Machine-translated from Chinese.
微分形式
回忆$\,\forall\,f\in C^\infty(M),$ $(df)_p\in T_p^\ast M,$
称为$f$在$p$处的微分. 局部上我们有
$$
df|_U=\sum_i \frac{\partial {} }{\partial {}x^i} fdx^i,\quad \frac{\partial {} }{\partial {}x^i}f\in C^\infty(U)\,\Rightarrow\,df\in \mathcal{A}^1(M).
$$
$\omega\in\Gamma(\Lambda^rT^\ast M)=\mathcal{A}^r(M),$ 称为$r$次外微分形式.
令$\mathcal{A}(M):=\bigoplus_{r=0}^m \mathcal{A}^r(M),$
称为微分形式空间. 它以外积$\wedge$为乘法结构, 是一个分层代数.
若$f\in C^\infty(M,N),$
则它诱导了拉回映照$f^\ast :\Gamma(T_r^0(N))\rightarrow \Gamma(T_r^0(M)),$
$$
f^\ast (\Phi)(v_1,\cdots,v_r):=\Phi(f_\ast (v_1),\cdots,f_\ast (v_r)).
$$
考察系数:
$$
(f^\ast \Phi)_{i_1\cdots i_r}=\Phi\left(f_\ast \frac{\partial {} }{\partial {}x_{i_1} },\cdots,f_\ast \frac{\partial {} }{\partial {}x_{i_r} }\right)=\frac{\partial {}y^{\alpha_1}\circ f}{\partial {}x^{i_1} }\cdots\frac{\partial {}y^{\alpha_r}\circ f}{\partial {}x^{i_r} }\cdot \Phi_{\alpha_1\cdots\alpha_r}\circ f.
$$
确实是光滑的, 因此$f^\ast \Phi\in \Gamma(T_r^0(M)).$
于是$f^\ast :\mathcal{A}(N)\rightarrow \mathcal{A}(M),$
$f^\ast \sigma=\sum_{r=0}^n f^\ast \sigma_r,$
且$f^\ast (\omega\wedge\sigma)=f^\ast \omega\wedge f^\ast \sigma,$
$\,\forall\,\omega,\sigma\in \mathcal{A}(N).$ 它是一个代数同态.
外微分
设$d:\mathcal{A}(M)\rightarrow \mathcal{A}(M)$为线性映照, 满足:
$d:\mathcal{A}^r(M)\rightarrow\mathcal{A}^{r+1}(M);$
$\omega\in \mathcal{A}^r(M),$ $\sigma\in \mathcal{A}(M),$
$d(\omega\wedge\sigma)=d\omega\wedge\sigma+(-1)^r\omega\wedge d\sigma.$
$f\in \mathcal{A}^0(M)=C^\infty(M)$时, $df$为普通微分, 且$d^2f=0.$
则称$d$为$M$上外微分算子, 或简称外微分.
引理 1. 外微分算子为局部算子, 即若$\omega_1|_U=\omega_2|_U,$ 则$(d\omega_1)|_U=(d\omega_2)|_U.$
证: 只需说明$\omega|_U=0\Rightarrow (d\omega)|_U=0.$
取截断函数$f|_V\equiv 1,$ $\operatorname{supp}f\subset U$即可.
这样$f\omega=0.$ 从而$0=d(f\omega)=df\wedge \omega+f(d\omega),$
在$V$上有$df=0,$ 从而$(d\omega)|_V=0.$
由$V$的任意性即有$(d\omega)|_U=0.$
定理 2. $C^\infty$-流形$M$上外微分算子是存在唯一的.
证: 先证唯一性. 由截断函数, $d$可以诱导$U$上的外微分算子.
取引理中的$f,$ 对$\omega\in \mathcal{A}(U),$
局部定义$(d\omega)|_V:=(d(f\omega))|_V.$ 现证$d$在坐标系里是唯一的:
取$(U,\varphi;x^i)\subset M,$ $\omega\in \mathcal{A}^r(M),$
$\omega|_U=adx^{i_1}\wedge\cdots\wedge dx^{i_r}.$ 由外微分算子性质,
$d(\omega|_U)=da\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r}.$ 由局部性,
$(d\omega)|_U=d(\omega|_U)=da\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r},$
$d\omega$在$U$内有确切表达式, 因此$d$在坐标系中唯一. 从而,
$d$在整体上也唯一.
