17.5 The Poincaré lemma

中文

The fact that every exact differential form is closed has a kind of local converse.

Theorem 17.5 · Poincaré lemma

(Poincaré lemma) On any open set $U\subseteq M$ homeomorphic to ${\mathbb{R}}^n$ , every closed differentialform of degree $k\ge 1$ is exact: if $\mathrm{d}\alpha =0$ on $U$ where $\alpha \in {\Lambda }_k(U)$ , then there exists a $(k-1)$-form $\beta$ on $U$ such that $\alpha =\mathrm{d}\beta$

Proof

We prove the theorem on ${\mathbb{R}}^n$ itself, with coordinates $x^1,\dots ,x^n$ , and set $\alpha ={\alpha }_{i_1{i}_2\dots i_k}(\mathbf{x})\mathrm{d}{x}^{i_1}\wedge \mathrm{d}{x}^{i_2}\wedge \cdot \cdot \cdot \wedge \mathrm{d}{x}^{i_k}$ . Let ${\alpha }^t\left(0\le t\le 1\right)$ ) be the one-parameter family of $k-$ forms

$${\alpha }^t={\alpha }_{i_1\dots i_k}(t\mathbf{x})d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}.$$

The map $h_k:{\Lambda }_k({\mathbb{R}}^n)\to {\Lambda }_{k-1}({\mathbb{R}}^n)$ defined by

$$h_k\alpha ={\int }_0^1{t}^{k-1}{\mathrm{i}}_X{\alpha }^t\mathrm{d}t\text{ where }X=x^i\frac{\partial }{\partial x^i}$$

satisfies the key identity

$$(\mathrm{d}\circ h_k+h_{k+1}\circ \mathrm{d})\alpha =\alpha \text{\hspace{1em}(17.4)}$$

for any $k$-form $\alpha$ on ${\mathbb{R}}^n$ . To prove Eq. (17.4) write out the left-hand side,

$$(\mathrm{d}\circ h_k+h_{k+1}\circ \mathrm{d})\alpha ={\int }_0^1{t}^{k-1}{\mathrm{di}}_X{\alpha }^t+t^k{\mathrm{i}}_X(\mathrm{d}\alpha {)}^t\mathrm{d}t,$$

and

$$\begin{array}{l}\mathrm{d}{\alpha }^t=\frac{\partial {\alpha }_{i_1\dots i_k}(t\mathbf{x})}{\partial x^j}\mathrm{d}{x}^j\wedge \mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_k}\\ =t(\mathrm{d}\alpha {)}^t.\end{array}$$

Using the Cartan identity, Eq. (16.13),

$$\begin{array}{rl}(\mathrm{d}\circ h_k+h_{k+1}\circ \mathrm{d})\alpha & ={\int }_0^1{t}^{k-1}(\mathrm{d}\circ {\mathrm{i}}_X+{\mathrm{i}}_X\circ \mathrm{d}){\alpha }^t\mathrm{d}t\\ & ={\int }_0^1{t}^{k-1}{\mathcal{L}}_X{\alpha }^t\mathrm{d}t\end{array}$$

and from the component formula for the Lie derivative (15.39),

$$\begin{array}{l}{\left({\mathcal{L}}_X{\alpha }^t\right)}_{i_1\dots i_k}=\frac{\partial {\alpha }_{i_1\dots i_k}(t\mathbf{x})}{\partial x^j}{x}^j+\frac{\partial x^j}{\partial x^{i_1}}{\alpha }_{j{i}_2\dots i_k}(t\mathbf{x})+\cdots +\frac{\partial x^j}{\partial x^{i_k}}{\alpha }_{i_1\dots j}(t\mathbf{x})\\ =t\frac{\partial {\alpha }_{i_1\dots i_k}(t\mathbf{x})}{\partial t{x}^j}\frac{\mathrm{d}t{x}^j}{\mathrm{d}t}+{\delta }_{i_1}^j{\alpha }_{j\dots i_k}(t\mathbf{x})+\cdots +{\delta }_{x^{i_k}}^j{\alpha }_{i_1\dots j}(t\mathbf{x})\\ =t\frac{\mathrm{d}{\alpha }_{i_1\dots i_k}(t\mathbf{x})}{\mathrm{d}t}+k{\alpha }_{i_1\dots i_k}(t\mathbf{x}).\end{array}$$

Hence

$${\mathcal{L}}_X{\alpha }^t=t\frac{\mathrm{d}{\alpha }^t}{\mathrm{d}t}+k{\alpha }^t,$$

and Eq. (17.4) follows from

$$(\mathrm{d}\circ h_k+h_{k+1}\circ \mathrm{d})\alpha ={\int }_0^1{t}^k\frac{\mathrm{d}{\alpha }^t}{\mathrm{d}t}+k{t}^{k-1}{\alpha }^t\mathrm{d}t={\int }_0^1\frac{\mathrm{d}{t}^k{\alpha }^t}{\mathrm{d}t}\mathrm{d}t=\alpha .$$

If $\mathrm{d}\alpha =0$ we have $\alpha =\mathrm{d}\beta$ where $\beta =h_k\alpha$ , and the theorem is proved.

