16.1 Differential forms and exterior derivative

中文

Let $M$ be a differentiable manifold of dimension $n$. At any point $p\in M$ let ${\left({\Lambda }_r\right)}_p(M)\equiv {\Lambda }^{\ast r}(T_p(M))$ be the space of totally antisymmetric tensors, or $r$-forms, generated by the tangent space $T_p(M)$ (see Chapter 8). Denote the associated exterior algebra

$${\Lambda }_p(M)\equiv {\Lambda }^{\ast }\left(T_p(M)\right)={\left({\Lambda }_0\right)}_p(M)\oplus {\left({\Lambda }_1\right)}_p(M)\oplus \cdots \oplus {\left({\Lambda }_n\right)}_p(M)$$

with graded exterior product $\wedge :{\left({\Lambda }_r\right)}_p(M)\times {\left({\Lambda }_s\right)}_p(M)\to {\left({\Lambda }_{r+s}\right)}_p(M).$

A differential $r$-form $\alpha$ on an open subset $U\subseteq M$ is an $r$-form field, or assignment of an $r-\mathrm{form}{\alpha }_p$ at every point $p\in U$ , such that the function $\alpha (X_1,X_2,\dots ,X_r)(p)\equiv {\alpha }_p((X_1{)}_p,(X_2{)}_p,\dots ,(X_r{)}_p)$ is differentiable for all smooth vector fields $X_1,X_2,\dots ,X_r$ on $U$. The set of all differential $r$-forms on $U$ is denoted ${\Lambda }_r(U)$ ) and the differential exterior algebra on $U$ is the direct sum

$$\Lambda (U)={\Lambda }_0(U)\oplus {\Lambda }_1(U)\oplus \cdots \oplus {\Lambda }_n(U)$$

with exterior product defined by $(\alpha \wedge \beta {)}_p={\alpha }_p\wedge {\beta }_p$ . This product is linear, associative and obeys the usual anticommutative rule:

$$\begin{array}{c}\alpha \wedge (\beta +\gamma )=\alpha \wedge \beta +\alpha \wedge \gamma \\ \alpha \wedge (\beta \wedge \gamma )=(\alpha \wedge \beta )\wedge \gamma \\ \alpha \wedge \beta =(-1{)}^{rs}\beta \wedge \alpha \text{ if }\alpha \in {\Lambda }_r(U),\beta \in {\Lambda }_s(U).\end{array}\text{\hspace{1em}(16.1)}$$

Differential 0-forms are simply scalar fields, smooth real-valued functions $f$ on $U$, ${\Lambda }_0(U)=\mathcal{F}(U)$

In a coordinate chart $(U,\varphi ;x^i)$ a basis of ${\Lambda }_r(U)$ is

$$\mathrm{d}{x}^{i_1}\wedge \mathrm{d}{x}^{i_2}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}=\mathcal{A}\left(\mathrm{d}{x}^{i_1}\otimes \mathrm{d}{x}^{i_2}\otimes \cdots \otimes \mathrm{d}{x}^{i_r}\right),$$

and every differential $r$-form on $U$ has a unique expansion

$$\alpha ={\alpha }_{i_1{i}_2\dots i_r}\mathrm{d}{x}^{i_1}\wedge \mathrm{d}{x}^{i_2}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\text{\hspace{1em}(16.2)}$$

where the components ${\alpha }_{i_1{i}_2\dots i_r}$ are smooth functions on $U$ and are antisymmetric in al indices,

$${\alpha }_{i_1{i}_2\dots i_r}={\alpha }_{[i_1{i}_2\dots i_r]}.$$

If $f$ is a scalar field then its gradient

$$f_{,i}=\frac{\partial f}{\partial x^i}$$

forms the components of a covariant vector field known as its differential $\mathrm{d}f$ (see Chapter 15). This concept may be extended to a map on all differential forms, $d$ : $\Lambda (M)\to \Lambda (M)$ called the exterior derivative, such that $d{\Lambda }_r(M)\subseteq {\Lambda }_{(r+1)}(M)$ and satisfying the following conditions:

(ED1) If $f$ is a differential 0-form, then $\mathrm{d}f$ is its differential, defined by $⟨\mathrm{d}f,X⟩=Xf$ for any smooth vector field $X$.

