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‚Ψ
= 0, 0, 1
‚z


(3) Plug the tangent vectors into ω at the point Ψ(θ, z). We get
’ sin θ cos θ cos θ 0
(cos2 θ + sin2 θ) +z
0 0 01
This simpli¬es to the function z cos θ.
6. SUMMARY: HOW TO INTEGRATE A DIFFERENTIAL FORM 55

(4) Integrate the resulting function over R.
1 2π

z cos θ dθ dz
0 0
Note that the integrand comes from Step 3 and the limits of integration come
from Step 1.
CHAPTER 4


Di¬erentiation of Forms.

1. The derivative of a di¬erential 1-form

The goal of this section is to ¬gure out what we mean by the derivative of a
di¬erential form. One way to think about a derivative is as a function which measures
the variation of some other function. Suppose ω is a 1-form on R2 . What do we mean
by the “variation” of ω? One thing we can try is to plug in Vp , a particular vector in
the tangent space at the point, p. We can then look at how ω(Vp ) changes as we vary
p. But p can vary in lots of ways, so we need to pick one. In calculus we learn how
to take another vector, Wp , and use it to vary p. Hence, the derivative of ω, which
we shall denote “dω”, is a function that acts on both Vp and Wp . In other words, it
must be a 2-form!

Vp+tWp


Vp

Wp


t




Figure 1. Using Wp to vary Vp .

Let™s see how to use Wp to calculate the variation in ω(Vp ) in a speci¬c example.
Suppose ω = y dx ’ x2 dy, p = (1, 1), Vp = 1, 2 (1,1) , and Wp = 2, 3 (1,1) . Notice that
57
58 4. DIFFERENTIATION OF FORMS.

Vp+tW = 1, 2 is a vector similar to Vp , but pushed away by t in the direction
(1+2t,1+3t)

of Wp . Hence, the variation of ω(Vp ), in the direction of Wp , can be calculated as
follows:

’ ω( 1, 2
ω( 1, 2 (1+2t,1+3t) ) (1,1) )
lim
t
t’0

[(1 + 3t)(1) ’ (1 + 2t)2 (2)] ’ [(1)(1) ’ (1)2 (2)]
= lim
t
t’0
2
’5t ’ 8t
= ’5
= lim
t
t’0

What about the variation of ω(Wp ) in the direction of Vp ? We calculate:

’ ω( 2, 3
ω( 2, 3 (1+t,1+2t) ) (1,1) )
lim
t
t’0

[(1 + 2t)(2) ’ (1 + t)2 (3)] ’ [(1)(2) ’ (1)2 (3)]
= lim
t
t’0
’2t ’ 3t2
= ’2
= lim
t
t’0

This is a small problem. We want dω to be a 2-form. Hence, dω(Vp , Wp ) should
equal ’dω(Wp , Vp ). How can we use the variations above to de¬ne dω so this is true?
Simple. We just de¬ne it to be the di¬erence in these variations:

ω(Wp+tV ) ’ ω(Wp ) ω(Vp+tW ) ’ ω(Vp)
’ lim
(4) dω(Vp, Wp ) = lim
t t
t’0 t’0

Hence, in the above example, dω( 1, 2 (1,1), 2, 3 (1,1)) = ’2 ’ (’5) = 3.
Before going further we introduce some notation from calculus to make Equation
4 a little more readable. Suppose f is a function from Rn to R and V ∈ Tp Rn . Then
the derivative of f at p, in the direction of V , can be written as “∇V f ”. That is, we
de¬ne

f (p + tV ) ’ f (p)
∇V f = lim
t
t’0
One may also recall from calculus that ∇V f = ∇f (p) · V , where ∇f (p) denotes
the gradient of f evaluated at p. Using this notation, we can rewrite Equation 4 as


dω(Vp, Wp ) = ∇Vp ω(Wp ) ’ ∇Wp ω(Vp )
1. THE DERIVATIVE OF A DIFFERENTIAL 1-FORM 59

