Lemma: Generator-level sign audit

For every field generator,

Moreover, $({\mathcal{H}}_s,{\mathcal{H}}_s)=0$ modulo divergences.

Proof. If $p=|{\Phi }^A|$ and $R^A=s{\Phi }^A$, the odd biderivation rule gives

The factor in (eq:H) therefore gives $R^A$ for both even and odd generators. In particular,

so an unsigned common antifield coupling would be wrong in this convention.

The cotangent-lift calculation gives $\frac{1}{2}({\mathcal{H}}_s,{\mathcal{H}}_s)$ linear in antifields, with coefficients $s^2{\Phi }^A$, after integration by parts. They vanish by Theorem thm:kinematic. There is no antifield-independent term.