Lemma: The candidate fiber map

For symmetric $r_{ab}$, ${\mathfrak{a}}^{\alpha \mu ,ab}{r}_{ab}=\frac{1}{2}{\eta }^{\alpha \mu }{\mathrm{tr}}_{\eta }r-r^{\alpha \mu }$. As an endomorphism $\mathfrak{a}$ of ${\mathrm{Sym}}^2$, ${\mathfrak{a}}^2=\mathrm{id}$; so $\mathfrak{a}$ is invertible of rank ten, and ${\eta }_{\alpha \mu }{\mathfrak{a}}^{\alpha \mu ,ab}{r}_{ab}={\mathrm{tr}}_{\eta }r=2b_1[r]$.

Proof. ${\eta }^{\alpha (a}{\eta }^{b)\mu }{r}_{ab}=r^{\alpha \mu }$ by symmetry. $\mathfrak{a}(\mathfrak{a}r)=\frac{1}{2}\eta \mathrm{tr}(\frac{1}{2}\eta \mathrm{tr}r-r)-(\frac{1}{2}\eta \mathrm{tr}r-r)=\frac{1}{2}\eta (2\mathrm{tr}r-\mathrm{tr}r)-\frac{1}{2}\eta \mathrm{tr}r+r=r$, using ${\mathrm{tr}}_{\eta }\eta =4$. The trace identity is $\frac{1}{2}\cdot 4\mathrm{tr}r-\mathrm{tr}r$.