A compact spacelike surface has a normal exponential tube

In a surface slice chart let $E_1,E_2$ be its smooth tangent basis and put $G_{ij}=g(E_i,E_j)$. Positivity makes $G$ invertible. Define $PV={\sum }_{i,j=1}^2{E}_i(G^{-1}{)}^{ij}g(E_j,V)$. Then $P$ is smooth, $P{\mathrm{\mid }}_{TS}$ is the identity, and $g(V-PV,E_i)=0$. Thus $TM{\mathrm{\mid }}_S=TS\oplus \ker P$, with $\ker P=NS$ of rank two; this also gives the smooth normal bundle since the projections agree on chart overlaps by uniqueness. The supplied $n,\nu$ are linearly independent by their displayed inner products, so span $NS$.

By the joint smoothness in [F1] and the smooth supplied frame, $\mathrm{\Xi }$ is smooth on an open subset of $S\times {\mathbb{R}}^2$ containing its zero section. Since $\mathrm{\Xi }(s,0,0)=s$ and its vertical differential is the identity on normal vectors, its differential there is $(X,u,v)\mapsto X+u{n}_s+v{\nu }_s$. Step 1.1 makes this an isomorphism. In local charts the determinant of this differential is continuous, so the locus $U$ where $\mathrm{\Xi }$ is defined and its differential invertible is open and contains $S\times \{(0,0)\}$. By [F2], $\mathrm{\Xi }$ is a local diffeomorphism throughout $U$.

For the boundary assertion use the full three-dimensional tangent spaces of ${\mathrm{\Sigma }}_{+}$, including at $S$. In a boundary chart their basis $E_1,E_2,E_3$ is smooth up to the boundary. With $H_{ij}=g(E_i,E_j)$, the same formula as in step 1.1, now summed from $1$ to $3$, defines the unique smooth orthogonal projection $P_{\mathrm{\Sigma }}$ onto $T{\mathrm{\Sigma }}_{+}$. Write $w=N-P_{\mathrm{\Sigma }}N$. Orthogonality gives $g(w,w)=g(N,N)-g(P_{\mathrm{\Sigma }}N,P_{\mathrm{\Sigma }}N)<0$ and $g(w,N)=g(w,w)<0$. Thus $n=w\mathrm{/}\sqrt{-g(w,w)}$ is the future unit normal, smooth up to $S$. The projection is intrinsic, so these formulas agree on overlaps. Inverse Gram matrices are smooth since their determinants are nonzero, using the adjugate formula. This construction uses only boundary tangent bundles and requires no extension of the hypersurface through its boundary.

Apply [F6] with $X=K=S$, $Z={\mathbb{R}}^2$, $z_0=(0,0)$ and $N=U$. It supplies an open $W$ about $(0,0)$ with $S\times W\subset U$. Choose ${ϵ}_0>0$ so that $D_{{ϵ}_0}=\{(a,b):\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }<{ϵ}_0\}\subset W$. Thus $\mathrm{\Xi }$ is defined and is a local diffeomorphism on the entire uniform domain $S\times D_{{ϵ}_0}$.

The surface $S$, with its slice-chart smooth structure and subspace topology, is a topological manifold. Apply [F7] under [F3] to give $S$ a compatible metric; its assumed topological compactness is compactness for that metric. Suppose no radius smaller than ${ϵ}_0$ makes $\mathrm{\Xi }$ injective. Using Choice, for each integer $k\ge 1$ select unequal data $(s_k,a_k,b_k)$ and $(s_k^{\mathrm{\prime }},a_k^{\mathrm{\prime }},b_k^{\mathrm{\prime }})$ in $S\times D_{{ϵ}_0\mathrm{/}(k+1)}$ with the same image. Sequential compactness first gives a subsequence with $s_k\to s$ and then a further subsequence with $s_k^{\mathrm{\prime }}\to s^{\mathrm{\prime }}$. Along it all four fiber coordinates tend to zero. Continuity and $\mathrm{\Xi }(s,0,0)=s$ show that the common images tend to both $s$ and $s^{\mathrm{\prime }}$ in the Hausdorff manifold $M$, so $s=s^{\mathrm{\prime }}$. Both data sequences therefore eventually lie in one inverse-function neighborhood of $(s,0,0)$ from step 2.1, contradicting its injectivity. Hence some $0<ϵ<{ϵ}_0$ gives injectivity on $S\times D_{ϵ}$.

A local diffeomorphism is open, because its local restrictions have open images. Hence the injective restriction from step 4.1 has open image $O$ containing $S$; its local smooth inverses agree by injectivity and give its smooth global inverse on $O$. If $(a,b)\in D_{ϵ}$ and $0\le r\le 1$, then $(ra,rb)\in D_{ϵ}$. The scaling law in [F1] therefore identifies the entire normal radial geodesic segment with $r\mapsto \mathrm{\Xi }(s,ra,rb)$ in $O$.

In each boundary chart of ${\mathrm{\Sigma }}_{+}$ put $\rho =x^3\ge 0$, so $S$ is $\rho =0$ locally and $d\rho \mathrm{\ne }0$ there. Set $q={\sum }_{i,j=1}^3{E}_i(H^{-1}{)}^{ij}d\rho (E_j)$; then $h(q,V)=d\rho (V)$ for every $V\in T{\mathrm{\Sigma }}_{+}$. The smooth vector $q$ is nonzero, is orthogonal to $TS$, and satisfies $d\rho (q)=h(q,q)>0$. Consequently $\nu =q\mathrm{/}\sqrt{h(q,q)}$ along $S$ is the smooth inward unit boundary normal. On overlaps it agrees because the orthogonal line to $TS$ in the positive definite space $T{\mathrm{\Sigma }}_{+}$ has exactly one inward unit vector, with inwardness chart independent by [F5]. Since $n\perp T{\mathrm{\Sigma }}_{+}$, one has $g(n,\nu )=0$. Both fields are normal to $S$, so step 1.1 shows they form the required smooth orthonormal normal frame. Together with step 5.1 this proves every assertion.