Theorem: Evaluated detecting coefficient

For (eq:witness-test),

The mass-vertex contribution to this test coefficient is zero. The formula holds for every $m\ge 0$.

Proof. In the positive seed chart, $r=-t$. At $q=0$,

Thus the two $t$-derivatives must both fall on $q^2/2$ for a contribution at the incidence. The remaining factor $|\xi {|}^2/2$ makes $T_{J,\beta \rho }=0$, so the mass term in (eq:positive-quadratic) vanishes.

Also $T_{J,\beta \rho \nu }=0$. Therefore the derivative of $f$ contributes zero, and the surviving kinetic term has one $s$-derivative and one output derivative on $|\xi {|}^2/2$. At $\Sigma$, with $c,t,y$ fixed during $s$-differentiation,

The scalar $\alpha /\beta$ here is the ratio of characteristic coordinates. Consequently

The explicit tensor (eq:Bkin) gives

Hence the bracketed kinetic derivative in (eq:positive-quadratic), after summing the output direction, becomes

Since $\mathsf{A}(\eta {)}^{\rho \mu }={\eta }^{\rho \mu }$,

Finally $\ell \cdot r=-ϵ(1-\mathbf{n}\cdot \omega )$. Substitute into (eq:positive-quadratic) to obtain (eq:witness-value).

No cutoff or tangential-amplitude derivative survives this evaluation: a nonzero incidence jet must use both $t$-derivatives on $q^2/2$ and the $s,y$-derivatives on $|\xi {|}^2/2$. This also verifies that no omitted localization derivative affects the coefficient.

Proof. The hypotheses and continuity permit a sufficiently small isolated tangential neighborhood where the continuous complex density

is nonzero and has values in a fixed open half-plane after multiplication by one complex phase. Choose a nonnegative nonzero $\psi$ there. All other factors in (eq:witness-value) are positive: $\alpha ,\beta ,ϵ>0$ and $1-\mathbf{n}\cdot \omega >0$. Thus $c_{\ast }=A_{-2,5+}({\Phi }_{\ast })\ne 0$.

Support isolation makes the complete pairing equal to the $5+$ pairing. The controlled non-tip expansion gives (eq:failure-main). Its inverse-quadratic term cannot be canceled by lower orders. The fixed complete-family pairing is unbounded. A distributional limit would have finite pairings on every fixed test, so no raw limit exists.