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

1$\begingroup$ What does "not collapsed" mean? Radius of injectivity uniformly bounded away from zero? $\endgroup$– YCorMay 1, 2014 at 10:47

2$\begingroup$ arxiv.org/abs/0903.2142 I guess Miles Simon's paper is exactly what you want. $\endgroup$– ChihWei ChenMay 1, 2014 at 11:17

2$\begingroup$ I checked in the given reference: "noncollapsed" means that the volume of an arbitrary ball of radius 1 is bounded away from 0. $\endgroup$– YCorMay 2, 2014 at 0:08

$\begingroup$ @ChihWeiChen Maybe you can make an answer of your comment. $\endgroup$– Thomas RichardMay 2, 2014 at 9:34

$\begingroup$ Salut! Thomas, you are much more familiar than me. I wish to see your answer if you have time to write it up. $\endgroup$– ChihWei ChenMay 2, 2014 at 10:37
1 Answer
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 GromovHausdorff sense, and for Hausdorff dimensions $\dim X=\dim M_i$ for large $i$, then $X$ is homeomorphic to smooth manifold (with nonsmooth metric in general, of course).
For the Perelman stability theorem see p.23 here https://www.math.psu.edu/petrunin/papers/alexandrov/perelmanASWCBFB2+.pdf .
Added: May I add another statement which is a combination of the above Perelman theorem and BuragoGromovPerelman theorem (see Corollary 10.10.11 in the book "Metric geometry" by BuragoBuragoIvanov).
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 GromovHausdorff 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 $i$.

$\begingroup$ This statement says that if $X$ has sectional curvature bounded below, then for every $(X_n)$ tending to $X$ GromovHausdorff, eventually $X_n$ is homeomorphic to $X$. This does not answer the question. $\endgroup$– YCorMay 2, 2014 at 16:55

$\begingroup$ (If $X_n$ is the hyperbolic 3fold obtained by modding out the hyperbolic 3space 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 GromovHausdorff to the real line. Hence some noncollapsing assumption is necessary.) $\endgroup$– YCorMay 2, 2014 at 16:58

$\begingroup$ @YvesCornulier: Sure, one assumes that $\dim X=\dim M_i$ for large $i$. This is the noncollapsing assumption. Here $\dim X$ is the Hausdorff dimension. $\endgroup$– asvMay 2, 2014 at 17:05

$\begingroup$ Still, I don't understand the stability theorem page 23 (an openness theorem) of the link and the question (a closedness theorem), nor can I see any collapsing hypothesis in the stability theorem. $\endgroup$– YCorMay 2, 2014 at 19:07

$\begingroup$ 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 GromovHausdorff (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 GHmetric, hence $M_i$ belongs to this neighborhood for large $i$, hence $M_i$ is homeomorphic to $X$ for large $i$. $\endgroup$– asvMay 3, 2014 at 7:06