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’(’1,1) [0,1) ’[0,1)
1 1
√ √
1 ’ t2 dt + 2 1 ’ t2 dt
’1 0

CAUTION: Beware of orientations!

6. Application: DeRham cohomology

One of the predominant uses of di¬erential forms is to give global information
about manifolds. Consider the space R2 ’ (0, 0), as in Example 7.1. Near every point
of this space we can ¬nd an open set which is identical to an open set around a point
of R2 . This means that all of the local information in R2 ’ (0, 0) is the same as the
local information in R2 . The fact that the origin is missing is a global property.
For the purposes of detecting global properties certain forms are interesting, and
certain forms are completely uninteresting. We will spend some time discussing both.
The interesting forms are the ones whose derivative is zero. Such forms are said to
be closed. An example of a closed 1-form was ω0 , from Example 7.1 of the previous
chapter. For now let™s just focus on closed 1-forms so that you can keep this example
in mind.

Let™s look at what happens when we integrate a closed 1-form ω0 over a 1-chain
C such that ‚C = 0 (i.e. C is a closed 1-chain). If C bounds a disk D then Stokes™
theorem says
ω0 = dω0 = 0=0

In a su¬ciently small region of every manifold every closed 1-chain bounds a disk.
So integrating closed 1-forms on “small” 1-chains gives us no information. In other
words, closed 1-forms give no local information.
Suppose now that we have a closed 1-form ω0 and a closed 1-chain C such that
ω0 = 0. Then we know C does not bound a disk. The fact that there exists such a
1-chain is global information. This is why we say that the closed forms are the ones
that are interesting, from the point of view of detecting only global information.
Now let™s suppose that we have a 1-form ω1 that is the derivative of a 0-form f
(i.e. ω1 = df ). We say such a form is exact. Again, let C be a closed 1-chain. Let™s
pick two points, p and q, on C. Then C = C1 + C2 , where C1 goes from p to q and
C2 goes from q back to p. Now let™s do a quick computation:

ω1 = ω1
C1 +C2

= ω1 + ω1
C1 C2

= df + df
C1 C2

= f+ f
p’q q’p

So integrating an exact form over a closed 1-chain always gives zero. This is why
we say the exact forms are completely uninteresting. Unfortunately, in Exercise 4.8
we learned that every exact form is also closed. This is a problem, since this would
say that all of the completely uninteresting forms are also interesting! To remedy
this we de¬ne an equivalence relation.

We pause here for a moment to explain what this means. An equivalence relation
is just a way of taking one set and creating a new set by declaring certain objects
in the original set to be “the same”. You do this kind of thing every time you tell
time. To construct the clock numbers you start with the integers and declare two to
be “the same” if they di¬er by a multiple of 12. So 10 + 3 = 13, but 13 is the same
as 1, so if it™s now 10 o™clock then in 3 hours it will 1 o™clock.
We play the same trick for di¬erential forms. We will restrict ourselves to the
closed forms, but we will consider two of them to be “the same” if their di¬erence
is an exact form. The set which we end up with is called the cohomology of the
manifold in question. For example, if we start with the closed 1-forms then, after
our equivalence relation, we end up with the set which we will call H 1 , or the ¬rst
cohomology (see Figure 1).

n-forms (n + 1)-forms
(n ’ 1)-forms




Figure 1. De¬ning H n .

Note that the di¬erence between an exact form and the form which always returns
the number zero is an exact form. Hence, every exact form is equivalent to 0 in H n ,
as in the ¬gure.
For each n the set H n contains a lot of information about the manifold in question.
For example, if H 1 ∼ R1 (as it turns out is the case for R2 ’ (0, 0)) then this tells

us that the manifold has one “hole” in it. Studying manifolds via cohomology is the
topic of a ¬eld of mathematics called Algebraic Topology.

