Proposition: Regression to XVI-J, including coincidence

Keep only $\mathsf{k}=\{0,0\}$ in (eq:UK). The result is exactly

the constant-slot distribution of XVI-J. Its smeared functional is ${{\mathcal{C}}^{\mathfrak{R}}}_{2e}(h,\phi )$ for every smooth $h,\phi$. These are equalities of distributions and functionals, not merely off-diagonal equalities.

Proof. Proposition prop:hess gives $L^{00}=L$, so (eq:Kpq) gives the same dual-basis tensor $\mathcal{K}$. The sign is positive and the unordered multiplicity is one, hence

For the numerical factor, use the same subtraction function $\varpi$, the same projection ${\mathsf{W}}_0\psi =\psi -\psi (0)\varpi$, and the same zero moment as XVI-J. Proposition prop:RK then gives ${\mathfrak{R}}_{\mathrm{K}}({\mathcal{S}}^2)={\bar{q}}^{\mathfrak{R}}$ globally. Their pullbacks and defined root products therefore coincide, including $y=w$, and so do their smooth pairings.