 BRST-exact covariant gauge fixing

For a left left-acting BRST differential with $sA=Dc$, $s\bar{c}=b$, $sb=0$, choose $\Psi ={\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded Leibniz rule gives $s\Psi =b\partial \cdot A+\xi b^2/2-\bar{c}\partial \cdot Dc$. Integrating the auxiliary Gaussian yields $-(\partial \cdot A{)}^2/(2\xi )-\bar{c}\partial \cdot Dc$. At $\xi =1$ this is Feynman gauge. Opposite antighost or exact-action sign conventions require a corresponding change of fermion.

// Target: relativistic-quantum-field.bigb