5.4 Differential forms

Before we define differential forms, we examine the symmetry property of tensors. The symmetry operation on a tensor $\omega \in {\mathcal{T}}_{r,p}^0(M)$ is defined by

$$P\omega \left(V_1,\dots ,V_r\right)\equiv \omega \left(V_{P(1)},\dots ,V_{P(r)}\right)\text{\hspace{1em}(5.59)}$$

where $V_i\in T_{p}M$ and $P$ is an element of $S_r$ , the symmetric group of order $r$ . Take the coordinate basis $\{e_{\mu }\}=\{\partial /\partial x^{\mu }\}$ . The component of $\omega$ in this basis is

$$\omega \left(e_{{\mu }_1},e_{{\mu }_2},\dots ,e_{{\mu }_r}\right)={\omega }_{{\mu }_1{\mu }_2\dots {\mu }_r}.$$

The component of $P\omega$ is obtained from (5.59) as

$$P\omega \left(e_{{\mu }_1},e_{{\mu }_2},\dots ,e_{{\mu }_r}\right)={\omega }_{{\mu }_{P(1)}{\mu }_{P(2)}\dots {\mu }_{P(r)}}.$$

For a general tensor of type $(q,r)$ , the symmetry operations are defined for $q$ indices and $r$ indices separately.

For $\omega \in {\mathfrak{T}}_{r,p}^0(M)$ , the symmetrizer $\mathcal{S}$ is defined by

$$\mathbb{S}\omega =\frac{1}{r!}\sum _{P\in S_r}P\omega \text{\hspace{1em}(5.60)}$$

while the anti-symmetrizer $\mathcal{A}$ is

$$\mathcal{A}\omega =\frac{1}{r!}\sum _{P\in S_r}\mathrm{sgn}(P)P\omega \text{\hspace{1em}(5.61)}$$

where $\mathrm{sgn}(P)=+1$ for even permutations and $-1$ for odd permutations. $\mathfrak{S}\omega$ is totally symmetric (that is, $P\mathfrak{S}\omega =\mathfrak{S}\omega$ for any $P\in S_r$ ) and $\mathcal{A}\omega$ is totally antisymmetric ( $P\mathcal{A}\omega =\mathrm{sgn}(P)\mathcal{A}\omega$ ).

5.4.1 Definitions

Definition 5.4

A differential form of order $r$ or an $r$ -form is a totally anti-symmetric tensor of type $(0,r)$ .

Let us define the wedge product $\wedge$ of $r$ one-forms by the totally anti-symmetric tensor product

$$\mathrm{d}{x}^{{\mu }_1}\wedge \mathrm{d}{x}^{{\mu }_2}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_r}=\sum _{P\in S_r}\mathrm{sgn}(P)\mathrm{d}{x}^{{\mu }_{P(1)}}\wedge \mathrm{d}{x}^{{\mu }_{P(2)}}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_{P(r)}}.\text{\hspace{1em}(5.62)}$$

For example,

$$\begin{array}{rl}& \mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }=\mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\nu }-\mathrm{d}{x}^{\nu }\otimes \mathrm{d}{x}^{\mu }\\ & \mathrm{d}{x}^{\lambda }\wedge \mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }=\mathrm{d}{x}^{\lambda }\otimes \mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\nu }+\mathrm{d}{x}^{\nu }\otimes \mathrm{d}{x}^{\lambda }\otimes \mathrm{d}{x}^{\mu }\\ & +\mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\nu }\otimes \mathrm{d}{x}^{\lambda }-\mathrm{d}{x}^{\lambda }\otimes \mathrm{d}{x}^{\nu }\otimes \mathrm{d}{x}^{\mu }\\ & -\mathrm{d}{x}^{\nu }\otimes \mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\lambda }-\mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\lambda }\otimes \mathrm{d}{x}^{\nu }.\end{array}$$

It is readily verified that the wedge product satisfies the following.

