Gr13 Convex Normal Neighborhoods

(Assume a compact set is visited near the endpoint.) Suppose the assertion fails for $K$: there are $t_k\in (\alpha ,\beta )$ with $t_k\to \beta$ and $\gamma (t_k)\in K$; passing to a subsequence we may assume $t_k$ is strictly increasing to $\beta$.

(A limit point and a local flow.) By sequential compactness of $K$ [L3] there is a subsequence, renamed $\gamma (t_k)$, converging to some $q\in K$. Apply [L1] at $q$: there are $h>0$, an open neighborhood $U$ of $q$ and a smooth $\mathrm{\Phi }:(-h,h)\times U\to M$ whose slices are integral curves. Since $\gamma (t_k)\to q$, for some $k$ both $\gamma (t_k)\in U$ and $\beta -t_k<h$ hold.

(Extension beyond the maximal time.) Define $\tilde{\gamma }:(\alpha ,t_k+h)\to M$ by $\tilde{\gamma }(t):=\gamma (t)$ for $t<t_k$ and $\tilde{\gamma }(t):=\mathrm{\Phi }(t-t_k,\gamma (t_k))$ for $t\in [t_k,t_k+h)$. The two pieces agree at $t_k$; on the overlap near $t_k$ they are both integral curves with the same value at $t_k$, so they agree by uniqueness of integral curves [L1], and $\tilde{\gamma }$ is an integral curve of $X$ through $p$. Since $t_k+h>\beta$, its domain strictly contains $(\alpha ,\beta )$, contradicting maximality [L2]. Hence no such sequence exists, which is the assertion; the past-end assertion is the same argument applied to $-X$ (the maximal integral curve of $-X$ through $p$ is $t\mapsto \gamma (-t)$, with endpoint $-\alpha$).

(Deterministic diagonal extraction after countable choice.) Enumerate $D=\{q_1,q_2,\dots \}$ [L1]. Compactness gives a finite cover of $B$ by balls of radius $1\mathrm{/}n$ for each $n\ge 1$; [L3] chooses one ordered finite list of centers for each $n$, from this family fixed in advance. Set $A_0=\mathbb{N}$. At stage $n$, classify the indices $m\in A_{n-1}$ by the tuple of least ball indices containing ${\sigma }_m(q_1),\dots ,{\sigma }_m(q_n)$ in the chosen radius-$1\mathrm{/}n$ cover. There are finitely many classes, so at least one is infinite; let $A_n$ be the infinite class with lexicographically least tuple. This defines nested infinite sets by deterministic recursion, without dependent choice. Put $m_0=0$ and let $m_n$ be the least member of $A_n$ greater than $m_{n-1}$. For fixed $i$ and $n\ge i$, every $m_j$ with $j\ge n$ lies in $A_n$, so the values ${\sigma }_{m_j}(q_i)$ lie in one ball of radius $1\mathrm{/}n$ and have pairwise distances less than $2\mathrm{/}n$. They are Cauchy and converge by [L2]. Write $\sigma (q_i)={\lim }_j{\sigma }_{m_j}(q_i)$. The only choice was the simultaneous selection of finite covers.

(The limit on the dense set is Lipschitz and extends to $I$.) For $q,q^{\mathrm{\prime }}\in D$, continuity of $d$ and the Lipschitz bound give $d(\sigma (q),\sigma (q^{\mathrm{\prime }}))={\lim }_{j}d({\sigma }_{m_j}(q),{\sigma }_{m_j}(q^{\mathrm{\prime }}))\le K\mathrm{\mid }q-q^{\mathrm{\prime }}\mathrm{\mid }$ [L2]. For $t\in I$ choose a sequence $q_n\in D$ with $q_n\to t$ [L1]; by the previous estimate $(\sigma (q_n){)}_n$ is Cauchy, hence convergent in the complete space $B$, and the limit does not depend on the chosen sequence (two sequences may be interleaved). Define $\sigma (t)$ as this limit; passing to the limit in the Lipschitz estimate shows $\sigma$ is $K$-Lipschitz on $I$.

