A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] \rightarrow X$ is the sup of $$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + \cdots + d(c(t_{N-1}), c(1)) $$ over all $0 < t_1 < t_2\cdots < t_{N-1} < 1$ and $N > 0$.

A geodesic space is a length space, where for each $x,y \in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.

A Riemannian manifold $M$ and its metric completion $\overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.

But is $\overline{M}$ necessarily a geodesic space? If not, what is a counterexample?

This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds

Also, note that if $\overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $\overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.


1 Answer 1


I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $\Sigma:=\{1,\tfrac{1}{2},\tfrac{1}{3},\ldots\}\cup \{-1,-\tfrac{1}{2},-\tfrac{1}{3},\ldots\}$. The set $\Sigma\cup \{0\}$ is closed in $\mathbb{R}$.

Let $(M,g)$ be the complement of $[0,1]\times (\Sigma\cup\{0\})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $\overline{M}$ of $(M,g)$ contains the following "extra points":

  • $\{0,1\}\times (\Sigma\cup\{0\})$

  • for each $(t,s)\in(0,1)\times\Sigma$, two points $(t,s)_\pm$, coming from the (two different) directional limits $\lim_{y\to s^\pm}(t,y)$.

Importantly, as far as I can tell, there is nothing in $\overline{M}$ corresponding to the points in the segment $(0,1)\times\{0\}$.

If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $\overline{M}$ connecting them.

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    $\begingroup$ Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/… $\endgroup$
    – Misha
    Jan 18, 2019 at 21:48
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    $\begingroup$ Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold. $\endgroup$
    – macbeth
    Jan 18, 2019 at 21:53
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    $\begingroup$ Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$. $\endgroup$
    – Misha
    Jan 18, 2019 at 21:55
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    $\begingroup$ @macbeth, thanks! $\endgroup$
    – Deane Yang
    Jan 18, 2019 at 23:36

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