[proofplan] We connect the two principal connections by the affine path $A_t=A_0+t(A_1-A_0)$ and compute the variation of the Chern-Weil form along this path. The difference $a=A_1-A_0$ is horizontal and equivariant, so every $A_t$ is again a principal connection. Differentiating the curvature gives $\frac{d}{dt}F_{A_t}=d_{A_t}a$, and the infinitesimal $\operatorname{Ad}$-invariance identity together with the Bianchi identity rewrites $\frac{d}{dt}P_0(F_{A_t})$ as an [exterior derivative](/theorems/1525). Integrating in $t$ gives a basic transgression form whose descent to $M$ has exterior derivative equal to the difference of the two endpoint Chern-Weil forms. [/proofplan]

[step:Handle the degree zero case separately] If $k=0$, then $P_0\in I^0(G)$ is a constant polynomial. Hence $P_0(F_{A_0})_M=P_0(F_{A_1})_M$ as degree-$0$ forms on $M$, and their de Rham cohomology classes are equal. Hence assume for the rest of the proof that $k\ge 1$. [/step]

[step:Build the affine path of principal connections]Define \begin{align*} a:=A_1-A_0\in\Omega^1(P;\mathfrak g). \end{align*} For each $t\in[0,1]$, define \begin{align*} A_t:=A_0+t a\in\Omega^1(P;\mathfrak g). \end{align*} We verify that $A_t$ is a principal connection. Let $\mathfrak X(P)$ denote the real [vector space](/page/Vector%20Space) of smooth vector fields on $P$. If $\xi\in\mathfrak g$ and $\xi_P\in\mathfrak X(P)$ denotes the fundamental vector field generated by $\xi$, then $A_0(\xi_P)=\xi$ and $A_1(\xi_P)=\xi$, so \begin{align*} a(\xi_P)=0. \end{align*} Therefore \begin{align*} A_t(\xi_P)=A_0(\xi_P)+t a(\xi_P)=\xi. \end{align*} For every $g\in G$, the connection equivariance identities give \begin{align*} R_g^*A_0=\operatorname{Ad}_{g^{-1}}A_0 \end{align*} and \begin{align*} R_g^*A_1=\operatorname{Ad}_{g^{-1}}A_1. \end{align*} Subtracting gives \begin{align*} R_g^*a=\operatorname{Ad}_{g^{-1}}a. \end{align*} Hence \begin{align*} R_g^*A_t=\operatorname{Ad}_{g^{-1}}A_t. \end{align*} Thus $A_t$ reproduces fundamental vector fields and is $G$-equivariant, so $A_t$ is a principal connection for every $t\in[0,1]$.[/step]

[guided]The affine formula $A_t=A_0+t(A_1-A_0)$ is useful only if it stays inside the space of principal connections. We check the two defining conditions directly. First define the difference form \begin{align*} a:=A_1-A_0\in\Omega^1(P;\mathfrak g). \end{align*} For each $t\in[0,1]$, define \begin{align*} A_t:=A_0+t a\in\Omega^1(P;\mathfrak g). \end{align*} Let $\mathfrak X(P)$ denote the real vector space of smooth vector fields on $P$. Let $\xi\in\mathfrak g$, and let $\xi_P\in\mathfrak X(P)$ be the fundamental vector field on $P$ generated by $\xi$. Since $A_0$ and $A_1$ are principal connections, both reproduce fundamental vector fields: \begin{align*} A_0(\xi_P)=\xi \end{align*} and \begin{align*} A_1(\xi_P)=\xi. \end{align*} Subtracting these two equalities gives \begin{align*} a(\xi_P)=0. \end{align*} Consequently \begin{align*} A_t(\xi_P)=A_0(\xi_P)+t a(\xi_P)=\xi. \end{align*} So the first connection axiom holds for $A_t$. Now we check equivariance. For every $g\in G$, the right action map $R_g:P\to P$ satisfies \begin{align*} R_g^*A_0=\operatorname{Ad}_{g^{-1}}A_0 \end{align*} and \begin{align*} R_g^*A_1=\operatorname{Ad}_{g^{-1}}A_1. \end{align*} Subtracting gives the transformation law for the difference form: \begin{align*} R_g^*a=\operatorname{Ad}_{g^{-1}}a. \end{align*} Therefore \begin{align*} R_g^*A_t=R_g^*A_0+tR_g^*a=\operatorname{Ad}_{g^{-1}}A_0+t\operatorname{Ad}_{g^{-1}}a=\operatorname{Ad}_{g^{-1}}A_t. \end{align*} Thus $A_t$ also satisfies the equivariance axiom. The point of this verification is that the path $t\mapsto A_t$ is not merely a path of $\mathfrak g$-valued $1$-forms; it is a path through genuine principal connections, so its curvatures are legitimate Chern-Weil inputs.[/guided]

