= Solution
Choose an auxiliary <null vector> $N$ with $U\cdot N=-1$, parallel-transported along each affine ray. The <screen-space projector> and its induced positive metric are
$$
\boxed{P^a{}_b=\delta^a_b+U^aN_b+N^aU_b,\qquad P_{ab}=g_{ab}+U_aN_b+N_aU_b.}
$$
They annihilate $U,N$, obey $P^2=P$, and have trace two in four <spacetime> dimensions. Since both $U$ and $N$ are parallel along $U$, $\nabla_UP=0$. Equivalently, the screen represents the two-dimensional quotient $U^\perp/\operatorname{span}(U)$.
Define the <optical matrix of a null congruence> by
$$
\widehat B_{ab}=P_a{}^cP_b{}^d\nabla_dU_c.
$$
For a transverse deviation, its screen projection $\zeta=PZ$ satisfies $\nabla_U\zeta^a=\widehat B^a{}_b\zeta^b$. Decompose the <matrix> as
$$
\boxed{\widehat B_{ab}=\frac12\theta P_{ab}+\widehat\sigma_{ab}+\widehat\omega_{ab},\quad\theta=P^{ab}\widehat B_{ab},\quad\widehat\sigma_{ab}=\widehat B_{(ab)}-\frac12\theta P_{ab},\quad\widehat\omega_{ab}=\widehat B_{[ab]}.}
$$
These are the <null expansion>, <null shear> and <null twist>. <Null expansion> gives the fractional change of a small transverse area, $\theta=d\log(dA)/d\lambda$. <Null shear> is symmetric trace-free distortion; <null twist> is the antisymmetric rotation part. This <null expansion> convention uses the full trace, not half the trace. The auxiliary screen fixes representatives, while the transverse optical content is intrinsic to the <null geodesic congruence>.
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