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for a ļ¬xed p. We can make this point a little clearer by re-examining the above

integrand. Note that it is of the form f (ā)Area(Ā·, Ā·). For a ļ¬xed point, ā, of R3 this

is an operator on Tā R3 Ć— Tā R3 , much like a 2-form is.

But so far all we have done is to deļ¬ne 2-forms at ļ¬xed points of R3 . To really

generalize the above integral we have to start considering entire families of 2-forms,

Ļp : Tp R3 Ć— Tp R3 ā’ R, where p ranges over all of R3 . Of course, for this to be useful

weā™d like such a family to have some ānicenessā properties. For one thing, we would

like it to be continuous. That is, if p and q are close then Ļp and Ļq should be similar.

An even stronger property that we will insist on is that the family, Ļp , is diļ¬er-

entiable. To see what this means recall that for a ļ¬xed p, a 2-form Ļp can always

be written as ap dx ā§ dy + bp dy ā§ dz + cp dx ā§ dz, where ap , bp , and cp are constants.

But if we let our choice of p vary over all of R3 then so will these constants. In

other words, ap , bp and cp are all functions from R3 to R. To say that the family,

Ļp , is diļ¬erentiable we mean that each of these functions is diļ¬erentiable. If Ļp is

33

34 3. DIFFERENTIAL FORMS

diļ¬erentiable then we will refer to it as a diļ¬erential form. When there can be no

confusion we will suppress the subscript, p.

Example 3.1. Ļ = x2 y dxā§dyā’xz dyā§dz is a diļ¬erential 2-form on R3 . On the

space T(1,2,3) R3 it is just the 2-form 2dxā§dy ā’3dy ā§dz. We will denote vectors in

T(1,2,3) R3 as dx, dy, dz (1,2,3). Hence, the value of Ļ( 4, 0, ā’1 (1,2,3), 3, 1, 2 (1,2,3))

is the same as the 2-form, 2dx ā§ dy + dy ā§ dz, evaluated on the vectors 4, 0, ā’1

and 3, 1, 2 , which we compute:

Ļ( 4, 0, ā’1 , 3, 1, 2 (1,2,3) )

(1,2,3)

= 2dx ā§ dy ā’ 3dy ā§ dz ( 4, 0, ā’1 , 3, 1, 2 )

0 ā’1

40

ā’3

=2 =5

31 1 2

Suppose Ļ is a diļ¬erential 2-form on R3 . What does Ļ act on? It takes a pair

of vectors at each point of R3 and returns a number. In other words, it takes two

vector ļ¬elds and returns a function from R3 to R. A vector ļ¬eld is simply a choice

of vector in Tp R3 , for each p ā R3 . In general, a diļ¬erential n-form on Rm acts on n

vector ļ¬elds to produce a function from Rm to R (see Fig. 1).

6 0

21

Ļ ā

9 ā’3

7

2 Ļ

3

Figure 1. A diļ¬erential 2-form, Ļ, acts on a pair of vector ļ¬elds, and

returns a function from Rn to R.

2. INTEGRATING DIFFERENTIAL 2-FORMS 35

is a vector ļ¬eld on R3 . For example, it

Example 3.2. V1 = 2y, 0, ā’x (x,y,z)

contains the vector 4, 0, ā’1 ā T(1,2,3) R3 . If V2 = z, 1, xy and Ļ is the

(x,y,z)

diļ¬erential 2-form, x2 y dx ā§ dy ā’ xz dy ā§ dz, then

Ļ(V1, V2 ) = x2 y dx ā§ dy ā’ xz dy ā§ dz( 2y, 0, x (x,y,z) , z, 1, xy (x,y,z))

0 ā’x

2y 0

= x2 y = 2x2 y 2 ā’ x2 z,

ā’ xz

z1 1 xy

which is a function from R3 to R.

