= 0, 0, 1

‚z

(3) Plug the tangent vectors into ω at the point Ψ(θ, z). We get

’ sin θ cos θ cos θ 0

(cos2 θ + sin2 θ) +z

0 0 01

This simpli¬es to the function z cos θ.

6. SUMMARY: HOW TO INTEGRATE A DIFFERENTIAL FORM 55

(4) Integrate the resulting function over R.

1 2π

z cos θ dθ dz

0 0

Note that the integrand comes from Step 3 and the limits of integration come

from Step 1.

CHAPTER 4

Di¬erentiation of Forms.

1. The derivative of a di¬erential 1-form

The goal of this section is to ¬gure out what we mean by the derivative of a

di¬erential form. One way to think about a derivative is as a function which measures

the variation of some other function. Suppose ω is a 1-form on R2 . What do we mean

by the “variation” of ω? One thing we can try is to plug in Vp , a particular vector in

the tangent space at the point, p. We can then look at how ω(Vp ) changes as we vary

p. But p can vary in lots of ways, so we need to pick one. In calculus we learn how

to take another vector, Wp , and use it to vary p. Hence, the derivative of ω, which

we shall denote “dω”, is a function that acts on both Vp and Wp . In other words, it

must be a 2-form!

Vp+tWp

Vp

Wp

t

Figure 1. Using Wp to vary Vp .

Let™s see how to use Wp to calculate the variation in ω(Vp ) in a speci¬c example.

Suppose ω = y dx ’ x2 dy, p = (1, 1), Vp = 1, 2 (1,1) , and Wp = 2, 3 (1,1) . Notice that

57

58 4. DIFFERENTIATION OF FORMS.

Vp+tW = 1, 2 is a vector similar to Vp , but pushed away by t in the direction

(1+2t,1+3t)

of Wp . Hence, the variation of ω(Vp ), in the direction of Wp , can be calculated as

follows:

’ ω( 1, 2

ω( 1, 2 (1+2t,1+3t) ) (1,1) )

lim

t

t’0

[(1 + 3t)(1) ’ (1 + 2t)2 (2)] ’ [(1)(1) ’ (1)2 (2)]

= lim

t

t’0

2

’5t ’ 8t

= ’5

= lim

t

t’0

What about the variation of ω(Wp ) in the direction of Vp ? We calculate:

’ ω( 2, 3

ω( 2, 3 (1+t,1+2t) ) (1,1) )

lim

t

t’0

[(1 + 2t)(2) ’ (1 + t)2 (3)] ’ [(1)(2) ’ (1)2 (3)]

= lim

t

t’0

’2t ’ 3t2

= ’2

= lim

t

t’0

This is a small problem. We want dω to be a 2-form. Hence, dω(Vp , Wp ) should

equal ’dω(Wp , Vp ). How can we use the variations above to de¬ne dω so this is true?

Simple. We just de¬ne it to be the di¬erence in these variations:

ω(Wp+tV ) ’ ω(Wp ) ω(Vp+tW ) ’ ω(Vp)

’ lim

(4) dω(Vp, Wp ) = lim

t t

t’0 t’0

Hence, in the above example, dω( 1, 2 (1,1), 2, 3 (1,1)) = ’2 ’ (’5) = 3.

Before going further we introduce some notation from calculus to make Equation

4 a little more readable. Suppose f is a function from Rn to R and V ∈ Tp Rn . Then

the derivative of f at p, in the direction of V , can be written as “∇V f ”. That is, we

de¬ne

f (p + tV ) ’ f (p)

∇V f = lim

t

t’0

One may also recall from calculus that ∇V f = ∇f (p) · V , where ∇f (p) denotes

the gradient of f evaluated at p. Using this notation, we can rewrite Equation 4 as

dω(Vp, Wp ) = ∇Vp ω(Wp ) ’ ∇Wp ω(Vp )

1. THE DERIVATIVE OF A DIFFERENTIAL 1-FORM 59

There are other ways to determine what dω is than by using Equation 4. Recall

that a 2-form acts on a pair of vectors by projecting them onto each coordinate

plane, calculating the area they span, multiplying by some constant, and adding.

