BV Density Bridge and Pushforward Master Defect

Put $f=\log \mu$. Because $f$ is field-only, ${\mathrm{\Delta }}_{0}f=0$ and $(f,f)=0$; the odd-Laplacian product identity therefore gives ${\mathrm{\Delta }}_{\rho }^2=0$ on this chart. The same semidensity has the two expressions $e^{-S_0\mathrm{/}\mathrm{\hslash }-I_1}\sqrt{{\rho }_{\mathrm{B}\mathrm{V}}}=e^{-S_0\mathrm{/}\mathrm{\hslash }-J_1}\sqrt{{\rho }_0}$ exactly when $J_1=I_1-f$. For even $S_0,f$, the shifted graded-skew rule makes $(S_0,f)=(f,S_0)$, and hence $(S_0,I_1)-{\mathrm{\Delta }}_{\rho }{S}_0=(S_0,J_1)-{\mathrm{\Delta }}_0{S}_0$. A metric derivative at fixed Darboux coordinates gives ${\delta }_{\sigma }{\mathrm{\Delta }}_{\rho }=({\delta }_{\sigma }f,-)$ and ${\delta }_{\sigma }{J}_1={\delta }_{\sigma }{I}_1-{\delta }_{\sigma }f$.

For an even full action $S$, set $A=\frac{1}{2}(S,S)-\mathrm{\hslash }\mathrm{\Delta }S$ and $\mathrm{\Psi }=e^{-S\mathrm{/}\mathrm{\hslash }}$. Applying the second-order product identity to the exponential gives $\mathrm{\Delta }\mathrm{\Psi }={\mathrm{\hslash }}^{-2}A\mathrm{\Psi }$. Split $\mathrm{\Delta }={\mathrm{\Delta }}_L+{\mathrm{\Delta }}_H$ in fixed product Darboux coordinates and integrate over the specified high cycle; because that cycle is fixed, ${\mathrm{\Delta }}_L$ commutes with the high integration. Define $e^{-S_L\mathrm{/}\mathrm{\hslash }}={\int }_H\mathrm{\Psi }$ and $A_L=\frac{1}{2}(S_L,S_L{)}_L-\mathrm{\hslash }{\mathrm{\Delta }}_L{S}_L$; the same exponential identity gives $A_L{e}^{-S_L\mathrm{/}\mathrm{\hslash }}={\mathrm{\hslash }}^2{\mathrm{\Delta }}_L{\int }_H\mathrm{\Psi }$. Therefore $A_L{e}^{-S_L\mathrm{/}\mathrm{\hslash }}={\int }_{H}A\mathrm{\Psi }-{\mathrm{\hslash }}^2{\int }_H{\mathrm{\Delta }}_H\mathrm{\Psi }$. The last integral is the exact boundary remainder; deleting it requires a separate zero-boundary hypothesis on the actual cycle. When it vanishes, $A_L=⟨A{⟩}_H$, where $⟨A{⟩}_H=({\int }_{H}A\mathrm{\Psi })\mathrm{/}({\int }_H\mathrm{\Psi })$.

Let the high Lagrangian be $y^{\ast }=0$ and suppose $S_0$ satisfies the classical master equation. On a branch where $H={\mathrm{\partial }}_y^2{S}_0$ is invertible, the formal implicit-function theorem gives a stationary section $y=y_{\ast }(x,x^{\ast })$ solving ${\mathrm{\partial }}_y{S}_0=0$. Define $S_{0,L}=S_0(x,y_{\ast };x^{\ast },0)$. Differentiating this composite in a low coordinate produces no term from $D{y}_{\ast }$ because its coefficient is ${\mathrm{\partial }}_y{S}_0=0$. In the full classical master equation, the high bracket term also contains ${\mathrm{\partial }}_y{S}_0$ and vanishes on the graph; the remaining low bracket is therefore $(S_{0,L},S_{0,L}{)}_L=0$. No induced low antifield vertex is discarded.

Let $I_{1,\mathrm{u}\mathrm{p}}$ be the upstream one-loop functional, ${\mu }_H$ the high density factor, and $H=D_y^2{S}_0{\mathrm{\mid }}_{y_{\ast }}$. Translating to $y=y_{\ast }+\eta$, the normalized formal Gaussian stationary-phase integral contributes $(\mathrm{sdet}H{)}^{-1\mathrm{/}2}$ and the high density contributes ${\mu }_H(x,y_{\ast })$; hence $J_{1,L}=I_{1,\mathrm{u}\mathrm{p}}(x,y_{\ast })+\frac{1}{2}\mathrm{STr}\log H-\log {\mu }_H(x,y_{\ast })$, up to the field-independent normalization absorbed in $I_{1,\mathrm{u}\mathrm{p}}$. Here $\mathrm{STr}$ counts even high coordinates positively and odd high coordinates negatively; all source derivatives use the fixed graded left/right ordering of the antibracket. Differentiating the stationarity equation gives $U_I:={\mathrm{\partial }}_{x_I^{\ast }}{y}_{\ast }=-H^{-1}{D}_y{\mathrm{\partial }}_{x_I^{\ast }}{S}_0$. The superdeterminant identity $D\log \mathrm{sdet}H=\mathrm{STr}(H^{-1}DH)$ then yields ${\mathrm{\partial }}_{x_I^{\ast }}{J}_{1,L}=D_{x_I^{\ast }}^{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}}(I_{1,\mathrm{u}\mathrm{p}}-\log {\mu }_H)+\frac{1}{2}\mathrm{STr}\text{\hspace{0.17em}⁣}\left[H^{-1}(D_y^2{\mathrm{\partial }}_{x_I^{\ast }}{S}_0+(D_y^3{S}_0)[U_I])\right]$, where $D_{x_I^{\ast }}^{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}}F={\mathrm{\partial }}_{x_I^{\ast }}F+D_{y}F[U_I]$ is the total derivative along the stationary section. Thus the cubic term and every explicit source/density derivative are included. If the upstream one-loop functional contains $S_{\mathrm{F}\mathrm{T}}$, its source derivative contributes ${\mathrm{\partial }}_{x_I^{\ast }}{S}_{\mathrm{F}\mathrm{T}}{\mathrm{\mid }}_y+D_y{S}_{\mathrm{F}\mathrm{T}}[U_I]$ inside the total derivative; neither term is promoted to a leading-action Hessian vertex.

In a fixed reference chart suppose $S_0$ obeys the classical master equation and write $s=(S_0,-)$ and $A_1=(S_0,J_1)-{\mathrm{\Delta }}_0{S}_0$. The graded Jacobi identity gives $s^2=0$. Applying ${\mathrm{\Delta }}_0$ to $(S_0,S_0)=0$ and using its second-order identity gives $s{\mathrm{\Delta }}_0{S}_0=0$; Jacobi also gives $s(S_0,J_1)=0$. Therefore $s{A}_1=0$. This is a consistency identity only. If a local $C_1$ is separately supplied with $s{C}_1=-A_1$, substitution into $S_0+\mathrm{\hslash }(J_1+C_1)$ cancels the coefficient of $\mathrm{\hslash }$ in $\frac{1}{2}(S,S)-\mathrm{\hslash }{\mathrm{\Delta }}_{0}S$, which is precisely the one-loop master equation. The lemma proves no existence or locality of such $C_1$.