An immediate corollary of this theorem and de Rham’s theorem is that all homology groups $H_k({\mathbb{R}}^n)$ are trivial for $k\ge 1$ ; that is, all Betti numbers for $k\ge 1$ vanish in Euclidear space, $b_k=b^k=0$ . Of course $b_0=b^0=1$ since there is a single connected component.

Example 17.7

In ${\mathbb{R}}^3$ let $\alpha$ be the 1-form $\alpha =A_1\mathrm{d}{x}^1+A_2\mathrm{d}{x}^2+A_3\mathrm{d}{x}^3$ . Its exterior derivative is

$$\mathrm{d}\alpha =(A_{2,1}-A_{1,2})\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2+(A_{1,3}-A_{3,1})\mathrm{d}{x}^3\wedge \mathrm{d}{x}^1+(A_{3,2}-A_{2,3})\mathrm{d}{x}^2\wedge \mathrm{d}{x}^3$$

and Poincaré’s lemma asserts that $\mathrm{d}\alpha =0$ if and only ifthere exists a function $f$ on ${\mathbb{R}}^3$ such that $\alpha =\mathrm{d}f$ . In components,

$$A_{2,1}-A_{1,2}=A_{1,3}-A_{3,1}=A_{3,2}-A_{2,3}=0⟺A_1=f_{,1},A_2=f_{,2},A_3=f_{,3},$$

or in standard 3-vector language, with $\mathbf{A}=(A_1,A_2,A_3)$ 1

$$\nabla \times \mathbf{A}=0⟺\mathbf{A}=\nabla f.$$

If $\alpha$ is the differential 2-form $\alpha =A_3\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2+A_2\mathrm{d}{x}^3\wedge \mathrm{d}{x}^1+A_1\mathrm{d}{x}^2\wedge \mathrm{d}{x}^3$ , then

$$\mathrm{d}\alpha =(A_{1,1}+A_{2,2}+A_{3,3})\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3.$$

The Poincaré lemma says

$$\mathrm{d}\alpha =0⟺\alpha =\mathrm{d}\beta \text{ where }\beta =B_1\mathrm{d}{x}^1+B_2\mathrm{d}{x}^2+B_3\mathrm{d}{x}^3,$$

or in components

$$A_{1,1}+A_{2,2}+A_{3,3}=0⟺A_1=B_{3,2}-B_{2,3},A_2=B_{1,3}-B_{3,1},A_3=B_{2,1}-B_{1,2},$$

which reduces to the familiar 3-vector statement

$$\nabla \cdot \mathbf{A}=0⟺\text{ there exists }\mathbf{B}\text{ such that }\mathbf{A}=\nabla \times \mathbf{B}.$$

Example 17.8

On the manifold $M={\mathbb{R}}^2-\{0\}$ with coordinates $x^1=x,x^2=y$ , let $\omega$ be the differential 1-form

$$\omega =\frac{-y\mathrm{d}x+x\mathrm{d}y}{x^2+y^2},$$

which cannot be extended smoothly to a 1-form on all of ${\mathbb{R}}^2$ because of the singular behaviour at the origin. On $M$, however, it is closed since

$$\begin{array}{c}\mathrm{d}\omega =\frac{-\mathrm{d}y\wedge \mathrm{d}x}{x^2+y^2}+\frac{2y^2\mathrm{d}y\wedge \mathrm{d}x}{(x^2+y^2{)}^2}+\frac{\mathrm{d}x\wedge \mathrm{d}y}{x^2+y^2}-\frac{2x^2\mathrm{d}x\wedge \mathrm{d}y}{(x^2+y^2{)}^2}\\ =\frac{2\mathrm{d}x\wedge \mathrm{d}y}{x^2+y^2}-\frac{2(x^2+y^2)}{(x^2+y^2{)}^2}\mathrm{d}x\wedge \mathrm{d}y=0.\end{array}$$

