《微分流形》第七章-流形上的积分 "Differential Manifolds" Chapter 7 - Integrals on Manifolds
DreamAR

外微分形式的积分

定义

我们熟知有积分变量代换公式: 设$f\in C_c(\mathbb{R}^m),$ $h:\mathbb{R}^m\approx \mathbb{R}^m$为微分同胚, 则

$$ \int_{\mathbb{R}^m}f(x)dx^1\cdots dx^m=\int_{\mathbb{R}^m}(f\circ h)(y)|\det Dh|dy^1\cdots dy^m. $$

若$h:\mathbb{R}^m\rightarrow \mathbb{R}^m$满足$\det Dh(y)>0,$ 则处处有$\det Dh>0.$ 称它是保定向的微分同胚.

由上节课内容, $\Omega={}dx^{1}\wedge\cdots\wedge{}dx^{ {}m}\in \mathcal{A}^m(\mathbb{R}^m)$给了$\mathbb{R}^m$一个自然的定向. 现$(\mathbb{R}^m,\mathrm{id};x),(\mathbb{R}^m,h^{-1};y)$为两个不同坐标系, 令$\omega=f(x)dx^1\wedge\cdots\wedge dx^m$为具紧支集连续$m$次外微分形式, 那么由于

$$ dx^1\wedge\cdots\wedge dx^m=\det(Dh)(y){}dy^{1}\wedge\cdots\wedge{}dy^{ {}m}, $$

我们知道$\omega=f\circ h(y)\det(Dh){}dy^{1}\wedge\cdots\wedge{}dy^{ {}m}$为另一表示. 上述公式提示:

$$ \int_{\mathbb{R}^m}\omega:=\int_{\mathbb{R}^m}f(x)dx^1\cdots dx^m. $$

设$(M^m,\Omega)$是定向$C^\infty$-流形, $\omega$为$M$上具紧支集$m$次外微分形式, 即$\operatorname{supp}\omega=\overline{\left\{q\mid\omega(q)\neq 0\right\}} \Subset M.$ 下面来定义$\omega$在$M$上的积分.

当$\operatorname{supp}\omega\subset (U,\varphi;x^i),$ 与定向$\Omega$相符的坐标系时, $\omega=f dx^1\wedge\cdots\wedge dx^m,$ $f\in C_c(U).$ 由前面的叙述, 定义

$$ \int_M \omega:=\int_U \omega=\int_{\varphi(U)}(f\circ\varphi^{-1})(x)dx^1\cdots dx^m. $$

一般地, 取$\{(U_i,\varphi_i)\}$为与定向相符的坐标图册, 以及从属于它的单位分解. 那么$\omega=\sum_i f_i\omega,$ $\operatorname{supp}(f_i\omega)\subset U_i.$ 由$\omega$具紧支集, $\{\operatorname{supp}f_i\}$局部有限, 和式总有意义. 于是我们定义

$$ \int_M \omega:=\sum_i \int_M f_i\omega, $$

其中$\int_M f_i\omega$是已定义好的.

良定性

我们需要验证该定义的良定性: 是否与坐标系/图册以及单位分解选取无关?

若$\operatorname{supp}\omega \subset (U,\varphi;x)\cap (V,\psi;y),$ $f{}dx^{1}\wedge\cdots\wedge{}dx^{ {}m}=\omega=g{}dy^{1}\wedge\cdots\wedge{}dy^{ {}m}.$ 需验证:

$$ \int_{\psi(U)}(g\circ\psi^{-1})(y)dy^1\cdots dy^m=\int_M\omega=\int_{\varphi(U)}(f\circ\varphi^{-1})(x)dx^1\cdots dx^m. $$

令$h=\varphi\circ\psi^{-1},$ $h(y)=x.$ 那么由积分变量代换公式, 以及$g=f\det(Dh),$ 结合坐标系都是与定向$\Omega$相符的, 即有等式成立.

设有两种单位分解$\{f_i\},\{g_j\},$ 那么$\{f_ig_j=g_jf_i\}$也是一个单位分解. 因此

$$ \sum_j\int_M g_j\omega=\sum_{i,j}\int_M f_ig_j\omega=\sum_i\int_M f_i\omega. $$

其它性质

容易看出, $\int_{(M,\Omega)}\omega=-\int_{(M,-\Omega)}\omega.$

定义$\mathcal{K}$为全体具紧支集连续的$m$次形式, 构成向量空间, 且$\int_M:\mathcal{K}\rightarrow \mathbb{R}$为线性泛函.

