6
$\begingroup$

Here I am considering the Gromov-Hausdorff convergence for metric spaces. I know two compact metric spaces are isometric if and only if $d_{GH}(X,Y)=0$, where $d_{GH}$ denotes the Gromov-Hausdorff (pseudo)distance. I was wondering if a weaker assumption (e.g. $d_{GH}(X,Y)$ small enough) could imply homeomorphism between $X$ and $Y$.

To be more concrete, I am especially interested in the following problem: let $(X,d)$ be a metric space and let $d_n$ a sequence of metrics on $X$ such that $(X,d_n)$ converges to $(X,d)$ in the Gromov-Hausdorff convergence. Can I say that the topology induced by $d_n$ is equivalent to the native topology (induced by $d$) of $X$, at least for $n$ large enough?

$\endgroup$
5
$\begingroup$

Consider the sequence of annuli $A=\{1\le |z|\le 1+1/n\}$ in the complex plane. Their limit in the GH-sense is the unit circle. Since the annuli and the circle have the same cardinality, you can assume (although it is very unnatural, GH-convergence is set up to avoid this) that the corresponding metrics are defined on the same set. If you want Riemannian manifolds as your examples, think of a sequence of tori converging to the unit circle. On the positive side, Perelman's stability theorem (used in the proof of his more famous theorem) states that under certain conditions ("Alexandrov", "curvature bounds", "noncollapsing"), if $X_n$'s GH-converge to $X$ then for large $n$ indeed $X$ is homeomorphic to $X_n$.

$\endgroup$

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.