2.4 Differential Forms

Some tensors are more interesting than others. A particularly interesting class are totally anti-symmetric $(0,p)$ tensors fields. These are called p-forms. The set of all p-forms over a manifold M is denoted ${\Lambda }^p(M)$

We’ve met some forms before. A 0-form is simply a function. Meanwhile, as we saw previously, a 1-form is another name for a covector. The anti-symmetry means that we can’t have any form of degree $p>n=\dim (M)$ . A p-form has $\left(\genfrac{}{}{0ex}{}{n}{p}\right)$ different components. Forms in ${\Lambda }^n(M)$ are called top forms.

Given a p-form $\omega$ and a q-form η, we can take the tensor product (2.24) to construct a $(p+q)$ -tensor. If we anti-symmetrise this, we then get a $(p+q)$ -form. This construction is called the wedge product, and is defined by

$$(\omega \wedge \eta {)}_{{\mu }_1\dots {\mu }_p{\nu }_1\dots {\nu }_q}=\frac{(p+q)!}{p!q!}{\omega }_{[{\mu }_1\dots {\mu }_p}{\eta }_{{\nu }_1\dots {\nu }_q]}$$

where the $\left[\dots \right]$ in the subscript tells us to anti-symmetrise over all indices. For example, given ω, $\eta \in {\Lambda }^1(M)$ , we can construct a 2-form

$$(\omega \wedge \eta {)}_{\mu \nu }={\omega }_{\mu }{\eta }_{\nu }-{\omega }_{\nu }{\eta }_{\mu }$$

For one forms, the anti-symmetry ensures that $\omega \wedge \omega =0$ . In general, if $\omega \in {\Lambda }^p(M)$ and $\eta \in {\Lambda }^q(M)$ , then one can show that

$$\omega \wedge \eta =(-1{)}^{pq}\eta \wedge \omega$$

This means that $\omega \wedge \omega =0$ for any form of odd degree. We can, however, wedge even degree forms with themselves. (Which you know already for 0-forms where the wedge product is just multiplication of functions.)

As a more specific example, consider $M={\mathbf{R}}^3$ and $\omega ={\omega }_{\mu }d{x}^{\mu }$ and $\eta ={\eta }_{\mu }d{x}^{\mu }$ . We then have

$$\begin{array}{rl}& \omega \wedge \eta =({\omega }_{1}d{x}^1+{\omega }_{2}d{x}^2+{\omega }_{3}d{x}^3)\wedge ({\eta }_{1}d{x}^1+{\eta }_{2}d{x}^2+{\eta }_{3}d{x}^3)\\ & =({\omega }_1{\eta }_2-{\omega }_2{\eta }_1)d{x}^1\wedge d{x}^2+({\omega }_2{\eta }_3-{\omega }_3{\eta }_2)d{x}^2\wedge d{x}^3+({\omega }_3{\eta }_1-{\omega }_1{\eta }_3)d{x}^3\wedge d{x}^1\end{array}$$

Notice that the components that arise are precisely those of the cross-product acting on vectors in ${\mathbf{R}}^3$ . This is no coincidence: what we usually think of as the cross-product between vectors is really a wedge product between forms. We’ll have to wait to Section 3 to understand how to map from one to the other.

It can also be shown that the wedge product is associative, meaning

$$\omega \wedge (\eta \wedge \lambda )=(\omega \wedge \eta )\wedge \lambda$$

We can then drop the brackets in any such product.

Given a basis $\{f^{\mu }\}$ of ${\Lambda }^1(M)$ , a basis of ${\Lambda }^p(M)$ can be constructed by wedge products $\{f^{{\mu }_1}\wedge ...\wedge f^{{\mu }_p}\}$ . We will usually work with the coordinate basis $\{d{x}^{\mu }\}$ . This means that any p-form ω can be written locally as

$$\omega =\frac{1}{p!}{\omega }_{{\mu }_1\dots {\mu }_p}d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_p}\text{\hspace{1em}(2.25)}$$

Although locally any $p\cdot$ -form can be written as (2.25), this may not be true globally. This, and related issues, will become of some interest in Section 2.4.3.

