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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 $\partial X \neq \mathbb{S}^4$:
I would still be interested to learn if a similar example is known for $n < 5$.

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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." –  Peter Samuelson Sep 2 '12 at 2:15
@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 –  Andy Putman Sep 2 '12 at 2:34
That is surprising! (At least to someone who doesn't often think about non-smooth manifolds.) –  Peter Samuelson Sep 2 '12 at 3:03
@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. –  Andy Putman Sep 2 '12 at 3:09

1 Answer 1

up vote 12 down vote accepted

For a PL manifold, the answer is YES. This is proved by M. Davis and T. Januszkiewicz in Davis, Michael W.(1-OHS); Januszkiewicz, Tadeusz(PL-WROC) Hyperbolization of polyhedra. J. Differential Geom. 34 (1991), no. 2, 347–388.

For a topological manifold the answer is NO, as shown in the same paper.

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PDF download link for the paper cited by Igor or (non-paywalled) –  Joseph O'Rourke Sep 2 '12 at 13:54
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. –  Igor Belegradek Sep 3 '12 at 1:34
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 by Bartels-Lueck-Weinberger. –  Igor Belegradek Sep 3 '12 at 1:37
@Igor: Thanks, especially for the reference to the 2010 paper, "On hyperbolic groups with spheres as boundary"---very apropos! –  Joseph O'Rourke Sep 3 '12 at 1:48
@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 $\mathbb R^n$. –  Igor Belegradek Sep 3 '12 at 2:43

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