(ED2) For any pair of differential forms, $\alpha ,\beta \in \Lambda (M),\mathsf{d}(\alpha +\beta )=\mathsf{d}\alpha +\mathsf{d}\beta$

(ED3) If $f$ is a differential 0-form then ${\mathrm{d}}^{2}f\equiv \mathrm{d}(\mathrm{d}f)=0.$

(ED4) For any $r$-form $\alpha$, and any $\beta \in \Lambda (M)$

$$\mathrm{d}(\alpha \wedge \beta )=(\mathrm{d}\alpha )\wedge \beta +(-1{)}^r\alpha \wedge \mathrm{d}\beta .\text{\hspace{1em}(16.3)}$$

Condition (ED4) says that it is an anti-derivation (see Section 8.4, Eq. (8.18)), and (ED3) will be shown to hold for differential forms of all orders. The general theory of differentia forms and exterior derivative may be found in [1–9]. Our aim in the following discussion is to show that the operator $d$ exists and is uniquely defined.

Lemma 16.1 · Local smooth cutoff functions

Let $U$ be an open subset of a differentiable manifold $M$. For any point $p\in U$ there exist open sets $W$ and $W^{\prime }$ where $W$ has compact closure with $p\in \overline{W}\subseteq W^{\prime }\subseteq U$ , and a smooth function $h\ge 0$ such that $h=1$ on $W$ and $h=0$ on $M-W^{\prime }$

Proof

Let $f:\mathbb{R}\to \mathbb{R}$ be the smooth non-negative function, defined by

$$f(t)=\{\begin{array}{ll}{\mathrm{e}}^{-1/t}& \text{ if }t>0,\\ 0& \text{ if }t\le 0.\end{array}$$

For every $a>0$ let $g_a:\mathbb{R}\to \mathbb{R}$ be the non-negative smooth function

$$g_a(t)=\frac{f(t)}{f(t)+f(a-t)}=\{\begin{array}{ll}0& \text{ if }t\le 0,\\ >0& \text{ if }0<t<a,\\ 1& \text{ if }t\ge a.\end{array}$$

If $b>a$ the smooth function $h_{a,b}:\mathbb{R}R$ defined by

$$h_{a,b}(t)=1-g_{b-a}(t-a)$$

has the value 1 for $t\le a$ and is 0 for $t\ge b$ . On the open interval $(a,b)$ it is positive with values between 0 and 1. Let $(V,\varphi ;x^i)$ be a coordinate chart at $p$ such that $x^i(p)=0$ and $V\subseteq U$ Let $b>0$ be any real number such that the open ball $B_b(0)\subset \varphi (V)$ , and let $0<a<b$ Set $W^{\prime }={\varphi }^{-1}(B_b(0))$ and $W={\varphi }^{-1}(B_a(0))$ ). The closure of $W$, being the homeomorphic image of a compact set, is compact and $\overline{W}\subset W^{\prime }.\mathrm{Let}\tilde{h}:{\mathbb{R}}^n\mathbb{R}$ be the smooth map

$$\tilde{h}\left(x^1,x^2,\dots ,x^n\right)=h_{a,b}(r)\text{ where }r=\sqrt{{\left(x^1\right)}^2+{\left(x^2\right)}^2+\cdots +{\left(x^n\right)}^2},$$

and the positive function $h:M\to \mathbb{R}$ defined by

$$h(p)=\{\begin{array}{ll}\tilde{h}\circ \varphi (p)& \text{ for }p\in W\\ 0& \text{ for }p\in M-W^{\prime }\end{array}$$

has all the desired properties.

Construction and uniqueness of the exterior derivative

If $\alpha$ is an $r$-form whose restriction to $U$ vanishes, $\alpha {|}_U=0.$ , then $h\alpha =0$ on all of $M$ where $h$ is the function defined in Lemma 16.1, and by property (ED4),

$$\mathrm{d}(h\alpha )=\mathrm{d}h\wedge \alpha +h\mathrm{d}\alpha =0.$$

Restricting this equation to $W$ we have ${\mathrm{d}\alpha |}_W=0$ , and in particula $(\mathrm{d}\alpha ){|}_p=0$ . Since $p$ is an arbitrary point of $U$ it follows that $d\alpha {|}_U=0$ . Hence, if $\alpha$ and $\beta$ are any pair of $r$-forms such that $\alpha {|}_U=\beta {|}_U$ , then $(\mathrm{d}\alpha ){|}_U=(\mathrm{d}\beta ){|}_U$ . Thus if $d$ exists, satisfying (ED1)–(ED4), then it has a local character and is uniquely defined everywhere.

