11. Integration of Forms

11.1Orientation

To discuss the application of the smooth linear algebra of vector bundles, we have the following proposition.

命题 11.1.1. Suppose $E\stackrel{\pi }{\to }M$ is a smooth vector bundle of rank $k$. Then the following are equivalent:

(1)	$E$ is orientable.
(2)	${\Lambda }^{k}E$ is orientable.
(3)	${\Lambda }^{k}E$ is trivial.

证明. ${\Lambda }^{k}E$ has rank $\left(\begin{array}{c}k\\ k\end{array}\right)=1$.

(2)$\iff$(3): we proved before for arbitrary rank $1$ bundle.

(1)$\iff$(2): By definition, $E$ is orientable if and only if $\exists$ system of local trivializations $(U_i,{\phi }_i)$,$${\phi }_i:{\pi }^{-1}(U_i)\to U_i\times {\mathbb{R}}^k$$for which all ${\phi }_j\circ {\phi }_i^{-1}$ are orientation-preserving on $\{p\}\times {\mathbb{R}}^k$ for all $p\in U_i\cap U_j$.$$\begin{array}{rl}{\phi }_j\circ {\phi }_i^{-1}:(U_i\cap U_j)\times {\mathbb{R}}^k& \to (U_i\cap U_j)\times {\mathbb{R}}^k\\ (p,v)& \mapsto (p,g_{ji}(p)\cdot v)\end{array}$$where $g_{ji}:U_i\cap U_j\to G{L}_k(\mathbb{R})$. So all $g_{ji}$ takes value in $G{L}_k^{+}(\mathbb{R})\subset G{L}_k(\mathbb{R})$ $\iff$ $\det g_{ji}(p)>0$ for all $p\in U_i\cap U_j$. $(U_i,{\Lambda }^k{\phi }_i)$ form a system of local trivializations for ${\Lambda }^{k}E$, whose transition maps are $\det g_{ji}(p)$. So if (1) holds, then (2) follows.

For the converse, choose an open covering of $M$ by $U_i$ such that both $E$ and ${\Lambda }^{k}E$ are trivial over all $U_i$.

证明. Over each $U_i$, we have trivializations$$\begin{array}{rl}{\phi }_i:{\pi }^{-1}(U_i)& \to U_i\times {\mathbb{R}}^k\\ {\psi }_i:({\pi }^{\prime }{)}^{-1}(U_i)& \to U_i\times {\mathbb{R}}^k\end{array}\begin{array}{rl}\pi :E& \to M\\ {\pi }^{\prime }:{\Lambda }^{k}E& \to M\end{array}$$If (2) holds, we may choose the ${\psi }_i$, so that all ${\psi }_j\circ {\psi }_i^{-1}$ are orientable-preserving on $\mathbb{R}$.$$\begin{array}{rl}{E}_p& \stackrel{{\phi }_i}{\to }\{p\}\times {\mathbb{R}}^k\\ {\Lambda }^k{E}_p& \stackrel{{\Lambda }^k{\phi }_i}{\to }\{p\}\times \mathbb{R}\\ {\Lambda }^k{E}_p& \stackrel{{\psi }_i}{\to }\{p\}\times \mathbb{R}\end{array}$$By composing ${\phi }_i$ with a reflection in a hyperplane in ${\mathbb{R}}^k$, we may assume that ${\Lambda }^k{\phi }_i$ and ${\psi }_i$ define the same orientation on ${\Lambda }^k{E}_p$.

Since the ${\psi }_i$ have orientation-preserving transition map by assumption, the same is now true for the ${\phi }_i$, so (2)$\Rightarrow$(1).

$\square$

$\square$

注 11.1.2. $E^{\ast }$ is (non-canonically) isomorphic to $E$.

If $⟨,⟩$ is a metric on $E$, then$$\begin{array}{rl}f:E& \to E^{\ast }\\ v& \mapsto ⟨v,-⟩\end{array}$$is a bundle homomorphism which is an isomorphism.

定义 11.1.3. A smooth manifold $M$ is orientable if $TM\to M$ is an orientable vector bundle.

定义 11.1.4. A volume form on $M$ is a differential form $\omega \in \Gamma ({\Lambda }^n{T}^{\ast }M)$ where $n=\dim M$, s.t. $\omega (p)\ne 0$, $\forall p\in M$.

推论 11.1.5. For a smooth $n$-dimensional manifold, the following are equivalent:

(1)	$M$ is orientable.
(2)	${\Lambda }^{n}TM$ is orientable.
(3)	${\Lambda }^{n}TM$ is trivial.
(4)	$M$ admits a volume form.

