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对合性
命题 1.1. 设$\mathcal{D}$是$M$上光滑分布. 若$\mathcal{D}$是可积的, 则$\,\forall\,X,Y\in \chi(\mathcal{D}),$ 即$\,\forall\,p\in M,$ $X_p,Y_p\in \mathcal{D}_p,$ 那么$[X,Y]\in \chi(\mathcal{D}).$
证: 任取含$p$坐标系$(U,\varphi;x^i),$
要证$[X,Y]_p\in \mathcal{D}_p.$
由可积性过$p$点存在积分子流形$i:N^k\hookrightarrow M,$ 局部来看由秩定理,
可以视其为切片. 从而$X,Y$只在前$k$个自然基上有光滑分量. 计算得到,
$[X,Y]$同样只在前$k$个自然基上有光滑分量,
因此$[X,Y]\in \chi(\mathcal{D}).$
注 1.2. 可以证明, 若$X_1,X_2$与$Y_1,Y_2$是$F$相关的, 即$F_\ast X_i=Y_i,$ 则$[X_1,X_2],[Y_1,Y_2]$也是$F$相关的. 由此可以证明在题设下, $[X|_N,Y|_N]$与$[X,Y]$是$i$相关的, 即$i_\ast [X|_N,Y|_N]=[X,Y],$ 从而$[X|_N,Y|_N]=[X,Y]|_N.$
定义 1.3. 设$\mathcal{D}$是$M$上光滑分布, 若$\,\forall\,X,Y\in \chi(\mathcal{D}),$ $[X,Y]\in \chi(D),$ 则称$\mathcal{D}$是对合的.
由前面的命题, 可积性$\Rightarrow$对合性. 事实上欲证对合性,
只需说明$\,\forall\,\mathcal{D}$的局部标架场$X_1,\cdots,X_k,$
$[X_i,X_j]\in \chi(\mathcal{D}),$ $\,\forall\,1\le i,j\le k.$
利用标架性质即可说明此时满足对合性的原始条件. 从而这两个条件是等价的.
Frobenius定理
定理 1.4 (Frobenius). 设$\mathcal{D}$是$M^m$上$k$维分布. 如果$\mathcal{D}$是对合的, 那么$\,\forall\,p\in M,$ $\,\exists\,$含$p$坐标系$(U,\varphi;x^i)$使得$k$维切片都是$\mathcal{D}$的积分子流形.
证: $k=1$时, 由Flow box定理, 每点局部都有坐标系,
使得任意$x^1$-曲线为$X$的积分曲线,
即$1$维切片都是$\mathcal{D}$的$1$维积分子流形.
归纳地, 若命题对$k-1$维分布成立,
那么设$\mathcal{D}$是$M$上对合的$k$维分布,
取局部标架场$X_1,\cdots, X_k,$ 那么由对合性,
$$[X_i,X_j]\equiv 0 \mod \{X_1,\cdots,X_k\}.$$
由Flow box定理,
可以取到坐标系使得$X_k=\frac{\partial {} }{\partial {}y_k},$
那么对$1\le \lambda,\mu,\nu\le k-1,$
令$X_\lambda'=X_\lambda-X_\lambda(y^k)X_k,$ 则$X_{\lambda}'(y_k)=0,$
$X_k(y_k)=1,$
从而$\mathcal{D}=\operatorname{span}\{X_1',\cdots,X_{k-1}',X_k\},$
仍是局部标架场.
计算得到,
光是$k-1$维分布$\mathcal{D}':=\operatorname{span}\{X_1',\cdots,X_{k-1}'\}$就满足对合性.
此时由归纳假设, $\mathcal{D}'$是可积分布,
有坐标系$(V,\psi;z^i)$使得$k-1$维切片都是$\mathcal{D}'$的积分子流形.
