Proposition: Failure of the characteristic-cone test

For every nonzero $f\in {C_c}^{\infty }(M_{+})$,

In particular, its full mixed kernel cannot be admitted merely by replacing the causal exclusion cones by the null exclusion cones.

Proof. Fix $\mathbf{n}\in S^2$, let $r>0$, and consider the smooth complexified configuration

A direct use of (eq:C) gives

Solving the zero-Cauchy equation yields

Choose $a\in {C_c}^{\infty }(M_{+})$ with $\int fa{\mathrm{d}}^{4}x\ne 0$, and set

Substitution into (eq:B) gives

Indeed, the nonoscillating term is obtained from

all terms containing $e^{-2irt}$ decrease rapidly after the $t$-integration, including the term with the prefactor $rt$. The term containing ${\partial }_{t}a$ is $O(r^{-1})$.

This is a Fourier test of the compact kernel $a(x)b_f(x,z)$, contracted with the constant tensor $P_{\alpha \beta 00}$, along a ray for which both slot covectors are null and have the same temporal orientation. If the characteristic-cone condition held, compactness and the definition of the wavefront set would imply rapid decrease along that ray. The nonzero limit contradicts it.

If desired, the plane wave in the graviton slot can be multiplied by a compact cutoff equal to one on a neighborhood of ${\mathcal{D}}_{\mathrm{supp}f}$, without changing the calculation.