Machine-translated from Chinese.
参考书: Warner《微分流形与李群基础》
流形上的微分形式
拉回映射
光滑函数$f:\mathbb{R}^n\rightarrow \mathbb{R}^m$可诱导拉回映射$f^\ast :\Omega^0(\mathbb{R}^m)\rightarrow\Omega^0(\mathbb{R}^n)$,
$f^\ast (g)=g\circ f$;
同时还可诱导$f^\ast :\Omega^1(\mathbb{R}^m)\rightarrow \Omega^1(\mathbb{R}^n)$,
$f^\ast (dy_i)=df_i=d(y_i\circ f)$.
容易延拓该定义得到$f^\ast :\Omega^\ast (\mathbb{R}^m)\rightarrow\Omega^\ast (\mathbb{R}^n)$,
$f^\ast (g_Idy_{i_1}\wedge ...\wedge dy_{i_q})$ =
$f^\ast (g_I)f^\ast (dy_{i_1}\wedge ...\wedge dy_{i_q})=(g_I\circ f)df_{i_1}\wedge ... \wedge df_{i_q}$.
$0$维上的定义比较自然, $1$维为何要对基$dy_i$如此定义需要想一下.
我们已知$d:\Omega^0\rightarrow \Omega^1$,
发现$df^\ast (g)=d(g\circ f)=\sum_i\frac{\partial {}g\circ f}{\partial {}f_i}df_i$,
$dg=\sum_i\frac{\partial {}g}{\partial {}y_i}dy_i$.
注意$f^\ast (\frac{\partial {}g}{\partial {}y_i})=\frac{\partial {}g\circ f}{\partial {}f_i}$,
所以我们定义$f^\ast (dy_i)=df_i$也是相当自然的.
同时由上分析我们得到如下命题:
命题 1.1 (可交换性). $df^\ast =f^\ast d$.
也可以说我们的定义就是从希望满足可交换性出发的.
一阶微分不变性
记$f:\mathbb{R}^n\rightarrow\mathbb{R}^n$为微分同胚,
$f_i=x_i\circ f$为新坐标, 称为由微分同胚$f$定义的坐标.
由链式法则,
$dg=\sum_{i=1}^{n}\frac{\partial {}g}{\partial {}f_i} df_i=\sum_{i,j=1}^{n}\frac{\partial {}g}{\partial {}f_i}\frac{\partial {}f_i}{\partial {}x_j} dx_j=\sum_{j=1}^n \frac{\partial {}g}{\partial {}x_j}dx_j,$
于是$dg$与坐标系无关, 这就是在说一阶微分不变性.
拉回映射与一阶微分不变性是流形上定义微分形式与微分算子$d$的基础.
流形上微分形式
记$M$为光滑$n$维流形,
$\{(U_\alpha, \phi_\alpha)\}_{\alpha\in\Lambda}$为坐标图册.
$M$上的微分形式$\omega$为一个集类$\{\omega_\alpha\in \Omega^\ast (U_\alpha)\}$,
该集类需要满足相容性条件：$\,\forall\,U_\alpha\cap U_\beta\neq \varnothing$,
记$U_\alpha\cap U_\beta=U_{\alpha\beta}$,
$\omega_\alpha|_{U_{\alpha\beta} }=\omega_\beta|_{U_{\alpha\beta} }$,
这样才能使得局部给出的微分形式能够拼成一个整体,
即集类$\{\omega_\alpha\in \Omega^\ast (U_\alpha)\}$拼成$\omega$.
这里$\omega_\alpha=\sum_I {f_\alpha}_I d{\phi_\alpha}_I$,
$\phi_\alpha:U_\alpha\approx \mathbb{R}^n$,
即将曲面片上的微分形式拉到$\mathbb{R}^n$中考虑.
$\omega_\alpha|_{U_{\alpha\beta} }$的定义是由嵌入$i:U_{\alpha\beta}\rightarrow U_\alpha$,
从$\Omega^\ast (U_\alpha)$拉回到$\Omega^\ast (U_{\alpha\beta})$中的,
即定义为$i^\ast \omega_\alpha$.
记$M$上的微分形式全体为$\Omega^\ast (M)$，类似地可以定义微分算子$d$与外积$\wedge$.
这里需注意由于流形上的微分形式有要求, 需要验证相容性条件.
由$d$与$i^\ast$的可交换性，$\{d\omega_\alpha\}$确实满足相容性条件,
可以拼成$d\omega$, 因此定义良好. $\wedge$同理.
同样的在流形上也有$\Omega^q(M)$, 可以按型分解.
