[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*}
[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]
[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]