Let ${\Omega }^k(M)=\Gamma ({\Lambda }^k{T}^{\ast }M)$. When $k=0$, ${\Omega }^0(M)=\Gamma (M\times \mathbb{R})={\mathcal{C}}^{\infty }(M)$. We define$$\begin{array}{rl}d:{\mathcal{C}}^{\infty }(M)& \to {\Omega }^k(M)\\ f& \mapsto df\end{array}$$where$$\begin{array}{rl}(df)(p):T_{p}M& \to \mathbb{R}\\ x& \mapsto D_{p}f(x)\end{array}$$

引理 11.1.6. Let $\phi :M\to N$ be a differetiable map, $f\in {\mathcal{C}}^{\infty }(N)$. Then ${\phi }^{\ast }(f)=f\circ \phi$ and ${\phi }^{\ast }(df)=d({\phi }^{\ast }(f))$. So ${\phi }^{\ast }\circ d=d\circ {\phi }^{\ast }$.

证明. Let $X\in T_{p}M$.$$\begin{array}{rl}({\phi }^{\ast }df)(X)& =df(D_p\phi (X))\\ & =(D_{\phi (p)}f)(D_p\phi (X))\\ & =D_p(f\circ \phi )(X)\\ & =D_p({\phi }^{\ast }(f))(X)\\ & =d({\phi }^{\ast }(f))(X)\end{array}$$

$\square$

Let $U\subset {\mathbb{R}}^n$ be open, $f\in {\mathcal{C}}^{\infty }$.$$df(X)=Df(X)$$At every point $p\in U$, $\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_n}$ form a basis of $T_{p}U$$$X=\sum _{i=1}^n{\lambda }_i\frac{\partial }{\partial x_i}$$Let $d{x}_1,\dots ,d{x}_n$ be the dual basis of $T_p^{\ast }M$.

Claim 11.1.7. $df=\sum _{i=1}^n\frac{\partial f}{\partial x_i}d{x}_i$.

证明. For all $X$, we must have $df(X)=Df(X)$. Take $X=\frac{\partial }{\partial x_i}$. Then $df\left(\frac{\partial }{\partial x_i}\right)=Df\left(\frac{\partial }{\partial x_i}\right)=\frac{\partial f}{\partial x_i}$.$$\left(\sum _{j=1}^n\frac{\partial f}{\partial x_j}d{x}_j\right)\left(\frac{\partial }{\partial x_i}\right)=\sum _{j=1}^n\frac{\partial f}{\partial x_j}d{x}_j\left(\frac{\partial }{\partial x_i}\right)=\frac{\partial f}{\partial x_i}$$

$\square$

$$\omega =\sum _{1⩽i_1<\cdots <i_k⩽n}{f}_{i_1,\dots ,i_k}d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\in {\Omega }^k(U)$$Define$$d\omega =\sum _{1⩽i_1<\cdots <i_k⩽n}(d{f}_{i_1,\dots ,i_k})\wedge d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\in {\Omega }^{k+1}(U)$$

Claim 11.1.8. $d^2\equiv 0$, so $d(d\omega )=0$.

证明. $$\begin{array}{rl}d(d\omega )& =d\sum _{i_j}\left(\sum _{\alpha =1}^n\frac{\partial f_{i_j}}{\partial x_{\alpha }}d{x}_{\alpha }\right)\wedge d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\\ & =\sum _{i_j}\sum _{\alpha ,\beta =1}^n\frac{\partial }{\partial x_{\beta }}\left(\frac{\partial f_{i_j}}{\partial x_{\alpha }}\right)d{x}_{\beta }\wedge d{x}_{\alpha }\wedge d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\\ & =\sum _{i_j}\sum _{\alpha <\beta }\cancel{\left(\frac{\partial f_{i_j}}{\partial x_{\beta }\partial x_{\alpha }}-\frac{\partial f_{i_j}}{\partial x_{\alpha }\partial x_{\beta }}\right)}d{x}_{\beta }\wedge d{x}_{\alpha }\wedge d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}=0\end{array}$$

$\square$

引理 11.1.9. Let $\phi :M\to N$ be a differentiable map, $\omega \in {\Omega }^k(N)$. Let $M,N\subset {\mathbb{R}}^n$ be open subsets. Then $d{\phi }^{\ast }\omega ={\phi }^{\ast }d\omega$.

证明. $\omega =\sum _{i_j}{f}_{i_1,\dots ,i_k}d{y}_{i_1}\wedge \cdots \wedge d{y}_{i_k}$.
$\begin{array}{rl}d{\phi }^{\ast }\omega & =d\sum _{i_j}{\phi }^{\ast }(f_{i_1,\dots ,i_k}){\phi }^{\ast }(d{y}_{i_1})\wedge \cdots \wedge {\phi }^{\ast }(d{y}_{i_k})\\ & =d\sum _{i_j}{\phi }^{\ast }(f_{i_1,\dots ,i_k})d({\phi }^{\ast }{y}_{i_1})\wedge \cdots \wedge d({\phi }^{\ast }{y}_{i_k})\\ & =\sum _{i_j}d{\phi }^{\ast }(f_{i_1,\dots ,i_k})\wedge d({\phi }^{\ast }{y}_{i_1})\wedge \cdots \wedge d({\phi }^{\ast }{y}_{i_k})\\ & ={\phi }^{\ast }\left(\sum _{i_j}d{f}_{i_1,\dots ,i_k}\wedge d{y}_{i_1}\wedge \cdots \wedge d{y}_{i_k}\right)={\phi }^{\ast }d\omega \end{array}$

$\square$

Claim 11.1.10. If $\omega \in {\Omega }^k(U)$ and $\eta \in {\Omega }^l(U)$, then $d(\omega \wedge \eta )=d\omega \wedge \eta +(-1{)}^k\omega \wedge (d\eta )$.

