A complete no-return extension of a strictly convex manifold

Use [F2] to attach the smooth cylinder $[0,\mathrm{\infty })\times \mathrm{\partial }M$ to $M$ along its boundary. By [F3], the metric coefficients extend smoothly across each boundary point to the double. Compactness of $\mathrm{\partial }M$ gives a finite such cover; [F4] patches these local extensions by a smooth partition of unity. After shrinking the collar, the patched tensor remains positive definite, since it agrees with $g$ on $M$ and positive definiteness is open. Solve the geodesic ODE for the extended metric from each boundary point in the outward unit-normal direction. At $(x,0)$ the derivative of $(x,s)\mapsto {\exp }_x(s{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}(x))$ sends $(w,a)$ to $w+a{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}(x)$, an isomorphism; [F15] therefore gives local inverse charts along the zero section. A finite subcover gives a common domain where this map is a local diffeomorphism. If no uniform collar width made it injective, there would be distinct pairs $(x_j,s_j)$ and $(y_j,t_j)$ with equal images and $0\le s_j,t_j\downarrow 0$; [F20], first for $(x_j)$ and then for $(y_j)$ along the resulting subsequence, gives a common subsequence converging to $(x,0)$ and $(y,0)$, equality of the images forces $x=y$, and both pairs then lie in one local inverse chart at $(x,0)$, a contradiction. Thus a uniform outward normal collar exists.

In this normal collar, use $s$ for outward normal distance and write the extended metric as $d{s}^2+{\tilde{h}}_s$, where ${\tilde{h}}_s$ is a smooth family of metrics on $\mathrm{\partial }M$. The Gauss-lemma orthogonality follows by differentiating $G({\mathrm{\partial }}_s,J)$ along a variation field $J$ of normal geodesics: $D_s{\mathrm{\partial }}_s=0$, $[{\mathrm{\partial }}_s,J]=0$, and metric compatibility give ${\mathrm{\partial }}_{s}G({\mathrm{\partial }}_s,J)=\frac{1}{2}J(G({\mathrm{\partial }}_s,{\mathrm{\partial }}_s))=0$, while the initial value is zero. Also $G({\mathrm{\partial }}_s,{\mathrm{\partial }}_s)=1$. For tangent fields $X,Y$ transported along this collar, metric compatibility and torsion freeness yield ${\mathrm{\partial }}_s{\tilde{h}}_s(X,Y)=G({\mathrm{\nabla }}_X{\mathrm{\partial }}_s,Y)+G(X,{\mathrm{\nabla }}_Y{\mathrm{\partial }}_s)$; hence $\frac{1}{2}{\mathrm{\partial }}_s{\tilde{h}}_s{\mathrm{\mid }}_{s=0}$ is the second fundamental form for the outward normal. By [F1], ${\mathrm{\partial }}_s{\tilde{h}}_s$ is positive definite at $s=0$, and compactness makes it positive definite for $0\le s<2\epsilon$ for some $\epsilon >0$.

Choose smooth nonincreasing ${\xi }_1:[0,\mathrm{\infty })\to [0,1]$ with ${\xi }_1=1$ on $[0,\epsilon ]$ and ${\xi }_1=0$ on $[2\epsilon ,\mathrm{\infty })$, and set ${\xi }_2=1-{\xi }_1$. Since the family ${\tilde{h}}_s\mathrm{/}s$ is bounded relative to the positive metric ${\tilde{h}}_0$ on the compact transition interval $[\epsilon ,2\epsilon ]\times \mathrm{\partial }M$, choose $C>0$ large enough that $sk-{\tilde{h}}_s$ is positive semidefinite there for $k=C{\tilde{h}}_0$. Define $h_s={\xi }_1(s){\tilde{h}}_s+s{\xi }_2(s)k$. This is positive definite: on $[0,\epsilon ]$ it equals ${\tilde{h}}_s$, on $[2\epsilon ,\mathrm{\infty })$ it equals $sk$, and on the transition both terms are positive semidefinite with positive-definite sum.