To show the existence of the operator $d$, let $(U;x^i)$ be a coordinate chart at any point $p.$ Expanding $\alpha$ according to Eq. (16.2) we have, using (ED1)–(ED4),

$$\begin{array}{l}\mathrm{d}\alpha =\mathrm{d}{\alpha }_{i_1\dots i_r}\wedge \mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}+{\alpha }_{i_1\dots i_r}{\mathrm{d}}^2{x}^{i_1}\wedge \mathrm{d}{x}^{i_2}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\\ -{\alpha }_{i_1\dots i_r}\mathrm{d}{x}^{i_1}\wedge {\mathrm{d}}^2{x}^{i_2}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}+\dots \\ =\mathrm{d}{\alpha }_{i_1\dots i_r}\mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\\ ={\alpha }_{i_1\dots i_r,j}\mathrm{d}{x}^j\wedge \mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}.\end{array}$$

Performing a cyclic permutation of indices, and using the total antisymmetry of the wedge product,

$$\mathrm{d}\alpha =(-1{)}^r{\alpha }_{[i_1\dots i_r,i_{r+1}]}\mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\wedge \mathrm{d}{x}^{i_{r+1}}.\text{\hspace{1em}(16.4)}$$

It still remains to verify that conditions (ED1)–(ED4) hold for Eq. (16.4). Firstly, this formula reduces to $\mathrm{d}f=f_{,i}\mathrm{d}{x}^i$ in the case of a 0-form, consistent with (ED1). Condition (ED2) follows trivially. To verify (ED3),

$${\mathrm{d}}^{2}f=\mathrm{d}(f_{,i}\mathrm{d}{x}^i)=f_{,ij}\mathrm{d}{x}^j\wedge \mathrm{d}{x}^i=f_{,[ij]}\mathrm{d}{x}^j\wedge \mathrm{d}{x}^i=0$$

since

$$f_{,[ij]}=\frac{1}{2}(\frac{{\partial }^{2}f}{\partial x^i\partial x^j}-\frac{{\partial }^{2}f}{\partial x^j\partial x^i})=0.$$

Finally Eq. (16.4) implies (ED4):

$$\begin{array}{l}\mathrm{d}(\alpha \wedge \beta )=\mathrm{d}({\alpha }_{i_1{i}_2...i_r}{\beta }_{j_1{j}_2...j_s}\mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\mathrm{d}{x}^{j_1}\wedge \cdots \wedge \mathrm{d}{x}^{j_s})\\ =\mathrm{d}\left({\alpha }_{i_1{i}_2...i_r}{\beta }_{j_1{j}_2...j_s}\right)\wedge \mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\mathrm{d}{x}^{j_1}\wedge \cdots \wedge \mathrm{d}{x}^{j_s}\\ =(\mathrm{d}{\alpha }_{i_1{i}_2...i_r}{\beta }_{j_1{j}_2...j_s}+{\alpha }_{i_1{i}_2...i_r}\mathrm{d}{\beta }_{j_1{j}_2...j_s})\wedge \mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\mathrm{d}{x}^{j_1}\wedge \cdots \wedge \mathrm{d}{x}^{j_s}\\ =(\mathrm{d}\alpha )\wedge \beta +(-1{)}^r\alpha \wedge \mathrm{d}\beta .\end{array}$$

The last step follows on performing the $r$ interchanges needed to bring the $\mathrm{d}{\beta }_{j_1{j}_2\dots j_s}$ term between $d{x}^{i_r}$ and $\mathrm{d}{x}^{j_1}$ . This shows the existence and uniqueness of the operator $d$ on every coordinate neighbourhood on $M$.

Exercise

For all differential forms $\alpha$ and $\beta ,$ , and any pair of real numbers $a$ and ${}^{b,}$ show that

$$\mathrm{d}(a\alpha +b\beta )=a\mathrm{d}(\alpha )+b\mathrm{d}(\beta ).$$

The property (ED3) extends to arbitrary differential forms

$${\mathrm{d}}^2\alpha =\mathrm{d}(\mathrm{d}\alpha )=0,\text{\hspace{1em}(16.5)}$$

for, applying the operator $d$ to Eq. (16.2) and using (ED3) give

$$\mathrm{d}(\mathrm{d}\alpha )=\left({\mathrm{d}}^2{\alpha }_{i_1\dots i_r}\right)\wedge \mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}=0.$$

Example 16.1

Let $x=x^1,y=x^2,z=x^3$ be coordinates on the three-dimensional manifold $M={\mathbb{R}}^3$ . The exterior derivative of any 0-form $\alpha =f$ is

$$\mathrm{d}f=f_{,i}\mathrm{d}{x}^i=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z.$$

