Lemma: Polynomial transversality

Every tensor $\mathsf{E},{\mathsf{Q}}_2,{\mathsf{Q}}_0$ is symmetric in each index pair and under exchange of the pairs. Each satisfies

as a polynomial identity on all of ${\mathbb{R}}^4$.

Proof. For ${\mathsf{Q}}_2,{\mathsf{Q}}_0$, use $p^a{L}_{ab}=0$. For $\mathsf{E}$, contraction of (eq:E) cancels the terms proportional to $p_b{\eta }_{cd}$, $p_c{\eta }_{bd}+p_d{\eta }_{bc}$, and $p_b{p}_c{p}_d$. The symmetries follow from the displayed formulas.