So the 2-form is completely determined by the constants that you multiply by after

projecting. In order to ¬gure out what these constants are we are free to examine

the action of the 2-form on any pair of vectors. For example, suppose we have two

vectors that lie in the x-y plane and span a parallelogram with area 1. If we run

these through some 2-form and end up with the number 5 then we know that the

multiplicative constant for that 2-form, associated with the x-y plane is 5. This, in

turn, tells us that the 2-form equals 5 dx § dy+?. To ¬gure out what “?” is, we can

examine the action of the 2-form on other pairs of vectors.

Let™s try this with a general di¬erential 2-form on R3 . Such a form always looks

like dω = a(x, y, z)dx § dy + b(x, y, z)dy § dz + c(x, y, z)dx § dz. To ¬gure out

what a(x, y, z) is, for example, all we need to do is determine what dω does to the

vectors 1, 0, 0 (x,y,z) and 0, 1, 0 (x,y,z). Let™s compute this using Equation 4, assuming

ω = f (x, y, z)dx + g(x, y, z)dy + h(x, y, z)dz.

dω( 1, 0, 0 (x,y,z), 0, 1, 0 (x,y,z))

’ ω( 0, 1, 0 (x,y,z))

ω( 0, 1, 0 (x+t,y,z))

= lim

t

t’0

ω( 1, 0, 0 (x,y+t,z)) ’ ω( 1, 0, 0 (x,y,z))

’ lim

t

t’0

g(x + t, y, z) f (x, y + t, z)

’ lim

= lim

t t

t’0 t’0

‚g ‚f

(x, y, z) ’

= (x, y, z)

‚x ‚y

Similarly, direct computation shows:

‚h ‚g

(x, y, z) ’

dω( 0, 1, 0 (x,y,z), 0, 0, 1 (x,y,z)) = (x, y, z)

‚y ‚z

and,

‚h ‚f

(x, y, z) ’

dω( 1, 0, 0 (x,y,z), 0, 0, 1 (x,y,z)) = (x, y, z)

‚x ‚z

Hence, we conclude that

60 4. DIFFERENTIATION OF FORMS.

‚g ‚f ‚h ‚g ‚h ‚f

’ )dx § dy + ( ’ )dy § dz + ( ’ )dx § dz

dω = (

‚x ‚y ‚y ‚z ‚x ‚z

Exercise 4.1. Suppose ω = f (x, y)dx + g(x, y)dy is a 1-form on R2 . Show that

dω = ( ‚x ’ ‚f )dx § dy.

‚g

‚y

Exercise 4.2. Suppose ω = xy 2 dx + x3 z dy ’ (y + z 9 ) dz, V = 1, 2, 3 (2,3,’1) , and

W = ’1, 0, 1 (2,3,’1)

(1) Compute ∇V ω(W ) and ∇W ω(V ), at the point (2, 3, ’1).

(2) Use your answer to the previous question to compute dω(V, W ).

Exercise 4.3. If ω = y dx ’ x2 dy, ¬nd dω. Verify that dω( 1, 2 (1,1) , 2, 3 (1,1) ) = 3.

2. Derivatives of n-forms

Before jumping to the general case let™s look at the derivative of a 2-form. A

2-form, ω, acts on a pair of vectors, Vp and Wp , to return some number. To ¬nd

some sort of variation of ω we can vary the vectors Vp and Wp and examine how

ω(Vp , Wp ) varies. As in the last section one way to vary a vector is to push it in the

direction of some other vector, Up . Hence, whatever dω turns out to be, it will be a

function of the vectors Up , Vp , and Wp . So, we would like to de¬ne it to be a 3-form.

