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This section requires some knowledge of Topology and Algebra. It is not essential
for the ¬‚ow of the text.
While we are on the subject of di¬erential forms on subsets of Rn , there is a very
common construction of a topological space for which it is very easy to de¬ne what
we mean by a di¬erential form. Let™s look again at the cylinder, L, of the previous
section. One way to construct L is to start with the plane, R2 , and “roll it up.” More
technically, we can consider the map, µ(θ, z) = (cos θ, sin θ, 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 µ to {(θ, z) ∈ R2 |0 ¤ θ < 2π}, however this
set is not open.
94 7. MANIFOLDS

Note that for each point, (θ, z) ∈ R2 , Dµ is a 1-1 map from T R2
(θ,z) to T Lµ(θ,z) .
This is all we need in order for µ— „ to make sense, where „ is the form on L de¬ned
in the previous section.

Exercise 7.6. Show that µ— „ = dθ.

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

Exercise 7.7. Suppose ω0 is the 1-form on R2 which we used to de¬ne the winding
number. Let µ(r, θ) = (r cos θ, r sin θ). Let U = {(r, θ)|r > 0}. Then µ : U ’
{R2 ’(0, 0)} is a covering map. Hence, there is a 1-1 correspondence between a quotient
of U and R2 ’ (0, 0). Compute the lift of ω0 to U.

Let™s go back to the cylinder, L. Another way to look at things is to ask: How
can we recover L from the θ-z plane? The answer is to view L as a quotient space.
Let™s put an equivalence relation, R, on the points of R2 : (θ1 , z1 ) ∼ (θ2 , z2 ) if and
only if z1 = z2 , and θ1 ’ θ2 = 2nπ, for some n ∈ Z. We will denote the quotient of
R2 under this relation as R2 /R. µ now induces a 1-1 map, µ, from R2 /R onto L.
¯
Hence, these two spaces are homeomorphic.
Let™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. Let™s try and see why. In general, if
µ : U ‚ Rn ’ M ‚ Rm is a many-to-one map, di¬erentiable at each point of U, then
the sets, {µ’1 (p)}, partition U. Hence, we can form the quotient space, U/µ’1 , under
this partition. For each x ∈ µ’1 (p), Dµx is a 1-1 map from T Ux to T Mp , and hence,
Dµx is well de¬ned. If x and y are both in µ’1 (p), then Dµ’1 —¦Dµx is a 1-1 map from
’1
y
T Ux to T Uy . We will denote this map as Dµxy . We say a k-form, ω, on Rn descends
to a k-form on U/µ’1 if and only if ω(Vx1 , ..., Vxk ) = ω(Dµxy (Vx1 ), ..., Dµxy (Vx1 )), for
all x, y ∈ U such that µ(x) = µ(y).

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

Now suppose that we have a k-form, ω , on U which descends to a k-form on
˜
U/µ’1 , where µ : U ‚ Rn ’ M ‚ 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, µ : U/µ’1 ’ M is a 1-1 map. Hence, we can
¯
use it to push forward the form, ω. In other words, we can de¬ne a k-form on M as
follows: Given k vectors in T Mp , we ¬rst choose a point, x ∈ µ’1 (p). We then de¬ne

µ— ω(Vp1 , ..., Vpk ) = ω (Dµ’1 (Vp1 ), ..., Dµ’1(Vpk ))
˜ x x

It follows from the fact that ω descends to a form on U/µ’1 that it did not matter
˜
which point, x, we chose in µ’1 (p). Note that although µ is not 1-1, Dµx is, so Dµ’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 “
be a group of transformations of U. We say “ acts discretely if for each p ∈ U, there
exists an « > 0 such that N« (p) does not contain γ(p), for any non-identity element,
γ ∈ “. If “ acts discretely, then we can form the quotient of U by “, denoted U/“,
as follows: p ∼ q if there exists γ ∈ “ such that γ(p) = q (The fact that “ acts
discretely is what guarantees a “nice” topology on U/“).
Now, suppose ω is a k-form on U. We say ω descends to a k-form, ω, on U/“, if
˜ ˜
and only if ω (Vp1 , ..., Vpk ) = ω (Dγ(Vp1 ), ..., Dγ(Vp1 )), for all γ ∈ “.
˜ ˜
Now that we have decided what a form on a quotient of U is, we still have to
de¬ne n-chains, and what we mean by integration of n-forms over n-chains. We say
˜ ˜ ˜
an n-chain, C ‚ U, descends to an n-chain, C ‚ U/“, if γ(C) = C, for all γ ∈ “.
The n-chains of U/“ 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 “. 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 ∈ U. Now, let D = {q ∈ U|d(p, q)) ¤
d(γ(p), q), for all γ ∈ “}.
˜
Now, let C be an n-chain on U which descends to an n-chain, C, on U/“, and let
ω be an n-form that descends to an n-form, ω. Let D be a fundamental domain for
˜
“ in U. Then we de¬ne


ω≡ ω
˜
˜
C C©D
96 7. MANIFOLDS

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



Example 7.3. Suppose “ is the group of transformations of the plane generated
by (x, y) ’ (x + 1, y), and (x, y) ’ (x, y + 1). The space R2 /“ 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 “ 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 ‚ T 2 , counts the number of times C wraps around the “hole” of the
donut.