2.4.1 The Exterior Derivative

We learned in Section 2.3.1 how to construct a one-form $df$ from a function $f.$ In a coordinate basis, this one-form has components (2.17),

$$df=\frac{\partial f}{\partial x^{\mu }}d{x}^{\mu }$$

We can extend this definition to higher forms. The exterior derivative is a map

$$d:{\Lambda }^p(M)\to {\Lambda }^{p+1}(M)$$

In local coordinates (2.25), the exterior derivative acts as

$$(d\omega )=\frac{1}{p!}\frac{\partial {\omega }_{{\mu }_1\dots {\mu }_p}}{\partial x^{\nu }}d{x}^{\nu }\wedge d{x}^{{\mu }_1}\wedge \cdots \wedge d{x}^{{\mu }_p}\text{\hspace{1em}(2.26)}$$

Equivalently we have

$$(d\omega {)}_{{\mu }_1\dots {\mu }_{p+1}}=(p+1){\partial }_{[{\mu }_1}{\omega }_{{\mu }_2\dots {\mu }_{p+1}]}\text{\hspace{1em}(2.27)}$$

Importantly, if we subsequently act with the exterior derivative again, we get

$$d(d\omega )=0$$

because the derivatives are anti-symmetrised and hence vanish. This holds true for any p-form, a fact which is sometimes expressed as

$$d^2=0$$

It can be shown that the exterior derivative satisfies a number of further properties,

$d(\omega \wedge \eta )=d\omega \wedge \eta +(-1{)}^p\omega \wedge$ dη, where $\omega \in {\Lambda }^p(M)$

$d({\phi }^{\ast }\omega )={\phi }^{\ast }(d\omega )$ where ${\phi }^{\ast }$ is the pull-back associated to the map between manifolds, $\phi :M\to N$

• Because the exterior derivative commutes with the pull-back, it also commutes with the Lie derivative. This ensures that we have $d({\mathcal{L}}_X\omega )={\mathcal{L}}_X(d\omega )$

A p-form $\omega$ is said to be closed if $d\omega =0$ everywhere. It is exact if $\omega =d\eta$ everywhere for some η. Because $d^2=0$ , an exact form is necessary closed. The question of when the converse is true is interesting: we’ll discuss this more in Section 2.4.3.

Examples

Suppose that we have a one-form $\omega ={\omega }_{\mu }d{x}^{\mu }$ , the exterior derivative gives a 2-form

$$(d\omega {)}_{\mu \nu }={\partial }_{\mu }{\omega }_{\nu }-{\partial }_{\nu }{\omega }_{\mu }\Rightarrow d\omega =\frac{1}{2}({\partial }_{\mu }{\omega }_{\nu }-{\partial }_{\nu }{\omega }_{\mu })d{x}^{\mu }\wedge d{x}^{\nu }$$

As a specific example of this example, suppose that we take the one-form to live on ${\mathbf{R}}^3$ , with

$$\omega ={\omega }_{1}d{x}^1+{\omega }_{2}d{x}^2+{\omega }_{3}d{x}^3$$

Since this is a field, each of the components ${\omega }_{\mu }$ is a function of $x^1,x^2$ and $x^3$ . The exterior derivative is given by

$$\begin{array}{rl}d\omega & ={\partial }_2{\omega }_{1}d{x}^2\wedge d{x}^1+{\partial }_3{\omega }_{1}d{x}^3\wedge d{x}^1+{\partial }_1{\omega }_{2}d{x}^1\wedge d{x}^2\\ & +{\partial }_3{\omega }_{2}d{x}^3\wedge d{x}^2+{\partial }_1{\omega }_{3}d{x}^1\wedge d{x}^3+{\partial }_2{\omega }_{3}d{x}^2\wedge d{x}^3\\ & =({\partial }_1{\omega }_2-{\partial }_2{\omega }_1)d{x}^1\wedge d{x}^2+({\partial }_2{\omega }_3-{\partial }_3{\omega }_2)d{x}^2\wedge d{x}^3+({\partial }_3{\omega }_1-{\partial }_1{\omega }_3)d{x}^3\wedge d{x}^1\end{array}\text{\hspace{1em}(2.28)}$$

Notice that there’s no term like ${\partial }_1{\omega }_1$ because this would come with a $d{x}^1$ $\wedge d{x}^1=0$

In the olden days (before this course), we used to write vector fields in ${\mathbf{R}}^3$ as $\omega =$ $({\omega }^1,{\omega }^2,{\omega }^3)$ and compute the curl $\nabla \times \omega$ . But the components of the curl are precisely the components that appear in dω. In fact, our “vector” ω was really a one-form and the curl turned it into a two-form. It’s a happy fact that in ${\mathbf{R}}^3$ , vectors, one-forms and two-forms all have three components, which allowed us to conflate them in our earlier courses. (In fact, there is a natural map between them that we will meet in Section 3.)

