Null focusing and a restricted smooth future area theorem

Define the one-form $\alpha$ on $H$ by ${\alpha }_p(X):=g_p(k_p,X)$ for $p\in H$ and $X\in T_{p}H$. Since $k$ spans the radical of $g{\mathrm{\mid }}_{TH}$, $\alpha =0$. For smooth tangent fields $X,Y$ on $H$, the exterior derivative on $H$ is defined by $d_H\alpha (X,Y):=X(\alpha (Y))-Y(\alpha (X))-\alpha ([X,Y])\mathrm{.}$ Thus $d_H\alpha =0$. Metric compatibility and torsion freedom give $\begin{array}{rl}{\displaystyle 0=d_H\alpha (X,Y)}& {\displaystyle =Xg(k,Y)-Yg(k,X)-g(k,[X,Y])}\\ {\displaystyle }& {\displaystyle =g({\mathrm{\nabla }}_{X}k,Y)-g({\mathrm{\nabla }}_{Y}k,X)+g(k,{\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y])}\\ {\displaystyle }& {\displaystyle =g({\mathrm{\nabla }}_{X}k,Y)-g({\mathrm{\nabla }}_{Y}k,X),}\end{array}$ so $B$ is symmetric. By [F8], it is a well-defined symmetric form on each positive two-dimensional screen. In a local $\bar{g}$-orthonormal screen frame $(e_1,e_2)$ write $B_{AB}=\frac{1}{2}\theta {\delta }_{AB}+{\sigma }_{AB}$, where ${\sum }_A{\sigma }_{AA}=0$. Therefore $B_{AB}{B}_{BA}=\frac{1}{2}{\theta }^2+{\sum }_{A,B}{\sigma }_{AB}^2\mathrm{.}$

Fix $p\in H$. By [F6], choose slice coordinates with $H=\{x^3=0\}$ and, after a linear change of the tangential coordinates, $k^0(p)\mathrm{\ne }0$. The two-dimensional slice $S=\{x^0=x^3=0\}$ lies in $H$, and the three-dimensional slice ${\mathcal{N}}_0=\{x^0=0\}$ is transverse to $k$ at $p$. Along $S$, set $n=N\mathrm{/}\sqrt{-g(N,N)}$, $\alpha =-g(k,n)>0$, and $u=k\mathrm{/}\alpha -n$; then $g(u,n)=0$ and $g(u,u)=1$. Extend $\alpha$ and the coordinate components of $u$ smoothly from $S$ to ${\mathcal{N}}_0$ by holding them constant in the slice coordinate $x^3$. Project the extended vector to $n^{\perp }$ by $w=u+g(u,n)n$; after shrinking the neighborhood $g(w,w)>0$, so $e=w\mathrm{/}\sqrt{g(w,w)}$ is smooth and unit spacelike. The field $K_0=\alpha (n+e)$ on ${\mathcal{N}}_0$ is null and future-directed, and equals $k$ on $S$. The map $F(t,x)={\gamma }_{x,K_0(x)}(t)$ is smooth by [F4]; at $(0,p)$ its differential is $(s,v)\mapsto s{k}_p+v$ for $v\in T_p{\mathcal{N}}_0$, an isomorphism because $k_p\mathrm{\notin }{T}_p{\mathcal{N}}_0$. By [F5], after restriction $F$ is a diffeomorphism onto a neighborhood of $p$. Define $K$ there by $K_{F(t,x)}={\mathrm{\partial }}_{t}F(t,x)$. Then $K$ is smooth and affinely geodesic. Metric compatibility gives ${\mathrm{\partial }}_{t}g(K,K)=2g({\mathrm{\nabla }}_{K}K,K)=0$, so $K$ remains null. By [F12], the restriction of $k$ to $H$ has a smooth local integral flow $\mathrm{\Psi }(t,x)$ for $x$ near $p$ in $S$, and these curves remain in $H$. Since ${\mathrm{\nabla }}_{k}k=0$, each such curve is an affinely parameterized geodesic with initial velocity $k_x$; uniqueness in [F4] therefore identifies it with ${\gamma }_{x,k_x}(t)$. Its differential at $(0,p)$ is $(s,v)\mapsto s{k}_p+v$ from $\mathbb{R}\times T_{p}S$ to $T_{p}H$, which is an isomorphism. Thus [F5] makes $\mathrm{\Psi }$ a local diffeomorphism onto a neighborhood of $p$ in $H$. For $x\in S$, the curves defining $\mathrm{\Psi }$ and $F{\mathrm{\mid }}_{\mathbb{R}\times S}$ have the same initial data, so uniqueness in [F4] makes them agree; hence $K=k$ on this neighborhood in $H$.

