[proofplan]
The key geometric input is the Second Bianchi Identity, which asserts that the induced connection on $\mathrm{End}(E)$ annihilates the curvature: $d_{\mathrm{End}(E)} F_A = 0$. Working locally in a trivialising chart, we verify that $d(\mathrm{tr}(F_A^m)) = \mathrm{tr}(d(F_A)_\alpha^m)$ where the right side is the ordinary de Rham differential of a matrix-valued form, and then replace $d(F_A)_\alpha$ using the local Bianchi identity $d(F_A)_\alpha = [(F_A)_\alpha, \theta_\alpha]$. The resulting expression is a sum of traces of graded commutators, each of which vanishes by the cyclic invariance of the trace combined with sign tracking for wedge products of even-degree forms. Globalising via the trace's coordinate-independence yields $d(\mathrm{tr}(F_A^m)) = 0$ on $M$, so $\mathrm{tr}(F_A^m)$ is closed and represents a well-defined de Rham class.
[/proofplan]
[step:Reduce the differential of the trace to the trace of the local differential]
Fix a trivialising chart $U_\alpha$ with local frame and connection matrix $\theta_\alpha \in \Omega^1(U_\alpha; \mathrm{End}(\mathbb{R}^r))$, so by the [Local Formula for Curvature](/theorems/1540),
\begin{align*}
(F_A)_\alpha = d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha \;\in\; \Omega^2(U_\alpha; \mathrm{End}(\mathbb{R}^r)).
\end{align*}
Let $\Omega_\alpha := (F_A)_\alpha$. The power $\Omega_\alpha^m \in \Omega^{2m}(U_\alpha; \mathrm{End}(\mathbb{R}^r))$ is well-defined using matrix multiplication on the $\mathrm{End}(\mathbb{R}^r)$ factor and wedge product on the form factor. The trace $\mathrm{tr}: \mathrm{End}(\mathbb{R}^r) \to \mathbb{R}$ extends to a $C^\infty(U_\alpha)$-linear map on $\Omega^k(U_\alpha; \mathrm{End}(\mathbb{R}^r)) \to \Omega^k(U_\alpha)$ by applying it to the matrix factor pointwise.
Since $\mathrm{tr}$ is $\mathbb{R}$-linear and commutes with the scalar exterior derivative,
\begin{align*}
d\big(\mathrm{tr}(\Omega_\alpha^m)\big) = \mathrm{tr}\big(d(\Omega_\alpha^m)\big),
\end{align*}
where $d$ on the right is the component-wise de Rham differential of a matrix-valued form.
[/step]
[step:Compute $d(\Omega_\alpha^m)$ using the graded Leibniz rule for matrix-valued forms]
The matrix-valued de Rham differential satisfies the graded Leibniz rule: for $\eta \in \Omega^p(U_\alpha; \mathrm{End}(\mathbb{R}^r))$ and $\zeta \in \Omega^q(U_\alpha; \mathrm{End}(\mathbb{R}^r))$,
\begin{align*}
d(\eta \wedge \zeta) = d\eta \wedge \zeta + (-1)^p\, \eta \wedge d\zeta,
\end{align*}
where $\wedge$ combines wedge product with matrix multiplication (and is not skew-symmetric as matrices). Since $\Omega_\alpha$ has form degree 2 (even), the sign $(-1)^2 = 1$ is unity in every step, so iterating the Leibniz rule:
\begin{align*}
d(\Omega_\alpha^m) = \sum_{k=0}^{m-1} \Omega_\alpha^k \wedge d\Omega_\alpha \wedge \Omega_\alpha^{m-1-k}.