Locally it is possible everywhere to find a function $f$ such that $\omega =df$ . For example, it is straightforward to verify that the pair of differential equations

$$\frac{\partial f}{\partial x}=\frac{-y}{x^2+y^2},\frac{\partial f}{\partial y}=\frac{x}{x^2+y^2}$$

has a solution $f=\arctan (y/x)$ . However $f$ is not globally defined on $M$, since it is essentially the polar angle given by $x$ $r$ $\cos$ $f$, $y$ $r$ $\sin$ $f$ and increases by $2\pi$ on any circuit of the origin beginning at the positive branch of the $y$-axis. This demonstrates that Poincaré’s lemma does not in general hold on manifolds not homeomorphic with ${\mathbb{R}}^n$

Electrodynamics

An electromagnetic field is represented by an antisymmetric 4-tensor field $F$ in Minkowski space, having components $F_{\mu \nu }(x^{\alpha })(\mu ,\nu =1,...,4)$ (see Chapter 9). Define the Maxwell 2-form $\phi$ as having components $F_{\mu \nu ;}$

$$\begin{array}{l}\phi =F_{\mu \nu }\mathrm{d}{x}^{\mu }\mathrm{d}{x}^{\nu }\\ =2(B_3\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2+B_2\mathrm{d}{x}^3\wedge \mathrm{d}{x}^1+B_1\mathrm{d}{x}^2\wedge \mathrm{d}{x}^3\\ +E_1\mathrm{d}{x}^1\wedge \mathrm{d}{x}^4+E_2\mathrm{d}{x}^2\wedge \mathrm{d}{x}^4+E_3\mathrm{d}{x}^3\wedge \mathrm{d}{x}^4)\end{array}$$

where $\mathbf{E}=(E_1,E_2,E_3)$ is the electric field, $\mathbf{B}=(B_1,B_2,B_3)$ the magnetic field and $x^1=x,x^2=y,x^3=z,x^4=ct$ are inertial coordinates. The source-free Maxwell equations (9.37) can be written

$$\begin{array}{rcl}\mathrm{d}\phi =0& ⟺& F_{\mu \nu ,\rho }+F_{\nu \rho ,\mu }+F_{\rho \mu ,\nu }=0\\ & ⟺& \nabla \cdot \mathbf{B}=0,\nabla \times \mathbf{E}+\frac{1}{c}\frac{\partial \mathbf{B}}{\partial t}=0.\end{array}$$

By the Poincaré lemma, there exists a 1-form $\alpha$, known as the 4-vector potential, such tha $\phi =\mathrm{d}\alpha$ . Writing the components of $\alpha$ as $(A_1,A_2,A_3,-\varphi )$ this equation reads

$$\mathbf{B}=\nabla \times \mathbf{A},\mathbf{E}=-\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t}-\nabla \varphi .$$

To express the equations relating the electromagnetic field to its sources in terms of differential forms we must define the dual Maxwell 2-form $\ast \phi =\ast F_{\mu \nu }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }$ where

* $F_{\mu \nu }$ is defined as in Example 8.8,

$$\begin{array}{l}\ast \phi =2(-E_3\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2-E_2\mathrm{d}{x}^3\wedge \mathrm{d}{x}^1-E_1\mathrm{d}{x}^2\wedge \mathrm{d}{x}^3\\ +B_1\mathrm{d}{x}^1\wedge \mathrm{d}{x}^4+B_2\mathrm{d}{x}^2\wedge \mathrm{d}{x}^4+B_3\mathrm{d}{x}^3\wedge \mathrm{d}{x}^4).\end{array}$$

The distribution of electric charge present is represented by a 4-current vector field $J=J^{\mu }{e}_{\mu }$ having components $J^{\mu }=\left(\mathbf{j},\rho c\right)$ where $\rho (\mathbf{r},t)$ is the charge density and $j$ the charge flux density (see Section 9.4).

$$\begin{array}{l}\vartheta =\ast J_{\mu \nu \rho }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }\wedge \mathrm{d}{x}^{\rho }\\ =-\frac{1}{3!}{ϵ}_{\mu \nu \rho \sigma }{J}^{\sigma }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }\wedge \mathrm{d}{x}^{\rho }\\ =-c\rho \mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3+J^1\mathrm{d}{x}^2\wedge \mathrm{d}{x}^3\wedge \mathrm{d}{x}^4\\ +J^2\mathrm{d}{x}^3\wedge \mathrm{d}{x}^1\wedge \mathrm{d}{x}^4+J^3\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^4.\end{array}$$

