Theorem: Nilpotent relative differential

On ${\mathfrak{A}}_W$ define $s_{\mathrm{ext}}F:=\left(S_{\mathrm{ext}},F\right)$. Then:

1. $s_{\mathrm{ext}}$ is odd, has ghost number $+1$, and is a graded derivation: $s_{\mathrm{ext}}(FG)=(s_{\mathrm{ext}}F)G+(-1{)}^{|F|}F{s}_{\mathrm{ext}}G$;

2. ${s_{\mathrm{ext}}}^2=0$ on ${\mathfrak{A}}_W$;

3. on fields and antifields,

   $$s_{\mathrm{ext}}{\Phi }^A=s{\Phi }^A,s_{\mathrm{ext}}{\gamma }^{\ast \mu \nu }=\frac{{\delta }^{\mathrm{R}}{S}_{\mathrm{ext}}}{\delta {\gamma }_{\mu \nu }}={e_{\mathrm{D}}}^{\mu \nu }+(\text{terms linear in antifields}),$$

   and similarly $s_{\mathrm{ext}}{\varphi }^{\ast }=E_{\varphi }+(\dots )$, with $s$ the kinematic rule (eq:kin).

Proof. (i) $S_{\mathrm{ext}}$ is even of ghost number zero, so the grading of (eq:antibracket) gives the stated parity and ghost number; the derivation property is the graded Leibniz rule of the antibracket in its second slot HT (Ch. 15). (ii) For even $S$ the graded Jacobi identity gives $\left(S,\left(S,F\right)\right)=\frac{1}{2}\left(\left(S,S\right),F\right)$. By the copied relative identity (eq:relative), $\left(S_{\mathrm{ext}},S_{\mathrm{ext}}\right)$ vanishes on ${\mathfrak{F}}_W$ together with its functional derivatives along ${\mathfrak{F}}_W$ (XIV-A's proof: the $\gamma$- and $\varphi$-derivatives retain the factor $c^{\nu }$, and the $c$-derivative is $\kappa {\mathcal{K}}_{\nu }$ restricted to $W$, which is zero). Hence $\left(\left(S_{\mathrm{ext}},S_{\mathrm{ext}}\right),F\right)=0$ for every $F\in {\mathfrak{A}}_W$ and ${s_{\mathrm{ext}}}^2=0$. (iii) With $G={\Phi }^A(x)$ only the second term of (eq:antibracket) survives: $s_{\mathrm{ext}}{\Phi }^A=-{\delta }^{\mathrm{R}}{S}_{\mathrm{ext}}/\delta {\Phi }_A^{\ast }$. For even ${\Phi }^A$ the antifield is odd, and the right derivative of $\int {\Phi }_A^{\ast }s{\Phi }^A$ with respect to ${\Phi }_A^{\ast }$ is $-s{\Phi }^A$, so $s_{\mathrm{ext}}{\Phi }^A=s{\Phi }^A$; for odd ${\Phi }^A$ the antifield is even and the same result holds without the intermediate sign. With $G={\Phi }_A^{\ast }(x)$ only the first term survives, giving $s_{\mathrm{ext}}{\Phi }_A^{\ast }={\delta }^{\mathrm{R}}{S}_{\mathrm{ext}}/\delta {\Phi }^A$, whose antifield-independent part is the Euler derivative of $S_{\mathrm{D}}$.