(Uniform convergence on compact subintervals.) Let $[a,b]\subseteq I$ be a compact subinterval and let $\epsilon >0$. Choose finitely many points $q_1,\dots ,q_N\in D$ such that every $t\in [a,b]$ satisfies $\mathrm{\mid }t-q_i\mathrm{\mid }\le \epsilon \mathrm{/}(3K+3)$ for some $i$; this is possible, including when $a=b$, because $D$ is dense in $I$ and a compact interval is totally bounded in the sense of admitting such finite nets, the case $K=0$ being immediate from the estimator below with $\epsilon \mathrm{/}(3K+3)$ read as any positive number. Since ${\sigma }_{m_j}(q_i)\to \sigma (q_i)$ for each $i$, there is $j_0$ such that $d({\sigma }_{m_j}(q_i),\sigma (q_i))<\epsilon \mathrm{/}3$ for all $j\ge j_0$ and all $i\le N$. For $t\in [a,b]$ pick $i$ with $\mathrm{\mid }t-q_i\mathrm{\mid }\le \epsilon \mathrm{/}(3K+3)$; then $d({\sigma }_{m_j}(t),\sigma (t))\le d({\sigma }_{m_j}(t),{\sigma }_{m_j}(q_i))+d({\sigma }_{m_j}(q_i),\sigma (q_i))+d(\sigma (q_i),\sigma (t))<\frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon$ for $j\ge j_0$, because both ${\sigma }_{m_j}$ and $\sigma$ are $K$-Lipschitz. Hence ${\sigma }_{m_j}\to \sigma$ uniformly on $[a,b]$.

(A countable cover adapted to the metric.) By [L1] choose a countable cover $\{U_i{\}}_{i\ge 1}$ of $M$ by relatively compact coordinate charts $(U_i,x_i)$ and a smooth partition of unity $\{{\rho }_i\}$ subordinate to it, so that ${\rho }_i$ has compact support in $U_i$, only finitely many ${\rho }_i$ are nonzero near any point, and ${\sum }_i{\rho }_i=1$. For each $i$ let ${\delta }_i$ be the density supported in $U_i$ whose coefficient in the chart $x_i$ is ${\rho }_i\cdot w$, where $w>0$ is the smooth metric volume coefficient of [L2]; then ${\delta }_i\ge 0$, it is a smooth density, and ${\delta }_i>0$ on $\{{\rho }_i>0\}$ whenever this set is nonempty.

(The masses are finite and the weights positive.) Fix $i$. Since $\mathrm{supp}{\delta }_i$ is compact, cover it by finitely many relatively compact coordinate balls and let $\{{\theta }_1,\dots ,{\theta }_m\}$ be a subordinate finite smooth partition of unity. Applying the chart formula of [L5] with that finite cover exhibits ${\mu }_{{\delta }_i}(M)$ as a finite sum of Lebesgue integrals of bounded Borel functions supported in boxes; each term is finite by [L6], so $d_i:={\mu }_{{\delta }_i}(M)<\mathrm{\infty }$. Zero partition terms are allowed, so $d_i$ may be zero. Put $c_i:=(1+d_i{)}^{-1}\in (0,1]$.

(The normalized density.) Define $r_0:={\sum }_{i\ge 1}{2}^{-i}{c}_i{\delta }_i\mathrm{.}$ Only finitely many ${\delta }_i$ are nonzero near any point, so $r_0$ is a smooth density by [L2]; it is strictly positive at every point because in the chart of a point $q$ its coefficient is $w(q){\sum }_i{2}^{-i}{c}_i{\rho }_i(q)$ with $w(q)>0$ and some ${\rho }_i(q)>0$. Applying the chart formula to the locally finite sum and using [L6], ${\mu }_{r_0}(M)={\sum }_{i\ge 1}{2}^{-i}{c}_i{d}_i\le {\sum }_{i\ge 1}{2}^{-i}=1,$ and ${\mu }_{r_0}(M)>0$ by [L4], since $M$ is nonempty and $r_0$ is strictly positive. Hence $C:={\mu }_{r_0}(M)\in (0,1]$. Define $r:=r_0\mathrm{/}C$ and $\mu :={\mu }_r$; by the homogeneity of [L5], $\mu (M)={\mu }_{r_0}(M)\mathrm{/}C=1$.

(The properties of $\mu$.) By step 3.1, $\mu (M)=1$; since $r$ is a positive smooth density, [L3] makes $\mu$ compact-finite, locally finite, $\sigma$-finite and regular Borel. This is (a). Property (b) is the positivity clause of [L4]. For (c), let $(V,y)$ be a chart and let $E\subseteq V$ be Borel with $y(E)$ Lebesgue null. By the chart formula of [L5] applied with a partition whose first chart is $(V,y)$ and using $0\cdot \mathrm{\infty }=0$, $\mu (E)$ is the sum of Lebesgue integrals of nonnegative Borel coefficients over $y(E)$ and over the images of the remaining charts, and each of those images is null in its own chart because the chart transitions are $C^1$ diffeomorphisms on overlaps and carry null sets to null sets [L6]; each integral vanishes by [L6], so $\mu (E)=0$.

