Let $M \subset \mathbb{R}^d$ be a compact smooth $k$-dimensional manifold embedded in $\mathbb{R}^d$. Let $\mathcal{N}(\varepsilon)$ denote the minimal cardinal of an $\varepsilon$-cover $P$ of $M$; that is for every point $x \in M$ there exists a $p \in P$ such that $\| x - p\|_{2}<\varepsilon$.

Is it the case that $\mathcal{N}(\varepsilon) \in \Theta\left(\frac{1}{\varepsilon^k}\right)$? If so, is there a reference?


It turns out that this statement can be rephrased as: The Minkowski-Bouligand dimension of $M$ is equal to $k$ (this is immediate from the definition). Another word for this is the box dimension, and we will write it as $\mathrm{dim}_\mathrm{B}$. Here's a proof sketch that $\mathrm{dim}_\mathrm{B}(M^k)=k$.

By the inverse function theorem and compactness, we can write $M$ as a collection of finitely many smooth graphs over its tangent spaces. More precisely, there exists a $\rho>0$, points $x_1,\dotsc, x_N\in M$, smooth functions $u_1,\dotsc,u_N$ with $u_i:B_{\rho}(x_i)\cap T_{x_i}M\to\Bbb R^{d-k}$ and $\mathrm{Lip}(u_i)\le L$ such that $$M=\bigcup_{i=1}^N\mathrm{graph}(u_i).$$ It suffices to check the claim for each graph.

So let $B$ now be the $k$-dimensional unit ball and $u:B\to\Bbb R^{d-k}$ a Lipschitz function. It is clear by considering grids of spacing $\varepsilon$ that $\mathrm{dim}_\mathrm{B}(B)=k$. Let $\mathcal N_B(\varepsilon)$ be the covering number of $B$. Then $\mathcal N_{\mathrm{graph}(u)}(L\varepsilon)\le \mathcal N_B(\varepsilon)$. Therefore, $\mathrm{dim}_\mathrm{B}(\mathrm{graph}(u))\le \mathrm{dim}_\mathrm{B}(B).$

For the other inequality, we may assume that our graphical parametrization is in fact bi-Lipschitz by making the original balls smaller (but still having finitely many). That is, we may assume that $B_{\rho}(x_i)\cap T_{x_i}M$ is a Lipschitz graph over a subset of $B_{\rho}(x_i)\cap M$. This gives $\mathrm{dim}_\mathrm{B}(\mathrm{graph}(u))\ge \mathrm{dim}_\mathrm{B}(B)$, which completes the proof.

| cite | improve this answer | |

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.