(i) $\mathrm{d}{x}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{x}^{{\mu }_r}=0$ if some index $\mu$ appears at least twice.
(ii) $\mathrm{d}{x}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{x}^{{\mu }_r}=\mathrm{sgn}(P)\mathrm{d}{x}^{{\mu }_{P(1)}}\wedge \dots \wedge \mathrm{d}{x}^{{\mu }_{P(r)}}.$
(iii) $\mathrm{d}{x}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{x}^{{\mu }_r}$ is linear in each $\mathrm{d}{x}^{\mu }$

If we denote the vector space of $r$ -forms at $p\in M$ by ${\Omega }_p^r(M)$ , the set of $r$ -forms (5.62) forms a basis of ${\Omega }_p^r(M)$ and an element $\omega \in {\Omega }_p^r(M)$ is expanded as

$$\omega =\frac{1}{r!}{\omega }_{{\mu }_1{\mu }_2\dots {\mu }_r}d{x}^{{\mu }_1}\wedge d{x}^{{\mu }_2}\wedge \cdots \wedge d{x}^{{\mu }_r}\text{\hspace{1em}(5.63)}$$

where ${\omega }_{{\mu }_1{\mu }_2\dots {\mu }_r}$ are taken totally anti-symmetric, reflecting the anti-symmetry of the basis. For example, the components of any second-rank tensor ${\omega }_{\mu \nu }$ are decomposed into the symmetric part ${\sigma }_{\mu \nu }$ and the anti-symmetric part ${\alpha }_{\mu \nu }$ :

$${\sigma }_{\mu v}={\omega }_{(\mu v)}\equiv \frac{1}{2}\left({\omega }_{\mu v}+{\omega }_{v\mu }\right)\text{\hspace{1em}(5.64a)}$$

$${\alpha }_{\mu \nu }={\omega }_{[\mu \nu ]}\equiv \frac{1}{2}\left({\omega }_{\mu \nu }-{\omega }_{\nu \mu }\right).\text{\hspace{1em}(5.64b)}$$

Observe that ${\sigma }_{\mu \nu }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }=0$ , while ${\alpha }_{\mu \nu }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }={\omega }_{\mu \nu }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }$

Since there are $\left(\genfrac{}{}{0ex}{}{m}{r}\right)$ choices of the set $({\mu }_1,{\mu }_2,\dots ,{\mu }_r)$ out of $(1,2,\dots ,m)$ in (5.62), the dimension of the vector space ${\Omega }_p^r(M)$ is

$$\left(\begin{array}{c}m\\ r\end{array}\right)=\frac{m!}{(m-r)!r!}.$$

For later convenience we define ${\Omega }_p^0(M)=\mathbb{R}$ . Clearly ${\Omega }_p^1(M)=T_p^{\ast }M$ . If $r$ in (5.62) exceeds $m$ , it vanishes identically since some index appears at least twice in the anti-symmetrized summation. The equality $\left(\genfrac{}{}{0ex}{}{m}{r}\right)=\left(\genfrac{}{}{0ex}{}{m}{m-r}\right)$ implies $\dim {\Omega }_p^r(M)=\dim {\Omega }_p^{m-r}(M)$ . Since ${\Omega }_p^r(M)$ is a vector space, ${\Omega }_p^r(M)$ is isomorphic to ${\Omega }_p^{m-r}(M)$ (see section 2.2).

Define the exterior product of a $q$ -form and an $r$ -form $\wedge :{\Omega }_p^q(M)\times {\Omega }_p^r(M)\to {\Omega }_p^{q+r}(M)$ by a trivial extension. Let $\omega \in {\Omega }_p^q(M)$ and $\xi \in {\Omega }_p^r(M)$ , for example. The action of the $(q+r)$ -form $\omega \wedge \xi$ on $q+r$ vectors is defined by

$$\begin{array}{rl}& (\omega \wedge \xi )(V_1,\dots ,V_{q+r})\\ & =\frac{1}{q!r!}\sum _{P\in S_{q+r}}\mathrm{sgn}(P)\omega \left(V_{P(1)},\dots ,V_{P(q)}\right)\xi \left(V_{P(q+1)},\dots ,V_{P(q+r)}\right)\end{array}\text{\hspace{1em}(5.65)}$$

where $V_i\in T_{p}M$ . If $q+r>m$ , $\omega \wedge \xi$ vanishes identically. With this product, we define an algebra

$${\Omega }_p^{\ast }(M)\equiv {\Omega }_p^0(M)\oplus {\Omega }_p^1(M)\oplus \cdots \oplus {\Omega }_p^m(M).\text{\hspace{1em}(5.66)}$$

${\Omega }_p^{\ast }(M)$ is the space of all differential forms at $p$ and is closed under the exterior product.

