Hypersurface and null geometry — Examples

On $S$, $q=R^{2}d{\mathrm{\Omega }}^2$. For tangent $X,Y$, flat-space differentiation gives ${\mathrm{\nabla }}_X{\mathrm{\partial }}_r=X\mathrm{/}R$ and ${\mathrm{\nabla }}_X{\mathrm{\partial }}_t=0$. Therefore ${\chi }_{\mathrm{\ell }}=q\mathrm{/}R$ and ${\chi }_k=-q\mathrm{/}R$, and since the screen dimension is two, ${\theta }_{\mathrm{\ell }}=2\mathrm{/}R$ and ${\theta }_k=-2\mathrm{/}R$. Direct inner products give $g(\mathrm{\ell },\mathrm{\ell })=g(k,k)=0$ and $g(\mathrm{\ell },k)=-1-1=-2$; both fields are future because their $t$ components are $1$.

In Cartesian coordinates the inward geodesic is ${\gamma }_{\omega }(\lambda )=(\lambda ,(R-\lambda )\omega )$. Varying $\omega$ in a tangent direction $X\in T_{R\omega }(R{\mathbb{S}}^2)$ gives the Jacobi field $J_X(\lambda )=(1-\lambda \mathrm{/}R)X$. Its screen map is therefore $A_{\lambda }=(1-\lambda \mathrm{/}R)\mathrm{id}$ for $0\le \lambda <R$, with first rank loss at $R$. Hence $B=-\mathrm{id}\mathrm{/}(R-\lambda )$, $\theta =-2\mathrm{/}(R-\lambda )$, $\sigma =0$, and $\mathrm{Ric}=0$. Differentiating $\theta$ verifies the displayed equation directly, and $2\mathrm{/}\mathrm{\mid }\theta (0)\mathrm{\mid }=2\mathrm{/}(2\mathrm{/}R)=R$. Thus the explicit focal parameter equals the comparison value; under its Countable Choice convention, [F2] gives the same sharp bound.

Fix $R<\lambda <3R$. The point $p_{-}=(0,-R\omega )\in S$ and $q_{\lambda }={\gamma }_{\omega }(\lambda )$ have time separation $\lambda$ and spatial separation $\mathrm{\mid }(R-\lambda )\omega -(-R\omega )\mathrm{\mid }=\mathrm{\mid }2R-\lambda \mathrm{\mid }<\lambda \mathrm{.}$ The straight segment from $p_{-}$ to $q_{\lambda }$ is future timelike because its time component is positive and its squared interval is $-{\lambda }^2+(2R-\lambda {)}^2<0$. Thus $q_{\lambda }\in I^{+}(S)$. Under Countable Choice and Dependent Choice this also agrees with [F3] for parameters sufficiently close to $R$; the displayed segment proves the stronger stated range $R<\lambda <3R$ directly, without either choice assumption.

Since $e^f>0$, the boost preserves the future null rays, and $g({\mathrm{\ell }}^{\mathrm{\prime }},k^{\mathrm{\prime }})=e^f{e}^{-f}g(\mathrm{\ell },k)=-2$. The boost law in [F1] gives ${\widehat{\theta }}_0=e^f{\theta }_0$, so ${\widehat{\theta }}_0$ has the same sign as ${\theta }_0$. The new tangent is ${\mathrm{\ell }}^{\mathrm{\prime }}=d\mathrm{/}d{\lambda }^{\mathrm{\prime }}$, while $\mathrm{\ell }=d\mathrm{/}d\lambda$; solving $d\lambda \mathrm{/}d{\lambda }^{\mathrm{\prime }}=e^f$ with both parameters zero at the initial section gives ${\lambda }^{\mathrm{\prime }}=e^{-f}\lambda$ because $f$ is constant along the generator.

On the open parameter domain where $e^{f(p)}{\lambda }^{\mathrm{\prime }}\in J$, put $\widehat{F}(p,{\lambda }^{\mathrm{\prime }}):=F(p,e^{f(p)}{\lambda }^{\mathrm{\prime }})$. For $X\in T_{p}S$, differentiation at fixed ${\lambda }^{\mathrm{\prime }}$ gives $d_p\widehat{F}(X)=d_{p}F(X)+e^{f(p)}{\lambda }^{\mathrm{\prime }}(Xf)\mathrm{\ell }\mathrm{.}$ The added term vanishes in the screen quotient, so at corresponding points ${\widehat{A}}_{{\lambda }^{\mathrm{\prime }}}=A_{e^{f(p)}{\lambda }^{\mathrm{\prime }}}$. The screen metric is unchanged because the null line is unchanged. Along the fixed generator, $f(p)$ is constant, and differentiation with the screen connection gives ${\widehat{A}}_{{\lambda }^{\mathrm{\prime }}}^{\mathrm{\prime }}=e^{f(p)}{A}_{e^{f(p)}{\lambda }^{\mathrm{\prime }}}^{\mathrm{\prime }},\widehat{B}={\widehat{A}}^{\mathrm{\prime }}{\widehat{A}}^{-1}=e^{f(p)}B\mathrm{.}$ Taking trace and trace-free part yields $\widehat{\theta }=e^f\theta$ and $\widehat{\sigma }=e^f\sigma$. Therefore $d\widehat{\theta }\mathrm{/}d{\lambda }^{\mathrm{\prime }}=e^{2f}d\theta \mathrm{/}d\lambda$ and $\mathrm{\mid }\widehat{\sigma }{\mathrm{\mid }}^2=e^{2f}\mathrm{\mid }\sigma {\mathrm{\mid }}^2$. Tensoriality gives $\mathrm{Ric}({\mathrm{\ell }}^{\mathrm{\prime }},{\mathrm{\ell }}^{\mathrm{\prime }})=e^{2f}\mathrm{Ric}(\mathrm{\ell },\mathrm{\ell })$. Multiplying the Raychaudhuri equation in [F2] by $e^{2f}$ proves its form is unchanged. For a generator from $p$ with ${\theta }_0(p)<0$ and $\mathrm{Ric}(\mathrm{\ell },\mathrm{\ell })\ge 0$ through the comparison interval, the bound rescales as ${\lambda }_{\ast }^{\mathrm{\prime }}(p)=\frac{2}{\mathrm{\mid }{\widehat{\theta }}_0(p)\mathrm{\mid }}=e^{-f(p)}\frac{2}{\mathrm{\mid }{\theta }_0(p)\mathrm{\mid }}=e^{-f(p)}{\lambda }_{\ast }(p)\mathrm{.}$

By step 2.1, ${\widehat{A}}_{{\lambda }^{\mathrm{\prime }}}=A_{e^{f(p)}{\lambda }^{\mathrm{\prime }}}$, so its rank-loss event is the same point of $M$ and its parameter value is multiplied by $e^{-f(p)}$. Under [A2], [F3] applies: the geometric statement that points just beyond that event lie in $I^{+}(S)$ is unchanged by relabeling the affine parameter. Thus only the numerical affine label changes; the causal-boundary conclusion does not.