接下来说明$d$的存在性. 首先由上段,
局部定义$d\omega:=da\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r}$即可,
$\omega=adx^{i_1}\wedge\cdots\wedge dx^{i_r}\in \mathcal{A}^r(U).$
这样它自然延拓为外微分算子. 由于它处处有局部定义, 整体上也有定义,
且由局部唯一性, 它与坐标系选取无关.
由$d$的局部定义, 我们自然得到如下引理:
引理 3 (Poincaré引理). $d^2=0.$ 即$\,\forall\,\omega\in \mathcal{A}(M),$ $d(d\omega)=0.$
设$\omega=Adx+Bdy+Cdz\in \mathcal{A}^1(\mathbb{R}^3)$与$X=(A,B,C)\in \mathfrak{X}(\mathbb{R}^3)$相应,
当$\omega=df$时$X$即为$\nabla f.$ 那么$d\omega$与$\nabla\times X$相应.
设$\omega=Ady\wedge dz+Bdz\wedge dx+Cdx\wedge dy,$
与$X=(A,B,C)\in \mathfrak{X}(\mathbb{R}^3)$相应.
那么$d\omega=(\nabla\cdot X)dx\wedge dy\wedge dz.$
因此$\nabla\times (\nabla f)=0$由$d(df)=0$保证,
$\nabla\cdot(\nabla\times X)$由$d(d\omega)=0$保证.
定理 4. 设$\omega\in \mathcal{A}^1(M),$ $X,Y\in \mathfrak{X}(M),$ 则$d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y]).$
证: 设$\omega=fdg,$ 则$d\omega=df\wedge dg.$
此时$df\wedge dg(X,Y)=X(f)Y(g)-Y(f)X(g).$ 计算右侧得到相同的结果,
定理得证.
文章最后更新于 2021-12-19 11:27:06
differential form
Memories$\,\forall\,f\in C^\infty(M),$ $(df)_p\in T_p^\ast M,$
is called the differential of $f$ at $p$. Locally we have
$$
df|_U=\sum_i \frac{\partial {} }{\partial {}x^i} fdx^i,\quad \frac{\partial {} }{\partial {}x^i}f\in C^\infty(U)\,\Rightarrow\,df\in \mathcal{A}^1(M).
$$
$\omega\in\Gamma(\Lambda^rT^\ast M)=\mathcal{A}^r(M),$ is called $r$ sub-exterior derivative form.
Order$\mathcal{A}(M):=\bigoplus_{r=0}^m \mathcal{A}^r(M),$
called differential formal space. It has an outer product $\wedge$ as a multiplication structure and is a hierarchical algebra.
If $f\in C^\infty(M,N),$
then it induces pullback mapping $f^\ast :\Gamma(T_r^0(N))\rightarrow \Gamma(T_r^0(M)),$
$$
f^\ast (\Phi)(v_1,\cdots,v_r):=\Phi(f_\ast (v_1),\cdots,f_\ast (v_r)).
$$
Examination coefficient:
$$
(f^\ast \Phi)_{i_1\cdots i_r}=\Phi\left(f_\ast \frac{\partial {} }{\partial {}x_{i_1} },\cdots,f_\ast \frac{\partial {} }{\partial {}x_{i_r} }\right)=\frac{\partial {}y^{\alpha_1}\circ f}{\partial {}x^{i_1} }\cdots\frac{\partial {}y^{\alpha_r}\circ f}{\partial {}x^{i_r} }\cdot \Phi_{\alpha_1\cdots\alpha_r}\circ f.
$$
is indeed smooth, so $f^\ast \Phi\in \Gamma(T_r^0(M)).$
So$f^\ast :\mathcal{A}(N)\rightarrow \mathcal{A}(M),$
$f^\ast \sigma=\sum_{r=0}^n f^\ast \sigma_r,$
And $f^\ast (\omega\wedge\sigma)=f^\ast \omega\wedge f^\ast \sigma,$
$\,\forall\,\omega,\sigma\in \mathcal{A}(N).$ It is an algebraic homomorphism.
