Null screen index form with a spacelike initial boundary

Use the smooth tangent-bundle convention in [A1]. For representatives $X,Y\in {\dot{\gamma }}^{\perp }$, the curvature symmetries [F2] give $\bar{q}(\mathcal{R}[X],[Y])=g(R(X,\dot{\gamma })\dot{\gamma },Y)=g(R(Y,\dot{\gamma })\dot{\gamma },X)=\bar{q}(\mathcal{R}[Y],[X])$. Shifting a representative by a multiple of $\dot{\gamma }$ leaves its screen class and the curvature contraction unchanged. The screen derivative and metric are well-defined by [F1], and ${\chi }_{\mathrm{\ell }}$ is symmetric by [F1]. Thus the integral and boundary terms are symmetric bilinear forms on the specified screen fields, independent of representatives; $I$ is well-defined and symmetric.

For a smooth variation $F_s$, put $T={\mathrm{\partial }}_{\lambda }{F}_s$ and $Z={\mathrm{\partial }}_s{F}_s$. Torsion-freeness gives $D_{s}T=D_{\lambda }Z$. The first variation is $E^{\mathrm{\prime }}(s)=[g(Z,T){]}_0^L-{\int }_0^{L}g(Z,D_{\lambda }T)d\lambda \mathrm{.}$ At $s=0$, $T=\mathrm{\ell }$, $D_{\lambda }T=0$, $Z(L)=0$, and $Z(0)$ is tangent to $S$ and orthogonal to $\mathrm{\ell }$, so $E^{\mathrm{\prime }}(0)=0$. Differentiating once more and using $[D_s,D_{\lambda }]T=R(Z,T)T$ gives, at $s=0$, $E^{\mathrm{\prime }\mathrm{\prime }}(0)=[g(D_{s}Z,\mathrm{\ell }){]}_0^L+{\int }_0^L(\mathrm{\mid }{D}_{\mathrm{\ell }}Z{\mathrm{\mid }}^2-g(R(Z,\mathrm{\ell })\mathrm{\ell },Z))d\lambda \mathrm{.}$ Here the boundary terms $[g(Z,D_{\mathrm{\ell }}Z){]}_0^L$ from differentiation and integration by parts cancel. The final endpoint contribution is zero because $F_s(L)$ is fixed. At the initial endpoint, let $X=Z(0)\in T_{p}S$; along $\alpha (s)=F_s(0)\in S$ one has $g({\alpha }^{\mathrm{\prime }}(s),\mathrm{\ell }(\alpha (s)))=0$. Differentiating this identity gives $g(D_{s}Z(0),\mathrm{\ell })=-{\chi }_{\mathrm{\ell }}(X,X)$, so the initial boundary contribution in the bracket is $+{\chi }_{\mathrm{\ell }}(X,X)$. Substitution proves $E^{\mathrm{\prime }\mathrm{\prime }}(0)=I(Z,Z)$.

It remains to realize every admissible screen field $V$. Let $V_E$ be its smooth representative in $E_{\lambda }$, and let $X\in T_{p}S$ be the unique vector with $[X]=V(0)$. Since $X-V_E(0)$ has zero screen class, it equals $a{\mathrm{\ell }}_0$ for a real number $a$. The lift $Z(\lambda ):=V_E(\lambda )+a(1-\lambda \mathrm{/}L){\mathrm{\ell }}_{\lambda }$ is smooth, orthogonal to $\mathrm{\ell }$, has screen class $V$, and satisfies $Z(0)=X$ and $Z(L)=0$. In the rest of this construction write $V$ for this lift $Z$. Because $D_{\lambda }\mathrm{\ell }=0$, adding a multiple of $\mathrm{\ell }$ does not change either integral contraction in the index form; its initial boundary term is evaluated on $X$. Thus step 2.1 applies to this lift and gives the claimed screen index form. By [F4], for sufficiently small $s$ the preliminary family ${\tilde{F}}_s(\lambda )={\exp }_{\gamma (\lambda )}(sV(\lambda ))$ is smooth on the compact interval; a finite subdivision supplies the local exponential domains. At $\lambda =L$, $V(L)=0$, so ${\tilde{F}}_s(L)=\gamma (L)$. At $\lambda =0$, use a slice chart $(y,z)$ from [F3] with $S=\{z=0\}$. Since $V(0)\in T_{p}S$, the normal coordinate of ${\tilde{F}}_s(0)$ has value $z=O(s^2)$ and zero first derivative at $s=0$. Choose a smooth cutoff $\eta (\lambda )$ equal to $1$ at $0$ and zero before the end of this chart. Near $\lambda =0$, replace the preliminary coordinates by $y_s(\lambda )+\eta (\lambda )(y(\alpha (s))-y_s(0)),z_s(\lambda )-\eta (\lambda )z_s(0),$ where $\alpha (s)\subset S$ is any smooth chart curve with $\alpha (0)=p$ and ${\alpha }^{\mathrm{\prime }}(0)=V(0)$. At $\lambda =0$ this is exactly $\alpha (s)$; its first $s$-derivative at $0$ is unchanged because the correction has zero first derivative. The cutoff vanishes before the final endpoint, so the endpoint remains fixed. This constructs the required variation with screen variational field $V$.

The field realization in step 3.1 and the second-variation calculation in step 2.1 establish the statement for every admissible $V$; symmetry and well-definedness were proved in step 1.1.