It is well-known that the completion of a metric space which is homeomorphic to a ball can be very wild; in fact, I think, every compact manifold is the closure of an open ball!

But CAT(0) spaces are very different from general manifolds. If I have a CAT(0) with the path metric, that is totally bounded, and that is homeomorphic to an open ball, will its completion be a closed ball?

  • $\begingroup$ I don't understand the question. What is "totally bounded"? "homeomorphic to an open ball", what does it mean? open ball of what? $\endgroup$
    – YCor
    Jun 3, 2017 at 9:01
  • $\begingroup$ I am not able to construct an example right now with bare hands, but I would expect that the topology on the closure could be strictly finer, maybe even discrete on the boundary in certain examples. $\endgroup$ Jun 3, 2017 at 9:52
  • $\begingroup$ @SebastianGoette thanks; so what's the difference with "bounded"? Also what's a CAT(0) with the path metric, since a CAT(0) space is by definition geodesic? $\endgroup$
    – YCor
    Jun 3, 2017 at 9:54
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    $\begingroup$ @YCor, it's polite not to criticize terminology before googling to see if it's standard. The first hit for "totally bounded" is the Wikipedia article giving the definition: en.m.wikipedia.org/wiki/Totally_bounded_space . $\endgroup$
    – HJRW
    Jun 4, 2017 at 5:44
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    $\begingroup$ For the record, the question can be restated as: Let a bounded CAT(0) metric space be homeomorphic to an open ball of some Euclidean space. Is its completion homeomorphic to a closed ball (in the same Euclidean space)? at least assuming if necessary that the completion is compact? $\endgroup$
    – YCor
    Jun 4, 2017 at 11:34

1 Answer 1


The answer is "no".

Let $\Sigma$ be the suspension over Poincaré homology sphere. It admits a polyhedral $\mathrm{CAT}[1]$-metric.

Let $B$ be the unit ball in the Euclidean cone $\mathrm{Cone}\,\Sigma$.

Note that $B$ is a compact $\mathrm{CAT}[0]$-space; its interior is homeomorphic to the ball in $\mathbb{E}^5$ but its boundary is homeomorphic to $\Sigma$, which is not a manifold.


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