Suppose instead that we start with a 2-form B in ${\mathbf{R}}^3$ , which we write as

$$B=B_{1}d{x}^2\wedge d{x}^3+B_{2}d{x}^3\wedge d{x}^1+B_{3}d{x}^1\wedge d{x}^2$$

Taking the exterior derivative now gives

$$\begin{array}{rl}& dB={\partial }_1{B}_{1}d{x}^1\wedge d{x}^2\wedge d{x}^3+{\partial }_2{B}_{2}d{x}^2\wedge d{x}^3\wedge d{x}^1+{\partial }_3{B}_{3}d{x}^3\wedge d{x}^1\wedge d{x}^2\\ & =({\partial }_1{B}_1+{\partial }_2{B}_2+{\partial }_3{B}_3)d{x}^1\wedge d{x}^2\wedge d{x}^3\end{array}\text{\hspace{1em}(2.29)}$$

This time there is just a single component, but again it’s something familiar. Had we written the original three components of the two-form in old school vector notation $\mathbf{B}=(B_1,B_2,B_3)$ , then the single component of dB is what we previously called $\nabla \cdot \mathbf{B}$

The Lie Derivative Yet Again

There is yet another operation that we can construct on $p\cdot$ -forms. Given a vector field $X\in \mathfrak{X}(M)$ , we can construct the interior product, a map ${\iota }_X:{\Lambda }^p(\mathcal{M})\to {\Lambda }^{p-1}(\mathcal{M})$ . If $\omega \in {\Lambda }^p(M)$ , we define a ${\iota }_X\omega \in {\Lambda }^{p-1}(M)$ by

$${\iota }_X\omega (Y_1,\dots ,Y_{p-1})=\omega (X,Y_1,\dots ,Y_{p-1})\text{\hspace{1em}(2.30)}$$

In other words, we just put X in the first argument of ω. Acting on functions $f,$ we simply define ${\iota }_{X}f=0$

The anti-symmetry of forms means that ${\iota }_X{\iota }_Y=-{\iota }_Y{\iota }_X$ . Moreover, you can check that

$${\iota }_X(\omega \wedge \eta )={\iota }_X\omega \wedge \eta +(-1{)}^p\omega \wedge {\iota }_X\eta$$

where $\omega \in {\Lambda }^p(M)$

Consider a 1-form ω. There are two different ways to act with ${\iota }_X$ and d to give us back a one-form. These are

$${\iota }_{X}d\omega ={\iota }_X\frac{1}{2}({\partial }_{\mu }{\omega }_{\nu }-{\partial }_{\nu }{\omega }_{\mu })d{x}^{\mu }\wedge d{x}^{\nu }=X^{\mu }{\partial }_{\mu }{\omega }_{\nu }d{x}^{\nu }-X^{\nu }{\partial }_{\mu }{\omega }_{\nu }d{x}^{\mu }$$

and

$$d{\iota }_X\omega =d({\omega }_{\mu }{X}^{\mu })=X^{\mu }{\partial }_{\nu }{\omega }_{\mu }d{x}^{\nu }+{\omega }_{\mu }{\partial }_{\nu }{X}^{\mu }d{x}^{\nu }$$

Adding the two together gives

$$(d{\iota }_X+{\iota }_{X}d)\omega =(X^{\mu }{\partial }_{\mu }{\omega }_{\nu }+{\omega }_{\mu }{\partial }_{\nu }{X}^{\mu })d{x}^{\nu }$$

But this is exactly the same expression we saw previously when computing the Lie derivative (2.22) of a one-form. We learn that

$${\mathcal{L}}_X\omega =(d{\iota }_X+{\iota }_{X}d)\omega \text{\hspace{1em}(2.31)}$$

This expression is sometimes referred to as Cartan’s magic formula. A similar calculation shows that (2.31) holds for any p-form $\omega$

2.4.2 Forms You Know and Love

There are a number of examples of differential forms that you’ve met already, but likely never called them by name.

The Electromagnetic Field

The electromagnetic gauge field $A_{\mu }=\left(\varphi ,\mathbf{A}\right)$ should really be thought of as the components of a one-form on spacetime ${\mathbf{R}}^4$ . (Here I’ve set $c=1.)$ We write

$$A=A_{\mu }(x)d{x}^{\mu }$$

Taking the exterior derivative yields a 2-form $F=dA$ , given by

$$F=\frac{1}{2}{F}_{\mu \nu }d{x}^{\mu }\wedge d{x}^{\nu }=\frac{1}{2}({\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu })d{x}^{\mu }\wedge d{x}^{\nu }$$

But this is precisely the field strength $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ that we met in our lectures on Electromagnetism. The components are the electric and magnetic fields, arranged as

$$F_{\mu \nu }=\left(\begin{array}{cccc}0& -E_1& -E_2& -E_3\\ E_1& 0& B_3& -B_2\\ E_2& -B_3& 0& B_1\\ E_3& B_2& -B_1& 0\end{array}\right)\text{\hspace{1em}(2.32)}$$

By construction, we also have $dF=d^{2}A=0$ . In this context, this is sometimes called the Bianchi identity; it yields two of the four Maxwell equations. In old school vector calculus notation, these are $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{E}+\partial \mathbf{B}/\partial t=0$ . We need a little more structure to get the other two as we will see later in this chapter.

