Machine-translated from Chinese.
积分曲线
定理 1.1 (Flow box). 设$X\in\chi(M),$ 若$X_p\neq 0,$ 则存在含$p$坐标系$(U,\varphi;x^i)$使得$X|_U=\frac{\partial {} }{\partial {}x^1},$ 即$x^1$-曲线就是$X$在$U$中的积分曲线.
证: 任取坐标系$(U,\varphi;u^i),$ $\varphi(p)=0.$
$X|_U=\sum_i\xi^i(u)\frac{\partial {} }{\partial {}u^i}.$
不妨设$\xi^1(p)\neq 0,$ 且进而设其处处非零.
设$u$可写为$v$的函数(坐标变换),
使得$X|_U=\xi^1(v) \frac{\partial {} }{\partial {}v^1}.$
两边作用$u^\alpha,$ 有 $$\left\{
\begin{aligned}
&\frac{\partial {}u^\alpha}{\partial {}v^1}=\frac{\xi^\alpha(u^1,\cdots,u^m)}{\xi^1(u^1,\cdots,u^m)},\quad \alpha=1,\cdots,m.\\
&u^\alpha(0,v^2,\cdots,v^m)=v^\alpha
\end{aligned}
\right.$$ 由ODE理论,
$\,\exists\,\delta>0$使得$C_\delta^m=\{|u^i|<\delta\}\subset U,$
$\,\forall\,$给定初值$(0,v^2,\cdots,v^m)\in C_\delta^m,$
在$v^1\in (-\widetilde{\delta},\widetilde{\delta})$内有唯一解$u^\alpha(v).$
记$\widehat{\delta}=\min\{\delta,\widetilde{\delta}\},$
那么对于$u^\alpha:C_{\widehat{\delta} }^m\rightarrow \varphi(U),$
$\det(Du^\alpha)|_0=1,$ 局部为微分同胚.
不妨设本身就是微分同胚(不然可将$\widetilde{\delta}$缩小), 记为$G.$
此时在坐标系$(\widetilde{U},G^{-1}\circ\varphi;v^i)$下,
$X|_{\widetilde{U} }=\xi^1(v)\frac{\partial {} }{\partial {}v^1}.$
最后再做变换 $$\left\{
\begin{aligned}
&w^1=\int_0^{v_1} \frac{dt}{\xi^1(t,v^2,\cdots,v^n)}\\
&w^\alpha=v^\alpha,\quad \alpha=2,\cdots,m
\end{aligned}
\right.$$
即有$X|_{\widetilde{U} }=\frac{\partial {} }{\partial {}w^1},$
$w^1$-曲线为$X$在$\widetilde{U}$中积分曲线.
分布
定义 1.2. $M^m$为$C^\infty$流形, 若$\,\forall\,p\in M,$ 指定$T_pM$中$k$-维子空间$D_p,$ 则称$\mathcal{D}:p\rightarrow D_p$为$M$上$k$维分布.
若$\,\forall\,p\in M,$ $\,\exists\,U\ni p,$
$X_1,\cdots, X_k\in \chi(U),$
使得$\mathcal{D}_q=\operatorname{span}\{X_1(q),\cdots,X_k(q)\},$
$\,\forall\,q\in U,$ 则称$\mathcal{D}$是$M$上$C^\infty$分布,
$\{X_1,\cdots,X_k\}$称为$\mathcal{D}$的局部标架场. $k\in(0,m)$时,
$\mathcal{D}$可看作$TM$的非平凡秩$k$子丛.
定义 1.3. 设$\mathcal{D}$是$M$上$k$维$C^\infty$-分布, $F:W\hookrightarrow M$是子流形, 满足${F_{\ast } }_p(T_pW)=D_{F(p)},$ $\,\forall\,p\in W.$ 则称$(W,F)$为$\mathcal{D}$的积分子流形.
自然的问题是, 给定$M^m$上一个$k$维$C^\infty$-分布$\mathcal{D},$
$\,\forall\,p\in M,$ 是否存在$\mathcal{D}$的过$p$点的积分子流形?
答案是否定的. 事实上积分子流形应有积分曲线网,
那么若在$\mathbb{R}^3$上取$X=\frac{\partial {} }{\partial {}x}+y\frac{\partial {} }{\partial {}z},$
$Y=\frac{\partial {} }{\partial {}y},$
$\mathcal{D}=\operatorname{span}\{X,Y\}.$ 那么若有过零点的积分子流形,
应包含$xoy$平面的一小部分. 但是在$x$轴外该流形又不是积分子流形, 矛盾.
定义 1.4. 设$\mathcal{D}$是$M$上$k$维$C^\infty$-分布, 若过每点$p\in M$都存在$\mathcal{D}$的积分子流形, 则我们称$\mathcal{D}$是可积的. 否则, 称之为不可积的.
$k=0,m$时平凡. $k=1$时, 由积分曲线存在定理, $1$维分布总是可积的,
积分子流形即为积分曲线. 但积分曲线不都是某个分布的积分子流形,
因为积分曲线对应的光滑向量场允许有奇点,
但有奇点的光滑向量场并不能作为标架张成分布.
文章最后更新于 2021-11-03 16:49:42
integral curve
Theorem 1.1 (Flow box). Assume $X\in\chi(M),$ if $X_p\neq 0,$, then there is a coordinate system $(U,\varphi;x^i)$ containing $p$ such that $X|_U=\frac{\partial {} }{\partial {}x^1},$ is $x^1$ - the curve is the integral curve of $X$ in $U$.