[step:Differentiate the curvature along the affine path] For a principal connection $A_t$, let \begin{align*} F_t:=F_{A_t}=dA_t+\frac{1}{2}[A_t\wedge A_t]\in\Omega^2(P;\mathfrak g) \end{align*} denote its curvature, where $[\alpha\wedge\beta]$ denotes the graded bracket of $\mathfrak g$-valued forms obtained by wedging the scalar form components and applying the Lie bracket $[\cdot,\cdot]:\mathfrak g\times\mathfrak g\to\mathfrak g$. Since $\frac{d}{dt}A_t=a$, differentiating the displayed curvature formula gives \begin{align*} \frac{d}{dt}F_t=da+[A_t\wedge a]. \end{align*} Let \begin{align*} d_{A_t}:\Omega^r(P;\mathfrak g)\to\Omega^{r+1}(P;\mathfrak g) \end{align*} denote the covariant exterior derivative defined by \begin{align*} d_{A_t}\beta:=d\beta+[A_t\wedge\beta] \end{align*} for $\beta\in\Omega^r(P;\mathfrak g)$. Then \begin{align*} \frac{d}{dt}F_t=d_{A_t}a. \end{align*} Also, expanding $d_{A_t}F_t$ from $F_t=dA_t+\frac{1}{2}[A_t\wedge A_t]$ and using the graded Jacobi identity for the bracket of $\mathfrak g$-valued forms gives the Bianchi identity \begin{align*} d_{A_t}F_t=0. \end{align*} [/step]

[step:Rewrite the variation of the invariant polynomial as an exterior derivative] Extend $P_0$ to $\mathfrak g$-valued differential forms by wedging the form parts and applying the symmetric $k$-[linear map](/page/Linear%20Map) $P_0$ to the $\mathfrak g$-components. For each $t\in[0,1]$, differentiating the $k$ curvature entries gives \begin{align*} \frac{d}{dt}P_0(F_t)=kP_0(d_{A_t}a,F_t,\dots,F_t). \end{align*} We compute the exterior derivative of the transgression integrand. Since $a$ has degree $1$ and $F_t$ has degree $2$, the graded Leibniz rule for $P_0$ gives \begin{align*} dP_0(a,F_t,\dots,F_t)=P_0(da,F_t,\dots,F_t)-\sum_{j=2}^{k}P_0(a,F_t,\dots,dF_t,\dots,F_t). \end{align*} The infinitesimal $\operatorname{Ad}$-invariance identity for $P_0$ applied to the $\mathfrak g$-valued $1$-form $A_t$, the $\mathfrak g$-valued $1$-form $a$, and the $\mathfrak g$-valued $2$-form $F_t$ in each of the remaining $k-1$ entries gives \begin{align*} P_0([A_t\wedge a],F_t,\dots,F_t)-\sum_{j=2}^{k}P_0(a,F_t,\dots,[A_t\wedge F_t],\dots,F_t)=0. \end{align*} The minus signs in the summation are the graded signs obtained when the degree-$1$ form $A_t$ is moved past the preceding degree-$1$ entry $a$ before acting on a later degree-$2$ curvature entry. Adding this identity to the preceding Leibniz formula yields \begin{align*} dP_0(a,F_t,\dots,F_t)=P_0(d_{A_t}a,F_t,\dots,F_t)-\sum_{j=2}^{k}P_0(a,F_t,\dots,d_{A_t}F_t,\dots,F_t). \end{align*} The Bianchi identity $d_{A_t}F_t=0$ removes every term in the sum, so \begin{align*} dP_0(a,F_t,\dots,F_t)=P_0(d_{A_t}a,F_t,\dots,F_t). \end{align*} Therefore \begin{align*} \frac{d}{dt}P_0(F_t)=d\bigl(kP_0(a,F_t,\dots,F_t)\bigr). \end{align*} Here $P_0(a,F_t,\dots,F_t)$ has one entry equal to $a$ and $k-1$ entries equal to $F_t$. [/step]

[step:Integrate the exact variation to obtain a basic transgression form]Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on the interval $[0,1]$. Define the transgression form on $P$ by \begin{align*} \widetilde T:=k\int_{[0,1]}P_0(a,F_t,\dots,F_t)\,d\mathcal L^1(t)\in\Omega^{2k-1}(P). \end{align*} Since exterior differentiation commutes with integration over the compact parameter interval $[0,1]$, the previous step gives \begin{align*} d\widetilde T=\int_{[0,1]}\frac{d}{dt}P_0(F_t)\,d\mathcal L^1(t). \end{align*} The [fundamental theorem of calculus](/theorems/632) applied to the smooth map \begin{align*} [0,1]\to\Omega^{2k}(P),\qquad t\mapsto P_0(F_t) \end{align*} gives \begin{align*} d\widetilde T=P_0(F_{A_1})-P_0(F_{A_0}). \end{align*} We now check that $\widetilde T$ is basic. The form $a$ is horizontal and $G$-equivariant by construction, and each curvature form $F_t$ is horizontal and $G$-equivariant for the connection $A_t$. Since $P_0$ is $\operatorname{Ad}$-invariant, $P_0(a,F_t,\dots,F_t)$ is horizontal and $G$-invariant for each $t$. Integration in $t$ preserves horizontality and $G$-invariance, so $\widetilde T$ is basic. By the descent property for basic forms, there is a unique form \begin{align*} T_M\in\Omega^{2k-1}(M) \end{align*} such that \begin{align*} \pi^*T_M=\widetilde T. \end{align*} The descended Chern-Weil forms exist by [citetheorem:9762], whose hypotheses are exactly the principal bundle $\pi:P\to M$, the principal connections $A_i$, and the invariant polynomial $P_0$ fixed above. Their closedness is the content of [citetheorem:9763], applied separately to $A_0$ and $A_1$ with the same $P_0$.[/step]

