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This section requires some knowledge of Topology and Algebra. It is not essential
for the ļ¬‚ow of the text.
While we are on the subject of diļ¬erential forms on subsets of Rn , there is a very
common construction of a topological space for which it is very easy to deļ¬ne what
we mean by a diļ¬erential form. Letā™s look again at the cylinder, L, of the previous
section. One way to construct L is to start with the plane, R2 , and āroll it up.ā More
technically, we can consider the map, Āµ(Īø, z) = (cos Īø, sin Īø, z). In general this is a
many-to-one map, so it is not a parameterization, in the strict sense. To remedy this,
one might try and restrict the domain of Āµ to {(Īø, z) ā R2 |0 ā¤ Īø < 2Ļ}, however this
set is not open.
94 7. MANIFOLDS

Note that for each point, (Īø, z) ā R2 , DĀµ is a 1-1 map from T R2
(Īø,z) to T LĀµ(Īø,z) .
This is all we need in order for Āµā— Ļ„ to make sense, where Ļ„ is the form on L deļ¬ned
in the previous section.

Exercise 7.6. Show that Āµā— Ļ„ = dĪø.

In this case, we say that Āµ is a covering map, R2 is a cover of L, and dĪø is the lift
of Ļ„ to R2 .

Exercise 7.7. Suppose Ļ0 is the 1-form on R2 which we used to deļ¬ne the winding
number. Let Āµ(r, Īø) = (r cos Īø, r sin Īø). Let U = {(r, Īø)|r > 0}. Then Āµ : U ā’
{R2 ā’(0, 0)} is a covering map. Hence, there is a 1-1 correspondence between a quotient
of U and R2 ā’ (0, 0). Compute the lift of Ļ0 to U.

Letā™s go back to the cylinder, L. Another way to look at things is to ask: How
can we recover L from the Īø-z plane? The answer is to view L as a quotient space.
Letā™s put an equivalence relation, R, on the points of R2 : (Īø1 , z1 ) ā¼ (Īø2 , z2 ) if and
only if z1 = z2 , and Īø1 ā’ Īø2 = 2nĻ, for some n ā Z. We will denote the quotient of
R2 under this relation as R2 /R. Āµ now induces a 1-1 map, Āµ, from R2 /R onto L.
ĀÆ
Hence, these two spaces are homeomorphic.
Letā™s suppose now that we have a form on U, an open subset of Rn , and we
would like to know when it descends to a form on a quotient of U. Clearly, if we
begin with the lift of a form, then it will descend. Letā™s try and see why. In general, if
Āµ : U ā‚ Rn ā’ M ā‚ Rm is a many-to-one map, diļ¬erentiable at each point of U, then
the sets, {Āµā’1 (p)}, partition U. Hence, we can form the quotient space, U/Āµā’1 , under
this partition. For each x ā Āµā’1 (p), DĀµx is a 1-1 map from T Ux to T Mp , and hence,
DĀµx is well deļ¬ned. If x and y are both in Āµā’1 (p), then DĀµā’1 ā—¦DĀµx is a 1-1 map from
ā’1
y
T Ux to T Uy . We will denote this map as DĀµxy . We say a k-form, Ļ, on Rn descends
to a k-form on U/Āµā’1 if and only if Ļ(Vx1 , ..., Vxk ) = Ļ(DĀµxy (Vx1 ), ..., DĀµxy (Vx1 )), for
all x, y ā U such that Āµ(x) = Āµ(y).

Exercise 7.8. If Ļ„ is a diļ¬erential k-form on M, then Āµā— Ļ„ (the lift of Ļ„ ) is a diļ¬erential
k-form on U which descends to a diļ¬erential k-form on U/Āµā’1 .

Now suppose that we have a k-form, Ļ , on U which descends to a k-form on
Ė
U/Āµā’1 , where Āµ : U ā‚ Rn ā’ M ā‚ Rm is a covering map. How can we get a k-form
3. FORMS ON QUOTIENTS OF Rn (OPTIONAL) 95

on M? As we have already remarked, Āµ : U/Āµā’1 ā’ M is a 1-1 map. Hence, we can
ĀÆ
use it to push forward the form, Ļ. In other words, we can deļ¬ne a k-form on M as
follows: Given k vectors in T Mp , we ļ¬rst choose a point, x ā Āµā’1 (p). We then deļ¬ne