Exercise 5.13

Take the Cartesian coordinates $(x,y)$ in ${\mathbb{R}}^2$ . The two-form $\mathrm{d}x\wedge \mathrm{d}y$ is the oriented area element (the vector product in elementary vector algebra). Show that, in polar coordinates, this becomes $r\mathrm{d}r\wedge \mathrm{d}\theta$ .

Exercise 5.14

Let $\xi \in {\Omega }_p^q(M)$ , $\eta \in {\Omega }_p^r(M)$ and $\omega \in {\Omega }_p^s(M)$ . Show that

$$\xi \wedge \xi =0\text{i f}q\text{i s o d d}\text{\hspace{1em}(5.67a)}$$

$$\xi \wedge \eta =(-1{)}^{qr}\eta \wedge \xi \text{\hspace{1em}(5.67b)}$$

$$(\xi \wedge \eta )\wedge \omega =\xi \wedge (\eta \wedge \omega ).\text{\hspace{1em}(5.67c)}$$

We may assign an $r$ -form smoothly at each point on a manifold $M$ . We denote the space of smooth $r$ -forms on $M$ by ${\Omega }^r(M)$ . We also define ${\Omega }^0(M)$ to be the algebra of smooth functions, $\mathcal{F}(M)$ . In summary we have table 5.1.

Table 5.1.

5.4.2 Exterior derivatives

Definition 5.5

The exterior derivative ${\mathrm{d}}_r$ is a map ${\Omega }^r(M)\to {\Omega }^{r+1}(M)$ whose action on an $r$ -form

$$\omega =\frac{1}{r!}{\omega }_{{\mu }_1\dots {\mu }_r}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}$$

is defined by

$${\mathrm{d}}_r\omega =\frac{1}{r!}\left(\frac{\partial }{\partial x^v}{\omega }_{{\mu }_1\dots {\mu }_r}\right)\mathrm{d}{x}^v\wedge \mathrm{d}{x}^{{\mu }_1}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_r}.\text{\hspace{1em}(5.68)}$$

It is common to drop the subscript $r$ and write simply d. The wedge product automatically anti-symmetrizes the coefficient.

Example 5.10

The $r$ -forms in three-dimensional space are:

(i) ${\omega }_0=f(x,y,z)$
(ii) ${\omega }_1={\omega }_x(x,y,z)\mathrm{d}x+{\omega }_y(x,y,z)\mathrm{d}y+{\omega }_z(x,y,z)\mathrm{d}z,$
(iii) ${\omega }_2={\omega }_{xy}(x,y,z)\mathrm{d}x\wedge \mathrm{d}y+{\omega }_{yz}(x,y,z)\mathrm{d}y\wedge \mathrm{d}z+{\omega }_{zx}(x,y,z)\mathrm{d}z\wedge \mathrm{d}x$ and
(iv) ${\omega }_3={\omega }_{xyz}(x,y,z)\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z.$

If we define an axial vector ${\alpha }^{\mu }$ by ${\epsilon }^{\mu \nu \lambda }{\omega }_{\nu \lambda }$ , a two-form may be regarded as a ‘vector’. The Levi-Civita symbol ${\epsilon }^{\mu \nu \lambda }$ is defined by ${\epsilon }^{P(1)P(2)P(3)}=\mathrm{sgn}(P)$ and provides the isomorphism between $\mathcal{X}(M)$ and ${\Omega }^2(M)$ . [Note that both of these are of dimension three.]