If $k=0$, then $\omega =f\in {\mathcal{C}}^{\infty }(U)$. This formula becomes$$d(f\eta )=df\wedge \eta +fd\eta$$

11.2Exterior Derivative

定义 11.2.1. An exterior derivative on a smooth manifold $M$ is a $\mathbb{R}$-linear map $d:{\Omega }^k(M)\to {\Omega }^{k+1}(M)$ for all $k$ with the following properties:

(1)	If $k=0$, then $df(X)=Df(X)$.
(2)	$d(\omega \wedge \eta )=d\omega \wedge \eta +(-1{)}^{\mathrm{deg}\omega }\omega \wedge d\eta$.
(3)	$d^2=0$.
(4)	$d$ commutes with pullback by differentiable maps.
(5)	If $U\subset M$ open, then $(d\omega ){|}_U$ depends only on $\omega {|}_U$.

定理 11.2.2. There exists a unique exterior derivative $d$ on smooth manifolds satisfying (1)-(5).

证明. First uniqueness, then existence. (“$0$-form wedge a $k$-form is just $0$-form (=functions) times that $k$-form.”)

Uniqueness: On $0$-forms (=functions), $d$ is determined by (1). Let $\omega \in {\Omega }^k$, $k>0$. Then by (5), we need only consider $\omega {|}_U$ for charts $(U,\phi )$. Then$$\omega {|}_U={\phi }^{\ast }\sum _{i_1<\cdots <i_k}{f}_{i_1,\dots ,i_k}d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}$$$$\begin{array}{rl}(d\omega ){|}_U& \stackrel{(5)}{=}d\left(\omega {|}_U\right)=d\left({\phi }^{\ast }\sum _{i_1<\cdots <i_k}{f}_{i_1,\dots ,i_k}d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\right)\\ & \stackrel{(4)}{=}{\phi }^{\ast }d\left(\sum _{i_1<\cdots <i_k}{f}_{i_1,\dots ,i_k}d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\right)\\ & =\sum _{i_1<\cdots <i_k}{\phi }^{\ast }(d(f_{i_1,\dots ,i_k}d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}))\\ & \stackrel{(2)}{=}\sum _{i_1<\cdots <i_k}{\phi }^{\ast }((d{f}_{i_1,\dots ,i_k}\wedge d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k})+f_{i_1,\dots ,i_k}d(d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}))\\ & \stackrel{(2)+(3)}{=}\sum _{i_1<\cdots <i_k}{\phi }^{\ast }\underset{\text{uniquely determined by (1)}}{\underbrace{(d{f}_{i_1,\dots ,i_k}\wedge d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k})}}\end{array}$$

Existence: Let $\{U_i\mid i\in I\}$ be an open covering of $M$ by domains of charts. Let ${\rho }_i$ be a subordinate partition of unity.$$\omega =1\cdot \omega =\sum _{i\in I}\underset{=:{\omega }_i}{\underbrace{{\rho }_i\omega }}=\sum _{i\in I}{\omega }_i$$where $\mathrm{supp}({\omega }_i)\subset U_i$. Define $d\omega =\sum _{i\in I}d{\omega }_i$, with $d{\omega }_i$ defined as follows: if $${\omega }_i={\phi }_i^{\ast }\left(\sum _{i_1<\cdots <i_k}{f}_{i_1,\dots ,i_k}d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\right)$$where it extended by $0$ outside of $U_i$, then$$d{\omega }_i={\phi }_i^{\ast }\left(\sum _{i_1<\cdots <i_k}d{f}_{i_1,\dots ,i_k}\wedge d{x}_{i_1}\wedge \cdots \wedge d{x}_{i_k}\right)$$where $d{f}_{i_1,\dots ,i_k}$ is defined by (1).

Well-definess: Suppose $\alpha \in {\Omega }^k(M)$, s.t. $\mathrm{supp}(\alpha )\subset U_i\cap U_j$. Then ${\phi }_i^{\ast }\beta =\alpha ={\phi }_j^{\ast }\gamma$. We want to define $d\alpha$ as ${\phi }_i^{\ast }d\beta$, so we need to check ${\phi }_i^{\ast }d\beta ={\phi }_j^{\ast }d\gamma$.$$\begin{array}{rl}\gamma & =({\phi }_j^{-1}{)}^{\ast }\alpha =({\phi }_j^{-1}{)}^{\ast }{\phi }_i^{\ast }\beta =({\phi }_i\circ {\phi }_j^{-1}{)}^{\ast }\beta \\ d\gamma & =d({\phi }_i\circ {\phi }_j^{-1}{)}^{\ast }\beta =({\phi }_i\circ {\phi }_j^{-1}{)}^{\ast }d\beta \end{array}$$Therefore,$${\phi }_i^{\ast }d\beta ={\phi }_j^{\ast }d\gamma$$

