Minkowski dimension of the subset

Question

Suppose $A\subset R^N$ is a manifold with topological dimension d.

(To be precise, the case I need is

A is the set of m by n matrices with rank r, N=mn.

It is known that its codimension is $(m-r)(n-r)$.)

Now can we say any (compact) subset of A has

1. Hausdorff dimension less than or equal to d?

2. Minkowski dimension(box covering dimension) less than or equal to d?

3. If not, is there any bounding inequality for the Minkowski dimension using d?

   (For example, $dim_{M} (B) \leq 10d + 100 $ holds for all $B \subset A$.

   If that upper bound is $O(d)$ then it would be fantastic)

Answer 1

The short answer is that charts on smooth manifold can be taken to be bi-Lipschitz and then they preserve dimension (see, for example, Hausdorff Dimension of a manifold of dimension n?).

In more details one could argue as follows:

If $A^d$ is a smooth submanifold of $\mathbb{R}^n$ then for every point $a\in A$ there exists a neighborhood $U$ of $a$, an open neighborhood $V$ of the origin in $\mathbb{R}^n=\{x_1, \ldots, x_n\}$ and diffeomorphism $f_a(U, U\cap A, a) \to (V, V\cap H, 0)$ where $H$ is the coordinate space $\{x_1, \ldots, x_n| x_{d+1}=0,\ldots, x_n=0\}$. This is either a definition of a submanifold, or basic theorem in the subject. Moreover, we can take $f(U)$ to be an open ball and this diffeomorphism can be taken to have derivative bounded in operator norm $|Df_u|<C, |Df^{-1}_u|<C$ for all $u\in U$. (Proof: Pick $V$ to be open ball inside closed ball inside $f(U)$ and take new $U$ to be $f^{-1}(V)$; then by compactness the bounds hold on the closed ball for some $C$, and hence on the open ball as well.)

Now pick a compact subset $K$ of $A$. The sets $W_a=U_a\cap K$ with $a\in K$ form a cover, so there is a finite subcover $W_1, \ldots, W_N$ with $W_j=U_{a_j}\cap K\subset U_{a_j}\cap A$. If we show that $\dim U_{a_j}\cap A \leq d$, then the same is true for $W_j$s, and hence for their union $K$, which is what we want.

We prove that upper box counting (Minkowski) dimension of $ A_j=U_{a_j}\cap A$ is at most $d$. The bound you want then follows for Hausdorff dimension as well since it is bounded by the box counting dimension.

But the fact that that upper box counting dimension of $A_j$ is at most $d$ follows from two observations: 1) the image of $A_j$ under $f$ being an open disc of $d$-dimensional plane has box counting dimension $d$ 2) if we have any covering of $f(A_j)$ by $\varepsilon$ balls centered at $h_l$s, then the $C\varepsilon$ balls centered at $f^{-1}(h_l)$ cover $A_j$. Indeed, suppose $a\in A_j$ is arbitrary. Then $h=f(a)$ is connected to some $h_j$ by a segment of length at most $\varepsilon$ (the length is the integral of velocity, and by our bound on $|Df^{-1}|$ the velocity is rescaled by at most $C$, so integrates to at most $C$ times orgonal lenght; i.e. we ensure $f^{-1}$ is Lipschitz). Then the preimage of this segment connects $a$ to $f^{-1}(h_l)$, and has length at most $C\varepsilon$. So the balls around $f^{-1}(h_j)$ of size $C\varepsilon$ cover $A_j$.

These two observations together imply that the upper box counting dimension of $A_j$ is at most $d$.

In fact, by taking charts where each $U$ is a ball (end ensuring $f$ is Lipschitz) and running similar argument one can show that lower box counting dimension of $A_j$s is at least $d$.