The action of $\mathrm{d}$ is

$$+\left(\frac{\partial {\omega }_x}{\partial z}-\frac{\partial {\omega }_z}{\partial x}\right)dz\wedge dx,$$

(i) $\mathrm{d}{\omega }_0=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z,$
(ii) $\mathrm{d}{\omega }_1=\left(\frac{\partial {\omega }_y}{\partial x}-\frac{\partial {\omega }_x}{\partial y}\right)\mathrm{d}x\wedge \mathrm{d}y+\left(\frac{\partial {\omega }_z}{\partial y}-\frac{\partial {\omega }_y}{\partial z}\right)\mathrm{d}y\wedge \mathrm{d}z$
(iii) $\mathrm{d}{\omega }_2=\left(\frac{\partial {\omega }_{yz}}{\partial x}+\frac{\partial {\omega }_{zx}}{\partial y}+\frac{\partial {\omega }_{xy}}{\partial z}\right)\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z$ and
(iv) $\mathrm{d}{\omega }_3=0$

Hence, the action of $d$ on ${\omega }_0$ is identified with ‘grad’, on ${\omega }_1$ with ‘rot’ and on ${\omega }_2$ with ‘div’ in the usual vector calculus.

Exercise 5.15

Let $\xi \in {\Omega }^q(M)$ and $\omega \in {\Omega }^r(M)$ . Show that

$$\mathrm{d}(\xi \wedge \omega )=\mathrm{d}\xi \wedge \omega +(-1{)}^q\xi \wedge \mathrm{d}\omega .\text{\hspace{1em}(5.69)}$$

A useful expression for the exterior derivative is obtained as follows. Let us take $X=X^{\mu }\partial /\partial x^{\mu },Y=Y^{\nu }\partial /\partial x^{\nu }\in \mathfrak{X}(M)$ and $\omega ={\omega }_{\mu }\mathrm{d}{x}^{\mu }\in {\Omega }^1(M)$ . It is easy to see that the combination

$$X[\omega (Y)]-Y[\omega (X)]-\omega ([X,Y])=\frac{\partial {\omega }_{\mu }}{\partial x^{\nu }}(X^{\nu }{Y}^{\mu }-X^{\mu }{Y}^{\nu })$$

is equal to $\mathrm{d}\omega (X,Y)$ , and we have the coordinate-free expression

$$\mathrm{d}\omega (X,Y)=X[\omega (Y)]-Y[\omega (X)]-\omega ([X,Y]).\text{\hspace{1em}(5.70)}$$

For an $r$ -form $\omega \in {\Omega }^r(M)$ , this becomes

$$\begin{array}{rl}& \mathrm{d}\omega \left(X_1,\dots ,X_{r+1}\right)\\ & =\sum _{i=1}^r(-1{)}^{i+1}{X}_i\omega \left(X_1,\dots ,{\hat{X}}_i,\dots ,X_{r+1}\right)\\ & +\sum _{i<j}(-1{)}^{i+j}\omega ([X_i,X_j],X_1,\dots ,{\hat{X}}_i,\dots ,{\hat{X}}_j,\dots ,X_{r+1})\end{array}\text{\hspace{1em}(5.71)}$$

where the entry below $\hat{}$ has been omitted. As an exercise, the reader should verify (5.71) explicitly for $r=2$ .

We now prove an important formula:

$${\mathrm{d}}^2=0(\text{o r}{\mathrm{d}}_{r+1}{\mathrm{d}}_r=0).\text{\hspace{1em}(5.72)}$$

Take

$$\omega =\frac{1}{r!}{\omega }_{{\mu }_1\dots {\mu }_r}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}\in {\Omega }^r(M).$$

The action of ${\mathrm{d}}^2$ on $\omega$ is

$${\mathrm{d}}^2\omega =\frac{1}{r!}\frac{{\partial }^2{\omega }_{{\mu }_1\dots {\mu }_r}}{\partial x^{\lambda }\partial x^{\nu }}\mathrm{d}{x}^{\lambda }\wedge \mathrm{d}{x}^{\nu }\wedge \mathrm{d}{x}^{{\mu }_1}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_r}.$$

This vanishes identically since ${\partial }^2{\omega }_{{\mu }_1\dots {\mu }_r}/\partial x^{\lambda }\partial x^{\nu }$ is symmetric with respect to $\lambda$ and $\nu$ while $\mathrm{d}{x}^{\lambda }\wedge \mathrm{d}{x}^{\nu }$ is anti-symmetric.