$\square$

引理 11.2.3. If $\alpha \in {\Omega }^1(M)$, then$$d\alpha (X,Y)=L_X(\alpha (Y))-L_Y(\alpha (X))-\alpha ([X,Y])$$

证明. It is enough to check the formula for $\alpha =f\cdot dg$, where $f,g\in {\mathcal{C}}^{\infty }(M)$.$$\begin{array}{rl}& d\alpha =df\wedge dg\\ & d\alpha (X,Y)=df\wedge dg(X,Y)\begin{array}{rl}[t]& =df(X)dg(Y)-df(Y)dg(X)\\ & =L_{X}f{L}_{Y}g-L_{Y}f{L}_{X}g\end{array}\end{array}$$$$\begin{array}{rl}& L_X(\alpha (Y))-L_Y(\alpha (X))-\alpha ([X,Y])\\ & =L_X(fdg(X))-L_Y(fdg(X))-fdg([X,Y])\\ & =L_X(f{L}_Y(g))-L_Y(f{L}_{X}g)-fdg([X,Y])\\ & =(L_{X}f)(L_{Y}g)+f{L}_X{L}_{Y}g-(L_{Y}f)(L_{X}g)-f{L}_Y{L}_{X}g-fdg([X,Y])\\ & =(L_{X}f)(L_{Y}g)-(L_{Y}f)(L_{X}g)+\cancel{f(L_X{L}_{Y}g-L_Y{L}_{X}g-L_{[X,Y]}g)}\end{array}$$

$\square$

定义 11.2.4. For $X\in \mathfrak{X}(M)$, let ${\phi }_t$ be the flow. Then for $\omega \in {\Omega }^k(M)$, define the Lie derivative of $\omega$ as$$L_X\omega =\frac{d}{dt}{\phi }_t^{\ast }\omega {|}_{t=0}$$

Take $\alpha \in {\Omega }^1(M)$.$$\begin{array}{rl}& (L_X\alpha )(Y)(p)=\left(\frac{d}{dt}{\phi }_t^{\ast }\alpha {|}_{t=0}(p)\right)(Y(p))\\ & =\underset{t\to 0}{\lim }\frac{\alpha ({\phi }_t(p))(D_p{\phi }_t(Y(p))-\alpha (p)(Y(p)))}{t}\\ & =\underset{t\to 0}{\lim }\frac{\alpha ({\phi }_t(p))(D_p{\phi }_t(Y(p))-Y({\phi }_t(p))+\alpha ({\phi }_t(p))(Y({\phi }_t(p)))-\alpha (p)(Y(p)))}{t}\\ & =\alpha (L_{-X}Y)(p)+L_X(\alpha (Y))(p)\\ & =L_X(\alpha (Y))(p)-\alpha ([X,Y])(p)\end{array}$$We have proved $\begin{array}{rl}[t](L_X\alpha )(Y)& =L_X(\alpha (Y))-\alpha ([X,Y])\\ & =\cancel{L_X(\alpha (Y))}+d\alpha (X,Y)\cancel{-L_X(\alpha (Y))}+L_Y(\alpha (X))\end{array}$$$\begin{array}{rlrlr}\Rightarrow & & d\alpha (X,Y)=(L_X\alpha )(Y)-L_Y(\alpha (X))& & \end{array}$$

定义 11.2.5. For $X\in \mathfrak{X}(M)$, define the contraction with $X$$$\begin{array}{rl}{i}_X:{\Omega }^k(M)& \to {\Omega }^{k-1}(M)\\ \omega & \mapsto \omega (X,\dots ,X_k)\end{array}$$by$$i_X\omega (X_1,\dots ,X_{k-1})=\omega (X,X_1,\dots ,X_k)$$

We have $i_X\equiv 0$ by skew-symmetry.

Moreover, $i_X(\omega \wedge \eta )=(i_X\omega )\wedge \eta +(-1{)}^{\mathrm{deg}\omega }\omega \wedge i_X\eta$. e.g. if $\mathrm{deg}\omega =\mathrm{deg}\eta =1$, then$$\begin{array}{rl}(i_X(\omega \wedge \eta ))(Y)& =(\omega \wedge \eta )(X,Y)\\ & =\omega (X)\eta (Y)-\omega (Y)\eta (X)\\ & =((i_X\omega )\wedge \eta )(Y)+(-1{)}^{\mathrm{deg}\omega }(\omega \wedge i_X\eta )(Y)\end{array}$$In general, $\eta \wedge \omega =(-1{)}^{\mathrm{deg}\eta \cdot \mathrm{deg}\omega }\omega \wedge \eta$.

定理 11.2.6 (Cartan Formula). On ${\Omega }^k(M)$, we have $L_X=d\circ i_X+i_X\circ d$.