$\{\Omega^\ast (M),d\}$称为流形$M$的de Rham复形.
进一步可以定义$Z^q(M),B^q(M),H^q(M)$.
现考虑流形间的光滑映射$f:M\rightarrow N$,
这诱导了$f^\ast :\Omega^\ast (N)\rightarrow\Omega^\ast (M)$.
由于$d,f^\ast$可交换, 这便诱导了
$f^\ast : H^\ast (N)\rightarrow H^\ast (M)$,
$f^\ast [\omega]=[f^\ast \omega]$. 特别地,
考虑$i:U\rightarrow M$为开子集的包含映射,
则有$i^\ast : H^\ast (M)\rightarrow H^\ast (U)$,
$i^\ast [\omega]=[i^\ast \omega]=[\omega|_U]$.
也把$i^\ast [\omega]$记作$[\omega]|_U$, 于是有$[\omega]|_U=[\omega|_U]$.
单位分解
单位分解的存在性是微分流形的最基本工具.
我们回忆流形是满足$T_2$分离性公理且仿紧的.
$M$上的单位分解是非负光滑函数集族$\{\rho_\alpha\}_{\alpha\in\Lambda}$,
满足$M$上每点有一邻域, 其上只有有限多个$\rho_\alpha$不为零.
这样$\sum\rho_\alpha$有意义, 进一步要求和处处为$1$.
由于流形是仿紧的, 其上总存在单位分解.
对于给定的开覆盖$\{U_\alpha\}$(往往便取坐标图册中的开覆盖),
存在从属于它的单位分解,
即$\operatorname{supp} \rho_\alpha\subset U_\alpha$.
流形的定向
记微分同胚$T:\mathbb{R}^n\rightarrow \mathbb{R}^n$,
$T(y)=x$，记$T_i=x_i\circ T$,
$J(T)=\det (\frac{\partial {}T_i}{\partial {}y_j})$. 由于$T$是微分同胚,
$J(T)$不为零, 所以要么恒正, 要么恒负. $J(T)>0$时称$T$保定向,
反之称为反定向.
一般的, 若是开集间的微分同胚, 则要求在所有连通分支上都恒正,
才称为保定向.
设$M$是光滑流形, $\{(U_\alpha, \phi_\alpha)\}$是坐标图册,
若$\,\forall\,U_{\alpha\beta}$非空时,
微分同胚$\phi_\beta\circ \phi_\alpha^{-1}: \phi_\alpha(U_{\alpha\beta})\rightarrow\phi_\beta(U_{\alpha\beta})$保定向,
则称坐标图册为定向坐标图册. 若$M$有一个定向坐标图册, 则称$M$可定向.
命题 1.2. $n$维流形$M$可定向当且仅当它有整体定义的处处非零的光滑$n$形式.
注意任意两个处处非零的$n$-形式$\eta,\eta'$,
他们之间只相差一个处处非零的函数$f$: $\eta=f\eta'$. 若$M$连通,
则$f$恒正或恒负. $f$为正时称$\eta,\eta'$等价.
于是$M$上每个处处非零的光滑$n$-形式都落在两个等价类中,
每个等价类称为$M$的一个定向, 记作$[M]$.
流形上的积分
设$\omega$为$\mathbb{R}^n$中具有紧支集的光滑$n$-形式:
$\omega=f(x)dx_1\wedge \cdots\wedge dx_n$,
定义积分$\int_{\mathbb{R}^n}\omega=\int_{\mathbb{R}^n}fdx_1\cdots dx_n$.
设$T:\mathbb{R}^n\rightarrow\mathbb{R}^n$, $T(y)=x$为微分同胚,
则可以证明变量代换公式:
命题 1.3 (变量代换). $dT_1\wedge ...\wedge dT_n=J(T)dy_1\wedge ...\wedge dy_n$.
从而$T^\ast \omega=T^\ast fdT_1\wedge ... \wedge dT_n=(f\circ T)J(T)dy_1\wedge ...\wedge dy_n,$
对上式积分得到
$$\int_{\mathbb{R}^n}T^\ast \omega=\int_{\mathbb{R}^n}(f\circ T)J(T)dy_1\wedge ...\wedge dy_n,$$
又由积分变换公式
$$\int_{\mathbb{R}^n} \omega=\int_{\mathbb{R}^n}(f\circ T)|J(T)|dy_1\wedge ...\wedge dy_n,$$
因此$\int_{\mathbb{R}^n}T^\ast \omega=\pm \int_{\mathbb{R}^n}\omega$,
符号由$J(T)$正负决定.