\end{align*}
[guided]
We compute the exterior derivative of the matrix power $\Omega_\alpha^m$ using the graded Leibniz rule. The rule for matrix-valued forms reads
\begin{align*}
d(\eta \wedge \zeta) = d\eta \wedge \zeta + (-1)^{\deg \eta}\, \eta \wedge d\zeta,
\end{align*}
and iterating over the $m$ factors of $\Omega_\alpha^m = \Omega_\alpha \wedge \cdots \wedge \Omega_\alpha$ produces
\begin{align*}
d(\Omega_\alpha^m) = \sum_{k=0}^{m-1} (-1)^{2k}\, \Omega_\alpha^k \wedge d\Omega_\alpha \wedge \Omega_\alpha^{m-1-k} = \sum_{k=0}^{m-1} \Omega_\alpha^k \wedge d\Omega_\alpha \wedge \Omega_\alpha^{m-1-k},
\end{align*}
where the sign $(-1)^{2k}$ comes from moving $d$ past the first $k$ copies of $\Omega_\alpha$ (each of degree 2), and always evaluates to $+1$.
The simplification here is the key combinatorial reason Chern-Weil works for the trace of powers of a **curvature** form (degree 2), but is more delicate for general Lie-algebra-valued forms: even form degree ensures no signs, so the formula looks exactly like the Leibniz rule for a commutative algebra.
[/guided]
[/step]
[step:Apply the local form of the Second Bianchi Identity to substitute $d\Omega_\alpha$]
The [Second Bianchi Identity](/theorems/???), in its local form, states
\begin{align*}
d\Omega_\alpha = \Omega_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha = [\Omega_\alpha, \theta_\alpha],
\end{align*}
where $[\cdot, \cdot]$ denotes the graded commutator: for $\eta \in \Omega^p(U_\alpha; \mathrm{End}(\mathbb{R}^r))$ and $\zeta \in \Omega^q(U_\alpha; \mathrm{End}(\mathbb{R}^r))$,
\begin{align*}
[\eta, \zeta] = \eta \wedge \zeta - (-1)^{pq}\, \zeta \wedge \eta,
\end{align*}
and here $p = 2$, $q = 1$, so $(-1)^{pq} = (-1)^2 = 1$ and $[\Omega_\alpha, \theta_\alpha] = \Omega_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha$.
Let us verify this local identity directly. From $\Omega_\alpha = d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha$,
\begin{align*}
d\Omega_\alpha = d(d\theta_\alpha) + d(\theta_\alpha \wedge \theta_\alpha) = 0 + d\theta_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge d\theta_\alpha,
\end{align*}
using $d^2 = 0$ and the graded Leibniz rule with $\deg \theta_\alpha = 1$ (yielding the sign $-1$). Now substitute $d\theta_\alpha = \Omega_\alpha - \theta_\alpha \wedge \theta_\alpha$:
\begin{align*}
d\Omega_\alpha &= (\Omega_\alpha - \theta_\alpha \wedge \theta_\alpha) \wedge \theta_\alpha - \theta_\alpha \wedge (\Omega_\alpha - \theta_\alpha \wedge \theta_\alpha) \\
&= \Omega_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge \theta_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha + \theta_\alpha \wedge \theta_\alpha \wedge \theta_\alpha \\
&= \Omega_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha = [\Omega_\alpha, \theta_\alpha].
\end{align*}
[/step]
[step:Substitute into the sum and regroup traces of graded commutators]
Combining Steps 2 and 3:
\begin{align*}
d(\Omega_\alpha^m) = \sum_{k=0}^{m-1} \Omega_\alpha^k \wedge [\Omega_\alpha, \theta_\alpha] \wedge \Omega_\alpha^{m-1-k} = \sum_{k=0}^{m-1} \Omega_\alpha^k \wedge (\Omega_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha) \wedge \Omega_\alpha^{m-1-k}.
\end{align*}
Distributing and using associativity of $\wedge$:
\begin{align*}
d(\Omega_\alpha^m) = \sum_{k=0}^{m-1} \Omega_\alpha^{k+1} \wedge \theta_\alpha \wedge \Omega_\alpha^{m-1-k} - \sum_{k=0}^{m-1} \Omega_\alpha^k \wedge \theta_\alpha \wedge \Omega_\alpha^{m-k}.
