Rigidity of isometric immersions of a connected Lorentzian manifold, and agreement of data-fixing embeddings on overlaps

(The coordinate identity for the first differential.) In the charts of [L1], write $f=y\circ \psi \circ x^{-1}$ and $J^{\alpha }{}_a={\mathrm{\partial }}_a{f}^{\alpha }$. The equation $g_{ab}=g_{\alpha \beta }^{\mathrm{\prime }}(f)J^{\alpha }{}_a{J}^{\beta }{}_b$ makes $J$ invertible: if $Jv=0$, then $g(v,w)=0$ for all $w$, so $v=0$. Put $H^{\alpha }{}_{bc}={\mathrm{\partial }}_b{\mathrm{\partial }}_c{f}^{\alpha }+{\mathrm{\Gamma }}^{\mathrm{\prime }\alpha }{}_{\beta \gamma }(f)J^{\beta }{}_b{J}^{\gamma }{}_c$ and $C^a{}_{bc}=(J^{-1}{)}^a{}_{\alpha }{H}^{\alpha }{}_{bc}$. The target coefficient formula gives ${\mathrm{\partial }}_{\beta }{g}_{\gamma \delta }^{\mathrm{\prime }}=g_{\alpha \delta }^{\mathrm{\prime }}{\mathrm{\Gamma }}^{\mathrm{\prime }\alpha }{}_{\beta \gamma }+g_{\gamma \alpha }^{\mathrm{\prime }}{\mathrm{\Gamma }}^{\mathrm{\prime }\alpha }{}_{\beta \delta }$. Differentiating the metric identity therefore gives ${\mathrm{\partial }}_b{g}_{cd}=g_{ad}{C}^a{}_{bc}+g_{ca}{C}^a{}_{bd}$. Since $C^a{}_{bc}=C^a{}_{cb}$, adding this identity with indices $(b,c,d)$ and $(c,b,d)$ and subtracting the one with indices $(d,b,c)$ yields $2g_{ad}{C}^a{}_{bc}={\mathrm{\partial }}_b{g}_{cd}+{\mathrm{\partial }}_c{g}_{bd}-{\mathrm{\partial }}_d{g}_{bc}$. Thus $C^a{}_{bc}={\mathrm{\Gamma }}^a{}_{bc}$ and ${\mathrm{\partial }}_b{J}^{\alpha }{}_c={\mathrm{\Gamma }}^d{}_{bc}{J}^{\alpha }{}_d-{\mathrm{\Gamma }}^{\mathrm{\prime }\alpha }{}_{\beta \gamma }(f)J^{\beta }{}_b{J}^{\gamma }{}_c$. This follows from finite algebra and the smooth metric identity, without an application of the choice-bearing connection or geodesic suppliers.

(The agreement set of (a).) Let $A:=\{q\in N:{\psi }_1(q)={\psi }_2(q)\text{ and }d{\psi }_1(q)=d{\psi }_2(q)\}$. Then $p\in A$ by hypothesis, so $A\mathrm{\ne }\mathrm{\varnothing }$.

($A$ is closed.) Let $q$ lie in the closure of $A$. Since $N^{\mathrm{\prime }}$ is Hausdorff, if ${\psi }_1(q)\mathrm{\ne }{\psi }_2(q)$ there are disjoint neighborhoods of these values; their inverse images have an open intersection containing $q$ and missing $A$, a contradiction. Thus ${\psi }_1(q)={\psi }_2(q)=r$. Choose a source chart $x:V\to x(V)\subseteq {\mathbb{R}}^4$ around $q$ and a target chart $y:W\to y(W)\subseteq {\mathbb{R}}^4$ around $r$, shrinking $V$ so both ${\psi }_i(V)\subseteq W$. The coordinate maps $f_i=y\circ {\psi }_i\circ x^{-1}$ are smooth. On $x(A\cap V)$ their values and their $4$ by $4$ derivative matrices agree. Each entry of $D{f}_1-D{f}_2$ is continuous, so a nonzero entry at $x(q)$ would have a neighborhood disjoint from $x(A\cap V)$, contradicting $q\in \bar{A}$. Hence $D{f}_1(x(q))=D{f}_2(x(q))$. The invertible chart differentials identify these matrices with $d{\psi }_{i,q}:T_{q}N\to T_r{N}^{\mathrm{\prime }}$, giving $d{\psi }_{1,q}=d{\psi }_{2,q}$. Thus $q\in A$ and $A$ is closed.

