Proposition: Transported differential

The operator ${s_0}^{H^{\prime }}={\bar{\alpha }}_d{s}_0{\bar{\alpha }}_d^{-1}$ is a nilpotent, ghost-number-one derivation and preserves ${\mathcal{S}}_{H^{\prime }}$. It equals the symbol rule $s_0$ exactly when $\delta d=0$, with $\delta$ defined in (eq:differentialcondition).

Proof. Conjugation gives nilpotence, the derivation property, and sector stability. The symbol rule is a derivation for $H^{\prime }$ exactly when $\delta H^{\prime }=\delta d=0$. In that case its commutator with the self-contraction generator vanishes. Conversely equality on the quotient would make the symbol rule a quantum derivation there; its defect on two regular linear fields is a scalar testing $\delta d$. Scalars are not killed by the shell quotient, so all these tests must vanish.