Notice that V2 contains the vector 3, 1, 2 (1,2,3). So, from the previous example

we would expect that 2x2 y 2 ā’ x2 z equals 5 at the point (1, 2, 3), which is indeed

the case.

Exercise 3.1. Let Ļ be the diļ¬erential 2-form on R3 given by

Ļ = xyz dx ā§ dy + x2 z dy ā§ dz ā’ y dx ā§ dz

Let V1 and V2 be the following vector ļ¬elds:

V1 = y, z, x2 (x,y,z) , V2 = xy, xz, y (x,y,z)

(1) What vectors do V1 and V2 contain at the point (1, 2, 3)?

(2) Which 2-form is Ļ on T(1,2,3) R3 ?

(3) Use your answers to the previous two questions to compute Ļ(V1, V2 ) at the

point (1, 2, 3).

(4) Compute Ļ(V1 , V2 ) at the point (x, y, z). Then plug in x = 1, y = 2, and z = 3

to check your answer against the previous question.

2. Integrating Diļ¬erential 2-Forms

Let us now examine more closely integration of functions on subsets of R2 , which

you learned in calculus. Suppose R ā‚ R2 and f : R ā’ R. How did we learn to deļ¬ne

the integral of f over R? We summarize the procedure in the following steps:

(1) Choose a lattice of points in R, {(xi , yj )}.

1 2

(2) For each i, j deļ¬ne Vi,j = (xi+1 , yj ) ā’ (xi , yj ) and Vi,j = (xi , yj+1) ā’ (xi , yj )

(See Fig. 2). Notice that Vi,j and Vi,j are both vectors in T(xi ,yj ) R2 .

1 2

36 3. DIFFERENTIAL FORMS

1 2

(3) For each i, j compute f (xi , yj )Area(Vi,j , Vi,j ), where Area(V, W ) is the func-

tion which returns the area of the parallelogram spanned by the vectors V

and W .

(4) Sum over all i and j.

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

to 0. This is the number that we deļ¬ne to be the value of f dx dy.

R

2

Vi,j

yj

1

Vi,j

xi

Figure 2. The steps toward integration.

1 2

Letā™s focus on Step 3. Here we compute f (xi , yj )Area(Vi,j , Vi,j ). Notice that this

is exactly the value of the diļ¬erential 2-form Ļ = f (x, y)dx ā§ dy, evaluated on the

1 2

vectors Vi,j and Vi,j at the point (xi , yj ). Hence, in step 4 we can write this sum

1 2 1 2

as f (xi , yj )Area(Vi,j , Vi,j ) = Ļ(xi ,yj ) (Vi,j , Vi,j ). It is reasonable, then, to

i j i j

adopt the shorthand ā Ļā to denote the limit in Step 5. The upshot of all this is

R

the following:

If Ļ = f (x, y)dx ā§ dy then Ļ= f dx dy.

R R

Since all diļ¬erential 2-forms on R2 are of the form f (x, y)dx ā§ dy we now know

how to integrate them.

2. INTEGRATING DIFFERENTIAL 2-FORMS 37

CAUTION! When integrating 2-forms on R2 it is tempting to always drop the

āā§ā and forget you have a diļ¬erential form. This is only valid with dx ā§ dy. It is

NOT valid with dy ā§ dx. This may seem a bit curious since

f dx ā§ dy = f dx dy = f dy dx

All of these are equal to ā’ f dy ā§ dx.

Exercise 3.2. Let Ļ = xy 2 dx ā§ dy be a diļ¬erential 2-form on R2 . Let D be the region

of R2 bounded by the graphs of x = y 2 and y = x ā’ 6. Calculate Ļ. Answer: 189.

D

What about integration of diļ¬erential 2-forms on R3 ? As remarked at the end of

Section 5 we do this only over those subsets of R3 which can be parameterized by

subsets of R2 . Suppose M is such a subset, like the top half of the unit sphere. To

deļ¬ne what we mean by Ļ we just follow the steps above:

M

(1) Choose a lattice of points in M, {pi,j }.

