Theorem: Relative off-shell nilpotence

The derivation (eq:nonlinear) satisfies $s^2=0$ on $\mathcal{P}(U)[[\kappa ]]$. Consequently its truncation through second order is nilpotent modulo ${\kappa }^3$, without imposing free or interacting equations.

Proof. For the odd vector field $c$, put $w^{\mu }=c^{\nu }{\partial }_{\nu }{c}^{\mu }$. Expansion of the tensor Lie derivative with the odd Leibniz rule gives

Terms containing symmetric second derivatives cancel against antisymmetric ghost coefficients. It follows from $sc=\kappa w$ and (eq:sq) that

Since $q=\bar{q}+\kappa r$ and $s\bar{q}=0$, this gives $s^{2}r=0$ coefficientwise.

For the scalar,

by the same cancellation. Direct expansion of $s(c^{\nu }{\partial }_{\nu }{c}^{\mu })$ gives zero after substitution of $sc=\kappa c\partial c$. The nonminimal pair is nilpotent by definition. The square is an even derivation, so the generator checks prove the assertion on the algebra.