Example 5.11

It is known that the electromagnetic potential $A=(\varphi ,A)$ is a one-form, $A=A_{\mu }\mathrm{d}{x}^{\mu }$ (see chapter 10). The electromagnetic tensor is defined by $F=\mathrm{d}A$ and has the components

$$\left(\begin{array}{cccc}0& -E_x& -E_y& -E_x\\ E_x& 0& B_z& -B_y\\ E_y& -B_z& 0& B_x\\ E_z& B_y& -B_x& 0\end{array}\right)\text{\hspace{1em}(5.73)}$$

where

$$\mathbf{E}=-\nabla \varphi -\frac{\partial }{\partial x^0}\mathbf{A}\text{a n d}\mathbf{B}=\nabla \times \mathbf{A}$$

as usual. Two Maxwell equations, $\nabla \cdot \mathbf{B}=0$ and $\partial \mathbf{B}/\partial t=-\nabla \times \mathbf{E}$ follow from the identity $\mathrm{d}F=\mathrm{d}(\mathrm{d}A)=0$ , which is known as the Bianchi identity, while the other set is the equation of motion derived from the Lagrangian (1.245).

A map $f:M\to N$ induces the pullback $f^{\ast }:T_{f(p)}^{\ast }N\to T_p^{\ast }M$ and $f^{\ast }$ is naturally extended to tensors of type $(0,r)$ ; see section 5.2. Since an $r$ -form is a tensor of type $(0,r)$ , this applies as well. Let $\omega \in {\Omega }^r(N)$ and let $f$ be a map $M\to N$ . At each point $f(p)\in N$ , $f$ induces the pullback $f^{\ast }:{\Omega }_{f(p)}^{r}N\to {\Omega }_p^{r}M$ by

$$\left(f^{\ast }\omega \right)\left(X_1,\dots ,X_r\right)\equiv \omega \left(f_{\ast }{X}_1,\dots ,f_{\ast }{X}_r\right)\text{\hspace{1em}(5.74)}$$

where $X_i\in T_{p}M$ and $f_{\ast }$ is the differential map $T_{p}M\to T_{f(p)}N$

Exercise 5.16

Let $\xi ,\omega \in {\Omega }^r(N)$ and let $f:M\to N$ . Show that

$$\mathrm{d}\left(f^{\ast }\omega \right)=f^{\ast }(d\omega )\text{\hspace{1em}(5.75)}$$

$$f^{\ast }(\xi \wedge \omega )=\left(f^{\ast }\xi \right)\wedge \left(f^{\ast }\omega \right).\text{\hspace{1em}(5.76)}$$

The exterior derivative ${\mathrm{d}}_r$ induces the sequence

$$0\stackrel{i}{\to }{\Omega }^0(M)\stackrel{{\mathrm{d}}_0}{\to }{\Omega }^1(M)\stackrel{{\mathrm{d}}_1}{\to }\cdots \stackrel{d_{m-2}}{\to }{\Omega }^{m-1}(M)\stackrel{{\mathrm{d}}_{m-1}}{\to }{\Omega }^m(M)\stackrel{{\mathrm{d}}_m}{\to }0\text{\hspace{1em}(5.77)}$$

where $i$ is the inclusion map $0\hookrightarrow {\Omega }^0(M)$ . This sequence is called the de Rham complex. Since ${\mathrm{d}}^2=0$ , we have $\mathrm{im}{\mathrm{d}}_r\subset \ker {\mathrm{d}}_{r+1}$ . [Take $\omega \in {\Omega }^r(M)$ . Then ${\mathrm{d}}_r\omega \in \mathrm{im}{\mathrm{d}}_r$ and ${\mathrm{d}}_{r+1}({\mathrm{d}}_r\omega )=0$ imply ${\mathrm{d}}_r\omega \in \ker {\mathrm{d}}_{r+1}$ .] An element of $\ker {\mathrm{d}}_r$ is called a closed $r$ -form, while an element of $\mathrm{im}{\mathrm{d}}_{r-1}$ is called an exact $r$ -form. Namely, $\omega \in {\Omega }^r(M)$ is closed if $\mathrm{d}\omega =0$ and exact if there exists an $(r-1)$ -form $\psi$ such that $\omega =\mathrm{d}\psi$ . The quotient space $\ker {\mathrm{d}}_r/\mathrm{im}{d}_{r-1}$ is called the $r$ th de Rham cohomology group which is made into the dual space of the homology group; see chapter 6.

