[proofplan]
We differentiate the curvature of the smooth path of connections and identify its variation with the covariant [exterior derivative](/theorems/1525) $d_{A_t}\dot A_t$. Multilinearity and symmetry of the polarised invariant polynomial reduce the derivative of the Chern-Weil form to $kP_0(d_{A_t}\dot A_t,F_{A_t},\dots,F_{A_t})$. Finally, the covariant exterior derivative formula for invariant polynomials rewrites this expression as an ordinary exterior derivative, and the Bianchi identity removes all terms containing $d_{A_t}F_{A_t}$.
[/proofplan]
[step:Differentiate the curvature along the path of connections]
Fix $t\in[0,1]$. Let
\begin{align*}
d_{A_t}:\Omega^r(M;\operatorname{ad}Q)\to \Omega^{r+1}(M;\operatorname{ad}Q)
\end{align*}
denote the covariant exterior derivative induced by the connection $A_t$. Since $(A_t)_{t\in[0,1]}$ is a smooth path of connections, its derivative
\begin{align*}
\dot A_t=\frac{dA_t}{dt}
\end{align*}
is a well-defined element of $\Omega^1(M;\operatorname{ad}Q)$.
The standard curvature variation identity for a path of connections gives
\begin{align*}
\frac{d}{dt}F_{A_t}=d_{A_t}\dot A_t
\end{align*}
in $\Omega^2(M;\operatorname{ad}Q)$.
For completeness, in any local trivialisation of $Q$ over an [open set](/page/Open%20Set) $U\subset M$, the connection is represented by a $\mathfrak g$-valued $1$-form, still denoted $A_t$, and the curvature is
\begin{align*}
F_{A_t}=dA_t+\frac{1}{2}[A_t\wedge A_t].
\end{align*}
Differentiating this identity with respect to $t$ gives
\begin{align*}
\frac{d}{dt}F_{A_t}=d\dot A_t+\frac{1}{2}[\dot A_t\wedge A_t]+\frac{1}{2}[A_t\wedge \dot A_t].
\end{align*}
By the graded bracket convention for Lie-algebra-valued forms, the final two terms combine to $[A_t\wedge \dot A_t]$. Hence the local expression is
\begin{align*}
\frac{d}{dt}F_{A_t}=d\dot A_t+[A_t\wedge \dot A_t]=d_{A_t}\dot A_t.
\end{align*}
These local identities are compatible under changes of trivialisation, so they define the global identity above.
[/step]
[step:Use multilinearity and symmetry of the invariant polynomial]
By definition of the Chern-Weil convention,
\begin{align*}
P_0(F_{A_t},\dots,F_{A_t})\in\Omega^{2k}(M)
\end{align*}
is obtained by evaluating the symmetric $k$-linear form $P_0$ on $k$ copies of $F_{A_t}$, with wedge product of differential form components.
Since $P_0$ is multilinear and symmetric, differentiating the $k$ entries gives
\begin{align*}
\frac{d}{dt}P_0(F_{A_t},\dots,F_{A_t})
=
\sum_{j=1}^{k}P_0(F_{A_t},\dots,F_{A_t},\frac{d}{dt}F_{A_t},F_{A_t},\dots,F_{A_t}).
\end{align*}
Each summand is equal because all entries $F_{A_t}$ have even degree $2$ and $P_0$ is symmetric. Using the curvature variation identity from the previous step, we obtain
\begin{align*}
\frac{d}{dt}P_0(F_{A_t},\dots,F_{A_t})
=
kP_0(d_{A_t}\dot A_t,F_{A_t},\dots,F_{A_t}).
\end{align*}
[guided]
We now differentiate the expression
\begin{align*}
P_0(F_{A_t},\dots,F_{A_t})
\end{align*}
as a product-like multilinear expression. The notation means that the same $\operatorname{ad}Q$-valued $2$-form $F_{A_t}$ is inserted into all $k$ arguments of the symmetric $k$-linear invariant form $P_0$.
