Proposition: Pure-trace data do not determine the smooth kernel or its free-shell functional

There is a nonzero smooth compact kernel symmetric under exchange of its tensor slots, invisible to every pure-trace pair. It can have a nonzero evaluation on a declared free metric–scalar solution family.

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

Every pure-trace evaluation vanishes, whereas $K\ne 0$.

For an explicit free-shell comparison, let each metric input have constant components $h_{12}=h_{21}=1$ and all other components zero. These are smooth solutions of $\square h_{\mu \nu }=0$. Let

so $(\square +m^2)\varphi =0$. Choose the time support of $c$ where $\cos (mt)>0$. The evaluation is

Thus zero and $K$ also define different functionals on this specified free solution family.