5.4.3 Interior product and Lie derivative of forms

Another important operation is the interior product ${\mathrm{i}}_X:{\Omega }^r(M)\to {\Omega }^{r-1}(M)$ where $X\in \mathcal{X}(M)$ . For $\omega \in {\Omega }^r(M)$ , we define

$${\mathrm{i}}_X\omega \left(X_1,\dots ,X_{r-1}\right)\equiv \omega \left(X,X_1,\dots ,X_{r-1}\right).\text{\hspace{1em}(5.78)}$$

For $X=X^{\mu }\partial /\partial x^{\mu }$ and $\omega =(1/r!){\omega }_{{\mu }_1\dots {\mu }_r}\mathrm{d}{x}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{x}^{{\mu }_r}$ we have

$$\begin{array}{rl}{\mathrm{i}}_X\omega & =\frac{1}{(r-1)!}{X}^v{\omega }_{v{\mu }_2\dots {\mu }_r}\mathrm{d}{x}^{{\mu }_2}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_r}\\ & =\frac{1}{r!}\sum _{s=1}^r{X}^{{\mu }_s}{\omega }_{{\mu }_1\dots {\mu }_s\dots {\mu }_r}(-1{)}^{s-1}d{x}^{{\mu }_1}\wedge \cdots \wedge \widehat{d{x}^{{\mu }_s}}\wedge \cdots \wedge d{x}^{{\mu }_r}\end{array}\text{\hspace{1em}(5.79)}$$

where the entry below $\hat{\mu }$ has been omitted. For example, let $(x,y,z)$ be the coordinates of ${\mathbb{R}}^3$ . Then

$${\mathrm{i}}_{e_x}(\mathrm{d}x\wedge \mathrm{d}y)=\mathrm{d}y,{\mathrm{i}}_{e_x}(\mathrm{d}y\wedge \mathrm{d}z)=0,{\mathrm{i}}_{e_x}(\mathrm{d}z\wedge \mathrm{d}x)=-\mathrm{d}z.$$

The Lie derivative of a form is most neatly written with the interior product. Let $\omega ={\omega }_{\mu }\mathrm{d}{x}^{\mu }$ be a one-form. Consider the combination

$$\begin{array}{rl}\left({\mathrm{di}}_X+{\mathrm{i}}_X\mathrm{d}\right)\omega & =\mathrm{d}\left(X^{\mu }{\omega }_{\mu }\right)+{\mathrm{i}}_X\left[\frac{1}{2}\left({\partial }_{\mu }{\omega }_{\nu }-{\partial }_{\nu }{\omega }_{\mu }\right)\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }\right]\\ & =\left({\omega }_{\mu }{\partial }_v{X}^{\mu }+X^{\mu }{\partial }_v{\omega }_{\mu }\right)d{x}^v+X^{\mu }\left({\partial }_{\mu }{\omega }_v-{\partial }_v{\omega }_{\mu }\right)d{x}^v\\ & =\left({\omega }_{\mu }{\partial }_v{X}^{\mu }+X^{\mu }{\partial }_{\mu }{\omega }_v\right)d{x}^v.\end{array}$$

Comparing this with (5.55), we find that

$${\mathcal{L}}_X\omega =\left({\mathrm{di}}_X+{\mathrm{i}}_X\mathrm{d}\right)\omega .\text{\hspace{1em}(5.80)}$$

For a general $r$ -form $\omega =(1/r!){\omega }_{{\mu }_1\dots {\mu }_r}\mathrm{d}{x}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{x}^{{\mu }_r}$ , we have

$$\begin{array}{rl}& {\mathcal{L}}_X\omega =\underset{\epsilon \to 0}{\lim }\frac{1}{\epsilon }\left({\left({\sigma }_{\epsilon }\right)}^{\ast }\omega {|}_{{\sigma }_{\epsilon }(x)}-\omega {|}_x\right)\\ & =X^v\frac{1}{r!}{\partial }_v{\omega }_{{\mu }_1\dots {\mu }_r}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}\\ & +\sum _{s=1}^r{\partial }_{{\mu }_s}{X}^{\nu }\frac{1}{r!}{\omega }_{{\mu }_1\dots \nu \dots {\mu }_r}^{s}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}.\end{array}\text{\hspace{1em}(5.81)}$$

