本章主要内容为以微分形式为对象的计算

以下计算均建立在开集 UU 上的 kk 形式: Ωk(U)\Omega^k(U)
对于 α,βΩk(U)\alpha, \beta \in \Omega^k(U),记

α=i1,,ik=1nfi1ikdxi1dxikβ=j1,,jk=1ngj1jkdxj1dxjk\alpha = \sum_{i_1,\dots,i_k=1}^n f_{i_1\cdots i_k} dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ \beta = \sum_{j_1,\dots,j_k=1}^n g_{j_1\cdots j_k} dx_{j_1} \wedge \cdots \wedge dx_{j_k}

# 外积

定义 UU 上的 k+lk+l 形式 αβ\alpha \wedge \beta

αβ:=i1,,ik=1n j1,,j=1n(fi1ikgj1j)  dxi1dxikdxj1dxj.\alpha \wedge \beta := \sum_{i_1, \dots, i_k=1}^n \ \sum_{j_1,\dots,j_\ell=1}^n \bigl( f_{i_1 \cdots i_k} \, g_{j_1 \cdots j_\ell} \bigr) \; dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_\ell}.

并称为 α\alphaβ\beta外积 (exterior product)

这个式子可能让人望而生畏,但是只要明白原理:乘法分配律
就非常浅显易懂了

示例

(x2dx+ydz)(dyxydz)=x2dxdyx3ydxdz+ydzdyxy2dzdz=x2dxdyydydz+x3ydzdx,\begin{aligned} (x^2 \, dx + y \, dz) \wedge (dy - x y \, dz) &= x^2 \, dx \wedge dy - x^3 y \,dx \wedge dz + y \, dz \wedge dy - x y^2 \, dz \wedge dz \\ &= x^2 \, dx \wedge dy - y \, dy \wedge dz + x^3 y \, dz \wedge dx, \end{aligned}

实际上就是把每一项都分配地去乘每一项,其中将函数值直接做积,将微分形式按顺序楔积做积

外积计算满足分配律

(α1+α2)β=α1β+α2β(\alpha_1 + \alpha_2) \wedge \beta = \alpha_1 \wedge \beta + \alpha_2 \wedge \beta

α(β1+β2)=αβ1+αβ2\alpha \wedge (\beta_1 + \beta_2) = \alpha \wedge \beta_1 + \alpha \wedge \beta_2

以及结合律

(αβ)γ=α(βγ)(\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma)

并且与微分形式的次数相同,继承了行列式的交代性
对于 kk 形式 α\alpha\ell 形式 β\beta,有

βα=(1)kαβ\beta \wedge \alpha = (-1)^{k \ell} \, \alpha \wedge \beta

# 外微分

提醒:请回忆数学分析中学习的多元函数的全微分计算方法,即

df=i=1nfxidxidf = \sum_{i=1}^n \frac{\partial f}{\partial x_i} \, dx_i

借由全微分的概念,对于 UU 上的 kk 形式

ω=i1,,ik=1nfi1ikdxi1dxik,\omega = \sum_{i_1,\dots,i_k=1}^n f_{i_1 \cdots i_k} \, dx_{i_1} \wedge \cdots \wedge dx_{i_k},

定义 UU 上的 (k+1)(k+1) 形式

dω=i1,,ik=1ndfi1ikdxi1dxik,d\omega = \sum_{i_1,\dots,i_k=1}^n df_{i_1 \cdots i_k} \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k},

称为 ω\omega外微分 (Exterior derivative)

00 形式的外微分等于其作为函数的全微分

同样,不要被复杂的式子吓退,只需要以实例看出分配法则是如何作用的

示例

d(yz2dx+xydydz)=d(yz2)dx+d(xy)dy+d(1)dz=(0dx+z2dy+2yzdz)dx+(ydx+xdy+0dz)dy+0dz=z2dydx+2yzdzdx+ydxdy=(yz2)dxdy+2yzdzdx,\begin{aligned} d(yz^2 \, dx + xy \, dy - dz) &= d(yz^2) \wedge dx + d(xy) \wedge dy + d(-1) \wedge dz \\ &= (0 \, dx + z^2 \, dy + 2yz \, dz) \wedge dx + (y \, dx + x \, dy + 0 \,dz) \wedge dy + 0 \wedge dz \\ &= z^2 \, dy \wedge dx + 2yz \, dz \wedge dx + y \, dx \wedge dy \\ &= (y - z^2) \, dx \wedge dy + 2yz \, dz \wedge dx, \end{aligned}


