[proofplan] We apply the [infinitesimal Chern-Weil variation formula](/theorems/9790) to the smooth path of connections $A_t$. That formula identifies the $t$-derivative of the Chern-Weil form $P(F_{A_t},\ldots,F_{A_t})$ with the [exterior derivative](/theorems/1525) of the integrand defining the transgression form. Integrating this identity over the compact parameter interval $[0,1]$, commuting the exterior derivative with parameter integration, and applying the [fundamental theorem of calculus](/theorems/632) gives the endpoint difference. [/proofplan]

[step:Define the endpoint form and the transgression integrand] Define a smooth map of differential forms \begin{align*} \Phi:[0,1]\to\Omega^{2k}(M) \end{align*} by \begin{align*} \Phi(t):=P(F_{A_t},\ldots,F_{A_t}). \end{align*} Define another smooth map \begin{align*} \Psi:[0,1]\to\Omega^{2k-1}(M) \end{align*} by \begin{align*} \Psi(t):=kP(\dot A_t,F_{A_t},\ldots,F_{A_t}), \end{align*} where $F_{A_t}$ occurs $k-1$ times. With this notation, the transgression form is \begin{align*} T_P(A_t)=\int_{[0,1]}\Psi(t)\,d\mathcal L^1(t). \end{align*} The smoothness of $t\mapsto A_t$ implies smoothness of $t\mapsto F_{A_t}$ and $t\mapsto \dot A_t$, so both $\Phi$ and $\Psi$ are smooth parameter-dependent differential forms. [/step]

[step:Apply the infinitesimal variation formula to the path $A_t$]The hypotheses of [citetheorem:9790] are satisfied: $A_t$ is a smooth path of connections, $\dot A_t\in\Omega^1(M;\operatorname{ad}Q)$ is its derivative, and $P\in I^k(\mathfrak g)$ is an invariant symmetric $k$-linear polynomial. Therefore, for every $t\in[0,1]$, \begin{align*} \frac{d}{dt}P(F_{A_t},\ldots,F_{A_t})=d\bigl(kP(\dot A_t,F_{A_t},\ldots,F_{A_t})\bigr). \end{align*} Equivalently, using the notation from the previous step, \begin{align*} \frac{d}{dt}\Phi(t)=d\Psi(t). \end{align*}[/step]

[guided]We want to turn the endpoint difference $\Phi(1)-\Phi(0)$ into an exact form. The mechanism is the infinitesimal Chern-Weil variation formula, which says that the derivative of a Chern-Weil form along a smooth path of connections is exact. We verify its hypotheses in the present setting. The theorem requires a smooth path of connections; this is exactly the given family $(A_t)_{t\in[0,1]}$. It requires the tangent vector to the path, and by definition this is \begin{align*} \dot A_t=\frac{d}{dt}A_t\in\Omega^1(M;\operatorname{ad}Q). \end{align*} It also requires an invariant symmetric $k$-linear polynomial on the [Lie algebra](/page/Lie%20Algebra), and this is precisely the given $P\in I^k(\mathfrak g)$. Hence [citetheorem:9790] applies and gives, for every $t\in[0,1]$, \begin{align*} \frac{d}{dt}P(F_{A_t},\ldots,F_{A_t})=d\bigl(kP(\dot A_t,F_{A_t},\ldots,F_{A_t})\bigr). \end{align*} In the notation already introduced, this is the compact identity \begin{align*} \frac{d}{dt}\Phi(t)=d\Psi(t). \end{align*} This is the whole geometric input of the proof: the variation formula itself is where the curvature variation identity, the invariance of $P$, and the Bianchi identity are used.[/guided]

[step:Integrate the infinitesimal identity over the parameter interval] Integrating the identity $\frac{d}{dt}\Phi(t)=d\Psi(t)$ over $[0,1]$ gives \begin{align*} \int_{[0,1]}\frac{d}{dt}\Phi(t)\,d\mathcal L^1(t)=\int_{[0,1]}d\Psi(t)\,d\mathcal L^1(t). \end{align*} Since $\Psi$ is a smooth parameter-dependent differential form on the compact interval $[0,1]$, exterior differentiation on $M$ commutes with integration over the parameter $t$. Therefore \begin{align*} \int_{[0,1]}d\Psi(t)\,d\mathcal L^1(t)=d\int_{[0,1]}\Psi(t)\,d\mathcal L^1(t). \end{align*} By the definition of $T_P(A_t)$, this becomes \begin{align*} \int_{[0,1]}d\Psi(t)\,d\mathcal L^1(t)=dT_P(A_t). \end{align*} [/step]

[step:Use the fundamental theorem of calculus to obtain the endpoint difference] The map $\Phi:[0,1]\to\Omega^{2k}(M)$ is smooth, so the fundamental theorem of calculus for smooth parameter-dependent differential forms gives \begin{align*} \int_{[0,1]}\frac{d}{dt}\Phi(t)\,d\mathcal L^1(t)=\Phi(1)-\Phi(0). \end{align*} Substituting the definition of $\Phi$ yields \begin{align*} \Phi(1)-\Phi(0)=P(F_{A_1},\ldots,F_{A_1})-P(F_{A_0},\ldots,F_{A_0}). \end{align*} Combining this identity with the previous step gives \begin{align*} P(F_{A_1},\ldots,F_{A_1})-P(F_{A_0},\ldots,F_{A_0})=dT_P(A_t). \end{align*} This is exactly the asserted transgression formula. [/step]