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Suppose a metric space $X$ is the Gromov--Hausdorff limit of a sequence of Riemannian three-manifolds $M_i$ with Ricci curvature bounded from below and not collapsed. Is $X$ a manifold? What references are there concerning this problem?

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What does "not collapsed" mean? Radius of injectivity uniformly bounded away from zero? – YCor May 1 '14 at 10:47
2 I guess Miles Simon's paper is exactly what you want. – Chih-Wei Chen May 1 '14 at 11:17
I checked in the given reference: "non-collapsed" means that the volume of an arbitrary ball of radius 1 is bounded away from 0. – YCor May 2 '14 at 0:08
@Chih-WeiChen Maybe you can make an answer of your comment. – Thomas Richard May 2 '14 at 9:34
Salut! Thomas, you are much more familiar than me. I wish to see your answer if you have time to write it up. – Chih-Wei Chen May 2 '14 at 10:37

This does not answer the question since lower bound on Ricci curvature is replaced by a stronger condition of lower bound on sectional curvature. Nevertheless perhaps it might be relevant.

Edit: By the Perelman stability theorem, if a sequence of smooth compact connected Riemannian manifolds $M_i$ has uniformly bounded from below sectional (rather than Ricci) curvature, converges to a compact metric space $X$ in the Gromov-Hausdorff sense, and for Hausdorff dimensions $\dim X=\dim M_i$ for large $i$, then $X$ is homeomorphic to smooth manifold (with non-smooth metric in general, of course).

For the Perelman stability theorem see p.2-3 here .

Added: May I add another statement which is a combination of the above Perelman theorem and Burago-Gromov-Perelman theorem (see Corollary 10.10.11 in the book "Metric geometry" by Burago-Burago-Ivanov).

Let a sequence of smooth compact connected Riemannian manifolds $M_i$ has uniformly bounded from below sectional curvature and converges to a compact metric space $X$ in the Gromov-Hausdorff sense. Assume in addition that for some $\delta>0$ one has $vol(M_i)>\delta$ for all $i$. Then $X$ is homeomorphic to $M_i$ for large $i$.

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This statement says that if $X$ has sectional curvature bounded below, then for every $(X_n)$ tending to $X$ Gromov-Hausdorff, eventually $X_n$ is homeomorphic to $X$. This does not answer the question. – YCor May 2 '14 at 16:55
(If $X_n$ is the hyperbolic 3-fold obtained by modding out the hyperbolic 3-space by a hyperbolic element of translation length $1/n$ and if we choose a basepoint in the closed geodesic of $X_n$, then the sequence $(X_n)$ tends Gromov-Hausdorff to the real line. Hence some non-collapsing assumption is necessary.) – YCor May 2 '14 at 16:58
@YvesCornulier: Sure, one assumes that $\dim X=\dim M_i$ for large $i$. This is the non-collapsing assumption. Here $\dim X$ is the Hausdorff dimension. – sva May 2 '14 at 17:05
Still, I don't understand the stability theorem page 2-3 (an openness theorem) of the link and the question (a closedness theorem), nor can I see any collapsing hypothesis in the stability theorem. – YCor May 2 '14 at 19:07
1) Stability theorem of the link says that any compact Alexandrov space $X$ (thus $X$ is a compact metric space with curvature bounded from below and finite Hausdorff dimension) has a neighborhood in the Gromov-Hausdorff (GH) metric such that any other compact Alexandrov space in this neighborhood with the same lower bounds on curvature and same dimension is homeomorphic to $X$. In our situation $M_i\to X$ in GH-metric, hence $M_i$ belongs to this neighborhood for large $i$, hence $M_i$ is homeomorphic to $X$ for large $i$. – sva May 3 '14 at 7:06

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