\end{align*}
Reindexing the first sum by $j = k + 1$:
\begin{align*}
\sum_{k=0}^{m-1} \Omega_\alpha^{k+1} \wedge \theta_\alpha \wedge \Omega_\alpha^{m-1-k} = \sum_{j=1}^{m} \Omega_\alpha^j \wedge \theta_\alpha \wedge \Omega_\alpha^{m-j}.
\end{align*}
The second sum, reindexed by $j = k$, reads $\sum_{j=0}^{m-1} \Omega_\alpha^j \wedge \theta_\alpha \wedge \Omega_\alpha^{m-j}$. Subtracting, a telescoping cancellation leaves only the boundary terms $j = m$ and $j = 0$:
\begin{align*}
d(\Omega_\alpha^m) = \Omega_\alpha^m \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha^m = [\Omega_\alpha^m, \theta_\alpha],
\end{align*}
where the last equality uses the graded commutator with $\deg \Omega_\alpha^m = 2m$ (even) and $\deg \theta_\alpha = 1$, so $(-1)^{2m \cdot 1} = 1$.
[guided]
We now assemble the computation. From Step 2 we have
\begin{align*}
d(\Omega_\alpha^m) = \sum_{k=0}^{m-1} \Omega_\alpha^k \wedge d\Omega_\alpha \wedge \Omega_\alpha^{m-1-k},
\end{align*}
and from Step 3, $d\Omega_\alpha = \Omega_\alpha \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha$. Substituting and distributing:
\begin{align*}
d(\Omega_\alpha^m) = \sum_{k=0}^{m-1} \Omega_\alpha^{k+1} \wedge \theta_\alpha \wedge \Omega_\alpha^{m-1-k} - \sum_{k=0}^{m-1} \Omega_\alpha^k \wedge \theta_\alpha \wedge \Omega_\alpha^{m-k}.
\end{align*}
This is a telescoping sum. Reindex the first sum with $j = k+1$ (so $j$ runs from 1 to $m$):
\begin{align*}
\sum_{j=1}^{m} \Omega_\alpha^j \wedge \theta_\alpha \wedge \Omega_\alpha^{m-j} - \sum_{j=0}^{m-1} \Omega_\alpha^j \wedge \theta_\alpha \wedge \Omega_\alpha^{m-j}.
\end{align*}
All interior $j = 1, \ldots, m-1$ terms appear in both sums with opposite signs and cancel. Only the $j = m$ term (from the first sum) and the $j = 0$ term (from the second, with a negative sign) survive:
\begin{align*}
d(\Omega_\alpha^m) = \Omega_\alpha^m \wedge \theta_\alpha \wedge \Omega_\alpha^0 - \Omega_\alpha^0 \wedge \theta_\alpha \wedge \Omega_\alpha^m = \Omega_\alpha^m \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha^m.
\end{align*}
This is the graded commutator $[\Omega_\alpha^m, \theta_\alpha]$ (with sign $+1$, since $\deg \Omega_\alpha^m = 2m$ is even).
This is a useful algebraic fact in its own right: for matrix-valued forms of even degree, $d(\eta^m) = [\eta^m, \omega]$ whenever $d\eta = [\eta, \omega]$.
[/guided]
[/step]
[step:Apply the trace and use graded cyclic invariance to conclude]
Taking the trace:
\begin{align*}
d\big(\mathrm{tr}(\Omega_\alpha^m)\big) = \mathrm{tr}\big(d(\Omega_\alpha^m)\big) = \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha^m) = \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha) - \mathrm{tr}(\theta_\alpha \wedge \Omega_\alpha^m).
\end{align*}
The graded cyclic property of the trace for matrix-valued forms says: for $\eta \in \Omega^p(U_\alpha; \mathrm{End}(\mathbb{R}^r))$ and $\zeta \in \Omega^q(U_\alpha; \mathrm{End}(\mathbb{R}^r))$,
\begin{align*}
\mathrm{tr}(\eta \wedge \zeta) = (-1)^{pq}\, \mathrm{tr}(\zeta \wedge \eta).
