Proposition: Extension, moments, and basis independence

The map ${\mathfrak{R}}_{\mathrm{K}}:\mathcal{V}\to {\mathcal{D}}^{\prime }({\mathbb{R}}^4)$ is linear, well defined, and satisfies

For $u\in {\mathcal{H}}_k$,

These conditions determine the extension uniquely.

Every constant-coefficient relation among the punctured kernels is preserved:

In particular, the extended version of (eq:trace) is zero.

Finally,

the selected square extension of XVI-J, with the same $\varpi$, order-zero projection, and zero moment.

Proof. Lemma lem:vanishing and the continuous projection give a continuous distributional action, linear in $u$. Tests supported away from zero have zero Taylor jet and recover the punctured distribution. The direct grading makes the definition on $\mathcal{V}$ unambiguous.

The construction is the zero-moment point extension with projection (eq:Wrho); the scaling bound follows from BF (Theorem 5.3 and equations (38)–(39)). The moments hold because ${\mathsf{W}}_{\rho }{w^{(\rho )}}_{\alpha }=0$. For a nonzero $u\in {\mathcal{H}}_k$, its restriction already has scaling degree $k$, so the extension has degree exactly $k$. For $u=0$, the selected extension is zero.

Differences of extensions with the stated bound are point-supported distributions

Pairing with the dual Taylor tests makes all their coefficients zero if the moments agree, proving uniqueness. Linearity proves (eq:coherence), independently of any basis chosen to compute the finite span.

For ${\mathcal{S}}^2$, the data are exactly (eq:W0) and $⟨{\mathfrak{R}}_{\mathrm{K}}({\mathcal{S}}^2),\varpi ⟩=0$. Uniqueness at scaling degree four therefore proves (eq:Jextension) on all of ${\mathbb{R}}^4$.