设$\varphi:N^k\rightarrow M$为$k$维子流形. 若$\varphi^\ast \omega$在$N$中具紧支集, 则

$$ \int_N \varphi^\ast \omega:=\int_{\varphi(N)} \omega. $$

Stokes公式

回忆$\mathbb{R}^m$上的多元微积分, 我们可将微积分基本定理, Green公式, Gauss公式, Stokes公式统一为:

$$ \int_D d\omega=\int_{\partial D}\omega,\quad D\subset \mathbb{R}^m. $$

设$M^m$为$m$维$C^\infty$流形. 如果$M$中子集$D$由下面两类点构成:

  1. 内点: $p\in D,$ $\,\exists\,V\ni p$为开邻域, $V\subset D.$

  2. 边界点: $p\in D,$ $\,\exists\,(U,\varphi;x^i)$使得$\varphi(p)=0,$ $\varphi(U\cap D)=\{q\in D|x^m(q)\ge 0\}.$ 称这样的坐标系为适应坐标系.

则称$D$为$M$中带边区域. 用$\partial D$表示边界全体.

易见内点全体$D^\circ$为$M$中开子集, $U\cap \partial D=\{x^m(q)=0\}$为$(m-1)$维切片. 从而$\partial D$为$M$中正则超曲面.

全体适应坐标系构成了$\partial D$的一个覆盖, 构成了$\partial D$的坐标图册. 它确定了$\partial D$的$C^\infty$结构. 现$M$是定向流形, 则可要求$\partial D$的每个适应坐标系与定向$\Omega$相符.

在每个$(U_\alpha\cap \partial D;x_{\alpha}^1,\cdots,x_{\alpha}^{m-1})$上, 局部定义$(-1)^m{}dx_{(\alpha)}^{1}\wedge\cdots\wedge{}dx_{(\alpha)}^{ {}m-1}.$ 下面证明它与$(-1)^m{}dx_{(\beta)}^{1}\wedge\cdots\wedge{}dx_{(\beta)}^{ {}m-1}$定向相符, 因此$\partial D$上有一诱导定向.

定理 1. 设$D$是$C^\infty$-流形$M$中带边区域, 如果$(M,\Omega)$是可定向的, 则$\partial D$也是可定向的.

证: 取$(U,\varphi;x^i),(V,\psi;y^j)$为$\partial D$的适应坐标系, 与定向$\Omega$相符, 需考虑$\det\frac{D(y^1,\cdots,y^{m-1})}{D(x^1,\cdots,x^{m-1})}$的符号. 已知$\det D(\psi\circ \varphi^{-1})>0.$ 由于$\frac{\partial {}y^m}{\partial {}x^{i} }=0,$ $\,\forall\,i\neq m;$ $\frac{\partial {}y^m}{\partial {}x^m}\ge 0,$ 立即推得$\det\frac{D(y^1,\cdots,y^{m-1})}{D(x^1,\cdots,x^{m-1})}>0.$ 因此$\partial D$是可定向的.

特别地, 我们选取$(-1)^m {}dx_{(\alpha)}^{1}\wedge\cdots\wedge{}dx_{(\alpha)}^{ {}m-1}$与$\partial D$定向相符, 使Stokes公式成立.

定理 2 (Stokes公式). 设$(M^m,\Omega)$是定向$C^\infty$流形, $D$是$M$中带边区域, 从$M$中自然诱导定向. $\omega$是$M$一个具紧支集$C^1$的$(m-1)$次外微分形式, 则$\int_D d\omega=\int_{\partial D}\omega.$ 如果$\partial D=\varnothing,$ 则右端为零.

