One word of caution here...beware of orientations! If p ∈ Rn then we use the

notation “’p” to denote p with the negative orientation. So if p = ’3 ∈ R1 then ’p

is not the same as the point, 3. ’p is just the point, ’3, with a negative orientation.

So, if f (x) = x2 then ’p f = ’f (p) = ’9.

Exercise 4.6. If f is the 0-form x2 y 3 , p is the point (’1, 1), q is the point (1, ’1),

and r is the point (’1, ’1), then compute the integral of f over the points ’p, ’q, and

’r, with the indicated orientations.

Let™s go back to our exploration of derivatives of n-forms. Suppose f (x, y) dx is

a 1-form on R2 . Then we have already shown that d(f dx) = ‚f dy § dx. We now

‚y

compute:

‚f ‚f

df § dx = dy § dx

dx +

‚x ‚y

‚f ‚f

dx § dx + dy § dx

=

‚x ‚y

‚f

dy § dx

=

‚y

= d(f dx)

Exercise 4.7. If ω is an n-form, and f is a 0-form, then d(f dω) = df § dω.

Exercise 4.8. d(dω) = 0.

Exercise 4.9. If ω is an n-form, and µ is an m-form, then d(ω § µ) = dω § µ +

(’1)n ω § dµ.

4. ALGEBRAIC COMPUTATION OF DERIVATIVES 63

4. Algebraic computation of derivatives

In this section we break with the spirit of the text brie¬‚y. At this point we have

amassed enough algebraic identities that computing derivatives of forms can become

quite routine. In this section we quickly summarize these identities and work a few

examples.

4.1. Identities involving § only. Let ω be an n-form and ν be an m-form.

ω§ω = 0

ω § ν = (’1)nm ν § ω

ω § (ν + ψ) = ω § ν + ω § ψ

(ν + ψ) § ω = ν § ω + ψ § ω

4.2. Identities involving “d”. Let ω be an n-form, µ an m-form, and f a

0-form.

d(dω) = 0

d(ω + µ) = dω + dµ

d(ω § µ) = dω § µ + (’1)n ω § dµ

d(f dω) = df § dω

‚f ‚f ‚f

df = dx1 + dx2 + ... + dxn

‚x1 ‚x2 ‚xn

4.3. Some examples.

Example 4.1.

d xy dx ’ xy dy + xy 2 z 3 dz

64 4. DIFFERENTIATION OF FORMS.

= d(xy) § dx ’ d(xy) § dy + d(xy 2 z 3 ) § dz

= (y dx + x dy) § dx ’ (y dx + x dy) § dy

+(y 2z 3 dx + 2xyz 3 dy + 3xy 2 z 2 dz) § dz

= + x dy § dx ’ y dx § dy ’

y dx § dx x dy § dy

(

+y 2z 3 dx § dz + 2xyz 3 dy § dz + ((2z(dz( dz

3xy (2 ( §

(

= x dy § dx ’ y dx § dy + y 2z 3 dx § dz + 2xyz 3 dy § dz

= ’x dx § dy ’ y dx § dy + y 2 z 3 dx § dz + 2xyz 3 dy § dz

= (’x ’ y) dx § dy + y 2 z 3 dx § dz + 2xyz 3 dy § dz

Example 4.2.

d x2 (y + z 2 ) dx § dy + z(x3 + y) dy § dz

= d(x2 (y + z 2 )) § dx § dy + d(z(x3 + y)) § dy § dz

= 2x2 z dz § dx § dy + 3x2 z dx § dy § dz

= 5x2 z dx § dy § dz

Exercise 4.10. For each di¬erential n-form, ω, ¬nd dω.

(1) sin y dx + cos x dy

(2) xy 2 dx + x3 z dy ’ (y + z 9 ) dz

(3) xy 2 dy § dz + x3 z dx § dz ’ (y + z 9 ) dx § dy

(4) x2 y 3 z 4 dx § dy § dz

Exercise 4.11. If f is the 0-form x2 y 3 and ω is the 1-form x2 z dx + y 3z 2 dy (on

R3 ) then use the identity d(f dω) = df § dω to compute d(f dω). Answer: (3x4 y 2 ’

2xy 6 z)dx § dy § dz.

Exercise 4.12. Let f, g, and h be functions from R3 to R. If ω = f dy § dz ’ g dx §

dz + h dx § dy then compute dω.

CHAPTER 5

Stokes™ Theorem

1. Cells and Chains

Up until now we have not been very speci¬c as to the types of subsets of Rm on

which one integrates a di¬erential n-form. All we have needed is a subset that can

be parameterized by a region in Rn . To go further we need to specify what types of

regions.

Definition. Let I = [0, 1]. An n-cell, σ, is the image of di¬erentiable map,

φ : I n ’ Rm , with a speci¬ed orientation. We denote the same cell with opposite

orientation as ’σ. We de¬ne a 0-cell to be a point of Rm .

Example 5.1. Suppose g1 (x) and g2 (x) are functions such that g1 (x) < g2 (x)

for all x ∈ [a, b]. Let R denote the subset of R2 bounded by the graphs of the

equations y = g1 (x) and y = g2 (x), and by the lines x = a and x = b. In

Example 3.8 we show that R is a 2-cell (assuming the induced orientation).