证明. For $k=0$, the formula reduces to $L_X=i_X\circ d$. Apply $L_X$ to $f\in {\mathcal{C}}^{\infty }(M)$: $L_{X}f=i_{X}df=df(X)$ true! For $k=1$, let $\alpha \in {\Omega }^1(M)$, then$$\begin{array}{rl}(L_X\alpha )(Y)& =d\alpha (X,Y)+L_Y(\alpha (X))\\ & =d\alpha (X,Y)+d(\alpha (X))(Y)\\ & =((i_X\circ d)\alpha )(Y)+((d\circ i_X)\alpha )(Y)\end{array}$$$$\begin{array}{rlrlr}\Rightarrow & & L_X=i_{X}d+d{i}_X\text{on}1\text{-forms}& & \end{array}$$In general, Cartan formula is local and $\mathbb{R}$-linear, so it is enough to check it on $\omega ={\alpha }_1\wedge \cdots \wedge {\alpha }_k$, where ${\alpha }_i\in {\Omega }^1(M)$.$$\begin{array}{rl}{L}_X\omega & =\sum _{j=1}^k{\alpha }_1\wedge \cdots \wedge L_X{\alpha }_j\wedge \cdots \wedge {\alpha }_k\\ & =\sum _{j=1}^k({\alpha }_1\wedge \cdots \wedge i_{X}d{\alpha }_j\wedge \cdots \wedge {\alpha }_k+{\alpha }_1\wedge \cdots \wedge d{\alpha }_i(X)\wedge \cdots \wedge {\alpha }_k)\\ & \stackrel{!}{=}{i}_{X}d\omega +d(i_X\omega )\end{array}$$$$\begin{array}{rl}{i}_X\omega & =\sum _{j=1}^k(-1{)}^{i-1}{\alpha }_1\wedge \cdots \wedge {\alpha }_j(X)\wedge \cdots \wedge {\alpha }_k\\ d\left(i_X\omega \right)& =\sum _{j=1}^k(-1{)}^{j-1}d\left({\alpha }_j(X)\right)\wedge {\alpha }_1\wedge \cdots \wedge \widehat{{\alpha }_j}\wedge \cdots \wedge {\alpha }_k\\ & +\sum _{j=1}^k(-1{)}^{j-1}{\alpha }_j(X)d\left({\alpha }_1\wedge \cdots \wedge \widehat{{\alpha }_j}\wedge \cdots \wedge {\alpha }_k\right)\\ d\omega & =\sum _{j=1}^k(-1{)}^{j-1}{\alpha }_1\wedge \cdots \wedge d{\alpha }_j\wedge \cdots \wedge {\alpha }_k.\\ i_{X}d\omega & =i_X\sum _{j=1}^k{\alpha }_1\wedge \cdots \wedge d{\alpha }_j\wedge \cdots \wedge {\alpha }_k\end{array}$$where the hat ($\widehat{{\alpha }_j}$) means that the $j$th factor is omitted.

$\square$

11.3Manifolds with Boundary

We look at the half space $${\mathbb{H}}^n=\{x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n\mid x_n⩾0\}\subset {\mathbb{R}}^n$$The boundary is$$\partial {\mathbb{H}}^n=\{x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n\mid x_n=0\}={\mathbb{R}}^{n-1}\subset {\mathbb{H}}^n$$Then the interior is $$\mathrm{int}{\mathbb{H}}^n=\{x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n\mid x_n>0\}\underset{\text{open}}{\subset }{\mathbb{R}}^n$$

定义 11.3.1. A differentiable manifold with boundary is a topological space $M$ which is Hausdorff and has a countable basis for its topology and has an atlas $(U_i,{\phi }_i)$, $i\in I$, where $U_i\subset M$ are open, $M=\bigcup _{i\in I}{U}_i$, $${\phi }_i:U_i\to {\mathbb{H}}^n\text{are homeomorphisms onto their images}$$and, whenever $U_i\cap U_j\ne \varnothing$, $${\phi }_j\circ {\phi }_i^{-1}:{\phi }_i(U_i\cap U_j)\to {\phi }_j(U_i\cap U_j)\text{is a diffeomorphism}$$

If $U\subset {\mathbb{H}}^n$ is open, a map $f:U\to N$ is differentiable if it admits a differentiable extension to open set in ${\mathbb{R}}^n$.

$M$ manifold with boundary$$\begin{array}{rl}\partial M& :=\{p\in M\mid \exists (U_i,{\phi }_i),\text{s.t.}{\phi }_i(p)\in \partial {\mathbb{H}}^n\}\\ \mathrm{int}M& :=\{p\in M\mid \exists (U_i,{\phi }_i),\text{s.t.}{\phi }_i(p)\in \mathrm{int}{\mathbb{H}}^n\}\end{array}$$

引理 11.3.2. $\partial M$, $\mathrm{int}M$ are well-defined, $M=\partial M\cup \mathrm{int}M$.

