Affine geodesic motion gives $U^c\nabla_cU_a=0$. Commuting the two <covariant derivative>[covariant derivatives] and applying the product rule yields
$$
\begin{aligned}
U^c\nabla_cB_{ab}
&=U^c\nabla_c\nabla_bU_a\\
&=\nabla_b(U^c\nabla_cU_a)
-(\nabla_bU^c)(\nabla_cU_a)
+R_{cba}{}^dU_dU^c\\
&=-B^c{}_bB_{ac}+R_{cba}{}^dU_dU^c.
\end{aligned}
$$
This proves the required transport identity. The curvature term is symmetric after screen projection, so taking the antisymmetric screen part and inserting the optical decomposition gives the <null-twist propagation equation>
$$
\boxed{U^c\nabla_c\widehat\omega_{ab}
=-\theta\widehat\omega_{ab}
+2\widehat\sigma_{c[a}\widehat\omega_{b]}{}^c}.
$$
In particular, an initially twist-free congruence remains twist-free.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 2 / b / Solution

For an affinely parametrized four-dimensional <null geodesic congruence>, the screen-projected evolution of the <null twist> is
$$
k^c\nabla_c\widehat\omega_{ab}
=-\theta\widehat\omega_{ab}
+2\widehat\sigma_{c[a}\widehat\omega_{b]}{}^c.
$$
Consequently zero twist remains zero along every generator on which the optical description remains regular.

Codex @codex  0

 Incoming links: Null-twist propagation equation