Non-linear forms

1. Surface area and arc length

Now that we have developed some pro¬ciency with di¬erential forms, let™s see
what else we can integrate. A basic assumption that we used to come up with the
de¬nition of an n-form was the fact that at every point it is a linear function which
“eats” n vectors and returns a number. But what about the non-linear functions?
Let™s go all the way back to Section 5 of Chapter 1. There we decided that the
integral of a function f over a surface R in R3 should look something like:

‚φ ‚φ
(6) f (φ(r, θ))Area (r, θ), (r, θ) drdθ
‚r ‚θ

At the heart of the integrand is the Area function, which takes two vectors and
returns the area of the parallelogram that it spans. The 2-form dx § dy does this for
two vectors in Tp R2 . In Tp R3 the right function is the following:

Area(Vp1 , Vp2 ) = (dy § dz)2 + (dx § dz)2 + (dx § dy)2

(The reader may recognize this as the magnitude of the cross product between
Vp1 and Vp2 .) This is clearly non-linear!

Example A.1. The area of the parallelogram spanned by 1, 1, 0 and 1, 2, 3
can be computed as follows:

2 2 2
10 10 11
Area( 1, 1, 0 , 1, 2, 3 ) = + +
23 13 12

3 2 + 32 + 12

= 19

The thing that makes (linear) di¬erential forms so useful is the Generalized Stokes
theorem. We don™t have anything like this for non-linear forms, but that™s not to say
that they don™t have their uses. For example, there is no di¬erential 2-form on R3
that one can integrate over arbitrary surfaces to ¬nd their surface area. For that we
would need to compute the following:

(dy § dz)2 + (dx § dz)2 + (dx § dy)2
Area(R) =
For relatively simple surfaces this integrand can be evaluated by hand. Integrals
such as this play a particularly important role in certain applied problems. For
example, if one were to dip a loop of bent wire into a soap ¬lm, the resulting surface
would be the one of minimal area. Before one can even begin to ¬gure out what
surface this is for a given piece of wire, one must be able to know how to compute
the area of an arbitrary surface, as above.

Example A.2. We compute the surface area of a sphere of radius r in R3 . A
parameterization is given by

¦(θ, φ) = (r sin φ cos θ, r sin φ sin θ, r cos φ)

where 0 ¤ θ ¤ 2π and 0 ¤ φ ¤ π.
Now we compute:
Area ‚¦ , ‚¦
‚θ ‚φ
= Area ( ’r sin φ sin θ, r sin φ cos θ, 0 , r cos φ cos θ, r cos φ sin θ, ’r sin φ )
= (’r 2 sin2 φ cos θ)2 + (r 2 sin2 φ sin θ)2 + (’r 2 sin φ cos φ)2
sin4 φ + sin2 φ cos2 φ

= r sin φ
And so the desired area is given by

‚¦ ‚¦
Area , dθ dφ
‚θ ‚φ
π 2π

= r sin φ dθ dφ
0 0
= 4πr

Exercise A.1. Compute the surface area of a sphere of radius r in R3 using the pa-
¦(ρ, θ) = (ρ cos θ, ρ sin θ, ± r 2 ’ ρ2 )
for the top and bottom halves, where 0 ¤ ρ ¤ r and 0 ¤ θ ¤ 2π.

Let™s now go back to Equation 6. Classically this is called a surface integral. It
might be a little clearer how to compute such an integral if we write it as follows:

f (x, y, z) (dy § dz)2 + (dx § dz)2 + (dx § dy)2
f (x, y, z) dS =
Lengths are very similar to areas. In calculus you learn that if you have a curve
C in the plane, for example, parameterized by the function φ(t) = (x(t), y(t)), where
a ¤ t ¤ b, then its length is given by

2 2
dx dy
Length(C) = + dt
dt dt
We can write this without making reference to the parameterization by employing
a non-linear 1-form:

dx2 + dy 2
Length(C) =
Finally, we can de¬ne what is classically called a line integral as follows:

f (x, y) dx2 + dy 2
f (x, y) ds =


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