- #1
HughBennet
- 2
- 0
I have problems understanding the proof of this lemma:
$$\lambda \in \Lambda (m, n), \ \ \text{this means that it is an increasing function} \ \ \lambda: \{1,2,...,m\} \rightarrow \{1,...,n\}, \ \ \text{so} \ \ \lambda(1) < ... < \lambda(m)$$
$$p_{\lambda} : \mathbb{R}^m \ni (x_1, ..., x_m) \rightarrow (x_{\lambda(1)}, ..., x_{\lambda(m)}) \in \mathbb{R}^n$$
Assumptions:
$N$ is an $n-$ dimensional, orientable $C^1$ subvariety of $\mathbb{R}^m$, $1 \le n \le m$, $\lambda \in \Lambda(m,n)$, $h: \mathbb{R}^m \rightarrow \mathbb{R}$ is continuous, $\omega = hd_{\lambda}$ is summable on
$N$, $Z$ is a closed subset of $\mathbb{R}^n$ such that $\mathcal{L}^n(Z)=0$ (Lebesgue measure)
Then $$\int_{N \setminus p_{\lambda}^{-1}(Z)} \omega = \int_N \omega$$ with the induced orientation on $N \setminus p_{\lambda}^{-1}(Z)$Proof:First, without loss of generality, we assume that $\lambda(i) =i$, so $p_{\lambda} : \mathbb{R}^m = \mathbb{R}^{n} \times \mathbb{R}^{m-n} \ni (x_1, ..., x_m) \rightarrow (x_1, ..., x_n) \in \mathbb{R}^n$.
Why can we do that?
Now we note that $p^{-1} _{\lambda} (Z)$ is closed but doesn't necessarily have measure $0$.
(I see that this set is closed due to continuity of $p_{\lambda}$.)We define $A = N \cap (p^{-1}_{\lambda}(Z)) = N \cap (Z \times \mathbb{R}^{m-n})$
and $C = \{x \in A \ : \ T_xN \cap (\{0\} \times \mathbb{R}^{m-n}) \neq 0\}$
in other words,
$C=\{x \in A \ : \ \xi(x) \wedge e_{n+1} \wedge ... \wedge e_m = 0\}$
where $$\xi : N \rightarrow \xi (x) \in \Lambda_n \mathbb{R}^m$$ is such that $$1) \forall x \in N : \xi (x) \ \ \text{is simple and } \ \ T_{\xi (x)} (def= \{ v \in \mathbb{R}^m \ | \ v \wedge \xi (x) =0\}) = T_xN$$
$$2) \forall x \in N : \ || \xi (x) || = 1 \ \ \text{inner product on the exterior power is induced by the inner product on } \ \mathbb{R}^m$$
$$3) \xi \text{is continuous}$$Now, we continue with the proof:
We note that $C$ is closed - is it because $\xi$ is continuous, the vectors of the canonical basis "don't move" and $\{0\}$ is a closed set?
We then define $B:= A \setminus C$
and note that :
$$1) \mathcal{H}^n (B)=0 $$
(Hausdorff measure) I do not see why that measure is zero
$$2) \omega(x; \xi (x)) = 0 \ \ \text{for } \ x \in C$$
The space of $p$-linear antisymmetric maps $f: V^p \rightarrow W$ is identified with $(\Lambda_pV)^* = \Lambda_pV^* $, but why is this value zero?
So $$\int_A \omega(x; \ \xi (x)) d \mathcal{H}^n(x)=0$$
The integrated function has vale\ue zero on the set $C$. So we can narrow down the integration to $A \setminus C$ which has measure zero, and so the integral is zero. Is that correct?
Could you explain to me the things that I do not understand?
$$\lambda \in \Lambda (m, n), \ \ \text{this means that it is an increasing function} \ \ \lambda: \{1,2,...,m\} \rightarrow \{1,...,n\}, \ \ \text{so} \ \ \lambda(1) < ... < \lambda(m)$$
$$p_{\lambda} : \mathbb{R}^m \ni (x_1, ..., x_m) \rightarrow (x_{\lambda(1)}, ..., x_{\lambda(m)}) \in \mathbb{R}^n$$
Assumptions:
$N$ is an $n-$ dimensional, orientable $C^1$ subvariety of $\mathbb{R}^m$, $1 \le n \le m$, $\lambda \in \Lambda(m,n)$, $h: \mathbb{R}^m \rightarrow \mathbb{R}$ is continuous, $\omega = hd_{\lambda}$ is summable on
$N$, $Z$ is a closed subset of $\mathbb{R}^n$ such that $\mathcal{L}^n(Z)=0$ (Lebesgue measure)
Then $$\int_{N \setminus p_{\lambda}^{-1}(Z)} \omega = \int_N \omega$$ with the induced orientation on $N \setminus p_{\lambda}^{-1}(Z)$Proof:First, without loss of generality, we assume that $\lambda(i) =i$, so $p_{\lambda} : \mathbb{R}^m = \mathbb{R}^{n} \times \mathbb{R}^{m-n} \ni (x_1, ..., x_m) \rightarrow (x_1, ..., x_n) \in \mathbb{R}^n$.
Why can we do that?
Now we note that $p^{-1} _{\lambda} (Z)$ is closed but doesn't necessarily have measure $0$.
(I see that this set is closed due to continuity of $p_{\lambda}$.)We define $A = N \cap (p^{-1}_{\lambda}(Z)) = N \cap (Z \times \mathbb{R}^{m-n})$
and $C = \{x \in A \ : \ T_xN \cap (\{0\} \times \mathbb{R}^{m-n}) \neq 0\}$
in other words,
$C=\{x \in A \ : \ \xi(x) \wedge e_{n+1} \wedge ... \wedge e_m = 0\}$
where $$\xi : N \rightarrow \xi (x) \in \Lambda_n \mathbb{R}^m$$ is such that $$1) \forall x \in N : \xi (x) \ \ \text{is simple and } \ \ T_{\xi (x)} (def= \{ v \in \mathbb{R}^m \ | \ v \wedge \xi (x) =0\}) = T_xN$$
$$2) \forall x \in N : \ || \xi (x) || = 1 \ \ \text{inner product on the exterior power is induced by the inner product on } \ \mathbb{R}^m$$
$$3) \xi \text{is continuous}$$Now, we continue with the proof:
We note that $C$ is closed - is it because $\xi$ is continuous, the vectors of the canonical basis "don't move" and $\{0\}$ is a closed set?
We then define $B:= A \setminus C$
and note that :
$$1) \mathcal{H}^n (B)=0 $$
(Hausdorff measure) I do not see why that measure is zero
$$2) \omega(x; \xi (x)) = 0 \ \ \text{for } \ x \in C$$
The space of $p$-linear antisymmetric maps $f: V^p \rightarrow W$ is identified with $(\Lambda_pV)^* = \Lambda_pV^* $, but why is this value zero?
So $$\int_A \omega(x; \ \xi (x)) d \mathcal{H}^n(x)=0$$
The integrated function has vale\ue zero on the set $C$. So we can narrow down the integration to $A \setminus C$ which has measure zero, and so the integral is zero. Is that correct?
Could you explain to me the things that I do not understand?