ńņš. 15 |

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 āroll 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 Ļ0 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 Ļ0 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 ) ā¼ (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 ā¼ q if there exists Ī³ ā Ī“ such that Ī³(p) = q (The fact that Ī“ acts

discretely is what guarantees a āniceā 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

Ļā” Ļ

Ė

Ė

C Cā©D

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 āunluckyā points has measure zero.

That is, if we were to choose the point at random, then the odds of picking an

āunluckyā 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 āholeā 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, {Ļi : Ui ā’ M},

such that {Ļi (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 Ļi (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, Ļ1 (t) = (t, 1 ā’ t 2 3

ā

and Ļ4 (t) = (ā’ 1 ā’ t2 , t). Then {Ļi (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, {Ļi : Ui ā’ M}, such that {Ļi (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 Ļā— 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

Ļā— Ļ = ā dt , Ļā— Ļ = ā dt

1 2

1 ā’ t2 1 ā’ t2

ā’1

1

Ļā— Ļ = ā dt , Ļā— 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 Ļi(Ui ), for some i, then we could deļ¬ne the value of

C

Ļā— Ļ. But it may be that part of C lies in both

this integral to be the value of i

Ļā’1 (C)

i

Ļi (Ui ) and Ļj (Uj ). If we deļ¬ne Ļ to be the sum of the two integrals we get when

C

we pull-back Ļ under Ļi and Ļj , then we end up ādouble countingā the integral of Ļ

on C ā© Ļi (Ui ) ā© Ļj (Uj ). Somehow, as we move from Ļi (Ui ) into Ļj (Uj ), we want the

eļ¬ect of the pull-back of Ļ under Ļi to āfade outā, and the eļ¬ect of the pull back

under Ļj to āfade inā. This is accomplished by a partition of unity.

The technical deļ¬nition of a partition of unity subordinate to the cover, {Ļi (Ui )}

is a set of diļ¬erentiable functions, fi : M ā’ [0, 1], such that fi (p) = 0 if p ā Ļi (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.

Ļā— (fi Ļ)

Ļā” i

i

C Ļā’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 Ļi : Ui ā’ M be the identity map. Let {fi }

be a partition of unity, subordinate to the cover, {Ļi(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

Ļā— (fi Ļ)

Ļā” = fi Ļ = (fi Ļ) = fi Ļ= Ļ

i

i=1 i=1 C i=1 i=1

C C C C

Ļā’1 (C)

i

as one would expect!

6. APPLICATION: DERHAM COHOMOLOGY 99

Example 7.7. Let S 1 , Ui , Ļi , and Ļ be deļ¬ned as in Examples 7.4 and 7.5. A

partition of unity subordinate to the cover {Ļi (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

Ļā— (fi Ļ)

Ļā” i

i=1

Ļ Ļā’1 (Ļ)

i

ā ā ā

ā’ 1 ā’ t2 dt + 0 + t2 ā’ 1 ā’ t2 dt

1ā’

= dt +

ńņš. 15 |