The gauge field A is not unique. Given any function α, we can always shift it by a gauge transformation

$$A\to A+d\alpha \Rightarrow A_{\mu }\to A_{\mu }+{\partial }_{\mu }\alpha$$

This leaves the field strength invariant because $F\to F+d(d\alpha )=F$

Phase Space and Hamilton’s Equations

In classical mechanics, the phase space is a manifold M parameterised by coordinates $(q^i,p_j)$ where $q^i$ are the positions of particles and $p_j$ the momenta. Recall from our lectures on Classical Dynamics that the Hamiltonian $H(q,p)$ is a function on M, and Hamilton’s equations are

$${\dot{q}}^i=\frac{\partial H}{\partial p_i}\text{ and }{\dot{p}}_i=-\frac{\partial H}{\partial q^i}\text{\hspace{1em}(2.33)}$$

Phase space also comes with a structure called a Poisson bracket, defined on a pair of functions f and g as

$$\{f,g\}=\frac{\partial f}{\partial q^j}\frac{\partial g}{\partial p_j}-\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial q^j}$$

Then the time evolution of any function f can be written as

$$\dot{f}=\{f,H\}$$

which reproduces Hamilton’s equations if we take $f=q^i$ or $f=p_i$

Underlying this story is the mathematical structure of forms. The key idea is that we have a manifold M and a function H on M. We want a machinery that turns the function H into a vector field $X_H$ . Particles then follow trajectories in phase space that are integral curves generated by $X_H$

To achieve this, we introduce a symplectic two-form ω on an even-dimensional manifold M. This two form must be closed, $d\omega =0$ , and non-degenerate, which means that the top form $\omega \wedge \cdots \wedge \omega \ne 0$ . We’ll see why we need these requirements as we go along. A manifold M equipped with a symplectic two-form is called a symplectic manifold.

Any 2-form provides a map $\omega :T_p(M)\to T_p^{\ast }(M)$ . Given a vector field $X\in \mathfrak{X}(M)$ we can simply take the interior product with ω to get a one-form ${\iota }_X\omega$ . However, we want to go the other way: given a function H, we can always construct a one-form dH, and we’d like to exchange this for a vector field. We can do this if the map ω : $T_p(M)\to T_p^{\ast }(M)$ is actually an isomorphism, so the inverse exists. This turns out to be true provided that $\omega$ is non-degenerate. In this case, we can define the vector field $X_H$ by solving the equation

$${\iota }_{X_H}\omega =-dH\text{\hspace{1em}(2.34)}$$

If we introduce coordinates $x^{\mu }$ on the manifold, then the component form of this equation is

$$X_H^{\mu }{\omega }_{\mu \nu }=-{\partial }_{\nu }H$$

We denote the inverse as ${\omega }^{\mu \nu }=-{\omega }^{\nu \mu }$ such that ${\omega }^{\mu \nu }{\omega }_{\nu \rho }={\delta }_{\rho }^{\mu }$ . The components of the vector field are then

$$X_H^{\mu }=-{\omega }^{\nu \mu }{\partial }_{\nu }H={\omega }^{\mu \nu }{\partial }_{\nu }H$$

The integral curves generated by $X_H$ obey the differential equation (2.9)

$$\frac{d{x}^{\mu }}{dt}=X_H^{\mu }={\omega }^{\mu \nu }{\partial }_{\nu }H$$

These are the general form of Hamilton’s equations. They reduce to our earlier form (2.33) if we write $x^{\mu }=(q^i,p_j)$ and choose the symplectic form to have block diagonal form ${\omega }^{\mu \nu }=\left(\genfrac{}{}{0ex}{}{01}{-10}\right)$

To define the Poisson structure, we first note that we can repeat the map (2.34) to turn any function f into a vector field $X_f$ obeying ${\iota }_{X_f}\omega =-df$ . But we can then feed these vector fields back into the original 2-form $\omega .$ . This gives us a Poisson bracket,

$$\{f,g\}=\omega (X_g,X_f)=-\omega (X_f,X_g)$$

Or, in components,

$$\{f,g\}={\omega }^{\mu \nu }{\partial }_{\mu }f{\partial }_{\nu }g$$

There are many other ways to write this Poisson bracket structure in invariant form. For example, backtracking through various definitions we find

$$\{f,g\}=-{\iota }_{X_f}\omega (X_g)=df(X_g)=X_g(f)$$

The equation of motion in Poisson bracket structure is then

$$\dot{f}=\{f,H\}=X_H(f)={\mathcal{L}}_{X_H}f$$

which tells us that the Lie derivative along $X_H$ generates time evolution.

