MSE crosspost

It's known (due to Perelman) that in class of Alexandrov spaces of fixed dimension and bounded from below curvature Gromov-Hausdorff distance separates homeomorphism types — every $\epsilon$-close to $X$ space will be homeomorphic to $X$ for some $\epsilon$.

Well, if we have some finite metric space $X_{\delta}$ which is $\epsilon/2$-close to $X$, then $(X_{\delta}, n, C)$ define homeomorphism type of, say, compact Riemannian manifold, where $n$ is dimension and $C$ is lower curvature bound.

Now let's fix $C$ once for all (take $-1$, for example) and call finite metric space $X_{\delta}$ a model of a manifold $X$ if for some $\epsilon$ the only manifold $\epsilon$-close to $X_{\delta}$ is $X$ with some metric with curvature bounded below by $-1$. We can define two functions on homeomorphism (diffeo, if dim > 4, thanks to Grove-Peterson-Wu) classes of $n$-dimensional manifolds: $min \, |X_{\delta}|$ and $min \, k: X_{\delta} \to \Bbb R^k$ for isometric embedding into real space with some norm, where minimum is taken over all models. It seems appropriate to me to call first one metric complexity $mCom(X)$ and second one — essential dimension $edim(X)$.

  1. Can $edim(X)$ be strictly less than dimension of $X$ [Edit after @Sergio answer: if $X$ is not contractible or sphere]?

  2. Are there some bounds on $mCom$ in terms of something like LS category or topological complexity (i. e. minimal cardinality of open cover over which $eval: X^I \to X \times X$ has local sections?

  3. What is, for example, $mCom(S^1 \times S^1)$ — or something else $\geq 2$-dimensional — and what is the model? (I guess that for all surfaces answer should be derivable from known results about triangulations et cetera).

(I'm totally not an expert in this area, so maybe those questions are either very easy or hopelessly hard; if it's so, I'll gladly accept as an answer putting them into one of these two categories.)

  • $\begingroup$ Now that I am reading it again, what I interpreted as your definition of model and what I think you meant is this: call finite metric space $X_{\delta}$ model of an $n$-dimensional manifold $X$ if for some $ϵ$, $X$ and $X_{\delta} $ are $\epsilon$ close, and the only $n$-dimensional mnifold $2ϵ$-close to $X_{\delta}$ is $X$ with some metric with curv bound below by −1. With this defntn, it may be also ntural to ask, related to $edim$, for an $n$-dimensional manifold $X$, which is lowest integer $k$ sch that there is model $X_{\delta}$ of $X$ whch is at the same time a model of $k$-dim mnfld. $\endgroup$ – Sergio Zamora Jul 31 '18 at 15:38

The cigar gives both a positive answer for 1 and a negative answer for an upper bound in 2.

enter image description here

If it is thin enough, a sequence of aligned points along it is a model and it can be embedded in $\mathbb{R}$. Giving a positive answer for 1.

If it is thick enough, but very long, you will need a lot of points in your model (there is no upper bound) to differentiate it from the version with a hole in it. Giving a negative answer to an upper bound in 2.

enter image description here

For lower bounds of $mCom$ in terms of the topology I would expect a positive answer, since keeping track of the geometry requires more information than keeping track of the topology.

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    $\begingroup$ Thanks for example! It's (un)fortunately less complicated than I expected. But your "counterexample for bounds" is aimed at wrong statement — we want to fix topology, and find best metric (and model) for it. If I'm not mistaken, $1/4$-grid on flat torus of unit circumference is a model for $T^2$ (among curvature $\geq -1$ surfaces), so complexity of torus is at most 16; even $1/3$-grid will also probably work. (If you want, I'll accept your answer in few days if nothing else appears.) $\endgroup$ – Denis T. Jul 30 '18 at 21:52
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    $\begingroup$ Actually, this is more complicated than I thought. If you take a sequence of aligned points along the cigar, they will also be close to the projective plane obtained by doubling the length of the cigar and identifying antipodal points. Then my example is not correct. My apologies. $\endgroup$ – Sergio Zamora Jul 31 '18 at 19:57

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