Equations (9.38) may then be written as

$$\mathrm{d}\ast \phi =-\vartheta ⟺\nabla \cdot \mathbf{E}=4\pi \rho ,-\frac{1}{c}\frac{\partial \mathbf{E}}{\partial t}+\nabla \times \mathbf{B}=\frac{4\pi }{c}\mathbf{j}.$$

Charge conservation follows from

$$\mathrm{d}\vartheta =-{\mathrm{d}}^2\ast \phi =0⟺\nabla \cdot \mathbf{j}+\frac{1}{c}\frac{\partial \rho }{\partial t}=0.$$

Example 17.9

Although Maxwell’s vacuum equations take on the deceptively symmetrica form

$$\mathrm{d}\phi =0,\mathrm{d}\ast \phi =0$$

we cannot assume that $\ast \phi =\mathrm{d}\beta$ for a globally defined 1-form $\beta .$ . For example, the coulomb field

$$\mathbf{B}=0,\mathbf{E}=\left(\frac{qx}{r^3},\frac{qy}{r^3},\frac{qz}{r^3}\right)$$

corresponds to the Maxwell 2-form

$$\phi =\frac{2q}{r^3}\left(xdx\wedge d{x}^4+ydy\wedge d{x}^4+zdz\wedge d{x}^4\right)=\frac{2q}{r^2}dr\wedge d{x}^4$$

where $r^2=x^2+y^2+z^2$ , with dual 2-form

$$\ast \phi =-\frac{2q}{r^3}(zdx\wedge dy+ydz\wedge dx+xdy\wedge dz).$$

This 2-form is, however, only defined on the subspace $M={\mathbb{R}}^3-\{0\}$ . A short calculation in spherical polar coordinates results in

$$\ast \phi =-2q\sin \theta d\theta \wedge d\varphi =d(2q\cos \theta d\varphi )=d(2q(\cos \theta -1)d\varphi ).$$

Either of the choices $\beta =2q$ $\cos$ $\theta$ $d\varphi$ or ${\beta }^{\prime }=2q(\cos \theta -1)\mathrm{d}\varphi$ will act as a potential 1-form for $\ast \phi ,$ , but neither is defined on all of $M$ since the angular coordinate $\varphi$ is not well-defined on the $z$-axis where $\theta =0\mathrm{or}\pi$ . The 1-form $\beta$ is not well-defined on the entire $z$-axis, but the potential 1-form ${\beta }^{\prime }$ vanishes on the positive $z$-axis and has a singularity along the negative $z$-axis. It is sometimes called a Dirac string – a term commonly reserved for solutions representing magnetic monopoles.

The impossibility of a global potential $\beta$ can be seen by integrating $\phi$ over the uni 2-sphere

$${\int }_{S^2}\ast \phi ={\int }_0^{\pi }{\int }_0^{2\pi }-2q\sin \theta d\theta d\varphi =-8\pi q,$$

and using Stokes’ theorem (note that $S^2$ has no boundary)

$${\int }_{S^2}\ast \phi ={\int }_{S^2}d\beta ={\int }_{\partial S^2}\beta =0.$$

Problems

Problem 17.13

Let

$$\alpha =\frac{x\mathrm{d}y-y\mathrm{d}x}{x^2+y^2}.$$

Show that $\alpha$ is a closed 1-form on ${\mathbb{R}}^2-\{0\}$ . Compute its integral over the unit circle $S^1$ and show that it is not exact. What does this tell us of the de Rham cohomology of ${\mathbb{R}}^2-\{0\}$ and $S^1$?

Problem 17.14

Prove that every closed 1-form on $S^2$ is exact. Show that this statement does not extend to 2-forms by showing that the 2-form

$$\alpha =r^{-3/2}(x\mathrm{d}y\wedge \mathrm{d}z+y\mathrm{d}z\wedge \mathrm{d}x+z\mathrm{d}x\wedge \mathrm{d}y)$$

is closed, but has non-vanishing integral on $S^2$

Problem 17.15

Show that the Maxwell 2-form satisfies the identities

$$\phi \wedge \ast \phi =\ast \phi \wedge \phi =4({\mathbf{B}}^2-{\mathbf{E}}^2)\Omega$$ $$\phi \wedge \phi =-\ast \phi \wedge \ast \phi =8\mathbf{B}\cdot \mathbf{E}\Omega$$

where $\Omega =\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3\wedge \mathrm{d}{x}^4$