证明. Suppose $p\in U_i$, and ${\phi }_i(p)\subset \mathrm{int}{\mathbb{H}}^n$. If $p\in U_j$, then ${\phi }_j\circ {\phi }_i^{-1}$ maps ${\phi }_i(U_i\cap U_j)$ diffeomorphically to ${\phi }_j(U_i\cap U_j)$. If this touches $\partial {\mathbb{H}}^n$, shrink $U_j$, to get an open neighborhood of $p$ in $M$. which maps to $\mathrm{int}M$.

$\square$

Considering $U_i\cap \mathrm{int}M$ and restricting ${\phi }_i$, we obtain a smooth atlas for $\mathrm{int}M$, showing that $\mathrm{int}M$ is an $n$-dimensional manifold in the usual sense.

If $\partial M\ne \varnothing$, then considering $U_i\cap \partial M$ and restricting ${\phi }_i$, we obtain a smooth atlas for $\partial M$, showing that $\partial M$ is a $(n-1)$-dimensional smooth manifold in the usual sense.$$\{\text{manifolds}\}\subset \{\text{manifolds with}\partial \}$$

$(p,i,v)$, $p\in M$
$i\in I$, s.t. $p\in U_i$
$v\in {\mathbb{R}}^n=T_{{\phi }_i(p)}{\mathbb{R}}^n$

The usual definition of $TM\to W$ works for manifolds with boundary.

例 11.3.3.  

(0)	$M={\mathbb{H}}^n$, $\partial M=\partial {\mathbb{H}}^n$.
(1)	$\overline{B_{\epsilon }(p)}=\{x\in {\mathbb{R}}^n\mid d(x,p)⩽\epsilon \}$, $\partial \overline{B_{\epsilon }(p)}=S^{n-1}$.
(2)	$\begin{array}{r}M\text{manifold with boundary}\partial M\\ N\text{manifold (without boundary)}\end{array}\}\begin{array}{r}M\times N\text{is a manifold with boundary}\\ \partial (M\times N)=\partial M\times N\end{array}$

①	$M$ is orientable if $TM\to M$ is an orientable vector bundle.
②	$M$ is orientable if there is an atlas $(U_i,{\phi }_i)$, $i\in I$, s.t. all ${\phi }_j\circ {\phi }_i^{-1}$ are orientable-preserving.

①$\iff$②: Both definitions also apply to manifolds with boundary, and are still equivalent.

引理 11.3.4. If $M$ is an orientable manifold with boundary, then $\partial M$ is an orientable manifold (in the usual sense).

证明. $M$ is orientable $\Rightarrow$ $\exists$ atlas, s.t. all ${\phi }_j\circ {\phi }_i^{-1}$ are orientable-preserving. Suppose $p\in \partial M$, and $p\in U_i\cap U_j$.

$$\begin{array}{rl}{D}_{{\phi }_i(p)}(\phi )& =\left(\frac{\partial {\phi }_k}{\partial y_l}\right)\\ & =\left(\begin{array}{cc}\frac{\partial {\phi }_k}{\partial y_l},k,l⩽n-1& \begin{array}{c}\ast \\ ⋮\\ ⋮\\ \ast \end{array}\\ 0\cdots \cdots 0& \frac{\partial {\phi }_n}{\partial y_n}\end{array}\right)\end{array}$$where $\frac{\partial {\phi }_n}{\partial y_n}>0$. Since $D_{{\phi }_i(p)}\phi$ is orientation-preserving, the restriction $\phi {|}_{\partial \mathbb{H}}$ is also orientation-preserving.

$\square$

Let $x_1,\dots ,x_n$ be the linear coordinates on ${\mathbb{R}}^n$, $\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_n}$ is an oriented basis for ${\mathbb{R}}^n=T_0{\mathbb{R}}^n$. We want to choose a basis $v_1,\dots ,v_{n-1}$ for ${\mathbb{R}}^{n-1}\subset {\mathbb{R}}^n$, so that $-\frac{\partial }{\partial x_n}$, $v_1,\dots ,v_{n-1}$ is positively oriented in ${\mathbb{R}}^n$, i.e. it defines the same orientation as $\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_n}$.

$$\begin{array}{rl}{v}_1& =(-1{)}^n\frac{\partial }{\partial x_1}\\ v_2& =\frac{\partial }{\partial x_2}\\ & ⋮\\ v_{n-1}& =\frac{\partial }{\partial x_{n-1}}\end{array}$$

定义 11.3.5. If $M$ is oriented, so that $\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_n}$ give the orientation in a chart, then $v_1,\dots ,v_{n-1}$ define an orientation for $\partial M$, called the induced orientation on the boundary.

例 11.3.6. $\begin{array}{rl}M:=& ([0,1]\times [0,1])/\sim \\ & (0,t)\sim (1,1-t)\end{array}$ is a manifold with boundary $\partial M\cong S^1$.

注 11.3.7. If $M$ is orientable, so is $\mathrm{int}M$.

11.4Stokes’ Theorem

$\omega \in {\Omega }^n({\mathbb{R}}^n)$ $\Rightarrow$ $\omega =f\cdot d{x}_1\wedge \cdots \wedge d{x}_n$, $f\in {\Omega }^0({\mathbb{R}}^n)$. $\omega$ has compact support $\iff$ $f$ has compact support.

