Theorem: Exact off-shell field nilpotence

The operator $s$ is a ghost-number-one odd derivation, preserves the stated local and multilocal classical domains, and satisfies $s^2=0$.

Its zero-coupling field operator is exactly $s_0$.

Proof. For the odd vector field $c$, set

The graded coordinate calculation gives ${{\mathcal{L}}_c}^2={\mathcal{L}}_w$. Terms with symmetric second derivatives cancel against antisymmetric ghost coefficients. Therefore

Since $\bar{q}$ is fixed and formal multiplication by $\kappa$ is injective before truncation, $s^{2}r=0$.

Similarly,

Expanding $s(c^{\nu }{\partial }_{\nu }{c}^{\mu })$ gives zero after substituting $sc=\kappa c\partial c$. The nonminimal pair is nilpotent by definition.

Every fixed coefficient uses finitely many field factors and derivatives. Compactly smeared local terms remain compactly smeared local terms under $s$, and the Leibniz rule preserves finite multilocal products. The leading coefficient is (eq:s0).