Proposition: Gauge fixing by XI-B's fermion

Let $\Psi =-i\int {\bar{c}}_{\mu }(2D^{\mu }\gamma -b^{\mu }){\mathrm{d}}^{4}x$ and $S_{\Psi }[\Phi ,{\Phi }^{\ast }]:=S_{\mathrm{ext}}[\Phi ,{\Phi }^{\ast }+\delta \Psi /\delta \Phi ]$. Then $(S_{\Psi },S_{\Psi })=\mathfrak{K}$ on $\mathfrak{F}$ and $(S_{\Psi },S_{\Psi })=0$ on ${\mathfrak{F}}_W$. At antifield degree zero,

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 shift ${\Phi }^{\ast }\mapsto {\Phi }^{\ast }+\delta \Psi /\delta \Phi$ is a canonical transformation of the antibracket HT (Ch. 18), so $(S_{\Psi },S_{\Psi })$ is $(S_{\mathrm{ext}},S_{\mathrm{ext}})$ evaluated at shifted antifields. Since $\mathfrak{K}$ does not depend on antifields, both statements follow from Theorems thm:breaking and thm:relative. For (eq:Sgf), $S_{\mathrm{ext}}$ at ${\Phi }^{\ast }=\delta \Psi /\delta \Phi$ equals $S_{\mathrm{D}}+s\Psi$, and with $DK=\square$, $s(2D^{\mu }\gamma -b^{\mu })=\square c^{\mu }+\kappa D^{\mu }({\mathcal{L}}_c\gamma )$, so $s\Psi =\int [b\cdot (2D\gamma -b)+i\bar{c}\cdot (\square c+\kappa D{\mathcal{L}}_c\gamma )]$. The quadratic part is XI-B's $S_{\mathrm{FP}}+S_{\varphi }+s_0\Psi$, since the cubic-only modification does not change the free Hessian (XII-C).