In coordinates, expanding $d\alpha$ and the <Lie bracket of vector fields> shows
$$
d\alpha(X,Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X,Y]).
$$
The terms differentiating components of $X,Y$ cancel against the bracket, leaving $(\partial_i\alpha_j-\partial_j\alpha_i)X^iY^j$. This is the <exterior derivative of a one-form evaluated on vector fields>.

A <connection on a vector bundle> is a linear map $\nabla:\Gamma(E)\to\Omega^1(E)$ satisfying $\nabla(fs)=df\otimes s+f\nabla s$; evaluation on $X$ defines $\nabla_Xs$. Its <curvature form of a connection> is $d_A^2$, locally $F_A=dA+A\wedge A$. Direct expansion, using the displayed exterior-derivative identity, gives
$$
\boxed{F_A(X,Y)s=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s.}
$$
The derivative terms on a scalar multiplying $s$ cancel, and the expression is also linear over smooth functions in $X,Y$. Thus it is an alternating tensor with values in $\operatorname{End}(E)$, namely an element of $\Omega^2(\operatorname{End}(E))$.

The <dual connection> and <tensor product connection> induce the <endomorphism bundle connection> on $E^*\otimes E$:
$$
(\widetilde\nabla_X\phi)(s)=\nabla_X(\phi s)-\phi(\nabla_Xs).
$$
Expanding twice cancels the cross terms and gives
$$
([\widetilde\nabla_X,\widetilde\nabla_Y]\phi)(s)
=[\nabla_X,\nabla_Y](\phi s)-\phi([\nabla_X,\nabla_Y]s).
$$
Here $\phi,s$ are local smooth sections; derivatives are not defined for isolated fiber elements without extensions. Subtract the corresponding $\widetilde\nabla_{[X,Y]}$ identity to get the <curvature of an endomorphism bundle connection>:
$$
\boxed{F_{\operatorname{End}(A)}(X,Y)\phi=[F_A(X,Y),\phi].}
$$
This vanishes for all $\phi$ exactly when each $F_A(X,Y)$ is central in the full matrix algebra, hence scalar. For rank $r>0$, $\omega=r^{-1}\operatorname{tr}F_A$ is a smooth two-form and
$$
\boxed{F_{\operatorname{End}(A)}=0\quad\Longleftrightarrow\quad F_A=\omega\,\operatorname{id}_E.}
$$
Conversely scalar curvature commutes with every endomorphism. This is the <scalar-curvature criterion for a flat endomorphism connection>; it permits a nonflat connection on $E$ itself.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 115 / 3 / Solution

The <endomorphism bundle connection> of a rank-$r>0$ bundle is flat if and only if the original curvature satisfies $F_A=\omega\,\operatorname{id}$. By the <curvature of an endomorphism bundle connection>, flatness means each $F_A(X,Y)$ commutes with every fiber endomorphism. The centre of the full matrix algebra consists of scalar matrices, and $\omega=r^{-1}\operatorname{tr}F_A$ is the required smooth two-form. The original connection can still have nonzero scalar curvature.

Scalar-curvature criterion for a flat endomorphism connection

{title2=$F_{\operatorname{End}(A)}\phi=[F_A,\phi]$}

For the <endomorphism bundle connection>, $(\widetilde\nabla_X\phi)(s)=\nabla_X(\phi s)-\phi(\nabla_Xs)$. Expanding two derivatives and subtracting $\widetilde\nabla_{[X,Y]}$ gives
$$
F_{\operatorname{End}(A)}(X,Y)\phi
=F_A(X,Y)\circ\phi-\phi\circ F_A(X,Y).
$$
Thus its <curvature form of a connection> is the commutator action of the original curvature.

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 Incoming links: Curvature of an endomorphism bundle connection