We haven’t yet explained why the symplectic two-form must be closed, $d\omega =0$ . You can check that this is needed so that the Poisson bracket obeys the Jacobi identity. Alternatively, it ensures that the symplectic form itself is invariant under Hamiltonian flow, in the sense that ${\mathcal{L}}_{X_H}\omega =0$ . To see this, we use (2.31)

$${\mathcal{L}}_{X_H}\omega =(d{\iota }_{X_H}+{\iota }_{X_H}d)\omega ={\iota }_{X_H}d\omega$$

The second equality follows from the fact that $d{\iota }_{X_H}\omega =-d(dH)=0$ . If we insist that $d\omega =0$ then we find ${\mathcal{L}}_{X_H}\omega =0$ as promised.

Thermodynamics

The state space of a thermodynamic system is a manifold M. For the ideal gas, this is a two-dimensional manifold with coordinates provided by, for example, the pressure p and volume V. More complicated systems can have a higher dimensional state space, but the dimension is always even since, like in classical mechanics, thermodynamical variables come in conjugate pairs.

When we first learn the laws of thermodynamics, we have to make up some strange new notation, đ, which then never rears its head again. For example, the first law of thermodynamics is written as

$$dE=\vec{d}Q+\vec{d}W$$

Here dE is the infinitesimal change of energy in the system. The first law of thermodynamics, as written above, states that this decomposes into the heat flowing into the system $\overrightarrow{dQ}$ and the work done on the system đW.

Why the stupid notation? Well, the energy $E(p,q)$ is a function over the state space M and this means that we can write the change of the energy $\begin{array}{r}dE=\frac{\partial E}{\partial p}dp+\frac{\partial E}{\partial V}dV\end{array}$ But there is no such function $Q(p,q)$ or $W(p,q)$ and, correspondingly, đQ and đW are not exact differentials. Indeed, we have $\#W=-pdV$ and later, after we introduce the second law, we learn that $\mathbf{\Psi \Psi \Psi }dQ=TdS$ , with T the temperature and S the entropy, both of which are functions over M.

This is much more natural in the language of forms. All of the terms in the first law are one-forms. But the transfer of heat, đQ and the work đW are not exact one-forms and so can’t be written as d(something). In contrast, dE is an exact one-form. That’s what the đ notation is really telling us: it’s the way of denoting non-exact one-forms before we had a notion of differential geometry.

The real purpose of the first law of thermodynamics is to define the energy functional E. The $18^{\mathrm{th}}$ century version of the statement is something like: “the amount of work required to change an isolated system is independent of how the work is performed”. A more modern rendering would be: “the sum of the work and heat is an exact one-form”.

2.4.3 A Sniff of de Rham Cohomology

The exterior derivative is a map which squares to zero, $d^2=0$ . It turns out that one can have a lot of fun with such maps. We will now explore a little bit of this fun.

First a repeat of definitions we met already: a p-form is closed if $d\omega =0$ everywhere. A p-form is exact if $\omega =d\eta$ everywhere for some η. Because $d^2=0$ , exact implies closed. However, the converse is not necessarily true. It turns out that the way in which closed forms fail to be exact captures interesting facts about the topology of the underlying manifold.

We’ve met this kind of question before. In electromagnetism, we have a magnetic field B which obeys $\nabla \cdot \mathbf{B}=0$ . We then argue that this means we can write the magnetic field as $\mathbf{B}=\nabla \times \mathbf{A}$ . This is more properly expressed the language of forms. We we saw in the previous section, the magnetic field is really a 2-form

$$B=B_{1}d{x}^2\wedge d{x}^3+B_{2}d{x}^3\wedge d{x}^1+B_{3}d{x}^1\wedge d{x}^2$$

We computed the exterior derivative in (2.29); it is

$$dB=\left({\partial }_1{B}_1+{\partial }_2{B}_2+{\partial }_3{B}_3\right)d{x}^1\wedge d{x}^2\wedge d{x}^3$$

We see that the Maxwell equation $\nabla \cdot \mathbf{B}=0$ is really the statement that B is a closed two-form, obeying $dB=0$ . We also saw in (2.28) if we write $B=dA$ for some one-form $A_i$ , then the components are given by $\mathbf{B}=\nabla \times \mathbf{A}$ . Clearly writing $B=dA$ ensures that $dB=0$ . But when is the converse true? We have the following statement (which we leave unproven)

Theorem (The Poincar´e Lemma): On $M={\mathbf{R}}^n$ , closed implies exact.