(Conclusion.) The positive smooth density $r$ and its measure $\mu ={\mu }_r$ satisfy (a), (b) and (c) by step 4.1, and the construction used only countably many choices through the cover and partition of step 1.1 and the finite selections of step 2.1.

(Projection and full adapted shear.) Choose [L1]'s adapted chart centered at $p$ and write $e_a={\mathrm{\partial }}_{y^a}{\mathrm{\mid }}_p$, with $e_0=n_p$. The subspace $E=\mathrm{span}(e_1,e_2,e_3)$ complements $\mathbb{R}{n}_p$, so [L2]'s projection restricts to an isomorphism from $E$ onto $n_p^{\perp }$. In particular $T(n_p)=0$, not $n_p$. Put $w_k=e_k+g_p(e_k,n_p)n_p$. Finite Gram-Schmidt yields an orthonormal basis $v_i={\sum }_k{C}^k{}_i{w}_k$ of the rest space with $C$ invertible. Define new coordinates by the inverse linear transformation $y^k={\sum }_i{C}^k{}_i{x}^i$ and $y^0=x^0+{\sum }_i{a}_i{x}^i$, where $a_i={\sum }_k{C}^k{}_i{g}_p(e_k,n_p)$. Its block-triangular matrix is invertible. The chain rule gives ${\mathrm{\partial }}_{x^0}={\mathrm{\partial }}_{y^0}=n$ throughout the chart and ${\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p=a_i{n}_p+{\sum }_k{C}^k{}_i{e}_k=v_i$. Hence $g_{00}\equiv -1$, $g_{0i}(p)=0$ and $g_{ij}(p)={\delta }_{ij}$. A coordinate ball centered at zero inside this chart gives the adapted properties of (a).

(Positive time component on a compact temporal locus.) Let $K_0\subset P$ be compact and nonempty. In the $z$ coordinate trivialization define $\mathcal{Z}=\{(q,v):q\in K_0,\mathrm{\mid }v{\mathrm{\mid }}_{\mathrm{e}}=1,g_q(v,v)\le 0,g_q(v,n_q)\le 0\}$. It is closed in the compact product $K_0\times S^3$ by continuity. At each such vector the last inequality is strict by [L3], so $\mathcal{Z}$ consists exactly of normalized nonzero future causal vectors and is nonempty. For $w=\mathrm{\nabla }{z}^0$, set $\lambda =\sqrt{-g(w,w)}>0$ and $m=-w\mathrm{/}\lambda$, a future unit timelike vector on $K_0$. The exact sign is $v^0=d{z}^0(v)=g(w,v)=-\lambda g(m,v)>0$ by [L3]. Thus the continuous time-component function has a positive compact minimum $\beta ={\min }_{\mathcal{Z}}{v}^0>0$. No minimum over points where the gradient may be null is taken.

(Uniform compact cone bounds; (e).) For any nonzero future causal $v$ based in $K_0$, normalization $u=v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}_{\mathrm{e}}$ puts $(q,u)$ in step 1.2's compact cone. Consequently $v^0\ge \beta \mathrm{\mid }v{\mathrm{\mid }}_{\mathrm{e}}$ and $\mathrm{\mid }{v}_s\mathrm{\mid }\le \mathrm{\mid }v{\mathrm{\mid }}_{\mathrm{e}}\le {\beta }^{-1}{v}^0$. Choose $L=\max \{1,{\beta }^{-1}\}$, giving all bounds of (e). If $K_0$ is empty, take $L=1$. If the gradient is past timelike on the whole closure of $U_0⋐U^{\mathrm{\prime }}$, apply this result to $K_0=\overline{U_0}$. For merely pointwise temporal $U_0$, openness from [L4] provides local precompact neighbourhoods with closure in $P$, while each compact subset of $U_0$ is already an admissible $K_0$. This proves exactly the stated local and compact uniform scope.

(One adapted chart with uniform outer and inner cones.) At $p$ in step 1.1's chart, $\mathrm{\nabla }{x}^0=-n_p$. By [L4], choose a closed coordinate ball inside the chart on which this gradient remains past timelike. Also shrink the ball so the coordinate Euclidean operator norm of $g(q)-\eta$ is less than $1\mathrm{/}4$ on its closure. The compact bound of step 2.1 supplies $K\ge 1$ for both the full spatial norm and full coordinate norm on this ball, proving (a)-(b). If $\mathrm{\mid }e{\mathrm{\mid }}_{\mathrm{e}}=1$ and ${\eta }_{16}(e,e)\le 0$, then $\mathrm{\mid }{e}_s{\mathrm{\mid }}^2\le (e^0{)}^2\mathrm{/}16$, $(e^0{)}^2\ge 16\mathrm{/}17$, and $\eta (e,e)\le -15\mathrm{/}17$. Therefore $g_q(e,e)\le -15\mathrm{/}17+1\mathrm{/}4<0$ throughout the ball. Positive scaling proves every nonzero ${\eta }_{16}$-causal vector is $g$-timelike there. All properties hold on the same open coordinate ball $U=x^{-1}(B_r)$.

