Gromov compactness theorem states that in a class of Riemannian manifolds that have a uniformly bounded diameter and uniformly bounded below Ricci curvature every sequence of manifolds has a subsequence that has a limit in the Gromov-Hausdorff metric.

What are the counterexamples to this statement if I drop the boundedness of Ricci curvature? That is, I am looking for a sequence of Riemannian manifolds with arbitrary negative Ricci curvature of bounded diameter and such that no subsequence is converging in GH sense.

  • $\begingroup$ I think you can pick a torus and pinch one of the loop to a point. The curvature $\to -\infty$ around that point, and the metric converges to a cone metric. $\endgroup$ – user99914 Nov 2 '15 at 0:05
  • $\begingroup$ right, it does not converge to a manifold but it converges to something. can one make an example when there is no convergence at all? $\endgroup$ – Ivan Bodhidharma Nov 2 '15 at 0:46
  • $\begingroup$ It seems that you can always isometrically embeds all the family into a fixed $\mathbb R^N$ by Nash embedding. The diameter forces that they all stay inside a fixed ball, so a subsequence must converge in Gromov Hausdorff distance to a compact set. $\endgroup$ – user99914 Nov 2 '15 at 0:49
  • $\begingroup$ oh. so does Nash embedding work for sequences then? I can accept that every element can be embedded in a compact, but embedding all of them in the same compact is more. $\endgroup$ – Ivan Bodhidharma Nov 2 '15 at 0:52
  • 2
    $\begingroup$ @JohnMa Nash embedding doesn't help much, since the metric of the embedded manifold is intrinsic and may have little to do with the ambient Euclidean metric. I.e., the fact of being contained in a ball of $\mathbb R^N$ is of no material consequence. $\endgroup$ – user147263 Nov 3 '15 at 5:32

Without the Ricci curvature bound one does not have control on the doubling constant of the space (which is the only thing that the bound is used for). This means that the spaces may contain larger and larger sets of uniformly separated points (say, distance $\ge 1$ between any two points). This precludes being Cauchy in Gromov-Hausdorff metric, since if the GH distance between two spaces is $<\epsilon$ and one has a large $1$-separated subset, the other one must have a $(1-2\epsilon)$-separated subset of the same cardinality.

For a concrete example, take a sequence of open disks $D_n$ on the sphere $S^2$ with disjoint closures, and begin attaching a "needle" of length $3$ to each disk. Let $M_n$ be the space with $n$ such needles attached. The diameter of $M_n$ remains bounded by $10$. Also, if $m\gg n$, the distance $d_{GH}(M_m, M_n)$ cannot be small: there are only so many points at distance about $6$ from one another that one can fit into $M_n$ for a fixed $n$.

  • $\begingroup$ thank you! After discussing this with some people I have also found another example illustrating this: take unit balls in hyperbolic spaces of infinitely decreasing curvature. then one needs more and more balls of fixed radius $r$ to cover such balls, and there is no GH convegrence. $\endgroup$ – Ivan Bodhidharma Nov 3 '15 at 21:27
  • $\begingroup$ That also works, I just wanted closed manifolds for some reason. $\endgroup$ – user147263 Nov 3 '15 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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