# Is a weakly contractible connected metric space, uniquely geodesic?

A topological space is weakly contractible if all the homotopy groups are trivial.
It's connected if it's not the union of two disjoint nonempty open sets.
A metric space $(X,d)$ is uniquely geodesic if two points $x,y \in X$ are connected by a unique path of minimal length, precisely $d(x,y)$.

Question : Is a weakly contractible connected metric space, uniquely geodesic ?

In the case of a negative answer :
- What are the classical counter-examples ?
- Are there natural additive conditions for having an affirmative answer ?

## 1 Answer

The topological property of being weakly contractible does not say much about the metric.

The real line with the metric $d(x,y)=|x-y|^{1/2}$ is not geodesic; in fact it has no paths of finite length.

Pac-Man shape with the restriction metric from $\mathbb R^2$ gives another example: rectifiably connected, but not geodesic.

Yet another example: a non-strictly convex normed space such as $\ell_1$ or $\ell_\infty$. These are geodesic, but not uniquely geodesic.

A sufficient condition for being uniquely geodesic is the triangle comparison property dubbed $\mathrm{CAT}(0)$.

• I see, thank you very much ! The CAT(0) condition is not necessary, isn't it ? Do you have an example of uniquely geodesic non-CAT(0) contractible space ? – Sebastien Palcoux Aug 20 '13 at 18:23
• @SébastienPalcoux Yes, for example $\ell_p$ for $1<p<\infty$, $p\ne 2$, does not satisfy any curvature bounds, but is uniquely geodesic. The condition of being uniquely geodesic is so simple itself that I would not expect to have something related to it that's easier to verify that the condition itself. – user90090 Aug 20 '13 at 19:19