Let™s start by looking at the variation of ω(Vp , Wp ) in the direction of Up . We write

this as ∇Up ω(Vp , Wp ). If we were to de¬ne this as the value of dω(Up , Vp , Wp ) we would

¬nd that in general it would not be alternating. That is, usually ∇Up ω(Vp , Wp ) =

’∇Vp ω(Up , Wp ). To remedy this, we simply de¬ne dω to be the alternating sum of

all the variations:

dω(Up , Vp , Wp ) = ∇Up ω(Vp , Wp ) ’ ∇Vp ω(Up , Wp ) + ∇Wp ω(Up , Vp )

We leave it to the reader to check that dω is alternating and multilinear.

It shouldn™t be hard for the reader to now jump to the general case. Suppose ω

is an n-form and Vp1 , ..., Vpn+1 are n + 1 vectors. Then we de¬ne

n+1

dω(Vp1 , ..., Vpn+1 ) (’1)i+1 ∇Vpi ω(Vp1 , ..., Vpi’1 , Vpi+1 , ..., Vpn+1 )

=

i=1

3. INTERLUDE: 0-FORMS 61

In other words, dω, applied to n+1 vectors, is the alternating sum of the variation

of ω applied to n of those vectors in the direction of the remaining one. Note that

we can think of “d” as an operator which takes n-forms to (n + 1)-forms.

Exercise 4.4. dω is alternating and multilinear.

Exercise 4.5. Suppose ω = f (x, y, z) dx § dy + g(x, y, z) dy § dz + h(x, y, z) dx § dz.

Find dω (Hint: Compute dω( 1, 0, 0 , 0, 1, 0 , 0, 0, 1 )). Compute d(x2 y dx § dy +

y 2 z dy § dz).

3. Interlude: 0-forms

Let™s go back to Section 1, when we introduced coordinates for vectors. At that

time we noted that if C was the graph of the function y = f (x) and p was a point of

C then the tangent line to C at p lies in Tp R2 and has equation dy = m dx, for some

constant, m. Of course, if p = (x0 , y0 ) then m is just the derivative of f evaluated

at x0 .

Now, suppose we had looked at the graph of a function of 2-variables, z = f (x, y),

instead. At some point, p = (x0 , y0 , z0 ), on the graph we could look at the tangent

plane, which lies in Tp R3 . It™s equation is dz = m1 dx + m2 dy. Since z = f (x, y),

m1 = ‚f (x0 , y0 ), and m2 = ‚f (x0 , y0 ), we can rewrite this as

‚x ‚y

‚f ‚f

df = dx + dy.

‚x ‚y

Notice that the right-hand side of this equation is a di¬erential 1-form. This is a

bit strange; we applied the “d” operator to something and the result was a 1-form.

However, we know that when we apply the “d” operator to a di¬erential n-form we

get a di¬erential (n + 1)-form. So, it must be that f (x, y) is a di¬erential 0-form on

R2 !

In retrospect, this should not be so surprising. After all, the input to a di¬erential

n-form on Rm is a point, and n vectors based at that point. So, the input to a

di¬erential 0-form should be a point of Rm , and no vectors. In other words, a 0-form

on Rm is just another word for a real-valued function on Rm .

Let™s extend some of the things we can do with forms to 0-forms. Suppose f is

a 0-form, and ω is an n-form (where n may also be 0). What should we mean by

f § ω? Since the wedge product of an n-form and an m-form is an (n + m)-form, it

62 4. DIFFERENTIATION OF FORMS.

must be that f § ω is an n form. It™s hard to think of any other way to de¬ne this

as just the product, f ω.

What about integration? Remember that we integrate n-forms over subsets of

Rm that can be parameterized by a subset of Rn . So 0-forms get integrated over

things parameterized by R0 . In other words, we integrate a 0-form over a point. How

do we do this? We do the simplest possible thing; de¬ne the value of a 0-form, f ,

integrated over the point, p, to be ±f (p). To specify an orientation we just need to

say whether or not to use the ’ sign. We do this just by writing “’p” instead of