Smooth null screen and KN parameter family

Fix $p\in H$. Nondegeneracy of $g_p$ and $k_p\mathrm{\ne }0$ imply that the linear functional $v\mapsto g_p(k_p,v)$ is nonzero, so choose $v\in T_p\mathcal{M}$ with $g_p(k_p,v)=-1$. In a neighborhood in $H$, set $n:=N\mathrm{/}\sqrt{-g(N,N)}$ and $\alpha :=-g(k,n)>0$, so $n$ and $\alpha$ are smooth; define $v_0:=n\mathrm{/}\alpha$ and $\mathrm{\ell }:=v_0+\frac{1}{2}g(v_0,v_0)k$. Then $g(k,\mathrm{\ell })=g(k,v_0)=-1$, and $g(\mathrm{\ell },\mathrm{\ell })=g(v_0,v_0)+g(v_0,v_0)g(v_0,k)+\frac{1}{4}g(v_0,v_0{)}^{2}g(k,k)=0,$ since $g(v_0,k)=-1$ and $g(k,k)=0$. This construction is smooth locally and gives a null field $\mathrm{\ell }$ paired with $k$ by $-1$.

Let $P_p:={\mathrm{span}}_{\mathbb{R}}\{k_p,{\mathrm{\ell }}_p\}$. Its Gram matrix in the ordered basis $(k_p,{\mathrm{\ell }}_p)$ is $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$, with determinant $-1$ and eigenvalues $1,-1$, so it is nondegenerate with inertia $(1,1,0)$. Therefore $T_p\mathcal{M}=P_p\oplus P_p^{\perp }$; in a basis adapted to this orthogonal direct sum, the Gram matrix is block diagonal. The full inertia is $(3,1,0)$ by [F1], so [F3] makes the restriction to the two-dimensional $P_p^{\perp }$ have inertia $(2,0,0)$, hence be positive definite.

Every $X\in k_p^{\perp }$ has a unique decomposition $X=a{k}_p+b{\mathrm{\ell }}_p+z$ with $z\in P_p^{\perp }$. Pairing with $k_p$ gives $0=g(k_p,X)=-b$, so $b=0$. Hence $k_p^{\perp }=\mathbb{R}{k}_p\oplus P_p^{\perp }$, and the quotient map restricts to a linear isomorphism from $P_p^{\perp }$ to ${\mathcal{S}}_p$. Changing a representative by a multiple of $k_p$ does not change $g_p(X,Y)$ because every representative lies in $k_p^{\perp }$; under the isomorphism the quotient form is the positive-definite restriction to $P_p^{\perp }$ from step 2.1. This proves its dimension, well-definedness and positivity.

The local fields $k,\mathrm{\ell }$ are linearly independent and their $2\times 2$ Gram matrix is invertible. Choose two smooth local vector fields $w_1,w_2$ whose values at $p$ complete them to a basis. Solving the two linear equations $g(k,e_i)=g(\mathrm{\ell },e_i)=0$ for the $k,\mathrm{\ell }$ components of $w_i$ uses the inverse of that smooth invertible Gram matrix, so yields smooth fields $e_1,e_2\in P^{\perp }$; after shrinking the neighborhood they remain linearly independent and form a frame of $P^{\perp }$. By step 3.1 their quotient classes frame $\mathcal{S}$, so the quotient is a smooth rank-two vector bundle and $\bar{g}$ is a smooth positive-definite metric. Applying the two finite Gram–Schmidt operations to $e_1,e_2$ gives a local smooth $\bar{g}$-orthonormal screen frame: each denominator is the positive square root of a smooth positive squared norm, hence smooth by [F4].

The screen form is independent of representatives. Replacing $X$ by $X+fk$ changes $g({\mathrm{\nabla }}_{X}k,Y)$ by $fg({\mathrm{\nabla }}_{k}k,Y)=0$ because $k$ is affine. Replacing $Y$ by $Y+hk$ changes it by $hg({\mathrm{\nabla }}_{X}k,k)=\frac{h}{2}X(g(k,k))=0$, since $\mathrm{\nabla }$ is metric-compatible and $k$ is null on $H$. Thus $B$ descends to a smooth bilinear form on $\mathcal{S}$; smoothness follows from its smooth local-frame components.

Substituting $M=m$, $a=q=0$ into [F1] gives $\mathrm{\Sigma }=r^2$, $f_0=f=1-2m\mathrm{/}r$, the displayed ingoing metric and $N={\mathrm{\partial }}_v-\frac{1+f^2}{2}{\mathrm{\partial }}_r$; the potential vanishes. The inverse of the $(v,r)$ metric block $\left(\begin{array}{cc}-f& 1\\ 1& 0\end{array}\right)$ is $\left(\begin{array}{cc}0& 1\\ 1& f\end{array}\right)$, so $g^{-1}(dr,dr)=f$. At $r_H=2m>0$, $f(r_H)=0$ and ${\chi }^{\mathrm{♭}}=g({\mathrm{\partial }}_v,\cdot )=-fdv+dr=dr$, with $dr(\chi )=0$. Thus $H=\{r=2m\}$ is a null hypersurface and $\chi$ is tangent, null and normal there. Since every coefficient is independent of $v$, ${\mathcal{L}}_{\chi }g=0$, so [F3] makes $\chi$ Killing; $\chi$ never vanishes, so the nonzero-generator part is all of $H$. Finally $g(\chi ,N)=dr(N)=-1\mathrm{/}2<0$ on $H$, so its generator is future-directed. This verifies the exact horizon conditions in [F2].

The Killing norm is $g(\chi ,\chi )=-f$, so $d(g(\chi ,\chi ))=-f^{\mathrm{\prime }}(r_H)dr$. From $f(r)=1-2m\mathrm{/}r$ and [F5], $f^{\mathrm{\prime }}(r)=2m\mathrm{/}{r}^2$, so at $r_H=2m$ this is $-\frac{1}{2m}dr$. Comparing with [F2] and ${\chi }^{\mathrm{♭}}=dr$ gives $-2{\kappa }_{H}dr=-\frac{1}{2m}dr$, hence ${\kappa }_H=1\mathrm{/}(4m)$. A section at fixed $v$ and $r=2m$ has induced metric $(2m{)}^2\gamma$. In spherical coordinates $\gamma =d{\theta }^2+{\sin }^2\theta d{\psi }^2$, so its area is $(2m{)}^2{\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\psi =16\pi m^2$. Substitution in [F4] gives the same area and surface-gravity functions and gives $j={\mathrm{\Omega }}_{\mathrm{K}\mathrm{N}}={\mathrm{\Phi }}_{\mathrm{K}\mathrm{N}}=0$.

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 $A({\lambda }_2)\ge A({\lambda }_1)$.