Null Raychaudhuri and focal-point bound

For each $X\in T_{p}S$, the variation field $J_X(\lambda )=d{F}_{(p,\lambda )}(X,0)$ is a Jacobi field by [F3]. Its screen class satisfies the screen Jacobi equation $A_{\lambda }^{\mathrm{\prime }\mathrm{\prime }}+{\mathcal{R}}_{\mathrm{\ell }}{A}_{\lambda }=0,{\mathcal{R}}_{\mathrm{\ell }}[X]:=[R(X,\mathrm{\ell })\mathrm{\ell }],$ because [F2]-[F4] give $D_{\mathrm{\ell }}^2{J}_X=R(\mathrm{\ell },J_X)\mathrm{\ell }=-R(J_X,\mathrm{\ell })\mathrm{\ell }$. The screen projection is well-defined since changing $J_X$ by a multiple of $\mathrm{\ell }$ changes $R(J_X,\mathrm{\ell })\mathrm{\ell }$ by zero. Before the first focal instant $A_{\lambda }$ is invertible, so differentiating $B=A^{\mathrm{\prime }}{A}^{-1}$ gives $B^{\mathrm{\prime }}=A^{\mathrm{\prime }\mathrm{\prime }}{A}^{-1}-A^{\mathrm{\prime }}{A}^{-1}{A}^{\mathrm{\prime }}{A}^{-1}=-{\mathcal{R}}_{\mathrm{\ell }}-B^2\mathrm{.}$

The null hypersurface normal makes $B$ self-adjoint by [F1]. In two screen dimensions write $B=\frac{\theta }{2}\mathrm{i}\mathrm{d}+\sigma$; then $\mathrm{tr}\sigma =0$ and $\mathrm{tr}(B^2)=\frac{{\theta }^2}{2}+\mathrm{tr}({\sigma }^2)=\frac{{\theta }^2}{2}+\mathrm{\mid }\sigma {\mathrm{\mid }}_{\bar{q}}^2\mathrm{.}$ The last equality holds because $\sigma$ is self-adjoint on the positive screen.

At any point choose the local pair supplied by [F1] and set $n:=k\mathrm{/}2$, so $g(\mathrm{\ell },n)=-1$; complete it by a $\bar{q}$-orthonormal screen basis $E_1,E_2$. In this basis the inverse metric is $g^{-1}=-\mathrm{\ell }\otimes n-n\otimes \mathrm{\ell }+E_1\otimes E_1+E_2\otimes E_2\mathrm{.}$ Contracting the curvature in [F2], the two null-pair terms in $\mathrm{Ric}(\mathrm{\ell },\mathrm{\ell })$ have repeated first or last slots and vanish by [F4], leaving $\mathrm{Ric}(\mathrm{\ell },\mathrm{\ell })={\sum }_{A=1}^{2}g(R(E_A,\mathrm{\ell })\mathrm{\ell },E_A)=\mathrm{tr}{\mathcal{R}}_{\mathrm{\ell }}\mathrm{.}$

Taking the trace of the Riccati equation in step 1.1 and substituting steps 2.1 and 2.2 yields ${\theta }^{\mathrm{\prime }}=-\mathrm{tr}(B^2)-\mathrm{tr}{\mathcal{R}}_{\mathrm{\ell }}=-\frac{{\theta }^2}{2}-\mathrm{\mid }\sigma {\mathrm{\mid }}_{\bar{q}}^2-\mathrm{Ric}(\mathrm{\ell },\mathrm{\ell }),$ which is the stated evolution equation.

Put ${\lambda }_{\ast }=2\mathrm{/}\mathrm{\mid }\theta (0)\mathrm{\mid }$. While no focal instant has occurred, $A_{\lambda }$ is invertible and $B,\theta$ are smooth. By [A1], [A2] and step 3.1, ${\theta }^{\mathrm{\prime }}\le -{\theta }^2\mathrm{/}2<0$ whenever $\theta <0$, so $\theta$ remains negative. Therefore ${\left(\frac{1}{\theta }\right)}^{\mathrm{\prime }}=-\frac{{\theta }^{\mathrm{\prime }}}{{\theta }^2}\ge \frac{1}{2}\mathrm{.}$ If no focal instant occurred on $[0,{\lambda }_{\ast }]$, then $\theta ({\lambda }_{\ast })$ would be finite and negative, so $1\mathrm{/}\theta ({\lambda }_{\ast })<0$. Integrating the inequality instead gives $\frac{1}{\theta ({\lambda }_{\ast })}\ge \frac{1}{\theta (0)}+\frac{{\lambda }_{\ast }}{2}=0,$ a contradiction. Thus the screen Jacobi map loses rank at some ${\lambda }_f\le {\lambda }_{\ast }$, which is the claimed focal instant.