature by pullback to $Q$] Choose a principal connection form \begin{align*} A\in\Omega^1(Q;\mathfrak h) \end{align*} on the principal $H$-bundle $Q\to M$. Let \begin{align*} j:Q\to\widetilde Q \end{align*} be the smooth map defined by $j(q)=[q,e_G]$, where $e_G\in G$ is the identity element. On $Q\times G$, let $\operatorname{pr}_Q:Q\times G\to Q$ and $\operatorname{pr}_G:Q\times G\to G$ denote the two projections, and let $\theta_G\in\Omega^1(G;\mathfrak g)$ denote the left Maurer-Cartan form. Define a $\mathfrak g$-valued one-form $\widehat A\in\Omega^1(Q\times G;\mathfrak g)$ by \begin{align*} \widehat A=\operatorname{Ad}_{g^{-1}}\bigl(d\varphi_e(\operatorname{pr}_Q^*A)\bigr)+\operatorname{pr}_G^*\theta_G, \end{align*} where $g$ denotes the $G$-coordinate on $Q\times G$. Let $X\in\mathfrak h$, and let $X_{Q\times G}$ denote the fundamental vector field for the quotienting right $H$-action $(q,g)\cdot h=(qh,\varphi(h)^{-1}g)$. At $(q,g)$, the $Q$-component is the fundamental vector $X_Q(q)$, so $A(X_Q(q))=X$, and the $G$-component is the tangent vector generated by the curve $t\mapsto \varphi(\exp_H(tX))^{-1}g$. The left Maurer-Cartan form evaluates on this $G$-component as $-\operatorname{Ad}_{g^{-1}}(d\varphi_e(X))$. Hence \begin{align*} \widehat A(X_{Q\times G})=\operatorname{Ad}_{g^{-1}}(d\varphi_e(X))-\operatorname{Ad}_{g^{-1}}(d\varphi_e(X))=0. \end{align*} Thus $\widehat A$ is horizontal for the quotient $H$-orbits. The connection transformation law $(R_h^Q)^*A=\operatorname{Ad}^H_{h^{-1}}A$, the identity $d\varphi_e(\operatorname{Ad}^H_{h^{-1}}X)=\operatorname{Ad}^G_{\varphi(h)^{-1}}d\varphi_e(X)$, and left invariance of the left Maurer-Cartan form together give invariance of $\widehat A$ under the right $H$-action on $Q\times G$. Therefore $\widehat A$ is basic for the quotient map $Q\times G\to Q\times_\varphi G$, so it descends to a unique one-form \begin{align*} \widetilde A\in\Omega^1(\widetilde Q;\mathfrak g). \end{align*} Its value on fundamental vectors for the principal right $G$-action is the corresponding element of $\mathfrak g$, and its equivariance under that right $G$-action follows from the displayed formula for $\widehat A$; hence $\widetilde A$ is a principal $G$-connection. Restricting the descended form along $j$ sets $g=e_G$ and removes the Maurer-Cartan term on the constant second factor, hence \begin{align*} j^*\widetilde A=d\varphi_e\circ A. \end{align*} Let \begin{align*} F_A=dA+\frac{1}{2}[A\wedge A]\in\Omega^2(Q;\mathfrak h) \end{align*} and \begin{align*} F_{\widetilde A}=d\widetilde A+\frac{1}{2}[\widetilde A\wedge\widetilde A]\in\Omega^2(\widetilde Q;\mathfrak g) \end{align*} be the corresponding curvature forms. Since $d\varphi_e$ is a [Lie algebra](/page/Lie%20Algebra) homomorphism, exterior differentiation commutes with the constant linear map $d\varphi_e$, and the bracket is preserved. Hence \begin{align*} j^*F_{\widetilde A}=d\varphi_e(F_A). \end{align*} [guided] The induced connection is not merely asserted from the equation $j^*\widetilde A=d\varphi_e\circ A$; it is obtained by descending a concrete form from $Q\times G$. Let $\operatorname{pr}_Q:Q\times G\to Q$ and $\operatorname{pr}_G:Q\times G\to G$ be the projections, and let $\theta_G\in\Omega^1(G;\mathfrak g)$ be the left Maurer-Cartan form. On $Q\times G$ define \begin{align*} \widehat A=\operatorname{Ad}_{g^{-1}}\bigl(d\varphi_e(\operatorname{pr}_Q^*A)\bigr)+\operatorname{pr}_G^*\theta_G, \end{align*} where $g$ is the $G$-coordinate. We verify that $\widehat A$ is basic for the quotient by $H$. Let $X\in\mathfrak h$, and let $X_{Q\times G}$ be the fundamental vector field of the right $H$-action on $Q\times G$ generated by $X$. At a point $(q,g)$, this vector is obtained by differentiating the curve $t\mapsto (q\exp_H(tX),\varphi(\exp_H(tX))^{-1}g)$. The first component is the fundamental vector $X_Q(q)$ on $Q$, and the