Cartan curvature equation with input indices ($\Theta =d\theta -\theta \wedge \theta$)

Write $D{e}_i={\sum }_j{\theta }_{ij}\otimes e_j$, with the differentiated input index first. The graded Leibniz rule gives $D^2{e}_i={\sum }_j(d{\theta }_{ij}-{\sum }_k{\theta }_{ik}\wedge {\theta }_{kj})\otimes e_j$. For frames $e^{\prime }=ge$, the connection matrix transforms as ${\theta }^{\prime }=dg{g}^{-1}+g\theta g^{-1}$ and curvature by ${\Theta }^{\prime }=g\Theta g^{-1}$. Transposing to the usual coefficient-column convention gives $A={\theta }^t$ and $F=dA+A\wedge A={\Theta }^t$; the opposite sign is an index convention, not a different curvature.