(Absolutely continuous graph bound; (c).) Assume the declared Countable and Dependent Choice premises and let $\gamma :I\to U$ be future causal. On any compact $[s,t]\subset I$, smooth composition and [L5] make its coordinate functions absolutely continuous; by step 3.1, their almost-everywhere derivatives satisfy $v^0>0$ and $\mathrm{\mid }{v}_s\mathrm{\mid }\le K{v}^0$. The fundamental theorem then gives $x^0(\gamma (t))-x^0(\gamma (s))={\int }_s^t{v}^0>0$. For the spatial vector, the triangle inequality for its component integrals gives $\mathrm{\mid }{x}_s(\gamma (t))-x_s(\gamma (s))\mathrm{\mid }\le {\int }_s^t\mathrm{\mid }{v}_s\mathrm{\mid }\le K{\int }_s^t{v}^0$. These are the claimed bounds. Strict increase and continuity make $x^0(\gamma (I))$ an interval and give a unique spatial value for each of its time values; the increment estimate makes that function $K$-Lipschitz in the Euclidean norm. No smooth inverse time parametrization is assumed.

(Chronology from the flat minorant; (d).) For $q,q^{\mathrm{\prime }}\in U$ satisfying the stated strict ${\eta }_{16}$ inequality, put $d=x(q^{\mathrm{\prime }})-x(q)$. If $d^0>0$, the straight coordinate segment $x(q)+td$, $0\le t\le 1$, stays in the convex ball and its nonzero tangent is $g$-timelike by step 3.1. Its positive $d{x}^0$ component makes it future-directed, since a past timelike vector pairs negatively with the past timelike gradient $\mathrm{\nabla }{x}^0$ and so has negative $d{x}^0$ component. Its coordinate representative is affine, hence absolutely continuous directly; in any other chart it is smooth and has bounded derivative on each sufficiently small compact segment, giving a Lipschitz and hence absolutely continuous representative, with finite subdivision handling a larger compact segment. Thus it belongs to the stated causal-curve class and is a smooth future timelike curve from $q$ to $q^{\mathrm{\prime }}$. Reversing the segment when $d^0<0$ proves the other chronological direction. This establishes the rest of (d), without an absolutely continuous integration argument.

The chart and cone bounds are steps 1.1 and 3.1, the causal graph estimate is step 4.1, the minorant chronology is steps 3.1 and 4.2, and the general compact temporal bounds are step 2.1. They prove (a)-(e), with exactly the declared choice uses and temporal compactness hypotheses.

(Clause (a).) Let $q\in M$ and $v\in T_{q}M$ nonzero, and decompose $v=\alpha n_q+w$ as in [L2]. Then $g_q(v,n_q)=-\alpha$, so $h_q(v,v)=g_q(v,v)+2g_q(v,n_q{)}^2=-{\alpha }^2+g_q(w,w)+2{\alpha }^2={\alpha }^2+g_q(w,w)\mathrm{.}$ If $w\mathrm{\ne }0$ then $g_q(w,w)>0$ by positive definiteness on $n_q^{\perp }$, so $h_q(v,v)>0$; if $w=0$ then $\alpha \mathrm{\ne }0$ and $h_q(v,v)={\alpha }^2>0$. Symmetry of $h$ is inherited from $g$ and the symmetric product $n^{\mathrm{♭}}\otimes n^{\mathrm{♭}}$, and smoothness from [L1]. This proves (a).

