Theorem: Off-shell field nilpotence

The derivation $s$ satisfies $s^2=0$ on the local formal field algebra. Its leading field operator is exactly (eq:s0).

Proof. For the odd vector field $c$, set $w^{\mu }=c^{\nu }{\partial }_{\nu }{c}^{\mu }$. Expanding the Lie derivative with the graded Leibniz rule gives

The second-derivative terms cancel by antisymmetry of the ghost coefficients. Therefore

Since $\bar{q}$ is fixed, $s^{2}r=0$ coefficientwise.

Similarly,

Expansion proves $s(c^{\nu }{\partial }_{\nu }{c}^{\mu })=0$ after substituting $sc=\kappa c\partial c$. The nonminimal pair is nilpotent by definition. The generator calculation extends to the algebra.