There are other ways to determine what dω is than by using Equation 4. Recall
that a 2-form acts on a pair of vectors by projecting them onto each coordinate
plane, calculating the area they span, multiplying by some constant, and adding.
So the 2-form is completely determined by the constants that you multiply by after
projecting. In order to ¬gure out what these constants are we are free to examine
the action of the 2-form on any pair of vectors. For example, suppose we have two
vectors that lie in the x-y plane and span a parallelogram with area 1. If we run
these through some 2-form and end up with the number 5 then we know that the
multiplicative constant for that 2-form, associated with the x-y plane is 5. This, in
turn, tells us that the 2-form equals 5 dx § dy+?. To ¬gure out what “?” is, we can
examine the action of the 2-form on other pairs of vectors.
Let™s try this with a general di¬erential 2-form on R3 . Such a form always looks
like dω = a(x, y, z)dx § dy + b(x, y, z)dy § dz + c(x, y, z)dx § dz. To ¬gure out
what a(x, y, z) is, for example, all we need to do is determine what dω does to the
vectors 1, 0, 0 (x,y,z) and 0, 1, 0 (x,y,z). Let™s compute this using Equation 4, assuming
ω = f (x, y, z)dx + g(x, y, z)dy + h(x, y, z)dz.


dω( 1, 0, 0 (x,y,z), 0, 1, 0 (x,y,z))

’ ω( 0, 1, 0 (x,y,z))
ω( 0, 1, 0 (x+t,y,z))
= lim
t
t’0
ω( 1, 0, 0 (x,y+t,z)) ’ ω( 1, 0, 0 (x,y,z))
’ lim
t
t’0
g(x + t, y, z) f (x, y + t, z)
’ lim
= lim
t t
t’0 t’0
‚g ‚f
(x, y, z) ’
= (x, y, z)
‚x ‚y
Similarly, direct computation shows:

‚h ‚g
(x, y, z) ’
dω( 0, 1, 0 (x,y,z), 0, 0, 1 (x,y,z)) = (x, y, z)
‚y ‚z
and,

‚h ‚f
(x, y, z) ’
dω( 1, 0, 0 (x,y,z), 0, 0, 1 (x,y,z)) = (x, y, z)
‚x ‚z
Hence, we conclude that
60 4. DIFFERENTIATION OF FORMS.



‚g ‚f ‚h ‚g ‚h ‚f
’ )dx § dy + ( ’ )dy § dz + ( ’ )dx § dz
dω = (
‚x ‚y ‚y ‚z ‚x ‚z
Exercise 4.1. Suppose ω = f (x, y)dx + g(x, y)dy is a 1-form on R2 . Show that
dω = ( ‚x ’ ‚f )dx § dy.
‚g
‚y


Exercise 4.2. Suppose ω = xy 2 dx + x3 z dy ’ (y + z 9 ) dz, V = 1, 2, 3 (2,3,’1) , and
W = ’1, 0, 1 (2,3,’1)

(1) Compute ∇V ω(W ) and ∇W ω(V ), at the point (2, 3, ’1).
(2) Use your answer to the previous question to compute dω(V, W ).

Exercise 4.3. If ω = y dx ’ x2 dy, ¬nd dω. Verify that dω( 1, 2 (1,1) , 2, 3 (1,1) ) = 3.

2. Derivatives of n-forms

Before jumping to the general case let™s look at the derivative of a 2-form. A
2-form, ω, acts on a pair of vectors, Vp and Wp , to return some number. To ¬nd
some sort of variation of ω we can vary the vectors Vp and Wp and examine how
ω(Vp , Wp ) varies. As in the last section one way to vary a vector is to push it in the
direction of some other vector, Up . Hence, whatever dω turns out to be, it will be a
function of the vectors Up , Vp , and Wp . So, we would like to de¬ne it to be a 3-form.
Let™s start by looking at the variation of ω(Vp , Wp ) in the direction of Up . We write
this as ∇Up ω(Vp , Wp ). If we were to de¬ne this as the value of dω(Up , Vp , Wp ) we would
¬nd that in general it would not be alternating. That is, usually ∇Up ω(Vp , Wp ) =
’∇Vp ω(Up , Wp ). To remedy this, we simply de¬ne dω to be the alternating sum of
all the variations:


