# Orbifold vs Alexandrove space vs Limit of manifolds

I have a question about some defitions : Orbifold, Alexandrov space, limit of manifolds in Gromov-Hausdorff distance sense.

Consider following example.

Let $r> 0$

$L_c = \{ (x cos \theta, x sin \theta, cx) | 0 \leq x$ and $0 \leq \theta < 2\pi \}$

$S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$

$T$ : $(z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$

If $R$ is sufficiently large, then we have a two dimensional sphere $U_c$ enclosed by $L_c$, $S$, and $T$.

First notice the following. $lim_{r \rightarrow 0} U_c$ is an orbifold for some $c$

Question : Is any orbifold is a limit of manifolds in Gromov-Hausdorff sense ?

If this question is wide, we can restrict to the case of nonnegatively curved orbifolds. : Is a nonnegatively curved $n$-orbifold a limit of positively curved $n$-manifolds ?

Question 2: In the following paper, a space with curvature $\geq k$ is defined.

M. Gromov Y. Burago and G. Perelman, A.d. alexandrov spaces with curvature bounded below, Uspekhi Mat. Nauk 47 (2) (1992), 3–51.

Is a $n$-dimensional space with curvature $\geq k$, which is smooth except finite points, is a limit of $n$-manifolds of positive sectional curvature $\geq k$ ? I believe that this question is trivial and it is true.

I do not think that all orbifolds or spaces with curvature $\geq k$ are limits of manifolds.

However I can not deny it.

Since ${\bf R}^3={\bf R}^4 /S^1 = lim_{k \rightarrow \infty} {\bf R}^4/{\bf Z}_k$, orbifolds are different from spaces with curvature $\geq k$. But they are obtained from the sequences of manifolds.

MOTIVATION : Hsiang-Kleiner classified positively curved manifolds with $S^1$-action. I want to extend this result to positively curved orbifolds with $S^1$-action.

If orbifold is a limit of manifolds then the problem is simple.

Accordingly I want to know the questions. Thank you for your attention.

• Every compact length space is a Gromov-Hausdorff limit of two-dimensional Riemannian manifolds. You need to be more precise in your question. Do the approximating manifolds have to have the same lower curvature bound? Same dimension? – Sergei Ivanov Jun 8 '12 at 12:38
• @ IVanov : Thank you. Your suggestion is helpful for me. – Hee Kwon Lee Jun 8 '12 at 13:26
• One technical point: Orbifolds are different animals than topological spaces. An orbifold is a space plus some extra structure. That said, I think that your question is fairly self explanatory (at least in this regard). – Spice the Bird Jun 8 '12 at 19:53

Q1. Note that one oriented orthonormal frame bundle $FO$ over a smooth orbifold $O$ is a smooth manifold. This frame bundle admits a one-parameter family of metrics which collapse to the original manifold and its curvature can be made bounded from below.
Proof. Equip $FO$ with $SO(n)$-invariant metric, consider product space $[\varepsilon\cdot SO(n)]\times FO$ and factorize along diagonal action. (This often called Cheeger's trick, but I think it was known before Cheeger.)
Q2. It is an open question. It is expected that cones over some positively curved manifolds can not be approximated. Vitali Kapovitch has examples of such $n$ dimesional cones which can not be approximated by $m$-dimensional manifolds with $m< n+8$.
By Perelman's Stability Theorem, if a (compact) limit of $n$-dimensional Alexandrov spaces of curvature $\ge k$ has the same dimension, then the convergent spaces are eventually homeomorphic to the limit space. So in this context a limit of manifolds is always a topological manifold.
And there exist non-manifold examples of Alexandrov spaces, e.g., $\mathbb R^3/\{id,-id\}$ which is the cone over $\mathbb {RP}^2$. Or, if you want a compact example, take $S^3/\mathbb Z_2$ where $\mathbb Z_2$ acts on $S^3$ with two fixed points (think of $S^3$ lying in $\mathbb R^4$ and consider the reflection in some 1-dimensional axis). This is a compact orbifold of curvature $\ge 1$ and it is not a limit of manifolds of the same dimension with a uniform lower curvature bound.