并且,针对三维空间上的外微分计算,通过利用算子 \nabla 实现的梯度,散度和旋度,可以大幅简化计算过程
以下令

F=(f1f2f3),dx=(dxdydz)\boldsymbol F = \begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix},\quad d\boldsymbol x = \begin{pmatrix} dx \\ dy \\ dz \end{pmatrix}

00 形式 ff 的外微分

df=fxdx+fydy+fzdz=(gradf)(dxdydz)=(f)(dxdydz)\begin{aligned} df &= \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy + \frac{\partial f}{\partial z} \, dz \\ &= (\mathrm{grad} \, f) \cdot \begin{pmatrix} dx \\ dy \\ dz \end{pmatrix} \\ &= (\nabla f) \cdot \begin{pmatrix} dx \\ dy \\ dz \end{pmatrix} \end{aligned}

11 形式 α=f1dx+f2dy+f3dz=Fdx\alpha = f_1 \, dx + f_2 \, dy + f_3 \, dz = \boldsymbol F \cdot d\boldsymbol x 的外微分

dα=df1dx+df2dy+df3dz=(f1xdx+f1ydy+f1zdz)dx+(f2xdx+f2ydy+f2zdz)dy+(f3xdx+f3ydy+f3zdz)dz=(f2xf1y)dxdy+(f3yf2z)dydz+(f1zf3x)dzdx=(f2xf1yf3yf2zf1zf3x)(dxdydydzdzdx)=(rotF)dx=(×F)dx\begin{aligned} d\alpha &= df_1 \wedge dx + df_2 \wedge dy + df_3 \wedge dz \\ &= (f_{1x} \, dx + f_{1y} \, dy + f_{1z} \, dz) \wedge dx \\ &\quad + (f_{2x} \, dx + f_{2y} \, dy + f_{2z} \, dz) \wedge dy \\ &\quad + (f_{3x} \, dx + f_{3y} \, dy + f_{3z} \, dz) \wedge dz \\ &= (f_{2x} - f_{1y}) \, dx \wedge dy + (f_{3y} - f_{2z}) \, dy \wedge dz + (f_{1z} - f_{3x}) \, dz \wedge dx \\ &= \begin{pmatrix} f_{2x} - f_{1y} \\ f_{3y} - f_{2z} \\ f_{1z} - f_{3x} \end{pmatrix} \cdot \begin{pmatrix} dx \wedge dy \\ dy \wedge dz \\ dz \wedge dx \end{pmatrix} \\ &= (\mathrm{rot} \, \boldsymbol F) \cdot d\boldsymbol x \\ &= (\nabla \times \boldsymbol F) \cdot d\boldsymbol x \end{aligned}

22 形式 ω=f1dydz+f2dzdx+f3dxdy=Fdx\omega = f_1 \, dy \wedge dz + f_2 \, dz \wedge dx + f_3 \, dx \wedge dy = \boldsymbol F \cdot d\boldsymbol x 的外微分

dω=df1dydz+df2dzdx+df3dxdy=(f1xdx+f1ydy+f1zdz)dydz+(f2xdx+f2ydy+f2zdz)dzdx+(f3xdx+f3ydy+f3zdz)dxdy=(f1x+f2y+f3z)dxdydz=(divF)dxdydz=(F)dxdydz\begin{aligned} d\omega &= df_1 \wedge dy \wedge dz + df_2 \wedge dz \wedge dx + df_3 \wedge dx \wedge dy \\ &= (f_{1x} \, dx + f_{1y} \, dy + f_{1z} \, dz) \wedge dy \wedge dz \\ &\quad + (f_{2x} \, dx + f_{2y} \, dy + f_{2z} \, dz) \wedge dz \wedge dx \\ &\quad + (f_{3x} \, dx + f_{3y} \, dy + f_{3z} \, dz) \wedge dx \wedge dy \\ &= (f_{1x} + f_{2y} + f_{3z}) \, dx \wedge dy \wedge dz \\ &= (\mathrm{div} \, \boldsymbol F) \, dx \wedge dy \wedge dz \\ &= (\nabla \cdot \boldsymbol F) \, dx \wedge dy \wedge dz \end{aligned}

