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For Alexandrov manifold in the title we mean 3-dim Alexandrov apace which is also a topological. manifold. Shioya-Yamaguchi posted a conjecture on their paper "Collapsing 3-manifold with lower sectional curvature bound"

Any three-dimensional compact, simply connected, nonnegatively curved Alexandrov space without boundary which is a topological manifold is homeomorphic to a sphere.

It seems for me that Riemannian geometric tools do not apply here. So any progress in this directoin or it has been proved somewhere?

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The Poincare conjecture in dimension 3 tells us that any 3-dimensional compact simply connected topological manifold is diffeomorphic to the 3-sphere. Surely you don't want to assume that your space is a manifold? Is its dimension as a manifold perhaps larger than 3?

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  • $\begingroup$ @Ben, Yes, I know that, that's why I ask Alexandrov space. $\endgroup$
    – Ralph
    Commented Apr 16, 2013 at 20:39
  • $\begingroup$ Ralph: Forget about metric conditions, you have the condition of the space being simply-connected 3-dimensional topological manifold. I think, you are misstating their conjecture. $\endgroup$
    – Misha
    Commented Apr 16, 2013 at 21:46
  • $\begingroup$ @Misah, I copied the Conjecture 0.8. from their paper. The paper is in 2000, before the Perelman's proof. probably they thought there might be a geometric approach of Poincare Conjecture assuming nonnegative curvature? $\endgroup$
    – Ralph
    Commented Apr 17, 2013 at 8:55
  • $\begingroup$ @Ben, I think you are right, it follows from Poincare Conj now. I was confused by the statement at the beginning. $\endgroup$
    – Ralph
    Commented Apr 17, 2013 at 8:56

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