此时$\frac{\partial {}y^k}{\partial {}z^\lambda}=0,$
$\,\forall\,1\le \lambda\le k-1.$
$\mathcal{D}=\operatorname{span}\{\frac{\partial {} }{\partial {}z^1},\cdots,\frac{\partial {} }{\partial {}z^{k-1} },X_k\}.$
作用$y^k,$
计算得到$[\frac{\partial {} }{\partial {}z^\lambda},X_k]\equiv 0\mod \{\frac{\partial {} }{\partial {}z^1}\cdots\frac{\partial {} }{\partial {}z^{k-1} }\}.$
将$X_k$表示为$\sum_{A=1}^m \xi^A \frac{\partial {} }{\partial {}z^A},$
那么由前面的式子,
$\sum_{A=k}^m\frac{\partial {}\xi^A}{\partial {}z^\lambda}\frac{\partial {} }{\partial {}z^A}=0,$
这就说明$\xi^A=\xi^A(z^k,\cdots,z^m),$ $\,\forall\,A\ge k.$
令$X_k'=X_k-\sum_{\lambda=1}^{k-1}\xi^\lambda\frac{\partial {} }{\partial {}z^\lambda},$
只保留后面的分量. 由Flowbox定理,
可以取到新坐标系使得$X'_{k}=\frac{\partial {} }{\partial {}x^k},$
满足$x^\lambda=z^\lambda,$ $1\le \lambda\le k-1.$ 即前$k-1$个坐标不变.
那么此时即有$\mathcal{D}=\operatorname{span}\{\frac{\partial {} }{\partial {}x^1},\cdots,\frac{\partial {} }{\partial {}x^{k} }\},$
$k$维切片都是积分子流形.
单位分解
定理 1.5. 设光滑流形$M^m$上有紧集$\overline{V}\subset U,$ 那么存在$f\in C_c^\infty(M),$ 使得$f\in[0,1],$ $f|_V\equiv 1,$ $f|_{M\setminus U}\equiv 0.$
设$M$是拓扑空间, $\{A_\alpha\}$是$M$的一族子集.
如$\bigcup_{\alpha}A_\alpha=M,$
则称$\{A_\alpha\}$为$M$的一个覆盖.
设$\{A_\alpha\}$是$M$的覆盖, 如果$\,\forall\,p\in M,$
存在$U\ni p$使得$U$只与有限个$A_\alpha$相交,
则称$\{A_\alpha\}$是$M$的局部有限的覆盖.
设$\{U_\alpha\},$ $\{V_\beta\}$是$M$的两个覆盖.
若$\,\forall\,\beta,$
$\,\exists\,\alpha$使得$V_\beta\subset U_\alpha,$
则称$\{V_\beta\}$是$\{U_\alpha\}$的加细.
定义 1.6. $M$是拓扑空间, 若$M$的任意开覆盖$\{U_\alpha\}$有局部有限的加细, 则称$M$是仿紧的.
现设$M$是$m$维$C^\infty$流形, $\{A_\alpha\}$是$M$的开覆盖.
由第二可数公理, $\{A_\alpha\}$存在可数加细$\{U_j\},$
且每个开集都是坐标图$(U_j,\varphi_j).$ 坐标图可进一步改造为局部有限的,
且$\varphi_j(U_j)=B_1^m(0),$
$\{V_j=\varphi_j^{-1}(B_{1/2}^m(0))\}$也是$M$的开覆盖.
定义 1.7. 设$M$为光滑流形. 若$\{f_i\}\in C^\infty(M)$满足$f_i\in [0,1],$ $\sum_i f_i\equiv 1,$ 且$\{\operatorname{supp}f_i\}$是$M$的局部有限的覆盖, 那么称$\{f_i\}$为$M$的单位分解.
定理 1.8 (单位分解). 设$M^m$为光滑流形, 则存在单位分解$\{f_i\}.$
证: $\,\forall\,\bigcup_\alpha A_\alpha\supset M,$ 由上述引理,
存在局部有限加细$\{(U_i,\varphi_i)\},$ $\varphi_i(U_i)=B_1^m(0).$ 进而
$\,\exists\,\{\widetilde{f}_i\}\in C_c^\infty(U_i),$
$\widetilde{f}_i\in[0,1],$ $\widetilde{f}_i|_{V_i}\equiv 1.$
最后取$f_i=\frac{\widetilde{f}_i}{\sum_i \widetilde{f}_i}$即可.
事实上有$\operatorname{supp}f_i\subset U_i\subset A_\alpha,$
称$\{f_i\}$是从属于$\{A_\alpha\}$的单位分解.
因此任意开覆盖有从属于之的单位分解.