Since we’ve spent a lot of time mapping manifolds to ${\mathbf{R}}^n$ , this also has consequence for a general manifold M. It means that if ω is a closed p-form , then in any neighbourhood $\mathcal{O}\subset M$ it is always possible to find a $\eta \in {\Lambda }^{p-1}(M)$ such that $\omega =d\eta$ on $\mathcal{O}.$ . The catch is that it may not be possible to find such an η everywhere on the manifold.

An Example

Consider the one-dimensional manifold $M=\mathbf{R}$ . We can take a one-form $\omega =f(x)dx$ This is always closed because it is a top form. It is also exact. We introduce the function

$$g(x)={\int }_0^{x}d{x}^{\prime }f(x^{\prime })$$

Then $\omega =dg$

Now consider the topologically more interesting one-dimensional manifold ${\mathbf{S}}^1$ , which we can view as the phase $e^{i\theta }\in \mathbf{C}$ . We can introduce the form $\omega =d\theta$ on ${\mathbf{S}}^1$ . The way its written makes it look like its an exact form, but this is an illusion because, as we stressed in Section 2.1, θ is not a good coordinate everywhere on ${\mathbf{S}}^1$ because it’s not single valued. Indeed, it’s simple to see that there is no single-valued function $g(\theta )$ on ${\mathbf{S}}^1$ such that $\omega =dg$ . So on ${\mathbf{S}}^1$ , we can construct a form which, locally, can be written as $d\theta$ but globally cannot be written as d(something). So we have a form that is closed but not exact.

Another Example

On $M={\mathbf{R}}^2$ , the Poincar´e lemma ensures that all closed forms are exact. However, things change if we remove a single point and consider ${\mathbf{R}}^2-\{0,0\}$ . Consider the one-form defined by

$$\omega =-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$

This is not a smooth one-form on ${\mathbf{R}}^2$ because of the divergence at the origin. But removing that point means that $\omega$ becomes acceptable. We can check that $\omega$ is closed,

$$d\omega =-\frac{\partial }{\partial y}\left(\frac{y}{x^2+y^2}\right)dy\wedge dx+\frac{\partial }{\partial x}\left(\frac{x}{x^2+y^2}\right)dx\wedge dy=0$$

where the = 0 follows from a little bit of algebra. $\omega$ is exact if we can find a function $f,$ , defined everywhere on ${\mathbf{R}}^2-\{0,0\}$ such that $\omega =df$ , which means

$$\omega =\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\Rightarrow \frac{\partial f}{\partial x}=-\frac{y}{x^2+y^2}\text{and}\frac{\partial f}{\partial y}=\frac{x}{x^2+y^2}$$

We can certainly integrate these equations; the result is

$$f(x,y)={\tan }^{-1}\left(\frac{y}{x}\right)+\text{ constant }$$

But this is not a smooth function everywhere on ${\mathbf{R}}^2-\{0,0\}$ . This means that we can’t, in fact, write $\omega =df$ for a well defined function on ${\mathbf{R}}^2-\{0,0\}$ . We learn that removing a point makes a big difference: now closed no longer implies exact.

There is a similar story for ${\mathbf{R}}^3$ . Indeed, this is how magnetic monopoles sneak back into physics, despite being forbidden by the Maxwell equation $\nabla \cdot \mathbf{B}=0$ . You can learn more about this in the lectures on Gauge Theory.

Betti Numbers

We denote the set of all closed p-forms on a manifold M as $Z^p(M)$ . Equivalently, $Z^p(M)$ is the kernel of the map d $:{\Lambda }^p(M)\to {\Lambda }^{p+1}(M)$

We denote the set of all exact p-forms on a manifold M as $B^p(M)$ . Equivalently, $B^p(M)$ is the range of $d:{\Lambda }^{p-1}(M)\to {\Lambda }^p(M)$

The $p^{\mathrm{th}}$ de Rham cohomology group is defined to be

$$H^p(M)=Z^p(M)/B^p(M)$$

The quotient here is an equivalence class. Two closed forms $\omega ,{\omega }^{\prime }\in Z^p(M)$ are said to be equivalent if $\omega ={\omega }^{\prime }+\eta$ for some $\eta \in B^p(M)$ . We say that $\omega$ and ${\omega }^{\prime }$ sit in the same equivalence class [ω]. The cohomology group $H^p(M)$ is the set of equivalence classes; in other words, it consists of closed forms mod exact forms.

