Figure 2. De¬ning φ— ω.

Example 5.4. Suppose ω = y dx + z dy + x dz is a 1-form on R3 , and

φ(a, b) = (a + b, a ’ b, ab) is a map from R2 to R3 . Then φ— ω will be a 1-form on

2. PULL-BACKS 69

R2 . To determine which one, we can examine how it acts on the vectors 1, 0 (a,b)

and 0, 1 (a,b) .

φ— ω( 1, 0 (a,b) ) = ω(Dφ( 1, 0 (a,b)))

«®

1 1

1

= ω ° 1 ’1 »

0

b a (a,b)

= ω( 1, 1, b (a+b,a’b,ab) )

= (a ’ b) + ab + (a + b)b

= a ’ b + 2ab + b2

Similarly,

φ— ω( 0, 1 = ω( 1, ’1, a

(a,b) ) (a+b,a’b,ab) )

= (a ’ b) ’ ab + (a + b)a

= a ’ b + a2

Hence,

φ— ω = (a ’ b + 2ab + b2 ) da + (a ’ b + a2 ) db

Exercise 5.3. If ω = x2 dy § dz + y 2dz § dw is a 2-form on R4 , and φ(a, b, c) =

(a, b, c, abc), then what is φ— ω?

Exercise 5.4. If ω is an n-form on Rm and φ : Rn ’ Rm , then

‚φ ‚φ

φ— ω = ωφ(x1 ,...,xn) (x1 , ...xn ) dx1 § ... § dxn

(x1 , ...xn ), ...,

‚x1 ‚xn

In light of the preceding exercise Equation 3 can be re-written as

φ— ω

ω=

M R

70 5. STOKES™ THEOREM

Exercise 5.5. If σ is a k-cell in Rn , φ : Rn ’ Rm , and ω is a k-form on Rm then

φ— ω = ω

σ φ(σ)

Exercise 5.6. If φ : Rn ’ Rm and ω is a k-form on Rm then d(φ— ω) = φ— (dω).

3. Stokes™ Theorem

In calculus we learn that when you take a function, di¬erentiate it, and then

integrate the result, something special happens. In this section we explore what

happens when we take a form, di¬erentiate it, and then integrate the resulting form

over some chain. The general argument is quite complicated, so we start by looking

at forms of a particular type integrated over very special regions.

Suppose ω = a dx2 § dx3 is a 2-form on R3 , where a : R3 ’ R. Let R be the unit

cube, I 3 ‚ R3 . We would like to explore what happens when we integrate dω over

‚a

R. Note ¬rst that Exercise 4.7 implies that dω = ‚x1 dx1 § dx2 § dx3 .

Recall the steps used to de¬ne dω:

R

(1) Choose a lattice of points in R, {pi,j,k }. Since R is a cube, we can choose

this lattice to be rectangular.

1 2 3

(2) De¬ne Vi,j,k = pi+1,j,k ’ pi,j,k . De¬ne Vi,j,k and Vi,j,k similarly.

1 2 2

(3) Compute dωpi,j,k (Vi,j,k , Vi,j,k , Vi,j,k ).

(4) Sum over all i, j and k.

(5) Take the limit as the maximal distance between adjacent lattice points goes

to 0.

Let™s focus on Step 3 for a moment. Let t be the distance between pi+1,j,k and pi,j,k ,

)’a(pi,j,k )

a(p

‚a i+1,j,k

and assume t is small. Then ‚x1 (pi,j,k ) is approximately equal to .

t

This approximation gets better and better when we let t ’ 0, in Step 5.

1 3 1

The vectors, Vi,j,k through Vi,j,k , form a little cube. If we say the vector Vi,j,k

is “vertical”, and the other two are horizontal, then the “height” of this cube is t,

2 3

and the area of its base is dx2 § dx3 (Vi,j,k , Vi,j,k ), which makes its volume t dx2 §

2 3

dx3 (Vi,j,k , Vi,j,k ). Putting all this together, we ¬nd that

3. STOKES™ THEOREM 71

‚a

1 2 2 1 2 2

dx1 § dx2 § dx3 (Vi,j,k , Vi,j,k , Vi,j,k )

dωpi,j,k (Vi,j,k , Vi,j,k , Vi,j,k ) =

‚x1

a(pi+1,j,k ) ’ a(pi,j,k ) 2 3

≈ t dx2 § dx3 (Vi,j,k , Vi,j,k )

t

2 3 2 3

= ω(Vi+1,j,k , Vi+1,j,k ) ’ ω(Vi,j,k , Vi,j,k )

Let™s move on to Step 4. Here we sum over all i, j and k. Suppose for the moment

that i ranges between 1 and N. First, we ¬x j and k, and sum over all i. The result

2 3 2 3 2 3

is ω(VN,j,k , VN,j,k ) ’ ω(V1,j,k , V1,j,k ). Now notice that ω(VN,j,k , VN,j,k ) is a Riemann

j,k

2 3

sum for the integral of ω over the “top” of R, and ω(V1,j,k , V1,j,k ) is a Riemann

j,k

sum for ω over the “bottom” of R. Lastly, note that ω, evaluated on any pair of

vectors which lie in the sides of the cube, gives 0. Hence, the integral of ω over a

side of R is 0. Putting all this together, we conclude:

(5) dω = ω

R ‚R

Exercise 5.7. Prove that Equation 5 holds if ω = b dx1 § dx3 , or if ω = c dx1 § dx2 .

