[proofplan] We prove closedness after pulling the descended form back to the total space $P$. The pullback is the scalar form $P_0(F_A,\dots,F_A)$, so the problem becomes a computation with the curvature form on $P$. The [exterior derivative](/theorems/1525) is rewritten using the covariant exterior derivative $d_A$; the Bianchi identity kills the covariant derivative term, and infinitesimal $\operatorname{Ad}$-invariance kills the remaining bracket terms. [/proofplan]

[step:Reduce closedness on $M$ to closedness after pullback to $P$]If $k=0$, then $P_0(F_A)_M$ is a degree-$0$ constant characteristic form, hence its exterior derivative is zero. Assume from now on that $k\ge 1$. Define the scalar $2k$-form \begin{align*} \omega:=P_0(F_A,\dots,F_A)\in \Omega^{2k}(P). \end{align*} By the definition of the descended Chern-Weil form, \begin{align*} \pi^*P_0(F_A)_M=\omega. \end{align*} Naturality of the exterior derivative under pullback gives \begin{align*} \pi^*\bigl(dP_0(F_A)_M\bigr)=d\pi^*P_0(F_A)_M=d\omega. \end{align*} Since $\pi:P\to M$ is a surjective submersion, $\pi^*:\Omega^r(M)\to \Omega^r(P)$ is injective for every $r\ge 0$. Thus it is enough to prove $d\omega=0$.[/step]

[guided]The form whose closedness we want lives on the base $M$, but the connection form $A$ and curvature form $F_A$ naturally live on the principal bundle $P$. The descent hypothesis gives the bridge between the two levels: the descended form $P_0(F_A)_M$ is exactly the unique form on $M$ whose pullback is \begin{align*} \omega:=P_0(F_A,\dots,F_A)\in \Omega^{2k}(P). \end{align*} Therefore \begin{align*} \pi^*P_0(F_A)_M=\omega. \end{align*} We may test closedness after pullback because pullback commutes with the exterior derivative. Applying naturality of $d$ to the smooth map $\pi:P\to M$ gives \begin{align*} \pi^*\bigl(dP_0(F_A)_M\bigr)=d\bigl(\pi^*P_0(F_A)_M\bigr)=d\omega. \end{align*} The remaining point is injectivity. Since $\pi$ is a surjective submersion, for every $p\in P$ with $x=\pi(p)$ the [linear map](/page/Linear%20Map) \begin{align*} d\pi_p:T_pP\to T_xM \end{align*} is surjective. If a form $\eta\in \Omega^r(M)$ satisfies $\pi^*\eta=0$, then for any $x\in M$ choose $p\in P$ with $\pi(p)=x$, and for any $v_1,\dots,v_r\in T_xM$ choose lifts $\tilde v_1,\dots,\tilde v_r\in T_pP$ with $d\pi_p(\tilde v_i)=v_i$. Then \begin{align*} \eta_x(v_1,\dots,v_r)=(\pi^*\eta)_p(\tilde v_1,\dots,\tilde v_r)=0. \end{align*} Hence $\eta=0$. Applying this to $\eta=dP_0(F_A)_M$, it suffices to prove $d\omega=0$ on $P$.[/guided]

[step:Differentiate the invariant polynomial applied to curvature] The symmetric multilinear map $P_0:\mathfrak g^k\to \mathbb R$ extends to $\mathfrak g$-valued differential forms by wedging the form components and applying $P_0$ to the [Lie algebra](/page/Lie%20Algebra) components. Since $F_A$ has degree $2$, the graded Leibniz rule gives \begin{align*} d\omega=\sum_{j=1}^k P_0(F_A,\dots,F_A,dF_A,F_A,\dots,F_A), \end{align*} where $dF_A$ occurs in the $j$th slot. Let \begin{align*} d_A:\Omega^r(P;\mathfrak g)\to \Omega^{r+1}(P;\mathfrak g) \end{align*} denote the covariant exterior derivative determined by $A$, so that \begin{align*} d_A\alpha=d\alpha+[A\wedge \alpha] \end{align*} for every $\alpha\in \Omega^r(P;\mathfrak g)$, with the standard graded Lie-bracket convention. Applying this to $\alpha=F_A$ gives \begin{align*} dF_A=d_AF_A-[A\wedge F_A]. \end{align*} Substituting this identity into the preceding formula yields \begin{align*} d\omega=\sum_{j=1}^k P_0(F_A,\dots,d_AF_A,\dots,F_A)-\sum_{j=1}^k P_0(F_A,\dots,[A\wedge F_A],\dots,F_A). \end{align*} [/step]

[step:Use the Bianchi identity to remove the covariant derivative terms] The curvature form of a principal connection satisfies the Bianchi identity \begin{align*} d_AF_A=0. \end{align*} Therefore every term in the first sum vanishes, and we obtain \begin{align*} d\omega=-\sum_{j=1}^k P_0(F_A,\dots,[A\wedge F_A],\dots,F_A). \end{align*} [/step]

[step:Cancel the bracket terms by infinitesimal Ad-invariance] Because $P_0$ is $\operatorname{Ad}$-invariant, its infinitesimal invariance identity says that for every $X,Y_1,\dots,Y_k\in \mathfrak g$, \begin{align*} \sum_{j=1}^k P_0(Y_1,\dots,[X,Y_j],\dots,Y_k)=0. \end{align*} This is exactly [citetheorem:9759] applied to the symmetric polarisation of $P_0$. Apply this identity pointwise to the Lie algebra components of $A$ and $F_A$ in the extension of $P_0$ to differential forms. The wedge factors merely record the differential-form degrees, and since each curvature entry has even degree, no additional signs appear when the curvature slots are permuted. Hence \begin{align*} \sum_{j=1}^k P_0(F_A,\dots,[A\wedge F_A],\dots,F_A)=0. \end{align*} Consequently $d\omega=0$. [/step]

[step:Descend the vanishing to the base] From the first step, \begin{align*} \pi^*\bigl(dP_0(F_A)_M\bigr)=d\omega. \end{align*} The preceding steps prove $d\omega=0$, so \begin{align*} \pi^*\bigl(dP_0(F_A)_M\bigr)=0. \end{align*} Injectivity of pullback by the surjective submersion $\pi$ now gives \begin{align*} dP_0(F_A)_M=0. \end{align*} Thus the descended Chern-Weil form $P_0(F_A)_M$ is closed. [/step]