Exterior derivative
Assume $d:\mathcal{A}(M)\rightarrow \mathcal{A}(M)$ is a linear mapping, satisfying:
$d:\mathcal{A}^r(M)\rightarrow\mathcal{A}^{r+1}(M);$
$\omega\in \mathcal{A}^r(M),$ $\sigma\in \mathcal{A}(M),$
$d(\omega\wedge\sigma)=d\omega\wedge\sigma+(-1)^r\omega\wedge d\sigma.$
When $f\in \mathcal{A}^0(M)=C^\infty(M)$, $df$ is an ordinary differential, and $d^2f=0.$
Then $d$ is called $M$ on exterior derivative operator, or abbreviated Exterior derivative.
Lemma 1. Exterior derivative operators are local operators, that is, if $\omega_1|_U=\omega_2|_U,$ then $(d\omega_1)|_U=(d\omega_2)|_U.$
Certificate: Just explain $\omega|_U=0\Rightarrow (d\omega)|_U=0.$
Just take the truncation function $f|_V\equiv 1,$ $\operatorname{supp}f\subset U$.
Thus $f\omega=0.$ Thus $0=d(f\omega)=df\wedge \omega+f(d\omega),$
There is $df=0,$ on $V$ so $(d\omega)|_V=0.$
From the arbitrariness of $V$, we have $(d\omega)|_U=0.$
Theorem 2. There is a unique exterior derivative operator on $C^\infty$-manifold $M$.
Certificate: Proprove uniqueness. From the truncation function, $d$ can induce the exterior derivative operator on $U$.
Take $f,$ in the lemma to $\omega\in \mathcal{A}(U),$
Local definition $(d\omega)|_V:=(d(f\omega))|_V.$ shows that $d$ is unique in the coordinate system:
Take $(U,\varphi;x^i)\subset M,$ $\omega\in \mathcal{A}^r(M),$
$\omega|_U=adx^{i_1}\wedge\cdots\wedge dx^{i_r}.$ From the properties of exterior derivative operators,
$d(\omega|_U)=da\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r}.$ By locality,
$(d\omega)|_U=d(\omega|_U)=da\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r},$
$d\omega$ has an exact expression within $U$, so $d$ is unique in the coordinate system. Thus,
$d$ is also unique on the whole.
Next, we will explain the existence of $d$. First, from the above paragraph,
Just define $d\omega:=da\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r}$ locally,
$\omega=adx^{i_1}\wedge\cdots\wedge dx^{i_r}\in \mathcal{A}^r(U).$
In this way, it is naturally extended to the exterior derivative operator. Since it is locally defined everywhere, it is also defined overall,
And due to local uniqueness, it has nothing to do with the selection of the coordinate system.
From the local definition of $d$, we naturally get the following lemma:
Lemma 3 (Poincaré’s lemma). $d^2=0.$ is $\,\forall\,\omega\in \mathcal{A}(M),$ $d(d\omega)=0.$
Assume $\omega=Adx+Bdy+Cdz\in \mathcal{A}^1(\mathbb{R}^3)$ corresponds to $X=(A,B,C)\in \mathfrak{X}(\mathbb{R}^3)$,
When $\omega=df$, $X$ is $\nabla f.$, then $d\omega$ corresponds to $\nabla\times X$.
Let $\omega=Ady\wedge dz+Bdz\wedge dx+Cdx\wedge dy,$
Corresponds to $X=(A,B,C)\in \mathfrak{X}(\mathbb{R}^3)$.
Then $d\omega=(\nabla\cdot X)dx\wedge dy\wedge dz.$
Therefore $\nabla\times (\nabla f)=0$ is guaranteed by $d(df)=0$,
$\nabla\cdot(\nabla\times X)$ is guaranteed by $d(d\omega)=0$.
Theorem 4. Assume $\omega\in \mathcal{A}^1(M),$ $X,Y\in \mathfrak{X}(M),$ then $d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y]).$
Certificate: Assume $\omega=fdg,$ then $d\omega=df\wedge dg.$
At this time, the same result is obtained by calculating the right side of $df\wedge dg(X,Y)=X(f)Y(g)-Y(f)X(g).$,
The theorem is proved.
The article was last updated on 2021-12-19 11:27:06