现给出$M$上的积分.
选定定向$[M]$与相应的定向坐标图册$\{(U_\alpha,\phi_\alpha)\}$.
设$\{\rho_\alpha\}$是从属于$\{U_\alpha\}$的单位分解,
对$\tau\in H^n_{c}(M)$有分解$\tau=\sum_\alpha \rho_\alpha \tau$.
我们希望有 $$\int_{[M]}\tau=\sum_\alpha \int_{[M]}\rho_\alpha\tau,$$
进而研究右式又希望
$$\int_{[M]}\rho_\alpha\tau=\int_{U_\alpha} (\rho_\alpha\tau)|_{U_\alpha},$$
利用变量代换, 将积分拉回到$\mathbb{R}^n$进行
$$\int_{U_\alpha} (\rho_\alpha\tau)|_{U_\alpha}=\int_{\mathbb{R}^n}(\phi_\alpha^{-1})^\ast (\rho_\alpha\tau)|_{U_\alpha},$$
注意这里将被积形式零延拓到$\phi(U_\alpha)$外.
于是我们将$\int_{[M]}\tau:=\sum_\alpha \int_{\mathbb{R}^n}(\phi_\alpha^{-1})^\ast (\rho_\alpha\tau)|_{U_\alpha}$,
若方向明了则可改记$\int_{[M]}$为$\int_{M}$.
需要验证的是这个定义与定向坐标图册与单位分解无关.
关于这点我们可以基于两个不同的划分做更细的划分, 再将其做和,
从而得到两种不同划分得到的积分式是相等的.
带边流形
带边流形$M$的定向坐标图册诱导了边界处的定向坐标图册,
先定义好上半平面是如何诱导的,
再将任意$M$的边界处打到上半平面来决定诱导定向.
定理 1.1 (Stokes定理). $\int_Md\omega=\int_{\partial M}\omega|_{\partial M}$.
文章最后更新于 2021-09-19 19:06:22
Reference book: Warner "Basics of Differential Manifolds and Lie Groups"
Differential forms on manifolds
Pull back mapping
The smooth function $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ can induce the pullback map $f^\ast :\Omega^0(\mathbb{R}^m)\rightarrow\Omega^0(\mathbb{R}^n)$,
$f^\ast (g)=g\circ f$;
It can also induce $f^\ast :\Omega^1(\mathbb{R}^m)\rightarrow \Omega^1(\mathbb{R}^n)$,
$f^\ast (dy_i)=df_i=d(y_i\circ f)$.
It is easy to extend this definition to get $f^\ast :\Omega^\ast (\mathbb{R}^m)\rightarrow\Omega^\ast (\mathbb{R}^n)$,
$f^\ast (g_Idy_{i_1}\wedge ...\wedge dy_{i_q})$ =
$f^\ast (g_I)f^\ast (dy_{i_1}\wedge ...\wedge dy_{i_q})=(g_I\circ f)df_{i_1}\wedge ... \wedge df_{i_q}$.
The definition on the $0$ dimension is more natural. Why the $1$ dimension has such a definition for the base $dy_i$ needs to be thought about.
We already know $d:\Omega^0\rightarrow \Omega^1$,
Discover$df^\ast (g)=d(g\circ f)=\sum_i\frac{\partial {}g\circ f}{\partial {}f_i}df_i$,
$dg=\sum_i\frac{\partial {}g}{\partial {}y_i}dy_i$.
Note$f^\ast (\frac{\partial {}g}{\partial {}y_i})=\frac{\partial {}g\circ f}{\partial {}f_i}$,
So it is quite natural for us to define $f^\ast (dy_i)=df_i$.
At the same time, from the above analysis we get the following proposition:
Proposition 1.1 (Exchangeability). $df^\ast =f^\ast d$.
It can also be said that our definition is based on the hope of satisfying exchangeability.
First-order differential invariance
Denote $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ as diffeomorphism,
$f_i=x_i\circ f$ is the new coordinate, which is called the coordinate defined by diffeomorphism $f$.
According to the chain rule,
$dg=\sum_{i=1}^{n}\frac{\partial {}g}{\partial {}f_i} df_i=\sum_{i,j=1}^{n}\frac{\partial {}g}{\partial {}f_i}\frac{\partial {}f_i}{\partial {}x_j} dx_j=\sum_{j=1}^n \frac{\partial {}g}{\partial {}x_j}dx_j,$
So $dg$ has nothing to do with the coordinate system, which means First-order differential invariance.