Differentiate the definition to obtain ${\mathrm{\partial }}_s{h}_s={\xi }_1{\mathrm{\partial }}_s{\tilde{h}}_s+{\xi }_{2}k+{\xi }_2^{\mathrm{\prime }}(sk-{\tilde{h}}_s)$. Each term is positive semidefinite on the transition interval, and at least one of the first two is positive definite at every point; outside the transition, ${\mathrm{\partial }}_s{h}_s={\mathrm{\partial }}_s{\tilde{h}}_s>0$ near $s=0$ and ${\mathrm{\partial }}_s{h}_s=k>0$ for $s\ge 2\epsilon$. Thus ${\mathrm{\partial }}_s{h}_s$ is positive definite for every $s\ge 0$. Gluing $d{s}^2+h_s$ to $g$ gives a smooth Riemannian metric $G$ on the cylinder attachment, because $h_s={\tilde{h}}_s$ near $s=0$.

Let $p:N\to \mathbb{R}$ equal $s$ on the cylinder and be smooth and nonpositive on $M$, agreeing with the collar coordinate near $\mathrm{\partial }M$; such an extension is obtained by a smooth one-variable cutoff on the compact interior side of the collar. The sublevel set $\{p\le R\}$ is closed by continuity of $p$ and contained in the union of compact $M$ and $[0,\max (R,0)]\times \mathrm{\partial }M$, which is compact by [F16, F17]. Therefore the sublevel set is compact by [F19]. The differential of $p$ has bounded metric norm: it has norm $1$ on the cylinder and is bounded on compact $M$. If $C$ is such a bound, then on each $C^1$ piece of a path $\alpha$, [F18] and the chain rule give $\mathrm{\mid }(p\circ \alpha {)}^{\mathrm{\prime }}\mathrm{\mid }\le C\mathrm{\mid }\dot{\alpha }{\mathrm{\mid }}_G$; integrating by [F14] on each piece and summing gives $\mathrm{\mid }p(\alpha (b))-p(\alpha (a))\mathrm{\mid }\le C{L}_G(\alpha )$. Taking the infimum over paths in [F7] shows that $p$ is $C$-Lipschitz for $d_G$. Hence every $d_G$-Cauchy sequence has bounded $p$ and lies in one compact sublevel set. The metric topology agrees locally with the manifold topology by [F9]: the upper coordinate bound makes nearby coordinate points close in $d_G$, and the lower bound on a smaller chart ball makes any path leaving that ball have a fixed positive length. Thus the sublevel is compact as a metric subspace by [F11], and complete by [F12]. Every Cauchy sequence in $N$ therefore converges in $N$, so $(N,G)$ is complete.

The manifold $N$ is connected, nonempty, and boundaryless, so [F13] extends every geodesic from its initial data to all $t\in \mathbb{R}$. Let $\gamma$ start at $\mathrm{\partial }M$ with unit outward or tangent velocity, and put $r(t)=s(\gamma (t))$ while it is in the exterior. From $G=d{s}^2+h_s$, the Levi–Civita formula gives ${\mathrm{Hess}}_{G}s(\dot{\gamma },\dot{\gamma })=\frac{1}{2}({\mathrm{\partial }}_s{h}_s)({\dot{\gamma }}_T,{\dot{\gamma }}_T)\ge 0$, where ${\dot{\gamma }}_T$ is the component tangent to the level set $\{s=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\}$. Thus $r^{\mathrm{\prime }\mathrm{\prime }}(t)\ge 0$. If $r^{\mathrm{\prime }}(0)>0$, then $r^{\mathrm{\prime }}(t)\ge r^{\mathrm{\prime }}(0)$ and $r(t)>0$ grows without bound. If $r^{\mathrm{\prime }}(0)=0$, then $\dot{\gamma }(0)$ is a nonzero boundary-tangent vector and strict convexity gives $r^{\mathrm{\prime }\mathrm{\prime }}(0)>0$; continuity makes $r^{\mathrm{\prime }}(t)>0$ for some sufficiently small $t>0$, after which the same convexity estimate keeps $r^{\mathrm{\prime }}$ positive and $r(t)\to \mathrm{\infty }$. In either case the geodesic never returns to $M$ and eventually leaves every compact subset of $N$.