Proposition: A smooth kernel invisible to all pure-trace pairs

All pure-trace evaluations fail to determine an unrestricted compact smooth three-slot kernel, even when symmetry under exchange of the two tensor slots is required. They also fail to determine its functional on a specified free solution family.

Proof. Let $A^{12}=A^{21}=1$, all other components zero. Then $A^{\mu \nu }{\eta }_{+,\mu \nu }=0$. For nonnegative nonzero $b,c\in {C_c}^{\infty }({\mathbb{R}}^4)$, define

This is nonzero and tensor-slot symmetric. Every contraction with

vanishes.

For a free comparison, choose both metric inputs with constant $h_{12}=h_{21}=1$ and all other components zero. They satisfy the componentwise free wave equation. Let $\varphi (t)=\cos (mt)$, $m>0$, which satisfies $({{\partial }_t}^2-\Delta +m^2)\varphi =0$. Choose the support of $c$ where $\cos (mt)>0$. The kernel evaluation is

Thus zero and $K$ agree on all pure-trace tests but differ on that free solution family.