Let $(M^n,g)$ be an $n$-dimensional compact and connected Riemannian manifold with sectional curvature bounded above and below by $c,C$. Is it possible/known how to express the external covering number of $M^n$ as a function of $c,C$, $n$, $\operatorname{diam}(M^n)$, and possibly some other numerical data associated to $(M^n,g)$?

This must be known, but I have no idea where to look.

Definition: For $\varepsilon>0$, the $\varepsilon$-external covering number of $M^n$, denoted by $N_{\varepsilon}(M^n)$, is the smallest number of points $p_1,\dots,p_N\in M^n$ which are $\varepsilon$-dense wrt. $(M^n,g)$'s Riemannian distance $d_g$. That is, if $p\in M^n$ then there is some $p_i$, as above, such that $$ d_g(p,p_i)\le \varepsilon . $$

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    $\begingroup$ For lower bound you can use the Bishop-Gromov volume comparison, not sure about an upper bound. $\endgroup$ Feb 4, 2022 at 5:41
  • $\begingroup$ @DmytroYeroshkin How can this be done? I feel that I'm missing something. Also, for the upper bound I found the following unpublished article but it only holds for small balls. arxiv.org/pdf/1305.1529.pdf $\endgroup$ May 16, 2022 at 7:36
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    $\begingroup$ The simplest lower bound can be achieved as follows: By Bishop-Gromov, we know that the volume of an $\epsilon$-ball in $M$ is at most the volume of an $\epsilon$-ball in the model space with constant curvature $c$. Therefore in order for $N$ $\epsilon$-balls to cover $M$, $N$ must be at least $Vol(M)$ divided by the volume of an $\epsilon$-ball in the model space. I suspect there are ways to improve the bound by understanding how big the overlaps have to be, but I've not actually looked at the details. $\endgroup$ May 16, 2022 at 11:25
  • $\begingroup$ @DmytroYeroshkin Thanks, the lower-bound is obvious now; thanks for the helping details. $\endgroup$ May 16, 2022 at 11:28


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