Perelman's stability theorem shows in particular that a finite dimensional compact Alexandrov space $(X,d)$ such that $X$ is not a topological manifold cannot be approximated in the Gromov-Hausdorf topology by Riemannian manifolds of the same dimension whose sectional curvature is bounded from below.

My questions are :

(1) Are there examples of compact Alexandrov spaces (say non-negatively curved) $(X,d)$ such that $X$ is a topological manifold but $(X,d)$ cannot be approximated by Riemannian manifolds of non-negative sectional curvature ?

(2) Does it change something if we allow the manifolds $(M_n,g_n)$ in the sequence to have a lower bound on the sectional curvature which is only going to zero as $n$ goes to infinity ?



1 Answer 1


Suspicious example: Take a "funny" manifold with sectional curvature $\ge 1$ say $X$; funny means Cayley flag or Aloff--Walach/Eschenburg/Bazaikin space, (not $S^n$ or $\mathbb{C}\mathrm{P}^n$ or $\mathbb{H}\mathrm{P}^n$). Consider spherical suspension $\Sigma$ over $X$.

The space $\Sigma$ has curvature $\ge 1$. It is expected that $\Sigma$ can not be approximated by any Riemannian manifolds with lower curvature bound, but no one can prove it. The question has a lot in common with stabilized version of the Soul Theorem.

Partial results:

  • Petersen--Wilhelm--Zhu: $\Sigma$ can not be approximated by Riemannian manifolds with curvature $\ge 1/4+\varepsilon$

  • Kapovitch: for Cayley flag, if $\Sigma$ can be approximated then it is collapsing and the dimension drops down by 15 at least.

P.S. It seems you forget to say "same dimension" in your questions. In this case, in addition to Perelman's result you have a result of V. Kapovitch, which says that space of directions, as well as iterated spaces of directions have to be homeomorphic to a sphere. This was conjectured by P. Petersen. (There are examples of Alexandrov spaces which homeomorphic to $\mathbb R^n$, but space of directions at some point are not homeomorphic to spheres.)

  • $\begingroup$ Maybe the title was not well chosen. I put "metrically singular" in the sense that the singularities can not be detected by the topology of the underlying space. By the way thanks for your answer. $\endgroup$ Jun 7, 2011 at 6:13
  • $\begingroup$ The non-collapsing case was indeed my question, I should havebstated it more explicitely. Where can the result and the Example you mention in your P.S. be found ? Thanks again. $\endgroup$ Jun 7, 2011 at 8:30
  • $\begingroup$ @Thomas, now a ref is included. $\endgroup$ Jun 7, 2011 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.