Assume this is the case. Define$$\int \omega =\int fd{x}_1\cdots d{x}_n$$Let $\phi$ be an orientation-preserving diffeomorphism.$$\int {\phi }^{\ast }\omega =\int (f\circ \phi )\det \left(\frac{\partial {\phi }_i}{\partial x_j}\right)d{x}_1\cdots d{x}_j$$We get $\underset{{\mathbb{R}}^n}{\int }{\phi }_{\ast }\omega =\underset{{\mathbb{R}}^n}{\int }\omega$ if $\det \left(\frac{\partial {\phi }_i}{\partial x_j}\right)>0$. So $\int$ is well-defined under orientation-preserving changes of coordinates. Let $M$ be an orientable manifold with fixed orientation. Let $(U_i,{\phi }_i)$ be a smooth orientated atlas.

定理 11.4.1. There is a well-defined $\mathbb{R}$-linear map (works for $M$ with boundary by ${\mathbb{R}}^n\mapsto {\mathbb{H}}^n$) $$\underset{M}{\int }:{\Omega }_0^n(M)\to \mathbb{R}$$s.t. if $\mathrm{supp}(\omega )\subset U_i$, then $\underset{M}{\int }\omega =\underset{{\mathbb{R}}^n}{\int }({\phi }_i^{-1}{)}^{\ast }\omega$.

证明. Let $\omega \in {\Omega }_0^n(M)$, and let ${\rho }_i$ be a smooth partition of unity subordinate to $U_i$. $$\omega =1\cdot \omega =\sum _{i\in I}({\rho }_i\omega )$$If $\underset{M}{\int }$ exists and is $\mathbb{R}$-linear, then $$\underset{M}{\int }\omega =\sum _i\underset{M}{\int }({\rho }_i\omega )=\sum _i\underset{{\mathbb{R}}^n}{\int }({\phi }_i^{-1}{)}^{\ast }({\rho }_i\omega )$$(11.1)This shows that $\underset{M}{\int }$ is unique. Use (11.1) to define $\underset{M}{\int }$. This is well-defined, because all transition maps are orientation preserving, so $$\underset{{\mathbb{R}}^n}{\int }({\phi }_i^{-1}{)}^{\ast }({\omega }_i)=\underset{{\mathbb{R}}^n}{\int }({\phi }_j^{-1}{)}^{\ast }({\omega }_i)$$if $\mathrm{supp}({\omega }_i)\subset U_i\cap U_j$.

$\square$

Let $M$ be a smooth $n$-dimensional manifold with boundary.

定理 11.4.2 (Stoke’s Theorem). If $i:\partial M\hookrightarrow M$ is the inclusion and $M$ is orientable, then$$\underset{M}{\int }d\omega =\underset{\partial M}{\int }{i}^{\ast }\omega \forall \omega \in {\Omega }_0^{n-1}(M)$$where $\partial M$ carries the orientation induced from $M$.

证明. Case 1: $M={\mathbb{H}}^n$. $$\omega =\sum _{i=1}^n{f}_{i}d{x}_1\wedge \cdots \wedge \widehat{d{x}_i}\wedge \cdots \wedge d{x}_n$$where the hat ($\widehat{d{x}_i}$) means that the $i$th factor is omitted. Since $i^{\ast }$, $d$, $\int$ are $\mathbb{R}$-linear, we prove stokes theorem for each summand. Without loss of generality, $\omega =fd{x}_1\wedge \cdots \wedge \widehat{d{x}_i}\wedge \cdots \wedge d{x}_n$.

Subcase 1a: $i<n$ $\Rightarrow$ $i^{\ast }\omega \equiv 0$ $\Rightarrow$ $\underset{\partial {\mathbb{H}}^n}{\int }{i}^{\ast }\omega =0$. $$\begin{array}{rl}\underset{{\mathbb{H}}^n}{\int }d\omega & =\underset{{\mathbb{H}}^n}{\int }\sum _{j=1}^n\frac{\partial f}{\partial x_j}d{x}_j\wedge d{x}_1\wedge \cdots \wedge \widehat{d{x}_i}\wedge \cdots \wedge d{x}_n\\ & =\underset{{\mathbb{H}}^n}{\int }\frac{\partial f}{\partial x_i}d{x}_i\wedge d{x}_1\wedge \cdots \wedge \widehat{d{x}_i}\wedge \cdots \wedge d{x}_n\\ & =(-1{)}^{i-1}\underset{{\mathbb{H}}^n}{\int }\frac{\partial f}{\partial x_i}d{x}_1\wedge \cdots \wedge d{x}_n\\ & =(-1{)}^{i-1}\int \cdots \int \frac{\partial f}{\partial x_j}d{x}_1\cdots d{x}_n\\ & =(-1{)}^{i-1}\int \cdots \int {\int }_{-\infty }^{+\infty }\frac{\partial f}{\partial x_i}d{x}_{i}d{x}_1\cdots \widehat{d{x}_i}\cdots d{x}_n=0\end{array}$$${\int }_{-\infty }^{+\infty }\frac{\partial f}{\partial x_i}d{x}_i=0$, since $f$ has compact support.