文章最后更新于 2021-12-22 18:57:10

Integration of Differential Forms

definition

We know that there are Integral variable substitution formula: Set $f\in C_c(\mathbb{R}^m),$ $h:\mathbb{R}^m\approx \mathbb{R}^m$ is diffeomorphism, then

$$ \int_{\mathbb{R}^m}f(x)dx^1\cdots dx^m=\int_{\mathbb{R}^m}(f\circ h)(y)|\det Dh|dy^1\cdots dy^m. $$

If $h:\mathbb{R}^m\rightarrow \mathbb{R}^m$ satisfies $\det Dh(y)>0,$ Then there is $\det Dh>0.$ everywhere and it is called Keep direction diffeomorphism of .

From the content of the previous lesson, $\Omega={}dx^{1}\wedge\cdots\wedge{}dx^{ {}m}\in \mathcal{A}^m(\mathbb{R}^m)$ gives $\mathbb{R}^m$ a natural orientation. Now $(\mathbb{R}^m,\mathrm{id};x),(\mathbb{R}^m,h^{-1};y)$ is two different coordinate systems, Let $\omega=f(x)dx^1\wedge\cdots\wedge dx^m$ be the continuous exterior derivative form of $m$ with compact support, Then due to

$$ dx^1\wedge\cdots\wedge dx^m=\det(Dh)(y){}dy^{1}\wedge\cdots\wedge{}dy^{ {}m}, $$

We know that $\omega=f\circ h(y)\det(Dh){}dy^{1}\wedge\cdots\wedge{}dy^{ {}m}$ is another representation. Tips for the above formula:

$$ \int_{\mathbb{R}^m}\omega:=\int_{\mathbb{R}^m}f(x)dx^1\cdots dx^m. $$

Let $(M^m,\Omega)$ be the oriented $C^\infty$-manifold, $\omega$ is the exterior derivative form of the compact support $m$ on $M$, That is $\operatorname{supp}\omega=\overline{\left\{q\mid\omega(q)\neq 0\right\}} \Subset M.$ Next, define the integral of $\omega$ on $M$.

When $\operatorname{supp}\omega\subset (U,\varphi;x^i),$ When the coordinate system matches the orientation $\Omega$, $\omega=f dx^1\wedge\cdots\wedge dx^m,$ $f\in C_c(U).$ From the previous description, defined

$$ \int_M \omega:=\int_U \omega=\int_{\varphi(U)}(f\circ\varphi^{-1})(x)dx^1\cdots dx^m. $$

Generally, $\{(U_i,\varphi_i)\}$ is taken as the coordinate atlas consistent with the orientation, And the partition of unity subordinate to it. Then $\omega=\sum_i f_i\omega,$ $\operatorname{supp}(f_i\omega)\subset U_i.$ has a compact support from $\omega$, $\{\operatorname{supp}f_i\}$ is locally finite, and the sum always makes sense. So we define

$$ \int_M \omega:=\sum_i \int_M f_i\omega, $$

Where $\int_M f_i\omega$ is defined.

good quality

We need to verify the soundness of this definition: Is it independent of the coordinate system/atlas and partition of unity selection?

If $\operatorname{supp}\omega \subset (U,\varphi;x)\cap (V,\psi;y),$ $f{}dx^{1}\wedge\cdots\wedge{}dx^{ {}m}=\omega=g{}dy^{1}\wedge\cdots\wedge{}dy^{ {}m}.$ Need to verify:

$$ \int_{\psi(U)}(g\circ\psi^{-1})(y)dy^1\cdots dy^m=\int_M\omega=\int_{\varphi(U)}(f\circ\varphi^{-1})(x)dx^1\cdots dx^m. $$

Let $h=\varphi\circ\psi^{-1},$ $h(y)=x.$ then substitute the formula by the integral variable, And the $g=f\det(Dh),$ combined coordinate system is consistent with the orientation $\Omega$, that is, the equation is established.

There are two partitions of unity $\{f_i\},\{g_j\},$ Then $\{f_ig_j=g_jf_i\}$ is also a partition of unity. Therefore

$$ \sum_j\int_M g_j\omega=\sum_{i,j}\int_M f_ig_j\omega=\sum_i\int_M f_i\omega. $$

Other properties

It is easy to see, $\int_{(M,\Omega)}\omega=-\int_{(M,-\Omega)}\omega.$

Define $\mathcal{K}$ as all consecutive $m$-order forms with compact supports, forming a vector space, And $\int_M:\mathcal{K}\rightarrow \mathbb{R}$ is a linear functional.

