Lemma: The fiber map is minus the trace reversal

For symmetric $r_{ab}$, with $\mathfrak{a}$ as in (eq:a1), ${\mathfrak{a}}^{\alpha \mu ,ab}{r}_{ab}=\frac{1}{2}{\eta }^{\alpha \mu }{\mathrm{tr}}_{\eta }r-r^{\alpha \mu }=-(\mathcal{P}r{)}^{\alpha \mu }$, where $\mathcal{P}$ is XII-B's trace reversal. ${\mathfrak{a}}^{\alpha \mu ,ab}$ is symmetric in $(\alpha \mu )$ and in $(ab)$, and $b_1[r]=\frac{1}{2}{\mathrm{tr}}_{\eta }r$.

Proof. ${\eta }^{\alpha (a}{\eta }^{b)\mu }{r}_{ab}=\frac{1}{2}(r^{\alpha \mu }+r^{\mu \alpha })=r^{\alpha \mu }$. The symmetries are read off from (eq:a1).