defining normalization of the principal connection $A$ gives $A(X_Q(q))=X$. The second component is generated in $G$ by left multiplication of $g$ by $\varphi(\exp_H(tX))^{-1}$; the left Maurer-Cartan form therefore evaluates on it as $-\operatorname{Ad}_{g^{-1}}(d\varphi_e(X))$. Consequently \begin{align*} \widehat A(X_{Q\times G})=\operatorname{Ad}_{g^{-1}}(d\varphi_e(X))-\operatorname{Ad}_{g^{-1}}(d\varphi_e(X))=0. \end{align*} This proves horizontality along the $H$-orbits. For invariance, fix $h\in H$. The $Q$-term transforms by $(R_h^Q)^*A=\operatorname{Ad}^H_{h^{-1}}A$. Applying $d\varphi_e$ and using the intertwining identity $d\varphi_e(\operatorname{Ad}^H_{h^{-1}}X)=\operatorname{Ad}^G_{\varphi(h)^{-1}}d\varphi_e(X)$ matches precisely the change of the $G$-coordinate from $g$ to $\varphi(h)^{-1}g$. The remaining Maurer-Cartan term is unchanged by left multiplication because $\theta_G$ is the left Maurer-Cartan form. Thus $\widehat A$ is invariant under the right $H$-action. Since it is both horizontal and invariant, it descends to a unique one-form \begin{align*} \widetilde A\in\Omega^1(\widetilde Q;\mathfrak g). \end{align*} The descended form is a principal $G$-connection form because the formula for $\widehat A$ gives the usual fundamental-vector normalization for the right $G$-action and the usual $\operatorname{Ad}^G$-equivariance under that action. Along $j(q)=[q,e_G]$, the second coordinate is constant, so the pullback of $\operatorname{pr}_G^*\theta_G$ vanishes and $\operatorname{Ad}_{e_G^{-1}}$ is the identity on $\mathfrak g$. Thus \begin{align*} j^*\widetilde A=d\varphi_e\circ A. \end{align*} Now compute curvature. Pullback commutes with exterior differentiation and with the bracket of Lie-algebra-valued forms, so \begin{align*} j^*F_{\widetilde A}=d(j^*\widetilde A)+\frac{1}{2}[j^*\widetilde A\wedge j^*\widetilde A]. \end{align*} Substituting $j^*\widetilde A=d\varphi_e\circ A$ gives \begin{align*} j^*F_{\widetilde A}=d(d\varphi_e(A))+\frac{1}{2}[d\varphi_e(A)\wedge d\varphi_e(A)]. \end{align*} Because $d\varphi_e:\mathfrak h\to\mathfrak g$ is a constant linear map, it commutes with exterior differentiation. Because $\varphi$ is a Lie group homomorphism, $d\varphi_e$ is a Lie algebra homomorphism, so it preserves brackets. Therefore \begin{align*} j^*F_{\widetilde A}=d\varphi_e\left(dA+\frac{1}{2}[A\wedge A]\right)=d\varphi_e(F_A). \end{align*} [/guided] [/step] [step:Compare the Chern-Weil representative forms] Recall the Chern-Weil evaluation convention: if $B\in\Omega^2(P;\mathfrak g)$ and $P_0:\mathfrak g^k\to\mathbb R$ is symmetric and $k$-linear, then $P_0(B)$ denotes the scalar $2k$-form obtained by applying $P_0$ to the $k$ curvature-form slots and wedging the resulting differential-form components. When $k=0$, this convention means that a constant $P_0\in\mathbb R$ is carried to the same constant $0$-form on the base. Let \begin{align*} \omega_{\widetilde Q}\in\Omega^{2k}(M) \end{align*} be the Chern-Weil form on $M$ determined by $P_0$ and $\widetilde A$, so that \begin{align*} \pi_{\widetilde Q}^*\omega_{\widetilde Q}=P_0(F_{\widetilde A}). \end{align*} Let \begin{align*} \omega_Q\in\Omega^{2k}(M) \end{align*} be the Chern-Weil form on $M$ determined by $\varphi^*P_0$ and $A$, so that \begin{align*} \pi_Q^*\omega_Q=(\varphi^*P_0)(F_A). \end{align*} Since $\pi_{\widetilde Q}\circ j=\pi_Q$, pulling the first identity back by $j$ gives \begin{align*} \pi_Q^*\omega_{\widetilde Q}=j^*P_0(F_{\widetilde A}). \end{align*} Using $j^*F_{\widetilde A}=d\varphi_e(F_A)$ and the definition of $\varphi^*P_0$, we obtain \begin{align*} j^*P_0(F_{\widetilde A})=P_0(d\varphi_e(F_A))=(\varphi^*P_0)(F_A). \end{align*} Therefore \begin{align*} \pi_Q^*\omega_{\widetilde Q}=\pi_Q^*\omega_Q. \end{align*} The projection $\pi_Q:Q\to M$ is a surjective submersion, so pullback by $\pi_Q$ is injective on differential forms. Hence \begin{align*} \omega_{\widetilde Q}=\omega_Q. \end{align*} [guided] The comparison happens on $Q$ because $Q$ maps naturally into the extended bundle by $j(q)=[q,e_G]$. If $k=0$, then $P_0\in\mathbb R$ is a constant and both Chern-Weil representatives are the same constant $0$-form, so assume in the computation below that $k\ge 1$. Let $\omega_{\widetilde Q}\in\Omega^{2k}(M)$ be the Chern-Weil form determined by $P_0$ and $\widetilde A$, so \begin{align*} \pi_{\widetilde Q}^*\omega_{\widetilde Q}=P_0(F_{\widetilde A}). \end{align*} Let $\omega_Q\in\Omega^{2k}(M)$ be the Chern-Weil form determined by $\varphi^*P_0$ and $A$, so \begin{align*} \pi_Q^*\omega_Q=(\varphi^*P_0)(F_A). \end{align*} The map $j$ lies over the identity map of $M$, meaning $\pi_{\widetilde Q}\circ j=\pi_Q$. Pulling back the defining equation for $\omega_{\widetilde Q}$ along $j$ gives \begin{align*} \pi_Q^*\omega_{\widetilde Q}=j^*P_0(F_{\widetilde A}). \end{align*} Pullback commutes with evaluating a multilinear polynomial on differential-form slots, so this is \begin{align*} j^*P_0(F_{\widetilde A})=P_0(j^*F_{\widetilde A}). \end{align*} From the curvature comparison already proved, $j^*F_{\widetilde A}=d\varphi_e(F_A)$. Therefore \begin{align*} j^*P_0(F_{\widetilde A})=P_0(d\varphi_e(F_A))=(\varphi^*P_0)(F_A). \end{align*} The last equality is exactly the definition of the pulled-back invariant polynomial applied to each curvature-form slot. Hence \begin{align*} \pi_Q^*\omega_{\widetilde Q}=\pi_Q^*\omega_Q. \end{align*} Finally, $\pi_Q:Q\to M$ is a surjective submersion. Differential forms can be checked after pullback by a surjective submersion, so $\pi_Q^*\eta=0$ implies $\eta=0$ for every form $\eta$ on $M$. Applying this to $\eta=\omega_{\widetilde Q}-\omega_Q$ gives \begin{align*} \omega_{\widetilde Q}=\omega_Q. \end{align*} [/guided] [/step] [step:Pass from equality of forms to equality of Chern-Weil classes] By the [Chern-Weil homomorphism theorem](/theorems/9765) [citetheorem:9765], the class $\operatorname{cw}_{\widetilde Q}(P_0)$ is represented by $\omega_{\widetilde Q}$, and the class $\operatorname{cw}_Q(\varphi^*P_0)$ is represented by $\omega_Q$. The previous step proves equality of these representative forms, so their de Rham cohomology classes are equal: \begin{align*} \operatorname{cw}_{Q\times_\varphi G}(P_0)=[\omega_{\widetilde Q}]=[\omega_Q]=\operatorname{cw}_Q(\varphi^*P_0). \end{align*} For an inhomogeneous invariant polynomial $P_0\in I(G)$, the argument applies to each homogeneous component and the Chern-Weil construction is additive over the finite sum of components. This proves the stated identity in $H^{\mathrm{even}}_{\mathrm{dR}}(M)$. [guided] The last step is to translate equality of differential forms into equality of the characteristic classes defined by Chern-Weil theory. By the Chern-Weil homomorphism theorem [citetheorem:9765], $\operatorname{cw}_{\widetilde Q}(P_0)$ is the de Rham cohomology class of the descended Chern-Weil representative $\omega_{\widetilde Q}$, and $\operatorname{cw}_Q(\varphi^*P_0)$ is the de Rham cohomology class of the descended Chern-Weil representative $\omega_Q$. The previous step proved the stronger statement that the representatives are equal as forms on $M$: \begin{align*} \omega_{\widetilde Q}=\omega_Q. \end{align*} Equal closed forms represent equal de Rham cohomology classes, so \begin{align*} \operatorname{cw}_{Q\times_\varphi G}(P_0)=[\omega_{\widetilde Q}]=[\omega_Q]=\operatorname{cw}_Q(\varphi^*P_0). \end{align*} If $P_0$ is not homogeneous, write it as the finite sum of its homogeneous components in $I(G)$. The preceding argument applies to each component separately, and the Chern-Weil homomorphism is additive over this finite [direct sum](/page/Direct%20Sum). Therefore the same identity holds for every $P_0\in I(G)$ in $H^{\mathrm{even}}_{\mathrm{dR}}(M)$. [/guided] [/step]