We would like to treat cells as algebraic objects which can be added and sub-

tracted. But if σ is a cell it may not at all be clear what “2σ” represents. One way

to think about it is as two copies of σ, placed right on top of each other.

Definition. An n-chain is a formal linear combination of n-cells.

As one would expect, we assume the following relations hold:

σ’σ = …

nσ + mσ = (n + m)σ

σ+„ =„ +σ

65

66 5. STOKES™ THEOREM

You can probably guess what the integral of an n-form, ω, over an n-chain is.

Suppose C = ni σi . Then we de¬ne

ω= ni ω

i σi

C

Exercise 5.1. If f is the 0-form x2 y 3 , p is the point (’1, 1), q is the point (1, ’1),

and r is the point (’1, ’1), then compute the integral of f over the following 0-chains:

(1) p ’ q; r ’ p

(2) p + q ’ r

Another concept that will be useful for us is the boundary of an n-chain. As a

warm-up, we de¬ne the boundary of a 1-cell. Suppose σ is the 1-cell which is the

image of φ : [0, 1] ’ Rm with the induced orientation. Then we de¬ne the boundary

of σ (which we shall denote “‚σ”) as the 0-chain, φ(1) ’ φ(0). In general, if the n-cell

σ is the image of the parameterization φ : I n ’ Rm with the induced orientation

then

n

(’1)i+1 φ|(x1 ,...,xi’1,1,xi+1 ,...,xn) ’ φ|(x1 ,...,xi’1,0,xi+1 ,...,xn)

‚σ =

i=1

So, if σ is a 2-cell then

‚σ = (φ(1, x2 ) ’ φ(0, x2 )) ’ (φ(x1 , 1) ’ φ(x1 , 0))

= φ(1, x2 ) ’ φ(0, x2 ) ’ φ(x1 , 1) + φ(x1 , 0)

If σ is a 3-cell then

‚σ = (φ(1, x2 , x3 ) ’ φ(0, x2 , x3 )) ’ (φ(x1 , 1, x3 ) ’ φ(x1 , 0, x3 ))

+ (φ(x1 , x2 , 1) ’ φ(x1 , x2 , 0))

= φ(1, x2 , x3 ) ’ φ(0, x2 , x3 ) ’ φ(x1 , 1, x3 ) + φ(x1 , 0, x3 )

+φ(x1 , x2 , 1) ’ φ(x1 , x2 , 0)

An example will hopefully clear up the confusion this all was sure to generate:

2. PULL-BACKS 67

θ y

x

r

Figure 1. Orienting the boundary of a 2-cell.

Example 5.2. Suppose φ(r, θ) = (r cos πθ, r sin πθ). The image of φ is the

2-cell, σ, depicted in Figure 1. By the above de¬nition,

‚σ = (φ(1, θ) ’ φ(0, θ)) ’ (φ(r, 1) ’ φ(r, 0))

= (cos πθ, sin πθ) ’ (0, 0) + (r, 0) ’ (’r, 0)

This is the 1-chain depicted in Figure 1.

Finally, we are ready to de¬ne what we mean by the boundary of an n-chain. If

C = ni σi , then we de¬ne ‚C = ni ‚σi .

Example 5.3. Suppose

√

1 ’ r 2 ),

φ1 (r, θ) = (r cos 2πθ, r sin 2πθ,

√

φ2 (r, θ) = (’r cos 2πθ, r sin 2πθ, ’ 1 ’ r 2 ),

σ1 = Im(φ1 ) and σ2 = Im(φ2 ). Then σ1 + σ2 is a sphere in R3 . One can check

that ‚(σ1 + σ2 ) = ….

Exercise 5.2. If σ is an n-cell then ‚‚σ = ….

2. Pull-backs

Before getting to the central theorem of the text we need to introduce one more

concept. Let™s reexamine Equation 3:

‚φ ‚φ

ω=± (x1 , ...xn ) dx1 § ... § dxn

ωφ(x1 ,...,xn) (x1 , ...xn ), ...,

‚x1 ‚xn

M R

68 5. STOKES™ THEOREM

The form in the integrand on the right was de¬ned so as to integrate to give

the same answer as the form on the left. This is what we would like to generalize.

Suppose φ : Rn ’ Rm is a parameterization, and ω is a k-form on Rm . We de¬ne

the pull-back of ω under φ to be the form on Rn which gives the same integral over

any k-cell, σ, as ω does when integrated over φ(σ). Following convention, we denote

the pullback of ω under φ as “φ—ω”.

So how do we decide how φ— ω acts on a k-tuple of vectors in Tp Rn ? The trick is

to use φ to translate the vectors to a k-tuple in Tφ(p) Rm , and then plug them into ω.

The matrix Dφ, whose columns are the partial derivatives of φ, is an n — m matrix.

This matrix acts on vectors in Tp Rn , and returns vectors in Tφ(p) Rm . So, we de¬ne

(see Figure 2):

φ— ω(Vp1 , ..., Vpk ) = ω(Dφ(Vp1 ), ..., Dφ(Vpk ))

Tp Rn Tp Rm

Dφ

Vp1 Dφ(Vp2 )

Dφ(Vp1 )

Vp2

φ— ω ω