\end{align*}
Applying this with $\eta = \theta_\alpha$ (degree 1) and $\zeta = \Omega_\alpha^m$ (degree $2m$):
\begin{align*}
\mathrm{tr}(\theta_\alpha \wedge \Omega_\alpha^m) = (-1)^{1 \cdot 2m}\, \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha) = \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha),
\end{align*}
since $2m$ is even. Substituting,
\begin{align*}
d\big(\mathrm{tr}(\Omega_\alpha^m)\big) = \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha) - \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha) = 0.
\end{align*}
[guided]
We now wrap up by taking traces and using cyclic invariance. Recall the graded cyclic identity for traces of matrix-valued forms
\begin{align*}
\mathrm{tr}(\eta \wedge \zeta) = (-1)^{pq}\, \mathrm{tr}(\zeta \wedge \eta), \quad \deg \eta = p, \; \deg \zeta = q,
\end{align*}
which reflects the combination of the matrix trace identity $\mathrm{tr}(AB) = \mathrm{tr}(BA)$ with the graded commutativity of wedge on scalar forms.
Applied to $\eta = \theta_\alpha$ ($p = 1$) and $\zeta = \Omega_\alpha^m$ ($q = 2m$), the sign is $(-1)^{2m} = 1$, so
\begin{align*}
\mathrm{tr}(\theta_\alpha \wedge \Omega_\alpha^m) = \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha).
\end{align*}
Applying the trace to the formula $d(\Omega_\alpha^m) = \Omega_\alpha^m \wedge \theta_\alpha - \theta_\alpha \wedge \Omega_\alpha^m$ from Step 4:
\begin{align*}
\mathrm{tr}(d(\Omega_\alpha^m)) = \mathrm{tr}(\Omega_\alpha^m \wedge \theta_\alpha) - \mathrm{tr}(\theta_\alpha \wedge \Omega_\alpha^m) = 0.
\end{align*}
Combined with Step 1's identity $d(\mathrm{tr}(\Omega_\alpha^m)) = \mathrm{tr}(d(\Omega_\alpha^m))$, we obtain $d(\mathrm{tr}(\Omega_\alpha^m)) = 0$ locally.
Why does this require $2m$ even? If $\Omega_\alpha$ had odd form degree, the cyclic sign would be negative and the two trace terms would add rather than cancel, and the argument would break. This is why Chern-Weil theory uses powers of the **curvature** (degree 2) rather than powers of the **connection** 1-form (degree 1) — the even degree is what guarantees closedness.
[/guided]
[/step]
[step:Globalise to $M$ and conclude the cohomology class is well-defined]
The local expression $\mathrm{tr}(\Omega_\alpha^m) \in \Omega^{2m}(U_\alpha)$ is invariant under change of trivialisation: on overlaps $U_\alpha \cap U_\beta$, the curvature transforms by $\Omega_\alpha = \psi_{\alpha\beta}\, \Omega_\beta\, \psi_{\alpha\beta}^{-1}$ (from the [Local Formula for Curvature](/theorems/1540) transformation), hence $\Omega_\alpha^m = \psi_{\alpha\beta}\, \Omega_\beta^m\, \psi_{\alpha\beta}^{-1}$, and cyclic invariance of the matrix trace gives
\begin{align*}
\mathrm{tr}(\Omega_\alpha^m) = \mathrm{tr}(\psi_{\alpha\beta}\, \Omega_\beta^m\, \psi_{\alpha\beta}^{-1}) = \mathrm{tr}(\Omega_\beta^m).
\end{align*}
So the local forms patch into a well-defined global 2$m$-form $\mathrm{tr}(F_A^m) \in \Omega^{2m}(M)$. By the previous step, $d\big(\mathrm{tr}(F_A^m)\big) = 0$ on every chart $U_\alpha$, hence globally on $M$. Therefore $\mathrm{tr}(F_A^m)$ is closed, and its de Rham cohomology class
\begin{align*}
[\mathrm{tr}(F_A^m)] \in H^{2m}_{dR}(M)
\end{align*}
is well-defined, completing the proof.
[/step]