Pullback mapping and first-order differential invariance are the basis for defining differential forms and differential operators $d$ on manifolds.
differential form on manifold
Let $M$ be a smooth $n$-dimensional manifold,
$\{(U_\alpha, \phi_\alpha)\}_{\alpha\in\Lambda}$ is the coordinate atlas.
The differential form $\omega$ on $M$ is a set class $\{\omega_\alpha\in \Omega^\ast (U_\alpha)\}$,
This set class needs to satisfy the compatibility condition: $\,\forall\,U_\alpha\cap U_\beta\neq \varnothing$,
Note$U_\alpha\cap U_\beta=U_{\alpha\beta}$,
$\omega_\alpha|_{U_{\alpha\beta} }=\omega_\beta|_{U_{\alpha\beta} }$,
In this way, the differential forms given locally can be assembled into a whole,
That is, the set class $\{\omega_\alpha\in \Omega^\ast (U_\alpha)\}$ is spelled into $\omega$.
Here $\omega_\alpha=\sum_I {f_\alpha}_I d{\phi_\alpha}_I$,
$\phi_\alpha:U_\alpha\approx \mathbb{R}^n$,
That is, the differential form on the surface patch is taken into consideration in $\mathbb{R}^n$.
$\omega_\alpha|_{U_{\alpha\beta} }$ is defined by embedding $i:U_{\alpha\beta}\rightarrow U_\alpha$,
Pull from $\Omega^\ast (U_\alpha)$ back to $\Omega^\ast (U_{\alpha\beta})$,
That is defined as $i^\ast \omega_\alpha$.
Let the differential form on $M$ be $\Omega^\ast (M)$. Similarly, the differential operator $d$ and the outer product $\wedge$ can be defined.
It should be noted here that due to the requirements for the differential form on the manifold, the compatibility conditions need to be verified.
Due to the interchangeability of $d$ and $i^\ast$, $\{d\omega_\alpha\}$ does satisfy the compatibility condition,
It can be spelled as $d\omega$, so it is well defined. The same goes for $\wedge$.
Similarly, there is $\Omega^q(M)$ on the manifold, which can be decomposed by type.
$\{\Omega^\ast (M),d\}$ is called the de Rham complex of the manifold $M$.
It can be further defined $Z^q(M),B^q(M),H^q(M)$.
Now consider the smooth mapping between manifolds $f:M\rightarrow N$,
This induces $f^\ast :\Omega^\ast (N)\rightarrow\Omega^\ast (M)$.
Since $d,f^\ast$ is commutative, this induces
$f^\ast : H^\ast (N)\rightarrow H^\ast (M)$,
$f^\ast [\omega]=[f^\ast \omega]$. In particular,
Consider $i:U\rightarrow M$ as the inclusion map of the open subset,
Then there is $i^\ast : H^\ast (M)\rightarrow H^\ast (U)$,
$i^\ast [\omega]=[i^\ast \omega]=[\omega|_U]$.
Also write $i^\ast [\omega]$ as $[\omega]|_U$, so we have $[\omega]|_U=[\omega|_U]$.
partition of unity
The existence of partition of unity is the most basic tool of differential manifolds.
We recall that the manifold satisfies the $T_2$ separability axiom and is quasi-compact.
The partition of unity on $M$ is a family of nonnegative smooth functions $\{\rho_\alpha\}_{\alpha\in\Lambda}$,
It satisfies that every point on $M$ has a neighborhood, and there are only a finite number of $\rho_\alpha$ on it that are not zero.
This way $\sum\rho_\alpha$ makes sense, and further requirements and everywhere are $1$.
Since the manifold is quasi-compact, there is always partition of unity on it.
For a given open coverage $\{U_\alpha\}$ (often the open coverage in the coordinate atlas is taken),
Being is subject to the decomposition of its units,
That is $\operatorname{supp} \rho_\alpha\subset U_\alpha$.
Orientation of the manifold
Note diffeomorphism $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$,
$T(y)=x$, note $T_i=x_i\circ T$,
$J(T)=\det (\frac{\partial {}T_i}{\partial {}y_j})$. Since $T$ is diffeomorphism,
$J(T)$ is not zero, so it is either always positive or always negative. When $J(T)>0$ is used, it is said that $T$ maintains the orientation.
The opposite is called reverse orientation.
Generally, if it is a diffeomorphism between open sets, it is required to be constant positive on all connected components.
That's called Bao Zhi.
