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Let $X$ be a topological manifold of dimension $n$, equipped with a compatible CAT(0) metric. Are sufficiently small metric spheres in $X$ homeomorphic to metric spheres in Euclidean space $\mathbb{E}^n$?

[In "Ideal boundary of CAT(0) spaces" (1998) by Myung-Jin Jeon, this was unclear to the author; see the bottom of Page 104. That paper examined geodesic completeness for CAT(0) manifolds.]

EDIT: It is well-known that sufficiently small metric balls in any CAT(0) space are contractible. Using the manifold property, I believe that my question reduces to a very basic one:

If $U$ is a contractible open subset of $\mathbb{R}^n$, then is $U$ homeomorphic to $\mathbb{R}^n$?

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The answer to the question in the edit is no. The Whitehead manifold embeds in $S^3$. See en.wikipedia.org/wiki/Whitehead_manifold –  Richard Kent Aug 21 '11 at 3:24
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The question in the edit has a negative answer in all dimensions $>2$; read math.niu.edu/~rusin/known-math/95/contractible. In dimensions $>3$ the point is that there are many compact contractible manifolds, and the double of a compact contractible manifold along the boundary is a homotopy sphere, hence it is homeomorphic to $S^n$, and removing a point gives $\mathbb R^n$. –  Igor Belegradek Aug 21 '11 at 13:35
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The answer is no. By the double suspension theorem of Cannon and Edwards, if $X^n$ is a homology $n$-sphere, then the double suspension $S^2X$ is a sphere. In particular, the cone $CSX$ on $SX$ will be a simply connected manifold homeomorphic to $R^{n+2}$, since it is obtained from the double suspension by deleting a point. If there is a simplicial complex structure on $SX$ which is flag (e.g. by barycentric subdivision), then the cone $CSX$ will have a CAT(0) metric. The point is that one may make each of the simplices of the triangulation of $SX$ into a right-angled $n+1$ spherical simplex, then take the metric cone on this to get a quadrant of $R^{n+2}$, and glue these quadrants together in the manner prescribed by the simplicial complex. The metric ball about the cone point of $CSX$ will not be homeomorphic to a ball, since its boundary will be $SX$ which is not a manifold. Check out the papers of Davis-Januskiewicz for constructions of CAT(0) complexes.

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Okay, sorry, I got confused by the different uses of $X$... for my original question, a counter-example is $X := CS\Sigma$ where $\Sigma$ is a closed $n$-manifold with the homology type, but not the homotopy type, of a sphere. By Cannon--Edwards, $X \approx \mathbb{R}^{n+2}$, and by Davis--Januskiewicz, $X$ is CAT(0). Thanks again! –  Qayum Khan Sep 5 '11 at 1:40
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