1.5-order formalism
= 1.5-order formalism
{title2=$\delta S/\delta\omega=0$}
Solve the algebraic connection equation for $\widehat\omega(e,\chi)$, but perform the remaining variations as though the connection were independent. The chain-rule contribution $(\delta S/\delta\omega)\delta\widehat\omega$ vanishes on that solution. The resulting equations are those of the connection-eliminated second-order <action>. This is especially useful for local <supersymmetry> calculations, where the gravitino-induced contorsion is quadratic in <fermions> and the omitted induced connection variation would otherwise be lengthy.