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Let $M \subset \mathbb{R}^d$ be a compact smooth $k$-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 a $p \in P$ such that $\| x - p\|_{2}$. My goal is to show that $\mathcal{N}(\epsilon) \in \Theta\left(\frac{1}{\epsilon^k}\right)$. Note that the bound depends on the dimension $k$, not the embedding space $d$.

To this end, I'd like to construct an $\epsilon$-cover in each coordinate chart $(U, \phi)$ of $M$, where $\phi: U \subset \mathbb{R}^k \rightarrow M$. A cover in $U$ can be transformed under $\phi$ into a cover in $M$. If $\phi$ is $L$-Lipschitz, that is $\|\phi(x) - \phi(y)\|_2 \leq L \|x - y\|_{2}$, then I can create an $\epsilon$-cover over a subset $\phi(U) \subset M$, with slightly larger balls than those in $U \subset \mathbb{R}^k$ as determined by the Lipschitz constant $L$.

To have each coordinate chart $(U, \phi)$ be $L$-Lipschitz I'm willing to assume $M$ has bounded curvature. Can I subdivided $M$ into a set of coordinate charts which are $L$-Lipschitz and how many such charts do you need as a function of the curvature of $M$?

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  • $\begingroup$ Is the Lipschitz-ness aspect of your question really relevant? The number of smooth charts needed to cover $M$ is finite and a topological invariant (see mathoverflow.net/questions/2615/… ), and the charts in a smooth atlas can be shrunk very slightly to give a new atlas in which all charts $\phi$ have $\sup |D\phi|<\infty$. Domain-rescaling these charts gives a third atlas in which all charts have $\sup |D\phi| = 1$. Now we have a 1-Lipschitz atlas whose size is a topological invariant. $\endgroup$
    – macbeth
    May 29, 2019 at 17:26
  • $\begingroup$ Yes, I believe the Lipschitz-ness is important. The manifold is embedding in $\mathbb{R}^d$, it has some predefined geometry. I want to construct a set of coordinate charts on $M$ such that each chart is Lipschitz. This must depend on the curvature; as $L$ becomes smaller I should need more charts to ensure that each satisfy this Lipschitz property since the curvature is constant. $\endgroup$
    – user141213
    Jun 1, 2019 at 4:49
  • $\begingroup$ I don't see why the Lipschitz constant $L$ should depend on curvature: domain-rescaling an $L$-Lipschitz chart gives a new chart which is 1-Lipschitz. Indeed, let $U\subseteq\mathbb{R}^k$ be open and let $\phi:U\to M$ be a $L$-Lipschitz chart for $M$, i.e., $\forall x,y\in U, |\phi(x)-\phi(y)|\leq L|x-y|$. Then, defining $\hat\phi$ by $\hat\phi(x):=\phi(L^{-1}x)$, we have that $\forall x,y\in LU, |\hat\phi(x)-\hat\phi(y)|=|\phi(L^{-1}x)-\phi(L^{-1}y)|\leq |x-y|$. $\endgroup$
    – macbeth
    Jun 2, 2019 at 7:22

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