Let \((M,g)\) be a Riemannian manifold, \(p,q\in M\) and \(0<r<\operatorname{inj}(p),d(p,q)\). Then there is an \(x\in M\) with \(d(p,x)=r\) such that
\begin{equation} d(p,x)+d(x,q)=d(p,q).\nonumber \\ \end{equation}Every piecewise \(C^{1}\)-curve between \(p,q\) has to intersect \(\exp_{p}(\{|v|=r\})\). So there is a sequence \(v_{i}\in T_{p}M\), \(|v_{i}|=r\) such that
\begin{equation} d(p,\exp_{p}(v_{i}))+d(\exp_{p}(v_{i}))\to d(p,q).\nonumber \\ \end{equation}By compactness, there is a \(v\in T_{p}M\), \(|v|=r\) such that
\begin{equation} d(p,\exp_{p}(v))+d(\exp_{p}(v))\to d(p,q).\nonumber \\ \end{equation}Set \(x=\exp_{p}(v)\). ∎
Let \((M,g)\) be a connected Riemannian manifold. Then the following are equivalent:
\((M,d)\) is complete.
\((M,g)\) is geodesically complete.
\(\exp_{p}\) is defined on all of \(T_{p}M\) for some \(p\in M\).
There is a sequence \(K_{1}\subset K_{2}\subset\ldots\) of compact subsets of \(M\) such that if \(x_{i}\not\in K_{i}\), then \(d(p,x_{i})\to\infty\) for any \(p\in M\).
Closed and bounded subsets (with respect to \(d\)) are compact.
All these statements imply that between any two points \(p,q\in M\) there is at least one minimizing geodesic.
Note that compact manifolds are always complete.
Let \(\gamma_{v}:(-l_{-v},l_{v})\to M\) be a geodesic and assume that \(l_{v}<\infty\). Then \(\lim_{t\to l_{v}}\gamma_{v}(t)\) does not exist. Note that
\begin{equation} d(\gamma_{v}(t_{1}),\gamma_{v}(t_{2}))\leq\ell(\gamma_{v}|_{[t_{1},t_{2}]})\leq|v|\cdot|t_{1}-t_{2}|.\nonumber \\ \end{equation}So \(\gamma_{v}(t)\) is a Cauchy “sequence”. So \(\gamma_{v}(t)\) has a limit.
Geodesically complete implies that \(\exp_{p}\) is defined everywhere.
Assume that \(\exp_{p}\) is defined everywhere. We will now show that for any point \(q\in M\) there is a minimizing geodesic between \(p,q\). Set \(d:=d(p,q)\) and let \(0<r<\operatorname{inj}(p)\). Determine \(x=\exp_{p}(rv)\), \(|v|=1\) as in the previous Lemma. We claim that \(q=\gamma_{v}(d)\) and that \(\gamma_{v}|_{[0,d]}\) is minimizing. Choose \(t_{0}\in[0,d]\) maximal such that
\begin{equation} t_{0}+d(\gamma_{v}(t_{0}),q)=d(p,q).\nonumber \\ \end{equation}Assume that \(t_{0}<d\). Apply the previous Lemma at \(p^{\prime}:=\gamma_{v}(t_{0})\) with some \(r^{\prime}>0\) and obtain \(x^{\prime}\in M\) and a minimizing geodesic \(\sigma:[0,r^{\prime}]\to M\) between \(p^{\prime},x^{\prime}\) such that
\begin{equation} d(p^{\prime},x^{\prime})+d(x^{\prime},q)=d(p^{\prime},q).\nonumber \\ \end{equation}It follows that
\begin{equation} t_{0}+r^{\prime}+d(x^{\prime},q)=t_{0}+d(p^{\prime},x^{\prime})+d(x^{\prime},q)=d(p,q).\nonumber \\ \end{equation}Now
\begin{equation} d(p,x^{\prime})+d(x^{\prime},q)\leq t_{0}+d(p^{\prime},x^{\prime})+d(x^{\prime},q)=d(p,q)\leq d(p,x^{\prime})+d(x^{\prime},q).\nonumber \\ \end{equation}So we have
\begin{equation} d(p,x^{\prime})=t_{0}+r^{\prime}=\ell(\gamma_{v}|_{[0,t_{0}]})+\ell(\sigma).\nonumber \\ \end{equation}So the concatenation of \(\gamma_{v}|_{[0,t_{0}]}\) and \(\sigma\) is a minimizing geodesic and hence smooth. So for \(t^{\prime}=t_{0}+r^{\prime}\)
\begin{equation} x^{\prime}=\gamma_{v}(t^{\prime}).\nonumber \\ \end{equation}By choice of \(t_{0}\), we must have \(t_{0}=d\). So
\begin{equation} d+d(\gamma_{v}(d),q)=d(p,\gamma_{v}(d))+d(\gamma_{v}(d))=d(p,q)=d.\nonumber \\ \end{equation}So \(\gamma_{v}(d)=d\). Also
\begin{equation} \ell(\gamma_{v}|_{[0,d]})=d.\nonumber \\ \end{equation}Assume that \(\exp_{p}\) is defined on \(T_{p}M\). Then \(M=\exp_{p}(T_{p}M)\). So we can set \(K_{i}:=\exp_{p}(\overline{B(0_{p},i)})\).
Assume that the \(K_{i}\) exist and that \(S\subset M\) is bounded and closed. Then \(S\subset K_{i}\) for large \(i\), so it’s compact.
Finally, let \(\{x_{i}\}\) be a Cauchy sequence. Then the closure of \(\{x_{i}\}\) is bounded and hence \(\{x_{i}\}\) subconverges. ∎