We also have

$$\begin{array}{rl}& \left({\mathrm{di}}_X+{\mathrm{i}}_X\mathrm{d}\right)\omega \\ & =\frac{1}{r!}\sum _{s=1}^r\left[{\partial }_v{X}^{{\mu }_s}{\omega }_{{\mu }_1\dots {\mu }_s\dots {\mu }_r}+X^{{\mu }_s}{\partial }_v{\omega }_{{\mu }_1\dots {\mu }_s\dots {\mu }_r}\right]\\ & \times (-1{)}^{s-1}\mathrm{d}{x}^{\nu }\wedge \mathrm{d}{x}^{{\mu }_1}\wedge \cdots \wedge \widehat{\mathrm{d}{x}^{{\mu }_s}}\wedge \mathrm{d}{x}^{{\mu }_r}\\ & +\frac{1}{r!}[X^{\nu }{\partial }_{\nu }{\omega }_{{\mu }_1\dots {\mu }_r}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}\\ & +\sum _{s=1}^r{X}^{{\mu }_s}{\omega }_{{\mu }_1\dots {\mu }_s\dots {\mu }_r}(-1{)}^{s}d{x}^v\wedge d{x}^{{\mu }_1}\wedge \cdots \wedge \widehat{d{x}^{{\mu }_s}}\wedge \cdots \wedge d{x}^{{\mu }_r}]\\ & =\frac{1}{r!}\sum _{s=1}^r[{\partial }_v{X}^{{\mu }_s}{\omega }_{{\mu }_1\dots {\mu }_s\dots {\mu }_r}(-1{)}^{s-1}d{x}^v\wedge d{x}^{{\mu }_1}\wedge \cdots \wedge \widehat{d{x}^{{\mu }_s}}\wedge \cdots \wedge d{x}^{{\mu }_r}\\ & +\frac{1}{r!}{X}^v{\partial }_v{\omega }_{{\mu }_1\dots {\mu }_r}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_r}.\end{array}$$

If we interchange the roles of ${\mu }_s$ and $\nu$ in the first term of the last expression and compare it with (5.81), we verify that

$$\left({\mathrm{di}}_X+{\mathrm{i}}_X\mathrm{d}\right)\omega ={\mathcal{L}}_X\omega \text{\hspace{1em}(5.82)}$$

for any $r$ -form $\omega$ .

Exercise 5.17

Let $X,Y\in \mathcal{X}(M)$ and $\omega \in {\Omega }^r(M)$ . Show that

$${\mathrm{i}}_{[X,Y]}\omega =X({\mathrm{i}}_Y\omega )-Y({\mathrm{i}}_X\omega ).\text{\hspace{1em}(5.83)}$$

Show also that ${\mathrm{i}}_X$ is an anti-derivation,

$${\mathrm{i}}_X(\omega \wedge \eta )={\mathrm{i}}_X\omega \wedge \eta +(-1{)}^r\omega \wedge {\mathrm{i}}_X\eta \text{\hspace{1em}(5.84)}$$

and nilpotent,

$${\mathrm{i}}_X^2=0.\text{\hspace{1em}(5.85)}$$

Use the nilpotency to prove

$${\mathcal{L}}_X{\mathrm{i}}_X\omega ={\mathrm{i}}_X{\mathcal{L}}_X\omega .\text{\hspace{1em}(5.86)}$$

Exercise 5.18

Let $t\in {\mathfrak{T}}_m^n(M)$ . Show that

$${\left({\mathcal{L}}_{X}t\right)}_{{\nu }_1\dots {\nu }_m}^{{\mu }_1\dots {\mu }_n}=X^{\lambda }{\partial }_{\lambda }{t}_{{\nu }_1\dots {\nu }_m}^{{\mu }_1\dots {\mu }_n}+\sum _{s=1}^n{\partial }_{{\nu }_s}{X}^{\lambda }{t}_{{\nu }_1\dots \lambda \dots {\nu }_m}^{{\mu }_1\dots {\mu }_n}-\sum _{s=1}^n{\partial }_{\lambda }{X}^{{\mu }_s}{t}_{{\nu }_1\dots {\nu }_m}^{{\mu }_1\dots \lambda \dots {\mu }_n}.\text{\hspace{1em}(5.87)}$$