(Clause (b).) Differentiating ${\psi }_i\circ j=j^{\mathrm{\prime }}$ makes the differentials equal on $T_{j(\sigma )}\mathrm{\Sigma }$. The orthogonal complement of this positive definite three-plane in a Lorentzian four-space is a negative definite line, so it has exactly one future unit vector $n$; the target plane likewise has a unique future unit vector $n^{\mathrm{\prime }}$. The invertible linear isometries $d{\psi }_i$ carry the source plane to the target plane and preserve future cones, hence carry $n$ to $n^{\mathrm{\prime }}$. Every tangent vector has a unique decomposition into a surface-tangent vector and a multiple of $n$; thus $d{\psi }_1=d{\psi }_2$ on $T_{j(\sigma )}U$. This is pointwise finite-dimensional linear algebra and uses no construction of a global normal field or connection.

(Every component of $U_1\cap U_2$ meets $\mathrm{\Sigma }$.) Let $x\in U_1\cap U_2$. By [L4] $x$ lies in one of $I^{-}(\mathrm{\Sigma })$, $\mathrm{\Sigma }$, $I^{+}(\mathrm{\Sigma })$; if $x\in \mathrm{\Sigma }$ its component already meets $\mathrm{\Sigma }$; if $x\in I^{+}(\mathrm{\Sigma })$, choose a future timelike curve $\lambda$ in $M$ from a point of $\mathrm{\Sigma }$ to $x$ ([L4]); by convexity of $U_1$ and of $U_2$ (A development sharing the initial Cauchy surface with a larger development is causally convex in it clauses (a) and (c), applied to the causal curve $\lambda$ and to any subsegment with endpoints in $U_i$) $\lambda \subseteq U_1\cap U_2$, so $x$ lies in the same component of $U_1\cap U_2$ as a point of $\mathrm{\Sigma }$; the case $x\in I^{-}(\mathrm{\Sigma })$ is the time reverse.

($A$ is open without geodesic existence.) Let $q\in A$. Choose charts $x$ and $y$ with $x(q)=0$ and $y({\psi }_i(q))=0$, and shrink to a source coordinate ball $B$ so both $f_i=y\circ {\psi }_i\circ x^{-1}$ are defined there. By continuity shrink $B$ further so the pairs $(f_i(z),D{f}_i(z))$ lie in a fixed compact convex ball $Q$ about their common pair at $0$, with the target-coordinate projection of $Q$ contained in $y$'s domain. For $z\in B$ set $u_i(t)=(f_i(tz),D{f}_i(tz))$, $0\le t\le 1$. Step 1.1 gives ${\dot{f}}_i^{\alpha }=z^b{J}_i^{\alpha }{}_b$ and ${\dot{J}}_i^{\alpha }{}_c=z^b({\mathrm{\Gamma }}^d{}_{bc}(tz)J_i^{\alpha }{}_d-{\mathrm{\Gamma }}^{\mathrm{\prime }\alpha }{}_{\beta \gamma }(f_i)J_i^{\beta }{}_b{J}_i^{\gamma }{}_c)$, a common system ${\dot{u}}_i=F_z(t,u_i)$ with $u_1(0)=u_2(0)$. The right side is smooth near $[0,1]\times Q$, so its derivative in $u$ has a finite bound $L$ there; integration along the straight segment in the convex set $Q$ gives $\mathrm{\mid }{F}_z(t,u)-F_z(t,v)\mathrm{\mid }\le L\mathrm{\mid }u-v\mathrm{\mid }$. If equality holds at the start $a$ of an interval $[a,b]$ with $L(b-a)<1$, the ordinary fundamental theorem of calculus implies $D\le L(b-a)D$, where $D={\max }_{[a,b]}\mathrm{\mid }{u}_1-u_2\mathrm{\mid }$; hence $D=0$. A finite subdivision of $[0,1]$ into such intervals proves $u_1=u_2$ throughout (if $L=0$ one interval suffices). In particular $f_1(z)=f_2(z)$ and $D{f}_1(z)=D{f}_2(z)$ for every $z\in B$. Thus $x^{-1}(B)\subseteq A$, and $A$ is open. Only uniqueness of these already supplied smooth coordinate curves was proved; no selections of solutions are needed.

(Conclusion of (a).) $N$ is connected and $A$ is nonempty, open and closed by the preceding steps; hence $A=N$, which is the assertion of (a).

(The agreement set of (c).) Let $A\subseteq U_1\cap U_2$ be the set of points at which ${\psi }_1={\psi }_2$ and $d{\psi }_1=d{\psi }_2$; on $j(\mathrm{\Sigma })$ the maps have equal values by [L5] and equal differentials by clause (b), so $\mathrm{\Sigma }\subseteq A$. The set $A$ is open in $U_1\cap U_2$ by the argument of step 2.1 applied to coordinate balls contained in the open set $U_1\cap U_2$, and closed by the argument of step 1.3.

(Conclusion of (c).) Since $A$ is open and closed in $U_1\cap U_2$ and meets every component of $U_1\cap U_2$ by step 1.5, $A=U_1\cap U_2$; hence ${\psi }_1={\psi }_2$ on $U_1\cap U_2$, which is clause (c).