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Alexandrov manifold means Alexandrov space which happens to be a manifold, i.e. the space of directions is homeomorphic to shpere. Sorry for introducing this new term.

For such a open manifold does the soul theorem holds? i.e. the manifold is homeomorphic to the disk bundle over it's soul?

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you say "Alexandrov space which happens to be a manifold, i.e. the space of directions is homeomorphic to shpere". These two things are not quite the same. Here is example from "Regularity of limits of noncollapsing sequences of manifolds" by V. Kapovitch: Let $\Sigma^3$ be the Poincare homology sphere with the metric of constant curvature 1. Look at the multiple spherical suspensions of this metric space. –  Anton Petrunin Feb 17 '10 at 20:19
    
Sorry for the misleading of the new term. I am not saying it's the limit of smooth Riemannian manifold. The only requirement is the space of direction is homeo. to sphere. Kapvoitch example is top manifold and Alex. space ofcourse it's not smoothable. –  John B Feb 17 '10 at 20:49
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Could you could explain why you are interested in the question? Suppose you know the answer, what would it give you? Also do you have an example of an "Alexandrov manifold" for which you do not know the answer? –  Igor Belegradek Feb 17 '10 at 21:02
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You did not get it: There is an example of Alexandrov space which is homeomorphic to a mnfld, but it has points with space of directions NOT homeomorphic to a sphere... –  Anton Petrunin Feb 17 '10 at 21:47
    
My bad, I made a mistake here... I know what you mean. Thanks –  John B Feb 17 '10 at 22:03

2 Answers 2

This is not an answer to your question but there is a soul theorem for Alexandrov spaces due to Perelman (see Section 6 in here) with the conclusion that that the space deformation retracts to a soul. As Perelman explains, for Alexandrov spaces it is not true that the space is homeomorphic to a tubular neighborhood of the soul. I am not sure what happens for Alexandrov manifolds.

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It seems that if the soul is a point, then it's homeomorphic to $\mathbb R^n$. Cause Whitehead's example is only for 3-dim. –  John B Feb 17 '10 at 20:53
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You seem to be saying that contractible n-manifolds that are not homeomorphic to Euclidean space exist only in dimension 3. This is not true. For example there are tons of homology spheres in higher dimensions and each of them bounds a contractible manifold. There are also many contractible manifolds that are not interiors of manifolds with boundary. –  Igor Belegradek Feb 17 '10 at 21:07

It might be true. If I would have to prove it I would try to mimic standard proof in Riemannian geometry --- Take $f$ to be distant from the soul. If $f$ has no critical points then you space is homeomorphic to a nbhd of soul (which is likely to be a disc bundle). If there is a critical point then you might try to perturb $f$ by the functions you used in constructing the soul...

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