The Betti numbers $B_p$ of a manifold M are defined as

$$B_p=\dim H^p(M)$$

It turns out that these are always finite. The Betti number $B_0=1$ for any connected manifold. This can be traced to the existence of constant functions which are clearly closed but, because there are no $p=-1$ forms, are not exact. The higher Betti numbers are non-zero only if the manifold has some interesting topology. Finally, the Euler character is defined as the alternating sum of Betti numbers,

$$\chi (M)=\sum _p(-1{)}^p{B}_p\text{\hspace{1em}(2.35)}$$

Here are some simple examples. We’ve already seen that the circle ${\mathbf{S}}^1$ has a closed, non-exact one-form. This means that $B_1=1$ and $\chi =0$ . The sphere ${\mathbf{S}}^n$ has only $B_n=1$ and $\chi =1+(-1{)}^n$ . The torus ${\mathbf{T}}^n$ has $B_p=\left(\genfrac{}{}{0ex}{}{n}{p}\right)$ and $\chi =0$

2.4.4 Integration

We have learned how to differentiate on manifolds by using a vector field X. Now it is time to learn how to integrate. It turns out that the things that we integrate on manifolds are forms.

Integrating over Manifolds

To start, we need to orient ourselves. A volume form, or orientation on a manifold of dimension dim $(M)=n$ is a nowhere-vanishing top form v. Any top form has just a single component and can be locally written as

$$v=v(x)d{x}^1\wedge \dots \wedge d{x}^n$$

where we require $v(x)\ne 0$ . If such a top form exists everywhere on the manifold, then M is said to be orientable.

The orientation is called right-handed if $v(x)>0$ everywhere, and left-handed if $v(x)<0$ everywhere. Given one volume form v, we can always construct another by multiplying by a function, giving $\tilde{v}=fv$ where $f(x)>0$ everywhere or $f(x)<0$ everywhere.

It’s not enough to just write down a volume form with $v(x)\ne 0$ locally. We must also ensure that we can patch these volume forms together over the manifold, without the handedness changing. Suppose that we have two sets of coordinates, $x^{\mu }$ and ${\tilde{x}}^{\mu }$ that overlap on some region. In the new coordinates, the volume form is given by

$$v=v(x)\frac{\partial x^1}{\partial {\tilde{x}}^{{\mu }_1}}d{\tilde{x}}^{{\mu }_1}\wedge \cdots \wedge \frac{\partial x^n}{\partial {\tilde{x}}^{{\mu }_n}}d{\tilde{x}}^{{\mu }_n}=v(x)\det \left(\frac{\partial x^{\mu }}{\partial {\tilde{x}}^{\nu }}\right)d{\tilde{x}}^1\wedge \cdots \wedge d{\tilde{x}}^n$$

which has the same orientation provided

$$\det \left(\frac{\partial x^{\mu }}{\partial {\tilde{x}}^{\nu }}\right)>0\text{\hspace{1em}(2.36)}$$

Non-orientable manifolds cannot be covered by overlapping charts such that (2.36) holds. Examples include the M¨obius strip and real projective space ${\mathbf{RP}}^n$ for n even. (In contrast ${\mathbf{RP}}^n$ is orientable for n odd, and ${\mathbf{CP}}^n$ is orientable for all $n.)$ In these lectures, we deal only with orientable manifolds.

Given a volume form v on M, we can integrate any function $f:M\to \mathbf{R}$ over the manifold. In a chart $\varphi :\mathcal{O}\to U$ , with coordinates $x^{\mu }$ , we have

$${\int }_{\mathcal{O}}fv={\int }_{U}d{x}_1\dots d{x}_{n}f(x)v(x)$$

On the right-hand-side, we’re just doing normal integration over some part of ${\mathbf{R}}^n$ . The volume form is playing the role of a measure, telling us how much to weight various parts of the integral. To integrate over the entire manifold, we divide the manifold up into different regions, each covered by a single chart. We then perform the integral over each region and sum the results.

Integrating over Submanifolds

We don’t have to integrate over the full manifold M. We can integrate over some lower dimensional submanifold.

A manifold Σ with dimension $k<n$ is a submanifold of M if we can find a map $\varphi :\Sigma \to M$ which is one-to-one (which ensures that Σ doesn’t intersect itself in M) and ${\varphi }_{\ast }:T_p(\Sigma )\to T_{\varphi (p)}(M)$ is one-to-one.

