Proposition: A nonzero open-index vertex block

Suppose ${\chi }_{\mathrm{I}}\not\equiv 0$. There are a smooth real background $h$, tensor contractions of the open indices of ${{\mathcal{V}}^{00}}_{IJ}[h]$, and a compact test supported in a spacelike subset of $\Omega$, for which the resulting pairing is nonzero.

This does not imply that every source-contracted kernel ${U_A}^{\circ }[h]$, or the root-attached sum, is nonzero.

Proof. Choose $w_0$ with ${\chi }_{\mathrm{I}}(w_0)\ne 0$, and let

At $w_0$, the field and its first derivatives vanish, while

Take constant tensor directions

We calculate the Hessian at this jet rather than infer it from scaling degree. For the metric

put $a=1+\kappa s$, $b=a-\kappa t$. At $w_0$,

The supplied Ricci convention gives

Therefore the coefficient of $\kappa$ in the signed density $-2{\kappa }^{-2}\sqrt{-q_{\kappa }}R$ is

Indeed,

Thus

Every Hessian coefficient with $|a|+|b|>0$ has its remaining field differentiated at most once. Those coefficients vanish at $w_0$, because $h(w_0)=\partial h(w_0)=0$. Consequently all vertex derivative-edge corrections vanish at that point, not merely in a scaling estimate.

Choose open-index contractions whose images under the invertible matrix $\mathbb{P}$ are $A,C$. Choose

At this spacelike pair the selected component of ${\mathcal{V}}^{00}$ is exactly

The graph is a smooth real function on a neighborhood of this spacelike pair. By continuity, a sufficiently small nonnegative test supported there has a nonzero pairing with that component.