Proposition: Nilpotence fails off

On $\mathfrak{F}$, for every local functional $F$ of fields and antifields,

For $F_v=\int c_{\mu }^{\ast }{v}^{\mu }{\mathrm{d}}^{4}x$ with real $v\in {C_c}^{\infty }({M_T}^{\circ };TM)$,

If $\mathrm{supp}v\subset W$, then ${s_{\mathrm{ext}}}^2{F}_v=0$ identically, consistently with Theorem thm:sext.

Proof. The first equality is the Jacobi identity with (eq:breaking); the second holds because $\mathfrak{K}$ is antifield-independent. For $F_v$, ${\delta }^{\mathrm{L}}{F}_v/\delta c_{\mu }^{\ast }=v^{\mu }$ and ${\delta }^{\mathrm{R}}\mathfrak{K}/\delta c^{\nu }=\kappa {\mathcal{K}}_{\nu }$ (the density ${\mathcal{K}}_{\nu }$ is even, so left and right derivatives agree). Evaluating at $\gamma =0$, $\varphi ={\varphi }_m$, where XIV-A computed ${\mathcal{K}}_{\nu }=({\partial }_{\mu }g){T_0}^{\mu }{}_{\nu }[{\varphi }_m]$ with ${T_0}^{00}=\frac{1}{2}$, ${T_0}^{0i}=0$, gives (eq:offFW-witness); the integral is strictly negative by the copied falling-switch lemma. For $\mathrm{supp}v\subset W$, ${\mathcal{K}}_{\nu }=0$ on $W$ by (eq:breaking).