We can then integrate a k-form ω on M over a k-dimensional submanifold Σ. We do this by pulling back the k-form to Σ and writing

$${\int }_{\varphi (\Sigma )}\omega ={\int }_{\Sigma }{\varphi }^{\ast }\omega$$

For example, suppose that we have a one-form A living over M. If C is a onedimensional manifold, then we can introduce a map $\sigma :C\to M$ which defines a non-intersecting, one-dimensional curve $\sigma (C)$ which is a submanifold of M. We can then pull-back A onto this curve and integrate to get

$${\int }_{\sigma (C)}A={\int }_C{\sigma }^{\ast }A$$

This probably looks more familiar in coordinates. If the curve traces out a path $x^{\mu }(\tau )$ in M, we have

$${\int }_C{\sigma }^{\ast }A=\int d\tau A_{\mu }(x)\frac{d{x}^{\mu }}{d\tau }$$

But this is precisely the way the worldline of a particle couples to the electromagnetic field, as we previously saw in (1.20).

2.4.5 Stokes’ Theorem

Until now, we have considered only smooth manifolds. There is a slight generalisation that will be useful. We define a manifold with boundary in the same way as a manifold, except the charts map $\varphi :\mathcal{O}\to U$ where U is an open subset of ${\mathbf{R}}^{n+}:=:\{\{x^1,\dots ,x^n\}$ such that $x^n\ge 0\}$ . The boundary has co-dimension 1 and is denoted ∂M: it is the submanifold with coordinates $x^n=0$

Consider a manifold M with boundary ∂M. If the dimension of the manifold is $\dim (M)=n$ then for any $(n-1)$ -form ω, we have the following simple result

$${\int }_{M}d\omega ={\int }_{\partial M}\omega \text{\hspace{1em}(2.37)}$$

This is Stokes’ theorem.

We do not prove Stoke’s theorem here. The proof is fairly tedious, and does not differ greatly from the proofs of other things that you’ve called Stokes’ theorem (or Gauss’ divergence theorem) in the past. (See, for example, the lectures on Vector Calculus.) However, the wonderful thing about (2.37) is the way in which it unifies many results in a single elegant formula. To see this, we simply need to look at a few examples.

The Mother of all Integral Theorems

First, consider $n=1$ with M the interval I. We introduce coordinates $x\in [a,b]$ on the interval. The 0-form $\omega =\omega (x)$ is simply a function and $d\omega =(d\omega /dx)dx$ . In this case, the two sides of Stokes’ theorem can be evaluated to give

$${\int }_{M}d\omega ={\int }_a^b\frac{d\omega }{dx}dx\mathrm{and}{\int }_{\partial M}\omega =\omega (b)-\omega (a)$$

Equating the two, we see that Stokes’ theorem is simply a restatement of the fundamental theorem of calculus.

Next, we take $M\subset {\mathbf{R}}^2$ to be a manifold with boundary. We introduce a one-form with coordinates

$$\omega ={\omega }_{1}d{x}^1+{\omega }_{2}d{x}^2\Rightarrow d\omega =\left(\frac{\partial {\omega }_2}{\partial x^1}-\frac{\partial {\omega }_1}{\partial x^2}\right)d{x}^1\wedge d{x}^2$$

In this case, the ingredients in Stokes’ theorem are

$${\int }_{M}d\omega ={\int }_M\left(\frac{\partial {\omega }_2}{\partial x^1}-\frac{\partial {\omega }_1}{\partial x^2}\right)d{x}^{1}d{x}^2\mathrm{and}{\int }_{\partial M}\omega ={\int }_{\partial M}{\omega }_{1}d{x}^1+{\omega }_{2}d{x}^2$$

Equating the two gives the result usually referred to as Green’s theorem in the plane.

Finally, consider $M\subset {\mathbf{R}}^3$ to be a manifold with boundary, with a 2-form

$$\omega ={\omega }_{1}d{x}^2\wedge d{x}^3+{\omega }_{2}d{x}^3\wedge d{x}^1+{\omega }_{3}d{x}^1\wedge d{x}^2$$

The right-hand-side of Stokes theorem is

$${\int }_{\partial M}{\omega }_{1}d{x}^{2}d{x}^3+{\omega }_{2}d{x}^{3}d{x}^1+{\omega }_{3}d{x}^{1}d{x}^2$$

Meanwhile, we computed the exterior derivative of a 2-form in (2.29). The left-handside of Stokes’ theorem then gives

$${\int }_{M}d\omega ={\int }_M\left({\partial }_1{\omega }_1+{\partial }_2{\omega }_2+{\partial }_3{\omega }_3\right)d{x}^{1}d{x}^{2}d{x}^3$$

This time, equating the two gives us Gauss’ divergence theorem.

We see that Stokes’ theorem, as written in (2.37), is the mother of all integral theorems, packaging many famous results in a single formula. We’ll revisit this in Section 3.2.4 where we relate Stokes’ theorem to a more explicit form of the divergence theorem.

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正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

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来源版本：michaelmas-2019-source-20260923

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