Āµā— Ļ(Vp1 , ..., Vpk ) = Ļ (DĀµā’1 (Vp1 ), ..., DĀµā’1(Vpk ))
Ė x x

It follows from the fact that Ļ descends to a form on U/Āµā’1 that it did not matter
Ė
which point, x, we chose in Āµā’1 (p). Note that although Āµ is not 1-1, DĀµx is, so DĀµā’1
x
makes sense.
If we begin with a form on U, there is a slightly more general construction of a
form on a quotient of U, which does not require the use of a covering map. Let Ī“
be a group of transformations of U. We say Ī“ acts discretely if for each p ā U, there
exists an Ē« > 0 such that NĒ« (p) does not contain Ī³(p), for any non-identity element,
Ī³ ā Ī“. If Ī“ acts discretely, then we can form the quotient of U by Ī“, denoted U/Ī“,
as follows: p ā¼ q if there exists Ī³ ā Ī“ such that Ī³(p) = q (The fact that Ī“ acts
discretely is what guarantees a āniceā topology on U/Ī“).
Now, suppose Ļ is a k-form on U. We say Ļ descends to a k-form, Ļ, on U/Ī“, if
Ė Ė
and only if Ļ (Vp1 , ..., Vpk ) = Ļ (DĪ³(Vp1 ), ..., DĪ³(Vp1 )), for all Ī³ ā Ī“.
Ė Ė
Now that we have decided what a form on a quotient of U is, we still have to
deļ¬ne n-chains, and what we mean by integration of n-forms over n-chains. We say
Ė Ė Ė
an n-chain, C ā‚ U, descends to an n-chain, C ā‚ U/Ī“, if Ī³(C) = C, for all Ī³ ā Ī“.
The n-chains of U/Ī“ are simply those which are descendants of n-chains in U.
Integration is a little more subtle. For this we need the concept of a fundamental
domain for Ī“. This is nothing more than a closed subset of U, whose interior does
not contain two equivalent points. Furthermore, for each equivalence class, there is
at least one representative in a fundamental domain. Here is one way to construct a
fundamental domain: First, choose a point, p ā U. Now, let D = {q ā U|d(p, q)) ā¤
d(Ī³(p), q), for all Ī³ ā Ī“}.
Ė
Now, let C be an n-chain on U which descends to an n-chain, C, on U/Ī“, and let
Ļ be an n-form that descends to an n-form, Ļ. Let D be a fundamental domain for
Ė
Ī“ in U. Then we deļ¬ne

Ļā” Ļ
Ė
Ė
96 7. MANIFOLDS

Technical note: In general, this deļ¬nition is invariant of which point was chosen
in the construction of the fundamental domain, D. However, a VERY unlucky choice
Ė
will result in C ā© D ā‚ ā‚D, which could give a diļ¬erent answer for the above integral.
Fortunately, it can be shown that the set of such āunluckyā points has measure zero.
That is, if we were to choose the point at random, then the odds of picking an
āunluckyā point are 0%. Very unlucky indeed!

Example 7.3. Suppose Ī“ is the group of transformations of the plane generated
by (x, y) ā’ (x + 1, y), and (x, y) ā’ (x, y + 1). The space R2 /Ī“ is often denoted
T 2 , and referred to as a torus. Topologists often visualize the torus as the surface
of a donut. A fundamental domain for Ī“ is the unit square in R2 . The 1-form,
dx, on R2 descends to a 1-form on T 2 . Integration of this form over a closed
1-chain, C ā‚ T 2 , counts the number of times C wraps around the āholeā of the
donut.

4. Deļ¬ning Manifolds

As we have already remarked, a diļ¬erential n-form on Rm is just an n-form on
Tp Rm , for each point p ā Rm , along with some condition about how the form varies
in a diļ¬erentiable way as p varies. All we need to deļ¬ne a form on a space other
than Rm is some notion of a tangent space at every point. We call such a space a
manifold. In addition, we insist that at each point of a manifold the tangent space
has the same dimension, n, which we then say is the dimension of the manifold.
How do we guarantee that a given subset of Rm is a manifold? Recall that we
deļ¬ned the tangent space to be the span of some partial derivatives of a parameteri-
zation. However, insisting that the whole manifold is capable of being parameterized
is very restrictive. Instead, we only insist that every point of a manifold lies in a
subset that can be parameterized. Hence, if M is an n-manifold in Rm then there is
a set of open subsets, {Ui } ā‚ Rn , and a set of diļ¬erentiable maps, {Ļi : Ui ā’ M},
such that {Ļi (Ui )} is a cover of M. (That is, for each point, p ā M, there is an i,
and a point, q ā Ui , such that Ļi (q) = p).
5. DIFFERENTIAL FORMS ON MANIFOLDS 97

Example 7.4. S 1 , the unit circle in R2 , is a 1-manifold. Let Ui = (ā’1, 1), for
ā ā ā
2 ), Ļ (t) = (t, ā’ 1 ā’ t2 ), Ļ (t) = ( 1 ā’ t2 , t),
i = 1, 2, 3, 4, Ļ1 (t) = (t, 1 ā’ t 2 3
ā
and Ļ4 (t) = (ā’ 1 ā’ t2 , t). Then {Ļi (Ui )} is certainly a cover of S 1 with the
desired properties.

Exercise 7.9. Show that S 2 , the unit sphere in R3 , is a 2-manifold.