4. De¬ning Manifolds

As we have already remarked, a di¬erential n-form on Rm is just an n-form on
Tp Rm , for each point p ∈ Rm , along with some condition about how the form varies
in a di¬erentiable way as p varies. All we need to de¬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
de¬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 } ‚ Rn , and a set of di¬erentiable maps, {φi : Ui ’ M},
such that {φi (Ui )} is a cover of M. (That is, for each point, p ∈ M, there is an i,
and a point, q ∈ Ui , such that φi (q) = p).
5. DIFFERENTIAL FORMS ON MANIFOLDS 97

Example 7.4. S 1 , the unit circle in R2 , is a 1-manifold. Let Ui = (’1, 1), for
√ √ √
2 ), φ (t) = (t, ’ 1 ’ t2 ), φ (t) = ( 1 ’ t2 , t),
i = 1, 2, 3, 4, φ1 (t) = (t, 1 ’ t 2 3

and φ4 (t) = (’ 1 ’ t2 , t). Then {φi (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. Di¬erential Forms on Manifolds

Basically, the de¬nition of a di¬erential n-form on an m-manifold is the same as
the de¬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 ∈ M.
Let™s say M is an m-manifold. Then we know there is a set of open sets, {Ui } ‚
Rm , and a set of di¬erentiable maps, {φi : Ui ’ M}, such that {φi (Ui )} covers M.
Now, let™s say that ω is a family of n-forms, de¬ned on Tp M, for each p ∈ M. Then
we say that the family, ω, is a di¬erentiable n-form on M if φ— ω is a di¬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 = ’ 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 de¬ne a 1-form on S 1 , ω = ’y dx + x dy. (Actually, ω is a 1-form
on all of R2 . To get a 1-form on just S 1 , we restrict the domain of ω to the
tangent lines to S 1 .) To check that this is really a di¬erential form, we must
compute all pull-backs:
’1 1
φ— ω = √ dt , φ— ω = √ dt
1 2
1 ’ t2 1 ’ t2
’1
1
φ— ω = √ dt , φ— ω = √ dt
3 4
1 ’ t2 1 ’ t2
Since all of these are di¬erentiable on Ui = (’1, 1), we can say that ω is a
di¬erential form on S 1 .

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

M. Let™s suppose that C is an n-chain in M, and ω is an n-form. Then how do we
de¬ne ω? If C lies entirely in φi(Ui ), for some i, then we could de¬ne the value of
C
φ— ω. But it may be that part of C lies in both
this integral to be the value of i
φ’1 (C)
i
φi (Ui ) and φj (Uj ). If we de¬ne ω to be the sum of the two integrals we get when
C
we pull-back ω under φi and φj , then we end up “double counting” the integral of ω
on C © φi (Ui ) © φj (Uj ). Somehow, as we move from φi (Ui ) into φj (Uj ), we want the
e¬ect of the pull-back of ω under φi to “fade out”, and the e¬ect of the pull back
under φj to “fade in”. This is accomplished by a partition of unity.
The technical de¬nition of a partition of unity subordinate to the cover, {φi (Ui )}
is a set of di¬erentiable functions, fi : M ’ [0, 1], such that fi (p) = 0 if p ∈ φi (Ui ),
/
fi (p) = 1, for all p ∈ M. We refer the reader to any book on di¬erential
and
i
topology for a proof of the existence of partitions of unity.
We are now ready to give the full de¬nition of the integral of an n-form on an
n-chain in an m-manifold.


φ— (fi ω)
ω≡ i
i
C φ’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) ‚ R. Let
Ui = (i, i + 2), for i = 1, ..., 8. Let φi : Ui ’ M be the identity map. Let {fi }
be a partition of unity, subordinate to the cover, {φi(Ui )}. Let ω 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
φ— (fi ω)
ω≡ = fi ω = (fi ω) = fi ω= ω
i
i=1 i=1 C i=1 i=1
C C C C
φ’1 (C)
i


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

Example 7.7. Let S 1 , Ui , φi , and ω be de¬ned as in Examples 7.4 and 7.5. A
partition of unity subordinate to the cover {φi (Ui )} is as follows:
y2 y ≥ 0 0 y>0
f1 (x, y) = , f2 (x, y) =
y2 y ¤ 0
0 y<0

x2 x ≥ 0 0 x>0
f3 (x, y) = , f4 (x, y) =
x2 x ¤ 0
0 x<0
(Check this!) Let µ : [0, π] ’ S 1 be de¬ned by µ(θ) = (cos θ, sin θ). Then the
image of µ is a 1-cell, σ, in S 1 . Let™s integrate ω over σ:

4
φ— (fi ω)
ω≡ i
i=1
σ φ’1 (σ)
i
√ √ √
’ 1 ’ t2 dt + 0 + t2 ’ 1 ’ t2 dt
1’
= dt +

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