1 2 1

(2) For each i, j deļ¬ne Vi,j = pi+1,j ā’ pi,j and Vi,j = pi,j+1 ā’ pi,j . Notice that Vi,j

and Vi,j are both vectors in Tpi,j R3 (see Fig. 3).

2

1 2

(3) For each i, j compute Ļpi,j (Vi,j , Vi,j ).

(4) Sum over all i and j.

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

to 0. This is the number that we deļ¬ne to be the value of Ļ.

M

Unfortunately these steps arenā™t so easy to follow. For one thing, itā™s not always

clear how to pick the lattice in Step 1. In fact there is an even worse problem. In

1 2 2 1 1

Step 3 why did we compute Ļpi,j (Vi,j , Vi,j ) instead of Ļpi,j (Vi,j , Vi,j )? After all, Vi,j

and Vi,j are two randomly oriented vectors in T R3i,j . There is no reasonable way to

2

p

decide which should be ļ¬rst and which second. There is nothing to be done about

this. At some point we just have to make a choice and make it clear which choice

we have made. Such a decision is called an orientation. We will have much more to

say about this later. For now, we simply note that a diļ¬erent choice will only change

our answer by changing its sign.

While we are on this topic we also note that we would end up with the same

1 2

number in Step 5 if we had calculated Ļpi,j (ā’Vi,j , ā’Vi,j ) in Step 4, instead. Similarly,

2 1

if it turns out later that we should have calculated Ļpi,j (Vi,j , Vi,j ) then we could have

38 3. DIFFERENTIAL FORMS

z

1

Vi,j

pi,j

2

Vi,j

y

x

Figure 3. The steps toward integrating a 2-form.

1 2

also gotten the right answer by computing Ļpi,j (ā’Vi,j , Vi,j ). In other words, there

1 2

are really only two possibilities: either Ļpi,j (Vi,j , Vi,j ) gives the correct answer or

1 2

Ļpi,j (ā’Vi,j , Vi,j ) does. Which one will depend on our choice of orientation.

Despite all the diļ¬culties with using the above deļ¬nition of Ļ, all hope is not

M

lost. Remember that we are only integrating over regions which can be parameterized

by subsets of R2 . The trick is to use such a parameterization to translate the problem

into an integral of a 2-form over a region in R2 . The steps are analogous to those in

Section 5 of Chapter 1.

Suppose Ļ : R ā‚ R2 ā’ M is a parameterization. We want to ļ¬nd a 2-form,

f (x, y) dx ā§ dy, such that a Riemann sum for this 2-form over R gives the same

result as a Riemann sum for Ļ over M. Letā™s begin:

(1) Choose a rectangular lattice of points in R, {(xi , yj )}. This also gives a

lattice, {Ļ(xi , yj )}, in M.

1 2

(2) For each i, j, deļ¬ne Vi,j = (xi+1 , yj ) ā’ (xi , yj ), Vi,j = (xi , yj+1) ā’ (xi , yj ),

1 2

Vi,j = Ļ(xi+1 , yj ) ā’ Ļ(xi , yj ), and Vi,j = Ļ(xi , yj+1) ā’ Ļ(xi , yj ) (see Fig. 4).

Notice that Vi,j and Vi,j are vectors in T(xi ,yj ) R2 and Vi,j and Vi,j are vectors

1 2 1 2

in TĻ(xi ,yj ) R3 .

2. INTEGRATING DIFFERENTIAL 2-FORMS 39

1 2 1 2

(3) For each i, j compute f (xi , yj ) dx ā§ dy(Vi,j , Vi,j ) and ĻĻ(xi ,yj ) (Vi,j , Vi,j ).

(4) Sum over all i and j.

z

y

Ļ(xi , yj )

Ļ

2

Vi,j 1

Vi,j

yj

1

Vi,j

2

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