Proposition: Gauge-fixed relative differential

Let $S_{\Psi }$ be the gauge-fixed action of Input statement input:XIVA and $s_{\Psi }:=\left(S_{\Psi },\cdot \right)$. Then $s_{\Psi }$ is an odd, ghost-number-one, nilpotent derivation of ${\mathfrak{A}}_W$; $s_{\Psi }{\Phi }^A=s{\Phi }^A$ for every field; and the antifield-degree-zero part of $S_{\Psi }$ is (eq:Sgf), whose quadratic part is XI-B's free gauge-fixed action with auxiliary equation $b=D\gamma$ and ghost term $i\bar{c}\square c$.

Proof. The antifield shift ${\Phi }^{\ast }\mapsto {\Phi }^{\ast }+\delta \Psi /\delta \Phi$ is a canonical transformation of (eq:antibracket) HT (Ch. 18); it preserves ${\mathfrak{F}}_W$ because $\Psi$ does not depend on $c$, so $c^{\ast }$ is unchanged, and the other shifts are compactly supported for compact $\bar{c}$. Hence $(S_{\Psi },S_{\Psi })$ is $(S_{\mathrm{ext}},S_{\mathrm{ext}})$ at shifted antifields, which is zero on ${\mathfrak{F}}_W$ by (eq:relative), and nilpotence follows as in Theorem thm:sext. The field rule is unchanged because $\delta S_{\Psi }/\delta {\Phi }^{\ast }=\delta S_{\mathrm{ext}}/\delta {\Phi }^{\ast }$ under a translation of the antifields. The display (eq:Sgf) is copied.