总结:
00 形式的外微分为梯度 grad\mathrm{grad}
11 形式的外微分为旋度 rot\mathrm{rot}
22 形式的外微分为散度 div\mathrm{div}
也就是说
微分形式的外微分做到了统一梯度,旋度和散度


命题 外微分的计算性质
UU 上的 kk 形式 α1,α2\alpha_1, \alpha_2
UU 上的 kk 形式 α\alpha\ell 形式 β\beta,有

  • 线性d(α1+α2)=dα1+dα2d(\alpha_1 + \alpha_2) = d\alpha_1 + d\alpha_2
  • 乘积法则d(αβ)=dαβ+(1)kαdβd(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \, \alpha \wedge d\beta
  • 闭合d(dα)=0d(d\alpha) = 0
证明

(1)
外微分的定义给出

d(α1+α2)=i1,,ik=1nd(f1i1ik+f2i1ik)dxi1dxik=i1,,ik=1n(df1i1ik+df2i1ik)dxi1dxik=i1,,ik=1ndf1i1ikdxi1dxik+i1,,ik=1ndf2i1ikdxi1dxik=dα1+dα2\begin{aligned} d(\alpha_1 + \alpha_2) &= \sum_{i_1,\dots,i_k=1}^n d(f_{1 i_1 \cdots i_k} + f_{2 i_1 \cdots i_k}) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ &= \sum_{i_1,\dots,i_k=1}^n \bigl( df_{1 i_1 \cdots i_k} + df_{2 i_1 \cdots i_k} \bigr) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ &= \sum_{i_1,\dots,i_k=1}^n df_{1 i_1 \cdots i_k} \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} + \sum_{i_1,\dots,i_k=1}^n df_{2 i_1 \cdots i_k} \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ &= d\alpha_1 + d\alpha_2 \end{aligned}

(2)
出于外微分上加法的交换性和外积的分配律,只需要证明

α=fdxi1dxik,β=gdxj1dxj\alpha = f \, dx_{i_1} \wedge \cdots \wedge dx_{i_k},\quad \beta = g \, dx_{j_1} \wedge \cdots \wedge dx_{j_\ell}

的情形即可。
那么,由于外积

αβ=fgdxi1dxikdxj1dxj,\alpha \wedge \beta = f g \, dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_\ell},

所以对其外微分

d(αβ)=d(fg)dxi1dxikdxj1dxj=(dfg+fdg)dxi1dxikdxj1dxj=dfgdxi1dxikdxj1dxj+fdgdxi1dxikdxj1dxj=dαβ+(1)kαdβ\begin{aligned} d(\alpha \wedge \beta) &= d(fg) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_\ell} \\ &= (df \, g + f \, dg) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_\ell} \\ &= df \wedge g \, dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_\ell} + f \, dg \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_\ell} \\ &= d\alpha \wedge \beta + (-1)^k \, \alpha \wedge d\beta \end{aligned}

(3)
证明

α=fdxi1dxik\alpha = f \, dx_{i_1} \wedge \cdots \wedge dx_{i_k}

的情形即可。先进行一次外微分

dα=dfdxi1dxik=i=1nfxidxidxi1dxik\begin{aligned} d\alpha &= df \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ &= \sum_{i=1}^n \frac{\partial f}{\partial x_i} \, dx_i \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \end{aligned}

所以得到

d(dα)=i=1nd(fxi)dxidxi1dxik=i=1nj=1n2fxjxidxjdxidxi1dxik=0\begin{aligned} d(d\alpha) &= \sum_{i=1}^n d\bigl( \frac{\partial f}{\partial x_i} \bigr) \wedge dx_i \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ &= \sum_{i=1}^n \sum_{j=1}^n \frac{\partial^2 f}{\partial x_j \partial x_i} \, dx_j \wedge dx_i \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ &= 0 \end{aligned}

\square