Endomorphism bundle connection ($\mathrm{End}(\nabla )$)

A connection with local matrix $A$ induces on an endomorphism-valued $r$-form $\mu$ the covariant exterior derivative

The endomorphism bundle connection is ${\tilde{D}}_X\alpha =D_X\circ \alpha -\alpha \circ D_X$ on sections. Expanding twice makes the two middle terms cancel, proving the displayed commutator identity. Subtracting the derivative along the Lie bracket of vector fields gives $\tilde{R}(X,Y)\alpha =[R(X,Y),\alpha ]$. The operators $D_X$ are differential operators, rather than fibrewise endomorphisms; the corrected curvature operator is fibrewise linear.

For the endomorphism bundle connection, $({\tilde{\nabla }}_X\varphi )(s)={\nabla }_X(\varphi s)-\varphi ({\nabla }_{X}s)$. Expanding two derivatives and subtracting ${\tilde{\nabla }}_{[X,Y]}$ givesThus its curvature form of a connection is the commutator action of the original curvature.

The endomorphism bundle connection of a rank-$r>0$ bundle is flat if and only if the original curvature satisfies $F_A=\omega \mathrm{id}$. 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 $\omega =r^{-1}\mathrm{tr}{F}_A$ is the required smooth two-form. The original connection can still have nonzero scalar curvature.