ADM section time and causal spatial speed

In the product splitting, the map $(a,W)\mapsto (a,W+aX)$ is invertible and carries the quadratic form of $g$ to $-N^2{a}^2+h(W,W)$. Smoothness, $N>0$ and positive definiteness of $h$ therefore make $g$ smooth and Lorentzian of the asserted signature. Direct substitution gives $g(n,n)=-1$ and $g(n,W)=0$ for spatial $W$, while the restriction to each slice is $h$; hence $n$ is the asserted smooth unit normal. For a nonzero future causal vector $Y=a{\mathrm{\partial }}_v+W$, its pairing with $n$ is $-Na<0$, so $a>0$. The metric identity gives $\mathrm{\mid }W+aX{\mathrm{\mid }}_h\le Na$. Applied almost everywhere to the original curve parameter $s$, these identities give $v_s>0$ and the coordinate estimate $\mathrm{\mid }{x}_s{\mathrm{\mid }}_{\mathrm{e}}\le B{v}_s$ on any relatively compact coordinate neighborhood, for some finite $B$. The bound follows from smoothness of $N,X,h$ and the metric comparison.

Integrating $v_s>0$ on any nondegenerate compact parameter interval shows that $v$ is strictly increasing. It therefore has a continuous inverse on its image interval. To justify the required vector estimate using scalar integration, let $z:[a,b]\to {\mathbb{R}}^3$ have absolutely continuous coordinates, let $w:[a,b]\to \mathbb{R}$ be absolutely continuous with $w_s\ge 0$ almost everywhere, and suppose $\mathrm{\mid }{z}_s{\mathrm{\mid }}_H\le C{w}_s$ almost everywhere for a fixed inner product $H$ and constant $C\ge 0$. Put $D=z(b)-z(a)$. If $D\mathrm{\ne }0$, fix $e=D\mathrm{/}\mathrm{\mid }D{\mathrm{\mid }}_H$ and the real function $\varphi (s)=H(e,z(s))$. This fixed linear combination of coordinates is absolutely continuous, and ${\varphi }_s=H(e,z_s)$ almost everywhere. The scalar fundamental theorem makes ${\varphi }_s,w_s$ integrable; Cauchy–Schwarz gives ${\varphi }_s\le C{w}_s$. Consequently $\mathrm{\mid }D{\mathrm{\mid }}_H=\varphi (b)-\varphi (a)={\int }_a^b{\varphi }_{s}ds\le C{\int }_a^b{w}_{s}ds=C(w(b)-w(a))$. If $D=0$, the same bound follows from $w(b)\ge w(a)$. On each compact chart subarc apply this estimate with $z=x$, $w=v$, $H$ the Euclidean inner product and $C=B$, obtaining $\mathrm{\mid }x(s_2)-x(s_1){\mathrm{\mid }}_{\mathrm{e}}\le B(v(s_2)-v(s_1))$. Thus $x$ as a function of $v$ is locally Lipschitz; each coordinate is therefore locally absolutely continuous. This argument does not assume absolute continuity of the inverse of $v$.

At an interior differentiability point $v_0$ of $x(v)$, put $p=x(v_0)$ and, in a spatial chart, freeze $H=h_p$, $N_0=N(p)$ and $c=X(p)$, identifying coordinate tangent vectors with ${\mathbb{R}}^3$. On a sufficiently short subarc about $p$, continuity of the coefficients and the coordinate speed bound imply $\mathrm{\mid }{x}_s+v_{s}c{\mathrm{\mid }}_H\le (N_0+ϵ)v_s$ almost everywhere, where $ϵ\to 0$ as the subarc shrinks. On each compact subarc the coordinates of $z(s)=x(s)+v(s)c$ are absolutely continuous, since they are fixed linear combinations of the original absolutely continuous coordinates. Their derivatives and $v_s$ are integrable by the scalar fundamental theorem. Apply the scalar-projection estimate proved in step 2.1 with this $z$, $w=v$, the fixed inner product $H$ and $C=N_0+ϵ$; it yields $\mathrm{\mid }x(s_2)-x(s_1)+(v(s_2)-v(s_1))c{\mathrm{\mid }}_H\le (N_0+ϵ)(v(s_2)-v(s_1))$. Take one endpoint at $v_0$, divide by the positive increase of $v$, and shrink the subarc to $v_0$. The differentiability of $x(v)$ gives $\mathrm{\mid }{x}_v(v_0)+X(p){\mathrm{\mid }}_{h_p}\le N(p)$. By local Lipschitz continuity and the scalar fundamental theorem applied to the three coordinates, such differentiability points have full measure in the $v$-interval. Hence $\mathrm{\mid }{x}_v+X{\mathrm{\mid }}_h\le N$ almost everywhere. Its time component is one, so the reparametrized velocity is nonzero future causal. This proves the reparametrized causal inequality.

The triangle inequality gives $\mathrm{\mid }{x}_v{\mathrm{\mid }}_h\le N+\mathrm{\mid }X{\mathrm{\mid }}_h$. For a smooth real $f$, the smooth-composition chain rule gives $(f\circ x{)}^{\mathrm{\prime }}=df(x_v)=df(x_v+X)-X(f)$ almost everywhere. Applying the gradient identity and Cauchy–Schwarz of [F8] to $x_v+X$ yields $\mathrm{\mid }(f\circ x{)}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }df{\mathrm{\mid }}_h\mathrm{\mid }{x}_v+X{\mathrm{\mid }}_h+\mathrm{\mid }X(f)\mathrm{\mid }\le N\mathrm{\mid }df{\mathrm{\mid }}_h+\mathrm{\mid }X(f)\mathrm{\mid }$. Finally the increasing homeomorphism of parameter intervals preserves all endpoint limits in either direction, proving the inextendibility assertion.