Because $P_0$ is multilinear, the derivative with respect to $t$ is obtained by differentiating one argument at a time:
\begin{align*}
\frac{d}{dt}P_0(F_{A_t},\dots,F_{A_t})
=
\sum_{j=1}^{k}P_0(F_{A_t},\dots,F_{A_t},\frac{d}{dt}F_{A_t},F_{A_t},\dots,F_{A_t}).
\end{align*}
There are $k$ summands. Since $P_0$ is symmetric and the curvature form has even degree, moving the differentiated curvature term from the $j$-th slot to the first slot introduces no graded sign. Therefore every summand equals
\begin{align*}
P_0(\frac{d}{dt}F_{A_t},F_{A_t},\dots,F_{A_t}).
\end{align*}
The previous step identified the curvature derivative as
\begin{align*}
\frac{d}{dt}F_{A_t}=d_{A_t}\dot A_t.
\end{align*}
Substituting this identity into the differentiated multilinear expression gives
\begin{align*}
\frac{d}{dt}P_0(F_{A_t},\dots,F_{A_t})
=
kP_0(d_{A_t}\dot A_t,F_{A_t},\dots,F_{A_t}).
\end{align*}
This is the infinitesimal variation before using the Bianchi identity; the remaining task is to recognize the right-hand side as an ordinary exterior derivative.
[/guided]
[/step]
[step:Apply the invariant-polynomial differential identity]
Let
\begin{align*}
\alpha_1:=\dot A_t\in\Omega^1(M;\operatorname{ad}Q)
\end{align*}
and, for $2\le j\le k$, let
\begin{align*}
\alpha_j:=F_{A_t}\in\Omega^2(M;\operatorname{ad}Q).
\end{align*}
The covariant exterior derivative identity for an invariant symmetric polynomial states that
\begin{align*}
dP_0(\alpha_1,\dots,\alpha_k)
=
\sum_{j=1}^{k}(-1)^{|\alpha_1|+\cdots+|\alpha_{j-1}|}
P_0(\alpha_1,\dots,d_{A_t}\alpha_j,\dots,\alpha_k),
\end{align*}
where $|\alpha_j|$ denotes the differential form degree of $\alpha_j$. This identity follows from the ordinary Leibniz rule for $d$, together with the infinitesimal $\operatorname{Ad}$-invariance of $P_0$, which cancels the connection-bracket terms.
Substituting the degrees $|\dot A_t|=1$ and $|F_{A_t}|=2$ gives
\begin{align*}
dP_0(\dot A_t,F_{A_t},\dots,F_{A_t})
=
P_0(d_{A_t}\dot A_t,F_{A_t},\dots,F_{A_t})
-
\sum_{j=2}^{k}P_0(\dot A_t,F_{A_t},\dots,d_{A_t}F_{A_t},\dots,F_{A_t}).
\end{align*}
The minus sign appears because the total degree before the $j$-th slot is $1+2(j-2)$, which is odd.
[/step]
[step:Use the Bianchi identity to eliminate the curvature derivative terms]
The Bianchi identity for the connection $A_t$ states that
\begin{align*}
d_{A_t}F_{A_t}=0
\end{align*}
in $\Omega^3(M;\operatorname{ad}Q)$. Therefore every term in the sum
\begin{align*}
\sum_{j=2}^{k}P_0(\dot A_t,F_{A_t},\dots,d_{A_t}F_{A_t},\dots,F_{A_t})
\end{align*}
vanishes. Hence
\begin{align*}
dP_0(\dot A_t,F_{A_t},\dots,F_{A_t})
=
P_0(d_{A_t}\dot A_t,F_{A_t},\dots,F_{A_t}).
\end{align*}
Combining this identity with the result of the multilinearity step gives
\begin{align*}
\frac{d}{dt}P_0(F_{A_t},\dots,F_{A_t})
=
k\,dP_0(\dot A_t,F_{A_t},\dots,F_{A_t}).
\end{align*}
This is the desired identity in $\Omega^{2k}(M)$, and the proof is complete.
[/step]
