 $$\begin{aligned}  {\frac{{\mathrm{d}}}{{\mathrm{d}}\Lambda}}\Sigma^\Lambda_{\alpha\beta} & = & \frac{0}{0\pi} \int{\mathrm{d}}\bar\omega \sum_{\mu\nu}    \mathcal{S}^\Lambda_{\mu\nu}(\bar\omega)    \Gamma^\Lambda_{\alpha\nu\beta\mu}