[proofplan]
We prove that $P_0(F_A)$ satisfies the two defining properties of a basic form on a principal bundle: horizontality and invariance under the right $G$-action. Horizontality follows because the curvature form $F_A$ is horizontal, so every term in the alternating Chern-Weil expression vanishes when one input vector is vertical. Invariance follows from the curvature equivariance identity $(R_g)^*F_A=\operatorname{Ad}_{g^{-1}}F_A$ and the $\operatorname{Ad}$-invariance of $P_0$. Finally, we recall and prove the descent criterion for basic forms using local sections, obtaining the unique form on $M$ whose pullback is $P_0(F_A)$.
[/proofplan]
[step:Write the Chern-Weil form using the alternating multilinear convention]
For each point $p\in P$, define the vertical subspace
\begin{align*}
V_pP:=\ker(d\pi_p)\subset T_pP.
\end{align*}
The curvature form $F_A\in \Omega^2(P;\mathfrak g)$ is horizontal, meaning that
\begin{align*}
F_A(p)(v,w)=0
\end{align*}
whenever $v\in V_pP$ and $w\in T_pP$.
For $k\ge 1$, the scalar $2k$-form $P_0(F_A)\in \Omega^{2k}(P)$ is defined by
\begin{align*}
P_0(F_A)_p(u_1,\dots,u_{2k})
=
\frac{1}{2^k k!}
\sum_{\sigma\in S_{2k}}
\operatorname{sgn}(\sigma)
P_0\bigl(
F_A(p)(u_{\sigma(1)},u_{\sigma(2)}),
\dots,
F_A(p)(u_{\sigma(2k-1)},u_{\sigma(2k)})
\bigr)
\end{align*}
for $p\in P$ and $u_1,\dots,u_{2k}\in T_pP$. For $k=0$, the expression is the constant function $P_0\in \Omega^0(P)$, and the same descent argument is immediate. We therefore treat $k\ge 1$ below.
[/step]
[step:Show that $P_0(F_A)$ is horizontal]
Let $p\in P$, let $u_1,\dots,u_{2k}\in T_pP$, and suppose that $u_j\in V_pP$ for some $j\in \{1,\dots,2k\}$. In each summand of the defining formula for $P_0(F_A)_p(u_1,\dots,u_{2k})$, the vector $u_j$ appears in exactly one argument of one curvature factor
\begin{align*}
F_A(p)(u_{\sigma(2\ell-1)},u_{\sigma(2\ell)})
\end{align*}
for some $\ell\in \{1,\dots,k\}$. Since $F_A$ is horizontal, that curvature factor is zero. By multilinearity of $P_0$, the whole summand is zero. Hence
\begin{align*}
P_0(F_A)_p(u_1,\dots,u_{2k})=0.
\end{align*}
Thus $P_0(F_A)$ is horizontal.
[guided]
We must check horizontality of the $2k$-form $P_0(F_A)$, so we take a point $p\in P$ and tangent vectors $u_1,\dots,u_{2k}\in T_pP$. Assume that one of them, say $u_j$, lies in the vertical subspace
\begin{align*}
V_pP:=\ker(d\pi_p).
\end{align*}
The Chern-Weil form is built by feeding pairs of tangent vectors into copies of the curvature form and then applying the $k$-[linear map](/page/Linear%20Map) $P_0$. In a typical summand, the vector $u_j$ occurs in exactly one of the paired curvature inputs:
\begin{align*}
F_A(p)(u_{\sigma(2\ell-1)},u_{\sigma(2\ell)})
\end{align*}
for a unique $\ell\in \{1,\dots,k\}$. The defining property of the curvature of a principal connection is that $F_A$ is horizontal: if either input is vertical, the value of $F_A$ is zero. Therefore the curvature factor containing $u_j$ vanishes.
Since $P_0:\mathfrak g^k\to \mathbb R$ is multilinear, if one of its $k$ inputs is zero, then the value of $P_0$ on that $k$-tuple is zero. Thus every summand in the alternating formula for $P_0(F_A)_p(u_1,\dots,u_{2k})$ is zero. Hence
\begin{align*}
P_0(F_A)_p(u_1,\dots,u_{2k})=0.