(Length and the distance for locally absolutely continuous curves.) Let $M_0$ be a connected component, let $H$ be a smooth Riemannian metric on it, and let $\gamma :[a,b]\to M_0$ be locally absolutely continuous. Its image is compact, so subdivide $[a,b]$ into finitely many pieces whose coordinate images lie in convex chart balls with compact closure. On each piece the smooth metric coefficients are bounded and the coordinate derivative is integrable; hence the integrand defining ${\mathrm{\ell }}_H$ is measurable and dominated by a constant times the Euclidean coordinate speed, so ${\mathrm{\ell }}_H(\gamma )<\mathrm{\infty }$. Fix $\epsilon \in (0,1)$. Refine the subdivision so that on each piece the metric matrix differs from a fixed positive form $H_i$ by relative bounds $(1-\epsilon )H_i\le H\le (1+\epsilon )H_i$, which is possible by continuity and compactness. For the coordinate endpoints $x_i,x_{i+1}$ of a piece, the straight coordinate segment stays in its convex chart ball and has $H$-length at most $\sqrt{1+\epsilon }\mathrm{\mid }{x}_{i+1}-x_i{\mathrm{\mid }}_{H_i}$. Since $x_{i+1}-x_i=\int x^{\mathrm{\prime }}(t)dt$, the triangle inequality for the norm induced by $H_i$ gives $\mathrm{\mid }{x}_{i+1}-x_i{\mathrm{\mid }}_{H_i}\le \int \mathrm{\mid }{x}^{\mathrm{\prime }}(t){\mathrm{\mid }}_{H_i}dt\le (1-\epsilon {)}^{-1\mathrm{/}2}\int \sqrt{H_{\gamma (t)}({\gamma }^{\mathrm{\prime }}(t),{\gamma }^{\mathrm{\prime }}(t))}dt\mathrm{.}$ The distance between the piece endpoints is no larger than the length of this piecewise-$C^1$ segment. Summing over pieces and using the triangle inequality for $d_H$ gives $d_H(\gamma (a),\gamma (b))\le \sqrt{(1+\epsilon )\mathrm{/}(1-\epsilon )}{\mathrm{\ell }}_H(\gamma )$. Letting $\epsilon \downarrow 0$ proves the asserted inequality.

(The conformal factor.) By [L4] choose a smooth nonnegative proper $f$ and set $h^{\mathrm{\prime }}:=(1+\mathrm{\mid }df{\mathrm{\mid }}_h^2)h$. By [L3], applied at each point with $H=h$, the field $h^{\mathrm{\prime }}$ is positive definite and smooth and satisfies $\mathrm{\mid }df{\mathrm{\mid }}_{h^{\mathrm{\prime }}}\le 1$ everywhere. This proves (b).

(The proper function is $1$-Lipschitz for $h^{\mathrm{\prime }}$.) Let $q,q^{\mathrm{\prime }}\in M_0$ and let $\epsilon >0$. By Riemannian distance on a connected manifold, choose a piecewise-$C^1$ curve $\gamma :[a,b]\to M_0$ from $q$ to $q^{\mathrm{\prime }}$ whose $h^{\mathrm{\prime }}$-length is at most $d_{h^{\mathrm{\prime }}}(q,q^{\mathrm{\prime }})+\epsilon$. On each smooth piece, [L5] and Cauchy--Schwarz give $\mathrm{\mid }f(\gamma (v))-f(\gamma (u))\mathrm{\mid }\le {\int }_u^v\mathrm{\mid }df{\mathrm{\mid }}_{h^{\mathrm{\prime }}}\mathrm{\mid }{\gamma }^{\mathrm{\prime }}{\mathrm{\mid }}_{h^{\mathrm{\prime }}}dt\le L_{h^{\mathrm{\prime }}}(\gamma {\mathrm{\mid }}_{[u,v]})$. Summing over the finitely many pieces yields $\mathrm{\mid }f(q^{\mathrm{\prime }})-f(q)\mathrm{\mid }\le L_{h^{\mathrm{\prime }}}(\gamma )\le d_{h^{\mathrm{\prime }}}(q,q^{\mathrm{\prime }})+\epsilon$. Letting $\epsilon \downarrow 0$ gives $\mathrm{\mid }f(q^{\mathrm{\prime }})-f(q)\mathrm{\mid }\le d_{h^{\mathrm{\prime }}}(q,q^{\mathrm{\prime }})$.

(Completeness.) Let $(x_k)$ be a Cauchy sequence in $(M_0,d_{h^{\mathrm{\prime }}})$. There is $k_0$ with $d_{h^{\mathrm{\prime }}}(x_k,x_{k_0})\le 1$ for all $k\ge k_0$; by step 3.1, $f(x_k)\le f(x_{k_0})+1$ for those $k$, so the tail lies in the closed sublevel $K:=f^{-1}([0,f(x_{k_0})+1])$. The set $K$ is compact in $M$ by [L4]. Every connected component is closed (the closure of a connected set is connected), so $K_0:=K\cap M_0$ is compact in $M_0$ and contains the tail. By [L7] that tail has a subsequence converging in $K_0$; the limit is a manifold-topology limit and therefore a $d_{h^{\mathrm{\prime }}}$-limit by [L6]. Since $(x_k)$ is Cauchy, the whole sequence converges to the same limit. Hence $(M_0,d_{h^{\mathrm{\prime }}})$ is complete, which is (c).