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0-forms : $f : \mathbb R^3 \rightarrow \mathbb R$ 1-forms : $\alpha =adx + bdy + cdz = w_{<a,b,c>}$ 2-forms : $\beta = ady\wedge dz + bdz \wedge dx + c dx \wedge dy = \Phi_{<a,b,c>}$ 3-forms : $\gamma = gdx\wedge dy \wedge dz$

1 . suppose $\alpha_1$, $\alpha_2$ both are 1-forms

$\alpha_1 \wedge \alpha_2 = - \alpha_2 \wedge \alpha_1$

2 . $w_\vec{A} \wedge w_\vec{B} = \Phi_{\vec{A}\times \vec{B}}$

3 . $\vec A \times (\vec B \times \vec C) \neq (\vec A \times \vec B) \times \vec C$

but $w_\vec A \wedge (w_ \vec B \wedge w _\vec C) = (w_\vec A \wedge w_ \vec B) \wedge w _\vec C$

4 . $\Phi _{\vec A} \wedge w_\vec B = (\vec A \cdot \vec B)dx \wedge dy \wedge dz$

5 . $w _\vec A \wedge w _\vec B \wedge w _\vec C = \Phi _{\vec A \times \vec B} \wedge w_\vec C = (\vec A \times \vec B) \cdot \vec C \ dx \wedge dy \wedge dz$

6 . Hodge duality

$* w_ \vec F = \Phi_ \vec F$

$* \Phi _ \vec F = w_ \vec F$ $*(g\ dx \wedge dy \wedge dz) = g$ $*(f) = f \ dx \wedge dy \wedge dz$

$p \overset{*}{\mapsto} (3-p) form$

7 . $\alpha = \displaystyle \sum_i \alpha_i dx_i$

8 . $d \alpha = \displaystyle \sum_i d\alpha_i \wedge dx_i$

$d :\text{p-forms} \mapsto \text{(p+1)-forms}$

9 . ex

0 : $f = x^2 + yz$ 1 : $df = 2xdx + zdy + ydz = w_{<2x, z, y>} = w_{\nabla f}$

所以 $df = w_{\nabla f}$

10 . ex

$\alpha = yz dx - zdy + z^2 dz$

$$\hspace{-430px}\begin{align} d \alpha &= d(yz) \wedge dx - dz \wedge dy + d(z^2) \wedge dz \\& = (zdy + ydz) \wedge dx - dz \wedge dy + 2z dz \wedge dz \\& = dy \wedge dz + y dz \wedge dx - z dx \wedge dy \\& = \Phi_{<1,y,-z>}\end{align}$$

$\nabla \times <yz, -z, z^2> = det \begin{vmatrix}\hat x & \hat y & \hat z \\ \partial _x & \partial _y & \partial _z \\ yz& -z &z^2\end{vmatrix} = <1,y,-z>$

所以

$dw_\vec F = \Phi _{\nabla \times \vec F}$

11 .

$\beta = x^2 dy \wedge dz + (y+z)dz \wedge dx + z dx \wedge dy$

$$\hspace{-430px}\begin{align} d\beta & = 2xdx\wedge dy \wedge dz + (dy + dz) \wedge dz \wedge dx + dz \wedge dx \wedge dy\\& = (2x + 1 + 1)\ dx \wedge dy \wedge dz\end{align}$$

$d \Phi_{<x^2, y+z, z>} = (2x + 1+ 1)\ dx \wedge dy \wedge dz = \nabla \cdot <x^2, y+z, z>\ dx \wedge dy \wedge dz$

所以

$d \Phi _\vec F = (\nabla \cdot \vec F)\ dx \wedge dy \wedge dz$

12 . Identities

假设 $\alpha$ 为 p-form(包括0-form)

$d(\alpha + \beta) = d \alpha + d \beta$

$d(\alpha \wedge \beta) = d \alpha \wedge \beta + (-1)^p \alpha \wedge d \beta$ $d(d \alpha) = 0$

13 . $d^2 = 0$

对于$\alpha$为0-form

$d(df) = 0$ $d(w_{\nabla f}) = 0$ $\Phi _{\nabla \times \nabla f} = 0 \Longrightarrow \nabla \times \nabla f = 0$

对于$\alpha$为1-form

$d(dw_\vec F) = 0$ $d \Phi _ {\nabla \times \vec F} = 0$ $\nabla \cdot (\nabla \times \vec F) \ dxdydz= 0 \Longrightarrow \nabla \cdot (\nabla \times \vec F) = 0$

14 .

$d(w_\vec F \wedge w _ \vec G) = d w _ \vec F \wedge w _ \vec G - w_ \vec F \wedge dw_\vec G$

$d \Phi _ {\vec F \times \vec G} = \Phi _{\nabla \times F} \wedge w _\vec G - w_ \vec F \wedge \Phi _{\nabla \times \vec G}$ $\nabla \cdot (\vec F \times \vec G) dx \wedge dy \wedge dz = ((\nabla \times \vec F) \cdot \vec G - \vec F \cdot (\nabla \times \vec G))dx \wedge dy \wedge dz$ $\Longrightarrow \nabla \cdot (\vec F \times \vec G) = ((\nabla \times \vec F) \cdot \vec G - \vec F \cdot (\nabla \times \vec G))$

15 . integral

0-form : ∫{

a,b}f=f(b)−f(a) $\displaystyle \int_{\{a,b\}}f = f(b) - f(a)$ 1-form : $\displaystyle \int_C w_\vec F = \int _C \vec F \cdot d\vec r$ 2-form : $\displaystyle \int_S \Phi _\vec F = \iint_S \vec F \cdot d\vec S$ 3-form : $\displaystyle \int _ \mathcal E g \ dx \wedge dy \wedge dz = \iiint _ \mathcal E g\ dxdydz$

16 . Generalized stokes’ theorems

对于任何 p-form $\alpha$

$\displaystyle \int _ M d\alpha = \int _ {\partial M} \alpha$

对于 $M = C$

$\displaystyle \int_C df = \int_{\{a,b\}}f = f(b) - f(a) ( = \int _C w_{\nabla f} = \int _C \nabla f \cdot d \vec r)$

对于 $M = S$

$\displaystyle \int _{\partial S} w_\vec F = \int_S d w_\vec F = \int_S \Phi_{\nabla \times \vec F}$

$\displaystyle \int _{\partial S} \vec F \cdot d\vec r = \iint _ S (\nabla \times \vec F) \cdot d\vec S$

对于 $M = \mathcal E$

$\displaystyle \int _{\partial \mathcal E} \Phi _ \vec F = \int _ \mathcal E d \Phi_\vec F$ $\displaystyle \int _{\partial \mathcal E} \vec F \cdot d \vec S = \iiint _ \mathcal E(\nabla \cdot \vec F)\ dxdydz$

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