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?

  • $\begingroup$ 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. $\endgroup$
    – Chris
    Jul 17, 2021 at 14:53
  • $\begingroup$ 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? $\endgroup$
    – Saimel
    Jul 17, 2021 at 15:06
  • $\begingroup$ one thing is the distance among points in $\mathbb R^2$ and another the distance among curves in $\mathbb R^2$ $\endgroup$
    – janmarqz
    Jul 17, 2021 at 15:25

1 Answer 1


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".


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