 Solution

Away from a bifurcation set, the normal to a Killing horizon is its nonzero Killing vector field $K$. Any affine tangent to the same generators has the form $k=hK$. In the optical tensor, derivatives of $h$ contribute a factor of $K_a$ and disappear under the screen-space projector. Consequentlyby the Killing equation. The trace and trace-free symmetric part are therefore zero: $\theta =0$ and ${\sigma }_{ab}=0$.

For the antisymmetric part, locally the horizon normal is proportional to a gradient. The restriction of its dual one-form to every tangent direction on the horizon is zero. Taking its exterior derivative and evaluating on two screen tangents therefore gives zero as well; equivalently this is the hypersurface-orthogonality consequence of the Frobenius theorem. Since both screen indices are tangential, $B_{[ab]}=0$. Thus the optical scalars on a Killing horizon satisfyNo affine normalization of the Killing vector field itself was assumed. At a regular bifurcation surface, $K$ vanishes but the smooth null generators do not; the conclusions for their optical quantities follow by continuity along each horizon branch.