Proposition: Termwise scaling bounds

The nonzero numerical products ${p_{\mu \nu }}^{\circ }$ of (eq:p) are homogeneous on the punctured domain of degree

and therefore have scaling degree

For zero products the corresponding inequality suffices.

The vertex-only blocks with $\alpha =\beta =0$ have local transverse scaling degree at most six. Including the insertion source derivatives gives

locally, uniformly on compact tangential sets. This is an upper bound before root attachment, not a sharp degree of the total tensor sum.

Proof. The massless $d$ is homogeneous of degree $-2$. Its derivative of order $|\mu |$ has degree $-2-|\mu |$. The punctured product is already defined by Proposition prop:product; rescaling respects that product. Hence its degree is the sum of the two degrees.

For $\alpha =\beta =0$, the Hessian satisfies $|a|+|b|\le 2$, so the numerical edge degree is at most six. With the insertion derivatives included,

This gives the bound eight.

In relative coordinates, all remaining coefficients are smooth functions of $(r,w)$. Their derivatives are bounded on the compact sets used in a local transverse scaling test. Smooth multiplication cannot increase the bound there. Finite summation completes the estimate.