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Let $M \subset \mathbb{R}^d$ be a $k$-dimensional manifold embedded in $\mathbb{R}^d$. Let $\mathcal{N}(\epsilon)$ denote the size of the minimum $\epsilon$-cover $P$ of $M$, that is for every point $x \in M$ there exists $p \in P$ such that $\|x - p\|_2\leq \epsilon$. What is known about lower bounds on $\mathcal{N}(\epsilon)$?

If $M$ were $d$-dimensional then there is a simple volume argument that shows that $\mathcal{N}(\epsilon) \geq \frac{\operatorname{vol}_{d}(M)}{\operatorname{vol}_{d}(B^d_{\epsilon})}$ in terms of the $d$-dimensional volume of $M$ and a $d$-dimensional ball $B^d_{\epsilon}$ of radius $\epsilon$. However I'm interested in the case where $M$ is $k$-dimensional for $k < d$. In particular I'd like a bound in terms of the $k$-dimensional volume $\operatorname{vol}_{k}(M)$ and the $k$-dimensional ball $B^{k}_{\epsilon}$. However due to the curvature of $\mathcal{M}$ it doesn't like the same bound applies with $B^{k}_{\epsilon}$, but using $B^{d}_{\epsilon}$ may lead to a less tight result.

I'd also be interested in any known lower bounds, even ones that do not use volume arguments.

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  • $\begingroup$ Is $M$ compact? Does $M$ have boundary? $\endgroup$
    – Ben McKay
    Commented Sep 23, 2018 at 14:17
  • $\begingroup$ Yes, M is compact. I don't want to assume with boundary, but if that's the only way I can get a bound, I'll take it. $\endgroup$
    – user122374
    Commented Sep 23, 2018 at 17:12
  • $\begingroup$ Sorry I meant to say, I don't want to assume without boundary. $M$ may have boundary. $\endgroup$
    – user122374
    Commented Sep 23, 2018 at 18:14

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