Assume $M$ is a smooth manifold, $\{(U_\alpha, \phi_\alpha)\}$ is a coordinate atlas,
If $\,\forall\,U_{\alpha\beta}$ is not empty,
Diffeomorphism $\phi_\beta\circ \phi_\alpha^{-1}: \phi_\alpha(U_{\alpha\beta})\rightarrow\phi_\beta(U_{\alpha\beta})$ orientation-preserving,
The coordinate atlas is called an oriented coordinate atlas. If $M$ has an oriented coordinate atlas, it is said that $M$ can be oriented.
Proposition 1.2. A $n$-dimensional manifold $M$ is orientable if and only if it has a globally defined smooth $n$ form that is non-zero everywhere.
Note that for any two everywhere non-zero $n$-forms $\eta,\eta'$,
The only difference between them is a function $f$ that is non-zero everywhere: $\eta=f\eta'$. If $M$ is connected,
Then $f$ is always positive or negative. When $f$ is positive, it is said that $\eta,\eta'$ is equivalent.
So every everywhere non-zero smooth $n$-form on $M$ falls into two equivalence classes,
Each equivalence class is called an orientation of $M$, denoted as $[M]$.
integrals on manifolds
Let $\omega$ be a smooth $n$-form with compact support in $\mathbb{R}^n$:
$\omega=f(x)dx_1\wedge \cdots\wedge dx_n$,
Define integral $\int_{\mathbb{R}^n}\omega=\int_{\mathbb{R}^n}fdx_1\cdots dx_n$.
Let $T:\mathbb{R}^n\rightarrow\mathbb{R}^n$, $T(y)=x$ be diffeomorphisms,
Then we can prove the variable substitution formula:
Proposition 1.3 (Variable substitution). $dT_1\wedge ...\wedge dT_n=J(T)dy_1\wedge ...\wedge dy_n$.
Thus $T^\ast \omega=T^\ast fdT_1\wedge ... \wedge dT_n=(f\circ T)J(T)dy_1\wedge ...\wedge dy_n,$
Integrate the above equation to get
$$\int_{\mathbb{R}^n}T^\ast \omega=\int_{\mathbb{R}^n}(f\circ T)J(T)dy_1\wedge ...\wedge dy_n,$$
And by the integral transformation formula
$$\int_{\mathbb{R}^n} \omega=\int_{\mathbb{R}^n}(f\circ T)|J(T)|dy_1\wedge ...\wedge dy_n,$$
Therefore $\int_{\mathbb{R}^n}T^\ast \omega=\pm \int_{\mathbb{R}^n}\omega$,
The sign is determined by the positive or negative $J(T)$.
Now give the integral on $M$.
Select the orientation $[M]$ and the corresponding orientation coordinates album $\{(U_\alpha,\phi_\alpha)\}$.
Let $\{\rho_\alpha\}$ be the partition of unity subordinate to $\{U_\alpha\}$,
There is decomposition $\tau=\sum_\alpha \rho_\alpha \tau$ for $\tau\in H^n_{c}(M)$.
We would like to have $$\int_{[M]}\tau=\sum_\alpha \int_{[M]}\rho_\alpha\tau,$$
Then study the right formula and hope that
$$\int_{[M]}\rho_\alpha\tau=\int_{U_\alpha} (\rho_\alpha\tau)|_{U_\alpha},$$
Use variable substitution to pull the integral back to $\mathbb{R}^n$.
$$\int_{U_\alpha} (\rho_\alpha\tau)|_{U_\alpha}=\int_{\mathbb{R}^n}(\phi_\alpha^{-1})^\ast (\rho_\alpha\tau)|_{U_\alpha},$$
Note that the integrand form zero is extended outside $\phi(U_\alpha)$.
So we will $\int_{[M]}\tau:=\sum_\alpha \int_{\mathbb{R}^n}(\phi_\alpha^{-1})^\ast (\rho_\alpha\tau)|_{U_\alpha}$,
If the direction is clear, $\int_{[M]}$ can be rewritten as $\int_{M}$.
What needs to be verified is that this definition has nothing to do with the directional coordinate atlas and partition of unity.
Regarding this point, we can make a more detailed division based on two different divisions, and then sum them up,
It follows that the integral formulas obtained by the two different divisions are equal.
Manifolds with Boundary
The oriented coordinate atlas of the edged manifold $M$ induces the oriented coordinate atlas at the boundary,
First define how the upper half plane is induced,
Then hit the boundary of any $M$ to the upper half plane to determine the induced orientation.
Theorem 1.1 (Stokes theorem). $\int_Md\omega=\int_{\partial M}\omega|_{\partial M}$.
The article was last updated on 2021-09-19 19:06:22