[guided]We have shown that the derivative in $t$ is an exterior derivative. The next task is to integrate that identity and make sure the resulting primitive lives on the base manifold $M$, not merely on the total space $P$. Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $[0,1]$. Define \begin{align*} \widetilde T:=k\int_{[0,1]}P_0(a,F_t,\dots,F_t)\,d\mathcal L^1(t)\in\Omega^{2k-1}(P). \end{align*} The integrand is a smooth one-parameter family of differential forms on $P$, and the parameter interval $[0,1]$ is compact. Therefore exterior differentiation in the $P$-variables commutes with integration in the parameter $t$. Using the identity from the previous step, we get \begin{align*} d\widetilde T=\int_{[0,1]}d\bigl(kP_0(a,F_t,\dots,F_t)\bigr)\,d\mathcal L^1(t). \end{align*} Substituting \begin{align*} d\bigl(kP_0(a,F_t,\dots,F_t)\bigr)=\frac{d}{dt}P_0(F_t) \end{align*} gives \begin{align*} d\widetilde T=\int_{[0,1]}\frac{d}{dt}P_0(F_t)\,d\mathcal L^1(t). \end{align*} Now apply the fundamental theorem of calculus to the smooth map \begin{align*} [0,1]\to\Omega^{2k}(P),\qquad t\mapsto P_0(F_t). \end{align*} This yields \begin{align*} d\widetilde T=P_0(F_1)-P_0(F_0)=P_0(F_{A_1})-P_0(F_{A_0}). \end{align*} It remains to descend $\widetilde T$ from $P$ to $M$. A differential form on a principal bundle descends exactly when it is basic, meaning horizontal and invariant under the right $G$-action. We verify both properties. The form $a=A_1-A_0$ is horizontal because it vanishes on fundamental vector fields, and it is $G$-equivariant because \begin{align*} R_g^*a=\operatorname{Ad}_{g^{-1}}a \end{align*} for every $g\in G$. For each $t$, the curvature $F_t$ of the principal connection $A_t$ is horizontal and satisfies \begin{align*} R_g^*F_t=\operatorname{Ad}_{g^{-1}}F_t. \end{align*} Since $P_0$ is invariant under the adjoint action, applying $P_0$ to one copy of $a$ and $k-1$ copies of $F_t$ eliminates the simultaneous adjoint action: \begin{align*} R_g^*P_0(a,F_t,\dots,F_t)=P_0(a,F_t,\dots,F_t). \end{align*} Horizontality is also preserved because inserting any vertical vector into the form forces either the $a$ entry or a curvature entry to vanish. Hence $P_0(a,F_t,\dots,F_t)$ is basic for each $t$. Integration over $t$ preserves both horizontality and invariance, so $\widetilde T$ is basic. By the standard descent criterion for basic forms on a principal bundle, a basic form on $P$ is the pullback of a unique form on $M$. Therefore there exists a unique differential form \begin{align*} T_M\in\Omega^{2k-1}(M) \end{align*} such that \begin{align*} \pi^*T_M=\widetilde T. \end{align*} The endpoint Chern-Weil forms descend by [citetheorem:9762], applied to the principal bundle $\pi:P\to M$, the connection $A_i$, and the invariant polynomial $P_0$ for each $i\in\{0,1\}$. They are closed by [citetheorem:9763], again applied separately to $A_0$ and $A_1$. This is exactly the setting needed to compare their de Rham cohomology classes on the base.[/guided]

[step:Descend the transgression identity to the base] By definition of the descended Chern-Weil forms, \begin{align*} \pi^*\bigl(P_0(F_{A_i})_M\bigr)=P_0(F_{A_i}) \end{align*} for $i\in\{0,1\}$. Since exterior differentiation commutes with pullback, \begin{align*} \pi^*(dT_M)=d(\pi^*T_M)=d\widetilde T=P_0(F_{A_1})-P_0(F_{A_0}). \end{align*} Using the previous displayed descent identities, this becomes \begin{align*} \pi^*(dT_M)=\pi^*\bigl(P_0(F_{A_1})_M-P_0(F_{A_0})_M\bigr). \end{align*} The pullback $\pi^*:\Omega^{2k}(M)\to\Omega^{2k}(P)$ is injective because $\pi:P\to M$ is a surjective submersion. Hence \begin{align*} P_0(F_{A_1})_M-P_0(F_{A_0})_M=dT_M. \end{align*} Thus the two closed forms differ by an exact form on $M$, and therefore \begin{align*} [P_0(F_{A_0})_M]=[P_0(F_{A_1})_M]\in H^{2k}_{\mathrm{dR}}(M). \end{align*} This proves the theorem. [/step]