I'm reading Hubbard's Vector Calculus, Linear Algebra, and Differential forms, and it defines a relaxed parametrization in the following way:
Definition 5.2.1 ($k$-dimensional volume 0 of a subset of $\Bbb{R}^{n}$)
the definition is given by two cases:
- A bounded subset $X \subset \Bbb{R}^n$ has $k$-dimensional volume 0 if $$ \lim _{N \rightarrow \infty} \sum_{C \in \mathcal{D}_{N}\left(\mathbb{R}^{n}\right) \atop C \cap X \neq \phi}\Big(\underbrace{\frac{1}{2^{N}}}_{\text {sidelength of $C$}}\Big)^k = 0 $$
- An arbitrary subset $X \subset \Bbb{R}^n$ has $k$-dimensional volume 0 if for all $r > 0$, the bounded set $X \cap B_r(0)$ has $k$-dimensional volume 0.
Here $\mathcal{D}_{N}(\Bbb{R}^n)$ is the $N$th {\it dyadic paving} of $\Bbb{R}^n$, defined by: $\mathcal{D}_{N}(\Bbb{R}^n) = \{C_{\mathbf{k},N}\}$, with:
$$ C_{\mathbf{k}, N} \stackrel{\text { def }}{=}\left\{\mathbf{x} \in \mathbb{R}^{n} \mid \frac{k_{i}}{2^{N}} \leq x_{i}<\frac{k_{i}+1}{2^{N}} \text { for } 1 \leq i \leq n\right\}, $$ where $\mathbf{k} \stackrel{\text { def }}{=}\left(\begin{array}{c} k_{1} \\ \vdots \\ k_{n} \end{array}\right)$ and $k_1, \dots, k_n$ are integers.
Definition 5.2.3 ("Relaxed" parametrization of a manifold).
Let $M\subset \mathbb{R}^n$ be a $k$-dimensional manifold and let $U\subset \mathbb{R}^k$ be a subset with boundary of $k$-dimensional volume $0$. Let $X\subset U$ be such that $U-X$ is open. Then a continuous mapping $\gamma: U \to \mathbb{R}^n$ parametrizes M if:
- $\gamma(U) \supset M;$
- $\gamma(U-X)\subset M;$
- $\gamma: (U-X) \to M$ is one to one, of class $C^1$ with locally Lipschitz derivative;
- the derivative $[D\gamma(\mathbf{u})]$ is one to one for all $\mathbf{u}$ in $U-X$;
- X has a $k$-dimensional volume 0, as does $\gamma(X)\cap C$ for any compact subset $C\subset M$.
(note: Hubbard uses volume to refer to volume in the sense of Riemann)
I copied the definition from a related question Showing that the manifold $U = \bigcup_{i = 1}^\infty \left(a_i - \frac{1}{2^{i+k}}, a_i + \frac{1}{2^{i+k}}\right)$ has a [relaxed] parametrization
And the definition of manifold in the context is "locally graph", the fourth definition in the question I asked: "Local parametrizations" implying "Local flat" in the two definitions of submanifolds of Euclidean spaces
Theorem 5.2.6Existence of parametrizations. All manifolds can be (relaxed) parametrized.
Hubbard is lefting the theorem as an exercise. The hints for proving the theorem is (the full detailed proof should be very fiddly, as Hubbard says in the exercise):
- Cover $M$ by a sequence of cubes $C_i$ such that $M \cap C_i$ is the graph $\Gamma_i$ of a map $\mathbf{f_i}$ expressing $n-k$ variables in terms of the other $k$.
- Then these $\Gamma_i$ will probably not have disjoint interiors: doctor up $C_i$ to remove the part of $\Gamma_i$ that is in the interior of some $\Gamma_j, j < i$.
- Spread out the domains of these $\mathbf{f}_i$ so they are well separated.
- (No hints for part 1-4) the first condition of part 5 is then easy (Hubbard says so), and the seconds follows from the compactness (open covering implying finite open covering)
Could anyone give some more hints for proving this please? The following is what I tried:
(the proof that I tried but dit not achieve)
Since $M$ is a $k$-dimensional manifold in $\Bbb{R}^n$, for any point $x \in M$, there is some open neighborhood $B_x$ of $x$ such that $B_x \cap M$ is the graph of some $g_x: U_x \rightarrow \Bbb{R}^{n-k}$, where $U_x$ is some open set in $\Bbb{R}^k$, which means $B_x \cap M = \{\sigma_x(u, g(u))|u \in U_x\}$ for some permutation $\sigma_x \in S_n$. There is some $N_x$, some $\mathbf{k}_x$, such that $x \in C_{\mathbf{k_x}, N_x} \subset \overline{C}_{\mathbf{k_x}, N_x} \subset B_x$(I use $\partial, \mathring{~}, \bar{~}$ to denote the boundary, interior, closure of a set respectively, here just the open/half-open/close cube. I didn't prove this result, but it should be true since all finite binary numbers is dense? Are terminating decimals dense in the reals?)