Caution! Beware of signs and orientations.

Exercise 5.8. Use the previous problem to conclude that if ω = a dx2 § dx3 + b dx1 §

dx3 + c dx1 § dx2 is an arbitrary 2-form on R3 then Equation 5 holds.

Exercise 5.9. If ω is an arbitrary (n ’ 1)-form on Rn and R is the unit cube in Rn

then show that Equation 5 still holds.

This exercise prepares us to move on to the general case. Suppose σ is an n-cell

in Rm , φ : I n ‚ Rn ’ Rm is a parameterization of σ, and ω is an (n ’ 1)-form on

Rm . Then we can combine Exercises 5.5, 5.6, and 5.9 to give us

φ— ω = d(φ—ω) = φ— (dω) =

ω= ω= dω = dω

In In

‚I n

φ(‚I n ) φ(I n ) σ

‚σ

In general, this implies that if C = ni σi is an n-chain, then

72 5. STOKES™ THEOREM

ω= dω

C

‚C

This equation is called the Generalized Stokes™ Theorem. It is unquestionably the

most crucial result of this text. In fact, everything we have done up to this point

has been geared toward developing this equation and everything that follows will be

applications of this equation.

Example 5.5. Let ω = x dy be a 1-form on R2 . Let σ be the 2-cell which is

the image of the parameterization φ(r, θ) = (r cos θ, r sin θ), where 0 ¤ r ¤ R

and 0 ¤ θ ¤ 2π. By the Generalized Stokes™ Theorem,

dx dy = Area(σ) = πR2

dx § dy =

ω= dω =

σ σ σ

‚σ

ω = πR2

Exercise 5.10. Verify directly that

‚σ

Example 5.6. Let ω = x dy + y dx be a 1-form on R2 , and let σ be any 2-cell.

Then ω = dω = 0

σ

‚σ

Exercise 5.11. Find a 1-chain in R2 which bounds a 2-cell and integrate the form

x dy + y dx over this curve.

Example 5.7. Let C be the curve in R2 parameterized by φ(t) = (t2 , t3 ), where

’1 ¤ t ¤ 1. Let f be the 0-form x2 y. We use the Generalized Stokes Theorem

to calculate df .

C

The curve C goes from the point (1,-1) , when t = ’1, to the point (1,1), when

t = 1. Hence, ‚C is the 0-chain (1, 1) ’ (1, ’1). Now we use Stokes:

x2 y = 1 ’ (’1) = 2

df = f=

C ‚C (1,1)’(1,’1)

3. STOKES™ THEOREM 73

Exercise 5.12. Calculate df directly.

C

Example 5.8. Let ω = (x2 + y)dx + (x ’ y 2 )dy be a 1-form on R2 . We wish

to integrate ω over σ, the top half of the unit circle. First, note that dω = 0,

so that if we integrate ω over the boundary of any 2-cell, we would get 0. Let

„ denote the line segment connecting (-1,0) to (1,0). Then the 1-chain σ ’ „

bounds a 2-cell. So ω = 0, which implies that ω= ω. This latter integral

σ’„ σ „

is a bit easier to compute. Let φ(t) = (t, 0) be a parameterization of „ , where

’1 ¤ t ¤ 1. Then

1

2

t2 dt =

ω= ω= ω(t,0) ( 1, 0 ) dt =

3

’1

σ „ [’1,1]

Exercise 5.13. Let ω = (x + y 3 ) dx + 3xy 2 dy be a di¬erential 1-form on R2 . Let Q

be the rectangle {(x, y)|0 ¤ x ¤ 3, 0 ¤ y ¤ 2}.

(1) Compute dω.

(2) Use the generalized Stokes Theorem to compute ω.

‚Q

(3) Compute ω directly, by integrating ω over each each edge of the boundary of

‚Q

the rectangle, and then adding in the appropriate manner. Answer: If L, R, T ,

and B represent the 1-cells that are the left, right, top, and bottom of Q then

1 1

ω’ ω’ ω = 24 ’ 0 ’ 28 + 4 = 0

ω= ω= ω+

2 2

R L T B

‚Q (R’L)’(T ’B)

(4) How does ω compare to ω?

R’T ’L B

(5) Let S be any curve in the upper half plane (i.e., the set {(x, y)|y ≥ 0}) that

connects the point (0, 0) to the point (3, 0). What is ω? Why?