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pair of vectors is based changes. In other words, Area(Ā·, Ā·) does not act on Tp R3 Ć—Tp R3
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,
Ļp : 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 ānicenessā properties. For one thing, we would
like it to be continuous. That is, if p and q are close then Ļp and Ļq should be similar.
An even stronger property that we will insist on is that the family, Ļp , is diļ¬er-
entiable. To see what this means recall that for a ļ¬xed p, a 2-form Ļp 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,
Ļp , is diļ¬erentiable we mean that each of these functions is diļ¬erentiable. If Ļp 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

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 Ļā 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
āā§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 Ļpi,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 Ļpi,j (Vi,j , Vi,j ) instead of Ļpi,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
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 Ļpi,j (ā’Vi,j , ā’Vi,j ) in Step 4, instead. Similarly,
2 1
if it turns out later that we should have calculated Ļpi,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 Ļpi,j (ā’Vi,j , Vi,j ). In other words, there
1 2
are really only two possibilities: either Ļpi,j (Vi,j , Vi,j ) gives the correct answer or
1 2
Ļpi,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 ĻĻ(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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