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Let $X$ be a complete CAT$(0)$ metric space, and $\partial X$ its boundary. One way to define $\partial X$ is as the equivalence class of geodesic rays $\gamma(t), \gamma'(t)$ that remain within a constant distance of one another for large $t$.

Under what conditions and for which $n$ is it known that the boundary of a complete CAT$(0)$ $n$-manifold is homeomorphic to the $(n{-}1)$-sphere $\mathbb{S}^{n-1}$ ?

I believe this is known if $X$ is a complete $n$-dimensional Riemannian manifold of nonpositive sectional curvature, but I have not found clear counterexamples otherwise. I am especially interested in $n{=}3$. Pointers would be appreciated, as this area is relatively new to me. Thanks!

Answered. Here is a snippet from the Davis-Januszkiewicz paper Igor cites, describing an $n{=}5$ example where $\partial X \neq \mathbb{S}^4$:
    Davis-Janu.
I would still be interested to learn if a similar example is known for $n < 5$.

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    $\begingroup$ If I remember right from "Metric spaces of nonpositive curvature" (a book by Bridson and Haeflinger), the Cartan-Hadamard theorem says that a Riemannian manifold is a CAT(0) space if and only if it has nonpositive sectional curvatures. So if I understand your question right the third paragraph implies the answer is "always." $\endgroup$
    – Peter Samuelson
    Sep 2, 2012 at 2:15
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    $\begingroup$ @Peter Samuelson : It is true that a Riemannian manifold is CAT(0) if and only if it has nonpositive curvature. However, for $n \geq 4$ there exist smooth manifolds that can be given metrics (in the sense of "metric spaces") which are CAT(0), but which cannot be given nonpositively curved Riemannian metrics. This was proven for $n \geq 5$ by Davis-Januszkiewicz (see Igor's answer below for the ref) and very recently for $n=4$ by Davis-Januszkiewicz-LaFont (see math.osu.edu/~lafont.1/DMJ161.pdf). $\endgroup$
    – Andy Putman
    Sep 2, 2012 at 2:34
  • $\begingroup$ That is surprising! (At least to someone who doesn't often think about non-smooth manifolds.) $\endgroup$
    – Peter Samuelson
    Sep 2, 2012 at 3:03
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    $\begingroup$ @Peter Samuelson : These aren't non-smooth manifolds, just geodesic metrics on smooth manifolds that are not induced by Riemannian metrics. I'd have to think it through a bit, but I bet that you can arrange it so that the square of the distance function $M \times M \rightarrow \mathbb{R}$ is even smooth. But I agree that it is a surprising result. $\endgroup$
    – Andy Putman
    Sep 2, 2012 at 3:09

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For a piecewise Euclidean or piecewise hyperbolic metric on a PL manifold, the answer is YES. This is proved (p348 in published version) by M. Davis and T. Januszkiewicz in M. Davis, T.Januszkiewicz, Hyperbolization of polyhedra. J. Differential Geom. 34 (1991), no. 2, 347–388. Link: projecteuclid or (non-paywalled) Davis' page.

For a topological manifold with a piecewise Euclidean or piecewise hyperbolic metric, the answer is NO, as shown in the same paper (Section 5).

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    $\begingroup$ PDF download link for the paper cited by Igor projecteuclid.org/euclid.jdg/1214447212 or (non-paywalled) people.math.osu.edu/davis.12/old_papers/djJDG.pdf $\endgroup$
    – Joseph O'Rourke
    Sep 2, 2012 at 13:54
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    $\begingroup$ I find the phrases "for a PL manifold" or "for a topological manifold" misleading. The version of Cartan-Hadamard proved on p348 of Davis-Januszkiewicz is for metrics that are piecewise Euclidean or piecewise hyperbolic on a PL manifolds. This is a very special class of metrics, which in a way is even more rigid than Riemannian metric of nonpositive curvature. In section 5 of the same paper they show that the theorem is false for piecewise Euclidean or piecewise hyperbolic on a topological manifolds. $\endgroup$
    – Igor Belegradek
    Sep 3, 2012 at 1:34
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    $\begingroup$ I see no reason to expect the universal cover to be $\mathbb R^n$ for metrics which are NOT piecewise Euclidean or piecewise hyperbolic. Also a more recent relevant work is this paper math.uchicago.edu/~shmuel/existanceborel.pdf by Bartels-Lueck-Weinberger. $\endgroup$
    – Igor Belegradek
    Sep 3, 2012 at 1:37
  • $\begingroup$ @Igor: Thanks, especially for the reference to the 2010 paper, "On hyperbolic groups with spheres as boundary"---very apropos! $\endgroup$
    – Joseph O'Rourke
    Sep 3, 2012 at 1:48
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    $\begingroup$ @Joseph: if you have not yet done so, you may want to read the proof of Davis-Januszkiewicz's PL Cartan-Hadamard theorem. It is very illuminating. Essenially, the boundary at infinity is the inverse limit of large metric spheres which are connected sums of links in the singular points that lie inside the metric sphere. In the PL case the links are spheres, and with enough work the same follows for the inverse limit. In the non PL case there are points with non-1-connected links, so the fundamental group at infinity is nontrivial, and the universal cover cannot be $\mathbb R^n$. $\endgroup$
    – Igor Belegradek
    Sep 3, 2012 at 2:43

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