Local null normal pairs and the positive screen quotient

Fix $p\in S$ and choose a smooth local frame $E_1,E_2$ for $TS$. Its Gram matrix $Q_{ij}=g(E_i,E_j)$ is smooth and positive definite, so $Q^{-1}$ is smooth on a sufficiently small neighborhood. The map $P_T(v):={\sum }_{i,j=1}^2{E}_i(Q^{-1}{)}^{ij}g(v,E_j)$ is a smooth projection onto $TS$; $P_{\perp }:=\mathrm{i}\mathrm{d}-P_T$ is a smooth rank-two projection with image $\mathcal{N}S$. The vector ${\nu }_0:=P_{\perp }N$ is timelike because $g({\nu }_0,{\nu }_0)=g(N,N)-g(P_{T}N,P_{T}N)<0,$ and it is future because $g({\nu }_0,N)=g({\nu }_0,{\nu }_0)<0$. Thus $\nu :={\nu }_0\mathrm{/}\sqrt{-g({\nu }_0,{\nu }_0)}$ is a smooth future unit timelike normal. Choose a smooth local section $W$ of $\mathcal{N}S$ independent of $\nu$ at $p$ and shrink the neighborhood so independence persists; then $e_0:=W+g(W,\nu )\nu$ is nonzero, normal to $S$, and orthogonal to $\nu$. By [F1] and [F5], ${\nu }^{\perp }\cap {\mathcal{N}}_{p}S$ is positive one-dimensional, so after shrinking, $e:=e_0\mathrm{/}\sqrt{g(e_0,e_0)}$ is a smooth unit spacelike normal. The fields $\mathrm{\ell }:=\nu +e$ and $k:=\nu -e$ are future null normals and satisfy $g(\mathrm{\ell },k)=-2$. This also proves that $\mathcal{N}S$ is a smooth rank-two Lorentzian bundle.

On an overlap, each normal plane has exactly two future null rays. If ${\mathrm{\ell }}^{\mathrm{\prime }}$ selects the same ray as $\mathrm{\ell }$, then ${\mathrm{\ell }}^{\mathrm{\prime }}=a\mathrm{\ell }$ for a smooth function $a>0$; the normalization $g({\mathrm{\ell }}^{\mathrm{\prime }},k^{\mathrm{\prime }})=-2$ forces $k^{\mathrm{\prime }}=a^{-1}k$. Writing $f=\log a$ gives the asserted boost law. If ${\mathrm{\ell }}^{\mathrm{\prime }}$ selects the other future ray, interchange the labels first.

Fix a point of $H$. Normalize the supplied future timelike field by $T=N\mathrm{/}\sqrt{-g(N,N)}$ and put $\alpha :=-g(\mathrm{\ell },T)>0$. The smooth auxiliary field $k:=\frac{2}{\alpha }T-\frac{1}{{\alpha }^2}\mathrm{\ell }$ is future null and satisfies $g(\mathrm{\ell },k)=-2$: directly, $g(k,k)=-4\mathrm{/}{\alpha }^2+4\mathrm{/}{\alpha }^2=0$ and $g(k,T)=-1\mathrm{/}\alpha <0$. This $k$ is not normal to $H$; its nonzero pairing with $\mathrm{\ell }$ makes it transverse to $T_{p}H={\mathrm{\ell }}_p^{\perp }$. Let $P:=\mathrm{span}\{\mathrm{\ell },k\}$. Its Gram matrix is $\left(\begin{array}{cc}0& -2\\ -2& 0\end{array}\right)$, so $P$ is nondegenerate with inertia $(1,1,0)$, and $T_{p}M=P\oplus P^{\perp }$. Diagonalizing $g{\mathrm{\mid }}_{P^{\perp }}$ and adjoining that basis to a diagonal basis of $P$ diagonalizes $g_p$; uniqueness in [F5] and the ambient inertia $(3,1,0)$ from [F1] give inertia $(2,0,0)$ on $P^{\perp }$. Thus $P^{\perp }$ is positive definite. Because $P$ is a smooth nondegenerate rank-two subbundle locally, its orthogonal projection is smooth and $P^{\perp }$ has local smooth frames. Every vector has a unique decomposition $a\mathrm{\ell }+bk+z$ with $z\in P^{\perp }$; pairing with $\mathrm{\ell }$ shows that it lies in ${\mathrm{\ell }}^{\perp }$ exactly when $b=0$. Hence ${\mathrm{\ell }}^{\perp }=\mathbb{R}\mathrm{\ell }\oplus P^{\perp }$, so the quotient map identifies ${\mathcal{S}}_p$ with $P^{\perp }$ and $\bar{q}$ is independent of representatives and positive definite. If $S$ is a spacelike section of $H$, then $T_{p}S\subseteq {\mathrm{\ell }}^{\perp }$ and $T_{p}H=T_{p}S\oplus \mathbb{R}{\mathrm{\ell }}_p$; therefore $T_{p}S\to {\mathcal{S}}_p$ is an isomorphism, and $\bar{q}([X],[Y])=g(X,Y)=q(X,Y)$ proves it is an isometry.

A local extension of a section $\mathrm{\ell }$ from an embedded surface defines ${\mathrm{\nabla }}_X\mathrm{\ell }$ independently of the extension: the difference of two extensions vanishes on $S$, so its derivative in every tangent direction to $S$ vanishes there. For tangent fields $X,Y$ on $S$, metric compatibility and normality give ${\chi }_{\mathrm{\ell }}(X,Y)-{\chi }_{\mathrm{\ell }}(Y,X)=-g(\mathrm{\ell },{\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X)=-g(\mathrm{\ell },[X,Y])=0,$ because the connection is torsion-free and $[X,Y]$ is tangent to $S$. For a boost, ${\mathrm{\nabla }}_X(e^f\mathrm{\ell })=e^f(Xf)\mathrm{\ell }+e^f{\mathrm{\nabla }}_X\mathrm{\ell }$ and $g(\mathrm{\ell },Y)=0$, hence ${\chi }_{e^f\mathrm{\ell }}=e^f{\chi }_{\mathrm{\ell }}$. On $H$, $[{\mathrm{\nabla }}_X\mathrm{\ell }]$ is in the screen since $g({\mathrm{\nabla }}_X\mathrm{\ell },\mathrm{\ell })=\frac{1}{2}Xg(\mathrm{\ell },\mathrm{\ell })=0$. Replacing $X$ by $X+a\mathrm{\ell }$ changes this class by $a[{\mathrm{\nabla }}_{\mathrm{\ell }}\mathrm{\ell }]=0$ when $\mathrm{\ell }$ is affine; therefore $B$ is well-defined. Likewise $[D_{\mathrm{\ell }}(X+a\mathrm{\ell })]=[D_{\mathrm{\ell }}X+a^{\mathrm{\prime }}\mathrm{\ell }+a{D}_{\mathrm{\ell }}\mathrm{\ell }]=[D_{\mathrm{\ell }}X]$, and metric compatibility gives $\mathrm{\ell }\bar{q}([X],[Y])=\bar{q}({\mathcal{D}}_{\mathrm{\ell }}[X],[Y])+\bar{q}([X],{\mathcal{D}}_{\mathrm{\ell }}[Y])$, so the screen derivative is well-defined and metric-compatible. Self-adjointness of $B$ follows from the same torsion-free calculation with $X,Y$ tangent to $H$, since $\mathrm{\ell }$ is normal to $H$ and $[X,Y]$ remains tangent to $H$.