At each point of $H$, choose an algebraic null companion $\mathrm{\ell }$ with $g(k,\mathrm{\ell })=-1$ and a $\bar{g}$-orthonormal screen basis $e_1,e_2$, whose existence follows from [F8]. Let $A(X):={\mathrm{\nabla }}_{X}K$ for the local extension in step 1.2. Since $K$ is null throughout its neighborhood, metric compatibility gives $g(K,A(X))=0$ for every $X$; since $K$ is affine geodesic, $A(K)=0$. The inverse metric in this null frame is $g^{-1}=-K\otimes \mathrm{\ell }-\mathrm{\ell }\otimes K+{\sum }_A{e}_A\otimes e_A$. These two annihilation identities imply ${\mathrm{\nabla }}_a{K}^a={\sum }_{A}g(A(e_A),e_A)=\theta$ and $({\mathrm{\nabla }}_a{K}^b)({\mathrm{\nabla }}_b{K}^a)={\sum }_{A,B}{B}_{AB}{B}_{BA}$: $A$ maps into $K^{\perp }=\mathrm{span}\{K,e_1,e_2\}$, kills $K$, and has no $\mathrm{\ell }$ component, so the null-frame trace and the trace of $A^2$ have only the screen block. To commute the derivatives, expand in a coordinate chart and use torsion freedom: $({\mathrm{\nabla }}_a{\mathrm{\nabla }}_{b}K{)}^c-({\mathrm{\nabla }}_b{\mathrm{\nabla }}_{a}K{)}^c=({\mathrm{\partial }}_a{\mathrm{\Gamma }}^c{}_{bd}-{\mathrm{\partial }}_b{\mathrm{\Gamma }}^c{}_{ad}+{\mathrm{\Gamma }}^c{}_{ae}{\mathrm{\Gamma }}^e{}_{bd}-{\mathrm{\Gamma }}^c{}_{be}{\mathrm{\Gamma }}^e{}_{ad})K^d=R^c{}_{dab}{K}^d\mathrm{.}$ The connection terms acting on the derivative index cancel because ${\mathrm{\Gamma }}^e{}_{ab}={\mathrm{\Gamma }}^e{}_{ba}$. Contracting $c=a$ gives ${\mathrm{\nabla }}_a{\mathrm{\nabla }}_b{K}^a-{\mathrm{\nabla }}_b{\mathrm{\nabla }}_a{K}^a={\mathrm{Ric}}_{db}{K}^d$ by [F3]. Since $K$ is geodesic, $0={\mathrm{\nabla }}_a(K^b{\mathrm{\nabla }}_b{K}^a)=({\mathrm{\nabla }}_a{K}^b)({\mathrm{\nabla }}_b{K}^a)+K^b{\mathrm{\nabla }}_a{\mathrm{\nabla }}_b{K}^a$. Substituting the contracted commutator and restricting to $H$, where $K=k$ and ${\mathrm{\nabla }}_a{K}^a=\theta$, gives $k^b{\mathrm{\nabla }}_b\theta =-B_{AB}{B}_{BA}-\mathrm{Ric}(k,k)=-\frac{1}{2}{\theta }^2-{\sum }_{A,B}{\sigma }_{AB}^2-\mathrm{Ric}(k,k)\mathrm{.}$

Pull the metric on each cut back to $S_0$: in a chart $(U,x^1,x^2)$ on $S_0$, write $X_A=d{\mathrm{\Phi }}_{\lambda }({\mathrm{\partial }}_A)$ and $q_{AB}(\lambda ,x)=g(X_A,X_B)$. Since $T{S}_{\lambda }\subseteq TH\subseteq k^{\perp }$ and positive definiteness gives $T{S}_{\lambda }\cap \mathbb{R}k=\{0\}$, the quotient map identifies $T{S}_{\lambda }$ with the two-dimensional screen of [F8]; its metric is exactly the induced cut metric. Thus $q^{AB}{B}_{AB}=\theta$. The map $(\lambda ,x)\mapsto \mathrm{\Phi }(\lambda ,x)$ is smooth, so these coordinate fields are smooth; their coordinate parameter fields commute, $[k,X_A]=0$, along the image of the flow. Torsion freedom therefore gives ${\mathrm{\nabla }}_k{X}_A={\mathrm{\nabla }}_{X_A}k$. Differentiating $q_{AB}$ and using metric compatibility yields ${\mathrm{\partial }}_{\lambda }{q}_{AB}=B_{AB}+B_{BA}=2B_{AB}$, since $B$ is symmetric by step 1.1. Write the positive-definite symmetric matrix as $q=\left(\begin{array}{cc}{q}_{11}& q_{12}\\ q_{12}& q_{22}\end{array}\right)$. Direct differentiation of $\det q=q_{11}{q}_{22}-q_{12}^2$, together with $q^{-1}=\frac{1}{\det q}\left(\begin{array}{cc}{q}_{22}& -q_{12}\\ -q_{12}& q_{11}\end{array}\right),$ gives ${\mathrm{\partial }}_{\lambda }\det q=(\det q)q^{AB}{\mathrm{\partial }}_{\lambda }{q}_{AB}$. Since $\det q>0$, the ordinary chain rule for the positive square root gives ${\mathrm{\partial }}_{\lambda }\sqrt{\det q}=\frac{1}{2}\sqrt{\det q}{q}^{AB}{\mathrm{\partial }}_{\lambda }{q}_{AB}=\theta (\mathrm{\Phi }(\lambda ,x))\sqrt{\det q}\mathrm{.}$ Thus the local coefficient of ${\mathrm{\partial }}_{\lambda }{\mu }_{q_{\lambda }}$ is the coefficient of $\theta (\mathrm{\Phi }(\lambda ,\cdot )){\mu }_{q_{\lambda }}$.