Let $\varphi:N^k\rightarrow M$ be the $k$-dimensional submanifold. If $\varphi^\ast \omega$ has compact support in $N$, then

$$ \int_N \varphi^\ast \omega:=\int_{\varphi(N)} \omega. $$

Stokes formula

Recalling the multivariable calculus on $\mathbb{R}^m$, we can convert the fundamental theorem of calculus, Green's formula, Gauss formula and Stokes formula are unified as:

$$ \int_D d\omega=\int_{\partial D}\omega,\quad D\subset \mathbb{R}^m. $$

Let $M^m$ be the $m$-dimensional $C^\infty$ manifold. If the subset $D$ in $M$ consists of the following two types of points:

  1. Interior points: $p\in D,$ $\,\exists\,V\ni p$ is an open neighborhood, $V\subset D.$

  2. Boundary points: $p\in D,$ $\,\exists\,(U,\varphi;x^i)$ makes $\varphi(p)=0,$ $\varphi(U\cap D)=\{q\in D|x^m(q)\ge 0\}.$ Call such a coordinate system Adapt to coordinate system.

Then $D$ is called $M$ band edge area. Use $\partial D$ to represent the entire boundary.

It is easy to see that all interior points $D^\circ$ are the open subset of $M$, $U\cap \partial D=\{x^m(q)=0\}$ is a $(m-1)$-dimensional slice. Therefore $\partial D$ is the regular hypersurface in $M$.

The overall adaptive coordinate system constitutes a coverage of $\partial D$, It constitutes the coordinate atlas of $\partial D$. It determines the $C^\infty$ structure of $\partial D$. Now $M$ is an oriented manifold, then each adaptive coordinate system of $\partial D$ can be required to be consistent with the orientation $\Omega$.

On each $(U_\alpha\cap \partial D;x_{\alpha}^1,\cdots,x_{\alpha}^{m-1})$, Local definition $(-1)^m{}dx_{(\alpha)}^{1}\wedge\cdots\wedge{}dx_{(\alpha)}^{ {}m-1}.$ The following proves that it is consistent with the $(-1)^m{}dx_{(\beta)}^{1}\wedge\cdots\wedge{}dx_{(\beta)}^{ {}m-1}$ orientation, Therefore there is an induced orientation on $\partial D$.

Theorem 1. Let $D$ be the edge region in $C^\infty$-manifold $M$. If $(M,\Omega)$ is orientable, then $\partial D$ is also orientable.

Certificate: Take $(U,\varphi;x^i),(V,\psi;y^j)$ as the adaptive coordinate system of $\partial D$, Consistent with orientation $\Omega$, The symbol $\det\frac{D(y^1,\cdots,y^{m-1})}{D(x^1,\cdots,x^{m-1})}$ needs to be considered. Known $\det D(\psi\circ \varphi^{-1})>0.$ Due to $\frac{\partial {}y^m}{\partial {}x^{i} }=0,$ $\,\forall\,i\neq m;$ $\frac{\partial {}y^m}{\partial {}x^m}\ge 0,$ Push immediately $\det\frac{D(y^1,\cdots,y^{m-1})}{D(x^1,\cdots,x^{m-1})}>0.$ Therefore $\partial D$ is orientable.

In particular, We select $(-1)^m {}dx_{(\alpha)}^{1}\wedge\cdots\wedge{}dx_{(\alpha)}^{ {}m-1}$ to match $\partial D$ orientation, Make Stokes formula hold.

Theorem 2 (Stokes formula). Assume $(M^m,\Omega)$ is the oriented $C^\infty$ manifold, $D$ is the edge region in $M$, and the orientation is naturally induced from $M$. $\omega$ is $M$ a $(m-1)$ sub-exterior derivative form with a compact support $C^1$, then $\int_D d\omega=\int_{\partial D}\omega.$ If $\partial D=\varnothing,$, the right end is zero.

The article was last updated on 2021-12-22 18:57:10

  • 本文标题:《微分流形》第七章-流形上的积分"Differential Manifolds" Chapter 7 - Integrals on Manifolds
  • 本文作者:DreamAR
  • 创建时间:2021-12-22 20:49:33
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