dω(Up , Vp , Wp ) = ∇Up ω(Vp , Wp ) ’ ∇Vp ω(Up , Wp ) + ∇Wp ω(Up , Vp )
We leave it to the reader to check that dω is alternating and multilinear.
It shouldn™t be hard for the reader to now jump to the general case. Suppose ω
is an n-form and Vp1 , ..., Vpn+1 are n + 1 vectors. Then we de¬ne

n+1
dω(Vp1 , ..., Vpn+1 ) (’1)i+1 ∇Vpi ω(Vp1 , ..., Vpi’1 , Vpi+1 , ..., Vpn+1 )
=
i=1
3. INTERLUDE: 0-FORMS 61

In other words, dω, applied to n+1 vectors, is the alternating sum of the variation
of ω applied to n of those vectors in the direction of the remaining one. Note that
we can think of “d” as an operator which takes n-forms to (n + 1)-forms.

Exercise 4.4. dω is alternating and multilinear.

Exercise 4.5. Suppose ω = f (x, y, z) dx § dy + g(x, y, z) dy § dz + h(x, y, z) dx § dz.
Find dω (Hint: Compute dω( 1, 0, 0 , 0, 1, 0 , 0, 0, 1 )). Compute d(x2 y dx § dy +
y 2 z dy § dz).

3. Interlude: 0-forms

Let™s go back to Section 1, when we introduced coordinates for vectors. At that
time we noted that if C was the graph of the function y = f (x) and p was a point of
C then the tangent line to C at p lies in Tp R2 and has equation dy = m dx, for some
constant, m. Of course, if p = (x0 , y0 ) then m is just the derivative of f evaluated
at x0 .
Now, suppose we had looked at the graph of a function of 2-variables, z = f (x, y),
instead. At some point, p = (x0 , y0 , z0 ), on the graph we could look at the tangent
plane, which lies in Tp R3 . It™s equation is dz = m1 dx + m2 dy. Since z = f (x, y),
m1 = ‚f (x0 , y0 ), and m2 = ‚f (x0 , y0 ), we can rewrite this as
‚x ‚y
‚f ‚f
df = dx + dy.
‚x ‚y
Notice that the right-hand side of this equation is a di¬erential 1-form. This is a
bit strange; we applied the “d” operator to something and the result was a 1-form.
However, we know that when we apply the “d” operator to a di¬erential n-form we
get a di¬erential (n + 1)-form. So, it must be that f (x, y) is a di¬erential 0-form on
R2 !
In retrospect, this should not be so surprising. After all, the input to a di¬erential
n-form on Rm is a point, and n vectors based at that point. So, the input to a
di¬erential 0-form should be a point of Rm , and no vectors. In other words, a 0-form
on Rm is just another word for a real-valued function on Rm .
Let™s extend some of the things we can do with forms to 0-forms. Suppose f is
a 0-form, and ω is an n-form (where n may also be 0). What should we mean by
f § ω? Since the wedge product of an n-form and an m-form is an (n + m)-form, it
62 4. DIFFERENTIATION OF FORMS.

must be that f § ω is an n form. It™s hard to think of any other way to de¬ne this
as just the product, f ω.
What about integration? Remember that we integrate n-forms over subsets of
Rm that can be parameterized by a subset of Rn . So 0-forms get integrated over
things parameterized by R0 . In other words, we integrate a 0-form over a point. How
do we do this? We do the simplest possible thing; de¬ne the value of a 0-form, f ,
integrated over the point, p, to be ±f (p). To specify an orientation we just need to
say whether or not to use the ’ sign. We do this just by writing “’p” instead of

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