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pair of vectors is based changes. In other words, Area(·, ·) does not act on Tp R3 —Tp R3
for a ¬xed p. We can make this point a little clearer by re-examining the above
integrand. Note that it is of the form f (⋆)Area(·, ·). For a ¬xed point, ⋆, of R3 this
is an operator on T⋆ R3 — T⋆ R3 , much like a 2-form is.
But so far all we have done is to de¬ne 2-forms at ¬xed points of R3 . To really
generalize the above integral we have to start considering entire families of 2-forms,
ωp : Tp R3 — Tp R3 ’ R, where p ranges over all of R3 . Of course, for this to be useful
we™d like such a family to have some “niceness” properties. For one thing, we would
like it to be continuous. That is, if p and q are close then ωp and ωq should be similar.
An even stronger property that we will insist on is that the family, ωp , is di¬er-
entiable. To see what this means recall that for a ¬xed p, a 2-form ωp can always
be written as ap dx § dy + bp dy § dz + cp dx § 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,
ωp , is di¬erentiable we mean that each of these functions is di¬erentiable. If ωp is
33
34 3. DIFFERENTIAL FORMS

di¬erentiable then we will refer to it as a di¬erential form. When there can be no
confusion we will suppress the subscript, p.



Example 3.1. ω = x2 y dx§dy’xz dy§dz is a di¬erential 2-form on R3 . On the
space T(1,2,3) R3 it is just the 2-form 2dx§dy ’3dy §dz. We will denote vectors in
T(1,2,3) R3 as dx, dy, dz (1,2,3). Hence, the value of ω( 4, 0, ’1 (1,2,3), 3, 1, 2 (1,2,3))
is the same as the 2-form, 2dx § dy + dy § dz, evaluated on the vectors 4, 0, ’1
and 3, 1, 2 , which we compute:



ω( 4, 0, ’1 , 3, 1, 2 (1,2,3) )
(1,2,3)

= 2dx § dy ’ 3dy § dz ( 4, 0, ’1 , 3, 1, 2 )
0 ’1
40
’3
=2 =5
31 1 2



Suppose ω is a di¬erential 2-form on R3 . What does ω act on? It takes a pair
of vectors at each point of R3 and returns a number. In other words, it takes two
vector ¬elds and returns a function from R3 to R. A vector ¬eld is simply a choice
of vector in Tp R3 , for each p ∈ R3 . In general, a di¬erential n-form on Rm acts on n
vector ¬elds to produce a function from Rm to R (see Fig. 1).


6 0
21
ω √
9 ’3
7

2 π
3



Figure 1. A di¬erential 2-form, ω, acts on a pair of vector ¬elds, and
returns a function from Rn to R.
2. INTEGRATING DIFFERENTIAL 2-FORMS 35

is a vector ¬eld on R3 . For example, it
Example 3.2. V1 = 2y, 0, ’x (x,y,z)

contains the vector 4, 0, ’1 ∈ T(1,2,3) R3 . If V2 = z, 1, xy and ω is the
(x,y,z)

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

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

0 ’x
2y 0
= x2 y = 2x2 y 2 ’ x2 z,
’ 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 ’ x2 z equals 5 at the point (1, 2, 3), which is indeed
the case.


Exercise 3.1. Let ω be the di¬erential 2-form on R3 given by

ω = xyz dx § dy + x2 z dy § dz ’ y dx § dz

Let V1 and V2 be the following vector ¬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 ω on T(1,2,3) R3 ?
(3) Use your answers to the previous two questions to compute ω(V1, V2 ) at the
point (1, 2, 3).
(4) Compute ω(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 Di¬erential 2-Forms

Let us now examine more closely integration of functions on subsets of R2 , which
you learned in calculus. Suppose R ‚ R2 and f : R ’ R. How did we learn to de¬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 de¬ne Vi,j = (xi+1 , yj ) ’ (xi , yj ) and Vi,j = (xi , yj+1) ’ (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 de¬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
Let™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 di¬erential 2-form ω = f (x, y)dx § 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 ) = ω(xi ,yj ) (Vi,j , Vi,j ). It is reasonable, then, to
i j i j
adopt the shorthand “ ω” to denote the limit in Step 5. The upshot of all this is
R
the following:

If ω = f (x, y)dx § dy then ω= f dx dy.
R R

Since all di¬erential 2-forms on R2 are of the form f (x, y)dx § 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
“§” and forget you have a di¬erential form. This is only valid with dx § dy. It is
NOT valid with dy § dx. This may seem a bit curious since

f dx § dy = f dx dy = f dy dx

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

Exercise 3.2. Let ω = xy 2 dx § dy be a di¬erential 2-form on R2 . Let D be the region
of R2 bounded by the graphs of x = y 2 and y = x ’ 6. Calculate ω. Answer: 189.
D

What about integration of di¬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
de¬ne what we mean by ω 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 de¬ne Vi,j = pi+1,j ’ pi,j and Vi,j = pi,j+1 ’ 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 ωpi,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 de¬ne to be the value of ω.
M
Unfortunately these steps aren™t so easy to follow. For one thing, it™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 ωpi,j (Vi,j , Vi,j ) instead of ωpi,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 ¬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 di¬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 ωpi,j (’Vi,j , ’Vi,j ) in Step 4, instead. Similarly,
2 1
if it turns out later that we should have calculated ωpi,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 ωpi,j (’Vi,j , Vi,j ). In other words, there
1 2
are really only two possibilities: either ωpi,j (Vi,j , Vi,j ) gives the correct answer or
1 2
ωpi,j (’Vi,j , Vi,j ) does. Which one will depend on our choice of orientation.
Despite all the di¬culties with using the above de¬nition of ω, 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 φ : R ‚ R2 ’ M is a parameterization. We want to ¬nd a 2-form,
f (x, y) dx § dy, such that a Riemann sum for this 2-form over R gives the same
result as a Riemann sum for ω over M. Let™s begin:
(1) Choose a rectangular lattice of points in R, {(xi , yj )}. This also gives a
lattice, {φ(xi , yj )}, in M.
1 2
(2) For each i, j, de¬ne Vi,j = (xi+1 , yj ) ’ (xi , yj ), Vi,j = (xi , yj+1) ’ (xi , yj ),
1 2
Vi,j = φ(xi+1 , yj ) ’ φ(xi , yj ), and Vi,j = φ(xi , yj+1) ’ φ(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 Tφ(xi ,yj ) R3 .
2. INTEGRATING DIFFERENTIAL 2-FORMS 39

1 2 1 2
(3) For each i, j compute f (xi , yj ) dx § dy(Vi,j , Vi,j ) and ωφ(xi ,yj ) (Vi,j , Vi,j ).
(4) Sum over all i and j.

z
y


φ(xi , yj )
φ
2
Vi,j 1
Vi,j
yj
1
Vi,j
2

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