Certificate: Take any coordinate system $(U,\varphi;u^i),$ $\varphi(p)=0.$
$X|_U=\sum_i\xi^i(u)\frac{\partial {} }{\partial {}u^i}.$
Let us assume $\xi^1(p)\neq 0,$ and further assume that it is non-zero everywhere.
Suppose $u$ can be written as the function (coordinate transformation) of $v$,
Make$X|_U=\xi^1(v) \frac{\partial {} }{\partial {}v^1}.$
Both sides act $u^\alpha,$ with $$\left\{
\begin{aligned}
&\frac{\partial {}u^\alpha}{\partial {}v^1}=\frac{\xi^\alpha(u^1,\cdots,u^m)}{\xi^1(u^1,\cdots,u^m)},\quad \alpha=1,\cdots,m.\\
&u^\alpha(0,v^2,\cdots,v^m)=v^\alpha
\end{aligned}
\right.$$ According to ODE theory,
$\,\exists\,\delta>0$ makes $C_\delta^m=\{|u^i|<\delta\}\subset U,$
$\,\forall\,$ Given initial value $(0,v^2,\cdots,v^m)\in C_\delta^m,$
There is a unique solution $u^\alpha(v).$ in $v^1\in (-\widetilde{\delta},\widetilde{\delta})$
Note$\widehat{\delta}=\min\{\delta,\widetilde{\delta}\},$
So for $u^\alpha:C_{\widehat{\delta} }^m\rightarrow \varphi(U),$
$\det(Du^\alpha)|_0=1,$ is locally diffeomorphic.
We might as well assume that it is a diffeomorphism (otherwise $\widetilde{\delta}$ can be reduced), recorded as $G.$
At this time, under the coordinate system $(\widetilde{U},G^{-1}\circ\varphi;v^i)$,
$X|_{\widetilde{U} }=\xi^1(v)\frac{\partial {} }{\partial {}v^1}.$
Finally do the transformation $$\left\{
\begin{aligned}
&w^1=\int_0^{v_1} \frac{dt}{\xi^1(t,v^2,\cdots,v^n)}\\
&w^\alpha=v^\alpha,\quad \alpha=2,\cdots,m
\end{aligned}
\right.$$
That is $X|_{\widetilde{U} }=\frac{\partial {} }{\partial {}w^1},$
$w^1$ - The curve is $X$ integrating the curve in $\widetilde{U}$.
Distribution
Definition 1.2. $M^m$ is the $C^\infty$ manifold. If $\,\forall\,p\in M,$ specifies the $k$-dimensional subspace $D_p,$ in $T_pM$, then $\mathcal{D}:p\rightarrow D_p$ is called the $k$ dimension on $M$. Distribution.
If $\,\forall\,p\in M,$ $\,\exists\,U\ni p,$
$X_1,\cdots, X_k\in \chi(U),$
Make$\mathcal{D}_q=\operatorname{span}\{X_1(q),\cdots,X_k(q)\},$
$\,\forall\,q\in U,$ then $\mathcal{D}$ is said to be the distribution of $C^\infty$ on $M$,
$\{X_1,\cdots,X_k\}$ is called $\mathcal{D}$ local frame field.$k\in(0,m)$,
$\mathcal{D}$ can be regarded as the non-trivial rank $k$ subbundle of $TM$.
Definition 1.3. Assume $\mathcal{D}$ is the $k$-dimensional $C^\infty$-distribution on $M$, $F:W\hookrightarrow M$ is a sub-manifold, and satisfies ${F_{\ast } }_p(T_pW)=D_{F(p)},$ $\,\forall\,p\in W.$, then $(W,F)$ is called $\mathcal{D}$ Integral manifold.
The natural question is, given $M^m$, the last $k$ dimension $C^\infty$-distribution $\mathcal{D},$
$\,\forall\,p\in M,$ Is there an integrator manifold of $\mathcal{D}$ passing through the point $p$?
The answer is no. In fact, the integrator manifold should have a network of integral curves,
Then if $X=\frac{\partial {} }{\partial {}x}+y\frac{\partial {} }{\partial {}z},$ is taken from $\mathbb{R}^3$
$Y=\frac{\partial {} }{\partial {}y},$
$\mathcal{D}=\operatorname{span}\{X,Y\}.$ Then if there is an integrator manifold crossing zero points,
It should contain a small part of the $xoy$ plane. However, the manifold outside the $x$ axis is not an integrator manifold, which is a contradiction.
Definition 1.4. Assume $\mathcal{D}$ is a $k$-dimensional $C^\infty$-distribution on $M$. If there is an integral manifold of $\mathcal{D}$ through every point $p\in M$, then we say $\mathcal{D}$ is integrable. Otherwise, call it non-integrable.
When $k=0,m$ is trivial. When $k=1$, according to the existence theorem of the integral curve, the $1$-dimensional distribution is always integrable,
The integrator manifold is the integral curve. But the integral curve is not always the integrator manifold of a certain distribution.
Because the smooth vector field corresponding to the integral curve allows singular points,
However, a smooth vector field with singular points cannot be used as a frame span distribution.
The article was last updated on 2021-11-03 16:49:42
