Proposition: Boundary identity, corrected orientation

With $B^{\mu }=\sqrt{-g}(g^{\mu \nu }{{\Gamma }^{\lambda }}_{\nu \lambda }-g^{\lambda \nu }{{\Gamma }^{\mu }}_{\lambda \nu })$,

Proof. (i) Orientation. The statement above, with $-{\partial }_{\mu }{B}^{\mu }$, replaces the p09/p10 statement; it is used in conventions.BOUNDARY_IDENTITY, the exports and all certificates. (ii) Nontrivial conformal control. For $g=f\eta$ with $f$ a generic function of all four coordinates, sympy gives exactly $\sqrt{-g}R=-3\square f+\frac{3}{2f}(\partial f{)}^2$, $L_{\Gamma \Gamma }=\frac{3}{2f}(\partial f{)}^2$, $B^{\mu }=3{\partial }^{\mu }f$. The identity holds with residual $0$; the p10 orientation leaves $-6\square f\ne 0$. (iii) Generic jet proof. Let $G^{ab}$ be ten free symmetric symbols (the inverse metric), and let the 40 first jets $g_{ab,c}$ and 100 second jets $g_{ab,cd}$ be free. Use ${\partial }_c{G}^{ab}=-G^{ar}{g}_{rs,c}{G}^{sb}$ and ${\partial }_c\sqrt{-g}=\frac{1}{2}\sqrt{-g}{G}^{ab}{g}_{ab,c}$ (Jacobi; both checked exactly via $\partial \det =\mathrm{tr}(\mathrm{adj}\partial g)$ and $\mathrm{adj}g=\det g1$). Then $(\sqrt{-g}R-L_{\Gamma \Gamma }+{\partial }_{\mu }{B}^{\mu })/\sqrt{-g}$ is a polynomial in 150 variables over $\mathbb{Q}$ (2406, 1764 and 2442 terms in the three pieces) and is identically zero for all ten components, first and second jets; with the p10 orientation it has 2736 nonzero terms. Because the identity holds for free $G$, it holds a fortiori for $G=g^{-1}$.