 Solution

The Frobenius theorem says that hypersurface orthogonality of the nonzero covector $U_a$ is equivalent to $U_{[a}{\nabla }_b{U}_{c]}=0$, or locally $U_a=h{\nabla }_{a}S$ for some nonzero $h$. This immediately makes the screen projection of ${\nabla }_{[b}{U}_{a]}$ zero, since its antisymmetric part is a wedge product with $\nabla S$, which is proportional to $U$. Thus hypersurface orthogonality gives zero null twist.

For the converse, put $B_{ab}={\nabla }_b{U}_a$. Nullness gives $U^a{B}_{ab}=\frac{1}{2}{\nabla }_b(U^2)=0$, and the affine geodesic equation gives $B_{ab}{U}^b=0$. Resolve both tensor slots in the basis consisting of $U,N$ and the screen. These two identities exclude covector factors of $N$. Consequentlyfor some covector $q$. If $\hat{\omega }=0$, wedging this identity with $U$ makes $U_{[a}{\nabla }_b{U}_{c]}=0$. Applying Frobenius theorem completes the proof:This is the null twist vanishes exactly for hypersurface-orthogonal geodesics criterion; the geodesic assumption ensures that the screen criterion implies the full Frobenius condition. The resulting hypersurfaces are null, and the congruence follows their null generators.