In the book by ''Metric spaces of non-positive curvature'' by Bridson and Haefliger we have the following definition for a geodesic in a metric space:
Let $(X,d)$ be a metric space. A map $c:[0,l]\longrightarrow X$ is a geodesic if for all $s,t \in [0,l]$ we have $d(c(s),c(t))=\vert t-s \vert$.
So far so good. In the Example underneath this Definition they state the following:
''We emphasize that the paths which are commonly called geodesics in differential geometry need not be geodesics in the metric sense; $\textbf{in general they will only be local geodesics}$.''
I assume they mean by ''metric sense'' the metric on our manifold that is induced by our riemannian metric. Otherwise I don't know what they mean?
But if this is true I'm quite confused about this, since $\gamma:[0,1] \longrightarrow \mathbb{R}^2, t \longmapsto 2t$ is a geodesic in the riemannian sense (if we consider the standard riemannian metric on $\mathbb{R}^2$ with induced connection). But this will never be a local geodesic in the metric sense, as $d(\gamma(s),\gamma(t))=2\vert t-s \vert$ for all $s,t \in [0,1]$.
Where fails my thinking? And if it does not fail, what is the connection between geodesics in the metric sense and riemannian sense?