# Is the metric completion of a Riemannian manifold always a geodesic space?

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.

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.

• Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/… – Misha Jan 18 '19 at 21:48
• Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold. – macbeth Jan 18 '19 at 21:53
• Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$. – Misha Jan 18 '19 at 21:55
• @macbeth, thanks! – Deane Yang Jan 18 '19 at 23:36