# Does a (local) uniqueness theorem exist for geodesics in metric spaces?

In Riemannian geometry, we have the next result.

Let $M$ be a smooth manifold with an affine connection. For any point $p\in M$ and for any vector $v\in\mathrm T_pM$, there is a unique geodesic $\gamma\colon I\to M$ such that $$\gamma(0)=p\qquad\text{and}\qquad\gamma'(0)=v$$ where $I$ is a maximal open interval in $\mathbb R$ containing $0$. [Wikipedia]

This can be shown by applying theory of ODEs to the geodesic equation.

One can generalize the notion of geodesics from smooth manifolds with an affine connection to metric spaces. Is there a similar result, possibly assuming additional conditions on the metric space?

The short answer is no. In practice, for a generic metric space one must make specific requiriments (like: goedesics exist, geodesics are unique and so on...)

You can build many counterexamples to the local uniqueness of geodesic by simple means.

Example 1) the $L^1$ metric. No local uniqueness of geodesics between two points.

Take $\mathbb R^2$ with the metric $$d((x,y),(a,b))=|x-a|+|y-b|$$

For any two points of $\mathbb R^2$ there are infinitely many geodesics. (For instance any "monotone curve" from $(0,0)$ to $(1,1)$ is a geodesic)

Example 2) The path metric on a tree. Bifurcation of geodesics.

Take a tree $T$, for instance a regular trivalent tree, and a point $O\in T$. Then you have only 3 directions emanating from $O$, but infinitely many geodesics.

Example 3) the $\sqrt~$ metric. No geodesics.

Take $\mathbb R$ with the metric $d(x,y)=\sqrt{|x-y|}$. There are no paths of finite length except the constant pahts.

One could continue...

• Thank you. I see it doesn't even hold for complete locally compact length spaces. Jul 17, 2014 at 9:23

On the positive side, if $X$ is a goedesic metric space with negative curvature in some sense, for instance if $X$ is locally a CAT$(0)$ space, http://en.wikipedia.org/wiki/CAT%28k%29_space then localy, given two points $x,y$ there is a unique geodesic between them.