Fix any point $\mathrm{\Phi }(\lambda ,x)$ and suppose $\theta (\lambda ,x)<0$. Along that generator put $y:=-\theta >0$. Since ${\mathrm{\partial }}_{\lambda }=k$, step 2.1 and [A1] give $y^{\mathrm{\prime }}\ge y^2\mathrm{/}2$. Thus $y$ remains positive and ${\left(\frac{1}{y}\right)}^{\mathrm{\prime }}=-\frac{y^{\mathrm{\prime }}}{y^2}\le -\frac{1}{2}\mathrm{.}$ Integrating from $\lambda$ to $\lambda +2\mathrm{/}y(\lambda )$ gives $1\mathrm{/}y(\lambda +2\mathrm{/}y(\lambda ))\le 0$, impossible because [A2] supplies a smooth embedded cut there and $\theta$ is finite. Hence $\theta \ge 0$ at every point on every future cut.

Fix ${\lambda }_{\ast }>{\lambda }_0$ and a compact interval $[a,b]\subset ({\lambda }_0,\mathrm{\infty })$ with ${\lambda }_{\ast }\in (a,b)$. Compactness of $S_0$ gives a finite cover by coordinate charts; by the partition theorem in [F9] choose a finite smooth partition of unity $({\rho }_i)$ subordinate to it. Each $\mathrm{supp}{\rho }_i$ is a compact subset of its chart. In chart coordinates let $K_i:=x_i(\mathrm{supp}{\rho }_i)$ and write ${\rho }_i{\mu }_{q_{\lambda }}=f_i(x,\lambda )\mathrm{\mid }d{x}^{1}d{x}^2\mathrm{\mid }$. The coefficient $f_i$ is smooth on a neighborhood of $K_i\times [a,b]$ and, because ${\rho }_i$ has compact support inside the chart, its zero extension in $x$ is smooth. The set $K_i$ is compact and has finite Lebesgue measure. The density integral formula in [F9] gives $A(\lambda )={\sum }_i{\int }_{K_i}{f}_i(x,\lambda )dx$. Applying [F10] to each of this finite collection of fixed compact sets and using 2.2 gives $A^{\mathrm{\prime }}({\lambda }_{\ast })={\sum }_i{\int }_{K_i}{\mathrm{\partial }}_{\lambda }{f}_i(x,{\lambda }_{\ast })dx={\sum }_i{\int }_{K_i}{\rho }_i(x)\theta (\mathrm{\Phi }({\lambda }_{\ast },x)){\mu }_{q_{{\lambda }_{\ast }}}(x)={\int }_{S_0}\theta (\mathrm{\Phi }({\lambda }_{\ast },x)){\mu }_{q_{{\lambda }_{\ast }}}(x)\mathrm{.}$ The last equality is the density integral formula and ${\sum }_i{\rho }_i=1$. Diffeomorphism invariance of density integration in [F9] identifies this with ${\int }_{S_{{\lambda }_{\ast }}}\theta dA$. As ${\lambda }_{\ast }$ was arbitrary, the derivative identity holds for every $\lambda >{\lambda }_0$.

By step 3.1 the integrand in step 3.2 is nonnegative, and the Riemannian volume density is positive, so $A^{\mathrm{\prime }}(\lambda )\ge 0$ for $\lambda >{\lambda }_0$. The same finite chart formula shows $A$ is continuous at ${\lambda }_0$: on each compact $K_i$, $f_i(x,\lambda )$ converges uniformly to $f_i(x,{\lambda }_0)$ as $\lambda \downarrow {\lambda }_0$ by continuity on the compact product, and the finite-measure integral difference is bounded by ${\lambda }_2(K_i){\sup }_{K_i}\mathrm{\mid }{f}_i(x,\lambda )-f_i(x,{\lambda }_0)\mathrm{\mid }$. Thus $A$ is continuous on every $[{\lambda }_1,{\lambda }_2]$ with ${\lambda }_0\le {\lambda }_1<{\lambda }_2$ and differentiable in its interior; [F11] implies $A({\lambda }_2)\ge A({\lambda }_1)$.