文章最后更新于 2021-11-10 16:15:52
Involutivity
Proposition 1.1. Let $\mathcal{D}$ be a smooth distribution on $M$. If $\mathcal{D}$ is integrable, then $\,\forall\,X,Y\in \chi(\mathcal{D}),$ is $\,\forall\,p\in M,$ $X_p,Y_p\in \mathcal{D}_p,$ then $[X,Y]\in \chi(\mathcal{D}).$
Certificate: Take any coordinate system $(U,\varphi;x^i),$ containing $p$
Proof$[X,Y]_p\in \mathcal{D}_p.$
According to the integrability, there exists an integrator manifold $i:N^k\hookrightarrow M,$ through the point $p$. From a local perspective, according to the rank theorem,
It can be regarded as a slice. Therefore, $X,Y$ only has smooth components on the first $k$ natural basis. Calculated,
$[X,Y]$ also has smooth components only on the first $k$ natural basis,
Therefore $[X,Y]\in \chi(\mathcal{D}).$
Note 1.2. It can be proved that if $X_1,X_2$ and $Y_1,Y_2$ are related to $F$, that is, $F_\ast X_i=Y_i,$, then $[X_1,X_2],[Y_1,Y_2]$ is also related to $F$. It can be proved that under the problem setting, $[X|_N,Y|_N]$ and $[X,Y]$ are related to $i$, that is, $i_\ast [X|_N,Y|_N]=[X,Y],$ and thus $[X|_N,Y|_N]=[X,Y]|_N.$
Definition 1.3. Assume $\mathcal{D}$ is a smooth distribution on $M$. If $\,\forall\,X,Y\in \chi(\mathcal{D}),$ $[X,Y]\in \chi(D),$, then $\mathcal{D}$ is said to be involution of.
From the previous proposition, integrability $\Rightarrow$ is involutive. In fact, if we want to prove the involutive property,
Just state the local frame field $X_1,\cdots,X_k,$ of $\,\forall\,\mathcal{D}$
$[X_i,X_j]\in \chi(\mathcal{D}),$ $\,\forall\,1\le i,j\le k.$
The original condition for satisfying involutivity can be explained by using the frame property. Therefore, these two conditions are equivalent.
Frobenius theorem
Theorem 1.4 (Frobenius). Let $\mathcal{D}$ be the $k$-dimensional distribution on $M^m$. If $\mathcal{D}$ is involutive, then $\,\forall\,p\in M,$ $\,\exists\,$ contains the $p$ coordinate system $(U,\varphi;x^i)$ such that the $k$-dimensional slices are all integrator manifolds of $\mathcal{D}$.
Certificate: When $k=1$, according to the Flow box theorem, each point has a local coordinate system,
Make any $x^1$-curve be the integral curve of $X$,
That is, $1$-dimensional slices are all $1$-dimensional integral molecular manifolds of $\mathcal{D}$.
Inductively, if the proposition holds for the $k-1$-dimensional distribution,
Then suppose $\mathcal{D}$ is the $k$-dimensional distribution involuted on $M$,
Taking the local standard field $X_1,\cdots, X_k,$, then according to the involution property,
$$[X_i,X_j]\equiv 0 \mod \{X_1,\cdots,X_k\}.$$
According to the Flow box theorem,
The coordinate system can be obtained such that $X_k=\frac{\partial {} }{\partial {}y_k},$
So right $1\le \lambda,\mu,\nu\le k-1,$
Let $X_\lambda'=X_\lambda-X_\lambda(y^k)X_k,$ then $X_{\lambda}'(y_k)=0,$
$X_k(y_k)=1,$
Thus $\mathcal{D}=\operatorname{span}\{X_1',\cdots,X_{k-1}',X_k\},$
It is still a local benchmark field.
Calculated,
The $k-1$-dimensional distribution $\mathcal{D}':=\operatorname{span}\{X_1',\cdots,X_{k-1}'\}$ alone satisfies the involution property.
At this time, according to the inductive assumption, $\mathcal{D}'$ is an integrable distribution,
There is a coordinate system $(V,\psi;z^i)$ such that $k-1$-dimensional slices are all integral manifolds of $\mathcal{D}'$.
