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微分形式性质
定理 1. 设$f\in C^\infty(M,N),$ 则$f^\ast \circ d=d\circ f^\ast .$
证: $\,\forall\,\omega\in \mathcal{A}^r(N).$ 当$r=0$时,
$$
(f^\ast du)(X)=(du)(f_\ast X)=(f_\ast X)u=X(u\circ f)=X(f^\ast u)=d(f^\ast u)(X).
$$
当$r>0$时, 做归纳. 不妨设$\omega=udy^{i_1}\wedge\cdots\wedge dy^{i_r},$
那么$d\omega=du\wedge dy^{i_1}\wedge\cdots\wedge dy^{i_r}.$ 我们有:
$$
f^\ast d\omega=f^\ast [du\wedge dy^{i_1}\wedge\cdots\wedge dy^{i_r}]=df^\ast u\wedge df^\ast y^{i_1}\wedge\cdots\wedge df^\ast y^{i_r}=d[(f^\ast u) df^\ast y^{i_1}\wedge\cdots\wedge df^\ast y^{i_r}]=df^\ast \omega.
$$
流形的定向
定义 2. 若$S$上存在连续单位法向量$\vec{n},$ 则称$S$是可定向的.
一般地, 给定抽象$C^\infty$-流形$M,$ 由于缺少外围空间, 不便谈论法向量.
但切平面总是存在的. 我们通过选取基来确定法向量. 对任意$m$维向量空间,
称两组基是等价的, 若转移矩阵行列式为正. 此时它们确定了同一个法向量.
易见全体基组落在两个等价类中. 选定一个等价类便给定了$V$一个定向.
为了讨论连续性, 我们任取$0\neq \Omega\in \Lambda^m(V^\ast ).$
此时对两组等价基, $\{\widetilde{e_j}\}=\{e_i\}\det(a^i_j),$
$\Omega(\widetilde{e_1},\cdots,\widetilde{e_m})=\Omega(e_1,\cdots,e_m)\det(a^i_j).$
因此$\Omega$作用在等价基上是同号的.
那么选定$0\neq \Omega\in \Lambda^m(V^\ast )$就相当于给定了$V$的定向.
定义 3. 设$M^m$为$m$维$C^\infty$-流形, 如果$M$上存在连续的处处非零的$m$次外微分形式$\Omega,$ 则称$M$是可定向的, $(M,\Omega)$称为定向流形.
$\Omega$是$M\rightarrow \Lambda^m T^\ast M$的连续映照.
由于$\dim \Lambda^m(T_p^\ast M)=1,$ $\Omega=fdx^1\wedge\cdots\wedge dx^m,$
$f\in C^0(U),$ $\,\forall\,(U,x^i)\subset M.$
若$\Omega_1,\Omega_2$都是$M$的定向, 则$\Omega_2=f\Omega_1,$
$f$处处非零连续. 它们定向相同当且仅当$f>0.$ 假设$M$是连通的,
那么$f$要么处处为正, 或处处为负. 这给出了$M$的两种定向.
坐标系$(U,x^i)$上的$dx^1\wedge\cdots\wedge dx^m\in \mathcal{A}^m(U)$给出了$U$上的局部定向.
设$(M,\Omega)$为定向流形, 那么$\Omega|_U=fdx^1\wedge\cdots\wedge dx^m,$
$f\in C^0(U).$ 若$f>0,$ 则称$(U,x^i)$的定向与$\Omega$是一致的.
定理 4. 设$M$是$m$维$C^\infty$流形, 则它是可定向的$\Leftrightarrow$存在坐标图册$\{(U_\alpha,\varphi_\alpha;x^i_{(\alpha)})\},$ 使得$\,\forall\,\alpha,\beta,$ 在$U_\alpha\cap U_\beta$上有$\det[D(\varphi_\alpha\circ\varphi_\beta^{-1})]>0.$
证: $\Rightarrow:$
任取坐标图册$\{(U_\alpha,\varphi_\alpha;x_{(\alpha)}^i)\}.$
那么$\Omega|_{U_\alpha}=f_\alpha dx^1_{(\alpha)}\wedge\cdots\wedge dx_{(\alpha)}^m.$
若$f_\alpha>0,$ 则坐标系不动; 若$f_\alpha<0,$
则可以调整坐标系使得$f_\alpha$取正.
