Trace curvature transgression
= Trace curvature transgression
{title2=$\operatorname{Tr}F_{D_1}-\operatorname{Tr}F_{D_0}=d\operatorname{Tr}(D_1-D_0)$}
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.