[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*}[/step]

[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: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]