此时$\,\forall\,p\in U_\alpha\cap U_\beta,$
$$
\Omega|_{U_\alpha\cap U_\beta}=f_\alpha{}dx_{(\alpha)}^{1}\wedge\cdots\wedge{}dx_{(\alpha)}^{ {}m}=f_\beta{}dx_{(\beta)}^{1}\wedge\cdots\wedge{}dx_{(\beta)}^{ {}m}.
$$
因此$\det[D(\varphi_\alpha\circ\varphi_\beta^{-1})]=\frac{f_\beta}{f_\alpha}>0.$
$\Leftarrow:$ 将坐标图册加细使得它是局部有限的, 且保持题设性质.
取$f_i$为单位分解.
那么令$\omega=\sum_i f_i{}dx_{(i)}^{1}\wedge\cdots\wedge{}dx_{(i)}^{ {}m}$即可.
此时$\,\forall\,p\in M,$
$$
\omega(p)=\sum_i f_i(p){}dx_{(i)}^{1}\wedge\cdots\wedge{}dx_{(i)}^{ {}m}=\sum_i f_i(p)\det[D(\varphi_i\circ \varphi_{j_0}^{-1})]{}dx_{(j_0)}^{1}\wedge\cdots\wedge{}dx_{(j_0)}^{ {}m}.
$$
由单位分解与题设性质, 即有$\omega\neq 0.$
由过程可见, 构造的$\omega\in \mathcal{A}^m(M).$
因此处处非零的外微分形式可要求为光滑的.
文章最后更新于 2021-12-21 19:01:28
Properties of Differential Forms
Theorem 1. Assume $f\in C^\infty(M,N),$ then $f^\ast \circ d=d\circ f^\ast .$
Certificate: $\,\forall\,\omega\in \mathcal{A}^r(N).$ When $r=0$,
$$
(f^\ast du)(X)=(du)(f_\ast X)=(f_\ast X)u=X(u\circ f)=X(f^\ast u)=d(f^\ast u)(X).
$$
When $r>0$, do the induction. Let's assume $\omega=udy^{i_1}\wedge\cdots\wedge dy^{i_r},$
Then $d\omega=du\wedge dy^{i_1}\wedge\cdots\wedge dy^{i_r}.$ we have:
$$
f^\ast d\omega=f^\ast [du\wedge dy^{i_1}\wedge\cdots\wedge dy^{i_r}]=df^\ast u\wedge df^\ast y^{i_1}\wedge\cdots\wedge df^\ast y^{i_r}=d[(f^\ast u) df^\ast y^{i_1}\wedge\cdots\wedge df^\ast y^{i_r}]=df^\ast \omega.
$$
Orientation of the manifold
Definition 2. If there is a continuous unit normal vector $\vec{n},$ on $S$, then $S$ is said to be Orientable.
In general, given the abstract $C^\infty$-manifold $M,$, it is inconvenient to talk about normal vectors due to the lack of peripheral space.
But the tangent plane always exists. We determine the normal vector by selecting a basis. For any $m$-dimensional vector space,
The two sets of basis are said to be equivalent if the determinant of the transfer matrix is positive. At this time, they determine the same normal vector.
It is easy to see that the entire basis set falls into two equivalence classes. Selecting an equivalence class gives an orientation to $V$.
In order to discuss continuity, we choose $0\neq \Omega\in \Lambda^m(V^\ast ).$ arbitrarily
At this time, for two sets of equivalent bases, $\{\widetilde{e_j}\}=\{e_i\}\det(a^i_j),$
$\Omega(\widetilde{e_1},\cdots,\widetilde{e_m})=\Omega(e_1,\cdots,e_m)\det(a^i_j).$
Therefore $\Omega$ acts with the same sign on the equivalent basis.
