= Solution
Work on a connected, finite-dimensional <Riemannian manifold> without boundary. It is <geodesically complete> if every maximal affinely parametrized <geodesic> has parameter domain $\mathbb R$. Equivalently, for every $p$ the <Riemannian exponential map> $\exp_p$ is defined on all of $T_pM$: any finite affine time can be rescaled to time one, and negative times correspond to reversing the initial velocity.
The <Hopf-Rinow lemma> in its minimizing-geodesic form says: \b[if $\exp_p$ is defined on all of $T_pM$ for one point $p$, then every $q$ can be joined to $p$ by a geodesic of length $d(p,q)$.] We first establish the local ingredient, the <distance splitting through a small geodesic sphere>. If $D=d(p,q)>0$, choose $0<\varepsilon<D$ sufficiently small that the closed normal ball about $p$ is compact and radial lengths give <Riemannian distance>, as proved using the <Gauss lemma>. Take paths from $p$ to $q$ of lengths tending to $D$. Their first intersections $x_j$ with its boundary sphere have a subsequence converging to some $x$ on the sphere. Their lengths are at least $\varepsilon+d(x_j,q)$, giving $D\geq\varepsilon+d(x,q)$. The <triangle inequality> gives the reverse inequality. Therefore
$$
\boxed{d(p,q)=\varepsilon+d(x,q),\qquad d(p,x)=\varepsilon.}
$$
This splitting uses only a compact local <geodesic sphere>, not completeness or the existence of a globally minimizing path.
We also need that <a minimizing broken geodesic has no corner>. To see this directly, let the incoming and outgoing unit tangents at a junction be $u$ and $v$. Choose one nearby point on each segment so that they and the junction lie in a common <convex normal neighbourhood>. Move the junction with velocity $z$, joining it to those two fixed points by the unique short <geodesics>. The <Gauss lemma>, or the first variation formula, gives the derivative of the sum of their lengths as $\langle u-v,z\rangle$. If $u\ne v$, setting $z=v-u$ makes this derivative $-|u-v|^2<0$, contradicting minimality. Thus the tangent vectors agree, and uniqueness for the <geodesic equation> makes the broken path one smooth <geodesic>.
Now choose the unit-speed radial <geodesic> $\gamma$ through the point $x$ obtained by the distance splitting, with $\gamma(0)=p$ and $\gamma(\varepsilon)=x$. By the hypothesis on $\exp_p$, it is defined at least up to time $D$. Define
$$
A=\{t\in[\varepsilon,D]:d(\gamma(t),q)=D-t\}.
$$
This is a nonempty closed set. If $t\in A$, the <triangle inequality> and the length of $\gamma|_{[0,t]}$ give
$$
D\leq d(p,\gamma(t))+d(\gamma(t),q)\leq t+(D-t)=D,
$$
so $d(p,\gamma(t))=t$: that geodesic segment minimizes. Let $t_* =\max A$. If $t_*<D$, apply the <distance splitting through a small geodesic sphere> at $\gamma(t_*)$ with radius $0<\delta<D-t_*$. There is a point $y$, joined to $\gamma(t_*)$ by a short minimizing radial <geodesic> $\sigma$, for which
$$
d(\gamma(t_*),q)=\delta+d(y,q).
$$
The broken path $\gamma|_{[0,t_*]}*\sigma$ has length $t_*+\delta$. Moreover,
$$
D\leq d(p,y)+d(y,q)\leq t_*+\delta+d(y,q)=D.
$$
Consequently that broken path minimizes. It has no corner, so $\sigma$ is the continuation of $\gamma$ and $y=\gamma(t_*+\delta)$. The last distance equality puts $t_*+\delta$ in $A$, a contradiction. Hence $D\in A$ and $\gamma(D)=q$. This proves the <Hopf-Rinow lemma>, with the case $p=q$ supplied by the constant <geodesic>.
The <Hopf-Rinow theorem> asserts the equivalence of
* completeness for the <Riemannian distance>;
* <geodesic completeness>;
* compactness of every closed bounded subset.
These conditions imply the existence of a <minimizing geodesic> between any two points. Here is the deduction, including the implication back to geodesic completeness. If the manifold is <geodesically complete>, the <Hopf-Rinow lemma> applies at each point. For every $R\geq0$,
$$
\overline B(p,R)=\exp_p\bigl(\{v\in T_pM:|v|\leq R\}\bigr).
$$
One inclusion follows from the length of the radial <geodesic>; the other follows by taking an initial velocity of a minimizing one. The tangent-space ball is compact, so the metric ball is compact as its continuous image. Every closed bounded set is a closed subset of such a ball and is compact. A <Cauchy sequence> is bounded and therefore has a convergent subsequence in a compact ball; the Cauchy property makes the whole sequence converge. This proves completeness of the <Riemannian distance>.
Conversely, assume metric completeness and let $\gamma:[0,b)\to M$ be a <geodesic> with $b<\infty$. Its speed $c$ is constant, so $d(\gamma(s),\gamma(t))\leq c|s-t|$. Completeness therefore gives a limit $q$ as $t\uparrow b$. Eventually $\gamma$ lies in a coordinate neighbourhood with compact closure on which the <Riemannian metric> is uniformly comparable to the Euclidean metric. Its coordinate velocity is bounded by the constant-speed estimate. The <Christoffel symbols> are bounded there, so the <geodesic equation>
$$
\ddot x^k=-\Gamma^k_{ij}(x)\dot x^i\dot x^j
$$
makes its coordinate acceleration bounded. Thus the coordinate velocity also has a limit $v$ at time $b$. The local existence and uniqueness theorem for smooth <ordinary differential equations> extends the solution from initial data $(q,v)$ past $b$, contradicting maximality. Reversing time treats a finite left endpoint. Hence metric completeness implies <geodesic completeness>, and the preceding argument supplies compactness of closed bounded sets. Compactness of closed bounded sets already implies metric completeness by the Cauchy-sequence argument, completing all equivalences.
In fact, the one-point hypothesis of the <Hopf-Rinow lemma> suffices for the theorem: the same closed-ball image argument at that point puts every Cauchy sequence in a compact ball, giving metric completeness and then geodesic completeness at every point. By contrast, the existence of a <minimizing geodesic> for every pair alone does not imply completeness: an open Euclidean ball has minimizing straight segments between all its points but has Cauchy sequences converging to its missing boundary.
\b[Geodesic completeness, metric completeness and compactness of closed bounded sets are equivalent; under these conditions every pair is joined by a minimizing geodesic.]
Back to article page