At this time $\frac{\partial {}y^k}{\partial {}z^\lambda}=0,$
$\,\forall\,1\le \lambda\le k-1.$
$\mathcal{D}=\operatorname{span}\{\frac{\partial {} }{\partial {}z^1},\cdots,\frac{\partial {} }{\partial {}z^{k-1} },X_k\}.$
Effect$y^k,$
Calculate $[\frac{\partial {} }{\partial {}z^\lambda},X_k]\equiv 0\mod \{\frac{\partial {} }{\partial {}z^1}\cdots\frac{\partial {} }{\partial {}z^{k-1} }\}.$
Represent $X_k$ as $\sum_{A=1}^m \xi^A \frac{\partial {} }{\partial {}z^A},$
Then from the previous formula,
$\sum_{A=k}^m\frac{\partial {}\xi^A}{\partial {}z^\lambda}\frac{\partial {} }{\partial {}z^A}=0,$
This means $\xi^A=\xi^A(z^k,\cdots,z^m),$ $\,\forall\,A\ge k.$
Order$X_k'=X_k-\sum_{\lambda=1}^{k-1}\xi^\lambda\frac{\partial {} }{\partial {}z^\lambda},$
Only the following components are retained. According to the Flowbox theorem,
The new coordinate system can be obtained such that $X'_{k}=\frac{\partial {} }{\partial {}x^k},$
Satisfies $x^\lambda=z^\lambda,$ $1\le \lambda\le k-1.$, that is, the first $k-1$ coordinates remain unchanged.
Then there is $\mathcal{D}=\operatorname{span}\{\frac{\partial {} }{\partial {}x^1},\cdots,\frac{\partial {} }{\partial {}x^{k} }\},$ at this time
$k$-dimensional slices are all integral manifolds.
partition of unity
Theorem 1.5. Suppose there is a compact set $\overline{V}\subset U,$ on the smooth manifold $M^m$, then there is $f\in C_c^\infty(M),$ such that $f\in[0,1],$ $f|_V\equiv 1,$ $f|_{M\setminus U}\equiv 0.$
Assume $M$ is a topological space, and $\{A_\alpha\}$ is a family of subsets of $M$.
Such as $\bigcup_{\alpha}A_\alpha=M,$
Then $\{A_\alpha\}$ is called one of $M$ Cover.
Let $\{A_\alpha\}$ be the cover of $M$, if $\,\forall\,p\in M,$
There is $U\ni p$ such that $U$ only intersects a limited number of $A_\alpha$,
Then $\{A_\alpha\}$ is said to be of $M$ locally limited of coverage.
Let $\{U_\alpha\},$ $\{V_\beta\}$ be two covers of $M$.
If$\,\forall\,\beta,$
$\,\exists\,\alpha$ makes $V_\beta\subset U_\alpha,$
Then $\{V_\beta\}$ is said to be of $\{U_\alpha\}$ Add detail.
Definition 1.6. $M$ is a topological space. If any open cover $\{U_\alpha\}$ of $M$ has local finite thinning, then $M$ is said to be Paracompact of.
Now suppose $M$ is the $m$-dimensional $C^\infty$ manifold, and $\{A_\alpha\}$ is the open cover of $M$.
According to the second countable axiom, $\{A_\alpha\}$ there are countable additions $\{U_j\},$
And each open set is a coordinate graph $(U_j,\varphi_j).$. The coordinate graph can be further transformed into a locally finite one,
And $\varphi_j(U_j)=B_1^m(0),$
$\{V_j=\varphi_j^{-1}(B_{1/2}^m(0))\}$ is also the open override of $M$.
Definition 1.7. Let $M$ be a smooth manifold. If $\{f_i\}\in C^\infty(M)$ satisfies $f_i\in [0,1],$ $\sum_i f_i\equiv 1,$ and $\{\operatorname{supp}f_i\}$ is a local finite cover of $M$, then $\{f_i\}$ is called $M$ partition of unity.
Theorem 1.8 (partition of unity). Assume $M^m$ is a smooth manifold, then there is partition of unity $\{f_i\}.$
Certificate: $\,\forall\,\bigcup_\alpha A_\alpha\supset M,$ From the above lemma,
There is local limited thinning $\{(U_i,\varphi_i)\},$ $\varphi_i(U_i)=B_1^m(0).$ and then
$\,\exists\,\{\widetilde{f}_i\}\in C_c^\infty(U_i),$
$\widetilde{f}_i\in[0,1],$ $\widetilde{f}_i|_{V_i}\equiv 1.$
Finally, just take $f_i=\frac{\widetilde{f}_i}{\sum_i \widetilde{f}_i}$.
In fact there is $\operatorname{supp}f_i\subset U_i\subset A_\alpha,$
Say $\{f_i\}$ is Subordinate Factored into units of $\{A_\alpha\}$.
Therefore any open covering has a partition of unity subordinate to it.
The article was last updated on 2021-11-10 16:15:52