Example 5.12

Let us reformulate Hamiltonian mechanics (section 1.1) in terms of differential forms. Let $H$ be a Hamiltonian and $(q^{\mu },p_{\mu })$ be its phase space. Define a two-form

$$\omega =\mathrm{d}{p}_{\mu }\wedge \mathrm{d}{q}^{\mu }\text{\hspace{1em}(5.88)}$$

called the symplectic two-form. If we introduce a one-form

$$\theta =q^{\mu }\mathrm{d}{p}_{\mu },\text{\hspace{1em}(5.89)}$$

the symplectic two-form is expressed as

$$\omega =\mathrm{d}\theta .\text{\hspace{1em}(5.90)}$$

Given a function $f(q,p)$ in the phase space, one can define the Hamiltonian vector field

$$X_f=\frac{\partial f}{\partial p_{\mu }}\frac{\partial }{\partial q^{\mu }}-\frac{\partial f}{\partial q^{\mu }}\frac{\partial }{\partial p_{\mu }}.\text{\hspace{1em}(5.91)}$$

Then it is easy to verify that

$${\mathrm{i}}_{X_f}\omega =-\frac{\partial f}{\partial p_{\mu }}\mathrm{d}{p}^{\mu }-\frac{\partial f}{\partial q^{\mu }}\mathrm{d}{q}^{\mu }=-\mathrm{d}f.$$

Consider a vector field generated by the Hamiltonian

$$X_H=\frac{\partial H}{\partial p_{\mu }}\frac{\partial }{\partial q^{\mu }}-\frac{\partial H}{\partial q^{\mu }}\frac{\partial }{\partial p_{\mu }}.\text{\hspace{1em}(5.92)}$$

For the solution $(q^{\mu },p_{\mu })$ to Hamilton’s equation of motion

$$\frac{\mathrm{d}{q}^{\mu }}{\mathrm{d}t}=\frac{\partial H}{\partial p_{\mu }}\frac{\mathrm{d}{p}_{\mu }}{\mathrm{d}t}=-\frac{\partial H}{\partial q^{\mu }},\text{\hspace{1em}(5.93)}$$

we also obtain

$$X_H=\frac{\mathrm{d}{p}_{\mu }}{\mathrm{d}t}\frac{\partial }{\partial p_{\mu }}\frac{\mathrm{d}{q}^{\mu }}{\mathrm{d}t}\frac{\partial }{\partial q^{\mu }}=\frac{\mathrm{d}}{\mathrm{d}t}.\text{\hspace{1em}(5.94)}$$

The symplectic two-form $\omega$ is left invariant along the flow generated by $X_H$

$$\begin{array}{rl}{\mathcal{L}}_{X_H}\omega & =\mathrm{d}({\mathrm{i}}_{X_H}\omega )+{\mathrm{i}}_{X_H}(\mathrm{d}\omega )\\ & =\mathrm{d}\left({\mathrm{i}}_{X_H}\omega \right)=-{\mathrm{d}}^{2}H=0\end{array}\text{\hspace{1em}(5.95)}$$

where use has been made of (5.82). Conversely, if $X$ satisfies ${\mathcal{L}}_X\omega =0$ , there exists a Hamiltonian $H$ such that Hamilton’s equation of motion is satisfied

along the flow generated by $X$ . This follows from the previous observation that ${\mathcal{L}}_X\omega =\mathrm{d}({\mathrm{i}}_X\omega )=0$ and hence by Poincaré’s lemma, there exists a function $H(q,p)$ such that

$${\mathrm{i}}_X\omega =-\mathrm{d}H.$$

The Poisson bracket is cast into a form independent of the special coordinates chosen with the help of the Hamiltonian vector fields. In fact,

$${\mathrm{i}}_{X_f}({\mathrm{i}}_{X_g}\omega )=-{\mathrm{i}}_{X_f}(\mathrm{d}g)=\frac{\partial f}{\partial q^{\mu }}\frac{\partial g}{\partial p_{\mu }}-\frac{\partial f}{\partial q^{\mu }}\frac{\partial g}{\partial p_{\mu }}=[f,g{]}_{\mathrm{PB}}.(5.96)$$

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：9ccf560be2dffc1c39f12dabd141340fe85bfad73a2a66c1036542c58479895a