22
$\begingroup$

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.

$\endgroup$

1 Answer 1

17
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @macbeth, thanks! $\endgroup$
    – Deane Yang
    Jan 18, 2019 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.