# geodesic in metric space and in manifolds

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?

• In general "differential geometry geodesics" are only locally length minimizing; consider on the sphere the path from the north pole to the south pole, and then go a little further along this path. This is a geodesic, but the quicker path would have been to go the other way around the sphere. Jul 17, 2021 at 14:53
• Yes I understand that. But what I am confused about is that the authors claim (at least I understand it like this) is : differential geometry geodesics are locally metric geodesics. But I think the curve $\gamma$ is a counterexample? Jul 17, 2021 at 15:06
• one thing is the distance among points in $\mathbb R^2$ and another the distance among curves in $\mathbb R^2$ Jul 17, 2021 at 15:25

I have seen definition where they declare geodesic in a metric space to be curves which satisfy $$d(c(s), c(t)) = v|t-s|,$$ for some $$v\ge 0$$. See here for example. The definition you used might be called "unit speed geodesic in $$(X, d)$$".
What the authors want to say is that geodesics in Riemannian geometry might not be geodesic in your sense, even if it is of unit speed. A simple example is the curve $$c (t) = [t]$$ in the one dimensional manifold $$\mathbb R/\mathbb Z$$ with the Euclidean metric. Since $$c(0) = c(1)$$,
$$d(c(0), c(1)) = 0 \neq 1 = |1-0|.$$
On the other hand, if $$c(t)$$ is a unit speed geodesic in a Riemannian manifold $$(M, g)$$, then for each $$t_0$$, there is $$\epsilon>0$$ so that $$c|_{[t_0-\epsilon, t_0+ \epsilon]}$$ is length minimizing and $$c|_{[t_0-\epsilon, t_0+ \epsilon]}$$ is a geodesic in the sense of metric space. Thus the term "local geodesic".