Lemma: Antifield-degree decomposition

For $S^{\prime }\in {\mathfrak{S}}_{\mathrm{kin}}$,

Proof. $S_{\mathrm{D}}$ depends only on the even fields $\gamma ,\varphi$, so its left and right derivatives coincide with the ordinary Euler derivatives $e_{\mathrm{D}},E_{\varphi }$. In (eq:antibracket) with $F=G=S_{\mathrm{ext}}$, the antifield-degree-zero part is produced only by pairing a derivative of $S_{\mathrm{D}}$ with ${\delta }^{\mathrm{L}}{S}_{\mathrm{ext}}/\delta {\Phi }_A^{\ast }=s{\Phi }^A$, or with ${\delta }^{\mathrm{R}}{S}_{\mathrm{ext}}/\delta {\Phi }_A^{\ast }$; moving the odd antifield past the odd $s{\Phi }^A$ for even ${\Phi }^A$ gives the sign that makes the two terms equal, yielding $2\int (e_{\mathrm{D}}\cdot s\gamma +E_{\varphi }s\varphi )$. The antifield-degree-one part is a sum of terms $\pm 2\int {\Phi }_A^{\ast }{s}^2{\Phi }^A$, which vanish by Lemma lem:nil. There is no degree-two part because $S_{\mathrm{ext}}$ is linear in antifields. Adding ${S^{\prime }}_{\ge 2}$ changes $(S^{\prime },S^{\prime })$ only in antifield degree $\ge 1$. This is the standard decomposition of HT (Ch. 17), GPS.