# Geodesics (Local vs Global)

Let $$M$$ be a Riemannian manifold, and $$p,q\in M$$ two points. Now, if $$M$$ is a complete metric space the Hopf–Rinow theorem ensures that there is a geodesic $$C\subset M$$ joining $$p$$ and $$q$$ that minimizes distance.

On the other hand, we always have the notion of geodesic as curve that locally in a neighborhood of a given point minimizes distance.

My question is the following: can the global geodesic $$C$$ be constructed by "smoothly prolonging" local geodesics? Here with "smoothly prolonging" I mean that we can find local geodesics $$C_1,\dots, C_r\subset M$$ with $$p\in C_1, q\in C_r$$ such that their union $$C_1\cup C_2\cup\dots\cup C_r = C$$ is smooth.

• If I understand your question correctly this is about uniqueness of solutions of ode’s. The answer is yes ( if the metric is sufficiently smooth) Jul 16, 2019 at 17:05
• @ThomasRot I am not even sure if it is about uniqueness; it seems to be about localization. I think the question (which has a positive answer by the way) is essentially: "if $\gamma : [0,1]\to M$ is a length-minimizing geodesic from $\gamma(0)$ to $\gamma(1)$, then for any $0 \leq s < t \leq 1$, is $\gamma|_{[s,t]}$ a length minimizing geodesic between the end-points?" For complete $(M,g)$ this is true. Jul 16, 2019 at 17:09
• completeness is not even required (slip of hand there when I typed the previous comment). No exactly trivial (since the proof is somewhat tedious), but it is done in most Riemannian geometry textbooks. Essentially if there is a shorter path from $\gamma(s)$ to $\gamma(t)$, you can form a piecewise smooth curve joining $\gamma(0)$ and $\gamma(1)$ with smaller length. Then you have to either (i) prove length minimizers are automatically $C^\infty$ or (ii) prove that for any $\epsilon$, a piecewise smooth curve can be smoothly approximated by a curve no more than $\epsilon$ longer. Jul 16, 2019 at 17:23
• What do you mean by local and global geodesics? In Riemannian geometry there is just one notion, simply a geodesic (a curve of zero acceleration). Are you asking if every geodesic is a finite union of minimizing geodesics? This is of course false. Jul 16, 2019 at 19:27
• Every minimizing curve will be geodesic, i.e. have zero acceleration, if you parameterize it by it's arc length. This is covered in any Riemannian geometry textbook. Jul 17, 2019 at 5:09