5. Diļ¬erential Forms on Manifolds

Basically, the deļ¬nition of a diļ¬erential n-form on an m-manifold is the same as
the deļ¬nition of an n-form on a subset of Rm which was given by a single parame-
terization. First and foremost it is just an n-form on Tp M, for each p ā M.
Letā™s say M is an m-manifold. Then we know there is a set of open sets, {Ui } ā‚
Rm , and a set of diļ¬erentiable maps, {Ļi : Ui ā’ M}, such that {Ļi (Ui )} covers M.
Now, letā™s say that Ļ is a family of n-forms, deļ¬ned on Tp M, for each p ā M. Then
we say that the family, Ļ, is a diļ¬erentiable n-form on M if Ļā— Ļ is a diļ¬erentiable
i
n-form on Ui , for each i.

Example 7.5. In the previous section we saw how S 1 , the unit circle in R2 , is
a 1-manifold. If (x, y) is a point of S 1 , then T S(x,y) is given by the equation
1

dy = ā’ x dx, in T R2 , as long as y = 0. If y = 0, then T S(x,y) is given by
1
(x,y)
y
dx = 0. We deļ¬ne a 1-form on S 1 , Ļ = ā’y dx + x dy. (Actually, Ļ is a 1-form
on all of R2 . To get a 1-form on just S 1 , we restrict the domain of Ļ to the
tangent lines to S 1 .) To check that this is really a diļ¬erential form, we must
compute all pull-backs:
ā’1 1
Ļā— Ļ = ā dt , Ļā— Ļ = ā dt
1 2
1 ā’ t2 1 ā’ t2
ā’1
1
Ļā— Ļ = ā dt , Ļā— Ļ = ā dt
3 4
1 ā’ t2 1 ā’ t2
Since all of these are diļ¬erentiable on Ui = (ā’1, 1), we can say that Ļ is a
diļ¬erential form on S 1 .

We now move on to integration of n-chains on manifolds. The deļ¬nition of an
n-chain is no diļ¬erent than before; it is just a formal linear combination of n-cells in
98 7. MANIFOLDS

M. Letā™s suppose that C is an n-chain in M, and Ļ is an n-form. Then how do we
deļ¬ne Ļ? If C lies entirely in Ļi(Ui ), for some i, then we could deļ¬ne the value of
C
Ļā— Ļ. But it may be that part of C lies in both
this integral to be the value of i
Ļā’1 (C)
i
Ļi (Ui ) and Ļj (Uj ). If we deļ¬ne Ļ to be the sum of the two integrals we get when
C
we pull-back Ļ under Ļi and Ļj , then we end up ādouble countingā the integral of Ļ
on C ā© Ļi (Ui ) ā© Ļj (Uj ). Somehow, as we move from Ļi (Ui ) into Ļj (Uj ), we want the
eļ¬ect of the pull-back of Ļ under Ļi to āfade outā, and the eļ¬ect of the pull back
under Ļj to āfade inā. This is accomplished by a partition of unity.
The technical deļ¬nition of a partition of unity subordinate to the cover, {Ļi (Ui )}
is a set of diļ¬erentiable functions, fi : M ā’ [0, 1], such that fi (p) = 0 if p ā Ļi (Ui ),
/
fi (p) = 1, for all p ā M. We refer the reader to any book on diļ¬erential
and
i
topology for a proof of the existence of partitions of unity.
We are now ready to give the full deļ¬nition of the integral of an n-form on an
n-chain in an m-manifold.

Ļā— (fi Ļ)
Ļā” i
i
C Ļā’1 (C)
i

We start with a very simple example to illustrate the use of a partition of unity.

Example 7.6. Let M be the manifold which is the interval (1, 10) ā‚ R. Let
Ui = (i, i + 2), for i = 1, ..., 8. Let Ļi : Ui ā’ M be the identity map. Let {fi }
be a partition of unity, subordinate to the cover, {Ļi(Ui )}. Let Ļ be a 1-form on
M. Finally, let C be the 1-chain which consists of the single 1-cell, [2, 8]. Then
we have
8 8 8 8
Ļā— (fi Ļ)
Ļā” = fi Ļ = (fi Ļ) = fi Ļ= Ļ
i
i=1 i=1 C i=1 i=1
C C C C
Ļā’1 (C)
i

as one would expect!
6. APPLICATION: DERHAM COHOMOLOGY 99

Example 7.7. Let S 1 , Ui , Ļi , and Ļ be deļ¬ned as in Examples 7.4 and 7.5. A
partition of unity subordinate to the cover {Ļi (Ui )} is as follows:
y2 y ā„ 0 0 y>0
f1 (x, y) = , f2 (x, y) =
y2 y ā¤ 0
0 y<0

x2 x ā„ 0 0 x>0
f3 (x, y) = , f4 (x, y) =
x2 x ā¤ 0
0 x<0
(Check this!) Let Āµ : [0, Ļ] ā’ S 1 be deļ¬ned by Āµ(Īø) = (cos Īø, sin Īø). Then the
image of Āµ is a 1-cell, Ļ, in S 1 . Letā™s integrate Ļ over Ļ:

4
Ļā— (fi Ļ)
Ļā” i
i=1
Ļ Ļā’1 (Ļ)
i
ā ā ā
ā’ 1 ā’ t2 dt + 0 + t2 ā’ 1 ā’ t2 dt
1ā’
= dt +
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