Subcase 1b: $i=n$, $\omega =fd{x}_1\wedge \cdots \wedge d{x}_{n-1}$. $i^{\ast }\omega =f{|}_{\partial {\mathbb{H}}^n}d{x}_1\wedge \cdots \wedge d{x}_{n-1}$. $$\underset{\partial {\mathbb{H}}^n}{\int }{i}^{\ast }\omega =(-1{)}^n{\int }_{-\infty }^{+\infty }\cdots {\int }_{-\infty }^{+\infty }f{|}_{{\mathbb{H}}^n}d{x}_1\cdots d{x}_{n-1}$$where the equality comes from induced orientation, since $(-1{)}^{n}d{x}_1\cdots d{x}_{n-1}$ is positive oriented. Then $$\begin{array}{rl}\underset{{\mathbb{H}}^n}{\int }d\omega & =\underset{{\mathbb{H}}^n}{\int }\frac{\partial f}{\partial x_n}d{x}_n\wedge d{x}_1\wedge \cdots \wedge d{x}_{n-1}\\ & =(-1{)}^{n-1}\int \frac{\partial f}{\partial x_n}d{x}_1\wedge \cdots \wedge d{x}_n\\ & =(-1{)}^{n-1}\underset{{\mathbb{H}}^n}{\int }\frac{\partial f}{\partial x_n}d{x}_1\cdots d{x}_n\\ & =(-1{)}^{n-1}{\int }_{-\infty }^{+\infty }\cdots {\int }_{-\infty }^{+\infty }{\int }_0^{+\infty }\frac{\partial f}{\partial x_n}d{x}_{n}d{x}_1\cdots d{x}_{n-1}\\ & =(-1{)}^n{\int }_{-\infty }^{+\infty }\cdots {\int }_{-\infty }^{+\infty }f{|}_{\partial {\mathbb{H}}^n}d{x}_1\cdots d{x}_{n-1}\end{array}$$where ${\int }_0^{+\infty }\frac{\partial f}{\partial x_n}d{x}_n=\underset{=0}{\underbrace{f(x_1,\dots ,x_{n-1},\infty )}}-f(x_1,\dots ,x_{n-1},0)=-f{|}_{{\mathbb{H}}^n}$.

Case 2: $M$ arbitrary, $\omega \in {\Omega }_0^{n-1}(M)$, $\omega =\sum _i{\rho }_i\omega$. $$\begin{array}{rl}\underset{M}{\int }d\omega & =\sum _i\underset{M}{\int }d({\rho }_i\omega )\\ & =\sum _i\underset{{\mathbb{H}}^n}{\int }({\phi }_i^{-1}{)}^{\ast }d({\rho }_i\omega )\\ & =\sum _i\underset{{\mathbb{H}}^n}{\int }d({\phi }_i^{-1\ast }({\rho }_i\omega ))\\ & \stackrel{\text{Case 1}}{=}\sum _i\underset{\partial {\mathbb{H}}^n}{\int }{i}^{\ast }({\phi }_i^{-1}{)}^{\ast }({\rho }_i\omega )\\ & =\sum _i\underset{\partial {\mathbb{H}}^n}{\int }({\phi }_i^{-1}{)}^{\ast }{i}^{\ast }({\rho }_i\omega )=\underset{\partial M}{\int }{i}^{\ast }\omega \end{array}$$

$\square$

推论 11.4.3. If $\partial M=\varnothing$, then $\underset{M}{\int }d\omega =0$.

推论 11.4.4. If $\partial M=\varnothing$, then $M$ does not have a volume form $\omega$, which is $d\alpha$, with $\alpha \in {\Omega }_0^{n-1}(M)$.

证明. $$0<\underset{M}{\int }\omega =\underset{M}{\int }d\alpha =0$$This leads to a contradiction.

$\square$

例 11.4.5. $M=B_1(0)\subset {\mathbb{R}}^2$, $\partial M=S^1$, $\omega =dx\wedge dy=d(xdy)$. Then $$\mathrm{Area}(B_1(0))=\underset{M}{\int }d(xdy)=\underset{S^1}{\int }xdy=\underset{S^1}{\int }\cos \phi d(\sin \phi )=\underset{S^1}{\int }{\cos }^2\phi d\phi$$Since $$(\sin \phi \cos \phi {)}^{\prime }={\cos }^2\phi -{\sin }^2\phi =2{\cos }^2\phi -1\Rightarrow {\cos }^2\phi =\frac{1}{2}+\frac{1}{2}(\sin \phi \cos \phi {)}^{\prime }$$Thus, $$\mathrm{Area}(B_1(0))=\underset{S^1}{\int }\frac{1}{2}+\frac{1}{2}\underset{S^1}{\int }(\sin \phi \cos \phi {)}^{\prime }d\phi ={\int }_0^{2\pi }\frac{1}{2}d\phi =\pi$$