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Are there important/ interesting/ natural examples of compact Alexandrov spaces with curvature bounded from below which are not Gromov-Hausdorff limits of smooth compact Riemannian manifolds with uniformly bounded from below sectional curvature? (The condition of compactness might be relaxed somehow.)

EDIT: The dimension of manifolds in the approximating sequence is supposed to be bounded from above.

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  • $\begingroup$ Can you get a tripod in such a way? (tripod is the 1-skeleton of the 4-vertices graph given by 1 vertex liked to the three others) $\endgroup$
    – YCor
    Commented Jun 3, 2015 at 8:25
  • $\begingroup$ @YCor: Tripod has curvature bounded from above but not from below. $\endgroup$
    – asv
    Commented Jun 3, 2015 at 8:33
  • $\begingroup$ Ah OK (I saw "bounded from below" only in the second part of the sentence). $\endgroup$
    – YCor
    Commented Jun 3, 2015 at 8:52
  • $\begingroup$ What about the metric suspension over $\mathbb{RP}^2$. $\endgroup$
    – foliations
    Commented Jun 3, 2015 at 14:44

2 Answers 2

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  • There is an Alexandrov space, say $A$, with curvature $\ge 1$ which can not obtained as a limit of Riemannian manifolds with curvature $\ge \kappa$, if $\kappa>\tfrac14$; see "Metric constraints on exotic spheres via Alexandrov geometry." by Grove and Wilhelm.

  • There is an Alexandrov space, say $A$, such that if it can appear as a limit of Riemannian manifolds $M_n$ with uniformly bounded curvature then $\dim M_n\ge \dim A+8$ for all large $n$; see "Regularity of limits of noncollapsing sequences of manifolds" by Kapovitch. It is expected that in this formula one can exchange $8$ to $\infty$.

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  • $\begingroup$ Is the space $A$ compact in each example? In the second example, the curvature of $M_n$ is uniformly bounded from below or from both sides? $\endgroup$
    – asv
    Commented Jun 3, 2015 at 13:56
  • $\begingroup$ (1) yes always, (2) the construction is local and one can assume $A$ is compact, $\endgroup$ Commented Jun 4, 2015 at 10:10
  • $\begingroup$ Sorry, I did not understand your answer. What is "yes always"? Is it related to the fact that the bounds on curvature are always lower, or that $A$ is always compact? Could you please elaborate. Thanks. $\endgroup$
    – asv
    Commented Jun 4, 2015 at 11:09
  • $\begingroup$ in (2) you get examples which are compact and which are not compact. $\endgroup$ Commented Jun 5, 2015 at 13:54
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I believe (I am not an expert in the subject) a simple example (though using a highly non-trivial theorem) is given by considering the following example: Let $$\mathbb{S}^3\subset \mathbb{R}^4$$ be the round sphere of curvature $1$ and consider the isometry $\phi:\mathbb{S}^3\to \mathbb{S}^3$ induced by $$ (x_1, x_2, x_3, x_4)\mapsto (x_1, -x_2, -x_3, -x_4) $$ and let $$ X=\mathbb{S}^3/\phi. $$ be the obvious quotient space with the natural quotient metric. This quotient has curvature $\geq 1$, but is not a topological manifold (as small balls about $(\pm 1 , 0, 0, 0)$ look like cones over $\mathbb{RP}^2$).

Hence, by Perelman's stability theorem, $X$ cannot be the GH limit of smooth manifolds of curvature $\geq k$ for any $k\leq 1$.

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    $\begingroup$ The stability theorem concerns the situation when there is no collapse. In the situation of my question collapse is possible. $\endgroup$
    – asv
    Commented Jun 3, 2015 at 15:14

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