\end{align*}
This is exactly the horizontality condition for the form $P_0(F_A)$.
[/guided]
[/step]
[step:Show that $P_0(F_A)$ is invariant under the principal right action]
Fix $g\in G$. The curvature of a principal connection satisfies the equivariance identity
\begin{align*}
(R_g)^*F_A=\operatorname{Ad}_{g^{-1}}F_A,
\end{align*}
meaning that, for every $p\in P$ and $u,v\in T_pP$,
\begin{align*}
((R_g)^*F_A)_p(u,v)=\operatorname{Ad}_{g^{-1}}\bigl(F_A(p)(u,v)\bigr).
\end{align*}
Using the alternating formula for $P_0(F_A)$ and the naturality of pullback with respect to multilinear evaluation, we obtain
\begin{align*}
(R_g)^*P_0(F_A)=P_0\bigl((R_g)^*F_A\bigr).
\end{align*}
Substituting the curvature equivariance identity gives
\begin{align*}
(R_g)^*P_0(F_A)=P_0(\operatorname{Ad}_{g^{-1}}F_A,\dots,\operatorname{Ad}_{g^{-1}}F_A).
\end{align*}
Since $P_0$ is $\operatorname{Ad}$-invariant and $g^{-1}\in G$, the right-hand side equals $P_0(F_A)$. Therefore
\begin{align*}
(R_g)^*P_0(F_A)=P_0(F_A)
\end{align*}
for every $g\in G$, so $P_0(F_A)$ is $G$-invariant.
[/step]
[step:Descend the basic form to the base]
Let $\alpha:=P_0(F_A)\in \Omega^{2k}(P)$. The preceding steps show that $\alpha$ is horizontal and $G$-invariant, hence basic.
We now recall the descent argument. Let $U\subset M$ be an [open set](/page/Open%20Set) admitting a smooth local section $s:U\to P$ of $\pi$, and define
\begin{align*}
\beta_U:=s^*\alpha\in \Omega^{2k}(U).
\end{align*}
If $s':U\to P$ is another smooth local section, then there is a smooth map $h:U\to G$ such that
\begin{align*}
s'(x)=R_{h(x)}(s(x))
\end{align*}
for every $x\in U$. The derivative of $s'$ differs from the derivative of $R_{h(x)}\circ s$ by terms tangent to the $G$-orbits coming from the variation of $h$. Since $\alpha$ is horizontal, those vertical terms do not contribute. Since $\alpha$ is $G$-invariant, the remaining terms give
\begin{align*}
(s')^*\alpha=s^*\alpha.
\end{align*}
Thus $\beta_U$ is independent of the chosen local section.
On overlaps, the locally defined forms agree by the same independence argument, so they glue to a unique form
\begin{align*}
P_0(F_A)_M\in \Omega^{2k}(M).
\end{align*}
For every local section $s:U\to P$,
\begin{align*}
s^*P_0(F_A)=P_0(F_A)_M|_U.
\end{align*}
Pulling this identity back along $\pi:\pi^{-1}(U)\to U$ gives
\begin{align*}
\pi^*\bigl(P_0(F_A)_M|_U\bigr)=P_0(F_A)|_{\pi^{-1}(U)}.
\end{align*}
Since such open sets $U$ cover $M$, we obtain
\begin{align*}
\pi^*P_0(F_A)_M=P_0(F_A)
\end{align*}
on all of $P$.
Finally, if $\gamma\in \Omega^{2k}(M)$ also satisfies $\pi^*\gamma=P_0(F_A)$, then for every local section $s:U\to P$,
\begin{align*}
\gamma|_U=s^*\pi^*\gamma=s^*P_0(F_A)=P_0(F_A)_M|_U.
\end{align*}
The local sections cover $M$, so $\gamma=P_0(F_A)_M$. This proves both existence and uniqueness of the descended form.
[/step]
