 Solution

Let another local frame be $e^{\prime }=eA$ with $A:U\to G{L}_r(\mathbb{R})$ smooth. Since the coefficient columns satisfy $u=A{u}^{\prime }$, the Leibniz rule gives the change of frame of a vector-bundle connection:The curvature form of a connection is linear over smooth functions in its section argument, so its matrix transforms homogeneously:ConsequentlyThe entries of $A$ are functions, of degree zero, so ordinary cyclic invariance of trace applies without a graded sign. The local forms therefore glue to a globally defined two-form $\alpha =\mathrm{tr}\Theta$, the trace of vector-bundle curvature.

Apply the Bianchi identity proved in part (b):For the first term,The middle interchange has sign $(-1{)}^{2\cdot 1}=1$, and the next equality simply renames $i,j$. ThusOne can also check closedness locally: the off-diagonal terms in $\mathrm{tr}(\theta \wedge \theta )={\sum }_{i,j}{\theta }_{ij}\wedge {\theta }_{ji}$ cancel in pairs, while the diagonal terms vanish. Hence $\alpha =d(\mathrm{tr}\theta )$ in each frame, and $d^2=0$ gives the same conclusion.