Since the elements of $\mathbf{k_x}$, and $N_x$ is integer, $\{C_{\mathbf{k_x}, N_x}|x \in M\}$ is countable. Denote it as $\{C_i|i=1,2,\cdots\}$. Since every $x \in M$ is contained in some $C_i$, $\{C_i | i=1,2,\cdots\}$ is a covering of $M$. And by the axiom of choice, there is open set $B_i \supset \overline{C}_{i}$ , open set $U_i \subset \Bbb{R}^k$, smooth function $g_i: U_i \rightarrow \Bbb{R}^n$, permutation $\sigma_i \in S_n$, such that $B_i \cap M = \{\sigma_i(u, g_i(u))| u \in U_i\}$
By the definition of $C_{\mathbf{k}, N}$, $C_i$ and $C_j$ have disjoint interior, or the interior of one of them is inside the interior of the other: $\mathring{C}_i \cap \mathring{C}_j = \varnothing$, or $\mathring{C}_i \subset \mathring{C}_j$, or $\mathring{C}_j \subset \mathring{C}_i$. Remove all those $C_i$ whose interior is contained in some others' interior (Is this correct? and is there a more rigorous to make the statement?) For convenience, suppose $\{C_i\}$ has disjoint interior.
For each $C_i$, define $U_i = \pi(\sigma_i^{-1}(C_i)) \subset \Bbb{R}^k$, where $\pi: \Bbb{R}^n \rightarrow \Bbb{R}^k$ is the projection function taken the first $k$ coordinates. The $\{U_i\}$ can be spread out by some translations. For convenience , suppose $\{\overline{U_i}\}$ is disjoint. Let $U = \cup U_i$, and define $\gamma: U \rightarrow \Bbb{R}^n$ by $\gamma(u) = \sigma_i(u, g_i(u))$ if $u \in U_i$. And let $X_i = \pi(\sigma_i^{-1}(C_i \sim \mathring{C_i}))$. Since $\{U_i\}$ can be spread out by some translation, suppose $\{X_i\}$ is disjoint too. Let $X = \cup X_i$.
Now check all the parts in definition of "relaxed" arametrization of a manifold:
$\partial U$ has boundary of $k$-dimensional volume 0. Since each $\partial U_i$ has boundary of $k$-dimensional volume 0, and $\{U_i\}$ is disjoint, and jordan measure is countably additive on volume 0 set(? I'm not sure with this statement, It should be false, since $\Bbb{Q} \cap [0,1]$ is a counterexample), $\partial U$ has boundary of $k$-dimensional volume 0.
$X \subset U$, and $U - X$ is open. By definition $X_i = \pi(\sigma_i^{-1}(C_i \sim \mathring{C_i})) \subset U_i$. Hence $X \subset U$. And $U - X = \cup (U_i - X_i)$. For each $i$, $U_i - X_i = \pi(\sigma_i^{-1}(\mathring{C_i}))$ is open. Hence $U - X$ is open.
$\gamma(U) \supset M$. By definition $\gamma(U_i) = C_i \cap M$. Hence $\gamma(U) = \cup C_i \cap M$. Since $\cup C_i \supset M$, $\gamma(U) = M$
$\gamma(U - X) \subset M$. Since $\gamma(U) = M$, obviously $\gamma(U - X) \subset M$.
$\gamma: (U-X) \rightarrow M$ is one to one, of class $C^1$. By $\gamma(U_i-X_i) = \mathring{C_i}$ and $\{\mathring{C_i}\}$ is disjoint and $g_i$ is $C^1$, and $(u_1, g_i(u_1)), (u_2, g_i(u_2))$ cannot be the same point unless $u_1 = u_2$
the derivative $[D\gamma(\mathbf{u})]$ is one to one for all $\mathbf{u}$ in $U-X$. Since for $u \in U_i - X_i$, $[D\gamma(\mathbf{u})]$ is $\left[ \begin{array}{c} I \\ Dg_i(\mathbf{u}) \end{array} \right] $ with some row permutation, where $I$ is the identity matrix of size $k$, $[D\gamma(\mathbf{u})]$ is one to one in $U_i - X_i$. Hence $[D\gamma(\mathbf{u})]$ is one to one in $U-X$.
$X$ has $k$-dimensional volume 0, as does $\gamma(X) \cap C$ for any compact subset $C \subset M$. $X_i \subset \partial U_i$, and $\partial U$ has $k$-dimensional volume 0, so is $X$. By definition $\gamma(X) = \cup \partial C_i \cap M$. Suppose $C \subset M$ is any compact set. Then $\gamma(X) \cap C= \cup \partial C_i \cap C$. How to prove the second condition? I tried to find out and use some finite open covering but still got no idea here?
(summarize the question that I have)
- I'm not sure if the sketch of the proof is like so
- Some detailed I'm not confirmed, like the definition of $C_i$, and the part that making them disjoint.
- I'm not quite confirmed the construction of $U$ and $X$ too
- The proof for showing $\partial U$ of $k$-dimensional volume 0 seems wrong. Since each raitional point in $[0, 1]$ has 1-dimensional volume 0, but $\Bbb{Q} \cap [0,1]$ has no volume (not Riemann integrable).
- My conclusion seems stronger: I showd that $\gamma(U) = M$. It makes me feeling weird too.
- I cannot prove that $\gamma(X) \cap C$ has $k$-dimensional volume 0 for any compact subset $C \subset M$.