The three components are commonly known as the gradient of the scalar field $f.$

If $\omega =w_i\mathrm{d}{x}^i=A\mathrm{d}x+B\mathrm{d}y+C\mathrm{cd}z$ is a differential 1-form then

$$\mathrm{d}\omega =\left(\frac{\partial C}{\partial y}-\frac{\partial B}{\partial z}\right)\mathrm{d}y\wedge \mathrm{d}z+\left(\frac{\partial A}{\partial z}-\frac{\partial C}{\partial x}\right)\mathrm{d}z\wedge \mathrm{d}x+\left(\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y}\right)\mathrm{d}x\wedge \mathrm{d}y.$$

The components of the exterior derivative are traditionally written as components ofa vector field, known as the curl of the three-component vector field $(A,B,C)$ . Notice, however, that the tensor components of $d\omega \mathbf{\Sigma }=-w_{[i,j]}\mathbf{d}{x}^i\wedge \mathbf{d}{x}^j$ are half the curl components,

$$(\mathrm{d}\omega {)}_{ij}=-w_{[i,j]}=\frac{1}{2}(w_{j,i}-w_{i,j}).$$

If $\alpha ={\alpha }_{ij}\mathrm{d}{x}^i\wedge \mathrm{d}{x}^j=P\mathrm{d}y\mathrm{d}zQ\mathrm{d}zdxRdx\mathrm{d}y$ is a 2-form then

$$\mathrm{d}\alpha =\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z.$$

The single component of this 3-form is known as the divergence of the three-component vector field $(P,Q,R)$ . Equation (16.5) applied to the 0-form $f$ and 1-form $\omega$ gives the following classical results:

$$\begin{array}{l}{\mathrm{d}}^{2}f=0⟹\text{ curl grad }=0,\\ {\mathrm{d}}^2\omega =0⟹\text{ div curl }=0.\end{array}$$

Exercise

$\mathrm{If}\alpha ={\alpha }_{ij}d{x}^i\wedge d{x}^j$ is a 2-form on a manifold $M$, show that

$$(\mathrm{d}\alpha {)}_{ijk}=\frac{1}{3}({\alpha }_{ij,k}+{\alpha }_{jk,i}+{\alpha }_{ki,j}).\text{\hspace{1em}(16.6)}$$

More generally, lumping together the permutations of the first $r$ indices in Eq. (16.4) we obtain the following formula for the tensor components of the exterior derivative of an $r$-form $\alpha$:

$$(\mathrm{d}\alpha {)}_{i_1\dots i_{r+1}}=\frac{(-1{)}^r}{r+1}\sum _{\text{ cyclic }\pi }(-1{)}^{\pi }{\alpha }_{i_{\pi (1)}\dots i_{\pi (r)},i_{\pi (r+1)}}.\text{\hspace{1em}(16.7)}$$

Problems

Problem 16.1

Let $x^1=x,x^2=y,x^3=z$ be coordinates on the manifold ${\mathbb{R}}^3$ . Write out the com ponents ${\alpha }_{ij}$ and $(\mathrm{d}\alpha {)}_{ijk}$ , etc. for each of the following 2-forms:

$$\alpha =\mathrm{d}y\wedge \mathrm{d}z+\mathrm{d}x\wedge \mathrm{d}y,$$ $$\beta =x\mathrm{d}z\wedge \mathrm{d}y+y\mathrm{d}x\wedge \mathrm{d}z+z\mathrm{d}y\wedge \mathrm{d}x,$$ $$\gamma =\mathrm{d}(r^2(x\mathrm{d}x+y\mathrm{d}y+z\mathrm{d}z))\text{ where }{r}^2=x^2+y^2+z^2.$$

Problem 16.2

On the manifold ${\mathbb{R}}^n$ compute the exterior derivative $d$ of the differential form

$$\alpha =\sum _{i=1}^n(-1{)}^{i-1}{x}^i\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{i-1}\wedge \mathrm{d}{x}^{i+1}\wedge \cdots \wedge \mathrm{d}{x}^n.$$

Do the same for $\beta =r^{-n}\alpha$ where $r^2=(x^1{)}^2+\cdot \cdot \cdot +(x^n{)}^2$

Problem 16.3

Show that the right-hand side of Eq. (16.6) transforms as a tensor field of type (0, 3). Generalize this result to the right-hand side of Eq. (16.7), to show that this equation could be used $a$ a local definition of exterior derivative independent of the choice of coordinate system.