Curvature difference formula ($F_{\nabla +a}=F_{\nabla }+d_{\nabla }a+a\wedge a$)

If two objects ${\nabla }_1$ and ${\nabla }_2$ are each a connection on a vector bundle and satisfy ${\nabla }_1={\nabla }_2+a$ for an $\mathrm{End}(E)$-valued one-form $a$, thenThis follows by expanding $({\nabla }_2+a{)}^2$ with the covariant exterior derivative on the endomorphism bundle connection.

The difference of two connections on a vector bundle is a global endomorphism-valued one-form. Taking the trace of the curvature difference formula cancels both the commutator terms and the trace of the square of that one-form. Consequently the trace curvatures represent the same de Rham cohomology class.

For a real or complex line bundle, its endomorphism bundle is canonically trivial by scalar multiplication. The Bianchi identity therefore reduces to $d{F}_A=0$. The difference of two connections on a vector bundle is a global endomorphism-valued one-form, which here is an ordinary scalar one-form $a$. Since scalar one-forms have vanishing wedge square, the curvature difference formula reduces to $F_{A+a}-F_A=da$. Thus the de Rham cohomology class is independent of the connection on a vector bundle.