Then selecting $0\neq \Omega\in \Lambda^m(V^\ast )$ is equivalent to giving the orientation of $V$.
Definition 3. Let $M^m$ be a $m$-dimensional $C^\infty$-manifold. If there is a continuous everywhere non-zero $m$ sub-exterior derivative form $\Omega,$ on $M$, then $M$ is said to be Orientable, $(M,\Omega)$ is called directional manifold.
$\Omega$ is a continuous reflection of $M\rightarrow \Lambda^m T^\ast M$.
Due to $\dim \Lambda^m(T_p^\ast M)=1,$ $\Omega=fdx^1\wedge\cdots\wedge dx^m,$
$f\in C^0(U),$ $\,\forall\,(U,x^i)\subset M.$
If $\Omega_1,\Omega_2$ are all orientations of $M$, then $\Omega_2=f\Omega_1,$
$f$ is non-zero continuous everywhere. They have the same orientation if and only if $f>0.$ Assume $M$ is connected,
Then $f$ is either positive everywhere, or negative everywhere. This gives two orientations of $M$.
$dx^1\wedge\cdots\wedge dx^m\in \mathcal{A}^m(U)$ on coordinate system $(U,x^i)$ gives the local orientation on $U$.
Assume $(M,\Omega)$ is a directional manifold, then $\Omega|_U=fdx^1\wedge\cdots\wedge dx^m,$
$f\in C^0(U).$ If $f>0,$, it is said that the orientation of $(U,x^i)$ is consistent with $\Omega$.
Theorem 4. Assume $M$ is a $m$-dimensional $C^\infty$ manifold, then it is orientable $\Leftrightarrow$ and exists in the coordinate atlas $\{(U_\alpha,\varphi_\alpha;x^i_{(\alpha)})\},$ such that $\,\forall\,\alpha,\beta,$ has $\det[D(\varphi_\alpha\circ\varphi_\beta^{-1})]>0.$ on $U_\alpha\cap U_\beta$
Certificate: $\Rightarrow:$
Any coordinate atlas $\{(U_\alpha,\varphi_\alpha;x_{(\alpha)}^i)\}.$
Then $\Omega|_{U_\alpha}=f_\alpha dx^1_{(\alpha)}\wedge\cdots\wedge dx_{(\alpha)}^m.$
If $f_\alpha>0,$, the coordinate system does not move; if $f_\alpha<0,$
Then the coordinate system can be adjusted to make $f_\alpha$ positive.
At this time $\,\forall\,p\in U_\alpha\cap U_\beta,$
$$
\Omega|_{U_\alpha\cap U_\beta}=f_\alpha{}dx_{(\alpha)}^{1}\wedge\cdots\wedge{}dx_{(\alpha)}^{ {}m}=f_\beta{}dx_{(\beta)}^{1}\wedge\cdots\wedge{}dx_{(\beta)}^{ {}m}.
$$
Therefore $\det[D(\varphi_\alpha\circ\varphi_\beta^{-1})]=\frac{f_\beta}{f_\alpha}>0.$
$\Leftarrow:$ Detail the coordinate atlas so that it is locally limited and maintains the nature of the problem.
Take $f_i$ as the unit to decompose.
Then let $\omega=\sum_i f_i{}dx_{(i)}^{1}\wedge\cdots\wedge{}dx_{(i)}^{ {}m}$ suffice.
At this time $\,\forall\,p\in M,$
$$
\omega(p)=\sum_i f_i(p){}dx_{(i)}^{1}\wedge\cdots\wedge{}dx_{(i)}^{ {}m}=\sum_i f_i(p)\det[D(\varphi_i\circ \varphi_{j_0}^{-1})]{}dx_{(j_0)}^{1}\wedge\cdots\wedge{}dx_{(j_0)}^{ {}m}.
$$
From the partition of unity and the properties of the question, there is $\omega\neq 0.$
Visible by process, constructed $\omega\in \mathcal{A}^m(M).$
Therefore, the exterior derivative form that is non-zero everywhere can be required to be smooth.
The article was last updated on 2021-12-21 19:01:28
