# How far can I extend a geodesic and still have it be a shortest path?

First, I will set the scene: suppose $M$ is a compact connected smooth manifold equipped with a Riemannian metric. The Hopf-Rinow theorem guarantees that any two points $x,y\in M$ can be joined by a geodesic, and of all the geodesics joining $x$ and $y$, at least one is a shortest path from $x$ to $y$.

Now here is what I am trying to prove. Let $p\in M$. I want to show that there is a neighborhood $U$ of $p$ such that if $x,y\in U$, I can extend a shortest-path geodesic from $x$ to $y$ to a third point $z$ so that the path is also a shortest path from $x$ to $z$. The Hopf-Rinow theorem guarantees that the geodesic from $x$ to $y$ can be extended past $y$, but how can I guarantee that the geodesic continues being the shortest path for even just a little bit?

• I'm not too sure if this leads to a relevant conclusion, but Bellman's Principle of Optimality asserts that if the optimal path from $x$ to $y$ is $\gamma$, then the optimal path from $x$ to any other point on $\gamma$ between $x$ and $y$ is also along $\gamma$. Combining this with the Hopf-Rinow theorem may lead to the conclusion that all extensions of $\gamma$ are also optimal to the new endpoint, but I'm not too sure. – jnez71 May 6 '17 at 17:12
• Unfortunately, no. If $x$ and $y$ are antipodal points on a sphere, there are lots of minimal geodesics from $x$ to $y$, but which ever one you would like to extend does not extend to an optimal path to a third point. – Alex S May 6 '17 at 17:16
• Well the optimal path would still be "on" that geodesic, just if you go the other way around haha :P I see what you're saying, although that could be due to the fact that there are multiple (nonetheless infinite) optimal paths, so Bellman's principle would have to behave a bit differently. Can you think of another counterexample but where there is only one minimal geodesic from $x$ to $y$? Maybe uniqueness is key to the guarantee you seek? – jnez71 May 6 '17 at 18:19

This is, e.g., a consequence of the fact that the injectivity radius (see here: https://en.wikipedia.org/wiki/Cut_locus_(Riemannian_manifold)) of a compact smooth manifold is bounded from below (it's upper semicontinuous, see, e.g. here: The continuity of injectivity radius, but you'll find this also in textbooks about differential geometry). If you look at a neighbourhood of a point $p$ of the form $$\{\exp_p(tv): 0\le t \le r_i(p),\, v\in T_pM, ||v||= 1)$$ with $r_i(p)$ being the injectivity radius of $M$ in $p$, then any geodesic starting in $p$ with length less than $r_i(p)$ will be minimizing.
You just have to make your neighbourhood $U$ small enough to get the result you want.